Force and Laws of Motion

Class 9 – NCERT Science Solutions

Intext Questions 1

Question 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-rupee coin and a one-rupee coin?

Answer

As inertia of a body is directly proportional to its mass, hence, more the mass more is the inertia. The following objects therefore, hold greater inertia because of their higher mass.

(a) Stone

(b) Train

(c) Five-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 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 :

1. When the football player kicks a football to another player,
2. When that player kicks the football towards the goal.
3. When the goalkeeper stops the football.
4. When the goalkeeper kicks the football towards his team player.

Hence, the velocity of the football changes four times.

Agent supplying the force:

(a) The First case is the First player

(b) The Second case is the Second player

(c) The Third case is Goalkeeper

(d) The Fourth case is Goalkeeper

Question 3

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

Answer

When the tree is at rest, the leaves are also at rest. Now when the tree is vigorously shaken, the tree is in motion while the leaves are still at rest, due to inertia of rest. The force acts on the leaves in changing directions. Therefore, the leaves that are weakly attached to the branch fall off whereas the leaves that are firmly attached to the branch remain attached.

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

With the application of brakes, the bus stops but our body tends to continue in the same state of motion because of its inertia. Hence, we fall forward.

An opposite experience is encountered when we are sitting in a bus and the bus begins to move suddenly. Now, we tend to fall backwards. This is because the sudden start of the bus brings motion to the bus. But, our body opposes this motion because of its inertia. Hence, we fall backwards.

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.

Answer

Yes, it is possible. According to Newton’s first law of motion, an object remains in a state of rest or of uniform motion in a straight line unless compelled to change that state by an applied force.
A net zero external unbalanced force means that the object has balanced force. Hence, if it had an initial velocity then the object will continue travelling with a non-zero velocity.
So, the conditions for an object to be traveling with a non-zero velocity while experiencing a net zero external unbalanced force are:

1. The object must have an initial velocity.
2. The object must have experienced a force in the past that set it in motion, and the net force acting on it is now balanced, resulting in a constant velocity.

Question 2

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

Answer

The part of the carpet where the stick strikes, comes in motion at once, while the dust particles settled on it’s fur, remain in position due to inertia of rest. Thus, the part of the carpet moves ahead with the stick, leaving behind the dust particles which fall down due to the earth’s pull.

Question 3

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

Answer

When some luggage is placed on the roof of a bus which is initially at rest, the acceleration of the bus in the forward direction will exert a force (in the backward direction) on the luggage due to luggage’s inertia of rest.

On the other hand, when a moving bus suddenly stops due to the application of brakes, a force (in the forward direction) is exerted on the luggage due to luggage’s inertia of motion. Hence, tying the luggage will secure its position and prevent it from falling off the bus.

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

1. the batsman did not hit the ball hard enough.
2. velocity is proportional to the force exerted on the ball.
3. there is a force on the ball opposing the motion.
4. there is no unbalanced force on the ball, so the ball would want to come to rest.

Answer

there is a force on the ball opposing the motion.

Reason — When the ball rolls on a level ground, its motion is opposed by the force of friction (the friction arises between the ground and the ball). This frictional force eventually stops the ball.

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 it’s mass is 7 tonnes.

(Hint: 1 tonne = 1000 kg)

Answer

Given,

Distance (s) = 400 m

Time (t) = 20 sec

Initial velocity (u) = 0

According to the second equation of motion,

S = ut + 1221​at²

Substituting we get,

400 = 0 + 1221​ x a x 20 x 20

a = 400×220×2020×20400×2​ = 2 ms^-2

∴ acceleration of the car = 2 ms^-2.

Now, mass = 7 tonnes

Converting tonne into kg

1 tonne = 1000 kg

∴ 7 tonnes = 7000 kg

Force = mass x acceleration

F = 7000 x 2 = 14000 N

Hence, Force acting = 14000 N

Question 6

A stone of 1 kg 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 50 m. What is the force of friction between the stone and the ice?

Answer

Given,

Mass (m) = 1 kg

Initial velocity (u) = 20 ms^-1

Final velocity (v) = 0

Distance travelled (s) = 50 m

According to the third equation of motion,

2as = v² – u²

or

a = v2−u22s2sv2−u2​

Substituting we get,

a = 0−(20)22×50=−4001002×500−(20)2​=−100400​ = -4 [retardation]

Now, as Force = mass x acceleration

Substituting we get,

F = 1 × (-4) = -4 N

Negative sign indicates the opposing force which is friction

Hence, force of friction between the stone and the ice = -4 N

Question 7

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:

(a) the net accelerating force and

(b) the acceleration of the train

Answer

(a) Given,

Force exerted (F) = 40,000 N

Force of friction = -5000 N

Net accelerating force = 40,000 + (-5000) = 35,000 N

Hence, net accelerating force = 35,000 N

(b) Total mass of the train = mass of engine + mass of each wagon = 8000 + (5 × 2000) = 8000 + 10000 = 18000 kg

Now, as Force = mass x acceleration

Substituting we get,

35000 = 18000 x a

a = 35000180001800035000​ = 1.94 ms^-2

Hence, acceleration of the train is 1.94 ms^-2

Question 8

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 ms^-2?

Answer

Given,

Mass (m) = 1500 kg

Acceleration (a) = -1.7 ms^-2

Now, as Force = mass x acceleration

F = 1500 × (-1.7) = -2550 N

Hence, force required = -2550 N. (Negative sign indicates opposite direction.)

Question 9

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

1. (mv)²
2. mv²
3. 1221​mv²
4. mv

Answer

mv

Reason — The momentum of a body is defined as the product of its mass and velocity. For a body of mass m, moving with velocity v, linear momentum p is expressed as : p = 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

As the velocity of the cabinet is constant, its acceleration must be zero. Therefore, the effective force acting on it is also zero. Hence, the magnitude of opposing frictional force is equal to the force exerted on the cabinet i.e., 200 N. So, the total friction force is 200 N.

Question 11

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

The student’s justification is incorrect since action-reaction force always apply to different bodies and hence cannot cancel each other. When two equal and opposite forces act on the same body, they cancel each other.

As, the truck has a very high mass, the static friction between the road and the truck is high. When pushing the truck with a small force, the frictional force cancels out the applied force and the truck does not move. Hence, the truck will only move if a force greater than the friction force is applied to it.

Question 12

A hockey ball of mass 200 g 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 magnitude of change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.

Answer

Given,

Mass (m) = 200 g

Initial velocity (u) = 10 ms^-1

Final velocity (v) = -5 ms^-1

Initial momentum of the ball = mu = 200 × 10 = 2000 gms^-1

Final momentum of the ball = mv = 200 × -5 = -1000 gms^-1

∴ The change in momentum (mv – mu) = -1000 – 2000 = -3000 gms^-1

Convert gms^-1 to kgms^-1 :

1000 gms^-1 = 1 kgms^-1

So, 3000 gms^-1 = 3000100010003000​ = -3 kgms^-1

Hence, momentum of the ball reduces by 3 kgms^-1 after being struck by the hockey stick.

Question 13

A bullet of mass 10 g 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

Given,

Mass of the bullet (m) = 10 g

Convert g into kg

1000 g = 1 kg

So, 10 g = 101000100010​ = 0.01 kg

Initial velocity of the bullet (u) = 150 ms^-1

Terminal velocity of the bullet (v) = 0

Time period (t) = 0.03 s

To find the distance of penetration, the acceleration of the bullet must be calculated

Let the distance of penetration be s

According to the first equation of motion,

v = u + at

or

a = v – uttv – u​

Substituting we get,

a = 0−1500.03=1500.03=−1500030.030−150​=0.03150​=3−15000​ = -5000 ms^-2

∴ Acceleration = -5000 ms^-2.

According to the third equation of motion,

2as = v² – u²

or

a = v2−u22s2sv2−u2​

Substituting we get,

s = 0−(150)22×−5000=−22500−10000=2×−50000−(150)2​=−10000−22500​= = 2.25 m.

Hence, the total distance travelled is 2.25 m

Now, as Force = Mass x Acceleration

Substituting we get,

F = 0.01 × -5000 = -50 N

Negative sign indicates direction of recoil of bullet is opposite to the direction of bullet.

Hence, magnitude of the force = 50 N

Question 14

An object of mass 1 kg travelling in a straight line with a velocity of 10 ms^-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

Given,

Mass of the object (m₁) = 1 kg

Mass of the block (m₂) = 5 kg

Initial velocity of the object (u₁) = 10 ms^-1

Initial velocity of the block (u₂) = 0

Mass of the resulting object = m₁ + m₂ = 6 kg

Velocity of the resulting object (v) = ?

Total momentum before the collision = m₁u₁ + m₂u₂ = (1 × 10) + 0 = 10 kgms^-1

According to the law of conservation of momentum,

total momentum before the collision = total momentum after the collision.

∴ the total momentum post the collision is also 10 kgms^-1

Hence, total momentum just before the impact = 10 kgms^-1 and just after the impact = 10 kgms^-1

Now, (m₁ + m₂) × v = 10 kgms^-1

6 x v = 10

v = 106610​ = 5335​ ms^-1

Hence, velocity of the combined object = 5335​ ms^-1

Question 15

An object of mass 100 kg 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

Given,

Mass of the object (m) = 100 kg

Initial velocity (u) = 5 ms^-1

Final velocity (v) = 8 ms^-1

Time period (t) = 6 s

Now, initial momentum (m × u) = 100 × 5 = 500 kgms^-1

Final momentum (m × v) = 100 × 8 = 800 kgms^-1

Hence, Initial momentum = 500 kgms^-1 and final momentum = 800 kgms^-1

As per the first equation of motion,

a = v – uttv – u​

Substituting we get,

a = 8−5668−5​ = 0.5 ms^-2

We know, Force = mass x acceleration

Substituting we get,

F = 100 x 0.5 = 50 N

Hence, magnitude of the force = 50 N

Question 16

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

Kiran’s statement is false. The change in momentum of the insect and the motorcar is equal by conservation of momentum.

Akhtar’s statement is false. Force exerted is directly proportional to the mass and change in momentum. It does not depend on velocity. Secondly, according to Newton’s third law of motion car and insect would apply equal and opposite force on each other.

Rahul’s statement is correct. As per the Newton’s third law of motion, the force exerted by the insect on the car is equal and opposite to the force exerted by the car on the insect. Hence, insect and car experience same force.

The change in momentum, however, depends on the mass and resulting acceleration. The motorcar, with its larger mass, experiences less acceleration and, therefore, less change in velocity. On the other hand, the mass of the insect is very small compared to the motorcar, it suffers a huge change of velocity as compared to the motorcar. Due to this, insect dies.

Question 17

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 ms^-2.

Answer

Given,

Mass of the dumb-bell (m) = 10 kg

Distance covered (s) = 80 cm

Convert cm to m

100 cm = 1 m

So, 80 cm = 8010010080​ = 0.8 m

Initial velocity (u) = 0

Acceleration (a) = 10 ms^-2

Terminal velocity (v) = ?

Momentum of the dumb-bell when it hits the ground = mv

As per the third law of motion,

v² – u² = 2as

Substituting we get,

v² – 0 = 2 x 10 x 0.8

v² = 16

v = 1616​ ms^-1

v = 4 ms^-1

Hence, dumb-bell strikes the ground with a velocity of 4 ms^-1

The momentum transferred by the dumb-bell to the floor = mv = 10 × 4 = 40 kgms^-1

Hence, momentum transferred = 40 kgms^-1

Additional Exercises

Question A1

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

Time in seconds	Distance in meters
0	0
1	1
2	8
3	27
4	64
5	125
6	216
7	343

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

(b) What do you infer about the forces acting on the object?

Answer

(a)

Time (t) in s	Distance (s) in m	Velocity (v = s/t) in m/s	Acceleration (a = ∆v/∆t) in m/s
0	0	0	0
1	1	1	1
2	8	4	3
3	27	9	5
4	64	16	7
5	125	25	9
6	216	36	11
7	343	49	13

We notice from the table above that the acceleration of the object is increasing.

(b) We know Force = mass x acceleration

As mass of the object remains same, the increasing acceleration implies that the force acting on the object is increasing as well.

Question A2

Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of 0.2 ms^-2. With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort)

Answer

Given,

Mass of the car (m) = 1200 kg

When two persons push the car, it moves with uniform velocity i.e., acceleration of the car is 0 and total force on the car is balanced.

Hence, the force that accelerates the car is given by the third person.

Acceleration when third person starts pushing the car (a) = 0.2 ms^-2

We know, Force = mass x acceleration

Substituting we get,

F = 1200 × 0.2 = 240 N

The force applied by third person = 240 N.

As all 3 people push with the same muscular effort, hence, force applied by each person = 240 N

Question A3

A hammer of mass 500 g, moving at 50 ms^-1, 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?

Answer

Given,

Mass of the hammer (m) = 500 g

Convert g into kg

1000 g = 1 kg

500 g = 50010001000500​ = 0.5 kg

Initial velocity of the hammer (u) = 50 ms^-1

Terminal velocity of the hammer (v) = 0

Time period (t) = 0.01 s

As per the first equation of motion,

a = v – uttv – u​

Substituting we get,

a = 0−500.010.010−50​ = -5000 ms^-2

We know, Force = mass x acceleration

Substituting we get,

F = 0.5 x 5000 = -2500 N

According to the third law of motion, the nail exerts an equal and opposite force on the hammer and as the force exerted on the nail by the hammer is -2500 N, hence, the force exerted on the hammer by the nail will be 2500 N.

Question A4

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.

Answer

Given,

Mass of the car (m) = 1200 kg

Initial velocity (u) = 90 kmh^-1

Convert kmh^-1 to ms^-1 : multiply by 518185​

90 x 518185​ = 25 ms^-1

Final velocity (v) = 18 kmh^-1

Convert kmh^-1 to ms^-1 : multiply by 518185​

18 x 518185​ = 5 ms^-1

Time period (t) = 4 sec

As per the first equation of motion,

a = v – uttv – u​

Substituting we get,

a = 5−25445−25​ = -5ms^-2

Hence, acceleration of the car is -5 ms^-2

Negative sign shows retardation.

Initial momentum of the car = m × u = 1200 × 25 = 30,000 kgms^-1

Final momentum of the car = m × v = 1200 x 5 = 6,000 kgms^-1

∴ Change in momentum = mv – mu = 6,000 – 30,000 = -24,000 kgms^-1

Hence, change in momentum = 24,000 kgms^-1

We know, Force = mass x acceleration

Substituting we get,

F = 1200 × -5 = -6000 N

Hence, the magnitude of force required to slow down the vehicle = 6000 N