1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Quadratic Graph & Bouncing Ball

  1. Feb 10, 2013 #1
    Hi guys. I've been a lurker for a while, but I've recently become super stumped on this physics question (physics is far from my forté). I've attached the graph of my bouncing ball. First, I had to identify the coefficients and what they mean. I understand that A is acceleration, B is initial velocity, and C is displacement.

    I understand that the graph is shifting, and B and C of the second bounce are not necessarily equal to those numbers. How do I find the true values of B and C for each subsequent bounce?

    1. The problem statement, all variables and given/known data
    This is my graph:
    http://i.imgur.com/5pnfJY5.png


    2. Relevant equations
    Y = Ax^2 + Bx + C
    y = y0 + vot + 1/2at^2

    3. The attempt at a solution
    For the first bounce, I was able to find the initial height by doing:
    Y = Ax^2 + Bx + C, with x = 0.6 seconds
    Y = -4.460(0.6)^2 + 5.425(0.6) - 0.7426
    Y = 0.9068 m

    And velocity:
    dy/dx = 2(-4.460)x + 5.425, with x = 0.6 seconds
    dy/dx = 0.073 m/s

    I believe I need to find these values in order to calculate the kinetic energy before and after impacts, and finding linear momentums - I have to fill out a chart like this one for seven bounces:
    MDqVQY8.png

    I also have no mass for the ball, p is basically equal to the velocity. And I am told that PE must be equal to KE of the ball at the beginning and at the end of each interval of free flight.
     
  2. jcsd
  3. Feb 11, 2013 #2
    y = Ax + Bx2
    y = y'
    x' = x + δ
    ∴ x = x' - δ

    Rewrite above equation using this relationship between the two variables
     

    Attached Files:

  4. Feb 11, 2013 #3
    So the B value would stay the same?
     
  5. Feb 12, 2013 #4
    It seems so yes.
    What type of detector did you use to get the data?
     
  6. Feb 13, 2013 #5
    I used a motion sensor hooked up to Logger Lite software, can't remember the exact specifics as I did the data collection some time ago.

    I just want to make sure I'm doing this right, my tutor hasn't responded to me yet. So for the second bounce, using your equation, this is what I did:

    Bounce 1 occurred between 0.2-0.9 seconds.
    δ = 0.9 - 0.7 = 0.2
    x = x' - δ
    = 0.9 - 0.7
    x = 0.2

    Bounce 2:
    y = Ax^2 + Bx
    = -4.320(0.2)^2 + 12.36(0.2)
    = 2.29 m ??? Should this not be lower than the first bounce? (and why am I not incorporating C?)

    Then
    dy/dx = 2(-4.320)x + 12.36
    = 2(-4.320)(0.2) + 12.36
    = 10.632 m/s

    I don't have a clue if I'm going about this correctly!
     
    Last edited: Feb 13, 2013
  7. Feb 14, 2013 #6
    So the ball was bouncing up and down while the motion sensor was looking down onto it?
    This means that you have time on the x-axis and distance from the motion sensor on the y?
    What are the units for each?
    If the variables on the axes are y(t) then one would expect a relationship of the form:

    y = yo - vot + 1/2 gt2

    which does not agree with the fitted coefficients - the signs are wrong. So maybe the motion sensor's measurements was used to calculate the ball's distance from the floor. In this case one would expect

    y = yo + vot - 1/2 gt2

    in which case the signs of the fitted coefficients agree, but still the meaning of C0 worries me.
     
    Last edited: Feb 14, 2013
  8. Feb 14, 2013 #7
    Sorry, I'll go back to the beginning. I have a feeling this is simpler than what it seems, but my tutor has yet to respond to me.

    Yes, the sensor was held vertically above the bouncing ball. The data collected produced this graph:

    http://i.imgur.com/QSxyot1.png

    y-axis is the vertical distance between the ball and the sensor, x-axis is time.

    This graph was analyzed using Graphical Analysis, the y-axis was altered so it would show the ball's position relative to the ground instead of the sensor.

    I did quadratic fits for each bounce interval on the new graph, which I posted earlier but I'll post it again:

    http://i.imgur.com/xme9ZBP.png

    After that, these were my instructions:
    "Fitting the data points corresponding to each interval, using the equation Y = A t⋀2 + B t + C (see Fig 5.3). Write a clear interpretation of the meaning of each parameter in this equation.

    I said that A represents acceleration, B represents initial velocity, and C represents initial position.

    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! Can you explain this apparent discrepancy?

    I'm guessing it's due to the shift, but I'm not understanding these values. I don't understand why A and C are negative, why C is becoming increasingly negative, as well as what is going on with the velocity. I'm not sure how to account for these in the analysis either. The example they gave produced a graph similar to mine, with similar patterns in how the A, B, and C values increase/decrease.

    And THEN I have to do this:

    Using the maximum height of the ball, it should be possible to calculate its maximum potential energy (PE) in the middle of each interval. From the principle of conservation of mechanical energy, this must be equal to the kinetic energy (KE) of the ball at the beginning and at the end of each interval of free flight. Therefore, you should be able to calculate the kinetic energy per unit mass (KE/m) just before and just after each impact with the ground. The momentum per unit mass (p/m) can also be calculated just before and just after each impact. Note that this quantity is basically equal to the ball’s velocity.

    After showing detailed sample calculations, construct a table similar to the one shown below, in which you include the changes in the kinetic energy and momentum as a result of each impact. Note that the momentum is a vector quantity, and therefore, the sign is very important.


    Du0jh00.png

    And this is where I'm trying to get the velocities... unless I don't have to and just use the B values. But I'm getting numbers that do not make sense to me.
     
  9. Feb 14, 2013 #8

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    If you mean the time shift, yes. Consider the second bounce as if it were just the top part of a single trajectory that started at time 0, i.e. about one second before any of its datapoints. Where would the trajectory have started at time 0, and with what speed?
    For A, what is the cause of the acceleration? Which way does that act? Which way are you measuring Y?
    For C, see my comment above on time shift.

    You can't use the B values for the reasons given: they do not represent velocities all at the same height.
     
  10. Feb 14, 2013 #9
    I knew it was simpler than what I was doing!

    Okay, so since I have max. heights for each of the bounces, I can find PE as g*h. So before impact, KE = PE. After impact, KE would be the PE of the following bounce, correct?

    And since momentum is the velocity in this case (no mass was recorded), it would be:
    1/2 mv^2 = mgh
    v = sqrt 2*g*h

    Am I correct in my thinking here?
     
  11. Feb 14, 2013 #10

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Yes, that's all correct. Btw, in the graphs, something odd seems to have happened between the third and fourth bounces. It looks like the sensor got knocked down a bit.
     
  12. Feb 14, 2013 #11
    Probably me! I was holding it above while simultaneously trying to make the ball bounce directly below. I'll just take note of the discrepancy in the write up, since it's basically saying the ball gained energy somehow, haha.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Quadratic Graph & Bouncing Ball
  1. Bouncing ball? (Replies: 2)

Loading...