Collisions in two dimensions

  • Thread starter rvnt
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Homework Statement



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?

Homework Equations


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



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
 

Answers and Replies

  • #2
kuruman
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How big is vA?
How are the trig functions of θA and θB related?
 
  • #3
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I think the angles are both 45degrees...and maybe vA= v'A Cos (45degrees)+v'B Cos(45degrees)= 2v'A Cos(45degrees)??
 
  • #4
kuruman
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I think the angles are both 45degrees...
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) = ______ ???
and maybe vA= v'A Cos (45degrees)+v'B Cos(45degrees)= 2v'A Cos(45degrees)??
What does vA represent? The velocity of what? Does it have a value that you can ascertain?
 
  • #5
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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?
 
  • #6
kuruman
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vA is the velocity of the electron before the collsion.
You mean the nucleus. There is no "electron" before the collision. What is a number for that? Read the problem.
 

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