A frictionless circular wire of unit radius is fixed on the...

Let the uniform angular speed of the particles be $\omega$.

The position vector of the particle on the right at an angle $\theta$ is given by $\vec{r}_1 = \cos\theta \hat{i} + \sin\theta \hat{j}$.

Since it moves from $\theta = \dfrac{\pi}{2}$ to $\theta = -\dfrac{\pi}{2}$, its angular velocity is $\dfrac{d\theta}{dt} = -\omega$.

The velocity of the first particle is $\vec{v}_1 = \dfrac{d\vec{r}_1}{dt} = (-\sin\theta \hat{i} + \cos\theta \hat{j})\dfrac{d\theta}{dt} = \omega\sin\theta \hat{i} - \omega\cos\theta \hat{j}$.

By symmetry, the position vector of the particle on the left is $\vec{r}_2 = -\cos\theta \hat{i} + \sin\theta \hat{j}$.

Its velocity is $\vec{v}_2 = \dfrac{d\vec{r}_2}{dt} = (\sin\theta \hat{i} + \cos\theta \hat{j})\dfrac{d\theta}{dt} = -\omega\sin\theta \hat{i} - \omega\cos\theta \hat{j}$.

The total linear momentum of the system is $\vec{P} = m\vec{v}_1 + m\vec{v}_2$.

Given $m = 1$, we have $\vec{P} = (\omega\sin\theta \hat{i} - \omega\cos\theta \hat{j}) + (-\omega\sin\theta \hat{i} - \omega\cos\theta \hat{j}) = -2\omega\cos\theta \hat{j}$.

The magnitude of the total linear momentum is $P = |\vec{P}| = 2\omega\cos\theta$ (since $\cos\theta \geq 0$ for $\theta \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$).

The graph of $P$ versus $\theta$ is a cosine curve which is zero at $\theta = \pm\dfrac{\pi}{2}$ and maximum at $\theta = 0$. This corresponds to the fourth graph.

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