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 restitution will be

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0.8

0.25

0.5

0.4

answer is B.

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Detailed Solution

From law of conservation of linear momentum

$m V + 4 m (0) = m \times (0) + 4 m V^{1}$

$m V = 4 m V^{1}$

$V^{1} = \frac{V}{4}$

$e = \frac{V^{1}}{V} = \frac{V}{4 V} = \frac{1}{4} = 0.25$

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