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Coefficient of restitution problem?

  1. Sep 1, 2014 #1
    The coefficient of restitution, e, of a ball hitting the floor is defined as the ratio of the speed of the ball after it rebounds from the impact to the speed of the ball right before it hits the floor.
    Derive a formula for the coefficient of restitution when a ball is released from an initial height H and rebounds to a final height h.

    The textbook solution is set up this way:
    Before: mgH=1/2mv2before
    vbefore=√2gH

    After: 1/2mv2after=mgh
    vafter=√2gh
    With the final answer being e= √(h/H).
    I'm confused on the set up, where do the before and after equations come from, particularly mgH and mgh? I know mgy is the gravitational potential energy at a given height, but why is it set equal to 1/2mv2 in the before and after equation?
     
  2. jcsd
  3. Sep 1, 2014 #2

    Nathanael

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    Their solution comes from conservation of energy. [itex]mgH[/itex] is set equal to [itex]\frac{1}{2}mv^2[/itex] because the gravitational potential energy is converted to kinetic energy (and vice versa, on the way up).
     
  4. Sep 1, 2014 #3

    andrevdh

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    The ball motions under the influence of a conservative force -gravity. So the mechanical energy is conserved both on the way down and on the way up. That does not include the collision with the floor, so energy before and after the collision is not the same. So PE + KE = constant for both up and down, although the mechanical energy up is less than that on the way down.
     
  5. Sep 1, 2014 #4
    That kind of makes sense to me, but in the chapter the equation is listed as ΔK= -ΔUgrav. So if 1/2mv2 represents the kinetic energy, wouldn't the equation be 1/2mv2= -mgH?
     
  6. Sep 1, 2014 #5

    Nathanael

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    When ΔUgrav is positive, is ΔK positive or negative?

    When ΔUgrav is negative, is ΔK positive or negative?


    The negative sign basically means this:

    "If you increase the gravitational potential energy, the kinetic energy decreases."

    Similarly,

    "If you decrease the gravitational potential energy, the kinetic energy increases."
     
  7. Sep 1, 2014 #6

    haruspex

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    Gravitational PE is, in general, ##-\frac{GMm}r##. For constant g, that's -gmr. The gain in PE, ΔUgrav, due to a gain Δr in r is therefore -gmΔr.
    Choose whether up is positive or negative. E.g., suppose we take up to be positive.
    If a ball falls a distance H, what is Δr? Is g positive or negative? So what sign do you get for ΔUgrav?
     
  8. Sep 2, 2014 #7

    andrevdh

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    Lets consider the ball falling down and use the subscript i (initial) for the top and f (final) for the bottom. Now since the mechanical energy is conserved we have that

    Ui + Ki = Uf + Kf

    so that

    Ui - Uf = Kf - Ki

    The change in a quantity, ∇, is the final value minus the intial value so that

    -∇U = ∇K
     
  9. Sep 2, 2014 #8

    haruspex

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    Δ is the usual notation.
     
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