Solution to Question 50
Physics Principle: Conservation of Energy and Momentum.
1. Analysis of Final Configuration:
Initially, the four particles form a regular tetrahedron with side length \(l\). The thread connecting A and B is cut. Due to electrostatic repulsion, A and B move apart. The system will expand until the potential energy is minimized (maximum kinetic energy). This minimum potential energy state corresponds to the maximum possible separation between A and B, which occurs when the tetrahedron flattens into a planar rhombus (since the other 5 threads remain intact).
In this flat rhombus configuration:
- The triangles \(\Delta ACD\) and \(\Delta BCD\) become coplanar.
- The distance \(CD = l\) (fixed by thread).
- The distance \(AB\) becomes the sum of the altitudes of the two equilateral triangles: \( r_{AB} = 2 \times (\frac{\sqrt{3}}{2}l) = \sqrt{3}l \).
2. Momentum and Velocity Symmetry:
The system starts from rest, so the net momentum is zero.
By symmetry, the pair A and B moves outwards along one axis (say x-axis), and the pair C and D moves ‘inward’ towards the plane along the perpendicular axis (z-axis) as the structure collapses.
At the instant the structure becomes flat (\(z=0\)), the potential energy is at a minimum, implying kinetic energy is at a maximum.
Crucially, at this instant of maximum speed, the rate of change of the variable separation \(x\) (distance AB) is momentarily zero (since it is at a maximum). Thus, the velocity of A and B is purely perpendicular to the plane (z-direction). Similarly, C and D move purely in the z-direction.
Due to conservation of momentum and the symmetric masses:
$$ |\vec{v}_A| = |\vec{v}_B| = |\vec{v}_C| = |\vec{v}_D| = v $$
All four particles have the same speed \(v\).
3. Conservation of Energy:
Initial Potential Energy (6 edges of length \(l\)): $$ U_i = 6 \frac{k q^2}{l} $$ Final Potential Energy (5 edges of length \(l\), 1 edge of length \(\sqrt{3}l\)): $$ U_f = 5 \frac{k q^2}{l} + \frac{k q^2}{\sqrt{3}l} $$ Total Kinetic Energy: $$ K_{total} = 4 \times \left( \frac{1}{2} m v^2 \right) = 2 m v^2 $$ Equating gain in KE to loss in PE: $$ 2 m v^2 = U_i – U_f $$ $$ 2 m v^2 = \frac{k q^2}{l} \left( 6 – 5 – \frac{1}{\sqrt{3}} \right) = \frac{k q^2}{l} \left( 1 – \frac{1}{\sqrt{3}} \right) $$ $$ v^2 = \frac{k q^2}{2 m l} \left( 1 – \frac{1}{\sqrt{3}} \right) $$
Substitute \(k = \frac{1}{4\pi\epsilon_0}\): $$ v = \sqrt{ \frac{q^2}{8\pi\epsilon_0 m l} \left( 1 – \frac{1}{\sqrt{3}} \right) } $$