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One Dimensional motion of particle in a potential field

  1. Sep 8, 2014 #1
    A classical particle constrained to move in one dimension (x) is in the potential field V(x) = V0(x – a)(x –b)/(x – c)^2, 0 < a < b < c < ∞.
    a. Make a sketch of V
    b. Discuss the possible motions, forbidden domains, and turning points. Specifically, if the
    particle is known to be at x → ∞ with E = 3V0(b – 4a + 3c)/(c – b), at which value of x
    does it reflect?

    I'm not sure how to approach this problem any tips or advice would be greatly appreciated.
     
  2. jcsd
  3. Sep 8, 2014 #2

    BvU

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    Hello Saban, and welcome to PF.
    Here's my tip:
    Start with a)
     
  4. Sep 8, 2014 #3
    Thanks I have it drawn and reviewed it with my professor who told me the sketch is correct. I'm still not sure as to how to complete part b though.
     
  5. Sep 8, 2014 #4

    BvU

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    I don't have your sketch, but I can kind of telepatically pick it up :smile:
    So you see a big peak at x=c. And a minimum between a and b. Right ?

    Good that you already did the first discussion in part b) and now want to complete it.

    In the completion of part b the particle comes from the right, with a given E.
    Is it clear to you that it doesn't have enough energy to get to x < c ?

    So what you know about the turning point is that at that point there is no more kinetic energy, just potential energy.
    Since the given E(x≈∞) = Ekin(x≈∞) + V(x≈∞) is the same as E at the turning point x, you obtain an equation in x.
     
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