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 the coefficient of restitution $\left(e\right)$ 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 law of conservation of linear momentum, we get

$mv=4m{v}^{\prime }$

$velocityofblockhavingmass4m,v^{\prime }=v/4$

$e=\frac{v/4}{v}=0.25$