Solution to Question 48
Physics Principle: Energy Conservation in Relative Motion (or Center of Mass frame).
Problem Statement:
One bead is projected with speed \(v\) towards another identical bead on a parallel rod (distance \(d\)). We need the minimum speed for the first bead to “pass over” (overtake/cross the x-coordinate of) the second bead.
Step 1: Condition for Passing
As the projected bead approaches, it repels the stationary bead. The first bead slows down, and the second bead accelerates.
The “most difficult” point to reach is the configuration of maximum potential energy, which occurs when the beads are side-by-side (distance between them is minimized to \(d\)).
If the system has enough kinetic energy to reach this state with a non-negative relative velocity, the first bead will pass the second.
Step 2: Conservation of Energy and Momentum
Let’s analyze this in the Laboratory Frame.
Initial State: Separation \(\infty\).
\(v_1 = v\), \(v_2 = 0\).
\(E_i = \frac{1}{2}mv^2\).
Momentum \(P = mv\).
Critical State (Side-by-side):
Separation \(d\).
By symmetry/conservation of momentum, at closest approach (or crossing point in the limit), the beads must move with the same velocity \(V\) to minimize kinetic energy loss to potential energy.
Conservation of Momentum: \(m v + 0 = m V + m V \implies 2mV = mv \implies V = v/2\).
Energy at Critical State:
$$ E_f = \frac{1}{2}m V^2 + \frac{1}{2}m V^2 + U_{elec} $$
$$ E_f = m (\frac{v}{2})^2 + \frac{kq^2}{d} = \frac{1}{4}mv^2 + \frac{kq^2}{d} $$
Step 3: Solve for Minimum Velocity
For the passing to occur, Initial Energy \(\ge\) Critical Final Energy. $$ E_i \ge E_f $$ $$ \frac{1}{2}mv^2 \ge \frac{1}{4}mv^2 + \frac{kq^2}{d} $$ $$ \frac{1}{4}mv^2 \ge \frac{kq^2}{d} $$ $$ v^2 \ge \frac{4kq^2}{md} $$ $$ v \ge \sqrt{\frac{4kq^2}{md}} $$
Substituting \( k = \frac{1}{4\pi\epsilon_0} \): $$ v \ge \sqrt{\frac{4q^2}{4\pi\epsilon_0 m d}} = \sqrt{\frac{q^2}{\pi\epsilon_0 m d}} $$