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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
Explanation
Momentum conservation gives $v/4$ for the heavy block; $e = (v/4)/v$.
Detailed Solution
According to the law of conservation of linear momentum,
$mv + 4m\times0 = 4mv' + 0 \Rightarrow v' = \frac{v}{4}$
$e = \frac{\text{Relative velocity of separation}}{\text{Relative velocity of approach}} = \frac{v/4}{v}$
$e = \frac{1}{4} = 0.25$