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 $$\left(\frac{m}{M}\right)$$ is $$\frac{1}{x}$$. The value of $$x$$ will be ___.

A particle of mass $$m$$ moving with velocity $$u$$ hits one end of a rod of mass $$M$$ and length $$L$$ perpendicularly, with the collision being completely elastic, and the particle comes to rest after the collision.

Let $$V_{cm}$$ be the velocity of the centre of mass of the rod and $$\omega$$ be its angular velocity after the collision. Conservation of linear momentum gives $$mu = MV_{cm}$$, so $$V_{cm} = \frac{mu}{M}$$.

Conservation of angular momentum about the centre of the rod gives $$mu\cdot\frac{L}{2} = I_{rod}\,\omega = \frac{ML^2}{12}\,\omega$$, so $$\omega = \frac{6mu}{ML}$$.

Conservation of kinetic energy (elastic collision): $$\frac{1}{2}mu^2 = \frac{1}{2}MV_{cm}^2 + \frac{1}{2}I_{rod}\,\omega^2$$.

Substituting: $$mu^2 = M\cdot\frac{m^2u^2}{M^2} + \frac{ML^2}{12}\cdot\frac{36m^2u^2}{M^2L^2} = \frac{m^2u^2}{M} + \frac{3m^2u^2}{M} = \frac{4m^2u^2}{M}$$.

Therefore $$m = \frac{M}{4}$$, giving $$\frac{m}{M} = \frac{1}{4} = \frac{1}{x}$$, so $$x = 4$$.