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 \( \vec{V_A} = V_0 \hat{AC}, \, \vec{V_B} = V_0 \hat{BA}, \, \vec{V_C} = V_0 \hat{CB} \).Here, \( \hat{AC}, \hat{CB}, \hat{BA} \) 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:
Show Hint
In problems involving angular momentum of a system of particles, remember to calculate the angular momentum of each particle and then sum them up. For systems with symmetry like an equilateral triangle, the center of mass can simplify the calculation.
Step 1: As the system forms an equilateral triangle, the net angular momentum is computed relative to its center of mass. Initially, determine the system's center of mass \( r \). For an equilateral triangle, the distance from any vertex to the center of mass is \( \frac{2r}{\sqrt{3}} \), with \( r \) representing the side length.
\[r = \frac{a}{\sqrt{3}}\]
Step 2: The angular momentum of each individual mass is expressed as:
\[L = mvr\]
Here, \( v \) denotes the velocity of each mass. The net angular momentum is obtained by summing the angular momentum of all masses.
\[L_{\text{total}} = 3 \times m \times v_0 \times \frac{a}{\sqrt{3}} = \frac{\sqrt{3}}{2} m v_0\]
Consequently, the magnitude of the system's net angular momentum at the point of collision is \( \frac{\sqrt{3}}{2} m v_0 \), aligning with option (3).
