AP Physics 1 Unit 1 Exercises

docx

School

University of Illinois, Urbana Champaign *

*We aren’t endorsed by this school

Course

Subject

Physics

Date

Jan 9, 2024

Type

docx

Pages

Uploaded by DeaconArtShark16

Akhil Morisetty PD:6 AP Physics 1 Exercise 1 Some wonderfully nerdy children decide to experiment with how full a basketball needs to be before it can bounce “perfectly.” Their definition of a perfect bounce is as follows: if you drop the ball on a hard surface (the students use an asphalt outdoor basketball court), the ball will bounce all the way back to the same height from which it was dropped. They start with a flat basketball and an electric air pump. Dropping the basketball at first results in no bounce whatsoever. After filling the basketball for 5 seconds, they then drop it again and measure the bounce height that results. They then fill the basketball for 5 more seconds, and they repeat the experiment. And so on. Every drop is made from an initial height of 1.0 meter above the ground. Their data table is shown below: Time spent filling the ball (seconds) Bounce height (cm) 0.0 0 5.0 0 10.0 0 15.0 2 20.0 5 25.0 15 30.0 30 35.0 50 40.0 70 45.0 75 50.0 78 55.0 80 60.0 80 The children stop their experiment there, because the basketball looks like it will explode if it’s filled any more. a) First, some basics: in this experiment, what were the independent and dependent variables, and what quantities were controlled? Independent Variable: Time spent filling the basketball Dependent Variable: Bounce Height Quantities Controlled: Asphalt bounced on, Height ball is bounced from, Air pump used, Same ball b) If the children were to plot a graph of Bounce Height as a function of Air Filling Time (that is, with bounce height on the vertical axis and air filling time on the horizontal axis), would the graph be linear? If so, describe its slope and intercepts in the form of an equation. If not, describe the graph shape with a sketch and/or words. (You don’t actually need to plot the graph to scale.) Graph is not linear. The graph starts out at being really steep and towards the end the graph starts to flatten out. The slope starts out as high and decreases towards the end

Akhil Morisetty PD:6 c) One of the children wants to keep filling the ball, because, they say, “The ball’s clearly only 80% full! Which means we need to fill it for…” Here they do a quick calculation and discover that 1.2 times 60 seconds is 72 seconds, and they continue, “…12 more seconds! Then the ball will bounce all the way to where it was dropped from!” The other kids look nervously at the ball, which really does look like it cannot handle 12 more seconds of filling. For safety’s sake, they do not add any more air to the ball. Your task: identify and discuss in depth, in a clear and coherent paragraph, at least two flaws with this student’s reasoning ( other than the safety concerns). For each flaw, use evidence from the data set to support your reasoning. Use the back of this page, or a new piece of paper, for your answer. The first flaw is that the student said that the ball is only 80% full. The amount the ball bounces has stopped increasing indicating that the ball has been filled to the max. At 55 seconds and 60 seconds the bounce height is the same showing the bounce height has reached its limit. The second flaw is that the student’s method would only work if the bounce height’s increase was linear but its not. The students thinks that increasing the filling time by 20% will also increase bounce height by 20%, but as the graph from question b shows the bounce height does not have a linear increase.

Akhil Morisetty PD:6 AP Physics 1 Exercise 2 There’s an old story about a tortoise (or turtle) and a hare (bunny rabbit). One version of it goes like this: A tortoise challenged a hare to a 1-km race. The hare knew he was faster than the tortoise, so he happily accepted the challenge. They started the race at the same time, with the hare running at 5 km/h. When the hare reached the halfway point, he looked back and saw the tortoise far off behind. So the hare laughed and decided to take a nap. He rested longer than intended, though, because when he woke up, he looked ahead and saw the tortoise about to cross the finish line! The hare raced ahead at 10 km/h and tried to catch up, but the tortoise won—just barely, by less than a second! The tortoise had walked at a constant speed of 0.5 km/h the entire time (note: this is way too fast for the average tortoise, but oh well; it makes the numbers in this problem work out, so just work with me, here). The moral of this story, of course, is, “Slow and steady wins the race.” a) Draw a well-labeled pair of motion diagrams representing the motion of both the tortoise and the hare. (Hint: when one dot represents a long time interval, such as when the hare takes a nap, you circle that dot and label it with how much time gets spent in that location. Feel free to save this step for after you’ve solved part b), if that’s more convenient.) b) Using only the motion diagram you made , solve for how long the hare must have been asleep. The format for showing your work should be as follows: to the left of the vertical line below, briefly explain in complete sentences how you are reasoning through the answer. (And remember: it must be clear that your reasoning is based off the motion diagram.) To the right of the vertical line, show all mathematical steps involved. Note: I know you could solve this without the aid of the diagram. The point is that, in the future, you might only be given the diagram and have to solve something similar. So that’s the skill we’re practicing here. The hare slept for 111 minutes before getting up and start running. First I found how much time the hare ran before stopping. Since it ran 2 dots worth of time and each dot is 3 min. it ran for 6 min. Then I looked how much the hare ran after it woke up. Since it ran 1 dot worth of time, it ran for 3 minutes at the end. When you add 3 min. and 6 min.(the time the hare ran) and subtract from the total time the hare took to finish which is 120 min the difference is 111 min. 111 minutes is how much the hare didn’t run or was asleep.

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Akhil Morisetty PD:6 c) Find and correct a flaw in the following statement: “The hare had a speed of 5 km/h, then 0 km/h, and then 10 km/h. So, on average, the hare’s velocity was 5 km/h toward the finish line, because (5 + 0 + 10) / 3 = 5.” We don’t know how long the hare ran at each of those speeds. We also don’t know how much the hare ran. The hare ran a total of 2 km. in 2 hours. So the average velocity would be .5 which is the same as the tortoise. Since the time intervals are uneven we can’t take the average of the three speed. The hare didn’t run the same amount of time at each of those speeds. That’s why the statement is flawed.

Akhil Morisetty PD:6 AP Physics 1 Exercise 3 Position as a Function of Time Position (m) 2.0 1.0 2.0 3.0 4.0 5.0 Time (s) The graph above shows the motion of a ball that rolls toward a wall and bounces directly backward after hitting the wall. Using the graph, answer the following questions. a) At what time does the ball collide with the wall? Briefly explain. 2 seconds because the ball has a negative velocity before 2 sec. and after 2 sec. has a positive velocity indicating a change in direction. b) Which direction: toward the wall, or away from the wall, is the positive direction? Briefly explain. Away from the wall because the slope of the line for the ball after hitting the wall is positive. c) Without doing any mathematical calculations, answer the following: is the ball’s speed after hitting the wall less than, greater than, or the same as, the ball’s speed before hitting the wall? Briefly explain. Less than because before hitting the wall the ball travels 2 meters in 2 seconds and after hitting the wall the ball travels the same distance in a longer time of 3 seconds indicating the ball’s speed is less than it was before hitting the wall d) Calculate the speed of the ball before hitting the wall, and the speed of the ball after hitting the wall. Then, verify that your answers are consistent with what you wrote in part c). (Hint: speed is, mathematically, the absolute value of velocity.) Speed before hitting the wall is 1 m/s Speed after hitting the wall is .67 m/s e) “Change in velocity” is a very useful concept in physics, which we’ll use many times in later units. Its symbol is  v , and it’s found by the following subtraction: “final velocity” minus “initial velocity.” Consider the timeframe before the ball hits the wall to be “initial,” and consider the timeframe after the ball hits the wall to be “final.” Solve for the ball’s  v . (HINT: this is change in velocity , not change in speed .) .67 m/s – (-1 m/s) = +1.67 m/s Change in Velocity = +1.67 m/s f) What is the average velocity of the ball… i) …from time t = 0 seconds to t = 5 seconds? Show your work. (2-2)/(5-0) = 0/5 = 0 m/s Average Velocity = 0/ms ii) …from time t = 1 second to t = 5 seconds? Show your work. (2-1)/(5-1) = .25 m/s Average Velocity +.25 m/s

Akhil Morisetty PD:6 AP Physics 1 Exercise 4 American sprinter Florence Griffith-Joyner holds set the women’s world record for the 100-meter dash in 1988. Her time was 10.49 seconds. Assume that she ran in a straight line from the starting line to the finish line, and that she maintained a constant speed throughout the race (not completely realistic, but just go with it for now). Furthermore, just to make things a little easier for this exercise, we’re going to estimate her race completion time to be 10.5 seconds, to 3 significant figures. Note: throughout the race, assume that the sprinters are running in the positive direction. a) Below, you will see a dividing line. On the left side of the line, draw a motion diagram for Griffith-Joyner’s sprint, and on the right side, sketch the shape of a graph of her position as a function of time. Use 0.5 s intervals for the motion diagram. For the graph, label both intercepts clearly, but you do not need to plot every point. Assume that the reference frame is that of a stationary observer watching the race from the starting line. Motion Diagram: Graph: b) Now repeat part a), but this time, from the reference frame of a stationary observer standing at the finish line. c) Now do the exercise again, but this time, from the reference frame of someone who is in the race and loses to Griffith-Joyner by exactly 1.0 meter—that is, when Griffith-Joyner crosses the finish line, this opponent is at the 99 m mark. (Note: you are still diagramming Griffith-Joyner’s motion; just from the losing sprinter’s reference frame.)

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Akhil Morisetty PD:6 d) Figure out however you like, but show your work: by how many seconds did the sprinter in part c) lose the race? The sprinter lost by .1 seconds. e) Sketch the shape of the position vs. time graph of the losing sprinter’s motion, from Griffith-Joyner’s reference frame. You don’t need to label any specific values on the axes.

Related Documents

Final-Exam-a-Fall-2021(1).docx

FINAL PROJECT PHYS.docx

Lab 2 Uniform acceleration.pdf

ES 1023-2123- Lab 8 Earthquake Seismology.pdf

Experiment 13 - Declan Rogers.docx

Review for Honors Physics final .pdf

Chapter15-DiscussionQuestion1.pdf

Rotational_Equilibrium_Lab_102-1.pdf

Physics1410_Chapter_14_practice_questions (1).docx

Phys 1410_Resonance Lab.docx

Physics1410_Chapter_16_practice_questions (1).docx

Physics1410_Chapter_17_practice_questions.docx

Recommended textbooks for you

Physics for Scientists and Engineers: Foundations...

Physics

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Cengage Learning

Physics for Scientists and Engineers with Modern ...

Physics

ISBN:9781337553292

Author:Raymond A. Serway, John W. Jewett

Publisher:Cengage Learning

Physics for Scientists and Engineers, Technology ...

Physics

ISBN:9781305116399

Author:Raymond A. Serway, John W. Jewett

Publisher:Cengage Learning

Physics for Scientists and Engineers

Physics

ISBN:9781337553278

Author:Raymond A. Serway, John W. Jewett

Publisher:Cengage Learning

Principles of Physics: A Calculus-Based Text

Physics

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Cengage Learning

College Physics

Physics

ISBN:9781938168000

Author:Paul Peter Urone, Roger Hinrichs

Publisher:OpenStax College

SEE MORE TEXTBOOKS

Recommended textbooks for you

Physics for Scientists and Engineers: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning
Physics for Scientists and Engineers with Modern ...
Physics
ISBN:9781337553292
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Physics for Scientists and Engineers, Technology ...
Physics
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Physics for Scientists and Engineers
Physics
ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
College Physics
Physics
ISBN:9781938168000
Author:Paul Peter Urone, Roger Hinrichs
Publisher:OpenStax College