Sunday, March 8, 2020
Details of the Experiment Essays
Details of the Experiment Essays Details of the Experiment Essay Details of the Experiment Essay The data which was provided was taken from the following experiment: An air track was set up at a slight angle and a small trolley was placed on it. The trolley had a 10cm long flag attached to the top of it, so that a light gate could time how long it took the trolley to pass a given point. There were two light gates set up exactly 10.0cm apart which gave 2 separate time readings for different starting positions up the sloping air track. Data taken from the Experiment: The experiment was completed and some data was extracted from it and given for analysis and calculation. The data which was taken from the experiment is given below: Ta/s Tb/s y/cm 0.268 0.206 10.0 0.205 0.173 20.0 0.173 0.152 30.0 0.174 0.154 40.0 0.153 0.139 50.0 0.139 0.128 60.0 0.130 0.120 70.0 0.119 0.113 80.0 0.112 0.106 90.0 Constant Value Length of flag on trolley 10.0 cm Mass of the Trolley 179 g Distance between gates 10.0 cm Acceleration of free fall 9.81 m.s-2 Original Calculations: The first few calculations to be applied to the data taken from the experiment was to calculate the velocity of the trolley at each of the two points. This was found by using the distance (10cm flag on trolley) and the time (reading at A or B) and putting them in the following formula: Distance = Velocity x Time 10cm = x 0.268s etc This gave all of the velocities and once the first had been calculated, the formula could be used in Microsoft (c) Excel and the Fill Down command used to calculate the velocity for every individual set of data. The Data table now looked like this with the new accelerations included: Ta/s Tb/s y/cm Velocity A (ms-1) Velocity B (ms-1) 0.268 0.206 10.0 0.373 0.485 0.205 0.173 20.0 0.488 0.578 0.173 0.152 30.0 0.578 0.658 0.174 0.154 40.0 0.575 0.649 0.153 0.139 50.0 0.654 0.719 0.139 0.128 60.0 0.719 0.781 0.130 0.120 70.0 0.769 0.833 0.119 0.113 80.0 0.840 0.885 0.112 0.106 90.0 0.893 0.943 In order to see what this data is representing it can all be plotted onto a graph showing the calculated velocities of the trolley at the different heights up the slope (y) given in the original data. This Graph has been plotted and included on the next page. By looking at the data on the graph a simple line of best fit was plotted on the Velocity at A data, this line shows how close the results are together and how well correlated they are. As can be seen, the data is all related very closely together as the line fits very well into the data. Trolley Acceleration: Another graph of the Velocity at A against the Velocity at B was plotted to check the data. This shows how the speed changed along the trolleys journey and due to the almost perfect straight line fit it seams obvious that the speed of the trolley is increasing at a steady, constant rate. This suggests that the acceleration of the trolley is constant as physics suggests it should be through Newtons Laws stating that F=ma therefore in order for the trolley the change its acceleration it would have to change either the force acting on it which is gravity and wont change or it has to change mass, which it also cant. Therefore Acceleration must stay constant. One of the general equations of motion can be used here: The formula above can be used to calculate the velocity on an individual trolley experiment. This will give the acceleration of the trolley because we know that v is the velocity at point B, u is the velocity at point A and s is the distance between A and B which is 10cm. From this information the following accelerations were calculated from the data: Ta/s Tb/s y/cm Acc/ms-2 0.268 0.206 10.0 0.482 0.205 0.173 20.0 0.481 0.173 0.152 30.0 0.494 0.174 0.154 40.0 0.457 0.153 0.139 50.0 0.452 0.139 0.128 60.0 0.464 0.130 0.120 70.0 0.514 0.119 0.113 80.0 0.385 0.112 0.106 90.0 0.464 The data gained from the experiment back up the theories stated earlier that acceleration would remain constant, all the data was plotted on a graph of distance up slope against Acceleration. The values stick at approximately 0.466 as a simple average, but this will not do scientifically, therefore another graph can be plotted so that the acceleration can be calculated that way. Using the equation earlier and re-arranging it so it fits a y = mx + c style graph equation, the acceleration can be calculated from the y intersect of the trend line. According to the equation found, the acceleration should be half of the value of the y intersect. The data was then plotted onto another graph and the software found the best trend line and gave the equation for it which on the printed graph (included on the next page) reads as: y =0.987x + 0.0991 This means that the acceleration should have been 0.0991 x 5 = 0.4955m/s2 The Angle of the slope: The trolley was released on a sloped air track set at a consistent angle. This angle was not given in the results of the experiment or in the original data. Therefore it must be calculated. One way of calculating the angle is to use trigonometry to work out the angle between the forces and movements at work. The force causing the trolley to accelerate is gravity. This acts vertically downwards and is a constant. Using the acceleration as calculated in the previous exercise you get the following equation: Because this equation uses the acceleration which was proved to be a constant previously and gravity which is also a constant, the angle should certainly be a constant too, but thinking the other way around. We know the angle was constant in the experiment and so that proves through the formula that acceleration also has to be constant. By plugging in all the data from the previous calculations on acceleration to provide an angle for each test you get the following: Ta/s Tb/s y/cm Angle (Acc) 0.268 0.206 10.0 2.816855682 0.205 0.173 20.0 2.809567951 0.173 0.152 30.0 2.88356838 0.174 0.154 40.0 2.668975616 0.153 0.139 50.0 2.64043687 0.139 0.128 60.0 2.710441452 0.130 0.120 70.0 3.001325343 0.119 0.113 80.0 2.248658736 0.112 0.106 90.0 2.711100835 As you can see the angles are all pretty similar showing that the theory on the angle being kept constant stands. The angle can also be calculated through Gravitational Potential Energy and Kinetic Energy by using the vertical height the trolley was released from as well as the distance up the slope. Vertical Height above the light gate: Because of the way the experiment was carried out, only data for the distance up the slope has been given, the actual vertical height is unknown. This vertical height is very useful in calculating the Kinetic energy of the trolley and can also be used to calculate the angle. Because of these two separate uses of the height, there are two ways of calculating it. the easiest way is to use the Gravitational Potential Energy Formula and the kinetic energy Formula: This gives the following calculated data for the different velocities at A: V A (ms-1) Height (m) 0.373 0.0071 0.488 0.0121 0.578 0.0170 0.575 0.0168 0.654 0.0218 0.719 0.0264 0.769 0.0302 0.840 0.0360 0.893 0.0406 The calculated heights do relate quite closely to the velocities at first glance, therefore another graph was plotted of the Velocity squared against the vertical height. The gradient of the line connecting the data should certainly have a gradient of 19.62 and should cross at the origin. The graph which was plotted does fit these predictions which helps to promote my findings as being correct. Problems with the results: By looking carefully at the data given to me in the beginning and by looking at some of the graphs plotted originally it appears that some if not most of the data is inaccurate. The readings start off fine up to the 30.0cm test, but the next test appears to have been exactly the same distance up the slope as the previous one, thus giving two readings for 30.0cm and skewing the data on the graph. This is the first obvious error on the data. The next area where error appears to have arisen is on the acceleration calculated. Each result should, according to physics, be the same, however the readings seam to vary a great deal around the calculated valve. This error has been discussed on the Graph of the different accelerations. It could be due to the errors in measurement on the original data of Ã ¯Ã ¿Ã ½0.0005s on each time reading and Ã ¯Ã ¿Ã ½0.2cm on each distance reading. However the first error discovered will have has a large effect on the 40.0cm reading and the next error may explain the rest of the data: By looking at the graph of distance released up the track against velocity, it looks like the data after the repeated result is all wrong. By looking at the trend of the data before and after the kink, is appears that they share a similar gradient but the two lines are offset from each other. If each of the readings after 30.0cm were shifted down by 10.0cm and re-plotted it would form a far smoother curve. It is my prediction that during the experiment, the data got out of line and each experiment after the 30.0cm reading was really the test before it and so instead of the y values being: 10, 20, 30, 40, 50, 60, 70, 80, 90 they should be: 10, 20, 30, 30, 40, 50, 60, 70, 80 This shows how the data has become out of line and therefore needs further investigation. In order to see whether the height readings are correct a formula can be constructed to use the acceleration and velocities between points A and B and the angle of slope, regardless of the actual distance up the slope to calculate what y was for that test. This equation comes from the Kinetic Energy equation: K.E. = 0.5 x m x v2 and the Gravitational potential energy equation used in the previous problem to form: This formula was then used on the data from the experiment to calculate what the actual distance y was: V A (ms-1) y MEASURED y CALCULATED 0.373 0.1000 0.140 0.488 0.2000 0.240 0.578 0.3000 0.337 0.575 0.4000 0.333 0.654 0.5000 0.431 0.719 0.6000 0.522 0.769 0.7000 0.597 0.840 0.8000 0.712 0.893 0.9000 0.804 The data in the table does help to highlight that the data does come out of line but it also highlights that the early results are approximately 5cm out. This could be because the velocity is measured over a 10.0cm flag which would average the velocity to the middle of the flag thus making the actual distance y 5.0 cm further than expected.