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Energy and charged particle collisions

  1. Feb 22, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the initial velocity of an alpha particle with a mass of ##6.64 \times 10^{-27} kg## and a charge of ##+3.2 \times 10^{-19} C##, if it undergoes a head on "collision" with a gold nucleus. You may assume the gold nucleus does not move at all during the interaction. The charge on the gold nucleus is ##+2.53 \times 10^{-17} C## and the distance of closest approach between the two is ##4.7 \times 10^{-15} m##.

    2. Relevant equations
    ##K_E\ =\ E_E##

    or

    ##K_E\ +\ E_E\ =\ E_E##?

    3. The attempt at a solution
    I've already calculated initial velocity using ##K_E\ =\ E_E## and got ##6.8 \times 10^{7} m/s## but I suspect this is incorrect.

    I'm having trouble understanding energy transformations during collisions of charged particles. I think that by "initial velocity" they mean just before the particles collide? In which case I should take into account EPE for the initial total energy. I think the final energy at the instant of closest approach must be entirely EPE because the gold nucleus is stationary and so the particles' combined velocity must be zero.
     
  2. jcsd
  3. Feb 22, 2016 #2

    PeroK

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    You're correct on the second point, but not the first. If two particles repel each other, then (unless one has an initial velocity towards the other), they will simply move further apart. Unlike gravity, however, where the velocity gets larger until the two objects actually physically collide, the velocity during repulsion will decrease until the smaller particle stops and then goes back the way it came.
     
    Last edited: Feb 22, 2016
  4. Feb 22, 2016 #3
    So initial velocity refers to the velocity of the particle when it's far away from the other, and EPE is negligible?
     
  5. Feb 22, 2016 #4

    PeroK

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    Yes, you have to assume EPE is negligible initially; otherwise, where it starts will affect the minimum distance.
     
  6. Feb 22, 2016 #5
    Okay, thanks for your help
     
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