Particle Dacay with four vectors (not too hard)

[Zeebo]

Problem:
Given that the $$\Lambda$$ has Q=0, B=1, S=-1 and the K(+) has Q=1, B=0, S=1 and the K− is its
antiparticle, show that the following reactions are suitable for making strange particles. Calculate
the threshold kinetic energy of the beam for producing these reactions in a ‘fixed target’ experiment.
(a) p + p --> $$\Lambda$$ + K(+) + p
(b) p + p --> K+ + K− + p + p

Attempt at a Solution:
I had no trouble showing that the two sides conserve Q,B and S. I'm simply having trouble setting up the four vectors for part 2. If someone could simply show me the setup for a), including the components of the four vectors, I should be fine.

 

[NateD]

I'm not sure what the momentum of the target is, so I'm not sure how to calculate it.Solution: Part (a): The four vectors for this reaction are: Beam: p + P --> \Lambda + K(+) + pInitial state: (p1,m1), (P2,m2)Final state: (\Lambda3,m3), (K(+)4,m4), (p5,m5)Conservation of four-momentum requires that: p1 + P2 = \Lambda3 + K(+)4 + p5The threshold kinetic energy of the beam can be found by equating the total four-momentum of the initial and final states and solving for E1, the kinetic energy of the beam. p1 + P2 = \Lambda3 + K(+)4 + p5 E1 + m1 + E2 + m2 = E3 + m3 + E4 + m4 + E5 + m5 E1 = (E3 + m3 + E4 + m4 + E5 + m5) - (m1 + E2)Part (b): The four vectors for this reaction are: Beam: p + P --> K+ + K− + p + p Initial state: (p1,m1), (P2,m2)Final state: (K(+)3,m3), (K(-)4,m4), (p5,m5), (p6,m6)Conservation of four-momentum requires that: p1 + P2 = K(+)3 + K(-)4 + p5 + p6 The threshold kinetic energy of the beam can be found by equating the total four-momentum of the initial and final states and solving for E1, the kinetic energy of the beam. p1 + P2 = K(+)3 + K(-)4 + p5 + p6 E1 + m1 + E2