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Phy 201
Your 'ball down ramp' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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A ball is timed as it rolls from rest to the end of a ramp. The slope of the ramp is varied. Preliminary conclusions are drawn about the motion and the effect of ramp slope. A subsequent lab exercise uses the data from this lab to reach additional conclusions.
Most students report completion times between 45 minutes and 75 minutes hour, with a few reporting times as short as 25 minutes or as long as 2 hours. Median time of completion is around 1 hour.
Timing Ball down Ramp
The picture below shows a ball near the end of a grooved steel track (this steel track is a piece of 'shelf standard'); the shelf standard is supported by a stack of two dominoes. Your lab materials package contains two pieces of shelf standard; the shelf standard shown in the figure is white, but the one in your kit might be colored black, gold, silver or any of a variety of other colors.
If a ball rolls from an initial state of rest down three ramps with different slopes, the same distance along the ramp each time, do you think the time required to roll the length of the ramp will be greatest or least for the steepest ramp, or will the interval on the steepest ramp be neither the greatest nor the least? Explain why you think you have correctly predicted the behavior of the system.
Your answer (start in the next line):
The time required to roll down the steepest ramp will be the least. It just seems like as long as the ramp is steeper, the ball will accelerate more quickly.
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If we write down the slopes from least to greatest, next to the time intervals observed for those slopes, would you expect the time intervals to be increasing or decreasing, or do you think there would be no clear pattern? Explain why you think you have correctly described the behavior of the numbers in the table.
Your answer (start in the next line):
I would imagine the intervals would decrease as the ball accelerates more and more when it rolls down the steeper ramps.
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Set up the shelf standard ramp on a reasonably level table, using a piece of 30-cm shelf standard and a single domino under the high end of the ramp. Position the dominoes so that the last .5 cm of the ramp extends beyond the point where the ramp contacts the domino,.and do the same in all subsequent setups.
Set the bracket on the table, touching the lower end of the ramp so that a ball rolling down the ramp will strike the bracket..
Mark a point about 3 cm below the top end of the ramp. Place a domino on the ramp to its high end is at this point, and place the ball just above the domino, so the domino is holding it back. Quickly pull the domino away from the ball so the ball begins to roll freely down the ramp. Allow the ball to roll until it strikes the bracket.
The bracket will probably move a little bit. Reset it at the end of the ramp.
Determine how far the ball rolled from release until it struck the bracket.
Now repeat, but this time use the TIMER. The first click will occur at the instant you release the ball, the second at the instant the ball strikes the bracket. Practice until you are as sure as you can be that you are clicking and pulling back the domino at the same instant, and that your second click is simultaneous with the ball striking the bracket.
When you are ready, do 5 trials 'for real' and record your time intervals.
Then reverse the system--without otherwise changing the position of the ramp, place the domino under the left end and position the bracket at the right end.
Time 5 trials with the ramp in this position.
In the space below, give the time interval for each trial, rounded to the nearest .001 second. Give 1 trial on each line, so that you will have a total of 10 lines, the first 5 lines for the first system, then 5 lines for the second system.
Beginning in 11th line give a short narrative description of what your data means and how it was collected.
Also describe what you were thinking, relevant to physics and the experiment, during the process of setting up the system and performing the trials.
Your answer (start in the next line):
1.961
1.844
1.828
1.719
1.828
2.016
2.109
2.063
2.102
2.102
These are the timed intervals for the ball to roll down a ramp, either from the left or right.
My surface isn’t entirely level, as you see the second set of trials are a little longer.
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Now place two dominoes under the right end and repeat the process, obtaining the time interval for each of 5 trials.
Then place the two dominoes under the left end and repeat once more.
Enter your 10 time intervals using the same format as before.
Your answer (start in the next line):
1.453
1.328
1.273
1.445
1.297
1.438
1.432
1.391
1.336
1.477
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Repeat the preceding using 3 dominoes instead of 2. Enter your 10 time intervals using the same format as before.
Your answer (start in the next line):
1.078
1.18
1.164
1.117
1.109
1.148
1.109
1.063
1.086
1.109
1.141
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Repeat the preceding again, still using the 3 domino setup, but this time place a CD or a DVD disk (or something of roughly similar thickness) on the 'low' end of the ramp. You need time only 5 intervals, but if you prefer you may use 10. Enter your 5 (or 10) time intervals using the same format as before.
Your answer (start in the next line):
1.125
1.078
1.195
1.156
1.117
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Repeat the preceding one last time, still using the 3 domino setup, but remove the disk and replace it with a piece of paper. You need time only 5 intervals, but if you prefer you may use 10. Enter your 5 (or 10) time intervals using the same format as before.
Your answer (start in the next line):
1.086
0.992
1
1.102
1.07
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Do your results support or fail to support the hypotheses you stated in the first two questions, regarding the relationship between time intervals and slopes? Explain.
Your answer (start in the next line):
They support my hypothesis that a steeper ramp will result in a smaller time interval.
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How do you think the average velocity of the ball is related to the slope of the ramp? Explain in as much detail as possible.
Your answer (start in the next line):
The average velocity of the ball should increase as the slope gets steeper. Without doing all the math, I’m assuming the relationship is proportional to how many dominoes are stacked at one end. and the two end experiments with the cd and paper underneath the other end should show a small change in that proportion.
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Speculate on what it is that causes the average velocity on these ramps to change with slope.
Your answer (start in the next line):
The steeper the ramp, the less resistance the ball has to fall downward at the force of gravity. When the ramp is more level, there is more “ramp in the way” of the balls falling straight downward.
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How might you verify whether your speculations are indeed valid explanations?
Your answer (start in the next line):
By doing a bunch of timed intervals of the ball rolling down a ramp I suppose.
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Do your data conclusively show that the disk made a difference?
Your answer (start in the next line):
It looks like a the cd did make a small difference, but there aren’t enough accurate trials to see.
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Do your data conclusively show that the piece of paper made a difference?
Your answer (start in the next line):
The piece of paper made no noticeable difference.
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Imagine that someone is placing different objects below the 'low' end of the ramp, and you are timing the ball. Assume that somehow the object placed below the 'low' end is hidden from you in a way that does not interfere with the timing process. Compared to the thickness of the DVD, how thin would the object have to be before you would be unable, using the TIMER, to observe a difference in times down the ramp?
Answer this question in the first line below. Express your answer in multiples or fractions of the thickness of a disk.
Starting in the second line, explain how you came to your conclusion, based on the results you obtained in this experiment. Also discuss how you could modify or refine the experiment, still using the TIMER, to distinguish the effect of the thinnest possible object placed under the 'low end.
Your answer (start in the next line):
The object would probably have to be ¾ the thickness of the cd to see a noticeable difference.
I came up with that number by just guessing that the object must be of some “noticeable” thickness, unlike a piece of paper, ¾ of a CD isn’t too thin. I don’t know how I would find the thinnest possible object to fit under the low end of the ramp. There is a large amount of human error here including how the ball is released, and precise clicking of the timer. Even then, the timer program isn’t entirely accurate either.
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Had you placed the disk below the 'low' end of the ramp in a 1-domino setup, do you think the difference in times would have been greater or less? Do you think you would be better able distinguish the presence of a thinner object using the 1-domino setup, or the 3-domino setup? Explain your reasoning below:
Your answer (start in the next line):
The difference in times would be greater than with 3 dominos on one end and a cd on the other. I think I could better distinguish the difference with the 1 domino setup, because it makes a more “significant” change in the slope. If the slope is 1, and the cd changes that by 0.1, then the slope is 0.9. If the slope is 3, and the cd changes that by 0.1, then the slope is 2.9, a larger steeper number to deal with making the intervals more accurate only to greater decimal places such as 0.000001.
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Does the ball's velocity change more or less quickly with the 3-domino setup or the 1-domino setup? Explain as best you can how you could use your results to support your answer.
Your answer (start in the next line):
The velocity should change more quickly with the 3 domino setup, as shown with the intervals it takes less time to travel down a steeper slope.
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Your instructor is trying to gauge the typical time spent by students on these experiments. Please answer the following question as accurately as you can, understanding that your answer will be used only for the stated purpose and has no bearing on your grades:
• Approximately how long did it take you to complete this experiment?
Your answer (start in the next line):
35 minutes
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This lab exercise is based on the observations you previously made of a ball rolling down ramps of various slopes. We further investigate the relationship between ramp slope and acceleration.
The mean time reported to complete this exercise is 2 hours. The most frequently reported times range from 1 hour to 3 hours, with some reports of shorter or longer times.
Note that there are a number of repetitive calculations in this exercise. You are encouraged to use a spreadsheet as appropriate to save you time, but be sure your results check out with a handwritten analysis of at least a few representative trials.
Document your data
For ramps supported by 1, 2 and 3 dominoes, in a previous exercise you reported time intervals for 5 trials of the ball rolling from right to left down a single ramp, and 5 trials for the ball rolling from left to right.
If in that experiment you were not instructed to take data for all three setups in both directions, report only the data you were instructed to obtain.
(Note: If you did the experiment using the short ramp and coins, specify which type of coin you used. In the instructions below you would substitute the word 'coins' for 'dominoes').
Go to your original data or to the 'readable' version that should have been posted to your access page, and copy your data as indicated in the boxes below:
Copy the 10 trials for the 1-domino setups, which you should have entered into your original lab submission in the format specified by the instruction
'In the box below, give the time interval for each trial, rounded to the nearest .001 second. Give 1 trial on each line, and give the 5 trials for the first system, then the 5 trials for the second system. You will therefore give 10 numbers on 10 lines.'
In the 'readable' posted version this data will follow the boldfaced heading
'5 trials each way 1 domino'
Enter your 10 numbers on 10 lines below, and on the first subsequent line briefly indicate the meaning of the data:
------>>>>>> ten trials for 1-domino setups
Your answer (start in the next line):
1.387
1.558
1.449
1.587
1.484
2.730
2.730
2.824
2.746
3.024
This is the 10 trials from a ball on the ramp from right to left being first five and left to right being the second five.
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Enter your data for the 2-domino setups in the same format, being sure to include your brief explanation:
On the 'readable' posted version this data will follow the boldfaced heading
'5 trials each way 2 dominoes'
------>>>>>> 2 domino results
Your answer (start in the next line):
1.215
1.246
1.234
1.137
1.199
1.480
1.703
1.625
1.667
1.578
First five is right to left and second five is left to right.
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Enter your data for the 3-domino setups in the same format, including brief explanation.
On the 'readable' posted version this data will follow the boldfaced heading
'5 trials each way 3 dominoes'
------>>>>>> 3 domino results
Your answer (start in the next line):
1.000
.934
1.077
1.016
.980
1.219
1.219
1.199
1.219
1.250
The first five is from right to left and second five is from left to right.
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Calculate mean time down ramp for each setup
In the previous hypothesis testing exercise, you calculated and reported the mean and standard deviation of times down each of the two 1-domino setups, one running right-left and the other left-right.
You may use any results obtained from that analysis (provided you are confident that your results follow correctly from your data), or you may simply recalculate this information, which can be done very quickly and easily using the Data Analysis Program at
http://www.vhcc.edu/dsmith/genInfo/labrynth_created_fall_05/levl1_15\levl2_51/dataProgram.exe\
In any case, calculate as needed and enter the following information, in the order requested, giving one mean and standard deviation per line in comma-delimited format:
Mean and standard deviation of times down ramp for 1 domino, right-to-left.
Mean and standard deviation of times down ramp for 1 domino, left-to-right.
Mean and standard deviation of times down ramp for 2 dominoes, right-to-left.
Mean and standard deviation of times down ramp for 2 dominoes, left-to-right.
Mean and standard deviation of times down ramp for 3 dominoes, right-to-left.
Mean and standard deviation fof times down ramp or 3 dominoes, left-to-right.
On the first subsequent line briefly indicate the meaning of your results and how they were obtained:
------>>>>>> mean, std dev each setup each direction
Your answer (start in the next line):
1.493, 0.08111
2.811, 0.1254
1.206,0.04265
1.611,0.08667
1.001, 0.05228
1.221,0.01828
The first number in the colum is the mean and the secon dis the standard deviation and I used the data program.
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Calculate average ball velocity for each setup
Assuming that the ball traveled 28 cm from release until the time it struck the bracket, determine each of the following, using the mean time required for the ball to travel down the ramp:
Average ball velocity for 1 domino, right-to-left.
Average ball velocity for 1 domino, left-to-right.
Average ball velocity for 2 dominoes, right-to-left.
Average ball velocity for 2 dominoes, left-to-right.
Average ball velocity for 3 dominoes, right-to-left.
Average ball velocity for 3 dominoes, left-to-right.
Report your six results in the box below, one result per line, in the order requested above.
Starting in the seventh line explain how you obtained your results, giving the details of how you obtained at least one of your results. These details should include the definition of the average velocity, and should explain how you used the mean time and the distance down the ramp to arrive at your result, and should show the numbers used and the numbers obtained in each step.
------>>>>>> ave velocities each of six setups
Your answer (start in the next line):
18.75cm/s
9.96cm/s
23.22cm/s
17.38cm/s
27.97cm/s
22.93cm/s
I divided the 28 m by the mean in seconds to get the average velocity. The average velocity is 'ds /'dt. For example: 28 m divided by 1.492s=18.75m/s.
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Calculate average ball acceleration for each setup
Assuming that the velocity of the ball changed at a constant rate in each trial, use the mean time interval and the 28 cm distance to determine the average rate of change of velocity with respect to clock time. You will determine your results in the following order:
Average rate of change of ball velocity with respect to clock time for 1 domino, right-to-left.
Average rate of change of ball velocity with respect to clock time for 1 domino, left-to-right.
Average rate of change of ball velocity with respect to clock time for 2 dominoes, right-to-left.
Average rate of change of ball velocity with respect to clock time for 2 dominoes, left-to-right.
Average rate of change of ball velocity with respect to clock time for 3 dominoes, right-to-left.
Average rate of change of ball velocity with respect to clock time for 3 dominoes, left-to-right.
Report your six results in the box below, one result per line, in the order requested above.
Starting in the seventh line explain how you obtained your results, giving the details of how you obtained at least one of your results. These details should include the definition of the average rate of change of velocity with respect to clock time and should explain, step by step, how you used the mean time and the distance down the ramp to arrive at your result, and should show the numbers used and the numbers obtained in each step.
------>>>>>> ave roc of vel each of six setups
Your answer (start in the next line):
12.56m/s^2
3.54m/s^2
19.25m/s^2
10.79m/s^2
27.94m/s^2
19.6m/s^2
The first line takes the vAve calculated early of 18.75cm/s dividing it by the mean time which is 1.493s then a=18.75m/s/1.492s=12.56cm/s^2, because aAve =vAve/'dt
@&
You don't get acceleration by dividing average velocity by change in clock time.
*@
@&
*@
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Average left-right and right-left velocities for each slope
For the 1-domino system you have obtained two values for the average rate of change of velocity with respect to clock time, one for the right-left setup and one for the left-right. Average those two values and note your result.
For the 2-domino system you have also obtained two values for the average rate of change of velocity with respect to clock time. Average those two values and note your result.
For the 3-domino system you have also obtained two values for the average rate of change of velocity with respect to clock time. Average those two values and note your result.
Report your results in the box below, giving one average rate of change of velocity with respect to clock time per line, in the order requested. Starting the the first subsequent line, briefly indicate how you obtained your results and what you think they mean.
------>>>>>> ave of right-left, left-right each slope
Your answer (start in the next line):
8.05cm/s^2
15.02cm/s^2
23.77cm/s^2
These are the average of each set up of dominoes and they were obtained by adding the two trials of left and right and right and left together and dividing 2 for the average.
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Find acceleration for each slope based on average of left-right and right-left times
Average the mean time required for the right-to-left run with the mean time for the left-to-right run.
Using this average mean time, recalculate your average rate of velocity change with respect to clock time for the 1-domino trials
Do the same for the 2-domino results, and for the 3-domino results.
Report your results in the box below, giving one average rate of change of velocity with respect to clock time per line, in the order requested. In the subsequent line explain how you obtained your results and what you think they mean.
------>>>>>> left-right, right-left each setup, ave mean times and give ave accel
Your answer (start in the next line):
14.355cm/s/2.152s=6.671cm/s^2
20.3cm/s/1.409s=14.406cm/s^2
25.45cm/s/1.111s=22.907cm/s^2
I averaged the two trials for the mean clocktime and averaged the average velocities for the two trials for each set.
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Compare acceleration results for the two different methods
You obtained data for three basic setups, each with a different slope. Each basic setup was done with a right-left and a left-right version.
You previously calculated a single average rate of change of velocity with respect to clock time for each slope, by averaging the right-left rate with the left-right rate.
You have now calculated a single average rate of change of velocity with respect to clock time for each slope, but this time by using the average of the mean times for the right-left and left-right versions.
Answer the following questions in the box below:
Since both methods give a single average rate of change of velocity with respect to clock time, would you therefore expect these two results to be the same for each slope?
Are the results you reported here, based on the average of the two mean times, the same as those you obtained previously by average the two rates? Are they nearly the same?
Why would you expect that they would be the same or nearly the same?
If they are not exactly the same, can you explain why?
------>>>>>> ave of mean vel, ave based on mean of `dt same, different, why
Your answer (start in the next line):
I would expect that the two rates would be the same or at least similar. Because they are both using the same data just two different ways of getting them.
They are different because they are different values being calculated the numbers are not the same when averaging the different rates.
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Associate acceleration with ramp slope
Your results will clearly indicate that, as expected, acceleration increases when ramp slope increases. We want to look further at just how the acceleration changes with ramp slope.
If you set up the ramps according to instructions, then the ramp slopes for 1-, 2- and 3-domino systems should have been approximately equal to .03, .06 and .09 (if you used coins and the 15 cm ramp instead of dominoes and the 30-cm ramp, your ramp slopes will be different; each dime will correspond to a ramp slope of about .007, each penny to a slope of about .010, each quarter to a slope of about .013).
For each slope you have obtained two values for the average rate of change of velocity with respect to clock time on that slope. You may use below the values obtained in the preceding box, or the values you obtained in the box preceding that one. Use the one in which you have more faith.
In the box below, report in the first line the ramp slope and the average rate of change of velocity with respect to clock time for the 1-domino system. Use comma-delimited format.
Using the same format report your results for the 2-domino system in the second line, and for the 3-domino system in the third.
In your fourth line specify the units of these quantities. Ramp slope is a unitless quantity; be sure you report this. Also briefly explain how you got your results and what they tell you about this system:
------>>>>>> ramp slope ave roc of vel each system
Your answer (start in the next line):
.03, 6.671cm/s^2
.06, 14.406cm/s^2
.09, 22.907cm/s^2
The rampslope is first number in the columns, the second number is the rate calculated in the previous box.
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Graph acceleration vs. ramp slope
A graph of acceleration vs. ramp slope will contain three data points. The graph will visually represent the way acceleration changes with ramp slope. A straight line through your three data points will have a slope and a y-intercept, each of which has a very significant meaning.
Your results constitute a table with three rows and two columns, representing rate of velocity change vs. ramp slope.
Sketch in your lab notebook a graph of the table you have just entered. The graph will be of rate of change of velocity with respect to clock time vs. ramp slope. Be sure to follow the y vs. x convention to put the right quantities on the horizontal and vertical axes (if it's y vs. x, then y is on the vertical, x on the horizontal axis).
Your graph might look something like the following. Note, however, that this graph is a little too long for its height. On a good graph the region occupied by the data points should be about as high as it is wide. To save space on the page, graphs depicted here are often not high enough for their width
Sketch the best possible straight line through your 3 data points. Unless the points lie perfectly along a straight line, which due to experimental uncertainty is very unlikely, the best possible line will not actually pass through any of these points. The best-fit line can be constructed reasonably well by sketching the line which passes as close as possible, on the average, to the 3 points.
For reference, other examples of 3-point graphs and best-fit lines are shown below.
Describe your best-fit line by giving the following:
On the first line, the horizontal intercept of your best-fit line. The horizontal intercept will be specified here by a single number, which will be the coordinate at which the line passes through the horizontal axis of your graph.
On the second line, the vertical intercept of your best-fit line. The horizontal intercept will be specified here by a single number, which will be the coordinate at which the line passes through the vertical axis of your graph.
On the third line, give the units of your horizontal intercept and the meaning of that intercept.
On the fourth line, give the units of your vertical intercept and the meaning of that intercept.
Starting in the fifth line, give a brief written description of your graph and an explanation of what you think it might tell you about the system:
------>>>>>> horiz int, vert int, units and meaning of horiz, then vert int
Your answer (start in the next line):
.015
-8
No units for slope
m/s^2 for acceleration
The line doesn't pass through every line but it is pretty close and seem that the acceleration seems close to being constant.
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Mark the point on your best-fit line which would correspond to a ramp slope of .10. Determine as accurately as you can the rate of velocity change that goes with this point, so that you have both the horizontal and vertical coordinates of the point.
Report the horizontal and vertical coordinates of that point on the first line below, in the specified order, in comma-delimited format. Starting at the second line, explain how you made your estimate and how accurate you think it might have been. Explain, briefly, what your numbers mean and how you got them.
------>>>>>> mark and report best fit line coord for ramp slope .10
Your answer (start in the next line):
(.10, 30)
The slope is accurate to +-.005 and acceleration is +-.05
The numbers are the slope and acceleration that meets it on the graph.
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Determine the slope of the best-fit line
We defined rise, run and slope between graph points:
The 'run' from one graph point to another is the change in the horizontal coordinate, from the first point to the second.
The 'rise' from one graph point to another is the change in the vertical coordinate, from the first point to the second.
The slope between the two graph points is the rise-to-run ratio, calculated as slope = rise / run.
As our first point we will use the horizontal intercept of your best-fit line, the point where that line goes through the horizontal axis.
As our second point we will use the point on that line corresponding to ramp slope .10.
In the box below give on the first line the run from the first point to the second.
On the second line give the rise from the first point to the second.
On the third line give the slope of your best-fit straight line.
Starting in the fourth line, give a brief explanation and an indication of what you think the slope might tell you about the system.
------>>>>>> slope of graph based on horiz int, ramp slope .10 point
Your answer (start in the next line):
.02
7.735cm/s/s
386.75cm/s/s
The slope is the change of rate between the accerlation and the ramp slope.
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Assess the uncertainties in your result
The rest of this exercise is optional for Phy 121 and Phy 201 students whose goal is a C grade
Calculate average of mean times and average of standard deviations for 1-domino ramp
Since there is uncertainty in the timing data on which the velocities and rates of velocity change calculated in this experiment have been based, there is uncertainty in the velocities and rates of velocity change.
We first estimate this uncertainty for the 1-domino case.
In the box below, report in the first line the right-to-left mean time, the left-to-right mean time and the average of these two mean times on the 1-domino ramp. This third number, which you also calculated previously, will be called 'the average of the mean times'.
In the second line report the standard deviation of right-to-left times, the standard deviation of left-to-right times and the average of these standard deviations for the 1-domino ramp. This third number will be called 'the average of the standard deviations'.
Starting in the next line give a brief explanation and speculate on the significance of these results.
------>>>>>> 1 dom ramp mean rt-left and left-rt, then std def of both
Your answer (start in the next line):
1.493,2.811, 2.152
0.08111, 0.1254, .103
The first line is the means and then average of the means. the second line is the standard deviation and the average of the two.
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Use average time and standard deviation to estimate minimum and maximum possible velocity and acceleration for first ramp
We will use the average of the mean times and the average of the standard deviations to estimate our error in the average velocity and in the acceleration on the 1-domino ramp.
We will assume that the actual time down the ramp is within in the interval defined by mean +- std dev, where 'mean' is in this case the average of the mean times, and 'std dev' is the average of the standard deviations.
Using these values for mean and std dev:
Sketch a number line and sketch the interval from mean - std dev to mean + std dev. The interval will be centered at the average of the mean times as you reported it in the previous box, and will extend a distance equal to the average of the standard deviations (as also reported in the previous box) on either side.
So for example if the average of the mean times was 1.93 seconds and the standard deviation .11 second, the interval would extend from 1.93 sec - .11 sec = 1.82 sec to 1.93 sec + .11 sec to 2.04 sec.. This interval would be bounded on the left by 1.82 sec and on the right by 2.04 sec..
Report in the first line of the box below the left and right boundaries of your interval. Starting in the second line explain briefly, in your own words, what these numbers represent.
------>>>>>> boundaries of intervals rt-left, left-rt
Your answer (start in the next line):
(2.409, 2.255)
I subtracted the std and added the std to get the high low boundaries to the mean.
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Instead of 'rate of velocity change with respect to clock time' we will now begin to use the word 'acceleration'. So 'average acceleration' means exactly the same thing as 'average rate of velocity change with respect to clock time', and vice versa.
Since we are assuming here that acceleration is constant on a straight ramp, in this context we can simply say 'acceleration' rather than 'average acceleration'.
Using this terminology:
If the time down the ramp is equal to that of the left-hand boundary of the interval you just sketched, then what would be the average velocity and the acceleration of the ball? Report in comma-delimited format on the first line below.
Find the same quantities for the right-hand boundary of your interval, and report in similar format on the second line.
In the third line report the resulting minimum and maximum possible values of acceleration on this interval, using comma-delimited format. Your results will just be a repeat of the results you just obtained.
Starting on the fourth line, explain what your numbers represent and why it is likely that the actual acceleration of the ball on a 1-domino ramp, if set up carefully so that right-left symmetry is assured, would be between the two results you have given.
------>>>>>> 1-dom vel and accel left boundary of interval, rt boundary, min and max possible accel
Your answer (start in the next line):
13.67m/s,6.67 m/s/s
12.41m/s, 5.50m/s/s
Min=2.049s, max=2.255s
The first line is the velocity and acceleration for the left-hand boundary and the second line is the velocity and acceleration for the right hand boundary. The third line is the minimum and maximum values of acceleration.
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Repeat for 2- and 3-domino ramps
Do the same for the 2-domino data, and report in identical format, including explanations:
------>>>>>> 2-dom vel and accel left boundary of interval, rt boundary, min and max possible accel
Your answer (start in the next line):
20.83cm/s, 15.50cm/s^2
18.99cm/s, 12.88cm/s^2
1.344, 1.474
The first line is the velocity and acceleration for the left-hand boundary and the second line is the velocity and acceleration for the right hand boundary. The third line is the minimum and maximum values of acceleration.
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Do the same for the 3-domino data, and report in identical format, including explanations:
------>>>>>> 3-dom vel and accel left boundary of interval, rt boundary, min and max possible accel
Your answer (start in the next line):
25.95cm/s, 24.05cm/s^2
24.37cm/s, 21.21cm/s^2
1.079,1.149
The first line is the velocity and acceleration for the left-hand boundary and the second line is the velocity and acceleration for the right hand boundary. The third line is the minimum and maximum values of acceleration.
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Now make a table of your results, as follows.
You will recall that slopes of .03, .06 and .09 correspond to the 1-, 2- and 3-domino ramps.
In the first line report the slope and the lower limit on acceleration for the 1-domino ramp.
In the second line report the slope and the lower limit on acceleration for the 2-domino ramp.
In the third line report the slope and the lower limit on acceleration for the 3-domino ramp.
In the fourth line report the slope and the upper limit on acceleration for the 1-domino ramp.
In the fifth line report the slope and the upper limit on acceleration for the 2-domino ramp.
In the sixth line report the slope and the upper limit on acceleration for the 3-domino ramp.
Starting in the seventh line give a brief explanation, in your own words, of what these numbers mean and what they tell you about the system:
------>>>>>> slope and lower limit 1, 2, 3 dom; slope and upper limit 1, 2, 3 dom
Your answer (start in the next line):
.03, 6.67cm/s^s
.06, 15.50cm/s^2
.09,24.05cm/s^2
.03, 5.50cm/s^2
.06, 12.88cm/s^2
.09, 21.21cm/s^2
These are the relation of slope and acceleration on upper and lower limits.
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Plot acceleration vs. ramp slope using vertical segments to represent velocity ranges
On your graph of acceleration vs. ramp slope, plot the points specified by this table.
When you are done you will have three points lying directly above the .03 label of your horizontal axis. Connect these three points with a line segment running vertically from the lowest to the highest.
You will also have three points above the .06 label, which you will similarly connect with a segment, and three points above the .09 label, which you will also connect.
Your graph will now contain the best-fit straight line you made earlier, and the three short vertical line segments you have just drawn. Your graph will look something like the one below, though your short vertical line segments will probably be a little thinner than the ones shown here, and unlike yours the graph shown here does not contain the best-fit line. And of course your points won't be the same as those used in constructing this graph:
Does your graph fit this description?
Does your best-fit straight line pass through the three short vertical segments?
Give your answer and be sure to include a couple of sentences of explanation.
------>>>>>>
best-fit line thru error bars?
Your answer (start in the next line):
Yes my graph looks like the one above and the line passes through the vertical segments.
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Determine max and min possible slopes of acceleration vs. ramp slope graph
It should be possible to draw a number of straight lines which pass through all three vertical segments. Some of these lines will have greater slopes than others. For example note that the figures below show two lines which pass through all three vertical segments, with the line in the second graph being steeper than the line in the first.
Draw the steepest possible straight line which passes through all three vertical segments on your graph.
Using the x-intercept of this line and the point on this line corresponding to ramp slope .10, determine the slope of the line.
In the first line below report the rise, run and slope of your new line. Use comma-delimited format.
Starting in the second line give a brief statement of what your numbers mean, including an explanation of how you obtained your slope. Be sure to include the coordinates of the two points you used and the resulting rise and run.
------>>>>>>
max possible graph slope
Your answer (start in the next line):
33m/s^2, .085, 388.2
The numbers are the rise, run, and slope of the best fit line.
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Now draw the least-steep possible straight line which passes through all three vertical segments.
Follow the same instructions as before, and report your results for this line in the same way, including a brief explanation:
------>>>>>> min possible graph slope
Your answer (start in the next line):
28m/s^2, .01, 280
Once again this is the rise, run, and slope of the least steep line.
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Your instructor is trying to gauge the typical time spent by students on these experiments. Please answer the following question as accurately as you can, understanding that your answer will be used only for the stated purpose and has no bearing on your grades:
Approximately how long did it take you to complete this experiment?
------>>>>>>
Your answer (start in the next line):
5hrs
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*#&!
@&
I believe you accelerations were obtained by dividing average velocity by change in clock time, which is incorrect.
You need to correct those results. This will also affect the slope of your graph, which after correction will come out very close to the expected value.
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#
*@