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Homework Help: Distance Of Closest Approach

  1. Jul 10, 2008 #1
    1. The problem statement, all variables and given/known data

    Two Protons are moving directly toward one another. When they are very far apart, their initial speeds are 2.1 x 10^6 m/s. What is the distance of closest approach?

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 10, 2008 #2

    Doc Al

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    Show what you've done so far.

    Hint: What's conserved?
     
  4. Jul 10, 2008 #3

    nicksauce

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    Your are expected to show an attempt at the problem. What are your thoughts on the problem?
     
  5. Jul 10, 2008 #4
    ok, will do
     
  6. Sep 28, 2008 #5
    Okay I have the same problem except my initial speed is 1.2*10^6

    So, I used the equation 1/2mVo^2 = Kq^2/r

    I am solving for r and I keep getting 1.92 * 10^-13, but it is wrong. What am I doing wrong?
     
  7. Sep 29, 2008 #6

    Doc Al

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    Realize that both protons are moving and thus have kinetic energy.
     
  8. Sep 29, 2008 #7
    So, I am using the wrong formula? Not getting it...
     
  9. Sep 29, 2008 #8

    Doc Al

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    You need to set total KE equal to PE.
     
  10. Sep 29, 2008 #9
    I thought that is what I did
     
  11. Sep 29, 2008 #10

    Doc Al

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    What's the KE of each proton? (Symbolically--no need for numbers yet.)
     
  12. Sep 29, 2008 #11
    I am not sure what you are looking for
     
  13. Sep 29, 2008 #12

    Doc Al

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    The basic expression for the KE of each proton, like you used in post #5.
     
  14. Sep 29, 2008 #13
    1/2mVo^2
     
  15. Sep 29, 2008 #14

    Doc Al

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    Good. That's the KE of one proton. So what's the total KE of both protons?
     
  16. Sep 29, 2008 #15
    so,

    1/2mVo^2 + 1/2mVo^2= Kq^2/r

    ?
     
  17. Sep 29, 2008 #16

    Doc Al

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    Good.
     
  18. Sep 29, 2008 #17
    Okay I got the correct answer 9.6e-14 .... finally :-)
     
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