#$&*
Phy 201
Your 'cq_1_09.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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A ball accelerates uniformly as it rolls 20 cm down a ramp, starting from rest, in 2 seconds.
What are its average velocity, final velocity and acceleration?
answer/question/discussion: ->->->->->->->->->->->-> :
vave = ds /dt = 20cm/ 2sec = 10cm/s
vf = 10cm/s (2) = 20cm/s
a ave = 10cm/s / 2sec = 5cm/s/s
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If the time interval is in error so that it is 3% longer than the actual time interval, then what are the actual values of the final velocity and acceleration?
answer/question/discussion: ->->->->->->->->->->->-> :
.03(2) = 2.06sec
Vf = 19.42 cm/s
A ave = 4.71cm/s/s
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What is the percent error in each?
answer/question/discussion: ->->->->->->->->->->->-> :
.03%, .06%
Your proportional errors are .03 and .06, which would be 3% and 6%
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If the percent error is the same for both velocity and acceleration, explain why this must be so.
answer/question/discussion: ->->->->->->->->->->->-> :
because they are calculated from one another.
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If the percent errors are different explain why it must be so.
answer/question/discussion: ->->->->->->->->->->->-> :
average velocity contains error therefore final velocity contains same error. But to derive acceleration the calculations have us use 2.06 again creating error (x2).
#$&*
________________________________________
Copy and paste your work into the box below and submit as indicated:
Your instructor is trying to gauge the typical time spent by students on these questions. 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 much time did you spend on this question? 30min
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30 min
** **
Copy this document, from this point down, into a word processor or text editor.
Follow the instructions, fill in your data and the results of your analysis in the given format.
Regularly save your document to your computer as you work.
When you have completed your work:
Copy the document into a text editor (e.g., Notepad; but NOT into a word processor or html editor, e.g., NOT into Word or FrontPage).
Highlight the contents of the text editor, and copy and paste those contents into the indicated box at the end of this form.
Click the Submit button and save your form confirmation.
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):
2.395
2.605
2.595
2.566
2.585
1.967
1.844
2.046
2.066
1.934
5 trials each way 1 domino
:
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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):
1.368
1.418
1.484
1.352
1.370
1.451
1.341
1.281
1.244
1.371
5 trials each way 2 domino
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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):
1.108
1.210
1.011
1.008
1.071
1.128
1.125
1.078
1.144
1.1
5 trials each way 3 dominoes'
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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):
2.149, .08739
1.971, .08967
1.398, .05384
1.338, .08055
1.082, .08321
1.115, .02599
:
#$&*
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):
13.029cm/s
14.206cm/s
20.029cm/s
20.927cm/s
25.878cm/s
25.112cm/s
Ds / dt = Vave
This mean times are used to derive the average velocity of the object on the ramp.
:
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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):
6.063cm/s/s
7.208cm/s/s
14.327cm/s/s
15.641cm/s/s
23.917cm/s/s
22.520cm/s/s
:
#$&*
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):
13.612, one domino
20.478, two domino
25.495, three domino
These times are calculated from data recorded from the ramp experiment and averaged together from left to right and right top left.
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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):
6.6355cm/s/s
14.984cm/s/s
23.2185cm/s/s
:
#$&*
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):
The averages are not the same, and I noticed this early in the experiment, that these velocities were gone to vary. The desk I conducted the experiment on is obviously out of level enough to make a difference. Another note worthy point is left to right setup took some real concentration, for I was conducted this with my not dominate hand.
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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):
One domino 6.6355 cm/s/s
Two domino 14.984 cm/s/s
Three domino 23.218 cm/s/s
These are the averaged accelerations, these are my more faithful measurements.
:
#$&*
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):
.11
27
Slope by height of domino
Cm/s/s
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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):
.1 is very close to my x-intersect, hope I havent done any wrong.
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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):
.1 and .9
0 and 23.22
slope of 28.75
I used the data recovered from the graph to calculate the slope of 28.75
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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):
2.149, 1.971, 2.06
.08739, 08964, .088515
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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):
1.971, 2.149
These are the lower and upper limits of the average of the times that it took to complete the ramp at one domino slope.
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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):
14.206cm/s, 7.208cm/s/s
13.029cm/s, 6.063cm/s/s
7.208cm/s/s, 6.063cm/s/s
These numbers represent the lower and upper boundaries of accelerations of the ball. Surprisingly, the lower limit is the first average that I calculated.
#$&*
Repeat for 2- and 3-domino ramps
Do the same for the 2-domino data, and report in identical format, including explanations:
------>>>>>> 2-dom vel and accel left boundary of interval, rt boundary, min and max possible accel
Your answer (start in the next line):
21.5cm/s, 16.5cm/s/s
19.512cm/s, 14.998cm/s/s
16.5cm/s/s, 14.998cm/s/s
These are the values found of the two domino slope.
#$&*
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):
:26.819cm/s, 25.689 cm/s/s
24.263cm/s, 21.025cm/s/s
25.689cm/s/s, 21 .025cm/s/s
These are the values of the steepest slope.
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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):
202.143
249.967
233.611
240.267
275.167
285.433
These numbers are slopes of accelerations and ramp heights, upper and lower limits.
#$&*
Plot acceleration vs. ramp slope using vertical segments to represent velocity ranges
On your graph of acceleration vs. ramp slope, plot the points specified by this table.
When you are done you will have three points lying directly above the .03 label of your horizontal axis. Connect these three points with a line segment running vertically from the lowest to the highest.
You will also have three points above the .06 label, which you will similarly connect with a segment, and three points above the .09 label, which you will also connect.
Your graph will now contain the best-fit straight line you made earlier, and the three short vertical line segments you have just drawn. Your graph will look something like the one below, though your short vertical line segments will probably be a little thinner than the ones shown here, and unlike yours the graph shown here does not contain the best-fit line. And of course your points won't be the same as those used in constructing this graph:
Does your graph fit this description?
Does your best-fit straight line pass through the three short vertical segments?
Give your answer and be sure to include a couple of sentences of explanation.
------>>>>>>
best-fit line thru error bars?
Your answer (start in the next line):
My graph does look similar to the graph above. The sloped line does go through the short bars, barely. It seems that my steepest slope at the lowest limit are the closest to not fitting.
:
#$&*
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):
:256.9
this is the slope of the steepest line graphed, it was obtained by the highest point in the lower/ upper bar plotted.
#$&*
Now draw the least-steep possible straight line which passes through all three vertical segments.
Follow the same instructions as before, and report your results for this line in the same way, including a brief explanation:
------>>>>>> min possible graph slope
Your answer (start in the next line):
210.25
This is the slope of the most shallow line situation, it was derived by the lowest of the points on the bars plotted.
:
#$&*
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):
2.5hrs
:
#$&*
Very well done, and your results are right in the middle of the expected range.
&#See any notes I might have inserted into your document, and before looking at the link below see if you can modify your solutions. If there are no notes, this does not mean that your solution is completely correct.
Then please compare your old and new solutions with the expanded discussion at the link
Solution
Self-critique your solutions, if this is necessary, according to the usual criteria. Insert any revisions, questions, etc. into a copy of this posted document. Mark any insertions with &&&& so they can be easily identified.If your solution is completely consistent with the given solution, you need do nothing further with this problem. &#