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Distance Of Closest Approach

 
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Jul10-08, 06:42 PM   #1
 

Distance Of Closest Approach


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
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Jul10-08, 06:45 PM   #2
 
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Show what you've done so far.

Hint: What's conserved?
Jul10-08, 06:47 PM   #3
 
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Your are expected to show an attempt at the problem. What are your thoughts on the problem?
Jul10-08, 06:49 PM   #4
 

Distance Of Closest Approach


ok, will do
Sep28-08, 08:44 PM   #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?
Sep29-08, 05:01 AM   #6
 
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Realize that both protons are moving and thus have kinetic energy.
Sep29-08, 11:25 AM   #7
 
So, I am using the wrong formula? Not getting it...
Sep29-08, 11:30 AM   #8
 
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You need to set total KE equal to PE.
Sep29-08, 11:49 AM   #9
 
I thought that is what I did
Sep29-08, 11:55 AM   #10
 
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What's the KE of each proton? (Symbolically--no need for numbers yet.)
Sep29-08, 11:57 AM   #11
 
I am not sure what you are looking for
Sep29-08, 12:04 PM   #12
 
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Quote by nckaytee View Post
I am not sure what you are looking for
The basic expression for the KE of each proton, like you used in post #5.
Sep29-08, 12:09 PM   #13
 
1/2mVo^2
Sep29-08, 12:11 PM   #14
 
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Quote by nckaytee View Post
1/2mVo^2
Good. That's the KE of one proton. So what's the total KE of both protons?
Sep29-08, 12:33 PM   #15
 
so,

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

?
Sep29-08, 12:35 PM   #16
 
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Good.
Sep29-08, 12:43 PM   #17
 
Okay I got the correct answer 9.6e-14 .... finally :-)
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