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Homework Help: Collision of a moving proton into a proton at rest (relativity)

  1. May 9, 2010 #1
    1. The problem statement, all variables and given/known data
    A high speed proton collides with a proton at rest, two proton-antiproton pairs were created. That is some of the accelerated protons kinetic energy was enough to be converted into the mass of the new particles. If the proton-antiproton pairs move along together as a single particle of rest mass M=4m show that hte kinetic energy of the accelerated proton before collision is:

    K=6mc2


    2. Relevant equations

    relativistic momentum: p=gamma(mv)
    Relativistic kinetic energy: K=(gamma-1)mc2
    energy as a function of momemtum: E2=(pc)2 + (mc2)2


    3. The attempt at a solution
    Heres my train of thought....So the kinetic energy of the moving proton must equal the mass energy of 2 additional particles in the product of the collision but those products ( the proton antiproton pairs) are also moving so thats additional kinetic energy. The reactancts have a mass of 2m and the products have a mass of 4m. The total momemtum must be the same since its conserved.

    I tried figuring out the speed of the intital moving proton and the speed of the blob of 4 particles and got :

    4u2 + v2=3c2

    where u is the velocity of the proton and the v is the velocity of the 4 particle blob.

    Then Iam not sure to procede from there. I am assuming that all the mass of the proton at rest was converted into energy but it doesnt really state that. It just seems wierd because the intital proton can moving at any speed greater than the least amount of speed required to create 4 particles that are moving...if that makes sense since it says that SOME of the accelerated protons kinetic energy was converted.

    Thank you for your precious time.
     
    Last edited: May 9, 2010
  2. jcsd
  3. May 10, 2010 #2

    vela

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    This actually can't happen because you have a total charge of +2 before the collision and a total charge of 0 after the collision. A proton-antiproton collision would work, though.
    It would help if you showed how you got this result.

    Generally, when you solve this type of problem, you want to work with the particles' energy and momentum, not their velocities.

    Are you familiar with working with four-vectors?
     
  4. May 10, 2010 #3
    Well if the proton is going fast enough to overcome the electric repulsive forces it would collide wouldnt it? thats how colliders work dont they?

    And i have never heard of 4 vectors sorry =/.

    And the velocities i got it from equating the total energies of the "reactancts" and "products"

    E1=gamma(2mc2) 2m because theres 2 protons initially

    E2=gamma(4mc2) equated both equations but obviously i used the definiition of gamma because they both have different velocities.
     
    Last edited: May 10, 2010
  5. May 10, 2010 #4

    vela

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    That's not the point. You can't have interactions where electric charge just simply appears or disappears.
    E1 isn't correct because only one proton is moving. Your expression would only be correct if both protons were moving with the same speed so they have the same gamma.

    Start by writing down the equations for conservation of energy and momentum in terms of the energies and momenta of the individual protons and the blob. Square the two equations, and then make use of the third "relevant equation" to combine and simplify your results.
     
  6. May 10, 2010 #5
    Well you have the proton at rest: E=mc2 K=0 p=0

    The moving proton E1=K1+ mc2 p=(gamma1)mu whre u is the speed of the proton

    The blob: E2=K2+4mc2 p=(gamma2)mv where v is the velocity of the blob

    Equating any of teh energy/momentum equations together just seems like a big mess because of the velocity that is also inside the gamma function...
     
    Last edited: May 10, 2010
  7. May 11, 2010 #6
    This is the only problem i have left unsolved before tomorrows test(which is verbatim from homework). Ive tried solving for the velocities using the previously mentioned equations but no luck =/.
     
  8. May 11, 2010 #7

    vela

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    To repeat...
     
  9. May 11, 2010 #8
    Sorry vela. I thought you mean start by energies and momentum to get velcoties.
    Ok i managed to obtain the mass in terms of the K1 using hte euation above and the third relavant equation.

    m=(pc)2 - K12/(2K1c2) i dont know if this will help.

    When you say the equations for conservation of energy and momentum are you refering to:

    Conservation of energy:K2+4mc2=K1+ mc2
    Conservation of momentum: p=(gamma1)mu = p=(gamma2)mv
    ??
     
    Last edited: May 11, 2010
  10. May 11, 2010 #9

    vela

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    The units don't work out in that equation. Stick with the total energy E and momentum p for each particle. Once you solve for the total energy of the moving proton, you can calculate its kinetic energy by subtracting out the rest energy.
     
  11. May 11, 2010 #10
    yes i know what your saying:

    Kinetic energy energy of moving proton is K1=(gamma1-1)mc2
    But this takes me back to the initial quetion i would have to solve for the velocity in the gamma to obtain the kinetic energy just in terms of mass and c2.

    Iam stupid sorry, iam not understanding at all. I dont see how you can solve this problem with just using the momentum and energies.
     
  12. May 11, 2010 #11

    vela

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    E1 = energy of moving proton
    E2 = energy of proton at rest
    E = energy of blob

    p1 = momentum of moving proton
    p2 = momentum of proton at rest
    p = momentum of blob

    Conservation equatlons are

    E1 + E2 = E
    p1 + p2 = p

    Plus you know p2=0 and E2=mc2.

    Square the first equation. Multiply the second equation by c and then square it. Subtract the second from the first.
     
  13. May 11, 2010 #12
    I feel like a complete retard now.....but i still cant see it. I did what you told me and i still cant see how the momentum cancels out.
     
  14. May 12, 2010 #13

    vela

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    What is E2-(pc)2 equal to?
     
  15. May 12, 2010 #14
    I failed , sorry vela. The my ******* proffesor put this one test and he knew it was the only one i didnt awnser. Out of 30 homework questions he always puts 6 on the test and he puts the only one i didnt have. Oh well i will still get an A but i wanted 100% cause it makes him mad when someone gets a 100%


    Thanks for helping me even though i failed :( really apreciate your time.
     
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