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course PHY241
9/27 9
I don't know if it's possible to develop a balance to weigh questions assigned to students via emails and webpages, but I suspect if it was we would be up to about a ton.Most of these questions are broken down into relatively small steps, and if you understand the ideas and read carefully you should find that they can be answered relatively quickly. Of course that's not the case for all of them, and a large number of quick answers can still take a significant amount of time.
The questions on a topic typically build in difficulty. My advice: When you start to get bogged down move on to the questions about another experiment or another concept. Let your brain process what you have done for a day or so, then come back to the questions that troubled you. This should give you a new perspective. And don't forget the Question Form at http://vhcc2.vhcc.edu/dsmith/forms/question_form.htm if you want to ask questions before the next class.
Given the delay in developing these questions and getting them to you, I don't expect them all to be done by Monday. But do try to get a good start by then, and should have most or all questions answered by Wednesday.
This document includes notes on revising the rubber band experiment. If you did it completely right the first time, which a few people did, no need to revise it. Otherwise you should do so.
There are pictures later in in this document that email won't handle. You can check the posted version for those pictures.
Inserting answers
You have been inserting your answers in the lines between the ends of the questions and the #$&* mark. Starting with this assignment, you will see **** in the line immediately following each question, and #$&* below that. Please insert your answers starting on a new line between the **** and #$&*. The **** and #$&* should each be alone on its own line, one before and one after your response. If you have already inserted your answers into a document lacking the ****, that's OK.
Brief at-home experiment
Is the magnitude of a ball's acceleration up an incline the same as the magnitude of its acceleration down the incline?
Test this.
One possible suggested method: Use the short ramp to get the ball started, and let it roll from the short ramp onto the longer ramp, with the longer ramp inclined so the ball rolls up, rather than down. Get the ball started on the short ramp, either by inclining it toward the long ramp or giving it a push (before it reaches the long ramp). Click the TIMER at the instant the ball hits the 'bump' between the two ramps, again when the ball comes to rest for an instant before accelerating back down the long ramp, and once more when the ball again hits the 'bump'.
Everyone will tend to anticipate their 'clicks', and to try to compensate for their anticipation. If the effect of anticipation is the same for all three timed events, then uncertainties in your results will all be due to the TIMER. If the degree of anticipation differs, as it inevitably will, then the uncertainties are compounded. If the degree of anticipation (and/or compensation) tends to be either greater or less for the second event (the ball stopping for an instant) than for the first and third (the 'clicks' made when the ball hits the 'bump'), then a systematic error is introduced (you will have a tendency to 'short' one interval while extending the other).
If there exists a difference between the times up and down the ramp, and if the difference is great enough to show through the uncertainties and systematic errors, then you might get a useful result (e.g., either the actual times are the same, or they differ).
Give a brief report of your data and your results:
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Displacement and force vector for a rubber band
The points (2, 1) and (4, 6) have the following characteristics:
* From the first point to the second the 'run' is 4 - 2 = 2 and the 'rise' is 6 - 1 = 5.
* The slope of the line segment connecting the points is therefore 5 / 2 = 2.5.
* A right triangle can be constructed whose hypotenuse is the line segment from (2, 1) to (4, 6), and whose legs are parallel to the coordinate axes. The leg parallel to the x axis has length 2, and the leg parallel to the y axis has length 5. The hypotenuse therefore has length sqrt(2^2 + 5^2) = sqrt(29) = 5.4, approx..
* The area of the trapezoid formed by projecting the two points onto the x axis, whose sides are the projection lines, the segment [2, 4] of the x axis and the line segment between the two points, has 'graph altitudes' 1 and 5, giving it average 'graph altitude' (1 + 6) / 2 = 3.5, and width 5 - 1 = 4, so its area is 14.
* The vector from the first point to the second has x component 2 and y component 5. Its length is sqrt(20) and its direction is specified by the ratio of the y to the x component.
Any time you see two points on a graph you should be aware that you can easily construct the right triangle and use it to calculate the slope of the segment joining them, the distance between the points, the area of the associated trapezoid and the components of the vector. Depending on the nature of the graph, some of these quantities will sense and be have useful interpretations, while others probably will not.
In the particular example of a rubber band stretched between the two points, with the coordinates in centimeters, the most important quantities are the components and length of the vector. The vector from the first point to the second is in this case a displacement vector. The displacement vector has x component 2 cm, y component 5 cm and length sqrt( 2 cm)^2 + (5 cm)^2 ) = sqrt( 29 cm^2) = sqrt(29) cm, about 5.4 cm.
* We denote the displacement vector from (2 cm, 1 cm) to (4 cm, 6 cm) as <2 cm, 5 cm>
If we divide the this vector 2 we get a vector of length 2.15 cm. The x component will be 1 cm (half of the original 2 cm) and the y component 5 cm /2 = 2.5 cm (half of the original 5 cm). The ratio of the components is the same, so this vector will be in the same direction as the original vector. If we multiply this vector by 2 we get a vector of length 8.6 cm. Its components will be double those of the original vector, and it will also be in the same direction as the original vector.
If we divide our vector by 5.4 cm, then the resulting vector has length 5.3 cm / (5.3 cm) = 1. The x and y components will be 2 cm / (5.4 cm) = .37 and 5 cm / (5.4 cm) = .92. As before, the ratio of the components is the same as for the original vector (accurate to 2 significant figures),
The vector <.37, .92> has magnitude 1 (calculated to 2 significant figure), as you can easily verify using the Pythagorean Theorem. Its direction is the same as that of the displacement vector <2 cm, 5 cm>.
Now we invoke a rule to find the tension in the rubber band. Assume that for this rubber band the tension is .5 Newtons for each centimeter in excess of its 'barely-exerting-force' length of 4.8 cm.
* The 5.4 cm length of the stretched rubber band is therefore .6 cm in excess of the 'barely-exerting-force' length of 4.8 cm.
* According to the rule, we conclude that the tension is therefore .6 cm * .5 N / cm = .3 N.
The vector <.37, .92> has magnitude 1, and its direction is the same as that of the displacement vector. A rubber band can exert a force only in the direction of its displacement vector.
* If we multiply our vector <.37, .92> , which we again mention has length 1, by .3 N, we will obtain a vector whose magnitude is .3 N. (whatever quantity we multiply by a vector of length 1, we end up with a vector whose length is equal to that quantity, or more correctly whose length represents that quantity)
* The direction of this vector will be the same as that of our original vector.
* <.37, .92> * .3 N = <0.11 N, 0.28 N>
* This is the tension vector. It tells us that the tension of this rubber band has x component 0.11 N and y component 0.28 N.
Using your data from the rubber band experiment, check your previous calculations of the displacement vector, the unit vector and the force vector for each rubber band. If your original work on this experiment was not done correctly, please correct it and resubmit. Please acknowledge that you have read this instruction and understand it.
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ok
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Linear force function and work\energy
If the graph below indicates the tension in Newtons vs. length in cm for a rubber band:
At what length does the rubber band begin exerting a force?
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5cm
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What is the average force exerted between the lengths x = 5 cm and x = 7 cm?
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.5N
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As the rubber band is stretched from length 5 cm to length 7 cm, through what distance is the force exerted?
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1N
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How much work is therefore done in stretching the rubber band from 5 cm to 7 cm length?
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2N/cm
you need to show the calculation, including units at every step
your units here are not correct
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How much work is done in stretching the rubber band from the 5 cm length to the 8 cm length?
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4.5N/cm
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How much work is done in stretching the rubber band from the 6 cm length to the 8 cm length?
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2N/cm
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What are the areas beneath the graph between each of the following pairs of lengths:
* x = 5 cm and x = 7 cm
* x = 5 cm and x = 8 cm
* x = 6 cm and x = 8 cm
* x = 7 cm and x = 8 cm
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1cm^2
2.25cm^2
2cm^2
1.25cm^2
the numbers are mostly right but the units aren't
you need to show at least one calculation with units at every step
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Verify that the equation of the straight line in the given graph is F(x) = (x - 5) * .5, where F(x) is force in Newtons when x is position in centimeters.
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f(x)=(6-5)*.5=.5
f(x)=(7-5)*.5=1
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University Physics Students: Confirm your results for graph areas by integrating the force function over appropriate intervals. Suggested method: Integrate the function symbolically from x = a to x = b, then use the resulting expression for appropriate values of a and b.
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[(.25b-2.5)b]-[(.25a-2.5)a]
[(.25(7)-2.5)7]-[(.25(a)-2.5)a]=1
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Nonlinear force function and work\energy
The graph given previously was linear. That graph would be realistic for a well-made spring, but not for a rubber band.
The graph given below is more realistic.
Estimate the average force exerted by this rubber band between the x = 5 cm and x = 6 cm lengths. Give your estimate in the first line, and explain how you made the estimate in the second:
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.5N
looked on the graph for halfway between the two points
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Repeat, for the x = 6 cm to x = 7 cm length interval.
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.8N
same as above
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Repeat, for the x = 5 cm to x = 5.5 cm length interval.
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.25N
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Repeat, for the x = 5.5 cm to x = 6 cm length interval.
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.6N
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Estimate the work done in stretching the rubber band for each of these intervals. Give your four estimates in the first line below, separated by commas. Starting in the second line explain how you got your estimates:
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.5N/cm,.8N/cm,.125N/cm,.3N/cm
Used my estimates for the avg force and simply calculated work force*`dx
good except for the units
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How much work do you estimate is done between the x = 5 cm length and the x = 7 cm length? Give your estimate in the first line, your explanation starting in the second.
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1.3N/cm
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If a trapezoid was constructed by projecting the x = 5 and x = 7 points of the graph, what would be its area, and would this result be an overestimate or an underestimate of the actual area beneath the graph between these two points?
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0.9cm^2,underestimate
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If a trapezoid was constructed by projecting the x = 5 and x = 8 points of the graph, what would be its area, and would this result be an overestimate or an underestimate of the actual area beneath the graph between these two points?
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2.55cm^2,it looks like it would be almost exact. The 2 curves should even each other out
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University Physics Students Only:
Note that these questions are easy to answer, if you understand what to do. Understanding what to do is fairly challenging at this point. I expect that some will get this, and some will not, and I can't predict who will fall into which category. Of course I'd love it if you would make it easy on me, and everyone would get these:
Suppose the function for the tension is T(L) = (L-6.5)^3*.15 + .5 + (L-5)*.2. How much work is done between length L = 5 cm and L = 7 cm, according to this function?
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1.85T/cm
T isn't a unit
you need to show the details of the calculation, including units
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How much work is done between lengths L = L_1 and L = L_2? Apply your expression to the work done over each of the following intervals:
* L = 5 cm to L = 6 cm
* L = 6 cm to L = 7 cm
* L = 5.5 cm to L = 6 cm
* L = 5.5 cm to L = 7 cm
Give your four results in the first line below, separated by commas. Starting in the second line explain how you got your results.
.6875,.23125,.1156,.703125
I plugged in the legnths and found the difference in tension and multiplied by the change in x
Directions of force and displacement vectors matter for work\energy
If force F and displacement `ds are both along the x axis, then what is the sign of F * `ds in each of the following cases:
* F is in the positive direction and `ds is in the positive direction.
* F is in the positive direction and `ds is in the negative direction.
* F is in the negative direction and `ds is in the positive direction.
* F is in the negative direction and `ds is in the negative direction.
Give your answers in the given order, in the first line below, separated by commas. Starting in the second line explain how you got your results.
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+,-,-,+
well two negatives make a positive,two positives make a positive and a negative and positive make a negative
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If the x axis in the car-and-magnet experiment was your meter stick, then what were the directions of F and `dx, in each of the following situations:
* F is the force exerted on the 'car magnet' by the 'fixed magnet', `dx is the displacement of the car for 10 cm after its release.
* F is the force exerted on the 'car magnet' by the 'fixed magnet', `dx is the displacement of the car as you move it toward the last 10 cm toward the 'fixed magnet'.
* F is the force exerted by the 'car magnet' on the 'fixed magnet', `dx is the displacement of the car for 10 cm after its release.
* F is the force exerted by the 'car magnet' by the 'fixed magnet', `dx is the displacement of the car as you move it toward the last 10 cm toward the 'fixed magnet'.
* F is the force exerted on the car by friction, `dx is the displacement of the car after being released.
* F is the force exerted by the car against friction, `dx is the displacement of the car after being released.
Give your six answers in the first line below, separated by commas. Starting in the second line explain how you reasoned out your answers.
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The figure below represents two plausible force vs. separation models for two of the ceramic magnets used in the experiment. Force is in Newtons while separation is in centimeters.
If the two magnets were touching before release, which model appears to predict the greater kinetic energy at the x = 2 cm position, and which at the x = 9 cm position?
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red appears to be greater
blue appears to be greater
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&#Good responses. See my notes and let me know if you have questions. &#
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