Question

Two bars of masses $m_1$ and $m_2$ connected by a weightless spring of stiffness $\kappa$ (figure shown above) rest on a smooth horizontal plane. Bar 2 is shifted a small distance $x$ to the left and then released. If the velocity of the centre of inertia of the system after bar 1 breaks off the wall is given as $v_{cm}=\frac{sx\sqrt{m_{2}k}}{(m_1+m_2)}$. Find $s$.

Answer 1

After releasing the bar 2 acquires the velocity $v_2$, obtained by the energy, conservation
$\frac{1}{2}{m}_2{v}_2^2=\frac{1}{2}\kappa x^2$ or, $v_2=x\sqrt{\frac{\kappa }{m_2}}$ (1)
Thus the sought velocity of C.M.
$v_{cm}=\frac{0+m_{2}x\sqrt{\frac{\kappa }{m_2}}}{m_1+m_2}=\frac{x\sqrt{m_{2}k}}{(m_1+m_2)}$