Block A of mass m is hanging from a vertical spring of force constant k . Another identical block B strikes the block A with velocity v  and sticks to it. The value of −3π/4  for which the spring just attains natural length is

The initial extension of the spring y0=mgk ………(i)The velocity of combined block after collision   mv=(m+m)v ∴v'=v2 Now from conservation of mechanical energy, we have 12(2m)(v2)2+12ky02=2m  gy0 ………..(ii)After solving above equations, we get  v=6mkg