A uniform rod of length L lies on a smooth horizontal table. A particle moving on the table strikes the rod perpendicularly at an end and stops. Find the distance traveled by the centre of the rod by the time it turns through a right angle. Show that if the mass of the rod is four times that of the particle, the collision is elastic.

Let the mass of the particle = m and the mass of the rod = M Let the particles strike the rod with velocity V. If we take two body to be a system, Therefore, the net external torque and external force = 0

Therefore, Applying laws of conservation of linear momentum,

MV' = mV (V' = velocity of the rod after striking)

$\Rightarrow \frac{V^{\prime }}{V}=\frac{m}{M}$

Again applying laws of conservation of angular momentum,

$\Rightarrow \frac{mVR}{2}=\left(\frac{M{R}^2}{12}\right)\times \left(\frac{\pi }{2t}\right)$

$\Rightarrow t=\left(\frac{MR\pi }{m\times 12\times V}\right)$

$\frac{m{v}^2}{R}$

Therefore, distance travelled

$=V^{\prime }t=V^{\prime }\left(MR\pi \right)\times \left(\frac{R\pi }{12}\right)$

$=\left(\frac{m}{M}\right)\times \left(\frac{M}{m}\right)\times \left(\frac{R\pi }{12}\right)$

$=\frac{R\pi }{12}$