Coefficient of Restitution in Block Collision – Rankers Physics
Topic: Center of Mass , Momentum and Collision
Subtopic: Collision

Coefficient of Restitution in Block Collision

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)

$0.8$
$0.25$
$0.5$
$0.4$

Solution:

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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