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 $-3\pi /4$  for which the spring just attains natural length is

The initial extension of the spring $y_0=\frac{mg}{k}$ ………(i)The velocity of combined block after collision   $mv=\left(m+m\right)v$ $\therefore v\text{'}=\frac{v}{2}$ Now from conservation of mechanical energy, we have $\frac{1}{2}\left(2m\right){\left(\frac{v}{2}\right)}^2+\frac{1}{2}k{y}_0^2=2mg{y}_0$ ………..(ii)After solving above equations, we get  $v=\sqrt{\frac{6m}{k}}g$