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A moving block having mass m, collides with another stationary block having mass 4 m. 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

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detailed solution

Correct option is B

Let final velocity of the block of mass 4m=v'Initial velocity of block of mass 4 m=0 Final velocity of block of mass m=0According to law of conservation of linear momentum,mv+4m×0=4mv'+0v'=v4Coefficient of restitution,e= Relative velocity of separation  Relative velocity of approach =v/4v

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