Lab Report2

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Northeastern University *

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Physics

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Apr 3, 2024

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Report for Experiment #3 Motion in One and Two Dimensions Abstract In this times’ experiment, we used webcam to record the motion of a hover puck on a glass table. We have done four investigation for this experiment. The first one was to test the error ( δx: 0.0072&δy: 0.0058 ) when the puck stayed still in the middle of the glass table. Secondly, we used a wooden block to higher the glass table to create a oblique plane. After that, we put the hover puck at the top of the oblique glass table and let it free fall. Then, we record the speed of it and calculate the error of speed( δ V:2.4706) on the Y-axis. Thirdly, we gave a force to the puck to make it move and record the motion. Then, we get the error of speed (δ Vx:3.06696& δVy: 2.4706). Finally, we used the length(65.9cm) and height(8.9cm) to calculate the acceleration.

Introduction In this lab, we were repeating a famous experiment of Galileo. In Galileo’s experiment, he came to the conclusion that a cannonball shot from a tower in the horizontal direction would hit the ground at the same time as a ball dropped straight down from the same height. In this lab, we were trying to record the motion of a hover puck both in one dimension and two dimensions and calculate the acceleration of it to test the concept of Galileo that the vertical motion had to be independent of the horizontal motion. In this lab, our goals are: 1. To study the time dependence of displacement, velocity, and acceleration in one and two dimensions. 2.To show that motion in one direction is independent of motion in a perpendicular direction. In investigation 1, we mainly focused on testing the equipment and calculating the error of displacement. In investigation 2, we formally started to achieve our objects for this lab. Investigation 2 was about the displacement. In this investigation, we put the puck at the top of a inclined plane of the glass table and let it free fall. We recorded the displacement depended on time by using the webcam. In investigation 3, we tried to give the puck a force that points to upper right at the left corner of the inclined glass table and recorded the motion of the puck depended on time. After that, we used the equation: V(t) = ∆D ∆t (V(t) is the velocity depended on time, ∆D is the distance of the displacement within a period fo time, ∆ t is a period of time) to get the velocity on Y-axis and X-axis. Finally, in investigation 4, we used a ruler to test the height of the inclined plane to the desk and the length of the inclined plane. Then, we used the equation: Sinθ = D L = ma mg (in this equation, θ is the degree of the angle that the inclined plane with the table, D is the magnitude of the height, L is the length of the plane, m is the mass of the puck, a is the acceleration of the puck on the inclined plane, and g is the magnitude of the gravity that we need to calculate) to get the degree of the angle. Then, we used this equation again to calculate the magnitude of the gravity and compared the gravity we calculated to the real one.

Investigation 1 In investigation 1, the equipment that we used are glass table, hover puck, and webcam. The picture on the right shows how the webcam was installed on the glass table. Webcam was used to record the motion of the hover puck on the glass table. There was an app on the lab computer to help us to use the webcam. In this investigation, we firstly turn the puck on and put it on the glass table. Then, we adjusted the legs of the webcam and tried to make the glass table close to a plane which the hover puck will not move when it was placed on middle of the glass table. After that, we adjusted the setting of the webcam app to make the webcam was able to track the hover puck and started recording. After 10 seconds of recording, the webcam app will record the data of the x positions and y position of the hover puck each 0.0333 second. The hover puck stayed off during the whole recording. When we finished the recording, we copied the data into excel. Then, we started to analyse the data. Table 1-Displacement, time measurements (with absolute error) There were too much data. Thus, I picked the time from 0s to 0.3s. In table, 1 the first three rows are the data that we collected from the webcam app. t(s) is the time, x1 is the x-position, and y1 is the y-position. We tested for 10 seconds and these are the first ten times of recording. Than, we used excel to calculate the standard deviation of x and y t(s) x1(cm) y1(cm) δ x σ x δ y σ y 0.0334 33.17 34.921 0.138088 0.013809 0.110253 0.011025 0.0667 33.158 34.917 0.138056 0.013806 0.11038 0.011038 0.1 33.157 34.927 0.138279 0.013828 0.110408 0.011041 0.1333 33.158 34.924 0.138508 0.013851 0.110594 0.011059 0.1667 33.143 34.936 0.138732 0.013873 0.110765 0.011077 0.2 33.149 34.926 0.138784 0.013878 0.110834 0.011083 0.2333 33.149 34.924 0.138971 0.013897 0.11102 0.011102 0.2667 33.166 34.93 0.139159 0.013916 0.111194 0.011119 0.3 33.16 34.903 0.139257 0.013926 0.111372 0.011137

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automatically. The absolute error of x and y were δ = 10 σ based on what we knew in the lab manual. In these two graphs that we created, the x-axis for the graph, X positions, is the x position of the puck same for the graph, Y position.The y-axis for both of the graphs are how many times that the puck were in certain intervals of position. The positions of x and y were obtained by the FWHM FWHM

webcam app which can record the position of the puck each 0.333 second. The positions in these graphs were recording for 10 seconds. Based on the equation bellow, δn = W 2 2ln2 we can estimate the uncertainty δn which were 0.0098521for Y position and 0.0091727 for X position. These were really small values. Thus, we used 10 times standard deviation as the uncertainty for the rest of the investigations. The glass table might cause a systematic error. Since the glass table was not completely a horizontal plane and there had to have some angle which made the glass table a inclined plane, the data of position that we collected were based on the inclined plane which may cause the data away from the real value. Investigation 2 In investigation 2, the setup were glass table, hover puck, webcam, and wooden block. Different from the first investigation, this time, we put a wooden block under one of the legs of the glass table to create an inclined plane. After we had an inclined plane, we turned the hover puck on and place it in the middle top of the tilted table. We waited for one second after we clicked on start at the webcam app and release the puck without give a push. Then, we were recording the motion of the hover puck for 7 seconds and stopped. After we got the data, we plugged them into excel. Table 2-Displacement, time, velocity measurements (with absolute error) t(s) x1(cm) y1(cm) vy(cm/s) t avg (s) δ vy 0.6997 33.207 60.183 -0.4451 0.71655 4.696297 0.7334 33.206 60.168 -0.81325 0.75 4.696297 0.7666 33.231 60.141 -1.68675 0.7832 4.696297 0.7998 33.265 60.085 -5.3012 0.8164 4.696297 0.833 33.296 59.909 -5.63798 0.84985 4.696297 0.8667 33.326 59.719 -8.04217 0.8833 4.696297 0.8999 33.354 59.452 -10.512 0.9165 4.696297 0.9331 33.394 59.103 -11.0241 0.9497 4.696297

0.9663 33.416 58.737 -14.2136 0.98315 4.696297 1 33.422 58.258 -17.1988 1.0166 4.696297 1.0332 33.433 57.687 -16.747 1.0498 4.696297 1.0664 33.409 57.131 -21.4759 1.083 4.696297 1.0996 33.422 56.418 -21.543 1.11645 4.696297 1.1333 33.417 55.692 -23.1928 1.1499 4.696297 1.1665 33.398 54.922 -27.3193 1.1831 4.696297 1.1997 33.331 54.015 -25.9644 1.21655 4.696297 1.2334 33.305 53.14 -30.6325 1.25 4.696297 1.2666 33.31 52.123 -32.1687 1.2832 4.696297 1.2998 33.287 51.055 -33.2229 1.3164 4.696297 1.333 33.216 49.952 -35.1039 1.34985 4.696297 1.3667 33.195 48.769 -37.7711 1.3833 4.696297 1.3999 33.135 47.515 -38.4036 1.4165 4.696297 1.4331 33.081 46.24 -41.3253 1.4497 4.696297 1.4663 33.01 44.868 -42.819 1.48315 4.696297 1.5 32.971 43.425 -43.1627 1.5166 4.696297 1.5332 32.904 41.992 -48.2229 1.5498 4.696297 1.5664 32.827 40.391 -47.6506 1.583 4.696297 1.5996 32.744 38.809 -48.9318 1.61645 4.696297 1.6333 32.688 37.16 -52.5904 1.6499 4.696297 1.6665 32.594 35.414 -52.3193 1.6831 4.696297

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1.6997 32.502 33.677 -55.4896 1.71655 4.696297 1.7334 32.411 31.807 -56.8976 1.75 4.696297 1.7666 32.318 29.918 -58.7651 1.7832 4.696297 1.7998 32.201 27.967 -61.0542 1.8164 4.696297 1.833 32.109 25.94 -62.0475 1.84985 4.696297 1.8667 32.011 23.849 -63.6145 1.8833 4.696297 1.8999 31.875 21.737 -67.0181 1.9165 4.696297 1.9331 31.788 19.512 -67.9518 1.9497 4.696297 1.9663 31.632 17.256 -68.0415 1.98315 4.696297 2 31.521 14.963 -73.6446 2.0166 4.696297 2.0332 31.405 12.518 -72.5602 2.0498 4.696297 2.0664 31.256 10.109 -75.0301 2.083 4.696297 There were too much data. Thus, I picked a specific part of data which all the value of the velocity in y direction were negative. The average velocity in the y direction is defined as V(t avg ) = ∆y ∆t we knew that ∆y equals to y2-y1 and ∆t equals to t2-t1. The subscripts 1 and 2 represent any two consecutive data points in your data set. Based on the equation, we can calculate the average velocity at y direction in each 0.033 second. The t avg equaled to (t1+t2)/2. By using the error propagation equation, δ∆y = δy 1 2 + δy 2 2 since the δy 1 and δy 2 were the same, we can find that δ∆y = 2 δy =0.155917. Since time did not have error, we could get a equation δv y = δ∆y ∆t

Based on the δ∆y that we found which is 0.155917, we gained δv y =4.696297. In this graph, the hover puck would firstly stay still from 0s to 7s. Then, the hover puck was released. It started to move downward. Until about 2.066s, it hit the lower edge of the glass table and rebounded up. During the time range between 2.133s to 3.466s, the hover puck went upward until all its’ kinetic energy switched to the gravitational potential energy and went downward. The hover puck was repeating the up and down motion until run out of energy. There were also some extra value in this graph. We selected two specific data series to draw straight lines. The slope for the two straight lines were -55.716cm/ � 2 and -60.879cm/ � 2 . Since the plot was the relationship between speed and time, the slope of the straight lines were the acceleration. The first data series which between 0.6997s to 2.0664s had an uncertainty of 39.504 and the second data series which between 2.133s to 3.466s had an uncertainty of 178.49. The � 2 represented a statistical measure of how close the data are to the fitted trendline. Both of the � 2 were 0.9979 which is closed to 1 means all the data points are almost exactly on the trendline. By using the IPL Straight Line Fit Calculator, I gained slope m=-55.7159 ± 1.79358 and uncertainty b=39.5036 ±2.58449 for first series and m=-60.8795±1.85967, b=178.490±5.25817. The value both for m and b which gained from the IPL Straight Line Fit Calculator were same as the value that we gained from the graph. This situation was under the expectation. Since both of them were calculated by the machine and we only got uncertainty of speed, it was possible that we gain the same answer. The webcam might cause some random error. As we could see from the graph, there are some strange points which appeared in weird places. The reason why this happened was the webcam lost the hover puck which may cause the speed of it appear at weird places.

Investigation 3 The setup for investigation 3 were glass table, hover puck, webcam, and wooden block. Same with the investigation 2, we put a wooden block under one of the legs of the glass table to create an inclined plane. In this investigation, we firstly turned on the puck and placed it in the left corner. After we clicked the start recording, we waited for a second and pushed the puck upper right to make a parabola that went all the way up and came back down in the right corner. Table 3-Displacement, time, velocity measurements (with absolute error) t(s) x1(cm) y1(cm) � 푎�± (s) Vx(cm/s) δ Vx Vy(cm/s) δ Vy 1.0664 8.943 17.401 1.08325 16.35015 5.794918 50.35608 4.626617 1.1001 9.494 19.098 1.1167 23.5241 5.882191 62.92169 4.696295 1.1333 10.275 21.187 1.1499 23.19277 5.882191 62.28916 4.696295 1.1665 11.045 23.255 1.1831 23.49398 5.882191 59.03614 4.696295 1.1997 11.825 25.215 1.21655 23.23442 5.794918 56.20178 4.626617 1.2334 12.608 27.109 1.25 23.34337 5.882191 54.75904 4.696295 1.2666 13.383 28.927 1.2832 23.07229 5.882191 52.40964 4.696295 1.2998 14.149 30.667 1.31665 22.49258 5.794918 50.32641 4.626617 1.3335 14.907 32.363 1.3501 22.53012 5.882191 48.73494 4.696295 1.3667 15.655 33.981 1.3833 22.8012 5.882191 46.0241 4.696295 There were too much data. Thus, I picked the first ten values after the puck started moving. The t avg equaled to (t1+t2)/2. The subscripts 1 and 2 represent any two consecutive data points in your data set. The velocity could be calculated by the follow equation V = ∆d ∆t for the velocity of x,Vx, the ∆d were the difference between two x position after each 0.0332s. ∆ t was 0.0332s for the all data series. For the velocity of y, the ∆d were the difference between two y position after each 0.0332s. Based on the error propagation equation that we used before δ∆y = δy 1 2 + δy 2 2

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since the uncertainty were the same, we can find that δ∆y = 2 δy =0.155917 and δ∆x = 2 δx =0.195289. Since time did not have error, we could get a equation δv y = δ∆y ∆t Based on the δ∆y that we found which is 0.155917 and δ∆x that we found which is 0.195289 , we gained δv y =4.696295, δv x =5.882191. In this graph, we had two straight lines which shows the slope, uncertainty and � 2 . For the velocity of y, we gained the slope of -57.354, uncertainty of 123.3, and the � 2 of 0.9981. For the velocity of x, we gained the slope of -5.2318, uncertainty of 30.036, and the � 2 of 0.9597. The slope represented the acceleration of the hover puck and � 2 represented a statistical measure of how close the data are to the fitted trendline. The � 2 of Vy is larger than Vx which means Vy has more data points on the trendline than Vx. By using the IPL Straight Line Fit Calculator, what we got for Vy is the slope m=- 57.3473 ± 0.848409, the uncertainty b=123.286 ± 1.97675, for Vx is m=-5.23091 ± 0.835507, b=30.0319 ± 2.13473. I expected these value to be the same or very closed to the value that we gained from the graph. The slope of Vy in this investigation which is -57.3473 is closed to what I found in investigation 2 which is -55.716. However, the uncertainty 39.5036 in investigation 2 is larger than this investigation, 0.848409. The webcam might cause some random error. As we could see from the graph, there are some strange points which appeared in weird places. The reason why this happened was the webcam lost the hover puck which may cause the speed of it appear at weird places.

Investigation 4 In investigation 4, the setup were glass table, wooden block and ruler. We used ruler to measure the height h and the length of glass table. Then, we used them to calculate the acceleration due to gravity of investigation 2 and 3. Based on the graphs and the trigonometric function, sinθ = h L h was 8.2cm and L was 65.9cm based on the measurement of ruler. Thus, we gained sin θ which is 0.124431. Due to the Newton ’ s second law F = ma the force of gravity Fg equals to mg and the force of acceleration Fa equals to ma in which m is the mass of the hover puck, g is the gravity, and a is the acceleration. Based on the graph, we can also know that sinθ = ma mg = a g Thus, g=a/ sinθ . Due to the previous investigations, we knew that acceleration equals to the slope of velocity. In this case, a equals to the slope of velocity in y direction. For investigation 2 and 3, we totally got 3 slopes of y velocity (a1=-55.7159cm/ � 2 , a2= -60.8795 cm/ � 2 , a3=-57.3473cm/ � 2 ). Since the sinθ is a constant which is 0.124431, we can calculate g1=a1/ sinθ =447.7654cm/ � 2 , g2=a2/ sinθ =487.9483cm/ � 2 , and g3=a3/ sinθ =460.8763cm/ � 2 . Since the sinθ is a constant which is 0.124431, we gained an equation L h Fg N Fg h θ θ Fa

δg = δa sinθ Based on the uncertainty we gained from the previous investigations which were δa 1=39.5036, δa 2=178.490, δa 3=123.286 ,we can calculate δg 1= δa 1/ sinθ =317.4731, δg 2= δa 2/ sinθ =1434.4496, δg 3= δa 3/ sinθ =990.7981. ± 푎� = ± 1 + ± 2 + ± 3 3 = 465.53 cm/ � 2 δ± 푎� = δg 1 2 + δg2 2 + δg3 2 3 = 580.679 The accepted value of g is 980.665cm/ � 2 and the ± 푎� in our experiment is 465.53 cm/ � 2 which almost a half of the accepted value. This huge difference was due to the very large uncertainty which is 580.679 even larger than the value of the ± 푎� . The reason why this happen maybe was the error of webcam. Since the webcam could not perfectly lock the motion of the puck, there may have some extreme value during the recording which may cause the huge difference. Conclusion During this lab, we used the webcam to record the motion of a hover puck in both one dimension and two dimensions, and calculated the gravity. The goals we had in this lab were to study the time dependence of displacement, velocity, and acceleration in one and two dimensions and showed that motion in two perpendicular direction were independent with each other. In investigation 1, we tested the webcam. In investigation 2, we recorded the free fall motion of the hover puck in an inclined plane. In investigation 3, we pushed the hover puck to make it move in two directions. In investigation 4, we calculated the gravity based on the trigonometric function. During the investigation 2 and 3, we learned the time dependence of displacement, velocity, and acceleration in one and two dimensions based on the understanding of the graphs. Since the slope of investigation 3 in y direction was similar to one of the slopes in investigation 2. I learned that whether the x direction which is perpendicular to the y direction move or not, the motion in y direction will not be interfered. The uncertainty of the acceleration and ± 푎� were too large. The reason of it maybe the error of webcam which it could not perfectly lock the motion of the puck. What I can do to improve in next lab is to try to remember the equations of error which may reduce the time that I spent on finding the equation and make the procedure of experiment more smoothly.

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Questions 1. Time in investigation 2 is 1.3667s. Time in investigation 3 is 1.0337s. The time of two investigation were not the same because they were not start at the same position. In investigation 2, the puck started at the top of the plane. However, in investigation 3, the puck started at the bottom of the plane. 2. a 2θ is the acceleration of a puck falling down an incline of angle 2θ 3. The meaning of the intercept of your line with the v-axis is the velocity at start. If the recording is started at the same time the puck started to move, the intercept with the v-axis should be close to the origin. 4. If I performed Investigations 2 and 3 again with an incline of θ =0, the trajectories of Investigations 2 will stay still because the net force that acted on the puck is zero. The trajectories of Investigations 3 will be like billiards. It will hit the edge of the glass table and rebound until there is not energy. 5. Since the table is a square, the puck goes up the same distance in the y direction as it travels in the x direction. We get an equation that ∆x = ∆y . Since the acceleration for both x and y are constant, ∆y = Vyt − 1/2gt 2 , ∆x = Vxt . Based on the equation that we found in the experiment sinθ = ma mg = a g g=asin θ . Thus, Viyt − 1/2( asin θ)t 2 = Vixt . θ =arcsin(a t 2 /2t(Vix-Viy).

Lab Report2

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