 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ia/paper-4/10a/c/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

B

. Two-body decay kinematics gives the massless-particle energies

E_{1} = \frac{( m _{A}^{2} - m _{B}^{2} ) c ^{2}}{2 m _{B}}, E_{2} = \frac{( m _{B}^{2} - m _{C}^{2} ) c ^{2}}{2 m _{B}} .

θ

is the angle between their momenta, then

(Q_{1} + Q_{2})^{2} = 2 Q_{1} \cdot Q_{2} = \frac{2 E _{1} E _{2}}{c ^{2}} (1 - cos θ) .

Since

0 \leq 1 - cos θ \leq 2

0 \leq (Q_{1} + Q_{2})^{2} \leq \frac{( m _{A}^{2} - m _{B}^{2} ) ( m _{B}^{2} - m _{C}^{2} ) c ^{2}}{m _{B}^{2}} .

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