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Dan Ni JOURNAL OF HS SCIENCE INVESTIGATIONS October 17, 2011 AP Physics C: Energy on the Air Track Lab Dan Ni, Taehyun Kim, Kevin Jou Department of Science, Shanghai American School, Pudong, Shanghai Links Executive Community, 1600 Ling Bai Lu, San Jia Gang, Pudong, Shanghai, China 201201 (Submitted for review, October 21, 2011) Abstract: This lab investigated how the distance of compression of a spring-­‐like ring structure on an air track glider affected the velocity of the air track glider. In turn, the apparatus is used to study the conservation of mechanical energy, namely from elastic potential energy due to Hooke’s Law and the kinetic energy of the glider. The ring on the air track glider was compressed in increments of 4 mm, after which a photogate measured the amount of time taken for the glider to pass through. The velocity was calculated and a relationship between its square and the square of the distance compressed from equilibrium was determined. The results showed a positive linear relationship, indicating that the ring on the air track glider was shown to obey Hooke’s Law and that the system could be modeled with the theoretical mathematical model of the conservation of total mechanical energy. Research Question How does the distance that the ring structure is compressed from equilibrium affect the velocity of the air track glider? By extension, is the total mechanical energy of the apparatus conserved? Expectations Independent variable: distance of compression of the ring structure of the air track glider, in the following values (cm): 0.4, 0.8, 1.2, 1.6, 2.0 Dependent variable: the time taken for the air track glider to travel through the photogate and in turn its velocity Controlled variables: •
Mass of the air track glider •
Length of the air track glider •
Same ring structure used •
Same air track used Dan Ni JOURNAL OF HS SCIENCE INVESTIGATIONS October 17, 2011 Assumptions Over the course of this experiment, a number of simplifying assumptions were made. These include: •
The air track is assumed to be frictionless so a resistant friction force does not have to be taken into account. Therefore, after the initial acceleration from the spring the velocity can be assumed to be constant. Also, the system is isolated because of the absence of the nonconservative force of friction. The total mechanical energy of the system is therefore conserved. •
Similarly, the velocity of the air track glider is assumed to be constant during the time it travels through the photogate. The velocity can then be calculated using the simplified glider and •
, where is the length of the air track is the time taken to travel through the photogate. The ring structure on the air track glider is assumed to obey Hooke’s law in that the force it exerts on the glider is proportional to the distance of compression from equilibrium. In equation form, , where F is the force the ring exerts on the glider, x is the distance of compression and k is the spring constant of the ring. If the ring structure obeys Hooke’s law, then its elastic potential energy can be modeled with the function . Mathematical Model Since the system is assumed to be isolated and total mechanical energy of the system constant, a model that can represent the transformation of energy during the experiment is: where k is the spring constant of the ring structure, x is the distance of compression of the ring structure, m is the mass of the air track glider and v is the velocity of the air track glider. In terms of the dependent variable, Dan Ni JOURNAL OF HS SCIENCE INVESTIGATIONS October 17, 2011 Thus, the expectation is of a positive linear relationship between v2 and x2 with a slope of . Method 1. Set up the apparatus as shown in the picture and diagrams. Figure 1: Photo of the Air Track Apparatus and Glider
Figure 2: Top View of the Apparatus with a Compressed Ring Structure Dan Ni JOURNAL OF HS SCIENCE INVESTIGATIONS October 17, 2011 2. Push the air track glider back 4 mm (measured on the meterstick), compressing the ring structure. 3. Let go of the glider, trying not to interfere with its forward motion. 4. Subtract the two times recorded on the photogate to find the time the air track glider took to travel through. 5. Repeat two more times for a total of three trials at 4 mm. 6. Repeat steps 2-­‐5 with distances of compression of 8 mm, 12 mm, 16 mm, and 20 mm. 7. Find the average of the times taken to travel through the photogate. 8. Divide the length of the air track glider by these averages to find the velocities of the air track glider. 9. Square the distances of compression and the velocities and plot them on an XY scatter plot, with the square of the distances of compression on the x-­‐axis and the square of the velocities of the y-­‐axis. Use the Microsoft Trend Line function to find the equation of the line and its R2 value. Results Mass of the air track glider: 352.6 g (±0.1 g); 0.3526 kg Length of the air track glider: 15.2 cm (±0.2 cm); 0.152 m Dan Ni JOURNAL OF HS SCIENCE INVESTIGATIONS October 17, 2011 Table 1: The Time Taken to Travel Through the Photogate at a Range of Distances Compressed from Equilibrium Time Taken to Travel Through the Photogate Distance Compressed from Equilibrium (±0.1 cm) Trial 1 (±0.01 s) Trial 2 (±0.01 s) Trial 3 (±0.01 s) Average 0.4 0.99 0.94 0.98 0.97 0.8 0.58 0.58 0.62 0.59 1.2 0.49 0.45 0.43 0.46 1.6 0.31 0.36 0.39 0.35 2.0 0.29 0.23 0.29 0.27 Average Time Taken to Travel Through the Photogate = Sum of Trials ÷ 3 = (0.99 + 0.94 + 0.98) ÷ 3 = 0.97 s Table 2: The Square of the Distance Compressed from Equilibrium and the Square of the Velocity of the Air Track Glider Distance Compressed Velocity of Distance Velocity from Equilibrium Average Time Glider Squared Squared (±0.001 m) (s) (m/s) (m2) (m2/s2) 0.004 0.97 0.16 1.6 × 10-­‐5 0.025 0.008 0.59 0.26 6.4 × 10-­‐5 0.070 0.012 0.46 0.34 1.4 × 10-­‐4 0.12 0.016 0.35 0.45 2.6 × 10-­‐4 0.20 0.020 0.27 0.58 4.0 × 10-­‐4 0.33 Velocity of the Glider = Length of the air track glider ÷ Average Time = 0.152 ÷ 0.97 = 0.16 m/s Dan Ni JOURNAL OF HS SCIENCE INVESTIGATIONS October 17, 2011 Figure 1: The Velocity of the Air Track Glider Squared vs. Distance Compressed Velocity of the Air Track Car Squared (m2/
s2) from Equilibrium Squared 0.35 0.3 0.25 0.2 0.15 0.1 0.05 0 0 0.00005 0.0001 0.00015 0.0002 0.00025 0.0003 0.00035 0.0004 0.00045 Distance Compressed from Equilibrium Squared (m2) Derived from Microsoft Excel Trendline Function: Equation of the line: y = 741.2x + 0.010 R² = 0.991 Determining the value of k: From the mathematical model, From figure 1, slope = 741.2 = and the slope is . = 261.3 kg/s2 Discussion The trend in figure 1 and the R2 value that is very close to 1 indicate that the experimental data are very close to linear with little random error. Because of this linear relationship, the total mechanical energy of the system can be said to follow the relationship and was therefore conserved. Consequently, the simplifying assumption that the velocity was constant did not Dan Ni JOURNAL OF HS SCIENCE INVESTIGATIONS October 17, 2011 seem to significant affect the results and the air track did its job of eliminating friction. Also, the simplifying assumption that the ring structure obeyed Hooke’s Law was fairly accurate; the spring constant of the ring was found using the slope of the line and determined to be 261.3 kg/s2. A possible follow-­‐up question could involve collisions and the transfer of momentum. Since the air track is assumed to be frictionless and the system is isolated, the total momentum before and after the collision should remain constant. The experiment could use the same independent variable of the distance of compression of the ring structure but measure a dependent variable of the velocity of a second air track glider that collides with the first. Thus, the apparatus could be tested to see if it produced collisions that were elastic or inelastic. An elastic collision would have the same total kinetic energy before and after while an inelastic collision would lose energy afterwards.