A block having mass m collides with an another stationary block having mass 2m. The lighter block comes to rest after collision. If the velocity of first block is v, then the value is coefficient of restitution will must be

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`