phys 223 lab 4 report

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Apr 3, 2024

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January 30, 2024 PHYS 223-02 Experiment 3 Jasmine Warren, Mia Hardy, Ella Dozer Introductions and Objectives For this experiment the objective was to determine acceleration during free fall motion and up and down motion. To start we used a lighter ball about 10 cm below the sensor, started data acquisition on the computer, and dropped the ball multiple times to get some options. We then would copy the results of distances and times into a note and the find the multiple times where the distance was at a min/max. We would copy those times/distances from the min to the max into the program and run to see it plotted. We would try this with our multiple drops to pick the best set of data and then adjust the acceleration if necessary to fit the curve. Once we found our best data set, we would copy that into the table and calculate velocity. We then repeated those steps with a heavier ball. Next, we went back to the lighter ball and repeated those steps but instead we started at the bottom and threw the ball up and then let it fall. We also took note of the accelerations from each setup and the velocity from the up and down motion. We then had to plot distance vs time to find the best fit linear and quadratic coefficient. Data Sheets and Graphs Recorded Time t (s) Time t (s) Distance (cm) Velocity (cm/s) 1 3.222 0 17.2 0 2 3.277 0.055 22.6 98.182
3 3.332 0.11 24.1 27.273 4 3.387 0.165 32.4 150.909 5 3.443 0.221 44.0 207.143 6 3.561 0.339 78.2 289.831 7 3.620 0.398 98.9 350.847 8 3.680 0.458 122.5 393.333 9 3.741 0.519 132.3 160.656 10 3.803 0.581 132.7 6.452 Table 1: Data from the light ball during motion down. 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0 20 40 60 80 100 120 140 f(x) = 134.18 x² + 154.33 x + 10.72 Time t (s) Distance (cm) Graph 1: Time vs distance for the light ball during motion down.
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0 50 100 150 200 250 300 350 400 450 f(x) = 282.3 x + 88.12 Time t (s) Velocity (cm/s) Graph 2: Time vs Velocity for the light ball during motion down. Recorded Time t (s) Time t (s) Distance (cm) Velocity (cm/s) 1 16.775 0 10.2 0 2 16.830 0.055 11.5 23.636 3 16.884 0.109 11.2 -5.556 4 16.938 0.163 12.4 22.222 5 16.993 0.218 16.5 74.545 6 17.048 0.273 21.8 96.364 7 17.113 0.338 204.7 2813.846 8 17.178 0.403 205.1 6.154 Table 2: Data from the heavy ball during motion down.
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0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0 50 100 150 200 250 f(x) = 2507.59 x² − 500.54 x + 21.24 Time t (s) Distance (cm) Graph 3: Time vs distance for the heavy ball during motion down. 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 -500 0 500 1000 1500 2000 2500 3000 f(x) = 3004.51 x − 206.6 Time t (s) Velocity (cm/s) Graph 4: Time vs velocity for the heavy ball during motion down. Recorded Time t (s) Time t (s) Distance (cm) Velocity (cm/s) 1 24.447 0 92.3 0 2 24.505 0.058 72.8 -336.207 3 24.561 0.114 55.2 -314.286
4 24.617 0.170 40.9 -255.357 5 24.845 0.398 14.3 -116.667 6 24.901 0.454 14.6 5.357 7 24.957 0.510 19.1 80.357 8 25.012 0.565 23.4 78.182 9 25.244 0.797 82.9 256.466 10 25.303 0.856 107.7 420.339 11 25.717 1.270 207.0 239.855 Table 3: Data from the light ball during motion up then down. 0 0.2 0.4 0.6 0.8 1 1.2 1.4 0 50 100 150 200 250 f(x) = 277.94 x² − 241.34 x + 79.98 Time t (s) Distance (cm) Graph 5: Time vs distance for the light ball during motion up then down.
0 0.2 0.4 0.6 0.8 1 1.2 1.4 -400 -300 -200 -100 0 100 200 300 400 500 f(x) = 515.92 x − 238.24 Time t (s) Velocity (cm/s) Graph 6: Time vs velocity for the light ball during motion up then down. Analysis Before graphing I took the times we got from the computer and set them where the initial time is 0 and then going up from there by taking the time the computer showed subtracted by the first time to see how much time had elapsed if it stated at 0 rather than what it did. I plotted the distance vs time for the light ball during motion down and fit with a quadratic dependence (Graph 1). The linear coefficient was v 0 = 154.33 m s and the quadratic coefficient was a 2 = 134.18 m s 2 . I plotted the distance vs time for the heavy ball during motion down and fit with a quadratic dependence (Graph 3). The linear coefficient was v 0 = 500 . 54 m s and the quadratic coefficient was a 2 = 2507.6 m s 2 . The acceleration for the heavy ball was much greater than the light ball. This could be because the heavy ball was much smaller and harder for the sensor to detect. We also
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had to adjust the acceleration on the computer from 9.81 to 4.00 m/s 2 which could also contribute to error. I plotted the distance vs time for the light ball during motion up then down and fit with a quadratic dependence (Graph 5). The linear coefficient was v 0 = 241.34 m s and the quadratic coefficient was a 2 = 277.94 m s 2 . This acceleration is closer to the light ball during motion down but is still a little higher. I then calculated the velocity for all my data points from all 3 experimental setups. I did this by using the equation v = ∆h ∆t . I found h by taking the position at one point in time subtracted by the distance from before, all the way down the chart. t was calculated the same way but with times. I plotted the velocity vs time for the light ball during motion down and fit with a linear dependence (Graph 2). The intercept was v 0 = 88.119 m s and the slope was a = 282.3 m s 2 . The 0 was much lower than what I got from the distance plot, but the a was very close once you divide it by 2. I plotted the velocity vs time for the heavy ball during motion down and fit with a linear dependence (Graph 4). The intercept was v 0 =− 206.6 m s and the slope was a = 3004.5 m s 2 . The 0 was lower in magnitude and negative compared to what I got from the distance plot, and the a was also much smaller once you divide it by 2. I plotted the velocity vs time for the light ball during motion up then down and fit with a linear dependence (Graph 6). The intercept was v 0 =− 238.24 m s and the slope was
a = 515.92 m s 2 . The 0 was close to the same magnitude but negative compared to what I got from the distance plot, but the a was close once you divide it by 2. The acceleration at the turning point in motion up then down can be found by using the velocity at the highest point divided by the time at that point. The highest point is 14.3 cm, and the acceleration is calculated a = v t = − 116.667 0.398 =− 293.133 . This is negative because it was moving up then changing direction, so the velocity was negative being opposite of the gravitational acceleration which is down. Discussion The data collection for this experiment was somewhat difficult. It was hard to get a good throw where we weren’t interfering with the sensor by standing under it while also making sure the ball was thrown straight up so it was being sensed. We also had to do multiple throws for each setup because it was so hard to get good data. Our computer was also messing up for a while where it was deleting our data before I could copy it over, but we dd some troubleshooting and got that fixed. The data we ended up getting didn’t look super great no matter how many times we tried. Especially for the heavy ball trial because we had to adjust the acceleration on the computer to be 4.00 to help the line fit the data. For the light ball in both scenarios, we were able to keep the acceleration 9.81 and we noted the initial velocity from the computer during the motion up then down to be -4.00. Conclusion
The objective of this experiment was to find acceleration during free fall motion and during motion up then down. We were able to keep the acceleration at 9.81 on the computer for the setups with the light ball but not the heavy ball. When I graphed it separately and used the equations of the best fit quadratic and linear lines, the acceleration varied. This can be from the trouble we had with keeping the ball under the sensor when collecting data, giving us inconsistent results.
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