Question

Head-on collisions are observed between protons each moving with velocity (relative to the fixed observer) corresponding to 10¹⁰ eV. If one of the protons were to be at rest relative to the observer, what would the energy of the other need to be so as to produce the same collision energy as before. The rest energy of the proton is 10⁹eV.

Answer 1

When the protons travel toward each other with equal energy, and therefore with the same speed, their net momentum is zero. In that case the Lab system is reduced to the C-M system, and the observer is watching the events sitting in the C.M. system. The total energy is then, 

E^∗ = 10¹⁰eV + 10¹⁰eV + 10⁹ eV + 10⁹ eV = 22 × 10⁹ or 22 GeV 

Let E₁ be the energy of a proton in the lab system moving toward the target proton originally at rest, then if the total energy available in the CMS has to be the same as E^∗ = 22 GeV, 

(m₁² + m₂² + 2E₁ M₁)^1/2 

= E^∗ = 22 GeV 

(1² + 1² + 2E₁ × 1)^1/2 = 22 

Or E₁ = 241 GeV 

Therefore required kinetic energy = 241 − 1 = 240GeV