1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Bouncing Electrongs

  1. Feb 16, 2008 #1
    [SOLVED] Bouncing Electrongs

    Two electrons, each with mass m and charge q, are released from positions very far from each other. With respect to a certain reference frame, electron A has initial nonzero speed v toward electron B in the positive x direction, and electron B has initial speed 3v toward electron A in the negative x direction. The electrons move directly toward each other along the x axis (very hard to do with real electrons). As the electrons approach each other, they slow due to their electric repulsion. This repulsion eventually pushes them away from each other.

    What is the minimum separation (r_min) that the electrons reach?
    Express your answer in term of q, m, v, and k.

    All I understand so far is that both electrons are moving at the same nonzero speed in the same direction.

    I need help starting this question. I understand that U1 + K1 = U2 + K2 for conservation of energy.

    I also understand that U = kqQ/r for a point charge. So in this case I believe U1 = -ke^2/r, and then K1 = mv^2/r. Also I believe U2 = -ke^2/r and K2 = -3mv^2/r.

    Please tell me if these are correct assumptions, and what I need to do next.

    Thanks!
     
  2. jcsd
  3. Feb 16, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    If you mean at the moment of minimum separation: good! You'll have to find that speed.

    Good.

    Two problems: (1) the potential energy should be positive, since the charges are the same; (2) KE is always positive.

    Hint: What else is conserved?
     
  4. Feb 16, 2008 #3
    ok error on the U1 should be ke^2/r, and K1 = mv^2 and both of these added together must equal U2 = ke^2/r added with K2= 3mv^2/r, where we can take r as minimum.

    So since these must be conserved I could just solve for r_min?
     
  5. Feb 17, 2008 #4

    Doc Al

    User Avatar

    Staff: Mentor

    KE = 1/2 mv^2, not mv^2.

    Not until you determine the common speed of the electrons when they reach r_min. Hint: What else is conserved?
     
  6. Feb 17, 2008 #5
    I am not sure what else must be conserved...given the potential and kinetic energy, I can only think of charge and mass remaining conserved.
     
  7. Feb 17, 2008 #6

    Doc Al

    User Avatar

    Staff: Mentor

    Nope, there's one more conservation law that you're forgetting. :wink:
     
  8. Feb 17, 2008 #7
    conservation of linear momentum?
     
  9. Feb 17, 2008 #8

    Doc Al

    User Avatar

    Staff: Mentor

    Absolutely!
     
  10. Feb 17, 2008 #9
    ok so kinetic energy and momentum must be conserved in this system.

    so (1/2)mv^2 + (3/2)mv^2 = mv^2 (since both electrons move at the same speed in the same direction when they are at their minimum separation)?

    change in momentum final must be equal to change in momentum initial.

    i think i am doing something wrong
     
  11. Feb 17, 2008 #10

    Doc Al

    User Avatar

    Staff: Mentor

    Total mechanical energy is conserved, not kinetic energy.

    Use conservation of momentum to figure out the speed the electrons must have at minimum separation. Then you can use conservation of energy to figure out the separation.
     
  12. Feb 17, 2008 #11
    ok this is what I got so far:

    m1v1 + m2v2 = m1v1_f + m2v2_f

    based on the information given in the problem:

    m1=m2, v1 = v, v2 = 3v, and at their minimum separation they are going the same speed in the same direction so...

    mv + 3mv = mv + mv
    4mv = 2mv; this equality doesnt make much sense to me, i believe i am still making a mistake somewhere
     
  13. Feb 17, 2008 #12

    Doc Al

    User Avatar

    Staff: Mentor

    direction counts!

    They are moving towards each other--opposite directions; thus they should have opposite signs.
     
  14. Feb 17, 2008 #13
    ok so redoing it once again...

    mv - 3mv = 2mv_2

    v_2 = -v, speed at minimum separation?
     
  15. Feb 17, 2008 #14

    Doc Al

    User Avatar

    Staff: Mentor

    Sounds good.
     
  16. Feb 17, 2008 #15
    ok now that I have the speed during minimum separation, i use the conservation of mechanical energy laws...

    as written before

    U1 + K1 = U2 + K2; K1 = (mv^2/2) + (m(-3v)^2/2); K2 = (mv^2/2) + (mv^2/2)

    (ke^2/r) + (10mv^2/2) = (ke^2/r) + (mv^2)

    but in my equation above, I will not be able to solve for r...so once again I think I am missing something
     
  17. Feb 17, 2008 #16

    Doc Al

    User Avatar

    Staff: Mentor

    Since the electrons start from very far away, you can say U1 = 0 (r_1 = infinity). Then just solve for r_2 (which is r_min) in terms of k,q,m, & v.
     
  18. Feb 17, 2008 #17
    ok thank you so much for your help so far Doc Al

    this is what I know come up with:


    U1 + K1 = U2 = K2

    U1 = 0

    K1 = 5mv^2; using kinetic energy of both electrons

    U2= kq^2/r

    K2 = mv^2


    Solving for r_min I get (kq^2)/(4mv^2). Does this make sense?
     
  19. Feb 17, 2008 #18

    Doc Al

    User Avatar

    Staff: Mentor

    How did you get this? What's the speed of each electron at r_min?

    You're almost done.
     
  20. Feb 17, 2008 #19
    I got K2 by taking -v as the speed of the electron at r_min; and since there are two electrons I added their kinetic energy together. Is this ok to do? so (mv^2/2) + (mv^2/2) = mv^2?
     
  21. Feb 17, 2008 #20

    Doc Al

    User Avatar

    Staff: Mentor

    You're right. (You caught me sleeping!)

    Your answer is correct.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?