just after collision of $\mathrm{B}$ with $\mathrm{A}$ the speed of combined mass is $\frac{\mathrm{v}}{2}$

For spring to just attain natural length the combined mass must rise up by $:$

$\mathrm{x}_{0}=\frac{\mathrm{mg}}{\mathrm{k}}$ (see figure) and comes to rest.

Now, applying conservation of energy

$\frac{1}{2}(2 m)\left(\frac{v}{2}\right)^{2}+\frac{1}{2} k\left(x_{0}^{2}\right)=(2 m) g x_{0}$

$\& x_{0}=\frac{mg}{k}$

se get, $\mathrm{v}=\mathrm{g} \sqrt{\frac{6 \mathrm{m}}{\mathrm{k}}}$

Alternate solving by $SHM:$

In $\mathrm{SHM} \vee=\omega \sqrt{\mathrm{A}^{2}-\mathrm{x}^{2}}$

we can write, $\frac{\mathrm{v}}{2}=\sqrt{\frac{\mathrm{k}}{2 \mathrm{m}}} \sqrt{\left(\frac{2 \mathrm{mg}}{\mathrm{k}}\right)^{2}-\left(\frac{\mathrm{mg}}{\mathrm{k}}\right)^{2}}$

$v=g \sqrt{\frac{6 m}{k}}$