course Phy 201
ic_class_090909vvvv
Rubber band measurements
You have data that allows you to determine the length of the thin rubber bands and the length of the thick rubber band, for each setup.
Report the span of your hand on the 'ruler' you used to measure the rubber band system, as well as your height in inches. Report as two numbers separated by commas:
&&&& (report your numbers starting on the next line):
23cm, 75
Report you raw data below (this will be the quantities you actually measured in class; note that you didn't measure the lengths of the rubber bands but the positions of their ends, so your raw data will not include the lengths). :
&&&& Your raw data from class should be reported starting in the next line:
20 cm
A data report needs to be followed by an explanation of how the data was obtained and what it means.
Beginning in the line below, report the lengths of the 'thin' rubber band and the length of the 'thick' rubber band, as determined from your raw data. Each line should be reported as two numbers, separated by a comma:
&&&& (lengths of 'thin' then 'thick' bands, first trial (least stretch)): 7, 7.5
&&&& (lengths of 'thin' then 'thick' bands, second trial (medium stretch)): 9, 10
&&&& (lengths of 'thin' then 'thick' bands, third trial (greatest stretch observed)): 12, 11
These results make perfect sense.
However the only data you reported was in the single number in the single line
20 cm
Your raw data do not support the numbers you reported here.
Now sketch a graph of y vs. x, where y = length of 'thin' rubber band and x = length of 'thick' band. Your graph will consist of three points.
&&&& What is the slope between the first and second point on your graph (give the slope in the line below, then starting in a new line explain how you calculated the slope)?
1.25
I used this equation m=(y2-y1)/(x2-x1)= slope. m=(10-7.5)/(9-7)= 1.25
&&&& What is the slope of the graph between the second and third point on your graph (give the slope in the line below, then starting in a new line explain how you calculated the slope)?
1/3
I used the equation for slope. m=(11-10)/(12-9)= 1/3
&&&& Do you believe that the difference in the slopes is due mainly to unavoidable errors in measurement, or to the actual behavior of the rubber bands? Support your answer with the best arguments you can.
I believe it was due to the behavior of the rubberbands because the thin rubberband is not as strong as the large rubber band and can be stretched farther with less force.
* In the second setup you had three 'thin' rubber bands, all stretched between the same two paper clips, opposing the single 'thicker' rubber band. You observed this system for three different degrees of stretch.
Beginning in the line below, report the lengths of the 'thin' rubber bands and the length of the 'thick' rubber band, as determined from your raw data. Each line should be reported as two numbers, separated by a comma:
&&&& (lengths of 'thin' then 'thick' bands, first trial (least stretch)): 6.5, 7.7
&&&& (lengths of 'thin' then 'thick' bands, second trial (medium stretch)): 8, 9
&&&& (lengths of 'thin' then 'thick' bands, third trial (greatest stretch observed)): 8.9, 10.3
Now sketch a graph of y vs. x, where y = common length of 'thin' rubber bands and x = length of 'thick' band. Your graph will consist of three points.
&&&& What is the slope between the first and second point on your graph (give the slope in the line below, then starting in a new line explain how you calculated the slope)?
1.15
I used the slope fromula. m=(8-6.5)/(9-7.7)= 1.15
&&&& What is the slope between the second and third point on your graph (give the slope in the line below, then starting in a new line explain how you calculated the slope)?
.9/1.3
I used the slope formula. y is the 3 rubberbands and x is the 1 thick band.
&&&& Do you believe that the difference in the slopes is due mainly to unavoidable errors in measurement, or to the actual behavior of the rubber bands? Support your answer with the best arguments you can.
I think that it is due to the actual behavior of the rubberband because the strength of the 3 thick rubberbands exceeds the strength of the 1 thick band. Because of the difference in strength I was able to use less force to stretch out the thick rubberband compared to the greater amount of force it would have taken to stretch the 3 thin bands to the same length.
Question to think about (and answer as best you can): You you think that during the measurement process, the rubber bands were being held apart by force, power, energy or something else, or perhaps by all three? Answer beginning in the line below, and support your answers with your best reasoning. Don't worry about being wrong, but do give it some thought. &&&&
I feel that they were being held together by all three. However, the force, power, and energy that is pulling the rubber band is also trying to pull it back together. When you're pulling agianst the rubber band it is taking your force, power, and energy to fight the force, power, and energy that comes from the rubberband itself. So in order to pull the rubberband apart it requires an increasingly larger amount of all force, power, and energy.
Acceleration of Toy Cars in Two Opposite Directions
You also measured the motion of a couple of toy cars, moving in two opposite directions. You were asked to obtain information that will tell you the acceleration of each car in each direction.
&&&& Give your raw data in starting in the line below. Be sure to include all information necessary to interpret your data.
Pick 'north' or 'south' for your positive direction. State your choice:
North
&&&&
For the first trial in which the car moved in the northerly direction, explain how you reason out the acceleration. Show how you reason out your results, starting with the raw data, based on the definitions of rate of change, average velocity and average acceleration. You may assume that the acceleration is uniform, so that the v vs. t graph is in fact trapezoidal. Give you explanation starting in the line below:
The car went 39.5 cm. before stopping and it took 2.5 oscilations using a short pendulum, A=(39.5-0)/ 2.5= 15.8 cm./t
&&&&
Repeat for the first trial in the southerly direction:
The car went 33.4 cm. before stopping and it took 2.5 oscilations using a short pendulum, A=(33.4-0)/ 2.5= 13.36 cm./t
&&&&
Find the acceleration for the next trial; however you may abbreviate your calculations and don't need to repeat the verbal explanations:
Trial 1: A=(39.5-0)/ 2.5= 15.8 cm./t
Trial 2: A=(33.4-0)/ 2.5= 13.36 cm./t
It appears that you've calculated average velocities, but not accelerations.
&&&&
(you may copy the two lines above as many times as necessary to account for all the trials you think it necessary to report your results; for these additional lines your work may be further abbreviated)
See my notes. Overall you did pretty well here, but your data report should have been more complete, and you still need to calculate the accelerations of the car.
.