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course PHY 241
12/28; 12:30
Analysis: Each roll up the long incline corresponds to an increase in the PE of the ball, which can be calculated given the mass of the ball (we will assume mass 60 grams) and the change in its vertical position while on that incline.
Assume that the thickness of the two-domino stack supporting the long ramp is 1.8 cm, so that a ball which rolls all the way up the ramp from the bottom to the top would rise 1.8 cm while traveling 30 cm. Just to be sure you understand what this means, answer the following:
How much higher does the ball go if it travels 30 cm (yes, the answer should be really obvious)?
****
1.8 cm. Lol
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How much higher would it go if it only traveled 10 cm along the ramp?
**** 1.8cm / 30cm * 10cm = .6 cm
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How much higher would it go if it only traveled 3 cm along the ramp?
**** same as above except varying hypotenuse = .18 cm
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How much higher would it go if it only traveled 1 cm along the ramp?
**** .06 cm, this is intuitive when examining the 10 cm `dist along the ramp.
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How could you use that last result to figure out how much higher the ball goes if it travels 7 cm along the ramp?
**** multiply the result by seven = .42 cm; this is about right simply by looking at the 10 cm and 3 cm `d(ramp).
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Your answers to that last series of questions should have been .6 cm, .18 cm, .06 cm and 7 * .06 cm = .42 cm. If you didn't get those, you should let the instructor know, because you're going to need to be able to figure out how far the ball rises along that ramp for each of your various trials.
You have observed how far the ball travels along longer ramp, from various starting positions on the shorter ramp.
For each trial, figure out how far the ball traveled along the longer ramp, and how much higher it was at its 'turnaround point' than when it first contacted the ramp. If you need to estimate anything in order to do your calculation, make a reasonable estimate.
If you're not sure of your calculations you can enter them here and submit this for further guidance. Otherwise you can leave the space below blank and move on.
****
How far `dx along the ramp / trail: [the length of the whole ramp was 29 cm.]
Only the edge: 3.1 cm
Radius of the ball: 5.0 cm
2D thick: [1domino’s thickness = 0.9 cm] 6.59 cm
1D wide: [2.55 cm] `dx=6.8 cm
1D long: [5.15] 10.9 cm
1D(thick + long): 11.5 cm
1D (wide + long):14.75 cm
1D (2l): 18.5 cm
1D (2l+1w): 20.9 cm
`dy (vertical):
[Calculation for a `dy]: 2D thick: 1.8 cm/29cm * 6.59 cm= 0.409 cm
Only the edge [0 cm]: 0.192414 cm
Radius of the ball [1.25 cm**]: 0.310 cm
2D thick: 1.8 cm/29cm * 6.59 cm= 0.409 cm
1D wide: [2.55 cm] `dy = 0.422 cm
1D long: [5.15] 0.676 cm
1D(thick + long): 0.713793 cm
1D (wide + long): 0.915517 cm
1D (2l): 1.148276 cm
1D (2l+1w): 1.297241 cm
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Assuming that the dominoes are .9 cm thick, 2.5 cm wide and 5.0 cm long, find the distance the ball rolled along the first incline before rolling off the edge. For your very first trial, where the domino was released just at the edge, you would report distance 0.
**** I actually accounted for the radius of the ball when receiving the lab data.
0
**I think the diameter of the ball was 2.5 cm. I do not have the ball any more.
=>1.25 = `d(ramp)
2D thick: 1.8 cm
1D wide: [2.5 cm]
1D long: [5.0]
2D(thick + long): 5.9 cm
2D (wide + long): 7.5 cm
2D (2l): 10 cm
3D (2l+1w): 15 cm
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Now make a table with three columns. The first column will be the rolling distances you just calculated. The second column will be the corresponding distances the ball rolled up the longer incline. The third column will be the change in the ball's height as it rolled along the longer incline.
Give your table:
**** all measured in cm
0 3.1 0.192
1.25 5 0.314
1.8 6.59 0.409
2.5 6.8 0.422
5 10.9 0.676
5.9 11.5 0.714
7.5 14.75 0.916
10 18.5 1.15
15 20.9 1.3
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Assume that on the first ramp, the ball descended .1 cm for every centimeter it rolled. Assume also a ball mass of 60 grams.
For each trial, figure out how much gravitational PE the ball lost while rolling down the shorter incline (not counting the fall from the edge of the first incline to the longer incline), and how much gravitational PE it gained while rolling up the longer incline.
Report your results in the form of a table showing PE gained on longer incline vs. PE lost on shorter incline.
****
PE gained PE lost
0.00113 J 0 J
0.001848 J 0.000736 J
0.002407 J 0.001059 J
0.002484 J 0.001472 J
0.003979 J 0.002943 J
0.004203 J 0.003473 J
0.005392 J 0.004415 J
0.006769 J 0.005886 J
0.007652 J 0.008829 J
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At what position on the shorter incline did the PE gained on the longer incline begin to exceed the PE lost on the shorter?
**** I do not understand this question though simple there seems to be something I am missing. It asks, “At position on the shorter incline did the PE gained on the longer incline begin to exceed the PE lost on the shorter?”
By looking at the data and intuition says the PE gained on the longer incline is greater up to a certain point due to the edge effect. So the PE gained is greater from 0 up to somewhere between 10 and 15 cm on the shorter ramp.
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Plot each line of your table as a point on a graph of y = PE gained vs. x = PE lost. Sketch the smooth curve that seems to best indicate the trend of your data. Sketch also the line y = x.
Describe your graph.
**** the graph increases at a slower and slower rate: this is due to the edge effect decreasing due to faster and faster speed of the ball traveling down the ramp.
The function appears to be about f(x) = x^.75 [very rough estimate.]
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How is it possible that the ball could have more KE when it reaches the longer ramp than the PE it lost on the shorter? Give your best explanation of how that might have occurred on some of your trials.
**** For one the angle of the short will always have at least a small effect on the ball. Also as the ball goes over the edge the ball is speeding up due to gravity. Bc of this the ball speeds up in the x direction of the ramp. Due to this increase in speed the ball is able to travel up the longer ramp. The reason I say the importance of focus is the x direction [parallel with the horizontal] is due to the impact for example if we dropped a ball straight down and it hit a perfectly level surface it would bounce straight up again and not go any where horizontally.
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You appear to have good data. See the appended instructions for analysis.
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