Chapter 9 : Force And Laws Of Motion - Ncert Solutions for Class 9 Physics CBSE

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Chapter 9 - Force And Laws Of Motion Excercise 118

Solution 1
Mass is a measure of the inertia of a body. The greater the mass of a body; the greater is its inertia.
 
(a) Mass of a stone is more than the mass of a rubber ball of same size. Hence, inertia of a stone is greater than that of a rubber ball of same size.
(b) Mass of a train is more than the mass of a bicycle. Hence, inertia of a train is greater than that of a bicycle.
 
(c) Mass of a five rupee coin is more than that of a one-rupee coin. Hence, inertia of a five rupee coin is greater than that of a one-rupee coin.

Concept Insight: More mass means more inertia.

Solution 2

The velocity of the ball changes four times.

(i) 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 force is applied by the kick of the player.

(ii) When the ball reaches another player, he kicks the ball towards the goal post. As a result, the direction of the ball and its speed both get changed. Therefore, its velocity also changes. In this case, the force is applied by the kick of the player.

(iii) When the goalkeeper collects the ball, the ball comes to rest, i.e. its speed reduces to zero from a certain value. Thus, the velocity of the ball changes. In this case, the force is applied by the hands of the goalkeeper.

(iv) The goalkeeper then kicks the stationary ball towards his team player, i.e., the speed of the ball increases from zero to a certain value. Hence, its velocity changes once again. In this case, the force is applied by the kick of the goalkeeper.

 Concept Insight:- Velocity is a vector quantity. It has both magnitude and direction.

Solution 3

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

 
Concept Insight:- Inertia resists change in state of motion.
Solution 4

When a moving bus brakes to a stop, we fall in the forward direction because though the lower part of our body comes to a stop when the bus stops but the upper part of the body continues to be in motion in the forward direction due to its inertia, thus making us fall in the forward direction.

When a bus accelerates from rest, we fall backwards because though the lower part of our body starts moving with the bus but the upper part of the body tries to remain at rest due to its inertia, thus making us fall in the backward direction.

Concept Insight:- Inertia resists any change in the state of motion of a body.

Chapter 9 - Force And Laws Of Motion Excercise 126

Solution 1

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

Concept Insight:- Action and reaction forces act on two different bodies and that too in opposite directions.

Solution 2

Due to the backward reaction of the water ejecting from the hose pipe.

When a fireman 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 hose pipe tends to go backward and slips from the hands of fireman which makes it difficult for the fireman to hold the hose pipe.

 
Concept Insight:- Action and reaction forces act on two different bodies and that too in opposite directions.

Solution 3
Mass of the rifle,  m1 = 4 kg

Mass of the bullet,  m2 = 50 g = 0.05 kg

Recoil velocity of the rifle = v1

Bullet is fired with an initial velocity, v2 = 35 m/s

Initially, the rifle is at rest.

Thus, its initial velocity, v = 0

Total initial momentum (before firing) of the rifle and bullet system  = (m1 + m2)v =0

Total final momentum (after firing) of the rifle and bullet system

= m1v1 + m2v2
= 4(v1) + 0.05 × 35
= 4v1 + 1.75

According to the law of conservation of momentum:

Concept Insight:- Total momentum after the firing = Total momentum before the firing

4v1 + 1.75 = 0
4v1 = − 1.75

begin mathsize 14px style straight v subscript 1 space equals space minus fraction numerator 1.75 over denominator 4 end fraction equals negative 0.4375 space straight m divided by straight s end style
The negative sign indicates that the rifle recoils backwards with a velocity of 0.4375 m/s.

Chapter 9 - Force And Laws Of Motion Excercise 127

Solution 1

Mass of first object, m1 = 100 g = 0.1 kg 

Mass of second object, m2 = 200 g = 0.2 kg

Velocity of first object before collision, v1 = 2 m/s

Velocity of second object before collision, v2 = 1 m/s

Velocity of first object after collision, v3 = 1.67 m/s

Velocity of second object after collision = v4

According to the law of conservation of momentum:

Concept Insight:- Total momentum before collision = Total momentum after collision

begin mathsize 14px style therefore straight m subscript 1 straight v subscript 1 space plus space straight m subscript 2 straight v subscript 2 space equals space straight m subscript 1 straight v subscript 3 space plus space straight m subscript 2 straight v subscript 4
left parenthesis 0.1 right parenthesis cross times 2 space plus space left parenthesis 0.2 right parenthesis cross times 1 space equals space left parenthesis 0.1 right parenthesis cross times 1.67 space plus space left parenthesis 0.2 right parenthesis cross times straight v subscript 4
0.2 space plus space 0.2 space equals space 0.167 space plus space 0.2 straight v subscript 4
0.4 space equals space 0.167 plus space 0.2 straight v subscript 4
0.4 minus 0.167 space equals space 0.2 straight v subscript 4
0.233 space equals space 0.2 straight v subscript 4
straight v subscript 4 equals fraction numerator 0.233 over denominator 0.2 end fraction
therefore straight v subscript 4 equals 1.165 space straight m divided by straight s end style

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

Chapter 9 - Force And Laws Of Motion Excercise 128

Solution 1

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 the same non-zero velocity.

Concept Insight:- To change the state of motion, a net non-zero external unbalanced force must be applied on the object.
Solution 2

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 retain their state of rest. Hence, the dust particles come out of the carpet.

 
Concept Insight:- Inertia resists change in state of motion.

Solution 3

When the bus suddenly accelerates from rest 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 and hence may fall down from the roof of the bus.

Similarly, when the moving bus stops suddenly, then due to its inertia of motion, the luggage kept on the roof of the bus tends to remain in motion and hence may fall down from the roof of the bus.

Hence, it is advised to tie the luggage kept on the roof of a bus with a rope so that it does not fall down when the bus starts or stops suddenly.

 
Concept Insight:- Inertia resists change in state of motion

Solution 4

(c) There is a force on the ball opposing the motion.

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.

Solution 5
Initial velocity, u = 0 (since the truck is initially at rest)
 
Distance travelled, s = 400 m
Time taken, t = 20 s
Acceleration, a=?
 
According to the second equation of motion:
 
begin mathsize 14px style straight s equals ut plus 1 half at squared end style
begin mathsize 14px style 400 equals 0 plus 1 half cross times straight a cross times left parenthesis 20 right parenthesis squared
400 equals 1 half cross times straight a cross times 400
400 equals 200 straight a
straight a equals 400 over 200
therefore straight a equals 2 space straight m divided by straight s squared end style
 
Given: 1 tonne = 1000 kg
 
Therefore, 7 tonnes = 7000 kg
 
Mass of truck, m = 7000 kg
 
From Newton's second law of motion:
 
Concept Insight:- 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.

Solution 6
Initial velocity of the stone, u = 20 m/s
 
Final velocity of the stone, v = 0 (finally the stone comes to rest)
 
Distance covered by the stone, s = 50 m
 
According to the third equation of motion:
 
v2 = u2 + 2as
 
where, a = acceleration
 
(0)2 = (20)2 + 2 × a × 50
0 = 400 + 100 a
-400 = 100 a

begin mathsize 14px style negative 400 over 100 equals straight a end style
begin mathsize 14px style therefore end stylea = - 4 m/s2

Concept Insight:- 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.

Solution 7
(a) Force exerted by the engine, F = 40000 N
 
Frictional force offered by the track, Ff = 5000 N
 
Hence, net accelerating force, Fa = F - Ff = 40000 - 5000 = 35000 N
 
 
(b) Let acceleration of the train be a.
 
Net accelerating force on the wagons, Fa = 35000 N
 
Mass of the wagons, M = Mass of a wagon × Number of wagons = 2000 x 5 = 10000 kg

From Newton's second law of motion:

Concept Insight:- Force = Mass x Acceleration
 
Fa = Ma
 
begin mathsize 14px style straight a equals straight F subscript straight a over straight M equals 35000 over 10000 equals 3.5 space straight m divided by straight s squared end style
Hence, the acceleration of the train is 3.5 m/s2.

Mass of the four wagons behind the first wagon = 4 2000 = 8000 kg

Acceleration of the wagons = 3.5 m/s2

Thus, force of wagon 1 on remaining four wagons behind it = 8000 3.5 = 28000 N

Hence, the force exerted by wagon 1 on wagon 2 is 28000 N.

Solution 8
Mass of the automobile vehicle, m = 1500 kg
 
Final velocity, v = 0 (finally 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.

Solution 9
(d)  mv
 
Mass of the object = m
 
Velocity = v
 
Concept Insight:- Momentum = Mass × Velocity
 
Momentum = mv

Solution 10

Force applied, P = 200 N

Force of friction, F = ?

As the wooden cabinet is to move across the floor with a constant velocity, no force (f) is spent in accelerating the cabinet, i.e.,

f = P-F = 0

or, F = P = 200 N


Concept Insight:- For a non-accelerated motion, no net force is required.

Solution 11

Mass of first object, m1 = 1.5 kg

Mass of second object, m2 = 1.5 kg

Velocity first object before collision, v1 = 2.5 m/s

Velocity of second object 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:

Concept Insight:- Total momentum before collision = Total momentum after collision

m1 v1 + m2 v2 = (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.

Chapter 9 - Force And Laws Of Motion Excercise 129

Solution 1

When we push a massive truck parked along the roadside, it does not move. The justification given by the student that the two opposite and equal forces cancel each other is totally wrong. This is because force of action and reaction never act on one body. There is no question of their cancellation. The truck does not move because the push applied is far less than the force of friction between the truck and the road.

Concept Insight:- Action and reaction forces act on different objects.

Solution 2
Mass of the hockey ball, m = 200 g = 0.2 kg

Hockey ball travels with velocity, v1 = 10 m/s

Initial momentum = mv1

After being struck by the stick, the hockey ball travels in the opposite direction with velocity, v2 = -5 m/s

Final momentum = mv2

Concept Insight:- Change in momentum = Final momentum - Initial momentum

Change in momentum = mv2 - mv1  = m (v2 - v1) =  0.2 [-5-10] = 0.2 (-15) = -3 kg ms-1

Hence, the change in momentum of the hockey ball is -3 kg ms-1.

Solution 3

Mass of the bullet, m = 10 g = 0.01 kg

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 finally comes to rest)

Time taken to come to rest, t = 0.03 s

According to the first equation of motion,

v = u + at

where , a is the acceleration of the bullet

0 = 150 + (a × 0.03)

begin mathsize 14px style 0 equals 150 plus 0.03 straight a
minus 150 equals 0.03 straight a
minus fraction numerator 150 over denominator 0.03 end fraction equals straight a
therefore straight a equals negative 5000 space straight m divided by straight s squared end style

 
Concept Insight:- Negative sign indicates that the velocity of the bullet is decreasing.

According to the third equation of motion:

v2 = u2 + 2 as

0 = (150)2 + 2×(-5000)×s

0 = 22500 + 2×(-5000)×s 

0=22500 - 10000 s

10000 s = 22500

begin mathsize 14px style straight s equals 22500 over 10000 equals 2.25 space straight m end style

Hence, the distance of penetration of the bullet into the block is 2.25 m.

From Newton's second law of motion:

Concept Insight:- Force, F = Mass Acceleration

Mass of the bullet, m = 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.

Solution 4

Mass of the object, m1 = 1 kg

Velocity of the object before collision, v1 = 10 m/s

Mass of the stationary wooden block, m2 = 5 kg

Velocity of the wooden block before collision, v2 = 0 m/s

Total momentum before collision = m1v1+m2v= (1×10)+(5×0) = 10 kg m s-1

According to the law of conservation of momentum, the total momentum just after the impact will be the same as the total momentum just before the impact.

i.e., the total momentum just after the impact will be 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 = m+ m2

Velocity of the combined system = v

According to the law of conservation of momentum:

Concept Insight:- Total momentum before collision = Total momentum after collision

m1v1+m2v2 = (m1+m2)v

(1×10)+(5×0) = (1+5)v

10+0 = 6v

10 = 6v

begin mathsize 14px style straight v equals 10 over 6 equals 5 over 3 straight m divided by straight s end style

v = 1.67 m/s

Hence, velocity of the combined object after collision will be 1.67 m/s.

 
Solution 5

The suggestion made by Kiran that the insect suffered a greater change in momentum as compared to the change in momentum of the motor car is wrong. The suggestion made by Akhtar that the motor car exerted a larger force on the insect because of large velocity of motor car is also wrong. The explanation put forward by Rahul is correct. On collision of insect with motor car, both experience the same force as action and reaction are always equal and opposite. Further, changes in their momenta are also the same. Only the signs of changes in momenta are opposite, i.e., change in momenta of the two occur in opposite directions, though magnitude of change in momentum of each is the same.

Solution 6

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 floor) = v

Concept Insight Choose the equation of motion wisely out of the three, to minimize the number of steps in calculations.

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 floor = mv = 10×4 = 40 kg m s-1

Momentum transferred to the floor is 40 kg m/s.

Chapter - Excercise

Solution 1

Chapter 9 - Force And Laws Of Motion Excercise 130

Solution 1

(a) A careful observation of the distance-time table shows that

begin mathsize 14px style straight s proportional to straight t cubed end style

It is known that

(i) for motion with uniform velocity (zero acceleration)

begin mathsize 14px style straight s equals ut
straight i. straight e. comma space straight s proportional to straight t end style

 

(ii) for motion with uniform acceleration

begin mathsize 14px style straight s equals ut plus 1 half at squared
straight i. straight e. comma space straight s proportional to straight t squared end style

 

In the present case, begin mathsize 14px style straight s proportional to straight t cubed end style.  Therefore, we conclude in this case that acceleration must be increasing uniformly with time.


(b) As F = ma, therefore, F  a. Hence, the force must also be increasing uniformly with time.

Solution 2

Here, mass of motorcar, m = 1200 kg

Let each person exert a push F on the motorcar.

Total push of two persons = F + F = 2F

As this push gives a uniform velocity to the motorcar along a level road, it must be a measure of the force of friction (f) between the motorcar and the road,

i.e., f = 2F.

When three person push, total force applied = F + F + F = 3F

Force that produces acceleration (a=0.2 m/s2),

i.e., ma = 3F-f = 3F-2F = F

or, F = ma = 1200 × 0.2 = 240 N

Solution 3

begin mathsize 14px style Given colon
Mass space of space the space hammer comma space straight m equals 500 space straight g equals 500 over 1000 equals 0.5 space kg
Initial space velocity comma space straight u equals 50 space straight m divided by straight s
Final space velocity comma space straight v equals 0 space straight m divided by straight s
Time comma space straight t equals 0.01 space straight s
Force space of space the space nail space on space the space hammer comma space straight F equals ?
straight F equals ma equals fraction numerator straight m left parenthesis straight v minus straight u right parenthesis over denominator straight t end fraction
space space space equals fraction numerator 0.5 left parenthesis 0 minus 50 right parenthesis over denominator 0.01 end fraction
space space space equals fraction numerator 0.5 left parenthesis negative 50 right parenthesis over denominator 0.01 end fraction
space space space equals negative fraction numerator 25 over denominator 0.01 end fraction
straight F equals negative 2500 space straight N end style

Thus, the force of the nail on the hammer is 2500 N. Negative sign indicates the opposing force.

Solution 4

begin mathsize 14px style Given colon
Mass comma space straight m equals 1200 space kg
Initial space velocity comma space straight u equals 90 space km divided by straight h equals fraction numerator 90 cross times 1000 space straight m over denominator 60 cross times 60 space straight s end fraction equals 25 space straight m divided by straight s
Final space velocity comma space straight v equals 18 space km divided by straight h equals fraction numerator 18 cross times 1000 space straight m over denominator 60 cross times 60 space straight s end fraction equals 5 space straight m divided by straight s
Time comma space straight t equals 4 space straight s
Acceleration comma space straight a equals ?
By space equation space of space motion comma space we space get comma
straight v equals straight u plus at
straight a equals fraction numerator straight v minus straight u over denominator straight t end fraction
space space space equals fraction numerator 5 minus 25 over denominator 4 end fraction
straight a equals negative 5 straight m divided by straight s squared end style

Change in momentum = mv-mu = m(v-u) = 1200×(5-25)= -24000 kg m/s

Force required , F = ma = 1200×(-5) = -6000 N

Magnitude of force required = 6000 N

Negative sign shows that force is opposing the motion.

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