In an accelerator, two protons with equal kinetic energies collide head-on. The following reaction takes place: $\mathrm{p}+\mathrm{p} \rightarrow \mathrm{p}+\mathrm{p}+\mathrm{p}+\overline{\mathrm{p}} .$ What is the minimum possible kinetic energy of each of the incident proton beams?

Initially, we have two protons colliding head-on: \(p_1 + p_2\) After the collision, we have three protons and one antiproton: \(p_3 + p_4 + p_5 + \overline{p_6}\)

For the final state, we have an additional proton-antiproton pair. The rest mass of a proton (and an antiproton) is approximately \(938 \,\text{MeV}/c^2\). Therefore, the total rest mass energy of the additional proton-antiproton pair is: \(E_{rest} = 2 \times (938 \,\text{MeV}/c^2) = 1876 \,\text{MeV}\)

The minimum kinetic energy required for the collision to happen requires the final state particles to be at rest. In this case, the conservation of momentum states that the initial momenta must sum up to zero: \(p1_{initial} + p2_{initial} = 0\)

The minimum possible kinetic energy is when the final particles are all at rest, and thus the initial kinetic energy of both protons must be at least equal to the total rest mass energy of the additional proton-antiproton pair: \(\frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 \geq E_{rest}\) Since the initial protons have equal kinetic energies and their momenta sum up to zero, we have: \(2 \times \frac{1}{2}m_1v^2 \geq E_{rest}\) From the rest mass energy of the proton-antiproton pair, we determined that \(E_{rest} = 1876 \,\text{MeV}\). Therefore: \(mv^2 \geq 1876\,\text{MeV}\) To find the minimum kinetic energy of each proton, divide both sides by 2: \(\frac{1}{2}mv^2 \geq 938 \,\text{MeV}\) So the minimum possible kinetic energy of each of the incident proton beams is approximately \(938 \,\text{MeV}\).