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Relativistic particle problem.

  1. Jun 25, 2003 #1
    An unstable particle at rest breaks up into two fragments of unequal mass. The rest mass of the lighter particle is 2.5 x 10-28 kg, and that of the heavier fragment is 1.67 x 10-27 kg. If the lighter fragment has a speed of 0.983c after the breakup, what is the speed of the heavier fragment?


    What is the idea here? Where does the energy come from, from an external source, or from the mass in the particles? (note: lower case m corresponds to the mass of the light particle and upper case M corresponds to the mass of the heavy particle)

    v1: speed of the lighter particle
    v2: speed of the heavier particle

    We know that the total relativistic energy is:

    E = KE + moc2

    KE = 1/2mv12, m is the relativistic mass

    So if relativistic energy of the light particle is conserved we get this equation.

    mc2 = 1/2mv2 + moc2

    We can also conserve the relativistic mass of both particles:

    (m + M)c2 = 1/2mv12 + 1/2Mv22 + (mo + Mo)c2

    Is this the correct way to setup the problem?

    I tried solving for v2 without any luck. So I hope there is an easier way--the correct way.

    Thanks
     
    Last edited: Jun 25, 2003
  2. jcsd
  3. Jun 25, 2003 #2

    Tom Mattson

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    The idea is to apply conservation of energy and momentum to a relativistic problem.

    It comes from the mass of the inital body. When you are done solving the problem, you can verify that the combined masses of the fragments is less than the mass of the whole. Rest mass energy was converted to kinetic energy.

    That's a bad choice, in my opinion. I would use capital M for the mass of the body before disintegration, and mL, mH for the masses of the light and heavy fragments, respectively. Descriptive variable names can help you track which quantity is which throughout the problem.

    OK

    No, that's nonrelativistic KE. Look up relativistic KE in your book. It is:

    KE=(γ-1)mc2

    and if you add that to the rest mass energy to get the total energy of a free particle, you get:

    E=mc2+KE=mc2+(γ-1)mc2
    E=γmc2

    This should all be in your book.

    No, even if you were right about KE you wouldn't get that. The energy of the light particle by itself is not conserved.

    No, on the left you have the total mass equal to (m+M), which is not true.

    I am going to use the notation I suggested.

    The total energy prior to disintegration is:

    Ei=Mc2

    The total energy after disintegration is:

    Ef=γLmLc2+γHmHc2

    Let Ei=Ef, and you have the equation for conservation of energy. Since you have two unknowns (M and vH), you need another equation. Luckily, conservation of momentum saves the day.

    Total momentum before disintegration:

    pi=0

    Total momentum after disintegration:

    pf=γLmLvL+γHmHvH

    Let pi=pf, and you have the equation for conservation of momentum.

    edit: fixed variable subscripts
     
    Last edited: Jun 25, 2003
  4. Jun 26, 2003 #3
    I got vH = .3265*c

    However the correct answer should be .285*c

    I founs these equations:

    From: pL = pH I got:

    VH = (YL/YH)*(mLo/mHo)*VL

    From Ei = Ef:

    I found:

    YH = (mLo + mHo - YL*mLo)/mHo

    I then substited YH in VH = (YL/YH)*(mLo/mHo)*VL and calculated my answer.

    What went wrong?

    Thanks.
     
  5. Jun 27, 2003 #4

    Tom Mattson

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    How did you eliminate the mass M of the initial particle?
     
  6. Jun 27, 2003 #5
    Uhhhhh... By making it go to zero.

    Nevertheless you can solve it strickly with ph=pl.
     
  7. Jun 27, 2003 #6

    Tom Mattson

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    Yes, that's true--I didn't notice that at first.

    So you got it then?
     
  8. Jun 27, 2003 #7
    Yeah, I fiquered it out right before I turned it in. It was a Newton's Third Law problem with a Special Relativity twist.
     
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