A mass 2m is at rest and another mass m is moving with a velocity. An elastic collision takes place between them. Show that the mass m loses `8/9` part of its initial kinetic energy in this collision.

Explanation

Let the velocity of mass m before collision =u and the velocity of mass m after collision =`v_1`. The velocity of mass 2m after collision =`v_2`.
From the law of conservation of momentum ,
`mu=mv_1+2mv_2 or, u=v_1+2v_2 or, u-v_1 =2v_2` ………..(1)
From the law of conservation of energy,
`1/2 mu^2=1/2mv_1^2+1/2(2m)v_2^2`
or, `u^2 =v_1^2+2v_2^2 or, u^2-v_1^2=2v_2^2`
`or, (u+v_1)(u-v_1)=2v_2^2`........(2)
Dividing (2) by (1) ,we get,
`u+v_1=v_2` .........(3)
From (1) and (3) we get , `v_2=2/3u therefore v_1=-1/3 u`
Initial kinetic energy of the mass m`=1/2 mu^2`.
Its kinetic energy after collision `=1/2 mv_1^2=1/2m*1/9u^2=1/(18)mu^2.`
Hence, decreases in kinetic energy
`=1/2 mu^2-1/(18)mu^2=8/9xx1/2mu^2`
`=8/9` of its initial kinetic energy.