Ball A of mass 2 kg moving with a velocity of 2 m/s, strikes directly on a ball B of mass 4 kg at rest. The ball A, after striking comes to rest. Find the coefficient of restitution after the collision.

Concept:

Collision:

Collision means two objects coming into contact with each other for a very short period. In other words, collision is a reciprocating interaction between two masses for a very short interval wherein the momentum and energy of the colliding masses changes. While playing carroms, you might have noticed the effect of a striker on coins when they both collide.

For a collision between two objects, the coefficient of restitution is the ratio of the relative speed after to the relative speed before the collision. The coefficient of restitution is a number between 0 (perfectly inelastic collision) and 1 (elastic collision) inclusive.

m₁ = 2 kg, m₂ = 4 kg

u₁ = 2 m/s, u₂ = 0

According to Law of con conservation of Linear momentum:-

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

m₁ u₁ = m₂ v₂ [u₂ = 0, v₁ = 0]

2 × 2 = 4 × v₂

v₂ = 1 m/s

coefficient of restitution (e):-

e = \(\rm \frac{Relative\ velocity\ after \ collasion}{Relative \ velocity\ before\ collasion}\)

= \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)\(\rm \frac{v_2-v_1}{u_1-u_2}\)

\(\rm \frac{1-0}{2-0}\)

= 0.5

So the correct answer is Option 2