Copy of Linear Motion Lab 4

docx

School

University Of Connecticut *

*We aren’t endorsed by this school

Course

4098

Subject

Physics

Date

Dec 6, 2023

Type

docx

Pages

Uploaded by bam721

AP I Physics Mr. Hafner Name Hugo Onghai Lab Group Members: Annie Maier, Santiago Burgos-Fallon, Kasumi Layne-Stasik Linear Motion Discovery – Lab 4 Learning Objectives 1. Observe objects moving at a constant speed and object moving with changing speed. 2. Graph the relationships between distance and time for moving objects. 3. Interpret graphs relating distance and time for moving objects. Part I - Constant Speed Lab Procedure I 1. Somewhere on the floor (either inside or out in the hall), take two carts and come up with a way, before testing, to see where they will intersect. Write the thought process/procedure in the space below: Align the two cars on either end of the meter stick and launch them simultaneously towards each other. Timing how long a car will take to get from one end of a meter stick to the other will help us calculate the velocity the car traveled at. Considering that a meter stick is 100cm, and we determine time traveled with the stopwatch, we can calculate the velocity of both cars relatively accurately. 2. Draw out a diagram of what you have found (with labels) before you test it 3. What is the distance that they intersect? Show work below Velocities: Red Car = 100cm/4sec = +25cm/s Blue Car = 100cm/11sec = +9.09cm/s Time each car would travel before intersecting: 100cm = 9.09*t + 25*t t = 2.933 seconds Meter Stick (100 cm) 100cm_________ Time to travel the meter stick Meter Stick (100 cm) 100cm_________ Time to travel the meter stick

Red Car: 2.933*25cm/s = 73.33cm Blue Car: 2.933*9.09cm/s = 26.66cm Estimation: The red car will travel 73.33 cm before intersecting, and the blue car 26.66 cm before intersecting. 4. Place a piece of tape on the floor and test your results. Was your estimation correct? What was your percent error? The blue car traveled 30 cm and the red car traveled 70 cm until they collided/intersected. Percent Error: (| o - a | / a)*100% o = obtained a = actual Blue: |26.66cm - 30cm|/30cm = 11.13% off for blue car Red: |73.33cm - 70cm|/cm = 4.73% off for red car Lab Procedure II 1. Outside in the hallway you will need a straight stretch of 5 meters on the floor that is clear of obstructions. You will choose a starting point for your car and mark it with masking tape on the floor. 2. Start the car, and place it on the starting point. Release the car (your lab partner should start their stopwatch at the same time). Let the car move in a straight line for 2.0 s. Notice exactly where the car is after 2.0s. Repeat this for several trials, until you find the point that the car consistently crosses after 2.0s. Mark this point with masking tape, and label it “0.00 m”. Throughout this lab, you will start the car at the original starting point, but you will begin to measure the distance and time of the car’s motion when the car crosses the 0.00 m mark. 3. Start the car, and place it on the floor at the starting point. Observe the car as it moves. Start your stopwatch as the car crosses the 0.00 m mark. 4. After the car has traveled 5 seconds mark this position with masking tape. 5. Repeat steps #3 and #4 for 4.0 s, 3.0s, 2.0s, and 1.0s. Label each time point with masking tape on the ground. Conduct 3 trials for each second to ensure your masking tape is in an accurate location.

6. Measure the distance from the 0.00 m mark to each timed position mark. (Do not include the distance from the starting point in any measurements). Create a chart for the position and time of each piece of masking tape below. Time (s) 0s 1s 2s 3s 4s 5s Positio n (m) blue 0cm red 0cm blue 20cm red 51cm blue 30cm red 79cm blue 54cm red 127cm blue 66cm red 160cm 77cm red 192 Graphs & Analysis I 1. Describe in words what is meant by constant speed. When speed is constant, that implies the velocity and speed stay at set values. This means that acceleration is 0, i.e. the body of motion doesn’t accelerate. Therefore, the body travels a distance at a fixed rate. 2. Why is it necessary to create the “0.00 m” mark 2 seconds after the cart has been set into motion? The 0.00 m mark is created after 2 seconds so that the car is already at constant velocity. From the very start of the car to 2 seconds into its motion it is still accelerating. 3. Construct a graph with time on the x-axis and distance on the y-axis. Utilize the data from Procedure #6.

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

4. What does this graph represent (pretend like you didn’t just conduct this experiment and you are looking at this graph independently) and allow you to state about your car in any moment of time? The graph represents the distance from the state and distance between the cars. One could also find the average velocity between calculated points by determining the slope of the line. 5. Calculate the slope of your best fit line/curve. What is the physical significance of the slope of your best fit line? Assuming that error above and below the best fit line/curve cancel out, then the line of best fit represents distance over time more accurately/ideally than the graph of points. This would make the slope the average velocity. Blue: y = 15.629x + 2.095 Red: y = 38.143x + 6.143 6. How far did the car travel in 1.0s time interval from 0.0s – 1.0s, 1.0 – 2.0s, 2.0 – 3.0s, 3.0s – 4.0s, 4.0s – 5.0s. Record these values in a chart below. What do you notice when comparing these distances? Why do you think this relationship exists? For the BLUE CAR: 0.0s – 1.0s 1.0 – 2.0s 2.0 – 3.0s 3.0s – 4.0s 4.0s – 5.0s 20cm 10cm 24cm 12cm 11cm TIME (s) DISTANCE (cm) Red = red car Blue = blue car

From 2.0-3.0s, 3.0-4.0s, and 4.0-5.0s, the decreases in the amount travelled but in a way that looks like it may level out. To be honest,it is hard to find a definitive trend because it seems that our data has some error. I would think that the car would be accelerating until it reached top speed, at which point the distance traveled per second would stay constant. I think one problem that could have led to this relationship is the slightly veering towards the right of the car. It can be inconsistent sometimes, as evident by the first few segments of time, but overall tends to become more significant as the car continues to drive. 7. Predict the position of the car at 7.0 s. Explain how you arrived at this prediction. Using the lines of best fit… Blue: y = 15.629x + 2.095 y = 15.629(7.0) + 2.095 y = 111.498 cm 8. Utilizing the v = d/t formula create a graph of velocity vs. time from 0s – 5s. Time should be on the x-axis and velocity on the y-axis. Compare this graph to the one you created in question 2. What similarities do these two graphs have? Both graphs have 5 line segments. Here, the slope of each line segment in the distance vs. time graph corresponds to the y-value/velocity of the line segment. Time (s) 0s - 1s 1s - 2s 2s - 3s 3s - 4s 4s - 5s velocity 20cm/s 10cm/s 24cm/s 12cm/s 11cm/s TIME (s) 0 5 10 20 VELOCITY (cm/s) BLUE CAR

(m/s) Part II - Moving at an Increasing Speed Lab Procedure III 1. One end of the track needs to be supported with 3 AP I Physics textbook. The track needs to be secured so it does not move. The base of the track should be resting on the floor/lab table. Your starting point will be the top of the ramp. 2. The ball should be placed at the starting point and held in place with a ruler. To release the ball you need to quickly swing the ruler out of way. Start your stopwatch the instant the ball is released. Stop the watch once the ball reaches the base of the track (right before it strikes the floor). 3. Place your ball back at the top of the ramp and hold it in place with a ruler. Allow the ball to roll down the ramp by quickly lifting the meter stick. After 0.25s, 0.50s, 0.75s, 1.0 s, and 1.25s you will mark the position on the ramp with a piece of masking tape. This will take a minimum of 3 trials for each separate time interval. The use of video could be very helpful in this experiment. 4. Measure the distance from the 0.00 m mark to each timed position mark. Create a chart for the position and time of each piece of masking tape below. 1.25=> 1.37, 1.35, 1.33 73.25cm 1.0=> .96, 1.07, 1.11, 53cm, .75=> .81, .71, .75 22.5cm .50=> .51, .46, .51, 13cm .25=> .33, .28, .25, 9cm Time Marks Position (distance from 0.00cm starting point) 1.35 (1.25) 73.25cm 1.05 (1.00) 53.0cm .76 (.75) 22.5cm .49 (.50) 13.0cm .29 (.25) 9.0cm Graphs & Analysis II

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

1. Why is it necessary to complete multiple trials in order to determine where the .25s, .5s, .75s, etc. marks are on the ramp? How does this process cause possible experimental error in this lab procedure? This guess-and-check process is full of potential errors. First, due to the nature of our stopwatch and human error, it is impossible to exactly ever get .25s, .5s, etc.. If you only used one trial to determine the marks, then your mark would basically be based solely on a guess, which is a very inaccurate way to measure something. At least with multiple trials, you can adjust your mark based upon the previous trials to hopefully get closer to your actual times. We need to know how long it takes to travel to a mark, the distance the ball travels, to calculate the velocity. Displacement/Time = Velocity. 2. Describe the motion of the ball as it travels down the ramp. How can you tell from your data and observations? How is this different from the car in the first experiment? Why is there a difference? Unlike the car, the motion here is not at a constant speed. As the ball goes down the ramp, it accelerates. I can tell because as time goes on, the distance between marks increases. 3. Construct a graph with time on the x-axis and distance on the y-axis. Utilize the data from Procedure #7. 4. Construct a best fit line/curve on the data points that you have plotted above. Which seems more appropriate in this case a best fit curve or a best fit line? A best fit curve seems more appropriate, since distance vs. time is not linear. There is acceleration. y = 39.21x^2 - 0.48x + 4.51

5. Why would the graph from the first experiment (Graph & Analysis I, Question #3) be a best fit line and the graph above be a best fit curve? What are the differences in motion between the car on the and the steel ball on the ramp? The graph with the car should have a best fit line. In the car graph, the motion is pretty much linear because there is no or very little acceleration. Here, there is a lot of acceleration because we take into account the ball traveling from a stopped position. 6. What does the graph above (Graph & Analysis II, Question 3) represent in terms of the motion of the steel ball? Describe the motion. The motion is not at a constant speed; the steel ball is accelerating/speeding up towards the end. 7. How far did the steel ball travel in 0.25s time interval from 0.0s – 0.25s, 0.25s – 0.5s, 0.5s – 0.75s, 0.75s – 1.0s, 1.0s – 1.25s and so on. Record these values in a chart below. What do you notice when comparing these distances? Why do you think this relationship exists? 0.0s - 0.29s 0.29s - 0.49s 0.49 - 0.76s 0.76 - 1.05s 1.05 - 1.35s 9.0cm 4cm 9.5cm 30.5cm 22.25cm 8. Predict the position of the steel ball at 2.0 s. Explain how you arrived at this prediction. This can be done in words or with a formula. y = 39.21x^2 - 0.48x + 4.51 let x = 2.0s and solve for y y = 160.39cm 9. Utilizing your answers from Graph & Analysis II, Questions 7 construct a graph with “Time” on the x-axis and “Change in Position” on the y-axis. This should be thought of as a bar graph, not a typical Distance vs. Time graph.

10. Compare the shape of the graphs you made in Graph & Analysis II, Questions 3 & 9. What differences are there between these graphs. Why does this occur? The bar graph does not have as steep of an increase, because each bar represents only the distance travelled within the time interval, not the cumulative distance as in the dot graph. Otherwise, they are very similar in shape, both increasing non-linearly. Because of error (discussed below), the bars aren’t perfectly increasing each time. 11. Describe in words the difference between the motion of the car in Experiment I and the motion of the steel ball in Experiment II. Why does this difference occur? Because we started the time when the ball began motion, acceleration to bring the ball to top speed was taken into account. This is why less distance was traveled early on than later on. This is unlike the car, which stayed at a relatively constant speed and traveled the same distance amount throughout each interval. 12. If I were to drop a tennis ball out of a building that is 10 meters tall its motion would be most closely related to the car or the steel ball? How does the motion differ though from its closely related companion. The tennis ball is more closely related to the steel ball. Like the steel ball, the tennis ball took time to accelerate. 13. Describe the motion of a bicyclist who starts from rest and gets a gentle nudge down a steep hill that is 15 m long, he then reaches a flat stretch of road which is 25 m long (what effect would air resistance have in this 25 meter stretch of road), and then finally he approaches a steep hill going upwards that is 150 m long. All of the time the bicyclist is not pedaling. You can make some reasonable assumptions throughout your description of the motion. From rest, the bicyclist accelerates very quickly down the steep hill, has relatively constant velocity along the flat stretch of road (only slowing down a bit due to air resistance, friction, and other losses of energy), and

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

finally decelerates a lot while going up hill, until it stops. After stopping on the hill, he will begin to gain velocity in the opposite direction as well as accelerate in that same opposite direction. Part III Sources of Error – Think about how you could have obtained more accurate data. 1. Equipment error - motion was not completely straight, the car veered to the right. The stopwatch for very small times (in the ball experiment) was difficult to accurately pause/measure time. The rulers/meter sticks used could only measure so precisely. 2. Human error - Especially for the steel ball experiment, it was hard to perfectly time and synchronize starting the stopwatch and stopping the stopwatch. Improvements to the Lab – How could we have made this lab more efficient? What would you change about the ramp? What would you change about the distance the steel ball rolls down the ramp? 1. I would increase the length of the ramp, or I would automate the time recording method with a laser stopwatch. It proved to be very difficult to measure .25 of a second with human reaction times/physical ability. 2. It would have been better if all lab partners were in person at the same time. We would have been able to get more done and our chromebooks would have been running faster without the google meet in the background.

Copy of Linear Motion Lab 4

Related Documents