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Relativistic energy/momentum conservation problem

  1. Mar 14, 2013 #1
    1. The problem statement, all variables and given/known data

    A particle of rest mass m0 moving at a speed of 3/4 c collides with a same-mass particle at rest and they stick together to form a composite particle. What is the rest mass of the composite particle and what is its speed?

    2. Relevant equations

    [tex]E = \gamma mc^2[/tex]
    [tex]p = \gamma mu[/tex]
    [tex]\gamma = \frac{1}{\sqrt{1- \beta ^2}}[/tex]

    3. The attempt at a solution

    So far I have written out equations for initial energy and momentum and final energy and momentum. I rearranged both for [itex]\beta_{final}[/itex] and set them equal, with the aim of solving for [itex]v_{final}[/itex]. I got a velocity which was larger than the velocity of the initial particle which surely can't be right.

    I won't write out all my workings as it would take an age but here is what I got for [itex]\beta_{final}[/itex] for the energy equation after cancelling/simplifying:

    [tex]\beta_{final} = -3 -4 \beta_{initial} ^2[/tex]

    And for the momentum equation:

    [tex]\beta_{final} = 1 - \frac{(4 v_{final} ^2)(1 - \beta_{initial} ^2)}{v_{initial} ^2}[/tex]

    As I said, I then set them equal and solved for [itex]v_{final}[/itex]. Any ideas?

    Cheers!
     
    Last edited: Mar 14, 2013
  2. jcsd
  3. Mar 14, 2013 #2

    mfb

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    A small β_initial would lead to β_final close to -3. This cannot be right.
    I don't see why you use both v and β, they just differ by a factor of c and it is convenient to use β everywhere. The second equation is wrong, too.

    Do you know 4-vectors? It is possible to solve it without them, but this would be easier.
     
  4. Mar 14, 2013 #3

    Doc Al

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    Why don't we start with this one. How did you get this? What happened to the masses?
     
  5. Mar 14, 2013 #4
    Do you mean that the equation should be rest mass not mass? If so, the rest mass is given as [itex]m_0[/itex] but yeah I should have written it out fully. Nope I don't know 4-vectors. This is a coursework assignment and at this stage we haven't covered all the material so I am guessing I am supposed to solve using the methods we have covered.

    Using the first equation, for energy.

    [tex]\frac{m_0 c^2}{\sqrt{1- \beta_{initial} ^2}} = \frac{2m_0 c^2}{\sqrt{1- \beta_{final} ^2}}[/tex]

    Then I rearranged for [itex]\beta_{final} ^2[/itex]. [itex]m_0[/itex] and [itex]c[/itex] cancel out. Surely that is essential for solving as both final mass are unknowns?
     
    Last edited: Mar 14, 2013
  6. Mar 14, 2013 #5

    Doc Al

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    You have assumed that the rest mass of the composite particle is 2m0. Do not assume that! (That's one of the things you are trying to solve for.)

    Edit: You also ignored the rest energy of one of the masses.
     
    Last edited: Mar 14, 2013
  7. Mar 14, 2013 #6

    vela

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    How did you come up with that equation? It makes no sense to me.

    I suggest you stick with energies and momenta in your calculations. Write your equations in terms of E's and p's; don't write them in terms of velocities. It'll simplify the algebra quite a bit. Once you have the energy and momentum of the composite particle, then you can solve for its velocity.
     
  8. Mar 14, 2013 #7
    I had a feeling that was too simple to be true, thanks. I went through it again this time with [itex](m_0 + m_1)[/itex] instead of [itex]2m_0[/itex] and that gave me the final velocity equal to the initial velocity which can't be right? Am I missing something here?

    EDIT: In the equation you quoted?
     
  9. Mar 14, 2013 #8

    Doc Al

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    I'm still not sure what you are doing. The rest mass of each particle before the collision is m0. Write the total energy of the system before the collision.

    After the collision you have one combined particle. Call its mass m1. Write its total energy.

    That's how you'd get your energy equation. Then do the same for momentum.
     
  10. Mar 14, 2013 #9
    Ahh bugger me I completely forgot to include the energy of the stationary particle! Will have a go at it and report back.
     
  11. Mar 15, 2013 #10

    mfb

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    "Rest mass" and "mass" are the same.
    The concept of a "relativistic mass" is not used in physics any more.
     
  12. Mar 15, 2013 #11
    I see, then what's wrong with the equation?

    I've spent about 6 hours on this question now. I have tried to express the mass of the composite particle in terms of [itex]m_0[/itex] and then substitute into the energy equation but I just can't get a sensible answer out :/. And it's due in in 2 hours!
     
  13. Mar 15, 2013 #12

    Doc Al

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    Please do as I asked in post #8 above. (Do not try to guess the mass of the composite particle. The equations will tell you that.)
     
  14. Mar 15, 2013 #13

    Doc Al

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    Here's the idea:

    Energy of particle 1 (mass = m0) + Energy of particle 2 (mass = m0) = Energy of composite particle (mass = m1)
     
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