Three equal masses m are kept at vertices (A, B, C) of an equilateral triangle of side a in free space. At t=0, they are given an initial velocity V_A⃗=V₀ A C⃗, V_B⃗=V₀ B A⃗ and V_C⃗=V₀ C B⃗. Here, A C⃗, C B⃗ and B A⃗ are unit vectors along the edges of the triangle. If the three masses interact gravitationally, then the magnitude of the net angular momentum of the system at the point of collision is :

Solution

Physics sir · 4 steps

Gravity here is internal, so L never changes; find it at t = 0 about the centroid and you are done.

1. Centre of mass. The three velocities point along AC⃗, CB⃗ and BA⃗, which go round the triangle and add to zero. So the total momentum is zero and the centre of mass, the centroid, stays at rest.
2. L is conserved. Gravitational forces between the masses are internal and act along the lines joining them, so they give no net torque. The angular momentum at the moment of collision equals its value at t = 0.
3. Lever arm. Each velocity runs along one side, so its lever arm about the centroid is the inradius r = a/2√3. The motion goes A → C → B → A, so all three contributions have the same sense and add.

   L = 3 × m V₀ × a/2√3 = √3/2 a m V₀

4. So. The net angular momentum at collision is √3/2 a m V₀, option (d).

The idea to keep. With only internal central forces, L is conserved, so evaluate it at the easiest instant. If total momentum is zero, L is the same about every point.

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