Question

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 of coefficient of restitution will must be

• A. 0.5
• B. 0.4
• C. 0.6
• D. 0.8

Answer 1

Answer: (A)

0.5

Since there is no external force acting on the system, the linear momentum of the system must be conserved.

Hence momentum before collision = momentum after collision

$⟹mv+0=0+\left(2m\right)v^{\prime }$

$v^{\prime }=\frac{v}{2}$

The coefficient of restitution = Velocity of separation/Velocity of approach

$=\frac{v^{\prime }}{v}$

$=\frac{1}{2}$