Physics - Center of Mass Question with Solution | TestHub

$V_{CM}=\frac{m{V}_0+m{V}_0}{3m}=\frac{2V_0}{3}$ Let $u$ be the velocity of beads relative to the hoop just before collision. In $y$-direction, the beads and the hoop will have same velocity equal to $V_{CM}=\frac{2V_0}{3}$. In ground frame, speed of each bead is $=\sqrt{u^2+v_{CM}^2}=\sqrt{u^2+\frac{4v_0^2}{9}}$ Energy conservation gives : $\frac{1}{2}m\left(u^2+\frac{4v_0^2}{9}\right)\times 2+\frac{1}{2}m{\left(\frac{2v_0}{3}\right)}^2=\frac{1}{2}m{v}_0^2\times 2$ $\Rightarrow u=\frac{V_0}{\sqrt{3}}$

∴ Speed of eachbead $=\sqrt{u^2+\frac{4v_0^2}{9}}=\frac{\sqrt{7}}{3}{v}_0$