Two particles of equal mass m start moving simultaneously from the...

Let the particles start from the position $(0, R)$ on the circular track. At any instant, let each particle have covered an angular distance $\phi$.

The velocity of the first particle (moving clockwise) is $\vec{v}_1 = v\cos\phi \hat{i} - v\sin\phi \hat{j}$.

The velocity of the second particle (moving counter-clockwise) is $\vec{v}_2 = -v\cos\phi \hat{i} - v\sin\phi \hat{j}$.

The total linear momentum of the system is $\vec{P} = m\vec{v}_1 + m\vec{v}_2 = -2mv\sin\phi \hat{j}$.

The magnitude of the total momentum is $P = 2mv\sin\phi$.

We are given that $P$ equals the momentum of a single particle, which is $mv$.

Therefore, $2mv\sin\phi = mv$

$\sin\phi = \dfrac{1}{2}$

$\phi = \dfrac{\pi}{6}$.

A common mistake is assuming the angle between the velocity vectors is $2\phi$ such that $P = 2mv\cos\phi$, which incorrectly leads to $\phi = \pi/3$.

Answer: $\dfrac{\pi}{6}$