Question

Two protons, each having a speed of 0.930c in the laboratory, are moving toward each other.

Part A Determine the momentum of each proton in the laboratory. Enter your answers numerically separated by a comma. |p1|, |p2| =   GeV/c   Part B Determine the total momentum of the two protons in the laboratory. |p1+p2| = GeV/c   Part C Determine the momentum of one proton as seen by the other proton. |p12| = GeV/c

Answer 1

Part A) P = Mv = mv/sqrt(1 - (v/c)^2) = m .93c/sqrt(1 - .93^2) = 1836e/c^2 .93c/sqrt() = (.93*1836*.511E6/299.79E6)/sqrt(1 - .93^2) = 7.92 eV.s/m

Note m = 1836e/c^2 where e = .511E6 eV for electron rest mass equivalent. That is, the proton rest mass is roughly 1836 times more massive than an electron.

Part B) 2*7.92 eV.s/m = 15.84 eV.s/m

Part C) We have W = (2*.93)/(1 + .93^2)*C = .997C is the relative speed; so P' = = (.997*1836*.511E6/299.79E6)/sqrt(1 - .997^2) = 40.3 eV.s/m