Concepts in Mechanics

Since v = 60 km h-I, a = 1,5 m s - ~ ,

Terminal Questions
1. We have learnt from Sec. 1.5 that the terms 'rest' and 'motion' are relative. So,
whenever I say that 'I am moving' or 'I am at rest', I am supposed to mention about
the observer with respect to whom I am talking about my state. That is why the
statement "I am moving" is meaningless.
2. Refer to Fig. 1.29. This is a modified form of Fig. 1.24 where the Cartesian x and
y-axes are along OB and OA, respectively. (a) Let r be the position vector of the
midpoint M ofAB. Let the coordinates of B and A be (x, 0) and (0, y), respectively,
at any time t.

- - . -
a n d r = O M = O A + A M = O A + 2 A B = y ~ + 2 ( x i - y j") = 21( x"i t y j ) .
I " I " A

Fig. 1.29
The position vector of M is r = 21 (xi + yj^),where x2 + y2 = L~ = a constant equaI
to the square of the length of the ladder.
Now, using Eq. 1.3b, we get r = L/2 (o: x2 + y2 = L2). This means that the point M
is always at a distance L/2from 0.In other words, it'describes a circle of radius L/2
with 0 as centre,
(b) The velocity of M =*r- =- '(&3+4f).
df 2 dl dt
dx
Now ;i;= a constant = vo (given). Again as x2 t y 2 = L2. we have

When B is at a distance b from 0,we get x = OB = b and y = OA = ~ L - Zb 2 .

Correspondingly, 3 = - .Ir;.-h.
bv

So, the velocity ofJ4 at this instant is given by

3. Refer to Fig. 1.30. We first select a three-dimensional Caqtesian coordinate system.

Its origin is at the centre of the circle along which electron 2 executes uniform
circular motion. Its z-axis is taken along the direction of the magnetic field, such
that electron 1 moves along it. Let the position vectors of 1 and 2 at any time t be
rl and r2, respectively. Let the uniform angular speed of 2 be o and the radius of the
circle a. Then,
4 4 A
r, = v1t k , r 2 = a (COSw t i + sin o t j )
/ The relative velocity of 2 with respect to 1 is given by
A
d d ?r
v, = T ~ J(r2 - r l ) = ;jl( a (cos o t i + sin o t J) - v,t k )
A

,
P. A A

or v,, =-om (sin m t i - cos m t j ) - v k, d d the acceleration of 2

Fig. 1.30 -
with respect to 1 is given by a,, = dvdlj2 = - nm2(cos wr l + sin mt ) )
UNIT 2 FORCE AND MOMENTUM
Structure
2.1 Introduction
Objectives
2.2 Causes of Motion
Newton's Laws of Motion
Applications of Newton's Laws
Equilibrium of Forces
2.3 Linear Momentum
Conservation of Linear Momentum
Impulsc
Motion with Variable Mass
2.4 Summary
2.5 Terminal Questions
2.6 Answers

2.1 INTRODUCTION
In Unit 1, we learnt how to describe the motion of a particle in terms of displacement,
velocity and acceleration. We did not ask what caused the motion, In this unit we shall study
the factors affecting motion. For this we shall recall Newton's laws 0.f motion and apply
them to a variety of situations. Using Newton's laws we shall also establish the condition for
a particle's equilibrium, when it is acted on by several coplanar forces.
We will use the familiar concept of linear momentum to study the motion of systems having
more than one particle. In this process we shall establish the principle of conservation of
linear momentum and apply it to solve problems in which a knowledge of the forces acting
on the system is not needed. Finally, we shall recall the concept of impulse and use it to
study the motion of variable mass systems. Any change o i motion of an object is
accompanied by perfoimance of work and expenditure of energy. Therefore, in the next unit
we shall study the concepts of work and energy.
Objectives
After studying this unit you should be able to:
@ apply Newton's laws of motion
@ solve problems using conditions for equilibrium of forces
@ apply the law of conservation of linear momentum
@ solve problems concerning impulse and variable mass systems.

2.2 CAUSES OF MOTION

What makes things move? An answer to this question was suggested by Aristotle, way back
in the fourth century B.C.For nearly 2,000 years following the work of Aristotle, most
people believed in his answer, that a force-a push or a pull - was needed to keep
something moving. And the motion ceased when the force was removed. This idea made a
lot of common sense. When an ox stopped pulling an ox-cart, the cart quickly came to a stop.
But these ideas were first critically examined by Galileo who carried out a series of
0
\
'..- 4'

(b)
experiments to show that no cause or force is needed to maintain the motion of an object. -,
,,, ,
, -
Study Fig. 2.1 carefully to understand this. (c)

What do you think actually happens in the case of (c)?

Fig.2.1: (a) The ball rolling down
The ball does stop on the flat surface after some time. But it is seen that the smoother the a frictionless incline will rise
surface, the longer it takes for the ball to come to rest. Moreover, if the surface is reasonably approximately t0 its starting
smooth and flat, the ball moves more or less in a straight line. So if the element of friction heightOn aJecOnd
indine; (b) making the second
can be completely removed,'the ball would move indefinitely with a constant velocity as it
more
would never be able to reach the starting height. Galileo concluded that any object in in the ball's tra?elling further In
motion, if not obstructed will continue,to move with a constant speed along a horizontal line. the hgrlzontal direction to atbin
So, there would be no change in the motion of an object, unless an external agent acted on it the sake height; (c) what happens
to cause the change, In thb case? 27
Concepts In Mechenlcs That was Galileo's version of inertia. Inertia resists changes, not only from the state of rest,
but also from motion with a constant speed along a straight line. So the interest shifted from
the causes of motion to the causes for changes in motion.
>
Galilee's work set the stage for centuries of progress in mechanics, beginning with the
achievements of Isaac Newton. Newton's laws of motion are the basis of mechanics. We
will now briefly discuss these laws.

2.2.1 Newton's Laws of Motion

Galileo's version of inertia was formalised by Newton in a form that has come to be known
as Newton's first law of motion.
Newton's first law of motion
Stated in Newton's words, the first law of motion is:
"Every body continues in its state of rest or of uniform motion in a straight line
unless it is compelled to change that state by forces impressed upon it."
Newton's first law is also known as law ofinertia and the motion of a body not subject to
the action of other forces is said to be inertial motion. With the help of this law we can
dejine force as an external cause which changes or tends to change the state of rest or of
uniform motion of a body.
Have you noted that the first law does not tell you anything about the observer? But we
know from Sec. 1.5 that the description of motion depends very much on the observer. So it
would be worthwhile to know: For what kind of observer does Newton's first law of motion
hold?
Fig. 2.2: (a) The observer 0 and Suppose that an object P is at rest with respect to an observer 0 who is also at rest
the object P are at rest with (Fig. 2.2a). Let another observer O'be accelerating with respect to 0.P will appear to 0'to
respect to each other; (b) 0 ' is be accelerating in a direction opposite to the acceleration of 0' (Fig. 2.2b). According to
accelerating with respect to 0.
Newton's first law the cause of the acceleration is some force. So 0' will infer that P is
being acted upon by a force. But 0 knows that no force is acting on P. It only appears to be
accelerated to 0'. Hence, the first law does not hold good for 0'. It holds good for 0 .
An observer like 0 who is at rest or is moving with a constant velocity is called an inertial
observer and the one like 0', a non-inertial observer.
But hbw do we know whether an observer is inertial or not. For this, we need to measure the
observer's velocity with respect to some standard. It is a common practice to consider the
earth as a standard. Now the place where one is performing one's experiment has an
acceleration (as discussed in Sec. 1.5) towards the polar axis due to the daily rotation of the
earth. Again the centre of.the earth has an acceleration towards the sun owing to its yearly
motion around the sun. The sun also has an acceleration towards the centre of the Galaxy,
and so on. ~ e n d ethe
, search for an absolute inertial frame is unending.
So we modify the definition of an inertial observer. We say two observers are inertial with
respect to one another when they are either at rest or in uniform motion with respect to one
another.. Ifan observer has an acceleration with respect to another then they are non-
inertial with respect to one another. Thus, a car moving with a constant velocity and a man
standing on a road are inertial with respect to one another while a car in the process of
gathering speed, and the man, are non-inertial with respect to each other.
The first law tells you how to detect the presence or absence of force on a body. In a sense,
it tells you what a force does - it produces acceleration (either positive or negative) in a
body. But the first law does not give a quantitative, measurable definition of force. This is
what the second law does.
Newton's second iaw of motion
If you are struck by a very fast moving cricket ball you get injured but if you are hit by a
flower moving with the same velocity as that of the ball you do not at all feel perturbed.
However, if you are struck by a slower ball the injury is less serious. This indicates that any
kind of impact made by an object depends on two things- its mass and velocity. Hence,
Newton felt the necessity of defining the product of mass and velocity which later came to
be known as linear momentum. Mathematically speaking, linear momentum
P = mv. (2.1)
Thus, p is a vector quantity in the direction of velocity. The introduction of the above
 quantity paved the way for stating the second law, w h i ~ hin the words of Newton is as
follows:
'The change of motion of an object is proportional to the force impressed; and is
made in the direction of the straight line in which the force is impressed."
By "change of motion", Newton meant the rate of change of momentum with time. So
mathematically we have
* ~s~p%-wu

where F is the impressed force and k is a constant of proportionality. The differential

d
operator - indicates the rate of change with time. Now, if the mass of the body remains
A+
U L
constant (i.e. neither the body is gaining in mass like a conveyer belt nor is it disintegrating
like a rocket), then
dp = -
- d (nzv) = nl -
clv = nra,
dt dt dt
dv
where a = -= the acceleration of the body. Thus, from Eq. 2.2, we get
dt
F=kma,and
F = kmn,
We had seen earlier that the need for a second law was fell in order to provide a quantitative
definition of force. So something must be done with the constant k, We have realised that
the task of a force F acting on a body of mass m is to produce in it an acceleration a. Hence,
anything appearing in the expression for force other than m and a must be a pure number,
i.e, k is a pure number. So we can afford to make a choice for its numerical value.
We define unit force as one which produces unit acceleration in its direction when it acts on
a unit mass. So, we obtain from Eq. 2.3b that 1 = k. 1.1 or k = 1. Thus, Eqs. 2.2 and 2.3 take
the form
F=- dp ,and
dt
F = ma, for constant mass.
Now we know from Sec. 1.3 that if the position vector of a particlejs r at a time t then its
velocity v and acceleration a are given by Eqs. 1.24a and 1.26a. Substituting for a and v in
Eq. 2.4, we get

Eq. 2.5 is a second order differential equation in r. If we know the force F acting on a body
of mass m, we can integrate Eq. 2.5 to determine r as a function of t. The function r ( t )
would give us the path of the particle. Since Eq. 2.5 is of second ord& we shall c6me across
two constants of integration. So we require two initial conditions to work out a solution of
this equation. Conversely, if we know the path or trajectory of an accelerating particle, we
can use Eq. 2.5 to determine the force acting on the body. Eq. 2.5 also enables us to
determine unknown masses from measured forces and accelerations.
So far, we have considered only one force acting on the body. But often several forces act on
the same body. For example, the force of gravity, the force of air on the wi.ngs and body of
the plane and the force associated with engine thrust act on a flying jet (Fig. 2.3).

Fig. 2.3: Forces on a jet: Fe,the thrust of the engine, PII,the force of the air provides both lift and drag, FK
the force of gravity.
29
Conceptsin Mechanics
In such cases, we add the individual forces vectorially, to find the net force acting on the
object. The object's mass and acceleration are related to this net force by Newton's second
law. You may now like to apply Newton's second law to a simple situation.
s a n~
Astronailta 011 the Skylab mission of the 1970s found their masses by using a chair on which
a k~aownforce was exerted by a spring. With an astronaut strapped in the chair, the 15 kg
an acceleration of 2.04 x 10-' IA s-' when the spring force was 2.07 N.
cliair u~~tle~.went
What w;lS Lhc astronaut's mass?
Newton's third law of motion
So far we have been trying to understand how and why a single body moves. We have
identified force as the cause of change in the motio~iof a body. But how does one exert a
force on this body? Inevitably, there is an agent that makes this possible. Very often, your
hands or feet are the agents. In football, your feet bring the ball into motion. Thus, forces
arise from interactions between systems. This fact is made clear in Newton's third law of
motion. To put it in his own words:
"To every uction there is m~equal and opposite reaction."
Here the words 'action' and 'reaction' mean forces as defined by the first and second laws.
If a body A exerts a force FA,on a body B, then the body B in turn exerts a force F, on A,
such that
FAD = -F,,.

So, we have FA,+ F,, = 0. (2.6)

Notice that Newton's third Law deals with fwo forces, each acting on a different body. You
may now like to work out an SAQ based on the third law.

SAQ 2
a) Whcn a footbullcr kicks the ball, the ball and the man experience forces of the same
1n:lgnitude but in opposite directions according to the third law. The ball moves but the.
man does 1101 move. Why'?
h) The earth attracts an apple with a force of magnitude F. What is the magnitude of the
force with which the apple attracts the earth? The apple moves towards the earth. Whv
does not the reveae happen?
Newton's laws of motion give us the means to understand most aspects of motion. Let us
now apply them to a variety of physical situations involving objects in motion.

2.2.2 Applications of Newton's Laws

To apply Newton's laws, we must identify the body whose motion interests us. Then we
should identify all the forces acting on the body, draw them on a vector diagram and find
the net force acting on the body. Newton's second law can then be used to determine the
body's acceleration. We will now use this basic method to solve a few examples.
Example 1: Projectile Motion
The motion of a shot fired by a gun and that of a ball thrown by a fieldsman to another are
all examples of projectiles. Let us consider such a projectile of mass m (Fig. 2.4).It is
thrown from a point 0 with a velocity v , along OA making an angle 8 with the horizontal.
Letthe particle be at a point P (OP = r) at time t. If we neglect air resistance, then the only
force acting on the particle is a constant force, F = ntg, due to gravity. Let us determine the
particle's path. Eq. 2.5 gives
d 2r
m,=mg, (2.7)
dt
d 2r
or ~ = g ,
dt

On integrating with respect to t we get

dr (2.8a)
-=gt+A,
dt
where A is a constant of integration. As the other two factors in Eq. 2.8a are vectors having Force and Momentum
dimensions of velocity, A must also be a vector having the dimension of velocity. To
dr
-
determine A, we use the initial condition that velocity = = vowhen t = 0.
dt
SO A = v,,.Hence,

On integrating with respect to t again, we get

r = v,t + gt
2
+ B,
where B, like A in Eq. 2.8a, is a constant vector of integration, but it has the dimension of
length. To determine B, we need another initial condition. Letting r = 0 at t = 0, we get B =O.
Hence,
r = V 0 t + 71f gt2 (2.10)
dr
We have essentially used two initial conditions: -
. dt
= vo and r = 0 at t = 0.Since vois
along OA and t is scalar, we understand that votis along OA. Again g is directed vertically
1 2
downwards and t is a scalar, so gt2 is directed vertically downwards, i.e. along AP
(Fig. 2.4).
We use the law of vector addition to get
QP = OA + AP. (2.11)
Thus, we get the location of the particle. As time advances OA is lengthened and so is AP,
and we get the location of the particle by adding QA and AP.
Example 2: Friction
A heavy-block is kept on a rough floor. You apply a force by pulling on a rope attached to it,
but it still does not move. Is it a contradiction of Newton's laws? Discuss the motion of the
block.
Refer to Fig. 2%. Let us first find out all the forces that act on the heavy block. There is a
force of gravity mg acting downwards. The block exerts this force on the floor. Therefore, 1;;
the floor exerts an equal and opposite normal force of reaction N on the block. N is normal
to the surface of the floor. The third force results from your pull on thg rope. Let F, be the
force that you exert on the rope. The rope exerts a force of reaction F,'on you and a force of
action, say F, on the block. Let Fi be the force that the block exerts on the rope. Then
according to Newton's third law of motion (a)
F, = -F,'; F, = -F;. (2.1 2)
Let us assume that the rope is massless. Then, from Newton's second law, the net force
acting on the rope is zero and we have,

or F, = F, , from Eq. 2.12. (b)

So a massless rope transmits the force you exert on it to the block without any change. The Fig, 2.5
three forces mg, N and F, acting on the block do not add up to zero. N and mg cancel each
other, leaving a net force F, shown in Fig. 2.5b. Since the block remains at rest, the net force
acting on it must be zero according to the first law. There must, therefore, be another force
which acts on the block. This force must also be horizontal, directed opposite to F, and
equal in magnitude. Actually, there is such a force which is the contact force between the
floor and the block, known as the force of friction. It is shown in Fig. 2.5b by a dotted line.
Friction is a force that acts between two surfaces to oppose their relative motion (see Fig.
2.6). The force of static friction f, acts between surfaces at rest with respect to each other.
The maximum force of static friction f,,,is the same as the smallest force necessary to start
motion. Once motion has started, the force of friction usually decreases, so that a smaller
force is required to maintain a uniform motion.

The force acting between surfaces in relative motion is called the force of kinetic friction
fk.fkis less than fVDi
The ratio of the magnitude of maximum force of static frictionJn, to the 31
~unceprsm Mechanics magnitude of normal force of reaction N between the two surfaces is called the coefficient of
static friction y ire.
?,

(a)
rI=P IN*

where y, is the coefficient of kinetic friction,

The discussion on friction brings us to an important class of problems in which an object
undergoes motion against resistive forces. Another example of resistive force is air
resistance to projectile motion. The motion of raindrops, or cars is also affected by air
resistance. So let us discuss an example on motion where resistive forces are present.

(b)
Example 3: Motion against resistive forces
Suppose an object moves under the influence of a constant force F,,, with a resistive force R
Fig. 2,6: Friction acts two opposing its motion. Let R always act in a direction opposite to the object's instantaneous
surfaces to oppose their relative velocity. In general the resistive force is a function of speed, so that Newton's second law
motion. Even the smoothest becomes:
surface is actually rough on a F - R ( v ) = m - dv
microscopicscale. (n) When two o dr (2.1 3)
surfaces are in their
The resistive force of dry friction (Fig. 2.7a) is almost independent of v, so that
irregularities adhere because of
electrical forces between the R(v) = F , = constant.
molecules. This gives rise to a
force that opposes their relative In this case Eq. 2.13 reduces to the simple case of acceleration under a constant net force.
motion; (b) when the normal force
between the surfaces increases,
the irregularities are crushed
together and the contact area
between the surfaces increases.
This increases the force of
friction.
-FA F A
U *

4
/"
sr E,
f ?
0 v 0 t

Pig. 2.7: (a) Resistive force for an object resisted by dry friction and (b) by fluid friction; (c) terminal speed I;,,
of an object in a fluid resistive medium. F,,/nr is the slope of the curve at 0.

In the case of air resistance or fluid resistance, R(v) increases with v (Fig. 2.7b). It is usually
described by the relation \

For the sake of simplicity, let us consider only one-dimensional motion under the resistive
force of Eq. 2.14. So we can use the scalar form of Eq. 2.13, which is

Eq. 2.15 is not very easy to solve and we do not intend to go into its formal mathematical
solution. Let us, however,'consider some qualitative features of the possible solution.
Let the object start" moving under a constant force F,. Its initial acceleration will have almost
a constant yalue 2,
since v is very small. Thus, v will be a linear function oft (Fig. 2.7 c).
m
As v increases, R (v) will increase and the net driving force is reduced to a value below F,,
giving a steadily decreasing slope in the graph of v(t). When R(v) approaches F,, the net
force acting on the body tends to be zero. Then the object's velocity acquires a limiting
32 constant magnitude vm. The value of vmis the positive solution of the quadratic equation
 Force and Momentum

In such a situation, the body moves with zero acceleration under zero net force. It is not the
unaccelerated motion of objects moving under no force at all. So every time we see a car
moving along a straight road at a steady speed, a jet plarie flying through the air at a constant
speed, or raindrops falling with a uniform terminal velocity, wk see bodies moving under . -
zero net force. Their motion at a constant speed does not mean that no force is acting on

,-dT
them. Now you may like to work out an SAQ based on this concept.

SAQ 3 .
A boli of mass t?l is being pulled across a rough floor by 11-teanls of a massless rope that
s angle 0 with the horizontal (Fig. 2.8). The coefficiea~tof kinetic Mction between-
m d i ~ ail
in tllc ~.opc\ ~ h e nthe h)x[]loves at a conritani
the hox and the tli:)i)l is y,. W h n ~is t l 1cnsior.1
~
- - - - -- ---
vrlocity?
Fig. 2.8
One simple but important application of Newton's laws is the study of bodies in equilibrium.
A large number of situations may be reduced to problems concerning the equilibrium of
forces on a particle. For example, the construction of buildings and suspension bridges,
,
design of aircrafts and ships, loading or unloading operations, involve forces in equilibrium.
So let usnaw study equilibrium of forces acting on a particle.

2.2.3 Equilibrium of Forces

We say that aparticle is in equilibrium, when the resultant of ull the forces acting on it is
zero. It then follows from Newton's firs1 law of motion that a particle in equilibrium is either
at rest or is moving in a straight line with constant speed. It is found that for a large number
of problems, we have to deal with equilibrium of forces lying in a plane. Therefore, we shall
restrict our discussion to the case when a particle is in equilibrium under the influence of a
number of coplanar forces, F,, F, ,F, , ..... . The req~ired~condition is given by
F, + F2+ F3+ .....= 0. (2.16a)
Since the forces are coplanar, we can resolve them along two mutually perpendicular
directions of x and y-axes (Fig. 2.9), 0 being the particle. So Eq, 2.16a can be rewritten as
6 CI h

(F,,T+ F,,j) -I- (F2,,i+. F2?j) 0,

or (6, + F,, +.-.)l+ (F,, + F,, +...)j= 0 .
or F , x + F 2 r + .... = 0 ,
Fig. 2.9
and F,y+ F2?+ .... = 0. (2.16b)
Eqs. 2.16b can be expressed in a concise foim as
CFx=o, CFy= 0,
I
where I:denotes sumination of the x- or y-components of the forces. We shall now apply Eq.
4 2.16~to work out an example.
--
Example 4
I A particle of mass m is hung by two light strings as showmin Fig. 2.10a. The ends A and B
are held by hands. The strings OA and OB make angles 0 with the vertical. Find the values
of T and T ' in terms of m and 0.
Through 0, we consider two mutually perpendicular directions of x and y-axes, the latter
being along the vertical.
From Eq. 2 . 1 6 ~we have
-T cos (900- e) + T cos (90" - a) = o, (2.17)
and Tcos0+I:'cos0-mg=O (2.18)
Hence, from Eq. 2.17, we get
T ' = T, 0 + 0").
(a;

Thus, from Eq. 2.1 8, we have

2T cos 0 = mg,

Fig. 2.10
If 8 is increased, cos 0 will decrease, thereby increasing the tension. This may lead to the 33