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Relating quadratic graphs & bouncing movement

  1. Feb 22, 2015 #1
    1. The problem statement, all variables and given/known data
    Hi everyone,
    I am doing a lab currently (and it looks like a few other people on here have had similar questions...) and I'm having trouble with one of the concepts. I had to drop a ball from a height and measure the movement with a motion sensor, which was graphed with the Logger Pro program. I had to create a quadratic equation for a few of the curves created in the graph, which the program does automatically. I have attached a picture (hopefully it is clear enough) of the resulting graph and the quadratic equations for 4 of the curves.

    2. Relevant equations
    Ax^2 + Bx + C
    y = yo + vo(t) - 1/2 g(t)^2

    3. The attempt at a solution
    The first part of the question is:
    "Write a clear interpretation of the meaning of each parameter in this equation."
    Which I understand well enough. I do understand how the quadratic equation relates to the kinematic equation y = yo + vo(t) - 1/2 g(t)^2.
    A = 0.5g
    B = vo
    C = yo

    What I'm having problems with is the following part:
    "From the fit results of each interval, you should notice that the B (parameter) increases as the ball makes a new bounce. If B is interpreted as the initial velocity of the ball for the corresponding bounce, this seems to contradict the observed loss of mechanical energy after each bounce. Provide an explanation of this apparent discrepancy."

    I don't understand why, for example in the first interval, B = 15.31, and C = -11.85, yet A seems fairly close to the actual value of 0.5g. Furthermore, I don't understand why B and C are in fact getting larger in the later intervals when C should be 0 (if it really is y0) and B should be getting smaller because the ball, in real life, is losing energy. I think I may have worked out from other posts on here that B and C are not the actual v0 and y0 values (although still not too sure why), but I think I may have worked out how to find them. For example, for the first interval, the bounce actually begins at t = 1.25 and the max height (y = 0.742) is reached at t = 1.65 (this is in the table on the left), so I worked out that, for the bounce to actually begin at t = 0, then max height is reached at t = 0.4, so (t' = t-t0) and y0 = 0:

    y = y0 + v0(t) - 1/2g(t)^2
    v0 = (y + 1/2g(t)^2 ) / t
    v0 = 3.71

    Does this make sense? If so, why does the graph tell me that B (or v0) = 15.31? What is the significance of this number? The only thing that I can think of is that y0 = -11.85, literally... so the intercept occurs at y = -11.85 (which obviously is impossible since the ground is at 0m but maybe the computer doesn't know that?) and at this negative intercept Fig 3.png , the velocity is 15.31, even though this isn't the actual moment the interval begins. Can anyone confirm/deny/correct/comment on this?

    Any help is appreciated!! :)
     
  2. jcsd
  3. Feb 22, 2015 #2
    If you just consider the function y(t) = At2 + Bt + C, then, in terms of its graph, A and B have a relationship to the location of the vertex of the parabola, and C has a relationship to the intercept with the y-axis (see here). Does that show you why B must increase and C decrease?
     
  4. Feb 22, 2015 #3
    I'm sorry, I'm not quite sure if I understand... From the link you sent:

    "The coefficient b alone is the declivity of the parabola as y-axis intercepts.
    The coefficient c controls the height of the parabola, more specifically, it is the point where the parabola intercept the y-axis."

    So, does this mean that, in the quadratic equation, C is increasingly negative because the graphs are shifting away from the y-axis over time? I'm not sure why B is increasingly positive, though.
     
  5. Feb 22, 2015 #4

    haruspex

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    Think of the parabola for the second bounce as being just part of a longer parabola that started at t=0. What would that correspond to in terms of the throwing of the ball?
     
  6. Feb 22, 2015 #5
    Exactly. If you imagine extending the limits of the y-axis in the graph, so you also see the intercept of the parabola with the y-axis, then, if the parabola has the same "shape", translating it "to the right" would have to lower the intercept point.

    If the ##t##-coordinate of the vertex of the parabola is given by ##t = -\frac{B}{2A}##, and ##A## is a negative constant, then what does that tell you about ##B## as ##t## increases?
     
  7. Feb 22, 2015 #6
    I suppose it would mean that B = t*(2A), so it increases by quite a bit
    So, just to understand the context:
    a7a7b7b2fda741b65e73c3e189fe954f.png

    These refer to coordinates of the vertex/max height of the parabola, so the coordinate for x (which happens to be time in my graph) is -b/2a and y (which is height in m) could be found with -(delta)/4a? (Not that the question is asking for that in particular, but just so that I make sure I am understanding properly :]).

    So, C is referring to an intercept along the y-axis, but since the graph is moving further away from the y-axis (as t increases), C becomes increasingly negative because the intercept is further down.

    And B becomes increasingly larger because it refers to a the point along the x-axis associated with the highest point in the parabola, and as t increases, this value increases as well (even though the parabola is getting smaller over time).

    Am I getting it?
    (PS thank you everyone for your help!)
     
  8. Feb 22, 2015 #7
    B = -2At, but yes.

    Yes, but 'max height of the parabola' is probably not a good description, since you'd usually associate that with a certain shape of the graph. The parabola could open upwards instead, for instance. It's the coordinates for its vertex.

    This might be a bit nitpicky, but I'd just state B's relationship to the location of the vertex of the parabola. 'highest point in the parabola' and 'parabola is getting smaller over time' is ambiguous. It's clear that you get it, though.
     
  9. Feb 22, 2015 #8
    Awesome, thank you so much for your help!! :smile::smile:
     
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