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Disintegration, special relativity

  1. Mar 24, 2013 #1
    1. The problem statement, all variables and given/known data

    Hello everybody,
    A particle of mass at rest Mo in the laboratory is at rest when it disintegrates into three particles of mass at rest mo. Two of them have velocity and direction as indicated.

    a. Find the expression of the velocity and angle θ of particle #3.
    b. Find the ratio Mo/mo.


    2. Relevant equations



    3. The attempt at a solution

    In the scheme, we have particle #1 going in the direction <---- with velocity 4c/5, particle #2 with direction perpendicular to #1 going downward with velocity 3c/5, and particle #3 is going upward in the right direction with angle θ.

    From conservation of momentum:
    Pi=Pf and P=gamma*m*v I took c=1
    P1+P2+P3=0
    P1=(4/5)*m/((1-(-4/5)^2)^(1/2))
    P2=(3/5)m/((1-(-3/5)^2)^(1/2))
    P3=mv/((1-(v)^2)^(1/2))

    As, P1+P2+P3=0 ==> P1=-P2-P3 and solve for v ?
    How to find theta?
    Thank you!
     
    Last edited by a moderator: Mar 25, 2013
  2. jcsd
  3. Mar 25, 2013 #2

    Simon Bridge

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    Remember that momentum is a vector.
     
  4. Mar 25, 2013 #3
    So I have to write that
    ||P3||= ((P1)^2+(P2)^2)^1/2 ?
    And how to find Mo/mo ?
     
  5. Mar 25, 2013 #4

    vela

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    Why don't you start by trying to find the x and y components of ##\vec{p}_3## first?
     
  6. Mar 25, 2013 #5
    I found the velocity and the angle theta, but not the ratio Mo/mo
     
  7. Mar 25, 2013 #6

    vela

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    Show your work. We can't help without seeing what you've done.
     
  8. Mar 25, 2013 #7
    P1=-4mo/5(1-(-4/5)²)^1/2 i=-4mo/3 i
    P2=-3mo/5(1-(-3/5)²)^1/2 j=-3mo/4 j

    P3=-P1-P2=4mo/3 i+3mo/4 j
    P3=mo(337/144)^1/2
    And P3=mo*v/(1-v²) ==> v=(P3²/(mo²+P3²))^1/2
    I replaced P3 by its balue i found v=0,83c

    tanθ= 3/5 / 4/5 = 3/4
    θ= arctan (3/4) ≈ 37°
     
  9. Mar 25, 2013 #8

    Simon Bridge

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    Find the ratio of masses from conservation of energy.
     
  10. Mar 26, 2013 #9
    γMo=E1+E2+E3
    1mo+γ2mo+γ3mo
    γMo = Mo because the particle is at rest when it decays
    Mo=mo(γ123)
    Mo/mo=γ123
    =5/3+5/4+√481/12
    =35/12+√481/12

    Is it correct !?
    Thanks
     
  11. Mar 26, 2013 #10

    Simon Bridge

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    That's what I was thinking of - does that make it correct?
    Can you think of any way to check it?

    Expect
    Mo > mo since we are told that mo is bigger - so Mo/mo > 1
    We can do better: Mo > 3mo - since some of the rest energy of the parent mass becomes kinetic energy in the daughters. If Mo = 3mo, then the daughters would be stationary too.
    Should Mo be a lot bigger or just a bit bigger than 3mo?
    Is that what you've got?
     
  12. Mar 26, 2013 #11
    If i understoof well, I'd say that Mo sould be a bit bigger than 3mo. And I found that Mo/mo=4.7 so thats's correct
     
  13. Mar 26, 2013 #12

    Simon Bridge

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    Why would you expect that Mo would not be much bigger than mo?

    Get used to showing your reasoning for your answers with the answers.
    It's counter-intuitive: if you provide reasons it means people are more likely to be able to tell where you went wrong. Nobody likes being wrong. Yet, in science, being wrong is essential.

    Notice how thinking about the answer like that gives you confidence in the answer?
    That way you don't need to rely on some authority figure ;)
     
  14. Mar 26, 2013 #13
    Since some of the rest energy of the first particle becomes kinetic energy in the three particles and they have relativistic velocities so their kinetic energies are very large, so Mo would'nt be much bigger than mo ?
     
  15. Mar 26, 2013 #14

    Simon Bridge

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    Now you are thinking - at what speed would the kinetic energy be equal to the rest-mass energy?
    How does that compare to the speeds here?
     
  16. Mar 26, 2013 #15
    The kinetic energy would be equal to the rest mass energy if v=0.86c
    And the velocity of #1 is 0.8, #2 = 0.6 and #3=0.83 so we are not that far from 0.86c
     
  17. Mar 26, 2013 #16

    Simon Bridge

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    ...
    So you'd expect the total energies for two of the particles to be about 2mo and the other to be somewhat less ... back of envelope, you'd get 4 < Mo/m0 < 6. Your value is pretty much in that ballpark. When you get used to the values, because of the way gamma changes with speed, you'd see that you'd expect a number close to 5. (If, however, one of the speeds was 0.9998c, that would be a different story.)

    It is useful to get a feel for the speeds at different kinetic-energies.
    When is the KE half the rest energy? When is it double? Wat is the ratio for 0.99c, 0.999c, 0.9999c? That sort of thing.
    But you should get the idea now :)
     
  18. Mar 26, 2013 #17
    Thank you for your help and explanations ! :smile:
     
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