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Force (potential, I think) to Kinetic Energy - Two charges

  • Thread starter gulz
  • Start date
  • #1
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Two particles each have a mass of 7.7 x 10-5 kg. One has a charge of +5.9 x 10-6 C, and the other has a charge of -5.9 x 10-6 C. They are initially held at rest at a distance of 0.91 m apart. Both are then released and accelerate toward each other. How fast is each particle moving when the separation between them is one-half its initial value?

mass particle = 7.7x10^-5 kg
q1 = 5.9x10^-6 C
q2 = -5.9x10^-6 C
r0 = .91m
r1 = .455m
v0 = 0 m/s
v1 = ???


Homework Equations



F = kq1*q2/(r^2)
KE = .5 * m * v^2

The Attempt at a Solution



I believe the first thing to do would be to gather the force between the charges, using
F = k*q1*q2/(.91^2).

The next thing I think I should do is substitute F in for KE, in the equation
KE = .5 * m * v^2

One of the problems I'm not sure about is how the KE equation factor in two charges moving in opposite directions.

By substituting F for KE, am I already factoring in all the vectors I need?

edit: I think another way of describing what I'm feeling is: Is the velocity in the KE equation the velocity of each charge, regardless of direction?
 

Answers and Replies

  • #2
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Oh, and if it helps explain, the charges are set up like:


<-----q1...q2----->
 

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