Let r be the radius of the groove.

As P and Q lie at opposite ends of a diameter, therefore distance Covered $=\frac{1}{2}\left(2\pi r\right)=\pi r$

$\therefore \text{Speed of P before impact,}u=\frac{\pi r}{t}$

Speed of Q before impact = 0.

Let v and v' be velocities of P and Q respectively after impact.

$\therefore v\text{'}–v=e(u–0)=eu$

Thus the velocity of Q relative to P after the first impact = $eu$

Let the next impact occur after $t\text{'}$ seconds. This would happen when Q completes one round more than P in time $t\text{'}$.

Distance travelled in one round = $2\pi r$

$\therefore t=\frac{\text{Relative distance}}{\text{Relative velocity}}=\frac{2\pi r}{{\text{v}}^{\prime }-\text{v}}$

$t^{\prime }=\frac{2\pi r}{\text{eu}}=\frac{2\left(ut\right)}{\mathrm{eu}}=\frac{2t}{e}$