Two blocks `A` and `B` of masses `m` and `2m`, respectively are connected by a spring of force constant `k`. The masses are moving to the right with uniform velocity `v` each, the heavier mass leading the lighter one. The spring is in the natural length during this motion. Block `B` collides head on with a third block `C` of mass `m`, at rest, the collision being completely inelastic. Calculate the maximum compression of the spring.

Answer

`x=sqrt((mv^(2))/(12k))`

Explanation

First considering an inelastic collision between `B` and `C`.
`2mv(m+2m)v_(B)`
`v_(B)=2/3v` velocity of (`B` and `C`)
Then velocity of centre of mass during maximum compression
`v_(CM)=(3mv)/(4m)=3/4v`

`1/2mv_(4)^(2)+1/23mv_((b+c))^(2)=1/24mv_(CM)^(2)+1/2kx^(2)`
Then substituting the value `x=sqrt((mv^(2))/(12k))`