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**Summary:**Finding the KE of a two proton collision that creates Kaons. Given the rest KE of protons and kaons, what is the minimum KE of one proton that can create the two kaons.

In high-energy physics, new particles can be created by collisions of fast-moving projectile particles with stationary particles. Some of the kinetic energy of the incident particle is used to create the mass of the new particle. A proton-proton collision can result in the creation of a negative kaon (K−) and a positive kaon (K+):

p+p→p+p+K−+K+

Part A:

Calculate the minimum kinetic energy of the incident proton that will allow this reaction to occur if the second (target) proton is initially at rest. The rest energy of each kaon is 493.7 MeV, and the rest energy of each proton is 938.3 MeV. (

*Hint:*It is useful here to work in the frame in which the total momentum is zero. Note that here the Lorentz transformation must be used to relate the velocities in the laboratory frame to those in the zero-total-momentum frame.)

Part C: Suppose that instead the two protons are both in motion with velocities of equal magnitude and opposite direction. Find the minimum combined kinetic energy of the two protons that will allow the reaction to occur.

So I'm trying to solve part A before doing part C. The answer given for part A is 2494 MeV, but I can't seem to get the right answer. I believe my conceptual understanding of the solution is not correct.

I start with conservation of energy.

E(p) + E(p) = E(p) + E(p) + E(k) + E(k)

Since, only one proton is moving, I can say:

νmc^2 + 938.3 = 938.3 + 938.3 + 493.7 + 493.7

Solving for lorentz transformation gives me ν = 2.05.

Using ν, I plug it into the KE equation:

K=(ν-1)mc^2 * (1 MeV / 1.6*10^-13 J) = 983.4 MeV

983.4 MeV ≠ 2494 MeV

What am I doing wrong?

**[Moderator's note: Moved from a technical forum and thus no template.]**

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