Question

A moving block having mass $m$, collides with another stationory block having mass 2$m$. The lighter block comes to rest after collision. When initial velocity of the lighter block is $v$, then the value of coefficient of restitution ($e$) will be

• A. 0.8
• B. 0.25
• C. 0.5
• D. 0.4

Answer 1

The correct option is C 0.5

Using law of law of conservation of linear momentum, we get
$mv=2mv\text{'}$
velocity of block having mass $2m,=v\text{'}=v/2$
$e=\frac{\text{velocity of seperation after collision}}{\text{velocity of approach before collision}}e=\frac{v/2}{v}=\frac{1}{2}=0.5$