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Homework Help: Question envolves the centre-of-mass and momentum.

  1. Feb 28, 2010 #1
    1. The problem statement, all variables and given/known data

    An electron collides elastically with a hydrogen atom that is initially at rest. Assume all the motion occurs along a straight line. What fraction of the electron's initial kinetic energy is transferred to the atom? (Take the mass of the hydrogen atom to be 1840 times the mass of an electron.)

    2. Relevant equations

    KE = (1/2)mv^2

    uapp = vsep

    p = mv

    3. The attempt at a solution

    I used the formulas above, and came up with 3 equations:

    p-cons

    u1x = v1x + 1840v2x

    u1x = v2x - v1x

    KE

    u1x^2 = v1x^2 + 1840v2x^2

    But i keep trying and can't use them to find any one velocity.
     
  2. jcsd
  3. Feb 28, 2010 #2

    Doc Al

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    Just combine these to get V1 in terms of U1. That's all you need to compare initial and final KE of the electron.

    (Note that the second of these two equations comes from combining conservation of momentum with conservation of energy.)
     
  4. Feb 28, 2010 #3

    vela

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    You can't because you have three velocities and really only two equations. The equation u1x = v2x - v1x was derived by combining the conservation of momentum and conservation of kinetic energy equations.

    In any case, it doesn't matter because the problem is asking for the fraction of the electron's initial kinetic energy transferred to the atom. If you write that down mathematically, you'll see what you need is a ratio of two of the velocities, which is a quantity you can solve for.
     
  5. Feb 28, 2010 #4
    I have tried combining the two equations, but that only gives me a new equation with two unknown velocities, and I need all three velocities to find the inital and final KE.

    How would I use an equation with 2 unknown variables to compare the initial and final KE??

    Help??
     
  6. Feb 28, 2010 #5

    Doc Al

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    As I said earlier, just solve for V1 in terms of U1. Then you can compare the ratio of the final and initial KE of the electron. (You don't need all three velocities.)
     
  7. Feb 28, 2010 #6
    So your saying that when you add the equaitons and get:

    V1i = 1841v2f

    How can this be used to compare the two KE?

    since KEi = (1/2)mv^2

    KEi = (1/2)(1)(1841v2f)^2

    It doesn't give an actual number, and again there ends up with two variables

    and then KEf would be:

    KEf = (1/2)(1)(v1f)^2 + (1/2)(1840)(v2f)^2

    neither of these can really be compared to give a percentage??
     
  8. Feb 28, 2010 #7

    Doc Al

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    I didn't say anything about adding the equations. Solve for V1 in terms of U1. (Hint: Eliminate V2.)
     
  9. Feb 28, 2010 #8
    You did say to combine the equations. But, when you do solve for u1, there are still 2 variables, and you can't find the KEf or KEi without at least one of these, and therefore can't compare the two.
     
  10. Feb 28, 2010 #9

    Doc Al

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    You have two equations and three unknowns. Eliminate V2. Then express V1 in terms of U1.

    Do it.
     
  11. Feb 28, 2010 #10
    Ok:

    u1x = v1x + 1840v2x

    u1x = v2x - v1x

    so...

    v2x= u1x + v1x

    and then...

    u1x = v1x + 1840(u1x + v1x)

    u1x = v1x + 1840u1x + 1840v1x

    -1839u1x = 1841v1x

    u1x = -1841v1x/1839

    there.

    What is this even showing me. I have still have 2 unknowns?
     
  12. Feb 28, 2010 #11

    Doc Al

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    Good. (I meant to say, Express V1 in terms of U1.)

    Now:
    What's the initial KE of the electron? (This will be in terms of U1, of course.)
    What's the final KE of the electron? (Express it in terms of U1.)

    Compare! (Find the ratio: KEf/KEi)
     
  13. Feb 28, 2010 #12
    So, I did what you said, and its starting to look like an answer, thanks for that.

    But now I have:

    u1x = (u1x)^2 ((1839/1841)^2 + (1840)(2/1840)^2)

    with the left side being inital and the right side being the final KE's

    I was wondering though, how would I convert that into a percentage?
     
    Last edited: Feb 28, 2010
  14. Feb 28, 2010 #13

    vela

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    That equation couldn't possibly be correct. You have a velocity on the LHS, not a KE, and the RHS side is a velocity squared, also not a KE.

    Write down KEi in terms of u1x and KEf in terms of v1x. Divide one by the other. You'll end up with a ratio which you can evaluate given your earlier result.
     
  15. Feb 28, 2010 #14

    Doc Al

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    Not sure what you did here.

    KEi = 1/2m(U1)^2
    KEf = 1/2m(V1)^2

    Express V1 in terms of U1, then rewrite KEf in terms of U1.
     
  16. Feb 28, 2010 #15
    Oh i think its because i used KEf = 1/2m1(v1f)^2 + 1/2m2(v2f)^2

    Wouldnt that be right?
     
  17. Feb 28, 2010 #16

    vela

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    No, that ratio would come out to be 1 because it's an elastic collision, i.e. total KE(initial)=total KE(final). Go back and see what the question was actually asking for.
     
  18. Mar 1, 2010 #17

    Doc Al

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    No. Just compare the initial and final KE of the electron, not the total KE.
     
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