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:
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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: Since the system is an equilateral triangle, the net angular momentum is calculated with respect to the center of mass. First, find the center of mass \( r \) of the system. For an equilateral triangle, the distance from each vertex to the center of mass is \( \frac{2r}{\sqrt{3}} \), where \( r \) is the side length.
\[
r = \frac{a}{\sqrt{3}}
\]
Step 2: The angular momentum of each mass is given by:
\[
L = mvr
\]
Where \( v \) is the velocity of each mass. The net angular momentum is the sum of the angular momentum of each mass.
\[
L_{\text{total}} = 3 \times m \times v_0 \times \frac{a}{\sqrt{3}} = \frac{\sqrt{3}}{2} m v_0
\]
Thus, the magnitude of the net angular momentum of the system at the point of collision is \( \frac{\sqrt{3}}{2} m v_0 \), so the correct answer is option (3).
