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Homework Help: Two protons collding

  1. Oct 31, 2007 #1
    1. The problem statement, all variables and given/known data
    I have to protons coliding - they have same magnitude of velocity and momentum and they collide head-on:

    p + p -> p + (p + p_) (the "p_" is an anti-proton).

    I have to find the minimum combined kinetic energy of the two protons for this process to run.

    3. The attempt at a solution

    I use P_total = P_1 + P_2 <=>

    (all the masses squared ...) * c^2 = (P_1 + P_2)^2.

    But this is where I encounter my problem. The 4-vector for e.g. P_1 is [E/c ; gamma*m*v] - I don't know v?

    Is there another way of doing this?

    - another thing: When I have found the total energy, do I have to subtract it with the rest mass of a proton x 2 (2*m_p*c^2) to find combined kinetic energy ?
     
    Last edited: Oct 31, 2007
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  3. Oct 31, 2007 #2

    Avodyne

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    You don't know v because that's what you're trying to find: if you knew v, you would know the total energy (and yes you have to subtract the rest energy to get the kinetic energy).

    You've written P_1 in components; what is P_2 ? Then, what is P_1 + P_2 ?
     
  4. Oct 31, 2007 #3
    I want to find E - finding E and v is overkill, isn't it?

    1) P_2 = [E/c ; -gamma*m*v]

    2) P_1 + P_2 = 2E/c?
     
    Last edited: Oct 31, 2007
  5. Oct 31, 2007 #4

    Avodyne

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    Yes, finding v is overkill. And more precisely, P_1 + P_2 = [2E/c; 0].
     
  6. Nov 1, 2007 #5

    clem

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    "p + p -> p + (p + p_) (the "p_" is an anti-proton)"

    You must mean -->p+p+(p+p_)
    You need a cm energy W=4M.
    The initial cm energy is W=(T+2M).
    Solve for T.
     
  7. Nov 1, 2007 #6
    So I get that T = 2*M?

    I have another question, related to the first post. If one of the protons are at rest, I have to find the kinetic energy for the process to begin:

    I would use that P_1 + P_2 = P_f (of the 4 protons) - so

    2*m^2*c^2 + 2*E*m = 4m^2 * c^2 (I have squred it)

    Two questions for that equation:

    1) Is it correct?

    2) When I have 4m^2 * c^2, how do I do this? (938 MeV/c^2)^2 * c^2 or what? So 4m^2 * c^2 = 938 ^2?
     
    Last edited: Nov 1, 2007
  8. Nov 1, 2007 #7
    If I do what Clem did in his post for the second scenario (where one proton is in rest, other is moving), I get that:

    W = 4M

    W_cm = (T+M), so T = 3M? Does that make sence?
     
  9. Nov 1, 2007 #8
    Ok, the two questions in my post are solved (#1 and #2). Now the original question - where the 2 protons have equal magnitude momentum - still remains.

    This is what I have so far:

    (P_1 + P_2)^2 = P_f^2 <=>

    (2E/c;0)^2 = 16*m^2*c^2 <=>

    4E^2/c2^2 = 16*m^2*c^2, where I find the total energy E and use: E - 2*m*c^2 = E_kinetic.

    Is this correct? I can use this method for all particles as long as I change the masses, right?
     
  10. Nov 1, 2007 #9
    Guys, I solved the problem.. I forgot that I found the total energy for EACH proton, not the total total-energy for both particles.. that's why my result didn't add up.

    Sorry for posting so many times, but all my problems are solved now - thanks.
     
  11. Nov 1, 2007 #10

    clem

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    "I have another question, related to the first post. If one of the protons are at rest, I have to find the kinetic energy for the process to begin:"

    If p_2=0 (at rest), then
    W^2=(E_1+M)^2-(p_1)^2=2M^2+2ME_1.
    This equals (4M)^2.
    Solve for E_1. Then T_1=E_1-M=6M.
     
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