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Kinetic Enrgey according to another viewer

  1. Sep 20, 2004 #1

    ori

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    2 particles with mass M and Kinetic Engergy T moving toward each other.
    what is the kinetic energy T' as it measured by the other particle
    given data: T=...
    m=...

    thanks
     
  2. jcsd
  3. Sep 20, 2004 #2
    Why you proposed this question?
     
  4. Sep 20, 2004 #3

    ori

    User Avatar

    training for test

    that's what i did, could anyone tell what's my prob here:
    for first particle
    T=mc^2(r-1)
    since we're working with Mev , c=1
    T=m(r-1)
    r=T/M+1
    therefore
    (1) V^2=T/(T+M)
    (V is speed of particle one according to "the lab")

    therefore the speed of particle2 is -V (simetry)
    now we use additional formula
    u1=(u1'+u3)/(1+u1'u3)
    whille u1 is velocity of particle1 according to the lab (u1=V)
    u1' is velocity of particle 1 according to particle 2
    u3 is the velocity between lab to particle2 (u3=-V)
    therefore
    u1'=2V/(V^2+1)

    now i look on kinetic energy of particle 1 according to particle 2:
    T'=m(r-1)
    while r=1/sqrt(1-u1'^2)
    i assigned u1' & V and got
    T'=4/3M
    this isn't correct..
     
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