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Relativity and Momentum

  1. Aug 7, 2008 #1
    1. The problem statement, all variables and given/known data
    What is the momentum of a proton with a kinetic energy of 750MeV?


    2. Relevant equations
    E = m0c2 + K
    E2 = p2c2 + m0c2
    Mp = 1.67 x 10-21

    3. The attempt at a solution

    I think whats confusing me is K = 750MeV, but I tried to solve for p

    First, I converted 750MeV to 750x106 eV

    Then solved for E in the first equation
    E = 750x106 eV

    Then used that to solve for p in the second equation
    But i get a wrong answer.

    The correct answer is 1404MeV/c
     
    Last edited: Aug 7, 2008
  2. jcsd
  3. Aug 7, 2008 #2

    hage567

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    In your second equation, [tex]m_oc^2 [/tex] needs to be squared. So it should be [tex] E^2 = p^2c^2 + (m_oc^2)^2[/tex]

    You also need to express the proton mass in electronvolts. Are you doing that?
     
  4. Aug 8, 2008 #3

    Redbelly98

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    In other words, you get that E = K. But that's only true when m0=0, which is not the case for a proton.

    You did mean to say proton, and not photon?
     
  5. Aug 8, 2008 #4
    Hey,

    So I fixed the equation and figured it out

    E = m0c2 + KE

    m0c2 = 1.5x10-10j = 939.4MeV

    E = 939 + 750 = 1689MeV

    E2 = p2c2 + m02c2

    16892 = p2c2 + 9392

    p2c2 = 191000

    pc = 1404MeV

    p = 1404MeV/c

    Thanks!
     
  6. Aug 9, 2008 #5
    is it okay to solve like this??

    1/2(Mp)v2=K
    1/2(p2)/Mp=K
    p=square root of 2K(Mp)
    p=6.34x10^-19 kg m/s
     
  7. Aug 9, 2008 #6

    Andrew Mason

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    Your arithmetic is not correct. It should be: p2c2 = 1,970,000

    AM
     
  8. Aug 9, 2008 #7

    Andrew Mason

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    Not for relativistic speeds. A proton with 750 Mev of kinetic energy is moving at relativistic speed. At non-relativistic speed you can use:

    [tex]\text{KE} = \frac{1}{2}mv^2 = \frac{p^2}{2m}[/tex]

    [tex]p = \sqrt{2m\text{KE}}[/tex]

    You had it right if you were using Mp for proton mass and p for momentum, but it is confusing.

    AM
     
  9. Aug 9, 2008 #8
    How can i know whether it is in non-relativistic or relativistic speed???
     
  10. Aug 9, 2008 #9

    Redbelly98

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    For non-relativistic speeds, KE has to be a lot less than mc2. Equivalently, this means the speed is a lot less than c.

    In this problem, KE is 750 MeV. That is comparable to mc2 = 940 MeV, so it is relativistic in this case.
     
  11. Aug 9, 2008 #10
    So for relativistic speed wouldn't you just substitute that equation,

    K = 1/2((lorenz factor) - 1)mv2

    substitute for p

    K/((lorenz factor) - 1) = p2/2m

    and you could figure out how fast the proton is moving given the kinetic energy?
     
  12. Aug 9, 2008 #11

    Redbelly98

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    No, the relation is

    K = ( (Lorenz factor) - 1 ) m c2

    Given K, solve for the Lorenz factor, from which you can get v.
     
  13. Aug 10, 2008 #12
    So can i use E=hf to calculate the momentum as it's relativistic??
     
  14. Aug 10, 2008 #13

    Redbelly98

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    Why would you think that? Momentum does not appear anywhere in your equation ???
     
  15. Aug 11, 2008 #14
    i thought the wavelength in the equation E=hc/lambda is the same as the de Broglie's wavelength......
    is the de Broglie's wavelength of a electron with non-relativistic speed couldnt be sub into the E=hf equation??
     
  16. Aug 11, 2008 #15

    Redbelly98

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    Sometimes, not always. There actually 3 "speed regimes", not 2, to think about.

    1. Non-relativistic: K is much less than mc2
    2. Moderately relativistic: K is comparable to mc2
    3. Extremely relativistic: K is much larger than mc2

    That equation, E = hf = hc/lambda = pc , is valid in the extremely relativistic case. Another way of seeing this is from two equations that are true in all 3 cases:

    E2 = (pc)2 + (mc2)2
    and
    E = K + mc2

    In case 3 (and only then) we can neglect the mc2 terms because it is much smaller than the other quantities, and we get

    E = K = pc (and also E = hc/lambda = hf)

    Hope that helps.
     
  17. Aug 12, 2008 #16
    o,
    finally i understood!!
    thnx ya!!
     
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