1. Key idea: angular momentum about the center of mass is conserved

Since the masses interact only through internal gravitational forces, the net external torque on the system is zero. Hence, the total angular momentum of the system about the center of mass remains constant.

At the instant of collision, all three particles meet at one point. The angular momentum of the system about that collision point is the same as the angular momentum about the center of mass, because the collision point is the center of mass of the system.

So we only need to calculate the initial angular momentum about the center of mass.

2. Geometry of the equilateral triangle

Let the center of mass (which is also the centroid for equal masses) be $O$.

For an equilateral triangle of side $a$, the distance from centroid to each vertex is

Place the triangle so that the vertices are at angular positions $0^\circ,120^\circ,240^\circ$ around $O$.

Each mass has speed $V_0$, directed along the side:

• at $A$: along $AC$
• at $B$: along $BA$
• at $C$: along $CB$

By symmetry, each velocity makes the same angle with the corresponding radius vector from $O$.

3. Find the angle between radius vector and velocity

In an equilateral triangle, the line from a vertex to the centroid bisects the $60^\circ$ angle there. So the radius vector $\overrightarrow{OA}$ makes $30^\circ$ with each side through $A$.

But the position vector of mass at $A$ relative to the center is $\overrightarrow{OA}$, while its velocity is along $AC$. Therefore the angle between $\overrightarrow{OA}$ and $\overrightarrow{V_A}$ is

Similarly for the other two masses.

Hence angular momentum magnitude of each particle about $O$ is

4. Direction of angular momentum of each mass

Now check whether the three angular momenta add or cancel.

The directions of velocities are cyclic: $A\to C$, $C\to B$, $B\to A$. This corresponds to the same sense of rotation for all three particles about the center. Therefore, all three angular momentum vectors are parallel and add directly.

So total angular momentum is

5. Angular momentum at collision

Since angular momentum is conserved, the magnitude of the net angular momentum at the point of collision is the same:

6. Option check

• A: $\frac12 amV_0$ ❌
• B: $3amV_0$ ❌
• C: $\frac32 amV_0$ ❌
• D: $\frac{\sqrt{3}}{2}amV_0$ ✅

Therefore, the correct answer is Option D.