Ball+on+Incline+-+Ellen+Apple+and+Raul+Ramirez

Title of Lab: Ball on Incline

Researchers: Ellen Apple and Raul Ramirez

Research Question: How does the time it takes for a tennis ball to roll down an incline relate to the distance that it travels?

Research: theta = 15º Where d is displacement, vi is initial velocity, a is acceleration, and t is the time elapsed, d = vi * t + 1/2 * at^2. In this case, the initial velocity is 0, so d = 1/2 *at^2. Solve for t^2 to show the relationship between time and displacement. 2d = at^2 2d/a = t^2

Here, acceleration can be written in terms of gravity and the angle. a = gsin(theta) g = 9.81 m/s^2

Hypothesis: t^2 = 2d / gsin(theta) t^2 = 2d / 9.81*sin(theta)

Materials: tennis ball, wooden board, stopwatch, meter stick, textbooks, protractor with string and weight attached

Procedure: 1. Set up the board so that the angle of inclination is 15 degrees. (Use the textbooks to prop up one side of the board, and check the angle with the protractor.) -This angle should remain constant throughout the experiment. 2. Place the meter stick on the board so that the bottom of the meter stick is approximately at the end of the board. (It should also be parallel to the board.) 3. Hold the ball at a certain distance along the incline from the bottom. 4. Release the ball, and immediately afterwards, start the stopwatch. (The person holding the ball should also be the one timing with the stopwatch to reduce error.) 5. As the ball reaches the bottom of the incline, stop the stopwatch. 6. Record the the time it took for the ball to reach the bottom of the incline. 7. Reset the stopwatch. 8. Repeat steps 3-7 at different distances along the incline.

Data:


 * Displacement in cm || Time in s ||
 * 0 || 0 ||
 * 5 || 0.22 ||
 * 10 || 0.31 ||
 * 15 || 0.35 ||
 * 20 || 0.44 ||
 * 25 || 0.47 ||
 * 30 || 0.53 ||
 * 35 || 0.59 ||
 * 40 || 0.67 ||
 * 45 || 0.69 ||
 * 50 || 0.70 ||
 * 55 || 0.71 ||
 * 60 || 0.72 ||
 * 65 || 0.74 ||
 * 70 || 0.81 ||
 * 75 || 0.85 ||
 * 80 || 0.91 ||
 * 85 || 0.94 ||
 * 90 || 1.06 ||
 * 95 || 1.25 ||
 * 100 || 1.95 ||


 * Displacement in cm || Time^2 in s ^2 ||
 * 0 || 0 ||
 * 5 || 0.0484 ||
 * 10 || 0.0961 ||
 * 15 || 0.1225 ||
 * 20 || 0.1936 ||
 * 25 || 0.2209 ||
 * 30 || 0.2809 ||
 * 35 || 0.3481 ||
 * 40 || 0.4489 ||
 * 45 || 0.4761 ||
 * 50 || 0.49 ||
 * 55 || 0.5041 ||
 * 60 || 0.5184 ||
 * 65 || 0.5476 ||
 * 70 || 0.6561 ||
 * 75 || 0.7225 ||
 * 80 || 0.8281 ||
 * 85 || 0.8836 ||
 * 90 || 1.1236 ||
 * 95 || 1.5625 ||
 * 100 || 3.8025 ||

Data Analysis:

The normal relationship between time and displacement is a square root curve, and for most of the graph, it follows that. However, the later displacements don't follow this, and the spike in the data was caused by experimental error. (Most likely to do with the stopwatch) The slope of the best fit line is 0.0113. Using the equation 2/gsin(theta), we found our experimental g to be 7.73 m/s^2.

Conclusion: Our experimental g was approximately 7.73 m/s^2. This gave us a percentage error of about 21.2% since the force due to gravity is 9.81 m/s^2. There was a lot of error, mainly experimental. The stopwatch wasn't very accurate, and starting and stopping the stopwatch at the exact moments we needed was difficult (if not nearly impossible). Recording the times using a more accurate instrument would give better data. Running more trials for each distance would also give better data. Because our data was inconclusive, our hypothesis was not supported by the experiment.