A block A of mass m is in equilibrium after being suspended from the ceiling with the help of a spring of force constant k. The block B of mass m, strikes the block A with a speed v and sticks to it. The value of v for which the spring just attains its natural length is

 Let the velocity of the combined system just after the collision ${\mathrm{v}}^{\text{'}}$, then using conservation of linear momentum we get $\mathrm{mv}=2{\mathrm{mv}}^{\text{'}}, \Rightarrow {\mathrm{v}}^{\text{'}}=\frac{\mathrm{v}}{2}$

Initial elongation of the spring is $\mathrm{x}=\frac{\mathrm{mg}}{\mathrm{k}}$.

Now, using the conservation of mechanical energy we get $\frac{1}{2}2\mathrm{m}{\left({\mathrm{v}}^{\text{'}}\right)}^2+\frac{1}{2}{\mathrm{kx}}^2=2\mathrm{mgx}$

$m\left(\frac{{\mathrm{v}}^2}{4}\right)+\frac{{\mathrm{m}}^2{\mathrm{g}}^2}{2\mathrm{k}}=\frac{2{\mathrm{m}}^2{\mathrm{g}}^2}{\mathrm{k}}$

$\mathrm{v}=\sqrt{\frac{6{\mathrm{mg}}^2}{\mathrm{k}}}$