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. 0.4
• B. 0.5
• C. 0.8
• D. 0.25

Answer 1

Answer: (D)

According to law of conservation of linear momentum,
$es0=4m{v}^{\prime }+0$
$v^{\prime }=\frac{v}{4}$
$e=\frac{\text{ Relative velocity of separation }}{\text{ Relative velocity of approach }}=\frac{\frac{v}{4}}{v}$
$e=\frac{1}{4}=0.25$