A uniform thin stick of mass M = 24kg and length L rotates on a friction less horizontal plane, with its centre of mass stationary. A particle of mass mis placed on the plane at a distance $\mathrm{x}=\frac{\mathrm{L}}{3}$  from the centre of the stick . This stick hits the particle elastically 
If the value of m so that after the collision, there is no rotational motion of the stick is $m_0$
and the minimum value of x can we get a value of ‘m’ so that the rod has no rotational motion after elastic collision, then
 

Let the velocity of particle after collision be V1 and that of centre of the stick be V₂ (as shown)
 

Momentum conservation gives $:{\mathrm{mV}}_1={\mathrm{mV}}_2$ …..(i)
Angular momentum conservation about $\mathrm{C}:\frac{{\mathrm{ML}}^2}{12}\mathrm{\omega }={\mathrm{mV}}_1\frac{\mathrm{L}}{3}\Rightarrow {\mathrm{V}}_1=\frac{\mathrm{ML\omega }}{4\mathrm{m}}\dots \dots \dots \text{ (ii) }$
Collision is elastic, i.e.,$\mathrm{e}=1$

$\frac{{\mathrm{V}}_1-\left(-{\mathrm{V}}_2\right)}{\mathrm{\omega }\frac{\mathrm{L}}{3}-0}=1\Rightarrow {\mathrm{V}}_1+{\mathrm{V}}_2=\frac{\mathrm{\omega L}}{3}$
Substituting for V₁ and V₂ from(i) and (ii) 
$\frac{\mathrm{M\omega L}}{4\mathrm{m}}+\frac{\mathrm{\omega L}}{4}=\frac{\mathrm{\omega L}}{3}$ ……(iii)
$\Rightarrow \frac{\mathrm{M}}{\mathrm{m}}=\frac{1}{3}\Rightarrow \mathrm{m}=3\mathrm{M}$
In above equation $\left(iii\right)$  if we replace $\frac{\mathrm{L}}{3}$by x and repeat the steps, we get equation (iii) as $\text{ as }\frac{\mathrm{M}}{\mathrm{m}}\frac{{\mathrm{L}}^2}{12\mathrm{x}}\mathrm{\omega }+\frac{{\mathrm{L}}^2\mathrm{\omega }}{12\mathrm{x}}=\mathrm{\omega x}$
$\Rightarrow \frac{\mathrm{M}}{\mathrm{m}}\frac{{\mathrm{L}}^{\mathrm{\prime }}}{12\mathrm{x}}=\mathrm{x}-\frac{{\mathrm{L}}^{\mathrm{\prime }}}{12\mathrm{x}}$
this number shall be greater than zero
$\therefore \mathrm{x}>\frac{{\mathrm{L}}^2}{12\mathrm{x}}\Rightarrow \mathrm{x}>\frac{\mathrm{L}}{\sqrt{12}}$