## Chapter 9 Force and Laws of Motion

Intext Questions

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 rupees coin and a one-rupee coin?

Answer: Heaver or more massive objects offer larger inertia. So,

a) stone has more inertia

b) train has more inertia

c) five rupees coin has more inertia.

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 football changes 4 times.

The first player kicks to second player who kicks to goalkeeper (third person) who stops it and passes to his team mate (fourth person).

Agent supplying force:

• Case I: First Player

• Case II: Second Player

• Case III: Goal Keeper

• Case IV: Goal Keeper

Question 3: 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 branch vigorously. This is because when the branches of a tree are shaken, it moves to and fro, but its leaves tend to continue inertia of rest. Due to this, the leaves fall down from the tree when shaken vigorously.

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: Initially, when the bus accelerates in a forward direction from a state of rest, the passengers experience a force exerted on them in the backward direction due to their inertia opposing the forward motion. Once, the bus starts moving, the passengers are in a state of motion in the forward direction. When the brakes are applied, the bus moves towards a position of rest. Now, a force in the forward direction is applied on the passengers because their inertia resists the change in the motion of the bus. This causes the passengers to fall forwards when the brakes are applied.

Question 5: 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, action force is equal to reaction but acts on two different bodies and in opposite direction. When a horse with cart push the ground in backward direction, the ground reacts and exerts an equal force on horse and cart in forward direction. This is how horse can pull a cart.

Question 6: Explain, why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity.

Answer: When a fireman holds a hose, which ejects large amounts of water at a high velocity, then a reaction force is exerted on him by ejecting water in the backward direction, by Newton’s third law of motion. As a result of the backward force, the stability of the fireman decreases. Therefore, it is difficult for him to hold a hose, which ejects large amount of water at a high velocity.

Question 7: From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 m s⁻¹. Calculate the initial recoil velocity of the rifle.

Answer:

Given

Mass of rifle = 4 kg (4000 g)

Mass of bullet = 50 g

Initial velocity of bullet = 35 m s⁻¹

Recoil velocity of the rifle = ?

Since, the rifle was initially at rest, the initial momentum of the rifle will be 0. (p = mv = 4000 × 0 = 0)

The total momentum of the rifle and bullet after firing = m₁v₁ + m₂v₂.

As per Law of Conservation of Momentum, total momentum of the rifle and bullet after firing = 0.

So, m₁v₁ + m₂v₂ = 0.

v₂ = -m₁v₁/ m₂

v₂ = -50 × 35/4000

v₂ = -7/16

v₂ = -0.4375 m s⁻¹

The negative sign indicates that recoil velocity is opposite to the bullet’s motion.

Question 8: 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 first object moves at a velocity of 1.67 m s⁻¹. Determine the velocity of the second object.

Answer:

Given

Mass of object 1 = 100 g (0.1 kg)

Mass of object 2 = 200 g (0.2 kg)

Velocity of object 1 (before collision) = 2 m s⁻¹

Velocity of object 2 (before collision) = 1 m s⁻¹

Velocity of object 1 (after collision) = 1.67 m s⁻¹

Velocity of object 2 (after collision) = ?

We know that, m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ by Law of Conservation of Momentum

Substituting the values, we get

(0.1 × 2) + (0.2 × 1) = (0.1 × 1.67) + (0.2 × x)

0.2 + 0.2 = 0.167 + 0.2x

0.4 - 0.167 = 0.2x

0.233 = 0.2x

x = 0.233/0.2= 1.165 m s⁻¹

Exercise Questions

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: When a net zero external unbalanced force is applied on the body, it is possible for the object to be travelling with non-zero velocity. In fact, once an object comes into motion and there is a condition, in which is imposed by any external force; the object will continue to remain in motion. It is necessary that the object moves at a constant velocity and in a particular direction.

Question 2: When a carpet is beaten with a stick, dust comes out of it. Explain.

Answer: When a carpet is beaten with a stick, the stick exerts a force on the carpet which sets it in motion. Initially the dust particles are at rest along with the carpet. Beating the carpet with the stick makes the carpet move but the dust particles remain at rest due to inertia of rest. Hence, the dust gets 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: While the bus is moving, luggage tends to remain in inertia of motion state. When the bus stops, the luggage tends to resist the change and due to inertia of motion it moves forward and may fall off. That’s why it is advised to tie any luggage kept on the roof of the bus with a rope.

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 want to 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 tonnes (Hint: 1 tonne = 1000 kg.)

Answer:

Given

Initial velocity of truck = 0 m s⁻¹

Distance travelled = 400 m

Time taken = 20 s

Acceleration = ?

Mass = 7 tonnes (7000 kg)

Force = ?

We know that, s = ut + 1/2 at²

Substituting the values, we get

400 = 1/2 × a × (20)²

400 = 200a

a = 2 m s⁻²

We know that, F = ma

Substituting the values, we get

F = 7000 × 2

F = 14000 N

Question 6: 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?

Answer:

Given

Mass of stone = 1 kg

Initial velocity = 20 m s⁻¹

Final velocity = 0 m s⁻¹

Distance travelled = 50 m

Force of friction between stone and ice = ?

We know that, 2as = v² - u²

Substituting the values, we get

2 × a × 50 = 0² - 20²

100a = -400

a = -4 m s⁻²

We know that, F = ma

Substituting the values, we get

F = 1 × -4

F = -4 N

The negative sign indicates the force applied in opposite to the motion.

Question 7: 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;

b) the acceleration of the train

Answer:

Given

Mass of engine = 8000 kg

Mass of 5 wagons = 10000 kg

Force exerted by engine = 40000 N

Force exerted by track (in opposite direction) = 5000 N

a) Net Accelerating Force = ?

Since, the frictional force is acting in opposite direction the force will be negative.

= 40000 N - 5000 N

= 35000 N

b) We know that, F = ma

Substituting the values, we get

35000 = 18000 × a

a = 35000/18000

a = 1.94 m s⁻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 m s⁻²?

Answer:

Given

Mass of automobile = 1500 kg

Negative acceleration = 1.7 m s⁻²

Force between vehicle and road = ?

We know that, F = ma

Substituting the values, we get

F = 1500 × (-1.7)

F = -2550 N

The negative sign implies that the force applied is opposite to the motion of an object.

Question 9: What is the momentum of an object of mass m, moving with a velocity v?

a) (mv)²

b) mv²

c) 1/2 mv²

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: For an object to move with constant velocity, net external force experienced by it must be zero. Thus a friction of 200 N must be exerted on the cabinet so that it moves with a constant velocity under the applied horizontal force of 200 N.

Question 11: 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:

Given

Mass of first object = 1.5 kg

Mass of second object = 1.5 kg

Initial velocity of first object = 2.5 m s⁻¹

Initial velocity of second object = -2.5 m s⁻¹

Final velocity of first object = Final velocity of second object = v

We know that, total momentum before collision = total momentum after collision

m₁u₁ + m₂u₂ = (m₁ + m₂)v

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

3.75 - 3.75 = 3v

0 = 3v

v = 0 m s⁻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: When we push a truck parked along the roadside, the road applies a static frictional force in the opposite direction of force applied on the truck to oppose the motion of the truck. The road offers only the required amount of frictional force to keep the truck at rest. If more force is applied, the road applies more friction. Hence, the friction force cancels out the force applied. Student’s justification is that two opposite and equal force cancel each other and hence the truck doesn’t move. However, the action and reaction forces do not cancel each other as they act on the different bodies whereas the truck didn’t move due to zero net force which is balanced by the static friction of the truck.

Question 13: 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 change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.

Answer:

Given

Mass of hockey ball = 200 g (0.2 kg)

Initial velocity of ball = 10 m s⁻¹

Final velocity of ball = -5 m s⁻¹

Change in momentum = ?

We know that, change in momentum = m(v - u)

Substituting the values, we get

p = 0.2(-5 - 10)

p = 0.2 × -15

p = -3 kg m s⁻¹

Question 14: 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:

Given

Mass of a bullet = 10 g (0.01 kg)

Initial velocity of bullet = 150 m s⁻¹

Final velocity of bullet = 0 m s⁻¹

Time taken = 0.03 s

We know that, v = u + at

Substituting the values, we get

0 = 150 + a(0.03)

-150 = 0.03a

a = -150/0.03

a = -5000 m s⁻²

We know that, 2as = v² - u²

Substituting the values, we get

2 × -5000 × s = 0² - 150²

-10000s = -22500

s = 22500/10000

s = 2.25 m

We know that, F = ma

Substituting the values, we get

F = 0.01 × -5000

F = -50 N

Question 15: 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:

Given

Mass of object = 1 kg

Initial velocity of object = 10 m s⁻¹

Mass of wooden block = 5 kg

Initial velocity of block = 0 m s⁻¹

Final velocity of object = Final velocity of block = v

We know that, total momentum before collision = total momentum after collision

m₁u₁ + m₂u₂ = (m₁ + m₂)v

(1 × 10) + (5 × 0) = (1 + 5)v

10 = 6v

v = 10/6 = 1.66 m s⁻¹

Question 16: 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 final momentum of the object. Also, find the magnitude of the force exerted on the object.

Answer:

Given

Mass of an object = 100 kg

Initial velocity of the object = 5 m s⁻¹

Final velocity of the object = 8 m s⁻¹

Time taken = 6 s

Initial momentum = mu = 100 × 5 = 500 kg m s⁻¹

Final momentum = mv = 100 × 8 = 800 kg m s⁻¹

Force = ?

We know that, F = m(v - u)/t

Substituting the values, we get

F = 100(8 - 5)/6

F = 300/6

F = 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: As per the law of conservation of momentum, the total momentum before the collision between the insect and the car is equal to the total momentum after the collision. Therefore, the change in the momentum of the insect is much greater than the change in momentum of the car (since force is proportional to mass).

Akhtar’s assumption is partially right. Since the mass of the car is very high, the force exerted on the insect during the collision is also very high.

Kiran’s statement is false. The change in momentum of the insect and the motorcar is equal by conservation of momentum. The velocity of insect changes accordingly due to its mass as it is very small compared to the motorcar. Similarly, the velocity of motorcar is very insignificant because its mass is very large compared to the insect.

Rahul’s statement is completely right. As per the 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. Rahul’s suggestion that the change in the momentum is the same as that in stated in the law of conservation of momentum.

Question 18: 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 s⁻².

Answer:

Given

Mass of dumb-bell = 10 kg

Distance covered = 80 cm (0.8 m)

Downward acceleration = 10 m s⁻²

Initial velocity = 0 m s⁻¹

Final velocity = ?

We know that, 2as = v² - u²

Substituting the values, we get

2 × 10 × 0.8 = v²

16 = v²

v = 4 m s⁻¹

The momentum transferred by the dumb-bell to the floor = 10 × 4 = 40 kg m s⁻¹.

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