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Two charged beads and maximum velocity.

  1. Jan 24, 2012 #1
    I apologize for the brevity but the forums logged me out and ate my entire post. I just want to get the information written down this time.

    1. The problem statement, all variables and given/known data
    We have two charged beads, 12.0cm apart. What are their maximum velocities when released?
    Bead A: -5.0nC, 15.0g
    Bead B: -10.0nC, 25.0g

    2. Relevant equations

    kqq/r

    Ui +Ki = Uf + Kf

    3. The attempt at a solution

    I have already found the potential energy of the system, but my answer for the first bead is incorrect - I am getting about 2.23 for the first bead and the book indicates ~1.77.

    I thought I just needed to use Vf = sqrt(2Ui/r), but I am apparently missing part of the process. I have been at this for five hours and really have no idea how to proceed. Any help would be greatly appreciated!
     
  2. jcsd
  3. Jan 24, 2012 #2

    tiny-tim

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    Welcome to PF!

    Hi Szechuan! Welcome to PF! :smile:

    hmm :rolleyes: … another equation might help …

    have you tried conservation of momentum? :wink:
     
  4. Jan 24, 2012 #3
    Re: Welcome to PF!

    Hi tim, thanks for the quick reply! I have absolutely no idea how that would apply here. Our book starts at chapter 26 and only deals with electrical stuff.
     
  5. Jan 24, 2012 #4

    tiny-tim

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    Re: Welcome to PF!

    uhh? :redface: I didn't know you could get books that started at chapter 26 :confused:

    Anyway, the total momentum (mass times velocity) starts at 0, and so it has to stay 0 for ever. :wink:
     
  6. Jan 25, 2012 #5
    Hi tim,

    I'm afraid I still don't understand. I grasp that momentum is conserved, and that this means the different masses will move at different speeds under an identical force.

    I thought momentum was already accounted for by the mass component of the velocity equation. Since that isn't the case, I do not know how to relate these two concepts to properly assign the right proportions of the final speed. I tried splitting them based on their proportion to the total mass but that still doesn't work. I don't want anyone to do the calculations for me but if you could clarify how I incorporate momentum to solve this I would be grateful!

    PS: The textbook was divided into sections - mechanics was last term but I have not completed that yet. There were some other issues last term preventing me from taking a full course load and I had to drop it.
     
  7. Jan 25, 2012 #6

    tiny-tim

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    Hi Szechuan! :smile:
    ok, call the speeds vA and vB

    then your conservation of momentum equation is:

    mAvA = mBvB

    Now combine that with your energy equation, and solve. :smile:
     
  8. Jan 25, 2012 #7
    Do you mean I should multiply the final velocity by the bead mass? I tried that and it did not work out.

    I am not sure I understand how to combine the two equations. I am posting from my phone but I will test a few things out after class and see what I get.
     
  9. Jan 25, 2012 #8
    Sorry, I've been rearranging things but I just don't understand how those are supposed to fit together; I don't even know where or how I am supposed to combine these equations.

    I don't even definitively know what my "energy equation" is.
     
  10. Jan 25, 2012 #9

    tiny-tim

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    Hi Szechuan! :smile:

    Your energy equation is (KE + PE)before = (KE + PE)after.

    It will include vA and vB.

    Together with the momentum equation, that makes two equations with vA and vB

    so you'll have two equations and two unknowns (vA and vB), so you should be able to solve them (and find the unknowns). :smile:
     
  11. Jan 25, 2012 #10
    OH! So by combine you mean rearrange the equations to be equivalent to V and thus equivalent to each other? I'm not sure how that lets me solve for v but I'll try it!
     
  12. Jan 26, 2012 #11

    tiny-tim

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    How's it going?

    What equations do you have? :smile:
     
  13. Jan 27, 2012 #12
    I haven't had a chance to try out the new info just yet. The class average for last week's quiz* on charge + fields was quite low, so our professors opted to do some review of that material over the past couple days. Since they wouldn't be able to finish the material this week, we were granted an extension on the assignment; of course, it was granted so late in the week that it didn't really give us time to work on other things. :grumpy:

    Tomorrow I am going to sit down with your hints (and a copy of last term's mechanics text from the class I haven't taken) and try to suss out the applicable relationship between momentum, conservation of energy, and electrical potential. I'll report back once I have my results.

    Thanks!

    *PS: My mark was quite good (~B-) given the class has a ~40% failure rate. I would have done better, but I used the wrong formula for an infinite line of charge (Which was a very silly mistake! I chose to go from memory instead of using the crib sheet they provided!)
     
  14. Jan 28, 2012 #13
    Update:

    I'm still working on it, but so far I'm just ending up with equations that simplify back to the equivalent of Va = Va.

    PS: Is there a reason the forums are timing me out after ten minutes? I don't have time to compose a reply before it starts throwing errors.
     
  15. Jan 28, 2012 #14

    tiny-tim

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    Hi Szechuan! :wink:


    Show us your equations (one for energy, one for momentum), and then we can see where you're going wrong. :smile:
    I don't know: try asking in the Feedback forum.
     
  16. Jan 28, 2012 #15
    Conservation:
    1/2mava2 + 1/2mbvb2 = Ui

    Momentum:
    MaVa = MbVb

    Tautology:
    Ek = p2/m = 1/2mv2

    I'm sure the fact that I haven't taken physics in a decade is probably contributing to my problems...
     
  17. Jan 28, 2012 #16

    tiny-tim

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    Looks ok …

    now put vb = mava/mb (from the second equation) into the first equation, and solve. :smile:
     
  18. Jan 28, 2012 #17
    I'm subbing that in, but I am still getting the wrong answer.

    (1/2)*(MaVa2) + (1/2)*(MbVb2/Mb) = Ui

    Va = root(2MbUi/Ma+Ma2)


    Comes to ~50 cm/s when I need 1.77.

    I actually tried subbing in momentum earlier and nothing would work out correctly, which is why I asked.

    Edit: Correcting typos.
     
  19. Jan 28, 2012 #18

    tiny-tim

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    How did you get that? :confused:

    (it's not even dimensionally correct … you cant have mass and mass2 added together)

    Show us your intermediate steps.
     
  20. Jan 28, 2012 #19
    Math and I don't get along. I had 90% on all of my theoretical stuff in calculus, but barely passed because I make so many arithmetic errors. It's not that I rush, I just can't see them. I did those calculations three times and it looked perfect to me.

    Process:
    Ka + Kb = Ui

    Vb = (MaVa)/Mb

    Kb = MbVb2/2

    = (Mb/2)*(MaVa/Mb)2

    = Mb/2*(Ma2Va2/Mb2)

    = Ma2Va2/2Mb

    Ka = (MaVa2)/2

    Ka + Kb = (MaVa2)/2 + (Ma2Va2)/2Mb = Ui

    = Va2(Ma/2 + Ma2/2Mb) = Ui

    = Va2 = Ui/(Ma/2 + Ma2/2Mb)

    = root(Ui/(Ma/2 + Ma2/2Mb))

    = root((Ui/2)*(Ma + Ma2/Mb))
     
  21. Jan 28, 2012 #20

    tiny-tim

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    It's only the last line that's wrong …
    at the last (crucial!) moment, you've got very confused as to what's above the line, and what's below! :wink:

    (btw, a slight short-cut would be to write mv2 = (mv)2/m :wink:)
     
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