Past exam of the physics course of the year 4 of the natural sciences course of the University of Cambridge / 2015 / Paper 1 / Section A / A6 / Solution

For total relativistic energy $E=\gamma m{c}^2$, the speed is $v=c\sqrt{1-(m{c}^2/E{)}^2}$. Both beam energies are enormously greater than the rest energy, so expanding the square root avoids obscuring the small difference:With all energies in GeV, $m{c}^2=0.9383$, $E_1=4000$ and $E_2=6500$, givingThe deficits below the speed of light are about $8.25\mathrm{m}{\mathrm{s}}^{-1}$ and $3.12\mathrm{m}{\mathrm{s}}^{-1}$. The large energy increase therefore produces only a small speed increase; the energy mainly changes the Lorentz factor. These numbers use the quoted beam energies as total energies. Interpreting them as kinetic energies requires adding the rest energy, with only a tiny correction at these scales.