In the figure shown, initially the spring is in relaxed position. A bullet of mass m hits the block 'A' with horizontal speed `v_(0)` and gets embedded into it. Find the maximum compression in the spring. Stiffness of the spring is K. Neglect any friction and deformation in the spring during collision.

Explanation

Since no external force is associated with the system `(M+M+m)` conservation of horizontal momentum of the system yields.
`mV_(0)=(M+m)v_(1)=(2M+m)v_(2)`
Where `v_(1)=` velocity of (M+m) just after the impact & `v_(2)=` velocity of (M+M+m) just of the maximum compression of the spring.
`rArr v_(1)=(mv_(0))/(M+m) & V_(2)=(mv_(0))/(2M+m)`
The loss of K.E. between the two portion (a) and (b) is given as,
`Delta KE=(1)/(2)(M+m)v_(1)^(2)-(1)/(2)(2M+m)v_(2)^(2)`

`rArr Delta KE=(m^(2)v_(0)^(2))/(2)[(1)/((M+m))-(1)/((2M+m))]=(m^(2)Mv_(0)^(2))/(2(M+m)(2M+m))`
From conservation of mechanical energy
`rArr (1)/(2)kx^(2)=(m^(2)Mv_(0)^(2))/(2(M+m)(2M+m))`
`rArr x = sqrt((m^(2)Mv_(0)^(2))/(2k(M+m)(2M+m)))=mv_(0)sqrt((M)/(2k(M+m)(2M+m)))`