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

(i) If `sigma` be the mass/unit area of the disc,
mass of the whole disc, `m_(1) = pi R^(2) sigma`
radius of the hold `=R//2`
mass of the hold, `m_(2) = pi(R//2)^(2) sigma`
distance of centre of mass of the hold from O, i.e., `x_(2) = R/2`
Since, the hole has been taken out, its mass is taken as negative.
If O is the origin,
`x_(cm) = (m_(1)x_(1)-m_(2)x_(2))/(m_(1) + m_(2)) =((piR^(2) sigma) xx 0-pi(R//2)^(2) sigma xx (R //2))/(pi R^(2) sigma - pi(R//2)^(2) sigma)`
`=-(pi R^(2) sigma//8)/((3//4)pi R^(2) sigma) = -R/6`
The negative sign indicates that the CM is to the left of O.
(ii) When the block-2 is shifted to the left by a small distance x,Potential energy of compression of the spring,`U=1/2 kx^(2)`
As the block-1 break off the wall and the spring is unstretched, its potential energy is converted to kinetic energy of block-2 which acquires a velocity `v_(2)` , where
`1/2m_(2)v_(2)^(2) = 1/2 kx^(2)` or `v_(2) = sqrt(k/m_(2))x`
Since, both the block-I is initially at rest, `v_(1)=0`.
If `v_(cm)` is the velocity of the centre of mass of the system,
`v_(cm) = (m_(1)v_(1) xx m_(2) v_(2))/(m_(1) + m_(2))`
`=(m_(1) xx + m_(2)(k//m_(2)) x)/(m_(1) + m_(2)) = (xsqrt(km_(2))/(m_(1) + m_(2)))`