Two blocks of masses `m_(1)` and `m_(2)`, connected by a weightless spring of stiffness `k` rest on a smooth horizontal plane as shown in Fig. Block 2 is shifted a small distance `x` to the left and then released. Find the velocity of centre of mass of the system after block 1 breaks off the wall.

Explanation

If `m_(2)` is shifted by a distance x and released, the mass `m_(1)` will break off from the wall when the spring restores its natural length and `m_(2)` will start going towards right. At the time of breaking `m_(1), m_(2)` will be going towards right with a velocity `v`, which is given as
`1/2kx^(2)=1/2m_(2)v^(2)impliesv=(sqrt(k/m_(2)))x`
and the velocity of the centre of mass at this instant is
`v_(CM)=(m_(1)xx0+m_(2)xxv)/(m_(1)+m_(2))=((sqrt(m_(2)k)x))/(m_(1)+m_(2))`