Block $A$ of mass $m$ is hanging from a vertical spring having stiffness $k$ and is at rest. Block $B$ of identical mass strikes the block $A$ with velocity $v$ and sticks to it. Then the value of $v$ for which the spring just attains natural lengths is

The correct option is B $\sqrt{\frac{6m{g}^2}{k}}$
The initial extension in spring is $x_0=\frac{mg}{k}$
Let $v^{\prime }$ be the final velocity of the combined two blocks.
Conserving linear momentum; $m\times v+0=2m{v}^{\prime }\Rightarrow v^{\prime }=\frac{v}{2}$
Just after collision of block $B$ with block $A$, the speed of combined mass is $\frac{v}{2}$.
For the spring to just attain natural length, the combined mass must rise up by $x_0=\frac{mg}{k}$ and comes to rest.


Applying conservation of mechanical energy between initial and final states,
$\frac{1}{2}\times 2m\times {\left(\frac{v}{2}\right)}^2+\frac{1}{2}\times k\times {\left(\frac{mg}{k}\right)}^2=2mg\times \left(\frac{mg}{k}\right)+0$
On solving, we get $v=\sqrt{\frac{6m{g}^2}{k}}$