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Collisions in two dimensions

  1. Aug 7, 2010 #1
    1. The problem statement, all variables and given/known data

    A radioactive nucleus at rest decays into a second nucleus, an electron and a neutrino.The electron and neutrino are emitted at right angles and have momenta of 9.3x10^23kg*m/s and 5.40x10^23kg*m/s, respectively. What are the magnitude and direction of the momentum of the second (recoiling) nucleus?

    2. Relevant equations
    px conserved: mAvA=mAv'ACosΘ'A+mBv'BCosΘ'B
    py conserved: mAvA=mAv'AsinΘ'A+mBv'BSinΘ'B



    3. The attempt at a solution
    Really confused on how to approach this one? Tried to draw a picture first of what's going on but not really working...please help
     
  2. jcsd
  3. Aug 7, 2010 #2

    kuruman

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    How big is vA?
    How are the trig functions of θA and θB related?
     
  4. Aug 8, 2010 #3
    I think the angles are both 45degrees...and maybe vA= v'A Cos (45degrees)+v'B Cos(45degrees)= 2v'A Cos(45degrees)??
     
  5. Aug 8, 2010 #4

    kuruman

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    There is nothing in the problem that says that they are. The problem says that the neutrino and the electron are emitted at right angles. Let

    θn = neutrino angle
    θe = electron angle

    then θn + θe = 90o

    in which case
    cos(θn) = cos(90o - θe) = ______ ???
    sin(θn) = sin(90o - θe) = ______ ???
    What does vA represent? The velocity of what? Does it have a value that you can ascertain?
     
  6. Aug 8, 2010 #5
    vA is the velocity of the electron before the collsion. p=mv the mass of an electron is 9.11*10^-31kg so v=(9.30*10^-23)/(9.11*10^-31)=1.0208*10^8m/s so I think va= 2(1.0208*10^8m/s) CosΘ?
     
  7. Aug 8, 2010 #6

    kuruman

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    You mean the nucleus. There is no "electron" before the collision. What is a number for that? Read the problem.
     
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