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

  1. Feb 15, 2012 #1
    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 = ???


    2. Relevant equations

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

    3. 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?
     
  2. jcsd
  3. Feb 15, 2012 #2
    Oh, and if it helps explain, the charges are set up like:


    <-----q1...q2----->
     
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