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Solving system of equations to find angle and energy of Au atom and α

  1. Mar 12, 2013 #1
    1. The problem statement, all variables and given/known data

    An alpha particle traveling with a kinetic energy of 5.5 MeV and a rest-mass of 3727.8 MeV/c^2 strikes a gold atom with a rest-mass of 183,476 MeV/c^2.

    -The gold atom is initially at rest
    -The alpha particle deflects perpendicular to the horizontal in the after state


    2. Relevant equations

    Conservation of Energy:


    Conservation of momentum:



    3. The attempt at a solution

    Total Energy Before Collison:


    Momentum before (in x-direction):


    ==> (3727.8+5.5)^2=P^2c^2+(3727.8)^2

    ==> P=202.6 MeV/c


    Conservation of momentum:

    in x: P(Au)cosθ=202.6MeV/c (1)

    in y: P(Au)sinθ-P(α)=0 (2)

    Conservation of Energy:


    ==> √(P(Au)^2+183476^2)+√(P(α)^2+3727.8^2)=187,209 MeV (3)

    Solve the system of equations to find θ, P(Au), and P(α):

    Solve (3) for P(α):


    Plug back into Eq. (2):


    Solve (1) for θ:


    Plug into (2) and solve for Au:


    ==> P(Au)=26,172.7 MeV/c

    Solve for θ:


    ==> θ=89.56°

    Solve for P(α):


    ==> P(α)=26,171.9 MeV/c

    However, when I plug these numbers back into equation (3) the solution doesn't come out right. I have been looking over this for hours and can't figure out where my error might be.

  2. jcsd
  3. Mar 13, 2013 #2

    Simon Bridge

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    Possibly algebra problems?
    Go through each step carefully - without going too deep, the following stood out:
    Are you missing a c^2 in there? Typo?
    #3 was:
    ... is there a c^2 missing there too?
    How did you get from this to your conclusion
    Have I read that right:
    ##p_{Au}\sin(\arccos(202.6/p_{Au}))=\sqrt{1.36998\times 10^9 - p_{Au}} \\ \Rightarrow p_{Au}=26172.7\text{MeV/c}##
    ... how did you manage to ull out the ##p_{Au}## ?

    ... I'm not saying these are the mistakes - only that this is what stands out.
  4. Mar 13, 2013 #3
    I looked back over each step carefully. I must have made my mistake when solving for Pα in Equation 3. I ended up writing a script on MATLAB that solved for the three unknowns and got the answers that way. Thanks for the tips!
  5. Mar 13, 2013 #4

    Simon Bridge

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    That's where I think it first went pear shaped too ... the c^2 thing you can get away with by adopting units so that c=1. Many people do that implicitly.

    A good discipline is to do all the algebra in the symbols, leaving the numbers to last.
    Worst case is, you'll have a clear path to backtrack and troubleshoot - in this case, a lot was hidden when you combined numbers so I couldn't tell what you'd done.

    Still... no worries aye ;)
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