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Physics · PYQ NEET UG 2018
हिन्दीA 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
Correct answer · Option 2
\( 0.25 \)
Explanation
From conservation of momentum: \( mv = 4m v' \) \( \Rightarrow v' = v/4 \). The coefficient of restitution is \( e = \frac{\text{relative velocity of separation}}{\text{relative velocity of approach}} = \frac{v/4 - 0}{v - 0} = 0.25 \).
