Consider two identical metallic spheres of radius R each having charge Q and mass m. Their centers have an initial separation of 4R. Both the spheres are given an initial speed of u towards each other. The minimum value of u, so that they can just touch each other is:
(Take k=1/4πε₀ and assume kQ²>Gm² where G is the Gravitational constant)

Solution

Physics sir · 4 steps

Both spheres stop together at the moment they touch, so all their kinetic energy goes into potential energy.

1. Why both stop. The spheres start with equal and opposite momenta, so the total momentum is zero throughout. For the minimum u, they reach contact with zero speed. The centres are then 2R apart.
2. Net potential energy. At separation r, the electric repulsion gives +kQ²/r and the gravitational attraction gives -Gm²/r. So the total is U = (kQ² - Gm²)/r, which is positive because kQ² > Gm².

   2 × 1/2mu² = (kQ² - Gm²)(1/2R - 1/4R) = (kQ² - Gm²)/4R

3. Solve for u. This gives u² = (kQ² - Gm²)/4mR = kQ²/4mR(1 - Gm²/kQ²).
4. So. u = √(kQ²/4mR(1 - Gm²/kQ²)), which is option (d).

The idea to keep. When two bodies approach each other from rest-frame symmetry, total momentum is zero. Count the kinetic energy of both bodies, and add every interaction, including gravity, into the potential energy.

Answer from the NTA final answer key. Solution by PrepWiser.