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HELP! Finding Initial Separation

  1. Jul 7, 2011 #1
    One particle has a mass of 3.00 10-3 kg and a charge of +7.50 µC. A second particle has a mass of 6.00 10-3 kg and the same charge. The two particles are initially held in place and then released. The particles fly apart, and when the separation between them is 0.100 m, the speed of the 3.00 10-3 kg particle is 130 m/s. Find the initial separation between the particles.


    So, I used conservation of energy and figured out this equation;

    kq^2/r = kq^2/d +1/2m1v1 + 1/2m2v2
    where r is initial separation and m1 is 3.00e-3 and m2 is 6.00e-3.

    By F1=F2, m1a1=m2a2 and I found out that v2 is v1/2. Am I right so far?
    Then I solve the equation above for r and I got 0.0928 but it was wrong.
    Can anybody help me please? I don't have any clue where I screwed up..

    Thanks!
     
  2. jcsd
  3. Jul 7, 2011 #2

    tiny-tim

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    Welcome to PF!

    Hi Vettel! Welcome to PF! :smile:

    (try using the X2 and X2 icons just above the Reply box :wink:)
    Yes. :smile:

    (But conservation of momentum would get you there quicker. :wink:)
    your v should be squared. :redface:
     
  4. Jul 7, 2011 #3
    oh wow.. what a stupid mistake... no wonder I keep getting the wrong answer! Thank you so much! :)
     
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