NCERT Solutions for Class 9 Science Chapter 8 Force and Laws of Motion

NCERT
SOLUTIONS
CLASS - 9th

aglase .co

Book : Science Ncert Solutions | Chapter-9 Science

Class : 9th
Subject : Science
Chapter : 9
Chapter Name : FORCE AND LAWS OF MOTION

Q1 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 �ve-rupees coin and a one-rupee coin?

Answer. Inertia is the measure of the mass of the body. The greater is the mass of the
body; the greater is its inertia and vice-versa.
(a) Mass of a stone is more than the mass of a rubber ball for the same size. Hence,
inertia of the stone is greater than that of a rubber ball.
(b) Mass of a train is more than the mass of a bicycle. Hence, inertia of the train is
greater than that of the bicycle.
(c) Mass of a �ve rupee coin is more than that of a one-rupee coin. Hence, inertia of the
�ve rupee coin is greater than that of the one-rupee coin.

Page : 118 , Block Name : Questions

Q2 In the following example, try to identify the number of times the velocity of the ball
changes: “A football player kicks a football to another player of his team who kicks the
football towards the goal. The goalkeeper of the opposite team collects the football and
kicks it towards a player of his own team”.
Also identify the agent supplying the force in each case.

Answer. The velocity of the ball changes four times.
As a football player kicks the football, its speed changes from zero to a certain value.
As a result, the velocity of the ball gets changed. In this case, the player applied a force
to change the velocity of the ball. Another player kicks the ball towards the goal post. As
a result, the direction of the ball gets changed. Therefore, its velocity also changes. In
this case, the player applied a force to change the velocity of the ball. The goalkeeper
collects the ball. In other words, the ball comes to rest. Thus, its speed reduces to zero
from a certain value. The velocity of the ball has changed. In this case, the goalkeeper
applied an opposite force to stop/change the velocity of the ball. The goalkeeper kicks
the ball towards his team players. Hence, the speed of the ball increases from zero to a
certain value. Hence, its velocity changes once again. In this case, the goalkeeper applied
a force to change the velocity of the ball.

Page : 118 , Block Name : Questions

Page 1 of 12 Aglasem Schools

Book : Science Ncert Solutions | Chapter-9 Science

Q3 Explain why some of the leaves may get detached from a tree if we vigorously shake
its branch.

Answer. Some leaves of a tree get detached when we shake its branches vigorously. This
is because when the branches of a tree are shaken, it moves to and fro, but its leaves tend
to remain at rest. This is because the inertia of the leaves tend to resist the to and fro
motion. Due to this reason, the leaves fall down from the tree when shaken vigorously.

Page : 118 , Block Name : Questions

Q4 Why do you fall in the forward direction when a moving bus brakes to a stop and fall
backwards when it accelerates from rest?

Answer. Due to the inertia of the passenger Every body tries to maintain its state of
motion or state of rest. If a body is at rest,
then it tries to remain at rest. If a body is moving, then it tries to remain in motion. In a
moving bus, a passenger moves with the bus. As the driver applies brakes, the bus comes
to rest. But, the passenger tries to maintain his state of motion. As a result, a forward
force is exerted on him. Similarly, the passenger tends to fall backwards when the bus
accelerates from rest. This is because when the bus accelerates, the inertia of the
passenger tends to oppose the forward motion of the bus. Hence, the passenger tends to
fall backwards when the bus accelerates forward.

Page : 118 , Block Name : Questions

Q1 If action is always equal to the reaction, explain how a horse can pull a cart.

Answer. A horse pushes the ground in the backward direction. According to Newton’s
third law of motion, a reaction force is exerted by the Earth on the horse in the forward
direction. As a result, the cart moves forward.

Page : 126 , Block Name : Questions

Q2 Explain, why is it dif�cult for a �reman to hold a hose, which ejects large amounts of
water at a high velocity.

Answer. Due to the backward reaction of the water being ejected When a �reman holds a
hose, which is ejecting large amounts of water at a high velocity, then a reaction force is
exerted on him by the ejecting water in the backward direction. This is because of
Newton’s third law of motion. As a result of the backward force, the stability of the

Page 2 of 12 Aglasem Schools

Book : Science Ncert Solutions | Chapter-9 Science

�reman decreases. Hence, it is dif�cult for him to remain stable while holding the hose.

Page : 126 , Block Name : Questions

Q3 From a ri�e of mass 4 kg, a bullet of mass 50 g is �red with an initial velocity of 35 m
s–1 . Calculate the initial recoil velocity of the ri�e.

Answer. Mass of the ri�e, m1=41kg
Mass of the bullet, m2=50g
Recoil velocity of the ri�e = v1
Bullet is �red with an initial velocity v2 = 35m/s
Initially, the ri�e is at rest.
Thus, its initial velocity, v = 0
Total initial momentum of the ri�e and bullet system (m1+m2)v=0
Total momentum of the ri�e and bullet system after �ring:
=m1v1 = m2v2 = 4v1 = 0.005×35
=4v1 + 1.75
According to the law of conservation of momentum:
Total momentum after the �ring = Total momentum before the �ring
4v1 + 1.75 = 0
v1 = – 1.75 / 4 = -0.4375m/s
The negative sign indicates that the ri�e recoils backwards with a velocity of 0.4375 m/s.

Page : 126 , Block Name : Questions

Q4 Two objects of masses 100 g and 200 g are moving along the same line and direction
with velocities of 2 m s–1 and 1 m s–1, respectively They collide and after the collision,
the �rst object moves at a velocity of 1.67 m s–1 . Determine the velocity of the second
object.

Answer. Mass of one of the objects, m1 = 100 g = 0.1 kg
Mass of the other object, m2 = 200 g = 0.2 kg
Velocity of m1 before collision, v1=2 m/s
Velocity of m2 before collision, v2=1 m/s
Velocity of m1 after collision,v3=1.67m/s
Velocity of m2 after collision = v4
According to the law of conservation of momentum:
Total momentum before collision = Total momentum after collision
Therefore m1v1 + m2v2 = m3v3 + m4v4
(0.1)2 + (0.2)1 = (0.1)1.67 + (0.2)v4
0.4 = 0.167 + 0.2v4

Page 3 of 12 Aglasem Schools

Book : Science Ncert Solutions | Chapter-9 Science

Therefore v4 = 1.165m/s
Hence, the velocity of the second object becomes 1.165 m/s after the collision.

Page : 127 , Block Name : Questions

Q1 An object experiences a net zero external unbalanced force. Is it possible for the
object to be travelling with a non-zero velocity? If yes, state the conditions that must be
placed on the magnitude and direction of the velocity. If no, provide a reason.

Answer. Yes. Even when an object experiences a net zero external unbalanced force, it is
possible that the object is travelling with a non-zero velocity. This is possible only when
the object has been moving with a constant velocity in a particular direction. Then, there
is no net unbalanced force applied on the body. The object will keep moving with a non-
zero velocity. To change the state of motion, a net non-zero external unbalanced force
must be applied on the object.

Page : 128 , Block Name : Exercise

Q2 . When a carpet is beaten with a stick, dust comes out of it. Explain.

Answer. Inertia of an object tends to resist any change in its state of rest or state of
motion. When a carpet is beaten with a stick, then the carpet comes to motion. But, the
dust particles try to resist their state of rest. According to Newton’s �rst law of motion,
the dust particles stay in a state of rest, while the carpet moves. Hence, the dust particles
come out of the carpet.

Page : 128 , Block Name : Exercise

Q3 Why is it advised to tie any luggage kept on the roof of a bus with a rope?

Answer. When the bus accelerates and moves forward, it acquires a state of motion.
However, the luggage kept on the roof, owing to its inertia, tends to remain in its state of
rest. Hence, with the forward movement of the bus, the luggage tends to remain at its
original position and ultimately falls from the roof of the bus. To avoid this, it is advised
to tie any luggage kept on the roof of a bus with a rope.

Page : 128 , Block Name : Exercise

Q4 . A batsman hits a cricket ball which then rolls on a level ground. After covering a
short distance, the ball comes to rest. The ball slows to a stop because
(a) the batsman did not hit the ball hard enough.

Page 4 of 12 Aglasem Schools

Book : Science Ncert Solutions | Chapter-9 Science

(b) velocity is proportional to the force exerted on the ball.
(c) there is a force on the ball opposing the motion.
(d) there is no unbalanced force on the ball, so the ball would want to come to rest.

Answer. (c) A batsman hits a cricket ball, which then rolls on a level ground. After
covering a short distance, the ball comes to rest because there is frictional force on the
ball opposing its motion. Frictional force always acts in the direction opposite to the
direction of motion.
Hence, this force is responsible for stopping the cricket ball.

Page : 128 , Block Name : Exercise

Q5 A truck starts from rest and rolls down a hill with a 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
tonnes (Hint: 1 tonne = 1000 kg.)

Answer. Initial velocity, u = 0 (since the truck is initially at rest)
Distance travelled, s = 400 m
Time taken, t = 20 s
According to the second equation of motion:
s=u + 1/2 at square
Where
Acceleration = a
400 = 0 + 1/2 a (20)sq
400 = 1/2 a (400)
a = 2m/s sq
1 metric tonne = 1000kg
Therefore 7 metric tonnes = 7000kg
Mass of truck, m = 7000 kg
From Newton’s second law of motion:
Force, F = Mass × Acceleration
F = ma = 7000 × 2 = 14000 N
Hence, the acceleration of the truck is 2 m/s2 and the force acting on the truck is 14000
N.

Page : 128 , Block Name : Exercise

Q6 A stone of 1 kg is thrown with a velocity of 20 m s–1 across the frozen surface of a
lake and comes to rest after travelling a distance of 50 m. What is the force of friction
between the stone and the ice?

Page 5 of 12 Aglasem Schools

Book : Science Ncert Solutions | Chapter-9 Science

Answer. Initial velocity of the stone, u = 20 m/s
Final velocity of the stone, v = 0 (�nally the stone comes to rest)
Distance covered by the stone, s = 50 m
According to the third equation of motion:
v2 = u2 + 2as
Where,
Acceleration, a
(0)2 = (20)2 + 2 × a × 50
a = −4 m/s2
The negative sign indicates that acceleration is acting against the motion of the
stone.
Mass of the stone, m = 1 kg
From Newton’s second law of motion:
Force, F = Mass × Acceleration
F = ma
F = 1 × (− 4) = −4 N
Hence, the force of friction between the stone and the ice is −4 N.

Page : 128 , Block Name : Exercise

Q7 A 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track.
If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then
calculate:
(a) the net accelerating force and
(b) the acceleration of the train.

Answer. (a) 35000 N (b) 1.944 m/s2 (c) 28000 N
(a)Force exerted by the engine, F = 40000 N
Frictional force offered by the track, Ff = 5000 N
Net accelerating force, Fa = F − Ff = 40000 − 5000 = 35000 N
Hence, the net accelerating force is 35000 N.
(b)Acceleration of the train = a
The engine exerts a force of 40000 N on all the �ve wagons.
Net accelerating force on the wagons, Fa = 35000 N
Mass of the wagons, m = Mass of a wagon × Number of wagons
Mass of a wagon = 2000 kg
Number of wagons = 5
∴ m = 2000 × 5 = 10000 kg
Mass of the engine, m′ = 8000 kg
Total mass, M = m + m′ = 18000 kg
From Newton’s second law of motion:

Page 6 of 12 Aglasem Schools

Book : Science Ncert Solutions | Chapter-9 Science

Fa = Ma
a = F(a)/M = 35000/18000 = 1.944m/s sq
Hence, the acceleration of the wagons and the train is 1.944 m/s2.
(c)Mass of all the wagons except wagon 1 is 4 × 2000 = 8000 kg
Acceleration of the wagons = 3.5 m/s2
Thus, force exerted on all the wagons except wagon 1
= 8000 × 3.5 = 28000 N
Therefore, the force exerted by wagon 1 on the remaining four wagons is 28000 N.
Hence, the force exerted by wagon 1 on wagon 2 is 28000 N.

Page : 128 , Block Name : Exercise

Q8 An automobile vehicle has a mass of 1500 kg. What must be the force between the
vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 m s–2?

Answer. Mass of the automobile vehicle, m = 1500 kg
Final velocity, v = 0 (�nally the automobile stops)
Acceleration of the automobile, a = −1.7 ms−2
From Newton’s second law of motion:
Force = Mass × Acceleration = 1500 × (−1.7) = −2550 N
Hence, the force between the automobile and the road is −2550 N, in the direction
opposite to the motion of the automobile.

Page : 128 , Block Name : Exercise

Q9 What is the momentum of an object of mass m, moving with a velocity v?
(a) (mv)2
(b) mv2
(c) ½ mv2
(d) mv

Answer. (d) mv
Mass of the object = m
Velocity = v
Momentum = Mass × Velocity
Momentum = mv

Page : 128 , Block Name : Exercise

Q10 Using a horizontal force of 200 N, we intend to move a wooden cabinet across a �oor
at a constant velocity. What is the friction force that will be exerted on the cabinet?

Page 7 of 12 Aglasem Schools

Book : Science Ncert Solutions | Chapter-9 Science

Answer. A force of 200 N is applied in the forward direction. Thus, from Newton’s third
law of motion, an equal amount of force will act in the opposite direction. This opposite
force is the frictional force exerted on the cabinet. Hence, a frictional force of 200 N is
exerted on the cabinet.

Page : 128 , Block Name : Exercise

Q11 Two objects, each of mass 1.5 kg, are moving in the same straight line but in
opposite directions. The velocity of each object is 2.5 m s-1 before the collision during
which they stick together. What will be the velocity of the combined object after
collision?

Answer. Mass of one of the objects, m1 = 1.5 kg
Mass of the other object, m2 = 1.5 kg
Velocity of m1 before collision, v1 = 2.5 m/s
Velocity of m2, moving in opposite direction before collision, v2 = −2.5 m/s
(Negative sign arises because mass m2 is moving in an opposite direction)
After collision, the two objects stick together.
Total mass of the combined object = m1 + m2
Velocity of the combined object = v
According to the law of conservation of momentum:
Total momentum before collision = Total momentum after collision
m1v1 + m2 v1 = (m1 + m2) v
1.5(2.5) + 1.5 (−2.5) = (1.5 + 1.5) v
3.75 − 3.75 = 3 v
v=0
Hence, the velocity of the combined object after collision is 0 m/s.

Page : 128 , Block Name : Exercise

Q12 According to the third law of motion when we push on an object, the object pushes
back on us with an equal and opposite force. If the object is a massive truck parked along
the roadside, it will probably not move. A student justi�es this by answering that the two
opposite and equal forces cancel each other. Comment on this logic and explain why the
truck does not move.

Answer. The truck has a large mass. Therefore, the static friction between the truck and
the road is also very high. To move the car, one has to apply a force more than the static
friction. Therefore, when someone pushes the truck and the truck does not move, then it
can be said that the applied force in one direction is cancelled out by the frictional force

Page 8 of 12 Aglasem Schools

Book : Science Ncert Solutions | Chapter-9 Science

of equal amount acting in the opposite direction. Therefore, the student is right in
justifying that the two opposite and equal cancel each other.

Page : 129 , Block Name : Exercise

Q13 A hockey ball of mass 200 g travelling at 10 m s-1 is struck by a hockey stick so as to
return it along its original path with a velocity at 5 m s-1. Calculate the magnitude of
change of momentum occurred in the motion of the hockey ball by the force applied by
the hockey stick.

Answer. Mass of the hockey ball, m = 200 g = 0.2 kg
Hockey ball travels with velocity, v1 = 10 m/s
Initial momentum = mv1
Hockey ball travels in the opposite direction with velocity, v2 = −5 m/s
Final momentum = mv2
Change in momentum = mv1 − mv2 = 0.2 [10 − (−5)] = 0.2 (15) = 3 kg m s−1
Hence, the change in momentum of the hockey ball is 3 kg m s−1

Page : 129 , Block Name : Exercise

Q14 A bullet of mass 10 g travelling horizontally with a velocity of 150 m s-1 strikes a
stationary wooden block and comes to rest in 0.03 s. Calculate the distance of
penetration of the bullet into the block. Also calculate the magnitude of the force
exerted by the wooden block on the bullet.

Answer. Now, it is given that the bullet is travelling with a velocity of 150 m/s.
Thus, when the bullet enters the block, its velocity = Initial velocity, u = 150 m/s
Final velocity, v = 0 (since the bullet �nally comes to rest)
Time taken to come to rest, t = 0.03 s
According to the �rst equation of motion, v = u + at
Acceleration of the bullet, a
0 = 150 + (a ×0.03 s)
a = -150/0.03 = -5000m per second sq
(Negative sign indicates that the velocity of the bullet is decreasing.)
According to the third equation of motion:
v2 = u2 + 2as
0 = (150)2 + 2 (−5000) s
s = -(150 sq)/=2(5000) = 22500/10000 = 2.25m

Hence, the distance of penetration of the bullet into the block is 2.25 m.
From Newton’s second law of motion:

Page 9 of 12 Aglasem Schools

Book : Science Ncert Solutions | Chapter-9 Science

Force, F = Mass × Acceleration
Mass of the bullet, m = 10 g = 0.01 kg
Acceleration of the bullet, a = 5000 m/s2
F = ma = 0.01 × 5000 = 50 N
Hence, the magnitude of force exerted by the wooden block on the bullet is 50 N.

Page : 129 , Block Name : Exercise

Q15 An object of mass 1 kg travelling in a straight line with a velocity of 10 m s-1 collides
with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off
together in the same straight line. Calculate the total momentum just before the impact
and just after the impact. Also, calculate the velocity of the combined object

Answer. Mass of the object, m1 = 1 kg
Velocity of the object before collision, v1 = 10 m/s
Velocity of the wooden block before collision, v2 = 0 m/s
∴ Total momentum before collision = m1 v1 + m2 v2
= 1 (10) + 5 (0) = 10 kg m s−1
It is given that after collision, the object and the wooden block stick together.
Total mass of the combined system = m1 + m2
Velocity of the combined object = v
According to the law of conservation of momentum:
Total momentum before collision = Total momentum after collision
m1 v1 + m2 v2 = (m1 + m2) v
1 (10) + 5 (0) = (1 + 5) v
v = 10/6 = 5/3 m per second sq
The total momentum after collision is also 10 kg m/s.
Total momentum just before the impact = 10 kg m s−1
Total momentum just after the impact = (m1+m2)v = 6×5/3 = 10kg m/s
Hence, Velocity of the combined object after collision = 5/3 m/s

Page : 129 , Block Name : Exercise

Q16 An object of mass 100 kg is accelerated uniformly from a velocity of 5 m s–1 to 8 m
s–1 in 6 s. Calculate the initial and �nal momentum of the object. Also, �nd the
magnitude of the force exerted on the object.

Answer. Initial velocity of the object, u = 5 m/s
Final velocity of the object, v = 8 m/s
Mass of the object, m = 100 kg
Time take by the object to accelerate, t = 6 s

Page 10 of 12 Aglasem Schools

Book : Science Ncert Solutions | Chapter-9 Science

Initial momentum = mu = 100 × 5 = 500 kg m s−1
Final momentum = mv = 100 × 8 = 800 kg m s−1
Force exerted on the object F = (mv=mu)/t
= m(v-u)/t = (800-500)/6 = 50N
Initial momentum of the object is 500 kg m s−1.
Final momentum of the object is 800 kg m s−1.
Force exerted on the object is 50 N.

Page : 129 , Block Name : Exercise

Q17 Akhtar, Kiran and Rahul were riding in a motorcar that was moving with a high
velocity on an expressway when an insect hit the windshield and got stuck on the
windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that
the insect suffered a greater change in momentum as compared to the change in
momentum of the motorcar (because the change in the velocity of the insect was much
more than that of the motorcar). Akhtar said that since the motorcar was moving with a
larger velocity, it exerted a larger force on the insect. And as a result the insect died.
Rahul while putting an entirely new explanation said that both the motorcar and the
insect experienced the same force and a change in their momentum. Comment on these
suggestions.

Answer. According to the law of conservation of momentum:
Momentum of the car and insect system before collision = Momentum of the car and
insect system after collision
Hence, the change in momentum of the car and insect system is zero.
The insect gets stuck on the windscreen. This means that the direction of the insect is
reversed. As a result, the velocity of the insect changes to a great amount. On the other
hand, the car continues moving with a constant velocity. Hence, Kiran’s suggestion that
the insect suffers a greater change in momentum as compared to the car is correct. The
momentum of the insect after collision becomes very high because the car is moving at a
high speed. Therefore, the momentum gained by the insect is equal to the momentum
lost by the car.
Akhtar made a correct conclusion because the mass of the car is very large as compared
to the mass of the insect. Rahul gave a correct explanation as both the car and the insect
experienced equal forces caused by the Newton’s action-reaction law. But, he made an
incorrect statement as the system suffers a change in momentum because the
momentum
before the collision is equal to the momentum after the collision.

Page : 129 , Block Name : Exercise

Page 11 of 12 Aglasem Schools

Book : Science Ncert Solutions | Chapter-9 Science

Q18 How much momentum will a dumb-bell of mass 10 kg transfer to the �oor if it falls
from a height of 80 cm? Take its downward acceleration to be 10 m s–2 .

Answer. Mass of the dumbbell, m = 10 kg
Distance covered by the dumbbell, s = 80 cm = 0.8 m
Acceleration in the downward direction, a = 10 m/s2
Initial velocity of the dumbbell, u = 0
Final velocity of the dumbbell (when it was about to hit the �oor) = v
According to the third equation of motion:
v2 = u2 + 2as
v2 = 0 + 2 (10) 0.8
v = 4 m/s
Hence, the momentum with which the dumbbell hits the �oor is
= mv = 10 × 4 = 40 kg m s−1

Page : 129 , Block Name : Exercise

Page 12 of 12 Aglasem Schools