Two blocks A(5kg) and B(5kg) attached to the ends of a spring of spring constant 1000N/m are place on a horizontal smoothplane with the spring undeformed. Simultaneously velocities of 10m/s and 4m/s along the line of the spring in the same direction are imparted to A and B.Find Amplitude of SHM.

spring force is an internal force...so momentum conservation is possible....now in any collision there is a point where the maximum deformation takes place and and there is max loss in kinetic energy...in this case the kinetic energy will convert in spring potential energy and there will be max extension..so we conclude that when a situation similar to perfect inelastic collision will take place there will be maximum extension in the spring..i.e. when the velocities of the two block will be same..

m₁v₁+m₂v₂=(m₁+m₂)v

5x10+5x4=10xv

v=6m/s

loss in KE =gain in spring potential energy

1/2x5x4²+1/2x5x10²-1/2x(5+5)x6²=1/2x1000xX²

80+500-360=1000xX²

0.22=X²

X=0.47m