Question

A moving block having mass $m$ collides with another stationary block having mass $4m$. The lighter block comes to rest after collision. When the initial velocity of the lighter block is $v$, then the value of coefficient of restitution ($e$) will be

• A. $0.8$
• B. $0.5$
• C. $0.4$
• D. $0.25$

Answer 1

The correct option is D $0.25$

Using law of conservation of linear momentum, we get
$mv=4m{v}^{\prime }$
Velocity of block having mass $4m$,
$v^{\prime }=v/4$
$e=\frac{v/4}{v}=0.25$