A wooden block of mass ' m ' moves with velocity ' v ' and collides with another block of mass ' 4 ~m ', which is at rest. After collision the block of mass ' m ' comes to rest. The coefficient of restitution will be

Step-by-step solution

aligned & M ₁= m , u ₁= v , v ₁=0 & M ₂=4 ~m , u ₂=0, v ₂=? aligned By law of conservation of momentum, we have aligned & m ₁ u ₁+ m ₂ u ₂= m ₁ v ₁+ m ₂ v ₂ & mv +0=0+4 mv ₂ & v =4 v ₂ or v ₂= v 4 aligned Coefficient of restitution e = v ₂- v ₁ u ₁- u ₂ = v 4 -0 v -0 = 1 4 =0.25