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Homework Help: Conservation of energy-Moving Charges

  1. Mar 23, 2010 #1
    1. The problem statement, all variables and given/known data

    If a helium nucleus with a mass of 6.68x10^-27 kg, a charge of +2e, and an initial velocity of 1.50x10^7 m/s is projected head-on toward a gold nucleus with a charge of +79e, how close will the helium atom come to the gold nucleus before it stops and turns around? (Assume the gold nucleus is held in place by other gold atoms and does not move.)


    2. Relevant equations

    Ki+Ui=Kf+Uf

    E=DeltaV/d a=Fe/m=eE/m=edeltaV/md


    Vf=square root [Vi^2+2ad]





    3. The attempt at a solution

    I know this is a conservation of energy for moving charges problem but I just can't seem to figure out how to calculate the distance with these equations and the given information. I feel like I am completely missing something on this one! Help please :)
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 23, 2010 #2

    kuruman

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    It is conservation of energy indeed. Which ones (if any) of the initial and final quantities in the energy conservation equation is (are) zero?
     
  4. Mar 23, 2010 #3
    The final velocity is zero.
     
  5. Mar 23, 2010 #4

    kuruman

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    You have four quantities in the energy conservation equation, Ki, Ui, Kf and Uf. Which one(s) do you think are zero?
     
  6. Mar 23, 2010 #5
    Kf right?
     
  7. Mar 23, 2010 #6

    kuruman

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    Yes, and what else?
     
  8. Mar 23, 2010 #7
    Ui is also zero
     
  9. Mar 24, 2010 #8

    kuruman

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    Correct because initially the particles are very far from each other. Now can you put the energy conservation equation together?
     
  10. Mar 24, 2010 #9
    (1.50x10^7m/s)+0=0+Uf

    Uf=1.50x10^7
     
  11. Mar 24, 2010 #10

    kuruman

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    Where and how did you get this number? Energy is expressed in Joules. What equation did you use and what numbers did you plug in? I have to know exactly what you did in order to help you.
     
  12. Mar 24, 2010 #11
    This is where I'm having trouble. The 1.50x10^7 m/s is the Vi. I'm not sure how to get the energy.
     
  13. Mar 24, 2010 #12

    kuruman

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    What is the expression that gives the potential energy of two charges separated by some distance r apart? Look it up if you don't remember.
     
  14. Mar 24, 2010 #13
    UE = k(q1q2/r) k= 9 x 10^9 N m2/C2
     
  15. Mar 24, 2010 #14

    kuruman

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    That's the one. As we said before, you need to say

    Uf = Ki

    You already have that Uf = kq1q2/r

    What expression can you write down for Kf? Look up "kinetic energy" if you have to.
     
  16. Mar 24, 2010 #15
    Kf=Ki+(Ui-Uf)

    Kf=1/2mv^2
     
  17. Mar 24, 2010 #16

    kuruman

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    OK. Can you find a number for this kinetic energy in Joules? You know everything that goes in the equation.
     
  18. Mar 24, 2010 #17
    Kf=(1/2)(6.68x10^-27kg)(1.50x10^7m/s)^2=

    =7.52x10^-13 J
     
  19. Mar 24, 2010 #18

    kuruman

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    What do you think you should do next?
     
  20. Mar 24, 2010 #19
    Well I'm a little confused as to why the equation is equal to Kf and not Ki because isn't Kf=0?
     
  21. Mar 25, 2010 #20

    kuruman

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    You are correct, 7.52x10-13 J is Ki not Kf. Based on all that is said above, what is the next step?
     
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