Distance Of Closest Approach

  • Thread starter kdrobey
  • Start date
  • #1
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Homework Statement



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?

Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
Doc Al
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Show what you've done so far.

Hint: What's conserved?
 
  • #3
nicksauce
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5
Your are expected to show an attempt at the problem. What are your thoughts on the problem?
 
  • #4
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ok, will do
 
  • #5
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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?
 
  • #6
Doc Al
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Realize that both protons are moving and thus have kinetic energy.
 
  • #7
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So, I am using the wrong formula? Not getting it...
 
  • #8
Doc Al
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You need to set total KE equal to PE.
 
  • #9
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I thought that is what I did
 
  • #10
Doc Al
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What's the KE of each proton? (Symbolically--no need for numbers yet.)
 
  • #11
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I am not sure what you are looking for
 
  • #12
Doc Al
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I am not sure what you are looking for
The basic expression for the KE of each proton, like you used in post #5.
 
  • #13
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1/2mVo^2
 
  • #14
Doc Al
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1/2mVo^2
Good. That's the KE of one proton. So what's the total KE of both protons?
 
  • #15
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so,

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

?
 
  • #16
Doc Al
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Good.
 
  • #17
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Okay I got the correct answer 9.6e-14 .... finally :-)
 

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