A rod of mass ‘m’ and length l is placed on a smooth table. An another particle of same mass ‘m’ strikes the rod with velocity ${\mathrm{v}}_0$ in a distance perpendicular to the rod at distance x(<l/2) from its centre. Particle sticks to the rod. Let $\mathrm{\omega }$ be the angular speed of system after collision. Then the maximum possible value of impulses (by varying x) that can be imparted to the particle during collision. Particle still sticks to the rod

C is the center of mass of rod and O the center of mass of system. Conserving angular momentum about point O. ${\mathrm{mv}}_0\frac{\mathrm{x}}{2}={\mathrm{I}}_0\mathrm{\omega }=\left[\mathrm{m}{\left(\frac{\mathrm{x}}{2}\right)}^2+\frac{{\mathrm{ml}}^2}{12}+\mathrm{m}{\left(\frac{\mathrm{x}}{2}\right)}^2\right]\mathrm{\omega }$

$\therefore \mathrm{\omega }=\frac{6{\mathrm{v}}_0}{\left(6\mathrm{x}+{\mathrm{l}}^2/\mathrm{x}\right)}$

For $\mathrm{\omega }$ to be maximum or minimum quantity $\left(6\mathrm{x}+\frac{{\mathrm{l}}^2}{\mathrm{x}}\right)$ should be minimum or maximum respectively.

$\frac{\mathrm{d}}{\mathrm{dx}}\left(6\mathrm{x}+\frac{{\mathrm{l}}^2}{\mathrm{x}}\right)=0 \mathrm{or} 6-\frac{{\mathrm{l}}^2}{{\mathrm{x}}^2}=0\mathrm{ }\therefore \mathrm{x}=\left(\frac{\mathrm{l}}{\sqrt{6}}\right)$

Further $\frac{{\mathrm{d}}^2\left(6\mathrm{x}+\frac{{\mathrm{l}}^2}{\mathrm{x}}\right)}{{\mathrm{dx}}^2} \mathrm{at} \mathrm{x}=\frac{\mathrm{l}}{\sqrt{6}}$ is positive, Hence at $\mathrm{x}=\frac{\mathrm{l}}{\sqrt{6}}$

Denominator is minimum $\Rightarrow \mathrm{\omega }$ is maximum.

Further at $\mathrm{x}=0,\mathrm{\omega }=0$, therefore $\mathrm{\omega }$ will first increase and then it will decreases as x varies from x=0 to x=l/2.

Impulse $\left|\mathrm{J}\right|=\mathrm{\Delta P}=\left|{\mathrm{P}}_{\mathrm{f}}-{\mathrm{P}}_{\mathrm{i}}\right|={\mathrm{P}}_{\mathrm{i}}-{\mathrm{P}}_{\mathrm{f}}$ (as $\left.{\mathrm{P}}_{\mathrm{i}}>{\mathrm{P}}_{\mathrm{f}}\right)$ 

will be maximum when the change in momentum of the ball is maximum.

$\Rightarrow$the final velocity of the ball must be minimum.

This is possible when $\mathrm{r\omega }$ is zero(r is distance from axis of rotation).

So for maximum impulse there must only be linear momentum transfer.

$|\mathrm{J}{|}_{\max }=\mathrm{m}\left({\mathrm{v}}_0-\frac{{\mathrm{v}}_0}{2}\right)=\frac{{\mathrm{mv}}_0}{2}$