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Conservation of Energy - Max height WITH Air resistance

  1. Apr 28, 2015 #1
    1. The problem statement, all variables and given/known data
    A ball [itex](m=0.7kg)[/itex] is dropped from a height [itex](h=30m)[/itex]. The air resistance force is given by the formula [itex]F=kv^2[/itex] where [itex]k=0.0228[/itex] and [itex]v[/itex] is the ball's speed. The ball bounces after a perfectly elastic collision.
    • Calculate the maximum height the ball reaches after the collision!

    2. Relevant equations
    Since we are talking about a perfectly elastic collision, the kinetic energy before and after the collision is the same.
    [itex]E_k = E_k'[/itex].

    Also the law of conservation of energy holds.

    3. The attempt at a solution

    I am not sure what I'm doing wrong. Here's what I tried:
    • DURING FALL
    First I tried to write the energy state during the fall. Left side of the equation is the state at [itex]t=0[/itex], and the right side is at the moment just before the collision.
    [tex] E_k + E_p + W = E_{k}' + E_{p}'[/tex]
    [itex]W[/itex] is the work that the air resistance force does on the ball.
    Kinetic energy at start equals 0, and potential energy at the moment of collision equals 0.
    That leads to (*):
    [tex] E_p + W = E_{k}'[/tex]
    The speed of the rigid body during the free fall is given by [itex]v^2=2gh[/itex], which leads to [tex]F(v)=kv^2 \Rightarrow F(s)=2kgs[/tex]
    The work of the air resistance along the fall
    [tex]W=\int F(s) \, ds=\int 2kgs\, ds=2kg\int s\,ds=2kg\frac{s^2}{2}=kgs^2[/tex]
    I repeat one of the previous equations (*), expanded:
    [tex]mgh - kgh^2 = \frac{mv^2}{2}[/tex]
    Here is the thing I am not sure about: should the Work be negative (like I have written) since the direction of the force that does the work is opposite of the movement?

    Entering the given values in this equation I get [itex]v_{collision}=3.668 [m/s][/itex].
    • DURING RISE
    Then I tried to write the energy state after collision and during the rise. Left side of the equation is the state at the moment just after the collision and the right side is the moment in which the height is at maximum [itex](y_{max})[/itex] (after the collision).
    [tex] E_k + E_p + W = E_{k}' + E_{p}'[/tex]
    [itex]W[/itex] is the work that the air resistance force does on the ball during its rise.
    Potential energy at the moment of collision equals 0, and kinetic energy at the highest point equals 0.
    That leads to (**):
    [tex] E_k + W = E_{p}'[/tex]
    I used the previously calculated formula for work, only this time, the height is not [itex]h[/itex], but [itex]y_{max}[/itex].
    Now I have:
    [tex]\frac{mv^2}{2} - kgy_{max}^2 = mgy_{max}[/tex]

    This gives me a quadratic equation for [itex]y_{max}[/itex].

    The roots of this quadratic equation are ([itex]v[/itex] entered as [itex]v_{collision}[/itex]):
    [itex]y_{max1} = 0.671[m],y_{max2}=-31.372[m][/itex].

    The latter has no sense for height, and the former is incorrect. The correct answer is [itex]9.15[m][/itex].

    What do I do incorrectly?
    Any hints?
     
  2. jcsd
  3. Apr 28, 2015 #2

    BvU

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    Hello KP, welcome to PF :smile: !
    ##
    v^2=2gh## gets spoilt by the loss of kinetic energy due to air resistance, so you'll have to revert to a differential equation. Would that work for you ?
     
  4. Apr 28, 2015 #3
    It probably would, but I am not sure what you mean. You mean diff equation for [itex]v[/itex]? Unfortunatelly, I am not sure I know what I need to do.
     
  5. Apr 28, 2015 #4

    BvU

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    Yes, I mean ##F = ma## which is indeed a differential equation for ##v##. Fill in ##ma = m{dv\over dt}## and the two forces acting on ##m##.
     
  6. Apr 28, 2015 #5

    haruspex

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    This is a valiant attempt to account for energy loss due to air resistance, but it doesn't work.
    You start by computing the speed it would have after a fall of a certain distance with no resistance, then find the drag at that point. But it's too late, since it would not have reached that speed. As a result, you overestimate the loss.
    As BvU says, you need to start with forces and obtain a differential equation.
     
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