Coefficient of Restitution in Block Collision – Rankers Physics

Collision: Practice Problem & Solution

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 (2018)
$0.8$
$0.25$
$0.5$
$0.4$

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Collision. Understanding the underlying formula is key to arriving at the correct answer below:

By momentum conservation, $mv = 4m v_2 \implies v_2 = v/4$. Coefficient of restitution $e = \frac{v_2 - 0}{v - 0} = \frac{v/4}{v} = 0.25$.

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