2017 USA Physics Olympiad Exam Part A 13 proton is given by mpc2=938 MeV and the mass of the A is given by mAc2=1232 MeV. Solution Restoring the factors of c and plugging in the numbers gives Ep=1.4×1020eV. This is known as the GZK bound for cosmic rays. The following relationships may be useful in solving this problem: velocity parameter B=吕 Lorentz factor y=√1- 1 relativistic momentum p=YBmc relativistic energy E=mc2 relativistic doppler shift 卡=V隔 Copyright C2017 American Association of Physics Teachers2017 USA Physics Olympiad Exam Part A 13 proton is given by mpc 2 = 938 MeV and the mass of the ∆ is given by m∆c 2 = 1232 MeV. Solution Restoring the factors of c and plugging in the numbers gives Ep = 1.4 × 1020 eV. This is known as the GZK bound for cosmic rays. The following relationships may be useful in solving this problem: velocity parameter β = v c Lorentz factor γ = √ 1 1−β2 relativistic momentum p = γβmc relativistic energy E = γmc2 relativistic doppler shift f f0 = q1−β 1+β Copyright c 2017 American Association of Physics Teachers