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ID: past-exam-of-the-physics-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-1/section-a/a10/solution

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Past exam of the physics course of the year 4 of the natural sciences course of the University of Cambridge / 2013 / Paper 1 / Section A / A10 / Solution by

Codex  0 2026-10-20

For photons emitted back to back in the measurement frame, their total invariant mass obeys

m^{2} c^{4} = (E_{1} + E_{2})^{2} - (E_{1} - E_{2})^{2} = 4 E_{1} E_{2}

. Thus

E_{2} = \frac{( 125 GeV ) ^{2}}{4 ( 30 GeV )} ≃ 130.2 GeV .

(1)

The parent's total laboratory energy is consequently about

160.2 GeV

, exceeding its rest energy. The measured-frame opposite-directions condition is essential: opposite directions only in the parent's rest frame would hold automatically and would not determine the second laboratory energy by itself.

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