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CBSE Physics | Conceptual Questions on Force and Laws of Motion (with Solutions)

August 31st, 2015

Conceptual Questions on Force and Laws of Motion (with Solution)

Conceptual Questions (with solution)

Question 1. A cricket player lowers his hand while catching the ball. Why?

Answer.

When a cricket player lowers his hands, the time to stop the ball increases, so the rate of change of momentum decrease and hence less force has to be applied to stop the ball, due to less force the hands of the player are not injured.

That is why, a cricket player lowers his hands while catching the ball.

Question 2. A person falling on a cemented floor gets injured but a person falling on a heap of sand is not injured. Why?

Answer.

When a person falls on a cemented floor, the rate of change of momentum is very high because the person is abruptly stopped i.e. stopped in a very small interval of time.

Hence the person gets injured due to the application of a large force on his body by to the floor. On the other hand, when a person falls on a heap of sand, the sand yields under the weight of the person.

As a result of this, the person comes to rest in a longer period of time. So the change in momentum takes place at longer interval of time. Hence a small force is exerted on the body of the person when he falls on a heap of sand. Therefore, he does not get injured.

Question 3. A bullet fired from a gun makes hole in the window pane while passing through it, but the stone striking the window pane breaks it into pieces. Why?

Answer.

In case of bullet, only a small portion of the window pane where the bullet strikes comes in motion because the bullet makes contact with the pane for a very short time due to its high speed. On the other hand, the remaining portion of the window pane remains at rest due to inertia of rest. Thus, a small hole is made by the bullet in the window pane.

In case of a stone, the speed of the stone is very low compared to the speed of the bullet. So the stone makes contact with the window pane for longer period of time. During this long time, the whole window pane comes in motion and hence breaks into pieces.

Question 4. Our hand hurts more when we hit the wall than when we hit a sponge seat of a car. Why?

Answer.

When we hit the wall with our hand, then the hand comes to rest in small interval of time. So large change in momentum of our hand takes place in small interval of time. Hence, large force is exerted on our hand. On the other hand, when we hit the sponge seat of a car, the seat yields and the hand comes to rest after long interval of time. So, the change in momentum of the hand takes place in large interval of time. Hence, less force is exerted on our hand.

Question 5. A karate player breaks a pile of tiles or bricks with a single blow. How?

Answer.

When a karate player strikes the pile of tiles with his hands, he des so as fast as possible. In other words, the time taken to strike the pile of tiles is very small. As the momentum of the hand of a karate player reduces to zero when his hand strikes the pile of tiles in a very small interval of time, therefore, the rate of change of momentum is very large and hence a very large force is exerted on the pile of tiles. This force is enough to break the pile of tiles.

Question 6. “Action and reaction are equal and opposite but even then they do not cancel each other.” Explain, Why?

Answer.

Two equal and opposite forces can cancel each other if they act on the same body. But action and reaction do not act on the same body. Action acts on one body and the reaction acts on another body. Hence they cannot cancel each other.

Question 7. It is difficult to balance our body when we accidently step on a peel of banana. Explain. Why?

Answer.

When we walk on the ground, our foot pushes the ground in the backward direction (Action). On the other hand, ground pushes our foot in the forward direction (Reaction). This reaction of the ground helps us to move in the forward direction. But, when our foot falls on a peel of banana, then our foot slips on the peel of banana and cannot push the round in the backward direction. Consequently, no reaction force acts on our foot. Hence we lose the balance and fall down.

Question 8. When a quick jerk is given to a smooth thick cardboard placed on a tumbler with a small coin placed on the cardboard the coin falls in the tumbler. Whey?

Answer.

The coin was initially at rest. When the cardboard moves because of the jerk, the coin tends to remain at rest due to inertia of rest. When the cardboard leaves contact with the coin, the coin falls in the tumbler on account of gravity.

Question 9. Which has more inertia, a bicycle or a train and why?

Answer.

A train because of its large mass.

Question 10. What type of force is acting on the body whose motion is shown in the following v–t graph.

clip_image002

Answer.

(a) Since the velocity is constant, acceleration is zero. Therefore no force is acting on the body.

(b) This is a case of constant retardation.

Therefore, a constant retarding force is acting on the body.

(c) This is a case of constant acceleration.

Therefore, a constant accelerating force is acting on the body.

Question 11. A truck and car moving with same velocity come to halt after head on collision. If the collision lasts for 10 s.

(a) Which vehicle experiences the greater force of impact?

(b) Which vehicle experiences the greater acceleration?

(c) Which vehicle experiences the greater change in momentum?

(d) Why is the car damaged more than the truck?

Answer.

(a) Since momentum of truck is more than the momentum of the car, because of its greater mass so the car will experience greater force of impact exerted by the truck.

(b) Since, clip_image004so both will have same acceleration. Where a = acceleration, v = final velocity, u = initial velocity, t = time taken to change in velocity

(c) The truck experiences a large change in momentum because its initial momentum is more than the initial momentum of the car.

(d) The car gets damaged more than the truck because the truck transfer more momentum to the car.

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CBSE Physics | Force and Laws of Motion: Law of Conservation of Momentum

August 31st, 2015

Force and Laws of Motion: Part 3

Law of Conservation of momentum: –

According to this law, the total momentum of an object remains constant if no net external force is applied on the object.

Illustration of the law of conservation of momentum: –

Let a moving ball collides with a stationary ball lying on the ground, observe what happens after collision. The moving ball will slow down i.e. its velocity decreases after colliding with the stationary ball.

On the other hand, the stationary ball beings to move i.e. its velocity increase after collision. We know, momentum of a body = mass of the body clip_image002velocity of the body

Therefore, the momentum of moving ball decrease after collision and the momentum of the stationary ball increase after collision.

Thus, we find that when two balls collide with each other, then moving ball loses momentum and the stationary ball gains momentum. The loss of momentum of one ball is equal to the gain of momentum of other ball.

However, the total momentum of these colliding balls before and after the collision remains the same. This is the law of conservation of momentum.

Proof of the law of conservation of momentum: –

Consider a system consisting of two bodies A and B of masses m1 and m2 respectively. suppose these bodies are moving with velocities u1and u2 as shown in fig below Let u1 > u2clip_image004

These bodies collide with each other for a small interval of time t. At the time of collision FAB is the force on A due to B and FBA is the force on B due to A. According to Newton’s third law of motion FAB = – FBA.

Due to these forces, the momentum of the bodies changes. Let V1 and V2 is the velocities of body

A and body B respectively after collision.

Now, initial momentum of body A = m1u1

Initial momentum of body B = m2u2

Final momentum of body A = m1v1

Final momentum of body B = m2 v2

Total momentum of the system before collision = m1 u1 + m2 u2

Total momentum of the system after collision = m1v1 + m2 v2

Change in momentum of the body A = final momentum of body A – Initial momentum of body A

= m1v1 – m1u1

and change in momentum of the body B = final momentum of body B – initial momentum of body B

= m2 v2 – m2 u2

According to Newton’s second law of motion

clip_image006

and clip_image008

According to Newton’s third law of motion FAB = –F BA

clip_image010 clip_image012

or clip_image014

or clip_image016

or clip_image018

or clip_image020

Hence, total momentum of the system before collision = Total momentum of the system after collision.

This is the law of conservation of momentum.

Application of conservation of Momentum

(i) Rocket Propulsion (Movement of A rocket in the upward direction)

The movement of a rocket in the upward direction can also be explained with the help of the law of conservation of momentum.

The momentum of rocket before it is fired is zero. When the rocket is fired, gases are produced in the combustion chamber of the rocket due to the burning of fuel. These gases come out of the rear of the rocket with high speed. The direction of the momentum of the gases coming out of the rocket is in the downward direction. To conserve the momentum of the system (rocket + gases), the rocket moves upward with a momentum equal to the momentum of gases. The rockets continue to move upward as long as the gases are ejected out of the rocket.

(ii) Inflated balloon lying on the surface of a floor moves forward when pierced with a pin: –

The momentum of the inflated balloon before it is pierced wit a pin is zero. When it is pierced with a pin, air in it comes out with in the backward direction. To conserve the momentum, the balloon moves in the forward direction.

(iii) Recoil of gun:

The recoil of a gun can also be explained with the help of the law of conservation of momentum. Before firing, the gun and the bullet are at rest, therefore, momentum of the system before firing is zero.

When the bullet is fired, it leaves the gun in the forward direction with certain momentum. Since no external force acts on the system, so the momentum of the system must be zero after firing. This is possible only if the gun moves backward with a momentum equal to the momentum of the bullet. That is why gun recoils or moves backward.

Recoil velocity of a gun

Let, mass of the bullet = m

Velocity of the bullet after firing = v

Mass of the gun = M.

Velocity of the gun after firing = V

Since the system is at rest before firing, so moment of the system (gun + bullet) before firing = 0

According to the law of conservation of momentum,

Momentum of system after firing = Moment of system before firing

i.e. MV + mv = 0

or, MV = – mv

or, clip_image022

Negative sign shows that the direction of the velocity of the gun after firing is opposite to the direction of the velocity of the bullet.

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