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Asked in Re-NEET 2026 · Discrete systems and motion of the centre of mass
Take θ as the angle of particle 1 from the horizontal diameter; by symmetry particle 2 is at π-θ.
Velocities: ⃗v₁=v(sin θ,-cos θ) and ⃗v₂=v(-sin θ,-cos θ).
The components across AB cancel and those along it add: P=2mv cos θ.
P=0 at θ=±π/2 and P=2mv at θ=0: a single hump, graph (3).
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