Correct Answer - B
Common velocity after collision be v than by COLM
`2Mv = Mu rArr v = (u)/(2)`
Hence, kinetic energy
`= 1/2` energy
`= 1/2(2M)(u/2)^(2) = 1/4Mu^(2)`
It is also the total energy of vibration because the spring is unstretched at this moment, hence if A is the amplitude, then
`1/2 KA^(2) = 1/4 Mu^(2) rArr A = (sqrt((M)/(2K)))u`