Question

A body A of mass 2M is placed on a smooth horizontal surface and a small block B of mass M is placed at the top of the body, starting from rest, block B slides down on smooth surface of A, at instance when block B leaves A, the speed v of A

Answer 1

$Mv= 2Mu+(M+2M)v,$

$\frac{Mv^2}2=\frac{2Mu^2}2+\frac{(M+2M)v^2}2,$

$u=v\frac{2M}{2(M+2M)}=\frac 13v.$