A uniform rod of length L rests on a smooth horizontal table. A...

Let the mass of the particle be $m and its initial velocity be$v$. Let the mass of the rod be$M$.<br><br>Since the particle stops after the collision, by conservation of linear momentum, the velocity of the centre of mass of the rod$v_c$is given by:<br><br>$mv = Mv_c \Rightarrow v_c = \dfrac{mv}{M}$<br><br>By conservation of angular momentum about the centre of the rod:<br><br>$mv\dfrac{L}{2} = I\omega$<br><br>Substituting the moment of inertia of the rod$I = \dfrac{ML^2}{12}$:<br><br>$mv\dfrac{L}{2} = \dfrac{ML^2}{12}\omega \Rightarrow \omega = \dfrac{6mv}{ML}$<br><br>The time$t$taken by the rod to turn through a right angle$\left(\theta = \dfrac{\pi}{2}\right)$is:<br><br>$t = \dfrac{\theta}{\omega} = \dfrac{\pi/2}{\dfrac{6mv}{ML}} = \dfrac{\pi ML}{12mv}$<br><br>The distance$d$travelled by the centre of the rod during this time is:<br><br>$d = v_c t = \left(\dfrac{mv}{M}\right)\left(\dfrac{\pi ML}{12mv}\right) = \dfrac{\pi L}{12}