A uniform rod of mass 1 kg and length 1 m hangs at rest from a smooth horizontal hinge at its upper end. A lump of putty of mass 2 kg moving…

Angular momentum about the hinge is conserved in the collision (hinge force has no torque about it). With the putty at distance $x=0.5\ \mathrm{m}$: $m v x=I\omega$, where $I=\dfrac{ML^2}{3}+mx^2=\dfrac13+0.5=\dfrac56\ \mathrm{kg\,m^2}$. After impact, energy is conserved as the system swings up $90^\circ$. The rod's centre of mass rises $0.5\ \mathrm{m}$ and the putty at the midpoint rises $0.5\ \mathrm{m}$, so $\tfrac12I\omega^2=Mg(0.5)+mg(0.5)=5+10=15\ \mathrm{J}$. Hence $\omega^2=\dfrac{2\times15}{5/6}=36$ and $\omega=6\ \mathrm{rad\,s^{-1}}$. Then $v=\dfrac{I\omega}{mx}=\dfrac{(5/6)(6)}{2\times0.5}=5\ \mathrm{m\,s^{-1}}$. A common slip is to treat the collision as conserving linear momentum, ignoring the impulse at the hinge.