Class 11 NEET Physics System of Particles and Rotational Motion

Concept:The net linear momentum is the vector sum of momenta of the two particles, which depends on their positions on the circular path.Explanation:Both particles have mass $m=1$ and uniform speed $v$.Let one particle move clockwise and the other anticlockwise from $\theta=\pi/2$ to $\theta=-\pi/2$.At an angle $\theta$, their positions are symmetric about the vertical axis.The velocity vectors are tangential to the circle.For the particle on the right side, velocity components: $v_x = -v\sin\theta$, $v_y = -v\cos\theta$.For the particle on the left side, velocity components: $v_x = v\sin\theta$, $v_y = -v\cos\theta$.So net momentum: $\vec{P}_{\text{net}} = (-v\sin\theta + v\sin\theta)\hat{i} + (-v\cos\theta - v\cos\theta)\hat{j} = -2v\cos\theta\,\hat{j}$.Magnitude: $P = 2v\,|\cos\theta|$.At $\theta=\pi/2$ and $\theta=-\pi/2$, $\cos\theta=0$, so $P=0$.At $\theta=0$, $\cos\theta=1$, so $P=2v$ (maximum).So the graph of $P$ vs $\theta$ is symmetric, starting at zero, rising to a peak at $\theta=0$, then falling back to zero.Answer:The graph that shows this shape (symmetric, zero at both ends, maximum at $\theta=0$) is option B.