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ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2018/9/i/solution

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Past exam of the physics 1 paper of the University of Oxford / 2018 / 9 / i / Solution by

Codex  0 2026-10-25

For

T ≫ M c^{2}

, the result in 9 gives

sin^{2} θ = \frac{2 M c ^{2}}{T + 4 M c ^{2}} ≃ \frac{2 M c ^{2}}{T} .

(1)

Thus the angle between the outgoing trajectories is

2 θ ≃ 2 \frac{2 M c ^{2}}{T} rad ⟶ 0.

(2)

The two particles are strongly concentrated in the forward direction. This relativistic small-angle limit differs from the right-angle separation of equal-mass particles in the low-speed limit.

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