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Completely Elastic Proton Collision

  1. Dec 10, 2007 #1
    [SOLVED] Completely Elastic Proton Collision

    In my physics class we tried to solve this problem but nobody in the class could figure it out and the teacher was having difficulty with it. It is possible that the book put in a problem that isn't possible to solve...

    i am not very good with latex coding so feel free to ask about my sub/superscript system

    A proton (atomic mass 1.01 u) with a speed of 518 m/s collides elastically with another proton at rest. The original proton is scattered 64.0 degrees from its initial direction.
    (A) What is the direction of the velocity of the target proton after the collision?
    (B) What are the speeds of the two protons after the collision?

    1. The problem statement, all variables and given/known data

    A proton (atomic mass 1.01 u) with a speed of 518 m/s collides elastically with another proton at rest. The original proton is scattered 64.0 degrees from its initial direction.

    Variables
    m[tex]_{1}[/tex]=1.01 u
    v[tex]_{1}[/tex]=518 m/s
    m[tex]_{2}[/tex]=1.01 u
    v[tex]_{2}[/tex]=0 m/s
    v[tex]^{'}_{1}[/tex]=?
    v[tex]^{'}_{2}[/tex]=?
    [tex]\vartheta[/tex]=64 degrees (b/n path of first proton before and after collision)
    [tex]\varphi[/tex]=? (b/n path of second proton after collision and path of first proton before collision)

    2. Relevant equations

    Total momentum before equals total momentum after
    P[tex]_{T}[/tex]=P[tex]^{'}_{T}[/tex]
    momentum = mass X velocity
    P=mv

    3. The attempt at a solution

    we set up a force diagram so that the path of proton #1 is a horizontal line
    vvvv total momentum in X direction vvvv
    P[tex]_{T,X}[/tex]= 518 u[tex]\times[/tex]m/s=V[tex]^{'}_{1}[/tex]cos(64)+V[tex]^{'}_{2}[/tex]cos([tex]\varphi[/tex])
    we have three variables in this equation and this makes a problem to solve for just one variable

    Total momentum in Y direction is zero prior to collision and therefore is zero after collision
    P[tex]_{T,Y}[/tex]=0 u[tex]\times[/tex]m/s=V[tex]^{'}_{1}[/tex]sin(64)+V[tex]^{'}_{2}[/tex]sin([tex]\varphi[/tex])

    this is the point where I have absolutely no idea what to do and apparently my teacher and the rest of the class as well.....
    Any help/hint(s) would be very greatly appreciated
    Thanks,
    Akatz
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 10, 2007 #2
    Are we supposed to ignore the Coulumb repulsion? Could we replace the protons with billiard balls of the same mass?
     
  4. Dec 10, 2007 #3
    yes you can ignore couloumbs law
    good idea....think of the protons as two billiard balls
     
  5. Dec 10, 2007 #4
    Ok, so right now you have 2 equations with 3 unknowns. You need one more equation in order to solve this problem. Try using conservation of energy (only accounting for kinetic energy). You'll end up with an equation for each p(x-direction), p(y-direction), and energy with unknowns, v1, v2 and the angle the 2nd ball makes with the original trajectory. Does this help?
     
  6. Dec 10, 2007 #5
    thats what i did
    i have the equation for conservation of momentum in the y axis
    and i have conservation of momentum in the x axis
    and that is where i am stuck
    thanks for the help
    i could isolate the unknown i am trying to find but it would be defined in terms of other unknowns
    guh...
     
  7. Dec 10, 2007 #6
    Use conservation of energy. Only kinetic energy terms will appear. This will give you 3 equations with 3 unknowns. Have you seen conservation of energy before?
     
  8. Dec 10, 2007 #7
    yes
    ok i think i get what your saying now
    Thanks a ton!
     
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