Bal+on+Incline-+Aaron+Floyd

Title of Lab: Ball on an Incline

Researchers: Aaron & Daniel

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

Research: Source: ([]) If we roll a ball down a hill halfway, it will take time (t) to make that journey. If we roll it sown the entire hill, it will take the time (t + b), b depends on the acceleration of the ball, to roll the length of the hill. Based off of this knowledge, we know that it takes a ball longer to roll a longer distance than it does a shorter distance. We can use the kinematics equation Δx = vit + 0.5at^2 to help us discover the relation between distance, or displacement (Δx), and time (t).

Hypothesis: The time it takes for the ball to roll down the incline will increase as the distance increases. The time will be directly related to the square root of the distance. x=.5(9.81/sin(theta))t^2.

Materials: The materials used in this project include: 1. Stopwatch 2. Wooden Incline 3. Protractor with weighted string attached 4. Tennis Ball 5. Meter Stick

Procedure: 1) To start out, set up the wooden incline by putting one end on the table and another on a chair. 2) Use books to elevate it to desired spot. 3) Measure the angle to make sure it falls between 10 and 20 degrees from horizontal and keep that angle consistent throughout the experiment. 4) Add the meter stick on the incline, towards one side. You are now ready to collect data. 5) Hold the ball at the desired distance and drop it as you are starting the stopwatch. 6) Let the stop watch run until it reaches the bottom of the incline. 7) Record the data 8) Repeat procedure numbers 5-7, occasionally checking to make sure the angle of the incline has not changed.

Data:

ϴ = 15° || || ||  ||
 * Distance rolled down incline (cm) || Time (s) - Trial 1 || Time (s) - Trial 2 || Time (s) - Trial 3 || Time (s) - Trial 4 || Time (s) - Trial ||
 * 100 || 1.06 || 1.03 || 1.03 || 1.09 || 1.12 ||
 * 90 || 1.00 || 0.96 || 0.97 || 0.94 || 0.96 ||
 * 80 || 0.93 || 0.87 || 0.97 ||
 * 70 || 0.81 || 0.84 || 0.81 ||
 * 60 || 0.79 || 0.81 || 0.81 || 0.75 || 0.75 ||
 * 50 || 0.75 || 0.75 || 0.72 || 0.75 || 0.78 ||
 * 40 || 0.68 || 0.66 || 0.60 || 0.68 || 0.63 ||
 * 30 || 0.41 || 0.50 || 0.47 || 0.46 || 0.59 ||
 * 20 || 0.44 || 0.44 || 0.40 || 0.47 || 0.44 ||
 * 10 || 0.19 || 0.25 || 0.18 || 0.18 || 0.22 ||  ||
 * 0 || 0.00 || 0.00 || 0.00 || 0.00 || 0.00 ||  ||

Data Analysis:
 * Distance (cm) || Average Time (s) ||
 * 100 || 1.1 ||
 * 90 || 1.0 ||
 * 80 || 0.9 ||
 * 70 || 0.8 ||
 * 60 || 0.8 ||
 * 50 || 0.8 ||
 * 40 || 0.7 ||
 * 30 || 0.5 ||
 * 20 || 0.4 ||
 * 10 || 0.2 ||
 * 0 || 0.0 ||

The slope of the best fit line = 0.11 cm/s^2 The slope represents the acceleration of the ball. Based off of the equation, Δx = vit + 0.5at^2, the acceleration should have been 1.8 cm/s^2. From this theoretical answer, we have concluded that our percent error is 94%.

Possible causes of this large amount of error could include inaccurate timing with the stopwatch or inaccurate measurements with the meter stick.

Conclusion: In conclusion part of our hypothesis was correct. The time was directly proportionate to the square root of the distance. Our formula that we thought would work is not correct.