A uniform rod of mass $m$ aaaand length l, lies on a horizontal frictionless table. A particle of same mass moving with a speed $v_0$ sticks to one end of the rod at $t=0$. At $t=t$

Consider the motion of the $c.m$. of the particle and the rod.

Conserving angular momentum about the centre of mass of the rod -particle system, 

    [$\frac{1}{12}m{l}^2+m{\left(\frac{{\displaystyle \frac{6v}{5l}}l}{4}\right)}^2]\omega +m{\left(\frac{l}{4}\right)}^2\omega = mv\frac{l}{4}\Rightarrow \omega =\frac{6v}{5l}$

So force exerted by the particle on the rod = m${\omega }^2\left(\frac{l}{4}\right)=\frac{9m{\left(v_0\right)}^2}{25l}$

Path of the centre of the rod is a cycloid .