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Find velocity of an accelerated proton using kinetic energy

  1. Oct 24, 2015 #1
    1. The problem statement, all variables and given/known data

    So I'm trying to find the final velocity of a proton that is being accelerated by a particle accelerator, just before it collides with a particle. All I have is its kinetic energy just before collision: 4.7066×10^(-13) J.

    I thought I should use KE = 1/2 mv^2 but then realised it should be travelling near the speed of light, so I used Einstein's equation for special relativity to find the relative mass

    I.e. I subbed in "m (rest)/sqrt(1-(v^2/c^2))" for mass

    However, when I solved for 'v' (which involved lots of algebra and logs), I got 1.713×10^31 m/s

    THAT'S WAY TOO FAST!
    I have checked my working a few times so I suspect there is something intrinsically wrong- should I not use relativistic mass?


    2. Relevant equations
    KE = 1/2 mv^2
    m (relative) = m (rest)/sqrt(1-(v^2/c^2))

    3. The attempt at a solution
    Attempt is attached as a pdf- to preserve formatting

    Thanks for any thoughts or help! :)
     

    Attached Files:

  2. jcsd
  3. Oct 24, 2015 #2

    mfb

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    You used the wrong classical answer for velocity and plugged it into the relativistic formula? That does not work.
     
  4. Oct 24, 2015 #3
    You certainly need to use relativistic energy.
    Ek = E - E0 is the given KE.
    Note that Ek = m0 c^2 (1 / A - 1) where m0 may be referred to as the rest mass of the proton.
    where A is the relativistic factor (1 - v^2 / c^2)^1/2.
     
  5. Oct 24, 2015 #4

    ehild

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    You made a very basic error. You applied logarithm to a sum. log(a+b) is not log(a) + log(b)!

    Determine the speed of the proton with the classical formula. With v obtained, calculate the Lorentz factor and see if it appreciably deviates from 1. If the difference is less than 0.01, you do not need the relativistic formula.
     
  6. Oct 24, 2015 #5
    @J Hann thanks- I'll re-calculate it with that- we only ever learned relativistic mass, length and time in class so I assumed there weren't other equations (whoops)

    @ehild I applied the multiplication rule... log (a) + log (b) = log (a*b)
    But thanks for recommending the Lorentz factor- the more maths and equation knowledge I can put in my solution the better! :)
     
  7. Oct 24, 2015 #6
    @J Hann Wait a sec- shouldn't I just use E= m(r)*c^2 - m0*c^2 ? Or is your equation derived from that?
     
  8. Oct 24, 2015 #7

    mfb

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    I would start here.
     
  9. Oct 24, 2015 #8
    Ok- I did so, and got roughly 2.4*10^7 m/s, which gave a Lorentz factor of 1.003

    So I'll have a go with the relativistic equation now...

    Does 1.003 represent that there is a dilation of 3%? I'm just trying to figure out how to explain that calculation, and the internet generally describes the Lorentz factor as time dilation, which doesn't really make sense in this case...

    Thanks for all the help C:
     
  10. Oct 24, 2015 #9

    mfb

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    Which means you can probably neglect special relativity. If you don't want to do that:

    The Lorentz factor depends on velocity only, and you can directly calculate it based on the given energy and the proton mass. Solve for velocity, done.
     
  11. Oct 24, 2015 #10
    Yep I used special relativity, because my teacher keeps emphasising that we should be doing "complicated" calculations.
    Now I'll have to see if my teacher likes what I've done :)
     
  12. Oct 24, 2015 #11
    Some people will say (as per Einstein) that the total energy whether at rest or moving is
    m c^2 / (1 - v^2 / c^2)^1/2
    and that the relativistic quantity involved is momentum not mass - mass is mass at rest or moving.
    If that's confusing, don't worry because most textbooks use m = m0 / (1 - v^2 / c^2)^1/2
    and the results are the same.
    The 1.003 that you found involves 1 / (1 - v^2 / c^2)^1/2 and allows you to calculate v as
    a percentage of c.
    The binomial theorem is commonly used expand (1 - v^2 / c^2)^-1/2 when v nearly equals c.
     
    Last edited: Oct 24, 2015
  13. Oct 24, 2015 #12

    ehild

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    If your teacher wants you to apply SR, use the formula
    [tex]KE=mc^2 \left( \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1 \right)[/tex]
    to get the Lorentz factor, and the speed v from that. You will get the same v with three significant digits as you have got with the classical method.
    Avoid to use logarithms. You have a calculator, don't you?
     
    Last edited: Oct 24, 2015
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