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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

1

0.8

2

0.25

3

0.5

4

0.4

Step-by-Step Solution

According to law of conservation of linear momentum, mv+4m(0)=m(0)+4mv′mv + 4m(0) = m(0) + 4mv', which gives v′=v/4v' = v/4. The coefficient of restitution e=Relative velocity of separationRelative velocity of approach=v/4v=1/4=0.25e = \frac{\text{Relative velocity of separation}}{\text{Relative velocity of approach}} = \frac{v/4}{v} = 1/4 = 0.25.

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