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Velocity of pions after moving far away from each other?

  1. Feb 26, 2015 #1
    1. The problem statement, all variables and given/known data
    A proton and an antiproton, each with an initial speed of 5.90 multiply.gif 107 m/s when they are far apart. When these two particles collide, they react to form two new particles: a positive pion (π+, charge +e) and a negative pion (π−, charge −e). Each pion has a rest mass of 2.5 multiply.gif 10-28 kg. These pions have enough energy that they move away from each other. When these two pions have moved very far away from each other, how fast is each pion going, v?

    2. Relevant equations
    E of system = kinetic + rest + potential energies
    kinetic = 1/2mv^2
    rest= mc^2
    electric potential = (9x10^9)(q1 * q2)/radius

    3. The attempt at a solution
    Esys1 = Esys3
    (2) (1/2) (1.7x10^-27) (5.9x10^7)^2 + (2) (1.7x10^-27) (3x10^8)^2 = (2) (1/2) (2.5x10^-28)(v final)^2 + (2) (2.5x10^-28) (3x10^8)^2
    v = 1.03328 x 10^9
    This answer turned out to be wrong! Can anyone see where I went wrong?
     
  2. jcsd
  3. Feb 26, 2015 #2

    haruspex

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    Look at the magnitudes of the terms in your equation. The rest masses dominate. So when you take the difference in the rest masses you get a relatively small difference between two large numbers. That can turn a small numerical error into a much more significant one.
    I would try using a more accurate value for c.
    (What is the supposed answer?)
     
  4. Feb 26, 2015 #3
    I don't know what the answer is, but I know it is supposed to be close to the speed of light... I tried plugging in 2.99792x10^8, and now my answer is 1.03258x10^9. Do you think this is correct? I only have one submission left!
     
  5. Feb 26, 2015 #4

    haruspex

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    With such great speeds, you should be using the generic relativistic mass, not adding rest mass to Newtonian KE, maybe? But that would yield a slightly smaller number.
     
  6. Feb 26, 2015 #5
    Oh I forgot about that! I now have v = 2.99792x10^8! Do you think it's right?
     
  7. Feb 26, 2015 #6

    haruspex

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    Still seems too close to c. Please post your working.
     
  8. Feb 26, 2015 #7
  9. Feb 26, 2015 #8

    haruspex

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    Rather a lot of arithmetic operations doing it that way. Do it all algebraically until the final step. You should get ##c^2-v_2^2 = (c^2-v_1^2)(\frac {m_2}{m_1})^2##, which gives me about 2.967E8.
     
  10. Feb 26, 2015 #9
    Oh ok! That certainly makes it simpler. Thank you!
     
  11. Feb 26, 2015 #10
    If you don't mind me asking, how did you get to that point? I tried it algebraically and got v = c at the end.
     
  12. Feb 26, 2015 #11

    haruspex

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    Post your algebra and I'll check it.
     
  13. Feb 26, 2015 #12
  14. Feb 26, 2015 #13

    haruspex

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    Two different gammas?
     
  15. Feb 26, 2015 #14
    Ohhhhh..... :mad:
     
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