Inelastic Collision Questions

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Very Short Answers Questions [1 Mark Questions]

Ques. Define inelastic collision.

Ans. In a particular collision, if there is dissipation of energy i.e. the total kinetic energy of the objects before and after the collision is not conserved, then the collision is known as the inelastic collision.

Ques. In an inelastic collision,

1. Momentum is not conserved but kinetic energy is conserved
2. Momentum is conserved but kinetic energy is not conserved
3. Both momentum and kinetic energy is conserved
4. Neither momentum nor kinetic energy is conserved

Ans. The correct option is b. Momentum is conserved but kinetic energy is not conserved.

Explanation: In an inelastic collision, due to the loss in energy, the total kinetic energy of the objects before and after the collision is not conserved.

Since no external force supported the collision, therefore momentum is conserved.

Ques. A bullet strikes a wooden block resting on a frictionless horizontal surface and stops in it. Later the combined system moves. In this case

1. Only kinetic energy is conserved
2. Only momentum is conserved
3. Both kinetic energy and momentum are conserved
4. Neither kinetic energy nor momentum is conserved

Ans. The correct option is b. Only momentum is conserved.

Explanation: Kinetic energy is not conserved because this is the case of perfectly inelastic collision.

Momentum is conserved because there is no external force acting on the system.

Ques. When two bodies collide, there is no loss in the overall kinetic energy. Which type of collision is this?

1. Elastic collision
2. Inelastic collision
3. Perfectly elastic collision
4. None of the options

Ans. The correct option is c. Perfectly elastic collision

Explanation: In a perfectly elastic collision, both linear momentum and kinetic energy are conserved. Therefore, after the collision, there is no loss in the overall kinetic energy.

Ques. In which type of collision does no conversion of energy take place?

1. Inelastic collision
2. Perfectly inelastic collision
3. Elastic collision
4. None of the options

Ans. The correct option is c. Elastic collision

Explanation: In an elastic collision there is no loss of energy i.e. the total kinetic energy before and after the collision is the same. Hence, no conversion of energy takes place.

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Short Answers Questions [2 Marks Questions]

Ques. Give three examples of inelastic collision.

Ans. Examples of inelastic collisions are

• A ball dropped from a certain height unable to rise to its original height.
• The accident of two vehicles.
• When a soft mudball is thrown against the wall, it will stick to the wall.

Ques. Define perfectly inelastic collision.

Ans. A collision is said to be a perfectly inelastic collision if two bodies after collision stick together and move as one body.

In this collision, the linear momentum of the system remains conserved but the kinetic energy is not conserved.

Ques. Define Collision. How it occurs?

Ans. The interaction between two bodies due to which the direction and magnitude of the velocity of the colliding bodies change is called Collision.

A collision occurs when two bodies come in direct contact with each other. It is a situation in which two or more bodies exert forces on each other in about a relatively short time.

Ques. Define elastic collision. What is the value of the coefficient of restitution in case of a perfectly elastic collision?

Ans. In a particular collision, if there is no dissipation of energy, the total kinetic energy of the object before the collision is equal to the total kinetic energy of the objects after the collision, then the collision is said to be an elastic collision.

An elastic collision is said to be perfectly elastic if both the linear momentum and kinetic energy of the system remain conserved.

In this situation, the value of the coefficient of restitution is e = 1.

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Long Answers Questions [3 Marks Questions]

Ques. A bullet weighing 10 g and moving with a velocity of 800 ms^-1 strikes a 10 kg block resting on a frictionless surface. What is the speed of the block after the perfectly inelastic collision?

Ans. In a perfectly inelastic collision, the two colliding bodies stick together and move as one body with velocity

v = (m₁u₁ + m₂u₂) / (m₁ + m₂)

Given

• Mass of the bullet, m₁ = 10 g = 10 x 10^-3 kg = 10^-2 kg
• Mass of the block, m₂ = 10 kg
• The initial velocity of the bullet, u₁ = 800 m/s
• The initial velocity of the block, u₂ = 0

Therefore, the final velocity after the collision is given by

v = (10^-2 x 800 + 10 x 0) / (10^-2 + 10)

⇒ v = 8/10.01 = 0.799 m/s ≅ 8 m/s

Ques. Define the coefficient of restitution.

Ans. The coefficient of restitution is an experimental quantity used to represent the nature of collision between two bodies.

The formula of the coefficient of restitution is given by the ratio of the velocity of separation to the velocity of the approach. i.e. 

Coefficient of restitution, e = v_separation/v_approach = |(v₂ - v₁) / (u₂ - u₁)|

Where u₁ and v₁ are the initial and final velocity of the first body respectively and, u₂ and v₂ are the initial and final velocity of the second body respectively.

Ques. What is the value of the coefficient of restitution in case of a perfectly inelastic collision? Write the expression for the final velocity of the two colliding objects undergoing a perfectly inelastic collision.

Ans. The coefficient of restitution in the case of a perfectly inelastic collision is zero. i.e. e = 0.

Let a body of mass m₁ moving with velocity v₁ collides with a body of mass m₂ moving velocity v₂. If the collision is perfectly inelastic, then after the collision both the body will stick together and move as one body with velocity

v = (m₁u₁ + m₂u₂) / (m₁ + m₂)

Ques. What are the differences between elastic and inelastic collisions?

Ans. The difference between elastic and inelastic collisions are as follows

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Very Long Answers Questions [5 Marks Questions]

Ques. Two particles of mass 1 kg and 2 kg are moving with speeds of 2 m/s and 5 m/s respectively. They collide perfectly inelastically. Find

1. Their final velocities
2. Loss in kinetic energy

Ans. In a perfectly inelastic collision, the two colliding bodies will stick together and move as one body with velocity

v = (m₁u₁ + m₂u₂) / (m₁ + m₂)

Given

• m₁ = 1 kg
• m₂ = 2 kg
• u₁ = 2 m/s
• u₂ = - 5 m/s

1. Let the body of mass m₁ is moving towards the positive X-axis and the body of mass m₂ is moving towards the negative X-axis. After the collision, both the body will stick together and moves as one body, therefore the velocity of both bodies will be

v = (1 x 2 - 2 x 5) / (1 + 2) = - 8/3 m/s towards negative X-axis

2. Loss in kinetic energy is given by

ΔK.E. = 1/2 (m₁m₂ / m₁ + m₂) (u₁ - u₂)²

ΔK.E. = 1/2 (1 x 2 / 1 + 2) (2 + 5)² = 49/3 J

Ques. A small ball is dropped from rest from a height of 10 m on a horizontal floor. If the coefficient of restitution between the ground and the body is 0.5, then find the maximum height it can rise after the collision.

Ans. Coefficient of restitution, e = vseparation/vapproach

Where

• v_separation = velocity of the ball just leaving the ground
• v_approach = velocity of the ball just before hitting the ground

But v_approach = √(2gh₁)

Where h₁ is the height from where the ball is dropped.

Also v_separation = √(2gh₂)

Where h₂ is the maximum height of the ball after collision from the ground.

Therefore, the coefficient of restitution, e = √(2gh₂) / √(2gh₁)

⇒ e = √(h₂ / h₁)

⇒ h₂ = e₂h₁

Given

• e = 0.5
• h₁ = 10 m

Therefore, the maximum height the ball can rise after collision

h₂ = 0.52 x 10 = 2.5 m

Ques. A moving particle of mass m makes a head-on perfectly inelastic collision with a particle of mass 2m which is initially at rest. Find the fractional loss in energy of the colliding particle after the collision.

Ans. After the head-on perfectly inelastic collision, the two bodies will stick together and moves as one body with velocity

v = (m₁u₁ + m₂u₂) / (m₁ + m₂)

Given

• m₁ = m
• m₂ = 2m

Let the particle of mass m is moving with a velocity u and the particle of mass 2m is at rest, therefore,

• u₁ = u
• u₂ = 0

On substituting the values, the final velocity of the bodies will be

v = (mu + 0) / (m + 2m) = u/3

Now, initial kinetic energy, K_i = 1/2 m₁u₁² + 1/2 m₂u₂² 

⇒ K_i = mu²/2

Final kinetic energy, K_f = 1/2 (m₁ + m₂)v² 

⇒ K_f = 1/2 (m + 2m)(u/3)²

⇒ K_f = 1/2 (3m)(u²/9) = mu²/6

Loss in kinetic energy, ΔK = K_f - K_i = mu²/6 - mu²/2 = - mu²/3

Fractional loss in kinetic energy = (K_f - K_i )/K_i = (- mu²/3) / (mu²/2) = -2/6 = - 1/3 

Therefore, after collision fractional loss in kinetic energy is 1/3 of its initial kinetic energy.

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