JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Conservation of Momentum

The three masses sit at the vertices of an equilateral triangle of side $a$, each launched with speed $V_0$ directed along one of the triangle's edges, in a cyclic pattern ($A$ along $AC$, $B$ along $BA$, $C$ along $CB$). By this three-fold symmetry, the mutual gravitational forces keep the configuration symmetric at all times, and all three masses converge on the centroid at the same instant.

Angular momentum about the centroid $G$ is conserved, since the gravitational forces between the particles are internal and produce no net external torque on the system. To find it at $t=0$ (which then holds all the way to the collision), note that the velocity of each mass lies exactly along the corresponding side of the triangle — the perpendicular distance from $G$ to that line is the triangle's inradius,


Each mass contributes angular momentum of magnitude $m{V}_{0}r$ about $G$, and by the cyclic symmetry all three contributions add in the same rotational sense:


This is the magnitude of the net angular momentum of the system at the point of collision.