# NCERT Solutions for Class 9 Science Chapter 9 Force and Laws of Motion Questions and Answers

## Force and Laws of Motion Questions and Answers Class 9

Page : 118

Q1. Which of the following has more inertia?

(i) A stone and a rubber ball of same size

(ii) A bicycle and a train

(iii) A five-rupees coin and a one-rupee coin.

Ans: Inertia of an object is the measurement of its mass, objects with more mass have more inertia.

(a) A stone of the same size

(b) a train

(c) a five-rupees coin

Q2. In the following example, try to identify the number of times at which 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 ball and kicks it towards a player of his own team. Also, identify the agent supplying the force in each case.

Ans: (i) When the first player kicks the ball toward another player of his team, then the velocity of the ball will change from 0 to ‘u’ because the first player applies some force on the ball.

(ii) When another player kicks the ball towards the goal, then the velocity of the ball will change, here again, the force is applied to the ball.

(iii) When the goalkeeper of the opposite team collects the ball. its velocity becomes zero.

(iv) The velocity of the ball changes when the goalkeeper applies some force on the ball.

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

Ans: Due to the inertia the leaves tend to remain in the state of rest when the position of the branch changes. therefore leaves get attached to a tree if we vigorously shake its branch.

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?

Ans: We fall in the forward direction when on a moving bus when brakes are applied because of inertia.

As our upper body continues to move while the bus gets stopped. The same goes when a bus is accelerated from rest our upper body tends to be at rest while the bus starts moving.

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

Ans: The third law of motion states that ‘action is always equal to the reaction’ but this will be implied in two different bodies.

In this case : The horse pushes the ground in the backward direction and the ground exerts equal and opposite force on the horse and cart system to push them forward.

This causes the force of friction between the wheels of the cart and the ground, then the cart is pushed forward.

Q2. Explain, why is it difficult for a fireman to hold a hose-pipe, which ejects large amount of water with a high velocity?

Ans: A fireman finds it difficult to hold a hose pipe which is ejecting a large amount of water at high velocity.

Because the stream of water rushing out of the pipe in the forward direction exerts a large force on the pipe in the backward direction.

Therefore, the fireman struggles to keep the hose pipe at rest.

Q3. From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 m/s. Calculate the initial recoil velocity of the rifle.

Ans: Given, Mass of the gun, M = 4 kg,

Mass of bullet, m = 50 g = 50 × 10-3 kg,

Initial velocity of bullet, u = 35 m/s

Let the recoil velocity of the gun be, v m/s.

Before firing the bullet both gun and bullet were at rest, so the total momentum of the gun and bullet is zero.

After firing, the momentum of the bullet = mu

The momentum of gun = Mv

The total momentum of bullet and gun after firing

⇒ mu + Mv

Since there is no external force applied to the system.

So, total momentum after firing = total momentum before firing I for

⇒ mu + Mv = 0

⇒ Mv = – mu

Here, a negative sign shows that the recoil velocity of the gun is in the direction opposite to the velocity of the bullet.

Page : 127

Q1. Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 m/s and 1m / s respectively. They collide and after collision, the first object moves at a velocity of 1.67 m/s. Determine the velocity of the second object.

Ans: Before collision,

m1 = 100g = 100/1000 = 0.1 kg [:1 kg=1000 g]

m2 = 200g = 200/1000 = 0.2kg

u1 = 2m / s ,

u2 = 1m / s

After collision,

m1 = 100g = 0.1 kg,

m2 = 200g = 0.2kg

v1 = 1.67m / s

Let the velocity of the second object after collision be v2 m/s Since, there is no external force on the system.

.. Total momentum before collision = Total momentum after collision

⇒ m1 x u1 + m2 x u2 = m1 x v1 + m2 x v2

⇒ 0.1 x 2 + 0.2 x 1 = 0.1 x 1.67 + 0.2 x v2

⇒ 0.2 + 0.2 = 0.167 + 0.2v2

⇒ 1.2v2 = 0.4 – 0.167

⇒ 0.2v2 = 0.233

⇒ v2 = 0.233/0.2 = 1.16 m/s

Exercises

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.

Ans: Yes it is possible for the object to be travelling with a non-zero velocity.

When an object experiences a net zero external unbalanced force, then it can move with a non-zero velocity.

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

Ans: When a carpet is beaten with a stick, the stick exerts a force on the carpet. due to which the fibres of the carpet attain the state of motion but the dust particles remain at rest due to inertia of rest. Therefore, the dust particles fall down.

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

Ans: It is advised to tie any luggage kept on the roof of a bus with a rope because when the bus comes into the state of rest due to the inertia of motion the luggage remains in the state of motion.

When the bus comes into the state of motion, the luggage due to the inertia of rest remains in the state of rest. and may fall down. to avoid this it is advised to tie the luggage on the roof of a bus.

Q4. A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the ball comes to rest. Why does the ball slows down to stop because:

(a) the batsman did not hit the ball hard enough

(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.

Ans: (c) there is a force on the ball opposing the motion.

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 tonne. (Hint: 1 tonne = 1000 kg).

Ans: The truck starts from rest, so initial velocity, u = 0

Distance, s = 400m , Time, t = 20s

Mass, m = 7 tonne = 7 X 1000 = 7000 kg

⇒ 400 = 200a

⇒ a = 2 m/s2

From Newton’s second law of motion, the force acting on the truck,

⇒ F = ma = 7000 X 2 = 14000N = 1.4 X 104

Thus, the acceleration of the truck is 2 m/s2 and the force acting on it is 1.4 X 104 N

Q6. A stone of size 1 kg is thrown with a velocity of 20 m/s 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?

Ans: Given, Mass of stone, m = 1kg,

Initial velocity, u = 20 m/s

Final velocity, v = 0

Distance covered, s = 50m

From Newton’s third law of motion, v2 = u2 + 2as

⇒ (0)2 = (20)2 + 2a(50)

⇒ 100 a = – 400

⇒ a = – 4 m/s2

Here, the negative sign shows that there is a retardation in the motion of stone,

Force of friction between stone and ice

= Force required to stop the stone = ma = 1 x (- 4) = – 4N or 4N

Q7. An 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:

(i) the net acceleration force.

(ii) the acceleration of the train and

(iii) the force of wagon 1 on wagon 2.

Ans: (i) Net accelerating force = Force exerted by engine – Friction force

= 40000-5000 = 35000 = 3.5 x 104 N

(ii) From Newton’s second law of motion,

accelerating force = mass of the train X acceleration of train

a = F/m

Mass of train = 5 x Mass of one wagon

= 5 x 2000=10000 kg

Acceleration = 35000/10000 = 3.5 m/s²

(iii) Force of wagon 1 on wagon 2 =(35000-3.5 x 2000) N = 28000 N

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 1.7m / (s2)?

Ans: Given, mass, m = 1500kg, acceleration, a = – 1.7 m/s2

From Newton’s second law of motion,

F = ma = 1500x(- 1.7) = – 2550N

Q9. What is the momentum of an object of mass m moving with a velocity v?

(a) m2v2

(b) mv2

(c) mu2

(d) mv

Ans: (d) mv

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

Ans: The cabinet will move across the floor with constant velocity if there is no net external force applied to it.

So, an external force of 200 N should be applied to the cabinet in the opposite direction.

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 before the collision during which they stick together. What will be the velocity of the combined object after collision?

Ans: Mass of the first object, m₁ =1.5 kg

Mass of second object, m₂ = 1.5 kg

Velocity of first object before collision,u1 = 2.5 m/s

The velocity of the second object before the collision,

u2=-2.5 m/s

[Here, the negative sign is taken because the second object is moving in the direction opposite to the direction of motion of the first object.]

Mass of combined objects after the collision,

m = m₁ + m₂

= 1.5+1.5=3 kg

Let the velocity of the combined object after collision be v m/s.

Here, there is no external force, so from the law of conservation of momentum, momentum after collision = momentum before the collision

⇒ mv = m1u1 + m₂u2

⇒ 3 x v = 1.5 x 2.5+1.5 (-2.5)= 0

⇒ v=0

Q12. According to the third law of motion when we push 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 justifies 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.

Ans: two equal and opposite forces cancel each other only, if they act on the same body.

But according to Newton’s third law: action and reaction force always act on two different bodies, so they cannot cancel each other.

show the logic given by the student is not correct.

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

Ans: Given, Mass of hockey ball,

m = 200g = 0.2kg

Initial velocity, u = 10m / s

Final velocity, v = – 5m / s

Change in the momentum of the ball = Final momentum – Initial momentum

= mv – mu = m(v – u)

= 0.2(-5-10)

= 0.2 x (- 15)

= – 3kg – m / s

Q14. A bullet of mass 10 g travelling horizontally with a velocity of 150 m/s 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.

Ans: Given, mass of bullet,

m = 10 g = 10/1000 kg = 0.01 kg

Initial velocity, u = 150 m/s

Final velocity, v = 0

Time, t = 0.03s

From Newton’s first law of motion,

⇒ v = u + at

⇒ 0 = 150 + a x 0.03

⇒ a = – 150/0.03 = – 5000 m / s2

Distance covered by the bullet before coming to rest is given by

⇒ v2 = u2 + 2as

⇒ 0 = (150)2 + 2(- 5000) s

The magnitude of the force applied by the bullet on the block,

⇒ F = ma = 0.01 x – 5000 = – 50N

Q15. An object of mass 100 kg is accelerated uniformly from a velocity of 5 m/s to 8 m/s in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.

Ans: Given, Mass of the object, m = 100kg

Initial velocity, u = 5 m/s

Final velocity, v = 8 m/s,

Time, t = 6s

(i) Initial momentum = mu

=100 x 5 = 500 kg-m/s

and final momentum = mv

=100 x 8 = 800 kg-m/s

(ii) From Newton’s second law, the force exerted on the object = rate of change of momentum

= Change in momentum/Time

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

Ans: Given, m = 100 kg, u = 5 m/s, v = 8m/s

Initial momentum = mu = 100 x 5

= 500 kg-m/s

and final momentum = mv = 100 × 8

= 800 kg-m/s

Force exerted on the object,

F = ma = 100 X ½  = 50N

Q17. Akhtar, Kiran and Rahul were riding on a motorcar that was moving with a high velocity on an expressway when an insect hit 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. Akhtar said that, since the motorcar was moving with 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 motorcar and the insect experienced the same force and change in their momentum. Comment on these suggestions.

Ans: According to the law of conservation of momentum,

the momentum of the car and insect system before collision = momentum of the car and insect system after a collision

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.

Q18. How much momentum will a dumb-bell of mass 10 kg transfer to the floor, if it falls from a height of 80 cm? Take, its downward acceleration to be 10 m/s².

Ans: Given, Mass of dumb-bell, m = 10kg,

Initial velocity, u = 0,

Acceleration, a = 10 m / s2

Distance covered, s = 80cm = 0.8m

⇒ v2 = u2 + 2as

⇒ v2 = 0 + 2 x 10 x 0.8 = 16

⇒ v = √16 = 4 m/s

The momentum of dumb-bell just before it touches the floor is given by

⇒ p = mv = 10 x 4 = 40 kg-m/s

When the dumb-bell touches the floor, then its velocity becomes zero.

Thus, the total momentum of the dumb-bell is transferred to the floor.

So, the momentum transferred to the floor is 40 kg- m/s.

Q19. The following is the distance-time table of an object in motion:

(i) What conclusion can you draw about the acceleration? Is it constant, increasing and decreasing or zero?

(ii) What do you infer about the force acting on the object?

Ans: (i) Initial velocity, u = 0

by Newton’s 2nd law = a = 2st2

Acceleration is increasing.

(ii) net unbalance force is acting on the object.

Q20. Two persons manage to push a motorcar of mass 1200 kg at an uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of 0.2 m/s². With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort.)

Ans: Given:

Mass of a motorcar, m = 1200kg,

Acceleration produced, a = 0.2 m/s²

Force applied on the car by three persons,

⇒ F = ma

⇒ 1200 x 0.2

⇒ 240 N

Force applied on the car by one person = 240/3 = 80N

Each person pushes the motorcar with a force of 80 N.

Q21. A hammer of mass 500 g moving at 50 m/s strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer?

Ans: Given, Mass of the hammer = 500 g = 0.5 kg

Initial velocity of hammer, u = 50 m/s

Final velocity of hammer, v = 0 m/s

Time, t = 0.01s

According to Newton’s second law of motion,

the force of the nail on the hammer = rate of change of momentum of hammer

The force of the nail on the hammer is equal and opposite.

Q22. A motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90 km/h. Its velocity is slowed down to 18 km/h in 4 s by an unbalanced external force. Calculate the acceleration and change in momentum. Also, calculate the magnitude of the force required.

Ans: Given, mass, m = 1200 kg,

Initial velocity, u = 90 km/h

(ii). Change in momentum

= Final momentum – Initial momentum

=mv – mu

= m(v – u)

= 1200x(5-25)

=1200x(- 20)

= – 24000 kg-m/s

(iii). Magnitude of the force required

= Rate of change of momentum

= Change in momentum/Time

= – 24000/4 = – 6000N