Physics Help Forum Impulse without mass/height

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 Feb 12th 2018, 09:58 PM #1 Junior Member   Join Date: Feb 2018 Posts: 2 Impulse without mass/height I have a graph of a jumper who is bungee jumping, and I'm asked to find their impulse during their first bounce. However, I'm not given their mass or their original height. Using the equation d = 1/2at^2, I found the bungee cord's natural length to be 20m and using the equation vf^2 = vi^2 + 2ad, I found the max. speed to be 20 m/s. How can I go further? Attached Thumbnails
 Feb 13th 2018, 04:21 AM #2 Member   Join Date: Oct 2017 Location: Glasgow Posts: 67 The impulse, $\displaystyle J$, is $\displaystyle J = \int^{t_2}_{t_1} F(t) dt$ But since, in your specific case, $\displaystyle F(t) = m a(t)$ Then, $\displaystyle J = m \int^{t_2}_{t_1} a(t) dt$ You should be able to use your data set to evaluate the integral numerically and by selecting values of $\displaystyle t_2$ and $\displaystyle t_1$ to correspond to the collision data from the first bounce. topsquark likes this.
Feb 13th 2018, 06:06 PM   #3
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 Originally Posted by benit13 The impulse, $\displaystyle J$, is $\displaystyle J = \int^{t_2}_{t_1} F(t) dt$ But since, in your specific case, $\displaystyle F(t) = m a(t)$ Then, $\displaystyle J = m \int^{t_2}_{t_1} a(t) dt$ You should be able to use your data set to evaluate the integral numerically and by selecting values of $\displaystyle t_2$ and $\displaystyle t_1$ to correspond to the collision data from the first bounce.
But he's not given the mass.

 Feb 14th 2018, 05:31 AM #4 Member   Join Date: Oct 2017 Location: Glasgow Posts: 67 Ahhh, yes... I think it's possible to derive the mass from your dataset, but only if you know the string constant of the bungee cord. Force balance, neglecting air resistance, suggests that during the stretching part of the bungee jump, $\displaystyle F_{net} = F_{bungee} - F_{weight}$ $\displaystyle F = kx(t) - mg = ma(t)$ Therefore, $\displaystyle m = \frac{kx(t)}{a(t) + g}$ Since you know $\displaystyle a(t)$, you just need to know $\displaystyle x(t)$ to obtain the mass. To get $\displaystyle x(t)$ you need to perform numerical integration. There are numerous techniques for this (I can't remember any off the top of my head). Once you have $\displaystyle x(t)$, you can then calculate the mass at each point and take an average: $\displaystyle m = \bar{m} = \frac{1}{n} \sum_{i=1}^n \frac{kx_i}{a_i + g}$ where i iterates over all points between $\displaystyle t_1$ and $\displaystyle t_2$.
 Feb 14th 2018, 04:28 PM #5 Junior Member   Join Date: Feb 2018 Posts: 2 I'm thinking of just finding an integral and multiplying it by the variable m. To find an equation for the function a(t) (to use in the integral), could I calculate the slope of a portion of the acceleration graph and put it in y = mx + b form? Certain values of t would be the upper and lower bounds of the integral.
 Feb 15th 2018, 02:37 AM #6 Senior Member     Join Date: Jun 2016 Location: England Posts: 377 Noisy Data Looking at your graph, I would suggest that the actual signal underlying all the noise is a sinusoid, with a decay function. This also matches what I would expect for a mass on a spring. You could try fitting such a function through the data and then work with that function rather than with the noisy data. I would suggest that since you are only looking at 2 cycles, you can get away with using a very simple (linear) decay function or perhaps even just ignoring it completely. excel provides the "linest" function for least squares curve fitting, it is rather awkward to use, {it uses array formulas} but I have found that (once you have figured out how to use it) it produces very good results. Alternatively use the old "eyeball method". Estimate the amplitude and period of a sinusoid that you think will fit then tweak it manually until it looks correct. You can derive the sum of the squares of the residuals quite easily and just adjust the amplitude and the period to minimise this value. (excel provides the "solver" add-in method which could be used to automate this process) studiot and benit13 like this. __________________ ~\o/~ Last edited by Woody; Feb 15th 2018 at 08:43 AM.
 Feb 15th 2018, 04:12 AM #7 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 747 Good suggestion Woody. I think there is enough information for harmonic analysis or an FFT.

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