Two bars of masses `.m_(1).` and `.m_(2).` connected by a weightles spring of stiffness constant K as shown in figure, rest on a smooth horizontal plane. Bar 2 is shifted through a small distance x to the left and then released. Find the velocity of the centre of mass of the system after bar 1 breaks off the wall.

Explanation

Velocity of the bar 2 at the instant of break off will be obtained as `(1)/(2)m_(2)v_(2)^(2)=(1)/(2)kx^(2)`
`v_(2)= x sqrt((k)/(m_(2)))`
The velocity of the bar 1 at that instant is zero `(v_(1)=0)`
The velocity of the centre of mass is given by
`V_(c )=((m_(1)v_(1)+m_(2)v_(2)))/((m_(1)+m_(2)))`
After the substitution of the value of `v_(1)` and `v_(2)`, we get
`V_(c )= x sqrt((km_(2))/((m_(1)+m_(2))))`