Question

Two particles, each of mass $M$ and speed $v$, move as shown. They simultaneously strike the ends of a uniform rod of mass $M$ and length $d$ which is pivoted at its center. The particles stick to the ends of the rod. Find the ratio of total initial kinetic energy of the two particles to the total loss in kinetic energy in the collision of the two particles with the rod?

Answer 1

Angular momentum conservation
$2\left(M v \frac{d}{2}\right)=\left[\frac{M d^{2}}{12}+2\left(M\left(\frac{d}{2}\right)^{2}\right)\right] \omega $
$\Rightarrow \omega=\frac{12 v}{7 d}$
$K E_{i}=2\left(\frac{1}{2} M v^{2}\right) $
$\Rightarrow K E_{f}=\frac{1}{2} I \omega^{2}=\frac{1}{2}\left(\frac{7 M d^{2}}{12}\right) \omega^{2} $
$\Delta K E=\frac{1}{7} M v^{2}$
$ \Rightarrow \frac{K E_{i}}{\Delta K E}=7$