Physics Solution – Question 49

Solution to Question 49

1. System Analysis

We consider a system of three small non-conducting balls A, B, and C, each of mass $m$ and charge $q$. Balls A and B are rigidly attached to the ends of a light rod. Based on the initial configuration shown:

Distance between A and C is $3d$.
Distance between C and B is $d$.
Therefore, the total length of the rod (distance AB) is $4d$.

Ball C slides on the rod. The system is released from rest. We need to find the maximum speed of ball C.

Figure 1: Initial configuration of the system.

2. Energy Considerations

By conservation of mechanical energy, the maximum kinetic energy occurs when the potential energy is minimum. Since charges A and B repel C, the electrostatic potential energy is minimized when C is at the equilibrium position between them. Due to symmetry (equal charges on A and B), this occurs at the midpoint of the rod.

Initial Position: $r_{AC} = 3d$, $r_{BC} = d$.
Final (Min PE) Position: $r_{AC} = 2d$, $r_{BC} = 2d$.

The change in potential energy $\Delta U = U_i – U_f$ is:

$$U_i = k\frac{q^2}{4d} + k\frac{q^2}{3d} + k\frac{q^2}{d} \quad (\text{Interaction AB, AC, BC})$$ $$U_f = k\frac{q^2}{4d} + k\frac{q^2}{2d} + k\frac{q^2}{2d}$$ $$\Delta U = \left( k\frac{q^2}{3d} + k\frac{q^2}{d} \right) – \left( k\frac{q^2}{2d} + k\frac{q^2}{2d} \right)$$ $$\Delta U = \frac{kq^2}{d} \left[ \left(\frac{1}{3} + 1\right) – 1 \right] = \frac{kq^2}{3d}$$

3. Momentum Conservation

The system is released from rest, so the initial momentum is zero. No external forces act along the rod axis. Let $v_C$ be the velocity of ball C and $v_{AB}$ be the velocity of the rigid rod (balls A and B combined).

$$P_i = 0$$ $$P_f = m v_C + (2m) v_{AB} = 0$$ $$v_{AB} = -\frac{v_C}{2}$$

4. Calculation of Maximum Speed

The loss in potential energy converts to the kinetic energy of the system.

$$\Delta U = K_{total}$$ $$\frac{kq^2}{3d} = \frac{1}{2}m v_C^2 + \frac{1}{2}(2m) v_{AB}^2$$ Substitute $v_{AB} = \frac{v_C}{2}$: $$\frac{kq^2}{3d} = \frac{1}{2}m v_C^2 + m \left( \frac{v_C^2}{4} \right)$$ $$\frac{kq^2}{3d} = \frac{3}{4} m v_C^2$$

Solving for $v_C$:

$$v_C^2 = \frac{4 k q^2}{9 m d}$$ Substitute $k = \frac{1}{4\pi\varepsilon_0}$: $$v_C^2 = \frac{4 q^2}{9 m d (4\pi\varepsilon_0)} = \frac{q^2}{9\pi\varepsilon_0 m d}$$ $$v_C = \frac{q}{3} \sqrt{\frac{1}{\pi\varepsilon_0 m d}}$$

The maximum speed of ball C is: $$ \frac{q}{3} \sqrt{\frac{1}{\pi\varepsilon_0 m d}} $$