Motion in One and Two Dimensions

Report for Experiment #3 Motion in One and Two Dimensions Abstract This experiment explores the relationship between motions in different dimensions. A frictionless puck will be used to perform different motion on the table, and its change in position would be recorded by using a video tracker. The experiment would prove that the puck’s motion in one dimension is independent from its perpendicular dimension. Furthermore, the data retrieved will be used to calculate the acceleration of gravity. The calculated value of 𝑔𝑎𝑣𝑔 = −11.05 ± 0.157 𝑚/𝑠 2. Introduction The purpose of this lab was to study the relationship between position, velocity, and acceleration in both one and two dimensions. Position refers to the location of an object (in this case a point) at the specific moment of time. Velocity is the rate at which an object changes its position. While acceleration is the rate at which an object changes its velocity. They are all vector quantities, which means that they have a magnitude and direction. Velocity can be derived from change in position, or the derivative of x 𝑣= 𝛥𝑥 𝑥2 − 𝑥1 = 𝛥𝑡 𝑡2 − 𝑡1 𝑣 is the velocity, x is the position, t is the time, and 𝛥 is the change. The same applies to acceleration, which is the derivative of velocity: 𝑎= 𝛥𝑣 𝑣2 − 𝑣1 = 𝑡2 − 𝑡1 𝛥𝑡 a is the acceleration, v is the velocity, t is the time, and 𝛥 is the change. One thing can be noticed from these equations is that they are all time dependent, which means that without the progression of time, none of the quantities would change. Therefore, they are computed as changes occur with respect to time. The experiment explains how the motions of each dimension are independent from each other. In order to describe the motion of an object, a set of equations are being used called kinematics equations and assume the object has a constant acceleration. 𝑣𝑥 = 𝑣0 + 𝑎𝑡 𝑥 = 𝑥 0 + 𝑣0𝑥 𝑡 + 1 2 𝑎𝑥 𝑡 𝑣𝑥2 = 𝑣0𝑥2 + 2𝑎𝑥 (𝑥 − 𝑥0 ) 𝑣 stands for velocity, 𝑣0 is the initial velocity, x is the position, t is the time, and a is the acceleration. The equations used the x-dimension as an example, the same applies to motion in the y-dimension. Galileo’s experiment on motions has led to the discovery that motions in one dimension are independent from the motions in a perpendicular direction. Two balls with the same mass and density, one fired at the horizontal direction and the other straight down, will reach the ground at the same time despite the difference between the distance of two balls. This proves that the two dimensions, x and y, can be treated separately. The distance between two balls was caused by the extra force on the horizontal dimension on one of the balls, while gravity has the same effect on both balls, therefore making them land at the same time. Therefore, the objective of this experiment is to understand the relationship between time and displacement, velocity, and acceleration. Furthermore, the experiment will prove how motions in two dimensions are independent of each other. Investigation 1 would be the set-up of the puck and the calibration of the video tracker while estimating the random error of the experiment. Investigation 2 would be measuring the puck motion in one dimension, finding the velocity of the puck. Investigation 3 would incorporate the other dimension of motion, calculating the velocity of both x and y-dimension. Investigation 4 would look for the calculation of gravity based on the data retrieved from the previous investigations. Investigation 1 The materials that would be needed in the first investigation would be the hover puck, the glass table, webcam, ruler, and wooden block. In the first investigation, the live video tracker software would be first calibrated in order to track the motion of the puck. Turn the puck on to adjust the table legs so that the puck will stay stationary at the center. Place the wooden block on the side of the glass table with one leg to be used as an elevation. Start the IPL tracker program and have the puck in focus. Then, increase the exposure as long as the frame rate is at around 30 frames/sec. Then adjust the settings of the color and other qualities so that the screen is bright enough without oversaturation. Next, define the outer corner of the table, measure the length between the selected corner. Go to the color masking tab and adjust the setting so that everything is black beside the white spot that is indicating the position of the puck. place the puck in the center position, then start recording for 10 seconds and record the data. The data retrieved would be the position of the puck over time. Table 1: puck’s position over time on the x and y plain. t(s) 0.0334 0.0667 0.1 0.1334 0.1667 0.2 0.2333 0.2667 0.3 0.3333 0.3667 0.4 0.4333 0.4667 0.5 0.5334 0.5667 0.6 0.6334 0.6667 0.7 0.7333 0.7667 0.8 0.8333 0.8667 0.9 0.9333 0.9667 x1(cm) y1(cm) 32.506 32.24 32.476 32.225 32.503 32.225 32.49 32.209 32.5 32.225 32.496 32.242 32.502 32.23 32.496 32.222 32.508 32.224 32.487 32.24 32.487 32.228 32.486 32.24 32.477 32.22 32.503 32.234 32.503 32.214 32.49 32.239 32.494 32.241 32.506 32.243 32.502 32.22 32.504 32.221 32.508 32.225 32.487 32.237 32.487 32.229 32.493 32.23 32.512 32.232 32.504 32.216 32.521 32.239 32.502 32.228 32.504 32.224 t(s) 1.4333 1.4667 1.5 1.5334 1.5667 1.6 1.6334 1.6667 1.7 1.7333 1.7667 1.8 1.8333 1.8667 1.9 1.9333 1.9667 2 2.0334 2.0667 2.1 2.1334 2.1667 2.2 2.2333 2.2667 2.3 2.3333 2.3667 x1(cm) 32.506 32.482 32.499 32.486 32.506 32.487 32.503 32.494 32.487 32.494 32.524 32.504 32.487 32.491 32.512 32.493 32.485 32.517 32.5 32.493 32.499 32.494 32.511 32.479 32.498 32.48 32.501 32.477 32.487 y1(cm) 32.228 32.236 32.244 32.221 32.227 32.218 32.214 32.247 32.221 32.231 32.224 32.225 32.234 32.232 32.228 32.214 32.231 32.254 32.245 32.231 32.22 32.215 32.228 32.232 32.227 32.223 32.238 32.23 32.226 t(s) 2.8333 2.8667 2.9 2.9333 2.9667 3 3.0334 3.0667 3.1 3.1334 3.1667 3.2 3.2333 3.2667 3.3 3.3333 3.3667 3.4 3.4333 3.4667 3.5 3.5334 3.5667 3.6 3.6334 3.6667 3.7 3.7333 3.7667 x1(cm) 32.496 32.491 32.492 32.488 32.502 32.506 32.476 32.504 32.508 32.52 32.497 32.485 32.502 32.495 32.479 32.483 32.505 32.505 32.512 32.488 32.497 32.495 32.502 32.485 32.515 32.499 32.499 32.495 32.501 y1(cm) 32.214 32.221 32.229 32.239 32.223 32.23 32.227 32.222 32.228 32.239 32.223 32.224 32.227 32.238 32.226 32.232 32.229 32.223 32.236 32.229 32.248 32.214 32.226 32.221 32.238 32.219 32.223 32.222 32.225 1 1.0334 1.0667 1.1 1.1334 1.1667 1.2 1.2333 1.2667 1.3 1.3333 1.3667 1.4 t(s) 4.2333 4.2667 4.3 4.3333 4.3667 4.4 4.4333 4.4667 4.5 4.5334 4.5667 4.6 4.6334 4.6667 4.7 4.7333 4.7667 4.8 4.8333 4.8667 4.9 4.9333 4.9667 5 5.0334 5.0667 5.1 5.1334 5.1667 32.505 32.5 32.487 32.475 32.482 32.501 32.503 32.503 32.515 32.488 32.482 32.492 32.505 x1(cm) 32.49 32.496 32.504 32.492 32.506 32.499 32.503 32.508 32.495 32.489 32.5 32.48 32.481 32.504 32.497 32.5 32.494 32.514 32.478 32.503 32.496 32.493 32.503 32.489 32.48 32.495 32.527 32.515 32.504 32.224 32.227 32.228 32.228 32.232 32.246 32.227 32.226 32.239 32.257 32.222 32.237 32.212 y1(cm) 32.23 32.236 32.223 32.235 32.23 32.22 32.228 32.221 32.225 32.24 32.225 32.231 32.216 32.217 32.229 32.233 32.231 32.234 32.243 32.207 32.244 32.216 32.23 32.223 32.227 32.215 32.24 32.223 32.212 2.4 2.4333 2.4667 2.5 2.5334 2.5667 2.6 2.6334 2.6667 2.7 2.7333 2.7667 2.8 t(s) 5.6 5.6334 5.6667 5.7 5.7333 5.7667 5.8 5.8333 5.8667 5.9 5.9333 5.9667 6 6.0334 6.0667 6.1 6.1334 6.1667 6.2 6.2333 6.2667 6.3 6.3333 6.3667 6.4 6.4333 6.4667 6.5 6.5334 32.489 32.5 32.489 32.488 32.513 32.515 32.507 32.496 32.492 32.488 32.51 32.513 32.514 x1(cm) 32.522 32.499 32.491 32.51 32.484 32.491 32.489 32.499 32.495 32.498 32.49 32.495 32.514 32.506 32.495 32.504 32.491 32.506 32.493 32.489 32.503 32.492 32.507 32.485 32.497 32.488 32.499 32.498 32.486 32.232 32.221 32.215 32.241 32.222 32.235 32.231 32.232 32.23 32.233 32.235 32.229 32.234 y1(cm) 32.238 32.223 32.241 32.225 32.226 32.216 32.228 32.245 32.229 32.232 32.22 32.247 32.226 32.241 32.205 32.229 32.208 32.227 32.218 32.238 32.226 32.223 32.225 32.231 32.24 32.211 32.247 32.205 32.242 3.8 3.8333 3.8667 3.9 3.9333 3.9667 4 4.0334 4.0667 4.1 4.1334 4.1667 4.2 t(s) 6.9667 7 7.0334 7.0667 7.1 7.1334 7.1667 7.2 7.2333 7.2667 7.3 7.3333 7.3667 7.4 7.4333 7.4667 7.5 7.5334 7.5667 7.6 7.6334 7.6667 7.7 7.7333 7.7667 7.8 7.8333 7.8667 7.9 32.504 32.504 32.483 32.498 32.491 32.494 32.497 32.502 32.528 32.494 32.487 32.511 32.482 x1(cm) 32.501 32.505 32.498 32.495 32.5 32.49 32.488 32.495 32.492 32.498 32.485 32.488 32.504 32.499 32.5 32.487 32.483 32.495 32.495 32.488 32.493 32.493 32.506 32.501 32.491 32.495 32.493 32.508 32.51 32.232 32.228 32.239 32.226 32.228 32.225 32.242 32.218 32.245 32.213 32.242 32.219 32.238 y1(cm) 32.242 32.211 32.249 32.212 32.253 32.211 32.235 32.211 32.219 32.236 32.23 32.237 32.212 32.241 32.208 32.248 32.209 32.246 32.203 32.233 32.212 32.24 32.227 32.208 32.244 32.218 32.238 32.211 32.238 5.2 5.2333 5.2667 5.3 5.3333 5.3667 5.4 5.4333 5.4667 5.5 5.5334 t(s) 5.5667 8.3333 8.3667 8.4 8.4333 8.4667 8.5 8.5334 8.5667 8.6 8.6334 8.6667 8.7 8.7333 8.7667 8.8 8.8333 8.8667 8.9 8.9333 8.9667 9 9.0334 9.0667 9.1 9.1334 9.1667 9.2 9.1334 32.486 32.501 32.488 32.486 32.487 32.497 32.493 32.499 32.494 32.516 32.501 x1(cm) 32.511 32.502 32.515 32.491 32.493 32.489 32.501 32.496 32.492 32.497 32.492 32.5 32.502 32.503 32.517 32.5 32.485 32.496 32.489 32.501 32.513 32.514 32.496 32.479 32.492 32.493 32.488 32.522 32.493 32.214 32.229 32.232 32.239 32.23 32.215 32.216 32.218 32.244 32.232 32.235 y1(cm) 32.216 32.227 32.217 32.235 32.213 32.233 32.222 32.228 32.241 32.204 32.237 32.21 32.24 32.216 32.238 32.222 32.234 32.24 32.235 32.237 32.217 32.233 32.229 32.234 32.227 32.241 32.23 32.245 32.241 9.2333 9.2667 32.49 32.501 32.224 32.242 6.5667 6.6 6.6334 6.6667 6.7 6.7333 6.7667 6.8 6.8333 6.8667 6.9 t(s) 6.9333 9.7667 9.8 9.8333 9.8667 9.9 9.9333 9.9667 10 32.508 32.5 32.499 32.496 32.504 32.485 32.509 32.498 32.483 32.497 32.499 x1(cm) 32.48 32.506 32.496 32.5 32.483 32.512 32.5 32.492 32.497 32.218 32.235 32.218 32.234 32.234 32.231 32.239 32.206 32.228 32.206 32.247 y1(cm) 32.212 32.224 32.229 32.24 32.217 32.242 32.204 32.252 32.212 7.9333 7.9667 8 8.0334 8.0667 8.1 8.1334 8.1667 8.2 8.2333 8.2667 32.495 32.51 32.51 32.508 32.506 32.499 32.5 32.477 32.485 32.485 32.512 32.205 32.245 32.217 32.245 32.217 32.218 32.223 32.228 32.239 32.224 32.233 9.3 9.3333 9.3667 9.4 9.4333 9.4667 9.5 9.5334 9.5667 9.6 9.6334 9.6667 32.511 32.49 32.5 32.502 32.497 32.498 32.511 32.496 32.478 32.485 32.516 32.502 32.223 32.217 32.229 32.221 32.215 32.221 32.227 32.215 32.233 32.234 32.237 32.238 Figure 1: Histogram of the horizontal position of the puck and relative uncertainty for the position Figure 2: Histogram of the vertical position of the puck and its relative uncertainty for the position The width of the graph is related to the uncertainty of the measurements. The uncertainty of x and y is calculated by using the standard deviation equation 𝜎=√ 𝛴(𝑥1 − 𝜇)2 𝑁 where 𝜎 is the standard deviation, 𝑥𝑖 is the value of each position, N is the size of the data, and 𝜇 is the mean of the data. The computed value of the standard deviation for δx is 0.010121 m, and for δy is 0.010861m. Due to the puck being at a resting position, the uncertainty would be relatively much smaller compared to the puck in motion. It can be determined that the random error increases by a factor of 10 when in motion. Therefore, the error of the position regards x and y would be 10 times the standard deviation. Another source of error could be the systematic error. This can be caused by an incorrect set up or calibration of the video tracker, or it could be caused by the unnoticed change of the position of the table, causing an incorrect alignment from the live video, therefore affecting the entire experiment data. Investigation 2 For investigation 2, all the equipment mentioned in Investigation 1 will be used in investigation 2. For this part, the puck would be turned on then placed and held at the top center of the glass table. Then start the recording, wait for about a second and release the puck. Record for about 8 seconds. The data of x and y would be recorded and would indicate the position of the puck over time. For the second investigation, since the puck was released straight down, this means the only change of motion would occur in the y-dimension, which will be used to calculate the velocity of the puck. The kinematic equation would be used to derive the velocity from the change in position over change in time, which is 𝑣𝑦 (𝑡𝑎𝑣𝑔 ) = 𝛥𝑦 𝛥𝑡 𝑣𝑦 is the velocity in y-dimension, 𝛥𝑦 is change in the y position and 𝛥𝑡 is change in time. Since the investigation look for the average velocity for a time interval, the time coordinate of the average velocity will be the midpoint of the time interval, computed by using the equation 𝑡𝑎𝑣𝑔 = 𝑡2 + 𝑡1 2 Table 2: Partial data of the puck’s position, velocity, and velocity error over time t(s) 1.8334 1.8667 1.9 1.9333 1.9667 2 2.0334 2.0668 2.1001 2.1334 2.1667 2.2001 2.2334 2.2667 2.3 2.3334 2.3667 2.4 2.4333 2.4667 2.5 2.5334 2.5668 2.6001 2.6334 2.6667 2.7001 2.7334 2.7667 𝑥1 (cm) 29.146 29.1 29.091 29.066 29.066 29.001 28.977 28.95 28.917 28.863 28.829 28.768 28.73 28.666 28.627 28.553 28.499 28.418 28.381 28.256 28.197 28.114 28.044 27.927 27.863 27.758 27.656 27.555 27.442 𝑦1 (cm) 5.394 7.808 10.071 12.042 14.071 15.88 17.594 19.054 20.444 21.745 22.863 23.913 24.728 25.531 26.041 26.615 26.906 27.184 27.217 27.192 26.989 26.755 26.308 25.757 25.076 24.239 23.341 22.278 21.073 𝑇𝑎𝑣𝑔 1.85005 1.88335 1.91665 1.95 1.98335 2.0167 2.0501 2.08345 2.11675 2.15005 2.1834 2.21675 2.25005 2.28335 2.3167 2.35005 2.38335 2.41665 2.45 2.48335 2.5167 2.5501 2.58345 2.61675 2.65005 2.6834 2.71675 2.75005 2.78335 𝑣𝑦 72.49249 67.95796 59.18919 60.7485 54.32432 51.31737 43.71257 41.74174 39.06907 33.57357 31.43713 24.47447 24.11411 15.31532 17.18563 8.738739 8.348348 0.990991 -0.7485 -6.0961 -7.00599 -13.3832 -16.5465 -20.4505 -25.1351 -26.8862 -31.9219 -36.1862 -39.3393 𝛿𝑣𝑦 4.612545 4.612545 4.612545 4.598735 4.612545 4.598735 4.598735 4.612545 4.612545 4.612545 4.598735 4.612545 4.612545 4.612545 4.598735 4.612545 4.612545 4.612545 4.598735 4.612545 4.598735 4.598735 4.612545 4.612545 4.612545 4.598735 4.612545 4.612545 4.612545 2.8 2.8334 2.8667 2.9 2.9333 2.9667 3 3.0334 3.0668 3.1001 3.1334 3.1667 3.2001 3.2334 3.2667 3.3 3.3334 3.3667 3.4 3.4333 3.4667 3.5 3.5334 3.5668 3.6001 3.6334 3.6667 3.7001 3.7334 3.7667 3.8 3.8334 3.8667 3.9 27.368 27.262 27.123 27.046 26.904 26.813 26.694 26.586 26.514 26.367 26.231 26.091 25.945 25.849 25.709 25.612 25.481 25.388 25.275 25.198 25.107 25.036 24.951 24.886 24.87 24.806 24.757 24.745 24.706 24.692 24.699 24.702 24.718 24.725 19.763 18.325 16.773 15.056 13.239 11.285 9.212 6.952 4.636 3.443 5.098 6.574 8.045 9.174 10.258 11.2 12.047 12.667 13.184 13.558 13.806 13.951 13.888 13.805 13.435 13.086 12.511 11.9 11.031 10.133 9.07 7.91 6.625 5.15 2.8167 2.85005 2.88335 2.91665 2.95 2.98335 3.0167 3.0501 3.08345 3.11675 3.15005 3.1834 3.21675 3.25005 3.28335 3.3167 3.35005 3.38335 3.41665 3.45 3.48335 3.5167 3.5501 3.58345 3.61675 3.65005 3.6834 3.71675 3.75005 3.78335 3.8167 3.85005 3.88335 3.91665 -43.0539 -46.6066 -51.5616 -54.5646 -58.503 -62.2523 -67.6647 -69.3413 -35.8258 49.6997 44.32432 44.04192 33.9039 32.55255 28.28829 25.35928 18.61862 15.52553 11.23123 7.42515 4.354354 -1.88623 -2.48503 -11.1111 -10.4805 -17.2673 -18.2934 -26.0961 -26.967 -31.9219 -34.7305 -38.5886 -44.2943 -44.3544 4.598735 4.612545 4.612545 4.612545 4.598735 4.612545 4.598735 4.598735 4.612545 4.612545 4.612545 4.598735 4.612545 4.612545 4.612545 4.598735 4.612545 4.612545 4.612545 4.598735 4.612545 4.598735 4.598735 4.612545 4.612545 4.612545 4.598735 4.612545 4.612545 4.612545 4.598735 4.612545 4.612545 4.612545 T is the time, x is the horizontal position, y is the vertical position, 𝑇𝑎𝑣𝑔 is the average time calculated using the above equation, 𝑣𝑦 is the velocity in y-dimension, 𝛿𝑣𝑦 is the error of the velocity. The yellow highlighted portion is the data points used for the first trendline, the green highlighted portion is the data for the second trendline. The unit of the velocity would be cm/s, and for time would be second. The error of the velocity δv is also obtained by using the error propagation 𝛿𝑣𝑦 = 1 √2(𝛿𝑦)2 𝑡2 − 𝑡1 𝛿𝑦 = 10 ∗ 𝑆𝑇𝐷 𝛿𝑣𝑦 is the error of the velocity in cm/s, t is the time in second, and 𝛿𝑦 is the error of the position of the puck in the y-dimension in meter, which is retrieved by using the standard deviation of the y position in investigation 1 times 10. Figure 3: Scatter plot of velocity of the y-dimension vs time average. The velocity of the puck has a constant change over time due to no addition force, i.e., only a constant acceleration due to gravity is acting on the puck. The sharp change from negative velocity sudden to a positive velocity, for example at from t = 1.7s to t = 1.8s, was caused by the puck hitting the bottom of the glass table, resulting a change in direction that is reflection by the graph. Trend Line 1 80 60 40 Velocity 20 0 0 0.5 1 1.5 2 2.5 -20 -40 y = -117.07x + 286.23 R² = 0.9986 -60 -80 Time Figure 4: 1st Trend line derived from the velocity vs. time graph. 3 3.5 Trend Line 2 60 40 Velocity 20 0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 -20 y = -119.62x + 420.96 R² = 0.9965 -40 -60 Time Figure 5: 2nd Trendline of derived from the velocity vs. time graph The trendlines above are the representation of downward velocity of the puck. The value of 𝑅2 are 0.9986 and 0.9965, which are very close to 1, indicating that the data points are very close to the trendline, meaning that the velocity change over time is a constant number, indicating a constant acceleration. The slope here represents the velocity of the puck in motion, starting from the moment after the puck has hit the bottom till it comes back down again. Using the [1]IPL 𝑐𝑚 𝑐𝑚 Straight Line, the slope of the 1st trendline is -117.075 2 with an error of ±2.128 𝑠2 . The slope of the second line is -119.62 𝑐𝑚 𝑠2 𝑠 with an error of ±3.835 𝑐𝑚 𝑠2 . Comparing the magnitude of both slopes, they are close to each together which is expected because the puck is going through the same motion with a constant acceleration, therefore, the change of velocity over time should remain the same. One reason that may explain the difference between the two values could be the random error, the puck might be slightly shifted when it is being hit from the bottom. Investigation 3 For investigation 3, the same equipment from the last investigations will be used. This time, the puck will start at the left bottom corner of the table, then launched in a parabola shape that goes up then down to the bottom right corner. Record the entire process. Table 3: Partial data of the puck’s position, velocity, error in both dimensions over time 𝑡𝑎𝑣𝑔 0.2499 0.2832 0.31655 0.35 0.3833 0.4166 0.44995 0.4833 0.5166 0.54995 0.5833 0.6166 0.6499 0.68325 0.7166 0.7499 0.7832 0.81655 0.85 0.8833 0.9166 0.94995 0.9833 1.0166 1.04995 1.0833 1.1166 1.1499 1.18325 1.2166 1.2499 1.2832 1.31655 1.35 𝑣𝑦 𝛿𝑣𝑦 𝑣𝑥 𝛿𝑣𝑥 82.75449 77.13855 70.71642 66.88623 63.91566 57.96407 54.11411 50.65868 44.87952 42.74627 36.56627 33.53293 28.94578 25.43284 21.08434 16.85629 13.34337 9.164179 5.628743 1.024096 -2.24551 -7.41742 -9.43114 -13.9157 -18.6866 -22.4699 -25.988 -28.3735 -34 -36.5964 -40.6886 -45.0301 -48.6567 -52.4551 4.598735 4.626438 4.585007 4.598735 4.626438 4.598735 4.612545 4.598735 4.626438 4.585007 4.626438 4.598735 4.626438 4.585007 4.626438 4.598735 4.626438 4.585007 4.598735 4.626438 4.598735 4.612545 4.598735 4.626438 4.585007 4.626438 4.598735 4.626438 4.585007 4.626438 4.598735 4.626438 4.585007 4.598735 38.47305 37.8012 38.08955 37.81437 39.42771 38.17365 39.36937 39.58084 38.73494 39.58209 40.09036 40.2994 39.21687 40 41.14458 41.13772 40.87349 39.76119 41.79641 40.87349 41.73653 42.19219 41.40719 40.69277 42.1791 41.65663 42.36527 41.80723 42.65672 42.46988 41.79641 43.31325 42.29851 43.50299 4.285406 4.311222 4.272614 4.285406 4.311222 4.285406 4.298275 4.285406 4.311222 4.272614 4.311222 4.285406 4.311222 4.272614 4.311222 4.285406 4.311222 4.272614 4.285406 4.311222 4.285406 4.298275 4.285406 4.311222 4.272614 4.311222 4.285406 4.311222 4.272614 4.311222 4.285406 4.311222 4.272614 4.285406 𝑡(𝑠) 0.2332 0.2666 0.2998 0.3333 0.3667 0.3999 0.4333 0.4666 0.5 0.5332 0.5667 0.5999 0.6333 0.6665 0.7 0.7332 0.7666 0.7998 0.8333 0.8667 0.8999 0.9333 0.9666 1 1.0332 1.0667 1.0999 1.1333 1.1665 1.2 1.2332 1.2666 1.2998 1.3333 𝑥1 (𝑐𝑚) 4.89 6.175 7.43 8.706 9.969 11.278 12.553 13.864 15.186 16.472 17.798 19.129 20.475 21.777 23.117 24.483 25.857 27.214 28.546 29.942 31.299 32.693 34.098 35.481 36.832 38.245 39.628 41.043 42.431 43.86 45.27 46.666 48.104 49.521 𝑦1 (𝑐𝑚) 9.148 11.912 14.473 16.842 19.076 21.198 23.134 24.936 26.628 28.118 29.55 30.764 31.884 32.845 33.697 34.397 34.96 35.403 35.71 35.898 35.932 35.857 35.61 35.295 34.833 34.207 33.461 32.593 31.651 30.512 29.297 27.938 26.443 24.813 1.3833 1.4166 1.44995 1.4833 1.5166 1.54995 1.5833 -55.0904 -61.0778 -64.0541 -67.9341 -70.6325 -77.0149 -78.8253 4.626438 4.598735 4.612545 4.598735 4.626438 4.585007 4.626438 43.25301 43.47305 43.81381 44.04192 43.8253 44.29851 44.54819 4.311222 4.285406 4.298275 4.285406 4.311222 4.272614 4.311222 1.3667 1.3999 1.4333 1.4666 1.5 1.5332 1.5667 50.974 52.41 53.862 55.321 56.792 58.247 59.731 23.061 21.232 19.192 17.059 14.79 12.445 9.865 The velocity of the puck in both dimensions are derived using the equation discussed in introduction, and the error of the velocities are derived using the equation discussed in investigation 2. The data presented in the chart are the data being used to create the trendline of 𝑣𝑦 and 𝑣𝑥 . Vy vs. Time Scatter Plot 100 80 60 velocity 40 20 0 -20 0 0.5 1 1.5 2 2.5 3 3.5 -40 -60 -80 -100 time Figure 6: Velocity in y-dimension over time The graph above displays the velocity of the puck in the y-dimension over time. Compares to the previous graph in investigation 2, the two graphs look very similar in terms of shape and pattern despite the puck is going through a different motion. Same analysis of figure 3 can be apply to this graph, the puck’s velocity at t = 0.233s and t = 1.567s experienced a sharp turn due to the puck hitting the bottom of the table, causing a sudden change in direction. Vx vs. Time Scatter Plot 50 40 30 velocity 20 10 0 -10 0 0.5 1 1.5 2 2.5 3 3.5 -20 -30 -40 -50 time Figure 7: Velocity in x-dimension over time The graph displays the change in velocity of the puck in the x-dimension over time. The sudden change of the positive velocity to negative velocity at t = 1.567s was caused by the puck hitting the far-right corner, resulting a change in direction of the puck in the horizontal dimension. Compares to the change of the velocity in the y-dimension, the velocity change in x-dimension is relatively smaller. This indicates that the change in x position is the same over time, resulting a near constant velocity, therefore no acceleration occurred in the x-dimension. Relate back to the y-dimension, where a clear change in velocity over time has presented, it can be concluded that the two dimensions are independent of each other, therefore the acceleration due to gravity in the y-dimension does not affect the motion in the x-dimension. Vy trend line 100 80 60 velocity 40 20 0 -20 0 0.2 0.4 0.6 0.8 1 1.2 1.4 -40 -60 -80 -100 y = -119.09x + 107.76 R² = 0.9991 time 1.6 1.8 Figure 8: Velocity in y-dimension trendline The slope of the graph represents the acceleration of the puck in the y-dimension. Using the IPL Straight line fit the slope of the line is -119.096 with an error of ±1.824. Compares to figure 5 in investigation 2, the slopes of both lines are very close to each other despite their difference in the motion of the puck. This indicates that the acceleration for both motions are the same in vertical direction. Therefore, it can be concluded that the motion in the y-dimension does not depend on the motion in the x-dimension. Vx Trend line 45 y = 4.6418x + 36.954 R² = 0.9243 44 velocity 43 42 41 40 39 38 37 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 time Figure 8: Velocity in x-dimension trendline The slope of the trendline is 4.642 with an error of ±1.700, derived from the IPL straight line fit calculator. Compares to the slope of the y-dimension, the velocity in x-dimension is relatively much smaller, indicating a small, close to zero acceleration. Theoretically the change in velocity in x-dimension should be zero, but the actual data did not reflect it. This might be caused by the unaccounted systematic error. Since the puck was launched by hand without any additional help on precision, it is likely that extra push might be added upon the puck, causing a slight variation as a result of a random error. Investigation 4 For this investigation, the data derived from previous investigations will be used to derive the value of g, the gravitational acceleration. ՜ ՜ 𝑔 ՜ Figure 9: Free-body diagram of the puck for investigation 2 and 3 First, find sin (𝜃), 𝜃 is the angle at which the table is tilted. To find the angle, measure the length of the table and height of both the table and the wooden block. Then using the Pythagorean 𝐻 theorem to find sin(𝜃) = 𝐿 . The height would be 6.3 ± 0.05cm, and the length would be 58.7 ± 0.05 cm. The relative error of sin(𝜃) is δ𝐻 2 δ𝐿 2 δsin(𝜃) ) = √( ) +( 𝐿 sin(𝜃) 𝐻 Where H stands for height, L stands for length, resulting δsin(𝜃 ) sin(𝜃) = 0.00798 Next, g is being calculated using the value of sin (𝜃), 𝑚𝑔 sin(𝜃) = 𝑚𝑎𝑦 𝑎𝑦 sin(𝜃) 𝑔= Which 𝑎𝑦 is the slope of the graph. 𝑔1 = −10.91 𝑚/𝑠 2 , derived from figure 4 in investigation 2. 𝑔2 = −11.15 𝑚/𝑠 2, derived from figure 5 in investigation 2. 𝑔3 = −11.01 𝑚/𝑠 2, derived from figure 8 in investigation 3. Using the above three g value, the 𝑔𝑎𝑣𝑔 = 𝑔1 +𝑔2 +𝑔3 3 = −11.02 𝑚/𝑠 2 Their uncertainties are calculated by using the error propagation δsin(𝜃) 2 δ𝑎𝑦 2 δ𝑔 ) +( ) = √( sin(𝜃) 𝑎𝑦 𝑔 δ𝑔 is the error of the gravity, 𝑎𝑦 is the slope of the trendlines calculated using the IPL straight line fit calculator. The value of δ𝑔1 = 0.224 𝑚/𝑠 2, δ𝑔2 = 0.368 𝑚/𝑠 2 , δ𝑔3 = 0.190 𝑚/𝑠 2 . The average of the errors is δ𝑔𝑎𝑣𝑔 = 0.157 𝑚/𝑠 2, calculated by using the equation δ𝑔𝑎𝑣𝑔 = √(δ𝑔1 )2 + (δ𝑔2 )2 + (δ𝑔3 )2 3 The final measured 𝑔𝑎𝑣𝑔 = −11.05 ± 0.157 𝑚/𝑠 2 . Compares to the acceptable g value of 9.8 𝑚/𝑠 2 , the experimental value did not fall within the range of the value. This means that an error most likely occurred during the experiment. One source error could be the systematic error that took place during the experiment. An example would be unleveled table due to limited human capacity as it can be difficult for human to level the table at exact without additional help of tools. This could impact the entire experiment, causing a shift in data from desired values. Conclusion In conclusion, the goals of the experiment are to prove that motion in one dimension is independent from the perpendicular dimension and to determine the value of g using experimental data. The puck is going through different motions in a frictionless setting and the position of the puck over time is being recorded to determine the velocity of the puck. The slope of the velocity was analyzed to understand the relationships between the motion of the puck in different dimensions, which indicates how the motion in one dimension is independent from the perpendicular dimension, that the change in position, velocity, or acceleration in one dimension would not affect the other dimension. Investigation 1 focused on the setting up of the experiment and determining the error of the puck’s position. Investigation 2 start with the puck’s motion in one-dimension, the y-dimension. The acceleration of the puck was determined to be constant since the slope of the puck is a 𝑐𝑚 𝑐𝑚 straight line, with m = -117.075 ± 2.128 𝑠2 and m = -119.628 ± 3.836 𝑠2 . Investigation 3 put an emphasis on the motion in both dimension, showing a similar value in the change of velocity of y-dimension over time, indicating the independence of motions in x and y-dimension, since the velocity of x did not have much variation over time due to no additional force occurred in the horizontal dimension, while gravity continue to act on the vertical dimension, producing a 𝑐𝑚 similar slope m = -119.096 ± 1.824 2 . Investigation 4 asks to derive the gravity acceleration 𝑠 based on the collected data from the experiment. The g value in respect to each measured acceleration in y-dimension are 𝑔1 = −10.91 ± 0.224 𝑚/𝑠 2, 𝑔2 = −11.15 ± 0.368 𝑚/𝑠 2 , 𝑔3 = −11.01 ± 0.190 𝑚/𝑠 2 . The calculated 𝑔𝑎𝑣𝑔 = −11.02 ± 0.157 𝑚/𝑠 2, while the actual value of gravity = 9.8 𝑚/𝑠 2 does not fall within the range. Some of the possible unaccounted error could be having an uneven surface of the wooden block, creating a height difference that would affect the calculation of the gravity acceleration. Another error could be a delay or malfunction that occur with the video tracker that goes unnoticed. To improve these potential draw backs, the wooden block should have an even surface to have a unified height level. Furthermore, double check the video tracker and make sure that all category has been adjusted. Questions 1. The time that the puck reached the bottom of the air table in investigation 2 happens at t = 2. 3. 4. 5. 1.767, for investigation 3 it happens at t = 1.567. There is some minor difference between the two-value due to random error, but the two numbers are close enough to conclude that just like the picture of the canon of the ball hitting the ground at the same time, the puck hit the bottom of the table at the same time. This is because although the motion of the puck in both investigation is different, their change in velocity over time are the same, indicating that the acceleration is the same. The only acceleration that is acting upon the puck is the gravitational acceleration. The acceleration of 𝑎𝜃 = 𝑔 sin(𝜃), then 𝑎2𝜃 = 𝑔 sin(2𝜃 ). The intercept of the line means that the velocity is at zero at the specific moment of the time. At the point, the position of the puck is at the highest point, therefore velocity switched from positive to negative. It also indicates the initial velocity of the puck being zero at t = 0. The value can be expected to be close to origin since when at t = 0, v = 0. For investigation 2, the puck will remain at the same position since the normal force acting on the puck would be the same as the gravity force resulting a net force of zero. For investigation 3, the puck will proceed in a straight-line motion rather than creating a parabola shape because the gravitation acceleration will be cancelled by the normal force, no additional acceleration will caused the puck to change its direction. Since value of the position of x and y-dimension would be the same, use the Pythagorean theorem, the angle would be 45 degrees. Honor Questions: a. There would not be a significant change of the result for both investigations. b. The drop time would increase because gravity would create less force acting on the puck, therefore decrease the acceleration, indicating a smaller change in velocity or greater change in time. c. No, because the puck released first will gain more speed at a faster rate than the second puck. This cause the distance between the two pucks to be varied over time. d. The average acceleration and velocity would not change, the travel time to the highest point would be 1.2 second.