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Distance Of Closest Approach |
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| Jul10-08, 06:42 PM | #1 |
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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 |
| 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?
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| Jul10-08, 06:49 PM | #4 |
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Distance Of Closest Approach
ok, will do
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| Sep28-08, 08:44 PM | #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? |
| Sep29-08, 05:01 AM | #6 |
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Realize that both protons are moving and thus have kinetic energy.
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| Sep29-08, 11:25 AM | #7 |
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So, I am using the wrong formula? Not getting it...
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| Sep29-08, 11:49 AM | #9 |
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I thought that is what I did
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| Sep29-08, 11:55 AM | #10 |
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What's the KE of each proton? (Symbolically--no need for numbers yet.)
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| Sep29-08, 11:57 AM | #11 |
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I am not sure what you are looking for
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| Sep29-08, 12:04 PM | #12 |
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| Sep29-08, 12:09 PM | #13 |
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1/2mVo^2
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| Sep29-08, 12:11 PM | #14 |
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| Sep29-08, 12:33 PM | #15 |
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so,
1/2mVo^2 + 1/2mVo^2= Kq^2/r ? |
| Sep29-08, 12:43 PM | #17 |
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Okay I got the correct answer 9.6e-14 .... finally :-)
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