EXPLORE

PHYSICS

O-LEVEL

IMENDA T. M
NKUMBI INTERNATIONAL SCHOOL
MOBILE NO: 0978728265
Email:titusimenda@mail.com

2010
What is physics?
Physics is the branch of science which deals with the properties and interaction of matter
and energy.

Properties of matter
The properties of matter are called physical quantities.

Physical quantities are measurable features or properties of objects.

Types of physical quantities

There are two types of physical quantities:

 Base quantities
 Derived quantities

1. Base quantities
These are quantities with only one SI unit.
SI units. (International system of units). This is a system of units which is universally
agreed to be used in measurements of quantities worldwide.

Base unit Symbol For measuring

Metre M Length
Kilogram Kg Mass
Second S Time
Ampere A Electric current
Kelvin K Temperature
Mole Mol Amount of substance
Candela Cd Luminous intensity

2. Derived quantities
These are quantities which are expressed by combining two or more base units.

Derived quantity SI unit Symbol

Speed Metre per second m/s
Acceleration Metre per second squared m/s2
Density Kilogram per cubic metre Kg/m3
Force Newton Kgm/s2
Energy Joule Kgm2/s2
Electricity Coulomb As

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Conversion of units

Measure of distance
10mm = 1cm
100cm = 1m
1000m = 1Km
1Km = 100000cm = 1000000 mm

Measure of mass
1Kg = 1000g
1tonne = 1000Kg = 1000000g

Measure of time
60 seconds = 1 minute
60 minutes = 1 hour
24 hours = 1 day-night
7 days = 1 week
4 weeks = 1 month
12 months = 1 year = 360 days
Examples
1. Convert the following to the stated units.
(a) 200 kg to g,
(b) 30cm to m

Solution
(a) 1Kg → 1000g
200Kg → x

2 ooKg x 1000 g
x=
1 Kg

x = 200000g

(b) 100cm → 1m
30cm → x

30 cm x 1 m
x=
100 cm

x = 0.3m
Activity one
1. Convert the following to the stated units
(a) 8.0Km to m
(b) 0.8cm to m
(c) 500m to Km
(d) 13m to mm

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Rounding off numbers
When considering whole numbers:
 Zeros at the end of the number are not significant. Not that zeros at the end of the
whole number are place holders so that the other digits do not lose their place
values
 Zeros between non – zero digits are significant
When rounding off decimal numbers:
 Zeros at the end of a decimal number are significant
 Zeros between non – zero digits are significant
 Zeros at the beginning at of a decimal number are not significant.
When decimal numbers are rounded off, the number of decimal places to be
rounded off must be specified

Scientific notation
Scientific notation is also called standard form.
Scientific notation is a method of expressing a number in the form: a x 10n, where 1
< a < 10 and n is an integer.
This is where numbers are expressed in the power of ten

Examples
1. Express the following in standard form
(a) 3000000
(b) 4200
(c) 600
(d) 0.0016
(e) 0.235
(f) 0.2001
(g) 0.2000

Solution
(a) 3 x 106
(b) 4.2 x 103
(c) 6 x 102
(d) 1.6 x 10-3
(e) 2.35 x 10-1
(f) 2.001 x 10-1
(g) 2 x 10-1

Activity two
1. Write down the standard form of;
(a) 6423
(b) 5200
(c) 60003
(d) 0.03

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 (e) 0.3002
(f) 0.004010

Examples
1. Round off the following numbers according to the specifications:
(a) 683 to the nearest ten
(b) 683 to nearest hundred
(c) 786 to the nearest ten
(d) 9.3 to the nearest whole number
(e) 5.7 to the nearest whole number
(f) 9.9 to the nearest whole number

Solution
(a) 680
(b) 700
(c) 790
(d) 9
(e) 6
(f) 10

Rounding off decimal numbers

.

Examples
1. Round off the following according to the decimal places specified
(a) 6.83 correct to one decimal place
(b) 1.057 correct two decimal places
(c) 0.0863648 correct to two decimal places
(d) 0.95 correct to one decimal place
Solution
(a) 6.8
(b) 1.06
(c) 0.09
(d) 1.0

Activity three
1. Round off the following according to the decimal places specified.
(a) 4.38 correct to one decimal place
(b) 2.065 correct to two decimal places
(c) 0.004689 correct to three decimal places.

Fundamental quantities

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There are three fundamental quantities upon which all measurements are based. These
are;
 Length
 Time
 Mass

Length
Symbol: L
SI unit: metre, m
Definition: Length is distance between two or more points.

Instruments used to measure length

 Rule
 Vernier calipers
 Micrometer screw gauge

The rule
Accuracy: 1mm
Quantity measured: Length

Common types of rules

 metre rule (100cm rule)
 30cm rule
 15cm rule

Meter rule
The metre rule is used to measure length of more than 1mm.
It is usually graduated in centimeters.
It has sub- divisions in millimeters.

Correct use of a rule

1. The eye should be placed vertically above the point to be measured to avoid
parallax error.

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 2. If a rule has no zero edge, it means you cannot use this point. Therefore, to take a
reading, start slightly inwards say at 1cm and remember to subtract from the final
reading.

Example
1. A piece of cotton is measured between two points on a ruler.

When the length of cotton is wound closely around a pen, it goes round six times.

What is the length of the cotton?

Solution
Length of cotton = 15.6cm – 2.4cm
=13.2cm

Vernier Calipers

Accuracy: 0.01cm
Quantity measured: Length
Use: It is used to measure the length of solids where an ordinary rule cannot be used.
The Vernier calipers can also be used to measure the diameter of balls and cylinders.

Structure of the vernier calipers

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Main scale
The main scale is on the stem and fixed.
It is marked in centimeters, cm.

Vernier scale
The vernier scale is movable and slides on the main scale.
It is marked in millimeters, mm.
1
It has an accuracy of up to th of a millimeter
10
The vernier scale has ten divisions that correspond to nine divisions of the main scale.

Internal jaws
They measure internal diameter of objects.

External jaws
They measure external diameter of objects.

How to read vernier calipers

1. Find the value on the main scale that appears just before the zero of the vernier
scale in centimeters, cm.
2. Find the value of the line on the vernier scale that coincides with a line on the
main scale and multiply it by 0.01cm in order to convert it into cm.
3. Add main scale reading and vernier scale reading.

Precautions when using vernier calipers

1. Zero the instrument before taking a reading
2. Clean the instrument so that it is free from dust particles.

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Example
1. State the readings shown in the diagram of Vernier calipers below.

Main scale reading = 0.10cm

Vernier scale reading = 3 x 0.01cm = 0.03cm
Vernier calipers reading = 0.10cm + 0.03cm
= 0.13cm

Activity four
1. Find the readings registered by the vernier calipers below.
(a)

(b)

Micrometer screw gauge

Accuracy: 0.01mm
Quantity measured: Length
Use: It is used to measure smallest size of length such as;
 the thickness of a hair,
 the diameter of a wire,
 the thickness of a piece of paper,
 the thickness of a coin,
 the thickness of a razor blade.

Structure of the micrometer screw gauge

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Important parts of the micrometer screw gauge
1. Sleeve
This is the part that bears the sleeve scale.
The sleeve scale is graduated in millimeters, mm.
The sleeve scale measures correct to 0.5mm
2. Thimble
This is the part that bears the thimble scale.
The thimble scale measures correct to a 100th of a millimeter or 0.01mm.
A thimble scale has 50 divisions and each division represents 0.01mm.
3. Anvil and Spindle
These two parts hold the object that is being measured by the instrument.
4. Ratchet
This is a part used to move the spindle towards or away from the anvil in order to hold
the object.
Measurement using the micrometer screw gauge
The two parts or scales are considered, namely;
 thimble scale
 sleeve scale
How to read the micrometer screw gauge
1. Find the value on the sleeve scale which appears just before the edges of the
thimble. The value above the horizontal line gives the whole numbers.
The value below the horizontal line but in front of the whole number obtained is a
mark of 0.5mm and is added to the whole number.
2. Find the value on the thimble scale which is in line with the horizontal line of the
sleeve scale and multiply it by 0.01mm.
Note
If there isn’t any mark in line, but the horizontal line or point is in between the
mark, the highest mark is taken and then multiplied by 0.01mm.
3. Add the sleeve scale reading and thimble scale reading.
Precautions when using a micrometer screw gauge.
1. Zero the instrument before making any measurement.
2. Clean the anvil and spindle before making any measurement.
3. Turn the ratchet gently.
Example

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 1. State the measurement shown in the diagram of the micrometer screw below.

Sleeve reading = 6.5mm,

Thimble reading = 32 x 0.01mm
= 0.32mm
Micrometer reading = 6.5mm + 0.32mm
= 6.82mm .

Activity five
1. State the measurement shown in the diagram of the micrometer below

Time
Symbol: t
SI unit: Second, s
Definition: It is the measure of how long matter occupies a given space
A time measurement enables us to determine the interval between the beginning and the
end of an event.

Instruments for measuring time

 Simple pendulum
 Stop watch
 Ticker tape timer
 Oscilloscope (C.R.O)

The simple pendulum

A simple pendulum is a small heavy bob suspended by a light inextensible string.
This consists of a string tied to a horizontal support. A bob is suspended at the lower end

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of the string.

Terms used to describe a simple pendulum

Oscillation
An oscillation is a complete to and fro movement of the bob. It is also called a cycle or
vibration or swing.

Note:
A swing from;
1
(a) A to B = or 0.25 oscillations.
4
1
(b) A to C = or 0.5 oscillations
2
3
(c) A to C and back to B = or 0.75 oscillations
4
(d) A to C and back to A = 1 complete oscillation

Amplitude

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Symbol: A
SI unit: metre, m
Definition: Amplitude is the maximum displacement of the bob from the rest position.

Length of the pendulum

Symbol: L
SI unit: metre, m
Definition: Length of the pendulum is the distance from the supporter (point of
suspension) to the centre of the bob.

Period of a pendulum
Symbol: T
SI unit: Second, s
Definition: Period of the pendulum is the time taken by the bob to make a complete
oscillation.

Factors affecting the period of the pendulum

 Length of the pendulum, L
 Acceleration due to gravity, g

(a) Length of the pendulum, L

Period of the pendulum becomes;
(i) longer if the length of the pendulum is increased
(ii) shorter if the period of the pendulum is reduced
(b) Acceleration due to gravity, g
When;
(i) gravity is high, period reduces
(ii) gravity is low period increases
Period does not depend on the mass or material of the bob.

Relationship between period and time

 t=nxT

t
 T=
n

t
 n=
T
Note
 t = time interval in seconds, s
 n = number of oscillations ( swings/cycles/times)
 T = period of the pendulum in seconds, s
Frequency
Symbol: f
SI unit: Hertz, Hz
Definition: Frequency is the number of oscillations in one second.

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Relationship between frequency and period

1
 period =
frequency

1
T=
f

1
 frequency =
period

1
f=
T

number of oscillations
 frequency =
time

n
f=
t

Displacement- time graph for a simple pendulum

Note
 Amplitude = 2cm
 Period of the pendulum = 1.0s
 When length of the pendulum increases, period also increases but frequency
reduces.
 When length of the pendulum reduces, period also reduces but frequency
increases.

Determining (measuring) period of the pendulum

 Set the pendulum oscillating
 Note the time, t, and the number of oscillations, n.
 Calculate the period, T, using the formula;
t
T=
n

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Measuring time interval using a simple pendulum.
 Set the pendulum oscillating.
 Note the number of oscillations, n.
 Calculate time by using the formula;
t=nxT
Note
 A number of runs are done and the average is taken to minimize error.

Experiment

Aim: To determine the relationship between the length (L) and period (T) of the
pendulum

Apparatus
 Bob
 Clamp and stand
 String
 Stop watch

Method
Measure and record the length of the string from the point of support to the centre of the
bob.
Pull the bob to one side with angular amplitude of less than 10o.
Release the bob so that it starts swinging.
When the bob reaches the maximum displacement, start the stop watch and start counting
Record the time taken for 20 complete oscillations
Repeat the experiment with different lengths (L). Record values in the table.

Results
Length of string (cm) Time taken for 20 complete oscillations (s) Period (s)
1 30cm
2 20cm
3 10cm

Conclusion
Period of the pendulum depends on the length of the pendulum and acceleration due to
gravity.

Examples

1. In an experiment to measure the period of the pendulum, the time taken for 50
complete oscillations was found to be one minute. What is the period of the
pendulum?

Data Solution

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 T =? t
T=
t = 60 seconds n
n = 50 60 s
T=
50
T = 1.2s

2. What is the period of a pendulum that makes 50 cycles in 9s?

Data Solution
T =? t
T=
t = 9s n
n = 50 9s
T=
50
T = 0.18s

3. A pendulum has period 0.6s. Calculate the time it takes to make 75 cycles?

Data Solution
t =? t=nxT
n = 75 t = 75 x 0.6s
T = 0.6s t = 45s

4. How many cycles are made by a pendulum whose period is 1.2s in 30s?
Data Solution
n =? t
n=
T = 1.2s T
t = 30s 30 s
n=
1.2 s
n = 25 cycles

5. A pendulum makes 96 cycles in 4.8s.What is its frequency?

Data Solution
f =? n
f=
n = 96 t
t = 4.8s 96
f=
4.8 s
f = 20Hz

6. A pendulum's frequency is 15Hz.How many cycles does it make in 3.6s?

Data Solution
n =? n=fxt
f = 15Hz n = 15Hz x 3.6s
t = 3.6s n = 54 cycles

7. What is the time taken for a pendulum of frequency 25Hz to make 40 cycles?

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 Data Solution
t =? n
t=
n = 40 f
f = 25Hz 40
t=
25 Hz
t = 1.6s

8. The figure below shows a simple pendulum that oscillates between position A and
C.

a) If it takes 2 seconds for the bob to move from A to C and back to B, find the
number of oscillations.
b) Calculate the period of the pendulum.
c) Calculate the frequency of the pendulum

Data Solution
a A to C back to B n = 0.75 oscillations
b T =? t
T=
t = 2s n
n = 0.75 2s
T=
0.75
T = 2.67s
c f =? 1
f=
T= 2.67s T
1
f=
2.67 s
f = 0.37Hz

9. The diagram below shows an oscillating pendulum.

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If the period of the pendulum is 0.4s, find the time taken for the pendulum to swing from;
a) A to C
b) A to B
c) A to C and back to B
Data Solution
a t =? t=nxT
n = 0.5 oscillations t = 0.5 x 0.4s
T = 0.4s t = 0.2s
b t =? t=nxT
n = 0.25 oscillations t = 0.25 x 0.4s
T = 0.4s t = 0.1s
c t =? t=nxT
n = 0.75 oscillations t = 0.75 x 0.4s
T = 0.4s t = 0.3s

10. The bob of a simple pendulum is pulled to one side and released. The motion
during its swing is shown in the graph.

(a) What is the value of the period of the pendulum?

(b) Calculate the frequency of the pendulum
(c) What would you do in order to change the periodic time of the same
pendulum to 1.5s?

Data Solution

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 a T = 2.0s
b f =? 1
f=
T = 2.0s T
1
f=
2.0 s
f = 0.5Hz
c By reducing the length of the pendulum.

Activity six
1. Find the period of the pendulum if it oscillates 15 times for 45 seconds.
2. The diagram below shows an oscillating pendulum.

a) If it takes 3 seconds for the bob to move from A to C, find the period of the
pendulum.
b) Find the time taken for 12 complete oscillations.
3. The bob of the pendulum shown below takes 0.25s to swing from A to C.

a) If A and C are extreme points, determine;

(i) the period of the pendulum
(ii) the frequency of the pendulum
b) State whether the frequency of oscillations will increase, decrease or remain the
same if;
(i) the length of the string is increased
(ii) the mass of the bob is increased
(iii) The distance between A and C is increased.

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 4 Briefly describe how the period of the pendulum would be measured.

5 Study the displacement- time graph for a simple pendulum.

(a) What is value of the period of the pendulum?

(b) State the maximum displacement of the pendulum.
(c) Naosa carried out an experiment to determine the time Kakula took to finish
drinking one litre of castle using a simple pendulum. The period of the
pendulum was 1.5 seconds and its length was 0.8m.
(i) Calculate the time taken for Kakula to finish drinking one litre of castle if
the number of oscillations were 50.
(ii) State what will happen to the frequency of pendulum if the length was;
(a) reduced to 0.5m
(b) Increased by 0.5m.

6 The graph below is for a pendulum bob which was pulled to one side and then
released to swing. Assume that there is no friction of any sort as the bob
swings.

(a) What do you understand by period of the pendulum?

(b) After how long does the pendulum bob reach the maximum distance of travel?
(c) If the pendulum bob swings at the rate of 5m/s, how far from the starting position
is it at 8 seconds later from the time it started swinging?
(d) Explain why this pendulum would be suitable for keeping or measuring time.

Scalar and vector quantities

Scalar quantity
It is a quantity which has magnitude (size) with no direction

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Scalar quantities can easily be added and subtracted.
Examples of scalar quantities
 Distance
 Speed
 Mass
 Volume
 Temperature

Vector quantity
It is a quantity which has both magnitude (size) and direction.
Vector quantities are mainly represented graphically or an arrow with a point (→)
Examples of Vector quantities
 Displacement
 Velocity
 Acceleration
 Force
 Weight
 Momentum
Kinematics
Kinematics is the science of describing the motion of objects using words, diagrams,
numbers, graphs, and equations. Kinematics is a branch of mechanics.
Mechanics is the study of the motion of objects.

Motion
Motion is the change of position of an object in a given direction.

Types of motion
1. Linear motion
This is the movement of an object along a straight line or path e.g. a car travelling along a
straight road.

2. Motion Circular (Rotational motion)

This is the movement in a circle about the centre or an axis e.g. a spinning wheel or
rotating fan.
3. Oscillatory motion
This is the movement where an object moves to and fro about a fixed position e.g. the
swinging of the bob of the pendulum.

4. Random motion
This is the movement of an object in a disorderly manner e.g. in the case of gaseous
particles.

Linear motion
Four parameters are required to describe motion in a straight line. These are;

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  Distance or displacement
 Speed or velocity
 Acceleration
 Time

Distance
Symbol: s
SI unit: metre, m
Definition: Distance is the length between two or more points. It can also be defined as
the actual path travelled by an object from its initial position to the final position.

Displacement
Symbol: s
SI unit: metre, m
Definition: Displacement is the distance travelled in a specified direction.

Similarity between distance and displacement

1. The SI unit for both distance and displacement is the metre, m.

Differences between distance and displacement

1. Distance is a scalar quantity while displacement is a vector quantity.
2. Distance is the length between two points while displacement is the distance
travelled in a specified direction.
Examples
1. A car moves 15Km to the East and 13Km to the North.

(a) Find the distance and displacement of the car.

Solution
(i)Distance = 15Km + 13Km
= 28Km
(ii)Displacement = 15Km East and 13Km North.
2. The circumference of a round bout is 50m and the car turns it once.
(a) Find the distance and displacement of the car.
Solution
(i)Distance = 50m
(ii)Displacement = 0m because the car came back to the starting point.
3. A boy walks forward 25m and backward 15m.
(a) Find the distance and displacement of the boy.
Solution
(i)Distance = 25m + 15m
= 40m
(ii)Displacement = 25m – 15m
= 10m forward
Speed
Symbol: V
SI unit: metre per second, m/s (or ms-1)

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Definition: Speed is the rate of change of distance with time.

total distance covered

Formula: Average speed =
total time taken
s
v=
t
Note
1m/s = 3.6Km/h

Examples
1. Express
(a) 72Km/h in m/s
(b) 10m/s in Km/h
Solution
( 72 x 1000 ) m
(a) =
( 60 x 60 ) s
= 20m/s

OR

3.6Km/h → 1m/s
72Km/h → x
72 Km /h X 1 m/s
x=
3.6 Km/h
x = 20m/s

(b) 1m/s → 3.6Km/h

10m/s→ x
10 m/ s X 3.6 Km/h
x=
1 m/ s
x = 36Km/h

2. A car travels from Lusaka to Mongu 600Km away in 8hours. Find the average
speed of the car in Km/h.
Data Solution
v =? s
v=
s = 600Km t
t = 8h 600 Km
v=
8h
v = 75Km/h

3. A cheetah runs at a speed of 20m/s in 50 seconds. Calculate how far it will travel
in this time.
Data Solution
s =? s=vxt
v= 20m/s s = 20m/s x 50s

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 t = 50s s = 100m

4. A bus takes 40 minutes to complete its 24Km route. Calculate its average speed in
m/s.
Data Solution
v =? s
v=
s = 24Km = 24000m t
t = 40min = 2400s 24000 m
v=
2400 s
v = 10m/s

Velocity
Symbol: V
SI unit: metre per second, m/s (or ms-1)
Definition: Velocity is the rate of change of displacement with time.

displacement
Formula: Average velocity =
time taken

s
v=
t

Similarity between speed and velocity

1. Both speed and velocity are measured in metres per second, m/s.

Differences between speed and velocity

1. Speed is a scalar quantity while velocity is vector quantity.
2. Speed is the rate of change of distance with time while velocity is the rate of
change of displacement with time.

Note
Speed is called velocity when it has direction and velocity is called speed when it has
no direction.

Example
1. Car 1 moves 10m/s east and car 2 moves 10m/s north. Find the speed and velocity
of the two cars.
Solution
(a) Both car 1 and 2 have the same speed of 10m/s
(b) Car 1 has a velocity of 10m/s east while car 2 has a velocity of 10m/s north.

Acceleration
Symbol: a
SI unit: metre per second squared, m/s2 (or ms-2)

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Definition: Acceleration is the rate of change of velocity with time.

Final velocity−initial velocity

Formula: Average acceleration =
time taken
V −U
a=
t
Types of acceleration
1. Positive acceleration
This is when velocity is increasing with time.
It is always given a positive sign.
2. Negative acceleration
This is when velocity is decreasing with time. It is also called retardation or deceleration
It is always given a negative sign
3. Uniform acceleration
This is when the rate of velocity is constant
Under uniform acceleration, velocity is changing continuously but at the same rate.
Uniform acceleration is also called constant acceleration
4. Non uniform acceleration
This is the acceleration in which the rate of change of velocity is not constant.
The rate of change of velocity keeps on changing.
Note
1. Negative acceleration is called retardation or deceleration
2. When speed or velocity is constant, acceleration, a = 0m/s2
3. From rest, initial velocity, u = 0m/s
4. Moving at/travelling at/moving with, initial velocity, u = given velocity in m/s
5. To rest, final velocity, v = 0m/s

Examples
1. A car starting from rest increases its velocity uniformly to 15m/s in 3 seconds.
What is the acceleration?
Data Solution
a =?
v−u
a=
v = 15m/s t
15 m/ s−0 m/s
u = 0m/s a=
3s
15 m/ s
t = 3s a=
3s
a = 5m/s2

2. A car slows down from 36m/s to rest in 12s. Calculate the retardation.

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 Data Solution
a =? 0 m/s−36 m/s
a=
v = 0m/s 12 s
u = 36m/s −36 m/s
t = 12s a=
12 s
a = -3m/s2

Equations of uniformly accelerated linear motion

1. v = u + at

1
2. s = ut + at2
2

3. v2 = u2 + 2as

( v +u ) t
4. s =
2
Examples
1. A car travelling at 10m/s accelerates at 2m/s2 for 3 seconds. What is its final
velocity?

Data Solution
v =? v = u + at
u = 10m/s v = 10m/s + (2m/s2 x 3s)
t = 3s v = 10m/s + 6m/s
a = 2m/s2 v = 16m/s

2. A car starts from rest accelerates at 3m/s2.How far does it travel in 4 seconds?

Data Solution
s =? 1
s = ut + at2
u = 0m/s 2
t = 4s 1
a = 3m/s2 s = 0 m/s x 4 s+ x 3m/s2 x (4s )2
2
s = 24m

3. A car accelerates from rest to a velocity of 8m/s over a distance of 200m. How
long does it take to accelerate from rest to 8m/s?

Data Solution
t =? 2s
t=
s = 200m v +u
v = 8m/s 2 x 200 m
u = 0m/s t=
8 m/s+ 0 m/s

26 | P a g e
 t = 50s

4. A car accelerates uniformly from rest until it reaches a velocity of 10m/s in 5s.
How far does it travel during the 5s?

Data Solution
s =? (v+ u) t
s=
v =10m/s 2
u = 0m/s ( 10 m/ s+0 m/ s ) 5 s
t = 5s s=
2
s = 25m

Activity seven
1. A car travelling at 20m/s accelerates at the rate of 2m/s2 for 30 seconds. Calculate;
(a) the final velocity of the car
(b) the distance travelled by the car.

Time graphs

Distance-time graphs
A distance time - graph is a graph where distance is plotted against time.
The diagrams below represent the distance time graphs for the motion of an object.

s (m) s (m)

time/s time/s
Description Description
The object was accelerating. The object was decelerating or retarding.

30
10

s (m) s (m)

0 3 6 time/s 0 3 6 time/s

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Description Description
The object stopped moving.(was at rest) The object was moving with constant
The horizontal straight line indicates zero velocity. It travelled a distance of 30m in
speed. 6s.
The slope on the distance-time graph
represents velocity.
distance
velocity =
time
s
v=
t

Example
1. An object travelled a distance of 40m in 4 seconds.
(a) Sketch the distance- time graph to interpret the information above.
(b) Calculate the velocity of an object.
Solution

(a) 40

Distance/ m

0 2 4 time/s
distance
(a) Velocity =
time

s
v=
t

40 m
v=
4s

v = 10m/s

Activity eight
1. The diagram below shows the distance-time graph of an object.

C D
Distance / m

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 B

A Time / s
Describe the motion of an object from;
(a) A to B
(b) B to C
(c) C to D

Velocity (speed)-time graphs

Velocity-time graph is a graph where velocity is plotted against time.
The diagrams below show the velocity-time graphs for the motion of an object.

12 12

v (m/s)
v (m/s)

0 1 2 time/s 0 1 2 3 4 time/s

Description Description
The object was moving from rest with The object was moving from rest with
constant acceleration to a velocity of 12m/s constant acceleration to a velocity of 12m/s
in 2s. in 2s. It then moved with constant velocity
The slope indicates constant acceleration. of 12m/s in 2s.

12 12

v (m/s) v (m/s)

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 0 1 2 3 4 5 6 time/s 0 1 2 3 4 5 6 7 8 9 time/s
Description Description
The object was moving at a constant The object was moving from rest with
velocity of 12m/s in 3s and then constant (uniform) acceleration to a
decelerated uniformly to rest in 3s. velocity of 12m/s in 2s and then it moved
with constant velocity of 12m/s in 4s and
finally decelerates uniformly to rest in 3s.

Summary of velocity-time graphs

v (m/s) v (m/s)

time/s
time/s

Description Description
Non-uniform acceleration Non-uniform deceleration or retardation

v (m/s) v (m/s)

Description time/s
Constant velocity Description time/s
Horizontal line represents zero acceleration Increasing (uniform) velocity
Constant or uniform acceleration

+4 4

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 v (m/s) v (m/s)

0 1 2 3 4 time/s 0 1 2 3 time/s

-4
Description Description
Negative velocity shows that the object was Non-uniform deceleration
dropping or falling.
Constant (uniform) deceleration

Note
1. The slope (gradient) on the velocity-time graph represents acceleration.

v−u
a=
t

2. The area under the velocity-time graph represents the distance covered.
For a;
(a) triangle,

h or h

b b

1
Distance, s = bh
2

(b) rectangle,
l

Distance, s = l x b

31 | P a g e
 (c) trapezium,
a

H
h h

1
Distance, s = ( a+b ) h
2

3. Velocity of a body must be changing when the body is accelerating

Example
1. The diagram below shows a speed versus time graph for an arrow which was shot
vertically upwards.

(a) At what speed was the arrow shot?

(b) How long did it take the arrow to reach its highest point?
(c) Determine how high the arrow rose.

Data Solution
(a) v = 100m/s
(b) t = 10s

32 | P a g e
(c) s =?
1
s = bh
b = 10s 2

h = 100m/s 1
s= x 10s x 100m/s
2

s = 500m

1. The figure below shows a velocity-time graph for a car travelling along a straight
road in 10s.

(a) Describe the motion of the car in 10s.

(b) Find the acceleration of the car in the first 2s.
(c) Find the acceleration of the car between 2 and 6 seconds of the journey.
(d) Calculate the acceleration of the car in the last 4s of its motion.
(e) Find the distance travelled by the car in the first 2s.
(f) Calculate the distance travelled by the car during the constant velocity
(between 2 and 6 seconds)
(g) Find the total distance travelled by the car.

33 | P a g e
1 Data Solution
(a) The car was moving from rest with constant acceleration
to a velocity of 10m/s in 2s and then it moved with
constant velocity of 10m/s in 4s and finally decelerates
uniformly to rest in 4s.
(b a =? v−u
a=
) v = 10m/s t
u = 0m/s 10 m/ s−0 m/s
t = 2s a=
2s
a = 5m/s2
(c) a =? a = 0m/s2 because velocity is constant.
v = 10m/s or
u = 10m/s v−u
a=
t = 4s t
10 m/ s−10 m/s
a=
4s
0 m/s
a=
4s
a = 0m/s2
(d a =? v−u
a=
) v = 0m/s t
u = 10m/s om/s−10 m/s
t = 4s a=
4s
a = -2.5m/s2
(e) s =? 1
s = bh
b = 2s 2
h = 10m/s 1
s = x 2 s x 10 m/s
2
s = 10m
(f) s =? s=lxb
l = 4s s = 4s x 10m/s
b = 10m/s s = 40m
(g s =? 1
s = (a + b)h
) a = 4s 2
b = 10s 1
h = 10m/s s = (4s + 10s)10m/s
2
1
s = x 14s x 10m/s
2
s = 70m

Activity nine
1. A car moving from rest acquires a velocity of 20m/s with uniform acceleration in
4s. It then moves with this velocity for 6s and again accelerates uniformly to
30m/s in 5s. It travels for 3s at this velocity and then comes to rest with uniform
deceleration in 12s.

34 | P a g e
 (a) Draw a velocity-time graph
(b) Calculate the total distance covered.
(c) Calculate the average speed.
2. A car starting from rest accelerates uniformly to 20m/s in 5s. And it accelerates
more to 40m/s in 2s and then decelerates until it stops 8s later.
(a) Draw the speed-time graph
(b) Calculate the retardation
(c) Calculate the total distance travelled
(d) Calculate the average speed.
3. A car accelerated uniformly from 10m/s to 20m/s. It travelled a distance of 50m
during this time.
(a) What the acceleration of the car?
(b) How long does it take to travel this distance?
4. A car stating from rest accelerates uniformly at 5m/s2 in 3s.
(a) Calculate the final velocity
(b) Calculate the distance covered.
5. A man drives a car at 5Km/h. He brakes and stops in 3s. Calculate the retardation.
6. A man rides a bicycle. He accelerates from rest to a velocity of 8m/s in 5s. What
is the acceleration?
7. An object moving at a velocity of 10m/s comes to rest in 4s.
(a) Sketch the velocity-time graph for the motion of this object.
(b) Using your graph, calculate the acceleration of the object.
8. The table below shows the readings obtained by a group of pupils performing an
experiment to determine variation of velocity with time for a car starting from
rest.

Velocity, m/s 0 10 20 20 20
Time, s 0 2 4 6 8

(a) On the axes above, draw the velocity-time graph.

(b) Calculate the acceleration of the car for the first 4 seconds.

Acceleration-time graphs
Acceleration-time graph is a graph where acceleration is plotted against time.
The diagrams below represent the acceleration - time graphs

35 | P a g e
 +1.25 Acceleration

Acceleration

(m/s2) 0 time/s (m/s2)

-1.25 time/s

Description Description
Constant deceleration Constant acceleration
Decreasing velocity Increasing velocity
Example
1. As it went past an observer standing by the road side, a bus decelerated at
1.25m/s2. Thirty seconds later, the bus stopped.
(a) How far from the observer has the bus moved when it stopped?
(b) What was the speed of the bus as it went past the observer?
(c) On the axis below, sketch an acceleration- time graph for the motion of the bus.

a (m/s2) 0 time/s

1 Data Solution
(a) s =? 1
s = ut + at2
u = 0m/s 2
t = 30s 1
a = -1.25m/s2 s = om/s x 30 s + x1.25m/s2x(30s)2
2
1
s = om/s x 30 s + x1.25m/s2x30s x 30s
2
s =562.5m
(b u =? v = u + at
) v = 0m/s u = v – at
t = 30s u = 0m/s – (-1.25m/s2) x 30s
a = -1.25m/s2 u = 37.5m/s
(c)

a (m/s2)

36 | P a g e
 0 30 time/s

-1.25

Activity ten
1. Starting from rest at t = 0s, a car moves in a straight line with an acceleration
given by the graph below.

10

a (m/s2)

0 1 2 3 4 5 6 7 time/s
(a) What is the speed of the car at t = 3s?

Acceleration due to gravity: free fall

Symbol: g
SI unit: metre per second squared, m/s2
Definition: It is the acceleration of free falling objects.
All objects accelerate uniformly downwards on the earth if air resistance is ignored. This
is called acceleration due to gravity.
Objects fall because of the gravitational attraction between the objects and the earth.
If an object is dropped from the top of the building, it accelerates uniformly downwards.
If an object is released without applying force, it starts from rest. This is called free fall.

Free fall (dropping), u = 0m/s

g = 10m/s2

If an object is thrown vertically upwards, it decelerates to the top. Then the object stops
momentarily on the top and then it starts to fall freely.

Throwing up, v = 0m/s

g = -10m/s2

37 | P a g e
The time taken for an object thrown vertically upwards to rise is equal to the time it will
take to drop, t1 = t2
t1: time taken from the ground to the top.
t2: time taken from the top to the ground.
Total time, t = t1 + t2
The equations of motion can be applied to free falling objects using “g” instead of “a”
and “h” instead of “s".
1. v = u + gt

1
2. h = ut+ gt2
2

3. v2 = u2 + 2gh

(v+ u) t
4. h =
2

Note
g = acceleration due to gravity [10 m/s2]
h = height or distance [m].
Examples
1. A stone is thrown upwards with an initial velocity of 20m/s. Air resistance is
ignored.
(a) How far does it reach to the top?
(b) How long does it take to the top?
(c) What is its velocity just before reaching the ground?
(d) How long does it take to the ground?

Data Solution

38 | P a g e
a v 2−u 2
h=
2g
h =?
u = 20m/s 02−202
v = 0m/s h=
2 x (−10)
g = -10m/s2
−400
h=
−20
h = 20m
b t1 =?
u = 20m/s v−u
t1 =
v = 0m/s g
g = -10m/s2 20−0
t1 =
−10
t1 = 2s
c u=v v = 20m/s (Final velocity is equal to initial velocity,
the velocity with which it was thrown)
d t =? t = t1 + t2
t1 = 2s t = 2s + 2s
t1 = t2 = 2s t = 4s

2. A body falls freely from rest. Air resistance is ignored.

(a) What is its velocity after 1s?
(b) How far does it reach in 1s?

2 Data Solution
a v =? v = u + gt
u = 0m/s v = 0m/s + (10m/s2 x 1s)
g = 10m/s2 v = 0m/s + 10m/s
t = 1s v = 10m/s
b h =? 1
h = ut + gt2
u = 0m/s 2
t = 1s 1
g = 10m/s2 h = om/s x 1 s+ x10m/s2x1sx1s
2
h = 5m

3. A ball is thrown vertically upwards with a velocity of 10m/s.

Calculate,
(a) the maximum height that the ball reaches.
(b) the total time the ball is in the air
(c) the velocity with which the ball hits the ground.

39 | P a g e
3 Data Solution
a
v 2−u 2
h=
2g
h =?
u = 10m/s 02−102
v = 0m/s h=
2 x (−10)
g = -10m/s2
−100
h=
−20
h = 5m
b 2h
t1 =
v +u
t1 =? 2 x5m
t1 = t2 t1 =
0 m/s+10 m/ s
t =? 10 m
h = 5m t1 =
10 m/ s
u = 10m/s t1 =1s
v = 0m/s t1 = t2 = 1s
t = t1 + t2
t = 1s + 1s
t = 2s
c u=v v = 10m/s (Final velocity is equal to initial velocity, the
velocity with which it was thrown.)

Activity eleven
1. A stone is released from the top of a building and takes 3s to reach the ground.
The air resistance is ignored.
(a) What was the final velocity of the stone?
(b) How tall is the building?
2. A ball is thrown vertically upwards with an initial velocity of 40m/s.
(a) Find the maximum height the ball reaches.
(b) How long does the ball remain in the air? (assuming air resistance is ignored)

Terminal velocity

40 | P a g e
Terminal velocity is constant velocity reached by a falling object when the air resistance
is equal to the weight of the object.
Every falling object experiences some air resistances which increase with speed. When a
falling object acquires a high speed such that air resistance becomes equal to the weight
of the object, the object stops accelerating and falls with constant velocity. This constant
velocity is called terminal velocity.

Factors (conditions) that affect terminal velocity

 Size of the object
 Shape of the object
 Weight of the object
 Mass of the object

An object of low density but large surface area reaches terminal velocity e.g. a feather.
A man who jumps out of a helicopter has a high terminal velocity, but when he opens the
parachute to his advantage, terminal velocity reduces due to increased air resistance.

[A] If a coin and a feather are enclosed in a long tube which contains air and the tube is
inverted, the coin falls much faster than a feather. A feather falls more slowly because it
has a low density and large surface area.
Terminal velocity is reached where there is air.

[B] If air is pumped out of the tube with a vacuum pump and the tube is inverted, both the
feather and the coin fall at the same time and the same acceleration called acceleration
due to gravity.
Terminal velocity is not reached in a vacuum.

41 | P a g e
Activity twelve
1. Give an example where a person uses terminal velocity to his advantage.
2. Explain a reason why a piece of paper falls more slowly than a stone, although
both of them are on earth and are supposed to have the same acceleration of
10m/s2.

3. The figure below shows a feather, dropped from the top of a building which
reaches terminal velocity at point B.

The velocity of the feather at B is 30m/s. If time taken for the feather to move from B to
C is 3s, what is its velocity at C?

Recording motion using a ticker tape timer

A ticker tape timer is a device that can be used to record motion of an object.
A ticker tape makes dots on a paper tape.
When a paper tape is pulled through the timer, a dot is marked on the tape every 0.02s.
1
In one second, 50 dots are made on the paper tape. This implies that a dot is made in
50 s
or 0.02s.

Implications (interpretations) of ticker tape dot pattern

Constant speed

Accelerating

Decelerating

42 | P a g e
Determining time from the ticker tape
Time = number of dot spaces x 0.02s
Example
1. Determine the time interval between x and y.

Solution
t = number of dot spaces x 0.02s
t = 7 x 0.02s
t = 0.14s

Determining speed from ticker tapes

distance
Speed =
time
s
v=
t

Example
1. From the ticker tape shown below, work out the speed.

Solution
t = number of dot spaces x 0.02s
t = 4 x 0.02s
t = 0.8s

distance
Speed =
time
s
v=
t
0.2 m
v=
0.8 s
v = 2.5m/s

43 | P a g e
Mass
Symbol: m
SI unit: Kilogram, Kg
Definition: Mass is the quantity of matter contained in a substance.
The mass of an object is also the measure of its inertia.
Measuring instruments: beam balance.
Mass of an object is constant (same) everywhere the object is taken e.g. if the stone on
earth is 75Kg, its mass on the moon will also be 75Kg.

Conversion of units
1 Kg = 1000g
1 tonne = 1000Kg
1 tonne = 1000000g

Measurement of mass
Comparing masses using a beam balance
When measuring the mass of a substance, we compare the mass of the measured object
with standard masses (known masses)

Procedure
1. Place the standard mass (e.g. 10kg) on one pan.(Standard mass of a substance of
mass 10kg is needed)
2. Place the measured object on the other pan until the object and standard mass
balances.
3. When the two balances, it means they have the same mass or weight.

Precautions
1. Clean the pans and beams

44 | P a g e
 2. Adjust the zeroing screw so that the pointer coincides with the zero mark.
3. Read the mass of the known mass object when the beam is balanced.
Determining the mass of a liquid

Experiment
Aim: To find the mass of the liquid, m

Apparatus
 Triple beam balance
 Beaker
 Liquid

Method
Place a dry empty beaker on the beam balance and record its mass, m1.
Pour the liquid into the beaker. Measure and record the mass of the liquid and beaker, m2.
Find the mass of the liquid using the formula; m = m2 – m1.

Conclusion
Mass of liquid = mass of beaker and liquid – mass of empty beaker

Precaution
1. The beaker should be cleaned and dried before the experiment.

Example
1. In an experiment to determine the mass of a certain volume of paraffin, the mass
of the beaker was found to be 20g. When the paraffin was poured into the beaker,
the mass increased to 42.5g. What was the mass of paraffin?

Solution
Mass of paraffin = mass of beaker and liquid – mass of empty beaker
m = m2 – m1
m = 42.5g – 20g
m = 22.5g

Determining the mass of air

Experiment

Aim: To find the mass of air, m.

Apparatus
 Bottle with air
 Beam balance
 Vacuum pump

45 | P a g e
Method
Place the bottle filled with air on the beam balance and record the mass, m1
Remove the air from the bottle using the vacuum pump. Measure and record the mass of
the empty bottle, m2
Find the mass of the air using the formula; m = m1 – m2
Conclusion
Mass of air = mass of bottle with air – mass of empty bottle

Example
1. The mass of the bottle filled with air is 50.65g. When the air is removed from the
bottle, the mass of the empty bottle is 50g. Calculate the mass of air.
Solution
Mass of air = mass of bottle filled with air – mass of empty bottle
m = m1 – m2
m = 50.65g – 50g
m = 0.65g

Activity thirteen
1. A bottle filled with air with mass 22g has a mass of 53.2g. Find the mass of the
empty bottle.

Weight
Symbol: W
SI unit: Newton. N
Definition: Weight is the force of gravity acting on an object.
Measuring instrument: Spring balance.

46 | P a g e
The weight of an object varies from place to place i.e. from the earth to the moon.
Weight is less on the moon and more on the earth.
There is no weight in the outer space. (Weight is equal to zero newtons)

Factors that affect the weight of an object

 Mass of an object
 Acceleration due to gravity, g
 Distance of an object from the centre of the earth.
The greater the mass of an object, the greater its weight.
When g is high, weight is also high and when g is low, weight is also low.
On the earth’s surface, g varies depending on how far the object is from the centre of the
earth.
Nearer to the centre of the earth, g is high and further away from the centre of the earth, g
is low.
As the object is moved further away from the centre of the earth, weight reduces because
g keeps reducing.
A place at the south or North Pole where g is zero, weight is also zero.
As the object is moved closer to the centre of the earth, g increases and weight also
increases.
Example
1. Explain why;
(a) The weight of the miner increases as he goes along a deep mine?
(b) The weight of the space craft reduces as it moves upwards?
(c) The rocket’s weight is zero?

Solution

47 | P a g e
 (a) Because g increases as the miner goes closer to the centre of the earth and this
results in an increase in the weight.
(b) Because g reduces as an object moves away from the centre of the earth and
weight also reduces.
(c) Because in the space g is equal to zero and also results in weight to be zero.

Differences between mass and weight

1. Mass is the quantity of matter contained in a substance while weight is the force
of gravity acting on an object.
2. Mass is a scalar quantity while weight is a vector quantity.
3. Mass is measured using a beam balance while weight is measured using a spring
balance.
4. Weight varies slightly from place to place while mass does not change.
5. The SI unit for weight is the Newton, N, while the SI unit for mass is the
kilogram, Kg. (A mass of 1Kg weighs approximately 10N)

Relationship between mass and weight

Weight = mass x acceleration due to gravity
W = mg
Where;
W = weight in Newtons, N
m = mass in kilograms, Kg
g = acceleration due to gravity in Newton per kilogram, N/Kg

Note
1. The value of g on earth is 10N/Kg
2. The value of g on the moon is 1.6N/Kg

Example
1. The mass of a man is 70kg. What is his weight on the moon?
Data Solution
W =? W = mg
m = 70kg W = 70kg x 10N/kg
g = 10N/kg W = 700N.

2. The weight of an on object is 300N.

(a) What is its mass on earth?
(b) What is its (i) mass and (ii) weight on the moon?
(c) What is its (i) mass and (ii) weight in the outer space?
Data Solution
a
m =? w
m=
W = 300N g
g = 10N/kg 300 N
m=
10 N /kg
m = 30Kg

48 | P a g e
 b(i) Mass does not m = 30kg
change
(ii) W =? W = mg
m = 30kg W = 30kg x 1.6N/kg
g = 1.6N/kg W = 48N
c (i) Mass does not m = 30kg
change
(ii) g = 0N/kg W = mg
m = 30kg W = 30kg x 0N/kg
W = 0N

Activity fourteen
1. A stone of mass 20kg is placed on earth where gravitational strength is 10N/kg.
(a) Find the weight of the stone on earth.
(b) What is the weight of the same stone on the moon?
2. A block of mass 5000g is found at a place on earth where g is 10N/kg.
(a) Find its weight at this place.
(b) What is its mass when it is taken down into the mine?
3. An astronaut with a mass of 75kg on earth travels to the moon whose gravitational
strength is 1.6N/kg.
(a) What is meant by mass?
(b) What is the mass of an astronaut on the moon?
(c) What is his weight on the moon?

Centre of gravity (Centre of mass)

Centre of gravity of an object is the point through which its whole weight appears to act.
Centre of gravity can also be defined as a point within an object where its total mass
seems to originate from.
How to determine the centre of gravity of an irregular object
Aim: To find out (locate) the centre of gravity of an irregularly shaped object (plane
lamina)

Apparatus
 String
 Plane lamina/ paper
 Pen/pencil/ruler with a knife edge
 Pin
 Bob

Method
Make a small hole near the edge of a flat plane lamina
Hang the plane lamina by a needle and make sure that it can swing freely.
Hang the plumb line from the same needle and again make sure that it is also free to turn
Mark the position of the plumb line on the plane lamina (to do this accurately, make a
point near the bottom edge of the plane lamina over which the string passes)

49 | P a g e
Draw a straight line from the needle to this point to represent the position of the plumb
line [the centre of gravity lies some where along this line]
Make two other holes near the edge of the plane lamina so that all the three holes are as
far as possible.
Repeat the experiment and draw two other lines.

Observation
The irregular shaped plane lamina balances at point C.

Conclusion
Since the centre of gravity lies on each of the lines, their intersection locates the centre of
gravity.

Stability
Stability of an object is defined as the ability of an object to regain its original position
after it has been displaced slightly.
Stability can also be defined as a condition in which an object is not moving and cannot
fall.
A stationary object can either be stable or unstable
Something stable is an object which cannot easily fall when slightly pushed or tilted.
Something unstable is an object which can easily fall when slightly pushed or tilted.

Conditions for stability

 Low centre of gravity
 Wide base
1. Low centre of gravity
The position of centre of gravity affects the stability of an object.
The centre of gravity should be as low as possible.

50 | P a g e
 2. Wide base
The base area should be as large as possible
The wider the base, the more stable an object will be.
The mass of an object should be concentrated at the base.

Equilibrium
Equilibrium is a condition of an object in which the sum of all forces acting on it is zero
e.g. resultant force is zero
Objects which are in equilibrium are;
(a) those that are stationary i.e.at rest
(b) those that are moving with constant velocity
A stationary object can either be in a stable equilibrium, unstable equilibrium or neutral
equilibrium.
1. Stable equilibrium
An object is said be in stable equilibrium if when slightly pushed or tilted goes back to its
original position.
Examples

The objects above are more stable because they have;

(a) Low centre of gravity and
(b) Wider base
Objects in stable equilibrium do not easily fall when slightly pushed because the vertical
line of force from the centre of gravity does not easily fall on the other side of base.

2. Unstable equilibrium
An object is said to be in unstable equilibrium if when slightly pushed or tilted falls off
i.e.it does not go back to its original position.

Examples
Topples over when tilted
Topples over when tilted

High centre of gravity

High centre of gravity

51 | P a g e
 Vertical line of force Vertical line of force
The objects above are unstable because they have;
(a) High centre of gravity
(b) Smaller or narrow base
Objects in unstable equilibrium fall off easily because when slightly pushed or tilted, the
vertical line of force easily falls off on the other side of the base.

3. Neutral equilibrium
An object is said to be in neutral equilibrium if it stays in its new position after it has
been pushed slightly.
Example

When a ball and a cylinder are rolled, they come to rest in a new stable equilibrium.

Note
It is not advisable to put a heavy luggage on the roof of a minibus because it can topple
over at the corner when it is moving fast.

Activity fifteen
1. The figure below shows a bus

52 | P a g e
 (a) State three modifications that should be made in the design of the bus to make it
more stable.

1. The diagram below shows two identical rectangular wooden blocks A and B.
Block B has a layer of lead attached to its base. The blocks were tilted about
edges PQ as shown in the diagram below.

Explain why A topples over at a smaller angle of tilt than B

(a) State two conditions which can help to prevent a truck toppling over when tilted.

Volume
Symbol: V
SI unit: cubic metre, m3
Definition: Volume is the amount of space occupied by an object.

Other units for volume

Cubic centimeters, cm3
Mililitres, ml
Litres, L

Relationship of units

53 | P a g e
1ml = 1cm3
1L = 1000ml = 1000cm3
1m3 = 1000L = 1000000cm3

Note
In the laboratory, we usually use cubic centimeters because the cubic metre is a very
large unit.

Instruments for measuring volume of liquids

Measuring cylinder
It measures various volumes of liquids

Pipette
It measures a fixed amount of volume of liquid according to its capacity

Burette
It measures the required volume up to its capacity

Flasks
They give or measure approximate fixed volumes

Volume of regular solids

An irregular solid is an object whose sides can be measured easily.

Procedure
Measure the length of an object using a ruler or vernier calipers or micrometer screw
gauge
Use the appropriate formula to find the volume.

Object Formula
Cuboid (rectangle) V=lxbxh
V=Axh
Cube (square) V = l3
Sphere (circle) 4
V = πr3
3
Cylinder (wire or pipe) V = πr2h
V=Axh
Cone 1
V = π2h
3
Pyramid 1
V = bh
3
Prism 1
V = bh
3

54 | P a g e
 Examples
1. Find the volume of the block which has the following measurements;
length = 10cm, breadth = 6cm, height = 3cm.
Data Solution
V =? V=lxbxh
l = 10cm V = 10cm x 6cm x 3cm
b = 6cm V = 180cm3
h = 3cm

2. Find the volume of a cube of sides 4cm.

Data Solution
V =? V = l3
V=lxlxl
l = 4cm V = 4cm x 4cm x 4cm
V = 64cm3

3. Calculate the volume of the sphere of radius 6cm.

Data Solution
V =? 4 3
22 V = πr
3
π=
7 4 22
V = x x 6cm x 6cm x 6cm
r = 6cm 3 7
V = 905cm3

Activity sixteen
1. Calculate the volume of the pipe of cross section area 30cm2 and 50cm long.
2. Find the volume of a wire of diameter 0.2cm and height 7cm.

Volume of liquids
Liquids take the shape of the container in which they are placed.
If a container is filled to its capacity, its volume can be determined by pouring the
contents into the measuring cylinder.

How to read volumes of liquids

When a liquid is poured into a measuring cylinder, it forms a curved surface on the upper
part of the liquid.
The curve could be concave or convex depending on the properties of the liquid.
The curved surface is called meniscus and is caused by the attraction between the liquid
particles and the container.
When the meniscus is convex (i.e. curving upwards) it is read from the top and when it is
concave (i.e. curving downwards) it is read from the bottom.

Precautions
1. Place the measuring cylinder on the horizontal flat surface

Volume of irregular solids

55 | P a g e
An irregular solid is an object whose sides cannot be measured easily.
An irregular solid has no specific dimensions e.g. a stone
The volume of small solids is measured by the displacement method using;
 A measuring cylinder
 An over flow can

(a) Using a measuring cylinder

Experiment

Aim: To find the volume of the stone, V

Apparatus
 Measuring cylinder
 Water
 Stone
 Thin string

Method
Pour water into a measuring cylinder and record the initial water level, V1
Tie a piece of thin string around a small stone and slowly lower the stone into the
measuring cylinder until it is fully submerged. Record the final water level, V2
Find the volume of the stone using the formula, V = V2 – V1

Conclusion
Volume of water displaced by the stone is equal to the volume of the stone.

(b) Using an over flow can

Experiment

Aim: To find the volume of the stone, V

Apparatus
 Over flow can
 Water
 Measuring cylinder
 Tripod stand
 Small stone
 Thin string

Method
Place an over flow can on a tripod stand
Pour water into an over flow can until it begins to flow from the spout.

56 | P a g e
Leave the can until the water stops over flowing (dripping)
Place an empty measuring cylinder under the spout
Tie a piece of thin string around a small stone and slowly lower the stone into the can
until it is fully submerged.
Water from the can is collected in a measuring cylinder. Water collected in the cylinder is
the volume of the stone.

Conclusion
The water collected in the measuring cylinder is called displaced water and its volume is
equal to the volume of the stone lowered in the can.

Precautions
1. Use a thin string to reduce the amount of water displaced by it.
2. Use a solid that does not react or dissolve in the liquid.
3. Lower the irregular solid gently to avoid the splashing of the liquid.
4. Place the measuring cylinder on the flat or horizontal surface
5. Tap the measuring cylinder to remove any amount of air bubbles.
6. Place the eye level with the flat surface of the liquid [in case of water, read from
the bottom of the meniscus]

Example
1. 100cm3 of water is poured into a measuring cylinder. A block of copper wire is
gently lowered into the measuring cylinder and the water level rises to the 183cm
mark.
(a) What is the volume of the copper block?
(b) If the height of the block is 10cm, what is the cross sectional area?

Data Solution
a V =? V = v2 – v1
V2 =183cm3 V = 183cm3 – 100cm3
V1 = 100cm3 V = 83cm3
b
A =? V
A=
h
V = 83cm3
83 cm 3
h = 10cm A=
10 cm

A = 8.3cm2

Volume of a small irregular floating solid

Experiment

57 | P a g e
Aim: To find the volume of an irregular floating solid, V

Apparatus
 Cork (floating object)
 Stone
 Water
 Thin string
 Measuring cylinder

Method
Pour water into the measuring cylinder.
Tie a thin string around a small stone and gently lower the stone into the measuring
cylinder until it is fully submerged. Record this initial water level, V1.
Then tie a floating object together with the stone and then lower them into the same
measuring cylinder. Water level rises and record the this final water level, V2
Find the volume of the floating object using the formula, V = V2 – V1

Conclusion
Volume of floating object = final volume – initial volume

Note
The stone is used to make the floating object to sink or submerge
Anything that sinks can be used in place of a stone.

Density
Symbol: ρ
SI unit: kilogram per cubic metre, kg/m3
Definition: Density is defined as mass per unit volume of a substance
mass
Formula: Density =
volume

m
ρ=
v

Relationship of units
1kg/m3 = 0.001g/cm3
1g/cm3 = 1000kg/m3

Example
1. Convert
(a) 3kg/m3 into g/cm3
(b) 5g/cm3 into kg/m3
Solution
(a) 0.001g/cm3 → 1kg/m3
x → 3kg/m3

58 | P a g e
 0.001 g /cm3 x 3 kg /m 3
x=
1kg / m 3

x = 0.003g/cm3

(b) 1g/cm3 → 1000kg/m3

5g/cm3 → x

5 g /cm3 x 1000 kg /m 3
x=
1 g/cm 3

x = 5000kg/m3

Simple determination of density

Find the mass of an object using a beam balance
Find the volume of an object
Find the density of an object using the formula;
mass
Density =
volume

m
ρ=
v
Density of irregular solids

Experiment

Aim: To find the density of an irregular object, ρ

Apparatus
 Measuring cylinder
 Beam balance
 Stone
 Water
 Thin string

Method
Measure and record the mass of the stone, m
Pour water in the measuring cylinder and record the initial volume of water, V1
Slowly, lower the stone into a measuring cylinder using a thin string and record the final
volume of water, V2.
Find the density of the stone by using the formula;
m
ρ =
V 2−V 1

59 | P a g e
Examples
1. A body of mass 500g was suspended in 100cm3 of water by a piece of cotton. The
level rises to 150cm3 . What is its density?

Data Solution

ρ =? m
ρ =
V 2−V 1
m = 500g 500 g
ρ =
150 cm3 −100 cm3
V1= 100cm3
500 g
V2 = 150cm3 ρ =
50 cm3

ρ = 10g/cm3

Activity seventeen
1. A material has density of 9.0g/cm3 and volume 50cm3. What is its mass?
2. A metal has mass of 225g and volume of 30cm3. What is its density?

Density of liquids

Experiment

Aim: To find the density of a liquid, ρ

Apparatus
 Measuring cylinder
 Beam balance
 Liquid

Method
Measure and record the mass of an empty cylinder, m1.
Pour the liquid into the measuring cylinder. Measure and record the mass of the cylinder
and water, m2.
Record the volume of the liquid in the beaker, V
Find the density of the liquid by using the formula;
m 2−m1
ρ =
V

60 | P a g e
Example
1. A container of mass 200g and contains 160cm3 of liquid. The total mass of the
container and liquid is 520g. What is the density of the liquid?
Data Solution

ρ =? m 2−m1
ρ =
V
m1 = 200g
520 g−200 g
m2 = 520g ρ =
160 cm 3
v = 160cm3 320 g
ρ =
160 cm3

ρ = 2.0g/cm3

Activity eighteen
1. A stone of mass 20g and density 0.5g/cm3 was immersed into water in a
measuring cylinder whose initial volume was 30cm3. Find the final volume of the
water in the measuring cylinder.

2. What is meant by the density of a substance? State constituent units in which the
various quantities you have mentioned could be measured.
(a) A tin containing 5000cm3 of paint has a mass of 7.0kg.
(i) If the mass of the empty tin including the lid is 0.5kg, calculate the density
of the paint.
(ii)If the tin is made of a metal which has a density of 7800kgm-3, calculate
. the volume of metal used to make the tin and the lid.

Relative density
Symbol: ρr
Definition: Relative density is the ratio of the mass of a substance to the mass of water.
It is also the ratio of the density of a substance to the density of water
mass of liquid
Formula: Relative density =
mass of water

density of a substance
Relative density =
density of water

Units: Relative density has no units.

Note
Density of water = 1g/cm3 or 1000kg/m3
Relative density is also called specific gravity.

61 | P a g e
Example
1. Find the relative density of a liquid of mass 300g if it has the same volume as
100g of water.
Data Solution
ρr =? mass of liquid
ρr =
mass of water
Mass of liquid = 300g
300 g
Mass of water = 100g =
100 g
=3

Activity nineteen
1. The density of mercury is 13600kg/m3. The density of water is 1000kg/m3.
Calculate the relative density of mercury.

Experiment
Aim: To find the relative density of a liquid using the density bottle.

Method
Measure and record the mass of the density bottle, m1.
Measure and record the mass of the density bottle containing the water, m2.
Measure and record the mass of the density bottle containing the liquid under
investigation, m3.
Find the relative density by using the formula;

m3 −m 1
ρr =
m 2−m 1

Conclusion
mass of liquid
Relative density of liquid =
mass of water

NB: The density of a liquid is then found by multiplying relative density of the liquid by
the density of water.

Density of liquid = relative density x density of water

Precautions when using a density bottle

1. The density bottle must be thoroughly dried.
2. The water outside the density bottle must be dried completely with a dry cloth
3. The density bottle must be held by the neck to avoid expansion of the liquid [if
the bottle itself is held in the hands, the heat will cause expansion to the liquid]
4. Remove the water from the top of the stopper with a blotting paper.

62 | P a g e
Example
1. An empty relative density bottle weighs 25g. It weighs 65g when filled with a
liquid and 75g when filled with water.
(a) Calculate the relative density of the liquid
(b) Calculate the density of the liquid
Solution
(a) Mass of liquid = 65g - 25g
= 40g
Mass of water = 75g – 25g
= 50g
mass of liquid
Relative density of the liquid =
mass of water

40 g
Relative density of the liquid =
50 g
= 0.8
(b) Density of liquid = relative density x density of water
= 0.8 x 1g/cm3
= 0.8g/cm3
Activity twenty
1. An empty relative density bottle has a mass of 25g. When filled with a liquid of
relative density 0.92, its mass becomes 85g.
Calculate
(a) The mass of the bottle when filled with water.
(b) The capacity of the bottle

Density of air
Experiment

Aim: To find the density of air, ρ

Apparatus
 Beam balance / top pan balance
 Bottle / container with a top and tube
Note
A tube of the container can be connected to a suction pump which draw air in or suck air
out of the container

Method
Measure and record the mass of the container filled with air, m1.
Remove all the air from the container using a suction pump and then close the tap.
Measure and record the mass of the container without air. (empty container), m2.
NB: The volume, V, of the container should be known.
Next, open the container and fill it with water. Close the container tightly and make sure
all the air has been replaced by water

63 | P a g e
Measure and record the mass of the container filled with water, m3.
Find the density of air by using the formula;

m1 −m2
ρ=
V

Conclusion
mass of container withair−mass of empty container
Density of air =
volume of the air

Note
Volume of container = Volume of air = Volume of water
The volume of air depends very much on the temperature and pressure of the
surrounding. It is therefore important to take note of the temperature and atmospheric
pressure during the experiment.

Example
1. Mr.Naosa D.K, a physics teacher at Kambule Technical High School did an
experiment to find the density of air and he obtained the following results:
Mass of container = 265.12g
Mass of container and air = 265.42g
Mass of container and water = 515.12g
Take density of water to be 1g/cm3
Calculate
(a) The mass of air
(b) The mass of water
(c) The volume of the container
(d) The density of air
Solution
(a) Mass of air = mass of container with air – mass of empty container
= 265.42g – 265.12g
= 0.3g
(b) Mass of water = mass of container with water – mass of empty container
= 515.42g – 265.12g
= 250g

(c) Volume of container = volume of water

mass
Volume =
density

250 g
=
1 g /cm3

= 250cm3

64 | P a g e
 (d) Volume of air = volume of container

mass
Density of air =
volume

0.3 g
=
250 cm3

= 0.0012g/cm3

Activity twenty one

1. An experiment was carried out by Darlington Naosa junior to determine the
density of air using the density bottle.
The following results were obtained:
Mass of empty bottle = 309g
Mass of bottle filled with air = 310g
Mass of bottle filled with water = 1050g
Take density of water to be 1g/cm3

(a) What was the mass of water?

(b) What was the initial volume of the bottle?
(c) What was the mass of air?
(d) Calculate the density of air.

Density of a mixture
Method
Add the mass of the components to find the total mass
Add the volume of the components to find the total volume
Find the density of the mixture by using the formula;

Total mass of mixture

Density =
Total volume of mixture

Example
1. 30g of alcohol of volume 38cm3 is mixed in a jug with water of volume 20cm3
with mass 20g. Find the density of the mixture.

Solution
Total mass of mixture = 30g + 20g
= 50g
Total volume of mixture = 38cm3 + 20cm3
= 58cm3

Total mass of mixture

Density =
Total volume of mixture

65 | P a g e
 50 g
=
58 cm3

= 0.86g/cm3
Activity twenty two
1. 32g of kerosene of density 0.80g/cm3 is mixed with 8g of water.
(a) Find the total mass of the mixture
(b) Find the volume of kerosene
(c) Find the volume of water
(d) Calculate the total volume of the mixture
(e) Calculate the density of the mixture
2. 300cm3 of water is mixed with 300cm3 of pure alcohol. Calculate the density of
the mixture if the relative density of alcohol is 0.79.
Note
 When impurities of pollutants are added to a substance, its density increases e.g.
the density of water is 1g/cm3, but when salt is added to it, density increases
depending on the amount of impurities.
 An egg sinks in pure water because an egg is denser than pure water and an egg
floats in salt water because salt water is denser than an egg.
Force
Symbol: F
SI unit: Newton, N
Definition: Force is the push or pull exerted on an object
Measuring instrument: Spring balance.

Examples of forces
 Weight
 Friction
 Tension
 Up thrust
 Magnetic force
 Electric force
 Constant force

Effects of force on an object

1. Force can change the size and shape of an object
2. Force can change the motion of an object. Force can change the motion of an
object in the following ways:
(a) It makes an object to start moving
(b) It makes an object accelerates either uniformly or non-uniformly i.e. makes an
object accelerates uniformly if the force is constant and makes an object
accelerates non uniformly if the force varies.
(c) It makes an object decelerates
(d) It makes an object change direction

66 | P a g e
 3. Force can make an object to turn about the point (pivot). It can also make an
object to rotate.

Newton’s laws of motion

There are three basic laws of motion given by Sir Isaac Newton

Newton’s first law of motion

The law states that: Any given body continues in its state of rest or uniform motion in a
straight line unless it is compelled to change that by an external force exerted on it.
Newton’s first law of motion is also called the law of inertia.
Inertia is the property of a body that resists a change to its motion.
Inertia is not a force but a property of an object.
Inertia depends on the mass of an object.
If something has a high resistance [high mass] to the change of motion, its inertia is said
to be high.

Note
1. A wire car is easier to start and easier to stop
2. A heavy truck has high inertia and it is difficult to start moving and difficult to
stop.

Every day effects of inertia

(a) When a fast moving bus stops suddenly, the passengers tend to be thrown forward
to maintain their forward motion. Similarly, when a bus suddenly starts moving,
the passengers are thrown backwards; they tend to remain behind due to inertia.
(b) When a block is placed on a smooth card on the table and the card is suddenly
pulled away horizontally, the block remains behind.

Newton’s second law of motion

The law states that: An unbalanced force acting on a body produces an acceleration in the
direction of the force.
This acceleration is directly proportional to the force but inversely proportional to the
mass of the body.

1
a∞F and a∞
m

Force = mass x acceleration

F = ma

Note
F
 a=
m

67 | P a g e
 F
 m=
a

F = force [N]
m = mass [kg]
a = acceleration [m/s2] or [N/kg]

Example
1. A horizontal force of 5N was applied to a brick of mass 2kg resting on a
frictionless table. What was the acceleration of the brick?
Data Solution

a =? F
a=
m
F = 5N
5N
m = 2kg a=
2 kg

a = 2.5N/kg

Activity twenty three

1. A man pushes an 8kg luggage on the smooth floor. It starts from rest and reaches
the final velocity of 15m/s in 5seconds.
(a) Calculate the acceleration
(b) What was the force acting on the luggage?
Newton’s third law of motion
The law states that: To every action, there is an equal and opposite reaction.

Resultant force
Symbol: Rf
Definition: Resultant force is the sum of all forces acting on a body.
Formula: Resultant force = sum of forward forces – sum of backward forces
Or Resultant force = horizontal force - friction
Example
1. Find the resultant force of each of the following:

Direction of motion

5N 8N
(a)

68 | P a g e
 4N

(b) 11N 12N

1N 2N
(c)
4N 3N

Solution
(a) Rf = 8N – 5N
= 3N
(b) Rf = (4N + 12N) – 11N
= 16N – 11N
= 5N
(c) Rf = (2N + 3N) – (1N + 4)
= 4N – 4N
= 0N

2. The figure below shows the total forces acting forwards and backwards on a car at
different times X, Y and Z during a journey.

Direction of motion

69 | P a g e
In each case, the car is moving forwards. The mass of the car is 1000kg.
(a) State the name of one of the forces that is acting in the opposite direction to the
motion of the car.
(b) State whether the speed of the car is changing at time X.
Explain your answer.
(c) State whether the speed of the car at time Y is increasing, decreasing or is
constant.
Explain your answer.
(d) Calculate the acceleration of the car at time Y.

Data Solution
(a) Friction
(b) The speed is not changing
Forward force = Backward force Reason: Because the resultant force is zero
(i.e. Rf = 3000N – 3000N = 0N)
(c) Increasing
Forward force > Backward force Reason: Because the forward force is greater than the
backward force.
(d) a =? F
a=
m
F = 5000N – 3000N = 2000N
2000 N
m = 1000kg a=
1000 kg

a = 2m/s2

Activity twenty four

1. The figure below shows an object of mass 0.7kg resting on a horizontal

surface.

70 | P a g e
 6.0 N 2.5 N

If the object is pulled to the left by a force of 6.0 N and to the right by a force of 2.5N and
assuming that no other forces act on the object,
Calculate
(a) The resultant force
(b) The acceleration produced by the resultant force in (a)
(c) Explain why in practice the actual acceleration for the object may be lower than
your answer in (b) above.

Friction
Friction is the force which opposes the motion of two touching surfaces.
Friction acts in the opposite direction to the motion of an object.

Application of friction
1. It enables us to walk without slipping
2. It enables us to hold or grip something
3. It helps a vehicle to run and stop.

Friction and its effects on motion

1. It causes an object to move with constant velocity. In this case, horizontal
force (forward force) is equal to friction (backward force).e.g.

Direction of motion

20N 20N

Rf = 20N – 20N
= 0N

2. It causes an object to come to rest or decelerates. In this case, friction is

greater than forward force. e.g.

Direction of motion

71 | P a g e
 250N 200N

Rf = 200N – 250N
= -50N
3. It causes the change in the direction of motion

Problems (consequences) of friction

1. It produces unnecessary heat and reduces the efficiency of machines
2. It causes the wearing and tearing of surfaces in contact
How to reduce friction
1. Lubrication of surfaces in contact using grease or oil
2. Putting ball bearings between movable surfaces in contact

Activity twenty five

1. A man pushes a packing car having a total mass of 400kg across a floor at a
constant speed of 0.5m/s by exerting a horizontal force of 100N.
(a) How big was the force of friction acting on the car?
(b) What was the resultant force on the car?

Force and motion in a circular path

There are a number of objects which move round in circular motion.

Examples
1. The moon goes round the earth
2. The earth goes round the sun in an orbit
3. In the laboratory, a mass tied to a string can be made to swing round.

Centripetal force
Centripetal force is a force where the direction of the force is always directed towards the
Centre of the circle.
The force of circular motion is always at right angles to the motion.
Object
Direction of centripetal force and acceleration
Direction of motion

The acceleration caused by the centripetal force is called centripetal acceleration

Effects of force on the shape and size of an object

Force changes the shape and size of an object
The change of shape and size of an object is called deformation
Force can change the size and shape in the following ways;
(a) It compresses the object, hence reduces it in size
(b) It stretches the object, hence makes it longer

72 | P a g e
 (c) It twists the object, hence changes its shape

Elastic material
It is a substance which regains its original shape and size when the force applied has been
removed

Examples of elastic materials

 Spring
 Rubber

Elasticity
It is the ability of an elastic material to regain its original shape and size after the applied
force has been removed

Elastic limit of the spring

It is the maximum force that can be applied to a spring without stretching it permanently

Original (neutral) length

It is the length of the spring before being stretched
Original length = new length – extension

New length
It is the length the spring reaches when it is stretched
New length = original length + extension

Extension
It is the difference between the new length and original length of the spring
Extension = new length – original length

Experiment

Title: Hooke’s law

Aim: To find the relationship between loads and extensions on a spring

Apparatus
 Spring
 Loads (standard masses)
 Clamp and stand

Method
Support the spring vertically by means of a clamp and stand. Place a pan on the lower end
of the spring.
Measure the original length of the spring
Hang a load (standard mass) on the lower end of the spring
Calculate the new length of the spring

73 | P a g e
Calculate the extension of the spring
Repeat the experiment by adding loads
Calculate the spring constant by using the formula;
Load F
Constant = , K=
Extension E

Plot a graph of load against extension

Elastic limit = 6N

Load
Constant =
Extension

6N
Constant =
10 cm

Constant = 0.6N/cm
Conclusion
The extension of the loaded spring is directly proportional to the force applied, provided
the elastic limit is not exceeded. This is called Hooke’s law.
Example
1. A load of 1N extends a spring by 5mm. What load extends it by 10mm?
Solution
1N → 5mm
x → 10mm

1 N x 10 mm
x=
5 mm
x = 2N
2. Calculate the extension of a spring that would be produced by a 20N load if a 15N
load extends the spring by 3cm?
Solution
15N → 3cm

74 | P a g e
20N → x

20 N x 3 cm
x=
15 N

x = 4cm

Activity twenty six

1. A load of 4N extends a spring by 10mm. What load would extend it by 15mm?
2. A steel spring obeys Hooke’s law. A force of 8N extends a spring by 10mm.
Calculate the extension of the spring that would be produced by a force of 10N
3. A spring of neutral length 3cm is extended by a force of 4N. What will be
(a) the stiffness of the spring
(b) its extension when a force applied is 12N
(c) its length when a force applied is 12N
4. Use the data below to answer this question;
Original length = 20cm
New length = 25cm
Load = 50N
(a) Find the extension of the spring
(b) Calculate the elastic limit
(c) Find the extension caused by the 100N load that the elastic is not exceeded
(d) Find the new length when the spring is stretched by the 100N force.
5. In an experiment to verify Hooke’s law, standard masses were placed on the pan
which was attached to a suspended spring at the lower end. The corresponding
lengths of the stretched spring were recorded as shown below.

Load/N 0.0 1.0 1.5 2.0 2.5 3.0 3.5 3.8 3.4 3.1
0.0
Length of the spring/mm 500 505 510 515 520 525 528 530 530 530
Extension/mm

(a) Complete the table by filling in the values of extension

(b) Plot the graph of load against extension
(c) Show clearly on the graph the elastic limit
(d) Use your graph to determine the spring constant

6.

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 12cm

14cm
10N

20N

In the figure above, the length of the spring with 10N force hang on it is 12cm and with
20N is 14cm. What would be the length of the spring with 12N hang on it if the spring
obeys Hooke’s law?

Moments
Symbol: Г
SI: Newton metre, Nm
Definition: Moment is the turning effect of the force about the pivot
Moment of a force
Moment of a force about a pivot is the product of the force and perpendicular distance
from the point to the line of action of the force.
Moment = force x perpendicular distance
Г=Fxd
Note
Г = moment [Nm]
F = force [N]
d = perpendicular distance [m]

Perpendicular distance must be distance from the pivot to the force

Perpendicular distance must be at right angle to the force
Force

[A] Bar

Perpendicular distance

In this case, there is a moment because the force is perpendicular to the bar.
The force can produce the turning effect.

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[B] Bar Force

Not perpendicular distance

In this case, there is no moment because force is in the same direction of distance
The force doesn’t produce the turning effect.

Example
1. Calculate the moment of the force at the pivot
3N

2m

Data Solution
Г =? Г= F x d
F =3N Г= 3N x 2m
d = 2m Г = 6Nm
Principle of moments
The law states that: For a body in equilibrium, the sum of clockwise moment is equal to
the sum of anticlockwise moment about the same point.

Total anticlockwise moment = Total clockwise moment

Experiment
Title: Moments
Aim: To verify the principle of moment
Apparatus
 Long ruler (30cm or more)
 3 string
 Loads

Method
Hang a ruler by a string at the centre of mass and make it balanced
Hang some loads at a certain point from the pivot
Find the position where other loads are hanging to balance the ruler and measure the
length from the pivot to the position.
Calculate the clockwise moment and the anticlockwise moment
Repeat the experiment with different pairs of loads and distances

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 d1 d2

F1 F2
Anticlockwise moment, Г1 Clockwise moment, Г2

Г1 = Г2
F1 x d1 = F2 x d2
Conclusion
If a body is balanced, then the total clockwise moment is equal to the total anticlockwise
moment.
Example

1. Find the force, F1, if the bar below is balanced

0.5m 0.4m

F1 100N

Data Solution
Г1 = Г2 F1 x d1 = F2 x d2
F1 =? F1 x 0.5m = 100N x 0.4m
d1 =0.5m F1 x 0.5m = 40Nm
F2 = 100N 40 Nm
F1 =
d2 = 0.4m 0.5 m
F1 = 80N

2. Find the distance, d1, if the bar below is balanced.

3m
d1 2m

5N 3N 2N

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 Data Solution
Г1 = Г2 + Г3 F1 x d1 = F2 x F2 + F3 x d3
d1 =? 5N x d1 = 3N x 2m + 2N x 3m
F1 = 5N 5N x d1 = 6Nm + 6Nm
d2 = 2m 5N x d1 = 12Nm
F2 = 3N 12 Nm
d1 =
d3 = 3m 5N
F3 = 2N d1 = 2.4m

3. Below is the diagram of a uniform beam suspended on a pivot. Four coins of

equal masses are put on points B and F as shown below.

Coins

A B C D E F G H

(a) What happens to the beam if left to move freely?

Give a reason for your answer
(b) Which position on the beam would you put one coin to balance the beam? Mark
the position with letter P
Solution
(a) The beam will tilt anticlockwise
Reason: The anticlockwise moments are more than the clockwise moments
(b) P should be at H

Determining mass using the principle of moments

Hint
1. Weight of a body acts through its centre of gravity
2. Centre of gravity of a body with a uniform cross section is at mid points
Example
1. A metre rule pivoted at the 40cm mark is balanced by an 80g placed at the 20cm
mark. Find the mass of the rule.
Solution

0cm 20cm 40cm 50cm 100cm

80g W

20cm x 0.8N = 10cm x W

20 cm X 0.8 N
W=
10 cm

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 W = 1.6N

W
m=
g

1.6 N
m=
10 N /kg

m = 0.16kg = 160g

Activity twenty seven

1. Calculate the force, F3, if the bar below is balanced

4m
2m 2m

5N 3N F3

2. The diagram below shows a uniform rule, weight, W, pivoted at the 75cm mark
and balanced by a force of 2N acting at the 95cm mark.

50cm 75cm 95cm

W 2N
(a) Calculate the moment of the 2N force about the pivot
(b) Use the principle of moments to calculate the value of W.
3. A metre rule hangs by a string at the 80cm mark and a mass of 140g hangs at
95cm mark. The weight of the ruler appears on the centre of mass.
(a) Where is the pivot?
(b) What is the weight of the 140g mass?
(c) Calculate the weight of the ruler, W
(d) Calculate the mass of the ruler
4. The diagram below shows the uniform metre rule balanced horizontally on a
knife-edge placed at the 58cm mark when a mass of 20g is suspended from the
end.

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 0cm 58cm 100cm

20g

(a) Find the mass of the rule

(b) What is the weight of the rule (taking g = 10m/s2)
(c) A candle stand has a wide heavy base. Explain why the base has both heavy mass
and wide area.

Simple machines
A machine is any device by means of which a force applied at one point can be used to
overcome a force at another point.
Effort
Symbol: E
Definition: Effort is the applied force
Load
Symbol: L
Definition: Load is the force which the effort overcomes
Load can also be defined as the force an object pulls or pushes on a machine.

Mechanical advantage,
Symbol: M.A
Definition: Mechanical advantage is the ratio of the load to the effort
Load
Formula: M.A =
Effort
Units: M.A has no units since it is a ratio whose units cancel each other.

Velocity ratio
Alternative term: Ideal mechanical advantage
Symbol: V.R
Definition: Velocity ratio is the ratio of the distance moved by the effort to the distance
moved by the load

Distance moved by effort

Formula: V.R =
Distance moved by load

Units: V.R has no units since it is a ratio whose units cancel each other.

V.R of a lever

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Efficiency of a machine
Symbol: η
Definition: Efficiency is the ratio of the useful energy output to the energy input
multiplied by 100%.
Energy out put
Formula: η = x 100%
Energy∈ put

M.A
η= x 100%
V .R
Note
Efficiency of a machine can never be more than 100% because the energy out put (work
done by a machine) is never more than energy in put (work done on the machine)
η < 100%
Efficiency of a machine cannot be 100%
Reasons:
 Some energy is used to overcome friction
 Some energy is used to move parts of the machine

M.A< V.R
Generally, in an ideal situation, the efficiency of any machine is equal to100% and this
just theoretical. This means that M.A = V.R or energy out put = energy in put.

1. Prove that M.A ≤ V.R

Solution

Efficiency of a machine is equal or less than 100%

Efficiency ≤ 100%

η ≤ 100%

M.A
η= x 100%
V .R

M.A
x 100% ≤ 100%
V .R

1 M.A 1
x x 100% ≤ 100% x
100 % V .R 100 %

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M.A
≤1
V .R

M.A ≤ V.R, hence proved

2. Find the efficiency of an electric motor that is capable of pulling a 50kg mass
through a height of 15m after consuming 30,000J of electric energy.

Solution
Energy output = mgh
= 50kg x 10N/kg x 15m
= 7,500J

Energy in put = 30,000J

Enrgy out put

η= x 100%
Energy∈ put

7,500 J
= x 100%
30,000 J

= 25%

Types of simple machines

A. Levers

Load
Effort

Pivot

¿
V.R = Distance ¿ pivot ¿ effort Distance ¿ pivot ¿ load ¿

Some examples of levers

 Wheel barrow
 Claw hammer
 Table knife
 Scissors
 Bore hole

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Example
1. Study the diagram below and answer the questions that follow

60N
20N

10cm 40cm
Calculate
(a) The mechanical advantage
(b) The velocity ratio
(c) The efficiency

Solution
Load
(a) M.A =
Effort

60 N
M.A =
20 N

M.A = 3

¿
(b) V.R = Distance ¿ pivot ¿ effort Distance ¿ pivot ¿ load ¿

40 cm
V.R =
10 cm

V.R = 4
M.A
(c) Efficiency = x 100%
V .R

3
Efficiency = x 100%
4

Efficiency = 75%

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 2. A load is to be moved using a wheelbarrow. The total mass of the load and
wheelbarrow is 60kg. The gravitational field strength is 10N/kg

. F

Wheelbarrow

Load [weight]

70cm 50cm

Pivot

What is the size of force, F, needed just to lift the loaded wheelbarrow?

Data Solution
F =? W = mg
W = 60kg x 10N/kg
m = 60kg W = 600N
600N x 70cm = F x 120cm
g = 10N/kg
600 N x 70 cm
F=
d1 = 70cm 120 cm

d2 = 120cm (70cm + 50cm) F = 350N

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 B. Pulleys
A pulley is a wheel with a grooved rim mounted on a block

Types of pulleys
1. Single fixed pulley

Rope

Effort
Load

Load = Effort
M.A = 1

2. Single moving pulley

Effort Effort

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 Load
Load is twice effort
M.A = 2

3. Block and tackle

Effort

Load

V.R = Number of lines or pulley wheels

V.R = 4

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Activity twenty eight
1. The diagram below shows the pulley system

1000N

88 | P a g e
 2250N

Find
(a) The mechanical advantage
(b) The velocity ratio
(c) The efficiency

C. Inclined plane

Effort
Length of slope, l (distance moved by effort)

. Distance moved
by load (height, h)
Load (weight)

.
.
Sin Ɵ

Distance moved by effort

V.R =
Distance moved by load

OR

Length of slope
V.R =
Height

1
V.R =
sin Ɵ
Activity twenty nine
1. The diagram below shows an inclined plane.

89 | P a g e
 20m
10N
1m
70N

Find
(a) The mechanical advantage
(b) The velocity ratio
(c) The efficiency
D. Gears

Rotations

Output shaft Input shaft

Driving wheel Driven wheel

Teeth Teeth

Number of teeth∈driving wheel

V.R =
Number of teeth∈drven wheel

Number of rotations ∈driven wheel

Number of rotations in driving wheel =
V.R

Activity thirty
1. The figure below shows the diagram of rotating gear wheels. The driving wheel
has 36 teeth and the driven wheel has 12 teeth.

Rotations

Driving wheel Driven wheel

36 teeth 12 teeth

(a) Find the velocity ratio

90 | P a g e
 (b) If the driven wheel makes 15 rotations, how many rotations would the driving
wheel make

Work
Symbol: W
SI unit: Joule, J
Definition: Work is the product of the force and the distance moved in the direction of the
force.
Formula: Work = force x distance
W=Fxd
W = weight x height
W = mgh

Note
W = work [J]
F = force [N]
d = distance [m]
h = height [m]
m = mass [kg]
g = acceleration due to gravity [10m/s2] or [10N/kg]

Joule is the work done when the point of application of a force of 1 Newton moves
through 1 metre in the direction of the force.
1 Newton-metre is equal to 1 joule of work.
Work is said to have been done when we push an object through a certain point or when
we lift an object from the ground.

Relationship of units
1KJ = 1000J
1MJ = 1000KJ = 1000000J = 106J

NB: KJ = kilo joule

MJ = mega joule

Example
1. A force of 5N acts on a 2kg brick, moving it 8m horizontally from rest. Find the
work done by the force.

Data Solution
W =? W=Fxd
F = 5N W = 5N x 8m
d = 8m W = 40Nm
W = 40J

91 | P a g e
 2. A hawk picks a 2kg chicken and lifts it up to a branch of a tree 15m from the
ground. How much work has it done on the chicken?
g = 10N/kg

Data Solution
W =? W = mgh
m = 2kg W = 2kg x 10N/kg x 15m
g = 10N/kg W = 300Nm
h = 15m W = 300J

Activity thirty one

1. A car of mass 1000kg is accelerated at 2m/s2 from rest in 20 seconds.
Calculate
(a) The force acting on the car.
(b) The final velocity
(c) The distance travelled by the car
(d) The work done by the car.

2. A crane lifts a weight of 200N through 50m. Find the work done by the crane.

3. A crane lifts a car of mass 500kg through 5m. Find the work done by the crane.

4. A person exerts a horizontal force of 500N on a box, which also experiences a

friction force of 100N.
Box

100N 500N

3m
How much work is done against friction when the box moves a horizontal distance of
3m?

Energy
Symbol: E
SI unit: Joule, J
Definition: Energy is the ability to do work.

Potential energy
Symbol: P.E
SI unit: Joule, J
Definition: Potential energy is the energy which the body possess by virtual of its
position.

92 | P a g e
Formula: P.E = mgh

Note
P.E = potential energy [J]
m = mass [kg]
g = acceleration due to gravity [10m/s2] or [10N/kg]
h = height [m]

Example
1. A 2kg object is raised to a height of 5m. What is its potential emery?

Data Solution
P.E =? P.E = mgh
m = 2kg P.E = 2kg x 10N/kg x 5m
g = 10N/kg P.E = 100Nm
h = 5m P.E = 100J

Activity thirty two

1. A book which has a mass of 1.2kg is put on the desk of height 0.8m. Calculate
the potential energy. (Take g to be 10N/kg)

2. A rock of mass 10kg is on top of the hill. Calculate the height of the hill if the
potential energy of the rock is 5000J. (Take g to be 10N/kg)

Kinetic energy
Symbol: K.E
SI unit: Joule, J
Definition: Kinetic energy is the energy the body has due to its motion.
1
Formula: K.E = mv2
2
Note
K.E = kinetic energy [J]
m = mass [kg]
v = velocity [m/s]

Example
1. A 2kg stone is thrown vertically with a velocity of 5m/s. What is the kinetic
energy?
Data Solution
K.E =? 1
K.E = mv2
2
m = 2kg
1
v =5m/s K.E = x 2kg x (5m/s)2
2

93 | P a g e
 1
K.E = x 2kg x 5m/s x 5m/s
2
K.E = 25J

Activity thirty three

1. A car of mass 500kg moves with a velocity of 20m/s. Find the kinetic energy.

2. A 60kg pupil runs 600m in one minute uniformly.

(a) Calculate his velocity
(b) Calculate his kinetic energy.

The law of conservation of energy

The law states that: Energy cannot be created or destroyed but can only be changed from
one form to another.

Energy transformations
Each energy can be changed but the total energy is constant
When there is only P.E and K.E, then;
P.E + K.E = Constant

Examples of energy transformations

(I) A ball falling freely from a certain height

A P.E = 20J Total energy = 20J

K.E = 0J

15m

30m P.E = 10J Total energy = 20J

B K.E = 10J

15m

C P.E = 0J Total energy = 20J

K.E = 20J

[A] Before a ball is released, its potential energy is 20J and the kinetic energy is 0J
because it is not moving

[B] At the mid point of its journey, the potential energy drops to 10J but the kinetic
energy increases to 10J. At height 15m, P.E becomes equal to K.E.
The total energy is still 20J.

[C] Just before hitting the ground, the potential energy becomes 0J but the kinetic energy
increases to 20J. All P.E becomes K.E
There is no change in the total energy through out its falling.

94 | P a g e
Example
1. A 2kg stone is dropped from the top of a 20m building.
(a) What potential energy does it posses?
(b) At what height does its potential energy becomes equal to its kinetic energy?
(c) What is its kinetic energy just before it hits the ground?
(d) With what velocity does it reach the ground?

1 Data Solution
(a) P.E =? P.E = mgh
m = 2kg P.E = 2kg x 10N/kg x 20m
g = 10N/kg P.E = 400Nm
h = 20m P.E = 400J
(b) At height 10m
(c) All P.E becomes K.E K.E = 400J
(d)
v =? 1
K.E = mv2
2
K.E = 400J 1
400 = x 2 x v2
2
m = 2kg 2
v = 400
v = √ 400
v = 20m/s

(II) Simple pendulum

A C
P.E = 20J P.E =20J
K.E = 0J B K.E = 0J
P.E = 0J
K.E = 20J

[A] The pendulum bob is pulled to position A. Before it is released, its potential energy is
20J and kinetic energy is 0J because it is at rest.

95 | P a g e
[B] As the bob moves from A to B, it loses potential energy and gains kinetic energy of
20J because of reducing the height and increasing the velocity. It has maximum velocity
at B

[C] Moving from B to C, the bob slows down losing kinetic energy but gaining potential
energy. If air resistance is ignored, the height of A is the same as the height of C because
the potential energies must be the same.

Example
1. A pendulum bob of mass 0.1kg is raised to a height of 0.4m above its lowest
point. It is then released.
(a) What is its potential energy at this height?
(b) What is its kinetic energy at its lowest height?
(c) What is its maximum velocity?

1 Data Solution
(a) P.E =? P.E = mgh
m = 0.1kg P.E = 0.1kg x 10N/kg x 0.4m
g = 10N/kg P.E = 0.4Nm
h = 0.4m P.E = 0.4J
(b All P.E becomes K.E K.E = 0.4J
)
(c)
1
K.E = mv2
v =? 2

1
0.4 = x 0.1 x v2
2

m = 0.1kg v2 x 0.1= 0.4 x 2

v2 x 0.1 = 0.8

0.8
K.E = 0.4J v2 =
0.1

v2 = 8

v= √8
v = 2.83m/s

Activity thirty four

96 | P a g e
 1. A 25kg bag of mealie meal is lifted from the ground to the top of a wall 1.8m high
in 0.6 seconds.
(a) What type of energy has the mealie meal bag gained?
(b) If the bag is released from the wall, with what velocity does it strike the
ground?
(c) Calculate the power which developed
(d) On striking the ground, into what form is the energy of the bag converted?

2. A tin of mass 64g fell from a height of 11.25m.

(a) Work out the speed of the tin at the moment it struck the ground.
(b) Calculate the kinetic energy of the tin when it was just hitting the ground.

Power
Symbol: P
SI unit: watt, W
Definition: Power is the rate of doing work.

Work done
Formula: Power =
time taken
W
P=
t

mgh
P=
t

P = power [W]
W = wok [J]
t = time [s]
m = mass [kg]
g = acceleration due to gravity [10m/s2] or [10N/kg]
h = height [m]

Example
1. A machine can lift 200kg to a height of 100m in 20 seconds. Find the useful
power of the machine.
Data Solution
P =? mgh
P=
m = 200kg t
g = 10N/kg
h = 100m 200 kg x 10 N /kg x 100 m
t = 20s P =
20 s

P = 10,000W

97 | P a g e
 2. A boy whose mass is 40kg finds that he can ran up a flight of 45 steps each
16cm high in 5 seconds. Calculate the power.

Data Solution
mgh
P=
t
P =?
m = 40kg 40 kg X 10 N /kg X 7.2 m
g = 10N/kg P=
5s
h = 45 x 16cm = 720cm = 7.2m
t = 5s 2880 J
P=
5s

P = 576W

Activity thirty five

1. A force of 1000N is needed to push a mass of 30kg through a distance of 40m
to raise an inclined plane to a height of 5m.
Calculate
(a) The weight of the object
(b) The mechanical advantage
(c) The velocity ratio
(d) The efficiency of the inclined plane.
(e) The energy at the height of 5m
(f) The work done by the force of 1000N.

2. A pupil of mass 50kg runs up a flight of 20 stairs each 5cm high in a time of
20 seconds. [ Take g = 10N/kg]
Calculate
(a) The pupil’s gain in potential energy
(b) The useful power developed by the pupil in climbing the stairs.

Transformers
Definition: A transformer is a device which is used to change the voltage of an appliance
(load) by mutual induction
Structure of a transformer
A transformer consists of two coils (primary coil and secondary coil) wound on a soft
iron core.
The coil that is connected to the alternating current input is called primary coil and the
coil that provides the alternating current output is called the secondary coil.

98 | P a g e
Types of transformers
There are two types of transformers
1. Step - up transformer
This is a transformer which increases the voltage of an appliance
The voltage in the primary coil (input) is lower than the voltage in the secondary coil
(output)
The number of turns in the primary coil is less than the number of turns in the secondary
Coil

2. Step - down transformer

This is a transformer which reduces the voltage of an appliance
The voltage in the primary coil (input) is higher than the voltage in the secondary coil
(output)
The number of turns in the primary coil is greater than the number of turns in the
secondary coil

Principle of operation of a basic iron - cored transformer

A transformer functions by mutual induction. That is;

99 | P a g e
An alternating voltage applied to the primary coil causes an alternating current to flow in
the coil. The alternating current induces a changing magnetic field.
The changing magnetic field induces an alternating voltage in the secondary coil. This
causes flow of alternating current in the secondary coil

Circuit symbols
(a) Step - up transformer

Np < Ns
Vp < Vs
Ip > I s

(b) Step - down transformer

Np > Ns
Vp > Vs
Ip < I s
Note
 Np = Number of turns in primary coil

100 | P a g e
  Ns = Number of turns in secondary coil
 Vp = Voltage of primary coil
 Vs = Voltage of secondary coil
 Ip = Current in primary coil
 Is = Current in secondary coil
 A transformer will not operate using a direct current input because direct current
produces a steady magnetic field which cannot induce a voltage in the secondary
coil.
 Transformers are used to transmit electricity because they can easily convert the
type of voltage needed. For domestic purposes, a step down transformer can be
used to drop a very high voltage to a suitable voltage in our homes. A step up
transformer can be used to amplify the voltage so that industrial areas can utilize
such high voltages

Factors that cause energy loses in a transformer and how this can be minimized
If a transformer has efficiency 100%, it is called ideal transformer.
However, no transformer is ideal. This means that a transformer cannot be 100% perfect.
It has energy loses. The following are factors that can cause energy loses in a transformer
and how they can be minimized
1. The resistance of the coils. As the coils have resistance, they give off heat when
current flows through. Coil resistance and energy loses can be minimized by
making the coils from thick copper because thick copper does not heat up easily.
2. Magnetization and demagnetization of the core. Work has to be done to alter
sizes and direction of domains and heat is released in the process. These energy
loses are reduced by making the core from soft iron because soft iron is easy to
magnetize and easy to demagnetize
3. Eddy currents in the core. Eddy currents are small currents produced within the
iron. These occur because the core itself is a conductor in a changing magnetic
field. The energy loses are reduced by laminating the iron core.
Advantage of transmitting electrical energy using high voltage
This can reduce energy loses due to long distances since the energy is transmitted from
long distances, the wires offer resistance. Some energy will be lost in the cables due to
heating effect.

Advantage of transmitting electrical energy using alternating current

This is because alternating current can be transformed to higher voltage; which is
efficient to transmit

Transformer equations

Np Vp
1. =
Ns V s

2. IpVp = IsVs

Note

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Np = Number of turns in primary coil
Ns = Number of turns in secondary coil
Vp = Voltage of primary coil
Vs = Voltage of secondary coil
Ip = Current in the primary coil
Is = Current in secondary coil

Transformer calculations
Examples

1. The figure below represents a transformer with a primary coil of 400 turns and a
secondary coil of 200 turns

(a) If the primary coil is connected to a 240V a.c mains supply, calculate the
secondary voltage
(b) Distinguish between the step-down and step-up transformers
(c) Explain carefully how a transformer works
(d) Why is the core made of iron?

Solution

Np Vp
(a) =
Ns V s

V p x Ns
Vs =
Np

240V x 200
Vs =
400

Vs = 120V
(b) A step down transformer reduces the voltage of an appliance while a step up
transformer increases the voltage of an appliance

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 In the step down transformer, the voltage in the primary coil is higher than the
voltage in the secondary coil while in the step up transformer the voltage in the
primary coil is lower than the voltage in the secondary coil
In the step down transformer, the number of turns in the primary coil is greater
than the number of turns in the secondary coil while in the step up transformer;
the number of turns in the primary coil is less than the number of turns in the
secondary coil.
(c) An alternating voltage applied to the primary coil causes an alternating current to
flow in the coil. The alternating current induces a changing magnetic field.
The changing magnetic field induces an alternating voltage in the secondary coil.
This causes flow of alternating current in the secondary coil
(d) Because soft iron can magnetize and demagnetize easily.

2. The primary coil of a transformer is connected to a 240V a.c mains and a current
of 5A passes through. If the voltage at the secondary coil is 12V, calculate the
secondary current.
Data Solution
Is =? IsVs = IpVp

Ip = 5A I p xV p
Is =
Vs
Vs = 20V
5 A x 240V
Vp = 240V Is =
12V

Is = 100A

Exercise
1. The primary coil of a transformer has 800 turns; its secondary coil has 2400
turns. Voltage in the primary coil is 50V;
(a) Calculate voltage in the secondary coil;
(b) Given that the current in the secondary coil is 12A, determine the current
in the primary coil.
2. A step down transformer is required to transform 240V a.c to 12V a.c for a
model railway. If the primary coil has 1000 turns. How many turns should the
secondary coil have?

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