Problem Overview: We have a system of four identical charged sleeves (A, B, C, D) on frictionless rods. B, C, and D are initially in an unstable equilibrium. Sleeve A is launched with velocity u towards the cluster. We must determine the final velocities of all sleeves.

1. Defining the Velocity Parameter v_0

The cluster B-C-D is in unstable equilibrium. To solve the problem, we first determine the velocity with which the cluster would explode if disturbed from rest.

Energy Balance

The electrostatic potential energy available in the cluster (interaction of C with B and D) is: $$U = \frac{kq^2}{d} + \frac{kq^2}{d} = \frac{2kq^2}{d}$$ Let the explosion velocity of C be $V_{exp}$. By conservation of momentum, B and D would recoil at $-V_{exp}/2$. $$\frac{2kq^2}{d} = \frac{1}{2}mV_{exp}^2 + 2\left(\frac{1}{2}m\left(\frac{V_{exp}}{2}\right)^2\right) = \frac{3}{4}mV_{exp}^2$$ $$V_{exp} = \sqrt{\frac{8kq^2}{3md}}$$

Matching the Parameter $v_0$

Substituting $k = \frac{1}{4\pi\epsilon_0}$, we get: $$V_{exp} = \sqrt{\frac{8}{3md} \cdot \frac{q^2}{4\pi\epsilon_0}} = \sqrt{\frac{2q^2}{3\pi\epsilon_0 md}} = 2\sqrt{\frac{q^2}{6\pi\epsilon_0 md}}$$ We define the parameter $v_0$ as per the answer key: $$v_0 = \sqrt{\frac{q^2}{6\pi\epsilon_0 md}}$$ Therefore, the “Explosion Velocity” of the cluster is exactly $2v_0$.

2. Analysis of Motion Cases

We analyze the interaction by comparing the incoming velocity u to the explosion velocity 2v_0. The situation is similar to collision. In further explanation “hitting” or “striking” refers to the electrostatic interaction. The particles never physically touch each other.

Case I: $u = 2v_0$ (Exchange)

Sleeve A arrives with exactly the velocity C would have during an explosion.

• A strikes C elastically. Since masses are equal, they exchange velocities.
• $v_C$ becomes $u = 2v_0$.
• $v_A$ becomes 0. A replaces C at the unstable equilibrium point.
• Since A is at rest at the equilibrium point, the system (A-B-D) remains stable. B and D do not move.

Case II: $u > 2v_0$ (Transmission)

Sleeve A hits C, exchanging velocity. $v_C = u$. A momentarily takes the place of C with velocity $\approx 0$, but the system (A-B-D) still contains the potential energy $U$.

Since A enters the cluster, the cluster “explodes” using its potential energy. However, unlike Case I, A is part of this explosion dynamics. The potential energy converts into kinetic energy for A, B, and D.

Based on the explosion derivation in Section 1:
Center particle (now A) moves at the explosion velocity: $2v_0$.
Side particles (B and D) recoil at: $-v_0$.

Case III: $u < 2v_0$ (Reflection)

Sleeve A is too slow. The repulsive field triggers the breakup of the B-C-D cluster before A can replace C.
1. C escapes at $2v_0$.
2. The B-D pair recoils towards A with velocity $-v_0$.

Collision: Sleeve A (mass $m$, velocity $u$) collides with the pair B+D (mass $2m$, velocity $-v_0$).

Using standard 1D elastic collision formulas for $v_A$: $$v_A = \left( \frac{m – 2m}{m + 2m} \right) u + \left( \frac{2(2m)}{m + 2m} \right) (-v_0)$$ $$v_A = -\frac{1}{3}u – \frac{4}{3}v_0 = -\frac{1}{3}(u + 4v_0)$$

For the pair B+D: $$v_{BD} = \left( \frac{2m – m}{m + 2m} \right) (-v_0) + \left( \frac{2(m)}{m + 2m} \right) u$$ $$v_{BD} = -\frac{1}{3}v_0 + \frac{2}{3}u = \frac{1}{3}(2u – v_0)$$