A bullet of mass m strikes a block of mass M connected to a light spring of stiffness k, with a speed `v_(0)` and gets into it. Find the loss of K.E. of the bullet.

Explanation

The process of impact of bullet and block is transient. Within a very short time of impact, the compression of the spring is negligible. Therefore, the corresponding spring force is negligible. Even though it is external to the system (M+m), we can conserve its momentum just before and after the impact (impact force is internal). Conservation of linear momentum of bullet plus block just before and after the impact (impact force is internal). Conservation of linear momentum of bullet plus block just after and before impact yields
`(M+m)v=mv_(0)`
`rArr v = (mv_(0))/(M+m)`
where v = common velocity of block and bullet.
Therefore the loss of K.E. of the system
`Delta KE=(1)/(2)mv_(0)^(2)-(1)/(2)(M+m)v^(2)`
Putting `v=(mv_(0))/(M+m)` are obtain.
`Delta KE = (Mn v_(0)^(2))/(2(M+m))`.