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

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