JEE Physics 2025 Question with Solution

Given: Three equal masses $m$ are placed at the vertices of an equilateral triangle of side $a$. Their initial velocities are along the sides as given. By symmetry, the particles move toward the centroid and collide there.

Find: The magnitude of the net angular momentum of the system at the point of collision.

Angular momentum of the system is conserved because the gravitational interaction is internal and central. So it is enough to calculate the initial angular momentum about the center of mass.

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

$$r = \frac{a}{\sqrt{3}}$$

Each velocity is along a side, and the angle between the radius vector from centroid to a vertex and the corresponding velocity is $30^\circ$. Therefore the perpendicular distance from the centroid to the line of motion is

$$r_\perp = r \sin 30^\circ = \frac{a}{\sqrt{3}} \cdot \frac{1}{2} = \frac{a}{2\sqrt{3}}$$

So angular momentum of one particle about the centroid is

$$L_1 = m V_0 r_\perp = m V_0 \frac{a}{2\sqrt{3}}$$

All three angular momenta are in the same sense, hence

$$L_{\text{net}} = 3L_1 = 3 \times m V_0 \frac{a}{2\sqrt{3}} = \frac{\sqrt{3}}{2} a m V_0$$

Therefore, the magnitude of the net angular momentum at collision is $\frac{\sqrt{3}}{2} a m V_0$. The correct option is C.

Given: The three masses start from the vertices of an equilateral triangle and interact only through mutual gravitation.

Find: Net angular momentum when they collide.

Because the forces are internal, the total torque about the center of mass is zero. Hence total angular momentum about the center of mass remains constant throughout the motion, including at collision.

The center of mass of three equal masses at the vertices of an equilateral triangle is the centroid. Its distance from each vertex is

$$r = \frac{\sqrt{3}}{3}a = \frac{a}{\sqrt{3}}$$

Now consider one particle. Its speed is $V_0$ and its velocity is directed along a side of the triangle. The angle between the radius from centroid to the vertex and that side is $30^\circ$, so only the component perpendicular to the radius contributes:

$$V_\perp = V_0 \sin 30^\circ = \frac{V_0}{2}$$

Thus the angular momentum of one mass is

$$L_1 = m r V_\perp = m \cdot \frac{a}{\sqrt{3}} \cdot \frac{V_0}{2}$$

Therefore,

$$L_1 = \frac{a m V_0}{2\sqrt{3}}$$

For three identical masses, these contributions add:

$$L_{\text{net}} = 3 \cdot \frac{a m V_0}{2\sqrt{3}} = \frac{\sqrt{3}}{2} a m V_0$$

Hence, the net angular momentum at the point of collision is $\frac{\sqrt{3}}{2} a m V_0$.