Past exam of the mathematics course of the University of Cambridge / 2018 / ia / Paper 4 / 10A / a / Solution

The four-momentum isThusMultiplication by $c$ identifies the first two terms as the rest energy $m{c}^2$ and the Newtonian kinetic energy $m|\mathbf{u}{|}^2/2$.

For the fixed-target reaction, the initial invariant isAt threshold the four final particles are at rest in their centre-of-momentum frame, giving invariant $(4mc{)}^2$. Hence $\gamma =7$ and the least laboratory speed is