Question

A ball of mass m moving at a speed v makes a
head on collision with an identical ball at rest. The
kinetic energy of the ball after the collision is 4%
of the original. Find the coefficient of restitution.​

Answer 1

Hi, just considr initial vel. of I ball = u final vel. of I ball = v1 final vel. of II ball =v2 & coeff. of restitution = e now, e = vel. of separaion / vel. of approach e = v2 -v1 / u => v2-v1 = eu ....(1) momentum conservation => mu = mv1 +mv2 => u = v1 + v2 ...(2) energy eqn 3/4((mu^2)/2)= 1/2*m (v1^2 + v2^2) => 3/4*(u^2) = (v1^2 + v2^2) 2*(v1^2 + v2^2)=(v1 + v2)^2 + (v2 -v1)^2 = u^2 + (eu)^2 {FROM 1& 2 } => 3/2 (u^2) = u^2 + (eu)^2 => 1/2 (u^2) = (eu)^2 => e = root(1/2)