Elastic Collision Questions

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Very Short Answers Questions [1 Mark Questions]

Ques. In which type of collision the total kinetic energy is conserved?

1. Elastic collision
2. Inelastic collision

Ans. The correct answer is a. Elastic collision

Explanation: An elastic collision is one in which the system suffers no net loss of kinetic energy as a result of the collision. In elastic collisions, both momentum and kinetic energy are conserved.

Ques. The collision of two cars is an example of ______

1. Elastic collision
2. Inelastic collision

Ans. The correct answer is b. Inelastic collision

Explanation: An example of an inelastic collision is the collision between two cars. The total amount of momentum remains unchanged in this case, whereas the kinetic energy of the cars varies. The kinetic energy is used to cause the two cars to collide or stick together.

Ques. Which among the following is a collision in two dimensions?

1. Head-on Collision
2. Oblique Collision

Ans. The correct answer is b. Oblique Collision

Explanation: When one of the two bodies has a velocity that is at an angle with the collision line, an oblique collision occurs. The component of velocity perpendicular to the collision line remains unchanged in the case of an oblique collision. This can be regarded as a two-dimensional collision.

Ques. Does conversion of energy take place in an elastic collision?

1. Yes
2. No

Ans. The correct answer is a. No

Explanation: In an elastic collision, both momentum and kinetic energy are conserved. In the case of an elastic collision, the kinetic energy before and after the collision remains constant and is not transformed into any other type of energy.

Ques. What is the elastic collision formula of kinetic energy?

Ans. Let m₁ and m₂ be the masses of the two colliding bodies. Let u₁ and u₂ be their velocities before collision and v₁ and v₂ be their velocities after the collision, then the elastic collision formula of kinetic energy is given by

1/2 m₁u₁² + 1/2 m₂u₂² = 1/2 m₁v₁² + 1/2 m₂v₂²

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Short Answers Questions [2 Marks Questions]

Ques. What is meant by collision?

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. Explain elastic collision with an example.

Ans. An elastic collision is one in which the system suffers no net loss of kinetic energy as a result of the collision. Momentum and kinetic energy are both conserved in inelastic collisions.

For example, if two comparable carts are traveling at the same speed toward one other, they will collide, bouncing off each other with no loss of momentum. Because no energy was lost, this collision is perfectly elastic.

Ques. Explain inelastic collision with an example.

Ans. An inelastic collision occurs when two moving objects collide and lose kinetic energy and momentum as a result of their collision. 

For example, putting a mound of clay on the ground or witness a vehicle accident. The mound of clay will not return to its original location, and the vehicle will not continue on its original path.

Ques. Define the coefficient of restitution.

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

For a purely one-dimensional collision, the coefficient of restitution is defined as the ratio of velocity of separation to velocity of approach.

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Long Answers Questions [3 Marks Questions]

Ques. What are the applications of elastic collision?

Ans. The following are the applications of elastic collision

• The amount of force experienced by the body during the collision is affected by the amount of time involved in the collision. In other words, the lower the force acting on the body, the longer the time involved in the collision. As a result, in order to maximize the force, the time involved in the collision must be decreased.
• However, in order to minimize the force, the collision duration must be increased. There are several examples of this. One of them is that airbags in automobiles increase the time engaged in a collision while minimizing the effect of force on an object. The phenomenon behind this process is that the airbags reduce the force while increasing the object's collision time.
• An elastic collision can be considered in a variety of places or circumstances, but an inelastic collision is more realistic and happens in the majority of cases.

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

Ans. The differences between elastic and inelastic collisions are

Ques. What are the characteristics of elastic collision?

Ans. The following are the characteristics of the elastic collision

• An object's linear momentum is conserved in an elastic collision.
• The object's total energy is conserved in an elastic collision.
• The kinetic energy of the system is also conserved.
• During an elastic collision, conservative forces are involved.
• Mechanical energy is not transformed into heat in an elastic collision.

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Very Long Answers Questions [5 Marks Questions]

Ques. Two balls of mass 4 kg and 2 kg are moving with speeds of 10 m/s and 8 m/s respectively with the ball of heavier mass behind the lighter ball. The two balls collide with each other elastically. Find the maximum potential energy stored in the system of two balls during the collision.

Ans. Given

• Mass of the heavier ball, m₁ = 4 kg
• Speed of the heavier ball, v₁ = 10 m/s
• Mass of the lighter ball, m₂ = 2 kg
• Speed of the lighter ball, v₂ = 8 m/s

During collision for some time, the two bodies will stick together and move as one body of mass,

M = m₁ + m₂ = 4 + 2 = 6 kg

Let V be the common velocity of the system of balls during the collision.

Applying the law of conservation of momentum, we get

Initial momentum = Final momentum

⇒ m₁v₁ + m₂v₂ = MV

⇒ (4 x 10) + (2 x 8) = (6V)

⇒ 56 = 6V

⇒ V = 56/6 = 9.3 m/s

The maximum potential energy stored in the system is equal to the change in kinetic energy of the system.

(P.E)_max = ΔK.E

⇒ (P.E)_max = (K.E)_final - (K.E)_initial

⇒ (P.E)_max = (1/2 x 4 x 10² + 1/2 x 2 x 8²) - (1/2 x 10 x 9.3²)

⇒ (P.E)_max = 2.6 J

Ques. Two particles of equal mass moving with speed u₁ and u₂ in opposite directions collide perfectly elastically. Find their velocities just after the collision.

Ans. Given

• The initial velocity of the first particle is u₁
• The initial velocity of the second particle is u₂

Let m₁ and m₂ be the masses of the two particles and v₁ and v₂ respectively be their final velocities after collision then, we have

The final velocity of the first particle, 

v₁ = \((\frac{m_1 – m_2 e}{m_1 + m_2})\)u₁ + \((\frac{m_2(1 + e)}{m_1 + m_2})\)u₂

Where e is the coefficient of restitution. 

For perfectly elastic collision, e = 1

⇒ v₁ = \((\frac{m_1-m_2}{m_1+m_2})\) u₁ + \((\frac{2m_2}{m_1+m_2})\) u₂

Since m₁ = m₂ = m, therefore

v₁ = 0 + \((\frac{2m}{2m})\) u₂

⇒ v₁ = u₂

Now, the final velocity of the second particle, 

v₂ = \((\frac{m_2 – m_1 e}{m_1 + m_2})\) u₂ + \((\frac{m_1(1 + e)}{m_1 + m_2})\) u₁

For perfectly elastic collision, e = 1

⇒ v₂ = \((\frac{m_2-m_1}{m_1+m_2})\) u₂ + \((\frac{2m_1}{m_1+m_2})\) u₁

Since m₁ = m₂ = m, therefore

v₂ = 0 + \((\frac{2m}{2m})\) u₁

⇒ v₂ = u₁

It can be noted that if bodies of equal mass collide head-on and perfectly elastically, then their velocities get exchanged after the collision.

Ques. A very light particle moving with a speed of u₁ hits a very heavy block moving with a speed of u₂. If the collision is perfectly elastic, then find the velocities of the particle and block after the collision.

Ans. Given

• The initial speed of the light particle is u₁
• The initial speed of the heavy block is u₂

Let m₁ and m₂ be the masses of the particle and the block and v₁ and v₂ respectively be their final velocities after collision then, we have

The final velocity of the light particle, 

v₁ = \((\frac{m_1 – m_2 e}{m_1 + m_2})\) u₁ + \((\frac{m_2(1 + e)}{m_1 + m_2})\) u₂

For perfectly elastic collision, e = 1

⇒ v₁ = \((\frac{m_1-m_2}{m_1+m_2})\)u₁ + \((\frac{2m_2}{m_1+m_2})\) u₂

Since m₂ >> m₁, therefore the mass m₁ should be neglected in the above equation.

⇒ v₁ = \((\frac{– m_2}{m_2})\)u₁ + \((\frac{2m_2}{m_2})\)u₂

⇒ v₁ = – u₁ + 2u₂

Now, the final velocity of the heavy block, 

v₂ =\((\frac{m_2 – m_1 e}{m_1 + m_2})\) u₂ + \((\frac{m_1(1 + e)}{m_1 + m_2})\) u₁

For perfectly elastic collision, e = 1

⇒ v₂ = \((\frac{m_2-m_1}{m_1+m_2})\) u₂+ \((\frac{2m_1}{m_1+m_2})\) u₁

Since m₂ >> m₁, therefore the mass m₁ should be neglected in the above equation.

⇒ v₂ = \((\frac{m_2}{m_2})\) u₂+0

⇒ v₂ = u₂

Thus, the heavy block will keep moving approximately at the same speed.

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