## MP Board Class 9th Science Solutions Chapter 9 Force and Laws of Motion

### Force and Laws of Motion Intext Questions

**Force and Laws of Motion Intext Questions Page No. 118**

Question 1.

Which of the following has more inertia:

- a rubber ball and a stone of the same size?
- a bicycle and a train?
- a five – rupees coin and a one – rupee coin?

Answer:

As we know, inertia is the calculated value for the mass of the body. It is proportional to mass of the body:

- Inertia of the stone is greater than that of a rubber ball as mass of a stone is more than the mass of a rubber ball for the same size.
- Inertia of the train is greater than that of the bicycle. As mass of a train is more than the mass of a bicycle.
- Mass of a five rupee coin is more than that of a one – rupee coin. Hence, inertia of the five – rupee coin is greater than that of the one – rupee coin.

Question 2.

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 kick it towards a player of his own team”. Also, identify the agent supplying the force in each case.

Answer:

Four times:

- First, when a football player kicks to another player. Agent supplying the force: First case – First player. Second when that player kicks the football to the goalkeeper. Agent supplying the force. Second case – Second player.
- Third when the goalkeeper stops the football. Agent supplying the force: Third case – Goalkeeper.
- Fourth when the goalkeeper kicks the football towards a player of his own team. Agent supplying the force: Fourth case – Goalkeeper.

Question 3.

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

Answer:

When we shake any tree’s branches vigorously some leaves of that tree get detached because branches comes in motion while the leaves tend to remain at rest due to inertia of rest.

Question 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?

Answer:

Due to inertia of motion, we fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest.

- Case I: Since the driver applies brakes and bus comes to rest. But, the passenger tries to maintain its inertia of motion. As a result, a forward force is exerted on him.
- Case II: The passenger tends to fall backwards when the bus accelerates from rest because when the bus accelerates, the inertia of rest of the passenger tends to oppose the forward motion of the bus.

**Force and Laws of Motion Intext Questions Page No. 126**

Question 1.

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

Answer:

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

Question 2.

Explain, why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity.

Answer:

According to Newton’s third law of motion, a reaction force is exerted over fireman by the ejecting water in the backward direction when a fireman holds a hose, which is ejecting large amounts of water at a high velocity. As a result of the backward force, the stability of the fireman get affected. Hence, it is difficult for him to remain stable while holding the hose.

Question 3.

From a rifle of mass 4kg, a bullet of mass 50g is fired with an initial velocity of 35 ms^{-1}. Calculate the initial recoil velocity of the rifle.

Answer:

Given,

Mass of the rifle, m_{1 }= 4kg

Mass of the bullet, m_{2} = 50g = 0.05kg

Recoil velocity of the rifle = v_{1}

Bullet is fired with an initial velocity, v_{2}= 35 m/s

Condition:

Initially, the rifle is at rest.

Thus, its initial velocity, v = 0

Total initial momentum of the rifle and bullet system = (m_{1}+ m_{2})v = 0

Total momentum of the rifle and bullet system after firing = m_{1}v_{1 }+ m_{2}v_{2
}= 0.05 × 35 = 4v_{1} + 1.75

As per law of conservation of momentum:

Total momentum after the firing = Total momentum before the firing

⇒ 4v_{1 }+ 1.75 = 0

v_{1} = –\(\frac { 1.75 }{ 4 }\) = -0.4375 m/s

So, the rifle recoils backwards with a velocity of 0.4375 m/s because value is negative.

Question 4.

Two objects of masses 100g and 200g are moving along the same line and direction with velocities of 2 ms^{-1} and 1 ms^{-1}, respectively. They collide and after the collision, the first object moves at a velocity of 1.67 ms^{-1}. Determine the velocity of the second object.

Answer:

Given,

m_{1 }= 100g = 0.1kg

m_{2} = 200g = 0.2kg

Velocity of m_{1} before collision, v_{1} = 2 m/s

Velocity of m_{2} before collision, v_{2 }= 1 m/s

Velocity of m_{1} after collision, v_{3} = 1.67 m/s

Velocity of m_{2} after collision = v_{4}

As per the law of conservation of momentum:

Total momentum before collision = Total momentum after collision

Hence,

m_{1}v_{1 }+ m_{2}v_{2} = m_{1}v_{3}+ m_{2}v_{4}

Putting values

2(0.1) + 1(0.2) = 1.67(0.1) + v_{4}(0.2)

0.4 = 0.167 + 0.2v_{4}

v_{4 }= 1.165 m/s

Velocity of the second object = 1.165 m/s.

### Force and Laws of Motion NCERT Textbook Exercises

Question 1.

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, an object may travel with a non – zero velocity even when the net external force on it is zero. A rain drop falls down with a constant velocity. The weight of the drop is balanced by the upthrust and the viscosity of air. The net force on the drop is zero.

Question 2.

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

Answer:

When we beat the carpet with a stick, it comes into motion. But the dust particles continue to be at rest due to inertia and get detached from the carpet.

Question 3.

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

Answer:

Due to sudden jerks or due to the bus taking sharp turn on the road, the luggage may fall down from the roof because of its tendency to continue moving in the original direction. To avoid this, the luggage is tied with a rope on the roof.

Question 4.

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.

(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 come to rest.

Answer:

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

Question 5.

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

Answer:

Here, u = 0, s = 400m, t = 20 s, a = ?, F = ?.

m = 7 tonnes

= 7 × 1000kg

= 7000kg

⇒ s = ut + \(\frac { 1 }{ 2 }\)at^{2}

400 = (0 + 20) + \(\frac { 1 }{ 2 }\)a(20)^{2}

a = \(\frac { 400\times 2 }{ { 20 }^{ 2 } } \)

∴ a = 2 m/s^{2}

Force,

F = ma

= 7000 × 2 = 14,000 N.

Question 6.

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

Answer:

m = 1kg, u = 20 m/s, s = 50m, v = 0, F = ? a = ?.

⇒ v^{2} – u^{2} = 2as

(0)^{2} – (20)^{2} = 2a (50)

∴ – 400 = 100a

= \(\frac { 400 }{ 100 }\) – 4 m/s^{2}

Force of friction, F = m × a

= 1kg × -4 m/s^{2} = -4 N

Question 7.

A 8000kg engine pulls a train of 5 wagons, each of 2000kg, along a horizontal track. If the engine exerts a force of 40,000 N and the track offers a friction force of 5000 N, then calculate:

(a) the net accelerating force;

(b) the acceleration of the train; and

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

Answer:

m = 8,000 + 5 × 2,000 = 18,000kg

(a) The net accelerating force,

F = Engine force – friction force

= 40,000 – 5,000 = 35,000 N.

(b) The acceleration of the train,

a = \(\frac { F }{ m }\) = \(\frac { 35,000 }{ 18,000 }\) = \(\frac { 35 }{ 18 }\) = 1.94 ms^{-2}

(c) The force of wagon 1 on wagon 2

= The net accelerating force – mass of wagon × acceleration

= 35,000 – 2,000 × \(\frac { 35 }{ 18 }\)

= 35,000 – 3888.8 = 31,111.2 N.

Question 8.

An automobile vehicle has a mass of 1500kg. What must be the force between the vehicle and road if the vehicle is stopped with a negative acceleration of 1.7 ms^{-2}?

Answer:

Here, mass = 1500kg

a = -1.7 ms^{2
}F = m × a

= 1500 × (-1.7)

= -2550 N

The force between the vehicle and the road is 2,550 is, m a direction opposite to the direction of the vehicle.

Question 9.

What is the momentum of an object of mass m, moving with a velocity v? Choose correct option.

(a) (mv)^{2}

(b) mv^{2}

(c) \(\frac { 1 }{ 2 }\) × mv^{2}

(d) mv.

Answer:

(d) mv.

Question 10.

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?

Answer:

The cabinet will move with constant velocity only when the net force on it is zero.

Force of friction on the cabinet = 200 N, in a direction opposite to the direction of motion of the cabinet.

Question 11.

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

Answer:

Here, m_{1} = m_{2} = 1.5kg,

u_{1} = 2.5 ms^{-1}, u_{2} = – 2.5 ms^{-1}

Let v be the velocity of the combined object after the collision.

By conservation of momentum,

Total momenta after collision = Total momenta before collision

= (m_{1} + m_{2}) v = m_{1}u_{1} + m_{2}u_{2}

= (1.5 + 1.5) v = 1.5 × 2.5 + 1.5 × (-2.5)

= 3.0 v = 0

or

⇒ v = 0 ms^{-1}

Question 12.

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 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.

Answer:

Action and reaction always act on different bodies, so they cannot cancel each other. When we push a massive truck, the force of friction between its tyres and the road is very large and so, the truck does not move.

Question 13.

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

Answer:

Here, m = 200g = 0.2kg,

u = 10 ms^{-1},

v = -5 ms^{-1}

change in momentum = m (v – u)

= 0.2 (- 5 – 10) = -3kg ms^{-1}.

Question 14.

A bullet of mass 10g travelling horizontally with a velocity of 150 ms^{-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:

Here m = 10g = 0.01kg,

u = 150 ms^{-1},

v = 0, t = 0.03 s

a = \(\frac { v-u\quad }{ t } \) = \(\frac { 0-150\quad }{ 0.03 } \) = -5,000 ms^{-1
}The distance of penetration of the bullet into the block,

s = ut + \(\frac { 1 }{ 2 }\)at^{2}

150 × 0.03 + \(\frac { 1 }{ 2 }\) × (-5,000) × (0.03)

= 4.5 – 2.25 = 2.25

The magnitude of the force exerted by the wooden block on the bullet

= ma = 0.01 × 5,000 = 50 N.

Question 15.

An object of mass 1kg travelling in a straight line with a velocity of 10 ms^{-1} collides with, and sticks to, a stationary wooden block of mass 5kg. 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:

Here, m_{1} = 1kg, u_{1} = 10 ms^{-1}, m_{2} = 5kg, u_{2} = 0

Let v be the velocity of the combined object after the collision.

Total momentum just before the impact

= m_{1}u_{1} + m_{2}u_{2} = 1 × 10 + 5 × 0 = 10kg ms^{-1}

Total momentum just after the impact = (m_{1} + m_{2})v = (1 + 5)v

= 6v kg ms^{-1} by conservation of momentum,

6v = 10

= v = \(\frac { 10 }{ 6 }\) = \(\frac { 5 }{ 3 }\) ms^{-1}

∴ Total momentum just after the impact

= 6 × \(\frac { 5 }{ 3 }\) = 10 ms^{-1}

Question 16.

An object of mass 100kg is accelerated uniformly from a velocity of 5 ms^{-1} to 8 ms^{-1} in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.

Answer:

Here, m = 100kg,

u = 5 ms^{-1},

v = 8 ms^{-1},

t = 6 s

Initial momentum, P_{1 }= mu = 100 × 5 = 500kg ms^{-1}

Final momentum, P_{2} = mu = 100 × 8 = 800kg ms^{-1}

The magnitude of the force exerted on the object.

F = \(\frac { { P }_{ 2 }-{ P }_{ 1 } }{ t } \) = \(\frac { 800-500 }{ 6 }\) = \(\frac { 300 }{ 6 }\) = 50 N.

Question 17.

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:

Both, the motorcar and insect experience the equal force and hence, a same change in their momentum. So, we agree with Rahul. But due to smaller mass or inertia, the insect dies.

Question 18.

How much momentum will a dumb – bell of mass 10kg transfer to the floor if it falls from a height of 80cm? Take its downward acceleration to be 10 ms^{-2}.

Answer:

Here, m = 10kg, u = 0,

s = 80cm = 0.80m,

a = 10 m/s^{-2}

Let v be the velocity gained by the dumb-bell as it reaches the floor.

As v^{2} – u^{2} = 2as

v^{2} – 0^{2} = 2 × 10 × 0.80 = 16

or

v = 4 ms^{-1}

Momentum transferred by the dumb-bell to the floor

p = mv = 10 × 4 = 40kg ms^{-1}

### Force and Laws of Motion Additional Questions

**Force and Laws of Motion Multiple Choice Questions**

Question 1.

Which of the following statements is not correct for an object moving along a straight path in an accelerated motion?

(a) Its speed keeps changing.

(b) Its velocity always changes.

(c) It always goes away from the earth.

(d) A force is always acting on it.

Answer:

(c) It always goes away from the earth.

Question 2.

According to the third law of motion, action and reaction ________ .

(a) Always act on the same body.

(b) Always act on different bodies in opposite directions.

(c) Have same magnitude and directions.

(d) Act on either body at normal to each other.

Answer:

(b) Always act on different bodies in opposite directions.

Question 3.

A goalkeeper in a game of football pulls his hands backwards after holding the ball shot at the goal. This enables the goal keeper to ________ .

(a) Exert larger force on the ball.

(b) Reduce the force exerted by the ball on hands.

(c) Increase the rate of change of momentum.

(d) Decrease the rate of change of momentum.

Answer:

(b) Reduce the force exerted by the ball on hands.

Question 4.

The inertia of an object tends to cause the object ________ .

(a) To increase its speed.

(b) To decrease its speed.

(c) To resist any change in its state of motion.

(d) To decelerate due to friction.

Answer:

(c) To resist any change in its state of motion.

Question 5.

A passenger in a moving train tosses a coin which falls behind him. It means that motion of the train is ________ .

(a) Accelerated

(b) Uniform

(c) Retarded

(d) Along circular tracks.

Answer:

(a) Accelerated

Question 6.

An object of mass 2kg is sliding with a constant velocity of 4 ms^{-1} on a frictionless horizontal table. The force required to keep the object moving with the same velocity is ________ .

(a) 32 N

(b) 0 N

(c) 2 N

(d) 8 N.

Answer:

(b) 0 N

Question 7.

Rocket works on the principle of conservation of ________ .

(a) Mass

(b) Energy

(c) Momentum

(d) Velocity.

Answer:

(c) Momentum

Question 8.

A water tanker filled up to \(\frac { 2 }{ 3 }\) of its height is moving with a uniform speed. On sudden application of the brake, the water in the tank would ________ .

(a) Move backward

(b) Move forward

(c) Be unaffected

(d) Rise upwards.

Answer:

(b) Move forward

Question 9.

Which of the following represents example(s) of potential energy?

(a) A moving car

(b) A moving fan

(c) A book resting on the table

(d) Both (a) and (c).

Answer:

(d) Both (a) and (c).

Question 10.

Unit of force is ________ .

(a) Ampere

(b) Volt

(c) Joule

(d) Hertz.

Answer:

(c) Joule

Question 11.

Product of mass and acceleration of a body is called ________ .

(a) Acceleration

(b) Work

(c) Power

(d) Energy.

Answer:

(b) Work

Question 12.

Which of the following is correct about energy?

(a) Energy is not required to do work.

(b) Work can be expressed as Force × Displacement.

(c) Unit of power is joule.

(d) Power is the amount of work done per unit time.

Answer:

(c) Unit of power is joule.

Question 13.

An object of mass 3kg is falling from the height of 1m. The kinetic energy of the body will be when it touches the ground ________ .

(a) 29.4 N

(b) 29.4 J

(c) 30 N

(d) 15 J

Answer:

(b) 29.4 J

Question 14.

Two objects with masses 1kg and 9kg, and equal momentum. Calculate the ratios of their kinetic energies ________ .

(a) 3 : 1

(b) 9 : 1

(c) 1 : 1

(d) 1 : 2

Answer:

(b) 9 : 1

Question 15.

Considering air resistance negligible, the sum of potential and kinetic energies of the free falling body would be ________ .

(a) zero

(b) infinite

(c) would decrease

(d) remains fixed.

Answer:

(d) remains fixed.

**Force and Laws of Motion Very Short Answer Type Questions**

Question 1.

What do we call to the product of mass and velocity of an object?

Answer:

Momentum.

Question 2.

Define inertia.

Answer:

The property by which an object tends to remain in the state of rest or of uniform motion unless acted upon by some force is called inertia.

Question 3.

Which property has S.I. unit kilogram metres per second i.e., 1kg m/s?

Answer:

Momentum.

Question 4.

Give an example of scalar quantity.

Answer:

Mass.

Question 5.

Give an example of vector quantity.

Answer:

Momentum.

Question 6.

Calculate the total momentum of the bullet and the gun before firing.

Answer:

For both, it would be zero because both of them are at rest.

Question 7.

Which force slows down a moving bicycle when we try to stop?

Answer:

The force of friction.

Question 8.

Which kind of force of gravity work when an object is under free fall?

Answer:

Unbalanced force.

Question 9.

Which property of an object resist a change in their state of rest or motion?

Answer:

Inertia.

Question 10.

Which law of Newton is also known as Galileo’s law of inertia?

Answer:

First law.

Question 11.

Is force a vector quantity?

Answer:

Yes.

Question 12.

Which force of motion opposes motion of an object?

Answer:

Force of friction.

Question 3.

When action and reaction forces act on two different bodies, what kind of magnitude they have?

Answer:

Action and reaction forces act on two different bodies but they are equal in magnitude.

Question 14.

When gun moves in the backward direction, which kind of velocity is generated?

Answer:

Recoil.

Question 15.

Which factor of body is dependent on its mass?

Answer:

Inertia of a body depends on its mass.

**Force and Laws of Motion Short Answer Type Questions**

Question 1.

There are three solids made up of aluminium, steel and wood, of the same shape and same volume. Which of them would have highest inertia?

Answer:

Steel has highest inertia because it has greatest density and greatest mass, therefore, it has highest inertia.

Question 2.

Two balls of the same size but of different materials, rubber and iron are kept on the smooth floor of a moving train. The brakes are applied suddenly to stop the train. Will the balls start rolling? If so, in which direction? Will they move with the same speed? Give reasons for your answer.

Answer:

If the breaks are applied suddenly then, the balls will start rolling in the direction in which the train was moving. Due to the application of the brakes, the train comes to rest but due to inertia the balls try to remain in motion, therefore, they begin to roll. Direction and speed of all balls will not be same because the masses of the balls are not the same, therefore, the inertial forces are not same on both the balls. Thus, the balls will move with different speeds.

Question 3.

Two identical bullets are fired, one by a light rifle and another by a heavy rifle with the same force. Which rifle will hurt the shoulder more and why?

Answer:

According to law of conservation of momentum or explanation by Newton’s laws of motion, light rifle will hurt the shoulder more.

Question 4.

A horse continues to apply a force in order to move a cart with a constant speed. Explain why?

Answer:

The force applied by the horse balances the force of friction

Question 5.

Suppose a ball of mass m is thrown vertically upward with an initial speed v, its speed decreases continuously till it becomes zero. Thereafter, the ball begins to fall downward and attains the speed v again before striking the ground. It implies that the magnitude of initial and final momentums of the ball are same. Yet, it is not an example of conservation of momentum. Explain.

Answer:

Law of conservation of momentum is applicable to isolated system (no external force is applied). In this case, the change in velocity is due to the gravitational force of earth.

Question 6.

In which of the following conditions work done will be equal to zero?

Answer:

In the absence of any one of the two conditions given below, work done will be equal to zero, that is work is not considered to be executed:

- Force should act on the object.
- Object must be displaced.

Question 7.

Define energy and explain its forms.

Answer:

- Energy: Energy is the capacity of doing work. More the power, more will be energy and vice – versa. For example, a motorcycle has more energy than a bicycle.
- Forms of energy: There are many forms of energy, such as kinetic energy, potential energy, mechanical energy, chemical energy, electrical energy etc.

**Force and Laws of Motion Long Answer Type Questions**

Question 1.

Give the formulation of work. In which conditions work can occur?

Answer:

Work = Force × Displacement

or W = F × s

where, W is work ‘F’ is force and ‘s’ is displacement.

If force, F = 0

Therefore, work done, W = 0, s = 0

If displacement, s = 0

Therefore, Work done, W = F × 0 = 0

It proves that, there are two conditions for work to occur or be done:

- Force should act on the object.
- Object must be displaced.

Question 2.

Give the conditions when work done become positive and negative.

Answer:

When force is applied in the direction of displacement, the work done is considered as positive.

i.e., W = F × s

When force is applied in opposite direction of displacement, the work done is considered as negative.

i.e., W = -F × s = -Fs.

Question 3.

Explain positive and negative work.

Answer:

1. Positive work:

If a force displaces the object in its direction, then the work done is positive.

Here,

W = Fd

Example:

Motion of ball falling towards ground where displacement of ball is in the direction of force of gravity.

2. Negative work. If the force and the displacement are in opposite directions, then the work is said to be negative.

Here,

W = -Fd.

Example:

If a ball is thrown in upward direction but the force due to earth’s gravity is in the downward direction.

Question 4.

A cyclist moving along a circular path of radius 63m completes three rounds in 3minutes.

1. The total distance covered by him during this time.

2. Net displacement of cyclist.

3. The speed of the cyclist

Answer:

1. Total distance covered

s = 2πr × t

s = 2πr × 3

= 2 × \(\frac { 22 }{ 7 }\) × 63 × 3 = 1188m

2. Displacement = Zero

3. Speed = \(\frac { Distance }{ Time }\)

= \(\frac { 1188 }{ 180 }\)

= 6.6 m/s.

**Force and Laws of Motion Higher Order Thinking Skills (HOTS)**

Question 1.

As per Newton’s third law, every force is accompanied by equal and opposite force. How then can anything move?

Answer:

According to the Newton’s third law, action and reaction are two equal and opposite forces but they act on different bodies. This make the motion of a body possible.

Question 2.

The passengers travelling in a bus fall ahead when a speeding bus stops suddenly. Why?

Answer:

When the speeding bus stops suddenly lower part of the body, a long with the bus comes to rest while the upper part tends to remain in motion due to inertia of motion. That is why passengers fall ahead.

Question 3.

A player always runs some distance before taking a jump. Why?

Answer:

A player always runs for some distance before taking a jump because inertia of motion helps him to take a longer jump.

**Force and Laws of Motion Value Based Question**

Question 1.

Sushil saw his karate expert breaking a slate. He tried to break the slate but Sushil’s friend stopped him from doing so and told him that it would hurt, one needs lot of practice in doing such activity.

- How can a karate expert break the slate without any injury to his hand?
- What is Newton’s third law of motion?
- What value of Sushil’s friend is seen in the above case?

Answer:

- A karate expert Sushil applies the blow with large velocity in a very short interval of time on the slate, therefore large force is exerted on the slate and it breaks.
- To every action, there is an equal and opposite reaction, both act on different bodies.
- Sushil’s friend showed the value of being responsible and caring friend.