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): : 1.703 1.906 1.982 1.876 1.826 1.061 1.762 1.073 1.687 1.625 All these are within 0.2 of one another. This means the timing of the experiment was both fairly precise and accurate. #$&* Enter your data for the 2-domino setups in the same format, being sure to include your brief explanation: On the 'readable' posted version this data will follow the boldfaced heading '5 trials each way 2 dominoes' ------>>>>>> 2 domino results Your answer (start in the next line): 1.250 1.178 1.250 1.312 1.281 1.375 1.328 1.359 1.296 1.453 These are also very close to one another, but are shorter than the data for the 1-domino setup. This means the 2-domino setup caused the ball to travel the ramp in a shorter time period, meaning the acceleration was greater. #$&* 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): 1.093 1.125 1.109 1.140 1.140 1.062 1.062 1.063 1.062 1.093 Again these are shorter, meaning the ramp with the 3-domino setup has the greatest velocity. #$&* Calculate mean time down ramp for each setup In the previous hypothesis testing exercise, you calculated and reported the mean and standard deviation of times down each of the two 1-domino setups, one running right-left and the other left-right. You may use any results obtained from that analysis (provided you are confident that your results follow correctly from your data), or you may simply recalculate this information, which can be done very quickly and easily using the Data Analysis Program at http://www.vhcc.edu/dsmith/genInfo/labrynth_created_fall_05/levl1_15\levl2_51/dataProgram.exe\ In any case, calculate as needed and enter the following information, in the order requested, giving one mean and standard deviation per line in comma-delimited format: • Mean and standard deviation of times down ramp for 1 domino, right-to-left. • Mean and standard deviation of times down ramp for 1 domino, left-to-right. • Mean and standard deviation of times down ramp for 2 dominoes, right-to-left. • Mean and standard deviation of times down ramp for 2 dominoes, left-to-right. • Mean and standard deviation of times down ramp for 3 dominoes, right-to-left. • Mean and standard deviation of times down ramp or 3 dominoes, left-to-right. On the first subsequent line briefly indicate the meaning of your results and how they were obtained: ------>>>>>> mean, std dev each setup each direction Your answer (start in the next line): 1.85, .1038 1.67, .0609 1.25, .0497 1.36, .0509 1.12, .0204 1.06, .0136 The mean time of travel down the ramp gets shorter with each domino added, while the standard deviation seems to get smaller as well. #$&* 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): 15.13 cm/s 16.76 cm/s 22.40 cm/s 20.58 cm/s 25 cm/s 26.4 cm/s Results were obtained by taking the distance traveled (28 cm) and dividing them by the mean time interval for each domino setup in each direction #$&* Calculate average ball acceleration for each setup Assuming that the velocity of the ball changed at a constant rate in each trial, use the mean time interval and the 28 cm distance to determine the average rate of change of velocity with respect to clock time. You will determine your results in the following order: • Average rate of change of ball velocity with respect to clock time for 1 domino, right-to-left. • Average rate of change of ball velocity with respect to clock time for 1 domino, left-to-right. • Average rate of change of ball velocity with respect to clock time for 2 dominoes, right-to-left. • Average rate of change of ball velocity with respect to clock time for 2 dominoes, left-to-right. • Average rate of change of ball velocity with respect to clock time for 3 dominoes, right-to-left. • Average rate of change of ball velocity with respect to clock time for 3 dominoes, left-to-right. Report your six results in the box below, one result per line, in the order requested above. Starting in the seventh line explain how you obtained your results, giving the details of how you obtained at least one of your results. These details should include the definition of the average rate of change of velocity with respect to clock time and should explain, step by step, how you used the mean time and the distance down the ramp to arrive at your result, and should show the numbers used and the numbers obtained in each step. ------>>>>>> ave roc of vel each of six setups Your answer (start in the next line): 16.35 cm/s^2 20.07 cm/s^2 35.84 cm/s^2 30.26 cm/s^2 44.64 cm/s^2 49.81 cm/s^2 Since vf = vAve*2 when v0 is 0, I calculated vf and then divide by the change in time interval which here is the time interval since the time started at 0 as well. As seen below 15.13*2 = 30.26 cm/s / 1.85 s = 16.35 cm/s^2. #$&* 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 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): 18.21 cm/s^2 33.05 cm/s^2 47.23 cm/s^2 Rate of change of velocity is acceleration, so I took the reported accelerations from above and average then for 1 d0mino, 1 domino and 3 domino setups. They get fast3er as they go, meaning that the steeper the ramp, the more acceleration will occur. #$&* Find acceleration for each slope based on average of left-right and right-left times Average the mean time required for the right-to-left run with the mean time for the left-to-right run. Using this average mean time, recalculate your average rate of velocity change with respect to clock time for the 1-domino trials Do the same for the 2-domino results, and for the 3-domino results. Report your results in the box below, giving one average rate of change of velocity with respect to clock time per line, in the order requested. In the subsequent line explain how you obtained your results and what you think they mean. ------>>>>>> left-right, right-left each setup, ave mean times and give ave accel Your answer (start in the next line): 9.15 cm/s^2 16.47 cm/s^2 23.58 cm/s^2 I averaged both the clock time and the velocities for each domino setup. Using these averaged numbers, I calculated the average acceleration for each domino setup. #$&* 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 are both averaged results, but they were averaged from different numbers, so I wouldn’t expect them to be the same. The first set of results are more accurate in my opinion. #$&* Associate acceleration with ramp slope Your results will clearly indicate that, as expected, acceleration increases when ramp slope increases. We want to look further at just how the acceleration changes with ramp slope. If you set up the ramps according to instructions, then the ramp slopes for 1-, 2- and 3-domino systems should have been approximately equal to .03, .06 and .09 (if you used coins and the 15 cm ramp instead of dominoes and the 30-cm ramp, your ramp slopes will be different; each dime will correspond to a ramp slope of about .007, each penny to a slope of about .010, each quarter to a slope of about .013). For each slope you have obtained two values for the average rate of change of velocity with respect to clock time on that slope. You may use below the values obtained in the preceding box, or the values you obtained in the box preceding that one. Use the one in which you have more faith. In the box below, report in the first line the ramp slope and the average rate of change of velocity with respect to clock time for the 1-domino system. Use comma-delimited format. Using the same format report your results for the 2-domino system in the second line, and for the 3-domino system in the third. In your fourth line specify the units of these quantities. Ramp slope is a unitless quantity; be sure you report this. Also briefly explain how you got your results and what they tell you about this system: ------>>>>>> ramp slope ave roc of vel each system Your answer (start in the next line): .03, 18.21 cm/s^2 .06, 33.05 cm/s^2 .09, 47.23 cm/s^2 These results would have units of cm/s^2 because ramp slope is unit less and the other entity in this graph is acceleration, or change of rate of velocity. #$&* 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) 0.01 35 Slope Cm/s^2 #$&* Mark the point on your best-fit line which would correspond to a ramp slope of .10. Determine as accurately as you can the rate of velocity change that goes with this point, so that you have both the horizontal and vertical coordinates of the point. Report the horizontal and vertical coordinates of that point on the first line below, in the specified order, in comma-delimited format. Starting at the second line, explain how you made your estimate and how accurate you think it might have been. Explain, briefly, what your numbers mean and how you got them. ------>>>>>> mark and report best fit line coord for ramp slope .10 Your answer (start in the next line): .01, 0 This point is actually the 0.01 slope of the line as well as the x-intercept. I’m not very sure of how accurate this is, as I graphed the line, but am not sure how welkl the best fit line actually fits the data. #$&* 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): :At this point I’m pretty sure something has messed up along the way. My intercept point with the x axis is at the 0.01 slope mark, meaning the slope at this point is 0.