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

The given situation can be shown as

Before collision,

As the collision is perfectly elastic, therefore momentum is conserved, i.e.

${\mathrm{p}}_{\text{initially }}={\mathrm{p}}_{\text{finally }}$

$\Rightarrow \mathrm{mu}=\mathrm{Mv}$          ...(i)

Angular momentum will also be conserved about point O.

$\begin{array}{l}\Rightarrow \mathrm{mv}\cdot \frac{\mathrm{L}}{2}=\frac{{\mathrm{ML}}^2}{12}\mathrm{\omega }\\ \Rightarrow \mathrm{\omega }=\frac{6\mathrm{mv}}{\mathrm{ML}} ...\left(\mathrm{ii}\right)\end{array}$

$\because$ Coefficient of restitution,

$\mathrm{e}=\frac{\text{ Relative velocity after collision }}{\text{ Relative velocity before collision }}$

$\begin{array}{l}\Rightarrow 1=\frac{\mathrm{v}+\frac{\mathrm{\omega L}}{2}}{\mathrm{u}}\\ \Rightarrow \mathrm{v}+\frac{\mathrm{\omega L}}{2}=\mathrm{u} ...\left(\mathrm{iii}\right)\end{array}$

From Eqs. (ii) and (iii), we get

$\begin{array}{c}\mathrm{v}+\frac{3\mathrm{mu}}{\mathrm{M}}=\mathrm{u}\\ \Rightarrow \frac{\mathrm{mu}}{\mathrm{M}}+\frac{3\mathrm{mu}}{\mathrm{M}}=\mathrm{u} [\mathrm{using} \mathrm{Eq}.(\mathrm{i}\left)\right]\end{array}$

$\Rightarrow \frac{4\mathrm{mu}}{\mathrm{M}}=\mathrm{u}\Rightarrow \frac{\mathrm{m}}{\mathrm{M}}=\frac{1}{4}$        ...(iv)

According to question,

ratio of masses $\left(\frac{\mathrm{m}}{\mathrm{M}}\right)=\frac{1}{\mathrm{x}}$.

Comparing it with Eq. (iv), we get

x = 4