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Conservation of momentum (neuclear decay)

  1. Oct 25, 2008 #1
    1. The problem statement, all variables and given/known data
    A certain nucleus at rest suddenly decays into three particles, two of which are charged and can be easily detected. The data gathered for these two particles is:
    - Particle 1 has mass 2.93e-25 kg moving 5.69e+06 m/s at 69.5o.
    - Particle 2 has mass 1.33e-25 kg moving 9.31e+06 at 221o.

    Find the Momentum of the third particle
    _____i km m/s + _____ j km m/s

    3. The attempt at a solution
    I tried first calculating the Momentum of each of the particles and then split the 2D momentum vectors into i and j

    and then i set that the sum of all the i's = 0 and that the sum of all the j's = 0 since there aren't any external forces and momentum is conserved and then i tried solving for the missing i and the missing j.

    I keep getting the answer wrong...

    Thanks for your help
    Aditya
     
  2. jcsd
  3. Oct 25, 2008 #2
    You method sounds OK. I cannot tell where is your mistake unless you show your work (or at least results)
     
  4. Oct 25, 2008 #3
    Thanks for your reply...

    This is what i have done:
    my results came out to be
    -2.38e-20 i kg m/s + -1.741e-18 j kg m/s
     
  5. Oct 25, 2008 #4
    I don't get your results. As I said, if you show your work I could figure out where is your mistake.
    Maybe you calculator is set on the "wrong" mode (radians instead of degrees)
     
  6. Oct 25, 2008 #5
    well this is what i got
    2.55e-18 j + 8.06e-19 j + yj = 0
    y= -1.74e-18 j kg m/s

    9.52e-19 i + (-9.28e-19) i + xi = 0
    x= -2.37e-20 i kg m/s
     
  7. Oct 25, 2008 #6
    I don't know how you calculate your components but I can say this much:
    Something is not right. The second momentum (221 Deg) is in the third quadrant. Both x and y components must be negative. I don't see this in your components.
    And your y component of the first momentum (2.55 e-18) seems to be larger than the whole momentum...
     
  8. Oct 25, 2008 #7
    this is how im getting the 1D values
    m3v3 i = - (m1v1cosß +m2v2cosΘ)
    m3v3 j= - (m1v1sinß +m2v2sinΘ)
     
  9. Oct 25, 2008 #8
    hey i re-did the calculations and is this the right answer:
    -2.11e-19 i kg m/s + -1.72e-18 j kg m/s
     
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