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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 the coefficient of restitution (e) will be:

0.8

0.5

0.4

0.25

Solution:

Using law of conservation of linear momentum, we get
mv = 4mv'
velocity of block having mass 4m, v' = v/4
e = v'/v = (v/4)/v = 0.25