JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Conservation of Momentum

Let each body have mass $m$ and speed $v_0$, with the angle between their velocities being $2\theta$. The collision is perfectly inelastic, so they move off together, and by symmetry the combined velocity lies along the bisector of the original directions. Conserving momentum along that bisector, $2(m{v}_0\cos \theta )=(2m)\left(\frac{v_0}{2}\right)$, since the final common speed is half the initial. This gives $\cos \theta =\frac{1}{2}$, so $\theta =60^{\circ }$ and the angle between the initial velocities is $2\theta =120^{\circ }$.