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?

Two blocks of masses m_(1) = 2 kg and m_(2) = 5 kg are moving in the s

1.	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?
Answer» SOLUTION :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`