Question

Two identical spheres $P$ and $Q$ lie on a smooth horizontal circular groove at opposite ends of a diameter. $P$ is projected along the groove and at the end of time $t$ , impinges on $Q$ . If $e$ is the coefficient of restitution, then the second impact will occur after the shortest time of

• A. $\frac{2 \pi t}{e ⁡}$
• B. $\frac{ \pi t}{e ⁡}$
• C. $\frac{t}{e ⁡}$
• D. $\frac{2 t}{ e }$

Answer 1

Answer: (D)

Let $r$ be the radius of the groove.
As $P$ and $Q$ lie at opposite ends of a diameter, therefore distance covered $= \frac{1}{2} \left(2 \pi r \right) = \pi r ⁡$
$\therefore SpeedofPbeforeimpact,u=\frac{\pi r}{t}$
Speed of $Q$ before impact $=0$ .
Let $v$ and $v'$ be velocities of $P$ and $Q$ respectively after impact.
$\therefore v'-v=e\left(\right.u-0\left.\right)=eu$
Thus the velocity of $Q$ relative to $P$ after the first impact = $eu$
Let the next impact occur after $t'$ seconds. This would happen when $Q$ completes one round more than $P$ in time $t'$ .
Distance travelled in one round = $2\pi r$
$\therefore t = \frac{\text{Relative distance}}{\text{Relative velocity}} = \frac{2 \pi r }{\text{v}^{'} - \text{v}}$
$t^{'} = \frac{2 \pi r }{\text{eu}} = \frac{2 \left( u ⁡ t\right)}{ eu ⁡} = \frac{2 t}{ e ⁡}$