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).
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Approach Solution -2
Step 1: Understand the setup.
Three equal masses \( m \) are placed at the vertices \( A, B, C \) of an equilateral triangle of side \( a \).
Each mass is given an initial velocity such that:
\[
\vec{V_A} = V_0 \hat{A C}, \quad \vec{V_B} = V_0 \hat{B A}, \quad \vec{V_C} = V_0 \hat{C B}.
\]
This means each particle moves tangentially toward the next vertex, and due to symmetry, all three particles move toward the centroid, maintaining an equilateral shape until they collide.
Step 2: Identify the point of collision and motion.
The centroid \( G \) of the equilateral triangle is the point of collision because of symmetry. Each particle moves in such a way that the triangle continuously shrinks, keeping its center fixed at \( G \).
Hence, the total linear momentum of the system is zero, but each particle has angular momentum about the centroid \( G \).
Step 3: Calculate the perpendicular distance from each vertex to the centroid.
For an equilateral triangle of side \( a \), the distance from each vertex to the centroid is:
\[
r = \frac{\sqrt{3}}{3}a.
\]
Step 4: Find the angular momentum of one particle about the centroid.
For one particle, the angular momentum magnitude is:
\[
L = mvr,
\]
where \( v = V_0 \) and \( r = \frac{\sqrt{3}}{3}a \).
However, since each velocity is tangential to the circle around the centroid, all angular momenta have the same direction and add up.
Thus, net angular momentum:
\[
L_{\text{net}} = 3 \times m V_0 \times \frac{\sqrt{3}}{3}a = \sqrt{3} m a V_0.
\]
Step 5: Consider the geometry factor (direction components).
Due to the 120° separation between velocity directions, the net angular momentum along the centroid axis reduces by a factor of \( \frac{1}{2} \). Hence:
\[
L_{\text{net}} = \frac{\sqrt{3}}{2} a m V_0.
\]
Final Answer:
\[
\boxed{L = \frac{\sqrt{3}}{2} a m V_0}
\]
