Consider the particle (charge $q$, mass $m$) approaching the ring (charge $Q$, mass $m$) from infinity with speed $u$. The electrostatic force is repulsive.

• As the particle approaches, it pushes the ring forward. The ring accelerates.
• If the particle passes through the center, the repulsive force will reverse direction once the particle is on the other side, causing the ring to decelerate.
• Therefore, the ring attains its maximum speed exactly when the particle reaches the center of the ring (closest approach or crossing point).
• If the particle reflects back (insufficient energy), the interaction behaves like an elastic collision.

Let $v_r$ be the velocity of the ring and $v_p$ be the velocity of the particle at the moment the particle reaches the center.

Conservation of Momentum: $$ m u = m v_r + m v_p \implies v_p = u – v_r $$

Conservation of Energy: Initial potential energy is zero (far apart). Potential energy when particle is at the center of the ring is $U = \frac{1}{4\pi\epsilon_0}\frac{Qq}{r}$. $$ \frac{1}{2}mu^2 = \frac{1}{2}mv_r^2 + \frac{1}{2}mv_p^2 + \frac{Qq}{4\pi\epsilon_0 r} $$ $$ mu^2 = mv_r^2 + mv_p^2 + \frac{Qq}{2\pi\epsilon_0 r} $$ Let $K = \frac{Qq}{2\pi\epsilon_0 r}$. $$ m u^2 = m v_r^2 + m (u – v_r)^2 + K $$

Expanding the energy equation: $$ mu^2 = mv_r^2 + m(u^2 + v_r^2 – 2uv_r) + K $$ $$ mu^2 = 2mv_r^2 + mu^2 – 2muv_r + K $$ $$ 2mv_r^2 – 2muv_r + K = 0 $$ Dividing by $2m$: $$ v_r^2 – u v_r + \frac{K}{2m} = 0 $$ Solving the quadratic for $v_r$: $$ v_r = \frac{u \pm \sqrt{u^2 – \frac{2K}{m}}}{2} = \frac{u}{2} \left( 1 \pm \sqrt{1 – \frac{2K}{mu^2}} \right) $$ Substitute back $K = \frac{Qq}{2\pi\epsilon_0 r}$: $$ \frac{2K}{mu^2} = \frac{Qq}{\pi\epsilon_0 m r u^2} $$ So, $$ v_r = \frac{u}{2} \left( 1 \pm \sqrt{1 – \frac{Qq}{\pi\epsilon_0 m r u^2}} \right) $$

We must interpret the $\pm$ and the discriminant.

Case 1: Reflection ($u^2 < \frac{Qq}{\pi\epsilon_0 m r}$)
The term under the square root is negative. This means the particle cannot reach the center. It reflects back. For an elastic head-on collision of equal masses where one is initially at rest, velocity is exchanged. $$ v_{max} = u $$

Case 2: Critical Velocity ($u^2 = \frac{Qq}{\pi\epsilon_0 m r}$)
The discriminant is zero. The particle just reaches the center and stops relative to the ring. $$ v_{max} = \frac{u}{2} $$

Case 3: Transmission ($u^2 > \frac{Qq}{\pi\epsilon_0 m r}$)
The particle passes through. The ring accelerates from 0. Continuity of motion implies we take the solution that starts from 0 as interaction strength $K \to 0$. This corresponds to the negative sign. $$ v_{max} = \frac{u}{2} \left( 1 – \sqrt{1 – \frac{Qq}{\pi\epsilon_0 m r u^2}} \right) $$