Report for Experiment #3Motion in One and Two Dimensions EeswarAdluruLab Partner: RaymondLinTA: Abraham Tishelman-Charny 01/30/2018 Introduction Anymotion can be described through a breakdown of its x and y components. Thisapplies to both the velocity and acceleration of an object in motion.

Somemotions are one-dimensional, meaning that they only have either an x or ycomponent. An example of this is a ball being dropped free-falling from theroof of a building. There is no x component of acceleration present becausethere is no horizontal motion, however there is clearly a y component presentas the ball is accelerating vertically. Not all objects in motion are as simpleand clearly defined as this example, however. If a cannon shoots a ball up into the air, the ball will be moving bothhorizontally and vertically until it reaches some maximum height, and then theball will continue to move in the x direction and decrease height in the ydirection. The object in this scenario is considered to be moving in twodimensions.

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As one studies motion in two dimensions, the relationship betweenthe two components becomes apparent. In fact, Galileo conducted manyexperiments and came to the conclusion that when firing a cannon ball from atower in a horizontal direction, the ball would hit the ground at the same timeas a ball dropped straight down from the same height. This lab tests the theoryconcluded from Galileo’s experiment, in that downward motion under theinfluence of gravity is independent from the horizontal motion. Throughthe use of one-dimensional modern equations, the objective of Investigation 1was to analyze one dimensional motion. Using a frictionless air table tiled onan incline plane, a puck is released from the top of the table and makes itsway to the bottom. Marks are made by the puck via electric sparks and a sparktimer, every 0.1 seconds. Using the distance between the two dots and the timeit takes, velocity is determined.

This measurement is then applied to modernkinematic equations to determine what the one-dimensional acceleration due togravity is. Investigation 2 is concerned with velocity and acceleration in bothx and y components. Using the same setup as in Investigation 1, the onlydifference is that the puck is given a horizontal push as it makes its way tothe bottom of the air table. The same sequence of calculations are completed,and through Investigation 3, the acceleration due to gravity (a known constant,g = 9.8 m/s2) is derived from both a one dimensional, andtwo-dimensional motion. These quantities are then compared to their error.Ultimately, this lab will prove to familiarize students with the relationshipbetween objects in motion in one and two dimensions and enable them to see therelationship between x and y components of objects in motion.

Investigation 1 Toset up the first investigation, the air table is leveled flat by turning theair on and ensuring that the single puck being used doesn’t fall to onespecific edge of the table. Then a white piece of paper is placed on the table,and the x and y axes are labeled. The table is propped onto a wooden block tocreate the tilt which the investigation is based off of. The spark timer, whichis set to 100ms, is plugged in and when the puck is released from the top ofthe table, the timer switch is manually held and then released just before thepuck hits the opposite end of the table. The spark timer leaves a mark every100ms (?t), and the marks form a perpendicular line to the bottom of the table.Eight dots are then labeled, and ?y, or the distance between dots 1 and 3, 2and 4, 3 and 5, etc. till 8 is recorded into Table 3.

1. This information wasused to find the average velocity through the equation v= distance/time, or , where ?y occurs over two-time divisions so thetime interval between sparks, ?t, ismultiplied by 2. Additionally, errorsand relative errors for ?y wasdetermined and entered into the table. Table 3.1 Measurements and Calculations for Investigation 1 range ?y(cm) ??y(cm) ??y/?y t(ms) ?t ??t/?t v(cm/ms) ?v (cm/ms) ?v/v 1 to 3 1.9 0.05 0.

02631 100 0.1414 0.002 0.0095 0.0002507 0.02639 2 to 4 2.8 0.05 0.

01785 200 0.1414 0.002 0.014 0.0002515 0.01796 3 to 5 3.

8 0.05 0.01315 300 0.1414 0.002 0.019 0.

0002528 0.01330 4 to 6 4.7 0.

05 0.01063 400 0.1414 0.002 0.0235 0.0002543 0.

01082 5 to 7 5.7 0.05 0.00877 500 0.1414 0.002 0.0285 0.

0002564 0.00899 6 to 8 6.9 0.05 0.00724 600 0.1414 0.

002 0.0345 0.0002593 0.

00751 After plotting Velocity vs. Time, the slope of the trend line was determined to be5e-05,which is equal to the acceleration, a.Additionally, after using the IPL website, the graphical Error in Slope ( was determined tobe 6.09083e-7.Errors in time and change in time are neglected due to low relative errorpercentages, meaning that the values obtained are very close to the true value.The relative error of time is found to be +/- 0.

002. Using error propagation,the relative error can be found for the velocity of each data point, which canthen be used to find the uncertainty in velocity as well. An example of thiscalculation can be seen below for dots 1,3: (1) == 0.02638This value is equal to , so in order to find , the value of the relative error is multiplied by the velocity. Thismeans multiplying (0.02638) (0.

0095), to get 2.5066e-05 as the uncertainty ofvelocity for those data points.**Side Note: the errorbars are present in the graph above; however, they are too small to be seenwith the scale shown on the y axis(2) (3) (4) (5) In the end through plotting position vs.

time, the slope, oracceleration of the puck in motion in one dimension is determined to be 5e-5.The angle of the air table has not yet been factored in to this number but willbe later in Investigation 3. Investigation 2 Thegoal of investigation 2 was to understand motion in two-dimensions, and tocollect more data to test the theory that downward motion under the influenceof gravity is independent of horizontal motion. In theory, means that once ahorizontal component is added to the falling puck on the air table, theacceleration due to gravity should remain unaffected.

Just like Galileo notedin his canon ball experiment, the two balls hit the ground at the same time, asthey experienced the same vertical acceleration due to gravity. This shouldmean that if the puck is given both horizontal and vertical components, ifstudied over the same time interval as in the one-dimensional motion fromInvestigation 1, the acceleration in the vertical direction (y) should be thesame. This idea is ultimately what was being tested for this investigation. Toset up, the same piece of white paper was returned to the surface of the airtable, and the spark timer was turned on and set to 100ms. One lab partnerpressed the spark timer switch, as the other simultaneously launched the puckwith a horizontal push, and the spark timer was released just before the puckhit the opposite end of the table. This time, the series of dots created by thespark timer created a parabolic pattern, compared to the linear pattern inInvestigation 1. Starting around the third point, they were labeled one througheight.

The distance between marks 1 and 3 in the x direction and y directionwere recorded into tables 3.2 and 3.3, as well as from 2 to 4, 3 to 5, etc.(just like in Investigation 1, the distances were recorded over two-timedivisions). The determined errors werealso recorded and added to the tables below.

Four sets of graphs were createdfrom the data collected; x vs. t, y vs. t, vx vs. t, and vy vs.t. Velocity components were determined by using the same formula as inInvestigation1, or . Table 3.

2: Data in the HorizontalDirection (x) Dot range ?x(cm) ??x(cm) ??x/?x vx (cm/ms) ?vx (cm/ms) 1 to 3 7.3 0.05 0.

006849 0.0365 0.0002604 2 to 4 7.1 0.05 0.007042 0.0355 0.

0002598 3 to 5 6.8 0.05 0.007352 0.034 0.

0002590 4 to 6 6.5 0.05 0.007692 0.0325 0.

0002583 5 to 7 6.4 0.05 0.007812 0.032 0.

0002580 6 to 8 6.4 0.05 0.007812 0.032 0.0002580 Table 3.3: Data in the VerticalDirection (y) Dot range ?y(cm) ??y(cm) ??y/?y vy (cm/ms) ?vy (cm/ms) 1 to 3 8 0.05 0.

00625 0.04 0.0002624 2 to 4 8.8 0.05 0.005681 0.044 0.

0002650 3 to 5 9.6 0.05 0.005208 0.

048 0.0002677 4 to 6 10.1 0.05 0.004950 0.0505 0.

0002696 5 to 7 10.8 0.05 0.004629 0.054 0.0002723 6 to 8 11.7 0.

05 0.004273 0.0585 0.0002760 Table 3.4: Velocity and time Dot range t(ms) ?t ??t/?t v (cm/ms) ?v (cm/ms) ?v/v 1 to 3 100 0.1414 0.

002 0.0365 0.0002604 0.

007135 2 to 4 200 0.1414 0.002 0.0355 0.0002598 0.007321 3 to 5 300 0.

1414 0.002 0.034 0.0002590 0.

00762 4 to 6 400 0.1414 0.002 0.0325 0.

0002583 0.007948 5 to 7 500 0.1414 0.002 0.

032 0.0002580 0.008064 6 to 8 600 0.1414 0.002 0.032 0.

0002580 0.008064 **Error Bars are Present in ALL GRAPHS;however, they are too small to be seen along the increments of the y axes. Froman inspection of x vs.

t and y vs. t, the two are different because the slopeof the x vs. t graph is basically a straight line meaning that the puck did notmove nearly as much over the time period, as the puck did in the y direction,as shown by the increasing graph, showing that it was moving at a constantvelocity. As time increased so did the distance covered. Looking at the plot of vy vs t,the acceleration due to gravity (the slope of the trend line), ay, is 4e-05 m/s2. Theerror of this slope is 0.005976, meaning that thecalculated acceleration due to gravity doesn’t agree within errors.

Looking atvx vs. t, the error in slope is 0.005976. The velocity is constantbecause the slope of vx is within the error number of zero, as thetrue value of horizontal acceleration is supposed to be zero. Investigation 3 Themain purpose of the third investigation was to calculate the accelerations dueto gravity using the results obtained from Investigations 1 and 2. A formulafor acceleration due to gravity, g,was derived using the angle at which the air table was tilted, and eventuallythis was used to calculate g1 and g2, as well as their uncertainties.

G1 and g2 were then averaged to find gavg, andadditionally the uncertainty in that quantity was found. The relationshipbetween the angle of the air table and acceleration due to gravity can beexpressed by the equation , and can be found without actually measuring the angle of the air tablethrough the equation , where h2 is the height of the tallest vertical point of thetable, and h1 is the height of the lowest vertical point of thetable, and L is the horizontal length of the air table. The net height of thetable was found to be 3.38 cm, and the length of the tablewas found to be 66.6 cm. This meant that was equal to 0.0508. This numericalvalue was then plugged into the equation above in order to solve for theacceleration due to gravity, by plugging in the value for and the value of ay.

Values from bothinvestigations were used accordingly in order to solve for two accelerationsdue to gravity, g1 and g2. G1 was found to be 9.39 m/s2 and g2 was found to be 9.97 m/s2.Gavg was found by adding those two quantities together and dividingby two, to get 9.68 m/s2 as the average acceleration due togravity.

This however is not the end of the investigation. Next,uncertainty in g1 and g2 were to be found. First,uncertainty in was found to be 1.48x 10-4, which was calculated using the propagated errorformula shown in Investigation 1. Using the same propagated error formula, andthe equation , the uncertainty associated with each g1 and g2were found. was found to be 0.

120192by multiplying g1 by its relative error. was found to be 0.12127and was also derived by multiplying the value for g2 by its relativeerror. Lastly, using the formula , the average uncertainty of acceleration due to gravity was found to be0.

0854. Ourvalue of gavg, 9.68 m/s2, is lower thanthe accepted value which is 9.81 m/s2, and unfortunately theuncertainty of the average acceleration due to gravity does not agree with ourvalue, however if our error was doubled, then the value calculated fromInvestigations 1 and 2 would work. This discrepancy could be due to a number orreasons, but to name a few; underestimation of initial measurements, or friction. Conclusion Theoverarching goal of this lab was to prove the relationship between objects inmotion in one, versus two-dimensions. The x component of the puck moving in twodimensions had little to no effect on the motion in the vertical direction. Byusing an air table, a virtually “frictionless” surface was created in order tohelp simulate a scenario where air resistance and friction don’t exist, and theonly forces acting on the puck are either pure gravity, or pure gravity plushorizontal push.

Inthe end, the value for average acceleration due to gravity was in the ballparkof the actual acceleration due to gravity, which is 9.81 m/s2. Thecalculated value does not agree within the error; however, this does not changethe fact that the values from both Investigations were relatively close to oneanother, despite the additional horizontal component added to the puck inInvestigation 2. This Lab was therefore successful in proving Galileo’s theory,and showing that what happens in the x direction does not affect an objectsacceleration due to gravity. Somepossible errors which may have impacted the results of this lab, are that theair table could’ve not been 100% leveled.

In our first trial in experiment 1,the dots moved slightly horizontally, even though they were only supposed tohave moved only vertically, indicating that the balance was off. Additionally,if we were to do this lab again, we would shift the points picked to closer tothe release of the puck, to use as data points in Investigation 2 in order tohave our graph of y vs. t appear to be more parabolic in shape, as it issupposed to be. Questions1) The total time to reach the bottomof the air table for Investigations 1 and 2 are found to be around 1.

8seconds.These times will all be the same regardless of the mass of the object, as theyare all accelerating in the same vertical direction due to the samegravitational force, so it will take them the same amount of time to reach the bottom,whether there is a horizontal component involved or not.2) . The accelerations are the same.3) The meaning of the intercept of myline with the v-axis, is that this is the initial velocity of the puck at time0 s.

We expect this value to be close to the origin because it occurs at time0.4) With an incline of , the trajectories would look very different. First, there would be novertical drop, so the puck would not move, meaning that the position vs. timegraphs would be horizontal lines.5) If there was insufficient air flow throughthe table, this would mean friction is more likely to occur and oppose theaccelerations. This would cause deceleration and cause the slope of my graphsto become less slanted, and more horizontal.