 Solution

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

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

P = \sum_{i} Q_{i}

P \cdot P = M^{2} c^{2}

. Therefore

M = \frac{1}{c} (i \sum Q_{i}) \cdot (j \sum Q_{j}) .

In the parent rest frame,

M c^{2} = \sum_{i} E_{i}

E_{i} \geq m_{i} c^{2}

. Thus

M \geq i = 1 \sum N m_{i} .

 New to topics? Read the docs here!