Physics - W.E.P. Question with Solution | TestHub

The velocity of centre of mass $v_{\mathrm{cm}}=\frac{m_1{v}_1+m_2{v}_2}{m_1+m_2}$ When $v_1=0$ and $v_2=v_0$, then $v_{cm}=\frac{m_2{v}_0}{m_1+m_2}$ Now, let $x$ be the elongation in the spring. Change in potential energy $=$ potential energy stored in spring $\begin{array}{c}\Rightarrow \frac{1}{2}{m}_2{v}_0^2-\frac{1}{2}\left(m_1+m_2\right)\\ {\left(\frac{m_2{V}_0}{m_1+m_2}\right)}^2=\frac{1}{2}k{x}^2\end{array}$ This gives, $x={\left(\frac{m_1{m}_2}{m_1+m_2}\right)}^{\frac{1}{2}}{v}_0$