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What is the relative velocity of the two reference frames?

  1. Oct 8, 2006 #1
    c) What is the relative velocity of the two reference frames?
     
    Last edited: Oct 9, 2006
  2. jcsd
  3. Oct 8, 2006 #2

    OlderDan

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    Maybe just try part a) again, but make sure you account for the total energy and the momentum in both frames.

    Not sure where you went wrong. I got 5.66
     
  4. Oct 8, 2006 #3
    I probably didn't account for both frames. How did you use the 4 GeV/c?
     
  5. Oct 8, 2006 #4

    OlderDan

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    E^2-(cp)^2 = (E.)^2 = (m.c^2)^2 is invariant

    (I assume you are using the dot for the zero subscript)

    From your orignal data,
    (E.)^2 = 25GeV^2 - 9GeV^2 = 16GeV^2

    4GeV/c is the p in the second frame. Substitute that into your energy equation and solve for E
     
  6. Oct 8, 2006 #5
    Yes, I was using the dot for 0 subscript.

    E^2-(16 GeV^2)=16 GeV^2
    E= 5.66 GeV


    I could also use help on the rest of the problem, b and c.
     
    Last edited: Oct 8, 2006
  7. Oct 8, 2006 #6

    OlderDan

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    If you look back at the equations used, you should get part b) quite easily. Give it a try.
     
  8. Oct 8, 2006 #7

    OlderDan

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    What is the 16Gev^2 on the right hand side of your earlier equation?
     
  9. Oct 8, 2006 #8
    16Gev^2=E.^2
    4 GeV= E.

    931.49 MeV/c^2 = 1 amu
     
    Last edited: Oct 8, 2006
  10. Oct 8, 2006 #9

    OlderDan

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    So.. solve it for the rest mass and you've got it.
     
  11. Oct 8, 2006 #10

    OlderDan

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    Check your original post for conversion from MeV/c^2 to amu

    Part c) is the hardest part. Do you know how to find the velocity of the particle in each frame from the known energy and momentum? Once you have that, you need to use the relitivistic velocity addition formula. I'll let you work on that a bit.
     
    Last edited: Oct 8, 2006
  12. Oct 8, 2006 #11
    4 x 10^9 eV / 931.49 x 10^6 eV = 4.3 amu (should there not be a /c^2 in the conversion information?)

    I'm trying c now.

    v/c = cp/E
    v=c^2p/E

    v. = c^2(3 GeV/c) / 4 GeV = 2.25x10^8 m/sec
    v=2.12x10^8 m/sec

    I got u= 2.86x10^8 m/sec...
     
    Last edited: Oct 8, 2006
  13. Oct 9, 2006 #12
    v= c^2p/E

    ux=ux'+v/(1+vux'/c^2)

    I got something for part c but it was off (wrong).
     
    Last edited: Oct 9, 2006
  14. Oct 9, 2006 #13

    OlderDan

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    You left out parentheses in your numerator, but I expect you used them. You just have a sign problem. You know the velocity of the object in two frames. You are looking for the relativistic difference between the two. Try replacing the + in your equation with -
     
  15. Oct 9, 2006 #14
    ux'=(ux-v)/(1-vux/c^2)

    I got 0.185c. I used 1.8x10^8 m/sec for one v and 2.12x10^8 m/sec for another v.
     
    Last edited: Oct 9, 2006
  16. Oct 9, 2006 #15

    OlderDan

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    I see you just changed it.. I was trying to figure out where you got the first v :smile:

    I also got the .185c. I think you got it.
     
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