A frictionless circular wire of unit radius is fixed on the horizontal plane. Two point particles of unit mass start moving simultaneously from point $A \left( \theta = \frac{\pi}{2} \right)$ with identical uniform angular speeds in opposite directions, and meet again at point $B \left( \theta = -\frac{\pi}{2} \right)$. During this time, which of the following figures schematically represent the magnitude of the total linear momentum $\vec{P}$ of the system, as a function of $\theta$? 
To find the magnitude of the total linear momentum \(\vec{P}\) of the system as a function of \(\theta\), we first need to understand the motion of the two particles.
- Both particles start from point \(A \left( \theta = \frac{\pi}{2} \right)\) and move with identical uniform angular speeds but in opposite directions. They meet at point \(B \left( \theta = -\frac{\pi}{2} \right)\).
- Since both particles have unit mass and move with a uniform speed, the velocity magnitudes are equal. Let the linear speed of each particle be \(v\). Then, at any position \(\theta\), the particles are diametrically opposite each other.
- The linear momentum of a single particle is \(\vec{p}_i = m\cdot \vec{v}_i\), where \(m=1\) for both particles.
- Using trigonometric components, for a particle at angle \(\theta\):
- The velocity \(\vec{v}_1\) for one particle will have components along the tangent to the circle as \((-v \sin \theta, v \cos \theta)\).
- The opposite particle at \(\theta + \pi\) will have velocity \(\vec{v}_2\) as \((v \sin \theta, -v \cos \theta)\).
- The total linear momentum \(\vec{P} = \vec{p}_1 + \vec{p}_2\) will have components that add up to zero because the contributions in each direction cancel out.
- Thus, the magnitude of the total linear momentum, \(|\vec{P}|\), remains constant as the particles move, since the momenta perfectly cancel each other. Therefore, \(|\vec{P}| = 2 \times \text{(constant component on line of action)}\).
From the explanation, the magnitude of the total linear momentum does not vary with \(\theta\); it remains constant throughout the motion. Hence, the correct graph representing this scenario is a horizontal line at a constant value.
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