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A moving block of mass $m$ collides with another stationary block of mass $4m$. The lighter block comes to rest after the collision. If the initial velocity of the lighter block is $v$, then the coefficient of restitution $e$ is:

1. 0.5
2. 0.4
3. 0.8
4. 0.25

Correct answer: 0.25

Solution

Using conservation of momentum, the speed of the heavier block after collision is found first. Since the lighter block comes to rest, the relative speed of separation is just the heavier block’s final speed, and dividing by the initial relative speed gives $e=0.25$.

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