Solution
Given:
- Initial separation $r_0 = 60 \text{ cm}$.
- Angle of velocity vectors with the line joining them: $\theta = 45^\circ$.
- Condition: Total Kinetic Energy ($K_0$) = Potential Energy ($U_0$).
We need to find the distance of closest approach ($r_{min}$).
Step 1: Energy Analysis using Radial Component
The particles are approaching each other. The velocity can be resolved into two components:
- Radial component (along the line joining): $v_r = v \cos 45^\circ$
- Tangential component (perpendicular): $v_t = v \sin 45^\circ$
At the distance of closest approach ($r_{min}$), the radial velocity becomes momentarily zero. The kinetic energy associated with the radial component is converted entirely into electrostatic potential energy.
Initial State:
The problem states: $K_{total} = U_{total}$.
$$ 2 \left( \frac{1}{2} m v^2 \right) = \frac{k q^2}{r_0} \implies m v^2 = \frac{k q^2}{r_0} $$Step 2: Conservation of Energy
We can write the energy conservation equation focusing on the change in radial energy:
$$ U_{initial} + K_{radial} = U_{final} $$Note: The tangential kinetic energy is often treated as conserved (associated with angular momentum barrier) or non-participating in the “approach” in simplified collision approximations. Here, the energy associated with closing the gap comes from the radial velocity.
$$ \frac{k q^2}{r_0} + 2 \left( \frac{1}{2} m (v \cos 45^\circ)^2 \right) = \frac{k q^2}{r_{min}} $$Substitute $\cos 45^\circ = 1/\sqrt{2}$:
$$ \frac{k q^2}{r_0} + m v^2 \left( \frac{1}{2} \right) = \frac{k q^2}{r_{min}} $$From Step 1, we know $m v^2 = \frac{k q^2}{r_0}$. Substitute this into the equation:
$$ \frac{k q^2}{r_0} + \frac{1}{2} \left( \frac{k q^2}{r_0} \right) = \frac{k q^2}{r_{min}} $$Cancel $k q^2$ from both sides:
$$ \frac{1}{r_0} + \frac{1}{2 r_0} = \frac{1}{r_{min}} $$ $$ \frac{3}{2 r_0} = \frac{1}{r_{min}} $$ $$ r_{min} = \frac{2}{3} r_0 $$Given $r_0 = 60 \text{ cm}$:
$$ r_{min} = \frac{2}{3} \times 60 = 40 \text{ cm} $$Correct Option: (c) 40 cm