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

1. 0.5

2. 0.8

3. 0.25

4. 0.4

Doubtly Asked 30 Oct 2026 NEET UG 2018 Practice test neet Save WhatsApp Report

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Answer: 3. 0.25

Momentum conservation: $mv=0+4mv_2\Rightarrow v_2=\dfrac v4$ (lighter block at rest, $v_1=0$).

Coefficient of restitution = relative speed of separation / relative speed of approach:

$$e=\frac{v_2-v_1}{u_1-u_2}=\frac{v/4-0}{v-0}=0.25$$

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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