Given below are two statements: one is labelled as Assertion (A)...

Let the particles start from $\theta = 0$ and move with angular speed $\omega$ on a track of radius $R$.

The position vectors of the two particles at time $t$ are $\vec{r}_1 = R\cos(\omega t)\hat{i} + R\sin(\omega t)\hat{j}$ and $\vec{r}_2 = R\cos(\omega t)\hat{i} - R\sin(\omega t)\hat{j}$.

The position of the center of mass is $\vec{r}_{cm} = \dfrac{m\vec{r}_1 + m\vec{r}_2}{2m} = R\cos(\omega t)\hat{i}$.

The acceleration of the center of mass is $\vec{a}_{cm} = \dfrac{d^2\vec{r}_{cm}}{dt^2} = -\omega^2 R\cos(\omega t)\hat{i}$.

The particles meet again when they have each covered an angle of $\pi$, i.e., $\omega t = \pi$. At this instant, the magnitude of the acceleration of the center of mass is $|\vec{a}_{cm}| = \omega^2 R |\cos(\pi)| = \omega^2 R$, which is its maximum possible value. Alternatively, the net external force on the system is the vector sum of the centripetal forces provided by the track. When the particles meet, these forces are parallel and add up to a maximum value, leading to maximum acceleration of the center of mass. Thus, Assertion (A) is true.

The velocity of the center of mass is $\vec{v}_{cm} = \dfrac{d\vec{r}_{cm}}{dt} = -\omega R\sin(\omega t)\hat{i}$.

At the instant they meet ($\omega t = \pi$), $\vec{v}_{cm} = 0$, so the total linear momentum of the system is zero. Thus, Reason (R) is true.

However, the total linear momentum being zero (a kinematic condition of velocity) is not the cause of the acceleration being maximum. The maximum acceleration is due to the alignment of the external centripetal forces. Therefore, Reason (R) does not correctly explain Assertion (A).

Answer: Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A)