Two blocks of masses `m_(1) = 2 kg` and `m_(2) = 5` kg are moving in the same direction along a frictionless surface with speeds `10 m//s` and `3 m//s`, respectively, `m_(2)` being ahead of `m_(1)`. An ideal spring with `k = 1120 N//m` is attached to the back side of `m_(2)`. Find the maximum compression of the spring when the blocks collide. What are the final velocities of the blocks when they separate?

Explanation

Form conservation of momentum
`v=(m_(1)v_(1)+m_(2)v_(2))/(m_(1)+m_(2))=5m//s`
conservationof energy
`1/2m_(1)v_(1)^(2)+1/2m_(2)v_(2)^(2)=1/2(m_(1)+m_(2))v^(2)+1/2kx^(2)`
`implies kx^(2)=(m_(1)v_(1)^(2)+m_(2)v_(2)^(2)-(m_(1)-m_(2))v^(2))/k`
`implies x=sqrt(40/1120)=0.25m`
when the blocks have equa speeds, spring has maximum commpression. after his instasnt,the spring againexpands, and after sometimes `m_(1)` loses contact with the spring.
Let `v_(1)` and `v_(2)` be the velocities of the blocks after they lose contact.
`m_(1)u_(1)+m_(2)u_(2)=m_(1)v_(1)+m_(2)u_(2)`
and `v_(1)-v_(2)=-1(u_(1)-u_(2))`
`[e=1,` because there is no loss in `KE`]
Solving for `v_(1)` and `v_(2)` we get
`v_(1)=0m//s` and `v_(2)=7m//s`