Two blocks `A` and `B` of masses `m` and `2m` are connected by a massless spring of natural length `L` and spring constant `k`. The blocks are intially resting on a smooth horizontal floor with the spring at its natural length , as shown . A third identical block `C` of mass `m` moves on the floor with a speed `v_(0)` along the line joining `A` and `B` and collides elastically with `A`. FInd `(a)` the velocity of `c.m.` of system `(block A + B + spri ng)` and `(b)` the minimum compression of spring.

Explanation

Since the collision is head - on elastic , there will be exchange of velocity , block `C` comes to rest and block `A` acquires speed `v_(0)` towards right
`v_(c.m.) = (mv_(0) + 2m xx 0)/(m + 2m) = (v_(0))/(3)`
Since the external force is zero , `v_(c.m.) = constant`. As the blocks move , the spring will be compressed.
Due to the spring force , `kx` , the speed of `m` decreases and that of `2m` increases , until the spring is compressed to maximum. At the time of maximum compression (or maximum elongation , to be shown in other problem) blocks will move with same velocity .
`x_(0)` : maximum compression of the spring
By the momentum conservation between `(1)` and `(2)`
`mv_(0) + 2m xx 0 = mv + 2 mv rArr v = (v_(0))/(3)`
By the energy conservation between `(1)` and (2)
`(1)/(2) mv_(0)^(2) = (1)/(2) mv^(2) + (1)/(2) xx 2 mv^(2) + (1)/(2) k x_(0)^(2)`
` = (1)/(2) xx 3m((v_(0))/(3))^(2) + (1)/(2) kx_(0)^(2)`
`mv_(0)^(2) = (mv_(0)^(2))/(3) + kx_(0)^(2)`
`x_(0) = sqrt((2mv_(0)^(2))/(3k)) = v_(0) sqrt((2m)/(3k))`