Physics - SHM Question with Solution | TestHub

The initial extension in spring is$x_0=\frac{\mathit{m}\mathit{g}}{\mathit{k}}$. Just after the collision of$\mathrm{B}$with$\mathrm{A}$, the speed of combined mass is$\frac{\mathrm{v}}{2}$.

For the spring to just attain natural length the combined mass must rise up by$x_0=\frac{\mathit{m}\mathit{g}}{\mathit{k}}$(see figure) and comes to rest.



Applying conservation of energy between the initial and final state$\frac{1}{2}2m\mathrm{ }{\left(\frac{\mathrm{v}}{2}\right)}^2+\frac{1}{2}k{\left(\frac{mg}{k}\right)}^2=2mg \left(\frac{\mathit{m}\mathit{g}}{\mathit{k}}\right)$

Solving we get$\mathrm{v}=\sqrt{\frac{6m{g}^2}{k}}$