Siyavula Textbooks Grade 10 Physical Science 3.1

Siyavula textbooks: Grade 10 Physical
Science
Collection Editor:
Free High School Science Texts Project
Siyavula textbooks: Grade 10 Physical
Science
Collection Editor:
Authors:
Rory Adams
Heather Williams
Wendy Williams
Online:
< http://cnx.org/content/col11245/1.3/ >
CONNEXIONS
Rice University, Houston, Texas

This selection and arrangement of content as a collection is copyrighted by Free High School Science Texts Project.
It is licensed under the Creative Commons Attribution 3.0 license (http://creativecommons.org/licenses/by/3.0/).
Collection structure revised: August 29, 2011
PDF generated: October 29, 2012
For copyright and attribution information for the modules contained in this collection, see p. 322.
Table of Contents
1 Classication of matter
1.1 Mixtures, compounds and elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Giving names and formulae to substances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.3 Classication using the properties of matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
2 What are the objects around us made of
2.1 Atoms and molecules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
2.2 Intermolecular and intramolecular forces and the kinetic theory of matter . . . . . . . . . . . . . . . . . . . 21
2.3 The properties of matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
3 The atom
3.1 Models and atomic size . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
3.2 Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
3.3 Isotopes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
3.4 Energy quantisation and electron conguration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
3.5 Ionisation energy and the periodic table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
4 Physical and chemical change

4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
4.2 Energy changes and conservation of matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
5 Representing chemical change
5.1 Introduction, chemical symbols and chemical formulae . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
5.2 Balancing chemical equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
5.3 State symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
6 The water cycle

6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
6.2 Properties of water . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80
6.3 Water conservation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
7 The nitrogen cycle
7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
7.2 Human inuences and industry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93
8 The hydrosphere
8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99
8.2 Ions in aqueous solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
8.3 Electrolytes, ionisation and conductivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106
8.4 Precipitation reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 109
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116
9 Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117
10 Motion in one dimension
10.1 Frames of reference and reference point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 129
10.2 Displacement and distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132
10.3 Speed and velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
10.4 Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138
10.5 Description of motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140
10.6 Equations of motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158
iv
11 Gravity and mechanical energy

11.1 Weight and acceleration due to gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167
11.2 Potential energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173
11.3 Kinetic energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174
11.4 Mechanical energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183
12 Transverse pulses
12.1 Introduction and key concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 187
12.2 Graphs of particle motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190
12.3 Boundary conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201
13 Transverse waves
13.2 Graphs of particle motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211
13.3 Boundary conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226
14 Geometrical optics
14.1 Introduction and light rays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229
14.2 Reection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232
14.3 Refraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236
14.4 Mirrors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245
14.5 Total internal reection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264
15 Magnetism
15.1 Magnetic elds and permanent magnets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267
15.2 The Earth's magnetic eld . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275
16 Electrostatics
16.2 Forces between charges . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278
16.3 Conductors and insulators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 280
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286
17 Electric circuits
17.1 Key concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 287
17.2 Potential dierence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 293
17.3 Current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297
17.4 Resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301
17.5 Measuring devices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 308
Glossary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311
Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 319
Attributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322
Available for free at Connexions <http://cnx.org/content/col11245/1.3>

Chapter 1
Classication of matter
1.1 Mixtures, compounds and elements 1
1.1.1 Introduction
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Figure 1.1
All the objects that we see in the world around us, are made of matter. Matter makes up the air we
breathe, the ground we walk on, the food we eat and the animals and plants that live around us. Even our
own human bodies are made of matter!
Dierent objects can be made of dierent types of matter, or materials. For example, a cupboard (an
object) is made of wood, nails and hinges (the materials ). The properties of the materials will aect the
properties of the object. In the example of the cupboard, the strength of the wood and metals make the
cupboard strong and durable. In the same way, the raincoats that you wear during bad weather, are made
of a material that is waterproof. The electrical wires in your home are made of metal because metals are
a type of material that is able to conduct electricity. It is very important to understand the properties of
materials, so that we can use them in our homes, in industry and in other applications. In this chapter, we
will be looking at dierent types of materials and their properties.
The diagram below shows one way in which matter can be classied (grouped) according to its dierent
properties. As you read further in this chapter, you will see that there are also other ways of classifying
materials, for example according to whether or not they are good electrical conductors.
1 This content is available online at <http://cnx.org/content/m39993/1.1/>.
1
2 CHAPTER 1. CLASSIFICATION OF MATTER
Figure 1.2: The classication of matter
1.1.2 Mixtures
We see mixtures all the time in our everyday lives. A stew, for example, is a mixture of dierent foods such
as meat and vegetables; sea water is a mixture of water, salt and other substances, and air is a mixture of
gases such as carbon dioxide, oxygen and nitrogen.
Denition 1.1: Mixture

A mixture is a combination of two or more substances, where these substances are not bonded
(or joined) to each other.
In a mixture, the substances that make up the mixture:
• are not in a xed ratio Imagine, for example, that you have a 250 ml beaker of water. It doesn't
matter whether you add 20 g, 40 g, 100 g or any other mass of sand to the water; it will still be called
a mixture of sand and water.
• keep their physical properties In the example we used of the sand and water, neither of these substances
has changed in any way when they are mixed together. Even though the sand is in water, it still has
the same properties as when it was out of the water.
• can be separated by mechanical means To separate something by 'mechanical means', means that there
is no chemical process involved. In our sand and water example, it is possible to separate the mixture
by simply pouring the water through a lter. Something physical is done to the mixture, rather than
something chemical.
Some other examples of mixtures include blood (a mixture of blood cells, platelets and plasma), steel (a
mixture of iron and other materials) and the gold that is used to make jewellery. The gold in jewellery is not
pure gold but is a mixture of metals. The amount of gold in the jewellery is measured in karats (24 karat
would be pure gold, while 18 karat is only 75% gold).
We can group mixtures further by dividing them into those that are heterogeneous and those that are
homogeneous.
1.1.2.1 Heterogeneous mixtures

A heterogeneous mixture does not have a denite composition. Think of a pizza, that has a topping of
cheese, tomato, mushrooms and peppers (the topping is a mixture). Each slice will probably be slightly
dierent from the next because the toppings (the tomato, cheese, mushrooms and peppers) are not evenly
distributed. Another example would be granite, a type of rock. Granite is made up of lots of dierent mineral
substances including quartz and feldspar. But these minerals are not spread evenly through the rock and so
some parts of the rock may have more quartz than others. Another example is a mixture of oil and water.
Although you may add one substance to the other, they will stay separate in the mixture. We say that these
heterogeneous mixtures are non-uniform, in other words they are not exactly the same throughout.
Denition 1.2: Heterogeneous mixture
A heterogeneous mixture is one that is non-uniform and the dierent components of the mixture
can be seen.

3
1.1.2.2 Homogeneous mixtures

A homogeneous mixture has a denite composition, and specic properties. In a homogeneous mixture,
the dierent parts cannot be seen. A solution of salt dissolved in water is an example of a homogeneous
mixture. When the salt dissolves, it will spread evenly through the water so that all parts of the solution
are the same, and you can no longer see the salt as being separate from the water. Think also of a powdered
drink that you mix with water. Provided you give the container a good shake after you have added the
powder to the water, the drink will have the same sweet taste for anyone who drinks it, it won't matter
whether they take a sip from the top or from the bottom. The air we breathe is another example of a
homogeneous mixture since it is made up of dierent gases which are in a constant ratio, and which can't
be distinguished from each other.
Denition 1.3: Homogeneous mixture

A homogeneous mixture is one that is uniform, and where the dierent components of the mixture
cannot be seen.
An alloy is a homogeneous mixture of two or more elements, at least one of which is a metal, where the
resulting material has metallic properties. Alloys are usually made to improve the properties of the elements
that make them up. For example steel is much stronger than iron (which is the main component of steel).
1.1.2.3 Separating mixtures

Sometimes it is important to be able to separate a mixture. There are lots of dierent ways to do this. These
are some examples:
• Filtration A piece of lter paper in a funnel can be used to separate a mixture of sand and water.
• Heating / evaporation Heating a solution causes the liquid (normally water) to evaporate, leaving the
other (solid) part of the mixture behind. You can try this using a salt solution.
• Centrifugation This is a laboratory process which uses the centrifugal force of spinning objects to
separate out the heavier substances from a mixture. This process is used to separate the cells and
plasma in blood. When the test tubes that hold the blood are spun round in the machine, the heavier
cells sink to the bottom of the test tube. Can you think of a reason why it might be important to have
a way of separating blood in this way?
• Dialysis This is an interesting way of separating a mixture because it can be used in some important
applications. Dialysis works using a process called diusion. Diusion takes place when one substance
in a mixture moves from an area where it has a high concentration to an area where its concentration
is lower. When this movement takes place across a semi-permeable membrane it is called osmosis.
A semi-permeable membrane is a barrier that lets some things move across it, but not others. This
process is very important for people whose kidneys are not functioning properly, an illness called renal
failure.
note: Normally, healthy kidneys remove waste products from the blood. When a person has renal
failure, their kidneys cannot do this any more, and this can be life-threatening. Using dialysis, the
blood of the patient ows on one side of a semi-permeable membrane. On the other side there will
be a uid that has no waste products but lots of other important substances such as potassium
+
ions (K ) that the person will need. Waste products from the blood diuse from where their
concentration is high (i.e. in the person's blood) into the 'clean' uid on the other side of the
membrane. The potassium ions will move in the opposite direction from the uid into the blood.
Through this process, waste products are taken out of the blood so that the person stays healthy.

1.1.2.3.1 Investigation : The separation of a salt solution

Aim:
To demonstrate that a homogeneous salt solution can be separated using physical methods.
Apparatus:
glass beaker, salt, water, retort stand, bunsen burner.
Method:
1. Pour a small amount of water (about 20 ml) into a beaker.
2. Measure a teaspoon of salt and pour this into the water.
3. Stir until the salt dissolves completely. This is now called a salt solution. This salt solution is a
homogeneous mixture.
4. Place the beaker on a retort stand over a bunsen burner and heat gently. You should increase the heat
until the water almost boils.
5. Watch the beaker until all the water has evaporated. What do you see in the beaker?
Figure 1.3
Results:
The water evaporates from the beaker and tiny grains of salt remain at the bottom. (You may also
observe grains of salt on the walls of the beaker.)
Conclusion:
The salt solution, which is a homogeneous mixture of salt and water, has been separated using heating
and evaporation.
1.1.2.3.2 Discussion : Separating mixtures

Work in groups of 3-4
Imagine that you have been given a container which holds a mixture of sand, iron lings (small pieces of
iron metal), salt and small stones of dierent sizes. Is this a homogeneous or a heterogeneous mixture? In
your group, discuss how you would go about separating this mixture into the four materials that it contains.
1.1.2.3.3 Mixtures
1. Which of the following substances are mixtures ?
a. tap water
b. brass (an alloy of copper and zinc)
c. concrete
d. aluminium
e. Coca cola
f. distilled water
2. In each of the examples above, say whether the mixture is homogeneous or heterogeneous
2
Click here for the solution
2 See the le at <http://cnx.org/content/m39993/latest/http://www.fhsst.org/llm>

5
1.1.3 Pure Substances: Elements and Compounds

Any material that is not a mixture, is called a pure substance. Pure substances include elements and
compounds. It is much more dicult to break down pure substances into their parts, and complex chemical
methods are needed to do this.
1.1.3.1 Elements
An element is a chemical substance that can't be divided or changed into other chemical substances by any
ordinary chemical means. The smallest unit of an element is the atom.
Denition 1.4: Element

An element is a substance that cannot be broken down into other substances through chemical
means.
There are 112 ocially named elements and about 118 known elements. Most of these are natural, but
some are man-made. The elements we know are represented in the Periodic Table of the Elements, where
each element is abbreviated to a chemical symbol. Examples of elements are magnesium (Mg), hydrogen
(H), oxygen (O) and carbon (C). On the Periodic Table you will notice that some of the abbreviations do
not seem to match the elements they represent. The element iron, for example, has the chemical formula
Fe. This is because the elements were originally given Latin names. Iron has the abbreviation Fe because its
Latin name is 'ferrum'. In the same way, sodium's Latin name is 'natrium' (Na) and gold's is 'aurum' (Au).
1.1.3.2 Compounds
A compound is a chemical substance that forms when two or more elements combine in a xed ratio.
Water (H2 O), for example, is a compound that is made up of two hydrogen atoms for every one oxygen
atom. Sodium chloride (NaCl) is a compound made up of one sodium atom for every chlorine atom. An
important characteristic of a compound is that it has a chemical formula, which describes the ratio in
which the atoms of each element in the compound occur.
Denition 1.5: Compound

A substance made up of two or more elements that are joined together in a xed ratio.
Figure 1.4 might help you to understand the dierence between the terms element, mixture and com-
pound. Iron (Fe) and sulphur (S) are two elements. When they are added together, they form a mixture of
iron and sulphur. The iron and sulphur are not joined together. However, if the mixture is heated, a new
compound is formed, which is called iron sulphide (FeS). In this compound, the iron and sulphur are joined
to each other in a ratio of 1:1. In other words, one atom of iron is joined to one atom of sulphur in the
compound iron sulphide.
Figure 1.4: Understanding the dierence between a mixture and a compound
1.1.3.2.1 Elements, mixtures and compounds

1. In the following table, tick whether each of the substances listed is a mixture or a pure substance. If
it is a mixture, also say whether it is a homogeneous or heterogeneous mixture.

Substance Mixture or pure Homogeneous or heterogeneous mixture

zzy colddrink
steel
oxygen
iron lings
smoke
limestone (CaCO3 )
Table 1.1
3
2. In each of the following cases, say whether the substance is an element, a mixture or a compound.
a. Cu
b. iron and sulphur
c. Al
d. H2 SO4
e. SO3
4
1.2 Giving names and formulae to substances 5
1.2.1 Giving names and formulae to substances

It is easy to describe elements and mixtures. But how are compounds named? In the example of iron
sulphide that was used earlier, which element is named rst, and which 'ending' is given to the compound
name (in this case, the ending is -ide)?
The following are some guidelines for naming compounds:
1. The compound name will always include the names of the elements that are part of it.
• A compound of iron (Fe) and sulphur (S) is iron sulphide (FeS)
• A compound of potassium (K) and bromine (Br) is potassium bromide (KBr)
• A compound of sodium (Na) and chlorine (Cl) is sodium chlor ide (NaCl)
2. In a compound, the element that is on the left of the Periodic Table, is used rst when naming the
compound. In the example of NaCl, sodium is a group 1 element on the left hand side of the table,
while chlorine is in group 7 on the right of the table. Sodium therefore comes rst in the compound
name. The same is true for FeS and KBr.
3. The symbols of the elements can be used to represent compounds e.g. FeS, NaCl, KBr and H2 O.
These are called chemical formulae. In the rst three examples, the ratio of the elements in each
compound is 1:1. So, for FeS, there is one atom of iron for every atom of sulphur in the compound. In
the last example (H2 O) there are two atoms of hydrogen for every atom of oxygen in the compound.
4. A compound may contain compound ions. An ion is an atom that has lost (positive ion) or gained
(negative ion) electrons. Some of the more common compound ions and their formulae are given
below.
3 See the le at <http://cnx.org/content/m39993/latest/http://www.fhsst.org/lly>
4 See the le at <http://cnx.org/content/m39993/latest/http://www.fhsst.org/llV>

7
Name of compound ion Formula

2−
Carbonate CO3
2−
Sulphate SO4
−
Hydroxide OH
+
Ammonium NH4
−
Nitrate NO3
−
Hydrogen carbonate HCO3
3−
Phosphate PO4
−
Chlorate ClO3
−
Cyanide CN
2−
Chromate CrO4
−
Permanganate MnO4
Table 1.2
5. When there are only two elements in the compound, the compound is often given a sux (ending)
of -ide. You would have seen this in some of the examples we have used so far. For compound ions,
when a non-metal is combined with oxygen to form a negative ion (anion) which then combines with a
−
positive ion (cation) from hydrogen or a metal, then the sux of the name will be ...ate or ...ite. NO3
for example, is a negative ion, which may combine with a cation such as hydrogen (HNO3 ) or a metal
like potassium (KNO3 ). The NO3
−
anion has the name nitr ate. SO3
2−
ite, e.g.
in a formula is sulph
sodium sulphite (Na2 SO3 ).
SO4
2−
is sulph ate and PO
is phosphate.
4
3−
6. Prexes can be used to describe the ratio of the elements that are in the compound. You should know
the following prexes: 'mono' (one), 'di' (two) and 'tri' (three).
• CO (carbon monoxide) - There is one atom of oxygen for every one atom of carbon
• NO2 (nitrogen dioxide) - There are two atoms of oxygen for every one atom of nitrogen
• SO3 (sulphur trioxide) - There are three atoms of oxygen for every one atom of sulphur
tip: When numbers are written as 'subscripts' in compounds (i.e. they are written below and to
the right of the element symbol), this tells us how many atoms of that element there are in relation
to other elements in the compound. For example in nitrogen dioxide (NO2 ) there are two oxygen
atoms for every one atom of nitrogen. In sulphur trioxide (SO3 ), there are three oxygen atoms for
every one atom of sulphur in the compound. Later, when we start looking at chemical equations,
you will notice that sometimes there are numbers before the compound name. For example, 2H2 O
means that there are two molecules of water, and that in each molecule there are two hydrogen
atoms for every one oxygen atom.
1.2.1.1 Naming compounds

1. The formula for calcium carbonate is CaCO3 .
a. Is calcium carbonate a mixture or a compound? Give a reason for your answer.

b. What is the ratio of Ca:C:O atoms in the formula?
6
6 See the le at <http://cnx.org/content/m39999/latest/http://www.fhsst.org/llp>

2. Give the name of each of the following substances.
a. KBr
b. HCl
c. KMnO4
d. NO2
e. NH4 OH
f. Na2 SO4
7
3. Give the chemical formula for each of the following compounds.
a. potassium nitrate
b. sodium iodide
c. barium sulphate
d. nitrogen dioxide
e. sodium monosulphate
8
4. Refer to the diagram below, showing sodium chloride and water, and then answer the questions that
follow.
Figure 1.5
a. What is the chemical formula for water?

b. What is the chemical formula for sodium chloride?
c. Label the water and sodium chloride in the diagram.
d. Give a description of the picture. Focus on whether there are elements or compounds and if it is
a mixture or not.
9
5. What is the formula of this molecule?
Figure 1.6
a. C6 H2 O
b. C2 H6 O
c. 2C6HO
d. 2 CH6 O
10
7 See the le at <http://cnx.org/content/m39999/latest/http://www.fhsst.org/lld>

8 See the le at <http://cnx.org/content/m39999/latest/http://www.fhsst.org/llv>
9 See the le at <http://cnx.org/content/m39999/latest/http://www.fhsst.org/llL>
10 See the le at <http://cnx.org/content/m39999/latest/http://www.fhsst.org/llf>

9
1.3 Classication using the properties of matter 11
1.3.1 Metals, Semi-metals and Non-metals

The elements in the Periodic Table can also be divided according to whether they are metals, semi-metals
or non-metals. On the right hand side of the Periodic Table you can draw a 'zigzag' line (This line starts
with Boron (B) and goes down to Polonium (Po)). This line separates all the elements that are metals from
those that are non-metals. Metals are found on the left of the line, and non-metals are those on the right.
Along the line you nd the semi-metals. Metals, semi-metals and non-metals all have their own specic
properties.
1.3.1.1 Metals
Examples of metals include copper (Cu), zinc (Zn), gold (Au) and silver (Ag). On the Periodic Table, the
metals are on the left of the zig-zag line. There are a large number of elements that are metals. The following
are some of the properties of metals:
• Thermal conductors Metals are good conductors of heat. This makes them useful in cooking utensils
such as pots and pans.
• Electrical conductors Metals are good conductors of electricity. Metals can be used in electrical con-
ducting wires.
• Shiny metallic lustre Metals have a characteristic shiny appearance and so are often used to make
jewellery.
• Malleable This means that they can be bent into shape without breaking.
• Ductile Metals (such as copper) can be stretched into thin wires, which can then be used to conduct
electricity.
• Melting point Metals usually have a high melting point and can therefore be used to make cooking
pots and other equipment that needs to become very hot, without being damaged.
You can see how the properties of metals make them very useful in certain applications.
1.3.1.1.1 Group Work : Looking at metals

1. Collect a number of metal items from your home or school. Some examples are listed below:
• hammer
• wire
• cooking pots
• jewellery
• nails
• coins
2. In groups of 3-4, combine your collection of metal objects.

3. What is the function of each of these objects?
4. Discuss why you think metal was used to make each object. You should consider the properties of
metals when you answer this question.
1.3.1.2 Non-metals
In contrast to metals, non-metals are poor thermal conductors, good electrical insulators (meaning that they
do not conduct electrical charge) and are neither malleable nor ductile. The non-metals are found on the
right hand side of the Periodic Table, and include elements such as sulphur (S), phosphorus (P), nitrogen
(N) and oxygen (O).

1.3.1.3 Semi-metals
Semi-metals have mostly non-metallic properties. One of their distinguishing characteristics is that their
conductivity increases as their temperature increases. This is the opposite of what happens in metals. The
semi-metals include elements such as silicon (Si) and germanium (Ge). Notice where these elements are
positioned in the Periodic Table.
1.3.2 Electrical conductors, semi-conductors and insulators

An electrical conductor is a substance that allows an electrical current to pass through it. Electrical
conductors are usually metals. Copper is one of the best electrical conductors, and this is why it is used to
make conducting wire. In reality, silver actually has an even higher electrical conductivity than copper, but
because silver is so expensive, it is not practical to use it for electrical wiring because such large amounts
are needed. In the overhead power lines that we see above us, aluminium is used. The aluminium usually
surrounds a steel core which adds tensile strength to the metal so that it doesn't break when it is stretched
across distances. Occasionally gold is used to make wire, not because it is a particularly good conductor,
but because it is very resistant to surface corrosion. Corrosion is when a material starts to deteriorate at
the surface because of its reactions with the surroundings, for example oxygen and water in the air.
An insulator is a non-conducting material that does not carry any charge. Examples of insulators
would be plastic and wood. Do you understand now why electrical wires are normally covered with plastic
insulation? Semi-conductors behave like insulators when they are cold, and like conductors when they are
hot. The elements silicon and germanium are examples of semi-conductors.
Denition 1.6: Conductors and insulators

A conductor allows the easy movement or ow of something such as heat or electrical charge
through it. Insulators are the opposite to conductors because they inhibit or reduce the ow of
heat, electrical charge, sound etc through them.
1.3.2.1 Experiment : Electrical conductivity

Aim:
To investigate the electrical conductivity of a number of substances
Apparatus:
• two or three cells
• light bulb
• crocodile clips
• wire leads
• a selection of test substances (e.g. a piece of plastic, aluminium can, metal pencil sharpener, magnet,
wood, chalk).
Figure 1.7
Method:
1. Set up the circuit as shown above, so that the test substance is held between the two crocodile clips.
The wire leads should be connected to the cells and the light bulb should also be connected into the
circuit.

11
2. Place the test substances one by one between the crocodile clips and see what happens to the light
bulb.
Results:
Record your results in the table below:
Test substance Metal/non-metal Does the light bulb glow? Conductor or insulator
Table 1.3
Conclusions:
In the substances that were tested, the metals were able to conduct electricity and the non-metals were
not. Metals are good electrical conductors and non-metals are not.
The following simulation allows you to work through the above activity. For this simulation use the grab
bag option to get materials to test. Set up the circuit as described in the activity.
Figure 1.8
12
run demo
1.3.3 Thermal Conductors and Insulators

A thermal conductor is a material that allows energy in the form of heat, to be transferred within the
material, without any movement of the material itself. An easy way to understand this concept is through
a simple demonstration.
1.3.3.1 Demonstration : Thermal conductivity

Aim:
To demonstrate the ability of dierent substances to conduct heat.
Apparatus:
You will need two cups (made from the same material e.g. plastic); a metal spoon and a plastic spoon.
Method:
• Pour boiling water into the two cups so that they are about half full.
• At the same time, place a metal spoon into one cup and a plastic spoon in the other.
• Note which spoon heats up more quickly
12 http://phet.colorado.edu/sims/circuit-construction-kit/circuit-construction-kit-dc_en.jnlp

Results:
The metal spoon heats up faster than the plastic spoon. In other words, the metal conducts heat well,
but the plastic does not.
Conclusion:
Metal is a good thermal conductor, while plastic is a poor thermal conductor. This explains why cooking
pots are metal, but their handles are often plastic or wooden. The pot itself must be metal so that heat
from the cooking surface can heat up the pot to cook the food inside it, but the handle is made from a poor
thermal conductor so that the heat does not burn the hand of the person who is cooking.
An insulator is a material that does not allow a transfer of electricity or energy. Materials that are poor
thermal conductors can also be described as being good thermal insulators.
note: Water is a better thermal conductor than air and conducts heat away from the body about
20 times more eciently than air. A person who is not wearing a wetsuit, will lose heat very
quickly to the water around them and can be vulnerable to hypothermia (this is when the body
temperature drops very low). Wetsuits help to preserve body heat by trapping a layer of water
against the skin. This water is then warmed by body heat and acts as an insulator. Wetsuits
are made out of closed-cell, foam neoprene. Neoprene is a synthetic rubber that contains small
bubbles of nitrogen gas when made for use as wetsuit material. Nitrogen gas has very low thermal
conductivity, so it does not allow heat from the body (or the water trapped between the body and
the wetsuit) to be lost to the water outside of the wetsuit. In this way a person in a wetsuit is able
to keep their body temperature much higher than they would otherwise.
1.3.3.2 Investigation : A closer look at thermal conductivity

Look at the table below, which shows the thermal conductivity of a number of dierent materials, and then
answer the questions that follow. The higher the number in the second column, the better the material is
at conducting heat (i.e. it is a good thermal conductor). Remember that a material that conducts heat
eciently, will also lose heat more quickly than an insulating material.
Material Thermal Conductivity (W · m−1 · K −1 )

Silver 429
Stainless steel 16
Standard glass 1.05
Concrete 0.9 - 2
Red brick 0.69
Water 0.58
Snow 0.25 - 0.5
Wood 0.04 - 0.12
Polystyrene 0.03
Air 0.024
Table 1.4
Use this information to answer the following questions:
1. Name two materials that are good thermal conductors.

2. Name two materials that are good insulators.

13
3. Explain why:
a. cooler boxes are often made of polystyrene

b. homes that are made from wood need less internal heating during the winter months.
c. igloos (homes made from snow) are so good at maintaining warm temperatures, even in freezing
conditions.
note: It is a known fact that well-insulated buildings need less energy for heating than do buildings
that have no insulation. Two building materials that are being used more and more worldwide, are
mineral wool and polystyrene. Mineral wool is a good insulator because it holds air still in the
matrix of the wool so that heat is not lost. Since air is a poor conductor and a good insulator, this
helps to keep energy within the building. Polystyrene is also a good insulator and is able to keep
cool things cool and hot things hot. It has the added advantage of being resistant to moisture,
mould and mildew.
Remember that concepts such as conductivity and insulation are not only relevant in the building, industrial
and home environments. Think for example of the layer of blubber or fat that is found in some animals. In
very cold environments, fat and blubber not only provide protection, but also act as an insulator to help the
animal keep its body temperature at the right level. This is known as thermoregulation.
1.3.4 Magnetic and Non-magnetic Materials

We have now looked at a number of ways in which matter can be grouped, such as into metals, semi-metals
and non-metals; electrical conductors and insulators, and thermal conductors and insulators. One way in
which we can further group metals, is to divide them into those that are magnetic and those that are
non-magnetic.
Denition 1.7: Magnetism
Magnetism is one of the phenomena by which materials exert attractive or repulsive forces on other
materials.
A metal is said to be ferromagnetic if it can be magnetised (i.e. made into a magnet). If you hold
a magnet very close to a metal object, it may happen that its own electrical eld will be induced and the
object becomes magnetic. Some metals keep their magnetism for longer than others. Look at iron and steel
for example. Iron loses its magnetism quite quickly if it is taken away from the magnet. Steel on the other
hand will stay magnetic for a longer time. Steel is often used to make permanent magnets that can be used
for a variety of purposes.
Magnets are used to sort the metals in a scrap yard, in compasses to nd direction, in the magnetic
strips of video tapes and ATM cards where information must be stored, in computers and TV's, as well as
in generators and electric motors.
1.3.4.1 Investigation : Magnetism

You can test whether an object is magnetic or not by holding another magnet close to it. If the object is
attracted to the magnet, then it too is magnetic.
Find some objects in your classroom or your home and test whether they are magnetic or not. Then
complete the table below:

Object Magnetic or non-magnetic
Table 1.5
1.3.4.2 Group Discussion : Properties of materials

In groups of 4-5, discuss how our knowledge of the properties of materials has allowed society to:
• develop advanced computer technology

• provide homes with electricity
• nd ways to conserve energy
1.3.5 Summary
• All the objects and substances that we see in the world are made of matter.
• This matter can be classied according to whether it is a mixture or a pure substance.
• A mixture is a combination of one or more substances that are not chemically bonded to each other.
Examples of mixtures are air (a mixture of dierent gases) and blood (a mixture of cells, platelets and
plasma).
• The main characteristics of mixtures are that the substances that make them up are not in a xed
ratio, they keep their individual properties and they can be separated from each other using mechanical
means.
• A heterogeneous mixture is non-uniform and the dierent parts of the mixture can be seen. An
example would be a mixture of sand and water.
• A homogeneous mixture is uniform, and the dierent components of the mixture can't be seen. An
example would be a salt solution. A salt solution is a mixture of salt and water. The salt dissolves in
the water, meaning that you can't see the individual salt particles. They are interspersed between the
water molecules. Another example is a metalalloy such as steel.
• Mixtures can be separated using a number of methods such as ltration, heating, evaporation, cen-
trifugation and dialysis.
• Pure substances can be further divided into elements and compounds.
• An element is a substance that can't be broken down into simpler substances through chemical means.
• All the elements are recorded in the Periodic Table of the Elements. Each element has its own
chemical symbol. Examples are iron (Fe), sulphur (S), calcium (Ca), magnesium (Mg) and uorine
(F).
• A compound is a substance that is made up of two or more elements that are chemically bonded to
each other in a xed ratio. Examples of compounds are sodium chloride (NaCl), iron sulphide (FeS),
calcium carbonate (CaCO3 ) and water (H2 O).
• When naming compounds and writing their chemical formula, it is important to know the elements
that are in the compound, how many atoms of each of these elements will combine in the compound
and where the elements are in the Periodic Table. A number of rules can then be followed to name the
compound.

15
• Another way of classifying matter is into metals (e.g. iron, gold, copper), semi-metals (e.g. silicon
and germanium) and non-metals (e.g. sulphur, phosphorus and nitrogen).
• Metals are good electrical and thermal conductors, they have a shiny lustre, they are malleable and
ductile, and they have a high melting point. These properties make metals very useful in electrical
wires, cooking utensils, jewellery and many other applications.
• A further way of classifying matter is into electrical conductors, semi-conductors and insulators.
• An electrical conductor allows an electrical current to pass through it. Most metals are good
electrical conductors.
• An electrical insulator is not able to carry an electrical current. Examples are plastic, wood, cotton
material and ceramic.
• Materials may also be classied as thermal conductors or thermal insulators depending on whether
or not they are able to conduct heat.
• Materials may also be either magnetic or non-magnetic.
1.3.5.1 Summary
1. For each of the following multiple choice questions, choose one correct answer from the list provided.
a. Which of the following can be classied as a mixture:
a. sugar
b. table salt
c. air
13
d. iron Click here for the solution
b. An element can be dened as:
a. A substance that cannot be separated into two or more substances by ordinary chemical (or
physical) means
b. A substance with constant composition
c. A substance that contains two or more substances, in denite proportion by weight
d. A uniform substance
14
2. Classify each of the following substances as an element, a compound, a solution (homogeneous mixture),
or a heterogeneous mixture : salt, pure water, soil, salt water, pure air, carbon dioxide, gold and bronze.
15
3. Look at the table below. In the rst column (A) is a list of substances. In the second column (B) is a
description of the group that each of these substances belongs in. Match up the substance in Column
A with the description in Column B.
Column A Column B
iron a compound containing 2 elements
H2 S a heterogeneous mixture
sugar solution a metal alloy
sand and stones an element
steel a homogeneous mixture
13 See the le at <http://cnx.org/content/m39998/latest/http://www.fhsst.org/ll6>

14 See the le at <http://cnx.org/content/m39998/latest/http://www.fhsst.org/llF>
15 See the le at <http://cnx.org/content/m39998/latest/http://www.fhsst.org/llG>

Table 1.6
16
4. You are given a test tube that contains a mixture of iron lings and sulphur. You are asked to weigh
the amount of iron in the sample.
a. Suggest one method that you could use to separate the iron lings from the sulphur.
b. What property of metals allows you to do this?
17
5. Given the following descriptions, write the chemical formula for each of the following substances:
a. silver metal
b. a compound that contains only potassium and bromine
c. a gas that contains the elements carbon and oxygen in a ratio of 1:2
18
6. Give the names of each of the following compounds:
a. NaBr
b. BaSO4
c. SO2
19
7. For each of the following materials, say what properties of the material make it important in carrying
out its particular function.
a. tar on roads
b. iron burglar bars
c. plastic furniture
d. metal jewellery
e. clay for building
f. cotton clothing
20

17 See the le at <http://cnx.org/content/m39998/latest/http://www.fhsst.org/llA>
18 See the le at <http://cnx.org/content/m39998/latest/http://www.fhsst.org/llo>
19 See the le at <http://cnx.org/content/m39998/latest/http://www.fhsst.org/lls>
20 See the le at <http://cnx.org/content/m39998/latest/http://www.fhsst.org/llH>

Chapter 2
What are the objects around us made of
2.1 Atoms and molecules 1
2.1.1 Introduction: The atom as the building block of matter

We have now seen that dierent materials have dierent properties. Some materials are metals and some are
non-metals; some are electrical or thermal conductors, while others are not. Depending on the properties of
these materials, they can be used in lots of useful applications. But what is it exactly that makes up these
materials? In other words, if we were to break down a material into the parts that make it up, what would
we nd? And how is it that a material's microscopic structure (the small parts that make up the material)
is able to give it all these dierent properties?
The answer lies in the smallest building block of matter: the atom. It is the type of atoms, and the way
in which they are arranged in a material, that aects the properties of that substance.
It is not often that substances are found in atomic form. Normally, atoms are bonded (joined) to other
atoms to form compounds or molecules. It is only in the noble gases (e.g. helium, neon and argon) that
atoms are found individually and are not bonded to other atoms. We will look at the reasons for this in a
later chapter.
2.1.2 Molecules
Denition 2.1: Molecule
A molecule is a group of two or more atoms that are attracted to each other by relatively strong
forces or bonds.
Almost everything around us is made up of molecules. Water is made up of molecules, each of which has
two hydrogen atoms joined to one oxygen atom. Oxygen is a molecule that is made up of two oxygen atoms
that are joined to one another. Even the food that we eat is made up of molecules that contain atoms of
elements such as carbon, hydrogen and oxygen that are joined to one another in dierent ways. All of these
are known as small molecules because there are only a few atoms in each molecule. Giant molecules are
those where there may be millions of atoms per molecule. Examples of giant molecules are diamonds, which
are made up of millions of carbon atoms bonded to each other and metals, which are made up of millions of
metal atoms bonded to each other.
2.1.2.1 Representing molecules

The structure of a molecule can be shown in many dierent ways. Sometimes it is easiest to show what
diagrams, but at other
a molecule looks like by using dierent types of times, we may decide to simply
represent a molecule using its chemical formula or its written name.
17
18 CHAPTER 2. WHAT ARE THE OBJECTS AROUND US MADE OF
1. Using formulae to show the structure of a molecule. A chemical formula is an abbreviated

(shortened) way of describing a molecule, or some other chemical substance. In the chapter on classi-
cation of matter, we saw how chemical compounds can be represented using element symbols from the
Periodic Table. A chemical formula can also tell us the number of atoms of each element that are in a
molecule and their ratio in that molecule. For example, the chemical formula for a molecule of carbon
dioxide is CO2 The formula above is called the molecular formula of that compound. The formula
tells us that in one molecule of carbon dioxide, there is one atom of carbon and two atoms of oxygen.
The ratio of carbon atoms to oxygen atoms is 1:2.
Denition 2.2: Molecular formula

This is a concise way of expressing information about the atoms that make up a particular
chemical compound. The molecular formula gives the exact number of each type of atom in
the molecule.
A molecule of glucose has the molecular formula: C6 H12 O6 . In each glucose molecule, there are six
carbon atoms, twelve hydrogen atoms and six oxygen atoms. The ratio of carbon:hydrogen:oxygen is
6:12:6. We can simplify this ratio to write 1:2:1, or if we were to use the element symbols, the formula
would be written as CH2 O. This is called the empirical formula of the molecule.
Denition 2.3: Empirical formula
This is a way of expressing the relative number of each type of atom in a chemical compound.
In most cases, the empirical formula does not show the exact number of atoms, but rather
the simplest ratio of the atoms in the compound.
The empirical formula is useful when we want to write the formula for a giant molecule. Since giant
molecules may consist of millions of atoms, it is impossible to say exactly how many atoms are in each
molecule. It makes sense then to represent these molecules using their empirical formula. So, in the
case of a metal such as copper, we would simply write Cu, or if we were to represent a molecule of
sodium chloride, we would simply write NaCl. Chemical formulae therefore tell us something about
the types of atoms that are in a molecule and the ratio in which these atoms occur in the molecule,
but they don't give us any idea of what the molecule actually looks like, in other words its shape. To
show the shape of molecules we can represent molecules using diagrams. Another type of formula that
can be used to describe a molecule is its structural formula. A structural formula uses a graphical
representation to show a molecule's structure (Figure 2.1).
Figure 2.1: Diagram showing (a) the molecular, (b) the empirical and (c) the structural formula of
isobutane
2. Using diagrams to show the structure of a molecule Diagrams of molecules are very useful
because they help us to picture how the atoms are arranged in the molecule and they help us to see
the shape of the molecule. There are two types of diagrams that are commonly used:
• Ball and stick models This is a 3-dimensional molecular model that uses 'balls' to represent
atoms and 'sticks' to represent the bonds between them. The centres of the atoms (the balls)
are connected by straight lines which represent the bonds between them. A simplied example is
shown in Figure 2.2.

19
Figure 2.2: A ball and stick model of a water molecule
• Space-lling model This is also a 3-dimensional molecular model. The atoms are represented by
spheres. Figure 2.3 and Figure 2.4 are some examples of simple molecules that are represented
in dierent ways.
Figure 2.3: A space-lling model and structural formula of a water molecule. Each molecule is made
up of two hydrogen atoms that are attached to one oxygen atom. This is a simple molecule.
Figure 2.4: A space-lling model and structural formula of a molecule of ammonia. Each molecule is
made up of one nitrogen atom and three hydrogen atoms. This is a simple molecule.
Figure 2.5 shows the bonds between the carbon atoms in diamond, which is a giant molecule. Each
carbon atom is joined to four others, and this pattern repeats itself until a complex lattice structure is
formed. Each black ball in the diagram represents a carbon atom, and each line represents the bond
between two carbon atoms. Note that the carbon atoms on the edges are actually bonded to four
carbon atoms, but some of these carbon atoms have been omitted.
Figure 2.5: Diagrams showing the microscopic structure of diamond. The diagram on the left shows
part of a diamond lattice, made up of numerous carbon atoms. The diagram on the right shows how
each carbon atom in the lattice is joined to four others. This forms the basis of the lattice structure.
Diamond is a giant molecule.
note: Diamonds are most often thought of in terms of their use in the jewellery industry. However,
about 80% of mined diamonds are unsuitable for use as gemstones and are therefore used in industry
because of their strength and hardness. These properties of diamonds are due to the strong covalent
bonds (covalent bonding will be explained later) between the carbon atoms in diamond. The most
common uses for diamonds in industry are in cutting, drilling, grinding, and polishing.
2
This website allows you to view several molecules. You do not need to know these molecules, this is simply
to allow you to see one way of representing molecules.
2 http://alteredqualia.com/canvasmol/

Figure 2.6: Ball-and-stick view of ethanol from http://alteredqualia.com/canvasmol/3

21
2.1.2.1.1 Atoms and molecules

1. In each of the following, say whether the chemical substance is made up of single atoms, simple
molecules or giant molecules.
a. ammonia gas (NH3 )

b. zinc metal (Zn)
c. graphite (C)
d. nitric acid (HNO3 )
e. neon gas (Ne)
4
2. Refer to the diagram below and then answer the questions that follow:
Figure 2.7
a. Identify the molecule.

b. Write the molecular formula for the molecule.
c. Is the molecule a simple or giant molecule?
5
3. Represent each of the following molecules using its chemical formula, structural formula and ball and
stick model.
a. Hydrogen
b. Ammonia
c. sulphur dioxide
6
2.2 Intermolecular and intramolecular forces and the kinetic theory

of matter 7
2.2.1 Intramolecular and intermolecular forces

When atoms join to form molecules, they are held together by chemical bonds. The type of bond, and
the strength of the bond, depends on the atoms that are involved. These bonds are called intramolecular
forces because they are bonding forces inside a molecule ('intra' means 'within' or 'inside'). Sometimes we
simply call these intramolecular forces chemical bonds.
Denition 2.4: Intramolecular force

The force between the atoms of a molecule, which holds them together.
Examples of the types of chemical bonds that can exist between atoms inside a molecule are shown below.
These will be looked at in more detail in Grade 11.
3 http://alteredqualia.com/canvasmol/
4 See the le at <http://cnx.org/content/m39951/latest/http://www.fhsst.org/li5>
5 See the le at <http://cnx.org/content/m39951/latest/http://www.fhsst.org/liN>
6 See the le at <http://cnx.org/content/m39951/latest/http://www.fhsst.org/liR>

• Covalent bond Covalent bonds exist between non-metal atoms e.g. There are covalent bonds between
the carbon and oxygen atoms in a molecule of carbon dioxide.
• Ionic bond Ionic bonds occur between non-metal and metal atoms e.g. There are ionic bonds between
the sodium and chlorine atoms in a molecule of sodium chloride.
• Metallic bond Metallic bonds join metal atoms e.g. There are metallic bonds between copper atoms
in a piece of copper metal.
Intermolecular forces are those bonds that hold molecules together. A glass of water for example, contains
many molecules of water. These molecules are held together by intermolecular forces. The strength of the
intermolecular forces is important because they aect properties such as melting point and boiling point.
For example, the stronger the intermolecular forces, the higher the melting point and boiling point for that
substance. The strength of the intermolecular forces increases as the size of the molecule increases.
Denition 2.5: Intermolecular force

A force between molecules, which holds them together.
The following diagram may help you to understand the dierence between intramolecular forces and
intermolecular forces.
Figure 2.8: Two representations showing the intermolecular and intramolecular forces in water: space-
lling model and structural formula.
It should be clearer now that there are two types of forces that hold matter together. In the case of water,
there are intramolecular forces that hold the two hydrogen atoms to the oxygen atom in each molecule of
water (these are the solid lines in the above diagram). There are also intermolecular forces between each of
these water molecules. These intermolecular forces join the hydrogen atom from one molecule to the oxygen
atom of another molecule (these are the dashed lines in the above gure). As mentioned earlier, these
forces are very important because they aect many of the properties of matter such as boiling point, melting
point and a number of other properties. Before we go on to look at some of these examples, it is important
that we rst take a look at the Kinetic Theory of Matter.
2.2.1.1 Intramolecular and intermolecular forces
1. Using ammonia gas as an example...
a. Explain what is meant by an intramolecular force or chemical bond.

b. Explain what is meant by an intermolecular force.
8
2. Draw a diagram showing three molecules of carbon dioxide. On the diagram, show where the in-
9
tramolecular and intermolecular forces are. Click here for the solution
3. Why is it important to understand the types of forces that exist between atoms and between molecules?
10
Try to use some practical examples in your answer. Click here for the solution
8 See the le at <http://cnx.org/content/m39944/latest/http://www.fhsst.org/lin>

9 See the le at <http://cnx.org/content/m39944/latest/http://www.fhsst.org/liQ>
10 See the le at <http://cnx.org/content/m39944/latest/http://www.fhsst.org/liU>

23
2.2.2 The Kinetic Theory of Matter

The kinetic theory of matter helps us to explain why matter exists in dierent phases (i.e. solid, liquid
and gas), and how matter can change from one phase to the next. The kinetic theory of matter also helps
us to understand other properties of matter. It is important to realise that what we will go on to describe is
only a theory. It cannot be proved beyond doubt, but the fact that it helps us to explain our observations
of changes in phase, and other properties of matter, suggests that it probably is more than just a theory.
Broadly, the Kinetic Theory of Matter says that:
• Matter is made up of particles that are constantly moving.

• All particles have energy, but the energy varies depending on whether the substance is a solid, liquid
or gas. Solid particles have the least amount of energy and gas particles have the greatest amount of
energy.
• The temperature of a substance is a measure of the average kinetic energy of the particles.
• A change in phase may occur when the energy of the particles is changed.
• There are spaces between the particles of matter.
• There are attractive forces between particles and these become stronger as the particles move closer
together. These attractive forces will either be intramolecular forces (if the particles are atoms) or
intermolecular forces (if the particles are molecules). When the particles are extremely close, repulsive
forces start to act.
Table 2.1 summarises the characteristics of the particles that are in each phase of matter.
Property of matter Solid Liquid Gas

Particles Atoms or molecules Atoms or molecules Atoms or molecules
Energy and movement Low energy - particles Particles have less en- Particles have high en-
of particles vibrate around a xed ergy than in the gas ergy and are constantly
point phase moving
Spaces between parti- Very little space be- Smaller spaces than in Large spaces because of
cles tween particles. Parti- gases, but larger spaces high energy
cles are tightly packed than in solids
together
Attractive forces be- Very strong forces. Stronger forces than in Weak forces because of
tween particles Solids have a xed gas. Liquids can be the large distance be-
volume. poured. tween particles
Changes in phase Solids become liquids if A liquid becomes a gas In general a gas be-
their temperature is in- if its temperature is in- comes a liquid when it is
creased. In some cases a creased. It becomes a cooled. (In a few cases
solid may become a gas solid if its temperature a gas becomes a solid
if the temperature is in- decreases. when cooled). Parti-
creased. cles have less energy and
therefore move closer to-
gether so that the at-
tractive forces become
stronger, and the gas
becomes a liquid (or a
solid.)
Table 2.1: Table summarising the general features of solids, liquids and gases.

The following presentation is a brief summary of the above. Try to ll in the blank spaces before clicking
onto the next slide.

<http://static.slidesharecdn.com/swf/ssplayer2.swf ?doc=kinetictheory-done-100510040904-
phpapp02&stripped_title=kinetic-theory-done>
Figure 2.9
Let's look at an example that involves the three phases of water: ice (solid), water (liquid) and water
vapour (gas). Note that in the Figure 2.10 below the molecules in the solid phase are represented by single
spheres, but they would in reality look like the molecules in the liquid and gas phase. Sometimes we represent
molecules as single spheres in the solid phase to emphasise the small amount of space between them and to
make the drawing simpler.
Figure 2.10: The three phases of matter
Taking water as an example we nd that in the solid phase the water molecules have very little energy
and can't move away from each other. The molecules are held closely together in a regular pattern called
a lattice. If the ice is heated, the energy of the molecules increases. This means that some of the water
molecules are able to overcome the intermolecular forces that are holding them together, and the molecules
move further apart to form liquid water. This is why liquid water is able to ow, because the molecules are
more free to move than they were in the solid lattice. If the molecules are heated further, the liquid water
will become water vapour, which is a gas. Gas particles have lots of energy and are far away from each other.
That is why it is dicult to keep a gas in a specic area! The attractive forces between the particles are
very weak and they are only loosely held together. Figure 2.11 shows the changes in phase that may occur
in matter, and the names that describe these processes.
Figure 2.11: Changes in phase

25
2.3 The properties of matter 11
2.3.1 The Properties of Matter

Let us now look at what we have learned about chemical bonds, intermolecular forces and the kinetic theory
of matter, and see whether this can help us to understand some of the macroscopic properties of materials.
1. Melting point
Denition 2.6: Melting point
The temperature at which a solid changes its phase or state to become a liquid. The process is
called melting and the reverse process (change in phase from liquid to solid) is called freezing.
In order for a solid to melt, the energy of the particles must increase enough to overcome the bonds that
are holding the particles together. It makes sense then that a solid which is held together by strong
bonds will have a higher melting point than one where the bonds are weak, because more energy (heat)
is needed to break the bonds. In the examples we have looked at metals, ionic solids and some atomic
lattices (e.g. diamond) have high melting points, whereas the melting points for molecular solids and
other atomic lattices (e.g. graphite) are much lower. Generally, the intermolecular forces between
molecular solids are weaker than those between ionic and metallic solids.
2. Boiling point
Denition 2.7: Boiling point
The temperature at which a liquid changes its phase to become a gas. The process is called
evaporation and the reverse process is called condensation
When the temperature of a liquid increases, the average kinetic energy of the particles also increases
and they are able to overcome the bonding forces that are holding them in the liquid. When boiling
point is reached, evaporation takes place and some particles in the liquid become a gas. In other words,
the energy of the particles is too great for them to be held in a liquid anymore. The stronger the bonds
within a liquid, the higher the boiling point needs to be in order to break these bonds. Metallic and
ionic compounds have high boiling points while the boiling point for molecular liquids is lower. The
data in Table 2.2 below may help you to understand some of the concepts we have explained. Not all
of the substances in the table are solids at room temperature, so for now, let's just focus on the boiling
points for each of these substances. What do you notice?
Substance Melting point (0 C ) Boiling point (0 C )

Ethanol (C2 H6 O) - 114,3 78,4
Water 0 100
Mercury -38,83 356,73
Sodium chloride 801 1465
Table 2.2: The melting and boiling points for a number of substances
You will have seen that substances such as ethanol, with relatively weak intermolecular forces, have
the lowest boiling point, while substances with stronger intermolecular forces such as sodium chloride
and mercury, must be heated much more if the particles are to have enough energy to overcome the
forces that are holding them together in the liquid.

2.3.1.1 Forces and boiling point

The table below gives the molecular formula and the boiling point for a number of organic compounds called
alkanes (more on these compounds in grade 12). Refer to the table and then answer the questions that
follow.
0
Organic compound Molecular formula Boiling point ( C)
Methane CH4 -161.6
Ethane C2 H6 - 88.6
Propane C3 H8 -45
Butane C4 H10 -0.5
Pentane C5 H12 36.1
Hexane C6 H14 69
Heptane C7 H16 98.42
Octane C8 H18 125.52
Table 2.3
Data from: http://www.wikipedia.com
1. Draw a graph to show the relationship between the number of carbon atoms in each alkane and its
boiling point. (Number of carbon atoms will go on the x-axis and boiling point on the y-axis).
2. Describe what you see.
3. Suggest a reason for what you have observed.
4. Why was it enough for us to use 'number of carbon atoms' as a measure of the molecular weight of the
12
molecules? Click here for the solution
Density and viscosity

Density is a measure of the mass of a substance per unit volume. The density of a solid is generally higher
than that of a liquid because the particles are held much more closely together and therefore there are more
particles packed together in a particular volume. In other words, there is a greater mass of the substance
in a particular volume. In general, density increases as the strength of the intermolecular forces increases.
Viscosity is a measure of how resistant a liquid is to owing (in other words, how easy it is to pour the
liquid from one container to another). Viscosity is also sometimes described as the 'thickness' of a uid.
Think for example of syrup and how slowly it pours from one container into another. Now compare this to
how easy it is to pour water. The viscosity of syrup is greater than the viscosity of water. Once again, the
stronger the intermolecular forces in the liquid, the greater its viscosity.
It should be clear now that we can explain a lot of the macroscopic properties of matter (i.e. the
characteristics we can see or observe) by understanding their microscopic structure and the way in which
the atoms and molecules that make up matter are held together.
2.3.1.2 Investigation : Determining the density of liquids:

Density is a very important property because it helps us to identify dierent materials. Every material,
depending on the elements that make it up and the arrangement of its atoms, will have a dierent density.
The equation for density is:
Density = Mass/Volume
Discussion questions:
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27
To calculate the density of liquids and solids, we need to be able to rst determine their mass and volume.
As a group, think about the following questions:
• How would you determine the mass of a liquid?

• How would you determine the volume of an irregular solid?
Apparatus:
Laboratory mass balance, 10 ml and 100 ml graduated cylinders, thread, distilled water, two dierent
liquids.
Method:
Determine the density of the distilled water and two liquids as follows:
1. Measure and record the mass of a 10 ml graduated cyclinder.

2. Pour an amount of distilled water into the cylinder.
3. Measure and record the combined mass of the water and cylinder.
4. Record the volume of distilled water in the cylinder
5. Empty, clean and dry the graduated cylinder.
6. Repeat the above steps for the other two liquids you have.
7. Complete the table below.
Liquid Mass (g) Volume (ml) Density (g · ml−1 )

Distilled water
Liquid 1
Liquid 2
Table 2.4
2.3.1.3 Investigation : Determining the density of irregular solids:

Apparatus:
Use the same materials and equpiment as before (for the liquids). Also nd a number of solids that have
an irregular shape.
Method:
Determine the density of irregular solids as follows:
1. Measure and record the mass of one of the irregular solids.

2. Tie a piece of thread around the solid.
3. Pour some water into a 100 ml graduated cylinder and record the volume.
4. Gently lower the solid into the water, keeping hold of the thread. Record the combined volume of the
solid and the water.
5. Determine the volume of the solid by subtracting the combined volume from the original volume of
the water only.
6. Repeat these steps for the second object.
7. Complete the table below.
Solid Mass (g) Volume (ml) Density (g · ml−1 )

Solid 1
Solid 2
Solid 3
Table 2.5

2.3.2 Summary
• atom. Atoms can combine to form molecules.
The smallest unit of matter is the
• A molecule is a group of two or more atoms that are attracted to each other by chemical bonds.
• A small molecule consists of a few atoms per molecule. A giant molecule consists of millions of
atoms per molecule, for example metals and diamonds.
• The structure of a molecule can be represented in a number of ways.
• The chemical formula of a molecule is an abbreviated way of showing a molecule, using the symbols
for the elements in the molecule. There are two types of chemical formulae: molecular and empirical
formula.
• The molecular formula of a molecule gives the exact number of atoms of each element that are in
the molecule.
• The empirical formula of a molecule gives the relative number of atoms of each element in the
molecule.
• Molecules can also be represented using diagrams.
• A ball and stick diagram is a 3-dimensional molecular model that uses 'balls' to represent atoms and
'sticks' to represent the bonds between them.
• A space-lling model is also a 3-dimensional molecular model. The atoms are represented by spheres.
• In a molecule, atoms are held together by chemical bonds or intramolecular forces. Covalent
bonds, ionic bonds and metallic bonds are examples of chemical bonds.
• A covalent bond exists between non-metal atoms. An ionic bond exists between non-metal and
metal atoms and a metallic bond exists between metal atoms.
• Intermolecular forces are the bonds that hold molecules together.
• The kinetic theory of matter attempts to explain the behaviour of matter in dierent phases.
• The kinetic theory of matter says that all matter is composed of particles which have a certain amount
of energy which allows them to move at dierent speeds depending on the temperature (energy).
There are spaces between the particles and also attractive forces between particles when they come
close together.
• Understanding chemical bonds, intermolecular forces and the kinetic theory of matter can help to
explain many of the macroscopic properties of matter.
• Melting point is the temperature at which a solid changes its phase to become a liquid. The reverse
process (change in phase from liquid to solid) is called freezing. The stronger the chemical bonds and
intermolecular forces in a substance, the higher the melting point will be.
• Boiling point is the temperature at which a liquid changes phase to become a gas. The stronger the
chemical bonds and intermolecular forces in a substance, the higher the boiling point will be.
• Density is a measure of the mass of a substance per unit volume.
• Viscosity is a measure of how resistant a liquid is to owing.
2.3.2.1 Summary exercise

1. Give one word or term for each of the following descriptions.
a. The property that determines how easily a liquid ows.

b. The change in phase from liquid to gas.
c. A composition of two or more atoms that act as a unit.
d. Chemical formula that gives the relative number of atoms of each element that are in a molecule.
13
2. For each of the following questions, choose the one correct answer from the list provided.
a. Ammonia, an ingredient in household cleaners, can be broken down to form one part nitrogen (N)
and three parts hydrogen (H). This means that ammonia...
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29
a. is a colourless gas
b. is not a compound
c. cannot be an element
d. has the formula N3 H
14
b. If one substance A has a melting point that is lower than the melting point of substance B, this
suggests that...
a. A will be a liquid at room temperature.

b. The chemical bonds in substance A are weaker than those in substance B.
c. The chemical bonds in substance A are stronger than those in substance B.
d. B will be a gas at room temperature.
15
3. Boiling point is an important concept to understand.
a. Dene 'boiling point'.

b. What change in phase takes place when a liquid reaches its boiling point?
c. What is the boiling point of water?
d. Use the kinetic theory of matter and your knowledge of intermolecular forces to explain why water
changes phase at this temperature.
16
4. Refer to the table below which gives the melting and boiling points of a number of elements and then
answer the questions that follow. ( Data from http://www.chemicalelements.com)
Element Melting point Boiling point (0 C)

copper 1083 2567
magnesium 650 1107
oxygen -218.4 -183
carbon 3500 4827
helium -272 -268.6
sulphur 112.8 444.6
Table 2.6
a. What state of matter (i.e. solid, liquid or gas) will each of these elements be in at room temper-
ature?
b. Which of these elements has the strongest forces between its atoms? Give a reason for your
answer.
c. Which of these elements has the weakest forces between its atoms? Give a reason for your answer.
17
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Chapter 3
The atom
3.1 Models and atomic size 1
3.1.1
The following video covers some of the properties of an atom.
Veritasium video on the atom - 1

<http://www.youtube.com/v/SeDaOigLBTU;rel=0>
Figure 3.1
We have now looked at many examples of the types of matter and materials that exist around us and
we have investigated some of the ways that materials are classied. But what is it that makes up these
materials? And what makes one material dierent from another? In order to understand this, we need to
take a closer look at the building block of matter - the atom. Atoms are the basis of all the structures
and organisms in the universe. The planets, sun, grass, trees, air we breathe and people are all made up of
dierent combinations of atoms.
3.1.2 Models of the Atom

It is important to realise that a lot of what we know about the structure of atoms has been developed over
a long period of time. This is often how scientic knowledge develops, with one person building on the ideas
of someone else. We are going to look at how our modern understanding of the atom has evolved over time.
The idea of atoms was invented by two Greek philosophers, Democritus and Leucippus in the fth century
BC. The Greek word ατ oµoν (atom) means indivisible because they believed that atoms could not be
broken into smaller pieces.
Nowadays, we know that atoms are made up of a positively charged nucleus in the centre surrounded
by negatively charged electrons. However, in the past, before the structure of the atom was properly
understood, scientists came up with lots of dierent models or pictures to describe what atoms look like.
31
32 CHAPTER 3. THE ATOM
Denition 3.1: Model

A model is a representation of a system in the real world. Models help us to understand systems
and their properties. For example, an atomic model represents what the structure of an atom could
look like, based on what we know about how atoms behave. It is not necessarily a true picture of
the exact structure of an atom.
3.1.2.1 The Plum Pudding Model

After the electron was discovered by J.J. Thomson in 1897, people realised that atoms were made up of even
smaller particles than they had previously thought. However, the atomic nucleus had not been discovered
yet and so the 'plum pudding model' was put forward in 1904. In this model, the atom is made up of
negative electrons that oat in a soup of positive charge, much like plums in a pudding or raisins in a fruit
cake (Figure 3.2). In 1906, Thomson was awarded the Nobel Prize for his work in this eld. However, even
with the Plum Pudding Model, there was still no understanding of how these electrons in the atom were
arranged.
Figure 3.2: A schematic diagram to show what the atom looks like according to the Plum Pudding
model
The discovery of radiation was the next step along the path to building an accurate picture of atomic
structure. In the early twentieth century, Marie Curie and her husband Pierre, discovered that some elements
(the radioactive elements) emit particles, which are able to pass through matter in a similar way to X-rays
(read more about this in Grade 11). It was Ernest Rutherford who, in 1911, used this discovery to revise
the model of the atom.
3.1.2.2 Rutherford's model of the atom

Radioactive elements emit dierent types of particles. Some of these are positively charged alpha (α)
particles. Rutherford carried out a series of experiments where he bombarded sheets of gold foil with these
particles, to try to get a better understanding of where the positive charge in the atom was. A simplied
diagram of his experiment is shown in Figure 3.3.
Figure 3.3: Rutherford's gold foil experiment. Figure (a) shows the path of the α particles after they
hit the gold sheet. Figure (b) shows the arrangement of atoms in the gold sheets and the path of the α
particles in relation to this.
Rutherford set up his experiment so that a beam of alpha particles was directed at the gold sheets.
Behind the gold sheets was a screen made of zinc sulphide. This screen allowed Rutherford to see where the

33
alpha particles were landing. Rutherford knew that the electrons in the gold atoms would not really aect
the path of the alpha particles, because the mass of an electron is so much smaller than that of a proton.
He reasoned that the positively charged protons would be the ones to repel the positively charged alpha
particles and alter their path.
What he discovered was that most of the alpha particles passed through the foil undisturbed and could
be detected on the screen directly behind the foil (A). Some of the particles ended up being slightly deected
onto other parts of the screen (B). But what was even more interesting was that some of the particles were
deected straight back in the direction from where they had come (C)! These were the particles that had
been repelled by the positive protons in the gold atoms. If the Plum Pudding model of the atom were true
then Rutherford would have expected much more repulsion, since the positive charge according to that model
is distributed throughout the atom. But this was not the case. The fact that most particles passed straight
through suggested that the positive charge was concentrated in one part of the atom only.
Rutherford's work led to a change in ideas around the atom. His new model described the atom as
a tiny, dense, positively charged core called a nucleus surrounded by lighter, negatively charged electrons.
Another way of thinking about this model was that the atom was seen to be like a mini solar system where
the electrons orbit the nucleus like planets orbiting around the sun. A simplied picture of this is shown in
Figure 3.4.
Figure 3.4: Rutherford's model of the atom
3.1.2.3 The Bohr Model

There were, however, some problems with this model: for example it could not explain the very interesting
observation that atoms only emit light at certain wavelengths or frequencies. Niels Bohr solved this problem
by proposing that the electrons could only orbit the nucleus in certain special orbits at dierent energy levels
around the nucleus. The exact energies of the orbitals in each energy level depends on the type of atom.
Helium for example, has dierent energy levels to Carbon. If an electron jumps down from a higher energy
level to a lower energy level, then light is emitted from the atom. The energy of the light emitted is the
same as the gap in the energy between the two energy levels. You can read more about this in "Energy
quantisation and electron conguration". The distance between the nucleus and the electron in the lowest
energy level of a hydrogen atom is known as the Bohr radius.
note: Light has the properties of both a particle and a wave! Einstein discovered that light
comes in energy packets which are called photons. When an electron in an atom changes energy
levels, a photon of light is emitted. This photon has the same energy as the dierence between the
two electron energy levels.
3.1.3 The size of atoms

It is dicult sometimes to imagine the size of an atom, or its mass, because we cannot see an atom and also
because we are not used to working with such small measurements.

3.1.3.1 How heavy is an atom?

It is possible to determine the mass of a single atom in kilograms. But to do this, you would need very
modern mass spectrometers and the values you would get would be very clumsy and dicult to use. The
mass of a carbon atom, for example, is about 1.99 x 10−26 kg, while the mass of an atom of hydrogen is
−27
about 1.67 x 10 kg. Looking at these very small numbers makes it dicult to compare how much bigger
the mass of one atom is when compared to another.
To make the situation simpler, scientists use a dierent unit of mass when they are describing the mass
of an atom. This unit is called the atomic mass unit (amu). We can abbreviate (shorten) this unit to
just 'u'. Scientists use the carbon standard to determine amu. The carbon standard assigns carbon an
atomic mass of 12 u. Using the carbon standard the mass of an atom of hydrogen will be 1 u. You can
check this by dividing the mass of a carbon atom in kilograms (see above) by the mass of a hydrogen atom
in kilograms (you will need to use a calculator for this!). If you do this calculation, you will see that the
mass of a carbon atom is twelve times greater than the mass of a hydrogen atom. When we use atomic
mass units instead of kilograms, it becomes easier to see this. Atomic mass units are therefore not giving us
the actual mass of an atom, but rather its mass relative to the mass of one (carefully chosen) atom in the
Periodic Table. Although carbon is the usual element to compare other elements to, oxygen and hydrogen
have also been used. The important thing to remember here is that the atomic mass unit is relative to one
(carefully chosen) element. The atomic masses of some elements are shown in the table (Table 3.1) below.
Element Atomic mass (u)

Carbon (C) 12
Nitrogen (N) 14
Bromine (Br) 80
Magnesium (Mg) 24
Potassium (K) 39
Calcium (Ca) 40
Oxygen (O) 16
Table 3.1: The atomic mass number of some of the elements
The actual value of 1 atomic mass unit is 1.67 x 10−24 g or 1.67 x 10−27 kg. This is a very tiny mass!
3.1.3.2 How big is an atom?

tip: pm stands for picometres. 1 pm = 10
−12
m
Atomic diameter also varies depending on the element. On average, the diameter of an atom ranges from
100 pm (Helium) to 670 pm (Caesium). Using dierent units, 100 pm = 1 Angstrom, and 1 Angstrom =
10−10 m. That is the same as saying that 1 Angstrom = 0,0000000010 m or that 100 pm = 0,0000000010
m! In other words, the diameter of an atom ranges from 0.0000000010 m to 0.0000000067 m. This is very
small indeed.
3.2 Structure 2
3.2.1 Atomic structure

As a result of the work done by previous scientists on atomic models (that we discussed in "Models of the
Atom"), scientists now have a good idea of what an atom looks like. This knowledge is important because

35
it helps us to understand why materials have dierent properties and why some materials bond with others.
Let us now take a closer look at the microscopic structure of the atom.
So far, we have discussed that atoms are made up of a positively charged nucleus surrounded by one or
more negatively charged electrons. These electrons orbit the nucleus.
3.2.1.1 The Electron

The electron is a very light particle. It has a mass of 9.11 x 10−31 kg. Scientists believe that the electron
can be treated as a point particle elementary particle meaning that it can't be broken down into
or
anything smaller. The electron also carries one unit of negative electric charge which is the same as 1.6 x
10−19 C (Coulombs).
3.2.1.2 The Nucleus

Unlike the electron, the nucleus can be broken up into smaller building blocks called protons and neutrons.
Together, the protons and neutrons are called nucleons.
3.2.1.2.1 The Proton
Each proton carries one unit of positive electric charge. Since we know that atoms are electrically neutral,
i.e. do not carry any extra charge, then the number of protons in an atom has to be the same as the number
of electrons to balance out the positive and negative charge to zero. The total positive charge of a nucleus is
equal to the number of protons in the nucleus. The proton is much heavier than the electron (10 000 times
heavier!) and has a mass of 1.6726 x 10−27 kg. When we talk about the atomic mass of an atom, we are
mostly referring to the combined mass of the protons and neutrons, i.e. the nucleons.
3.2.1.2.2 The Neutron

The neutron is electrically neutral i.e. it carries no charge at all. Like the proton, it is much heavier than
the electron and its mass is 1.6749 x 10−27 kg (slightly heavier than the proton).
note: Rutherford predicted (in 1920) that another kind of particle must be present in the nucleus
along with the proton. He predicted this because if there were only positively charged protons in
the nucleus, then it should break into bits because of the repulsive forces between the like-charged
protons! Also, if protons were the only particles in the nucleus, then a helium nucleus (atomic
number 2) would have two protons and therefore only twice the mass of hydrogen. However, it
is actually four times heavier than hydrogen. This suggested that there must be something else
inside the nucleus as well as the protons. To make sure that the atom stays electrically neutral,
this particle would have to be neutral itself. In 1932 James Chadwick discovered the neutron and
measured its mass.
proton neutron electron

Mass (kg) 1.6726 x 10 −27
1.6749 x
−27
10 9.11 x 10−31
Units of charge +1 0 -1
Charge (C) 1.6 x

−19
10 0 -1.6 x 10−19
Table 3.2: Summary of the particles inside the atom
note: Unlike the electron which is thought to be a point particle and unable to be broken up
into smaller pieces, the proton and neutron can be divided. Protons and neutrons are built up of
smaller particles called quarks. The proton and neutron are made up of 3 quarks each.

3.2.2 Atomic number and atomic mass number

The chemical properties of an element are determined by the charge of its nucleus, i.e. by the number of
protons. This number is called the atomic number and is denoted by the letter Z.
Denition 3.2: Atomic number (Z)
The number of protons in an atom
The mass of an atom depends on how many nucleons its nucleus contains. The number of nucleons, i.e.
the total number of protons plus neutrons, is called the atomic mass number and is denoted by the letter
A.
Denition 3.3: Atomic mass number (A)
The number of protons and neutrons in the nucleus of an atom
Standard notation shows the chemical symbol, the atomic mass number and the atomic number of an
element as follows:
Figure 3.5
For example, the iron nucleus which has 26 protons and 30 neutrons, is denoted as:
56
26 Fe (3.1)
where the atomic number is Z = 26 and the mass number A = 56. The number of neutrons is simply the
dierence N = A − Z.
tip: Don't confuse the notation we have used above with the way this information appears on the
Periodic Table. On the Periodic Table, the atomic number usually appears in the top lefthand
corner of the block or immediately above the element's symbol. The number below the element's
symbol is its relative atomic mass. This is not exactly the same as the atomic mass number.
This will be explained in "Isotopes". The example of iron is shown below.
Figure 3.6
You will notice in the example of iron that the atomic mass number is more or less the same as its atomic
mass. Generally, an atom that contains n nucleons (protons and neutrons), will have a mass approximately
equal to nu. For example the mass of a C-12 atom which has 6 protons, 6 neutrons and 6 electrons is 12u,
where the protons and neutrons have about the same mass and the electron mass is negligible.
3.2.2.1 The structure of the atom

1. Explain the meaning of each of the following terms:
a. nucleus
b. electron

37
c. atomic mass
3
2. Complete the following table: (Note: You will see that the atomic masses on the Periodic Table are
not whole numbers. This will be explained later. For now, you can round o to the nearest whole
number.)
Element Atomic Atomic Number of Number of Number of

mass number protons electrons neutrons
Mg 24 12
O 8
17
Ni 28
40 20
Zn
C 12 6
Table 3.3
4
3. Use standard notation to represent the following elements:
a. potassium
b. copper
c. chlorine
5
35
4. For the element 17 Cl, give the number of ...
a. protons
b. neutrons
c. electrons
6
... in the atom. Click here for the solution
5. Which of the following atoms has 7 electrons?
5
a. 2 He
13
b. 6 C
7
c. 3 Li
15
d. 7 N
7
6. In each of the following cases, give the number or the element symbol represented by 'X'.
40
a. 18 X
x
b. 20 Ca
31
c. x P
3 See the le at <http://cnx.org/content/m39954/latest/ http://www.fhsst.org/ll0>

6 See the le at <http://cnx.org/content/m39954/latest/ http://www.fhsst.org/llX>
7 See the le at <http://cnx.org/content/m39954/latest/ http://www.fhsst.org/llk>

8
7. Complete the following table:
A Z N
235
92 U
238
92 U
Table 3.4
In these two dierent forms of Uranium...
a. What is the same ?

b. What is dierent?
Uranium can occur in dierent forms, called isotopes. You will learn more about isotopes in "Isotopes".
9
3.3 Isotopes 10
3.3.1 Isotopes
3.3.1.1 What is an isotope?
The chemical properties of an element depend on the number of protons and electrons inside the atom. So if
a neutron or two is added or removed from the nucleus, then the chemical properties will not change. This
means that such an atom would remain in the same place in the Periodic Table. For example, no matter
how many neutrons we add or subtract from a nucleus with 6 protons, that element will always be called
carbon and have the element symbol C (see the Table of Elements). Atoms which have the same number of
protons, but a dierent number of neutrons, are called isotopes.
Denition 3.4: Isotope
The isotope of a particular element is made up of atoms which have the same number of protons
as the atoms in the original element, but a dierent number of neutrons.
The dierent isotopes of an element have the same atomic number Z but dierent mass numbers A
because they have a dierent number of neutrons N. The chemical properties of the dierent isotopes of
an element are the same, but they might vary in how stable their nucleus is. Note that we can also write
elements as X-A where the X is the element symbol and the A is the atomic mass of that element. For
example, C-12 has an atomic mass of 12 and Cl-35 has an atomic mass of 35 u, while Cl-37 has an atomic
mass of 37 u.
‘ ‘
note: In Greek, same place reads as ι σoςτ o πoς (isos topos). This is why atoms which have
the same number of protons, but dierent numbers of neutrons, are called isotopes. They are in
the same place on the Periodic Table!
The following worked examples will help you to understand the concept of an isotope better.
Exercise 3.3.1: Isotopes (Solution on p. 56.)

234
For the element 92 U (uranium), use standard notation to describe:
1. the isotope with 2 fewer neutrons
8 See the le at <http://cnx.org/content/m39954/latest/ http://www.fhsst.org/llK>

9 See the le at <http://cnx.org/content/m39954/latest/ http://www.fhsst.org/llB>

39
2. the isotope with 4 more neutrons

40
Which of the following are isotopes of 20 Ca?
40
• 19 K
42
• 20 Ca
40
• 18 Ar

33
For the sulphur isotope 16 S, give the number of...
a. protons
b. nucleons
c. electrons
d. neutrons
3.3.1.1.1 Isotopes
1. Atom A has 5 protons and 5 neutrons, and atom B has 6 protons and 5 neutrons. These atoms are...
a. allotropes
b. isotopes
c. isomers
d. atoms of dierent elements
11
32 34
2. For the sulphur isotopes, 16 S and 16 S, give the number of...
a. protons
b. nucleons
c. electrons
d. neutrons
12
35
3. Which of the following are isotopes of 17 Cl?
17
a. 35 Cl
35
b. 17 Cl
37
c. 17 Cl
13
4. Which of the following are isotopes of U-235? (X represents an element symbol)
238
a. 92 X
238
b. 90 X
235
c. 92 X
14

12 See the le at <http://cnx.org/content/m39960/latest/http://www.fhsst.org/llZ>
13 See the le at <http://cnx.org/content/m39960/latest/http://www.fhsst.org/llW>
14 See the le at <http://cnx.org/content/m39960/latest/http://www.fhsst.org/llD>

3.3.1.2 Relative atomic mass

It is important to realise that the atomic mass of isotopes of the same element will be dierent because they
have a dierent number of nucleons. Chlorine, for example, has two common isotopes which are chlorine-35
and chlorine-37. Chlorine-35 has an atomic mass of 35 u, while chlorine-37 has an atomic mass of 37 u. In
the world around us, both of these isotopes occur naturally. It doesn't make sense to say that the element
chlorine has an atomic mass of 35 u, or that it has an atomic mass of 37 u. Neither of these are absolutely
true since the mass varies depending on the form in which the element occurs. We need to look at how
much more common one is than the other in order to calculate the relative atomic mass for the element
chlorine. This is the number that you nd on the Periodic Table.
Denition 3.5: Relative atomic mass

Relative atomic mass is the average mass of one atom of all the naturally occurring isotopes of a
particular chemical element, expressed in atomic mass units.
note: The relative atomic mass of some elements depends on where on Earth the element is found.
This is because the isotopes can be found in varying ratios depending on certain factors such as
geological composition, etc. The International Union of Pure and Applied Chemistry (IUPAC) has
decided to give the relative atomic mass of some elements as a range to better represent the varying
isotope ratios on the Earth. For the calculations that you will do at high school, it is enough to
simply use one number without worrying about these ranges.
Exercise 3.3.4: The relative atomic mass of an isotopic element (Solution on p. 56.)
The element chlorine has two isotopes, chlorine-35 and chlorine-37. The abundance of these isotopes
when they occur naturally is 75% chlorine-35 and 25% chlorine-37. Calculate the average relative
atomic mass for chlorine.
3.3.1.2.1 Isotopes
1. Complete the table below:
Isotope Z A Protons Neutrons Electrons

Carbon-12
Carbon-14
Chlorine-35
Chlorine-37
Table 3.5
15
2. If a sample contains 90% carbon-12 and 10% carbon-14, calculate the relative atomic mass of an atom
16
in that sample. Click here for the solution
3. If a sample contains 22.5% Cl-37 and 77.5% Cl-35, calculate the relative atomic mass of an atom in
17
that sample. Click here for the solution
15 See the le at <http://cnx.org/content/m39960/latest/http://www.fhsst.org/llj>

16 See the le at <http://cnx.org/content/m39960/latest/http://www.fhsst.org/llb>
17 See the le at <http://cnx.org/content/m39960/latest/http://www.fhsst.org/llT>

41
3.3.1.2.2 Group Discussion : The changing nature of scientic knowledge

Scientic knowledge is not static: it changes and evolves over time as scientists build on the ideas of others
to come up with revised (and often improved) theories and ideas. In this chapter for example, we saw
how peoples' understanding of atomic structure changed as more information was gathered about the atom.
There are many more examples like this one in the eld of science. For example, think about our knowledge
of the solar system and the origin of the universe, or about the particle and wave nature of light.
Often, these changes in scientic thinking can be very controversial because they disturb what people
have come to know and accept. It is important that we realise that what we know now about science may
also change. An important part of being a scientist is to be a critical thinker. This means that you need to
question information that you are given and decide whether it is accurate and whether it can be accepted
as true. At the same time, you need to learn to be open to new ideas and not to become stuck in what you
believe is right... there might just be something new waiting around the corner that you have not thought
about!
In groups of 4-5, discuss the following questions:
• Think about some other examples where scientic knowledge has changed because of new ideas and
discoveries:
· What were these new ideas?

· Were they controversial? If so, why?
· What role (if any) did technology play in developing these new ideas?
· How have these ideas aected the way we understand the world?
• Many people come up with their own ideas about how the world works. The same is true in science.
So how do we, and other scientists, know what to believe and what not to? How do we know when
new ideas are 'good' science or 'bad' science? In your groups, discuss some of the things that would
need to be done to check whether a new idea or theory was worth listening to, or whether it was not.
• Present your ideas to the rest of the class.
3.4 Energy quantisation and electron conguration 18
3.4.1 Energy quantisation and electron conguration

3.4.1.1 The energy of electrons
You will remember from our earlier discussions that an atom is made up of a central nucleus, which contains
protons and neutrons and that this nucleus is surrounded by electrons. Although these electrons all have
the same charge and the same mass, each electron in an atom has a dierent amount of energy. Electrons
that have the lowest energy are found closest to the nucleus where the attractive force of the positively
charged nucleus is the greatest. Those electrons that have higher energy, and which are able to overcome
the attractive force of the nucleus, are found further away.
3.4.1.2 Energy quantisation and line emission spectra

If the energy of an atom is increased (for example when a substance is heated), the energy of the electrons
inside the atom can be increased (when an electron has a higher energy than normal it is said to be "excited").
For the excited electron to go back to its original energy (called the ground state), it needs to release energy.
It releases energy by emitting light. If one heats up dierent elements, one will see that for each element,
light is emitted only at certain frequencies (or wavelengths). Instead of a smooth continuum of frequencies,
we see lines (called emission lines) at particular frequencies. These frequencies correspond to the energy of
the emitted light. If electrons could be excited to any energy and lose any amount of energy, there would be

a continuous spread of light frequencies emitted. However, the sharp lines we see mean that there are only
certain particular energies that an electron can be excited to, or can lose, for each element.
You can think of this like going up a ight of steps: you can't lift your foot by any amount to go from the
ground to the rst step. If you lift your foot too low you'll bump into the step and be stuck on the ground
level. You have to lift your foot just the right amount (the height of the step) to go to the next step, and
so on. The same goes for electrons and the amount of energy they can have. This is called quantisation
of energy because there are only certain quantities of energy that an electron can have in an atom. Like
steps, we can think of these quantities as energy levels in the atom. The energy of the light released when
an electron drops down from a higher energy level to a lower energy level is the same as the dierence in
energy between the two levels.
3.4.1.3 Electron conguration

We will start with a very simple view of the arrangement or conguration of electrons around an atom. This
view simply states that electrons are arranged in energy levels (or shells) around the nucleus of an atom.
These energy levels are numbered 1, 2, 3, etc. Electrons that are in the rst energy level (energy level 1) are
closest to the nucleus and will have the lowest energy. Electrons further away from the nucleus will have a
higher energy.
In the following examples, the energy levels are shown as concentric circles around the central nucleus.
The important thing to know for these diagrams is that the rst energy level can hold 2 electrons, the second
energy level can hold 8 electrons and the third energy level can hold 8 electrons.
1. Lithium Lithium (Li) has an atomic number of 3, meaning that in a neutral atom, the number of
electrons will also be 3. The rst two electrons are found in the rst energy level, while the third
electron is found in the second energy level (Figure 3.7).
Figure 3.7: The arrangement of electrons in a lithium atom.
2. Fluorine Fluorine (F) has an atomic number of 9, meaning that a neutral atom also has 9 electrons.
The rst 2 electrons are found in the rst energy level, while the other 7 are found in the second energy
level (Figure 3.8).
Figure 3.8: The arrangement of electrons in a uorine atom.
3. Argon Argon has an atomic number of 18, meaning that a neutral atom also has 18 electrons. The
rst 2 electrons are found in the rst energy level, the next 8 are found in the second energy level, and
the last 8 are found in the third energy level (Figure 3.9).
Figure 3.9: The arrangement of electrons in an argon atom.

43
But the situation is slightly more complicated than this. Within each energy level, the electrons move in
orbitals. An orbital denes the spaces or regions where electrons move.
Denition 3.6: Atomic orbital
An atomic orbital is the region in which an electron may be found around a single atom.
There are dierent orbital shapes, but we will be mainly dealing with only two. These are the 's' and 'p'
orbitals (there are also 'd' and 'f ' orbitals). The rst energy level contains only one 's' orbital, the second
energy level contains one 's' orbital and three 'p' orbitals and the third energy level contains one 's' orbital
and three 'p' orbitals (as well as 5 'd' orbitals). Within each energy level, the 's' orbital is at a lower energy
than the 'p' orbitals. This arrangement is shown in Figure 3.10.
Figure 3.10: The positions of the rst ten orbits of an atom on an energy diagram. Note that each
block is able to hold two electrons.
This diagram also helps us when we are working out the electron conguration of an element. The
electron conguration of an element is the arrangement of the electrons in the shells and subshells. There
are a few guidelines for working out the electron conguration. These are:
• Each orbital can only hold two electrons. Electrons that occur together in an orbital are called an
electron pair.
• An electron will always try to enter an orbital with the lowest possible energy.
• An electron will occupy an orbital on its own, rather than share an orbital with another electron. An
electron would also rather occupy a lower energy orbital with another electron, before occupying a
higher energy orbital. In other words, within one energy level, electrons will ll an 's' orbital before
starting to ll 'p' orbitals.
• The s subshell can hold 2 electrons
• The p subshell can hold 6 electrons
In the examples you will cover, you will mainly be lling the s and p subshells. Occasionally you may get
an example that has the d subshell. The f subshell is more complex and is not covered at this level.
The way that electrons are arranged in an atom is called its electron conguration.
Denition 3.7: Electron conguration
Electron conguration is the arrangement of electrons in an atom, molecule or other physical
structure.
An element's electron conguration can be represented using Aufbau diagrams or energy level diagrams.
An Aufbau diagram uses arrows to represent electrons. You can use the following steps to help you to draw
an Aufbau diagram:
1. Determine the number of electrons that the atom has.

2. Fill the 's' orbital in the rst energy level (the 1s orbital) with the rst two electrons.
3. Fill the 's' orbital in the second energy level (the 2s orbital) with the second two electrons.
4. Put one electron in each of the three 'p' orbitals in the second energy level (the 2p orbitals) and then
if there are still electrons remaining, go back and place a second electron in each of the 2p orbitals to
complete the electron pairs.
5. Carry on in this way through each of the successive energy levels until all the electrons have been
drawn.

tip: When there are two electrons in an orbital, the electrons are called an electron pair. If
the orbital only has one electron, this electron is said to be an unpaired electron. Electron
pairs are shown with arrows pointing in opposite directions. You may hear people talking of the
Pauli exclusion principle. This principle says that electrons have a property known as spin and two
electrons in an orbital will not spin the same way. This is why we use arrows pointing in opposite
directions. An arrow pointing up denotes an electron spinning one way and an arrow pointing
downwards denotes an electron spinning the other way.
An Aufbau diagram for the element Lithium is shown in Figure 3.11.
Figure 3.11: The electron conguration of Lithium, shown on an Aufbau diagram
A special type of notation is used to show an atom's electron conguration. The notation describes the
energy levels, orbitals and the number of electrons in each. For example, the electron conguration of lithium
2 1
is 1s 2s . The number and letter describe the energy level and orbital and the number above the orbital
shows how many electrons are in that orbital.
Aufbau diagrams for the elements uorine and argon are shown in Figure 3.12 and Figure 3.13 respectively.
2 2 5
Using standard notation, the electron conguration of uorine is 1s 2s 2p and the electron conguration
2 2 6 2 6
of argon is 1s 2s 2p 3s 3p .
Figure 3.12: An Aufbau diagram showing the electron conguration of uorine
Figure 3.13: An Aufbau diagram showing the electron conguration of argon
3.4.1.4 Core and valence electrons

Electrons in the outermost energy level of an atom are called valence electrons. The electrons that are in
the energy shells closer to the nucleus are called core electrons. Core electrons are all the electrons in an
atom, excluding the valence electrons. An element that has its valence energy level full is more stable and
less likely to react than other elements with a valence energy level that is not full.

45
Denition 3.8: Valence electrons

The electrons in the outer energy level of an atom
Denition 3.9: Core electrons

All the electrons in an atom, excluding the valence electrons
3.4.1.5 The importance of understanding electron conguration

By this stage, you may well be wondering why it is important for you to understand how electrons are
arranged around the nucleus of an atom. Remember that during chemical reactions, when atoms come into
contact with one another, it is the electrons of these atoms that will interact rst. More specically, it is the
valence electrons of the atoms that will determine how they react with one another.
To take this a step further, an atom is at its most stable (and therefore unreactive ) when all its orbitals
are full. On the other hand, an atom is least stable (and therefore most reactive ) when its valence electron
orbitals are not full. This will make more sense when we go on to look at chemical bonding in a later
chapter. To put it simply, the valence electrons are largely responsible for an element's chemical behaviour
and elements that have the same number of valence electrons often have similar chemical properties.
One nal point to note about electron congurations is stability. Which congurations are stable and
which are not? Very simply, the most stable congurations are the ones that have full energy levels. These
congurations occur in the noble gases. The noble gases are very stable elements that do not react easily (if
at all) with any other elements. This is due to the full energy levels. All elements would like to reach the
most stable electron congurations, i.e. all elements want to be noble gases.
3.4.1.5.1 Energy diagrams and electrons

1. Draw Aufbau diagrams to show the electron conguration of each of the following elements:
a. magnesium
b. potassium
c. sulphur
d. neon
e. nitrogen
2. Use the Aufbau diagrams you drew to help you complete the following table:
Element No. of energy No. of core No. of valence Electron con-

levels electrons electrons guration
(standard no-
tation)
Mg
Ne
Table 3.6
3. Rank the elements used above in order of increasing reactivity. Give reasons for the order you give.
19
Click here for the answer

3.4.1.5.2 Group work : Building a model of an atom

Earlier in this chapter, we talked about dierent 'models' of the atom. In science, one of the uses of models
is that they can help us to understand the structure of something that we can't see. In the case of the atom,
models help us to build a picture in our heads of what the atom looks like.
Models are often simplied. The small toy cars that you may have played with as a child are models.
They give you a good idea of what a real car looks like, but they are much smaller and much simpler. A
model cannot always be absolutely accurate and it is important that we realise this so that we don't build
up a false idea about something.
In groups of 4-5, you are going to build a model of an atom. Before you start, think about these questions:
• What information do I know about the structure of the atom? (e.g. what parts make it up? how big
is it?)
• What materials can I use to represent these parts of the atom as accurately as I can?
• How will I put all these dierent parts together in my model?
As a group, share your ideas and then plan how you will build your model. Once you have built your model,
discuss the following questions:
• Does our model give a good idea of what the atom actually looks like?
• In what ways is our model inaccurate ? For example, we know that electrons move around the atom's
nucleus, but in your model, it might not have been possible for you to show this.
• Are there any ways in which our model could be improved?
Now look at what other groups have done. Discuss the same questions for each of the models you see and
record your answers.
The following simulation allows you to build an atom
20
run demo
Figure 3.14: Build an atom simulation
This is another simulation that allows you to build an atom. This simulation also provides a summary
of what you have learnt so far.
21
Run demo
Figure 3.15: Build an atom simulation 2
20 http://phet.colorado.edu/sims/build-an-atom/build-an-atom_en.jnlp
21 See the le at <http://cnx.org/content/m39967/latest/http://scratch.mit.edu/projects/beckerr/867623>

47
3.5 Ionisation energy and the periodic table 22
3.5.1 Ionisation Energy and the Periodic Table

3.5.1.1 Ions
In the previous section, we focused our attention on the electron conguration of neutral atoms. In a neutral
atom, the number of protons is the same as the number of electrons. But what happens if an atom gains or
loses electrons? Does it mean that the atom will still be part of the same element?
A change in the number of electrons of an atom does not change the type of atom that it is. However,
the charge of the atom will change. If electrons are added, then the atom will become more negative. If
electrons are taken away, then the atom will become more positive. The atom that is formed in either of
ion. Put simply, an ion is a charged atom.
these cases is called an
Denition 3.10: Ion

An ion is a charged atom. A positively charged ion is called a cation +
e.g. Na , and a negatively
charged ion is called an anion −
e.g. F . The charge on an ion depends on the number of electrons
that have been lost or gained.
But how do we know how many electrons an atom will gain or lose? Remember what we said about
stability? We said that all atoms are trying to get a full outer shell. For the elements on the left hand side
of the periodic table the easiest way to do this is to lose electrons and for the elements on the right of the
periodic table the easiest way to do this is to gain electrons. So the elements on the left of the periodic table
will form cations and the elements on the right hand side of the periodic table will form anions. By doing
this the elements can be in the most stable electronic conguration and so be as stable as the noble gases.
Look at the following examples. Notice the number of valence electrons in the neutral atom, the number
of electrons that are lost or gained and the nal charge of the ion that is formed.
Lithium
A lithium atom loses one electron to form a positive ion (gure).
Figure 3.16: The arrangement of electrons in a lithium ion.
+
In this example, the lithium atom loses an electron to form the cation Li .
Fluorine
A uorine atom gains one electron to form a negative ion ().
Figure 3.17: The arrangement of electrons in a uorine ion.

You should have noticed in both these examples that each element lost or gained electrons to make a full
outer shell.
3.5.1.1.1 Investigation : The formation of ions

1. Use the diagram for lithium as a guide and draw similar diagrams to show how each of the following
ions is formed:
2+
a. Mg
+
b. Na
−
c. Cl
2−
d. O
2. Do you notice anything interesting about the charge on each of these ions? Hint: Look at the number
of valence electrons in the neutral atom and the charge on the nal ion.
Observations:
Once you have completed the activity, you should notice that:
• In each case the number of electrons that is either gained or lost, is the same as the number of electrons
that are needed for the atoms to achieve a full outer energy level.
• If you look at an energy level diagram for sodium (Na), you will see that in a neutral atom, there is
only one valence electron. In order to achieve a full outer energy level, and therefore a more stable
state for the atom, this electron will be lost.
• In the case of oxygen (O), there are six valence electrons. To achieve a full energy level, it makes more
sense for this atom to gain two electrons. A negative ion is formed.
3.5.1.2 Ionisation Energy

Ionisation energy is the energy that is needed to remove one electron from an atom. The ionisation energy
will be dierent for dierent atoms.
The second ionisation energy is the energy that is needed to remove a second electron from an atom, and
so on. As an energy level becomes more full, it becomes more and more dicult to remove an electron and
the ionisation energy increases. On the Periodic Table of the Elements, a group is a vertical column of the
elements, and a period is a horizontal row. In the periodic table, ionisation energy increases across a period,
but decreases as you move down a group. The lower the ionisation energy, the more reactive the element
will be because there is a greater chance of electrons being involved in chemical reactions. We will look at
this in more detail in the next section.
3.5.1.2.1 The formation of ions

Match the information in column A with the information in column B by writing only the letter (A to I)
next to the question number (1 to 7)

49
2+
1. A positive ion that has 3 less electrons than its neutral atom A. Mg
−
2. An ion that has 1 more electron than its neutral atom B. Cl
2−
3. The anion that is formed when bromine gains an electron C. CO3
3+
4. The cation that is formed from a magnesium atom D. Al
2−
5. An example of a compound ion E. Br
+
6. A positive ion with the electron conguration of argon F. K
+
7. A negative ion with the electron conguration of neon G. Mg
2−
H. O
−
I. Br
Table 3.7
23
3.5.2 The arrangement of atoms in the periodic table

<http://static.slidesharecdn.com/swf/ssplayer2.swf ?doc=periodictableintroduction-100510044643-
phpapp01&stripped_title=periodic-table-introduction>
Figure 3.18
The periodic table of the elements is a method of showing the chemical elements in a table. Most of the
work that was done to arrive at the periodic table that we know, can be attributed to a man called Dmitri
Mendeleev in 1869. Mendeleev was a Russian chemist who designed the table in such a way that recurring
("periodic") trends in the properties of the elements could be shown. Using the trends he observed, he even
left gaps for those elements that he thought were 'missing'. He even predicted the properties that he thought
the missing elements would have when they were discovered. Many of these elements were indeed discovered
and Mendeleev's predictions were proved to be correct.
To show the recurring properties that he had observed, Mendeleev began new rows in his table so that
elements with similar properties were in the same vertical columns, called groups. Each row was referred to
as a period. One important feature to note in the periodic table is that all the non-metals are to the right
of the zig-zag line drawn under the element boron. The rest of the elements are metals, with the exception
of hydrogen which occurs in the rst block of the table despite being a non-metal.
Figure 3.19: A simplied diagram showing part of the Periodic Table
23 See the le at <http://cnx.org/content/m39969/latest/http://www.fhsst.org/lid>

3.5.2.1 Groups in the periodic table

A group is a vertical column in the periodic table and is considered to be the most important way of
classifying the elements. If you look at a periodic table, you will see the groups numbered at the top of each
column. The groups are numbered from left to right as follows: 1, 2, then an open space which contains the
transition elements, followed by groups 3 to 8. These numbers are normally represented using Roman
numerals. In some periodic tables, all the groups are numbered from left to right from number 1 to number
18. In some groups, the elements display very similar chemical properties and the groups are even given
separate names to identify them.
The characteristics of each group are mostly determined by the electron conguration of the atoms of
the element.
• Group 1: These elements are known as the alkali metals and they are very reactive.
Figure 3.20: Electron diagrams for some of the Group 1 elements, with sodium and potasium incom-
plete; to be completed as an excersise.
• Group 2: These elements are known as the alkali earth metals. Each element only has two valence
electrons and so in chemical reactions, the group 2 elements tend to lose these electrons so that the
energy shells are complete. These elements are less reactive than those in group 1 because it is more
dicult to lose two electrons than it is to lose one.
• Group 3 elements have three valence electrons.
tip: The number of valence electrons of an element corresponds to its group number on the
periodic table.
• Group 7: These elements are known as the halogens. Each element is missing just one electron from
its outer energy shell. These elements tend to gain electrons to ll this shell, rather than losing them.
These elements are also very reactive.
• Group 8: These elements are the noble gases. All of the energy shells of the halogens are full and so
these elements are very unreactive.
Figure 3.21: Electron diagrams for two of the noble gases, helium (He) and neon (Ne).
• Transition metals: The dierences between groups in the transition metals are not usually dramatic.
3.5.2.1.1 Investigation : The properties of elements

Refer to Figure 3.20.
1. Use a periodic table to help you to complete the last two diagrams for sodium (Na) and potassium
(K).
2. What do you notice about the number of electrons in the valence energy level in each case?

51
3. Explain why elements from group 1 are more reactive than elements from group 2 on the periodic table
(Hint: Think back to 'ionisation energy').
It is worth noting that in each of the groups described above, the atomic diameter of the elements increases
as you move down the group. This is because, while the number of valence electrons is the same in each
element, the number of core electrons increases as one moves down the group.
3.5.2.2 Periods in the periodic table

A period is a horizontal row in the periodic table of the elements. Some of the trends that can be observed
within a period are highlighted below:
• As you move from one group to the next within a period, the number of valence electrons increases by
one each time.
• Within a single period, all the valence electrons occur in the same energy shell. If the period increases,
so does the energy shell in which the valence electrons occur.
• In general, the diameter of atoms decreases as one moves from left to right across a period. Consider
the attractive force between the positively charged nucleus and the negatively charged electrons in an
atom. As you move across a period, the number of protons in each atom increases. The number of
electrons also increases, but these electrons will still be in the same energy shell. As the number of
protons increases, the force of attraction between the nucleus and the electrons will increase and the
atomic diameter will decrease.
• Ionisation energy increases as one moves from left to right across a period. As the valence electron shell
moves closer to being full, it becomes more dicult to remove electrons. The opposite is true when
you move down a group in the table because more energy shells are being added. The electrons that
are closer to the nucleus 'shield' the outer electrons from the attractive force of the positive nucleus.
Because these electrons are not being held to the nucleus as strongly, it is easier for them to be removed
and the ionisation energy decreases.
• In general, the reactivity of the elements decreases from left to right across a period.
You may see periodic tables labeled with s-block, p-block, d-block and f-block. This is simply another way
to group the elements. When we group elements like this we are simply noting which orbitals are being lled
in each block. This method of grouping is not very useful to the work covered at this level.
3.5.2.2.1 Trends in ionisation energy

−1
Refer to the data table below which gives the ionisation energy (in kJ·mol ) and atomic number (Z) for a
number of elements in the periodic table:
Z Ionisation energy Z Ionisation energy
1 1310 10 2072
2 2360 11 494
3 517 12 734
4 895 13 575
5 797 14 783
6 1087 15 1051
7 1397 16 994
8 1307 17 1250
9 1673 18 1540

Table 3.8
1. Draw a line graph to show the relationship between atomic number (on the x-axis) and ionisation
energy (y-axis).
2. Describe any trends that you observe.
3. Explain why...
a. the ionisation energy for Z=2 is higher than for Z=1

b. the ionisation energy for Z=3 is lower than for Z=2
c. the ionisation energy increases between Z=5 and Z=7
24
3.5.2.2.2 Elements in the periodic table

Refer to the elements listed below:
Lithium (Li); Chlorine (Cl); Magnesium (Mg); Neon (Ne); Oxygen (O); Calcium (Ca); Carbon (C)
Which of the elements listed above:
1. belongs to Group 1
2. is a halogen
3. is a noble gas
4. is an alkali metal
5. has an atomic number of 12
6. has 4 neutrons in the nucleus of its atoms
7. contains electrons in the 4th energy level
8. has only one valence electron
9. has all its energy orbitals full
10. will have chemical properties that are most similar
11. will form positive ions
25
3.5.3 Summary
• Much of what we know today about the atom, has been the result of the work of a number of scientists
who have added to each other's work to give us a good understanding of atomic structure.
• Some of the important scientic contributors include J.J.Thomson (discovery of the electron, which
led to the Plum Pudding Model of the atom), Ernest Rutherford (discovery that positive charge
is concentrated in the centre of the atom) and Niels Bohr (the arrangement of electrons around the
nucleus in energy levels).
• Because of the very small mass of atoms, their mass is measured in atomic mass units (u). 1 u =
−24
1,67 × 10 g.
• nucleus (containing protons and neutrons), surrounded by elec-
An atom is made up of a central
trons.
• The atomic number (Z) is the number of protons in an atom.
• The atomic mass number (A) is the number of protons and neutrons in the nucleus of an atom.
The standard notation that is used to write an element, is Z X, where X is the element symbol, A is
A
•
the atomic mass number and Z is the atomic number.
24 See the le at <http://cnx.org/content/m39969/latest/http://www.fhsst.org/liv>
25 See the le at <http://cnx.org/content/m39969/latest/http://www.fhsst.org/liw>

53
• The isotope of a particular element is made up of atoms which have the same number of protons as
the atoms in the original element, but a dierent number of neutrons. This means that not all atoms
of an element will have the same atomic mass.
• The relative atomic mass of an element is the average mass of one atom of all the naturally occurring
isotopes of a particular chemical element, expressed in atomic mass units. The relative atomic mass is
written under the elements' symbol on the Periodic Table.
• The energy of electrons in an atom is quantised. Electrons occur in specic energy levels around an
atom's nucleus.
• Within each energy level, an electron may move within a particular shape of orbital. An orbital
denes the space in which an electron is most likely to be found. There are dierent orbital shapes,
including s, p, d and f orbitals.
• Energy diagrams such as Aufbau diagrams are used to show the electron conguration of atoms.
• The electrons in the outermost energy level are called valence electrons.
• The electrons that are not valence electrons are called core electrons.
• Atoms whose outermost energy level is full, are less chemically reactive and therefore more stable, than
those atoms whose outer energy level is not full.
• An ion is a charged atom. A cation is a positively charged ion and an anion is a negatively charged
ion.
• When forming an ion, an atom will lose or gain the number of electrons that will make its valence
energy level full.
• An element's ionisation energy is the energy that is needed to remove one electron from an atom.
• period in the periodic table.
Ionisation energy increases across a
• Ionisation energy decreases down a group in the periodic table.

<http://static.slidesharecdn.com/swf/ssplayer2.swf ?doc=atomictheoryrevision-100512033251-
phpapp01&stripped_title=atomic-theory-revision>
Figure 3.22
3.5.3.1 Summary
1. Write down only the word/term for each of the following descriptions.
a. The sum of the number of protons and neutrons in an atom

b. The dened space around an atom's nucleus, where an electron is most likely to be found
26
2. For each of the following, say whether the statement is True or False. If it is False, re-write the
statement correctly.
20 22
a. 10 Ne and 10 Ne each have 10 protons, 12 electrons and 12 neutrons.
b. The atomic mass of any atom of a particular element is always the same.
c. It is safer to use helium gas rather than hydrogen gas in balloons.
d. Group 1 elements readily form negative ions.
27
26 See the le at <http://cnx.org/content/m39969/latest/http://www.fhsst.org/lif>
27 See the le at <http://cnx.org/content/m39969/latest/http://www.fhsst.org/liG>

3. Multiple choice questions: In each of the following, choose the one correct answer.
a. The three basic components of an atom are:
a. protons, neutrons, and ions

b. protons, neutrons, and electrons
c. protons, neutrinos, and ions
d. protium, deuterium, and tritium
28
b. The charge of an atom is...
a. positive
b. neutral
c. negative
29
c. If Rutherford had used neutrons instead of alpha particles in his scattering experiment, the neu-
trons would...
a. not deect because they have no charge

b. have deected more often
c. have been attracted to the nucleus easily
d. have given the same results
30
234
d. Consider the isotope 92 U. Which of the following statements is true ?
234
a. The element is an isotope of 94 Pu
b. The element contains 234 neutrons
238
c. The element has the same electron conguration as 92 U
d. The element has an atomic mass number of 92
31
e. The electron conguration of an atom of chlorine can be represented using the following notation:
2 8 7
a. 1s 2s 3s
2 2 6 2 5
b. 1s 2s 2p 3s 3p
2 2 6 2 6
c. 1s 2s 2p 3s 3p
2 2 5
d. 1s 2s 2p
32
4. The following table shows the rst ionisation energies for the elements of period 1 and
2.

29 See the le at <http://cnx.org/content/m39969/latest/http://www.fhsst.org/liA>
30 See the le at <http://cnx.org/content/m39969/latest/http://www.fhsst.org/lio>
31 See the le at <http://cnx.org/content/m39969/latest/http://www.fhsst.org/lis>
32 See the le at <http://cnx.org/content/m39969/latest/http://www.fhsst.org/liH>

55
First ionisation energy (kJ.mol

−1
Period Element )
1 H 1312
He 2372
Li 520
Be 899
B 801
C 1086
2 N 1402
O 1314
F 1681
Ne 2081
Table 3.9
a. What is the meaning of the term rst ionisation energy ?

b. Identify the pattern of rst ionisation energies in a period.
c. Which TWO elements exert the strongest attractive forces on their electrons? Use the data in
the table to give a reason for your answer.
d. Draw Aufbau diagrams for the TWO elements you listed in the previous question and explain
why these elements are so stable.
e. It is safer to use helium gas than hydrogen gas in balloons. Which property of helium makes it a
safer option?
f. 'Group 1 elements readily form positive ions'. Is this statement correct? Explain your answer by
referring to the table.
33

Solutions to Exercises in Chapter 3

Solution to Exercise 3.3.1 (p. 38)
Step 1. We know that isotopes of any element have the same number of protons (same atomic number) in
each atom, which means that they have the same chemical symbol. However, they have a dierent
number of neutrons, and therefore a dierent mass number.
Step 2. Therefore, any isotope of uranium will have the symbol:
U (3.2)
Also, since the number of protons in uranium isotopes is always the same, we can write down the
atomic number:
92 U (3.3)
234
Now, if the isotope we want has 2 fewer neutrons than 92 U, then we take the original mass number
and subtract 2, which gives:
232
92 U (3.4)
Following the steps above, we can write the isotope with 4 more neutrons as:
238
92 U (3.5)

Step 1. We know that isotopes have the same atomic number but dierent mass numbers.
Step 2. You need to look for the element that has the same atomic number but a dierent atomic mass number.
42
The only element is 20 Ca. What is dierent is that there are 2 more neutrons than in the original
element.

Step 1. Z = 16, therefore the number of protons is 16 (answer to (a)).
Step 2. A = 33, therefore the number of nucleons is 33 (answer to (b)).
Step 3. The atom is neutral, and therefore the number of electrons is the same as the number of protons. The
number of electrons is 16 (answer to (c)).
Step 4.
N = A − Z = 33 − 16 = 17 (3.6)
The number of neutrons is 17 (answer to (d)).

Step 1. Contribution of Cl-35 = (75/100 x 35) = 26,25 u
Step 2. Contribution of Cl-37 = (25/100 x 37) = 9,25 u
Step 3. Relative atomic mass of chlorine = 26,25 u + 9,25 u = 35,5 u
If you look on the periodic table, the average relative atomic mass for chlorine is 35,5 u. You will
notice that for many elements, the relative atomic mass that is shown is not a whole number. You
should now understand that this number is the average relative atomic mass for those elements that
have naturally occurring isotopes.

Chapter 4
Physical and chemical change
4.1 Introduction 1
4.1.1 Physical and Chemical Change - Grade 10

Matter is all around us. The desks we sit at, the air we breathe and the water we drink are all examples of
matter. But matter doesn't always stay the same. It can change in many dierent ways. In this chapter, we
are going to take a closer look at physical and chemical changes that occur in matter.
4.1.2 Physical changes in matter

A physical change is one where the particles of the substances that are involved in the change are not
broken up in any way. When water is heated for example, the temperature and energy of the water molecules
increases and the liquid water evaporates to form water vapour. When this happens, some kind of change
has taken place, but the molecular structure of the water has not changed. This is an example of a physical
change.
H2 O (l)→ H2 O (g)
Conduction (the transfer of energy through a material) is another example of a physical change. As
energy is transferred from one material to another, the energy of each material is changed, but not its
chemical makeup. Dissolving one substance in another is also a physical change.
Denition 4.1: Physical change

A change that can be seen or felt, but that doesn't involve the break up of the particles in the
reaction. During a physical change, the form of matter may change, but not its identity. A change
in temperature is an example of a physical change.
You can think of a physical change as a person who is standing still. When they start to move (start
walking) then a change has occurred and this is similar to a physical change.
There are some important things to remember about physical changes in matter:
1. Arrangement of particles
When a physical change occurs, the particles (e.g. atoms, molecules) may re-arrange themselves
without actually breaking up in any way. In the example of evaporation that we used earlier, the water
molecules move further apart as their temperature (and therefore energy) increases. The same would
be true if ice were to melt. In the solid phase, water molecules are packed close together in a very
ordered way, but when the ice is heated, the molecules overcome the forces holding them together and
they move apart. Once again, the particles have re-arranged themselves, but have not broken up.
H2 O (s) →H2 O(l)
57
58 CHAPTER 4. PHYSICAL AND CHEMICAL CHANGE
Figure 4.1 shows this more clearly. In each phase of water, the water molecule itself stays the same,
but the way the molecules are arranged has changed. Note that in the solid phase, we simply show
the water molecules as spheres. This makes it easier to see how tightly packed the molecules are. In
reality the water molecules would all look the same.
Figure 4.1: The arrangement of water molecules in the three phases of matter
In a physical change, the total mass, the number of atoms and the number of molecules will always
stay the same.
2. Energy changes
Energy changes may take place when there is a physical change in matter, but these energy changes
are normally smaller than the energy changes that take place during a chemical change.
3. Reversibility
Physical changes in matter are usually easier to reverse than chemical changes. Water vapour for
example, can be changed back to liquid water if the temperature is lowered. Liquid water can be
changed into ice by simply decreasing the temperature.
4.1.3 Chemical Changes in Matter

When a chemical change takes place, new substances are formed in a chemical reaction. These new
products may have very dierent properties from the substances that were there at the start of the reaction.
The breakdown of copper(II) chloride to form copper and chlorine is an example of chemical change. A
simplied diagram of this reaction is shown in Figure 4.2. In this reaction, the initial substance is copper(II)
chloride, but once the reaction is complete, the products are copper and chlorine.
The decomposition of copper(II) chloride to form copper and chlorine. We write this as:
Figure 4.2:
CuCl2 →Cu+Cl2
Denition 4.2: Chemical change

The formation of new substances in a chemical reaction. One type of matter is changed into
something dierent.
There are some important things to remember about chemical changes:
1. Arrangement of particles
During a chemical change, the particles themselves are changed in some way. In the example of copper
(II) chloride that was used earlier, the CuCl2 molecules were split up into their component atoms. The

59
number of particles will change because each CuCl2 molecule breaks down into one copper atom (Cu)
and one chlorine molecule (Cl2 ). However, what you should have noticed, is that the number of atoms
of each element stays the same, as does the total mass of the atoms. This will be discussed in more
detail in a later section.
2. Energy changes
The energy changes that take place during a chemical reaction are much greater than those that take
place during a physical change in matter. During a chemical reaction, energy is used up in order to
break bonds, and then energy is released when the new product is formed. This will be discussed in
more detail in "Energy changes in chemical reactions".
3. Reversibility
Chemical changes are far more dicult to reverse than physical changes.
We will consider two types of chemical reactions: decomposition reactions and synthesis reactions.
4.1.3.1 Decomposition reactions
A decomposition reaction occurs when a chemical compound is broken down into elements or smaller
compounds. The generalised equation for a decomposition reaction is:
AB → A + B
One example of such a reaction is the decomposition of hydrogen peroxide (Figure 4.3) to form hydrogen
and oxygen according to the following equation:
2H2 O2 → 2H2 O+ O2
Figure 4.3: The decomposition of H2 O2 to form H2 O and O2
The decomposition of mercury (II) oxide is another example.
4.1.3.1.1 Experiment : The decomposition of mercury (II) oxide

Aim
To observe the decomposition of mercury (II) oxide when it is heated.
Note: Because this experiment involves mercury, which is a poisonous substance, it should be done in a
fume cupboard, and all the products of the reaction must be very carefully disposed of.
Apparatus
Mercury (II) oxide (an orange-red product); two test tubes; a large beaker; stopper and delivery tube;
Bunsen burner; wooden splinter.
Figure 4.4
Method
1. Put a small amount of mercury (II) oxide in a test tube and heat it gently over a Bunsen burner. Then
allow it to cool. What do you notice about the colour of the mercury (II) oxide?
2. Heat the test tube again and note what happens. Do you notice anything on the walls of the test tube?
Record these observations.
3. Test for the presence of oxygen by holding a glowing splinter in the mouth of the test tube.
Results
1. During the rst heating of mercury (II) oxide, the only change that took place was a change in colour
from orange-red to black and then back to its original colour.
2. When the test tube was heated again, deposits of mercury formed on the inner surface of the test tube.
What colour is this mercury?
3. The glowing splinter burst into ame when it was placed in the test tube, meaning that oxygen is
present.
Conclusions
When mercury oxide is heated, it decomposes to form mercury and oxygen. The chemical decomposition
reaction that takes place can be written as follows:
2HgO → 2Hg + O2
4.1.3.2 Synthesis reactions

During a synthesis reaction, a new product is formed from elements or smaller compounds. The generalised
equation for a synthesis reaction is as follows:
A + B → AB
One example of a synthesis reaction is the burning of magnesium in oxygen to form magnesium ox-
ide(Figure 4.5). The equation for the reaction is:
2Mg + O2 → 2MgO
Figure 4.5: The synthesis of magnesium oxide (MgO) from magnesium and oxygen
4.1.3.2.1 Experiment: Chemical reactions involving iron and sulphur

Aim
To demonstrate the synthesis of iron sulphide from iron and sulphur.
Apparatus
5,6 g iron lings and 3,2 g powdered sulphur; porcelain dish; test tube; Bunsen burner
Figure 4.6
Method
61
1. Measure the quantity of iron and sulphur that you need and mix them in a porcelain dish.
2. Take some of this mixture and place it in the test tube. The test tube should be about 1/3 full.
3. This reaction should ideally take place in a fume cupboard. Heat the test tube containing the mixture
over the Bunsen burner. Increase the heat if no reaction takes place. Once the reaction begins, you
will need to remove the test tube from the ame. Record your observations.
4. Wait for the product to cool before breaking the test tube with a hammer. Make sure that the test
tube is rolled in paper before you do this, otherwise the glass will shatter everywhere and you may be
hurt.
5. What does the product look like? Does it look anything like the original reactants? Does it have any
of the properties of the reactants (e.g. the magnetism of iron)?
Results
1. After you removed the test tube from the ame, the mixture glowed a bright red colour. The reaction
is exothermic and produces energy.
2. The product, iron sulphide, is a dark colour and does not share any of the properties of the original
reactants. It is an entirely new product.
Conclusions
A synthesis reaction has taken place. The equation for the reaction is:
Fe + S → FeS
4.1.3.2.2 Investigation : Physical or chemical change?

Apparatus:
Bunsen burner, 4 test tubes, a test tube rack and a test tube holder, small spatula, pipette, magnet, a
birthday candle, NaCl (table salt), 0,1M AgNO3 , 6M HCl, magnesium ribbon, iron lings, sulphur.
Note: AgNO 3 stains the skin. Be careful when working with it or use gloves.
Method:
1. Place a small amount of wax from a birthday candle into a test tube and heat it over the bunsen burner
until it melts. Leave it to cool.
2. Add a small spatula of NaCl to 5 ml water in a test tube and shake. Then use the pipette to add 10
drops of AgNO3 to the sodium chloride solution. NOTE: Please be careful AgNO3 causes bad stains!!
3. Take a 5 cm piece of magnesium ribbon and tear it into 1 cm pieces. Place two of these pieces into a
test tube and add a few drops of 6M HCl. NOTE: Be very careful when you handle this acid because
it can cause major burns.
4. Take about 0,5 g iron lings and 0,5 g sulphur. Test each substance with a magnet. Mix the two
samples in a test tube and run a magnet alongside the outside of the test tube.
5. Now heat the test tube that contains the iron and sulphur. What changes do you see? What happens
now, if you run a magnet along the outside of the test tube?
6. In each of the above cases, record your observations.
Questions:
Decide whether each of the following changes are physical or chemical and give a reason for your answer
in each case. Record your answers in the table below:

Description Physical or chemical change Reason

melting candle wax
dissolving NaCl
mixing NaCl with AgNO3
tearing magnesium ribbon
adding HCl to magnesium ribbon
mixing iron and sulphur
heating iron and sulphur
Table 4.1
4.2 Energy changes and conservation of matter 2
4.2.1 Energy changes in chemical reactions

All reactions involve some change in energy. During a physical change in matter, such as the evaporation of
liquid water to water vapour, the energy of the water molecules increases. However, the change in energy is
much smaller than in chemical reactions.
When a chemical reaction occurs, some bonds will break, while new bonds may form. Energy changes
in chemical reactions result from the breaking and forming of bonds. For bonds to break, energy must be
absorbed. When new bonds form, energy will be released because the new product has a lower energy than
the ìn between' stage of the reaction when the bonds in the reactants have just been broken.
In some reactions, the energy that must be absorbed to break the bonds in the reactants is less than the
total energy that is released when new bonds are formed. This means that in the overall reaction, energy
is released. This type of reaction is known as an exothermic reaction. In other reactions, the energy
that must be absorbed to break the bonds in the reactants is more than the total energy that is released
when new bonds are formed. This means that in the overall reaction, energy must be absorbed from the
surroundings. This type of reaction is known as an endothermic reaction. Most decomposition reactions
are endothermic and heating is needed for the reaction to occur. Most synthesis reactions are exothermic,
meaning that energy is given o in the form of heat or light.
More simply, we can describe the energy changes that take place during a chemical reaction as:
Total energy absorbed to break bonds - Total energy released when new bonds form
So, for example, in the reaction...
2Mg+O2 → 2MgO
Energy is needed to break the O-O bonds in the oxygen molecule so that new Mg-O bonds can be formed,
and energy is released when the product (MgO) forms.
Despite all the energy changes that seem to take place during reactions, it is important to remember that
energy cannot be created or destroyed. Energy that enters a system will have come from the surrounding
environment and energy that leaves a system will again become part of that environment. This is known as
the conservation of energy principle.
Denition 4.3: Conservation of energy principle
Energy cannot be created or destroyed. It can only be changed from one form to another.
Chemical reactions may produce some very visible and often violent changes. An explosion, for example,
is a sudden increase in volume and release of energy when high temperatures are generated and gases are
released. For example, NH4 NO3 can be heated to generate nitrous oxide. Under these conditions, it is
highly sensitive and can detonate easily in an explosive exothermic reaction.

63
4.2.2 Conservation of atoms and mass in reactions

The total mass of all the substances taking part in a chemical reaction is conserved during a chemical
reaction. This is known as the law of conservation of mass. The total number of atoms of each element
also remains the same during a reaction, although these may be arranged dierently in the products.
We will use two of our earlier examples of chemical reactions to demonstrate this:
1. The decomposition of hydrogen peroxide into water and oxygen
2H2 O2 → 2H2 O + O2
Figure 4.7
Left hand side of the equation

Total atomic mass = (4 × 1) + (4 × 16) = 68 u
Number of atoms of each element = (4 × H) + (4 × O)
Right hand side of the equation
Total atomic mass = (4 × 1) + (2 × 16) + (2 × 16) = 68 u
Number of atoms of each element = (4 × H) + (4 × O)
Both the atomic mass and the number of atoms of each element are conserved in the reaction.
2. The synthesis of magnesium and oxygen to form magnesium oxide 2Mg+O2 →2MgO
Figure 4.8
Left hand side of the equation

Total atomic mass = (2 × 24,3) + (2 × 16) = 80,6 u
Number of atoms of each element = (2 × Mg) + (2 × O)
Right hand side of the equation
Total atomic mass = (2 × 24,3) + (2 × 16) = 80,6 u
Number of atoms of each element = (2 × Mg) + (2 × O)
Both the atomic mass and the number of atoms of each element are conserved in the reaction.
4.2.2.1 Demonstration : The conservation of atoms in chemical reactions

Materials:
1. Coloured marbles or small balls to represent atoms. Each colour will represent a dierent element.
2. Prestik
Method:
1. Choose a reaction from any that have been used in this chapter or any other balanced chemical reaction
that you can think of. To help to explain this activity, we will use the decomposition reaction of calcium
carbonate to produce carbon dioxide and calcium oxide. CaCO3 →CO2 + CaO

2. Stick marbles together to represent the reactants and put these on one side of your table. In this
example you may for example join one red marble (calcium), one green marble (carbon) and three
yellow marbles (oxygen) together to form the molecule calcium carbonate (CaCO3 ).
3. Leaving your reactants on the table, use marbles to make the product molecules and place these on
the other side of the table.
4. Now count the number of atoms on each side of the table. What do you notice?
5. Observe whether there is any dierence between the molecules in the reactants and the molecules in
the products.
Discussion
You should have noticed that the number of atoms in the reactants is the same as the number of atoms
in the product. The number of atoms is conserved during the reaction. However, you will also see that the
molecules in the reactants and products is not the same. The arrangement of atoms is not conserved during
the reaction.
4.2.3 Law of constant composition

In any given chemical compound, the elements always combine in the same proportion with each other. This
is the law of constant proportion.
The law of constant composition says that, in any particular chemical compound, all samples of that
compound will be made up of the same elements in the same proportion or ratio. For example, any water
molecule is always made up of two hydrogen atoms and one oxygen atom in a 2:1 ratio. If we look at the
relative masses of oxygen and hydrogen in a water molecule, we see that 94% of the mass of a water molecule
is accounted for by oxygen and the remaining 6% is the mass of hydrogen. This mass proportion will be the
same for any water molecule.
This does not mean that hydrogen and oxygen always combine in a 2:1 ratio to form H2 O. Multiple
proportions are possible. For example, hydrogen and oxygen may combine in dierent proportions to form
H2 O2 rather than H2 O. In H2 O2 , the H:O ratio is 1:1 and the mass ratio of hydrogen to oxygen is 1:16. This
will be the same for any molecule of hydrogen peroxide.
4.2.4 Volume relationships in gases

In a chemical reaction between gases, the relative volumes of the gases in the reaction are present in a ratio
of small whole numbers if all the gases are at the same temperature and pressure. This relationship is also
known as Gay-Lussac's Law.
For example, in the reaction between hydrogen and oxygen to produce water, two volumes of H2 react
with 1 volume of O2 to produce 2 volumes of H2 O.
2H2 + O2 →2H2 O
In the reaction to produce ammonia, one volume of nitrogen gas reacts with three volumes of hydrogen
gas to produce two volumes of ammonia gas.
N2 +3H2 → 2NH3
This relationship will also be true for all other chemical reactions.
4.2.5 Summary
The following video provides a summary of the concepts covered in this chapter.

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Physical and chemical change

<http://www.youtube.com/v/QL7V3L3dfDM?fs=1&hl=en_US>
Figure 4.9
1. Matter does not stay the same. It may undergo physical or chemical changes.
2. A physical change means that the form of matter may change, but not its identity. For example,
when water evaporates, the energy and the arrangement of water molecules will change, but not the
structure of the water molecules themselves.
3. During a physical change, the arrangement of particles may change but the mass, number of atoms
and number of molecules will stay the same.
4. Physical changes involve small changes in energy and are easily reversible.
5. A chemical change occurs when one or more substances change into other materials. A chemical reaction
involves the formation of new substances with dierent properties. For example, magnesium and
oxygen react to form magnesium oxide (MgO)
6. A chemical change may involve a decomposition or synthesis reaction. During chemical change, the
mass and number of atoms is conserved, but the number of molecules is not always the same.
7. Chemical reactions involve larger changes in energy. During a reaction, energy is needed to break bonds
in the reactants and energy is released when new products form. If the energy released is greater than
the energy absorbed, then the reaction is exothermic. If the energy released is less than the energy
absorbed, then the reaction is endothermic. Chemical reactions are not easily reversible.
endothermic and synthesis reactions are usually exothermic.
8. Decomposition reactions are usually
9. The law of conservation of mass states that the total mass of all the substances taking part in
a chemical reaction is conserved and the number of atoms of each element in the reaction does not
change when a new product is formed.
10. The conservation of energy principle states that energy cannot be created or destroyed, it can
only change from one form to another.
11. The law of constant composition states that in any particular compound, all samples of that
compound will be made up of the same elements in the same proportion or ratio.
12. Gay-Lussac's Law states that in a chemical reaction between gases, the relative volumes of the gases
in the reaction are present in a ratio of small whole numbers if all the gases are at the same temperature
and pressure.

1. Complete the following table by saying whether each of the descriptions is an example of a physical or
chemical change:

Description Physical or chemical

hot and cold water mix together
milk turns sour
a car starts to rust
food digests in the stomach
alcohol disappears when it is placed on your skin
warming food in a microwave
separating sand and gravel
reworks exploding
Table 4.2
3
2. For each of the following reactions, say whether it is an example of a synthesis or decomposition
reaction:
a. (NH4 )2 CO3 →2NH3 + CO2 + H2 O

b. 4Fe + 3O2 →2Fe2 O3
c. N2 (g) + 3H2 (g)→2NH3
d. CaCO3 (s)→CaO + CO2
4
3. For the following equation: CaCO3 → CO2 + CaO show that the 'law of conservation of mass' applies.
5
3 See the le at <http://cnx.org/content/m39985/latest/http://www.fhsst.org/l3q>

4 See the le at <http://cnx.org/content/m39985/latest/http://www.fhsst.org/l3l>
5 See the le at <http://cnx.org/content/m39985/latest/http://www.fhsst.org/l3i>

Chapter 5
Representing chemical change
5.1 Introduction, chemical symbols and chemical formulae 1
5.1.1 Introduction
As we have already mentioned, a number of changes can occur when elements react with one another. These
changes may either be physical or chemical. One way of representing these changes is through balanced
chemical equations. A chemical equation describes a chemical reaction by using symbols for the elements
involved. For example, if we look at the reaction between iron (Fe) and sulphur (S) to form iron sulphide
(FeS), we could represent these changes either in words or using chemical symbols:
iron + sulphur → iron sulphide
or
Fe+S → FeS
Another example would be:
ammonia + oxygen → nitric oxide + water
or
4NH3 + 5O2 → 4NO + 6H2 O
Compounds on the left of the arrow are called the reactants and these are needed for the reaction to
take place. In this equation, the reactants are ammonia and oxygen. The compounds on the right are called
the products and these are what is formed from the reaction.
In order to be able to write a balanced chemical equation, there are a number of important things that
need to be done:
1. Know the chemical symbols for the elements involved in the reaction
2. Be able to write the chemical formulae for dierent reactants and products
3. Balance chemical equations by understanding the laws that govern chemical change
4. Know the state symbols for the equation
We will look at each of these steps separately in the next sections.
5.1.2 Chemical symbols

It is very important to know the chemical symbols for common elements in the Periodic Table, so that you
are able to write chemical equations and to recognise dierent compounds.
67
68 CHAPTER 5. REPRESENTING CHEMICAL CHANGE
5.1.2.1 Revising common chemical symbols

• Write down the chemical symbols and names of all the elements that you know.
• Compare your list with another learner and add any symbols and names that you don't have.
• Spend some time, either in class or at home, learning the symbols for at least the rst twenty elements
in the periodic table. You should also learn the symbols for other common elements that are not in
the rst twenty.
• Write a short test for someone else in the class and then exchange tests with them so that you each
have the chance to answer one.
5.1.3 Writing chemical formulae

A chemical formula is a concise way of giving information about the atoms that make up a particular
chemical compound. A chemical formula shows each element by its symbol and also shows how many atoms
of each element are found in that compound. The number of atoms (if greater than one) is shown as a
subscript.
Examples:
CH4 (methane)
Number of atoms: (1 x carbon) + (4 x hydrogen) = 5 atoms in one methane molecule
H2 SO4 (sulphuric acid)
Number of atoms: (2 x hydrogen) + (1 x sulphur) + (4 x oxygen) = 7 atoms in one molecule of sulphuric
acid
A chemical formula may also give information about how the atoms are arranged in a molecule if it is
written in a particular way. A molecule of ethane, for example, has the chemical formula C2 H6 . This formula
tells us how many atoms of each element are in the molecule, but doesn't tell us anything about how these
atoms are arranged. In fact, each carbon atom in the ethane molecule is bonded to three hydrogen atoms.
Another way of writing the formula for ethane is CH3 CH3 . The number of atoms of each element has not
changed, but this formula gives us more information about how the atoms are arranged in relation to each
other.
The slightly tricky part of writing chemical formulae comes when you have to work out the ratio in which
the elements combine. For example, you may know that sodium (Na) and chlorine (Cl) react to form sodium
chloride, but how do you know that in each molecule of sodium chloride there is only one atom of sodium
for every one atom of chlorine? It all comes down to the valency of an atom or group of atoms. Valency is
the number of bonds that an element can form with another element. Working out the chemical formulae
of chemical compounds using their valency, will be covered in Grade 11. For now, we will use formulae that
you already know.
5.2 Balancing chemical equations 2
5.2.1 Balancing chemical equations

5.2.1.1 The law of conservation of mass
In order to balance a chemical equation, it is important to understand the law of conservation of mass.
Denition 5.1: The law of conservation of mass

The mass of a closed system of substances will remain constant, regardless of the processes acting
inside the system. Matter can change form, but cannot be created or destroyed. For any chemical
process in a closed system, the mass of the reactants must equal the mass of the products.

69
In a chemical equation then, the mass of the reactants must be equal to the mass of the products. In
order to make sure that this is the case, the number of atoms of each element in the reactants must be
equal to the number of atoms of those same elements in the products. Some examples are shown below:
Example 1:
Fe+S → FeS
Figure 5.1
Reactants
Atomic mass of reactants = 55.8 u + 32.1 u = 87.9 u
Number of atoms of each element in the reactants: (1 × Fe) and (1 × S)
Products
Atomic mass of product = 55.8 u + 32.1 u = 87.9 u
Number of atoms of each element in the products: (1 × Fe) and (1 × S)
Since the number of atoms of each element is the same in the reactants and in the products, we say that
the equation is balanced.
Example 2:
H2 + O2 → H2 O
Figure 5.2
Reactants
Atomic mass of reactants = (1 + 1) + (16 + 16) = 34 u
Number of atoms of each element in the reactants: (2 × H) and (2 × O)
Product
Atomic mass of product = (1 + 1 + 16) = 18 u
Number of atoms of each element in the products: (2 × H) and (1 × O)
Since the total atomic mass of the reactants and the products is not the same and since there are more
oxygen atoms in the reactants than there are in the product, the equation is not balanced.
Example 3:
NaOH + HCl → NaCl + H2 O
Figure 5.3
Reactants

Atomic mass of reactants = (23 + 16 + 1) + (1 + 35.4) = 76.4 u

Number of atoms of each element in the reactants: (1 × Na) + (1 × O) + (2 × H) + (1 × Cl)
Products
Atomic mass of products = (23 + 35.4) + (1 + 1 + 16) = 76.4 u
Number of atoms of each element in the products: (1 × Na) + (1 × O) + (2 × H) + (1 × Cl)
Since the number of atoms of each element is the same in the reactants and in the products, we say that
the equation is balanced.
We now need to nd a way to balance those equations that are not balanced so that the number of atoms
of each element in the reactants is the same as that for the products. This can be done by changing the
coecients of the molecules until the atoms on each side of the arrow are balanced. You will see later in
Grade 11 that these coecients tell us something about the mole ratio in which substances react. They
also tell us about the volume relationship between gases in the reactants and products.
tip: Coecients
Remember that if you put a number in front of a molecule, that number applies to the whole molecule.
For example, if you write 2H2 O, this means that there are 2 molecules of water. In other words, there are
4 hydrogen atoms and 2 oxygen atoms. If we write 3HCl, this means that there are 3 molecules of HCl.
In other words there are 3 hydrogen atoms and 3 chlorine atoms in total. In the rst example, 2 is the
coecient and in the second example, 3 is the coecient.
Figure 5.4
3
run demo
5.2.1.2 Steps to balance a chemical equation

When balancing a chemical equation, there are a number of steps that need to be followed.
• STEP 1: Identify the reactants and the products in the reaction and write their chemical formulae.
• STEP 2: Write the equation by putting the reactants on the left of the arrow and the products on the
right.
• STEP 3: Count the number of atoms of each element in the reactants and the number of atoms of
each element in the products.
• STEP 4: If the equation is not balanced, change the coecients of the molecules until the number of
atoms of each element on either side of the equation balance.
• STEP 5: Check that the atoms are in fact balanced.
• STEP 6 (we will look at this a little later): Add any extra details to the equation e.g. phase.
Exercise 5.2.1: Balancing chemical equations 1 (Solution on p. 75.)

Balance the following equation:
Mg + HCl → MgCl2 + H2

CH4 + O2 → CO2 + H2 O
3 http://phet.colorado.edu/sims/balancing-chemical-equations/balancing-chemical-equations_en.jnlp

71

Nitrogen gas reacts with hydrogen gas to form ammonia. Write a balanced chemical equation for
this reaction.

In our bodies, sugar (C6 H12 O6 ) reacts with the oxygen we breathe in to produce carbon dioxide,
water and energy. Write the balanced equation for this reaction.
5.2.1.2.1 Balancing simple chemical equations

Balance the following equations:
1. Hydrogen fuel cells are extremely important in the development of alternative energy sources. Many
of these cells work by reacting hydrogen and oxygen gases together to form water, a reaction which
also produces electricity. Balance the following equation: H2 (g) + O2 (g)→ H2 O(l) Click here for the
4
solution
2. The synthesis of ammonia (NH3 ), made famous by the German chemist Fritz Haber in the early 20th
century, is one of the most important reactions in the chemical industry. Balance the following equation
5
used to produce ammonia: N2 (g) + H2 (g)→NH3 (g) Click here for the solution
6
3. Mg + P4 → Mg3 P2 Click here for the solution
7
4. Ca + H2 O→ Ca(OH)2 + H2 Click here for the solution
8
5. CuCO3 + H2 SO4 → CuSO4 + H2 O + CO2 Click here for the solution
9
6. CaCl2 + Na2 CO3 → CaCO3 + NaCl Click here for the solution
10
7. C12 H22 O11 + O2 + H2 O Click here for the solution
8. Barium chloride reacts with sulphuric acid to produce barium sulphate and hydrochloric acid. Click
11
here for the solution
12
9. Ethane (C2 H6 ) reacts with oxygen to form carbon dioxide and steam. Click here for the solution
10. Ammonium carbonate is often used as a smelling salt. Balance the following reaction for the decom-
position of ammonium carbonate: (NH4 )2 CO3 (s)→NH3 (aq) + CO2 (g) + H2 O(l) Click here for the
13
solution
5.3 State symbols 14
5.3.1 State symbols and other information

The state (phase) of the compounds can be expressed in the chemical equation. This is done by placing the
correct label on the right hand side of the formula. There are only four labels that can be used:
1. (g) for gaseous compounds

2. (l) for liquids
3. (s) for solid compounds
4 See the le at <http://cnx.org/content/m39977/latest/http://www.fhsst.org/lOg>

5 See the le at <http://cnx.org/content/m39977/latest/http://www.fhsst.org/lO4>
7 See the le at <http://cnx.org/content/m39977/latest/http://www.fhsst.org/lOT>
8 See the le at <http://cnx.org/content/m39977/latest/http://www.fhsst.org/lOb>
9 See the le at <http://cnx.org/content/m39977/latest/http://www.fhsst.org/lOj>
10 See the le at <http://cnx.org/content/m39977/latest/http://www.fhsst.org/lOD>
11 See the le at <http://cnx.org/content/m39977/latest/http://www.fhsst.org/lOW>
12 See the le at <http://cnx.org/content/m39977/latest/http://www.fhsst.org/lOZ>
13 See the le at <http://cnx.org/content/m39977/latest/http://www.fhsst.org/lOB>

4. (aq) for an aqueous (water) solution
Occasionally, a catalyst is added to the reaction. A catalyst is a substance that speeds up the reaction
without undergoing any change to itself. In a chemical equation, this is shown by using the symbol of the
catalyst above the arrow in the equation.
To show that heat was needed for the reaction, a Greek delta (∆) is placed above the arrow in the same
way as the catalyst.
tip: You may remember from chapter that energy cannot be created or destroyed during a chemical
reaction but it may change form. In an exothermic reaction, ∆H is less than zero and in an
endothermic reaction, ∆H is greater than zero. This value is often written at the end of a chemical
equation.

Solid zinc metal reacts with aqueous hydrochloric acid to form an aqueous solution of zinc chloride
(ZnCl2 )and hydrogen gas. Write a balanced equation for this reaction.
Exercise 5.3.2: Balancing chemical equations 5 (advanced) (Solution on p. 76.)

(NH4 )2 SO4 + NaOH → NH3 + H2 O + Na2 SO4
In this example, the rst two steps are not necessary because the reactants and products have
already been given.
The following video explains some of the concepts of balancing chemical equations.
Khan Academy video on balancing chemical equations

<http://www.youtube.com/v/RnGu3xO2h74&rel=0&hl=en_US&feature=player_embedded&version=3;rel=0>
Figure 5.5
5.3.1.1 Balancing more advanced chemical equations

Write balanced equations for each of the following reactions:
1. Al2 O3 (s) + H2 SO4 (aq) → Al2 (SO4 )3 (aq) + 3H2 O (l)

2. Mg(OH)2 (aq) + HNO3 (aq) → Mg(NO3 )2 (aq) + 2H2 O (l)
3. Lead(ll)nitrate solution reacts with potassium iodide solution.
4. When heated, aluminium reacts with solid copper oxide to produce copper metal and aluminium oxide
(Al2 O3 ).
5. When calcium chloride solution is mixed with silver nitrate solution, a white precipitate (solid) of
silver chloride appears. Calcium nitrate (Ca(NO3 )2 ) is also produced in the solution. Click here for
15
the solution
15 See the le at <http://cnx.org/content/m39995/latest/http://www.fhsst.org/lOK>

73

<http://static.slidesharecdn.com/swf/ssplayer2.swf ?doc=reactions-100511040544-
phpapp02&stripped_title=reactions-4046986>
Figure 5.6
5.3.2 Summary
• A chemical equation uses symbols to describe a chemical reaction.
• reactants are written on the left hand side of the equation and the products
In a chemical equation,
on the right. The arrow is used to show the direction of the reaction.
• When representing chemical change, it is important to be able to write the chemical formula of a
compound.
• In any chemical reaction, the law of conservation of mass applies. This means that the total atomic
mass of the reactants must be the same as the total atomic mass of the products. This also means
that the number of atoms of each element in the reactants must be the same as the number of atoms
of each element in the product.
• If the number of atoms of each element in the reactants is the same as the number of atoms of each
element in the product, then the equation is balanced.
• If the number of atoms of each element in the reactants is not the same as the number of atoms of
each element in the product, then the equation is not balanced.
• In order to balance an equation, coecients can be placed in front of the reactants and products
until the number of atoms of each element is the same on both sides of the equation.

Balance each of the following chemical equations:
1. NH4 + H2 O → NH4 OH
2. Sodium chloride and water react to form sodium hydroxide, chlorine and hydrogen.
3. Propane is a fuel that is commonly used as a heat source for engines and homes. Balance the following
equation for the combustion of propane: C3 H8 (l) + O2 (g)→CO2 (g) + H2 O(l) Click here for the
16
solution
4. Aspartame, an articial sweetener, has the formula C14 H18 N2 O5 . Write the balanced equation for its
17
combustion (reaction with O2 ) to form CO2 gas, liquid H2 O, and N2 gas. Click here for the solution
18
5. Fe2 (SO4 )3 + K(SCN) → K3 Fe(SCN)6 + K2 SO4 Click here for the solution
6. Chemical weapons were banned by the Geneva Protocol in 1925. According to this protocol, all chem-
icals that release suocating and poisonous gases are not to be used as weapons. White phosphorus, a
very reactive allotrope of phosphorus, was recently used during a military attack. Phosphorus burns
vigorously in oxygen. Many people got severe burns and some died as a result. The equation for this
spontaneous reaction is: P4 (s) + O2 (g) →P2 O5 (s)
a. Balance the chemical equation.
b. Prove that the law of conservation of mass is obeyed during this chemical reaction.
c. Name the product formed during this reaction.
16 See the le at <http://cnx.org/content/m39995/latest/http://www.fhsst.org/lOk>


d. Classify the reaction as endothermic or exothermic. Give a reason for your answer.
e. Classify the reaction as a synthesis or decomposition reaction. Give a reason for your answer.
19
(DoE Exemplar Paper 2 2007)

75

Step 1. Reactants: Mg = 1 atom; H = 1 atom and Cl = 1 atom
Products: Mg = 1 atom; H = 2 atoms and Cl = 2 atoms
Step 2. The equation is not balanced since there are 2 chlorine atoms in the product and only 1 in the reactants.
If we add a coecient of 2 to the HCl to increase the number of H and Cl atoms in the reactants, the
equation will look like this:
Mg + 2HCl → MgCl2 + H2
Step 3. If we count the atoms on each side of the equation, we nd the following:
Reactants: Mg = 1; H = 2; Cl = 2
Products: Mg = 1; H = 2; Cl = 2
The equation is balanced. The nal equation is:
Mg + 2HCl → MgCl2 + H2

Step 1. Reactants: C = 1; H = 4; O = 2
Products: C = 1; H = 2; O = 3
Step 2. If we add a coecient of 2 to H2 O, then the number of hydrogen atoms in the reactants will be 4,
which is the same as for the reactants. The equation will be:
CH4 + O2 → CO2 + 2H2 O
Step 3. Reactants: C = 1; H = 4; O = 2
Products: C = 1; H = 4; O = 4
You will see that, although the number of hydrogen atoms now balances, there are more oxygen atoms
in the products. You now need to repeat the previous step. If we put a coecient of 2 in front of O2 ,
then we will increase the number of oxygen atoms in the reactants by 2. The new equation is:
CH4 + 2O2 → CO2 + 2H2 O
When we check the number of atoms again, we nd that the number of atoms of each element in the
reactants is the same as the number in the products. The equation is now balanced.

Step 1. The reactants are nitrogen (N2 ) and hydrogen (H2 ) and the product is ammonia (NH3 ).
Step 2. The equation is as follows:
N2 + H2 → NH3
Step 3. Reactants: N = 2; H = 2
Products: N = 1; H = 3
Step 4. In order to balance the number of nitrogen atoms, we could rewrite the equation as:
N2 + H2 → 2NH3
Step 5. In the above equation, the nitrogen atoms now balance, but the hydrogen atoms don't (there are 2
hydrogen atoms in the reactants and 6 in the product). If we put a coecient of 3 in front of the
hydrogen (H2 ), then the hydrogen atoms and the nitrogen atoms balance. The nal equation is:
N2 + 3H2 → 2NH3

Step 1. Reactants: sugar (C6 H12 O6 ) and oxygen (O2 )
Products: carbon dioxide (CO 2) and water (H2 O)
Step 2. C6 H12 O6 + O2 → CO2 + H2 O
Step 3. Reactants: C=6; H=12; O=8;
Products: C=1; H=2; O=3;

Step 4. It is easier to start with carbon as it only appears once on each side. If we add a 6 in front of CO 2,
the equation looks like this:

C6 H12 O6 + O2 → 6CO2 + H2 O
Reactants: C=6; H=12; O=8;
Step 5. Let's try to get the number of hydrogens the same this time.
C6 H12 O6 + O2 → 6CO2 + 6H2 O
Step 6. C6 H12 O6 + 12O2 → 6CO2 + 6H2 O

Step 1. The reactants are zinc (Zn) and hydrochloric acid (HCl). The products are zinc chloride (ZnCl2 ) and
hydrogen (H2 ).
Step 2. Zn + HCl → ZnCl2 + H2
Step 3. You will notice that the zinc atoms balance but the chlorine and hydrogen atoms don't. Since there
are two chlorine atoms on the right and only one on the left, we will give HCl a coecient of 2 so that
there will be two chlorine atoms on each side of the equation.
Zn + 2HCl → ZnCl2 + H2
Step 4. When you look at the equation again, you will see that all the atoms are now balanced.
Step 5. In the initial description, you were told that zinc was a metal, hydrochloric acid and zinc chloride were
in aqueous solutions and hydrogen was a gas.
Zn(s) + 2HCl(aq) → ZnCl2 (aq) + H2 (g)

Step 1. With a complex equation, it is always best to start with atoms that appear only once on each side i.e.
Na, N and S atoms. Since the S atoms already balance, we will start with Na and N atoms. There
are two Na atoms on the right and one on the left. We will add a second Na atom by giving NaOH a
coecient of two. There are two N atoms on the left and one on the right. To balance the N atoms,
NH3 will be given a coecient of two. The equation now looks as follows:
(NH4 )2 SO4 + 2NaOH → 2NH3 + H2 O + Na2 SO4
Step 2. N, Na and S atoms balance, but O and H atoms do not. There are six O atoms and ten H atoms on
the left, and ve O atoms and eight H atoms on the right. We need to add one O atom and two H
atoms on the right to balance the equation. This is done by adding another H2 O molecule on the right
hand side. We now need to check the equation again:
(NH4 )2 SO4 + 2NaOH → 2NH3 + 2H2 O + Na2 SO4
The equation is now balanced.

Chapter 6
The water cycle
6.1 Introduction 1
6.1.1 Introduction
You may have heard the word 'cycle' many times before. Think for example of the word 'bicycle' or the
regular 'cycle tests' that you may have at school. A cycle is a series of events that repeats itself. In the
case of a bicycle, the wheel turns through a full circle before beginning the motion again, while cycle tests
happen regularly, normally every week or every two weeks. Because a cycle repeats itself, it doesn't have a
beginning or an end.
Our Earth is a closed system. This means that it can exchange energy with its surroundings (i.e. the
rest of the solar system), but no new matter is brought into the system. For this reason, it is important
that all the elements and molecules on Earth are recycled so that they are never completely used up. In the
next two sections, we are going to take a closer look at two cycles that are very important for life on Earth.
They are the water cycle and the nitrogen cycle.
6.1.2 The importance of water

For many people, it is so easy to take water for granted and yet life on Earth would not exist were it not for
this extraordinary compound. Not only is it believed that the rst forms of life actually started in water,
but most of the cells in living organisms contain between 70% and 95% water. Here in the cells, water acts
as a solvent and helps to transport vital materials such as food and oxygen to where they are needed, and
also removes waste products such as carbon dioxide and ammonia from the body. For many animals and
plants, water is their home. Think for example of sh and amphibians that live either all or part of the
time in rivers, dams and the oceans. In other words, if water did not exist, life as we know it would not be
possible.
Apart from allowing life to exist, water also has a number of other functions. Water shapes the landscape
around us by wearing away at rocks and also transports and deposits sediments on oodplains and along
coastal regions. Water also plays a very important role in helping to regulate Earth's climate. We will discuss
this again later in the chapter. As humans we use water in our homes, in industry, in mining, irrigation and
even as a source of electricitiy in hydro-electric schemes. In fact, if we were able to view Earth from space,
we would see that almost three quarters of our planet's surface is covered in water. It is because of this that
Earth is sometimes called the 'Blue Planet'. Most of this water is stored in the oceans, with the rest found
in ice (e.g. glaciers), groundwater (e.g. boreholes), surface water (e.g. rivers, lakes, estuaries, dams) and in
the atmosphere as clouds and water vapour.
77
78 CHAPTER 6. THE WATER CYCLE
note: In the search for life on other planets, one of the rst things that scientists look for is
water. However, most planets are either too close to the sun (and therefore too hot) for water to
exist in liquid form, or they are too far away and therefore too cold. So, even if water were to be
found, the conditions are unlikely to allow it to exist in a form that can support the diversity of
life that we see on Earth.
6.1.3 The movement of water through the water cycle

The water cycle is the continuous movement of water over, above and beneath the Earth's surface. As water
moves, it changes phase between liquid (water), solid (ice) and gas (water vapour). It is powered by solar
energy and, because it is a cycle, it has no beginning or end.
Denition 6.1: The Water Cycle

The water cycle is the continuous circulation of water across the Earth. The water cycle is
driven by solar radiation and it includes the atmosphere, land, surface water and groundwater. As
water moves through the cycle, it changes state between liquid, solid, and gas phases. The actual
movement of water from one part of the cycle to another (e.g. from river to ocean) is the result of
processes such as evaporation, precipitation, inltration and runo.
The movement of water through the water cycle is shown in Figure 6.1. In the gure, each process within
this cycle is numbered. Each process will be described below.
Figure 6.1: The water cycle
1. The source of energy The water cycle is driven by the sun, which provides the heat energy that is
needed for many of the other processes to take place.
2. Evaporation When water on the earth's surface is heated by the sun, the average energy of the water
molecules increases and some of the molecules are able to leave the liquid phase and become water
vapour. This is called evaporation. Evaporation is the change of water from a liquid to a gas as it
moves from the ground, or from bodies of water like the ocean, rivers and dams, into the atmosphere.
3. Transpiration Transpiration is the evaporation of water from the aerial parts of plants, especially the
leaves but also from the stems, owers and fruits. This is another way that liquid water can enter the
atmosphere as a gas.
4. Condensation When evaporation takes place, water vapour rises in the atmosphere and cools as the
altitude (height above the ground) increases. As the temperature drops, the energy of the water vapour
molecules also decreases, until the molecules don't have enough energy to stay in the gas phase. At this
point, condensation occurs. Condensation is the change of water from water vapour (gas) into liquid
water droplets in the air. Clouds, fog and mist are all examples of condensation. A cloud is actually a
collection of lots and lots of tiny water droplets. This mostly takes place in the upper atmosphere but
can also take place close to the ground if there is a signicant temperature change.
note: Have you ever tried breathing out on a very cold day? It looks as though you
are breathing out smoke! The moist air that you breathe out is much warmer than the air
outside your body. As this warm, moist air comes into contact with the colder air outside, its
temperature drops very quickly and the water vapour in the air you breathe out condenses.

79
The 'smoke' that you see is actually formed in much the same way as clouds form in the upper
atmosphere.
5. Precipitation Precipitation occurs when water falls back to the earth's surface in the form of rain or
snow. Rain will fall as soon as a cloud becomes too saturated with water droplets. Snow is similar to
rain, except that it is frozen. Snow only falls if temperatures in the atmosphere are around freezing.
0
The freezing point of water is 0 C.
6. Inltration If precipitation occurs, some of this water will lter into the soil and collect underground.
This is called inltration. This water may evaporate again from the soil at a later stage, or the
underground water may seep into another water body.
7. Surface runo This refers to the many ways that water moves across the land. This includes surface
runo such as when water ows along a road and into a drain, or when water ows straight across
the sand. It also includes channel runo when water ows in rivers and streams. As it ows, the
water may inltrate into the ground, evaporate into the air, become stored in lakes or reservoirs, or be
extracted for agricultural or other human uses.
tip: It is important to realise that the water cycle is all about energy exchanges. The sun is the
original energy source. Energy from the sun heats the water and causes evaporation. This energy
is stored in water vapour as latent heat. When the water vapour condenses again, the latent heat
is released and helps to drive circulation in the atmosphere. The liquid water falls to earth and will
evaporate again at a later stage. The atmospheric circulation patterns that occur because of these
exchanges of heat are very important in inuencing climate patterns.
6.1.3.1 Experiment : The Water Cycle

Materials:
Tile or piece of plastic (e.g. lid of ice-cream container) to make a hill slope; glass sh tank with a lid;
beaker with ice cubes; lamp; water
Set up a model of the water cycle as follows:
Figure 6.2
1. Lean the plastic against one side so that it creates a 'hill slope' as shown in the diagram.
2. Pour water into the bottom of the tank until about a quarter of the hill slope is covered.
3. Close the sh tank lid.
4. Place the beaker with ice on the lid directly above the hill slope.
5. Turn the lamp on and position it so that it shines over the water.
6. Leave the model like this for 20-30 minutes and then observe what happens. Make sure that you don't
touch the lamp as it will be very hot!
Observation questions:
1. Which parts of the water cycle can you see taking place in the model?
2. Which parts of the water cycle are not represented in the model?
3. Can you think of how those parts that are not shown could be represented?

4. What is the energy source in the model? What would the energy source be in reality?
5. What do you think the function of the ice is in the beaker?
2
This video provides a summary of the stages of the water cycle.
Figure 6.3
6.2 Properties of water 3
6.2.1 The microscopic structure of water

In many ways, water behaves very dierently from other liquids. These properties are directly related to the
microscopic structure of water and more specically to the shape of the molecule and its polar nature and
to the bonds that hold water molecules together.
6.2.1.1 The polar nature of water

Every water molecule is made up of one oxygen atom that is bonded to two hydrogen atoms. When atoms
bond, the nucleus of each atom has an attractive force on the electrons of the other atoms. This 'pull' is
stronger in some atoms than in others and is called the electronegativity of the atom. In a water molecule,
the oxygen atom has a higher electronegativty than the hydrogen atoms and therefore attracts the electrons
more strongly. The result is that the oxygen atom has a slightly negative charge and the two hydrogen atoms
each have a slightly positive charge. The water molecule is said to be polar because the electrical charge is
not evenly distributed in the molecule. One part of the molecule has a dierent charge to other parts. You
will learn more about this in Grade 11.
Figure 6.4: Diagrams showing the structure of a water molecule. Each molecule is made up of two
hydrogen atoms that are attached to one oxygen atom.
6.2.1.2 Hydrogen bonding in water molecules

In every water molecule, the forces that hold the individual atoms together are called intramolecular
forces. But there are also forces between dierent water molecules. These are called inter molecular
forces (Figure 6.5). You will learn more about these at a later stage, but for now it is enough to know that
in water, molecules are held together by hydrogen bonds. Hydrogen bonds are a much stronger type of
intermolecular force than those found in many other substances and this aects the properties of water.
2 http://www.teachersdomain.org/asset/ess05_vid_watercycle/

81
tip: Intramolecular and intermolecular forces
If you nd these terms confusing, remember that 'intra' means within (i.e. the forces within a molecule).
An intro vert is someone who doesn't express emotions and feelings outwardly. They tend to be quieter and
keep to themselves. 'Inter' means between (i.e. the forces between molecules). An inter national cricket
match is a match between two dierent countries.
Figure 6.5: Intermolecular and intramolecular forces in water. Note that the diagram on the left only
shows intermolecular forces. The intramolecular forces are between the atoms of each water molecule.
6.2.2 The unique properties of water

Because of its polar nature and the strong hydrogen bonds between its molecules, water has some special
properties that are quite dierent to those of other substances.
1. Absorption of infra-red radiation The polar nature of the water molecule means that it is able to
absorb infra-red radiation (heat) from the sun. As a result of this, the oceans and other water bodies
act as heat reservoirs and are able to help moderate the Earth's climate.
2. Specic heat
Denition 6.2: Specic heat
Specic heat is the amount of heat energy that is needed to increase the temperature of a
substance by one degree.
Water has a high specic heat, meaning that a lot of energy must be absorbed by water before its
temperature changes. Refer to Section 6.2.2.1 ( Demonstration : The high specic heat of water) for
activity.
You have probably observed this phenomenon if you have boiled water in a pot on the stove. The
metal of the pot heats up very quickly, and can burn your ngers if you touch it, while the water
may take several minutes before its temperature increases even slightly. How can we explain this in
terms of hydrogen bonding? Remember that increasing the temperature of a substance means that
its particles will move more quickly. However, before they can move faster, the bonds between them
must be broken. In the case of water, these bonds are strong hydrogen bonds, and so a lot of energy is
needed just to break these, before the particles can start moving faster. It is the high specic heat of
water and its ability to absorb infra-red radiation that allows it to regulate climate. Have you noticed
how places that are closer to the sea have less extreme daily temperatures than those that are inland?
During the day, the oceans heat up slowly, and so the air moving from the oceans across land is cool.
Land temperatures are cooler than they would be if they were further from the sea. At night, the
oceans lose the heat that they have absorbed very slowly, and so sea breezes blowing across the land
are relatively warm. This means that at night, coastal regions are generally slightly warmer than areas
that are further from the sea. By contrast, places further from the sea experience higher maximum
temperatures, and lower minimum temperatures. In other words, their temperature range is higher
than that for coastal regions. The same principle also applies on a global scale. The large amount of
water across Earth's surface helps to regulate temperatures by storing infra-red radiation (heat) from
the sun, and then releasing it very slowly so that it never becomes too hot or too cold, and life is
able to exist comfortably. In a similar way, water also helps to keep the temperature of the internal

environment of living organisms relatively constant. This is very important. In humans, for example,
a change in body temperature of only a few degrees can be deadly.
3. Melting point and boiling point The melting point of water is 00 C and its boiling point is 1000 C.
This large dierence between the melting and boiling point is mostly due to the strong intermolecular
forces in water (hydrogen bonds) and is very important because it means that water can exist as a
liquid over a large range of temperatures. The three phases of water are shown in Figure 6.6.
Figure 6.6: Changes in phase of water
4. High heat of vaporisation

Denition 6.3: Heat of vaporisation
Heat of vaporisation is the energy that is needed to change a given quantity of a substance
into a gas.
The strength of the hydrogen bonds between water molecules also means that it has a high heat of
vaporisation. 'Heat of vaporisation' is the heat energy that is needed to change water from the liquid
0
to the gas phase. Because the bonds between molecules are strong, water has to be heated to 100 C
before it changes phase. At this temperature, the molecules have enough energy to break the bonds
that hold the molecules together. The heat of vaporisation for water is 40,65 kJ · mol. It is very lucky
for life on earth that water does have a high heat of vaporisation. Can you imagine what a problem it
would be if water's heat of vaporisation was much lower? All the water that makes up the cells in our
bodies would evaporate and most of the water on earth would no longer be able to exist as a liquid!
5. Less dense solid phase Another unusual property of water is that its solid phase (ice) is less dense
than its liquid phase. You can observe this if you put ice into a glass of water. The ice doesn't sink
to the bottom of the glass, but oats on top of the liquid. This phenomenon is also related to the
hydrogen bonds between water molecules. While other materials contract when they solidify, water
expands. The ability of ice to oat as it solidies is a very important factor in the environment. If ice
sank, then eventually all ponds, lakes, and even the oceans would freeze solid as soon as temperatures
dropped below freezing, making life as we know it impossible on Earth. During summer, only the
upper few inches of the ocean would thaw. Instead, when a deep body of water cools, the oating ice
insulates the liquid water below, preventing it from freezing and allowing life to exist under the frozen
surface.
Figure 6.7: Ice cubes oating in water
note: Antarctica, the 'frozen continent', has one of the world's largest and deepest fresh-
water lakes. And this lake is hidden beneath 4 kilometres of ice! Lake Vostok is 200 km long
and 50 km wide. The thick, glacial blanket of ice acts as an insulator, preventing the water
from freezing.
6. Water as a solvent Water is also a very good solvent, meaning that it is easy for other substances to
dissolve in it. It is very seldom, in fact, that we nd pure water. Most of the time, the water that we
drink and use has all kinds of substances dissolved in it. It is these that make water taste dierent in

83
dierent areas. So why, then, is it important that water is such a good solvent? We will look at just a
few examples.
• Firstly, think about the animals and plants that live in aquatic environments such as rivers, dams
or in the sea. All of these living organisms either need oxygen for respiration or carbon dioxide for
photosynthesis, or both. How do they get these gases from the water in which they live? Oxygen
and carbon dioxide are just two of the substances that dissolve easily in water and this is how
plants and animals obtain the gases that they need to survive. Instead of being available as gases
in the atmosphere, they are present in solution in the surrounding water.
• Secondly, consider the fact that all plants need nitrogen to grow, and that they absorb this nitrogen
from compounds such as nitrates and nitrates that are present in the soil. The question remains,
however, as to how these nitrates and nitrites are able to be present in the soil at all, when most
of the Earth's nitrogen is in a gaseous form in the atmosphere. Part of the answer lies in the fact
that nitrogen oxides, which are formed during ashes of lightning, can be dissolved in rainwater
and transported into the soil in this way, to be absorbed by plants. The other part of the answer
lies in the activities of nitrogen-xing bacteria in the soil, but this is a topic that we will return
to in a later section.
It should be clear now, that water is an amazing compound and that without its unique properties, life on
Earth would denitely not be possible.
6.2.2.1 Demonstration : The high specic heat of water

1. Pour about 100 ml of water into a glass beaker.
2. Place the beaker on a stand and heat it over a bunsen burner for about 2 minutes.
3. After this time, carefully touch the side of the beaker (Make sure you touch the glass very lightly
because it will be very hot and may burn you!). Then use the end of a nger to test the temperature
of the water.
What do you notice? Which of the two (glass or water) is the hottest?
6.2.2.2 The properties of water

1. A learner returns home from school on a hot afternoon. In order to get cold water to drink, she adds
ice cubes to a glass of water. She makes the following observations:
• The ice cubes oat in the water.

• After a while the water becomes cold and the ice cubes melt.
a. What property of ice cubes allows them to oat in the water?

b. Briey explain why the water gets cold when the ice cubes melt.
c. Briey describe how the property you mentioned earlier aects the survival of aquatic life during
winter.
4
2. Which properties of water allow it to remain in its liquid phase over a large temperature range? Explain
5
why this is important for life on earth. Click here for the solution
4 See the le at <http://cnx.org/content/m39913/latest/http://www.fhsst.org/l3E>

5 See the le at <http://cnx.org/content/m39913/latest/http://www.fhsst.org/l3m>

6.3 Water conservation 6
6.3.1 Water conservation

Water is a very precious substance and yet far too often, earth's water resources are abused and taken for
granted. How many times have you walked past polluted rivers and streams, or seen the ow of water in
a river reduced to almost nothing because of its extraction for industrial and other uses? And if you were
able to test the quality of the water you see, you would probably be shocked. Often our water resources
are contaminated with chemicals such as pesticides and fertilisers. If water is to continue playing all the
important functions that were discussed earlier, it is vital that we reduce the impact of humans on these
resources.
6.3.1.1 Group work : Human impacts on the water cycle

Read the following extract from an article, entitled 'The Eects of Urbanisation on the Water Cycle' by
Susan Donaldson, and then answer the questions that follow.
As our communities grow, we notice many visible changes including housing developments, road
networks, expansion of services and more. These changes have an impact on our precious water
resources, with pollution of water being one of many such impacts. To understand these impacts
you will need to have a good knowledge of the water cycle!
It is interesting to note that the oceans contain most of earth's water (about 97%). Of the
freshwater supplies on earth, 78% is tied up in polar ice caps and snow, leaving only a very small
fraction available for use by humans. Of the available fresh water, 98% is present as groundwater,
while the remaining 2% is in the form of surface water. Because our usable water supply is so
limited, it is vitally important to protect water quality. Within the water cycle, there is no 'new'
water ever produced on the earth. The water we use today has been in existence for billions of
years. The water cycle continually renews and refreshes this nite water supply.
So how exactly does urbanisation aect the water cycle? The increase in hard surfaces (e.g.
roads, roofs, parking lots) decreases the amount of water that can soak into the ground. This
increases the amount of surface runo. The runo water will collect many of the pollutants that
have accumulated on these surfaces (e.g. oil from cars) and carry them into other water bodies
such as rivers or the ocean. Because there is less inltration, peak ows of stormwater runo are
larger and arrive earlier, increasing the size of urban oods. If groundwater supplies are reduced
enough, this may aect stream ows during dry weather periods because it is the groundwater
that seeps to the surface at these times.
Atmospheric pollution can also have an impact because condensing water vapour will pick up
these pollutants (e.g. SO2 , CO2 and NO2 ) and return them to earth into other water bodies.
However, while the eects of urbanisation on water quality can be major, these impacts can be
reduced if wise decisions are made during the process of development.
Questions
1. In groups, try to explain...
a. what is meant by 'urbanisation'

b. how urbanisation can aect water quality

85
2. Explain why it is so important to preserve the quality of our water supplies.

3. The article gives some examples of human impacts on water quality. In what other ways do human
activities aect water quality?
4. What do you think some of the consequences of these impacts might be for humans and other forms
of life?
5. Imagine that you are the city manager in your own city or the city closest to you. What changes would
you introduce to try to protect the quality of water resources in your urban area?
6. What measures could be introduced in rural areas to protect water quality?
Apart from the pollution of water resources, the overuse of water is also a problem. In looking at the water
cycle, it is easy sometimes to think that water is a never-ending resource. In a sense this is true because
water cannot be destroyed. However, the availability of water may vary from place to place. In South Africa
for example, many regions are extremely dry and receive very little rainfall. The same is true for many
other parts of the world, where the scarcity of water is a life and death issue. The present threat of global
warming is also likely to aect water resources. Some climate models suggest that rising temperatures could
increase the variability of climate and decrease rainfall in South Africa. With this in mind and remembering
that South Africa is already a dry country, it is vitally important that we manage our water use carefully. In
addition to this, the less water there is available, the more likely it is that water quality will also decrease.
A decrease in water quality limits how water can be used and developed.
At present, the demands being placed on South Africa's water resources are large. Table 6.1 shows the
water requirements that were predicted for the year 2000. The gures in the table were taken from South
Africa's National Water Resource Strategy, produced by the Department of Water Aairs and Forestry in
2004. In the table, 'rural' means water for domestic use and stock watering in rural areas, while 'urban'
means water for domestic, industrial and commercial use in the urban area. 'Aorestation' is included
because many plantations reduce stream ow because of the large amounts of water they need to survive.
Water Irrigation Urban Rural Mining Power Aorestation Total

man- and bulk generation
agement industrial
area
Limpopo 238 34 28 14 7 1 322
Thukela 204 52 31 46 1 0 334
Upper 114 635 43 173 80 0 1045

Vaal
Upper Or- 780 126 60 2 0 0 968

ange
Breede 577 39 11 0 0 6 633
Country 7920 2897 574 755 297 428 12871

total
Table 6.1: The predicted water requirements for various water management areas in South Africa for 2000
3
(million m /annum)
6.3.1.2 Case Study : South Africa's water requirements

Refer to Table 6.1 and then answer the following questions:
1. Which water management area in South Africa has the highest need for water...

a. in the mining and industry sector?

b. for power generation?
c. in the irrigation sector?
d. Suggest reasons for each of your answers above.
2. For South Africa as a whole...
a. Which activity uses the most water?

b. Which activity uses the least water?
3. Complete the following table, by calculating the percentage (%) that each activity contributes to the
total water requirements in South Africa for the year 2000.
Water use activity % of SA's total water requirements
Irrigation
Urban
Rural
Mining and bulk industry
Power generation
Aorestation
Table 6.2
Now look at Table 6.3, which shows the amount of water available in South Africa during 2000. In the table,
'usable return ow' means the amount of water that can be reused after it has been used for irrigation, urban
or mining.
Water man- Surface wa- Ground Irrigation Urban Mining and Total local
agement ter bulk indus- yield
area trial
Limpopo 160 98 8 15 0 281
Thukela 666 15 23 24 9 737
Upper Vaal 598 32 11 343 146 1130
Upper 4311 65 34 37 0 4447

Orange
Breede 687 109 54 16 0 866
Country 10240 1088 675 970 254 13227

total
Table 6.3: The available water yield in South Africa in 2000 for various water management areas (million
3
m /annum)
6.3.1.3 Case Study : Water conservation

Refer to Table 6.3 and then answer the following questions:
1. Explain what is meant by...

87
a. surface water
b. ground water
2. Which water management area has the...
a. lowest surface water yield?

b. highest surface water yield?
c. lowesttotal yield?
d. highesttotal yield?
3. Look at the country's total water requirements for 2000 and the total available yield.
a. Calculate what percentage of the country's water yield is already being used up.
b. Do you think that the country's total water requirements will increase or decrease in the coming
years? Give a reason for your answer.
4. South Africa is already placing a huge strain on existing water resources. In groups of 3-4, discuss
ways that the country's demand for water could be reduced. Present your ideas to the rest of the class
for discussion.
6.3.2 Summary
• Water is critical for the survival of life on Earth. It is an important part of the cells of living organisms
and is used by humans in homes, industry, mining and agriculture.
• Water moves between the land and sky in the water cycle. The water cycle describes the changes
in phase that take place in water as it circulates across the Earth. The water cycle is driven by solar
radiation.
• Some of the important processes that form part of the water cycle are evaporation, transpiration,
condensation, precipitation, inltration and surface runo. Together these processes ensure that water
is cycled between the land and sky.
• It is the microscopic structure of water that determines its unique properties.
• Water molecules are polar and are held together by hydrogen bonds. These characteristics aect
the properties of water.
• Some of the unique properties of water include its ability to absorb infra-red radiation, its high specic
heat, high heat of vaporisation and the fact that the solid phase of water is less dense that its liquid
phase.
• These properties of water help it to sustain life on Earth by moderating climate, regulating the internal
environment of living organisms and allowing liquid water to exist below ice, even if temperatures are
below zero.
• Water is also a good solvent. This property means that it is a good transport medium in the cells of
living organisms and that it can dissolve gases and other compounds that may be needed by aquatic
plants and animals.
• Human activities threaten the quality of water resources through pollution and altered runo patterns.
• As human populations grow, there is a greater demand for water. In many areas, this demand exceeds
the amount of water available for use. Managing water wisely is important in ensuring that there will
always be water available both for human use and to maintain natural ecosystems.
6.3.2.1 Summary Exercise

1. Give a word or term for each of the following phrases:
a. The continuous circulation of water across the earth.

b. The change in phase of water from gas to liquid.

c. The movement of water across a land surface.

d. The temperature at which water changes from liquid to gas.
7
2. In each of the following multiple choice questions, choose the one correct answer from the list provided.
a. Many of the unique properties of water (e.g. its high specic heat and high boiling point) are due
to:
i. strong covalent bonds between the hydrogen and oxygen atoms in each water molecule
ii. the equal distribution of charge in a water molecule
iii. strong hydrogen bonds between water molecules
iv. the linear arrangement of atoms in a water molecule
8
b. Which of the following statements is false?
i. Most of the water on earth is in the oceans.
ii. The hardening of surfaces in urban areas results in increased surface runo.
iii. Water conservation is important because water cannot be recycled.
iv. Irrigation is one of the largest water users in South Africa.
9
3. The sketch below shows a process that leads to rainfall in town X. The town has been relying only on
rainfall for its water supply because it has no access to rivers or tap water. A group of people told the
community that they will never run out of rainwater because it will never stop raining.
Figure 6.8
a. List the processes labelled P1 and P2 that lead to rainfall in town X.

b. Is this group of people correct in saying that town X will never run out of rainwater? Justify your
answer using the sketch. Recently, the amount of rainwater has decreased signicantly. Various
reasons have been given to explain the drought. Some of the community members are blaming
this group who told them that it will never stop raining.
c. What scientic arguments can you use to convince the community members that this group of
people should not be blamed for the drought?
d. What possible strategies can the community leaders adopt to ensure that they have a regular
supply of water.
10
7 See the le at <http://cnx.org/content/m39915/latest/http://www.fhsst.org/l3y>

8 See the le at <http://cnx.org/content/m39915/latest/http://www.fhsst.org/l3p>
9 See the le at <http://cnx.org/content/m39915/latest/http://www.fhsst.org/l3V>
10 See the le at <http://cnx.org/content/m39915/latest/http://www.fhsst.org/l3d>

Chapter 7
The nitrogen cycle
7.1 Introduction 1
7.1.1 Introduction
The earth's atmosphere is made up of about 78% nitrogen, making it the largest pool of this gas. Nitrogen
is essential for many biological processes. It is in all amino acids, proteins and nucleic acids. As you will see
in a later chapter, these compounds are needed to build tissues, transport substances around the body and
control what happens in living organisms. In plants, much of the nitrogen is used in chlorophyll molecules
which are needed for photosynthesis and growth.
So, if nitrogen is so essential for life, how does it go from being a gas in the atmosphere to being part of
living organisms such as plants and animals? The problem with nitrogen is that it is an 'inert' gas, which
means that it is unavailable to living organisms in its gaseous form. This is because of the strong triple bond
between its atoms that makes it dicult to break. Something needs to happen to the nitrogen gas to change
it into a form that it can be used. And at some later stage, these new compounds must be converted back
into nitrogen gas so that the amount of nitrogen in the atmosphere stays the same. This process of changing
nitrogen cycle (Figure 7.1).
nitrogen into dierent forms is called the
Denition 7.1: The nitrogen cycle

The nitrogen cycle is a biogeochemical cycle that describes how nitrogen and nitrogen-containing
compounds are changed in nature.
Very broadly, the nitrogen cycle is made up of the following processes:
• Nitrogen xation - The process of converting inert nitrogen gas into more useable nitrogen com-
pounds such as ammonia.
• Nitrication - The conversion of ammonia into nitrites and then into nitrates, which can be absorbed
and used by plants.
• Denitrication - The conversion of nitrates back into nitrogen gas in the atmosphere.
We are going to look at each of these processes in more detail.
89
90 CHAPTER 7. THE NITROGEN CYCLE
Figure 7.1: A simplied diagram of the nitrogen cycle
7.1.2 Nitrogen xation

Nitrogen xation is needed to change gaseous nitrogen into forms such as ammonia that are more useful to
living organisms. Some xation occurs in lightning strikes and in industrial processes, but most xation is
done by dierent types of bacteria living either in the soil or in parts of the plants.
1. Biological xation Some bacteria are able to x nitrogen. They use an enzyme called nitrogenase to
combine gaseous nitrogen with hydrogen to form ammonia. The bacteria then use some of this ammonia
to produce their own organic compounds, while what is left of the ammonia becomes available in the
soil. Some of these bacteria are free-living, in other words they live in the soil. Others live in the
root nodules of legumes (e.g. soy, peas and beans). Here they form a mutualistic relationship with
the plant. The bacteria get carbohydrates (food) from the plant and, in exchange, produce ammonia
which can be converted into nitrogen compounds that are essential for the survival of the plant. In
nutrient-poor soils, planting lots of legumes can help to enrich the soil with nitrogen compounds. A
simplied equation for biological nitrogen xation is:
+ −
N2 + 8H + 8e → 2NH3 + H2
−
In this equation the 8e means 8 electrons. Energy is used in the process, but this is not shown in the
above equation. Another important source of ammonia in the soil is decomposition. When animals
and plants die, the nitrogen compounds that were present in them are broken down and converted into
ammonia. This process is carried out by decomposition bacteria and fungi in the soil.
2. Industrial nitrogen xation In the Haber-Bosch process, nitrogen (N 2) is converted together with
hydrogen gas (H2 ) into ammonia (NH3 ) fertiliser. This is an articial process.
3. Lightning In the atmosphere, lightning and photons are important in the reaction between nitrogen
(N2 ) and oxygen (O2 ) to form nitric oxide (NO) and then nitrates.
note: It is interesting to note that by cultivating legumes, using the Haber-Bosch process to
manufacture chemical fertilisers and increasing pollution from vehicles and industry, humans have
more than doubled the amount of nitrogen that would normally be changed from nitrogen gas into
a biologically useful form. This has serious environmental consequences.
7.1.3 Nitrication
Nitrication involves two biological oxidation reactions: rstly, the oxidation of ammonia with oxygen to
−
form nitrite (NO2 ) and secondly the oxidation of these nitrites into nitrates.
1. NH3 + O2 → NO2
−
+ 3H
+
+ 2e
−
(production of nitrites )
2. NO2
−
+ H2 O→ NO3
−
+ 2H
+
+ 2e
−
(production of nitrates )
Nitrication is an important step in the nitrogen cycle in soil because it converts the ammonia (from the
nitrogen xing part of the cycle) into nitrates, which are easily absorbed by the roots of plants. This
absorption of nitrates by plants is called assimilation. Once the nitrates have been assimilated by the plants,

91
they become part of the plants' proteins. These plant proteins are then available to be eaten by animals.
In other words, animals (including humans) obtain their own nitrogen by feeding on plants. Nitrication is
performed by bacteria in the soil, called nitrifying bacteria.
7.1.3.1 Case Study : Nitrates in drinking water

Read the information below and then carry out your own research to help you answer the questions that
follow.
The negatively charged nitrate ion is not held onto soil particles and so can be easily washed out of the
soil. This is called leaching. In this way, valuable nitrogen can be lost from the soil, reducing the soil's
fertility. The nitrates can then accumulate in groundwater and eventually in drinking water. There are
strict regulations that control how much nitrate can be present in drinking water, because nitrates can be
reduced to highly reactive nitrites by microorganisms in the gut. Nitrites are absorbed from the gut and bind
to haemoglobin (the pigment in blood that helps to transport oxygen around the body). This reduces the
ability of the haemoglobin to carry oxygen. In young babies this can lead to respiratory distress, a condition
known as "blue baby syndrome".
1. How is nitrate concentration in water measured?

2. What concentration of nitrates in drinking water is considered acceptable? You can use drinking water
standards for any part of the world, if you can't nd any for South Africa.
3. What is 'blue baby syndrome' and what are the symptoms of the disease?
7.1.4 Denitrication
Denitrication is the process of reducing nitrate and nitrite into gaseous nitrogen. The process is carried
out by denitrication bacteria. The nitrogen that is produced is returned to the atmosphere to complete the
nitrogen cycle.
The equation for the reaction is:
− − +
2NO3 + 10e + 12H → N2 + 6H2 O
7.2 Human inuences and industry 2
7.2.1 Human Inuences on the Nitrogen Cycle

Humans have contributed signicantly to the nitrogen cycle in a number of ways.
• Atmospheric pollution is another problem. The main culprits are nitrous oxide (N2 O), nitric oxide
(NO) and nitrogen dioxide (NO2 ). Most of these gases result either from emissions from agricultural
soils (and particularly articial fertilisers), or from the combustion of fossil fuels in industry or motor
vehicles. The combustion (burning) of nitrogen-bearing fuels such as coal and oil releases this nitrogen
as NO2 or NO gases. Both NO2 and NO can combine with water droplets in the atmosphere to form
acid rain. Furthermore, both NO and NO contribute to the depletion of the ozone layer and some
2
are greenhouse gases. In high concentrations, these gases can contribute towards global warming.
• Both articial fertilisation and the planting of nitrogen xing crops, increase the amount of
nitrogen in the soil. In some ways this has positive eects because it increases the fertility of the soil
and means that agricultural productivity is high. On the other hand, however, if there is too much
nitrogen in the soil, it can run o into nearby water courses such as rivers or can become part of
the groundwater supply as we mentioned earlier. Increased nitrogen in rivers and dams can lead to a
problem called eutrophication. Eutrophication is the contamination of a water system with excess

nurtrients, which stimulates excessive algae growth at the expense of other parts of the ecosystem.
This occurs as eutrophication reduces oxygen levels in the water. Sometimes this can cause certain
plant species to be favoured over the others and one species may 'take over' the ecosystem, resulting in
a decrease in plant diversity. This is called a 'bloom'. Eutrophication also aects water quality. When
the plants die and decompose, large amounts of oxygen are used up and this can cause other animals
in the water to die.
7.2.1.1 Case Study : Fertiliser use in South Africa

Refer to the data table below, which shows the average fertiliser use (in kilograms per hectare or kg/ha)
over a number of years for South Africa and the world. Then answer the questions that follow:
1965 1970 1975 1980 1985 1990 1995 2000 2002

SA 27.9 42.2 57.7 80.3 66.6 54.9 48.5 47.1 61.4
World 34.0 48.9 63.9 80.6 86.7 90.9 84.9 88.2 91.9
Table 7.1
1. On the same set of axes, draw two line graphs to show how fertiliser use has changed in SA and the
world between 1965 and 2002.
2. Describe the trend you see for...
a. the world
b. South Africa
3. Suggest a reason why the world's fertiliser use has changed in this way over time.
4. Do you see the same pattern for South Africa?
5. Try to suggest a reason for the dierences you see in the fertiliser use data for South Africa.
6. One of the problems with increased fertiliser use is that there is a greater chance of nutrient runo
into rivers and dams and therefore a greater danger of eutrophication. In groups of 5-6, discuss the
following questions:
a. What could farmers do to try to reduce the risk of nutrient runo from elds into water systems?
Try to think of at least 3 dierent strategies that they could use.
b. Imagine you are going to give a presentation on eutrophication to a group of farmers who know
nothing about it. How will you educate them about the dangers? How will you convince them
that it is in their interests to change their farming practices? Present your ideas to the class.
7.2.2 The industrial xation of nitrogen

A number of industrial processes are able to x nitrogen into dierent compounds and then convert these
compounds into fertilisers. In the descriptions below, you will see how atmospheric nitrogen is xed to
produce ammonia, how ammonia is then reacted with oxygen to form nitric acid and how nitric acid and
ammonia are then used to produce the fertiliser, ammonium nitrate.
• Preparation of ammonia (NH3 ) The industrial preparation of ammonia is known as the Haber-
Bosch process. At a high pressure and a temperature of approximately 5000 C, and in the presence
of a suitable catalyst (usually iron), nitrogen and hydrogen react according to the following equation:
N2 + 3H2 → 2NH3 Ammonia is used in the preparation of artcial fertilisers such as (NH4 )2 SO4 and
is also used in cleaning agents and cooling installations.

93
note: Fritz Haber and Carl Bosch were the two men responsible for developing the Haber-
Bosch process. In 1918, Haber was awarded the Nobel Prize in Chemistry for his work. The
Haber-Bosch process was a milestone in industrial chemistry because it meant that nitrogenous
fertilisers were cheaper and much more easily available. At the time, this was very important
in providing food for the growing human population. Haber also played a major role in the
development of chemical warfare in World War I. Part of this work included the development of
gas masks with absorbent lters. He also led the teams that developed chlorine gas and other
deadly gases for use in trench warfare. His wife, Clara Immerwahr, also a chemist, opposed his
work on poison gas and committed suicide with his service weapon in their garden. During the
1920s, scientists working at his institute also developed the cyanide gas formulation Zyklon
B, which was used as an insecticide and also later, after he left the programme, in the Nazi
extermination camps. Haber was Jewish by birth, but converted from Judaism in order to be
more accepted in Germany. Despite this, he was forced to leave the country in 1933 because
he was Jewish 'by denition' (his mother was Jewish). He died in 1934 at the age of 65.
Many members of his extended family died in the Nazi concentration camps, possibly gassed
by Zyklon B.
• Preparation of nitric acid (HNO3 ) Nitric acid is used to prepare fertilisers and explosives. The
industrial preparation of nitric acid is known as the Ostwald process. The Ostwald process involves
the conversion of ammonia into nitric acid in various stages: Firstly, ammonia is heated with oxygen
in the presence of a platinum catalyst to form nitric oxide and water.
4NH3 (g) + 5O2 (g) → 4NO(g) + 6H2 O(g)
Secondly, nitric oxide reacts with oxygen to form nitrogen dioxide. This gas is then readily absorbed
by the water to produce nitric acid. A portion of nitrogen dioxide is reduced back to nitric oxide.
2NO(g) + O2 (g) → 2NO2 (g) 3NO2 (g) + H2 O(l)→ 2HNO3 (aq) + NO(g)
The NO is recycled and the acid is concentrated to the required strength by a process called distillation.
• Preparation of ammonium nitrate Ammonium nitrate is used as a fertiliser, as an explosive and

also in the preparation of 'laughing gas' which is used as an anaesthetic. Ammonium nitrate is prepared
by reacting ammonia with nitric acid:
NH3 + HNO3 → NH4 NO3
7.2.2.1 Debate : Fertiliser use

Divide the class into two groups to debate the following topic:
Increasing the use of articial fertilisers is the best solution to meet the growing food needs of the world's
human population.
One group should take the position of agreeing with the statement, and the other should disagree. In
your groups, discuss reasons why you have the opinion that you do, and record some notes of your discussion.
Your teacher will then explain to you how to proceed with the debate.
The following presentation shows other ways to represent the nitrogen cycle.

<http://static.slidesharecdn.com/swf/ssplayer2.swf ?doc=n2cycle-100511032952-
phpapp02&stripped_title=n2-cycle&userName=kwarne>
Figure 7.2

7.2.3 Summary
• Nitrogen is essential for life on earth, since it forms part of amino acids, proteins and nucleic
acids.
• The atmosphere is composed mostly of nitrogen gas, but the gas is inert, meaning that it is not
available to living organisms in its gaseous form.
• The nitrogen cycle describes how nitrogen and nitrogen-containing compounds are changed into
dierent forms in nature.
• The nitrogen cycle consists of three major processes: nitrogen xation, nitrication and denitri-
cation.
• Nitrogen xation is the conversion of atmospheric nitrogen into compounds such as ammonia that
are more easily used.
• biologically through the actions of bacteria, industrially through the Haber-
Nitrogen can be xed
Bosch process or by lightning.
• Nitrication converts ammonia into nitrites and nitrates, which can be easily assimilated by
plants.
• Denitrication converts nitrites and nitrates back into gaseous nitrogen to complete the nitrogen
cycle.
• Humans have had a number of impacts on the nitrogen cycle. The production of articial fer-
tilisers for example, means that there is a greater chance of runo into water systems. In some cases,
eutrophication may occur.
• Eutrophication is the contamination of a water system with excess nurtrients, which stimulates
excessive algae growth at the expense of other parts of the ecosystem. This occurs as eutrophication
reduces oxygen levels in the water.
• Many nitrogen gases such as NO, N2 O and NO2 are released by agricultural soils and articial fertilisers.
acid rain. Some of these
These gases may combine with water vapour in the atmosphere and result in
global warming.
gases are also greenhouse gases and may contribute towards
• A number of industrial processes are used to produce articical fertilisers.
• The Haber-Bosch process converts atmsopheric nitrogen into ammonia.
• The Ostwald process reacts ammonia with oxygen to produce nitric acid, which is used in the
preparation of fertilisers and explosives.
• If ammonia and nitric acid react, the product is ammonium nitrate, which is used as a fertiliser and
as an explosive.

1. Look at the diagram and the descriptions of the nitrogen cycle earlier in the chapter:
a. Would you describe the changes that take place in the nitrogen cycle as chemical or physical
changes? Explain your answer.
b. Are the changes that take place in the water cycle physical or chemical changes? Explain your
answer.
3
2. Explain what is meant by each of the following terms:
a. nitrogen xing
b. fertiliser
c. eutrophication
4
3 See the le at <http://cnx.org/content/m39924/latest/http://www.fhsst.org/lik>

95
3. Explain why the xing of atmospheric nitrogen is so important for the survival of life on earth. Click
5
4. Refer to the diagram below and then answer the questions that follow:
Figure 7.3
a. Explain the role of decomposers in the nitrogen cycle.

b. If the process taking place at (3) is nitrication, then label the processes at (1) and (5).
c. Identify the nitrogen products at (2) and (4).
d. On the diagram, indicate the type of bacteria that are involved in each stage of the nitrogen cycle.
e. In industry, what process is used to produce the compound at 2?

f. Does the diagram above show a 'cycle' ? Explain your answer.
6
5. NO and NO2 are both nitrogen compounds:
a. Explain how each of these compounds is formed?

b. What eect does each of these compounds have in the environment?
7
6. There are a number of arguments both 'for' and 'against' the use of articial fertilisers. Draw a table
8
to summarise the advantages and disadvantages of their use. Click here for the solution

7 See the le at <http://cnx.org/content/m39924/latest/http://www.fhsst.org/liX>
8 See the le at <http://cnx.org/content/m39924/latest/http://www.fhsst.org/lil>


Chapter 8
The hydrosphere
8.1 Introduction 1
8.1.1 Introduction
As far as we know, the Earth we live on is the only planet that is able to support life. Amongst other factors,
Earth is just the right distance from the sun to have temperatures that are suitable for life to exist. Also,
the Earth's atmosphere has exactly the right type of gases in the right amounts for life to survive. Our
planet also has water on its surface, which is something very unique. In fact, Earth is often called the 'Blue
Planet' because most of it is covered in water. This water is made up of freshwater in rivers and lakes, the
saltwater of the oceans and estuaries, groundwater and water vapour. Together, all these water bodies are
called the hydrosphere.
8.1.2 Interactions of the hydrosphere

It is important to realise that the hydrosphere interacts with other global systems, including the atmosphere,
lithosphere and biosphere.
• Atmosphere When water is heated (e.g. by energy from the sun), it evaporates and forms water
vapour. When water vapour cools again, it condenses to form liquid water which eventually returns
to the surface by precipitation e.g. rain or snow. This cycle of water moving through the atmosphere
and the energy changes that accompany it, is what drives weather patterns on earth.
• Lithosphere In the lithosphere (the ocean and continental crust at the Earth's surface), water is an
important weathering agent, which means that it helps to break rock down into rock fragments and then
soil. These fragments may then be transported by water to another place, where they are deposited.
This is called erosion. These two process, i.e. weathering and erosion, help to shape the earth's
surface. You can see this for example in rivers. In the upper streams, rocks are eroded and sediments
are transported down the river and deposited on the wide ood plains lower down. On a bigger scale,
river valleys in mountains have been carved out by the action of water, and clis and caves on rocky
beach coastlines are also the result of weathering and erosion by water.
• Biosphere In the biosphere, land plants absorb water through their roots and then transport this
through their vascular (transport) system to stems and leaves. This water is needed in photosynthesis,
the food production process in plants. Transpiration (evaporation of water from the leaf surface) then
returns water back to the atmosphere.
97
98 CHAPTER 8. THE HYDROSPHERE
8.1.3 Exploring the Hydrosphere

The large amount of water on our planet is something quite unique. In fact, about 71% of the earth is
covered by water. Of this, almost 97% is found in the oceans as saltwater, about 2.2% occurs as a solid in ice
sheets, while the remaining amount (less than 1%) is available as freshwater. So from a human perspective,
despite the vast amount of water on the planet, only a very small amount is actually available for human
consumption (e.g. drinking water). Before we go on to look more closely at the chemistry of the hydrosphere,
we are going to spend some time exploring a part of the hydrosphere in order to start appreciating what a
complex and beautiful part of the world it is.
8.1.3.1 Investigation : Investigating the hydrosphere

1. Choosing a study site: For this exercise, you can choose any part of the hydrosphere that you
would like to explore. This may be a rock pool, a lake, river, wetland or even just a small pond. The
guidelines below will apply best to a river investigation, but you can ask similar questions and gather
similar data in other areas. When choosing your study site, consider how accessible it is (how easy is
it to get to?) and the problems you may experience (e.g. tides, rain).
2. Collecting data: Your teacher will provide you with the equipment you need to collect the following
data. You should have at least one study site where you will collect data, but you might decide to have
more if you want to compare your results in dierent areas. This works best in a river, where you can
choose sites down its length.
a. Chemical data Measure and record data such as temperature, pH, conductivity and dissolved
oxygen at each of your sites. You may not know exactly what these measurements mean right
now, but it will become clearer later in the chapter.
b. Hydrological data Measure the water velocity of the river and observe how the volume of water
in the river changes as you move down its length. You can also collect a water sample in a clear
bottle, hold it to the light and see whether the water is clear or whether it has particles in it.
c. Biological data What types of animals and plants are found in or near this part of the hydrosphere?
Are they specially adapted to their environment?
Record your data in a table like the one shown below:
Site 1 Site 2 Site 3

Temperature
pH
Conductivity
Dissolved oxygen
Animals and plants
Table 8.1
3. Interpreting the data: Once you have collected and recorded your data, think about the following
questions:
• How does the data you have collected vary at dierent sites?
• Can you explain these dierences?
• What eect do you think temperature, dissolved oxygen and pH have on animals and plants that
are living in the hydrosphere?
2+ 2+ −
• Water is seldom 'pure'. It usually has lots of things dissolved (e.g. Mg , Ca and NO3 ions)
or suspended (e.g. soil particles, debris) in it. Where do these substances come from?
• Are there any human activities near this part of the hydrosphere? What eect could these
activities have on the hydrosphere?

99
8.1.4 The Importance of the Hydrosphere

It is so easy sometimes to take our hydrosphere for granted and we seldom take the time to really think about
the role that this part of the planet plays in keeping us alive. Below are just some of the very important
functions of water in the hydrosphere:
• Water is a part of living cells Each cell in a living organism is made up of almost 75% water, and
this allows the cell to function normally. In fact, most of the chemical reactions that occur in life,
involve substances that are dissolved in water. Without water, cells would not be able to carry out
their normal functions and life could not exist.
• Water provides a habitat The hydrosphere provides an important place for many animals and plants
− −
to live. Many gases (e.g. CO2 , O2 ), nutrients e.g. nitrate (NO3 ), nitrite (NO2 ) and ammonium
+ 2+ 2+
(NH4 ) ions, as well as other ions (e.g. Ca and Mg ) are dissolved in water. The presence of these
substances is critical for life to exist in water.
• Regulating climate One of water's unique characteristics is its high specic heat. This means that
water takes a long time to heat up and also a long time to cool down. This is important in helping
to regulate temperatures on earth so that they stay within a range that is acceptable for life to exist.
Ocean currents also help to disperse heat.
• Human needs Humans use water in a number of ways. Drinking water is obviously very important,
but water is also used domestically (e.g. washing and cleaning) and in industry. Water can also be
used to generate electricity through hydropower.
These are just a few of the very important functions that water plays on our planet. Many of the functions
of water relate to its chemistry and to the way in which it is able to dissolve substances in it.
8.2 Ions in aqueous solution 2
8.2.1 Ions in aqueous solution

As we mentioned earlier, water is seldom pure. Because of the structure of the water molecule, it is able
to dissolve substances in it. This is very important because if water wasn't able to do this, life would not
be able to survive. In rivers and the oceans for example, dissolved oxygen means that organisms (such as
sh) are still able to respire (breathe). For plants, dissolved nutrients are also available. In the human body,
water is able to carry dissolved substances from one part of the body to another.
Many of the substances that dissolve are ionic and when they dissolve they form ions in solution. We
are going to look at how water is able to dissolve ionic compounds, how these ions maintain a balance in the
human body, how they aect water hardness and how specic ions determine the pH of solutions.
8.2.1.1 Dissociation in water

Water is a polar molecule (Figure 8.1). This means that one part of the molecule has a slightly positive
charge (positive pole) and the other part has a slightly negative charge (negative pole).
Figure 8.1: Water is a polar molecule

It is the polar nature of water that allows ionic compounds to dissolve in it. In the case of sodium
+
chloride (NaCl) for example, the positive sodium ions (Na ) will be attracted to the negative pole of the
−
water molecule, while the negative chloride ions (Cl ) will be attracted to the positive pole of the water
molecule. In the process, the ionic bonds between the sodium and chloride ions are weakened and the water
molecules are able to work their way between the individual ions, surrounding them and slowly dissolving the
compound. This process is calleddissociation. A simplied representation of this is shown in Figure 8.2.
Denition 8.1: Dissociation

Dissociation in chemistry and biochemistry is a general process in which ionic compounds separate
or split into smaller molecules or ions, usually in a reversible manner.
Figure 8.2: Sodium chloride dissolves in water
The dissolution of sodium chloride can be represented by the following equation:

+ −
NaCl(s) → Na (aq) + Cl (aq)
The symbols s (solid), l (liquid), g (gas) and aq (material is dissolved in water) are written after the
chemical formula to show the state or phase of the material. The dissolution of potassium sulphate into
potassium and sulphate ions is shown below as another example:
+ 2−
K2 SO4 (s)→ 2K (aq) + SO4 (aq)
Remember that molecular substances (e.g. covalent compounds) may also dissolve, but most will not
form ions. One example is sugar.
C6 H12 O6 (s)
C6 H12 O6 (aq)
There are exceptions to this and some molecular substances will form ions when they dissolve. Hydrogen
chloride for example can ionise to form hydrogen and chloride ions.
+ −
HCl(g) → H (aq) + Cl (aq)
note: The ability of ionic compounds to dissolve in water is extremely important in the human
body! The body is made up of cells, each of which is surrounded by a membrane. Dissolved ions are
found inside and outside of body cells in dierent concentrations. Some of these ions are positive
2+ −
(e.g. Mg ) and some are negative (e.g. Cl ). If there is a dierence in the charge that is inside
and outside the cell, then there is a potential dierence across the cell membrane. This is called
the membrane potential of the cell. The membrane potential acts like a battery and aects
the movement of all charged substances across the membrane. Membrane potentials play a role
in muscle functioning, digestion, excretion and in maintaining blood pH to name just a few. The
movement of ions across the membrane can also be converted into an electric signal that can be
transferred along neurons (nerve cells), which control body processes. If ionic substances were not
able to dissociate in water, then none of these processes would be possible! It is also important to
realise that our bodies can lose +
ions such as Na , K , Ca
+ 2+ 2+
, Mg
−
, and Cl , for example when we
sweat during exercise. Sports drinks such as Lucozade and Powerade are designed to replace these
lost ions so that the body's normal functioning is not aected.
8.2.1.1.1 Ions in solution

1. For each of the following, say whether the substance is ionic or molecular.
a. potassium nitrate (KNO3 )

101
b. ethanol (C2 H5 OH)

c. sucrose (a type of sugar) (C12 H22 O11 )
d. sodium bromide (NaBr)
3
2. Write a balanced equation to show how each of the following ionic compounds dissociate in water.
a. sodium sulphate (Na2 SO4 )

b. potassium bromide (KBr)
c. potassium permanganate (KMNO4 )
d. sodium phosphate (Na3 PO4 )
4
8.2.1.2 Ions and water hardness

Denition 8.2: Water hardness
Water hardness is a measure of the mineral content of water. Minerals are substances such as
calcite, quartz and mica that occur naturally as a result of geological processes.
Hard water is water that has a high mineral content. Water that has a low mineral content is known
as soft water. If water has a high mineral content, it usually contains high levels of metal ions, mainly
calcium (Ca) and magnesium (Mg). The calcium enters the water from either CaCO3 (limestone or chalk)
or from mineral deposits of CaSO4 . The main source of magnesium is a sedimentary rock called dolomite,
CaMg(CO3 )2 . Hard water may also contain other metals as well as bicarbonates and sulphates.
note: The simplest way to check whether water is hard or soft is to use the lather/froth test. If
the water is very soft, soap will lather more easily when it is rubbed against the skin. With hard
water this won't happen. Toothpaste will also not froth well in hard water.
A water softener works on the principle of ion exchange. Hard water passes through a media bed,
usually made of resin beads that are supersaturated with sodium. As the water passes through the beads,
the hardness minerals (e.g. calcium and magnesium) attach themselves to the beads. The sodium that was
originally on the beads is released into the water. When the resin becomes saturated with calcium and
magnesium, it must be recharged. A salt solution is passed through the resin. The sodium replaces the
calcium and magnesium and these ions are released into the waste water and discharged.
8.2.1.3 The pH scale

The concentration of specic ions in solution aects whether the solution is acidic or basic. You will learn
about acids and bases in Grade 11. Acids and bases can be described as substances that either increase or
decrease the concentration of hydrogen (H
+ +
or H3 O ) ions in a solution. An acid increases the hydrogen ion
concentration in a solution, while a base decreases the hydrogen ion concentration. pH is used to measure
whether a substance is acidic or basic (alkaline).
Denition 8.3: pH
pH is a measure of the acidity or alkalinity of a solution. The pH scale ranges from 0 to 14.
Solutions with a pH less than seven are acidic, while those with a pH greater than seven are basic
(alkaline). pH 7 is considered neutral.
pH can be calculated using the following equation:
pH = −log H +

(8.1)
3 See the le at <http://cnx.org/content/m39937/latest/http://www.fhsst.org/l33>

4 See the le at <http://cnx.org/content/m39937/latest/http://www.fhsst.org/l3q>

or
pH = −log H3 O+

(8.2)
The brackets in the above equation are used to show concentration in mol·dm−3 .
Exercise 8.2.1: pH calculations (Solution on p. 116.)
Calculate the pH of a solution where the concentration of hydrogen ions is
−7 −3
1 × 10 mol·dm .
Exercise 8.2.2: pH calculations (Solution on p. 116.)

In a solution of ethanoic acid (or acetic acid), the following equilibrium is established:
− +
CH3 COOH (aq) + H2 O
CH3 COO (aq) + H3 O
− −3
The concentration of CH3 COO ions is found to be 0,003 mol·dm . Calculate the pH of the
solution.
Understanding pH is very important. In living organisms, it is necessary to maintain a constant pH so that

chemical reactions can occur under optimal conditions.
tip: It may also be useful for calculations involving the pH scale, to know that the following
+ − −14
equation can also be used: [H3 O ][OH ] = 1 × 10
note: A build up of acid in the human body can be very dangerous. Lactic acidosis is a condition
caused by the buildup of lactic acid in the body. It leads to acidication of the blood (acidosis) and
can make a person very ill. Some of the symptoms of lactic acidosis are deep and rapid breathing,
vomiting and abdominal pain. In the ght against HIV, lactic acidosis is a problem. One of the
antiretrovirals (ARV's) that is used in anti-HIV treatment is Stavudine (also known as Zerit or
d4T). One of the side eects of Stavudine is lactic acidosis, particularly in overweight women. If it
is not treated quickly, it can result in death.
In agriculture, farmers need to know the pH of their soils so that they are able to plant the right kinds of
crops. The pH of soils can vary depending on a number of factors such as rainwater, the kinds of rocks and
materials from which the soil was formed and also human inuences such as pollution and fertilisers. The
pH of rain water can also vary and this too has an eect on agriculture, buildings, water courses, animals
and plants. Rainwater is naturally acidic because carbon dioxide in the atmosphere combines with water to
form carbonic acid. Unpolluted rainwater has a pH of approximately 5,6. However, human activities can
alter the acidity of rain and this can cause serious problems such as acid rain.
8.2.1.3.1 Calculating pH
1. Calculate the pH of each of the following solutions:
−3
a. A 0,2 mol·dm KOH solution
−3
b. A 0,5 mol·dm HCl solution
5
−3 +
2. What is the concentration (in mol·dm ) of H3 O ions in a NaOH solution which has a pH of 12?
6
+ −
3. The concentrations of hydronium (H3 O ) and hydroxyl (OH ) ions in a typical sample of seawater
−8 −3 −6 −3
are 10 mol·dm and 10 mol·dm respectively.
a. Is the seawater acidic or basic?

b. What is the pH of the seawater?
c. Give a possible explanation for the pH of the seawater.


103
(IEB Paper 2, 2002)

7
The following simulation allows you to test the pH of various substances.
Figure 8.3
8
run demo
8.2.1.4 Acid rain

The acidity of rainwater comes from the natural presence of three substances (CO2 , NO, and SO2 ) in the
lowest layer of the atmosphere. These gases are able to dissolve in water and therefore make rain more
acidic than it would otherwise be. Of these gases, carbon dioxide (CO2 ) has the highest concentration and
therefore contributes the most to the natural acidity of rainwater. We will look at each of these gases in
turn.
Denition 8.4: Acid rain

Acid rain refers to the deposition of acidic components in rain, snow and dew. Acid rain occurs
when sulphur dioxide and nitrogen oxides are emitted into the atmosphere, undergo chemical trans-
formations and are absorbed by water droplets in clouds. The droplets then fall to earth as rain,
snow, mist, dry dust, hail, or sleet. This increases the acidity of the soil and aects the chemical
balance of lakes and streams.
1. Carbon dioxide Carbon dioxide reacts with water in the atmosphere to form carbonic acid (H 2 CO3 ).
CO2 + H2 O → H2 CO3
The carbonic acid dissociates to form hydrogen and hydrogen carbonate ions. It is the presence of
hydrogen ions that lowers the pH of the solution making the rain acidic.
+ −
H2 CO3 →H + HCO3
2. Nitric oxide Nitric oxide (NO) also contributes to the natural acidity of rainwater and is formed
during lightning storms when nitrogen and oxygen react. In air, NO is oxidised to form nitrogen
dioxide (NO2 ). It is the nitrogen dioxide which then reacts with water in the atmosphere to form
nitric acid (HNO 3 ).
3NO2 (g) + H2 O→2HNO3 (aq) + NO(g)

The nitric acid dissociates in water to produce hydrogen ions and nitrate ions. This again lowers the
pH of the solution making it acidic.
+ −
HNO3 → H + NO3
3. Sulphur dioxide Sulphur dioxide in the atmosphere rst reacts with oxygen to form sulphur trioxide,
sulphuric acid.
before reacting with water to form
2SO2 + O2 → 2SO3
SO3 + H2 O→ H2 SO4
Sulphuric acid dissociates in a similar way to the previous reactions.
− +
H2 SO4 → HSO4 + H
7 See the le at <http://cnx.org/content/m39937/latest/http://www.fhsst.org/l3T>

8 http://phet.colorado.edu/sims/ph-scale/ph-scale_en.jnlp

Although these reactions do take place naturally, human activities can greatly increase the concentration of
these gases in the atmosphere, so that rain becomes far more acidic than it would otherwise be. The burning
of fossil fuels in industries, vehicles etc is one of the biggest culprits. If the acidity of the rain drops to below
5, it is referred to as acid rain.
Acid rain can have a very damaging eect on the environment. In rivers, dams and lakes, increased
acidity can mean that some species of animals and plants will not survive. Acid rain can also degrade soil
minerals, producing metal ions that are washed into water systems. Some of these ions may be toxic e.g.
3+
Al . From an economic perspective, altered soil pH can drastically aect agricultural productivity.
Acid rain can also aect buildings and monuments, many of which are made from marble and limestone.
A chemical reaction takes place between CaCO3 (limestone) and sulphuric acid to produce aqueous ions
which can be easily washed away. The same reaction can occur in the lithosphere where limestone rocks are
present e.g. limestone caves can be eroded by acidic rainwater.
H2 SO4 + CaCO3 → CaSO4 . H2 O + CO2
8.2.1.4.1 Investigation : Acid rain

You are going to test the eect of 'acid rain' on a number of substances.
Materials needed:
samples of chalk, marble, zinc, iron, lead, dilute sulphuric acid, test tubes, beaker, glass dropper
Method:
1. Place a small sample of each of the following substances in a separate test tube: chalk, marble, zinc,
iron and lead
2. To each test tube, add a few drops of dilute sulphuric acid.
3. Observe what happens and record your results.
Discussion questions:
• In which of the test tubes did reactions take place? What happened to the sample substances?
• What do your results tell you about the eect that acid rain could have on each of the following:
buildings, soils, rocks and geology, water ecosystems?
• What precautions could be taken to reduce the potential impact of acid rain?
8.3 Electrolytes, ionisation and conductivity 9
8.3.1 Electrolytes, ionisation and conductivity

Conductivity in aqueous solutions, is a measure of the ability of water to conduct an electric current. The
more ions there are in the solution, the higher its conductivity.
Denition 8.5: Conductivity

Conductivity is a measure of a solution's ability to conduct an electric current.
8.3.1.1 Electrolytes
An electrolyte is a material that increases the conductivity of water when dissolved in it. Electrolytes can
be further divided into strong electrolytes and weak electrolytes.
Denition 8.6: Electrolyte

An electrolyte is a substance that contains free ions and behaves as an electrically conductive
medium. Because they generally consist of ions in solution, electrolytes are also known as ionic
solutions.

105
1. Strong electrolytes A strong electrolyte is a material that ionises completely when it is dissolved in
water:
AB (s, l, g) → A+ (aq) + B − (aq) (8.3)
This is a chemical change because the original compound has been split into its component ions
and bonds have been broken. In a strong electrolyte, we say that the extent of ionisation is high. In
other words, the original material dissociates completely so that there is a high concentration of ions
in the solution. An example is a solution of potassium nitrate:
KN O3 (s) → K + (aq) + N O3− (aq) (8.4)
2. Weak electrolytes A weak electrolyte is a material that goes into solution and will be surrounded
by water molecules when it is added to water. However, not all of the molecules will dissociate into
ions. The extent of ionisation of a weak electrolyte is low and therefore the concentration of ions in
the solution is also low.
AB (s, l, g) → AB (aq)
A+ (aq) + B − (aq) (8.5)
The following example shows that in the nal solution of a weak electrolyte, some of the original
compound plus some dissolved ions are present.
C2 H3 O2 H (l) → C2 H3 O2 H
C2 H3 O2− (aq) + H + (aq) (8.6)
8.3.1.2 Non-electrolytes
A non-electrolyte is a material that does not increase the conductivity of water when dissolved in it.
The substance goes into solution and becomes surrounded by water molecules, so that the molecules of the
chemical become separated from each other. However, although the substance does dissolve, it is not changed
in any way and no chemical bonds are broken. The change is a physical change. In the oxygen example
below, the reaction is shown to be reversible because oxygen is only partially soluble in water and comes out
of solution very easily.
C2 H5 OH (l) → C2 H5 OH (aq) (8.7)
O2 (g)
O2 (aq) (8.8)
8.3.1.3 Factors that aect the conductivity of water

The conductivity of water is therefore aected by the following factors:
• The type of substance that dissolves in water. Whether a material is a strong electrolyte (e.g.
potassium nitrate, KNO3 ), a weak electrolyte (e.g. acetate, CH3 COOH) or a non-electrolyte (e.g.
sugar, alcohol, oil) will aect the conductivity of water because the concentration of ions in solution
will be dierent in each case.
• The concentration of ions in solution. The higher the concentration of ions in solution, the higher
its conductivity will be.
• Temperature. The warmer the solution, the higher the solubility of the material being dissolved and
therefore the higher the conductivity as well.

8.3.1.3.1 Experiment : Electrical conductivity

Aim:
To investigate the electrical conductivities of dierent substances and solutions.
Apparatus:
Solid salt (NaCl) crystals; dierent liquids such as distilled water, tap water, seawater, benzene and
alcohol; solutions of salts e.g. NaCl, KBr; a solution of an acid (e.g. HCl) and a solution of a base (e.g.
NaOH); torch cells; ammeter; conducting wire, crocodile clips and 2 carbon rods.
Method:
Set up the experiment by connecting the circuit as shown in the diagram below. In the diagram, 'X'
represents the substance or solution that you will be testing. When you are using the solid crystals, the
crocodile clips can be attached directly to each end of the crystal. When you are using solutions, two carbon
rods are placed into the liquid and the clips are attached to each of the rods. In each case, complete the
circuit and allow the current to ow for about 30 seconds. Observe whether the ammeter shows a reading.
Figure 8.4
Results:
Record your observations in a table similar to the one below:
Test substance Ammeter reading
Table 8.2
What do you notice? Can you explain these observations?

Remember that for electricity to ow, there needs to be a movement of charged particles e.g. ions. With
the solid NaCl crystals, there was no ow of electricity recorded on the ammeter. Although the solid is made
up of ions, they are held together very tightly within the crystal lattice and therefore no current will ow.
Distilled water, benzene and alcohol also don't conduct a current because they are covalent compounds and
therefore do not contain ions.
The ammeter should have recorded a current when the salt solutions and the acid and base solutions
were connected in the circuit. In solution, salts dissociate into their ions, so that these are free to move in
the solution. Acids and bases behave in a similar way and dissociate to form hydronium and oxonium ions.
Look at the following examples:
KBr → K + + Br−
NaCl → Na+ + Cl−
HCl + H2 O → H3 O+ + Cl−
NaOH → Na+ + OH−
Conclusions:
Solutions that contain free-moving ions are able to conduct electricity because of the movement of charged
particles. Solutions that do not contain free-moving ions do not conduct electricity.

107
note: Conductivity in streams and rivers is aected by the geology of the area where the water is
owing through. Streams that run through areas with granite bedrock tend to have lower conduc-
tivity because granite is made of materials that do not ionise when washed into the water. On the
other hand, streams that run through areas with clay soils tend to have higher conductivity because
the materials ionise when they are washed into the water. Pollution can also aect conductivity.
A failing sewage system or an inow of fertiliser runo would raise the conductivity because of the
presence of chloride, phosphate, and nitrate (ions) while an oil spill (non-ionic) would lower the
conductivity. It is very important that conductivity is kept within a certain acceptable range so
that the organisms living in these water systems are able to survive.
8.4 Precipitation reactions 10
8.4.1 Precipitation reactions

Sometimes, ions in solution may react with each other to form a new substance that is insoluble. This is
precipitate.
called a
Denition 8.7: Precipitate

A precipitate is the solid that forms in a solution during a chemical reaction.
8.4.1.1 Demonstration : The reaction of ions in solution

Apparatus and materials:
4 test tubes; copper(II) chloride solution; sodium carbonate solution; sodium sulphate solution
Figure 8.5
Method:
1. Prepare 2 test tubes with approximately 5 ml of dilute Cu(II) chloride solution in each
2. Prepare 1 test tube with 5 ml sodium carbonate solution
3. Prepare 1 test tube with 5 ml sodium sulphate solution
4. Carefully pour the sodium carbonate solution into one of the test tubes containing copper(II) chloride
and observe what happens
5. Carefully pour the sodium sulphate solution into the second test tube containing copper(II) chloride
and observe what happens
Results:
1. A light blue precipitate forms when sodium carbonate reacts with copper(II) chloride
2. No precipitate forms when sodium sulphate reacts with copper(II) chloride
It is important to understand what happened in the previous demonstration. We will look at what happens
in each reaction, step by step.

1. Reaction 1: Sodium carbonate reacts with copper(II) chloride.

2+ − + 2−
When these compounds react, a number of ions are present in solution: Cu , Cl , Na and CO3 .
Because there are lots of ions in solution, they will collide with each other and may recombine in
dierent ways. The product that forms may be insoluble, in which case a precipitate will form, or the
product will be soluble, in which case the ions will go back into solution. Let's see how the ions in this
example could have combined with each other:
2+ 2−
Cu + CO3 → CuCO3
2+ −
Cu + 2Cl → CuCl2
+ −
Na + Cl → NaCl
+ 2−
Na + CO3 → Na2 CO3
You can automatically exclude the reactions where sodium carbonate and copper(II) chloride are the
products because these were the initial reactants. You also know that sodium chloride (NaCl) is soluble
in water, so the remaining product (copper carbonate) must be the one that is insoluble. It is also
possible to look up which salts are soluble and which are insoluble. If you do this, you will nd that
most carbonates are insoluble, therefore the precipitate that forms in this reaction must be CuCO3 .
The reaction that has taken place between the ions in solution is as follows:
+ 2− 2+ − + −
2Na + CO3 + Cu + 2Cl → CuCO3 + 2Na + 2Cl
2. Reaction 2: Sodium sulphate reacts with copper(II) chloride.
2+ − + 2−
The ions that are present in solution are Cu , Cl , Na and SO4 . The ions collide with each other
and may recombine in dierent ways. The possible combinations of the ions are as follows:
2+ 2−
Cu + SO4 → CuSO4
2+ −
Cu + 2Cl → CuCl2
+ −
Na + Cl → NaCl
+ 2−
Na + SO4 → Na2 SO4
If we look up which of these salts are soluble and which are insoluble, we see that most chlorides and
most sulphates are soluble. This is why no precipitate forms in this second reaction. Even when the
ions recombine, they immediately separate and go back into solution. The reaction that has taken
place between the ions in solution is as follows:
+ 2− 2+ − + 2− 2+ −
2Na + SO4 + Cu + 2Cl → 2Na + SO4 + Cu + 2Cl
Table 8.3 shows some of the general rules about the solubility of dierent salts based on a number of
investigations:
Salt Solubility
Nitrates All are soluble
Potassium, sodium and ammonium salts All are soluble
Chlorides All are soluble except silver chloride, lead(II) chlo-

ride and mercury(II) chloride
Sulphates All are soluble except lead(II) sulphate, barium sul-

phate and calcium sulphate
continued on next page

109
Carbonates All are insoluble except those of potassium, sodium

and ammonium
Table 8.3: General rules for the solubility of salts
8.4.2 Testing for common anions in solution

It is also possible to carry out tests to determine which ions are present in a solution.
8.4.2.1 Test for a chloride

Prepare a solution of the unknown salt using distilled water and add a small amount of silver nitrate
solution. If a white precipitate forms, the salt is either a chloride or a carbonate.
− + − −
Cl + Ag + NO3 → AgCl + NO3 (AgCl is white precipitate)
2− + − −
CO3 + 2Ag + 2NO3 → Ag2 CO3 + 2NO3 (Ag2 CO3 is white precipitate)
The next step is to treat the precipitate with a small amount of concentrated nitric acid. If the
precipitate remains unchanged, then the salt is a chloride. If carbon dioxide is formed, and the precipitate
disappears, the salt is a carbonate.
AgCl + HNO3 → (no reaction; precipitate is unchanged)
Ag2 CO3 + 2HNO3 → 2AgNO3 + H2 O + CO2 (precipitate disappears)
8.4.2.2 Test for a sulphate

Add a small amount of barium chloride solution to a solution of the test salt. If a white precipitate forms,
the salt is either a sulphate or a carbonate.
SO2−
4 + Ba
2+
+ Cl− → BaSO4 + Cl− (BaSO4 is a white precipitate)
2− 2+
CO3 + Ba + Cl− → BaCO3 + Cl− (BaCO3 is a white precipitate)
If the precipitate is treated with nitric acid, it is possible to distinguish whether the salt is a sulphate or
a carbonate (as in the test for a chloride).
BaSO4 + HNO3 → (no reaction; precipitate is unchanged)
BaCO3 + 2HNO3 → Ba(NO3 )2 + H2 O + CO2 (precipitate disappears)
8.4.2.3 Test for a carbonate

If a sample of the dry salt is treated with a small amount of acid, the production of carbon dioxide is a
positive test for a carbonate.
2−
Acid + CO3 → CO2
If the gas is passed through limewater and the solution becomes milky, the gas is carbon dioxide.
Ca(OH)2 + CO2 → CaCO3 + H2 O (It is the insoluble CaCO3 precipitate that makes the limewater go
milky)
8.4.2.4 Test for bromides and iodides

As was the case with the chlorides, the bromides and iodides also form precipitates when they are reacted
with silver nitrate. Silver chloride is a white precipitate, but the silver bromide and silver iodide precipitates
are both pale yellow. To determine whether the precipitate is a bromide or an iodide, we use chlorine water
and carbon tetrachloride (CCl4 ).
Chlorine water frees bromine gas from the bromide and colours the carbon tetrachloride a reddish brown.
Chlorine water frees iodine gas from an iodide and colours the carbon tetrachloride purple.

8.4.2.4.1 Precipitation reactions and ions in solution

1. Silver nitrate (AgNO3 ) reacts with potassium chloride (KCl) and a white precipitate is formed.
a. Write a balanced equation for the reaction that takes place.

b. What is the name of the insoluble salt that forms?
c. Which of the salts in this reaction are soluble?
11
2. Barium chloride reacts with sulphuric acid to produce barium sulphate and hydrochloric acid.
a. Write a balanced equation for the reaction that takes place.

b. Does a precipitate form during the reaction?
c. Describe a test that could be used to test for the presence of barium sulphate in the products.
12
3. A test tube contains a clear, colourless salt solution. A few drops of silver nitrate solution are added
to the solution and a pale yellow precipitate forms. Which one of the following salts was dissolved in
the original solution?
a. NaI
b. KCl
c. K2 CO3
d. Na2 SO4
13
(IEB Paper 2, 2005) Click here for the solution
8.4.3 Threats to the Hydrosphere

It should be clear by now that the hydrosphere plays an extremely important role in the survival of life
on Earth and that the unique properties of water allow various important chemical processes to take place
which would otherwise not be possible. Unfortunately for us however, there are a number of factors that
threaten our hydrosphere and most of these threats are because of human activities. We are going to focus
on two of these issues: overuse and pollution and look at ways in which these problems can possibly be
overcome.
1. Pollution
Pollution of the hydrosphere is also a major problem. When we think of pollution, we sometimes only
think of things like plastic, bottles, oil and so on. But any chemical that is present in the hydrosphere in
an amount that is not what it should be is a pollutant. Animals and plants that live in the hydrosphere
are specially adapted to surviving within a certain range of conditions. If these conditions are changed
(e.g. through pollution), these organisms may not be able to survive. Pollution then, can aect entire
aquatic ecosystems. The most common forms of pollution in the hydrosphere are waste products from
humans and from industries, nutrient pollution e.g. fertiliser runo which causes eutrophication (this
was discussed in chapter ref 7) and toxic trace elements such as aluminium, mercury and copper to
name a few. Most of these elements come from mines or from industries.
2. Overuse of water
We mentioned earlier that only a very small percentage of the hydrosphere's water is available as
freshwater. However, despite this, humans continue to use more and more water to the point where
water consumption is fast approaching the amount of water that is available. The situation is a serious
one, particularly in countries such as South Africa which are naturally dry and where water resources
are limited. It is estimated that between 2020 and 2040, water supplies in South Africa will no longer
be able to meet the growing demand for water in this country. This is partly due to population growth,
11 See the le at <http://cnx.org/content/m39941/latest/http://www.fhsst.org/l3c>

12 See the le at <http://cnx.org/content/m39941/latest/http://www.fhsst.org/l3O>
13 See the le at <http://cnx.org/content/m39941/latest/http://www.fhsst.org/l3x>

111
but also because of the increasing needs of industries as they expand and develop. For each of us, this
should be a very scary thought. Try to imagine a day without water...dicult isn't it? Water is so
much a part of our lives, that we are hardly aware of the huge part that it plays in our daily lives.
8.4.3.1 Discussion : Creative water conservation

As populations grow, so do the demands that are placed on dwindling water resources. While many people
argue that building dams helps to solve this water-shortage problem, the reality is that dams are only a
temporary solution and that they often end up doing far more ecological damage than good. The only
sustainable solution is to reduce the demand for water, so that water supplies are sucient to meet this.
The more important question then is how to do this.
Discussion:
Divide the class into groups, so that there are about ve people in each. Each group is going to represent
a dierent sector within society. Your teacher will tell you which sector you belong to from the following:
Farming, industry, city management or civil society (i.e. you will represent the ordinary 'man on the street').
In your groups, discuss the following questions as they relate to the group of people you represent: (Remember
to take notes during your discussions, and nominate a spokesperson to give feedback to the rest of the class
on behalf of your group)
• What steps could be taken by your group to conserve water?

• Why do you think these steps are not being taken?
• What incentives do you think could be introduced to encourage this group to conserve water more
eciently?
It is important to realise that our hydrosphere exists in a delicate balance with other systems and that
disturbing this balance can have serious consequences for life on this planet.
8.4.3.2 Group Project : School Action Project

There is a lot that can be done within a school to save water. As a class, discuss what actions could be
taken by your class to make people more aware of how important it is to conserve water.
8.4.4 Summary
• The hydrosphere includes all the water that is on Earth. Sources of water include freshwater (e.g.
rivers, lakes), saltwater (e.g. oceans), groundwater (e.g. boreholes) and water vapour. Ice (e.g. glaciers)
is also part of the hydrosphere.
• The hydrosphere interacts with other global systems, including the atmosphere, lithosphere and
biosphere.
• The hydrosphere has a number of important functions. Water is a part of all living cells, it provides a
habitat for many living organisms, it helps to regulate climate and it is used by humans for domestic,
industrial and other use.
• The polar nature of water means that ionic compounds dissociate easily in aqueous solution into
their component ions.
• Ions in solution play a number of roles. In the human body for example, ions help to regulate the
internal environment (e.g. controlling muscle function, regulating blood pH). Ions in solution also
determine water hardness and pH.
• Water hardness is a measure of the mineral content of water. Hard water has a high mineral
concentration and generally also a high concentration of metal ions e.g. calcium and magnesium. The
opposite is true for soft water.
• pH is a measure of the concentration of hydrogen ions in solution. The formula used to calculate pH
+ +
is as follows: pH = -log[H3 O ] or pH = -log[H ]. A solution with a pH less than 7 is considered acidic
and more than 7 is considered basic (or alkaline). A neutral solution has a pH of 7.

2−
• Gases such as CO2 , NO2 and SO4 dissolve in water to form weak acid solutions. Rain is naturally
acidic because of the high concentrations of carbon dioxide in the atmosphere. Human activities such
as burning fossil fuels, increase the concentration of these gases in the atmosphere, resulting in acid
rain.
• Conductivity is a measure of a solution's ability to conduct an electric current.
• An electrolyte is a substance that contains free ions and is therefore able to conduct an electric
current. Electrolytes can be divided into strong and weak electrolytes, based on the extent to which
the substance ionises in solution.
• A non-electrolyte cannot conduct an electric current because it dooes not contain free ions.
• The type of substance, the concentration of ions and the temperature of the solution aect its
conductivity.
• A precipitate is formed when ions in solution react with each other to form an insoluble product.
Solubility 'rules' help to identify the precipitate that has been formed.
• A number of tests can be used to identify whether certain anions are present in a solution.
• Despite the importance of the hydrosphere, a number of factors threaten it. These include overuse of
water, and pollution.

1. Give one word for each of the following descriptions:
a. the change in phase of water from a gas to a liquid

b. a charged atom
c. a term used to describe the mineral content of water
d. a gas that forms sulphuric acid when it reacts with water
14
2. Match the information in column A with the information in column B by writing only the letter (A to
I) next to the question number (1 to 7)
Column A Column B
1. A polar molecule A. H2 SO4
2. molecular solution B. CaCO3
3. Mineral that increases water hardness C. NaOH
4. Substance that increases the hydrogen ion concentration D. salt water
5. A strong electrolyte E. calcium
6. A white precipitate F. carbon dioxide
7. A non-conductor of electricity G. potassium nitrate
H. sugar water
I. O2
Table 8.4
15
3. For each of the following questions, choose the one correct answer from the list provided.
a. Which one of the following substances does not conduct electricity in the solid phase but is an
electrical conductor when molten?
14 See the le at <http://cnx.org/content/m39941/latest/http://www.fhsst.org/l3a>
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113
i. Cu
ii. PbBr2
iii. H2 O
iv. I2
16
b. The following substances are dissolved in water. Which one of the solutions is basic?
i. sodium nitrate
ii. calcium sulphate
iii. ammonium chloride
iv. potassium carbonate
17
−8 −6
4. The concentration of hydronium and hydroxyl ions in a typical sample of seawater are 10 and 10
respectively.
a. Is the seawater acidic or basic?

b. Calculate the pH of this seawater.
5. Three test tubes (X, Y and Z) each contain a solution of an unknown potassium salt. The following
observations were made during a practical investigation to identify the solutions in the test tubes: A: A
white precipitate formed when silver nitrate (AgNO3 ) was added to test tube Z. B: A white precipitate
formed in test tubes X and Y when barium chloride (BaCl2 ) was added. C: The precipitate in test
tube X dissolved in hydrochloric acid (HCl) and a gas was released. D: The precipitate in test tube Y
was insoluble in hydrochloric acid.
a. Use the above information to identify the solutions in each of the test tubes X, Y and Z.
b. Write a chemical equation for the reaction that took place in test tube X before hydrochloric acid
was added.
18
(DoE Exemplar Paper 2 2007) Click here for the solution

17 See the le at <http://cnx.org/content/m39941/latest/http://www.fhsst.org/l3r>
18 See the le at <http://cnx.org/content/m39941/latest/http://www.fhsst.org/l3Y>


−7 −3
Step 1. In this example, the concentration has been given and is 1 × 10 mol·dm
+
Step 2. pH = -log[H ]
−7
= -log(1 × 10 )
= 7

− +
Step 1. According to the balanced equation for this reaction, the mole ratio of CH3 COO ions to H3 O ions
is the same, therefore the concentration of these two ions in the solution will also be the same. So,
+ −3
[H3 O ] = 0,003 mol·dm .
+
Step 2. pH = -log[H3 O ]
= -log(0,003)
= 2,52

Chapter 9
Units 1
9.1 Introduction
Imagine you had to make curtains and needed to buy fabric. The shop assistant would need to know how
much fabric you needed. Telling her you need fabric 2 wide and 6 long would be insucient you have to
specify the unit (i.e. 2 metres wide and 6 metres long). Without the unit the information is incomplete and
the shop assistant would have to guess. If you were making curtains for a doll's house the dimensions might
be 2 centimetres wide and 6 centimetres long!
It is not just lengths that have units, all physical quantities have units (e.g. time, temperature, distance,
etc.).
Denition 9.1: Physical Quantity

A physical quantity is anything that you can measure. For example, length, temperature, distance
and time are physical quantities.
9.2 Unit Systems

9.2.1 SI Units
We will be using the SI units in this course. SI units are the internationally agreed upon units. Histori-
cally these units are based on the metric system which was developed in France at the time of the French
Revolution.
Denition 9.2: SI Units

The name SI units comes from the French Système International d'Unités, which means interna-
tional system of units.
There are seven base SI units. These are listed in Table 9.1. All physical quantities have units which can
be built from these seven base units. So, it is possible to create a dierent set of units by dening a dierent
set of base units.
These seven units are called base units because none of them can be expressed as combinations of the
other six. This is identical to bricks and concrete being the base units of a building. You can build dierent
things using dierent combinations of bricks and concrete. The 26 letters of the alphabet are the base units
for a language like English. Many dierent words can be formed by using these letters.
115
116 CHAPTER 9. UNITS
Base quantity Name Symbol

length metre m
mass kilogram kg
time second s
electric current ampere A
temperature kelvin K
amount of substance mole mol
luminous intensity candela cd
Table 9.1: SI Base Units
9.2.2 The Other Systems of Units

The SI Units are not the only units available, but they are most widely used. In Science there are three
other sets of units that can also be used. These are mentioned here for interest only.
9.2.2.1 c.g.s. Units

In the c.g.s. system, the metre is replaced by the centimetre and the kilogram is replaced by the gram. This
is a simple change but it means that all units derived from these two are changed. For example, the units of
force and work are dierent. These units are used most often in astrophysics and atomic physics.
9.2.2.2 Imperial Units

Imperial units arose when kings and queens decided the measures that were to be used in the land. All the
imperial base units, except for the measure of time, are dierent to those of SI units. This is the unit system
you are most likely to encounter if SI units are not used. Examples of imperial units are pounds, miles,
gallons and yards. These units are used by the Americans and British. As you can imagine, having dierent
units in use from place to place makes scientic communication very dicult. This was the motivation for
adopting a set of internationally agreed upon units.
9.2.2.3 Natural Units

This is the most sophisticated choice of units. Here the most fundamental discovered quantities (such as the
speed of light) are set equal to 1. The argument for this choice is that all other quantities should be built
from these fundamental units. This system of units is used in high energy physics and quantum mechanics.
9.3 Writing Units as Words or Symbols

Unit names are always written with a lowercase rst letter, for example, we write metre and litre. The
symbols or abbreviations of units are also written with lowercase initials, for example m for metre and ` for
litre. The exception to this rule is if the unit is named after a person, then the symbol is a capital letter.
For example, the kelvin was named after Lord Kelvin and its symbol is K. If the abbreviation of the unit
that is named after a person has two letters, the second letter is lowercase, for example Hz for hertz.

117
9.3.1 Naming of Units

For the following symbols of units that you will come across later in this book, write whether you think the
unit is named after a person or not.
1. J (joule)
2. ` (litre)
3. N (newton)
4. mol (mole)
5. C (coulomb)
6. lm (lumen)
7. m (metre)
8. bar (bar)
2
Click here for the solution.
9.4 Combinations of SI Base Units

To make working with units easier, some combinations of the base units are given special names, but it is
always correct to reduce everything to the base units. Table 9.2 lists some examples of combinations of SI
base units that are assigned special names. Do not be concerned if the formulae look unfamiliar at this stage
- we will deal with each in detail in the chapters ahead (as well as many others)!
newton (N) is
It is very important that you are able to recognise the units correctly. For instance, the
another name for the kilogram metre per second squared (kg·m·s−2 ), while the kilogram metre squared
) is called the joule (J).
2 −2
per second squared (kg·m ·s
Quantity Formula Unit Expressed in Base Units Name of Combination

Force ma kg·m·s
−2
N (newton)
1 −1
Frequency s Hz (hertz)
T
2 −2
Work Fs kg·m ·s J (joule)
Table 9.2: Some examples of combinations of SI base units assigned special names
tip: When writing combinations of base SI units, place a dot (·) between the units to indicate that
dierent base units are used. For example, the symbol for metres per second is correctly written as
−1 −1
m·s , and not as ms or m/s. Although the last two options will be accepted in tests and exams,
we will only use the rst one in this book.
9.5 Rounding, Scientic Notation and Signicant Figures

9.5.1 Rounding O
Certain numbers may take an innite amount of paper and ink to write out. Not only is that impossible, but
writing numbers out to a high precision (many decimal places) is very inconvenient and rarely gives better
answers. For this reason we often estimate the number to a certain number of decimal places. Rounding o
or approximating a decimal number to a given number of decimal places is the quickest way to approximate
a number. For example, if you wanted to round-o 2, 6525272 to three decimal places then you would rst
count three places after the decimal. 2, 652|5272 All numbers to the right of | are ignored after you determine
2 http://www.fhsst.org/lOX

whether the number in the third decimal place must be rounded up or rounded down. You round up the
nal digit (make the digit one more) if the rst digit after the | was greater or equal to 5 and round down
(leave the digit alone) otherwise. So, since the rst digit after the | is a 5, we must round up the digit in the
third decimal place to a 3 and the nal answer of 2, 6525272 rounded to three decimal places is 2,653.
Exercise 9.1: Rounding-o (Solution on p. 128.)

Round o π = 3, 141592654... to 4 decimal places.
Exercise 9.2: Rounding-o (Solution on p. 128.)

Round o 9, 191919... to 2 decimal places
9.5.2 Error Margins

In a calculation that has many steps, it is best to leave the rounding o right until the end. For example,
Jack and Jill walk to school. They walk 0,9 kilometers to get to school and it takes them 17 minutes. We
can calculate their speed in the following two ways.
Method 1:
17min
timeinhours = 60min
(9.1)
= 0, 283333333hr
Distance
speed = Time
0,9km
= 0,28333333hr
(9.2)
= 3, 176470588k · h−1
= 3, 18k · h−1
Method 2:
17min
timeinhours = 60min
(9.3)
= 0, 28hr
Distance
speed = Time
0,9km
= 0,28hr
(9.4)
= 3, 214285714k · h−1
= 3, 21k · h−1
You will see that we get two dierent answers. In Method 1 no rounding was done, but in Method 2, the
time was rounded to 2 decimal places. This made a big dierence to the answer. The answer in Method 1 is
more accurate because rounded numbers were not used in the calculation. Always only round o your nal
answer.
9.5.3 Scientic Notation

In Science one often needs to work with very large or very small numbers. These can be written more easily
in scientic notation, in the general form
d × 10e (9.5)
where d is a decimal number between 0 and 10 that is rounded o to a few decimal places. e is known as the
exponent and is an integer. If e > 0 it represents how many times the decimal place in d should be moved
to the right. If e < 0, then it represents how many times the decimal place in d should be moved to the

119
left. For example 3, 24 × 103 represents 3240 (the decimal moved three places to the right) and 3, 24 × 10−3
represents 0, 00324 (the decimal moved three places to the left).
If a number must be converted into scientic notation, we need to work out how many times the number
must be multiplied or divided by 10 to make it into a number between 1 and 10 (i.e. the value of e) and
what this number between 1 and 10 is (the value of d). We do this by counting the number of decimal places
the decimal comma must move.
For example, write the speed of light in scientic notation, to two decimal places. The speed of light is
299 792 458 m·s
−1
. First, nd where the decimal comma must go for two decimal places (to nd d) and then
count how many places there are after the decimal comma to determine e.
In this example, the decimal comma must go after the rst 2, but since the number after the 9 is 7,
d = 3, 00. e = 8 because there are 8 digits left after the decimal comma. So the speed of light in scientic
8 −1
notation, to two decimal places is 3,00 × 10 m·s .
9.5.4 Signicant Figures

In a number, each non-zero digit is a signicant gure. Zeroes are only counted if they are between two
non-zero digits or are at the end of the decimal part. For example, the number 2000 has 1 signicant gure
(the 2), but 2000,0 has 5 signicant gures. You estimate a number like this by removing signicant gures
from the number (starting from the right) until you have the desired number of signicant gures, rounding
as you go. For example 6,827 has 4 signicant gures, but if you wish to write it to 3 signicant gures it
would mean removing the 7 and rounding up, so it would be 6,83.
9.5.4.1 Using Signicant Figures

1. Round the following numbers:
a. 123,517 ` to 2 decimal places

−1
b. 14,328 km·h to one decimal place
c. 0,00954 m to 3 decimal places
3
2. Write the following quantities in scientic notation:
a. 10130 Pa to 2 decimal places

−2
b. 978,15 m·s to one decimal place
c. 0,000001256 A to 3 decimal places
4
3. Count how many signicant gures each of the quantities below has:
a. 2,590 km
b. 12,305 m`
c. 7800 kg
5
9.6 Prexes of Base Units

Now that you know how to write numbers in scientic notation, another important aspect of units is the
prexes that are used with the units.
3 http://www.fhsst.org/lOl
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5 http://www.fhsst.org/lON

Denition 9.3: Prex

A prex is a group of letters that are placed in front of a word. The eect of the prex is to
change meaning of the word. For example, the prex un is often added to a word to mean not, as
in unnecessary which means not necessary.
In the case of units, the prexes have a special use. The kilogram (kg) is a simple example. 1 kg is
equal to 1 000 g or 1 × 103 g. Grouping the 103 and the g together we can replace the 103 with the prex k
3
(kilo). Therefore the k takes the place of the 10 . The kilogram is unique in that it is the only SI base unit
containing a prex.
In Science, all the prexes used with units are some power of 10. Table 9.3 lists some of these prexes.
You will not use most of these prexes, but those prexes listed in bold should be learnt. The case of the
prex symbol is very important. Where a letter features twice in the table, it is written in uppercase for
6
exponents bigger than one and in lowercase for exponents less than one. For example M means mega (10 )
−3
and m means milli (10 ).
Prex Symbol Exponent Prex Symbol Exponent

yotta Y 1024 yocto y 10−24
zetta Z 1021 zepto z 10−21
exa E 1018 atto a 10−18
peta P 1015 femto f 10−15
tera T 1012 pico p 10−12
giga G 109 nano n 10−9
mega M 106 micro µ 10−6
kilo k 103 milli m 10−3
hecto h 102 centi c 10−2
deca da 101 deci d 10−1
Table 9.3: Unit Prexes
tip: There is no space and no dot between the prex and the symbol for the unit.
Here are some examples of the use of prexes:
• 40000 m can be written as 40 km (kilometre)

• 0,001 g is the same as 1 × 10−3 g and can be written as 1 mg (milligram)
• 2, 5 × 106 N can be written as 2,5 MN (meganewton)
• 250000 A can be written as 250 kA (kiloampere) or 0,250 MA (megaampere)
• 0,000000075 s can be written as 75 ns (nanoseconds)
• 3 × 10−7 mol can be rewritten as 0, 3 × 10−6 mol, which is the same as 0,3 µmol (micromol)
9.6.1 Using Scientic Notation

1. Write the following in scientic notation using Table 9.3 as a reference.
a. 0,511 MV
b. 10 c`
c. 0,5 µm
d. 250 nm

121
e. 0,00035 hg
6
2. Write the following using the prexes in Table 9.3.
a. 1,602 ×10−19 C
6
b. 1,992 ×10 J
4
c. 5,98 ×10 N
−4
d. 25 ×10 A
6
e. 0,0075 ×10 m
7
9.7 The Importance of Units

Without units much of our work as scientists would be meaningless. We need to express our thoughts clearly
and units give meaning to the numbers we measure and calculate. Depending on which units we use, the
numbers are dierent. For example if you have 12 water, it means nothing. You could have 12 ml of water,
12 litres of water, or even 12 bottles of water. Units are an essential part of the language we use. Units
must be specied when expressing physical quantities. Imagine that you are baking a cake, but the units,
like grams and millilitres, for the our, milk, sugar and baking powder are not specied!
9.7.1 Investigation : Importance of Units

Work in groups of 5 to discuss other possible situations where using the incorrect set of units can be to your
disadvantage or even dangerous. Look for examples at home, at school, at a hospital, when travelling and
in a shop.
9.7.2 Case Study : The importance of units

Read the following extract from CNN News 30 September 1999 and answer the questions below.
NASA: Human error caused loss of Mars orbiter November 10, 1999
Failure to convert English measures to metric values caused the loss of the Mars Climate Orbiter, a
spacecraft that smashed into the planet instead of reaching a safe orbit, a NASA investigation concluded
Wednesday. The Mars Climate Orbiter, a key craft in the space agency's exploration of the red planet,
vanished on 23 September after a 10 month journey. It is believed that the craft came dangerously close
to the atmosphere of Mars, where it presumably burned and broke into pieces. An investigation board
concluded that NASA engineers failed to convert English measures of rocket thrusts to newton, a metric
system measuring rocket force. One English pound of force equals 4,45 newtons. A small dierence between
the two values caused the spacecraft to approach Mars at too low an altitude and the craft is thought to
have smashed into the planet's atmosphere and was destroyed. The spacecraft was to be a key part of the
exploration of the planet. From its station about the red planet, the Mars Climate Orbiter was to relay
signals from the Mars Polar Lander, which is scheduled to touch down on Mars next month. The root
cause of the loss of the spacecraft was a failed translation of English units into metric units and a segment of
ground-based, navigation-related mission software, said Arthus Stephenson, chairman of the investigation
board. Questions:
1. Why did the Mars Climate Orbiter crash? Answer in your own words.
2. How could this have been avoided?
3. Why was the Mars Orbiter sent to Mars?
4. Do you think space exploration is important? Explain your answer.
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9.8 How to Change Units

It is very important that you are aware that dierent systems of units exist. Furthermore, you must be able
to convert between units. Being able to change between units (for example, converting from millimetres to
metres) is a useful skill in Science.
The following conversion diagrams will help you change from one unit to another.
×1000 ×1000
mm m km
÷1000 ÷1000
Figure 9.1: The distance conversion table
If you want to change millimetre to metre, you divide by 1000 (follow the arrow from mm to m); or if
you want to change kilometre to millimetre, you multiply by 1000×1000.
The same method can be used to change millilitre to litre or kilolitre. Use Figure 9.2 to change volumes:
×1000 ×1000
mℓ ℓ kℓ
dm3 m3
÷1000 ÷1000
Figure 9.2: The volume conversion table
Exercise 9.3: Conversion 1 (Solution on p. 128.)

Express 3 800 mm in metres.
Exercise 9.4: Conversion 2 (Solution on p. 128.)

Convert 4,56 kg to g.
9.8.1 Two other useful conversions

Very often in Science you need to convert speed and temperature. The following two rules will help you do
this:
Converting speed When converting km·h−1 to m·s−1 you divide by 3,6. For example 72 km·h
−1
÷ 3,6
−1
= 20 m·s .
−1 −1 −1 −1
When converting m·s to km·h , you multiply by 3,6. For example 30 m·s ×3,6 = 108 km·h .
Converting temperature Converting between the kelvin and celsius temperature scales is easy. To
convert from celsius to kelvin add 273. To convert from kelvin to celsius subtract 273. Representing the
kelvin temperature by TK and the celsius temperature by To C ,
TK = To C + 273 (9.6)

123
9.9 A sanity test

A sanity test is a method of checking whether an answer makes sense. All we have to do is to take a careful
look at our answer and ask the question Does the answer make sense?
Imagine you were calculating the number of people in a classroom. If the answer you got was 1 000 000
people you would know it was wrong it is not possible to have that many people in a classroom. That is
all a sanity test is is your answer insane or not?
It is useful to have an idea of some numbers before we start. For example, let us consider masses. An
average person has a mass around 70 kg, while the heaviest person in medical history had a mass of 635
kg. If you ever have to calculate a person's mass and you get 7 000 kg, this should fail your sanity check
your answer is insane and you must have made a mistake somewhere. In the same way an answer of 0.01 kg
should fail your sanity test.
The only problem with a sanity check is that you must know what typical values for things are. For
example, nding the number of learners in a classroom you need to know that there are usually 2050 people
in a classroom. If you get and answer of 2500, you should realise that it is wrong.
9.9.1 The scale of the matter... :

Try to get an idea of the typical values for the following physical quantities and write your answers into the
table:
Category Quantity Minimum Maximum

People mass
height
Transport speed of cars on freeways
speed of trains
speed of aeroplanes
distance between home and school
General thickness of a sheet of paper
height of a doorway
Table 9.4
9.10 Summary
1. You need to know the seven base SI Units as listed in Table 9.1. Combinations of SI Units can have
dierent names.
2. Unit names and abbreviations are written with lowercase letter unless it is named after a person.
3. Rounding numbers and using scientic notation is important.
4. Table 9.3 summarises the prexes used in Science.
5. Use gures Figure 9.1 and Figure 9.2 to convert between units.

9.11 End of Chapter Exercises

1. Write down the SI unit for the each of the following quantities:
a. length
b. time
c. mass
d. quantity of matter
8
2. For each of the following units, write down the symbol and what power of 10 it represents:
a. millimetre
b. centimetre
c. metre
d. kilometre
9
3. For each of the following symbols, write out the unit in full and write what power of 10 it represents:
a. µg
b. mg
c. kg
d. Mg
10
4. Write each of the following in scientic notation, correct to 2 decimal places:
a. 0,00000123 N
b. 417 000 000 kg
c. 246800 A
d. 0,00088 mm
11
5. Rewrite each of the following, accurate to two decimal places, using the correct prex where applicable:
a. 0,00000123 N
b. 417 000 000 kg
c. 246800 A
d. 0,00088 mm
12
6. For each of the following, write the measurement using the correct symbol for the prex and the base
unit:
a. 1,01 microseconds
b. 1 000 milligrams
c. 7,2 megameters
d. 11 nanolitre
13
−1
7. The Concorde is a type of aeroplane that ies very fast. The top speed of the Concorde is 2 172 km·hr .
−1 14
Convert the Concorde's top speed to m·s .Click here for the solution.
8 http://www.fhsst.org/lOQ
9 http://www.fhsst.org/lOU
10 http://www.fhsst.org/lOP
11 http://www.fhsst.org/lOE
12 http://www.fhsst.org/lOm
13 http://www.fhsst.org/lOy
14 http://www.fhsst.org/lOV

125
◦
8. The boiling point of water is 100 C. What is the boiling point of water in kelvin?Click here for the
15
solution.
15 http://www.fhsst.org/lOp


Solution to Exercise 9.1 (p. 120)
Step 1. π = 3, 1415|92654...
Step 2. The last digit of π = 3, 1415|92654... must be rounded up because there is a 9 after the |.
Step 3. π = 3, 1416 rounded to 4 decimal places.
Step 1. 9, 19|1919...
Step 2. The last digit of 9, 19|1919... must be rounded down because there is a 1 after the |.
Step 3. Answer = 9,19 rounded to 2 decimal places.

Step 1. Use Figure 9.1 . Millimetre is on the left and metre in the middle.
Step 2. You need to go from mm to m, so you are moving from left to right.
Step 3. 3 800 mm ÷ 1000 = 3,8 m
Step 1. Use Figure 9.1. Kilogram is the same as kilometre and gram the same as metre.
Step 2. You need to go from kg to g, so it is from right to left.
Step 3. 4,56 kg × 1000 = 4560 g

Chapter 10
Motion in one dimension
10.1 Frames of reference and reference point 1
10.1.1 Introduction
This chapter is about how things move in a straight line or more scientically how things move in one
dimension. This is useful for learning how to describe the movement of cars along a straight road or of trains
along straight railway tracks. If you want to understand how any object moves, for example a car on the
freeway, a soccer ball being kicked towards the goal or your dog chasing the neighbour's cat, then you have
to understand three basic ideas about what it means when something is moving. These three ideas describe
dierent parts of exactly how an object moves. They are:
1. position or displacement which tells us exactly where the object is,

2. speed or velocity which tells us exactly how fast the object's position is changing or more familiarly,
how fast the object is moving, and
3. acceleration which tells us exactly how fast the object's velocity is changing.
You will also learn how to use position, displacement, speed, velocity and acceleration to describe the motion
of simple objects. You will learn how to read and draw graphs that summarise the motion of a moving object.
You will also learn about the equations that can be used to describe motion and how to apply these equations
to objects moving in one dimension.
10.1.2 Reference Point, Frame of Reference and Position

The most important idea when studying motion, is you have to know where you are. The word position
describes your location (where you are). However, saying that you are here is meaningless, and you have to
specify your position relative to a known reference point. For example, if you are 2 m from the doorway, inside
your classroom then your reference point is the doorway. This denes your position inside the classroom.
Notice that you need a reference point (the doorway) and a direction (inside) to dene your location.
10.1.2.1 Frames of Reference

Denition 10.1: Frame of Reference
A frame of reference is a reference point combined with a set of directions.
A frame of reference is similar to the idea of a reference point. A frame of reference is dened as a
reference point combined with a set of directions. For example, a boy is standing still inside a train as it
pulls out of a station. You are standing on the platform watching the train move from left to right. To you
127
128 CHAPTER 10. MOTION IN ONE DIMENSION
it looks as if the boy is moving from left to right, because relative to where you are standing (the platform),
he is moving. According to the boy, and his frame of reference (the train), he is not moving.
A frame of reference must have an origin (where you are standing on the platform) and at least a positive
direction. The train was moving from left to right, making to your right positive and to your left negative.
If someone else was looking at the same boy, his frame of reference will be dierent. For example, if he was
standing on the other side of the platform, the boy will be moving from right to left.
For this chapter, we will only use frames of reference in the x-direction. Frames of reference will be
covered in more detail in Grade 12.
Figure 10.1: Frames of Reference
Figure 10.2

<http://static.slidesharecdn.com/swf/ssplayer2.swf ?doc=framesofreference-100511055617-
phpapp02&stripped_title=frames-of-reference&userName=kwarne>
Figure 10.3
10.1.2.2 Position
Denition 10.2: Position
Position is a measurement of a location, with reference to an origin.
A position is a measurement of a location, with reference to an origin. Positions can therefore be

negative or positive. The symbol x is used to indicate position. x has units of length for example cm, m
or km. Figure 10.4 shows the position of a school. Depending on what reference point we choose, we can
say that the school is 300 m from Joan's house (with Joan's house as the reference point or origin) or 500 m
from Joel's house (with Joel's house as the reference point or origin).

129
Figure 10.4: Illustration of position
The shop is also 300 m from Joan's house, but in the opposite direction as the school. When we choose a
reference point, we have a positive direction and a negative direction. If we choose the direction towards the
school as positive, then the direction towards the shop is negative. A negative direction is always opposite
to the direction chosen as positive.
Figure 10.5: The origin is at Joan's house and the position of the school is +300 m. Positions towards
the left are dened as positive and positions towards the right are dened as negative.
10.1.2.2.1 Discussion : Reference Points

Divide into groups of 5 for this activity. On a straight line, choose a reference point. Since position can have
both positive and negative values, discuss the advantages and disadvantages of choosing
1. either end of the line,

2. the middle of the line.
This reference point can also be called the origin".
10.1.2.2.2 Position
1. Write down the positions for objects at A, B, D and E. Do not forget the units.
Figure 10.6
2
2. Write down the positions for objects at F, G, H and J. Do not forget the units.
2 http://www.fhsst.org/laG

Figure 10.7
3
3. There are 5 houses on Newton Street, A, B, C, D and E. For all cases, assume that positions to the
right are positive.
Figure 10.8
a. Draw a frame of reference with house A as the origin and write down the positions of houses B,
C, D and E.
b. You live in house C. What is your position relative to house E?
c. What are the positions of houses A, B and D, if house B is taken as the reference point?
4
10.2 Displacement and distance 5
10.2.1 Displacement and Distance

Denition 10.3: Displacement
Displacement is the change in an object's position.
The displacement of an object is dened as its change in position (nal position minus initial position).
Displacement has a magnitude and direction and is therefore a vector. For example, if the initial position of
a car is xi and it moves to a nal position of xf , then the displacement is:
xf − xi (10.1)
However, subtracting an initial quantity from a nal quantity happens often in Physics, so we use the
shortcut ∆ to mean nal - initial. Therefore, displacement can be written:
∆x = xf − xi (10.2)
tip: The symbol ∆ is read out as delta. ∆ is a letter of the Greek alphabet and is used in
Mathematics and Science to indicate a change in a certain quantity, or a nal value minus an initial
value. For example, ∆x means change in x while ∆t means change in t.
3 http://www.fhsst.org/la7
4 http://www.fhsst.org/laA

131
tip: The words initial and nal will be used very often in Physics. Initial will always refer to
something that happened earlier in time and nal will always refer to something that happened
later in time. It will often happen that the nal value is smaller than the initial value, such that
the dierence is negative. This is ok!
Figure 10.9: Illustration of displacement
Displacement does not depend on the path travelled, but only on the initial and nal positions (Figure 10.9).
We use the word distance to describe how far an object travels along a particular path. Distance is the actual
distance that was covered. Distance (symbol d) does not have a direction, so it is a scalar. Displacement
is the shortest distance from the starting point to the endpoint from the school to the shop in the gure.
Displacement has direction and is therefore a vector.
shows the ve houses we discussed earlier. Jack walks to school, but instead of walking straight to school,
he decided to walk to his friend Joel's house rst to fetch him so that they can walk to school together. Jack
covers a distance of 400 m to Joel's house and another 500 m to school. He covers a distance of 900 m. His
displacement, however, is only 100 m towards the school. This is because displacement only looks at the
starting position (his house) and the end position (the school). It does not depend on the path he travelled.
To calculate his distance and displacement, we need to choose a reference point and a direction. Let's
choose Jack's house as the reference point, and towards Joel's house as the positive direction (which means
that towards the school is negative). We would do the calculations as follows:
Distance (d) = path travelled

= 400 m + 500 m (10.3)
= 900 m
Displacement (∆x) = xf − xi
= −100 m + 0 m (10.4)
= −100 m
Joel walks to school with Jack and after school walks back home. What is Joel's displacement and what
distance did he cover? For this calculation we use Joel's house as the reference point. Let's take towards the
school as the positive direction.
Distance (d) = path travelled

= 500 m + 500 m (10.5)
= 1000 m
Displacement (∆x) = xf − xi
= 0 m+0 m (10.6)
= 0 m

It is possible to have a displacement of 0m and a distance that is not 0 m. This happens when an object
completes a round trip back to its original position, like an athlete running around a track.
10.2.1.1 Interpreting Direction

Very often in calculations you will get a negative answer. For example, Jack's displacement in the example
above, is calculated as −100 m. The minus sign in front of the answer means that his displacement is 100 m
in the opposite direction (opposite to the direction chosen as positive in the beginning of the question).
When we start a calculation we choose a frame of reference and a positive direction. In the rst example
above, the reference point is Jack's house and the positive direction is towards Joel's house. Therefore Jack's
displacement is 100 m towards the school. Notice that distance has no direction, but displacement has.
10.2.1.2 Dierences between Distance and Displacement

Denition 10.4: Vectors and Scalars
A vector is a physical quantity with magnitude (size) and direction. A scalar is a physical quantity
with magnitude (size) only.
The dierences between distance and displacement can be summarised as:
Distance Displacement
1. depends on the path 1. independent of path taken
2. always positive 2. can be positive or negative
3. is a scalar 3. is a vector
4. does not have a direction 4. has a direction
Table 10.1
10.2.1.2.1 Point of Reference

1. Use to answer the following questions.
a. Jill walks to Joan's house and then to school, what is her distance and displacement?
b. John walks to Joan's house and then to school, what is his distance and displacement?
c. Jack walks to the shop and then to school, what is his distance and displacement?
d. What reference point did you use for each of the above questions?
6
2. You stand at the front door of your house (displacement, ∆x = 0 m). The street is 10 m away from
the front door. You walk to the street and back again.
a. What is the distance you have walked?

b. What is your nal displacement?
c. Is displacement a vector or a scalar? Give a reason for your answer.
7
6 http://www.fhsst.org/lao
7 http://www.fhsst.org/las

133
10.3 Speed and velocity 8
10.3.1 Speed, Average Velocity and Instantaneous Velocity

Denition 10.5: Velocity
Velocity is the rate of change of displacement.
Denition 10.6: Instantaneous velocity

Instantaneous velocity is the velocity of a body at a specic instant in time.
Denition 10.7: Average velocity

Average velocity is the total displacement of a body over a time interval.
Velocity is the rate of change of position. It tells us how much an object's position changes in time. This
is the same as the displacement divided by the time taken. Since displacement is a vector and time taken is
a scalar, velocity is also a vector. We use the symbol v for velocity. If we have a displacement of ∆x and a
time taken of ∆t, v is then dened as:
change in displacement (in m)

in m · s−1

velocity = change in time (in s)
(10.7)
∆x
v = ∆t
Velocity can be positive or negative. Positive values of velocity mean that the object is moving away from
the reference point or origin and negative values mean that the object is moving towards the reference point
or origin.
tip: An instant in time is dierent from the time taken or the time interval. It is therefore useful
to use the symbol t for an instant in time (for example during the 4
th
second) and the symbol ∆t
for the time taken (for example during the rst 5 seconds of the motion).
Average velocity (symbol v ) is the displacement for the whole motion divided by the time taken for the whole
motion. Instantaneous velocity is the velocity at a specic instant in time.
(Average) Speed (symbol s) is the distance travelled (d) divided by the time taken (∆t) for the journey.
Distance and time are scalars and therefore speed will also be a scalar. Speed is calculated as follows:
distance (in m)
speed in m · s−1 = (10.8)
time (in s)
d
s= (10.9)
∆t
Instantaneous speed is the magnitude of instantaneous velocity. It has the same value, but no direction.
Exercise 10.3.1: Average speed and average velocity (Solution on p. 158.)

James walks 2 km away from home in 30 minutes. He then turns around and walks back home
along the same path, also in 30 minutes. Calculate James' average speed and average velocity.
Figure 10.10
Exercise 10.3.2: Instantaneous Speed and Velocity (Solution on p. 158.)

A man runs around a circular track of radius 100 m. It takes him 120 s to complete a revolution
of the track. If he runs at constant speed, calculate:

1. his speed,
2. his instantaneous velocity at point A,
3. his instantaneous velocity at point B,
4. his average velocity between points A and B,
5. his average speed during a revolution.
6. his average velocity during a revolution.
Figure 10.11
10.3.1.1 Dierences between Speed and Velocity

The dierences between speed and velocity can be summarised as:
Speed Velocity
1. depends on the path taken 1. independent of path taken
2. always positive 2. can be positive or negative
3. is a scalar 3. is a vector
4. no dependence on direction and so is only posi- 4. direction can be guessed from the sign (i.e. pos-
tive itive or negative)
Table 10.2
Additionally, an object that makes a round trip, i.e. travels away from its starting point and then returns
to the same point has zero velocity but travels a non-zero speed.
10.3.1.1.1 Displacement and related quantities

1. Theresa has to walk to the shop to buy some milk. After walking 100 m, she realises that she does not
have enough money, and goes back home. If it took her two minutes to leave and come back, calculate
the following:
a. How long was she out of the house (the time interval ∆t in seconds)?
b. How far did she walk (distance (d))?
c. What was her displacement (∆x)?
−1
d. What was her average velocity (in m·s )?
−1
e. What was her average speed (in m·s )?
Figure 10.12

135
Figure 10.13
9
2. Desmond is watching a straight stretch of road from his classroom window. He can see two poles which
he earlier measured to be 50 m apart. Using his stopwatch, Desmond notices that it takes 3s for most
cars to travel from the one pole to the other.
∆x
a. Using the equation for velocity (v =
∆t ), show all the working needed to calculate the velocity
of a car travelling from the left to the right.
b. If Desmond measures the velocity of a red Golf to be −16, 67 m · s
−1
, in which direction was the
Gold travelling? Desmond leaves his stopwatch running, and notices that at t = 5, 0 s, a taxi
passes the left pole at the same time as a bus passes the right pole. At time t = 7, 5 s the taxi
passes the right pole. At time t = 9, 0 s, the bus passes the left pole.
c. How long did it take the taxi and the bus to travel the distance between the poles? (Calculate
the time interval (∆t) for both the taxi and the bus).
d. What was the velocity of the taxi and the bus?
e. What was the speed of the taxi and the bus?
f. What was the speed of taxi and the bus in km · h−1 ?
Figure 10.14
10
3. A rabbit runs across a freeway. There is a car, 100 m away travelling towards the rabbit.
Figure 10.15
a. If the car is travelling at 120 km · h−1 , what is the car's speed in m · s−1 .
b. How long will it take the a car to travel 100 m?
c. If the rabbit is running at 10 km · h−1 , what is its speed in m · s−1 ?
d. If the freeway has 3 lanes, and each lane is 3m wide, how long will it take for the rabbit to cross
all three lanes?
e. If the car is travelling in the furthermost lane from the rabbit, will the rabbit be able to cross all
3 lanes of the freeway safely?
9 http://www.fhsst.org/laH

11
10.3.1.1.2 Investigation : An Exercise in Safety

Divide into groups of 4 and perform the following investigation. Each group will be performing the same
investigation, but the aim for each group will be dierent.
1. Choose an aim for your investigation from the following list and formulate a hypothesis:
• Do cars travel at the correct speed limit?

• Is is safe to cross the road outside of a pedestrian crossing?
• Does the colour of your car determine the speed you are travelling at?
• Any other relevant question that you would like to investigate.
2. On a road that you often cross, measure out 50 m along a straight section, far away from trac lights
or intersections.
3. Use a stopwatch to record the time each of 20 cars take to travel the 50 m section you measured.
4. Design a table to represent your results. Use the results to answer the question posed in the aim of
the investigation. You might need to do some more measurements for your investigation. Plan in your
group what else needs to be done.
5. Complete any additional measurements and write up your investigation under the following headings:
• Aim and Hypothesis

• Apparatus
• Method
• Results
• Discussion
• Conclusion
6. Answer the following questions:
a. How many cars took less than 3s to travel 50 m?

b. What was the shortest time a car took to travel 50 m?
c. What was the average time taken by the 20 cars?
d. What was the average speed of the 20 cars?
e. Convert the average speed to km · h−1 .
10.4 Acceleration 12
10.4.1 Acceleration
Denition 10.8: Acceleration
Acceleration is the rate of change of velocity.
Acceleration (symbol a) is the rate of change of velocity. It is a measure of how fast the velocity of an
object changes in time. If we have a change in velocity (∆v ) over a time interval (∆t), then the acceleration
(a) is dened as:
change in velocity in m · s−1

in m · s −2

acceleration = (10.10)
change in time (in s)
∆v
a= (10.11)
∆t
11 http://www.fhsst.org/laF

137
Since velocity is a vector, acceleration is also a vector. Acceleration does not provide any information about
a motion, but only about how the motion changes. It is not possible to tell how fast an object is moving or
in which direction from the acceleration.
Like velocity, acceleration can be negative or positive. We see that when the sign of the acceleration and
the velocity are the same, the object is speeding up. If both velocity and acceleration are positive, the object
is speeding up in a positive direction. If both velocity and acceleration are negative, the object is speeding
up in a negative direction. If velocity is positive and acceleration is negative, then the object is slowing
down. Similarly, if the velocity is negative and the acceleration is positive the object is slowing down. This
is illustrated in the following worked example.
Exercise 10.4.1: Acceleration (Solution on p. 160.)

−1 1
A car accelerates uniformly from an initial velocity of 2 m·s to a nal velocity of 10 m·s in 8
−1
seconds. It then slows down uniformly to a nal velocity of 4 m·s in 6 seconds. Calculate the
acceleration of the car during the rst 8 seconds and during the last 6 seconds.
tip: Acceleration does not tell us about the direction of the motion. Acceleration only tells us how
the velocity changes.
tip: Deceleration
Avoid the use of the word deceleration to refer to a negative acceleration. This word usually means slowing
down and it is possible for an object to slow down with both a positive and negative acceleration, because
the sign of the velocity of the object must also be taken into account to determine whether the body is
slowing down or not.
10.4.1.1 Acceleration
−1 −1
1. An athlete is accelerating uniformly from an initial velocity of 0 m·s to a nal velocity of 4 m·s in
2 seconds. Calculate his acceleration. Let the direction that the athlete is running in be the positive
direction.
13
−1 −1
2. A bus accelerates uniformly from an initial velocity of 15 m·s to a nal velocity of 7 m·s in 4 seconds.
Calculate the acceleration of the bus. Let the direction of motion of the bus be the positive direction.
14
−1 −1
3. An aeroplane accelerates uniformly from an initial velocity of 200 m·s to a velocity of 100 m·s in 10
−1
seconds. It then accelerates uniformly to a nal velocity of 240 m·s in 20 seconds. Let the direction
of motion of the aeroplane be the positive direction.
a. Calculate the acceleration of the aeroplane during the rst 10 seconds of the motion.
b. Calculate the acceleration of the aeroplane during the next 14 seconds of its motion.
15
The following video provides a summary of distance, velocity and acceleration. Note that in this video a
dierent convention for writing units is used. You should not use this convention when writing units in
physics.
13 http://www.fhsst.org/l1k
14 http://www.fhsst.org/l10

Khan academy video on motion - 1

<http://www.youtube.com/v/8wZugqi_uCg&rel=0>
Figure 10.16
10.5 Description of motion 16
10.5.1 Description of Motion

The purpose of this chapter is to describe motion, and now that we understand the denitions of displacement,
distance, velocity, speed and acceleration, we are ready to start using these ideas to describe how an object
is moving. There are many ways of describing motion:
1. words
2. diagrams
3. graphs
These methods will be described in this section.

We will consider three types of motion: when the object is not moving (stationary object), when the
object is moving at a constant velocity (uniform motion) and when the object is moving at a constant
acceleration (motion at constant acceleration).
10.5.1.1 Stationary Object

The simplest motion that we can come across is that of a stationary object. A stationary object does not
move and so its position does not change, for as long as it is standing still. An example of this situation is
when someone is waiting for something without moving. The person remains in the same position.
Lesedi is waiting for a taxi. He is standing two metres from a stop street at t = 0 s. After one minute,
at t = 60 s, he is still 2 metres from the stop street and after two minutes, at t = 120 s, also 2 metres from
the stop street. His position has not changed. His displacement is zero (because his position is the same),
his velocity is zero (because his displacement is zero) and his acceleration is also zero (because his velocity
is not changing).
Figure 10.17
We can now draw graphs of position vs. time (x vs. t), velocity vs. time (v vs. t) and acceleration vs.
time (a vs. t) for a stationary object. The graphs are shown in Figure 10.18. Lesedi's position is 2 metres
from the stop street. If the stop street is taken as the reference point, his position remains at 2 metres for
120 seconds. The graph is a horizontal line at 2 m. The velocity and acceleration graphs are also shown.

139
They are both horizontal lines on the x-axis. Since his position is not changing, his velocity is 0 m · s−1 and
since velocity is not changing, acceleration is 0 m · s−2 .
Figure 10.18: Graphs for a stationary object (a) position vs. time (b) velocity vs. time (c) acceleration
vs. time.
Denition 10.9: Gradient

The gradient of a line can be calculated by dividing the change in the y -value by the change in the
x-value.
∆y
m =
∆x
Since we know that velocity is the rate of change of position, we can conrm the value for the velocity
vs. time graph, by calculating the gradient of the x vs. t graph.
tip: The gradient of a position vs. time graph gives the velocity.
If we calculate the gradient of the x vs. t graph for a stationary object we get:
∆x
v = ∆t
xf −xi
= tf −ti
(10.12)
2 m−2 m
= 120 s−60 s (initial position = final position)
= 0 m·s −1
(for the time that Lesedi is stationary)
Similarly, we can conrm the value of the acceleration by calculating the gradient of the velocity vs. time
graph.
tip: The gradient of a velocity vs. time graph gives the acceleration.
If we calculate the gradient of the v vs. t graph for a stationary object we get:
∆v
a = ∆t
vf −vi
= tf −ti
(10.13)
0 m·s−1 −0 m·s−1
= 120 s−60 s
= 0 m·s −2
Additionally, because the velocity vs. time graph is related to the position vs. time graph, we can use the
area under the velocity vs. time graph to calculate the displacement of an object.
tip: The area under the velocity vs. time graph gives the displacement.
The displacement of the object is given by the area under the graph, which is 0 m. This is obvious, because
the object is not moving.

10.5.1.2 Motion at Constant Velocity

Motion at a constant velocity or uniform motion means that the position of the object is changing at the
same rate.
Assume that Lesedi takes 100 s to walk the 100 m to the taxi-stop every morning. If we assume that
Lesedi's house is the origin, then Lesedi's velocity is:
∆x
v = ∆t
xf −xi
= tf −ti
(10.14)
100 m−0 m
= 100 s−0 s
= 1 m·s −1
Lesedi's velocity is 1 m·s

−1
. This means that he walked 1 m in the rst second, another metre in the second
second, and another in the third second, and so on. For example, after 50 s he will be 50 m from home. His
position increases by 1m every 1 s. A diagram of Lesedi's position is shown in Figure 10.19.
Figure 10.19: Diagram showing Lesedi's motion at a constant velocity of 1 m·s−1
We can now draw graphs of position vs.time (x vs. t), velocity vs.time (v vs. t) and acceleration vs.time
(a vs. t) for Lesedi moving at a constant velocity. The graphs are shown in Figure 10.20.
Figure 10.20: Graphs for motion at constant velocity (a) position vs. time (b) velocity vs. time (c)
acceleration vs. time. The area of the shaded portion in the v vs. t graph corresponds to the object's
displacement.
In the evening Lesedi walks 100 m from the bus stop to his house in 100 s. Assume that Lesedi's house
is the origin. The following graphs can be drawn to describe the motion.
Figure 10.21: Graphs for motion with a constant negative velocity (a) position vs. time (b) velocity
vs. time (c) acceleration vs. time. The area of the shaded portion in the v vs.t graph corresponds to the
object's displacement.

141
We see that the v vs. t graph is a horisontal line. If the velocity vs. time graph is a horisontal line,
it means that the velocity is constant (not changing). Motion at a constant velocity is known as uniform
motion.
We can use the x vs. t to calculate the velocity by nding the gradient of the line.
∆x
v = ∆t
xf −xi
= tf −ti
(10.15)
0 m−100 m
= 100 s−0 s
= −1 m · s −1
Lesedi has a velocity of −1 m · s−1 , or1 m · s−1 towards his house. You will notice that the v vs. t graph is
velocity of −1 m · s
−1
a horisontal line corresponding to a . The horizontal line means that the velocity stays
the same (remains constant) during the motion. This is uniform velocity.
We can use the v vs. t to calculate the acceleration by nding the gradient of the line.
∆v
a = ∆t
vf −vi
= tf −ti
(10.16)
1 m·s−1 −1 m·s−1
= 100 s−0 s
= 0 m·s −2
Lesedi has an acceleration of 0 m · s−2 . You will notice that the graph of a vs.t is a horisontal line
acceleration value of 0 m · s
−2
corresponding to an . There is no acceleration during the motion because his
velocity does not change.
We can use the v vs. t to calculate the displacement by nding the area under the graph.
v = Area under graph

= `× b
(10.17)
= 100 × (−1)
= −100 m
This means that Lesedi has a displacement of 100 m towards his house.
10.5.1.2.1 Velocity and acceleration

1. Use the graphs in Figure 10.20 to calculate each of the following:
a. Calculate Lesedi's velocity between 50 s and 100 s using the x vs. t graph. Hint: Find the gradient
of the line.
b. Calculate Lesedi's acceleration during the whole motion using the v vs. t graph.
c. Calculate Lesedi's displacement during the whole motion using the v vs. t graph.
17
2. Thandi takes 200 s to walk 100 m to the bus stop every morning. In the evening Thandi takes 200 s
to walk 100 m from the bus stop to her home.
a. Draw a graph of Thandi's position as a function of time for the morning (assuming that Thandi's
home is the reference point). Use the gradient of the x vs. t graph to draw the graph of velocity
vs. time. Use the gradient of the v vs. t graph to draw the graph of acceleration vs. time.

b. Draw a graph of Thandi's position as a function of time for the evening (assuming that Thandi's
home is the origin). Use the gradient of the x vs. t graph to draw the graph of velocity vs. time.
Use the gradient of the v vs. t graph to draw the graph of acceleration vs. time.
c. Discuss the dierences between the two sets of graphs in questions 2 and 3.
18
10.5.1.2.2 Experiment : Motion at constant velocity

Aim:
To measure the position and time during motion at constant velocity and determine the average velocity
as the gradient of a Position vs. Time" graph.
Apparatus:
A battery operated toy car, stopwatch, meter stick or measuring tape.
Method:
1. Work with a friend. Copy the table below into your workbook.
2. Complete the table by timing the car as it travels each distance.
3. Time the car twice for each distance and take the average value as your accepted time.
4. Use the distance and average time values to plot a graph of Distance vs. Time" onto graph paper.
Stick the graph paper into your workbook. (Remember that A vs. B" always means y vs. x").
5. Insert all axis labels and units onto your graph.
6. Draw the best straight line through your data points.
7. Find the gradient of the straight line. This is the average velocity.
Results:
Distance (m) Time (s)
1 2 Ave.
0,5
1,0
1,5
2,0
2,5
3,0
Table 10.3
Conclusions:
Answer the following questions in your workbook.
Questions:
1. Did the car travel with a constant velocity?
2. How can you tell by looking at the Distance vs. Time" graph if the velocity is constant?
3. How would the Distance vs. Time" look for a car with a faster velocity?
4. How would the Distance vs. Time" look for a car with a slower velocity?

143
10.5.1.3 Motion at Constant Acceleration

The nal situation we will be studying is motion at constant acceleration. We know that acceleration is the
rate of change of velocity. So, if we have a constant acceleration, this means that the velocity changes at a
constant rate.
Let's look at our rst example of Lesedi waiting at the taxi stop again. A taxi arrived and Lesedi got in.
The taxi stopped at the stop street and then accelerated as follows: After 1 s the taxi covered a distance of
2, 5 m, after 2 s it covered 10 m, after 3 s it covered 22, 5 m and after 4 s it covered 40 m. The taxi is covering
a larger distance every second. This means that it is accelerating.
Figure 10.22
To calculate the velocity of the taxi you need to calculate the gradient of the line at each second:
∆x
v1s = ∆t
xf −xi
= tf −ti
(10.18)
5 m−0 m
= 1,5 s−0,5 s
= 5 m·s −1
∆x
v2s = ∆t
xf −xi
= tf −ti
(10.19)
15 m−5 m
= 2,5 s−1,5 s
= 10 m · s −1
∆x
v3s = ∆t
xf −xi
= tf −ti
(10.20)
30 m−15 m
= 3,5 s−2,5 s
= 15 m · s −1
From these velocities, we can draw the velocity-time graph which forms a straight line.
The acceleration is the gradient of the v vs. t graph and can be calculated as follows:
∆v
a = ∆t
vf −vi
= tf −ti
(10.21)
15 m·s−1 −5 m·s−1
= 3 s−1 s
= 5 m·s −2
The acceleration does not change during the motion (the gradient stays constant). This is motion at constant
or uniform acceleration.
The graphs for this situation are shown in Figure 10.23.

Figure 10.23: Graphs for motion with a constant acceleration (a) position vs. time (b) velocity vs.
time (c) acceleration vs. time.
10.5.1.3.1 Velocity from Acceleration vs. Time Graphs

Just as we used velocity vs. time graphs to nd displacement, we can use acceleration vs. time graphs to
nd the velocity of an object at a given moment in time. We simply calculate the area under the acceleration
vs. time graph, at a given time. In the graph below, showing an object at a constant positive acceleration,
the increase in velocity of the object after 2 seconds corresponds to the shaded portion.
v = area of rectangle = a × ∆t
= 5 m · s−2 × 2 s (10.22)
= 10 m · s−1
The velocity of the object at t = 2s is therefore 10 m · s−1 . This corresponds with the values obtained in
Figure 10.23.
10.5.1.4 Summary of Graphs

The relation between graphs of position, velocity and acceleration as functions of time is summarised in
Figure 10.24.
Figure 10.24: Position-time, velocity-time and acceleration-time graphs.
tip: Often you will be required to describe the motion of an object that is presented as a graph
of either position, velocity or acceleration as functions of time. The description of the motion
represented by a graph should include the following (where possible):
1.whether the object is moving in the positive or negative direction

2.whether the object is at rest, moving at constant velocity or moving at constant positive
acceleration (speeding up) or constant negative acceleration (slowing down)
You will also often be required to draw graphs based on a description of the motion in words or from
a diagram. Remember that these are just dierent methods of presenting the same information. If
you keep in mind the general shapes of the graphs for the dierent types of motion, there should
not be any diculty with explaining what is happening.

145
10.5.1.5 Worked Examples

The worked examples in this section demonstrate the types of questions that can be asked about graphs.
Exercise 10.5.1: Description of motion based on a position-time graph (Solution on p.

161.)
The position vs. time graph for the motion of a car is given below. Draw the corresponding
velocity vs. time and acceleration vs. time graphs, and then describe the motion of the car.
Figure 10.25
Exercise 10.5.2: Calculations from a velocity vs. time graph (Solution on p. 162.)
The velocity vs. time graph of a truck is plotted below. Calculate the distance and displacement
of the truck after 15 seconds.
Figure 10.26
Exercise 10.5.3: Velocity from a position vs. time graph (Solution on p. 163.)
The position vs. time graph below describes the motion of an athlete.
1. What is the velocity of the athlete during the rst 4 seconds?

2. What is the velocity of the athlete from t=4 s to t=7 s?
Figure 10.27
Exercise 10.5.4: Drawing a v vs. t graph from an a vs. t graph (Solution on p. 163.)
The acceleration vs. time graph for a car starting from rest, is given below. Calculate the velocity
of the car and hence draw the velocity vs. time graph.
Figure 10.28

10.5.1.6 Graphs
1. A car is parked 10 m from home for 10 minutes. Draw a displacement-time, velocity-time and
acceleration-time graphs for the motion. Label all the axes.
19
2. A bus travels at a constant velocity of 12 m · s−1 for 6 seconds. Draw the displacement-time, velocity-
time and acceleration-time graph for the motion. Label all the axes.
20
3. An athlete runs with a constant acceleration of 1 m · s−2 for 4 s. Draw the acceleration-time, velocity-
time and displacement time graphs for the motion. Accurate values are only needed for the acceleration-
time and velocity-time graphs.
21
4. The following velocity-time graph describes the motion of a car. Draw the displacement-time graph
and the acceleration-time graph and explain the motion of the car according to the three graphs.
Figure 10.29
22
5. The following velocity-time graph describes the motion of a truck. Draw the displacement-time graph
and the acceleration-time graph and explain the motion of the truck according to the three graphs.
Figure 10.30
23
Figure 10.31
24
run demo
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24 http://phet.colorado.edu/sims/moving-man/moving-man_en.jnlp

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10.6 Equations of motion 25
10.6.1 Equations of Motion

In this chapter we will look at the third way to describe motion. We have looked at describing motion in
terms of graphs and words. In this section we examine equations that can be used to describe motion.
This section is about solving problems relating to uniformly accelerated motion. In other words, motion
at constant acceleration.
The following are the variables that will be used in this section:
vi initial velocity m · s−1 at t = 0s

=
vf = final velocity m · s−1 at time t

∆x = displacement (m)
(10.23)
t = time (s)
∆t = time interval (s)
a = acceleration m · s−1

vf = vi + at (10.24)
(vi + vf )
∆x = t (10.25)
2
1
∆x = vi t + at2 (10.26)
2
vf2 = vi2 + 2a∆x (10.27)
The questions can vary a lot, but the following method for answering them will always work. Use this when
attempting a question that involves motion with constant acceleration. You need any three known quantities
(vi , vf , ∆x, t or a) to be able to calculate the fourth one.
1. Read the question carefully to identify the quantities that are given. Write them down.
2. Identify the equation to use. Write it down!!!
3. Ensure that all the values are in the correct unit and ll them in your equation.
4. Calculate the answer and ll in its unit.
note: Galileo Galilei of Pisa, Italy, was the rst to determined the correct mathematical law
for acceleration: the total distance covered, starting from rest, is proportional to the square of the
time. He also concluded that objects retain their velocity unless a force often friction acts upon
them, refuting the accepted Aristotelian hypothesis that objects "naturally" slow down and stop
unless a force acts upon them. This principle was incorporated into Newton's laws of motion (1st
law).
10.6.1.1 Finding the Equations of Motion

The following does not form part of the syllabus and can be considered additional information.

10.6.1.1.1 Derivation of (10.24)

According to the denition of acceleration:
∆v
a= (10.28)
t
where ∆v is the change in velocity, i.e. ∆v = vf - vi . Thus we have
vf −vi
a = t
(10.29)
vf = vi + at
10.6.1.1.2 Derivation of (10.25)

We have seen that displacement can be calculated from the area under a velocity vs. time graph. For
uniformly accelerated motion the most complicated velocity vs. time graph we can have is a straight line.
Look at the graph below - it represents an object with a starting velocity of vi , accelerating to a nal velocity
vf over a total time t.
Figure 10.32
To calculate the nal displacement we must calculate the area under the graph - this is just the area of
the rectangle added to the area of the triangle. This portion of the graph has been shaded for clarity.
1
Area[U+25B5] = 2b ×h
1
2 t × (vf − vi )
= (10.30)
1 1
= 2 vf t − 2 vi t
Area = `×b
= t × vi (10.31)
= vi t
Displacement = Area + Area[U+25B5]

∆x = vi t + 21 vf t − 12 vi t (10.32)
(vi +vf )
∆x = 2 t
10.6.1.1.3 Derivation of (10.26)

This equation is simply derived by eliminating the nal velocity vf in (10.25). Remembering from (10.24)
that
vf = vi + at (10.33)

149
then (10.25) becomes
vi +vi +at
∆x = 2 t
2vi t+at2 (10.34)
= 2
∆x = vi t + 21 at2
10.6.1.1.4 Derivation of (10.27)

This equation is just derived by eliminating the time variable in the above equation. From (10.24) we know
v f − vi
t= (10.35)
a
Substituting this into (10.26) gives
2
+ 12 a f a i
vf −vi v −v
∆x = vi a
vf −2vi vf +vi2
2
vi2

vi v f 1
= a − a + 2 a a 2
vi vf vi2 vf2 vi vf vi2 (10.36)

= a − a + 2a − a + 2a
2a∆x = −2vi2 + vf2 + vi2
vf2 = vi2 + 2a∆x
This gives us the nal velocity in terms of the initial velocity, acceleration and displacement and is inde-
pendent of the time variable.
Exercise 10.6.1: Equations of motion (Solution on p. 164.)

A racing car is travelling north. It accelerates uniformly covering a distance of 725 m in 10 s. If
−1
it has an initial velocity of 10 m·s , nd its acceleration.
Exercise 10.6.2: Equations of motion (Solution on p. 165.)

A motorcycle, travelling east, starts from rest, moves in a straight line with a constant acceleration
and covers a distance of 64 m in 4 s. Calculate
1. its acceleration
2. its nal velocity
3. at what time the motorcycle had covered half the total distance
4. what distance the motorcycle had covered in half the total time.
10.6.1.1.4.1 Equations of motion

−1 −2
1. A car starts o at 10 m·s and accelerates at 1 m·s for 10 s. What is its nal velocity?
26
−2
2. A train starts from rest, and accelerates at 1 m·s for 10 s. How far does it move?
27
−1
3. A bus is going 30 m·s and stops in 5 s. What is its stopping distance for this speed?
28
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−1
4. A racing car going at 20 m·s stops in a distance of 20 m. What is its acceleration?
29
−1
5. A ball has a uniform acceleration of 4 m·s . Assume the ball starts from rest. Determine the velocity
and displacement at the end of 10 s.
30
−1
6. A motorcycle has a uniform acceleration of 4 m·s . Assume the motorcycle has an initial velocity of
−1
20 m·s . Determine the velocity and displacement at the end of 12 s.
31
−1
7. An aeroplane accelerates uniformly such that it goes from rest to 144 km·hr in 8 s. Calculate the
acceleration required and the total distance that it has traveled in this time.
32
10.6.2 Applications in the Real-World

What we have learnt in this chapter can be directly applied to road safety. We can analyse the relationship
between speed and stopping distance. The following worked example illustrates this application.
Exercise 10.6.3: Stopping distance (Solution on p. 166.)

−1
A truck is travelling at a constant velocity of 10 m·s when the driver sees a child 50 m in front
of him in the road. He hits the brakes to stop the truck. The truck accelerates at a rate of -1.25
−2
m·s . His reaction time to hit the brakes is 0,5 seconds. Will the truck hit the child?
10.6.3 Summary
• A reference point is a point from where you take your measurements.
• A frame of reference is a reference point with a set of directions.
• Your position is where you are located with respect to your reference point.
• The displacement of an object is how far it is from the reference point. It is the shortest distance
between the object and the reference point. It has magnitude and direction because it is a vector.
• The distance of an object is the length of the path travelled from the starting point to the end point.
It has magnitude only because it is a scalar.
• A vector is a physical quantity with magnitude and direction.
• A scalar is a physical quantity with magnitude only.
• Speed (s) is the distance covered (d) divided by the time taken (∆t):
d
s= (10.37)
∆t
• Average velocity (v ) is the displacement (∆x) divided by the time taken (∆t):
∆x
v= (10.38)
∆t
• Instantaneous speed is the speed at a specic instant in time.

• Instantaneous velocity is the velocity at a specic instant in time.
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151
• Acceleration (a) is the change in velocity (∆x) over a time interval (∆t):
∆v
a= (10.39)
∆t
• The gradient of a position - time graph (x vs. t) give the velocity.

• The gradient of a velocity - time graph (v vs. t) give the acceleration.
• The area under a velocity - time graph (v vs. t) give the displacement.
• The area under an acceleration - time graph (a vs. t) gives the velocity.
• The graphs of motion are summarised in .
• The equations of motion are used where constant acceleration takes place:
vf = vi + at
(vi +vf )
∆x = 2 t
(10.40)
1 2
∆x = vi t + 2 at
vf2 = vi2 + 2a∆x
10.6.4 End of Chapter Exercises: Motion in One Dimension

1. Give one word/term for the following descriptions.
a. The shortest path from start to nish.

b. A physical quantity with magnitude and direction.
c. The quantity dened as a change in velocity over a time period.
d. The point from where you take measurements.
e. The distance covered in a time interval.
f. The velocity at a specic instant in time.
33
2. Choose an item from column B that match the description in column A. Write down only the letter
next to the question number. You may use an item from column B more than once.
Column A Column B
a. The area under a velocity - time graph gradient
b. The gradient of a velocity - time graph area
c. The area under an acceleration - time graph velocity
d. The gradient of a displacement - time graph displacement
acceleration
slope
Table 10.4
34
3. Indicate whether the following statements are TRUE or FALSE. Write only 'true' or 'false'. If the
statement is false, write down the correct statement.
a. A scalar is the displacement of an object over a time interval.
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b. The position of an object is where it is located.

c. The sign of the velocity of an object tells us in which direction it is travelling.
d. The acceleration of an object is the change of its displacement over a period in time.
35
SC 2003/11 A body accelerates uniformly from rest for t0 seconds after which it continues with a constant velocity.
Which graph is the correct representation of the body's motion?
Figure 10.33
(a) (b)
Table 10.5
36
SC 2003/11 The velocity-time graphs of two cars are represented by P and Q as shown
Figure 10.37
The dierence in the distance travelled by the two cars (in m) after 4 s is ...
a. 12
b. 6
c. 2
d. 0
37
IEB 2005/11 HG The graph that follows shows how the speed of an athlete varies with time as he sprints for 100 m.
Figure 10.38
Which of the following equations can be used to correctly determine the time t for which he accelerates?
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153
a. 100 = (10) (11) − 12 (10) t

b. 100 = (10) (11) + 12 (10) t
c. 100 = 10t + 12 (10) t2
d. 100 = 21 (0) t + 12 (10) t2
38
SC 2002/03 HG1 In which one of the following cases will the distance covered and the magnitude of the displacement
be the same?
a. A girl climbs a spiral staircase.

b. An athlete completes one lap in a race.
c. A raindrop falls in still air.
d. A passenger in a train travels from Cape Town to Johannesburg.
39
SC 2003/11 A car, travelling at constant velocity, passes a stationary motor cycle at a trac light. As the car
overtakes the motorcycle, the motorcycle accelerates uniformly from rest for 10 s. The following
displacement-time graph represents the motions of both vehicles from the trac light onwards.
Figure 10.39
a. Use the graph to nd the magnitude of the constant velocity of the car.
b. Use the information from the graph to show by means of calculation that the magnitude of the
−2
acceleration of the motorcycle, for the rst 10 s of its motion is 7,5 m·s .
c. Calculate how long (in seconds) it will take the motorcycle to catch up with the car (point X on
the time axis).
d. How far behind the motorcycle will the car be after 15 seconds?
40
IEB 2005/11 HG Which of the following statements is true of a body that accelerates uniformly?
a. Its rate of change of position with time remains constant.
b. Its position changes by the same amount in equal time intervals.
c. Its velocity increases by increasing amounts in equal time intervals.
d. Its rate of change of velocity with time remains constant.
41
IEB 2003/11 HG1 The velocity-time graph for a car moving along a straight horizontal road is shown below.
Figure 10.40
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Which of the following expressions gives the magnitude of the average velocity of the car?
AreaA
a.
t
AreaA + AreaB
b.
t
AreaB
c.
t
AreaA − AreaB
d.
t
42
−1
SC 2002/11 SG A car is driven at 25 m·s in a municipal area. When the driver sees a trac ocer at a speed trap,
he realises he is travelling too fast. He immediately applies the brakes of the car while still 100 m away
from the speed trap.
a. Calculate the magnitude of the minimum acceleration which the car must have to avoid exceeding
−1
the speed limit, if the municipal speed limit is 16.6 m·s .
b. Calculate the time from the instant the driver applied the brakes until he reaches the speed trap.
−1
Assume that the car's velocity, when reaching the trap, is 16.6 m·s .
43
4. A trac ocer is watching his speed trap equipment at the bottom of a valley. He can see cars as they
enter the valley 1 km to his left until they leave the valley 1 km to his right. Nelson is recording the
times of cars entering and leaving the valley for a school project. Nelson notices a white Toyota enter
the valley at 11:01:30 and leave the valley at 11:02:42. Afterwards, Nelson hears that the trac ocer
−1
recorded the Toyota doing 140 km·hr .
a. What was the time interval (∆t) for the Toyota to travel through the valley?
b. What was the average speed of the Toyota?
−1
c. Convert this speed to km·hr .
−1
d. Discuss whether the Toyota could have been travelling at 140km·hr at the bottom of the valley.
e. Discuss the dierences between the instantaneous speed (as measured by the speed trap) and
average speed (as measured by Nelson).
44
IEB 2003/11HG A velocity-time graph for a ball rolling along a track is shown below. The graph has been divided up
into 3 sections, A, B and C for easy reference. (Disregard any eects of friction.)
Figure 10.41
a. Use the graph to determine the following:

i. the speed 5 s after the start
ii. the distance travelled in Section A
iii. the acceleration in Section C
b. At time t1 the velocity-time graph intersects the time axis. Use an appropriate equation of motion
to calculate the value of time t1 (in s).
c. Sketch a displacement-time graph for the motion of the ball for these 12 s. (You do not need to
calculate the actual values of the displacement for each time interval, but do pay attention to the
general shape of this graph during each time interval.)
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45
−1
5. In towns and cities, the speed limit is 60 km·hr . The length of the average car is 3.5 m, and the
width of the average car is 2 m. In order to cross the road, you need to be able to walk further than
the width of a car, before that car reaches you. To cross safely, you should be able to walk at least 2
m further than the width of the car (4 m in total), before the car reaches you.
−1 −1
a. If your walking speed is 4 km·hr , what is your walking speed in m·s ?
b. How long does it take you to walk a distance equal to the width of the average car?
−1
c. What is the speed in m·s of a car travelling at the speed limit in a town?
d. How many metres does a car travelling at the speed limit travel, in the same time that it takes
you to walk a distance equal to the width of car?
e. Why is the answer to the previous question important?
f. If you see a car driving toward you, and it is 28 m away (the same as the length of 8 cars), is it
safe to walk across the road?
g. How far away must a car be, before you think it might be safe to cross? How many car-lengths
is this distance?
46
−2
6. A bus on a straight road starts from rest at a bus stop and accelerates at 2 m·s until it reaches a
−1
speed of 20 m·s . Then the bus travels for 20 s at a constant speed until the driver sees the next bus
stop in the distance. The driver applies the brakes, stopping the bus in a uniform manner in 5 s.
a. How long does the bus take to travel from the rst bus stop to the second bus stop?
b. What is the average velocity of the bus during the trip?
47
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Step 1. The question explicitly gives
• the distance and time out (2 km in 30 minutes)

• the distance and time back (2 km in 30 minutes)
Step 2. The information is not in SI units and must therefore be converted.

To convert km to m, we know that:
1 km = 1 000 m
(10.41)
∴ 2 km = 2 000 m (multiplybothsidesby2, becausewewanttoconvert2kmtom.)
Similarly, to convert 30 minutes to seconds,
1 min = 60 s
(10.42)
∴ 30 min = 1 800 s (multiply both sides by30)
Step 3. James started at home and returned home, so his displacement is 0 m.
∆x = 0 m (10.43)
James walked a total distance of 4 000 m (2 000 m out and 2 000 m back).
d = 4 000 m (10.44)
Step 4. James took 1 800 s to walk out and 1 800 s to walk back.
∆t = 3 600 s (10.45)
Step 5.
s = d
∆t
4 000 m (10.46)
= 3 600 s
= 1, 11 m · s −1
Step 6.
∆x
v = ∆t
0 m (10.47)
= 3 600 s
= 0 m·s −1

Step 1. To determine the man's speed we need to know the distance he travels and how long it takes. We know
it takes 120 s to complete one revolution of the track.(A revolution is to go around the track once.)
Step 2. What distance is one revolution of the track? We know the track is a circle and we know its radius, so
we can determine the distance around the circle. We start with the equation for the circumference of
a circle
C = 2πr
= 2π (100 m) (10.48)
= 628, 32 m
Therefore, the distance the man covers in one revolution is 628, 32 m.

157
Step 3. We know that speed is distance covered per unit time. So if we divide the distance covered by the time
it took we will know how much distance was covered for every unit of time. No direction is used here
because speed is a scalar.
s = d
∆t
628,32 m
= (10.49)
120 s
= 5, 24 m · s−1
Step 4. Consider the point A in the diagram. We know which way the man is running around the track and
we know his speed. His velocity at point A will be his speed (the magnitude of the velocity) plus his
direction of motion (the direction of his velocity). The instant that he arrives at A he is moving as
indicated in the diagram. His velocity will be 5, 24 m · s−1 West.
Figure 10.42
Figure 10.43
Step 5. Consider the point B in the diagram. We know which way the man is running around the track and
we know his speed. His velocity at point B will be his speed (the magnitude of the velocity) plus his
direction of motion (the direction of his velocity). The instant that he arrives at B he is moving as
indicated in the diagram. His velocity will be 5, 24 m · s−1 South.
Figure 10.44

Figure 10.45
Step 6. To determine the average velocity between A and B, we need the change in displacement between A
and B and the change in time between A and B. The displacement from A and B can be calculated by
using the Theorem of Pythagoras:
2 2 2
(∆x) = (100m) + (100m)
= 20000m (10.50)
∆x = 141, 42135... m
1
The time for a full revolution is 120 s, therefore the time for a
4 of a revolution is 30 s.
∆x
vAB = ∆t
141,42...m
= (10.51)
30s
= 4.71 m · s −1
Figure 10.46
Velocity is a vector and needs a direction.

◦
Triangle AOB is isosceles and therefore angle BAO = 45 .
The direction is between west and south and is therefore southwest.
The nal answer is: v = 4.71 m · s−1 , southwest.
Step 7. Because he runs at a constant rate, we know that his speed anywhere around the track will be the
same. His average speed is 5, 24 m · s−1 .
Step 8. tip: Remember - displacement can be zero even when distance travelled is not!
To calculate average velocity we need his total displacement and his total time. His displacement is
zero because he ends up where he started. His time is 120 s. Using these we can calculate his average
velocity:
∆x
v = ∆t
0 m (10.52)
= 120 s
= 0 m·s −1

Step 1. Consider the motion of the car in two parts: the rst 8 seconds and the last 6 seconds.

159
For the rst 8 seconds:
vi = 2 m · s−1
vf = 10 m · s−1
(10.53)
ti = 0s
tf = 8s
For the last 6 seconds:
vi = 10 m · s−1
vf = 4 m · s−1
(10.54)
ti = 8s
tf = 14 s
Step 2. For the rst 8 seconds:
∆v
a = ∆t
10m·s −2m·s−1
−1
(10.55)
= 8s−0s
= 1 m·s −2
For the next 6 seconds:
∆v
a = ∆t
4m·s −10m·s−1
−1
(10.56)
= 14s−8s
= −1 m · s
−2
During the rst 8 seconds the car had a positive acceleration. This means that its velocity increased.
The velocity is positive so the car is speeding up. During the next 6 seconds the car had a negative
acceleration. This means that its velocity decreased. The velocity is negative so the car is slowing
down.

Step 1. The question gives a position vs. time graph and the following three things are required:
a. Draw a v vs. t graph.

b. Draw an a vs. t graph.
c. Describe the motion of the car.
To answer these questions, break the motion up into three sections: 0 2 seconds, 2 4 seconds and
4 6 seconds.
Step 2. For the rst 2 seconds we can see that the displacement remains constant - so the object is not moving,
thus it has zero velocity during this time. We can reach this conclusion by another path too: remember
that the gradient of a displacement vs. time graph is the velocity. For the rst 2 seconds we can see
that the displacement vs. time graph is a horizontal line, ie. it has a gradient of zero. Thus the velocity
during this time is zero and the object is stationary.
Step 3. For the next 2 seconds, displacement is increasing with time so the object is moving. Looking at the
gradient of the displacement graph we can see that it is not constant. In fact, the slope is getting steeper
(the gradient is increasing) as time goes on. Thus, remembering that the gradient of a displacement
vs. time graph is the velocity, the velocity must be increasing with time during this phase.

Step 4. For the nal 2 seconds we see that displacement is still increasing with time, but this time the gradient
is constant, so we know that the object is now travelling at a constant velocity, thus the velocity vs.
time graph will be a horizontal line during this stage. We can now draw the graphs:
So our velocity vs. time graph looks like this one below. Because we haven't been given any values
on the vertical axis of the displacement vs. time graph, we cannot gure out what the exact gradients
are and therefore what the values of the velocities are. In this type of question it is just important to
show whether velocities are positive or negative, increasing, decreasing or constant.
Figure 10.47
Once we have the velocity vs. time graph its much easier to get the acceleration vs. time graph as we
know that the gradient of a velocity vs. time graph is the just the acceleration.
Step 5. For the rst 2 seconds the velocity vs. time graph is horisontal and has a value of zero, thus it has a
gradient of zero and there is no acceleration during this time. (This makes sense because we know from
the displacement time graph that the object is stationary during this time, so it can't be accelerating).
Step 6. For the next 2 seconds the velocity vs. time graph has a positive gradient. This gradient is not changing
(i.e. its constant) throughout these 2 seconds so there must be a constant positive acceleration.
Step 7. For the nal 2 seconds the object is traveling with a constant velocity. During this time the gradient of
the velocity vs. time graph is once again zero, and thus the object is not accelerating. The acceleration
vs. time graph looks like this:
Figure 10.48
Step 8. A brief description of the motion of the object could read something like this: At t=0 s and object
is stationary at some position and remains stationary until t = 2 s when it begins accelerating. It
accelerates in a positive direction for 2 seconds until t=4 s and then travels at a constant velocity for
a further 2 seconds.

Step 1. We are asked to calculate the distance and displacement of the car. All we need to remember here
is that we can use the area between the velocity vs. time graph and the time axis to determine the
distance and displacement.
Step 2. Break the motion up: 0 5 seconds, 5 12 seconds, 12 14 seconds and 14 15 seconds.
For 0 5 seconds: The displacement is equal to the area of the triangle on the left:
1
Area[U+25B5] = 2 b×h
1
= 2 × 5 s × 4 m · s−1 (10.57)
= 10 m

161
For 5 12 seconds: The displacement is equal to the area of the rectangle:
Area = `×b
= 7 s × 4 m · s−1 (10.58)
2
= 28 m
For 12 14 seconds the displacement is equal to the area of the triangle above the time axis on the
right:
1
1
= 2 × 2 s × 4 m · s−1 (10.59)
= 4 m
For 14 15 seconds the displacement is equal to the area of the triangle below the time axis:
1
1
= 2 × 1 s × 2 m · s−1 (10.60)
= 1 m
Step 3. Now the total distance of the car is the sum of all of these areas:
∆x = 10 m + 28 m + 4 m + 1 m
(10.61)
= 43 m
Step 4. Now the total displacement of the car is just the sum of all of these areas. HOWEVER, because in the
last second (from t = 14 s to t = 15 s) the velocity of the car is negative, it means that the car was
going in the opposite direction, i.e. back where it came from! So, to nd the total displacement, we
have to add the rst 3 areas (those with positive displacements) and subtract the last one (because it
is a displacement in the opposite direction).
∆x = 10 m + 28 m + 4 m − 1 m
(10.62)
= 41 m in the positive direction

Step 1. The velocity is given by the gradient of a position vs. time graph. During the rst 4 seconds, this is
∆x
v = ∆t
4 m−0 m (10.63)
= 4 s−0 s
= 1 m·s −1
Step 2. For the last 3 seconds we can see that the displacement stays constant. The graph shows a horisontal
line and therefore the gradient is zero. Thus v = 0 m · s−1 .
Step 1. The motion of the car can be divided into three time sections: 0 2 seconds; 2 4 seconds and 4 6
seconds. To be able to draw the velocity vs. time graph, the velocity for each time section needs to be
calculated. The velocity is equal to the area of the square under the graph:

For 0 2 seconds:
Area = `×b
= 2 s × 2 m · s−2 (10.64)
= 4 m·s −1
−1
The velocity of the car is 4 m·s at t = 2s. For 2 4 seconds:
Area = `×b
= 2 s × 0 m · s−2 (10.65)
= 0 m · s−1
The velocity of the car is 0 m·s

−1
from t=2 s to t=4 s. For 4 6 seconds:
Area = `×b
= 2 s × −2 m · s−2 (10.66)
= −4 m · s−1
The acceleration had a negative value, which means that the velocity is decreasing. It starts at a
−1 −1
velocity of 4 m·s and decreases to 0 m·s .
Step 2. The velocity vs. time graph looks like this:
Figure 10.49

Step 1. We are given:
vi = 10 m · s−1
∆x = 725 m
(10.67)
t = 10 s
a = ?
Step 2. If you struggle to nd the correct equation, nd the quantity that is not given and then look for an
equation that has this quantity in it.
We can use equation (10.26)
1
∆x = vi t + at2 (10.68)
2
Step 3.
∆x = vi t + 12 at2
2
725 m 10 m · s−1 × 10 s + 12 a × (10 s)

=
(10.69)
725 m − 100 m 50 s2 a

=
a = 12, 5 m · s−2

163
−2
Step 4. The racing car is accelerating at 12,5 m·s north.

vi = 0m · s−1 (because the object starts from rest.)

∆x = 64m
t = 4s
a = ? (10.70)
vf = ?
t = ? at half the distance∆x = 32 m.
∆x = ? at half the time t = 2 s.
All quantities are in SI units.
Step 2. We can use (10.26)
1
∆x = vi t + at2 (10.71)
2
Step 3.
∆x = vi t + 12 at2
2
64 m 0m · s−1 × 4 s + 12 a × (4 s)

=
(10.72)
64 m 8 s2 a

=
a = 8 m · s−2 east
Step 4. We can use (10.27) - remember we now also know the acceleration of the object.
vf = vi + at (10.73)
Step 5.
vf = vi + at
vf 0m · s −1
+ 8m · s−2 (4 s)

= (10.74)
= 32 m · s −1
east
Step 6. We can use (10.26):
∆x = vi + 21 at2
2
32 m 0m · s−1 t + 12 8m · s−2 (t)

=
32 m 0 + 4m · s−2 t2

= (10.75)
2 2
8s = t
t = 2, 83 s
Step 7. Half the time is 2 s, thus we have vi , a and t - all in the correct units. We can use (10.26) to get the
distance:
∆x = vi t + 12 at2
1 2
= (0) (2) + (8) (2) (10.76)
2
= 16 m east

Step 8. a. The acceleration is 8 m · s−2 east

b. The velocity is 32 m · s−1 east
c. The time at half the distance is 2, 83 s
d. The distance at half the time is 16 m east

Step 1. It is useful to draw a timeline like this one:
Figure 10.50
We need to know the following:
• What distance the driver covers before hitting the brakes.

• How long it takes the truck to stop after hitting the brakes.
• What total distance the truck covers to stop.
Step 2. Before the driver hits the brakes, the truck is travelling at constant velocity. There is no acceleration
and therefore the equations of motion are not used. To nd the distance traveled, we use:
v = d
t
d (10.77)
10 = 0,5
d = 5m
The truck covers 5 m before the driver hits the brakes.
Step 3. We have the following for the motion between B and C:
vi = 10 m · s−1
vf = 0 m · s−1
(10.78)
a = −1, 25 m · s−2
t = ?
We can use (10.24)
vf = vi + at
0 = 10 + (−1, 25) t
(10.79)
−10 = −1, 25t
t = 8s
Step 4. For the distance we can use (10.25) or (10.26). We will use (10.25):
(vi +vf )
∆x = 2 t
10+0 (10.80)
∆x = s (8)
∆x = 40 m
Step 5. The total distance that the truck covers is dAB + dBC = 5 + 40 = 45 meters. The child is 50 meters
ahead. The truck will not hit the child.

Chapter 11
Gravity and mechanical energy
11.1 Weight and acceleration due to gravity 1
Veritasium video on falling objects - 1

<http://www.youtube.com/v/_mCC-68LyZM&rel=0>
Figure 11.1
11.1.1 Weight
Weight is the gravitational force that the Earth exerts on any object. The weight of an objects gives you an
indication of how strongly the Earth attracts that body towards its centre. Weight is calculated as follows:
W eight = mg
where m = mass of the object (in kg)
and g = the acceleration due to gravity (9, 8 m·s
−2
)
For example, what is Sarah's weight if her mass is 50 kg. Sarah's weight is calculated according to:
Weight = mg
(50 kg) 9, 8 m · s−2

=
(11.1)
= 490 kg · m · s−2
= 490 N
165
166 CHAPTER 11. GRAVITY AND MECHANICAL ENERGY
Figure 11.2
tip: Weight is sometimes abbreviated as Fg which refers to the force of gravity. Do not use the
abbreviation W for weight as it refers to `Work'.
Now, we have said that the value of g is approximately 9, 8 m · s−2 on the surface of the Earth. The actual
value varies slightly over the surface of the Earth. Each planet in our Solar System has its own value for g.
These values are listed as multiples of g on Earth in Table 11.1
Planet Gravitational Acceleration (multiples of g on Earth)
Mercury 0,376
Venus 0,903
Earth 1
Mars 0,38
Jupiter 2,34
Saturn 1,16
Uranus 1,15
Neptune 1,19
Pluto 0,066
Table 11.1: A list of the gravitational accelerations at the surfaces of each of the planets in our solar
system. Values are listed as multiples of g on Earth. Note: The "surface" is taken to mean the cloud tops
of the gas giants (Jupiter, Saturn, Uranus and Neptune).
Exercise 11.1.1: Determining mass and weight on other planets (Solution on p. 183.)
Sarah's mass on Earth is 50 kg. What is her mass and weight on Mars?

167
11.1.1.1 Dierences between Mass and Weight

Mass is measured in kilograms (kg) and is the amount of matter in an object. An object's mass does not
change unless matter is added or removed from the object.
The dierences between mass and weight can be summarised in the following table:
Mass Weight
1. is a measure of how much matter there is in an 1. is the force with which the Earth attracts an
object. object.
2. is measured in kilograms. 2. is measured in newtons
3. is the same on any planet. 3. is dierent on dierent planets.
4. is a scalar. 4. is a vector.
Table 11.2
11.1.1.1.1 Weight
1. A bag of sugar has a mass of 1 kg. How much does it weigh:
a. on Earth?
b. on Jupiter?
c. on Pluto?
2
2. Neil Armstrong was the rst man to walk on the surface of the Moon. The gravitational acceleration
1
on the Moon is
6 of the gravitational acceleration on Earth, and there is negligible gravitational
acceleration in outer space. If Neil's mass was 90 kg, what was his weight:
a. on Earth?
b. on the Moon?
c. in outer space?
3
3. A monkey has a mass of 15 kg on Earth. The monkey travels to Mars. What is his mass and weight
on Mars?
4
4. Determine your mass by using a bathroom scale and calculate your weight for each planet in the Solar
System, using the values given in Table 11.1
5
11.1.2 Acceleration due to Gravity

11.1.2.1 Gravitational Fields
A eld is a region of space in which a mass experiences a force. Therefore, a gravitational eld is a region
of space in which a mass experiences a gravitational force.
2 http://www.fhsst.org/lx5
3 http://www.fhsst.org/lxN
4 http://www.fhsst.org/lxR
5 http://www.fhsst.org/lxn

11.1.2.2 Free fall

tip: Free fall is motion in the Earth's gravitational eld when no other forces act on the object.
Free fall is the term used to describe a special kind of motion in the Earth's gravitational eld. Free fall
is motion in the Earth's gravitational eld when no other forces act on the object. It is basically an ideal
situation, since in reality, there is always some air friction which slows down the motion.
11.1.2.2.1 Experiment : Acceleration due to Gravity

Aim: Investigating the acceleration of two dierent objects during free fall.
Apparatus: Tennis ball and a sheet of A4 paper.
Method:
1. Hold the tennis ball and sheet of paper (horizontally) the same distance from the ground. Which one
would strike the ground rst if both were dropped?
Figure 11.3
2. Drop both objects and observe. Explain your observations.

3. Now crumple the paper into a ball, more or less the same size as the tennis ball. Drop the paper and
tennis ball again and observe. Explain your observations.
4. Why do you think the two situations are dierent?
5. Compare the value for the acceleration due to gravity of the tennis ball to the crumpled piece of paper.
6. Predict what will happen if an iron ball and a tennis ball of the same size are dropped from the same
height. What will the values for their acceleration due to gravity be?
If a metal ball and tennis ball (of the same size) were dropped from the same height, both would reach the
ground at the same time. It does not matter that the one ball is heavier than the other. The acceleration of
an object due to gravity is independent of the mass of the object. It does not matter what the mass of the
object is.
The shape of the object, however, is important. The sheet of paper took much longer to reach the ground
than the tennis ball. This is because the eect of air friction on the paper was much greater than the air
friction on the tennis ball.
If we lived in a world where there was no air resistance, the A4 sheet of paper and the tennis ball would
reach the ground at the same time. This happens in outer space or in a vaccuum.
Galileo Galilei, an Italian scientist, studied the motion of objects. The following case study will tell you
more about one of his investigations.
11.1.2.2.2 Case Study : Galileo Galilei

In the late sixteenth century, it was generally believed that heavier objects would fall faster than lighter
objects. The Italian scientist Galileo Galilei thought dierently. Galileo hypothesized that two objects
would fall at the same rate regardless of their mass. Legend has it that in 1590, Galileo planned out an
experiment. He climbed to the top of the Leaning Tower of Pisa and dropped several large objects to test his
theory. He wanted to show that two dierent objects fall at the same rate (as long as we ignore air resistance).

169
Galileo's experiment proved his hypothesis correct; the acceleration of a falling object is independent of the
object's mass.
A few decades after Galileo, Sir Isaac Newton would show that acceleration depends upon both force and
mass. While there is greater force acting on a larger object, this force is canceled out by the object's greater
mass. Thus two objects will fall (actually they are pulled) to the earth at exactly the same rate.
Questions: Read the case study above and answer the following questions.
1. Divide into pairs and explain Galileo's experiment to your friend.
2. Write down an aim and a hypothesis for Galileo's experiment.
3. Write down the result and conclusion for Galileo's experiment.
11.1.2.2.3 Research Project : Experimental Design

Design an experiment similar to the one done by Galileo to prove that the acceleration due to gravity of an
object is independent of the object's mass. The investigation must be such that you can perform it at home
or at school. Bring your apparatus to school and perform the experiment. Write it up and hand it in for
assessment.
11.1.2.2.4 Case Study : Determining the acceleration due to gravity 1

Study the set of photographs alongside showing the position of a ball being dropped from a height at constant
time intervals. The distance of the ball from the starting point in each consecutive image is observed to be:
x1 = 0 cm, x2 = 4, 9 cm, x3 = 19, 6 cm, x4 = 44, 1 cm, x5 = 78, 4 cm and x6 = 122, 5 cm. Answer the
following questions:
1. Determine the time between each picture if the frequency of the exposures were 10 Hz.
2. Calculate the velocity, v2 , of the ball between positions 1 and 3.
x3 − x1
v2 = (11.2)
t3 − t1
3. Calculate the velocity, v5 , of the ball between positions 4 and 6.
x6 − x4
v5 = (11.3)
t6 − t4
4. Calculate the acceleration the ball between positions 2 and 5.
v5 − v2
a= (11.4)
t5 − t2
5. Compare your answer to the value for the acceleration due to gravity (9, 8 m·s−2 ).
Figure 11.4

The acceleration due to gravity is constant. This means we can use the equations of motion under
6
constant acceleration that we derived in motion in one dimension to describe the motion of an object in
free fall. The equations are repeated here for ease of use.
vi initial velocity m · s−1 at t = 0s

=
vf = final velocity m · s−1 at time t

∆x = displacement (m)
(11.5)
t = time (s)
∆t = time interval (s)
g acceleration m · s−2

=
vf = vi + gt (11.6)
(vi + vf )
∆x = t (11.7)
2
1
∆x = vi t + gt2 (11.8)
2
vf2 = vi2 + 2g∆x (11.9)
11.1.2.2.5 Experiment : Determining the acceleration due to gravity 2

Work in groups of at least two people.
Aim: To determine the acceleration of an object in freefall.
Apparatus: Large marble, two stopwatches, measuring tape.
Method:
1. Measure the height of a door, from the top of the door to the oor, exactly. Write down the measure-
ment.
2. One person must hold the marble at the top of the door. Drop the marble to the oor at the same
time as he/she starts the rst stopwatch.
3. The second person watches the oor and starts his stopwatch when the marble hits the oor.
4. The two stopwatches are stopped together and the two times substracted. The dierence in time will
give the time taken for the marble to fall from the top of the door to the oor.
5. Design a table to show the results of your experiment. Choose appropriate headings and units.
6. Choose an appropriate equation of motion to calculate the acceleration of the marble. Remember that
the marble starts from rest and that it's displacement was determined in the rst step.
7. Write a conclusion for your investigation.
8. Answer the following questions:
a. Why do you think two stopwatches were used in this investigation?

b. Compare the value for acceleration obtained in your investigation with the value of acceleration
due to gravity (9, 8 m · s−2 ). Explain your answer.
Exercise 11.1.2: A freely falling ball (Solution on p. 183.)

A ball is dropped from the balcony of a tall building. The balcony is 15 m above the ground.
Assuming gravitational acceleration is 9, 8 m · s−2 , nd:
6 "Motion in One Dimension - Grade 10": Section Equations of Motion <http://cnx.org/content/m30867/latest/#cid10>

171
1. the time required for the ball to hit the ground, and
2. the velocity with which it hits the ground.
By now you should have seen that free fall motion is just a special case of motion with constant acceleration,
and we use the same equations as before. The only dierence is that the value for the acceleration, a, is
always equal to the value of gravitational acceleration, g. In the equations of motion we can replace a with
g.
11.1.2.2.6 Gravitational Acceleration

1. A brick falls from the top of a 5m high building. Calculate the velocity with which the brick reaches
the ground. How long does it take the brick to reach the ground?
7
2. A stone is dropped from a window. It takes the stone 1, 5 s to reach the ground. How high above the
ground is the window?
8
3. An apple falls from a tree from a height of 1, 8 m. What is the velocity of the apple when it reaches
the ground?
9
11.2 Potential energy 10
11.2.1 Potential Energy

The potential energy of an object is generally dened as the energy an object has because of its position
relative to other objects that it interacts with. There are dierent kinds of potential energy such as gravitional
potential energy, chemical potential energy, electrical potential energy, to name a few. In this section we will
be looking at gravitational potential energy.
Denition 11.1: Potential energy

Potential energy is the energy an object has due to its position or state.
Gravitational potential energy is the energy of an object due to its position above the surface of the
Earth. The symbol PE is used to refer to gravitational potential energy. You will often nd that the words
potential energy are used where gravitational potential energy is meant. We can dene potential energy (or
gravitational potential energy, if you like) as:
P E = mgh (11.10)
where PE = potential energy measured in joules (J)

m = mass of the object (measured in kg)
g = gravitational acceleration (9, 8 m · s−2 )
h = perpendicular height from the reference point (measured in m)
A suitcase, with a mass of 1 kg, is placed at the top of a 2m high cupboard. By lifting the suitcase
against the force of gravity, we give the suitcase potential energy. This potential energy can be calculated
using (11.10).
If the suitcase falls o the cupboard, it will lose its potential energy. Halfway down the cupboard, the
suitcase will have lost half its potential energy and will have only 9, 8 J left. At the bottom of the cupboard
the suitcase will have lost all it's potential energy and it's potential energy will be equal to zero.
7 http://www.fhsst.org/lxQ
8 http://www.fhsst.org/lxU
9 http://www.fhsst.org/lxP

Objects have maximum potential energy at a maximum height and will lose their potential energy as
they fall.
Figure 11.5
Exercise 11.2.1: Gravitational potential energy (Solution on p. 183.)

A brick with a mass of 1 kg is lifted to the top of a 4m high roof. It slips o the roof and falls to
the ground. Calculate the potential energy of the brick at the top of the roof and on the ground
once it has fallen.
11.2.1.1 Gravitational Potential Energy

1. Describe the relationship between an object's gravitational potential energy and its:
a. mass and
b. height above a reference point.
11
2. A boy, of mass 30 kg, climbs onto the roof of a garage. The roof is 2, 5 m from the ground. He now
jumps o the roof and lands on the ground.
a. How much potential energy has the boy gained by climbing on the roof ?
b. The boy now jumps down. What is the potential energy of the boy when he is 1m from the
ground?
c. What is the potential energy of the boy when he lands on the ground?
12
3. A hiker walks up a mountain, 800 m above sea level, to spend the night at the top in the rst overnight
hut. The second day he walks to the second overnight hut, 500 m above sea level. The third day he
returns to his starting point, 200 m above sea level.
a. What is the potential energy of the hiker at the rst hut (relative to sea level)?
b. How much potential energy has the hiker lost during the second day?
c. How much potential energy did the hiker have when he started his journey (relative to sea level)?
d. How much potential energy did the hiker have at the end of his journey?
13
11.3 Kinetic energy 14
11.3.1 Kinetic Energy

Denition 11.2: Kinetic Energy
Kinetic energy is the energy an object has due to its motion.
11 http://www.fhsst.org/lxE
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13 http://www.fhsst.org/lxy

173
Kinetic energy is the energy an object has because of its motion. This means that any moving object
has kinetic energy. The faster it moves, the more kinetic energy it has. Kinetic energy (KE ) is therefore
dependent on the velocity of the object. The mass of the object also plays a role. A truck of 2000 kg, moving
at 100 km · hr−1 , will have more kinetic energy than a car of 500 kg, also moving at 100 km · hr−1 . Kinetic
energy is dened as:
1
KE = mv 2 (11.11)
2
Consider the 1 kg suitcase on the cupboard that was discussed earlier. When the suitcase falls, it will gain
velocity (fall faster), until it reaches the ground with a maximum velocity. The suitcase will not have any
kinetic energy when it is on top of the cupboard because it is not moving. Once it starts to fall it will gain
kinetic energy, because it gains velocity. Its kinetic energy will increase until it is a maximum when the
suitcase reaches the ground.
Figure 11.6
Exercise 11.3.1: Calculation of Kinetic Energy (Solution on p. 184.)

A 1 kg brick falls o a 4m high roof. It reaches the ground with a velocity of 8, 85 m · s−1 . What
is the kinetic energy of the brick when it starts to fall and when it reaches the ground?
11.3.1.1 Checking units

According to the equation for kinetic energy, the unit should be kg · m2 · s−2 . We can prove that this unit
is equal to the joule, the unit for energy.
2
(kg) m · s−1 kg · m · s−2 · m

=
N·m becauseForce (N ) = mass (kg) × acceleration m · s−2

= (11.12)
= J (Work (J) = Force (N ) × distance (m))

We can do the same to prove that the unit for potential energy is equal to the joule:
(kg) m · s−2 (m) N·m

=
(11.13)
= J
Exercise 11.3.2: Mixing Units & Energy Calculations (Solution on p. 184.)

A bullet, having a mass of 150 g , is shot with a muzzle velocity of 960 m · s−1 . Calculate its kinetic
energy.
11.3.1.1.1 Kinetic Energy

1. Describe the relationship between an object's kinetic energy and its:
a. mass and
b. velocity

15
2. A stone with a mass of 100 g is thrown up into the air. It has an initial velocity of 3 m · s−1 . Calculate
its kinetic energy
a. as it leaves the thrower's hand.

b. when it reaches its turning point.
16
3. A car with a mass of 700 kg is travelling at a constant velocity of 100 km · hr−1 . Calculate the kinetic
energy of the car.
17
11.4 Mechanical energy 18
11.4.1 Mechanical Energy

tip: Mechanical energy is the sum of the gravitational potential energy and the kinetic energy.
Mechanical energy, U , is simply the sum of gravitational potential energy (P E ) and the kinetic energy (KE ).
Mechanical energy is dened as:
U = P E + KE (11.14)
U = P E + KE
(11.15)
U = mgh + 21 mv 2
11.4.1.1 Conservation of Mechanical Energy

The Law of Conservation of Energy states:
Energy cannot be created or destroyed, but is merely changed from one form into another.
Denition 11.3: Conservation of Energy

The Law of Conservation of Energy: Energy cannot be created or destroyed, but is merely changed
from one form into another.
So far we have looked at two types of energy: gravitational potential energy and kinetic energy. The sum
of the gravitational potential energy and kinetic energy is called the mechanical energy. In a closed system,
one where there are no external forces acting, the mechanical energy will remain constant. In other words,
it will not change (become more or less). This is called the Law of Conservation of Mechanical Energy and
it states:
The total amount of mechanical energy in a closed system remains constant.
Denition 11.4: Conservation of Mechanical Energy

Law of Conservation of Mechanical Energy: The total amount of mechanical energy in a closed
system remains constant.
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175
This means that potential energy can become kinetic energy, or vise versa, but energy cannot 'dissappear'.
The mechanical energy of an object moving in the Earth's gravitational eld (or accelerating as a result of
gravity) is constant or conserved, unless external forces, like air resistance, acts on the object.
We can now use the conservation of mechanical energy to calculate the velocity of a body in freefall and
show that the velocity is independent of mass.
Show by using the law of conservation of energy that the velocity of a body in free fall is independent of
its mass.
tip: In problems involving the use of conservation of energy, the path taken by the object can be
ignored. The only important quantities are the object's velocity (which gives its kinetic energy)
and height above the reference point (which gives its gravitational potential energy).
tip: In the absence of friction, mechanical energy is conserved and
Ubefore = Uafter (11.16)
In the presence of friction, mechanical energy is not conserved. The mechanical energy lost is
equal to the work done against friction.
∆U = Ubefore − Uafter = workdoneagainstfriction (11.17)
In general, mechanical energy is conserved in the absence of external forces. Examples of external forces are:
applied forces, frictional forces and air resistance.
In the presence of internal forces like the force due to gravity or the force in a spring, mechanical energy
is conserved.
The following simulation covers the law of conservation of energy.
19
run demo
Figure 11.7
11.4.1.2 Using the Law of Conservation of Energy

Mechanical energy is conserved (in the absence of friction). Therefore we can say that the sum of the PE
and the KE anywhere during the motion must be equal to the sum of the PE and the KE anywhere else
in the motion.
We can now apply this to the example of the suitcase on the cupboard. Consider the mechanical energy
of the suitcase at the top and at the bottom. We can say:
Figure 11.8
19 http://phet.colorado.edu/sims/energy-skate-park/energy-skate-park_en.jnlp

Utop = Ubottom
P Etop + KEtop = P Ebottom + KEbottom
1
mgh + 2 mv
2
= mgh + 21 mv 2
0 + 12 (1) v 2

(1) (9, 8) (2) + 0 = (11.18)
1 2
19, 6 J = 2v
2
39, 2 = v
v = 6, 26 m · s−1
The suitcase will strike the ground with a velocity of 6, 26 m · s−1 .
From this we see that when an object is lifted, like the suitcase in our example, it gains potential energy.
As it falls back to the ground, it will lose this potential energy, but gain kinetic energy. We know that energy
cannot be created or destroyed, but only changed from one form into another. In our example, the potential
energy that the suitcase loses is changed to kinetic energy.
The suitcase will have maximum potential energy at the top of the cupboard and maximum kinetic energy
at the bottom of the cupboard. Halfway down it will have half kinetic energy and half potential energy. As
it moves down, the potential energy will be converted (changed) into kinetic energy until all the potential
energy is gone and only kinetic energy is left. The 19, 6 J of potential energy at the top will become 19, 6 J
of kinetic energy at the bottom.
Exercise 11.4.1: Using the Law of Conservation of Mechanical Energy (Solution on p.

184.)
During a ood a tree truck of mass 100 kg falls down a waterfall. The waterfall is 5m high. If air
resistance is ignored, calculate
1. the potential energy of the tree trunk at the top of the waterfall.
2. the kinetic energy of the tree trunk at the bottom of the waterfall.
3. the magnitude of the velocity of the tree trunk at the bottom of the waterfall.
Figure 11.9
Exercise 11.4.2: Pendulum (Solution on p. 185.)

A 2 kg metal ball is suspended from a rope. If it is released from point A and swings down to the
point B (the bottom of its arc):
1. Show that the velocity of the ball is independent of it mass.

2. Calculate the velocity of the ball at point B.
Figure 11.10

177
11.4.1.2.1 Potential Energy

1. A tennis ball, of mass 120 g , is dropped from a height of 5 m. Ignore air friction.
a. What is the potential energy of the ball when it has fallen 3 m?

b. What is the velocity of the ball when it hits the ground?
20
2. A bullet, mass 50 g , is shot vertically up in the air with a muzzle velocity of 200 m · s−1 . Use the
Principle of Conservation of Mechanical Energy to determine the height that the bullet will reach.
Ignore air friction.
21
3. A skier, mass 50 kg, is at the top of a 6, 4 m ski slope.
a. Determine the maximum velocity that she can reach when she skies to the bottom of the slope.
b. Do you think that she will reach this velocity? Why/Why not?
22
4. A pendulum bob of mass 1, 5 kg, swings from a height A to the bottom of its arc at B. The velocity of
the bob at B is 4 m · s−1 . Calculate the height A from which the bob was released. Ignore the eects
of air friction.
23
5. Prove that the velocity of an object, in free fall, in a closed system, is independent of its mass.
24
11.4.2 Energy graphs

Let us consider our example of the suitcase on the cupboard, once more.
Figure 11.11
Let's look at each of these quantities and draw a graph for each. We will look at how each quantity
changes as the suitcase falls from the top to the bottom of the cupboard.
• Potential energy: The potential energy starts o at a maximum and decreases until it reaches zero
at the bottom of the cupboard. It had fallen a distance of 2 metres.
Figure 11.12
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• Kinetic energy: The kinetic energy is zero at the start of the fall. When the suitcase reaches the
ground, the kinetic energy is a maximum. We also use distance on the x-axis.
Figure 11.13
• Mechanical energy: The mechanical energy is constant throughout the motion and is always a
maximum. At any point in time, when we add the potential energy and the kinetic energy, we will get
the same number.
Figure 11.14
The following presentation provides a summary of some of the concepts covered in this chapter.

<http://static.slidesharecdn.com/swf/ssplayer2.swf ?doc=energy-100511053412-
phpapp02&stripped_title=energy-4047773>
Figure 11.15
11.4.3 Summary
• Mass is the amount of matter an object is made up of.
• Weight is the force with which the Earth attracts a body towards its centre.
• Newtons Law of Gravitation.
• A body is in free fall if it is moving in the Earth's gravitational eld and no other forces act on it.
• The equations of motion can be used for free fall problems. The acceleration (a) is equal to the
acceleration due to gravity (g).
• The potential energy of an object is the energy the object has due to his position above a reference
point.
• The kinetic energy of an object is the energy the object has due to its motion.
• Mechanical energy of an object is the sum of the potential energy and kinetic energy of the object.
• The unit for energy is the joule (J).
• The Law of Conservation of Energy states that energy cannot be created or destroyed, but can only
be changed from one form into another.
• The Law of Conservation of Mechanical Energy states that the total mechanical energy of an isolated
• The table below summarises the most important equations:

179
Weight Fg = m · g
Equation of motion vf = vi + gt
(vi +vf )
Equation of motion ∆x = 2 t
1 2
Equation of motion ∆x = vi t + 2 gt
Equation of motion vf2 = vi2 + 2g∆x
Potential Energy P E = mgh
Kinetic Energy KE = 12 mv 2
Mechanical Energy U = KE + P E
Table 11.3
11.4.4 End of Chapter Exercises: Gravity and Mechanical Energy

1. Give one word/term for the following descriptions.
a. The force with which the Earth attracts a body.

b. The unit for energy.
c. The movement of a body in the Earth's gravitational eld when no other forces act on it.
d. The sum of the potential and kinetic energy of a body.
e. The amount of matter an object is made up of.
25
2. Consider the situation where an apple falls from a tree. Indicate whether the following statements
regarding this situation are TRUE or FALSE. Write only 'true' or 'false'. If the statement is false,
write down the correct statement.
a. The potential energy of the apple is a maximum when the apple lands on the ground.
b. The kinetic energy remains constant throughout the motion.
c. To calculate the potential energy of the apple we need the mass of the apple and the height of
the tree.
d. The mechanical energy is a maximum only at the beginning of the motion.
e. The apple falls at an acceleration of 9, 8 m · s−2 .
26
IEB 2005/11 HG Consider a ball dropped from a height of 1m on Earth and an identical ball dropped from 1m on
the Moon. Assume both balls fall freely. The acceleration due to gravity on the Moon is one sixth
that on Earth. In what way do the following compare when the ball is dropped on Earth and on the
Moon.
Mass Weight Increase in kinetic energy

(a) the same the same the same
(b) the same greater on Earth greater on Earth
(c) the same greater on Earth the same
(d) greater on Earth greater on Earth greater on Earth
Table 11.4
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27
3. A man res a rock out of a slingshot directly upward. The rock has an initial velocity of 15 m · s−1 .
a. How long will it take for the rock to reach its highest point?
b. What is the maximum height that the rock will reach?
c. Draw graphs to show how the potential energy, kinetic energy and mechanical energy of the rock
changes as it moves to its highest point.
28
4. A metal ball of mass 200 g is tied to a light string to make a pendulum. The ball is pulled to the side
to a height (A), 10 cm above the lowest point of the swing (B). Air friction and the mass of the string
can be ignored. The ball is let go to swing freely.
a. Calculate the potential energy of the ball at point A.

b. Calculate the kinetic energy of the ball at point B.
c. What is the maximum velocity that the ball will reach during its motion?
29
5. A truck of mass 1, 2 tons is parked at the top of a hill, 150 m high. The truck driver lets the truck run
freely down the hill to the bottom.
a. What is the maximum velocity that the truck can achieve at the bottom of the hill?
b. Will the truck achieve this velocity? Why/why not?
30
6. A stone is dropped from a window, 3m above the ground. The mass of the stone is 25 g .
a. Use the Equations of Motion to calculate the velocity of the stone as it reaches the ground.
b. Use the Principle of Conservation of Energy to prove that your answer in (a) is correct.
31
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181

Step 1. m (on Earth) = 50 kg
m (on Mars) = ?
Weight (on Mars) = ?
Step 2. Sarah's mass does not change because she is still made up of the same amount of matter. Her mass on
Mars is therefore 50 kg.
Step 3.
Sarah' s weight = 50 × 0, 38 × 9, 8
(11.19)
= 186, 2 N

Step 1. It always helps to understand the problem if we draw a picture like the one below:
Figure 11.16
Step 2. We have these quantities:
∆x = 15 m
vi = 0 m · s−1 (11.20)
g = 9, 8 m · s−2
Step 3. Since the ball is falling, we choose down as positive. This means that the values for vi , ∆x and a will
be positive.
Step 4. We can use (11.8) to nd the time: ∆x = vi t + 12 gt2
Step 5.
∆x = vi t + 12 gt2
1 2
15 = (0) t + 2 (9, 8) (t)
15 = 4, 9 t2
(11.21)
t2 = 3, 0612...
t = 1, 7496...
t = 1, 75 s
Step 6. Using (11.6) to nd vf :
vf = vi + gt
vf = 0 + (9, 8) (1, 7496...) (11.22)
vf = 17, 1464...
Remember to add the direction: vf = 17, 15 m · s−1 downwards.

Step 1. • The mass of the brick is m = 1 kg

• The height lifted is h = 4m
Step 2. • We are asked to nd the gain in potential energy of the brick as it is lifted onto the roof.
• We also need to calculate the potential energy once the brick is on the ground again.
Step 3. Since the block is being lifted we are dealing with gravitational potential energy. To work out P E,
we need to know the mass of the object and the height lifted. As both of these are given, we just
substitute them into the equation for P E.
Step 4.
PE = mgh
= (1) (9, 8) (4) (11.23)
= 39, 2 J

Step 1. • The mass of the rock m = 1 kg
• The velocity of the rock at the bottom vbottom = 8, 85 m·s−1
These are both in the correct units so we do not have to worry about unit conversions.
Step 2. We are asked to nd the kinetic energy of the brick at the top and the bottom. From the denition
we know that to work out KE , we need to know the mass and the velocity of the object and we are
given both of these values.
Step 3. Since the brick is not moving at the top, its kinetic energy is zero.
Step 4.
1 2
KE = 2 mv
1
2
= 2 (1 kg) 8, 85 m · s−1 (11.24)
= 39, 2 J

Step 1. • We are given the mass of the bullet m = 150 g . This is not the unit we want mass to be in. We
need to convert to kg.
Mass in grams ÷ 1000 = Mass in kg

(11.25)
150 g ÷ 1000 = 0, 150 kg
• We are given the initial velocity with which the bullet leaves the barrel, called the muzzle velocity,
and it is v = 960 m · s−1 .
Step 2. • We are asked to nd the kinetic energy.
Step 3. We just substitute the mass and velocity (which are known) into the equation for kinetic energy:
1 2
KE = 2 mv
1 2
= (0, 150) (960) (11.26)
2
= 69 120 J

Step 1. • The mass of the tree trunk m = 100 kg
• The height of the waterfall h = 5 m. These are all in SI units so we do not have to convert.
Step 2. • Potential energy at the top
• Kinetic energy at the bottom
• Velocity at the bottom

183
Step 3.
PE = mgh
PE = (100) (9, 8) (5) (11.27)
PE = 4900 J
Step 4. The kinetic energy of the tree trunk at the bottom of the waterfall is equal to the potential energy it
had at the top of the waterfall. Therefore KE = 4900 J .
Step 5. To calculate the velocity of the tree trunk we need to use the equation for kinetic energy.
1 2
KE = 2 mv
1
(100) v 2

4900 = 2
98 = v2 (11.28)
v = 9, 899...
v = 9, 90 m · s−1 downwards

Step 1. • The mass of the metal ball is m = 2 kg
• The change in height going from point A to point B is h = 0, 5 m
• The ball is released from point A so the velocity at point, vA = 0 m · s−1 .
Step 2. • Prove that the velocity is independent of mass.
• Find the velocity of the metal ball at point B.
Step 3. As there is no friction, mechanical energy is conserved. Therefore:
UA = UB
P EA + KEA = P EB + KEB
1 2 2
mghA + 2 m(vA ) = mghB + 12 m(vB ) (11.29)
2
mghA + 0 = 0 + 21 m(vB )
1 2
mghA = 2 m(vB )
As the mass of the ball m appears on both sides of the equation, it can be eliminated so that the
equation becomes:
1 2
ghA = (vB ) (11.30)
2
2
2ghA = (vB ) (11.31)
This proves that the velocity of the ball is independent of its mass. It does not matter what its mass
is, it will always have the same velocity when it falls through this height.
Step 4. We can use the equation above, or do the calculation from 'rst principles':
2
(vB ) = 2ghA
2
(vB ) = (2) (9.8) (0, 5)
(11.32)
2
(vB ) = 9, 8
√
vB = 9, 8 m · s−1


Chapter 12
Transverse pulses
12.1 Introduction and key concepts 1
12.1.1 Introduction
This chapter forms the basis of the discussion into mechanical waves. Waves are all around us, even though
most of us are not aware of it. The most common waves are waves in the sea, but waves can be created in
any container of water, ranging from an ocean to a tea-cup. Waves do not only occur in water, they occur
in any kind of medium. Earthquakes generate waves that travel through the rock of the Earth. When your
friend speaks to you he produces sound waves that travel through the air to your ears. Light is made up of
electromagnetic waves. A wave is simply moving energy.
12.1.2 What is a medium?

In this chapter, as well as in the following chapters, we will speak about waves moving in a medium. A
medium is just the substance or material through which waves move. In other words the medium carries the
wave from one place to another. The medium does not create the wave and the medium is not the wave.
Therefore the medium does not travel with the wave as the wave propagates through it. Air is a medium
for sound waves, water is a medium for water waves and rock is a medium for earthquakes (which are also
a type of wave). Air, water and rock are therefore examples of media (media is the plural of medium).
Denition 12.1: Medium

A medium is the substance or material in which a wave will move.
In each medium, the atoms that make up the medium are moved temporarily from their rest position.
In order for a wave to travel, the dierent parts of the medium must be able to interact with each other.
12.1.3 What is a pulse ?

12.1.3.1 Investigation : Observation of Pulses
Take a heavy rope. Have two people hold the rope stretched out horizontally. Flick the rope at one end only
once.
185
186 CHAPTER 12. TRANSVERSE PULSES
Figure 12.1
What happens to the disturbance that you created in the rope? Does it stay at the place where it was
created or does it move down the length of the rope?
In the activity, we created a pulse. A pulse is a single disturbance that moves through a medium. In a
transverse pulse the displacement of the medium is perpendicular to the direction of motion of the pulse.
Figure 12.2 shows an example of a transverse pulse. In the activity, the rope or spring was held horizontally
and the pulse moved the rope up and down. This was an example of a transverse pulse.
Denition 12.2: Pulse

A pulse is a single disturbance that moves through a medium.
12.1.3.2 Pulse Length and Amplitude

The amplitude of a pulse is a measurement of how far the medium is displaced momentarily from a position
of rest. The pulse length is a measurement of how long the pulse is. Both these quantities are shown in
Figure 12.2.
Denition 12.3: Amplitude

The amplitude of a pulse is a measurement of how far the medium is displaced from rest.
Figure 12.2: Example of a transverse pulse
12.1.3.2.1 Investigation : Pulse Length and Amplitude

The graphs below show the positions of a pulse at dierent times.
Figure 12.3
Use your ruler to measure the lengths of a and p. Fill your answers in the table.

187
Time a p
t=0 s
t=1 s
t=2 s
t=3 s
Table 12.1
What do you notice about the values of a and p?

In the activity, we found that the values for how high the pulse (a) is and how wide the pulse (p) is the
same at dierent times. Pulse length and amplitude are two important quantities of a pulse.
12.1.3.3 Pulse Speed

Denition 12.4: Pulse Speed
Pulse speed is the distance a pulse travels per unit time.
2
In here we saw that speed was dened as the distance travelled per unit time. We can use the same
denition of speed to calculate how fast a pulse travels. If the pulse travels a distance d in a time t, then
the pulse speed v is:
d
v= (12.1)
t
Exercise 12.1.1: Pulse Speed (Solution on p. 201.)

A pulse covers a distance of 2m in 4s on a heavy rope. Calculate the pulse speed.
tip: The pulse speed depends on the properties of the medium and not on the amplitude or pulse
length of the pulse.
12.1.3.3.1 Pulse Speed

1. A pulse covers a distance of 5m in 15 s. Calculate the speed of the pulse.
3
2. A pulse has a speed of 5 cm · s−1 . How far does it travel in 2, 5 s?
4
3. A pulse has a speed of 0, 5 m · s−1 . How long does it take to cover a distance of 25 cm?
5
4. How long will it take a pulse moving at 0, 25 m · s−1 to travel a distance of 20 m?
6
5. The diagram shows two pulses in the same medium. Which has the higher speed? Explain your
answer.
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Figure 12.4
7
12.2 Graphs of particle motion 8
12.2.1 Graphs of Position and Velocity

When a pulse moves through a medium, there are two dierent motions: the motion of the particles of the
medium and the motion of the pulse. These two motions are at right angles to each other when the pulse is
transverse. Each motion will be discussed.
Consider the situation shown in Figure 12.6. The dot represents one particle of the medium. We see that
as the pulse moves to the right the particle only moves up and down.
12.2.1.1 Motion of a Particle of the Medium

First we consider the motion of a particle of the medium when a pulse moves through the medium. For the
explanation we will zoom into the medium so that we are looking at the atoms of the medium. These atoms
are connected to each other as shown in Figure 12.5.
Figure 12.5: Particles in a medium.
When a pulse moves through the medium, the particles in the medium only move up and down. We can
see this in Figure 12.6 which shows the motion of a single particle as a pulse moves through the medium.
Figure 12.6: Positions of a pulse in a rope at dierent times. The pulse moves to the right as shown by
the arrow. You can also see the motion of a point in the medium through which the pulse is travelling.
Each block is 1 cm.
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189
tip: A particle in the medium only moves up and down when a transverse pulse moves through
the medium. The pulse moves from left to right (or right to left). The motion of the particle is
perpendicular to the motion of a transverse pulse.
If you consider the motion of the particle as a function of time, you can draw a graph of position vs. time
and velocity vs. time.
12.2.1.1.1 Investigation : Drawing a position-time graph

1. Study Figure 12.6 and complete the following table:
time (s) 0 1 2 3 4 5 6 7 8 9
position (cm)
Table 12.2
2. Use your table to draw a graph of position vs. time for a particle in a medium.
The position vs. time graph for a particle in a medium when a pulse passes through the medium is shown
in Figure 12.7
Figure 12.7: Position against Time graph of a particle in the medium through which a transverse pulse
is travelling.
12.2.1.1.2 Investigation : Drawing a velocity-time graph

1. Study Figure 12.7 and complete the following table:
time (s) 0 1 2 3 4 5 6 7 8 9
velocity (cm.s−1 )
Table 12.3
2. Use your table to draw a graph of velocity vs time for a particle in a medium.
The velocity vs. time graph far a particle in a medium when a pulse passes through the medium is shown
in Figure 12.8.
Figure 12.8: Velocity against Time graph of a particle in the medium through which a transverse pulse
is travelling.

12.2.1.2 Motion of the Pulse

The motion of the pulse is much simpler than the motion of a particle in the medium.
tip: A point on a transverse pulse, eg. the peak, only moves in the direction of the motion of the
pulse.
Exercise 12.2.1: Transverse pulse through a medium (Solution on p. 201.)
Figure 12.9: Position of the peak of a pulse at dierent times (since we know the shape of the pulse
does not change we can look at only one point on the pulse to keep track of its position, the peak for
example). The pulse moves to the right as shown by the arrow. Each square is 0, 5 cm.
Given the series of snapshots of a transverse pulse moving through a medium, depicted in
Figure 12.9, do the following:
• draw up a table of time, position and velocity,

• plot a position vs. time graph,
• plot a velocity vs. time graph.
12.2.1.2.1 Travelling Pulse

1. A pulse is passed through a rope and the following pictures were obtained for each time interval:
Figure 12.10
a. Complete the following table for a particle in the medium:
time (s) 0,00 0,25 0,50 0,75 1,00 1,25 1,50 1,75 2,00
position (mm)
velocity (mm.s−1 )
Table 12.4
b. Draw a position vs. time graph for the motion of the particle at 3 cm.
c. Draw a velocity vs. time graph for the motion of the particle at 3 cm.
d. Draw a position vs. time graph for the motion of the pulse through the rope.
e. Draw a velocity vs. time graph for the motion of the pulse through the rope.
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191
12.3 Boundary conditions 10
12.3.1 Transmission and Reection of a Pulse at a Boundary

What happens when a pulse travelling in one medium nds that medium is joined to another?
12.3.1.1 Investigation : Two ropes

Find two dierent ropes and tie both ropes together. Hold the joined ropes horizontally and create a pulse
by icking the rope up and down. What happens to the pulse when it encounters the join?
When a pulse is transmitted from one medium to another, like from a thin rope to a thicker one, the
nature of the pulse will change where it meets the boundary of the two media (i.e. where the two ropes
are joined). Part of the pulse will be reected and part of it will be transmitted. Figure 12.11 shows the
general case of a pulse meeting a boundary. The incident pulse is the one that arrives at the boundary.
The reected pulse is the one that moves back, away from the boundary. The transmitted pulse is the
one that moves into the new medium, away from the boundary. The speed of the pulse depends on the mass
of the rope; the pulse is faster in the thinner rope and slower in the thick rope. When the speed of the pulse
increases, the pulse length will increase. If the speed decreases, the pulse length will decrease.
Figure 12.11: Reection and transmission of a pulse at the boundary between two media.
Consider a pulse moving from a thin rope to a thick rope. As the pulse crosses the boundary, the speed
of the pulse will decrease as it moves into the thicker rope. The pulse will move slower, so the pulse length
will decrease. The pulse will be reected and inverted in the thin rope. The reected pulse will have the
same length and speed but will have a smaller amplitude. This is illustrated in Figure 12.12.
When a pulse moves from a thick rope to a thin rope, the opposite will happen. As the pulse crosses the
boundary, the speed of the pulse will increase as it moves into the thinner rope. The pulse in the thin rope
will move faster, so the pulse length will increase. The pulse will be reected but not inverted in the thick
rope. The reected pulse will have the same length and speed but will have a smaller amplitude. This is
illustrated in Figure 12.13

12.3.1.2 Pulses at a Boundary I

1. Fill in the blanks or select the correct answer: A pulse in a heavy rope is traveling towards the boundary
with a thin piece of string.
a. The reected pulse in the heavy rope will/will not be inverted because .
b. The speed of the transmitted pulse will be greater than/less than/the same as the speed of
the incident pulse.
c. The speed of the reected pulse will be greater than/less than/the same as the speed of the
incident pulse.
d. The pulse length of the transmitted pulse will be greater than/less than/the same as the
pulse length of the incident pulse.
e. The frequency of the transmitted pulse will be greater than/less than/the same as the
frequency of the incident pulse.
11
2. A pulse in a light string is traveling towards the boundary with a heavy rope.
a. The reected pulse in the light rope will/will not be inverted because .
b. The speed of the transmitted pulse will be greater than/less than/the same as the speed of
the incident pulse.
c. The speed of the reected pulse will be greater than/less than/the same as the speed of the
incident pulse.
d. The pulse length of the transmitted pulse will be greater than/less than/the same as the
pulse length of the incident pulse.
12
12.3.2 Reection of a Pulse from Fixed and Free Ends

Let us now consider what happens to a pulse when it reaches the end of a medium. The medium can be
xed, like a rope tied to a wall, or it can be free, like a rope tied loosely to a pole.
12.3.2.1 Reection of a Pulse from a Fixed End

12.3.2.1.1 Investigation : Reection of a Pulse from a Fixed End
Tie a rope to a wall or some other object that cannot move. Create a pulse in the rope by icking one end
up and down. Observe what happens to the pulse when it reaches the wall.
11 http://www.fhsst.org/l1H

193
Figure 12.14: Reection of a pulse from a xed end.
When the end of the medium is xed, for example a rope tied to a wall, a pulse reects from the xed
end, but the pulse is inverted (i.e. it is upside-down). This is shown in Figure 12.14.
12.3.2.2 Reection of a Pulse from a Free End

12.3.2.2.1 Investigation : Reection of a Pulse from a Free End
Tie a rope to a pole in such a way that the rope can move up and down the pole. Create a pulse in the rope
by icking one end up and down. Observe what happens to the pulse when it reaches the pole.
When the end of the medium is free, for example a rope tied loosely to a pole, a pulse reects from the
free end, but the pulse is not inverted. This is shown in Figure 12.15. We draw the free end as a ring around
the pole. The ring will move up and down the pole, while the pulse is reected away from the pole.
Figure 12.15: Reection of a pulse from a free end.
tip: The xed and free ends that were discussed in this section are examples of boundary conditions.
You will see more of boundary conditions as you progress in the Physics syllabus.
12.3.2.2.2 Pulses at a Boundary II

1. A rope is tied to a tree and a single pulse is generated. What happens to the pulse as it reaches the
tree? Draw a diagram to explain what happens.
13
2. A rope is tied to a ring that is loosely tted around a pole. A single pulse is sent along the rope. What
will happen to the pulse as it reaches the pole? Draw a diagram to explain your answer.
14
The following simulation will help you understand the previous examples. Choose pulse from the options
(either manual, oscillate or pulse). Then click on pulse and see what happens. Change from a xed to a free
end and see what happens. Try varying the width, amplitude, damping and tension.
13 http://www.fhsst.org/l1F
14 http://www.fhsst.org/l1L

Phet simulation for transverse pulses

<wave-on-a-string.swf>
Figure 12.16
12.3.3 Superposition of Pulses

Two or more pulses can pass through the same medium at that same time. The resulting pulse is obtained
by using the principle of superposition. The principle of superposition states that the eect of the pulses
is the sum of their individual eects. After pulses pass through each other, each pulse continues along its
original direction of travel, and their original amplitudes remain unchanged.
Constructive interference takes place when two pulses meet each other to create a larger pulse. The
amplitude of the resulting pulse is the sum of the amplitudes of the two initial pulses. This is shown in
Figure 12.17.
Denition 12.5: Constructive interference is when two pulses meet, resulting in a

bigger pulse.
Figure 12.17: Superposition of two pulses: constructive interference.
Destructive interference takes place when two pulses meet and cancel each other. The amplitude of the
resulting pulse is the sum of the amplitudes of the two initial pulses, but the one amplitude will be a negative
number. This is shown in Figure 12.18. In general, amplitudes of individual pulses add together to give the
amplitude of the resultant pulse.
Denition 12.6: Destructive interference is when two pulses meet, resulting in a

smaller pulse.
Figure 12.18: Superposition of two pulses. The left-hand series of images demonstrates destructive
interference, since the pulses cancel each other. The right-hand series of images demonstrate a partial
cancelation of two pulses, as their amplitudes are not the same in magnitude.

195
Exercise 12.3.1: Superposition of Pulses (Solution on p. 201.)

The two pulses shown below approach each other at 1 m · s−1 . Draw what the waveform would
look like after 1 s, 2 s and 5 s.
Figure 12.19
tip: The idea of superposition is one that occurs often in physics. You will see much, much more
of superposition!
12.3.3.1 Superposition of Pulses

1. For the following pulse, draw the resulting wave forms after 1 s, 2 s, 3 s, 4 s and 5 s. Each pulse is
travelling at 1 m · s−1 . Each block represents 1 m. The pulses are shown as thick black lines and the
undisplaced medium as dashed lines.
Figure 12.20
15
Figure 12.21
16
15 http://www.fhsst.org.za/l1M
16 http://www.fhsst.org.za/l1e

Figure 12.22
17
Figure 12.23
18
Figure 12.24
19
Figure 12.25
20
7. What is superposition of waves?
21
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197
8. What is constructive interference?

22
9. What is destructive interference?
23
The following presentation provides a summary of the work covered in this chapter. Although the presenta-
tion is titled waves, the presentation covers pulses only.

<http://static.slidesharecdn.com/swf/ssplayer2.swf ?doc=g10wavesp1-100511062100-
phpapp01&stripped_title=g10-waves-p1&userName=kwarne>
Figure 12.26
12.3.4 Exercises - Transverse Pulses

1. A heavy rope is icked upwards, creating a single pulse in the rope. Make a drawing of the rope and
indicate the following in your drawing:
a. The direction of motion of the pulse

b. Amplitude
c. Pulse length
d. Position of rest
24
2. A pulse has a speed of 2, 5 m · s−1 . How far will it have travelled in 6 s?
25
3. A pulse covers a distance of 75 cm in 2, 5 s. What is the speed of the pulse?
26
4. How long does it take a pulse to cover a distance of 200 mm if its speed is 4 m · s−1 ?
27
5. The following position-time graph for a pulse in a slinky spring is given. Draw an accurate sketch
graph of the velocity of the pulse against time.
Figure 12.27
28
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26 http://www.fhsst.org.za/lr3
27 http://www.fhsst.org.za/lrO
28 http://www.fhsst.org.za/lrc

6. The following velocity-time graph for a particle in a medium is given. Draw an accurate sketch graph
of the position of the particle vs. time.
Figure 12.28
29
7. Describe what happens to a pulse in a slinky spring when:
a. the slinky spring is tied to a wall.

b. the slinky spring is loose, i.e. not tied to a wall.
(Draw diagrams to explain your answers.)

30
8. The following diagrams each show two approaching pulses. Redraw the diagrams to show what type
of interference takes place, and label the type of interference.
a.
Figure 12.29
b.
Figure 12.30
31
9. Two pulses, A and B, of identical shape and amplitude are simultaneously generated in two identical
wires of equal mass and length. Wire A is, however, pulled tighter than wire B. Which pulse will arrive
at the other end rst, or will they both arrive at the same time?
32
29 http://www.fhsst.org.za/lrx
30 http://www.fhsst.org.za/lra
31 http://www.fhsst.org.za/lrC

199

• the distance travelled by the pulse: d = 2m

• the time taken to travel 2 m: t = 4 s
We are required to calculate the speed of the pulse.
Step 2. We can use:
d
v= (12.2)
t
to calculate the speed of the pulse.
Step 3.
v = d
t
2m (12.3)
= 4s
= 0, 5 m · s−1
Step 4. The pulse speed is 0, 5 m · s−1 .

Step 1. Figure 12.9 shows the motion of a pulse through a medium and a dot to indicate the same position on
the pulse. If we follow the dot, we can draw a graph of position vs time for a pulse. At t = 0 s the dot
is at 0 cm. At t = 1s the dot is 1 cm away from its original postion. At t = 2s the dot is2 cm away
from its original postion, and so on.
Step 2.
time (s) 0 1 2 3 4 5 6 7 8 9
position (cm) 0 1 2 3 4 5 6 7 8 9
velocity (cm.s−1 ) 1 1 1 1 1 1 1 1 1 1
Table 12.5
Step 3.
Figure 12.31
Step 4.
Figure 12.32

Step 1. After 1 s, pulse A has moved 1m to the right and pulse B has moved 1m to the left.
Figure 12.33
Step 2. After 1s more, pulse A has moved 1m to the right and pulse B has moved 1m to the left.
Figure 12.34
Step 3. After 5 s, pulse A has moved 5m to the right and pulse B has moved 5m to the left.
Figure 12.35

Chapter 13
Transverse waves
13.1.1 Introduction
Waves occur frequently in nature. The most obvious examples are waves in water, on a dam, in the ocean, or
in a bucket. We are most interested in the properties that waves have. All waves have the same properties,
so if we study waves in water, then we can transfer our knowledge to predict how other examples of waves
will behave.
13.1.2 What is a transverse wave ?

2
We have studied pulses in Transverse Pulses , and know that a pulse is a single disturbance that travels
through a medium. A wave is a periodic, continuous disturbance that consists of a train of pulses.
Denition 13.1: Wave
A wave is a periodic, continuous disturbance that consists of a train of pulses.
Denition 13.2: Transverse wave
A transverse wave is a wave where the movement of the particles of the medium is perpendicular
to the direction of propagation of the wave.
13.1.2.1 Investigation : Transverse Waves

Take a rope or slinky spring. Have two people hold the rope or spring stretched out horizontally. Flick the
one end of the rope up and down continuously to create a train of pulses.
Figure 13.1
1. Describe what happens to the rope.

2. Draw a diagram of what the rope looks like while the pulses travel along it.

2 "Transverse Pulses - Grade 10" <http://cnx.org/content/m35714/latest/>
201
202 CHAPTER 13. TRANSVERSE WAVES
3. In which direction do the pulses travel?

4. Tie a ribbon to the middle of the rope. This indicates a particle in the rope.
Figure 13.2
5. Flick the rope continuously. Watch the ribbon carefully as the pulses travel through the rope. What
happens to the ribbon?
6. Draw a picture to show the motion of the ribbon. Draw the ribbon as a dot and use arrows to indicate
how it moves.
In the Activity, you have created waves. The medium through which these waves propagated was the rope,
which is obviously made up of a very large number of particles (atoms). From the activity, you would have
noticed that the wave travelled from left to right, but the particles (the ribbon) moved only up and down.
Figure 13.3: A transverse wave, showing the direction of motion of the wave perpendicular to the
direction in which the particles move.
When the particles of a medium move at right angles to the direction of propagation of a wave, the wave
is called transverse. For waves, there is no net displacement of the particles (they return to their equilibrium
position), but there is a net displacement of the wave. There are thus two dierent motions: the motion of
the particles of the medium and the motion of the wave.
The following simulation will help you understand more about waves. Select the oscillate option and
then observe what happens.
Phet simulation for Transverse Waves

<wave-on-a-string.swf>
Figure 13.4
13.1.2.2 Peaks and Troughs

Waves have moving peaks (or crests ) and troughs. A peak is the highest point the medium rises to and a
trough is the lowest point the medium sinks to.
Peaks and troughs on a transverse wave are shown in Figure 13.5.

203
Figure 13.5: Peaks and troughs in a transverse wave.
Denition 13.3: Peaks and troughs

A peak is a point on the wave where the displacement of the medium is at a maximum. A point
on the wave is a trough if the displacement of the medium at that point is at a minimum.
13.1.2.3 Amplitude and Wavelength

There are a few properties that we saw with pulses that also apply to waves. These are amplitude and
wavelength (we called this pulse length).
Denition 13.4: Amplitude

The amplitude is the maximum displacement of a particle from its equilibrium position.
13.1.2.3.1 Investigation : Amplitude
Figure 13.6
Fill in the table below by measuring the distance between the equilibrium and each peak and troughs in the
wave above. Use your ruler to measure the distances.
Peak/Trough Measurement (cm)
Table 13.1
1. What can you say about your results?

2. Are the distances between the equilibrium position and each peak equal?
3. Are the distances between the equilibrium position and each trough equal?
4. Is the distance between the equilibrium position and peak equal to the distance between equilibrium
and trough?

As we have seen in the activity on amplitude, the distance between the peak and the equilibrium position
is equal to the distance between the trough and the equilibrium position. This distance is known as the
amplitude of the wave, and is the characteristic height of wave, above or below the equilibrium position.
Normally the symbol A is used to represent the amplitude of a wave. The SI unit of amplitude is the metre
(m).
Figure 13.7
Exercise 13.1.1: Amplitude of Sea Waves (Solution on p. 226.)

If the peak of a wave measures 2 m above the still water mark in the harbour, what is the amplitude
of the wave?
13.1.2.3.2 Investigation : Wavelength
Figure 13.8
Fill in the table below by measuring the distance between peaks and troughs in the wave above.
Distance(cm)
Table 13.2
1. What can you say about your results?

2. Are the distances between peaks equal?
3. Are the distances between troughs equal?
4. Is the distance between peaks equal to the distance between troughs?
As we have seen in the activity on wavelength, the distance between two adjacent peaks is the same no
matter which two adjacent peaks you choose. There is a xed distance between the peaks. Similarly, we
have seen that there is a xed distance between the troughs, no matter which two troughs you look at. More
importantly, the distance between two adjacent peaks is the same as the distance between two adjacent
troughs. This distance is called the wavelength of the wave.

205
The symbol for the wavelength is λ (the Greek letter lambda) and wavelength is measured in metres (m).
Figure 13.9
Exercise 13.1.2: Wavelength (Solution on p. 226.)

The total distance between 4 consecutive peaks of a transverse wave is 6 m. What is the wavelength
of the wave?
13.1.2.4 Points in Phase

13.1.2.4.1 Investigation : Points in Phase
Fill in the table by measuring the distance between the indicated points.
Figure 13.10
Points Distance (cm)

A to F
B to G
C to H
D to I
E to J
Table 13.3
What do you nd?

In the activity the distance between the indicated points was the same. These points are then said to be
in phase. Two points in phase are separate by an integer (0,1,2,3,...) number of complete wave cycles. They
do not have to be peaks or troughs, but they must be separated by a complete number of wavelengths.
We then have an alternate denition of the wavelength as the distance between any two adjacent points
which are in phase.
Denition 13.5: Wavelength of wave
The wavelength of a wave is the distance between any two adjacent points that are in phase.

Figure 13.11
Points that are not in phase, those that are not separated by a complete number of wavelengths, are
called out of phase. Examples of points like these would be A and C, or D and E, or B and H in the
Activity.
13.1.2.5 Period and Frequency

Imagine you are sitting next to a pond and you watch the waves going past you. First one peak arrives, then
a trough, and then another peak. Suppose you measure the time taken between one peak arriving and then
the next. This time will be the same for any two successive peaks passing you. We call this time the period,
and it is a characteristic of the wave.
The symbol T is used to represent the period. The period is measured in seconds (s).
Denition 13.6: The period (T) is the time taken for two successive peaks (or troughs)
to pass a xed point.
Imagine the pond again. Just as a peak passes you, you start your stopwatch and count each peak going
past. After 1 second you stop the clock and stop counting. The number of peaks that you have counted in
the 1 second is the frequency of the wave.
Denition 13.7: The frequency is the number of successive peaks (or troughs) passing
a given point in 1 second.
The frequency and the period are related to each other. As the period is the time taken for 1 peak to
1
pass, then the number of peaks passing the point in 1 second is
T . But this is the frequency. So
1
f= (13.1)
T
or alternatively,
1
T = . (13.2)
f
1
For example, if the time between two consecutive peaks passing a xed point is
2 s, then the period of the
1
wave is
2 s. Therefore, the frequency of the wave is:
1
f = T
1
= 1 (13.3)
2 s
= 2 s−1
The unit of frequency is the Hertz (Hz) or s−1 .

Exercise 13.1.3: Period and Frequency (Solution on p. 226.)
What is the period of a wave of frequency 10 Hz?

207
13.1.2.6 Speed of a Transverse Wave

In Motion in One Dimension, we saw that speed was dened as
distance travelled
speed = . (13.4)
time taken
The distance between two successive peaks is 1 wavelength, λ. Thus in a time of 1 period, the wave will
travel 1 wavelength in distance. Thus the speed of the wave, v , is:
distance travelled λ
v= = . (13.5)
time taken T
1
However, f= T . Therefore, we can also write:
v = λ
T
1
= λ· T
(13.6)
= λ·f
We call this equation the wave equation. To summarise, we have that v =λ·f where
• v = speed in m · s−1
• λ = wavelength in m
• f = frequency in Hz
Exercise 13.1.4: Speed of a Transverse Wave 1 (Solution on p. 226.)

When a particular string is vibrated at a frequency of 10 Hz, a transverse wave of wavelength
0, 25 m is produced. Determine the speed of the wave as it travels along the string.
Exercise 13.1.5: Speed of a Transverse Wave 2 (Solution on p. 227.)

A cork on the surface of a swimming pool bobs up and down once every second on some ripples.
The ripples have a wavelength of 20 cm. If the cork is 2 m from the edge of the pool, how long does
it take a ripple passing the cork to reach the edge?
The following video provides a summary of the concepts covered so far.
Khan academy video on waves - 1

<http://www.youtube.com/v/tJW_a6JeXD8&rel=0>
Figure 13.12
13.1.2.6.1 Waves
1. When the particles of a medium move perpendicular to the direction of the wave motion, the wave is
called a ......... wave.
3
2. A transverse wave is moving downwards. In what direction do the particles in the medium move?
4
3 http://www.fhsst.org.za/liq
4 http://www.fhsst.org.za/li4

3. Consider the diagram below and answer the questions that follow:
Figure 13.13
a. the wavelength of the wave is shown by letter .

b. the amplitude of the wave is shown by letter .
5
4. Draw 2 wavelengths of the following transverse waves on the same graph paper. Label all important
values.
a. Wave 1: Amplitude = 1 cm, wavelength = 3 cm

b. Wave 2: Peak to trough distance (vertical) = 3 cm, peak to peak distance (horizontal) = 5 cm
6
5. You are given the transverse wave below.
Figure 13.14
Draw the following:
a. A wave with twice the amplitude of the given wave.

b. A wave with half the amplitude of the given wave.
c. A wave travelling at the same speed with twice the frequency of the given wave.
d. A wave travelling at the same speed with half the frequency of the given wave.
e. A wave with twice the wavelength of the given wave.
f. A wave with half the wavelength of the given wave.
g. A wave travelling at the same speed with twice the period of the given wave.
h. A wave travelling at the same speed with half the period of the given wave.
7
6. A transverse wave travelling at the same speed with an amplitude of 5 cm has a frequency of 15 Hz.
The horizontal distance from a crest to the nearest trough is measured to be 2,5 cm. Find the
a. period of the wave.

b. speed of the wave.
8
7. A y aps its wings back and forth 200 times each second. Calculate the period of a wing ap.
9
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209
8. As the period of a wave increases, the frequency increases/decreases/does not change.

10
9. Calculate the frequency of rotation of the second hand on a clock.
11
10. Microwave ovens produce radiation with a frequency of 2 450 MHz (1 MHz = 106 Hz) and a wavelength
of 0,122 m. What is the wave speed of the radiation?
12
11. Study the following diagram and answer the questions:
Figure 13.15
a. Identify two sets of points that are in phase.

b. Identify two sets of points that are out of phase.
c. Identify any two points that would indicate a wavelength.
13
12. Tom is shing from a pier and notices that four wave crests pass by in 8 s and estimates the distance
between two successive crests is 4 m. The timing starts with the rst crest and ends with the fourth.
Calculate the speed of the wave.
14
13.2 Graphs of particle motion 15
13.2.1 Graphs of Particle Motion

In Chapter , we saw that when a pulse moves through a medium, there are two dierent motions: the motion
of the particles of the medium and the motion of the pulse. These two motions are at right angles to each
other when the pulse is transverse. Since a transverse wave is a series of transverse pulses, the particle in
the medium and the wave move in exactly the same way as for the pulse.
When a transverse wave moves horizontally through the medium, the particles in the medium only move
up and down. We can see this in the gure below, which shows the motion of a single particle as a transverse
wave moves through the medium.
Figure 13.16
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13 http://www.fhsst.org.za/lrn
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tip: A particle in the medium only moves up and down when a transverse wave moves horizontally
through the medium.
16
As in here , we can draw a graph of the particles' position as a function of time. For the wave shown in
the above gure, we can draw the graph shown below.
Figure 13.17
The graph of the particle's velocity as a function of time is obtained by taking the gradient of the position
vs. time graph. The graph of velocity vs. time for the position vs. time graph above, is shown in the graph
below.
Figure 13.18
The graph of the particle's acceleration as a function of time is obtained by taking the gradient of the
velocity vs. time graph. The graph of acceleration vs. time for the position vs. time graph shown above, is
shown below.
Figure 13.19
As for motion in one dimension, these graphs can be used to describe the motion of the particle in the
medium. This is illustrated in the worked examples below.
Exercise 13.2.1: Graphs of particle motion 1 (Solution on p. 227.)

The following graph shows the position of a particle of a wave as a function of time.
Figure 13.20
1. Draw the corresponding velocity vs. time graph for the particle.

211
2. Draw the corresponding acceleration vs. time graph for the particle.
13.2.1.1 Mathematical Description of Waves

If you look carefully at the pictures of waves you will notice that they look very much like sine or cosine
functions. This is correct. Waves can be described by trigonometric functions that are functions of time
or of position. Depending on which case we are dealing with the function will be a function of t or x. For
example, a function of position would be:
x
y (x) = Asin 360◦ + φ (13.7)
λ
where A is the amplitude, λ the wavelength and φ is a phase shift. The phase shift accounts for the fact
that the wave at x=0 does not start at the equilibrium position. A function of time would be:
t

y (t) = Asin 360 ◦
+φ (13.8)
T
where T is the period of the wave. Descriptions of the wave incorporate the amplitude, wavelength, frequency
or period and a phase shift.
13.2.1.2 Graphs of Particle Motion

1. The following velocity vs. time graph for a particle in a wave is given.
Figure 13.21
a. Draw the corresponding position vs. time graph for the particle.
b. Draw the corresponding acceleration vs. time graph for the particle.
17
13.3 Boundary conditions 18
13.3.1 Standing Waves and Boundary Conditions

13.3.1.1 Reection of a Transverse Wave from a Fixed End
We have seen that when a pulse meets a xed endpoint, the pulse is reected, but it is inverted. Since a
transverse wave is a series of pulses, a transverse wave meeting a xed endpoint is also reected and the
reected wave is inverted. That means that the peaks and troughs are swapped around.
17 http://www.fhsst.org.za/lrU

Figure 13.22: Reection of a transverse wave from a xed end.
13.3.1.2 Reection of a Transverse Wave from a Free End

If transverse waves are reected from an end, which is free to move, the waves sent down the string are
reected but do not suer a phase shift as shown in Figure 13.23.
Figure 13.23: Reection of a transverse wave from a free end.
13.3.1.3 Standing Waves

What happens when a reected transverse wave meets an incident transverse wave? When two waves move
in opposite directions, through each other, interference takes place. If the two waves have the same frequency
and wavelength then standing waves are generated.
Standing waves are so-called because they appear to be standing still.
13.3.1.3.1 Investigation : Creating Standing Waves

Tie a rope to a xed object such that the tied end does not move. Continuously move the free end up and
down to generate rstly transverse waves and later standing waves.
We can now look closely how standing waves are formed. Figure 13.24 shows a reected wave meeting
an incident wave.
Figure 13.24: A reected wave (solid line) approaches the incident wave (dashed line).
When they touch, both waves have an amplitude of zero:

213
Figure 13.25: A reected wave (solid line) meets the incident wave (dashed line).
If we wait for a short time the ends of the two waves move past each other and the waves overlap. To
nd the resultant wave, we add the two together.
Figure 13.26: A reected wave (solid line) overlaps slightly with the incident wave (dashed line).
In this picture, we show the two waves as dotted lines and the sum of the two in the overlap region is
shown as a solid line:
Figure 13.27
The important thing to note in this case is that there are some points where the two waves always
destructively interfere to zero. If we let the two waves move a little further we get the picture below:
Figure 13.28
Again we have to add the two waves together in the overlap region to see what the sum of the waves
looks like.

Figure 13.29
In this case the two waves have moved half a cycle past each other but because they are completely out
of phase they cancel out completely.
When the waves have moved past each other so that they are overlapping for a large region the situation
looks like a wave oscillating in place. The following sequence of diagrams show what the resulting wave
will look like. To make it clearer, the arrows at the top of the picture show peaks where maximum positive
constructive interference is taking place. The arrows at the bottom of the picture show places where maximum
negative interference is taking place.
Figure 13.30
As time goes by the peaks become smaller and the troughs become shallower but they do not move.
Figure 13.31
For an instant the entire region will look completely at.
Figure 13.32
The various points continue their motion in the same manner.

215
Figure 13.33
Eventually the picture looks like the complete reection through the x-axis of what we started with:
Figure 13.34
Then all the points begin to move back. Each point on the line is oscillating up and down with a dierent
amplitude.
Figure 13.35
If we look at the overall result, we get a standing wave.
Figure 13.36: A standing wave
If we superimpose the two cases where the peaks were at a maximum and the case where the same waves
were at a minimum we can see the lines that the points oscillate between. We call this the envelope of the
standing wave as it contains all the oscillations of the individual points. To make the concept of the envelope
clearer let us draw arrows describing the motion of points along the line.

Figure 13.37
Every point in the medium containing a standing wave oscillates up and down and the amplitude of the
oscillations depends on the location of the point. It is convenient to draw the envelope for the oscillations
to describe the motion. We cannot draw the up and down arrows for every single point!
note: Standing waves can be a problem in for example indoor concerts where the dimensions of
the concert venue coincide with particular wavelengths. Standing waves can appear as `feedback',
which would occur if the standing wave was picked up by the microphones on stage and amplied.
13.3.1.4 Nodes and Anti-nodes

A node is a point on a wave where no displacement takes place at any time. In a standing wave, a node is a
place where two waves cancel out completely as the two waves destructively interfere in the same place. A
xed end of a rope is a node. An anti-node is a point on a wave where maximum displacement takes place.
In a standing wave, an anti-node is a place where the two waves constructively interfere. Anti-nodes occur
midway between nodes. A free end of a rope is an anti-node.
Figure 13.38
Denition 13.8: Node

A node is a point on a standing wave where no displacement takes place at any time. A xed end
of a rope is a node.
Denition 13.9: Anti-Node

An anti-node is a point on standing a wave where maximum displacement takes place. A free end
of a rope is an anti-node.
1
tip: The distance between two anti-nodes is only
2 λ because it is the distance from a peak to a
trough in one of the waves forming the standing wave. It is the same as the distance between two
adjacent nodes. This will be important when we work out the allowed wavelengths in tubes later.
We can take this further because half-way between any two anti-nodes is a node. Then the distance
from the node to the anti-node is half the distance between two anti-nodes. This is half of half a
1
wavelength which is one quarter of a wavelength,
4 λ.

217
13.3.1.5 Wavelengths of Standing Waves with Fixed and Free Ends

There are many applications which make use of the properties of waves and the use of xed and free ends.
Most musical instruments rely on the basic picture that we have presented to create specic sounds, either
through standing pressure waves or standing vibratory waves in strings.
The key is to understand that a standing wave must be created in the medium that is oscillating. There
are restrictions as to what wavelengths can form standing waves in a medium.
For example, if we consider a rope that can move in a pipe such that it can have
• both ends free to move (Case 1)

• one end free and one end xed (Case 2)
• both ends xed (Case 3).
Each of these cases is slightly dierent because the free or xed end determines whether a node or anti-node
will form when a standing wave is created in the rope. These are the main restrictions when we determine
the wavelengths of potential standing waves. These restrictions are known as boundary conditions and must
be met.
In the diagram below you can see the three dierent cases. It is possible to create standing waves with
dierent frequencies and wavelengths as long as the end criteria are met.
Figure 13.39
The longer the wavelength the less the number of anti-nodes in the standing waves. We cannot have
a standing wave with no anti-nodes because then there would be no oscillations. We use n to number the
anti-nodes. If all of the tubes have a length L and we know the end constraints we can nd the wavelength,
λ, for a specic number of anti-nodes.
13.3.1.5.1 One Node

Let's work out the longest wavelength we can have in each tube, i.e. the case for n = 1.
Figure 13.40
Case 1: In the rst tube, both ends must be anti-nodes, so we must place one node in the middle of the
1
tube. We know the distance from one anti-node to another is
2 λ and we also know this distance is L. So we

can equate the two and solve for the wavelength:
1
2λ = L
(13.9)
λ = 2L
Case 2: In the second tube, one end must be a node and the other must be an anti-node. Since we are
looking at the case with one node, we are forced to have it at the end. We know the distance from one node
1
to another is
2 λ but we only have half this distance contained in the tube. So :
1 1
2λ L

2 =
(13.10)
λ = 4L
Case 3: Here both ends are closed and so we must have two nodes so it is impossible to construct a case
with only one node.
13.3.1.5.2 Two Nodes

Next we determine which wavelengths could be formed if we had two nodes. Remember that we are dividing
the tube up into smaller and smaller segments by having more nodes so we expect the wavelengths to get
shorter.
Figure 13.41
Case 1: Both ends are open and so they must be anti-nodes. We can have two nodes inside the tube only
if we have one anti-node contained inside the tube and one on each end. This means we have 3 anti-nodes in
the tube. The distance between any two anti-nodes is half a wavelength. This means there is half wavelength
between the left side and the middle and another half wavelength between the middle and the right side so
there must be one wavelength inside the tube. The safest thing to do is work out how many half wavelengths
there are and equate this to the length of the tube L and then solve for λ.
1
2λ = L

2
(13.11)
λ = L
Case 2: We want to have two nodes inside the tube. The left end must be a node and the right end must
be an anti-node. We can have one node inside the tube as drawn above. Again we can count the number of
distances between adjacent nodes or anti-nodes. If we start from the left end we have one half wavelength
between the end and the node inside the tube. The distance from the node inside the tube to the right end
which is an anti-node is half of the distance to another node. So it is half of half a wavelength. Together

219
these add up to the length of the tube:
1 1 1
2λ 2λ L

+ 2 =
2
4λ + 14 λ = L
(13.12)
3
4λ = L
4
λ = 3L
Case 3: In this case both ends have to be nodes. This means that the length of the tube is one half
wavelength: So we can equate the two and solve for the wavelength:
1
2λ = L
(13.13)
λ = 2L
tip: If you ever calculate a longer wavelength for more nodes you have made a mistake. Remember
to check if your answers make sense!
13.3.1.5.3 Three Nodes

To see the complete pattern for all cases we need to check what the next step for case 3 is when we have an
additional node. Below is the diagram for the case where n = 3.
Figure 13.42
Case 1: Both ends are open and so they must be anti-nodes. We can have three nodes inside the tube
only if we have two anti-nodes contained inside the tube and one on each end. This means we have 4 anti-
nodes in the tube. The distance between any two anti-nodes is half a wavelength. This means there is half
wavelength between every adjacent pair of anti-nodes. We count how many gaps there are between adjacent
anti-nodes to determine how many half wavelengths there are and equate this to the length of the tube L
and then solve for λ.
1
2λ L

3 =
(13.14)
2
λ = 3L
Case 2: We want to have three nodes inside the tube. The left end must be a node and the right end must
be an anti-node, so there will be two nodes between the ends of the tube. Again we can count the number of
distances between adjacent nodes or anti-nodes, together these add up to the length of the tube. Remember
that the distance between the node and an adjacent anti-node is only half the distance between adjacent

nodes. So starting from the left end we count 3 nodes, so 2 half wavelength intervals and then only a node
to anti-node distance:
1 1 1
2λ 2λ L

2 + 2 =
λ + 14 λ = L
(13.15)
5
4λ = L
4
λ = 5L
Case 3: In this case both ends have to be nodes. With one node in between there are two sets of adjacent
nodes. This means that the length of the tube consists of two half wavelength sections:
1
2λ = L

2
(13.16)
λ = L
13.3.1.6 Superposition and Interference

If two waves meet interesting things can happen. Waves are basically collective motion of particles. So
when two waves meet they both try to impose their collective motion on the particles. This can have quite
dierent results.
If two identical (same wavelength, amplitude and frequency) waves are both trying to form a peak then
they are able to achieve the sum of their eorts. The resulting motion will be a peak which has a height
which is the sum of the heights of the two waves. If two waves are both trying to form a trough in the same
place then a deeper trough is formed, the depth of which is the sum of the depths of the two waves. Now in
this case, the two waves have been trying to do the same thing, and so add together constructively. This is
called constructive interference.
Figure 13.43
If one wave is trying to form a peak and the other is trying to form a trough, then they are competing to
do dierent things. In this case, they can cancel out. The amplitude of the resulting wave will depend on the
amplitudes of the two waves that are interfering. If the depth of the trough is the same as the height of the
peak nothing will happen. If the height of the peak is bigger than the depth of the trough, a smaller peak
will appear. And if the trough is deeper then a less deep trough will appear. This is destructive interference.
Figure 13.44

221
13.3.1.6.1 Superposition and Interference

1. For each labelled point, indicate whether constructive or destructive interference takes place at that
point.
Figure 13.45
Figure 13.46
19
2. A ride at the local amusement park is called "Standing on Standing Waves". Which position (a node
or an antinode) on the ride would give the greatest thrill?
20
3. How many nodes and how many anti-nodes appear in the standing wave below?
Figure 13.47
21
4. For a standing wave on a string, you are given three statements:
A. you can have any λ and any f as long as the relationship, v =λ·f is satised.
B. only certain wavelengths and frequencies are allowed
C. the wave velocity is only dependent on the medium
Which of the statements are true:
a. A and C only
b. B and C only
c. A, B, and C
d. none of the above
22
19 http://www.fhsst.org/lrP
20 http://www.fhsst.org/lrE
21 http://www.fhsst.org/lrm
22 http://www.fhsst.org/lry

5. Consider the diagram below of a standing wave on a string 9 m long that is tied at both ends. The
−1
wave velocity in the string is 16 m·s . What is the wavelength?
Figure 13.48
23
13.3.2 Summary
1. A wave is formed when a continuous number of pulses are transmitted through a medium.
2. A peak is the highest point a particle in the medium rises to.
3. A trough is the lowest point a particle in the medium sinks to.
4. In a transverse wave, the particles move perpendicular to the motion of the wave.
5. The amplitude is the maximum distance from equilibrium position to a peak (or trough), or the
maximum displacement of a particle in a wave from its position of rest.
6. The wavelength (λ) is the distance between any two adjacent points on a wave that are in phase. It is
measured in metres.
7. The period (T ) of a wave is the time it takes a wavelength to pass a xed point. It is measured in
seconds (s).
8. The frequency (f ) of a wave is how many waves pass a point in a second. It is measured in hertz (Hz)
or s−1 .
9. Frequency: f = T1
1
10. Period: T =
f
Speed: v = f λ or v =
λ
11.
T.
12. When a wave is reected from a xed end, the resulting wave will move back through the medium, but
will be inverted. When a wave is reected from a free end, the waves are reected, but not inverted.
13.3.3 Exercises
1. A standing wave is formed when:
a. a wave refracts due to changes in the properties of the medium

b. a wave reects o a canyon wall and is heard shortly after it is formed
c. a wave refracts and reects due to changes in the medium
d. two identical waves moving dierent directions along the same medium interfere
24
2. How many nodes and anti-nodes are shown in the diagram?
Figure 13.49
23 http://www.fhsst.org/lrG
24 http://www.fhsst.org/lrV

223
25
3. Draw a transverse wave that is reected from a xed end.
26
4. Draw a transverse wave that is reected from a free end.
27
5. A wave travels along a string at a speed of 1, 5m · s−1 . If the frequency of the source of the wave is 7,5
Hz, calculate:
a. the wavelength of the wave

b. the period of the wave
28
6. Water waves crash against a seawall around the harbour. Eight waves hit the seawall in 5 s. The
distance between successive troughs is 9 m. The height of the waveform trough to crest is 1,5 m.
Figure 13.50
a. How many complete waves are indicated in the sketch?

b. Write down the letters that indicate any TWO points that are:
i. in phase
ii. out of phase
iii. Represent ONE wavelength.
c. Calculate the amplitude of the wave.

d. Show that the period of the wave is 0,67 s.
e. Calculate the frequency of the waves.
f. Calculate the velocity of the waves.
29
25 http://www.fhsst.org/lrp
26 http://www.fhsst.org/lrd
27 http://www.fhsst.org/lrv
28 http://www.fhsst.org/lrw
29 http://www.fhsst.org/lrf


Step 1. The denition of the amplitude is the height of a peak above the equilibrium position. The still water
mark is the height of the water at equilibrium and the peak is 2m above this, so the amplitude is 2 m.
Step 1.
Figure 13.51
Step 2. From the sketch we see that 4 consecutive peaks is equivalent to 3 wavelengths.
Step 3. Therefore, the wavelength of the wave is:
3λ = 6m
6m
λ = 3
(13.17)
= 2m

Step 1. We are required to calculate the period of a 10 Hz wave.
Step 2. We know that:
1
T = (13.18)
f
Step 3.
1
T = f
1 (13.19)
= 10 Hz
= 0, 1 s
Step 4. The period of a 10 Hz wave is 0, 1 s.
Step 1. • frequency of wave: f = 10Hz
• wavelength of wave: λ = 0, 25m
We are required to calculate the speed of the wave as it travels along the string. All quantities are in
SI units.
Step 2. We know that the speed of a wave is:
v =f ·λ (13.20)
and we are given all the necessary quantities.

Step 3.
v = f ·λ
= (10 Hz) (0, 25 m) (13.21)
= 2, 5 m · s −1
Step 1. Available for free at Connexions <http://cnx.org/content/col11245/1.3>

225
Step 4. The wave travels at 2, 5 m · s−1 along the string.

We are given:
• frequency of wave: f = 1 Hz
• wavelength of wave: λ = 20 cm
• distance of cork from edge of pool: d = 2m
We are required to determine the time it takes for a ripple to travel between the cork and the edge of
the pool.
The wavelength is not in SI units and should be converted.
Step 2. The time taken for the ripple to reach the edge of the pool is obtained from:
d d

t= from v= (13.22)
v t
We know that
v =f ·λ (13.23)
Therefore,
d
t= (13.24)
f ·λ
Step 3.
20 cm = 0, 2 m (13.25)
Step 4.
t = d
f ·λ
2m (13.26)
= (1 Hz)(0,2 m)
= 10 s
Step 5. A ripple passing the leaf will take 10 s to reach the edge of the pool.

Step 1. The y vs. t graph is given. The vy vs. t and ay vs. t graphs are required.
Step 2. To nd the velocity of the particle we need to nd the gradient of the y vs. t graph at each time. At
point A the gradient is a maximum and positive. At point B the gradient is zero. At point C the
gradient is a maximum, but negative. At point D the gradient is zero. At point E the gradient is a
maximum and positive again.
Figure 13.52
Step 3. To nd the acceleration of the particle we need to nd the gradient of the vy vs. t graph at each time.
At point A the gradient is zero. At point B the gradient is negative and a maximum. At point C the
gradient is zero. At point D the gradient is positive and a maximum. At point E the gradient is zero.

Figure 13.53

Chapter 14
Geometrical optics
14.1 Introduction and light rays 1
14.1.1 Introduction
You are indoors on a sunny day. A beam of sunlight through a window lights up a section of the oor. How
would you draw this sunbeam? You might draw a series of parallel lines showing the path of the sunlight
from the window to the oor. This is not exactly accurate no matter how hard you look, you will not nd
unique lines of light in the sunbeam! However, this is a good way to draw light. It is also a good way to
model light geometrically, as we will see in this chapter.
These narrow, imaginary lines of light are called light rays. Since light is an electromagnetic wave, you
could think of a light ray as the path of a point on the crest of a wave. Or, you could think of a light ray
as the path taken by a miniscule particle that carries light. We will always draw them the same way: as
straight lines between objects, images, and optical devices.
We can use light rays to model mirrors, lenses, telescopes, microscopes, and prisms. The study of how
light interacts with materials is optics. When dealing with light rays, we are usually interested in the shape
of a material and the angles at which light rays hit it. From these angles, we can work out, for example,
the distance between an object and its reection. We therefore refer to this kind of optics as geometrical
optics.
14.1.2 Light Rays

In physics we use the idea of a light ray to indicate the direction that light travels. Light rays are lines with
arrows and are used to show the path that light travels. In Figure 14.1, the light rays from the object enters
the eye and the eye sees the object.
The most important thing to remember is that we can only see an object when light from the object
enters our eyes. The object must be a source of light (for example a light bulb) or else it must reect light
from a source (for example the moon), and the reected light enters our eyes.
tip: We cannot see an object unless light from that object enters our eyes.
Denition 14.1: Light ray

Light rays are straight lines with arrows to show the path of light.
227
228 CHAPTER 14. GEOMETRICAL OPTICS
Figure 14.1: We can only see an object when light from that object enters our eyes. We draw light as
lines with arrows to show the direction the light travels. When the light travels from the object to the
eye, the eye can see the object. Light rays entering the eye from the cart are shown as dashed lines. The
second wheel of the cart will be invisible as no straight, unobstructed lines exist between it and the eye.
tip: Light rays are not an exact description of a general source of light. They are merely used to
show the path that light travels.
14.1.2.1 Investigation : Light travels in straight lines

Apparatus:
You will need a candle, matches and three sheets of paper.
Method:
1. Make a small hole in the middle of each of the three sheets of paper.
2. Light the candle.
3. Look at the burning candle through the hole in the rst sheet of paper.
4. Place the second sheet of paper between you and the candle so that you can still see the candle through
the holes.
5. Now do the same with the third sheet so that you can still see the candle. The sheets of paper must
not touch each other.
Figure 14.2: Light travels in straight lines
6. What do you notice about the holes in the paper?
Conclusions:
In the investigation you will notice that the holes in the paper need to be in a straight line. This shows
that light travels in a straight line. We cannot see around corners. This also proves that light does not bend
around a corner, but travels straight.
14.1.2.2 Investigation : Light travels in straight lines

On a sunny day, stand outside and look at something in the distance, for example a tree, a ower or a car.
From what we have learnt, we can see the tree, ower or car because light from the object is entering our
eye. Now take a sheet of paper and hold it about 20 cm in front of your face. Can you still see the tree,
ower or car? Why not?
Figure 14.3 shows that a sheet of paper in front of your eye prevents light rays from reaching your eye.

229
Figure 14.3: The sheet of paper prevents the light rays from reaching the eye, and the eye cannot see
the object.
14.1.2.3 Shadows
Objects cast shadows when light shines on them. This is more evidence that light travels in straight lines.
The picture below shows how shadows are formed.
Figure 14.4
14.1.2.4 Ray Diagrams

A ray diagram is a drawing that shows the path of light rays. Light rays are drawn using straight lines and
arrow heads. The gure below shows some examples of ray diagrams.
Figure 14.5
Figure 14.6
14.1.2.4.1 Light Rays

1. Give evidence to support the statement: Light travels in straight lines. Draw a ray diagram to prove
this.
2
2 http://www.fhsst.org/lrt

14.2 Reection 3
14.2.1 Reection
When you smile into a mirror, you see your own face smiling back at you. This is caused by the reection
of light rays on the mirror. Reection occurs when a light ray bounces o a surface.
14.2.1.1 Terminology
4 5
In Transverse Pulses and Transverse Waves we saw that when a pulse or wave strikes a surface it is
reected. This means that waves bounce o things. Sound waves bounce o walls, light waves bounce o
mirrors, radar waves bounce o aeroplanes and it can explain how bats can y at night and avoid things as
thin as telephone wires. The phenomenon of reection is a very important and useful one.
We will use the following terminology. The incoming light ray is called the incident ray. The light
ray moving away from the surface is the reected ray. The most important characteristic of these rays
is their angles in relation to the reecting surface. These angles are measured with respect to the normal
of the surface. The normal is an imaginary line perpendicular to the surface. The angle of incidence,
θi is measured between the incident ray and the surface normal. The angle of reection, θr is measured
between the reected ray and the surface normal. This is shown in Figure 14.7.
When a ray of light is reected, the reected ray lies in the same plane as the incident ray and the normal.
This plane is called the plane of incidence and is shown in Figure 14.8.
Figure 14.7: The angles of incidence and reection are measured with respect to the surface normal.
Figure 14.8: The plane of incidence is the plane including the incident ray, reected ray, and the surface
normal.
14.2.1.2 Law of Reection

The Law of Reection states that the angles of incidence and reection are always equal and that the
reected ray always lies in the plane of incidence.

5 "Transverse Waves - Grade 10" <http://cnx.org/content/m32635/latest/>

231
Denition 14.2: Law of Reection

The Law of Reection states that the angle of incidence is equal to the angle of reection.
θi = θr (14.1)
The simplest example of the law of incidence is if the angle of incidence is 0◦ . In this case, the angle of
◦
reection is also 0 . You see this when you look straight into a mirror.
Figure 14.9: When a wave strikes a surface at right angles to the surface, then the wave is reected
directly back.
If the angle of incidence is not 0◦ , then the angle of reection is also not 0◦ . For example, if a light strikes
a surface at 60◦ to the surface normal, then the angle that the reected ray makes with the surface normal
is also 60◦ as shown in Figure 14.10.
Figure 14.10: Ray diagram showing angle of incidence and angle of reection. The Law of Reection
states that when a light ray reects o a surface, the angle of reection θr is the same as the angle of
incidence θi .
Exercise 14.2.1: Law of Reection (Solution on p. 264.)

◦
An incident ray strikes a smooth reective surface at an angle of 33 to the surface normal.
Calculate the angle of reection.
14.2.1.3 Types of Reection

The Law of Reection holds for every light ray. Does this mean that when parallel rays approach a surface,
the reected rays will also be parallel? This depends on the texture of the reecting surface.

(a)
(b)
Figure 14.11: Specular and diuse reection.
14.2.1.3.1 Specular Reection

Figure 14.11(a) shows a surface that is at and even. Parallel incident light rays hit the smooth surface and
parallel reected light rays leave the surface. This type of reection is called specular reection. Specular
reection occurs when rays are reected from a smooth, shiny surface. The normal to the surface is the
same at every point on the surface. Parallel incident rays become parallel reected rays. When you look in
a mirror, the image you see is formed by specular reection.
14.2.1.3.2 Diuse Reection

Figure 14.11(b) shows a surface with bumps and curves. When multiple rays hit this uneven surface, diuse
reection occurs. The incident rays are parallel but the reected rays are not. Each point on the surface
has a dierent normal. This means the angle of incidence is dierent at each point. Then according to the
Law of Reection, each angle of reection is dierent. Diuse reection occurs when light rays are reected
from bumpy surfaces. You can still see a reection as long as the surface is not too bumpy. Diuse reection
enables us to see all objects that are not sources of light.
14.2.1.3.2.1 Experiment : Specular and Diuse Reection

A bouncing ball can be used to demonstrate the basic dierence between specular and diuse reection.
Aim:
To demonstrate and compare specular and diuse reection.
Apparatus:
You will need:
1. a small ball (a tennis ball or a table tennis ball is perfect)

2. a smooth surface, like the oor inside the classroom
3. a very rough surface, like a rocky piece of ground
Method:
1. Bounce the ball on the smooth oor and observe what happens.
2. Bounce the ball on the rough ground oor and observe what happens.
3. What do you observe?
4. What is the dierence between the two surfaces?
Conclusions:
You should have seen that the ball bounces (is reected o the oor) in a predictable manner o the
smooth oor, but bounces unpredictably on the rough ground.

233
The ball can be seen to be a ray of light and the oor or ground is the reecting surface. For specular
reection (smooth surface), the ball bounces predictably. For diuse reection (rough surface), the ball
bounces unpredictably.
14.2.1.3.2.2 Reection
1. The diagram shows a curved surface. Draw normals to the surface at the marked points.
Figure 14.12
6
2. Which of the points, AH, in the diagram, correspond to the following:
a. normal
b. angle of incidence
c. angle of reection
d. incident ray
e. reected ray
Figure 14.13
7
3. State the Law of Reection. Draw a diagram, label the appropriate angles and write a mathematical
expression for the Law of Reection.
8
4. The diagram shows an incident ray I. Which of the other 5 rays (A, B, C, D, E) best represents the
reected ray of I?
Figure 14.14
9
6 http://www.fhsst.org/lrz
7 http://www.fhsst.org/lru
8 http://www.fhsst.org/lrJ
9 http://www.fhsst.org/lrS

◦
5. A ray of light strikes a surface at 15 to the surface normal. Draw a ray diagram showing the incident
ray, reected ray and surface normal. Calculate the angles of incidence and reection and ll them in
on your diagram.
10
◦
6. A ray of light leaves a surface at 65 to the surface. Draw a ray diagram showing the incident ray,
reected ray and surface normal. Calculate the angles of incidence and reection and ll them in on
your diagram.
11
7. Explain the dierence between specular and diuse reection.
12
8. We see an object when the light that is reected by the object enters our eyes. Do you think the
reection by most objects is specular reection or diuse reection? Explain.
13
9. A beam of light (for example from a torch) is generally not visible at night, as it travels through air.
Try this for yourself. However, if you shine the torch through dust, the beam is visible. Explain why
this happens.
14
14.3 Refraction 15
14.3.1 Refraction
In the previous sections we studied light reecting o various surfaces. What happens when light passes
through a medium? The speed of light, like that of all waves, is dependent on the medium in which it is
travelling. When light moves from one medium into another (for example, from air to glass), the speed of
light changes. The eect is that the light ray passing into a new medium is refracted, or bent. Refraction
is therefore the bending of light as it moves from one optical medium to another.
Denition 14.3: Refraction

Refraction is the bending of light that occurs because light travels at dierent speeds in dierent
materials.
When light travels from one medium to another, it will be bent away from its original path. When it
travels from an optically dense medium like water or glass to a less dense medium like air, it will be refracted
away from the normal (Figure 14.15). Whereas, if it travels from a less dense medium to a denser one, it
will be refracted towards the normal (Figure 14.16).
Figure 14.15: Light is moving from an optically dense medium to an optically less dense medium.
Light is refracted away from the normal.
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235
Figure 14.16: Light is moving from an optically less dense medium to an optically denser medium.
Light is refracted towards the normal.
Just as we dened an angle of reection in the previous section, we can similarly dene an angle of
refraction as the angle between the surface normal and the refracted ray. This is shown in Figure 14.17.
(a)
(b)
Figure 14.17: Light moving from one medium to another bends towards or away from the surface
normal. The angle of refraction θ is shown.
14.3.1.1 Refractive Index

Which is easier to travel through, air or water? People usually travel faster through air. So does light! The
speed of light and therefore the degree of bending of the light depends on the refractive index of material
through which the light passes. The refractive index (symbol n) is the ratio of the speed of light in a vacuum
to its speed in the material.
Denition 14.4: Refractive Index

The refractive index of a material is the ratio of the speed of light in a vacuum to its speed in the
medium.
note: The symbol c is used to represent the speed of light in a vacuum.
c = 299 792 485 m · s−1 (14.2)

8
For purposes of calculation, we use 3 × 10 m · s−1
. A vacuum is a region with no matter in it, not
even air. However, the speed of light in air is very close to that in a vacuum.
Denition 14.5: Refractive Index

The refractive index (symbol n) of a material is the ratio of the speed of light in a vacuum to
its speed in the material and gives an indication of how dicult it is for light to get through the
material.
c
n= (14.3)
v
where

n = refractive index (no unit)
c = speed of light in a vacuum (3, 00 × 108 m · s−1 )

v = speed of light in a given medium ( m · s−1 )
Table 14.1
14.3.1.1.1 Refractive Index and Speed of Light

Using
c
n= (14.4)
v
we can also examine how the speed of light changes in dierent media, because the speed of light in a
vacuum (c) is constant.
If the refractive index n increases, the speed of light in the material v must decrease. Light therefore
travels slowly through materials of high n.
Table 14.2 shows refractive indices for various materials. Light travels slower in any material than it does
in a vacuum, so all values for n are greater than 1.

237
Medium Refractive Index

Vacuum 1
Helium 1,000036
Air* 1,0002926
Carbon dioxide 1,00045
Water: Ice 1,31

◦
Water: Liquid (20 C) 1,333
Acetone 1,36
Ethyl Alcohol (Ethanol) 1,36
Sugar solution (30%) 1,38
Fused quartz 1,46
Glycerine 1,4729
Sugar solution (80%) 1,49
Rock salt 1,516
Crown Glass 1,52
Sodium chloride 1,54
Polystyrene 1,55 to 1,59
Bromine 1,661
Sapphire 1,77
Glass (typical) 1,5 to 1,9
Cubic zirconia 2,15 to 2,18
Diamond 2,419
Silicon 4,01
Table 14.2: Refractive indices of some materials. nair is calculated at STP.
14.3.1.2 Snell's Law

Now that we know that the degree of bending, or the angle of refraction, is dependent on the refractive index
of a medium, how do we calculate the angle of refraction?
The angles of incidence and refraction when light travels from one medium to another can be calculated
using Snell's Law.
Denition 14.6: Snell's Law

n1 sinθ1 = n2 sinθ2 (14.5)
where

n1 = Refractive index of material 1
θ1 = Angle of incidence
θ2 = Angle of refraction
Table 14.3
Remember that angles of incidence and refraction are measured from the normal, which is an imaginary
line perpendicular to the surface.
Suppose we have two media with refractive indices n1 and n2 . A light ray is incident on the surface
angle of incidence θ1 .
between these materials with an The refracted ray that passes through the second
medium will have an angle of refraction θ2 .
Exercise 14.3.1: Using Snell's Law (Solution on p. 264.)

◦
A light ray with an angle of incidence of 35 passes from water to air. Find the angle of refraction
using Snell's Law and Table 14.2. Discuss the meaning of your answer.
Exercise 14.3.2: Using Snell's Law (Solution on p. 264.)

A light ray passes from water to diamond with an angle of incidence of 75◦ . Calculate the angle
of refraction. Discuss the meaning of your answer.
If
n2 > n1 (14.6)
then from Snell's Law,
sinθ1 > sinθ2 . (14.7)
For angles smaller than 90◦ , sinθ increases as θ increases. Therefore,
θ1 > θ2 . (14.8)
This means that the angle of incidence is greater than the angle of refraction and the light ray is bent
toward the normal.
Similarly, if
n2 < n1 (14.9)
then from Snell's Law,
sinθ1 < sinθ2 . (14.10)
For angles smaller than 90 ◦

, sinθ increases as θ increases. Therefore,
θ1 < θ2 . (14.11)
This means that the angle of incidence is less than the angle of refraction and the light ray is away toward
the normal.
Both these situations can be seen in Figure 14.18.

239
(a)
(b)
Figure 14.18: Refraction of two light rays. (a) A ray travels from a medium of low refractive index to
one of high refractive index. The ray is bent towards the normal. (b) A ray travels from a medium with
a high refractive index to one with a low refractive index. The ray is bent away from the normal.
What happens to a ray that lies along the normal line? In this case, the angle of incidence is 0◦ and
sinθ2 = n2 sinθ1
n1
= 0 (14.12)
∴ θ2 = 0.
◦
This shows that if the light ray is incident at 0 , then the angle of refraction is also 0◦ . The ray passes
through the surface unchanged, i.e. no refraction occurs.
14.3.1.2.1 Investigation : Snell's Law 1

The angles of incidence and refraction were measured in ve unknown media and recorded in the table below.
Use your knowledge about Snell's Law to identify each of the unknown media A - E. Use Table 14.2 to help
you.
Medium 1 n1 θ1 θ2 n2 Unknown Medium

Air 1,000036 38 11,6 ? A
Air 1,000036 65 38,4 ? B
Vacuum 1 44 0,419 ? C
Air 1,000036 15 29,3 ? D
Vacuum 1 20 36,9 ? E
Table 14.4
14.3.1.2.2 Investigation : Snell's Law 2

Zingi and Tumi performed an investigation to identify an unknown liquid. They shone a beam of light into
the unknown liquid, varying the angle of incidence and recording the angle of refraction. Their results are
recorded in the following table:

Angle of Incidence Angle of Refraction

◦ ◦
0,0 0,00
◦ ◦
5,0 3,76
◦ ◦
10,0 7,50
◦ ◦
15,0 11,2
◦ ◦
20,0 14,9
◦ ◦
25,0 18,5
◦ ◦
30,0 22,1
◦ ◦
35,0 25,5
◦ ◦
40,0 28,9
◦ ◦
45,0 32,1
◦ ◦
50,0 35,2
◦ ◦
55,0 38,0
◦ ◦
60,0 40,6
◦ ◦
65,0 43,0
◦
70,0 ?
◦
75,0 ?
◦
80,0 ?
◦
85,0 ?
Table 14.5
1. Write down an aim for the investigation.

2. Make a list of all the apparatus they used.
3. Identify the unknown liquid.
◦ ◦ ◦ ◦
4. Predict what the angle of refraction will be for 70 , 75 , 80 and 85 .
Khan academy video on Snell's Law - 1

<http://www.youtube.com/v/HahjsBApxLE&rel=0>
Figure 14.19
14.3.1.3 Apparent Depth

Imagine a coin on the bottom of a shallow pool of water. If you reach for the coin, you will miss it because
the light rays from the coin are refracted at the water's surface.
Consider a light ray that travels from an underwater object to your eye. The ray is refracted at the
water surface and then reaches your eye. Your eye does not know Snell's Law; it assumes light rays travel

241
in straight lines. Your eye therefore sees the image of the at coin shallower location. This shallower location
is known as the apparent depth.
The refractive index of a medium can also be expressed as
real depth
n= . (14.13)
apparent depth
Figure 14.20
Exercise 14.3.3: Apparent Depth 1 (Solution on p. 264.)

A coin is placed at the bottom of a 40 cm deep pond. The refractive index for water is 1,33. How
deep does the coin appear to be?
Exercise 14.3.4: Apparent Depth 2 (Solution on p. 265.)

A R1 coin appears to be 7 cm deep in a colourless liquid known to be listed in Table 14.2. The
depth of the liquid is 10,43 cm.
1. Determine the refractive index of the liquid.

2. Identify the liquid.
14.3.1.3.1 Refraction
1. Explain refraction in terms of a change of wave speed in dierent media.
16
2. In the diagram, label the following:
a. angle of incidence
b. angle of refraction
c. incident ray
d. refracted ray
e. normal
Figure 14.21
17
3. What is refraction?
18
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4. Describe what is meant by the refractive index of a medium.

19
5. State Snell's Law.
20
6. In the diagram, a ray of light strikes the interface between two media.
Figure 14.22
Draw what the refracted ray would look like if:
a. medium 1 had a higher refractive index than medium 2.

b. medium 1 had a lower refractive index than medium 2.
21
◦
7. Light travels from a region of glass into a region of glycerine, making an angle of incidence of 40 .
a. Describe the path of the light as it moves into the glycerine.

b. Calculate the angle of refraction.
22
◦
8. A ray of light travels from silicon to water. If the ray of light in the water makes an angle of 69 to
the surface normal, what is the angle of incidence in the silicon?
23
9. Light travels from a medium with n = 1, 25 into a medium of n = 1, 34, at an angle of 27
◦
from the
interface normal.
a. What happens to the speed of the light? Does it increase, decrease, or remain the same?
b. What happens to the wavelength of the light? Does it increase, decrease, or remain the same?
c. Does the light bend towards the normal, away from the normal, or not at all?
24
10. Light travels from a medium with n = 1, 63 into a medium of n = 1, 42.
a. What happens to the speed of the light? Does it increase, decrease, or remain the same?
b. What happens to the wavelength of the light? Does it increase, decrease, or remain the same?
c. Does the light bend towards the normal, away from the normal, or not at all?
25
◦
11. Light is incident on a glass prism. The prism is surrounded by air. The angle of incidence is 23 .
Calculate the angle of reection and the angle of refraction.
26
12. Light is refracted at the interface between air and an unknown medium. If the angle of incidence is
◦ ◦
53 and the angle of refraction is 37 , calculate the refractive index of the unknown, second medium.
27
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243
13. A coin is placed in a bowl of acetone (n = 1,36). The coin appears to be 10 cm deep. What is the
depth of the acetone?
28
14. A dot is drawn on a piece of paper and a glass prism placed on the dot according to the diagram.
Figure 14.23
Use the information supplied to determine the refractive index of glass.

29
15. Light is refracted at the interface between a medium of refractive index 1,5 and a second medium of
◦
refractive index 2,1. If the angle of incidence is 45 , calculate the angle of refraction.
30
◦
16. A ray of light strikes the interface between air and diamond. If the incident ray makes an angle of 30
with the interface, calculate the angle made by the refracted ray with the interface.
31
17. Challenge Question: What values of n are physically impossible to achieve? Explain your answer.
The values provide the limits of possible refractive indices.
32
18. Challenge Question: You have been given a glass beaker full of an unknown liquid. How would you
identify what the liquid is? You have the following pieces of equipment available for the experiment:
a laser, a protractor, a ruler, a pencil, and a reference guide containing optical properties of various
liquids.
33
14.4 Mirrors 34
14.4.1 Mirrors
A mirror is a highly reective surface. The most common mirrors are at and are known as plane mirrors.
Household mirrors are plane mirrors. They are made of a at piece of glass with a thin layer of silver
nitrate or aluminium on the back. However, other mirrors are curved and are either convex mirrors or are
concave mirrors. The reecting properties of all three types of mirrors will be discussed in this section.
14.4.1.1 Image Formation

Denition 14.7: Image
An image is a representation of an object formed by a mirror or lens. Light from the image is seen.
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Figure 14.24: An object formed in a mirror is real and upright.
If you place a candle in front of a mirror, you now see two candles. The actual, physical candle is called
the object and the picture you see in the mirror is called the image. The object is the source of the incident
rays. The image is the picture that is formed by the reected rays.
The object could be an actual source that emits light, such as a light bulb or a candle. More commonly,
the object reects light from another source. When you look at your face in the mirror, your face does not
emit light. Instead, light from a light bulb or from the sun reects o your face and then hits the mirror.
However, in working with light rays, it is easiest to pretend the light is coming from the object.
An image formed by reection may be real or virtual. A real image occurs when light rays actually
intersect at the image. A real image is inverted, or upside down. A virtual image occurs when light rays do
not actually meet at the image. Instead, you "see" the image because your eye projects light rays backward.
You are fooled into seeing an image! A virtual image is erect, or right side up (upright).
You can tell the two types apart by putting a screen at the location of the image. A real image can be
formed on the screen because the light rays actually meet there. A virtual image cannot be seen on a screen,
since it is not really there.
To describe objects and images, we need to know their locations and their sizes. The distance from the
mirror to the object is the object distance, do .
The distance from the mirror to the image is the image distance, di .
14.4.1.2 Plane Mirrors
14.4.1.2.1 Investigation : Image formed by a mirror
1. Stand one step away from a large mirror
2. What do you observe in the mirror? This is called your image.
3. What size is your image? Bigger, smaller or the same size as you?
4. How far is your image from you? How far is your image from the mirror?
5. Is your image upright or upside down?
6. Take one step backwards. What does your image do? How far are you away from your image?
7. If it were a real object, which foot would the image of you right show t?
Figure 14.25: An image in a mirror is virtual, upright, the same size and inverted front to back.
When you look into a mirror, you see an image of yourself.

The image created in the mirror has the following properties:
1. The image is virtual.

245
2. The image is the same distance behind the mirror as the object is in front of the mirror.
3. The image is inverted front to back.
4. The image is the same size as the object.
5. The image is upright.
Virtual images are images formed in places where light does not really reach. Light does not really pass
through the mirror to create the image; it only appears to an observer as though the light were coming from
behind the mirror. Whenever a mirror creates an image which is virtual, the image will always be located
behind the mirror where light does not really pass.
Denition 14.8: Virtual Image

A virtual image is upright, on the opposite side of the mirror as the object, and light does not
actually reach it.
14.4.1.3 Ray Diagrams

We draw ray diagrams to predict the image that is formed by a plane mirror. A ray diagram is a geometrical
picture that is used for analyzing the images formed by mirrors and lenses. We draw a few characteristic
rays from the object to the mirror. We then follow ray-tracing rules to nd the path of the rays and locate
the image.
tip: A mirror obeys the Law of Reection.
The ray diagram for the image formed by a plane mirror is the simplest possible ray diagram. shows an
object placed in front of a plane mirror. It is convenient to have a central line that runs perpendicular to
the mirror. This imaginary line is called the principal axis.
tip: Ray diagrams
The following should be remembered when drawing ray diagrams:
1.Objects are represented by arrows. The length of the arrow represents the height of the object.
2.If the arrow points upwards, then the object is described as upright or erect. If the arrow
points downwards then the object is described as inverted.
3.If the object is real, then the arrow is drawn with a solid line. If the object is virtual, then
the arrow is drawn with a dashed line.
Method: Ray Diagrams for Plane Mirrors

Ray diagrams are used to nd the position and size and whether the image is real or virtual.
1. Draw the plane mirror as a straight line on a principal axis.
Figure 14.26

2. Draw the object as an arrow in front of the mirror.
Figure 14.27
3. Draw the image of the object, by using the principle that the image is placed at the same distance
behind the mirror that the object is in front of the mirror. The image size is also the same as the
object size.
Figure 14.28
4. Place a dot at the point the eye is located.

5. Pick one point on the image and draw the reected ray that travels to the eye as it sees this point.
Remember to add an arrowhead.
Figure 14.29
6. Draw the incident ray for light traveling from the corresponding point on the object to the mirror,
such that the law of reection is obeyed.
Figure 14.30
7. Continue for other extreme points on the object (i.e. the tip and base of the arrow).
Figure 14.31

247
Suppose a light ray leaves the top of the object traveling parallel to the principal axis. The ray will hit
the mirror at an angle of incidence of 0 degrees. We say that the ray hits the mirror normally. According to
the law of reection, the ray will be reected at 0 degrees. The ray then bounces back in the same direction.
We also project the ray back behind the mirror because this is what your eye does.
Another light ray leaves the top of the object and hits the mirror at its centre. This ray will be reected
at the same angle as its angle of incidence, as shown. If we project the ray backward behind the mirror, it
will eventually cross the projection of the rst ray we drew. We have found the location of the image! It is
a virtual image since it appears in an area that light cannot actually reach (behind the mirror). You can see
from the diagram that the image is erect and is the same size as the object. This is exactly as we expected.
We use a dashed line to indicate that the image is virtual.
Khan academy video on mirrors - 1

<http://www.youtube.com/v/nrOg85VPQgw&rel=0>
Figure 14.32
14.4.1.4 Spherical Mirrors

The second class of mirrors that we will look at are spherical mirrors. These mirrors are called spherical
mirrors because if you take a sphere and cut it as shown in Figure 14.33 and then polish the inside of one
and the outside of the other, you will get a concave mirror and convex mirror as shown. These two mirrors
will be studied in detail.
The centre of curvature is the point at the centre of the sphere and describes how big the sphere is.
Figure 14.33: When a sphere is cut and then polished to a reective surface on the inside a concave
mirror is obtained. When the outside is polished to a reective surface, a convex mirror is obtained.
14.4.1.5 Concave Mirrors

The rst type of curved mirror we will study are concave mirrors. Concave mirrors have the shape shown in
Figure 14.34. As with a plane mirror, the principal axis is a line that is perpendicular to the centre of the
mirror.

Figure 14.34: Concave mirror with principal axis.
If you think of light reecting o a concave mirror, you will immediately see that things will look very
dierent compared to a plane mirror. The easiest way to understand what will happen is to draw a ray
diagram and work out where the images will form. Once we have done that it is easy to see what properties
the image has.
First we need to dene a very important characteristic of the mirror. We have seen that the centre of
curvature is the centre of the sphere from which the mirror is cut. We then dene that a distance that is
half-way between the centre of curvature and the mirror on the principal axis. This point is known as the
focal point and the distance from the focal point to the mirror is known as the focal length (symbol f ).
Since the focal point is the midpoint of the line segment joining the vertex and the center of curvature, the
focal length would be one-half the radius of curvature. This fact can come in very handy, remember if you
know one then you know the other!
Denition 14.9: Focal Point

The focal point of a mirror is the midpoint of a line segment joining the vertex and the centre of
curvature. It is the position at which all parallel rays are focussed.
Why are we making such a big deal about this point we call the focal point? It has an important property
we will use often. A ray parallel to the principal axis hitting the mirror will always be reected through the
focal point. The focal point is the position at which all parallel rays are focussed.
Figure 14.35: All light rays pass through the focal point.
Figure 14.36: A concave mirror with three rays drawn to locate the image. Each incident ray is
reected according to the Law of Reection. The intersection of the reected rays gives the location of
the image. Here the image is real and inverted.
From Figure 14.36, we see that the image created by a concave mirror is real and inverted, as compared
to the virtual and erect image created by a plane mirror.

249
Denition 14.10: Real Image

A real image can be cast on a screen; it is inverted, and on the same side of the mirror as the
object.
14.4.1.5.1 Convergence
A concave mirror is also known as a converging mirror. Light rays appear to converge to the focal point of
a concave mirror.
14.4.1.6 Convex Mirrors

The second type of curved mirror we will study are convex mirrors. Convex mirrors have the shape shown
in Figure 14.37. As with a plane mirror, the principal axis is a line that is perpendicular to the centre of the
mirror.
We have dened the focal point as that point that is half-way along the principal axis between the centre
of curvature and the mirror. Now for a convex mirror, this point is behind the mirror. A convex mirror has
a negative focal length because the focal point is behind the mirror.
Figure 14.37: Convex mirror with principle axis, focal point (F) and centre of curvature (C). The
centre of the mirror is the optical centre (O).
To determine what the image from a convex mirror looks like and where the image is located, we need
to remember that a mirror obeys the laws of reection and that light appears to come from the image. The
image created by a convex mirror is shown in Figure 14.38.
Figure 14.38: A convex mirror with three rays drawn to locate the image. Each incident ray is reected
according to the Law of Reection. The reected rays diverge. If the reected rays are extended behind
the mirror, then their intersection gives the location of the image behind the mirror. For a convex mirror,
the image is virtual and upright.
From Figure 14.38, we see that the image created by a convex mirror is virtual and upright, as compared
to the real and inverted image created by a concave mirror.
14.4.1.6.1 Divergence
A convex mirror is also known as a diverging mirror. Light rays appear to diverge from the focal point of a
convex mirror.

14.4.1.7 Summary of Properties of Mirrors

The properties of mirrors are summarised in Table 14.6.
Plane Concave Convex

converging diverging
virtual image real image virtual image
upright inverted upright
image behind mirror image in front of mirror image behind mirror
Table 14.6: Summary of properties of concave and convex mirrors.
14.4.1.8 Magnication
In Figure 14.36 and Figure 14.38, the height of the object and image arrows were dierent. In any optical
system where images are formed from objects, the ratio of the image height, hi , to the object height, ho is
known as the magnication, m.
hi
m= (14.14)
ho
This is true for the mirror examples we showed above and will also be true for lenses, which will be introduced
in the next sections. For a plane mirror, the height of the image is the same as the height of the object, so
the magnication is simply m= hi
ho = 1. If the magnication is greater than 1, the image is larger than the
object and is said to be magnied . If the magnication is less than 1, the image is smaller than the object
so the image is said to be diminished.
Exercise 14.4.1: Magnication (Solution on p. 265.)
A concave mirror forms an image that is 4,8 cm high. The height of the object is 1,6 cm. Calculate
the magnication of the mirror.
14.4.1.8.1 Mirrors
1. List 5 properties of a virtual image created by reection from a plane mirror.
35
2. What angle does the principal axis make with a plane mirror?
36
3. Is the principal axis a normal to the surface of the plane mirror?
37
4. Do the reected rays that contribute to forming the image from a plane mirror obey the law of reection?
38
5. If a candle is placed 50 cm in front of a plane mirror, how far behind the plane mirror will the image
be? Draw a ray diagram to show how the image is formed.
39
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251
6. If a stool 0,5 m high is placed 2 m in front of a plane mirror, how far behind the plane mirror will the
image be and how high will the image be?
40
7. If Susan stands 3 m in front of a plane mirror, how far from Susan will her image be located?
41
8. Explain why ambulances have the word àmbulance' reversed on the front bonnet of the car?
42
9. Complete the diagram by lling in the missing lines to locate the image.
Figure 14.39
43
10. An object 2 cm high is placed 4 cm in front of a plane mirror. Draw a ray diagram, showing the object,
the mirror and the position of the image.
44
11. The image of an object is located 5 cm behind a plane mirror. Draw a ray diagram, showing the image,
the mirror and the position of the object.
45
12. How high must a mirror be so that you can see your whole body in it? Does it make a dierence if
you change the distance you stand in front of the mirror? Explain.
46
−1
13. If 1-year old Tommy crawls towards a mirror at a rate of 0,3 m·s , at what speed will Tommy and
his image approach each other?
47
14. Use a diagram to explain how light converges to the focal point of a concave mirror.
48
15. Use a diagram to explain how light diverges away from the focal point of a convex mirror.
49
16. An object 1 cm high is placed 4 cm from a concave mirror. If the focal length of the mirror is 2 cm,
nd the position and size of the image by means of a ray diagram. Is the image real or virtual? What
is the magnication?
50
17. An object 2 cm high is placed 4 cm from a convex mirror. If the focal length of the mirror is 4 cm,
nd the position and size of the image by means of a ray diagram. Is the image real or virtual? What
is the magnication?
51
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14.5 Total internal reection 52
14.5.1 Total Internal Reection and Fibre Optics

14.5.1.1 Total Internal Reection
14.5.1.1.1 Investigation : Total Internal Reection
Work in groups of four. Each group will need a raybox (or torch) with slit, triangular glass prism and
protractor. If you do not have a raybox, use a torch and stick two pieces of tape over the lens so that only
a thin beam of light is visible.
Aim:
To investigate total internal reection.
Method:
1. Place the raybox next to the glass block so that the light shines right through without any refraction.
See "Position 1" in diagram.
Figure 14.40
2. Move the raybox such that the light is refracted by the glass. See "Position 2".
Figure 14.41
3. Move the raybox further and observe what happens.
Figure 14.42
4. Move the raybox until the refracted ray seems to disappear. See "Position 4". The angle of the incident
light is called the critical angle.

253
Figure 14.43
5. Move the raybox further and observe what happens. See "Position 5". The light shines back into the
glass block. This is called total internal reection.
Figure 14.44
◦
When we increase the angle of incidence, we reach a point where the angle of refraction is 90 and the
refracted ray runs along the surface of the medium. This angle of incidence is called the critical angle.
Denition 14.11: Critical Angle

◦
The critical angle is the angle of incidence where the angle of reection is 90 . The light must
shine from a dense to a less dense medium.
If the angle of incidence is bigger than this critical angle, the refracted ray will not emerge from the
medium, but will be reected back into the medium. This is called total internal reection.
Total internal reection takes place when
• light shines from an optically denser medium to an optically less dense medium.
• the angle of incidence is greater than the critical angle.
Denition 14.12: Total Internal Reection

Total internal reection takes place when light is reected back into the medium because the angle
of incidence is greater than the critical angle.
Figure 14.45: Diagrams to show the critical angle and total internal reection.
◦
Each medium has its own unique critical angle. For example, the critical angle for glass is 42 , and that
◦
of water is 48,8 . We can calculate the critical angle for any medium.
14.5.1.1.2 Calculating the Critical Angle

Now we shall learn how to derive the value of the critical angle for two given media. The process is fairly
simple and involves just the use of Snell's Law that we have already studied. To recap, Snell's Law states:
n1 sinθ1 = n2 sinθ2 (14.15)

where n1 is the refractive index of material 1, n2 is the refractive index of material 2, θ1 is the angle of
incidence and θ2 is the angle of refraction. For total internal reection we know that the angle of incidence
is the critical angle. So,
θ1 = θc . (14.16)
◦
However, we also know that the angle of refraction at the critical angle is 90 . So we have:
θ2 = 90◦ . (14.17)
We can then write Snell's Law as:
n1 sinθc = n2 sin90◦ (14.18)
Solving for θc gives:
n1 sinθc = n2 sin90◦
sinθc = n2
n1 (1)
(14.19)

∴ θc = sin−1 nn12
tip: Take care that for total internal reection the incident ray is always in the denser medium.
Khan academy video on refraction - 1

<http://www.youtube.com/v/WRuatAcd2WY&rel=0>
Figure 14.46
Exercise 14.5.1: Critical Angle 1 (Solution on p. 265.)

Given that the refractive indices of air and water are 1 and 1,33, respectively, nd the critical
angle.
Exercise 14.5.2: Critical Angle 2 (Solution on p. 265.)

Complete the following ray diagrams to show the path of light in each situation.
Figure 14.47
Figure 14.48

255
Figure 14.49
Figure 14.50
14.5.1.2 Fibre Optics

Total internal reection is a powerful tool since it can be used to conne light. One of the most common
applications of total internal reection is in bre optics. An optical bre is a thin, transparent bre, usually
made of glass or plastic, for transmitting light. Optical bres are usually thinner than a human hair! The
construction of a single optical bre is shown in Figure 14.51.
The basic functional structure of an optical bre consists of an outer protective cladding and an inner
core through which light pulses travel. The overall diameter of the bre is about 125 µm (125 × 10−6 m) and
that of the core is just about 50 µm (10 × 10−6 m). The mode of operation of the optical bres, as mentioned
above, depends on the phenomenon of total internal reection. The dierence in refractive index of the
cladding and the core allows total internal reection in the same way as happens at an air-water surface. If
light is incident on a cable end with an angle of incidence greater than the critical angle then the light will
remain trapped inside the glass strand. In this way, light travels very quickly down the length of the cable.
Figure 14.51: Structure of a single optical bre.
14.5.1.2.1 Fibre Optics in Telecommunications

Optical bres are most common in telecommunications, because information can be transported over long
distances, with minimal loss of data. The minimised loss of data gives optical bres an advantage over
conventional cables.
Data is transmitted from one end of the bre to another in the form of laser pulses. A single strand is
capable of handling over 3000 simultaneous transmissions which is a huge improvement over the conventional
co-axial cables. Multiple signal transmission is achieved by sending individual light pulses at slightly dierent
◦
angles. For example if one of the pulses makes a 72,23 angle of incidence then a separate pulse can be sent
◦
at an angle of 72,26 ! The transmitted data is received almost instantaneously at the other end of the

cable since the information coded onto the laser travels at the speed of light! During transmission over
long distances repeater stations are used to amplify the signal which has weakened somewhat by the time
it reaches the station. The amplied signals are then relayed towards their destination and may encounter
several other repeater stations on the way.
14.5.1.2.2 Fibre Optics in Medicine

Optic bres are used in medicine in endoscopes.
note: Endoscopy means to look inside and refers to looking inside the human body for diagnosing
medical conditions.
The main part of an endoscope is the optical bre. Light is shone down the optical bre and a medical doctor
can use the endoscope to look inside a patient. Endoscopes are used to examine the inside of a patient's
stomach, by inserting the endoscope down the patient's throat.
Endoscopes allow minimally invasive surgery. This means that a person can be diagnosed and treated
through a small incision. This has advantages over open surgery because endoscopy is quicker and cheaper
and the patient recovers more quickly. The alternative is open surgery which is expensive, requires more
time and is more traumatic for the patient.
14.5.1.2.2.1 Total Internal Reection and Fibre Optics

1. Describe total internal reection, referring to the conditions that must be satised for total internal
reection to occur.
53
2. Dene what is meant by the critical angle when referring to total internal reection. Include a ray
diagram to explain the concept.
54
3. Will light travelling from diamond to silicon ever undergo total internal reection?
55
4. Will light travelling from sapphire to diamond undergo total internal reection?
56
5. What is the critical angle for light traveling from air to acetone?
57
◦
6. Light traveling from diamond to water strikes the interface with an angle of incidence of 86 . Calculate
the critical angle to determine whether the light be totally internally reected and so be trapped within
the water.
58
7. Which of the following interfaces will have the largest critical angle?
a. a glass to water interface

b. a diamond to water interface
c. a diamond to glass interface
59
8. If the bre optic strand is made from glass, determine the critical angle of the light ray so that the ray
stays within the bre optic strand.
60
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257
9. A glass slab is inserted in a tank of water. If the refractive index of water is 1,33 and that of glass is
1,5, nd the critical angle.
61
◦
10. A diamond ring is placed in a container full of glycerin. If the critical angle is found to be 37,4 and
the refractive index of glycerin is given to be 1,47, nd the refractive index of diamond.
62
11. An optical bre is made up of a core of refractive index 1,9, while the refractive index of the cladding
is 1,5. Calculate the maximum angle which a light pulse can make with the wall of the core. NOTE:
The question does not ask for the angle of incidence but for the angle made by the ray with the wall
◦
of the core, which will be equal to 90 - angle of incidence.
63

<http://static.slidesharecdn.com/swf/ssplayer2.swf ?doc=lightg10-100511072123-
phpapp01&stripped_title=light-g10&userName=kwarne>
Figure 14.52
14.5.2 Summary
1. We can see objects when light from the objects enters our eyes.
2. Light rays are thin imaginary lines of light and are indicated in drawings by means of arrows.
3. Light travels in straight lines. Light can therefore not travel around corners. Shadows are formed
because light shines in straight lines.
4. Light rays reect o surfaces. The incident ray shines in on the surface and the reected ray is the
one that bounces o the surface. The surface normal is the perpendicular line to the surface where the
light strikes the surface.
5. The angle of incidence is the angle between the incident ray and the surface, and the angle of reection
is the angle between the reected ray and the surface.
6. The Law of Reection states the angle of incidence is equal to the angle of reection and that the
reected ray lies in the plane of incidence.
7. Specular reection takes place when parallel rays fall on a surface and they leave the object as parallel
rays. Diuse reection takes place when parallel rays are reected in dierent directions.
8. Refraction is the bending of light when it travels from one medium to another. Light travels at dierent
speeds in dierent media.
9. The refractive index of a medium is a measure of how easily light travels through the medium. It is a
ratio of the speed of light in a vacuum to the speed of light in the medium.
c
v n=
10. Snell's Law gives the relationship between the refractive indices, angles of incidence and reection of
two media. n1 sinθ1 = n2 sinθ2
11. Light travelling from one medium to another of lighter optical density will be refracted towards the
normal. Light travelling from one medium to another of lower optical density will be refracted away
from the normal.
12. Objects in a medium (e.g. under water) appear closer to the surface than they really are. This is due
realdepth
to the refraction of light, and the refractive index of the medium. n= apparentdepth
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13. Mirrors are highly reective surfaces. Flat mirrors are called plane mirrors. Curved mirrors can be
convex or concave. The properties of the images formed by mirrors are summarised in Table 3.2.
14. A real image can be cast on a screen, is inverted and in front of the mirror. A virtual image cannot be
cast on a screen, is upright and behind the mirror.
15. The magnication of a mirror is how many times the image is bigger or smaller than the object.
imageheight(hi )
m= objectheight(h0 )
16. The critical angle of a medium is the angle of incidence when the angle of refraction is 90◦ and the
refracted ray runs along the interface between the two media.
17. Total internal reection takes place when light travels from one medium to another of lower optical
density. If the angle of incidence is greater than the critical angle for the medium, the light will be
reected back into the medium. No refraction takes place.
18. Total internal reection is used in optical bres in telecommunication and in medicine in endoscopes.
Optical bres transmit information much more quickly and accurately than traditional methods.
14.5.3 Exercises
1. Give one word for each of the following descriptions:
a. The image that is formed by a plane mirror.

b. The perpendicular line that is drawn at right angles to a reecting surface at the point of incidence.
c. The bending of light as it travels from one medium to another.

d. The ray of light that falls in on an object.
e. A type of mirror that focuses all rays behind the mirror.
64
2. State whether the following statements are TRUE or FALSE. If they are false, rewrite the statement
correcting it.
a. The refractive index of a medium is an indication of how fast light will travel through the medium.
b. Total internal refraction takes place when the incident angle is larger than the critical angle.
c. The magnication of an object can be calculated if the speed of light in a vacuum and the speed
of light in the medium is known.
d. The speed of light in a vacuum is about 3 × 108 m.s
−1
.
e. Specular reection takes place when light is reected o a rough surface.
65
3. Choose words from Column B to match the concept/description in Column A. All the appropriate
words should be identied. Words can be used more than once.
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259
Column A Column B
(a) Real image Upright
(b) Virtual image Can be cast on a screen
(c) Concave mirror In front
(d) Convex mirror Behind
(e) Plane mirror Inverted
Light travels to it
Upside down
Light does not reach it
Erect
Same size
Table 14.7
66
4. Complete the following ray diagrams to show the path of light.
Figure 14.53
Figure 14.54
Figure 14.55
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Figure 14.56
Figure 14.57
Figure 14.58
67
◦
5. A ray of light strikes a surface at 35 to the surface normal. Draw a ray diagram showing the incident
ray, reected ray and surface normal. Calculate the angles of incidence and reection and ll them in
on your diagram.
68
◦
6. Light travels from glass (n = 1,5) to acetone (n = 1,36). The angle of incidence is 25 .
a. Describe the path of light as it moves into the acetone.

b. Calculate the angle of refraction.
c. What happens to the speed of the light as it moves from the glass to the acetone?
d. What happens to the wavelength of the light as it moves into the acetone?
e. What is the name of the phenomenon that occurs at the interface between the two media?
69
7. A stone lies at the bottom of a swimming pool. The water is 120 cm deep. The refractive index of
water is 1,33. How deep does the stone appear to be?
70
◦
8. Light strikes the interface between air and an unknown medium with an incident angle of 32 . The
◦
angle of refraction is measured to be 48 . Calculate the refractive index of the medium and identify
the medium.
71
9. Explain what total internal reection is and how it is used in medicine and telecommunications. Why
is this technology much better to use?
72
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261
10. A candle 10 cm high is placed 25 cm in front of a plane mirror. Draw a ray diagram to show how the
image is formed. Include all labels and write down the properties of the image.
73
11. A virtual image, 4 cm high, is formed 3 cm from a plane mirror. Draw a labelled ray diagram to show
the position and height of the object. What is the magnication?
74
12. An object, 3 cm high, is placed 4 cm from a concave mirror of focal length 2 cm. Draw a labelled ray
diagram to nd the position, height and properties of the image.
75
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Step 1. We are given the angle between the incident ray and the surface normal. This is the angle of incidence.
We are required to calculate the angle of reection.
Step 2. We can use the Law of Reection, which states that the angle of incidence is equal to the angle of
reection.
◦ ◦
Step 3. We are given the angle of incidence to be 33 . Therefore, the angle of reection is also 33 .

Step 1. From Table 14.2, the refractive index is 1,333 for water and about 1 for air. We know the angle of
incidence, so we are ready to use Snell's Law.
Step 2. According to Snell's Law:
n1 sinθ1 = n2 sinθ2
1, 33sin35◦ = 1sinθ2
(14.20)
sinθ2 = 0, 763
θ2 = 49, 7◦ or 130, 3◦
Since 130, 3◦ is larger than 90◦ , the solution is:
θ2 = 49, 7◦ (14.21)
Step 3. The light ray passes from a medium of high refractive index to one of low refractive index. Therefore,
the light ray is bent away from the normal.

Step 1. From Table 14.2, the refractive index is 1,333 for water and 2,42 for diamond. We know the angle of
incidence, so we are ready to use Snell's Law.
Step 2. According to Snell's Law:
n1 sinθ1 = n2 sinθ2
◦
1, 33sin75 = 2, 42sinθ2
(14.22)
sinθ2 = 0, 531
θ2 = 32, 1◦ .
Step 3. The light ray passes from a medium of low refractive index to one of high refractive index. Therefore,
the light ray is bent towards the normal.

Step 1. n = 1,33
real depth = 40 cm
apparent depth = ?
Step 2.
real depth
n = apparent depth
40 (14.23)
1, 33 = x
40
x = 1,33 = 30, 08cm
The coin appears to be 30,08 cm deep.

263

Step 1. real depth = 7 cm
apparent depth = 10,43 cm
n = ?
Identify the liquid.
Step 2.
real depth
n = apparent depth
10,43
= (14.24)
7
= 1, 49
Step 3. Use Table 14.2. The liquid is an 80% sugar solution.

Step 1. Image height hi = 4,8 cm
Object heightho = 1,6 cm
Magnication m = ?
Step 2.
m = hi
ho
4,8
= (14.25)
1,6
= 3
The magnication is 3 times.

Step 1. We know that the critical angle is given by:
n2

θc = sin−1 (14.26)
n1
Step 2.

θc = sin−1 nn12

1
= sin−1 1,33 (14.27)
= 48, 8◦
Step 3. The critical angle for light travelling from water to air is 48, 8◦ .
Step 1. The critical angle for water is 48, 8◦ .
We are asked to complete the diagrams.
For incident angles smaller than 48, 8◦ refraction will occur.
◦
For incident angles greater than 48, 8 total internal reection will occur.
◦ ◦
For incident angles equal to 48, 8 refraction will occur at 90 .
The light must travel from a high optical density to a lower one.
Step 2.
Figure 14.59

Refraction occurs (ray is bent away from the normal)
Figure 14.60
Total internal reection occurs
Figure 14.61
θc = 48, 8◦
Figure 14.62
Refraction towards the normal (air is less dense than water)

Chapter 15
Magnetism
15.1 Magnetic elds and permanent magnets 1
15.1.1 Introduction
Magnetism is a force that certain kinds of objects, which are called `magnetic' objects, can exert on each
other without physically touching. A magnetic object is surrounded by a magnetic èld' that gets weaker
as one moves further away from the object. A second object can feel a magnetic force from the rst object
because it feels the magnetic eld of the rst object.
Humans have known about magnetism for many thousands of years. For example, lodestone is a magne-
tised form of the iron oxide mineral magnetite. It has the property of attracting iron objects. It is referred
to in old European and Asian historical records; from around 800 BCE in Europe and around 2 600 BCE
in Asia.
note: The root of the English word magnet is from the Greek word magnes, probably from
Magnesia in Asia Minor, once an important source of lodestone.
15.1.2 Magnetic elds

A magnetic eld is a region in space where a magnet or object made of magnetic material will experience a
non-contact force.
Electrons inside any object have magnetic elds associated with them. In most materials these elds
point in all directions, so the net magnetic eld is zero. For example, in the plastic ball below, the directions
of the magnetic elds of the electrons (shown by the arrows) are pointing in dierent directions and cancel
each other out. Therefore the plastic ball is not magnetic and has no magnetic eld.
Figure 15.1
In some materials (e.g. iron), called ferromagnetic materials, there are regions called domains, where
the electrons' magnetic elds line up with each other. All the atoms in each domain are grouped together
265
266 CHAPTER 15. MAGNETISM
so that the magnetic elds from their electrons point the same way. The picture shows a piece of an iron
needle zoomed in to show the domains with the electric elds lined up inside them.
Figure 15.2
In permanent magnets, many domains are lined up, resulting in a net magnetic eld. Objects made
from ferromagnetic materials can be magnetised, for example by rubbing a magnet along the object in one
direction. This causes the magnetic elds of most, or all, of the domains to line up in one direction. As a
result the object as a whole will have a net magnetic eld. It is magnetic. Once a ferromagnetic object has
been magnetised, it can stay magnetic without another magnet being nearby (i.e. without being in another
magnetic eld). In the picture below, the needle has been magnetised because the magnetic elds in all the
domains are pointing in the same direction.
Figure 15.3
15.1.2.1 Investigation : Ferromagnetic materials and magnetisation

1. Find 2 paper clips. Put the paper clips close together and observe what happens.
a. What happens to the paper clips?

b. Are the paper clips magnetic?
2. Now take a permanent bar magnet and rub it once along 1 of the paper clips. Remove the magnet and
put the paper clip which was touched by the magnet close to the other paper clip and observe what
happens. Does the untouched paper clip feel a force on it? If so, is the force attractive or repulsive?
3. Rub the same paper clip a few more times with the bar magnet, in the same direction as before. Put
the paper clip close to the other one and observe what happens.
a. Is there any dierence to what happened in step 2?

b. If there is a dierence, what is the reason for it?
c. Is the paper clip which was rubbed repeatedly by the magnet now magnetised?
d. What is the dierence between the two paper clips at the level of their atoms and electrons?
4. Now, nd a metal knitting needle, or a metal ruler, or other metal object. Rub the bar magnet along
the knitting needle a few times in the same direction. Now put the knitting needle close to the paper
clips and observe what happens.
a. Does the knitting needle attract the paper clips?

b. What does this tell you about the material of the knitting needle? Is it ferromagnetic?
5. Repeat this experiment with objects made from other materials. Which materials appear to be ferro-
magnetic and which are not? Put your answers in a table.
A ferromagnetic material is a substance that shows spontaneous magnetisation.

267
15.1.3 Permanent magnets

15.1.3.1 The poles of permanent magnets
Because the domains in a permanent magnet all line up in a particular direction, the magnet has a pair
of opposite poles, called north (usually shortened to N) and south (usually shortened to S). Even if the
magnet is cut into tiny pieces, each piece will still have both a N and a S pole. These magnetic poles always
occur in pairs. In nature, we never nd a north magnetic pole or south magnetic pole on its own.
Figure 15.4
Magnetic elds are dierent from gravitational and electric elds. In nature, positive and negative electric
charges can be found on their own, but you never nd just a north magnetic pole or south magnetic pole
on its own. On the very small scale, zooming in to the size of atoms, magnetic elds are caused by moving
charges (i.e. the negatively charged electrons).
15.1.3.2 Magnetic attraction and repulsion

Like (identical) poles of magnets repel one another whilst unlike (opposite) poles attract. This means that
two N poles or two S poles will push away from each other while a N pole and a S pole will be drawn towards
each other.
Denition 15.1: Attraction and Repulsion

Like poles of magnets repel each other whilst unlike poles attract each other.
Exercise 15.1.1: Attraction and Repulsion (Solution on p. 275.)

Do you think the following magnets will repel or be attracted to each other?
Figure 15.5
Exercise 15.1.2: Attraction and repulsion (Solution on p. 275.)

Do you think the following magnets will repel or be attracted to each other?
Figure 15.6

15.1.3.3 Representing magnetic elds

Magnetic elds can be represented using magnetic eld lines starting at the North pole and ending at the
South pole. Although the magnetic eld of a permanent magnet is everywhere surrounding the magnet (in all
three dimensions), we draw only some of the eld lines to represent the eld (usually only a two-dimensional
cross-section is shown in drawings).
Figure 15.7
In areas where the magnetic eld is strong, the eld lines are closer together. Where the eld is weaker,
the eld lines are drawn further apart. The number of eld lines drawn crossing a given two-dimensional
surface is referred to as the magnetic ux. The magnetic ux is used as a measure of the strength of the
magnetic eld over that surface.
tip:
1.Field lines never cross.

2.Arrows drawn on the eld lines indicate the direction of the eld.
3.A magnetic eld points from the north to the south pole of a magnet.
15.1.3.3.1 Investigation : Field around a Bar Magnet

Take a bar magnet and place it on a at surface. Place a sheet of white paper over the bar magnet and
sprinkle some iron lings onto the paper. Give the paper a shake to evenly distribute the iron lings. In
your workbook, draw the bar magnet and the pattern formed by the iron lings. Draw the pattern formed
when you rotate the bar magnet to a dierent angle as shown below.
Figure 15.8
As the activity shows, one can map the magnetic eld of a magnet by placing it underneath a piece of
paper and sprinkling iron lings on top. The iron lings line themselves up parallel to the magnetic eld.
15.1.3.3.2 Investigation : Field around a Pair of Bar Magnets

Take two bar magnets and place them a short distance apart such that they are repelling each other. Place
a sheet of white paper over the bar magnets and sprinkle some iron lings onto the paper. Give the paper
a shake to evenly distribute the iron lings. In your workbook, draw both the bar magnets and the pattern
formed by the iron lings. Repeat the procedure for two bar magnets attracting each other and draw what
the pattern looks like for this situation. Make a note of the shape of the lines formed by the iron lings, as
well as their size and their direction for both arrangements of the bar magnet. What does the pattern look
like when you place both bar magnets side by side?

269
Figure 15.9
As already said, opposite poles of a magnet attract each other and bringing them together causes their
magnetic eld lines to converge (come together). Like poles of a magnet repel each other and bringing them
together causes their magnetic eld lines to diverge (bend out from each other).
Figure 15.10
Figure 15.11
15.1.3.3.3 Ferromagnetism and Retentivity

Ferromagnetism is a phenomenon shown by materials like iron, nickel or cobalt. These materials can form
permanent magnets. They always magnetise so as to be attracted to a magnet, no matter which magnetic
pole is brought toward the unmagnetised iron/nickel/cobalt.
The ability of a ferromagnetic material to retain its magnetisation after an external eld is removed is
retentivity.
called its
Paramagnetic materials are materials like aluminium or platinum, which become magnetised in an
external magnetic eld in a similar way to ferromagnetic materials. However, they lose their magnetism
when the external magnetic eld is removed.
Diamagnetism is shown by materials like copper or bismuth, which become magnetised in a magnetic
eld with a polarity opposite to the external magnetic eld. Unlike iron, they are slightly repelled by a
magnet.
15.2 The Earth's magnetic eld 2
15.2.1 The compass and the earth's magnetic eld

A compass is an instrument which is used to nd the direction of a magnetic eld. A compass consists of
a small metal needle which is magnetised itself and which is free to turn in any direction. Therefore, when
in the presence of a magnetic eld, the needle is able to line up in the same direction as the eld.

Figure 15.12
note: Lodestone, a magnetised form of iron-oxide, was found to orientate itself in a north-south
direction if left free to rotate by suspension on a string or on a oat in water. Lodestone was
therefore used as an early navigational compass.
Compasses are mainly used in navigation to nd direction on the earth. This works because the earth itself
has a magnetic eld which is similar to that of a bar magnet (see the picture below). The compass needle
aligns with the earth's magnetic eld direction and points north-south. Once you know where north is, you
can gure out any other direction. A picture of a compass is shown below:
Figure 15.13
Some animals can detect magnetic elds, which helps them orientate themselves and navigate. Animals
which can do this include pigeons, bees, Monarch butteries, sea turtles and certain sh.
15.2.1.1 The earth's magnetic eld

In the picture below, you can see a representation of the earth's magnetic eld which is very similar to the
magnetic eld of a giant bar magnet like the one on the right of the picture. So the earth has two sets of
north poles and south poles: geographic poles and magnetic poles.
Figure 15.14
The earth's magnetic eld is thought to be caused by owing liquid metals in the outer core which causes
electric currents and a magnetic eld. From the picture you can see that the direction of magnetic north and
true north are not identical. The geographic north pole, which is the point through which the earth's
rotation axis goes, is about 11,5
o
away from the direction of the magnetic north pole (which is where a
compass will point). However, the magnetic poles shift slightly all the time.
Another interesting thing to note is that if we think of the earth as a big bar magnet, and we know
that magnetic eld lines always point from north to south, then the compass tells us that what we call the
magnetic north pole is actually the south pole of the bar magnet!
note: The direction of the earth's magnetic eld ips direction about once every 200 000 years!
You can picture this as a bar magnet whose north and south pole periodically switch sides. The
reason for this is still not fully understood.

271
The earth's magnetic eld is very important for humans and other animals on earth because it stops elec-
trically charged particles emitted by the sun from hitting the earth and us. Charged particles can also
damage and cause interference with telecommunications (such as cell phones). The stream of charged par-
ticles (mainly protons and electrons) coming from the sun is called the solar wind. These particles spiral in
the earth's magnetic eld towards the poles. If they collide with particles in the earth's atmosphere they
sometimes cause red or green lights or a glow in the sky which is called the aurora. This happens close to
the north and south pole and so we cannot see the aurora from South Africa.
This simulation shows you the Earth's magnetic eld and a compass.
Figure 15.15
3
run demo
15.2.2 Summary
1. Magnets have two poles - North and South.
2. Some substances can be easily magnetised.
3. Like poles repel each other and unlike poles attract each other.
4. The Earth also has a magnetic eld.
5. A compass can be used to nd the magnetic north pole and help us nd our direction.
This video provides a summary of the work covered in this chapter.
Khan academy video on magnets

<http://www.youtube.com/v/8Y4JSp5U82I&rel=0>
Figure 15.16
15.2.3 End of chapter exercises

1. Describe what is meant by the term magnetic eld.
4
2. Use words and pictures to explain why permanent magnets have a magnetic eld around them. Refer
to domains in your explanation.
5
3. What is a magnet?
6
3 http://phet.colorado.edu/sims/faraday/magnet-and-compass_en.jnlp
4 http://www.fhsst.org/llS
5 http://www.fhsst.org/lia
6 http://www.fhsst.org/lix

4. What happens to the poles of a magnet if it is cut into pieces?

7
5. What happens when like magnetic poles are brought close together?
8
6. What happens when unlike magnetic poles are brought close together?
9
7. Draw the shape of the magnetic eld around a bar magnet.
10
8. Explain how a compass indicates the direction of a magnetic eld.
11
9. Compare the magnetic eld of the Earth to the magnetic eld of a bar magnet using words and
diagrams.
12
10. Explain the dierence between the geographical north pole and the magnetic north pole of the Earth.
13
11. Give examples of phenomena that are aected by Earth's magnetic eld.
14
12. Draw a diagram showing the magnetic eld around the Earth.
15
7 http://www.fhsst.org/lic
8 http://www.fhsst.org/liO
9 http://www.fhsst.org/li3
10 http://www.fhsst.org/lii
11 http://www.fhsst.org/llu
12 http://www.fhsst.org/lil
13 http://www.fhsst.org/llh
14 http://www.fhsst.org/liJ
15 http://www.fhsst.org/llt

273

Step 1. We are required to determine whether the two magnets will repel each other or be attracted to each
other.
Step 2. We are given two magnets with the N pole of one approaching the N pole of the other.
Step 3. Since both poles are the same, the magnets will repel each other.

Step 1. We are required to determine whether the two magnets will repel each other or be attracted to each
other.
Step 2. We are given two magnets with the N pole of one approaching the S pole of the other.
Step 3. Since both poles are the dierent, the magnets will be attracted to each other.


Chapter 16
Electrostatics
16.1.1 Introduction
Electrostatics is the study of electric charge which is static (not moving).
16.1.2 Two kinds of charge

All objects surrounding us (including people!) contain large amounts of electric charge. There are two types
of electric charge: positive charge and negative charge. If the same amounts of negative and positive
charge are brought together, they neutralise each other and there is no net charge. Neutral objects are
objects which contain equal amouts of positive and negative charges. However, if there is a little bit more
of one type of charge than the other on the object then the object is said to be electrically charged. The
picture below shows what the distribution of charges might look like for a neutral, positively charged and
negatively charged object.
Figure 16.1
16.1.3 Unit of charge

Charge is measured in units called coulombs (C). A coulomb of charge is a very large charge. In electro-
statics we therefore often work with charge in microcoulombs (1 µC = 1 × 10−6 C) and nanocoulombs (1 nC
= 1 × 10−9 C).
16.1.4 Conservation of charge

Objects may become charged in many ways, including by contact with or being rubbed by other objects.
This means that they can gain extra negative or positive charge. For example, charging happens when you
rub your feet against the carpet. When you then touch something metallic or another person, you feel a
shock as the excess charge that you have collected is discharged.
275
276 CHAPTER 16. ELECTROSTATICS
tip: Charge, like energy, cannot be created or destroyed. We say that charge is conserved.
When you rub your feet against the carpet, negative charge is transferred to you from the carpet. The carpet
will then become positively charged by the same amount.
Another example is to take two neutral objects such as a plastic ruler and a cotton cloth (handkerchief ).
To begin, the two objects are neutral (i.e. have the same amounts of positive and negative charge).
Figure 16.2
Now, if the cotton cloth is used to rub the ruler, negative charge is transferred from the cloth to the ruler.
The ruler is now negatively charged and the cloth is positively charged. If you count up all the positive and
negative charges at the beginning and the end, there are still the same amount. i.e. total charge has been
conserved !
Figure 16.3
Note that in this example the numbers are made up to be easy to calculate. In the real world only a
tiny fraction of the charges would move from one object to the other, but the total charge would still be
conserved.
The following simulation will help you understand what happens when you rub an object against another
object.
2
run demo
Figure 16.4
16.2 Forces between charges 3
16.2.1 Force between Charges

The force exerted by non-moving (static) charges on each other is called the electrostatic force. The
electrostatic force between:
• like charges are repulsive

2 http://phet.colorado.edu/sims/balloons/balloons_en.jnlp

277
• opposite (unlike) charges are attractive.

In other words, like charges repel each other while opposite charges attract each other. This is dierent to
the gravitational force which is only attractive.
Figure 16.5
The closer together the charges are, the stronger the electrostatic force between them.
Figure 16.6
16.2.1.1 Experiment : Electrostatic Force

You can easily test that like charges repel and unlike charges attract each other by doing a very simple
experiment.
Take a glass rod and rub it with a piece of silk, then hang it from its middle with a piece string so that
it is free to move. If you then bring another glass rod which you have also charged in the same way next to
it, you will see the rod on the string turn away from the rod in your hand i.e. it is repelled. If, however,
you take a plastic rod, rub it with a piece of fur and then bring it close to the rod on the string, you will see
the rod on the string turn towards the rod in your hand i.e. it is attracted.
Figure 16.7
This happens because when you rub the glass with silk, tiny amounts of negative charge are transferred
from the glass onto the silk, which causes the glass to have less negative charge than positive charge, making
it positively charged. When you rub the plastic rod with the fur, you transfer tiny amounts of negative
charge onto the rod and so it has more negative charge than positive charge on it, making it negatively
charged.
Exercise 16.2.1: Application of electrostatic forces (Solution on p. 286.)
Two charged metal spheres hang from strings and are free to move as shown in the picture below.
The right hand sphere is positively charged. The charge on the left hand sphere is unknown.

Figure 16.8
The left sphere is now brought close to the right sphere.
1. If the left hand sphere swings towards the right hand sphere, what can you say about the
charge on the left sphere and why?
2. If the left hand sphere swings away from the right hand sphere, what can you say about the
charge on the left sphere and why?
aside: The electrostatic force determines the arrangement of charge on the surface of conductors.
This is possible because charges can move inside a conductive material. When we place a charge on
a spherical conductor the repulsive forces between the individual like charges cause them to spread
uniformly over the surface of the sphere. However, for conductors with non-regular shapes, there is
a concentration of charge near the point or points of the object. Notice in Figure 16.9 that we show
a concentration of charge with more − or + signs, while we represent uniformly spread charges
with uniformly spaced − or + signs.
Figure 16.9
This collection of charge can actually allow charge to leak o the conductor if the point is sharp
enough. It is for this reason that buildings often have a lightning rod on the roof to remove any
charge the building has collected. This minimises the possibility of the building being struck by
lightning. This spreading out of charge would not occur if we were to place the charge on an
insulator since charge cannot move in insulators.
note: The word 'electron' comes from the Greek word for amber. The ancient Greeks observed
that if you rubbed a piece of amber, you could use it to pick up bits of straw.
16.3 Conductors and insulators 4
16.3.1 Conductors and insulators

All atoms are electrically neutral i.e. they have the same amounts of negative and positive charge inside
them. By convention, the electrons carry negative charge and the protons carry positive charge. The basic
unit of charge, called the elementary charge, e, is the amount of charge carried by one electron.

279
All the matter and materials on earth are made up of atoms. Some materials allow electrons to move
relatively freely through them (e.g. most metals, the human body). These materials are called conductors.
Other materials do not allow the charge carriers, the electrons, to move through them (e.g. plastic,
glass). The electrons are bound to the atoms in the material. These materials are called non-conductors
or insulators.
If an excess of charge is placed on an insulator, it will stay where it is put and there will be a concentration
of charge in that area of the object. However, if an excess of charge is placed on a conductor, the like charges
will repel each other and spread out over the outside surface of the object. When two conductors are made
to touch, the total charge on them is shared between the two. If the two conductors are identical, then each
conductor will be left with half of the total charge.
aside: The basic unit of charge, namely the elementary charge is carried by the electron (equal
−19
to 1.602×10 C!). In a conducting material (e.g. copper), when the atoms bond to form the
material, some of the outermost, loosely bound electrons become detached from the individual
atoms and so become free to move around. The charge carried by these electrons can move around
in the material. In insulators, there are very few, if any, free electrons and so the charge cannot
move around in the material.
note: In 1909 Robert Millikan and Harvey Fletcher measured the charge on an electron. This
experiment is now known as Millikan's oil drop experiment. Millikan and Fletcher sprayed oil
droplets into the space between two charged plates and used what they knew about forces and in
particular the electric force to determine the charge on an electron.
Exercise 16.3.1: Conducting spheres and movement of charge (Solution on p. 286.)

I have 2 charged metal conducting spheres which are identical except for having dierent charge.
Sphere A has a charge of -5 nC and sphere B has a charge of -3 nC. I then bring the spheres together
so that they touch each other. Afterwards I move the two spheres apart so that they are no longer
touching.
1. What happens to the charge on the two spheres?

2. What is the nal charge on each sphere?
16.3.1.1 The electroscope

The electroscope is a very sensitive instrument which can be used to detect electric charge. A diagram of a
gold leaf electroscope is shown the gure below. The electroscope consists of a glass container with a metal
rod inside which has 2 thin pieces of gold foil attached. The other end of the metal rod has a metal plate
attached to it outside the glass container.
Figure 16.10
The electroscope detects charge in the following way: A charged object, like the positively charged rod
in the picture, is brought close to (but not touching) the neutral metal plate of the electroscope. This causes
negative charge in the gold foil, metal rod, and metal plate, to be attracted to the positive rod. Because

the metal (gold is a metal too!) is a conductor, the charge can move freely from the foil up the metal rod
and onto the metal plate. There is now more negative charge on the plate and more positive charge on the
gold foil leaves. This is called inducing a charge on the metal plate. It is important to remember that the
electroscope is still neutral (the total positive and negative charges are the same), the charges have just been
induced to move to dierent parts of the instrument! The induced positive charge on the gold leaves forces
them apart since like charges repel! This is how we can tell that the rod is charged. If the rod is now moved
away from the metal plate, the charge in the electroscope will spread itself out evenly again and the leaves
will fall down because there will no longer be an induced charge on them.
16.3.1.1.1 Grounding
If you were to bring the charged rod close to the uncharged electroscope, and then you touched the metal
plate with your nger at the same time, this would cause charge to ow up from the ground (the earth),
through your body onto the metal plate. Connecting to the earth so charge ows is called grounding. The
charge owing onto the plate is opposite to the charge on the rod, since it is attracted to the charge on the
rod. Therefore, for our picture, the charge owing onto the plate would be negative. Now that charge has
been added to the electroscope, it is no longer neutral, but has an excess of negative charge. Now if we
move the rod away, the leaves will remain apart because they have an excess of negative charge and they
repel each other. If we ground the electroscope again (this time without the charged rod nearby), the excess
charge will ow back into the earth, leaving it neutral.
Figure 16.11
16.3.2 Attraction between charged and uncharged objects

16.3.2.1 Polarisation of Insulators
Unlike conductors, the electrons in insulators (non-conductors) are bound to the atoms of the insulator and
cannot move around freely through the material. However, a charged object can still exert a force on a
neutral insulator due to a phenomenon called polarisation.
If a positively charged rod is brought close to a neutral insulator such as polystyrene, it can attract the
bound electrons to move round to the side of the atoms which is closest to the rod and cause the positive
nuclei to move slightly to the opposite side of the atoms. This process is called polarisation. Although it
is a very small (microscopic) eect, if there are many atoms and the polarised object is light (e.g. a small
polystyrene ball), it can add up to enough force to cause the object to be attracted onto the charged rod.
Remember, that the polystyrene is only polarised, not charged. The polystyrene ball is still neutral since no
charge was added or removed from it. The picture shows a not-to-scale view of the polarised atoms in the
polystyrene ball:
Figure 16.12

281
Some materials are made up of molecules which are already polarised. These are molecules which have a
more positive and a more negative side but are still neutral overall. Just as a polarised polystyrene ball can
be attracted to a charged rod, these materials are also aected if brought close to a charged object.
Water is an example of a substance which is made of polarised molecules. If a positively charged rod is
brought close to a stream of water, the molecules can rotate so that the negative sides all line up towards
the rod. The stream of water will then be attracted to the rod since opposite charges attract.
16.3.3 Summary
1. Objects can be positively charged, negatively charged or neutral.
2. Objects that are neutral have equal numbers of positive and negative charge.
3. Unlike charges are attracted to each other and like charges are repelled from each other.
4. Charge is neither created nor destroyed, it can only be transferred.
5. Charge is measured in coulombs (C).
6. Conductors allow charge to move through them easily.
7. Insulators do not allow charge to move through them easily.
The following presentation is a summary of the work covered in this chapter. Note that the last two slides
are not needed for exam purposes, but are included for general interest.

<http://static.slidesharecdn.com/swf/ssplayer2.swf ?doc=electrostatics-100511052714-
phpapp02&stripped_title=electrostatics-4047687>
Figure 16.13
16.3.4 End of chapter exercise

1. What are the two types of charge called?
5
2. Provide evidence for the existence of two types of charge.
6
3. Fill in the blanks: The electrostatic force between like charges is while the electrostatic force
between opposite charges is .
7
4. I have two positively charged metal balls placed 2 m apart.
a. Is the electrostatic force between the balls attractive or repulsive?

b. If I now move the balls so that they are 1 m apart, what happens to the strength of the electrostatic
force between them?
8
5. I have 2 charged spheres each hanging from string as shown in the picture below.
5 http://www.fhsst.org/lqs
6 http://www.fhsst.org/lqo
7 http://www.fhsst.org/lqA
8 http://www.fhsst.org/lqG

Figure 16.14
Choose the correct answer from the options below: The spheres will
a. swing towards each other due to the attractive electrostatic force between them.
b. swing away from each other due to the attractive electrostatic force between them.
c. swing towards each other due to the repulsive electrostatic force between them.
d. swing away from each other due to the repulsive electrostatic force between them.
9
6. Describe how objects (insulators) can be charged by contact or rubbing.
10
7. You are given a perspex ruler and a piece of cloth.
a. How would you charge the perspex ruler?

b. Explain how the ruler becomes charged in terms of charge.
c. How does the charged ruler attract small pieces of paper?
11
IEB 2005/11 HG An uncharged hollow metal sphere is placed on an insulating stand. A positively charged rod is brought
up to touch the hollow metal sphere at P as shown in the diagram below. It is then moved away from
the sphere.
Figure 16.15
Where is the excess charge distributed on the sphere after the rod has been removed?
a. It is still located at point P where the rod touched the sphere.

b. It is evenly distributed over the outer surface of the hollow sphere.
c. It is evenly distributed over the outer and inner surfaces of the hollow sphere.
d. No charge remains on the hollow sphere.
12
8. What is the process called where molecules in an uncharged object are caused to align in a particular
direction due to an external charge?
13
9. Explain how an uncharged object can be attracted to a charged object. You should use diagrams to
illustrate your answer.
14
9 http://www.fhsst.org/lqf
10 http://www.fhsst.org/lqW
11 http://www.fhsst.org/lqv
12 http://www.fhsst.org/lqd
13 http://www.fhsst.org/lqp
14 http://www.fhsst.org/lqP

283
10. Explain how a stream of water can be attracted to a charged rod.

15


Step 1. In the rst case, we have a sphere with positive charge which is attracting the left charged sphere. We
need to nd the charge on the left sphere.
Step 2. We are dealing with electrostatic forces between charged objects. Therefore, we know that like charges
repel each other and opposite charges attract each other.
Step 3. a. In the rst case, the positively charged sphere is attracting the left sphere. Since an electrostatic
force between unlike charges is attractive, the left sphere must be negatively charged.
b. In the second case, the positively charged sphere repels the left sphere. Like charges repel each
other. Therefore, the left sphere must now also be positively charged.

Step 1. We have two identical negatively charged conducting spheres which are brought together to touch each
other and then taken apart again. We need to explain what happens to the charge on each sphere and
what the nal charge on each sphere is after they are moved apart.
Step 2. We know that the charge carriers in conductors are free to move around and that charge on a conductor
spreads itself out on the surface of the conductor.
Step 3. a. When the two conducting spheres are brought together to touch, it is as though they become one
single big conductor and the total charge of the two spheres spreads out across the whole surface
of the touching spheres. When the spheres are moved apart again, each one is left with half of
the total original charge.
b. Before the spheres touch, the total charge is: -5 nC + (-3) nC = -8 nC. When they touch they
share out the -8 nC across their whole surface. When they are removed from each other, each is
left with half of the original charge:
−8 nC /2 = −4 nC (16.1)
on each sphere.

Chapter 17
Electric circuits
17.1 Key concepts 1
17.1.1 Electric Circuits

People all over the world depend on electricity to provide power for most appliances in the home and at work.
For example, ourescent lights, electric heating and cooking (on electric stoves), all depend on electricity to
work. To realise just how big an impact electricity has on our daily lives, just think about what happens
when there is a power failure or load shedding.
17.1.1.1 Discussion : Uses of electricity

With a partner, take the following topics and, for each topic, write down at least 5 items/appliances/machines
which need electricity to work. Try not to use the same item more than once.
• At home
• At school
• At the hospital
• In the city
Once you have nished making your lists, compare with the lists of other people in your class. (Save your
lists somewhere safe for later because there will be another activity for which you'll need them.)
When you start comparing, you should notice that there are many dierent items which we use in our
daily lives which rely on electricity to work!
tip: Safety Warning: We believe in experimenting and learning about physics at every opportu-
nity, BUT playing with electricity and electrical appliances can be EXTREMELY DANGER-
OUS! Do not try to build home made circuits alone. Make sure you have someone with you who
knows if what you are doing is safe. Normal electrical outlets are dangerous. Treat electricity with
respect in your everyday life. Do not touch exposed wires and do not approach downed power lines.
17.1.1.2 Closed circuits

In the following activity we will investigate what is needed to cause charge to ow in an electric circuit.
285
286 CHAPTER 17. ELECTRIC CIRCUITS
17.1.1.2.1 Experiment : Closed circuits

Aim:
To determine what is required to make electrical charges ow. In this experiment, we will use a lightbulb
to check whether electrical charge is owing in the circuit or not. If charge is owing, the lightbulb should
glow. On the other hand, if no charge is owing, the lightbulb will not glow.
Apparatus:
You will need a small lightbulb which is attached to a metal conductor (e.g. a bulb from a school electrical
kit), some connecting wires and a battery.
Method:
Take the apparatus items and try to connect them in a way that you cause the light bulb to glow (i.e.
charge ows in the circuit).
Questions:
1. Once you have arranged your circuit elements to make the lightbulb glow, draw your circuit.
2. What can you say about how the battery is connected? (i.e. does it have one or two connecting leads
attached? Where are they attached?)
3. What can you say about how the light bulb is connected in your circuit? (i.e. does it connect to one
or two connecting leads, and where are they attached?)
4. Are there any items in your circuit which are not attached to something? In other words, are there
any gaps in your circuit?
Write down your conclusion about what is needed to make an electric circuit work and charge to ow.
In the experiment above, you will have seen that the light bulb only glows when there is a closed circuit
i.e. there are no gaps in the circuit and all the circuit elements are connected in a closed loop. Therefore, in
order for charges to ow, a closed circuit and an energy source (in this case the battery) are needed. (Note:
you do not have to have a lightbulb in the circuit! We used this as a check that charge was owing.)
Denition 17.1: Electric circuit

An electric circuit is a closed path (with no breaks or gaps) along which electrical charges (electrons)
ow powered by an energy source.
17.1.1.3 Representing electric circuits

17.1.1.3.1 Components of electrical circuits
Some common elements (components) which can be found in electrical circuits include light bulbs, batteries,
connecting leads, switches, resistors, voltmeters and ammeters. You will learn more about these items in
later sections, but it is important to know what their symbols are and how to represent them in circuit
diagrams. Below is a table with the items and their symbols:

287
Component Symbol
light bulb
Figure 17.1
battery
Figure 17.2
switch
Figure 17.3
resistor
Figure 17.4
continued on next page

OR
Figure 17.5
voltmeter
Figure 17.6
ammeter
Figure 17.7
connecting lead
Figure 17.8
Table 17.1
17.1.1.3.2 Circuit diagrams

Denition 17.2: Representing circuits
A physical circuit is the electric circuit you create with real components.
A circuit diagram is a drawing which uses symbols to represent the dierent components in
the physical circuit.
We use circuit diagrams to represent circuits because they are much simpler and more general than
drawing the physical circuit because they only show the workings of the electrical components. You can see

289
this in the two pictures below. The rst picture shows the physical circuit for an electric torch. You can see
the light bulb, the batteries, the switch and the outside plastic casing of the torch. The picture is actually
a cross-section of the torch so that we can see inside it.
Figure 17.9: Physical components of an electric torch. The dotted line shows the path of the electrical
circuit.
Below is the circuit diagram for the electric torch. Now the light bulb is represented by its symbol, as
are the batteries, the switch and the connecting wires. It is not necessary to show the plastic casing of the
torch since it has nothing to do with the electric workings of the torch. You can see that the circuit diagram
is much simpler than the physical circuit drawing!
Figure 17.10: Circuit diagram of an electric torch.
17.1.1.3.3 Series and parallel circuits

There are two ways to connect electrical components in a circuit: in series or in parallel.
Denition 17.3: Series circuit
In a series circuit, the charge owing from the battery can only ow along a single path to return
to the battery.
Denition 17.4: Parallel circuit

In a parallel circuit, the charge owing from the battery can ow along multiple paths to return
to the battery.
The picture below shows a circuit with three resistors connected in series on the left and a circuit with
three resistors connected in parallel on the right. In the series circiut, the charge path from the battery goes
through every component before returning to the battery. In the parallel circuit, there is more than one
path for the charge to ow from the battery through one of the components and back to the battery.
Figure 17.11
This simulation allows you to experiment with building circuits.

Figure 17.12
2
run demo
Exercise 17.1.1: Drawing circuits I (Solution on p. 308.)

Draw the circuit diagram for a circuit which has the following components:
1. 1 battery
2. 1 lightbulb connected in series
3. 2 resistors connected in parallel
Exercise 17.1.2: Drawing circuits II (Solution on p. 308.)

Draw the circuit diagram for a circuit which has the following components:
1. 3 batteries in series
2. 1 lightbulb connected in parallel with 1 resistor
3. a switch in series with the batteries
17.1.1.3.3.1 Circuits
1. Using physical components, set up the physical circuit which is described by the circuit diagram below
and then draw the physical circuit:
Figure 17.13
3
2. Using physical components, set up a closed circuit which has one battery and a light bulb in series
with a resistor.
a. Draw the physical circuit.

b. Draw the resulting circuit diagram.
c. How do you know that you have built a closed circuit? (What happens to the light bulb?)
d. If you add one more resistor to your circuit (also in series), what do you notice? (What happens
to the light from the light bulb?)
e. Draw the new circuit diagram which includes the second resistor.
4
3. Draw the circuit diagram for the following circuit: 2 batteries and a switch in series, and 1 lightbulb
which is in parallel with two resistors.
2 http://phet.colorado.edu/sims/circuit-construction-kit/circuit-construction-kit-dc_en.jnlp
3 http://www.fhsst.org/lqZ
4 http://www.fhsst.org/lqB

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a. Now use physical components to set up the circuit.

b. What happens when you close the switch? What does does this mean about the circuit?
c. Draw the physical circuit.
5
17.1.1.3.3.2 Discussion : Alternative Energy

At the moment, most electric power is produced by burning fossil fuels such as coal and oil. In South Africa,
our main source of electric power is coal burning power stations. (We also have one nuclear power plant called
Koeberg in the Western Cape). However, burning fossil fuels releases large amounts of pollution into the
earth's atmosphere and contributes to global warming. Also, the earth's fossil fuel reserves (especially oil)
are starting to run low. For these reasons, people all across the world are working to nd alternative /other
sources of energy and on ways to conserve /save energy. Other sources of energy include wind power, solar
power (from the sun), hydro-electric power (from water, e.g. dammed rivers) among others.
With a partner, take out the lists you made earlier of the item/appliances/machines which used electricity
in the following environments. For each item, try to think of an alternative AND a way to conserve or save
power.
For example, if you had a ourescent light as an item used in the home, then:
• Alternative: use candles at supper time to reduce electricity consumption

• Conservation: turn o lights when not in a room, or during the day.
Topics:
• At home
• At school
• At the hospital
• In the city
Once you have nished making your lists, compare with the lists of other people in your class.
17.2 Potential dierence 6
17.2.1 Potential Dierence

17.2.1.1 Potential Dierence
When a circuit is connected and complete, charge can move through the circuit. Charge will not move unless
there is a reason, a force. Think of it as though charge is at rest and something has to push it along. This
means that work needs to be done to make charge move. A force acts on the charges, doing work, to make
them move. The force is provided by the battery in the circuit.
We call the moving charge "current" and we will talk about this later.
The position of the charge in the circuit tells you how much potential energy it has because of the force
being exerted on it. This is like the force from gravity, the higher an object is above the ground (position)
the more potential energy it has.
The amount of work to move a charge from one point to another point is how much the potential energy
has changed. This is the dierence in potential energy, called potential dierence. Notice that it is a
dierence between the value of potential energy at two points so we say that potential dierence is measured
between or across two points. We do not say potential dierence through something.
5 http://www.fhsst.org/lqK

Denition 17.5: Potential Dierence

Electrical potential dierence as the dierence in electrical potential energy per unit charge between
7
two points. The units of potential dierence are the volt (V).
The units are volt (V), which is the same as joule per coulomb, the amount of work done per unit charge.
Electrical potential dierence is also called voltage.
17.2.1.2 Potential Dierence and Parallel Resistors

When resistors are connected in parallel the start and end points for all the resistors are the same. These
points have the same potential energy and so the potential dierence between them is the same no matter
what is put in between them. You can have one, two or many resistors between the two points, the potential
dierence will not change. You can ignore whatever components are between two points in a circuit when
calculating the dierence between the two points.
Look at the following circuit diagrams. The battery is the same in all cases, all that changes is more
resistors are added between the points marked by the black dots. If we were to measure the potential
dierence between the two dots in these circuits we would get the same answer for all three cases.
Figure 17.14
Lets look at two resistors in parallel more closely. When you construct a circuit you use wires and you
might think that measuring the voltage in dierent places on the wires will make a dierence. This is not
true. The potential dierence or voltage measurement will only be dierent if you measure a dierent set of
components. All points on the wires that have no circuit components between them will give you the same
measurements.
All three of the measurements shown in the picture below (i.e. AB, CD and EF) will give you the
same voltage. The dierent measurement points on the left have no components between them so there is no
change in potential energy. Exactly the same applies to the dierent points on the right. When you measure
the potential dierence between the points on the left and right you will get the same answer.
Figure 17.15
17.2.1.3 Potential Dierence and Series Resistors

When resistors are in series, one after the other, there is a potential dierence across each resistor. The total
potential dierence across a set of resistors in series is the sum of the potential dierences across each of the
resistors in the set. This is the same as falling a large distance under gravity or falling that same distance
(dierence) in many smaller steps. The total distance (dierence) is the same.
Look at the circuits below. If we measured the potential dierence between the black dots in all of these
circuits it would be the same just like we saw above. So we now know the total potential dierence is the
7 named after the Italian physicist Alessandro Volta (17451827)

293
same across one, two or three resistors. We also know that some work is required to make charge ow
through each one, each is a step down in potential energy. These steps add up to the total drop which we
know is the dierence between the two dots.
Figure 17.16
Let us look at this in a bit more detail. In the picture below you can see what the dierent measurements
for 3 identical resistors in series could look like. The total voltage across all three resistors is the sum of the
voltages across the individual resistors.
Figure 17.17
Khan academy video on circuits - 1

<http://www.youtube.com/v/3o8_EARoMtg&rel=0>
Figure 17.18
17.2.1.4 Ohm's Law
Phet simulation for Ohm's Law

<ohms-law.swf>
Figure 17.19
The voltage is the change in potential energy or work done when charge moves between two points in the
circuit. The greater the resistance to charge moving the more work that needs to be done. The work done
or voltage thus depends on the resistance. The potential dierence is proportional to the resistance.

Denition 17.6: Ohm's Law

Voltage across a circuit component is proportional to the resistance of the component.
Use the fact that voltage is proportional to resistance to calculate what proportion of the total voltage
of a circuit will be found across each circuit element.
Figure 17.20
We know that the total voltage is equal to V1 in the rst circuit, to V1 + V2 in the second circuit and V1
+ V2 + V3 in the third circuit.
We know that the potential energy lost across a resistor is proportional to the resistance of the component.
The total potential dierence is shared evenly across the total resistance of the circuit. This means that the
potential dierence per unit of resistance is
Vtotal
Vper unit of resistance = (17.1)
Rtotal
Then the voltage across a resistor is just the resistance times the potential dierence per unit of resistance
Vtotal
Vresistor = Rresistor · . (17.2)
Rtotal
17.2.1.5 EMF
When you measure the potential dierence across (or between) the terminals of a battery you are measuring
the electromotive force (emf ) of the battery. This is how much potential energy the battery has to make
charges move through the circuit. This driving potential energy is equal to the total potential energy drops
in the circuit. This means that the voltage across the battery is equal to the sum of the voltages in the
circuit.
We can use this information to solve problems in which the voltages across elements in a circuit add up
to the emf.
EM F = Vtotal (17.3)
Exercise 17.2.1: Voltages I (Solution on p. 308.)

What is the voltage across the resistor in the circuit shown?
Figure 17.21
Exercise 17.2.2: Voltages II (Solution on p. 308.)

What is the voltage across the unknown resistor in the circuit shown?

295
Figure 17.22
Exercise 17.2.3: Voltages III (Solution on p. 309.)

What is the voltage across the unknown resistor in the circuit shown?
Figure 17.23
Exercise 17.2.4: Voltages IV (Solution on p. 309.)

What is the voltage across the parallel resistor combination in the circuit shown? Hint: the rest
of the circuit is the same as the previous problem.
Figure 17.24
17.3 Current 8
17.3.1 Current
17.3.1.1 Flow of Charge
We have been talking about moving charge. We need to be able to deal with numbers. How much charge
is moving, how fast is it moving? The concept that represents this information is called current. Current
allows us to quantify the movement of charge.
When we talk about current we talk about how much charge moves past a xed point in circuit in one
second. Think of charges being pushed around the circuit by the battery, there are charges in the wires but
unless there is a battery they won't move. When one charge moves the charges next to it also move. They
keep their spacing as if you had a tube of marbles like in this picture.

Figure 17.25
If you push one marble into the tube one must come out the other side. If you look at any point in the
tube and push one marble into the tube, one marble will move past the point you are looking at. This is
similar to charges in the wires of a circuit.
If one charge moves they all move and the same number move at every point in the circuit. This is due
to the conservation of charge.
17.3.1.2 Current
Now that we've thought about the moving charges and visualised what is happening we need to get back
to quantifying moving charge. I've already told you that we call moving charge current but we still need to
dene it precisely.
Denition 17.7: Current

Current is the rate at which charges moves past a xed point in a circuit. We use the symbol I to
show current and it is measured in amperes (A). One ampere is one coulomb of charge moving in
one second.
Q
I= (17.4)
∆t
When current ows in a circuit we show this on a diagram by adding arrows. The arrows show the
direction of ow in a circuit. By convention we say that charge ows from the positive terminal on a battery
to the negative terminal. We measure current with an ammeter
17.3.1.3 Series Circuits

In a series circuit, the charge has a single path from the battery, returning to the battery.
Figure 17.26
The arrows in this picture show you the direction that charge will ow in the circuit. They don't show
you much charge will ow, only the direction.
note: Benjamin Franklin made a guess about the direction of charge ow when rubbing smooth
wax with rough wool. He thought that the charges owed from the wax to the wool (i.e. from
positive to negative) which was opposite to the real direction. Due to this, electrons are said to
have a negative charge and so objects which Ben Franklin called negative (meaning a shortage
of charge) really have an excess of electrons. By the time the true direction of electron ow was
discovered, the convention of positive and negative had already been so well accepted in the
scientic world that no eort was made to change it.

297
tip: A battery does not produce the same amount of current no matter what is connected to it.
While the voltage produced by a battery is constant, the amount of current supplied depends on
what is in the circuit.
How does the current through the battery in a circuit with several resistors in series compare to the current
in a circuit with a single resistor (assuming all the resistors are the same)?
17.3.1.3.1 Experiment : Current in Series Circuits

Aim:
To determine the eect of multiple resistors on current in a circuit
Apparatus:
• Battery
• Resistors
• Light bulb
• Wires
Method:
1. Construct the following circuits
Figure 17.27
2. Rank the three circuits in terms of the brightness of the bulb.
Conclusions:
The brightness of the bulb is an indicator of how much current is owing. If the bulb gets brighter
because of a change then more current is owing. If the bulb gets dimmer less current is owing. You will
nd that the more resistors you have the dimmer the bulb.
Figure 17.28
17.3.1.4 Parallel Circuits
Figure 17.29

How does the current through the battery in a circuit with several resistors in parallel compare to the current
in a circuit with a single resistor?
17.3.1.4.1 Experiment : Current in Series Circuits

Aim:
To determine the eect of multiple resistors on current in a circuit
Apparatus:
• Battery
• Resistors
• Light bulb
• Wires
Method:
1. Construct the following circuits
Figure 17.30
2. Rank the three circuits in terms of the brightness of the bulb.
Conclusions:
The brightness of the bulb is an indicator of how much current is owing. If the bulb gets brighter
because of a change then more current is owing. If the bulb gets dimmer less current is owing. You will
nd that the more resistors you have the brighter the bulb.
Why is this the case? Why do more resistors make it easier for charge to ow in the circuit? It is because
they are in parallel so there are more paths for charge to take to move. You can think of it like a highway
with more lanes, or the tube of marbles splitting into multiple parallel tubes. The more branches there are,
the easier it is for charge to ow. You will learn more about the total resistance of parallel resistors later
but always remember that more resistors in parallel mean more pathways. In series the pathways come one
after the other so it does not make it easier for charge to ow.
Figure 17.31

299
17.4 Resistance 9
17.4.1 Resistance
17.4.1.1 What causes resistance?
We have spoken about resistors that reduce the ow of charge in a conductor. On a microscopic level,
electrons moving through the conductor collide with the particles of which the conductor (metal) is made.
When they collide, they transfer kinetic energy. The electrons therefore lose kinetic energy and slow down.
This leads to resistance. The transferred energy causes the resistor to heat up. You can feel this directly
if you touch a cellphone charger when you are charging a cell phone - the charger gets warm because its
circuits have some resistors in them!
Denition 17.8: Resistance

Resistance slows down the ow of charge in a circuit. We use the symbol R to show resistance and
it is measured in units called Ohms with the symbol Ω.
Volt
1 Ohm = 1 . (17.5)
Ampere
All conductors have some resistance. For example, a piece of wire has less resistance than a light bulb,
but both have resistance. A lightbulb is a very thin wire surrounded by a glass housing The high resistance
of the lament (small wire) in a lightbulb causes the electrons to transfer a lot of their kinetic energy in the
10
form of heat . The heat energy is enough to cause the lament to glow white-hot which produces light. The
wires connecting the lamp to the cell or battery hardly even get warm while conducting the same amount of
current. This is because of their much lower resistance due to their larger cross-section (they are thicker).
An important eect of a resistor is that it converts electrical energy into other forms of heat energy.
Light energy is a by-product of the heat that is produced.
note: There is a special type of conductor, called a superconductor that has no resistance,
but the materials that make up all known superconductors only start superconducting at very low
◦
temperatures (approximately -170 C).
17.4.1.1.1 Why do batteries go at?

A battery stores chemical potential energy. When it is connected in a circuit, a chemical reaction takes place
inside the battery which converts chemical potential energy to electrical energy which powers the electrons
to move through the circuit. All the circuit elements (such as the conducting leads, resistors and lightbulbs)
have some resistance to the ow of charge and convert the electrical energy to heat and, in the case of the
lightbulb, light. Since energy is always conserved, the battery goes at when all its chemical potential energy
has been converted into other forms of energy.
17.4.1.2 Resistors in electric circuits

It is important to understand what eect adding resistors to a circuit has on the total resistance of a circuit
and on the current that can ow in the circuit.
10 Flourescent lightbulbs do not use thin wires; they use the fact that certain gases glow when a current ows through them.
They are much more ecient (much less resistance) than lightbulbs.

17.4.1.2.1 Resistors in series

When we add resistors in series to a circuit, we increase the resistance to the ow of current. There is only
one path that the current can ow down and the current is the same at all places in the series circuit. Take
a look at the diagram below: On the left there is a circuit with a single resistor and a battery. No matter
where we measure the current, it is the same in a series circuit. On the right, we have added a second resistor
in series to the circuit. The total resistance of the circuit has increased and you can see from the reading on
the ammeter that the current in the circuit has decreased.
Figure 17.32

<http://www.youtube.com/v/7vHh1sfZ5KE&rel=0>
Figure 17.33
17.4.1.2.2 Resistors in parallel

In contrast to the series case, when we add resistors in parallel, we create more paths along which current
can ow. By doing this we decrease the total resistance of the circuit!
Take a look at the diagram below. On the left we have the same circuit as in the previous diagram with
a battery and a resistor. The ammeter shows a current of 1 ampere. On the right we have added a second
resistor in parallel to the rst resistor. This has increased the number of paths (branches) the charge can
take through the circuit - the total resistance has decreased. You can see that the current in the circuit has
increased. Also notice that the current in the dierent branches can be dierent.
Figure 17.34

301

<http://www.youtube.com/v/ZrMw7P6P2Gw&rel=0>
Figure 17.35
17.4.1.2.2.1 Resistance
11
1. What is the unit of resistance called and what is its symbol? Click here for the solution
2. Explain what happens to the total resistance of a circuit when resistors are added in series? Click here
12
for the solution
3. Explain what happens to the total resistance of a circuit when resistors are added in parallel? Click
13
14
4. Why do batteries go at? Click here for the solution
17.5 Measuring devices 15
17.5.1 Instruments to Measure voltage, current and resistance

As we have seen in previous sections, an electric circuit is made up of a number of dierent components such
as batteries, resistors and light bulbs. There are devices to measure the properties of these components.
These devices are called meters.
For example, one may be interested in measuring the amount of current owing through a circuit using
an ammeter or measuring the voltage provided by a battery using a voltmeter. In this section we will discuss
the practical usage of voltmeters, ammeters, and ohmmeters.
17.5.1.1 Voltmeter
A voltmeter is an instrument for measuring the voltage between two points in an electric circuit. In analogy
with a water circuit, a voltmeter is like a meter designed to measure pressure dierence. Since one is interested
in measuring the voltage between two points in a circuit, a voltmeter must be connected in parallel with the
portion of the circuit on which the measurement is made.
Figure 17.36: A voltmeter should be connected in parallel in a circuit.
11 http://www.fhsst.org/lqk
12 http://www.fhsst.org/lq0

Figure 17.36 shows a voltmeter connected in parallel with a battery. One lead of the voltmeter is connected
to one end of the battery and the other lead is connected to the opposite end. The voltmeter may also be
used to measure the voltage across a resistor or any other component of a circuit that has a voltage drop.
17.5.1.2 Ammeter
An ammeter is an instrument used to measure the ow of electric current in a circuit. Since one is interested
in measuring the current owing through a circuit component, the ammeter must be connected in series with
the measured circuit component (Figure 17.37).
Figure 17.37: An ammeter should be connected in series in a circuit.
17.5.1.3 Ohmmeter
An ohmmeter is an instrument for measuring electrical resistance. The basic ohmmeter can function much
like an ammeter. The ohmmeter works by suppling a constant voltage to the resistor and measuring the
current owing through it. The measured current is then converted into a corresponding resistance reading
through Ohm's Law. Ohmmeters only function correctly when measuring resistance over a component that
is not being powered by a voltage or current source. In other words, you cannot measure the resistance of a
component that is already connected to a live circuit. This is because the ohmmeter's accurate indication
depends only on its own source of voltage. The presence of any other voltage across the measured circuit
component interferes with the ohmmeter's operation. Figure 17.38 shows an ohmmeter connected with a
resistor.
Figure 17.38: An ohmmeter should be used when there are no voltages present in the circuit.
17.5.1.4 Meters Impact on Circuit

A good quality meter used correctly will not signicantly change the values it is used to measure. This
means that an ammeter has very low resistance to not slow down the ow of charge. A voltmeter has a very
high resistance so that it does not add another parallel pathway to the circuit for the charge to ow along.
17.5.1.4.1 Investigation : Using meters

If possible, connect meters in circuits to get used to the use of meters to measure electrical quantities. If
the meters have more than one scale, always connect to the largest scale rst so that the meter will not
be damaged by having to measure values that exceed its limits.

303
The table below summarises the use of each measuring instrument that we discussed and the way it
should be connected to a circuit component.
Instrument Measured Quantity Proper Connection

Voltmeter Voltage In Parallel
Ammeter Current In Series
Ohmmeter Resistance Only with Resistor
Table 17.2

<http://www.youtube.com/v/3NcIK0s3IwU&rel=0>
Figure 17.39
The following presentation summarizes the concepts covered in this chapter.

<http://static.slidesharecdn.com/swf/ssplayer2.swf ?doc=electriccurrent-100511051906-
phpapp02&stripped_title=electric-current-4047623&userName=kwarne>
Figure 17.40
17.5.2 Exercises - Electric circuits

1. Write denitions for each of the following:
a. resistor
b. coulomb
c. voltmeter
16
2. Draw a circuit diagram which consists of the following components:
a. 2 batteries in parallel
b. an open switch
c. 2 resistors in parallel
d. an ammeter measuring total current
16 http://www.fhsst.org/lqX

e. a voltmeter measuring potential dierence across one of the parallel resistors

17
3. Complete the table below:
Quantity Symbol Unit of meaurement Symbol of unit

e.g. Distance e.g. d e.g. kilometer e.g. km
Resistance
Current
Potential dierence
Table 17.3
18
SC 2003/11 The emf of a battery can best be explained as the ···
a. rate of energy delivered per unit current
b. rate at which charge is delivered
c. rate at which energy is delivered
d. charge per unit of energy delivered by the battery
19
IEB 2002/11 HG1 Which of the following is the correct denition of the emf of a battery?
a. It is the product of current and the external resistance of the circuit.

b. It is a measure of the cell's ability to conduct an electric current.
c. It is equal to the lost volts in the internal resistance of the circuit.
d. It is the power supplied by the battery per unit current passing through the battery.
20
IEB 2005/11 HG Three identical light bulbs A, B and C are connected in an electric circuit as shown in the diagram
below.
Figure 17.41
a. How bright is bulb A compared to B and C?

b. How bright are the bulbs after switch S has been opened?
c. How do the currents in bulbs A and B change when switch S is opened?
17 http://www.fhsst.org/lql
19 http://www.fhsst.org/lqN
20 http://www.fhsst.org/lqR

305
Current in A Current in B
(a) decreases increases
(b) decreases decreases
(c) increases increases
(d) increases decreases
Table 17.4
21
IEB 2004/11 HG1 When a current I is maintained in a conductor for a time of t, how many electrons with charge e pass
any cross-section of the conductor per second?
a. It
b. It/e
c. Ite
d. e/It
22
21 http://www.fhsst.org/lqn
22 http://www.fhsst.org/lqQ


Step 1.
Figure 17.42
Step 1.
Figure 17.43

Step 1. We have a circuit with a battery and one resistor. We know the voltage across the battery. We want
to nd that voltage across the resistor.
Vbattery = 2V (17.6)
Step 2. We know that the voltage across the battery must be equal to the total voltage across all other circuit
components.
Vbattery = Vtotal (17.7)
There is only one other circuit component, the resistor.
Vtotal = V1 (17.8)
This means that the voltage across the battery is the same as the voltage across the resistor.
Vbattery = Vtotal = V1 (17.9)
Vbattery = Vtotal = V1 (17.10)
V1 = 2V (17.11)

Step 1. We have a circuit with a battery and two resistors. We know the voltage across the battery and one
of the resistors. We want to nd that voltage across the resistor.
VA = 1V (17.13)

307
components that are in series.
The total voltage in the circuit is the sum of the voltages across the individual resistors
Vtotal = VA + VB (17.15)
Using the relationship between the voltage across the battery and total voltage across the resistors
Vbattery = V1 + Vresistor
2V = V1 + 1V (17.17)
V1 = 1V

Step 1. We have a circuit with a battery and three resistors. We know the voltage across the battery and two
of the resistors. We want to nd that voltage across the unknown resistor.
Vknown = VA + VC
(17.19)
= 1V + 4V
components that are in series.
The total voltage in the circuit is the sum of the voltages across the individual resistors
Vtotal = VB + Vknown (17.21)
Vbattery = VB + Vknown
7V = VB + 5V (17.23)
VB = 2V

Step 1. The circuit is the same as the previous example and we know that the voltage dierence between
two points in a circuit does not depend on what is between them so the answer is the same as above
Vparallel = 2V .

Step 2. We have a circuit with a battery and ve resistors (two in series and three in parallel). We know the
voltage across the battery and two of the resistors. We want to nd that voltage across the parallel
resistors, Vparallel .
Vknown = 1V + 4V (17.25)
components.
Voltages only add for components in series. The resistors in parallel can be thought of as a single
component which is in series with the other components and then the voltages can be added.
Vtotal = Vparallel + Vknown (17.27)
Vbattery = Vparallel + Vknown

7V = V1 + 5V (17.29)
Vparallel = 2V

GLOSSARY 309
Glossary
A Acceleration
Acceleration is the rate of change of velocity.
Acid rain
Acid rain refers to the deposition of acidic components in rain, snow and dew. Acid rain occurs
when sulphur dioxide and nitrogen oxides are emitted into the atmosphere, undergo chemical
transformations and are absorbed by water droplets in clouds. The droplets then fall to earth as
rain, snow, mist, dry dust, hail, or sleet. This increases the acidity of the soil and aects the
chemical balance of lakes and streams.
Amplitude
The amplitude is the maximum displacement of a particle from its equilibrium position.
Amplitude
The amplitude of a pulse is a measurement of how far the medium is displaced from rest.
Anti-Node
An anti-node is a point on standing a wave where maximum displacement takes place. A free
end of a rope is an anti-node.
Atomic mass number (A)

The number of protons and neutrons in the nucleus of an atom
Atomic number (Z)

The number of protons in an atom
Atomic orbital
An atomic orbital is the region in which an electron may be found around a single atom.
Attraction and Repulsion

Like poles of magnets repel each other whilst unlike poles attract each other.
Average velocity
Average velocity is the total displacement of a body over a time interval.
B Boiling point
The temperature at which a liquid changes its phase to become a gas. The process is called
evaporation and the reverse process is called condensation
C Chemical change
The formation of new substances in a chemical reaction. One type of matter is changed into
something dierent.
Compound
A substance made up of two or more elements that are joined together in a xed ratio.
Conductivity

310 GLOSSARY
Conductivity is a measure of a solution's ability to conduct an electric current.
Conductors and insulators

A conductor allows the easy movement or ow of something such as heat or electrical charge
through it. Insulators are the opposite to conductors because they inhibit or reduce the ow of
heat, electrical charge, sound etc through them.
Conservation of energy principle

Energy cannot be created or destroyed. It can only be changed from one form to another.
Conservation of Energy
The Law of Conservation of Energy: Energy cannot be created or destroyed, but is merely
changed from one form into another.
Conservation of Mechanical Energy

Law of Conservation of Mechanical Energy: The total amount of mechanical energy in a closed
Constructive interference is when two pulses meet, resulting in a bigger pulse.
Core electrons
All the electrons in an atom, excluding the valence electrons
Critical Angle
◦
The critical angle is the angle of incidence where the angle of reection is 90 . The light must
shine from a dense to a less dense medium.
Current
Current is the rate at which charges moves past a xed point in a circuit. We use the symbol I
to show current and it is measured in amperes (A). One ampere is one coulomb of charge
moving in one second.
Q
I= (17.4)
∆t
D Destructive interference is when two pulses meet, resulting in a smaller pulse.
Displacement
Displacement is the change in an object's position.
Dissociation
Dissociation in chemistry and biochemistry is a general process in which ionic compounds
separate or split into smaller molecules or ions, usually in a reversible manner.
E Electric circuit
An electric circuit is a closed path (with no breaks or gaps) along which electrical charges
(electrons) ow powered by an energy source.
Electrolyte
An electrolyte is a substance that contains free ions and behaves as an electrically conductive
medium. Because they generally consist of ions in solution, electrolytes are also known as ionic
solutions.
Electron conguration
GLOSSARY 311
Electron conguration is the arrangement of electrons in an atom, molecule or other physical

structure.
Element
An element is a substance that cannot be broken down into other substances through chemical
means.
Empirical formula
This is a way of expressing the relative number of each type of atom in a chemical compound. In
most cases, the empirical formula does not show the exact number of atoms, but rather the
simplest ratio of the atoms in the compound.
F Focal Point
The focal point of a mirror is the midpoint of a line segment joining the vertex and the centre of
curvature. It is the position at which all parallel rays are focussed.
Frame of Reference
A frame of reference is a reference point combined with a set of directions.
G Gradient
The gradient of a line can be calculated by dividing the change in the y -value by the change in
the x-value.
∆y
m =
∆x
H Heat of vaporisation
Heat of vaporisation is the energy that is needed to change a given quantity of a substance into a
gas.
Heterogeneous mixture
A heterogeneous mixture is one that is non-uniform and the dierent components of the mixture
can be seen.
Homogeneous mixture
A homogeneous mixture is one that is uniform, and where the dierent components of the
mixture cannot be seen.
I Image
An image is a representation of an object formed by a mirror or lens. Light from the image is
seen.
Instantaneous velocity
Instantaneous velocity is the velocity of a body at a specic instant in time.
Intermolecular force
A force between molecules, which holds them together.
Intramolecular force
The force between the atoms of a molecule, which holds them together.
Ion
An ion is a charged atom. A positively charged ion is called a cation e.g. +
Na , and a negatively
charged ion is called an anion e.g. F
−
. The charge on an ion depends on the number of
electrons that have been lost or gained.

312 GLOSSARY
Isotope
The isotope of a particular element is made up of atoms which have the same number of
protons as the atoms in the original element, but a dierent number of neutrons.
K Kinetic Energy
Kinetic energy is the energy an object has due to its motion.
L Law of Reection
The Law of Reection states that the angle of incidence is equal to the angle of reection.
θi = θr (14.1)
Light ray
Light rays are straight lines with arrows to show the path of light.
M Magnetism
Magnetism is one of the phenomena by which materials exert attractive or repulsive forces on
other materials.
Medium
A medium is the substance or material in which a wave will move.
Melting point
The temperature at which a solid changes its phase or state to become a liquid. The process is
called melting and the reverse process (change in phase from liquid to solid) is called freezing.
Mixture
A mixture is a combination of two or more substances, where these substances are not bonded
(or joined) to each other.
Model
A model is a representation of a system in the real world. Models help us to understand systems
and their properties. For example, an atomic model represents what the structure of an atom
could look like, based on what we know about how atoms behave. It is not necessarily a true
picture of the exact structure of an atom.
Molecular formula
This is a concise way of expressing information about the atoms that make up a particular
chemical compound. The molecular formula gives the exact number of each type of atom in the
molecule.
Molecule
A molecule is a group of two or more atoms that are attracted to each other by relatively strong
forces or bonds.
N Node
A node is a point on a standing wave where no displacement takes place at any time. A xed
end of a rope is a node.
O Ohm's Law
Voltage across a circuit component is proportional to the resistance of the component.

GLOSSARY 313
P Parallel circuit
In a parallel circuit, the charge owing from the battery can ow along multiple paths to return
to the battery.
Peaks and troughs

A peak is a point on the wave where the displacement of the medium is at a maximum. A point
on the wave is a trough if the displacement of the medium at that point is at a minimum.
pH
pH is a measure of the acidity or alkalinity of a solution. The pH scale ranges from 0 to 14.
Solutions with a pH less than seven are acidic, while those with a pH greater than seven are
basic (alkaline). pH 7 is considered neutral.
Physical change
A change that can be seen or felt, but that doesn't involve the break up of the particles in the
reaction. During a physical change, the form of matter may change, but not its identity. A
change in temperature is an example of a physical change.
Physical Quantity
A physical quantity is anything that you can measure. For example, length, temperature,
distance and time are physical quantities.
Position
Position is a measurement of a location, with reference to an origin.
Potential Dierence
Electrical potential dierence as the dierence in electrical potential energy per unit charge
23
between two points. The units of potential dierence are the volt (V).
Potential energy
Potential energy is the energy an object has due to its position or state.
Precipitate
A precipitate is the solid that forms in a solution during a chemical reaction.
Prex
A prex is a group of letters that are placed in front of a word. The eect of the prex is to
change meaning of the word. For example, the prex un is often added to a word to mean not,
as in unnecessary which means not necessary.
Pulse
A pulse is a single disturbance that moves through a medium.
Pulse Speed
Pulse speed is the distance a pulse travels per unit time.
R Real Image
A real image can be cast on a screen; it is inverted, and on the same side of the mirror as the
object.
Refraction
Refraction is the bending of light that occurs because light travels at dierent speeds in dierent
materials.
23 named after the Italian physicist Alessandro Volta (17451827)

314 GLOSSARY
Refractive Index
The refractive index (symbol n) of a material is the ratio of the speed of light in a vacuum to its
speed in the material and gives an indication of how dicult it is for light to get through the
material.
c
n= (14.3)
v
where
n = refractive index (no unit)
c = speed of light in a vacuum (3, 00 × 108 m · s−1 )

v = speed of light in a given medium ( m · s−1 )
Table 14.1
Refractive Index
The refractive index of a material is the ratio of the speed of light in a vacuum to its speed in
the medium.
Relative atomic mass

Relative atomic mass is the average mass of one atom of all the naturally occurring isotopes of a
particular chemical element, expressed in atomic mass units.
Representing circuits
A physical circuit is the electric circuit you create with real components.
A circuit diagram is a drawing which uses symbols to represent the dierent components in
the physical circuit.
Resistance
Resistance slows down the ow of charge in a circuit. We use the symbol R to show resistance
and it is measured in units called Ohms with the symbol Ω.
Volt
1 Ohm = 1 . (17.5)
Ampere
S Series circuit
In a series circuit, the charge owing from the battery can only ow along a single path to
return to the battery.
SI Units
The name SI units comes from the French Système International d'Unités, which means
international system of units.
Snell's Law
n1 sinθ1 = n2 sinθ2 (14.5)
where
θ1 = Angle of incidence
θ2 = Angle of refraction

GLOSSARY 315
Table 14.3
Specic heat
Specic heat is the amount of heat energy that is needed to increase the temperature of a
substance by one degree.
T The frequency is the number of successive peaks (or troughs) passing a given point in 1
second.
The law of conservation of mass

The mass of a closed system of substances will remain constant, regardless of the processes
acting inside the system. Matter can change form, but cannot be created or destroyed. For any
chemical process in a closed system, the mass of the reactants must equal the mass of the
products.
The nitrogen cycle

The nitrogen cycle is a biogeochemical cycle that describes how nitrogen and nitrogen-containing
compounds are changed in nature.
The period (T) is the time taken for two successive peaks (or troughs) to pass a xed
point.
The Water Cycle

The water cycle is the continuous circulation of water across the Earth. The water cycle is
driven by solar radiation and it includes the atmosphere, land, surface water and groundwater.
As water moves through the cycle, it changes state between liquid, solid, and gas phases. The
actual movement of water from one part of the cycle to another (e.g. from river to ocean) is the
result of processes such as evaporation, precipitation, inltration and runo.
Total Internal Reection

Total internal reection takes place when light is reected back into the medium because the
angle of incidence is greater than the critical angle.
Transverse wave
A transverse wave is a wave where the movement of the particles of the medium is perpendicular
to the direction of propagation of the wave.
V Valence electrons
The electrons in the outer energy level of an atom
Vectors and Scalars

A vector is a physical quantity with magnitude (size) and direction. A scalar is a physical
quantity with magnitude (size) only.
Velocity
Velocity is the rate of change of displacement.
Virtual Image
A virtual image is upright, on the opposite side of the mirror as the object, and light does not
actually reach it.
W Water hardness
316 GLOSSARY
Water hardness is a measure of the mineral content of water. Minerals are substances such as
calcite, quartz and mica that occur naturally as a result of geological processes.
Wave
A wave is a periodic, continuous disturbance that consists of a train of pulses.
Wavelength of wave
The wavelength of a wave is the distance between any two adjacent points that are in phase.

INDEX 317
Index of Keywords and Terms

Keywords are listed by the section with that keyword (page numbers are in parentheses). Keywords
do not necessarily appear in the text of the page. They are merely associated with that section. Ex.
apples, 1.1 (1) Terms are referenced by the page they appear on. Ex. apples, 1
A acceleration, 10.4(138), 138 Conservation of Mechanical Energy, 176

acceleration due to gravity, 11.1(167) constant composition, 4.2(62)
Acid rain, 105 Constructive interference is when two pulses
ammeter, 17.5(303) meet, resulting in a bigger pulse., 196
Amplitude, 188, 205 Core electrons, 45
Anti-Node, 218 Critical Angle, 255
aqueous solution, 8.2(101) current, 17.3(297), 298
D
Atomic mass number (A), 36
description of motion, 10.5(140)
atomic models, 3.1(31)
Destructive interference is when two pulses
Atomic number (Z), 36
meet, resulting in a smaller pulse., 196
Atomic orbital, 43
displacement, 10.2(132), 132
atomic size, 3.1(31)
Dissociation, 102
atoms, 2.1(17)
distance, 10.2(132)
Attraction and Repulsion, 269
Average velocity, 135
E Earth's magnetic eld, 15.2(271)
B balancing chemical equations, 5.2(68)

Electric circuit, 288
electric circuits, 17.1(287), 17.2(293),
Boiling point, 25
17.3(297), 17.4(301), 17.5(303)
boundary conditions, 12.3(193), 13.3(213)
Electrolyte, 106
C chemical change, 4.1(57), 58, 4.2(62) electrolytes, 8.3(106)
chemical equations, 5.2(68), 5.3(71) electron conguration, 3.4(41), 43
chemical formulae, 1.2(6), 5.1(67) electrons, 3.2(34)
chemical symbols, 5.1(67) electrostatics, 16.1(277), 16.2(278),
chemistry, 1.1(1), 1.2(6), 1.3(9), 16.3(280)
2.1(17), 2.2(21), 2.3(25), 3.1(31), Element, 5
3.2(34), 3.3(38), 3.4(41), 3.5(47), elements, 1.1(1)
4.1(57), 4.2(62), 5.1(67), 5.2(68), Empirical formula, 18
5.3(71), 6.1(77), 6.2(80), 6.3(84), energy changes, 4.2(62)
7.1(91), 7.2(93), 8.1(99), 8.2(101), energy quantisation, 3.4(41)
8.3(106), 8.4(109) equations of motion, 10.6(149)
F
classication of matter, 1.1(1), 1.2(6),
FHSST, 9(117)
1.3(9)
Focal Point, 250
compasses, 15.2(271)
force between charges, 16.2(278)
Compound, 5
frame of reference, 10.1(129), 129
compounds, 1.1(1)
conductivity, 8.3(106), 106
conductors, 16.3(280)
G geometrical optics, 14.1(229), 14.2(232),
14.3(236), 14.4(245), 14.5(254)
Conductors and insulators, 10
global cycle, 7.1(91), 7.2(93)
Conservation of Energy, 176
grade 10, 1.1(1), 1.2(6), 1.3(9), 2.1(17),
Conservation of energy principle, 62
2.2(21), 2.3(25), 3.1(31), 3.2(34),
conservation of matter, 4.2(62)

318 INDEX
3.3(38), 3.4(41), 3.5(47), 4.1(57), magnets, 15.1(267)

4.2(62), 5.1(67), 5.2(68), 5.3(71), mechanical energy, 11.1(167), 11.2(173),
6.1(77), 6.2(80), 6.3(84), 7.1(91), 11.3(174), 11.4(176)
7.2(93), 8.1(99), 8.2(101), 8.3(106), Medium, 187
8.4(109), 9(117), 10.1(129), 10.2(132), Melting point, 25
10.3(135), 10.4(138), 10.5(140), mirrors, 14.4(245)
10.6(149), 11.1(167), 11.2(173), mixture, 1.1(1), 2
11.3(174), 11.4(176), 12.1(187), Model, 32
12.2(190), 12.3(193), 13.1(203), Molecular formula, 18
13.2(211), 13.3(213), 14.1(229), Molecule, 17
14.2(232), 14.3(236), 14.4(245), molecules, 2.1(17)
14.5(254), 15.1(267), 15.2(271), motion in one dimension, 10.1(129),
16.1(277), 16.2(278), 16.3(280), 10.2(132), 10.3(135), 10.4(138),
17.1(287), 17.2(293), 17.3(297), 10.5(140), 10.6(149)
N
17.4(301), 17.5(303)
naming compounds, 1.2(6)
Gradient, 141
neutrons, 3.2(34)
graphs, 13.2(211)
nitrogen cycle, 7.1(91), 7.2(93)
gravity, 11.1(167), 11.2(173), 11.3(174),
Node, 218
11.4(176)
H Heat of vaporisation, 82
O objects composition, 2.1(17), 2.2(21),
2.3(25)
Heterogeneous mixture, 2
Ohm's Law, 296
Homogeneous mixture, 3
ohmmeter, 17.5(303)
human inuences, 7.2(93)
hydrosphere, 8.1(99), 8.2(101), 8.3(106),
8.4(109)
P Parallel circuit, 291
particle motion, 13.2(211)
I Image, 245
Peaks and troughs, 205
periodic table, 3.5(47)
industry, 7.2(93)
permanent magnets, 15.1(267)
Instantaneous velocity, 135
pH, 103
insulators, 16.3(280)
physical change, 4.1(57), 57, 4.2(62)
Intermolecular force, 22
Physical Quantity, 117
intermolecular forces, 2.2(21)
physics, 10.1(129), 10.2(132), 10.3(135),
Intramolecular force, 21
10.4(138), 10.5(140), 10.6(149),
intramolecular forces, 2.2(21)
11.1(167), 11.2(173), 11.3(174),
introduction, 6.1(77), 8.1(99), 12.1(187),
11.4(176), 12.1(187), 12.2(190),
13.1(203), 16.1(277), 17.1(287)
12.3(193), 13.1(203), 13.2(211),
Ion, 47
13.3(213), 14.1(229), 14.2(232),
ionisation, 8.3(106)
14.3(236), 14.4(245), 14.5(254),
ionisation energy, 3.5(47)
15.1(267), 15.2(271), 16.1(277),
ions, 8.2(101)
16.2(278), 16.3(280), 17.1(287),
Isotope, 38
17.2(293), 17.3(297), 17.4(301),
isotopes, 3.3(38)
17.5(303)
K kinetic energy, 11.3(174), 174 Position, 130
kinetic theory, 2.2(21) potential dierence, 17.2(293), 294
L
potential energy, 11.2(173), 173
Law of Reection, 233
Precipitate, 109
Light ray, 229
precipitation reactions, 8.4(109)
light rays, 14.1(229)
Prex, 122
M magnetic elds, 15.1(267) properties, 1.3(9)

properties of matter, 2.3(25)
Magnetism, 13, 15.1(267), 15.2(271)

INDEX 319
properties of water, 6.2(80) 11.2(173), 11.3(174), 11.4(176),

protons, 3.2(34) 12.1(187), 12.2(190), 12.3(193),
Pulse, 188 13.1(203), 13.2(211), 13.3(213),
Pulse Speed, 189 14.1(229), 14.2(232), 14.3(236),
R
14.4(245), 14.5(254), 15.1(267),
Real Image, 251
15.2(271), 16.1(277), 16.2(278),
reference point, 10.1(129)
16.3(280), 17.1(287), 17.2(293),
reection, 14.2(232)
17.3(297), 17.4(301), 17.5(303)
refraction, 14.3(236), 236
Specic heat, 81
Refractive Index, 237, 237
speed, 10.3(135)
Relative atomic mass, 40
state symbols, 5.3(71)
representing chemical change, 5.1(67),
structure, 3.2(34)
5.2(68), 5.3(71)
Representing circuits, 290 T the atom, 3.1(31), 3.2(34), 3.3(38),
resistance, 17.4(301), 301 3.4(41), 3.5(47)
S
The frequency is the number of successive
science, 1.1(1), 1.2(6), 1.3(9), 2.1(17),
peaks (or troughs) passing a given point in 1
2.2(21), 2.3(25), 3.1(31), 3.2(34),
second., 208
3.3(38), 3.4(41), 3.5(47), 4.1(57),
The law of conservation of mass, 68
4.2(62), 5.1(67), 5.2(68), 5.3(71),
The nitrogen cycle, 91
6.1(77), 6.2(80), 6.3(84), 7.1(91),
The period (T) is the time taken for two
7.2(93), 8.1(99), 8.2(101), 8.3(106),
successive peaks (or troughs) to pass a xed
8.4(109), 10.1(129), 10.2(132),
point., 208
10.3(135), 10.4(138), 10.5(140),
The Water Cycle, 78
10.6(149), 11.1(167), 11.2(173),
total internal reection, 14.5(254), 255
11.3(174), 11.4(176), 12.1(187),
transverse pulses, 12.1(187), 12.2(190),
12.2(190), 12.3(193), 13.1(203),
12.3(193)
13.2(211), 13.3(213), 14.1(229),
Transverse wave, 203
14.2(232), 14.3(236), 14.4(245),
transverse waves, 13.1(203), 13.2(211),
14.5(254), 15.1(267), 15.2(271),
13.3(213)
16.1(277), 16.2(278), 16.3(280),
17.1(287), 17.2(293), 17.3(297), U Units, 9(117)
V
17.4(301), 17.5(303)
Valence electrons, 45
Series circuit, 291
Vectors and Scalars, 134
SI Units, 117
velocity, 10.3(135), 135
Snell's Law, 239
Virtual Image, 247
South Africa, 1.1(1), 1.2(6), 1.3(9),
voltmeter, 17.5(303)
2.1(17), 2.2(21), 2.3(25), 3.1(31),
3.2(34), 3.3(38), 3.4(41), 3.5(47),
4.1(57), 4.2(62), 5.1(67), 5.2(68),
W water conservation, 6.3(84)
water cycle, 6.1(77), 6.2(80), 6.3(84)
5.3(71), 6.1(77), 6.2(80), 6.3(84),
Water hardness, 103
7.1(91), 7.2(93), 8.1(99), 8.2(101),
Wave, 203
8.3(106), 8.4(109), 9(117), 10.1(129),
Wavelength of wave, 207
10.2(132), 10.3(135), 10.4(138),
weight, 11.1(167)
10.5(140), 10.6(149), 11.1(167),

320 ATTRIBUTIONS
Attributions
Collection: Siyavula textbooks: Grade 10 Physical Science
Edited by: Free High School Science Texts Project
URL: http://cnx.org/content/col11245/1.3/
License: http://creativecommons.org/licenses/by/3.0/
Module: "Classication of matter: Mixtures, compounds and elements (Grade 10) [NCS]"
Used here as: "Mixtures, compounds and elements"
By: Free High School Science Texts Project
URL: http://cnx.org/content/m39993/1.1/
Pages: 1-6
Copyright: Free High School Science Texts Project
Module: "Classication of matter: Giving names and formulae to substances (Grade 10) [NCS]"
Used here as: "Giving names and formulae to substances"
Pages: 6-8
Module: "Classication of matter: Classication using the properties of matter (Grade 10) [NCS]"
Used here as: "Classication using the properties of matter"
Pages: 9-16
Module: "What are the objects around us made of: Atoms and molecules"
Used here as: "Atoms and molecules"
Pages: 17-21
Module: "What are the objects around us made of: Intermolecular and intramolecular forces and the kinetic
theory of matter"
Used here as: "Intermolecular and intramolecular forces and the kinetic theory of matter"
Pages: 21-24

ATTRIBUTIONS 321
Module: "What are the objects around us made of: The properties of matter"
Used here as: "The properties of matter"
Pages: 25-29
Module: "The atom: Models and atomic size"

Used here as: "Models and atomic size"
Pages: 31-34
Module: "The atom: Structure (Grade 10) [NCS]"

Used here as: "Structure"
Pages: 34-38
Module: "The atom: Isotopes (Grade 10) [NCS]"

Used here as: "Isotopes"
Pages: 38-41
Module: "The atom: Energy quantisation and electron conguration"

Used here as: "Energy quantisation and electron conguration"
Pages: 41-46
Module: "The atom: Ionisation energy and the periodic table (Grade 10) [NCS]"
Used here as: "Ionisation energy and the periodic table"
Pages: 47-55

322 ATTRIBUTIONS
Module: "Physical and Chemical change: Introduction (Grade 10) [NCS]"

Used here as: "Introduction"
Pages: 57-62
Module: "Physical and Chemical change: Energy changes and conservation of matter (Grade 10) [NCS]"
Used here as: "Energy changes and conservation of matter"
Pages: 62-66
Module: "Representing chemical change: Introduction, chemical symbols and chemical formulae (Grade 10)
[NCS]"
Used here as: "Introduction, chemical symbols and chemical formulae"
Pages: 67-68
Module: "Representing chemical change: Balancing chemical equations (Grade 10) [NCS]"
Used here as: "Balancing chemical equations"
Pages: 68-71
Module: "Representing chemical change: State symbols (Grade 10) [NCS]"

Used here as: "State symbols"
Pages: 71-74
Module: "The water cycle: Introduction"

Pages: 77-80

ATTRIBUTIONS 323
Module: "The water cycle: Properties of water"

Used here as: "Properties of water"
Pages: 80-83
Module: "The water cycle: Water conservation"

Used here as: "Water conservation"
Pages: 84-89
Module: "The nitrogen cycle: Introduction"

Pages: 91-93
Module: "The nitrogen cycle: Human inuences and industry"

Used here as: "Human inuences and industry"
Pages: 93-97
Module: "The hydrosphere: Introduction"

Pages: 99-101
Module: "The hydrosphere: Ions in aqueous solution"

Used here as: "Ions in aqueous solution"
Pages: 101-106

324 ATTRIBUTIONS
Module: "The hydrosphere: Electrolytes, ionisation and conductivity"

Used here as: "Electrolytes, ionisation and conductivity"
Pages: 106-109
Module: "The hydrosphere: Precipitation reactions"

Used here as: "Precipitation reactions"
Pages: 109-115
Module: "Units"
By: Rory Adams, Free High School Science Texts Project, Wendy Williams, Heather Williams
Pages: 117-128
Copyright: Rory Adams, Free High School Science Texts Project, Heather Williams
Module: "Motion in one dimension: Frames of reference and reference point (Grade 10) [NCS]"
Used here as: "Frames of reference and reference point"
Pages: 129-132
Module: "Motion in one dimension: Displacement and distance (Grade 10) [NCS]"
Used here as: "Displacement and distance"
Pages: 132-134
Module: "Motion in one dimension: Speed and velocity (Grade 10) [NCS]"
Used here as: "Speed and velocity"
Pages: 135-138
Module: "Motion in one dimension: Acceleration (Grade 10) [NCS]"

Used here as: "Acceleration"
Pages: 138-140

ATTRIBUTIONS 325
Module: "Motion in one dimension: Description of motion (Grade 10) [NCS]"

Used here as: "Description of motion"
Pages: 140-148
Module: "Motion in one dimension: Equations of motion (Grade 10) [NCS]"

Used here as: "Equations of motion"
Pages: 149-157
Module: "Gravity and mechanical energy: Weight and acceleration due to gravity (Grade 10) [NCS]"
Used here as: "Weight and acceleration due to gravity"
Pages: 167-173
Module: "Gravity and mechanical energy: Potential energy"

Used here as: "Potential energy"
Pages: 173-174
Module: "Gravity and mechanical energy: Kinetic energy"

Used here as: "Kinetic energy"
Pages: 174-176
Module: "Gravity and mechanical energy: Mechanical energy"

Used here as: "Mechanical energy"
Pages: 176-182

326 ATTRIBUTIONS
Module: "Transverse pulses: Introduction and key concepts (Grade 10) [NCS]"
Used here as: "Introduction and key concepts"
Pages: 187-190
Module: "Transverse pulses: Graphs of particle motion (Grade 10) [NCS]"

Used here as: "Graphs of particle motion"
Pages: 190-192
Module: "Transverse pulses: Boundary conditions (Grade 10) [NCS]"

Used here as: "Boundary conditions"
Pages: 193-200
Module: "Transverse waves: Introduction and key concepts (Grade 10) [NCS]"
Pages: 203-211
Module: "Transverse waves: Graphs of particle motion (Grade 10) [NCS]"

Used here as: "Graphs of particle motion"
Pages: 211-213
Module: "Transverse waves: Boundary conditions (Grade 10) [NCS]"

Used here as: "Boundary conditions"
Pages: 213-225

ATTRIBUTIONS 327
Module: "Geometrical optics: Introduction and light rays"

Used here as: "Introduction and light rays"
Pages: 229-231
Module: "Geometrical optics: Reection"

Used here as: "Reection"
Pages: 232-236
Module: "Geometrical optics: Refraction"

Used here as: "Refraction"
Pages: 236-245
Module: "Geometrical optics: Mirrors"

Used here as: "Mirrors"
Pages: 245-253
Module: "Geometrical optics: Total internal reection"

Used here as: "Total internal reection"
Pages: 254-263
Module: "Magnetism: Magnetic elds and permanent magnets (Grade 10) [NCS]"
Used here as: "Magnetic elds and permanent magnets"
Pages: 267-271

328 ATTRIBUTIONS
Module: "Magnetism: The Earth's magnetic eld (Grade 10) [NCS]"

Used here as: "The Earth's magnetic eld"
Pages: 271-274
Module: "Electrostatics: Introduction and key concepts (Grade 10) [NCS]"

Pages: 277-278
Module: "Electrostatics: Forces between charges (Grade 10) [NCS]"

Used here as: "Forces between charges"
Pages: 278-280
Module: "Electrostatics: Conductors and insulators (Grade 10) [NCS]"

Used here as: "Conductors and insulators"
Pages: 280-285
Module: "Electric circuits: Key concepts (Grade 10) [NCS]"

Used here as: "Key concepts"
Pages: 287-293
Module: "Electric Circuits: Potential dierence (Grade 10) [NCS]"

Used here as: "Potential dierence"
Pages: 293-297

ATTRIBUTIONS 329
Module: "Electric circuits: Current (Grade 10) [NCS]"

Used here as: "Current"
Pages: 297-300
Module: "Electric Circuits: Resistance (Grade 10) [NCS]"

Used here as: "Resistance"
Pages: 301-303
Module: "Electric Circuits: Measuring devices (Grade 10) [NCS]"

Used here as: "Measuring devices"
Pages: 303-307

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Siyavula textbooks: Grade 10 Physical

Rice University, Houston, Texas

4 Physical and chemical change

6 The water cycle

11 Gravity and mechanical energy

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1.1 Mixtures, compounds and elements 1

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1 This content is available online at <http://cnx.org/content/m39993/1.1/>.

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Figure 1.2: The classication of matter

Denition 1.1: Mixture

In a mixture, the substances that make up the mixture:

1.1.2.1 Heterogeneous mixtures

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1.1.2.2 Homogeneous mixtures

Denition 1.3: Homogeneous mixture

1.1.2.3 Separating mixtures

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1.1.2.3.1 Investigation : The separation of a salt solution

1.1.2.3.2 Discussion : Separating mixtures

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1.1.3 Pure Substances: Elements and Compounds

Denition 1.4: Element

Denition 1.5: Compound

Figure 1.4: Understanding the dierence between a mixture and a compound

1.1.3.2.1 Elements, mixtures and compounds

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Substance Mixture or pure Homogeneous or heterogeneous mixture

1.2 Giving names and formulae to substances 5

1.2.1 Giving names and formulae to substances

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Name of compound ion Formula

1.2.1.1 Naming compounds

a. Is calcium carbonate a mixture or a compound? Give a reason for your answer.

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2. Give the name of each of the following substances.

a. What is the chemical formula for water?

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1.3 Classication using the properties of matter 11

1.3.1 Metals, Semi-metals and Non-metals

1.3.1.1.1 Group Work : Looking at metals

2. In groups of 3-4, combine your collection of metal objects.

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1.3.2 Electrical conductors, semi-conductors and insulators

Denition 1.6: Conductors and insulators

1.3.2.1 Experiment : Electrical conductivity

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1.3.3 Thermal Conductors and Insulators

1.3.3.1 Demonstration : Thermal conductivity

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1.3.3.2 Investigation : A closer look at thermal conductivity

Material Thermal Conductivity (W · m−1 · K −1 )

Standard glass 1.05

Red brick 0.69

Snow 0.25 - 0.5

Wood 0.04 - 0.12

Use this information to answer the following questions:

Figure 1.2: The classication of matter

Denition 1.1: Mixture

Denition 1.3: Homogeneous mixture

Denition 1.4: Element

Denition 1.5: Compound

Figure 1.4: Understanding the dierence between a mixture and a compound

7 See the le at <http://cnx.org/content/m39999/latest/http://www.fhsst.org/lld>

1.3 Classication using the properties of matter 11

Denition 1.6: Conductors and insulators

a. Which of the following can be classied as a mixture:

b. An element can be dened as:

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Denition 2.2: Molecular formula

Denition 2.4: Intramolecular force

Denition 2.5: Intermolecular force

8 See the le at <http://cnx.org/content/m39944/latest/http://www.fhsst.org/lin>