Question

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 travelled by the centre of the rod by the time it turns through a right that of the particle, the collision is elastic

Answer 1

Let the mass of the particle $=m$ & the mass of the rod $=M$
Let the particle strikes the rod with a velocity $V$.
If we take the two body to be a system
Therefore the net external torque & net external forec $=0$
Therefore Applying laws of conservation of linear momentum
$M{V}^{\prime }=mV$ ($V=$ velocity of the rod after striking)
$\Rightarrow V^{\prime }/V=m/M$
Again applying laws of conservation of angular momentum
$\Rightarrow \frac{mVR}{2}=l\omega$
$\Rightarrow \frac{mVR}{2}=\frac{M{R}^2}{12}\times \frac{\pi }{2t}\Rightarrow t=\frac{MR\pi }{m12\times V}$
Therefore distance travelled:-
$V^{\prime }t=V^{\prime }\left(\frac{MR\pi }{m12\pi }\right)=\frac{m}{M}\times \frac{M}{m}\times \frac{R\pi }{12}=\frac{R\pi }{12}$