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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?

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- Thread starter kdrobey
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- #1

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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

Doc Al

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Show what you've done so far.

Hint: What's conserved?

Hint: What's conserved?

- #3

nicksauce

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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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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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The basic expression for the KE ofI am not sure what you are looking for

- #13

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1/2mVo^2

- #14

Doc Al

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Good. That's the KE of one proton. So what's the total KE of both protons?1/2mVo^2

- #15

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so,

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

?

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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