A rod of mass $M$ and length $L$ is lying on a horizontal frictionless surface. A particle of mass $m$ travelling along the surface hits at one end of the rod with a velocity $u$ in a direction perpendicular to the rod. The collision is completely elastic. After collision, particle comes to rest. The ratio of masses $\left(\dfrac{m}{M}\right)$ is $\dfrac1x$. The value of $x$ will be ______.
Numerical value type. Enter your answer.
Answer: 4
Linear momentum: $mu = Mv$.
Angular momentum about the centre: $mu\dfrac{L}{2} = \dfrac{ML^2}{12}\omega \Rightarrow \omega = \dfrac{6mu}{ML}$.
Energy (elastic): $\frac12mu^2 = \frac12Mv^2 + \frac12\cdot\frac{ML^2}{12}\omega^2 = \frac{m^2u^2}{2M} + \frac{3m^2u^2}{2M}$
$$mu^2 = \frac{4m^2u^2}{M} \Rightarrow \frac mM = \frac14 \Rightarrow x = 4$$
Solution by Sreeraj P, M.Sc Physics