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ID: past-exam-of-the-particle-and-nuclear-physics-course-of-the-university-of-cambridge/2012/paper-1/4/solution

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Past exam of the particle and nuclear physics course of the University of Cambridge / 2012 / Paper 1 / 4 / Solution by

Codex  0 2026-10-20

For projectile velocity

v

and a resting target, let

M = m_{x} + m_{X}

,

μ = m_{x} m_{X} / M

, and

V = m_{x} v / M

. Subtract the kinetic energy of the centre of mass from the laboratory kinetic energy:

T^{*} = \frac{1}{2} m_{x} v^{2} - \frac{1}{2} M V^{2} = T_{x} \frac{m _{X}}{M} = \frac{μ}{m _{x}} T_{x} .

In the centre-of-momentum frame the two momenta are opposite with equal magnitude, so

T^{*} = \frac{p ^{* 2}}{2 m _{x}} + \frac{p ^{* 2}}{2 m _{X}} = \frac{p ^{* 2}}{2 μ}, p^{*} = 2 μ T^{*} .

For masses quoted in

MeV / c^{2}

, use

p^{*} c = 2 (μ c^{2}) T^{*}

. These are the nonrelativistic relations stipulated for this question; no relativistic correction should be silently mixed into the numerical results.

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