Clarin Lab 2 bowling ball and air track motion-1.docx

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California State University, Monterey Bay *

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220

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Physics

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Apr 3, 2024

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Physics 220 Name: Branden Clarin Date: 08-31-2022 Bowling Ball and Air Track Purpose: During the lab you will come to understand average and instantaneous velocity. You will do this by using a bowling ball and air track to calculate the velocity. Understand the relationship between slope and velocity, well using excel or google sheets. Procedure: In part 1 you will use the air track to find the average velocity. My group would set the gate at a distance apart provided to find the time it takes to get from point A to point B. This would give us the time and since we had the distance you could solve for average velocity. Then we would repeat this process with a shorter distance, to difference in velocity. Next we would want to find the instantaneous velocity, so we would measure the length of the sail which would be 1 meters. Now setting up one gate at about 1 meters, then would measure the time it takes the sail to pass through the gate. Now we can solve for the instantaneous velocity, we found the time and distance which you can see in the sample calculation. Part 2 would be predicting if the speed would increase or decrease from different points. I would first set up on the gate at 30 cm and see how it takes the sail to pass though and predict if it was double in distance it would be faster or slower. I answer this in the question section of this lab report. Part 3 would be determining the average velocity of the bowling ball using a graph, because we know that slope is equal to average velocity. To set this up you will need to set a piece of tape measured out to 8 meters then at 2, 4, and 6 meter mark marks on the tape. So you can determine the time it takes the bowling ball to get to those points in one roll. We would do this so we can plug into my google sheet seen in the data section and question. Data: Part 1 A Distance Travel (m) Time Final (s) Average Velocity 1 1.5 03.2034 0.46825 m/s 2 1.5 03.1024 0.48350 m/s 3 1.5 03.5109 0.42724 m/s
Part 1 B Distance Travel (m) Time Final (s) Average Velocity 1 .5 02.6556 0.18828 m/s 2 .5 02.6352 0.18973 m/s 3 .5 02.6508 0.18862 m/s Part 1 C Length of Sail (m) Time Arrived (s) Instantaneous Velocity 1 1 00.2078 4.812 m/s 2 1 00.2064 4.845 m/s 3 1 00.2073 4.824 m/s Sample Calculations: 𝐴???𝑎?? 𝑉????𝑖?? = ? ? −? 𝑖 ? ? −? 𝑖 𝐼???𝑎??𝑎????? 𝑉????𝑖?? = ?? ?? 1.75 ?−.25 ? 03.2034 ?−0 ? = 0. 46825 ?/? 1 ? 00.2078 ? = 4. 812 ?/? Questions: Part 1 a. Distinguish in a complete sentence or short paragraph the difference between the average and instantaneous velocity. This is in general, not just in the activities you completed in this part of the lab. Please each person write their own definition. The difference between average and instantaneous is that the instantaneous velocity is the velocity at the moment the object passes through the point. Well the average velocity is the average velocity during the two points or more. b. Within the errors of your measurements, would you say that your results for all three parts (a through c) are the same? Do you expect that these should be the same? Why or why not? Please, each person answers this question in your own words. I don’t expect this to be the same due to the fact that the instantaneous velocity will be high, because you are calculating that a single point well average is several points to be taken for account. Part 2
a. Determine a position near the top of the air track from where you will consistently release a cart. Define a (downhill) point approximately 30 cm from this point. Time how long it takes the cart to reach this final point from the time of release. You may use the photocell gates or a stopwatch. 00.4593 s is the time it take the the sail to get through the gate at 30 cm b. Now, predict where the cart would be (if it were released from the top) when twice the time elapses from that measured in Part (a). Since you have not learned how to do this yet mathematically, I am not looking for an extensive calculation here. Will the cart move twice as far, less than twice as far, or more than twice as far? It should move twice as fast thought the gate, because it has more time to build speed a and accelerate. c. Now, determine a means to see if your prediction was correct. You may use photocell gates, stopwatches, or some creative method of your own. So we move the gate to 60 cm and as predicted more about twice as fast at 00.2954 s Part 3 a. Make a data table showing the times and distances you recorded for the bowling ball activity. On Excel, create a graph of the distance the ball traveled versus time (that means that distance is on the y-axis and time is on the x-axis). b. Determine the value of the slope, m, and find the value of the intercept, b, of the line you have sketched through the data. Don’t forget that these values have units! We know that the slope is the 0.601m/s x+(-0.248)
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c. What does the slope of the line represent? The slope is the average velocity and is used to find the average velocity in a graph c. Based on your graph , would you say that the bowling ball’s velocity is increasing, staying the same, or decreasing?. Give a clear reason for your answer (based completely on your observations and/or graph). Each person please answer this in her or his own words. From observing the graph you can see that the velocity of the bowling is increasing and you see well the ball is rolling and it is gaining speed. Conclusion: During this experiment I was able to see and do what we have been talking about in lecture. I have a better understanding of the relationship between slope and velocity. There may have been some error during the experiment, because we did have problems with the system we were using. There was for sure some human error during the bowling ball part due to having to be right when the bowl hits your point. Overall I would learn about the relationship and difference between average and instantaneous velocity and also slope.