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Force and Motion Problem

  1. Oct 23, 2008 #1
    1. The problem statement, all variables and given/known data
    An electron with a speed of 1.2 x 107 moves horizontally into a region where a constant vertical force of 4.5 x 10-16N acts on it. The mass of the electron is 9.11 x 10-31kg. Determine the vertical distance the electron is deflected during the time it has moved 30mm horizontally.

    2. Relevant equations

    3. The attempt at a solution
    4.5 x 10-16N
    | /|height=?
    |/ |
    |--|------------------> 1.2 x 107

    Honestly, I'm not exactly sure where to start. This seems like a rather simple problem but I don't know how to approach it...
  2. jcsd
  3. Oct 23, 2008 #2


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    Homework Helper

    Welcome to PF.

    First figure what the acceleration on the particle would be.

    Then it's just like a rock thrown horizontally off a building and affected by gravity - only using the numbers of the problem - 30mm, a, v.
  4. Oct 23, 2008 #3

    Doc Al

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    Staff: Mentor

    The force only acts vertically, so the horizontal motion is unaffected. Treat horizontal and vertical motion separately.

    Edit: Oops... LowlyPion beat me to it! :smile:
  5. Oct 23, 2008 #4
    Alright so...

    a=F/M = (4.5 x 10-16N) / (9.11 x 10-31kg) = 4.9x1014

    t = d/v = 30mm/(1.2 x 107) = 2.5 x 10-9

    Now that we have a and t, we can solve for vertical distance:

    D = vt-1/2at2

    = (1.2 x 107)*(2.5 x 10-9)-(1/2)*(4.9x1014)*(2.5 x 10-9)2
    = -1513.63 m
    = -1.5mm

    Is that right? Because that's very nearly the books answer.
    One more question, why is the answer negative? The books answer is positive 1.5mm.
  6. Oct 23, 2008 #5


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    Homework Helper

    Not quite. Because your initial velocity is normal to the deflection (the v in that direction is 0).

    This is your D
  7. Oct 23, 2008 #6
    Oh I see, so it should be like this:

    D = v0t+1/2at2

    = 0*(2.5 x 10-9)+(1/2)*(4.9x1014)*(2.5 x 10-9)2

    Got it. Thanks a lot.
    Last edited: Oct 23, 2008
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