1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Relativistic Momentum: Force Accelerating a proton

  1. Sep 25, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the force necessary to give a proton an acceleration of 10^19m/s^2 when the proton has a velocity ( along the same direction as the force) of 0.9c


    2. Relevant equations

    p=gamma * m * U

    F = d/dt [gamma * M * U]

    F=gamma^3 *m*a

    3. The attempt at a solution

    initial momentum = 1/(sqrt(1-0.9c^2/c^2)) * 1.67*10^-27kg * 0.9c = 1.03 *10^-18
     
  2. jcsd
  3. Sep 26, 2011 #2
  4. Sep 26, 2011 #3

    rude man

    User Avatar
    Homework Helper
    Gold Member

    hang on, relief is on its way, I will do it by tonight MST (but I stay up till at least 3 a.m.!).
     
  5. Sep 26, 2011 #4

    rude man

    User Avatar
    Homework Helper
    Gold Member

    OK, here we go, 2 glasses of Shiraz or not:

    We have F = dp/dt = (d/dt){m0v/(√(1 - v2/c2)}
    = m0(d/dt){v/√(1 - v2/c2)}
    = m0*du/dt

    where u = v/√(1 - v2/c2)

    So now du/dt = du/dv*dv/dt where dv/dt = 1e19m/s2

    Compute du/dv by standard way, then
    du/dt is known as a function of v.
    Let v = 0.9c and you have du/dt as a definite number.
    Finally F = m0*du/dt
    where m0 is the rest mass of a proton.

    Make sense?
     
  6. Sep 26, 2011 #5
    I think so, the answer I got was 2 * 10^-7 N

    Is that correct?
     
  7. Sep 27, 2011 #6

    rude man

    User Avatar
    Homework Helper
    Gold Member

    OK, I got
    du/dv = v(-1/2)(-2v/c2)[(1 - v2/c2)^(-3/2)] + (1 - v2/c2)^(-1/2)

    = 1/(1 - v2/c2)^(1/2) + 0.81/(1 - v2/c2)^(3/2)
    = 2.294 + 9.780 = 12.074
    Then, F = 12.074*1e19*1.67e-27 = 2.016e-7N
    Yay team!!
     
    Last edited: Sep 27, 2011
  8. Sep 27, 2011 #7
    Haha, Thank you so much :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Relativistic Momentum: Force Accelerating a proton
  1. Relativistic Momentum (Replies: 1)

  2. Relativistic momentum (Replies: 6)

Loading...