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Conservation of electric potential energy

  1. Oct 4, 2015 #1
    1. The problem statement, all variables and given/known data
    An alpha particle is a nucleus of helium. It has twice the charge and four times the mass of the proton. When they were very far away from each other, but headed toward directly each other, a proton and an alpha particle each had an initial speed of 3.6×10−3c, where c is the speed of light. What is their distance of closest approach? There are two conserved quantities. Make use of both of them. (c = 3.00 × 10^8 m/s, k = 1/4πε0 = 8.99 × 10^9 N · m2/C2, e = 1.60 × 10^-19 C, mproton = 1.67 x 10^-27kg)

    2. Relevant equations
    1/2 m v^2 + 1/2 m v^2 = q(kq / r)

    3. The attempt at a solution

    I used the above equation to get 9.46 * 10^-14 as the radius, but that is not the correct answer. I made the assumption that the electric potential far away is 0 and that when the distance is closest they have no velocity.
     
  2. jcsd
  3. Oct 4, 2015 #2

    mfb

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    That would violate momentum conservation.
     
  4. Oct 4, 2015 #3
    How would I incorporate the conservation of momentum into this? There don't seem to be any collisions.
     
    Last edited: Oct 4, 2015
  5. Oct 4, 2015 #4

    mfb

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    Momentum is always conserved, it does not matter if there are collisions or not. The particles will move at the point of closest approach, but what can you say about their relative motion? That will modify the energy balance, and conservation of momentum allows to find the additional parameter you have to consider.
     
  6. Oct 4, 2015 #5
    Shouldn't they get closer and closer, until they both have turned all of their kinetic energy into electric potential, then the electric energy converts back into velocity in the opposite direction? Like two magnets in a tube moving towards each other?

    I don't know much about this, my physics professor never teaches us anything that he puts onto the online homework. I already have the correct answer, because it was a multiple choice and I guessed, but I don't know how to get it.
     
  7. Oct 4, 2015 #6

    mfb

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    In the center of mass frame, that happens. In the lab frame, they cannot both stop at the same time, that would violate conservation of momentum.

    The helium nucleus is much heavier than the proton. Imagine a large truck hitting a car moving in the opposite direction. The car won't stop the truck. You do not have a direct contact here, but that does not matter - the conservation laws still apply.
     
  8. Oct 4, 2015 #7
    Ah, I see. So I would use m1v1 - m2v2 = -vf(m1+m2) to get the velocity where they are not moving relative to each other, then use their kinetic energies to find what the electric potential is and then the distance between them. Thanks!
     
  9. Oct 5, 2015 #8

    mfb

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    Right.
     
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