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Relativistic momentum

  1. Jun 24, 2010 #1
    1. The problem statement, all variables and given/known data

    A photon of momentum P strikes a nucleus at rest and s absorbed. If The final (excited)
    nucleus is M calculate its velocity.

    2. Relevant equations

    p = mv/(gamma)

    3. The attempt at a solution

    P = Mv/gamma

    P^2 - (Pv/c)^2 = (M^2)(v^2)

    P^2 = (Mv)^2 + (Pv/c)^2

    V = P / sqrt( (M^2) + ((P^2)/(C^2)) )

    The answer in the back is

    P(C^2) / sqrt((M^4)(C^2) + P(C^2))
  2. jcsd
  3. Jun 24, 2010 #2
    Your first formula is wrong. Should be [itex]p = \gamma mv[/itex].

    There are a couple of other equations to consider using also (though there is nothing wrong with your derivation):

    [tex]E^2 = p^2 c^2 + m^2 c^4[/tex]

    [tex]E = \gamma mc^2[/tex]

    [tex]pc^2 = Ev[/tex]
  4. Jun 24, 2010 #3
    I get the same answer as you, fys iks, starting out from [itex]p=Mv \gamma[/itex]. The units in the book's answer aren't consistent.
  5. Jun 24, 2010 #4
    In this problem, we don't know the energy, but we do know 3-momentum and mass, so it should be just a matter of solving [itex]p = \gamma mv[/itex] for v, shouldn't it?
  6. Jun 25, 2010 #5
    yes, energy isn't introduced until the next chapter. I found a similar question and tried out my equation and it worked so i just think the book made an error.

  7. Jun 25, 2010 #6
    It's the exponent of 4 that makes me think of the first equation I posted. I'm guessing the author of the solution was probably thinking like this:

    Divide numerator and denominator by c2 and you get your answer. A couple of typos in the denominator explains the mistake in the book (though that's no excuse). That's where my thinking was going when I provided the formulas. It's a bit quicker, but mathematically the same as your approach.
  8. Jun 25, 2010 #7
    Oh, I see, yeah, that's simpler than messing around with the gamma.
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