Center of MassHard

Question

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 resistitution (e) will be :-

Options

A.

0.5

B.

0.25

C.

0.8

D.

0.4

Solution

By conservation of linear momentum
mv = 4mv'⇒v'=v4
coefficient of restitution (e) =Velocity of separationVelocity of approach
=v4-ov-o=14=0.25

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