30 March 2009

Snell's Law in a prism

Snell's Law seems pretty easy when you're dealing with just a horizontal interface, and especially when the possibility of total internal reflection is taken out of the picture. A problem with angled surfaces, like those in the equilateral prism shown to the right, is worth assigning to see if your students are thinking about what the angles in Snell's Law really mean.

The goal of the problem is to determine angles X, Y, and Z made by light traveling through the equilateral prism. In stating the problem, I've been tricky in two ways. First of all, the 31o angle is NOT the angle of incidence at the left-hand surface. In optics, all angles are measured to the normal, not to a surface; so the incident angle to plug into Snell's Law is 59o. A student who plugs in 31o has neither paid attention in class nor collaborated with his classmates, and thus earns a red and blue VOID stamp.

Most students, though, figure out the correct incident angle. They plug in to get the refracted angle, 35o (from the normal, of course).

What's crazy here is the disconnect between geometry and the angles my students use for the right-hand interface. The vast majority of my class at this point makes two incompatible assertions:

1. Angles X and Y are each 60o, because the prism is equilateral.
2. The angle of incidence at the right interface is 35o, giving a refracted angle in air of 59o. What's the justification? Since the light entered the glass at 35o, it must make the same angle when it leaves.

This second assumption likely comes from class, where I show light incident on a water-air surface. If the light hits the water from the air at 45o, we show using Snell that the light refracts at 33o. Then we show again that light incident the other way, from water to air, can follow the same path: incident at 33o gives refraction into air at 45o. Perhaps, also, some have seen the calculation that light that hits a rectangular block of glass must exit the glass parallel to the original incident beam.* My class has incorrectly made the blanket interpretation that the angle of refraction in the air must, in all circumstances, be equal to the original angle of incidence. Oops.

The first assumption, though, is either lazy, or born of an inability to understand what a "refracted angle measured from the normal" means. Some careful but straightforward Euclidian geometry can show that angle X must be (90-35)=55o -- after all, angle X and the 35o angle of refraction make up the normal to the surface, which by definition is a 90o angle. Then, 180o in a triangle, so angle Y must be (180-55-60)=65o. The angle of incidence at the right-hand surface, therefore, is (90-65)=25o.

Snell's law shows that a 25o incident angle from glass to air give a refracted angle of 39o. Angle Z is the complement of 39o, and so is 51o. Done.

I assigned this problem in my general physics class after I had assigned a plug-and-chug-level problem set on Snell's Law. Sure enough, no one at all thought to trace through the geometry. That's partially my fault for giving them the plug-and-chug problems. I made them believe that Snell's Law didn't require any thinking. At the general physics level, I think it useful to do some plug-and-chug just to make the point that "Hey, this math is NOT too hard for you, so don't even think about using that as an excuse." Of course, then it is incumbant on me to guide the class a step beyond the plugging and the chugging. This problem is a good way to do that.

So, what did I do? I RE-assigned this problem, gave a brief hint on the order of "Don't assume that the light makes an equilateral triangle, actually do some geometry," and gave the quiz below. I won't grade the re-submitted problem, only the quiz.


P.S. This problem is good for AP physics, too... I probably would not draw in the normal lines, though.

* Check out Giancoli chapter 23 problem 38, which asks students to prove this fact geometrically.

1. What is the angle of incidence at the left-hand edge? Why?

2. Explain how to figure out angle X.

3. Explain how to figure out angle Y.

4. Explain how to figure out the angle of incidence at the right hand edge.

5. Calculate angle Z.

27 March 2009

Snel's Law demonstration

Let's get this out of the way right now: I'm not sure whether it's "Snel's Law" or "Snell's Law." My Serway credits the law to some dude (or dudette, I can't tell) named "Willebrørd Snell." However, I've seen two recent articles in The Physics Teacher and the American Journal of Physics which suggest that Snell should be spelled Snel. This is all an interesting debate, but in the end, who cares -- my purpose is to teach the law that bears the name.

To the right you see a fancy-pants refraction tank, sold by nada scientific for $200. If you have $200 lying around, it looks like a great tool -- the 6V light mounts anywhere around the circular tank, allowing you to send light either from air to water or from water to air. I don't know about the durability -- I had a similar tank whose stand broke off after only a few years of use.

Thing is, even though I do have $200 lying around, I prefer to spend money on other things. Refraction demonstrations can be done much more easily and cheaply. A fishbowl with a laser pointer works just fine -- use chalkdust to visualize the laser in the air, and a teeny drop of milk or coffee creamer to visualize the laser in the water.

The question becomes, how do you make quantitative measurements of incident and refracted angles for a laser entering water in a fishbowl? A dry erase marker will write and erase on glass. I have a student trace the incident and refracted rays with the dry erase marker. Then, I turn the lights back on, and use a protractor to measure angles of incidence and refraction. Snel's law always seems to be verified.


25 March 2009

Collaboration rules for AP problems assigned for homework

This time of year especially, I like to assign AP questions as homework. But when I do, the collaboration rules change.

Up until now, I've asked students merely to approach the problem individually, spending at least 5 minutes and showing me some kind of reasonable start. After that, collaboration is freely allowed and encouraged. I award full credit for a correct answer whether or not the individual work was on target.

Early in the year, I'm happy if someone merely copies the question and the diagram on his own; at least he's thought about the problem, and has context for the subsequent discussion with his friends. After a couple of months, I insist on a bit more... I start taking off unless the student does SOMETHING on his own, even if he's wrong.

Now, though, my students know how to learn physics. They no longer should be hesitant to approach a problem. Furthermore, an AP free response question always has several distinct parts -- if part (c) is hard, they should still be able to get full credit for parts (a), (b), and (d). Now is the time to practice this kind of approach to AP questions.

Throughout the school year, I nudge my class gradually toward becoming more independent physics students. Now it's time to take the final step. Many will be surprised how well they do on their own after relying on friends' input for much of the year. Others will expose blatant gaps in their knowledge; but they'll be forced to confront their lack of understanding, giving them incentive to figure out what they don't know.

Instructions for AP problems assigned as homework:

* Do the ENTIRE PROBLEM on your own before seeking assistance. Do NOT discuss the problem at all until you are completely finished. This should take no more than 15 minutes.

* If you make any changes, these must be in a DIFFERENT COLOR of ink or pencil. Cross out (do not erase) your original wrong work.

* You will earn full credit for your final answers, unless you fail to follow these collaboration rules.


23 March 2009

Secrets of grading homework

In order to get your students engaged on a nightly basis, you absolutely must make some pretense toward grading homework. If you expect your class to make an effort “for their own good,” without you checking their work at all, you’re deluding yourself – teenagers are not known for accurate evaluations of what’s good for them.

Let’s disregard philosophical issues of self-determinism vs. paternalism and note that even the most diligent student needs feedback on his efforts. Spending three hours doing 15 problems incorrectly is worse than useless; on the other hand, just 15 minutes of serious, careful thought about physics can be extraordinarily productive. A first-time physics student doesn’t know the difference between good effort and wasted time. Grading homework helps the student see that difference.

That said, there’s no need to grade every problem every night in tremendous detail. The purposes of grading homework are to inspire effort, and to provide useful feedback. Let’s say you assign two problems each night. Just knowing that one of the problems will be graded will be sufficient to inspire effort, as long as the students don’t know which problem will be graded. As for useful feedback, some of the most common mistakes in approaching physics problems are poor problem presentation, not checking the reasonability of answers, writing five paragraphs when one sentence would have been sufficient, or not explaining anything at all. Chances are, if the student is making these errors on one problem, he’s making them on both; there’s no need to grade every single assignment.

Near the beginning of the year, I find it important to grade a problem every night. Ideally, I grade it in time to send a brief comment to the class email folder:

"Good job explaining your reasoning on the normal force problem. The most common mistake was that a normal force requires a surface. You can’t label “normal force” for any old upward force!"

I’m establishing a tone. Those who are inclined to slack learn quickly that I will notice. My expectations become crystal clear right away.

As the class progresses, just the fact that I do some grading on a regular basis is enough to maintain the class’s focus. After a month or so, I can save up some problems to grade for my nights on duty. Or, I can grade the problems only every other night, or even every three nights. As long as they still know I’m watching, as long as I catch the egregious attempts at slacking1, I keep my class with me.
[1] And, because I graded a problem every night for a month, I know exactly who might make these egregious

What my homework grades mean

Different teachers take all kinds of different approaches to grading homework. Many are perfectly reasonable – please don’t take my ideas as gospel.

I grade each problem on (usually) a 10 points scale, just like an AP free response question. In principle, I have in mind a loose rubric. In reality, though, I am grading holistically.

On the top end of the scale, I want 10/10 to be reasonably tough to obtain. That means the student has presented the problem neatly and clearly, with explanations or annotations so that I know he understands what he did thoroughly.

I do occasionally give 11 or 12 points out of 10. Such bonus scores are given for outstanding presentation, for a clever explication of the meaning of a result, or for something out of the ordinary that I feel bears recognition.

The students’ overall homework grade is assigned based on a square root curve, where 81% is an A, 64% is a B, 49% is a C. Therefore, on an individual problem, 8 points is essentially an A, 6 or 7 points are a B.

Below that, the minimum passing score is 4. If I feel like the problem does not merit a passing score, I don’t even differentiate between a 0, 1, 2, or 3; I simply stamp VOID1 on the problem, and assign the student a time to come discuss the problem with me.

[2] Yes, I actually purchased a red and blue “VOID” stamp. It’s effective.

So what am I looking for when I grade?

It is critical that a stack of homework problems not take me all night to grade. I have to grade as fast as possible, which is not conducive to evaluating every aspect of the problem. If I know before I start what mistakes students might make, and what aspects of the problem require annotation or explanation, then I can move quickly.

For example, consider this problem, which I assigned last Thursday:

In a double-slit experiment it is found that blue light of wavelength 450 nm gives a second-order maximum at a certain location on the screen. What wavelength of visible light would have a minimum at the same location? (Hint: figure out which variable must be the same in each case!)

Here are the solution steps I expect to see:

• Writing the relevant equation, dsinθ = mλ.
• Recognizing that d and λ must stay the same
• Using a half-integer for m in the second case, because the light will be at a minimum
• Recognizing that while any half-integer m will work, the correct m will be the one that gives a wavelength in the visible spectrum

Now, what do I actually look for? I want to see some evidence that the student has not only done the steps above, but that he knows why he has done these steps.

For example, many folks start their work with the statement that m1λ1 = m2λ2. That’s technically and mathematically correct, and will lead to a correct solution. But that equation is not on the equation sheet, and so is not a legitimate way to begin a problem. Where did that equation come from?

If it’s the first equation I see, then most likely it came from a friend’s advice. That’s fine, as I encourage collaboration. Nonetheless, collaboration isn’t a substitute for individual thinking. It is critical that I catch students when they are using classmates’ ideas without an attempt to justify them with reasoning. So, this student will lose several points.

However, if this student says m1λ1 = m2λ2 “because d1sinθ1 = d2θ2, then I see what he’s doing. If he further mentions WHY the angle to the screen and the distance between slits does not change, then that’s worth full or extra credit.

Similarly, I expect to see some kind of annotation when it’s time to plug in m= 3/2: “Because of the destructive interference” or “because the light is at a minimum” works for me.

And finally, to earn full credit, I have to see some writing about the subtleties of choosing m. They should show that they tried other m’s besides 3/2, and that they gave wavelengths outside the visible spectrum; or, at least, they should tell me that they could have tried these other m’s. Saying that m= 3/2 for an illegitimate reason (for example, “because m was 2, so now it has to be less”) does not earn full credit. Failure to justify the choice of m earns the same points as if the justification were incorrect.

Beyond these four issues, I don’t really look at the solution. If the answer is wrong but the student explained the four bullet points clearly, I’ll probably just call is 9/10 and be done with it. After all, this is not a math class. It’s far more important to be able to explain the physics than to plug in correctly to the calculator. Throughout the year, my grading sends this message, as the class certainly shares stories of what they lost or gained points for. They see that I’m looking at their words far more than I’m looking at their numbers. And, since I know exactly what I’m looking for, it doesn’t take me very long to figure out whether I’ve seen the words I want to see – saving me time in the grading process.


18 March 2009

The Magic Tuner-Metronome

I learned a bunch today in the waves-in-a-musical instrument lab. (See yesterday's post for details.) For one thing, I learned that a tenor recorder is pitched in F. That means that when you think you're playing a C, you're actually playing an F. How did I learn that?

You might recall that I proposed to determine the correct octave being played by comparing the recorder's note to that played by a digital frequency generator. Three problems emerged:

(1) I only had an analog freqency generator available, that could give me a freqency to only about +/- 20 Hz; and

(2) My students were so bad at recorder playing that I'll bet THEIR frequency varied by at least that much, too.

(3) I used to have a pretty good ear for pitch, back when I was in choir and marching band. However, I discovered that matching a pitch with the frequency generator in a loud room was quite difficult.

After 20 minutes of putzing, someone asked if they could check their pitch with Charlie's tuner. What?!? Charlie has a tuner? Back when I was in band, a tuner was an enormous, bulky, expensive device. It might be able to tell you whether you were sharp or flat, but *you* had to tell *it* what note you were playing.

Amazing, the progress of technology. Charlie pulled a calculator-sized box out of his guitar case. We held up the box to the recorder, a student played -- or tried to play -- a note, and the box told us: That's a C4, which we know to be 262 Hz from our chart! When the student played the high C, the box registered as a C5, 523 Hz.

That's how I found out that the tenor recorder was pitched in F. The box told me so.

Anyway, I instantly went online to order one of these magic boxes. The item is a Boss TU-80C Chromatic Tuner & Metronome, $25.68 (including shipping) from amazon.com. I highly recommend you get one of these things if you do today's experiment.


17 March 2009

Lab -- waves on musical instruments

My general physics class is studying waves. So far, we've just done basic v=λf stuff. We're going to study standing waves, and harmonics in pipes and strings, soon.

For lab tomorrow, I've asked anyone who plays a musical instrument to bring it in. Groups that don't bring in a guitar or trumpet or something will get to use one of the recorders that I borrowed from the music department. (Does every third grader in America still learn to play recorder badly? I have interesting memories of Jingle Bells in the third grade's Arena of tonal Quality.)

Most instruments that I'll see tomorrow can be simplistically modeled as open pipes or fixed strings. And, except for the brass instruments, the wavelength of each note can be approximated as twice the length of the string or pipe. (Brass instruments are rarely playing the fundamental mode... I have to hazard a guess as to which harmonic they're playing. I always hope I get very few brass players in my class.)

So I'll have each group play a chromatic scale for one octave -- that's 12 notes. For each note, they'll approximate the frequency by looking at the chart I've printed on their lab sheet. I'll have a digital frequency generator hooked up in the back of the room so that they can try to differentiate between, say, the 262 Hz C and the 512 Hz C.

For each note, they'll estimate the wavelength by measuring the length of the pipe or string. Strings are straightforward; pipes are not. For the recorder, it's possible to estimate the pipe length as the distance from the mouthpiece to the first open hole, but even that's not quite right. We're using all kinds of crude estimates here.

But, crude or not, I usually see a nice, clear hyperbolic graph of frequency vs. wavelength. That allows me to teach these guys how to turn that into a linear graph, by plotting 1/wavelength on the horizontal axis. I'm happy for students at this level just to be exposed to the process of straightening a graph. AP students do this on a biweekly basis; this is, believe it or not, highly advanced math for the general class.

Anyway, the slope of the straight graph SHOULD be equal to the speed of waves on the string or in the pipe. For woodwinds, this should be the speed of sound in air, and will be unless they chose the wrong octave (i.e. they were playing the 262 Hz C, but they misidentified it as 512 Hz). For strings, they should have a wave speed in the high tens through the hundreds of m/s. It's possible to determine the wave speed by knowing the tension and linear density, but that's not a measurement I'd ask general students to make.

Below is the lab handout I'll give. This will most likely be a two-lab-period experiment -- I'll get the raw data and a start on the linear graph tomorrow, and next week we'll finish.

By the way, I welcome any comments or suggestions from people with better knowledge of the physics of music than I. I recognize how crazy some of these approximations are... but the point is, we get the hyperbolic relationship between frequency and wavelength. That's what I'm after.

Physics Experiment:
Waves in a Musical Instrument

Pre-lab preparation: Bring a musical instrument to class – a guitar, trombone, whatever. If you don’t bring an instrument, you’ll use a recorder.

Now, sketch in your notebook… what should a graph of frequency vs. wavelength look like? Make a sketch and show it to Mr. Jacobs. You will earn an extra point if this sketch turns out to be correct.

Goal of the experiment: You are to demonstrate that v = λf for waves in your instrument.

· You will find the wavelength and frequency for each note in one octave’s worth of a chromatic scale. (That’s 12 notes: C, C#, D, D#, E, F, F#, G, G#, A, A#, B.)

· The wavelength of the note should be twice the length of the pipe or string that produces the note.

· The frequency can be determined using the frequency table on the other side. Be careful – use the frequency generator in the back of the room to figure out what octave you’re playing. Mr. Jacobs will show you what this means.

· Graph frequency on the vertical axis, and the wavelength on the horizontal axis.

· Mr. Jacobs will show you how to convert this graph into a straight line graph.

Analysis: Answer the following questions thoroughly on a clean page in your lab notebook.

(1) Explain why the original f vs. λ graph looked the way it did.

(2) What is the meaning of the slope of the straight line graph?

(3) What was the slope of the straight-line graph?

(4) Comment on the reasonability of the slope you measured. No BS!

Frequencies of musical notes:
Note that multiplying a frequency by two gives the same note but an octave higher.

16 March 2009

Double Slit Interference -- First Assignments

Tomorrow I have to go back to work. Spring break is over, the guys have descended back upon campus, and it's time to talk wave interference. Double-slit interference is at first daunting, but quickly becomes straightforward. The nice thing is, at this stage of the year, the class can learn much of this chapter on their own based on a few hints and demonstrations.

Before we all left for break, I introduced the equation dsinθ = mλ. With a "take good notes" warning, I diagramed the typical double-slit situation, defining each variable, and explaining how the relationship comes about. I told the class how dsinθ was the path difference between light from the top and bottom slits. I told them that m represented the number of wavelengths in the path difference. But I am confident that none of this information sunk in. (That's okay -- it's in their notes, and they'll need to see the idea of "path difference" a bazillion times before they get it. Gotta start somewhere.)

Importantly, when I introduced the equation, I did several simple qualitative demonstrations. I used both red and green lasers through a couple of diffraction gratings, each time asking which variable I was changing, and what the pattern on the screen should look like. For example, I showed the red laser through the 600 lines / cm slit. I asked, "What will happen to the dots on the screen if I use the green laser through the same slits?" They had to figure out that d and m stay the same, while λ decreases... therefore, sinθ also decreases, and the dots will get closer together. Sure enough, that was what happened.

(Why did I use a diffraction grating rather than a true double slit? It's easier for the class to see, and the mathematics are identical. Once the students know what they're doing, I'll explain and demonstrate the differences between diffraction gratings, double slits, and single slits. As an introduction, though, screw subtlties: show the diffraction grating patterns, calculate for double slits.)

As I promised, the class's assignment for our first post-vacation meeting will be a couple of very simple problems with double slits. I keep the problems barely above the plug-and-chug level to start with, because I just want them to be able to identify the correct variables, and to plug properly into the calculator. When I graded a double slit problem on the 2004 AP Physics B exam, it broke my heart to see folks who wrote down the correct equation, then seemed to plug in values randomly for each variable. Half the battle is to know what the letters mean. The other half is recognizing that 500 nm is NOT equal to 5 x 10-9 m.

For tomorrow's quiz, I'm going to be sure that everyone gets the basics. I'll give the class a problem, but not ask them to solve for anything. Rather, they must simply identify variables and write the equation. Take a look:

You may use a calculator for today’s quiz.

A red laser with wavelength 620 nm in air shines through two slits which are separated by 0.50 mm. On a screen 2.0 m away from the slits, the laser makes an interference pattern. The brightest spot is located directly in front of the two slits. The next bright spot is located 0.25 cm from the brightest spot on the screen.

1. Assign a value to the following variables for the location of the next bright spot. USE UNITS OF METERS FOR ALL DISTANCE QUANTITIES!!! (You may want to make a sketch to help you with the trigonometry.)

d = (in units of meters)
m =
θ =
λ = (in units of meters)

2. What equation relates the variables above?

That's it! Tomorrow we'll go over subtleties, including the small angle approximation that allows for the equation x = mλL/d.

Oh, by the way... did anyone notice any problems with the picture at the top of the post?


15 March 2009

VIR charts and basic circuits

I got an email from Geoff Clarion the other day regarding VIR charts. Geoff graded AP physics exams with me for several years. He was known as the “cruise director” at the AP reading, because he didn’t just talk about how fun it would be to do something active and interesting – he actually made arrangements such that events happened. Geoff organized a yearly softball game*, volleyball nights, nights on the town, and so on.

I always say that the AP reading has been the best professional development I’ve ever been a part of. Many of my little tricks of the trade have been based on a snapshot of conversation from the reading. I mean, there I’m in the presence of 150 strong physics teachers, nerds like me who tend to talk shop when they’re not actually grading AP exams. I think it was Geoff who first suggested the utility of the green laser pointer as a demonstration tool. In turn, Geoff likes the VIR charts I use in my own class.

What is a VIR chart, you ask? It’s a straightforward way of organizing the answers to a circuit question, one which comes with a built-in reminder of when Ohm’s law is valid.

Everyone who takes physics can spit back Ohm’s law… V = IR. Wonderful. Problem is, too many students use this equation indiscriminately. Consider the basic circuit shown below. The goal is to determine the current through the 4 Ω resistor.

“Ah,” says the novice, “the voltage is 10 V, the resistance is 4 Ω, so by ohm’s law I divide to get a current of 2.5 A.”

“Boux,” says the teacher.** Ohm’s law cannot be used as a bludgeon. To teach the proper use of Ohm’s law, I organize my solution using a VIR chart like the one here.
The key to this chart: Ohm's law can ONLY be used when two of the three entries in a single row are known. Any other use of V=IR constitutes fraud.

Initially, I've only input the given values: the voltage of the battery, and the resistance of each individual resistor. Note that I know only one value in each row -- that indicates that there's more work to be done before Ohm's law is useful. The obvious approach is to find the equivalent resistance of the whole circuit. First I combine the parallel resistors to get this circuit:

Now, it's clear that the total resistance must be 5 Ω. I plug that into the VIR chart:

Voila, I have two of three entries in the "total" row completed. Ohm's law can be used to finish the row: I = V/R, giving a total current in the circuit of 2 A. Woo-hoo! A correct use of the law.

Warning: The VIR chart is not a magic square. It is not permissible just to add the resistances algebraically to get the total; we'll see that 2 A is NOT the sum of the "current" column. The chart is merely a method of organizing one's solution. It is only useful in conjunction with circuit diagrams.

With that reminder, I'll see what else I can figure out by looking at the circuit diagrams. By definintion, the "total" current in the circuit is the current coming out of the battery. Since R1 is in series with the battery, it must take the full current from the battery -- all 2 A. I put that in the chart:And now I have two entries in the top row, so I use Ohm's law to find the voltage across the first resistor:Almost done. Now the trick is to remember the rules about voltage in series and parallel resistors: The voltage across parallel resistors is the same, but the voltage across series resistors adds to the total. (That's just a simple statement of Kirchoff's loop rule.) Since the 2 Ω resistor takes 4 V, and the total is 10 V, the 3 Ω equivalent resistance must take the remaining 6 V. And, since voltage across parallel resistors is the same for each, I can fill in 6 V for BOTH R2 and R3. With two of the three entries in the rows, I can use Ohm's Law to find the missing currents.
Checking the reasonability of my chart, I see that Kirchoff's junction rule is satisfied. The two parallel resistors take different currents that add to the total, and the smaller resistor takes more of the current. My work here is finished.

* My stats: 2-4 with 2 runs scored, 3 assists and an error. Game MVP: Nebraska alumna Shelly Strand, my table leader at the time, whose defense at first base simply shut down the opposition (and saved me several more errors).

** “Boux” has sort of become a Jacobs catch phrase, as this is what I write on papers that show crazy mistakes like the one I’m describing here. The source of the phrase is an old Dave Barry column in which he asserts that Mets fans say, “Boo! You stupid bum!” while the more civilized Francophone Expos fans say, “Boux! Voux dumme bumme!” I’m paraphrasing, but that was the gist.

12 March 2009

AARGH! No internet. More posts coming soon.

I'm at the beautiful Jupiter (Florida) Waterfront Inn, but the internet isn't working in my room. Hence, the lack of posts. I've nearly finished a post about circuits and the VIR chart -- but that will have to go up on, probably, Sunday. Until then, I'll be enjoying the intercoastal waterway out my window, and then I'll be driving on I-95 for two straight days. Maybe an upcoming post will be the Physics of Interstate 95. Any suggestions?

06 March 2009

Bert, Ernie, Oscar, and nailing first-month misconceptions

I don’t care how good your lectures are, I don’t care how well your textbook presents the material, your students are going to internalize misconceptions about physics concepts.

Sure, some of these can be avoided by careful planning – for example, I don’t introduce Coulomb’s law for the force between electrical point charges at all. If I do, every problem that says anything about electricity automatically brings forth F = kQQ/r2. Instead, I start with the concept of an electric field; several days later, I explain that a point charge can CREATE an electric field by the equation E = kQ/r2; and then the equation for the force on a nearby charge follows from F = qE.

Nevertheless, even with my oh-so-careful presentation, I have to hammer the idea that the equation for the electric field created by a point charge only applies in that particular circumstance. They see questions on fundamentals quizzes asking when the equation is valid; they lose mongo points if they ever even write the equation in an illegal circumstance. My specially planned lectures, which evolved from 13 years of experience, still do not preempt misconceptions.

My attack strategy for misconceptions must necessarily take shape early in the course. In the very first month, we usually present one of the more conceptually difficult topics of the year: Newton’s second law. No matter how careful my presentations, students still try to set any old force (rather than the net force) equal to ma; they still assume that a force is required for motion to occur, assuming in their brains that Fnet = mv rather than ma.

I don’t believe in a magic technique to eliminate misconceptions. Rather, we must fight a war of attrition, converting one student at a time through a multi-faceted approach. Docking points on homework, asking the same question numerous different ways on fundamentals quizzes, multiple choice questions, in-class reminders, et cetera are all legitimate weapons in our attack on misconceptions. Today’s post demonstrates another part of the arsenal: make the student recognize the misconception when someone else writes it. Then, make that student write out an explanation for why this someone else has said something silly.

Below is a problem I use as a quiz in AP physics, or on a concepts test in general physics.

(The saddest part is that I often have to explain the use of the term “muppet.” RIP, Jim Henson.)

Bert, Ernie, and Oscar are discussing the gas mileage of cars. Specifically, they are wondering whether a car gets better mileage on a city street like Route 15 in Orange, or on a freeway like I 64. All agree (correctly) that the gas mileage of a car depends on the force that is produced by the car’s engine – the car gets fewer miles per gallon if the engine must produce more force.

Below is the statement made by each muppet – one statement is completely correct, and two contain errors. Identify the correct statement. For the others, explain thoroughly the error in physics (Free body diagrams may be useful).

Bert says: Gas mileage is better on I 64. On route 15 in town the car is always speeding up and slowing down because of the traffic lights, so since Fnet=ma and acceleration is large, the engine must produce a lot of force. However, on I 64, the car moves with constant velocity, and acceleration is zero. So the engine produces no force, allowing for better gas mileage.

Ernie says: Gas mileage is better on route 15. On route 15, the speed of the car is slower than the speed on the freeway. Acceleration is velocity divided by time, so the acceleration on route 15 is smaller. Since Fnet=ma, then, the force of the engine is smaller on route 15 giving better gas mileage.

Oscar says: Gas mileage is better on I 64. The force of the engine only has to be enough to equal the force of friction and air resistance – the engine doesn’t have to accelerate the car since the car maintains a constant speed. Whereas on route 15, the force of the engine must often be greater than the force of friction and air resistance in order to let the car speed up.

03 March 2009

Test - Correction - Recall Quiz

There's only one assignment you can give on which you can be sure of the class's full attention:

A test.

Thus, it's useful to sqeeze every last bit of usefulness you can out of a test. The cycle I try to use on every test all year is test - correction - recall quiz.

Consider the exam I gave last week. One of the questions was #3 from the 2007 AP Physics B exam. I can't post the question itself here, but you can get it at the College Board's archive of AP physics B exam questions. The problem shows a simple circuit and asks for a ranking of resistors in order of their current and voltage. Then, a capacitor replaces a resistor; students are asked to calculate current through the resistors and voltage across the capacitor.

The Test
On the test itself, most of my students get the ranking task and the calculation. The difficulty comes when the capacitor is added. Still, my class averaged 10.9 out of 15 points. That's 73%, or well into the range for an AP score of 5. (That's also a full standard deviation above the national average of 6.0 points. Statistics for many exam questions can be found at the archive page linked above.)

The Correction
In a test correction, students earn back half the points they originally missed by redoing the lettered parts they got wrong. Of course, I sometimes ask an additional question, or change the given values, so that they can't just parrot, they have to think about what they did wrong.

The most common mistake on the voltage ranking task is to assume that the first resistor takes the most voltage simply because it's first in line; or, to assume that the first resistor takes all 12 volts of the battery. Thus, for part (b) I ask the additional question:

First, draw me an example of a circuit in which the FIRST resistor does NOT experience the largest voltage across it. Explain your answer.

Now, justify the ranking you gave to the three voltages.

For the calculations in part (c) and (d), I simply change the R value from 100 ohms to 200 ohms so that they'll have to pay attention when they redo the circuit problem.
The most extensive part of the correction is for part (e), calculating the charge on the capacitor -- because that's the part that most people missed. I use the original question, but I add the following parts:
i. Diagram the new circuit.

ii. What does a capacitor do in a circuit?

iii. Explain how to figure out the voltage across the capacitor.

iv. Now calculate the charge on the capacitor.

The Recall Quiz
You might well be familiar with my "fundamentals quizzes," which test the basic facts that must be memorized before higher-level physics problem solving can happen. The third step in my testing process is to remind my class of the fundamental physcis facts behind each question. The questions on these recall quizzes are a bit too involved to ask on my regular fundamentals quizzes, because I assume familarity with the problem. The recall quizzes usually only ask about the most-missed issues on the original test -- there's no use beating a dead horse. Thus, the the recall quiz dealing with AP Physics B 2007 #3 only discusses the capacitor:

A capacitor is in parallel with RC in the circuit shown above.

i. Check one and explain briefly: After a long time, the voltage across the capacitor is

☐ Greater than e
☐ Less than e
☐ Equal to e

ii. Rank the current through the three items RA, RC, and C from greatest to least, with number 1 being greatest. If two items have the same current, give them the same ranking. Justify your ranking briefly.
____ RA ____ RC _____ C

iii. Rank the voltage across the three items RA, RC, and C from greatest to least, with number 1 being greatest. If two items have the same voltage, give them the same ranking. Justify your ranking briefly.
____ RA ____ RC _____ C