Two small particles of equal masses start moving in opposite directions from a point $$A$$ in a horizontal circular orbit. Their tangential velocities are $$v$$ and $$2v,$$  respectively, as shown in the figure. Between collisions, the particles move with constant speeds. After making how many elastic collisions, other than that at $$A,$$ these two particles will again reach the point $$A$$ ?

Solution :

Let the radius of the circle be $$r.$$ Then the two distance travelled by the two particles before first collision is $$2\pi r.$$   Therefore $$2v \times t + v \times t = 2\pi r$$     where $$t$$ is the time taken for first collision to occur.
$$\therefore t = \frac{{2\pi r}}{{3v}}$$
$$\therefore $$ Distance travelled by particle with velocity $$v$$ is equal to $$v \times \frac{{2\pi r}}{{3v}} = \frac{{2\pi r}}{3}.$$
Therefore the collision occurs at $$B.$$

As the collision is elastic and the particles have equal masses, the velocities will interchange as shown in the figure. According to the same reasoning as above, the 2^nd collision will take place at $$C$$ and the velocities will again interchange.
With the same reasoning the 3^rd collision will occur at the point $$A.$$ Thus there will be two elastic collisions before the particles again reach at $$A.$$