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Mass and Momentum Question

  1. Dec 30, 2008 #1
    1. The problem statement, all variables and given/known data
    Determine how accurately you will need to measure the speed of a particle that has energy 2 GeV to determine whether it is a pion (mass = 0.14 Gev/c^2) or a Kaon (mass = 0.49 Gev/C^2). *This question is in a relativity section, so I'm assuming relativistic equations are neccesary*

    2. Relevant equations
    [tex]
    E^2 = (p^2)(c^2) + (m^2)(c^4)
    [/tex]
    [tex]
    p = (\gamma)mv
    [/tex]

    3. The attempt at a solution
    I started by substituting [tex] p= (\gamma)mv [/tex] into the other equation so velocity can be related to mass, but I'm lost as to what the basic approach to this problem would be.
     
  2. jcsd
  3. Dec 30, 2008 #2

    Dick

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    Homework Helper

    That looks like a good start. I'd now try to solve for v in terms of E and m and get the value of v for a kaon and a pion and see what the difference is. It looks to me like it becomes a quadratic equation in v^2.
     
  4. Dec 31, 2008 #3
    use gamma r = E/(mc^2) = 1/sqrt(1 - v^2/c^2)
    so v/c = sqrt[1 - (mc^2/E)^2]
    for pion, mc^2/E = 0.14/2 = 0.07
    v/c = 0.9975
     
  5. Jan 2, 2009 #4
    Thanks, the only thing is when I solved it I didn't end up with a quadratic.
     
  6. Jan 3, 2009 #5

    Dick

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    What DID you get?
     
  7. Jan 4, 2009 #6
    I started by solving for p and substituing [tex] p= (\gamma mv) [/tex] for it.

    [tex]
    \frac{mv}{\sqrt{1 - v^2/c^2}} = \sqrt{\frac{E^2 - m^2c^4}{c^2}}
    [/tex]



    then I squared both sides and moved gamma and [tex] c^2 [/tex] to opposite sides

    [tex]
    m^2v^2c^2 = (E^2 - m^2c^4) (1 - v^2/c^2)
    [/tex]



    multiply out right hand side

    [tex]
    m^2v^2c^2 = E^2 + m^2c^2v^2 -m^2c^4 - \frac{E^2v^2}{c^2}
    [/tex]



    group [tex] v^2 [/tex] on same side and factor it out

    [tex]
    v^2 (c^2m^2 + E^2/c^2 - m^2c^2) = E^2 - m^2c^4
    [/tex]



    solve for v

    [tex]
    v = \sqrt{\frac{E^2 -m^2c^4}{c^2m^2 + E^2/c^2 - m^2c^2}}
    [/tex]
     
  8. Jan 5, 2009 #7

    Dick

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    That looks ok to me. You can cancel the two m^2*c^2 terms. It is even easier than I thought.
     
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