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Relativistic Momentum

  1. Mar 10, 2017 #1
    1. The problem statement, all variables and given/known data
    How much work is required to accelerate a proton from rest up to a speed of 0.999c?
    What would be the momentum of this proton?

    2. Relevant equations
    p=γmv

    3. The attempt at a solution
    I got part A, which was the momentum. I found that to be 20.1 GeV. Now for part B I have to find the momentum in units of GeV/c. I'm down to my last attempt on Mastering Physics. First, I tried going the simplest route and divided the 20.1 GeV from part A by c to get 6.7e-8 and that was wrong. I also tried using p=γmv to find an answer of 1.12e-5 J, which I converted to 7.01e-8 GeV, then divided that by c and got an incorrect figure of 2.34e-16.

    Any help would be greatly appreciated.
     
  2. jcsd
  3. Mar 10, 2017 #2

    PeroK

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    Can you find a formula that relates the energy to the momentum of a particle?
     
  4. Mar 10, 2017 #3
    Would that be E^2=p^2c^2 + m^2c^4? If that is the correct equation, I came up with an answer of 1.35e-6.
     
  5. Mar 10, 2017 #4

    PeroK

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    You could use that one. But, what about using ##E = \gamma mc^2## and ##p = \gamma mv##?
     
  6. Mar 10, 2017 #5
    Honestly, I'm not quite sure what I would do with those equations. I saw that a guy on Yahoo Answers did K/c=mvγ and that led me to an answer of 7.01e-8. I'm not confident that that is right though. Any idea if my answer of 1.35e-6 from my previous post is right?
     
  7. Mar 10, 2017 #6

    PeroK

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    I think your problem is probably understanding ##eV## units. The mass of a particle is normally given in ##MeV/c^2##. For example, the proton mass is about ##938 MeV/c^2##.

    The energy of a particle is, therefore, ##E = \gamma m## where ##E## is in ##MeV## and ##m## is the mass in ##MeV/c^2##.

    Momentum is given in units of ##MeV/c## and we have ##p = \gamma mv/c## in these units.

    In this case ##v/c = 0.999 \approx 1## so we have ##p \approx \gamma m = E##.

    These units take a bit of time and practice to get used to, but you can see how using them can simplify the calculations.
     
  8. Mar 10, 2017 #7
    Yes, thank you it makes sense now.
     
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