Three equal masses m$m$ are kept at vertices (A, B, C) of an equilateral triangle of side a$a$ in free space. At t = 0$t = 0$ , they are given an initial velocity vecV_mathrmA = V_0overrightarrowmathrmAC$\vec{V}_{\mathrm{A}} = V_{0}\overrightarrow{\mathrm{AC}}$ , vecV_mathrmB = V_0overrightarrowmathrmBA$\vec{V}_{\mathrm{B}} = V_{0}\overrightarrow{\mathrm{BA}}$ and vecV_mathrmC = V_0overrightarrowmathrmCB$\vec{V}_{\mathrm{C}} = V_{0}\overrightarrow{\mathrm{CB}}$ . Here, overrightarrowmathrmAC, overrightarrowmathrmCB$\overrightarrow{\mathrm{AC}}, \overrightarrow{\mathrm{CB}}$ and overrightarrowmathrmBA$\overrightarrow{\mathrm{BA}}$ 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:

### Related Formula vecL = sum left(vecr_i times m vecv_iright)$$\vec{L} = \sum \left(\vec{r}_i \times m \vec{v}_i\right)$$ vectau_textext = fracdvecLdt$$\vec{\tau}_{\text{ext}} = \frac{d\vec{L}}{dt}$$ ### Core Logic Since the three masses interact purely through mutual internal gravitational forces, the net external torque acting on the system about any central reference point is zero: vectau_textext = 0 implies vecL_textinitial = vecL_textfinal$$\vec{\tau}_{\text{ext}} = 0 \implies \vec{L}_{\text{initial}} = \vec{L}_{\text{final}}$$ Let us compute the total angular momentum about the centroid of the equilateral triangle:

From trigonometry, the perpendicular distance from the centroid to the velocity vector along any edge is: r_perp = fraca2sqrt3$$r_{\perp} = \frac{a}{2\sqrt{3}}$$ The initial angular momentum for one mass about the centroid is L_1 = m V_0 r_perp$L_1 = m V_0 r_{\perp}$. Since all three particles move cyclically in the same direction, their angular momenta reinforce cleanly: L_textnet = 3 cdot left(m V_0 fraca2sqrt3right) = frac32sqrt3 m V_0 a = fracsqrt32 a m V_0$$L_{\text{net}} = 3 \cdot \left(m V_0 \frac{a}{2\sqrt{3}}\right) = \frac{3}{2\sqrt{3}} m V_0 a = \frac{\sqrt{3}}{2} a m V_0$$ By conservation of angular momentum, this configuration value remains unchanged up to the point of collision. ### Pattern Recognition Mutual internal central forces can never alter the angular momentum of a system. Hence, the solution completely reduces to measuring the static configuration values at t=0$t=0$ about the center of mass symmetry. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion Class 11 Physics: Gravitation