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Hyatt
Physics Alexander Lab 124-021
February 21, 2007
Abstract
This experiment demonstrates the relationship between force, mass and acceleration. The lab was set up so that a glider was set on an air track, and a constant force would push the glider along the track. The first experiment involved an air track raised at different heights only on one side. The higher the track was raised, the greater the acceleration of the glider. The other experiments, differentiated only by differing glider masses, both used the air track at a level position, and used a hanging mass to pull the glider along the track. The acceleration was found experimentally. For the first experiment, the obtained data was used to find an experimental value for g, and then the % error was found for the difference between the experimental and actual values of g. The same was done for the other experiments except with respect to m1 instead of g.
Experiment Table 1
Data Table Exercise 1 D (cm) =100 H a (cm) sin(theta) (m/s^2) 1.29 0.0129 0.132 2.59 0.0259 0.263 3.89 0.0389 0.384 5.18 0.0518 0.516 Calculated G Percent Error 9.8148 0.05%
Table 1 deals with the data used to find the experimental value for gravity, or g. In this exercise, the air track was raised up a certain height, and then the acceleration was recorded at each height. Also, the angle of the incline was found by take arcsin of the Height (H) divided by the Distance (D). The g value was found by making a graph and making a line who's slope is: a/sin(theta). The experimental g value was found to be 9.8148m/s^2, while the actual value is 9.81m/s^2. The percent error between the two values is a negligible .05%.
Alexander Hyatt
Physics Lab 124-021
February 21, 2007
Table 2
Data Table Exercise 2 m1=.3308 1/m2 a 14.286 1.661 11.111 2.035 9.091 2.395 7.692 2.669 6.667 3.013 5.882 3.295 5.263 3.416 4.762 3.621 0.3393 2.54% 1.043 4.21%
m2 0.070 0.090 0.110 0.130 0.150 0.170 0.190 0.210
g/a 5.906 4.821 4.096 3.676 3.256 2.977 2.872 2.709
Calculated m1 Percent Difference Calculated yintercept Percent Difference
Table 2 is the data collected from exercise 2. Exercise 2 involved a glider with extra weight being moved across a level air track via a hanging mass pulling on it. Different acceleration values were found by adding 20g weights to the hanging mass each time a new run was done. A hundred grams were added to the glider to give it a mass of .3308g, this is the actual mass of the large glider, or m1. The calculated value for m1 was found by making a graph with g/a on the ordinate, and 1/m2 on the abscissa. The slope of the trendline of that data was the calculated value for m1. The calculated value for m1 turned out to be .3393g. The percent difference between the two values is 2.54%. Also, the equation used to find m1 has a y-intercept of 1. The calculated y-intercept for the trendline was 1.043. The percent difference between the two values is 4.21%.
Alexander Hyatt
Physics Lab 124-021
February 21, 2007
Table 3
Data Table Exercise 3 m1=.2308 1/m2 a 14.286 2.230 11.111 2.712 9.091 3.041 7.692 3.479 6.667 3.738 5.882 3.918 5.263 3.937 4.762 4.161 0.2148 7.18% 1.266 23.48%
m2 0.070 0.090 0.110 0.130 0.150 0.170 0.190 0.210
g/a 4.399 3.617 3.226 2.820 2.624 2.504 2.492 2.358
Calculated m1 Percent Difference Calculated yintercept Percent Difference
Table 3 is very similar to Table 2. This is because the only difference between exercise 2 and exercise 3 is the mass of the glider. Exercise 3 uses a small glider with a mass of .2308g, with no additional mass added to it. The goal of the exercise was still the same, to find the experimental value for m1 and take the percent difference between that value and the measured value as well as the percent difference between the y-intercept values. Again, a graph was created with the g/a values on the ordinate and the 1/m2 values on the abscissa. A trendline was created who's slope represented the experimental m1 value. This value was found to be .2148g. The percent difference between the calculated and measured values for m1 was 7.18%. Also, the percent difference between the calculated y-intercept, being 1.266, and the actual value was 23.48%.
Alexander Hyatt
Physics Lab 124-021
February 21, 2007
Graphs Graph 1
Experimental Determination of Gravity
0.6
0.5 y = 9.8148x 0.006 + R2 = 0.9997
Acceleration (a) [m/s^2]
0.4
0.3
0.2
0.1
0 0 0.01 0.02 0.03 sin(theta) [ang] 0.04 0.05 0.06
This is the Graph used to find the experimental value of gravity. The slope of the trendline of the data is the value for g. The experimental value was found to be 9.8148m/s^2, very close to the real value of 9.81m/s^2. The percent error in these two values is only .05%.
Alexander Hyatt
Physics Lab 124-021
February 21, 2007
Graph 2
Experimental Determination of m1 for the Large Glider
7.000
6.000
5.000 y = 0.3393x + 1.0429 R2 = 0.9986 4.000
g/a
3.000
2.000
1.000
0.000 0.000
2.000
4.000
6.000
8.000 1/m2 [1/kg]
10.000
12.000
14.000
16.000
This is the graph that was used to determine the experimental value for m1, for the large glider. The value was found to be .3393g, while the measured value was found to be .3308g. The percent difference between the two values is 2.54%. Also, the experimental y-intercept value was 1.043 while the actual value is 1. The percent difference between these two values is 4.21%.
Alexander Hyatt
Physics Lab 124-021
February 21, 2007
Graph 3
Experimental Determination of m1 for the Small Glider
5.000
4.500
4.000
3.500
3.000
g/a
y = 0.2148x + 1.266 R2 = 0.9899 2.500
2.000
1.500
1.000
0.500
0.000 0.000
2.000
4.000
6.000
8.000 1/m2 [1/kg]
10.000
12.000
14.000
16.000
Graph 3 is the experimental determination of m1 for the small glider. The experimental value was found to be .2148g, while the measured value was .2308g. The percent difference between these two values is 7.18%. Also, the experimental y-intercept was 1.266, while the actual y-intercept is 1. The percent difference between these two values is 23.48%.
Questions Calculation Details
Sin() = (H/D) = (1.29cm/100cm) = .0129 g = a/sin() = (.132m/s^2)/(.0129) = 10.233m/s^2 m1 = (g/a)/(1/m2) = (9.81/2.23)/(1/.070) = .3079g Percent Error = (2.70g/cm^3) = |measured actual|/actual * 100% = |9.81489.81|/9.81*100 = .05%
Alexander Hyatt
Physics Lab 124-021
February 21, 2007
Percent Difference = |measured2 measured1|/((measured1 + measured2)/2)*100 = |.2148.2308|/((.2308+.2148)/2)*100 = 7.18%
Questions
1. Newton's Second Law states that F=ma. If the mass of the glider in Exercise 1 were to double, how would the glider's acceleration be affected? What if the mass were halved? If the mass of the glider were to double, the acceleration would be halved, and if the mass of the glider were to be halved, then the acceleration would be doubled. 2. In Exercise 1, the computer calculated the slope and y-intercept of the velocity versus time plot. The slope is the average acceleration that the glider experienced. What does the y-intercept tell us? The y-intercept tells us the initial velocity of the glider when its picket fence starts to pass through the photogate. 3. In Exercise 1, you found the acceleration due to gravity by determining the slope of your graph. What was the y-intercept of your graph? Explain what a non-zero y-intercept would indicate about the experimental setup. My y-intercept was .0006. A non-zero y-intercept would indicate that there is some kind of flaw with the experimental setup, whether it be minor or major, because this would mean that there would be an initial acceleration other than zero, which is not what the experiment is supposed to show 4. Derive Equations 5.8 and 5.9 for the acceleration and tension in Exercise 2. m2g-m1a=m2a => m2g = m1a+m2a => m2g = a(m1+m2) => a = m2g/(m1+m2) a=T/m1 => a = m2g/(m1+m2) => T/m1 = m2g/(m1+m2) =>T =m1m2g/(m1+m2) 5. Examine Equation 5.8. What can be inferred about the mass of the glider, m1, if the acceleration of the glider equals that of gravity, a=g? Mass 1 equals 0. Because g = m2g/(0+m2) = m2g/m2 = g 6. Examine Equation 5.8. What can be inferred about the mass of the glider, m1, if the acceleration of the glider is zero, a=0? Mass 1 is infinitely large compared to m2. This is because the m2 value has a force weight acting on it. The only way for the m1 object not to move is for it to be infinitely larger than m2.
Alexander Hyatt
Physics Lab 124-021
February 21, 2007
7. Show that Equation 5.8 can be rewritten as Equation 5.10. a = m2g/(m1+m2) => 1/a = (m1+m2)/m2g => g/a = (m1+m2)/m2 => g/a = m1/m2 + m2/m2 => g/a = m1(1/m2) + 1

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