A uniform rod of mass 3.0 kg and length 1.2 m is hinged at one end and released from rest from the horizontal position. When the rod becomes vertical…

Energy conservation during the swing: $\tfrac12\cdot\tfrac13ML^2\omega^2=Mg\tfrac L2$, so $\omega^2=3g/L=25$ and $\omega=5\ \mathrm{rad\,s^{-1}}$; the free end has speed $u=\omega L=6\ \mathrm{m\,s^{-1}}$. For the elastic collision, angular momentum about the hinge and kinetic energy are conserved: $I\omega=I\omega'+mLv$ and $\tfrac12I\omega^2=\tfrac12I\omega'^2+\tfrac12mv^2$. With $I=\tfrac13ML^2$, the rod behaves at its tip like an effective mass $M_{\rm eff}=I/L^2=M/3=1.0\ \mathrm{kg}$. A one-dimensional elastic collision then gives the rod-tip velocity $u'=\dfrac{M_{\rm eff}-m}{M_{\rm eff}+m}u=\dfrac{1-2}{3}\times6=-2\ \mathrm{m\,s^{-1}}$, and the ball speed is $4\ \mathrm{m\,s^{-1}}$. Hence $\omega'=u'/L=-1.667\ \mathrm{rad\,s^{-1}}$; the rod's motion is reversed with magnitude $1.67$.