A block of mass m moving at speed `upsilon` collides with another block of mass 2 m at rest. The lighter block comes to rest after the collision. Find the coefficient of restitution.

Options

A. 0.5
B. 0.4
C. 0.6
D. 0.8

Answer

A

Explanation

Let the velocity of block of mass 2m after the collision be `v^(')`, then from conservation of momentum
`mv=2mv^(') rArr v^(')=v/2`
Now, the coefficient of restitution,
e = (Velocity of separation)/(Velocity of approach) = `(v^(')/v) = (v/2)/v = 1/2 = 0.5`