 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2001/ib/paper-4/18f/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

Conservation of momentum in the parent's rest frame gives

p_{A} + p_{B} = 0

. Equal daughter masses and the relativistic energy-momentum relation then give equal daughter energies. Energy conservation fixes

E_{A} = E_{B} = \frac{M c ^{2}}{2} .

Using

E^{2} = p^{2} c^{2} + m^{2} c^{4}

and the specified direction of A,

p_{A} = - \frac{c}{2} M^{2} - 4 m^{2}, p_{B} = + \frac{c}{2} M^{2} - 4 m^{2} .

The velocities follow from

v = p c^{2} / E

v_{A} = - V, v_{B} = V, V = c 1 - 4 (m / M)^{2} .

These are spatial momenta in the original inertial frame. For positive daughter rest mass,

m < M /2

gives

0 < V < c

 New to topics? Read the docs here!