A 2.00-GeV proton hits another 2.00-GeV proton in a headon collision. (a) Calculate v, β, p, K, and E for each of the initial protons. (b) What happens to the kinetic energy?
Strategy (a) By the convention just discussed, a 2.00-GeV proton has a kinetic energy of 2.00 GeV. We use Equation (2.65) to determine the total energy and Equation (2.70) to determine momentum if we know the total energy. To determine β and v, it helps to first determine the relativistic factor γ, which we can use Equation (2.65) to find. Then we use Equation (2.62) to find β and v. These are all typical calculations that are performed when doing relativistic computations.
\beta^{2}=\frac{\gamma^{2}-1}{\gamma^{2}}=\frac{(1.049)^{2}-1}{(1.049)^{2}}=0.091 (2.62)
E=\gamma m c^{2}=\frac{m c^{2}}{\sqrt{1-u^{2} / c^{2}}} (2.65)
E^{2}=p^{2} c^{2}+E_{0}^{2} (2.70)
