Force & Laws of Motion Class 9 Science Chapter- NCERT PDF Download

In the realm of physics, the study of motion and the forces that cause it forms the foundation for understanding the physical world. Class 9 Science introduces students to the fundamental concepts of force and the laws of motion, a crucial chapter that lays the groundwork for more advanced studies in physics. Let's delve into these concepts to gain a comprehensive understanding.

1. FORCE AND
LAWS OF MOTION
2.1 INTRODUCTION
In the preceding chapter we saw that uniform motion needs the
concept of velocity alone whereas non-uniform motion requires the
concept of acceleration in addition so far, we have not asked the
question as to what governs the motion of bodies. In this chapter, we
turn to this basic question.
Let us first guess the answer based on our common experience. To
move a football at rest, some one must kick it. To throw a stone
upwards, one has to give it an upward push. A clearly, some external
agency is needed to provide force to move a body from rest. Likewise,
an external force is needed also to retard or stop motion.
In short, a force is required to put a stationary body in motion or stop
a moving body and some external agency is needed to provide this
force. The external agency may or may not be in contact with the
body.
2.2 FORCE
“Force is a push or pull which changes or tends to change the
state of rest or of uniform motion, direction of motion, or the
shape and size of a body”
Example,
(i) When we push a ball lying on the ground, it starts rolling. The force
exerted has thus produced motion in the ball. However, when we
push a heavy stone, it does not move. The effort made in this case
has only tried to produce motion, but has not succeeded.
(ii) A ball falling downwards can be easily caught by our hands. The
motion of the ball has thus been destroyed. However, a big piece of
rock rolling down a hill cannot be stopped even when we try our best
to stop it.
Dimensions and Units of Force
As F = ma
F = (M) (LT
–2
) = [M
1
L
1
T
–2
]
This is the dimensional formula of force.
The units of force are of two types: Absolute units and Gravitational
units
2.1 Introduction
2.2 Force
2.3 Inertia
2.4 Newton’s first law of
motion
2.5 Momentum
2.6 Newton’s second law of
motion
2.7 Newton’s third law of
motion
2.8 Laws of conservation of
momentum
“IIT-JEE Foundation”
*2.9 Impulse
*2.10Friction
*2.11 Static Friction
*2.12Limiting Friction
*2.13Kinetic Friction
*2.14Laws of Limiting
Friction 9$9$&/$66(63+<6,&67+ $OOULJKWFRS\UHVHUYHG1RSDUWRIWKHPDWHULDOFDQEHSURGXFHGZLWKRXWSULRUSHUPLVVLRQ

2. (a) Absolute units
(i) The absolute unit of force on SI is newton (represented by N).
One Newton force is that much force which produces an acceleration of 1 ms
–2
in a body of mass 1 kg.
As F = ma
1 N = 1 kg × 1 ms
–2
= 1 kg ms
–2
(ii) The absolute unit of force on c.g.s. system is dyne.
One dyne force is that much force which produces an acceleration of 1 cms
–2
in a body of mass
one gram.
As F = ma
1 dyne = 1 g × 1 cms
–2
= 1 g cms
–2
Relation between newton and dyne
As 1N = 1 kg × 1 ms
–2
= 10
3
g × 10
2
cms
–2
1 N = 10
5
dyne
(b) Gravitational units
(i) The gravitational unit of force on SI is 1 kilogram weight (kg wt.) or 1 kilogram force (kg f). It is that much
force which produces on acceleration of 9.8 ms
–2
in a body of mass 1 kg. Thus
1 kg wt. or 1 kg f = 1 kg × 9.8 ms
–2
= 9.8 N
(ii) The gravitational unit of force on c.g.s. system is 1 gram weight (g wt.) or 1 gram force (1 g f). It is that
much force which produces an acceleration of 980 cm s
–2
in a body of mass 1 gram.
Thus,
1 g weight = 1 g f = 1 g × 980 cm s
–2
= 980 dyne
Note: The gravitational units of force are used to express weight of a body. For example, weight of a body
of mass 5 kg is 5 kg f or 5 kg wt. These units are, therefore, called the practical units.
Remember that in all numerical problems, we have to use only the absolute units of force. Gravitational
units are the practical units. They have to be converted into absolute units in all problems
2.3 INERTIA
“The property of a body to resist any change in its state of rest or of uniform motion in a straight
line”
2.3.1 Types of Inertia
(i) Inertia of rest: The property of the object that object remains at rest unless an external force is applied
on it is called inertia of rest.
For examples,
(a) When a bus starts suddenly, the passengers fall backwards.
(b) Beating of a hanging carpet.
(c) On shaking the branch of a tree, the fruits fall down.
(ii) Inertia of motion: The property of object that remains in uniform velocity unless a force is applied on it.
This inherent property of the object or body is called inertia of motion.
For examples,
(a) When a running bus stops suddenly, the passengers are jerked forward.
(b) The passenger getting out of a moving train or bus falls in the forward direction. 9$9$&/$66(63+<6,&67+ $OOULJKWFRS\UHVHUYHG1RSDUWRIWKHPDWHULDOFDQEHSURGXFHGZLWKRXWSULRUSHUPLVVLRQ

3. (iii) Inertia of direction: A body moving in a particular direction can never change its direction itself unless
a force is applied on it. This property of the body is called inertia of direction.
For example,
(a) A passenger experiences a sideway jerk when bus takes a sharp turn.
2.4 NEWTON ’S FIRST LAW OF MOTION
“A body continues in its state of rest or of uniform motion in a straight line unless an external force
acts on it.”
It define force and inertia.
2.5 MOMENTUM
“The product of the mass of a body and its velocity is called the momentum of the body.”
P = mv ; it is a vector quantity : SI unit : kg-m/s
Direction of momentum is along the direction of velocity.
Illustration 1
Find the linear momentum of a cricket ball of mass 250 g moving with a velocity of 72 km h
–1
.
Solution
m = 250 g = 25 × 10
–2
kg , v = 72 km h
–1
= 20 ms
–1
P = mv = 25 × 10
–2
× 20 = 5 kg ms
–1
2.6 NEWTON ’S SECOND LAW OF MOTION
The rate of change of momentum of a body is directly proportional to the force acting on it and takes place
in the direction of this force.
Suppose an object of mass, m is moving along a straight line with an initial velocity u. It is uniformly
accelerated to velocity, v in time, t by the application of a constant force F, throughout the time t. The initial
and final momentum of the object will be, p
1
= mu and p
2
= mv respectively.
The change in momentum =p
2
– p
1
= mv – mu
= m × (v–u)
The rate of change of momentum =
m (v u)
t
 
or, the applied force,
m (v u)
F
t
 

km (v u)
F
t
 
 = kma
Here
a [=(v–u)/t] is the acceleration, which is the rate of change of velocity. The quantity, k is a constant
of proportionally.
For k = 1,
F = m a
Force is a vector quantity 9$9$&/$66(63+<6,&67+ $OOULJKWFRS\UHVHUYHG1RSDUWRIWKHPDWHULDOFDQEHSURGXFHGZLWKRXWSULRUSHUPLVVLRQ

4. 2.6.1 Examples of Newton’ s Second Law of Motion
(a) A person falling on a cemented floor gets injured but a person falling on a heap of sand is not
injured.
(b) The vehicles are fitted with shockers (i.e. springs).
(c) A cricket player lowers his hands while catching the ball.
Illustration 2
A force of 1 N acts on a body of mass 1 g. Calculate the acceleration produced in the body?
Solution
Here, F = 1 N, m = 1 g = 10
–3
kg
Now, F = ma, a=
3
F 1
m
10

 = 10
3
m/s
2
Illu
stration 3
Calculate the force acting on a body which changes the momentum of body at the rate of
1 kg ms
–2
.
Solution
We know that F = rate of change of momentum
F = 1 kg ms
–2
= 1N
Try yourself
1 A mass of 5 kg is acted upon by a force of 1 N. Starting from rest, how much is distance covered by the
mass in 10 s?
2. A car of mass 1000 kg is moving with a speed of 36 km h
–1
on a level road. Calculate the retarding force
required to stop the car in a distance of 50 m.
2.7 NEWTON ’S THIRD LAW OF MOTION
“For every action there is an equal and opposite reaction, but action and reaction forces act on different
bodies.”
Let two bodies A and B press against each other as shown in figure. Let
1F

be the force exerted by body
A on body B and
2F

be the force exerted by body B on body A, then
F
1
F
2
(A)
(B)
1 2F F 
 
This relation shows that
1F

and
2F

are equal in magnitude but opposite in direction.
For examples,
(a) Book on the table (b) Walking of a man
(c) Recoil of a gun (d) Motion of rockets and jet planes
(e) The case of hose pipe (f) The case of a boat
2.8 LAWS OF CONSERVATION OF MOMENTUM
“If a group of bodies are exerting force on each other (i.e., interacting with each other), their total 9$9$&/$66(63+<6,&67+ $OOULJKWFRS\UHVHUYHG1RSDUWRIWKHPDWHULDOFDQEHSURGXFHGZLWKRXWSULRUSHUPLVVLRQ

5. momentum remains conserved before and after the interaction provided there is no external force acting
on them”.
Suppose two objects (two balls A and B, say) of masses m
A
and m
B
are travelling in the same direction
along a straight line at different velocities u
A
and u
B
respectively shown in figure (a). And there are no
other external unbalanced forces acting on them. Let u
A
> u
B
and the two balls collide with each other as
shown in figure (b). During collision which lasts for a time t, the ball A exerts a force F
AB
on ball B and the
ball B exerts a force F
BA
on ball A. Suppose v
A
and v
B
are the velocities of the two balls A and B after the
collision respectively in shown figure (c).
The momenta (plural of momentum) of ball A before and after the collision are m
A
u
A
and m
A
v
A
respectively.
The rate of change of its momentum (or F
AB
action) during the collision will be
A A
A
(v u )
m
t

.
Simiarly, the rate of change of momentum of ball B (=F
AB
or reaction) during the collision will be
B B
B
(v u )
m
t

.
According to the third law of motion, the force F
AB
exerted by ball A on ball B (action) and the force F
BA
exerted by the ball B on ball A (reaction) must be equal and opposite to each other. Therefore,
F
AB
= –F
BA
or
A A B B
A B
(v u ) (v u )
m m
t t
 
 
This gives,
m
A
u
A
+ m
B
u
B
=
m
A
v
A
+ m
B
v
B
or
m
1
u
1
+ m
2
u
2
= m
1
v
1
+ m
2
v
2
Total momentum before collision = Total momentum after collision
Illustration 4
A bullet of mass 20 g is fired by a gun of mass 20 kg. If the muzzle speed of the bullet is
100 m s
–1
, what is the recoil speed of gun?
Solution
Mass of gun, M = 20 kg
Mass of bullet, m =0.02 kg
Speed of bullet, v = 100 ms
–1
Then recoil speed of gun is given by 9$9$&/$66(63+<6,&67+ $OOULJKWFRS\UHVHUYHG1RSDUWRIWKHPDWHULDOFDQEHSURGXFHGZLWKRXWSULRUSHUPLVVLRQ

6. 0.02 100
20
mv
V
M

  = 0.1 ms
–1
Try
yourself
3. A bullet of mass 0.02 kg is fired from a gun weighing 7.5 kg. If the velocity of the bullet is 200 ms
–1
,
calculate the speed with which the gun recoils.
*2.9 IMPULSE
If a large force acts for a short duration then the effect on motion of the body is governed not by the force
alone but by the product of force and duration for which the force acts. Total (overall) effect of such a
force with time is called impulse. This is a vector quantity denoted by
I

. Direction of impulse is along the
direction of force. Its unit is N – s.
If during time interval  t the force
F

(or average force) remians constant, then –
impulse = force × time interval
I

=
F

×  t =
F

(t
2
– t
1
)
substituting value of
F

from Eq.
p
F
t





I

= 
P

=
F

 t
or
I

=
2
P

–
1
P

=
F

. t
Above Equations is called Impulse – Momentum Theorem. According to this, change in momentum of a
body is equal to impulse acquired by the body in the given time interval. Impulse can be positive, negative
or zero.
Illustration 5
A ball of mass 100 g moving with a velocity of 10 ms
–1
is brought to rest by a boy in 0.02 s.
Calculate (i) the impulse and (ii) the force applied by the boy.
Solution
Mass of ball, m = 100 g =
100
1000
= 0.1 kg
u = 10 ms
–1
; v = 0, t = 0.02s
(i) Impulse, Ft = Change in momentum
= mu – mv
= 0.1 × 10 – 0.1 × 0 = 1 Ns.
(ii) I = F t
1 = F × 0.02
F = 50 N
Try yourself
4. A bullet of mass 50 g moving with a speed of 500 ms
–1
is brought to rest in 0.01 s. Find the impulse and the
average force. 9$9$&/$66(63+<6,&67+ $OOULJKWFRS\UHVHUYHG1RSDUWRIWKHPDWHULDOFDQEHSURGXFHGZLWKRXWSULRUSHUPLVVLRQ

7. *2.10 FRICTION
The force which always opposes the motion of one body over the other body in contact with it is called the
frictional force or simply friction.
2.10.1 Cause of Friction Force
(i) A surface which may look smooth to the naked eye has a number of grooves or irregularities when
viewed under high power microscope (shown in figure). When one surface is placed over another surface,
humps of molecules press against each other and get interlocked. The pressure at points of contact are,
therefore, high and a type of cold welding takes place.
(ii) When one surface slides over the other, the bonds between the molecules are continuously broken and re-
build at other points. The force which is required to break the bonds between the molecules when one surface moves over another is called frictional force or force of friction.
Static Friction is a Self Adjusting Force
This means that neither the magnitude nor the direction of the force of static friction is fixed. Both these
things adjust themselves according to the applied force. We know that unless the block starts moving,
i.e. force of static friction is always equal to the applied force. When we change the applied force, the
force of direction of the force of static friction is always opposite to the direction of the applied force. For
example, when we try to move the body to the east, force of static friction is to the west and we try to
move the body to the north, force of static friction is to the south and so on.
Thus static friction adjusts itself so that its magnitude is equal to the magnitude of the applied force and its
direction is opposite to that of the applied force. Hence static friction is a self adjusting force.
Remember that friction arises only when body is actually sliding/rolling over the surface of another body
or the body is simply trying to slide/roll over the surface of the other.
Further, static friction alone is a self adjusting force.
*2.11 STATIC FRICTION
The friction that exists between the surfaces in contact when there is no relative motion is static friction..
*2.12 LIMITING FRICTION
The maximum frictional force present when a body just begins to slip over the surface of another body is
called the maximum static friction or force of limiting friction.
Rolling friction < Kinetic friction < Limiting friction
*2.13 KINETIC FRICTION
The force of friction which comes into play when one body moves over the surface of another body is
called kinetic friction. 9$9$&/$66(63+<6,&67+ $OOULJKWFRS\UHVHUYHG1RSDUWRIWKHPDWHULDOFDQEHSURGXFHGZLWKRXWSULRUSHUPLVVLRQ

8. 2.14.1 Types of Kinetic Friction
(a) Sliding Friction: The friction that exists between two surfaces in contact when the body slides on
the surface is known as the sliding friction.
(b) Rolling Friction: The friction that exists between two surfaces when the body rolls on the surface
is known as the rolling friction.
Cause of Rolling Friction
When a body rolls on a level track, the area of contact is very small. Therefore, pressure exerted which
is equal to weight/area is very large. This causes a depression in the surface below and a mount or bump
in front as shown in figure.
MOUNT
K
M
L
DEPRESSION
In turn, the surface of the rolling body in contact gets slightly compressed. Thus a rolling wheel (i) constantly pulls out of depression and goes uphill on the mount LM (ii) simultaneously detaches itself from the road KL, which is opposed by the forces of adhesion between the surfaces in contact. This causes rolling friction.
When a tyre is properly inflated, it becomes hard and gets compressed by the road to a much smaller
extent. Therefore, rolling friction reduces. Hence it is easier to drive a bicycle when its tyres are fully
inflated.
Note that the velocity of the point of contact of the wheel with respect to the floor remains zero all the
time, although the centre of the wheel moves forward. Therefore, rolling friction is often quite small
compared to the sliding friction. That is why heavy loads are transported by placing them on carts with
wheels. Thus sliding friction is converted into rolling friction. For example, rolling friction of steel on steel
is hardly 1% of sliding friction of steel on steel.
Coefficient of Friction
According to the first law of limiting friction,
F  R or F = µR
Where µ is a constant of proportionality and is called the coefficient of limiting friction between the two
surfaces in contact.
µ = F/R
Hence coefficient of limiting friction between any two surfaces in contact is defined as the ratio of the
force of limiting friction and normal reaction between them. The value of µ depends on
(i) Nature of the surfaces in contact i.e. whether dry or wet; rough or smooth; polished or non polished.
(ii) Material of this surface is contact.
*2.14 LAWS OF LIMITING FRICTION
(i) Friction always opposes motion.
(ii) Force of friction is directly proportional to the normal reaction between the two surfaces in contact.
(iii) The force of friction depends on the nature of surface in contact. 9$9$&/$66(63+<6,&67+ $OOULJKWFRS\UHVHUYHG1RSDUWRIWKHPDWHULDOFDQEHSURGXFHGZLWKRXWSULRUSHUPLVVLRQ

9. (iv) The force of friction is independent of the area of the surfaces in contact if normal reaction remains
constant.
Solved Examples
Example 1
A boy weighing 30 kg is sitting on a chair. How much reaction acts on the boy?
Solution
Action = Weight of boy on the chair in the downward direction
= mg = 30 × 9.8 = 294 N in downward direction
 Reaction on the boy = –Action = 294 N in the upward direction
Example 2
A body of mass 20 kg is moving with a speed of 5 ms
–1
. Calculate the distance travelled by the body
before coming to rest when a constant retarding force of 40 N is applied on it.
Solution
F = –40 N
(–ve sign shows that it is retarding force)
m = 20 kg, u = 5 ms
–1
, v= 0, t = ?
Step–1: Using relation F = ma, we have
F
a
m

2
40
2
20
ms

   i.e. retardation
Step–
2: Using v = u + at, we get
or 0 = 5 –2t or 2t = 5
t = 2.5 s
Step–3: S = ut + ½ at
2
= 5 × 2.5 + ½(–2) × (2.5)
2
= 6.25 m
Example 3
A body of mass 3 kg is moving with a velocity of 2 ms
–1
. Now a force is applied on the body so that
its velocity changes to 3.5 ms
–1
in 25 s. Calculate the direction and magnitude of the force acting
on the body.
Solution
Here, m = 3 kg, u = 2 ms
–1
, v = 3.5 ms
–1
, t = 25 s
v u
F ma m
t
 
 
 
 
(3.5 2) 4.5
3 0.18
25 25
N

   
The direction of force is along the direction of motion.
Examp
le 4
A child is standing in the middle of the road when a driver of a scooter moving with velocity 10 ms
–1
sees him. He applies breaks to save the child and brings his scooter to rest in 4s. Calculate the
retarding force on the scooter. Given, mass of scooter is 300 kg and that of a drive is 60 kg.
Solution
Here, m = 300 + 60 = 360 kg u = 10 ms
–1
, v = 0, t = 4s, F = ?
Step–1: Using v = u + at, we get 9$9$&/$66(63+<6,&67+ $OOULJKWFRS\UHVHUYHG1RSDUWRIWKHPDWHULDOFDQEHSURGXFHGZLWKRXWSULRUSHUPLVVLRQ

10. or
0 10 5
4 2
v u
a
t
 
   
= –2.5 ms
–2
Negative sign shows that it is retardation.
Step–2: Retarding force, F = ma
= –360 × 2.5 = –900 N
Example 5
A force of 10 N produces an acceleration of 2 ms
–2
in a body of mass m1 and 5 ms
–2
in a body of
mass m2. What will be the acceleration produced by the same force when both the bodies are tied
together?
Solution
F = 10 N, a1 = 2 ms
–2
 1
1
10
5
2
F
m kg
a
  
a2 = 5 ms
–2
2
2
10
2
5
F
m kg
a
  
Now, F = (m1 + m2)
a

1 2
10
7
F
a
m m
 

= 1.43 ms
–2
Example 6
A machine gun has a mass of 20 kg. It fires 35 g bullets at the rate of 4 bullets per second, with a speed of 400 ms
–1
. What force must be applied to the gun to keep it in position?
Solution
3
1
35 10 400
0.7
20
mv
V ms
M


 
  
Change in velocity of gun = 0.7 ms
–1
1
4
t s


2
0.7
2.8
1/ 4
a ms

 
Hence, F = ma = 98 × 10
–3
N
E
xample 7
A rubber ball of mass 200 g falls from a height of 1 m and rebounds to a height of 40 cm. Find the impulse and the average force between the ball and the ground, if time during which they are in contact was 0.1 s.
Solution
Velocity of ball just striking the ground
1
2 2 9.8 1 4.43
v gh ms

    
Velocity of the ball just rebounding
1
1
2 2 9.8 0.4 2.8
v gh ms

    
Impulse, Ft = mv1 –(–m
v) = m(v + v1)
= 0.2 × 7.23 = 1.446 Ns and
1.446
14.46
0.1
F N
 
***** 9$9$&/$66(63+<6,&67+ $OOULJKWFRS\UHVHUYHG1RSDUWRIWKHPDWHULDOFDQEHSURGXFHGZLWKRXWSULRUSHUPLVVLRQ

11. EXERCISE-I
1. The earth attracts on apple with a force of 1.5 N taking this as an action force, how much is the reaction
force, who exerts this reaction force? on which body does this reaction force act?
2. A coin falls towards the earth because the earth attracts the coin. Does the coin also attract the earth?
3. A body of mass m is placed on a table. The earth is pulling the body with a force mg. Taking this force to
be the action, what is the reaction?
4. You are riding in a car. The driver suddenly applies the breakes and you are pushed forward. Who pushed
you forward?
5. If you jump barefooted on a hard surface, your legs get injured, but they are not injured if you jump on a soft
surface like sand or pillow. Explain?
6. Define one newton force?
7. Give SI unit of momentum ?
8. Action and reaction are equal and opposite. Why cannot they cancel each other?
9. A jet engine works on the principle of newton’ s ....
10. Name the physical quantity that measures the inertia of a body?
11. One ...... is the force which produces an acceleration of 1 cm/s
2
when acting on a body of mass 1 gm.
12. Which is not the unit of force among the following?
(a) dyne (b) newton
(c) pound (d) kilogram weight
13. What do you mean by inertia?
14. On what factor does inertia of a body depend?
15. State Newton’ s second law of motion.
16. Name the unit of force in CGS system.
17. The rate of change of momentum of a particle is 5 kg ms
–2
. How much force acts on this body?
18. How much force acts on a body whose momentum is constant (i.e.
p

= constant)?
19. Dust can be removed from the carpet by beating it with a stick. Explain. why? 20. A stone tied to a string whirls in a horizontal circle It flies off tangentially when the string breaks suddenly.
Explain
21. When a person jumps out of a boat, the boat moves backward. Explain, why?
EXERCISE-II
1. A body of mass 1 Kg is kept at rest. A constant force of 1N starts acting on it. Find the time taken by the
body to move through a distance at 2 m.
2. A force at 4.0 N acts on a body of mass 2.0 Kg for 4 s. Assuming the body to be initially at rest, find.
(A) its velocity when the force stops acting, (B) the distance covered in 10 S after the force starts acting.
3. A feather of mass 0.2 Kg is dropped from a hight, it is found to fall down with a constant velocity. What is
net force acting on it?
4. A force produces an acceleration of 1.5 m/sec
2
in a disk such 3 disks are tied together and the same force
is applied on the combination. What will be the acceleration? 9$9$&/$66(63+<6,&67+ $OOULJKWFRS\UHVHUYHG1RSDUWRIWKHPDWHULDOFDQEHSURGXFHGZLWKRXWSULRUSHUPLVVLRQ

12. 5. A block of mass 120 g moves with a speed of 6.0 m/s on a fricitionaless horizontal surface towards another
block of mass 180 g kept at rest. They collide and the first block stops. Find the speed of the other block
after the collision.
6. A ball of mass 100 g and another ball of 120 g moves towards each other in the same direction with speed
6 m/s and 5 m/s respectively. If they stick to each other after colliding. What would be the velocity of the
combined mass after the collision?
7. A cart of 50 Kg is moving on a straight track with a speed of 12 m/s. A mass of 10 Kg is gently put into the
cart. What will be the velocity of the cart after this?
8. A javelin throw is marked foul if the athlete crosses over the line marked for the throw. Explain why
atheltes often fail to stop themselves before the line.
9. Why does a fielder pull his arms back while trying to stop or catch a cricket ball.
10. When you hit a wall and a piece of sponge with equal force, in which case will you be hurt more? Explain.
11. How do jet aeroplanes and rockets work.
12. Why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity?
13. A man pushes a box of mass 50 kg with a force of 80 N. What will be the acceleration of the box due to
this force? What would be the acceleration if the mass is halved?
14. A truck starts from rest and rolls down a hill with constant acceleration. It travels a distance of 400 m in
20 s. Find its acceleration. Find the force acting on it if its mass is 7 metric tonnes.
[Hint : 1 metric tonne = 1,000 kg]
15. The velocity-time graph of a ball moving on the surface of floor is shown in fig. Calculate the force acting
on the ball, if mass of the ball is 100 g.
16. A car of mass 1000 kg moving with a velocity of 45 km/h collides with a tree and comes to a stop in 5s.
What will be the force exerted by the car on the tree?
17. A cricket ball of mass 70 g moving with a velocity of 0.5 m/s is stopped by a player in 0.5 s. What is the
force applied by the player to stop the ball?
18. A force of 5 newton gives an acceleration of 8 m/s
2
to an object of mass m
1
, an acceleration of 24 m/s
2
. to
another object of mass m
2
What acceleration would it give if both the masses are tied together?
19. A bullet of mass 10 g is fired at a speed of 400 m/s from a gun of mass 4 kg. What is the recoil velocity of
the gun?
20. A man in a circus show jump from a height of 10m & is caught by a net spread below him. The net sags
down 2m due to impact. Calculate average force excrted by the net on the man to stop his fall. Takes mass
of the man = 60 kg & acceleration during free fall = 10 m/s
2
.
21. A bullet of mass 5 g travelling at a speed of 120 m/s penetrates deeply into a fixed target and is brought to
rest in 0.01 s. Calculate (i) the distance penetrated by the bullet into the target, and (ii) the average force
exerted on the bullet.
22. State Newton’ s second law of motion. Prove that F = ma, where symbols have their usual meanings.
23. State the law of conservation of momentum. Prove the law of conservation of momentum. 9$9$&/$66(63+<6,&67+ $OOULJKWFRS\UHVHUYHG1RSDUWRIWKHPDWHULDOFDQEHSURGXFHGZLWKRXWSULRUSHUPLVVLRQ

13. EXERCISE-III
SECTION-A
Fill in the blanks
1. The forces of action and reaction have ___________ magnitudes but ___________ directions.
2. A body kept at rest will remain ___________ if no unbalanced force acts on it.
3. The linear momentum of a system remains constant if no___________ force acts on it.
SECTION-B
Multiple choice question with one correct answers
1. A force of 100 N acts on 50 kg for 2 seconds. The same force acts on 25 kgs for 2 seconds. The ratio of
the momenta produced and the accelerations caused in two bodies respectively are –
(A) 1 : 1, 2 : 1 (B) 1 : 1, 1 : 2 (C) 1 : 2, 1 : 1 (D) none of these
2. The driver of a three-wheeler moving with a speed of 36 km/h sees a child standing in the middle of the
road and brings his vehicle to rest in 4.0 s just in time to save the child. The average retarding force on the
vehicle? The mass of the three-wheeler is 335 kg and mass of the driver is 65 kg.
(A) 1025 N (B) 10 N (C) 1000 N (D) none of these
3. A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 m/s, the
recoil speed of the gun –
(A) 14 m/s (B) 0.012 m/s (c) 0.016 m/s (D) 100 m/s
4. A horizontal force of 600 N pulls two masses 10 kg and 20 kg (lying on a frictionless table) connected by a
light string. The tension in the string , if the 20 kg mass is pulled
(A) 200 N (B) 400 N (C) 100 N (D) 50 N
5. Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a
frictionless pulley. The acceleration of the masses, when the masses are released is (g = 10 m/s
2
)
(A) 6 m/s
2
(B) 4 m/s
2
(C) 10 m/s
2
(D) 2 m/s
2
6. Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6 m/s collide and rebound
with the same speed. The magnitude of impulse imparted to each ball due to the other –
(A) 0.6 kg m/s (B) zero (C) 6 kg m/s (D) none of these
SECTION-C
Assertion & Reason
Instructions: In the following questions as Assertion (A) is given followed by a Reason (R). Mark your
responses from the following options.
(A) Both Assertion and Reason are true and Reason is the correct explanation of ‘Assertion’
(B) Both Assertion and Reason are true and Reason is not the correct explanation of ‘ Assertion’
(C) Assertion is true but Reason is false
(D) Assertion is false but Reason is true 9$9$&/$66(63+<6,&67+ $OOULJKWFRS\UHVHUYHG1RSDUWRIWKHPDWHULDOFDQEHSURGXFHGZLWKRXWSULRUSHUPLVVLRQ

14. 1.Assertion: Large force is required to move a body uniformly along a straight path
Reason: A body moving with uniform acceleration has a constant force acting on it.
2.Assertion: Linear momentum of a body changes when the body is moving in a circle.
Reason: In uniform circular motion the velocity changes because of change in direction.
3.Assertion: A player lowers his hands while catching a ball.
Reason: Impulse is the time rate of change of momentum.
4.Assertion: Mass is a measure of inertia of the body in linear motion.
Reason: Smaller the mass, smaller is the force required to change its state.
SECTION-D
Match the following (one to one)
Column-I and column-II contains four entries each. Entries of column-I are to be matched with some
entries of column-II. Only One entries of column-I may have the matching with the same entries of column-
II and one entry of column-II Only one matching with entries of column-I
1. Column I Column II
(A) Measure of inertia (P) Force × time
(B) Impulse (Q) Mass × velocity
(C) Momentum (R) Mass × acceleration
(D) Newton’ s second law of motion (S) Mass
EXERCISE-IV
SECTION-A
Multiple choice question with one correct answers
1. If a body is in equilibrium under a set of non-collinear forces, the minimum number of forces has to be:
(A) Four (B) Three (C) Two (D) Five
2. Bullets of 0.03 kg mass each hit a plate at the rate of 200 bullets per second, with a velocity of 50 m/sec.
reflect back with a velocity of 30 m/sec. The average force acting on the plate, in newtons is:
(A) 120 (B) 180 (C) 300 (D) 480
3. A force of 5 N making an angle  with the horizontal acting on an object displaces it by 0.4 m along the
horizontal direction.If the object gains kinetic energy of 1 J, the horizontal component of the force is:
(A) 1.5 N (B) 2.5 N (C) 3.5 N (D) 4.5 N
4. A force of 50 dynes is acted on a body of mass 5g which is at rest for an interval of 3sec. The impluse is
(A) 0.15 × 10
–3
Ns (B) 0.98 × 10
–3
Ns (C) 1.5 × 10
–3
N.s (D) 2.5 × 10
–3
5. Swimming is possible by the:
(A) first law of motion (B) second law of motion
(C) third law of motion (D) Newtons law of gravition
6. A cricket player catches a ball of mass 0.1 kg, moving with a speed 10 m/s in 0.1 second. Froce exerted by
him is (N)
(A) 4 (B) 2 (C) 1 (D) 10 9$9$&/$66(63+<6,&67+ $OOULJKWFRS\UHVHUYHG1RSDUWRIWKHPDWHULDOFDQEHSURGXFHGZLWKRXWSULRUSHUPLVVLRQ

15. SECTION-B
Multiple choice question with one or more than one correct answers
1. Newton’ s first law of motion defines
(A) Force (B) Inertia (C) Momentum (D) Acceleration
2. Which of the following statements is/are not correct ?
(A) A force is needed to keep a particle moving along a straight line with uniform velocity.
(B) The same force for the same time causes the same change in momentum for different bodies.
(C) Frictional force opposes both motion as well as relative motion.
(D) It is not motion, but relative motion that the frictional force opposes.
3. Which of the following statements is/are correct ?
(A) Momentum has both direction and magnitude
(B) Momentum is a scalar quantity
(C) The SI unit of momentum is kilogram-metre per second.
(D) The rate of change of momentum of an object is in the direction of force.
4. Action-Reaction forces act
(A) on the same body (B) on the different bodies
(C) along the same lines (D) in the same direction
5. A constant force acts on an object of mass 4 kg for a duration of 1 second. It increases the object velocity
from 4 ms
–1
to 6 ms
–1
. Select the correct options.
(A) Magnitude of the applied force is 8 N
(B) Final velocity of the object if the force applied for 5 sec is 24 m/s
(C) Change in momentum in 1 sec is 8 kg m sec
–1
(D) Change in momentum of object during 5 sec is 56 kg m/sec
6. A girl of mass 30 kg jumps with a horizontal velocity of 4 m sec
–1
onto a stationary cart with frictionless
wheels. The mass of cart is 2 kg. Select the correct options
(Assuming that there is no external unbalanced force working in the horizontal direction)
(A) The total momentum of the girl and the cart before the interaction is 120 kg m sec
–1
(B) The total momentumof the girl and the cart after the interaction is 120 kg m sec
–1
(C) The girl on cart will move with a velocity 15 m sec
–1
in the direction in which the girl jumped
(D) The girl on cart will move with a velocity 15/4 m sec
–1
in the opposite direction in which the girl jumped
SECTION-C
Comprehension
A truck is hauling a trailer along a level road as figure illustrates. The mass of the truck is m
1
= 8500 kg and
that of the trailer is m
2
= 27000 kg.
T T´
m = 27000 kg
2
m = 8500 kg
1
a = 0.78 m/s
2
D
Drawbar
The two move with an acceleration of a = 0.78 m/s
2
. Ignore the retarding forces of friction and air resis-
tances. 9$9$&/$66(63+<6,&67+ $OOULJKWFRS\UHVHUYHG1RSDUWRIWKHPDWHULDOFDQEHSURGXFHGZLWKRXWSULRUSHUPLVVLRQ

16. 1. The magnitude of the tension in the horizontal drawbar between the trailer and the truck
(A) 21000 N (B) 28000 N (C) 60000 N (D) None of these
2. The force D that propels the truck forward
(A) 21000 N (B) 28000 N (C) 60000 N (D) None of these
3. The action and reaction forces are
(A) D and T (B) T and T´ (C) D and T´ (D) None of these
SECTION-D
Match the following (one to many)
Column-I and column-II contains four entries each. Entries of column-I are to be matched with some
entries of column-II. One or more than one entries of column-I may have the matching with the some entries
of column-II and one entry of column-II may have one or more than one matching with entries of column-I
1. Column I Column II
(A) A rocket works on the principle of(P) First law of motion
(B) Recoiling of gun can be explained by (Q) Second law of motion
(C) The law which defines the force is(R) Third law of motion
(D) The law which gives the quantitative (S) Conservation of momentum
measurement of force is
EXERCISE-V
1. A constant force acts on an object of mass 5 kg for a duration of 2 s. It increases the object’s velocity from
3 m s
–1
to 7 m s
–1
. Find the magnitude of the applied force. Now, if the force was applied for a duration of
5 s, what would be the final velocity of the object.
2. Which would require a greater force–accelerating a 2 kg mass at 5 m s
–2
or a 4 kg mass at 2 m s
–2
?
3. A motorcar is moving with a velocity of 108 km/h and it takes 4 s to stop after the brakes are applied.
Calculate the force exerted by the brakes on the motorcar if its mass along with the passengers is 1000 kg.
4. A force of 5 N gives a mass m
1
, an accleration of 10 m s
–2
and a mass m
2
, an acceleration of 20 m s
–2
.
What acceleration would it give if both the masses were tied together?
5. A bullet of mass 20 g is horizontally fired with a velocity 150 m s
–1
from a pistol of mass 2 kg. what is the
recoil velocity of the pistol?
6. A girl of mass 40 kg jumps with a horizontal velocity of 5 m
–1
onto a stationary cart with frictionless wheels.
The mass of the cart is 3 kg. What is her velocity as the cart starts moving? Assume that there is no
external unbalanced force working in the horizontal direction.
EXERCISE-VI
1. Which of the following has more inertia: (a) a rubber ball and a stone of the same size? (b) a bicycle and a
train? (c) a five ruppes coin and a one rupee coin?
2. In the following example, try to identify the number of times the velocity of the ball changes: “A football
player kicks of a football to another player of his team who kicks the football towards the opposite team collects the football and kicks it towards a player of his won team”. Also identify the agent supplying the
force in each case. 9$9$&/$66(63+<6,&67+ $OOULJKWFRS\UHVHUYHG1RSDUWRIWKHPDWHULDOFDQEHSURGXFHGZLWKRXWSULRUSHUPLVVLRQ

17. 3. Explaing why some of the leaves may get detached from a tree if we vigorously shake its branch.
4. Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it
accelerates from rest?
5. If action is always equal to the reaction, explain how a horse can pull a cart.
6. Explain, why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high
velocity.
7. From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 m s
–1
. Calculate the
initial recoil velocity of the rifle.
8. Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2
ms
–1
and 1m s
–1
, respectively. They collide and after the collision, the first object moves at a velocity of 1.67
m s
–1
. Determine the velocity of the second object.
*****
Answers
TRY YOURSELF
1. 10 m 2. –1000 N 3. 0.53 ms
–1
4.–2500 N
EXERCISE-I
7. kg m/s
8. Because force of action & reaction act always on two different bodies.
9. Third law of motion 10. Mass
11. Dyne 12. (c) pound
13. The tendency of a body to oppose any change in its state of rest or uniform motion is called inertia of
the body.
14. Inertia of a body is directly proportional to its mass.
15. The rate of change of momentum of a body is directly proportional to the unbalanced force acting on
it.
16. dyne
17. F = rate of change of momentum = 5N
18. F = dp/dt. If p = constant, then no force acts on the body.
EXERCISE-II
1. 2 S 2. (a) 8 m/sec (b) 64 m
3. Zero 4. 0.5 m/sec
2
5. 4 m/sec 6. Zero
7. 10 m/sec
13. 1.6 m/s
2
, 3.2 m/s
2
14. a = 2m/s
2
, F = 140000 N
15. 0.5 N 16. 2500 N
17. .07 N 18. 6 m/s
2
. 9$9$&/$66(63+<6,&67+ $OOULJKWFRS\UHVHUYHG1RSDUWRIWKHPDWHULDOFDQEHSURGXFHGZLWKRXWSULRUSHUPLVVLRQ

18. 19. 1 m/s 20. (F = – 3000 N)
21. (i) 0.6 m, (ii) 60 N
EXERCISE-III
SECTION-A
1. Equal, Opposite 2. At rest 3. Unbalanced
SECTION-B
1. (B) 2. (C) 3. (C) 4. (A)
5. (D) 6. (A)
SECTION-C
1. (D) 2. (A) 3. (C) 4. (A)
SECTION-D
1. (A)-(S), (B)-(P), (C)-(Q), (D)-(R)
EXERCISE-IV
SECTION-A
1. (B) 2. (D) 3. (B) 4. (C) 5. (C)
6. (D)
SECTION-B
1. (A,B) 2. (A,C) 3. (A,C,D) 4. (B,C) 5. (A,C)
6. (A,B)
SECTION-C
1. (A) 2. (B) 3. (B)
SECTION-D
1. (A)- (R,S), (B)-(R,S), (C)-(P), (D)-(Q)
EXERCISE-V
1.–13 m/sec2. 2 kg mass 3. F = –7500 N 4.a = 6.67 m/sec
2
5. V = –1.5 m/sec6.V= +4.65 m/sec
EXERCISE-VI
3.–7/16 m/sec4. 1.98 m/sec
***** 9$9$&/$66(63+<6,&67+ $OOULJKWFRS\UHVHUYHG1RSDUWRIWKHPDWHULDOFDQEHSURGXFHGZLWKRXWSULRUSHUPLVVLRQ