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.)