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## 27 November 2010

### Corrections to the AP Physics Sunken Ocean Liner Problem

 Picture of the Andrea Doria from the Associated Press
The 2004 AP Physics B exam includes a clever problem combining kinematics and static fluids.  That's problem 2, linked here.  I assigned this problem on my trimester exam, and got reasonably good results.

Anytime you're teaching a class of 37 students, though, you're going to have frustrations grading their exams.  SOMEONE is going to make the silly mistake you warned them about a million times.  Someone, somewhere, is going to lose five points out of fifteen because they didn't think properly under pressure, even though that same person will later slap his head and say "d'oh!"

I make my students correct their mistakes on all tests and exams.  The corrections process is not only pedagogically sound, it is cathartic to the physics teacher's soul, as well.  Some of the questions I ask on corrections are to point out the most common misconceptions, so that students can eliminate said misconceptions from their brains.  Other questions are designed less as a teaching tool than as a deterrent to future silliness.  When I grade the exams, and I scream to an empty room, "How can you fail to put units on an answer after three months of AP Physics?!!!!!" I find that I feel much better by designing a "deterrent" style test correction.  Here's what I mean.

Part (a) of 2004 B2 asked students to find the gauge pressure at the bottom of the ocean, given an absolute pressure of 413 atmospheres.  Those who missed this question usually couldn't recall the meaning of "gauge pressure."  So, the correction is pretty simple:

(a) Don't solve, just tell me briefly what is the difference between gauge and absolute pressure.

Part (b) required students to use the absolute pressure and the equation for pressure in a static fluid column to find the depth of the sunken ocean liner.  The most common mistake was to fail to convert properly from atmospheres to pascals.  That's not a huge deal to me.  Of greater consequence are the numerous students who get an answer of 41 m for the depth of the ocean, and don't recognize why that's so silly.  (The ocean liner itself is probably longer than that!)  So all I ask is:

(b) Explain in no more than two sentences why the answer “41 m” is physically unreasonable.

But the other major problem with parts (a) and (b), and with ALL parts of this question, is the students who write down 8 significant digits, or who fail to put units on their answers.  My students certainly know how to use proper sig figs, and they have been called out numerous times for lack of units.  On an exam, units and signifcant figures are a matter of focus, not of true understanding.  So, here's the correction I use.  Imagine reading this correction with an evil cackle:

If you lost points for units or significant figures here, or anywhere in this problem, write out in your own handwriting:  From now on I will be sure to write units on every numerical answer, and I will limit my answer to two or three significant figures unless the problem requires more significant figures.  And, the value “41,205,100 Pa” has six significant figures, which is too many.”

Ah, I feel better now.  And it works -- I've heard students saying to one another, "Hey, don't worry, I had to write that after our earlier test, and I made sure to put units on every answer this time."

The only other additional question I ask students in correcting this problem is in the last part, which asks for the time for the ocean liner to fall to the bottom.  Some students set up a single kinematics chart, with an initial velocity of 0 m/s, a final velocity of 10 m/s, and a distance equal to the depth of the ocean.  Problem is, they were told that 10 m/s is the terminal velocity of the ocean liner, which was reached after falling for 30 s.  So it's a two-step problem -- find how far the liner falls in the first 30 s while it's accelerating, then use constant-speed kinematics to find the time to fall the rest of the way.  I ask students to re-solve this problem, but first, they answer:

To find the time of sinking, why can’t you just make a kinematics chart with vo = 0, a = 0.3 m/s2, and Dx = 4200 m?

The goal, of course, is for students who got a part wrong to figure out their mistakes.  Just asking students to redo the same questions without acknowledging their misconceptions or articulating in words their mistakes is, while better than NOT doing corrections, insufficient.  These kinds of additional questions make corrections engaging and effective.

## 20 November 2010

### The blessing and curse of an online timer

Whenever I give my classes a test or a quiz, I put a digital countdown of the time remaining up on the projector screen.  I found a nice web timer at

This timer is customizable as to the size and color of the digits, whether it counts up or down, even the sound or music that it plays at expiration. I like to set it for the hockey horn (HOOOOOOOOOOOOOOONK!) but often my students change to something a bit wimpier, like Pachelbel.  Go figure.

Particularly in my AP class, in which managing time on tests is a major issue, the timer display is a great teaching tool -- my guys have a clear knoweldge of the time remaining all the, well, time.  They get a kinesthetic sense of just how fast 80-seconds-per-multiple-choice-item is.  If nothing else, the timer saves arguments.  No one EVER says, "Can I have just one more minute?"  or "I thought we still had two minutes left."

Even on occasions when I've given a test or quiz with a substitute present, the students take it upon themselves to put the timer up on screen.  It giveth them comfort.  And forgetting that fact was where I screwed up today.

We took our first trimester exam, but in an unfamiliar setting -- all of my students can't fit in my classroom.  I couldn't project the online timer.  "So what," I thought.  These guys are used to taking timed tests, and most have watches.  I reminded them before the start that the multiple choice section was 30 minutes long, and that they should manage their time to get to every problem for sure.

When the 30 minutes were over, even some of my best students looked bewildered.  They explained that they had lost track of time, because the screen didn't count down.  Thing is, EVEN SOME OF THE STUDENTS WHO HAD BEEN LOOKING AT THEIR WATCHES said they were confused about the time.  I've trained these folks to expect a scoreboard-style timer; they did not function well with a different sort of scoreboard.

For the free response section, they quickly arranged for a reliable student with a watch to call out time in ten-minute intervals.  I think they still wanted the projected timer. I mean, I'm glad the cutesy little timer is so useful, but I had never expected how inseparably reliant my class had become upon it.

GCJ

## 16 November 2010

### Buoyant Force problem and demonstration with a crazy answer

A fluid mechanics problem I often assign is based on a problem from, I think, one of the older Serway editions.  It shows a beaker full of oil sitting on a platform scale.  A bar of iron is suspended in the oil from a rope which is attached to a spring scale.  The problem asks:

A 1.0 kg beaker containing 2.0 kg of oil (density = 916 kg/m3) rests on a platform scale. A 2.0 kg block of iron is suspended from a spring scale and is completely submerged in the oil.

(a) Which scale reading should be larger (or should they be the same)? Explain conceptually.

(b) When the iron is in equilibrium, what is the reading in the spring scale?

(c) When the iron is in equilibrium, what is the reading on the platform scale?

Most everyone gets the idea that the platform scale reads a bigger force -- after all, even without considering anything tricky (like fluid mechanics), the spring scale seems to read just the 20 N weight of the iron, while the platform scale seems to read the 30 N weight of the beaker/oil.  A bit more logic with buoyant forces convinces the students that the spring scale must read LESS THAN 20 N, because of the upward buoyant force on the iron.  No problem.

Part (b) is similar to a demonstration from class, and numerous example and practice problems in texts.  They know to draw a free body, calculate the buoyant force using Archimides' principle, and use the free body to calculate the tension in the string connected to the scale.  The only halfway tricky part is finding the volume of the iron, which is easily done once the density of iron is looked up.  The buoyant force is about 2 N in this case.

Part (c) is the part that causes trouble.  Most of the class, at least initially, says that the reading on the platform scale is just 30 N -- the weight of the oil plus the weight of the beaker.  Others get the right answer of 32 N, but for crazy reasons.  Some come to the conclusion that since the oil "lost" the 2 N buoyant force, that we must return these 2 N to the oil through the reading on the platform scale by conservation of force.  Others simply draw the buoyant force acting down directly on the beaker.  Many make no argument whatsoever, but just add in 2 N, presumably because their friends told them to and they couldn't quite explain it.

I'm glad that so many students have the physics instincts to recognize that 30 N can't be right.  A few will say they made a lucky guess, but I consider such a guess good physics intuition.  However, only a very few students get the justification for why the platform scale reads 32 N.  Do you know?

It's Newton's Third Law.

The buoyant force is the upward force of the oil on the iron.  Therefore, there must be a downward force of the iron on the oil.  When we consider the oil-beaker system, the downward forces sum to 32 N, including the weights of the oil and beaker, and the third law companion force to the buoyant force.

Do you believe me?

My students don't, at least not if they didn't get the answer right in the first place. So I set up a similar situation.  The picture at the top (credit to Frederic Lamontagne, WFS class of '11, for the photography) shows a beaker containing a submerged aluminum weight, just like in the problem.  When I remove the weight from the water, the reading in the spring scale increases, but the balance scale goes out of balance!  I have to rebalance the scale to make up for the removal of that downward force of the aluminum on the water.  Since the spring scale reading increased by 0.2 N, I had to add about 20 g to the balance scale reading.  Physics works.

## 10 November 2010

### HOW MUCH? The heck you say.

We're studying circular motion in regular physics, and I'm preparing my laboratory activity for the week after Thanksgiving break.  (Not sooner -- we have our trimester exams next week.)  I want to do the "swing a stopper on a string in a horizontal circle above your head" experiment, a classic developed in the PSSC era.  I discovered my setups for this experiment to be a tangled mess, with numerous missing pieces, broken strings, and not enough hollow tubes for the string to go through in any case.  Now you can tell why I haven't done this experiment with my class in about five years.

I looked on the PASCO site, hoping to find a reasonably priced set of hollow tubes with stoppers and the light, low friction thread that leads to quality data.  And what to my eyes did appear:  \$39 big ones for a set of five stoppers, two tubes, ten zip ties, and some regular old string.

Okay, my department's budget is nearly unlimited.  I can -- and do -- order any equipment I want or need for my class.  Nevertheless, there's something to be said for intelligent use of resources.  \$39 for items available in the storeroom?  Neither frugal nor intelligent.  This is why a former Chief Reader for the AP exam defined "Pasco" as a Latin verb meaning "to rob."*

* Before the Pasco Police come my way, please note that I am a HUGE customer, and a huge supporter of the company in general.  They sent me two loaner heat engines for my summer institutes -- no charge, no hassle, no problem.  (Of course, I probably garnered them 5-10 orders for said heat engines, so they got their money's worth.)  I tell anyone who will listen how reliable PASCO's products are, and how good their technical support is.  But the downside:  they're expensive.  And in this case, obnoxiously expensive.

I had no trouble finding stoppers, thread, and zip ties.  I'm going to use thread from Burrito Girl's* sewing kit rather than regular string; the chemistry department has stoppers of all sizes.  The trick was finding the hollow tubes without a trip to the hardware store.

* Burrito Girl is my wife and sidekick.

My classroommate Alex Tisch looked at the picture, and offered up a suggestion that would make the editors of the Tightwad Gazette croon:  what about a BIC pen with the ink part removed?  Two decades ago I used to take apart these pens when I was bored in class... now I could use that experience to save my department some dough.

In the event, I used a papermate brand pen.  The pen-tip is connected to a thin tube of ink, all of which can be removed from the pen casing easily; the cap on the other side took some wedging, but I got it out with a fingernail in less than one minute.  Voila, a "hollow tube," at a cost of about a quarter.

(Oh, you want to know about the actual experiment?  Attach the stopper to the string with the zip-tie, thread the string through the tube, and hang a mass from the bottom end of the string.  Hold the tube and swing the stopper in a horizontal circle at constant speed such that the hanging mass hangs in equilibrium.  The radius of circular motion can be measured with a ruler.  The speed of the mass can be determined with a stopwatch, knowing that speed is circumference divided by the time for one revolution.  A graph of speed squared on the vertical axis and radius on the horizontal axis yields a line whose slope is the centripetal acceleration of the stopper.  This acceleration can be shown to be equal to g times the ratio of the hanging mass to the stopper mass.)

GCJ

## 05 November 2010

### General Physics Lab: Time of Flight for a Projectile

I've been working this year on developing laboratory exercises for general physics that involve simple data collection and straight lines.  I want to drill the idea of taking a slope of a best-fit line, and interpreting the physical meaning of that slope.

I've already discussed the classic ball-off-of-a-table projectile lab, for which David Moore recommended using a photogate to measure the ball's horizontal speed.  The traditional experiment controls for the horizontal speed, and asks students to predict the landing spot on the floor.

I tried something different, because I wanted a graph.  I had the students measure the initial horizontal speed with the photogates... but then I had them measure the horizontal distance that the ball travels after flying off of the table.  They could make these measurements for a wide range of initial ball speeds.  I asked them to graph the landing distance on the veritical axis of a graph, and the velocity read from the photogates on the horizontal axis.

My class did a pretty good job of data collection. At this point, they're already quite good at drawing a best-fit line, at calculating the slope of the best-fit using two points on the line that are not data points.  Some even put proper units on the slope.  (The idea that a slope has units is, for some reason, a difficult concept to get across.)

What we're NOT yet good at is understanding the physical meaning of the slope of an experimental graph.

When I ask someone what the slope of this graph means, he invariably says "the slope is the change in the distance divided by the change in the speed."  Well, sure, but that's a mathematical answer.  I know that a slope is rise over run.  What's interesting and exciting is that the slope has a meaning beyond rise over run, which can only be determined with reference to the relevant equation.

The horizontal speed is constant; so speed = distance / time.  Some algebra rearranges this to time = distance / speed.  Well, distance / speed in this case is rise / run -- The slope is the time for the ball to fall to the ground.

One student's data is shown above.  The slope of his graph was 0.37 s.  The table from which he launched the ball was 72 cm high, predicting a time of fall of 0.38 s -- not bad, eh?

GCJ

## 03 November 2010

### Circular motion demonstration: turntable

When I arrived at Woodberry Forest eleven years ago, I discovered several 1970s-vintage turntables in the storeroom.  I turned one into a nice, quantitative, circular motion demonstration.

In the picture you see the turntable, with five identical brass masses placed on top.  Before I start the demonstration, I show the class that a brass mass placed right near the center stays put when the table rotates; but the same mass placed near the outer edge flies off.  The goal of the demonstration is to predict the maximum radius for which the mass will not fly off.

The friction force on the brass mass acts toward the center of the rotation, and is a centripetal force equal to mv2/r.  The variable r represents the radius of the circle -- that's what we're looking for here.

Problem is, the speed v itself depends on r, because a mass near the edge covers a larger distance in the same time as a mass near the center.  So we write v in terms of the period of revolution, Tv = 2πr/T.  The period of revolution is measured with a stopwatch.

The friction force can be set equal to μFn.  The normal force on the mass is simply its weight.  The coefficient of (static) friction must be measured... but I can do that with a spring scale or a force probe.  I pull a stack of brass masses with the scale.  I divide the reading in the scale just as the masses start to move by the weight of the stack of masses.

(Here the class can be asked whether the coefficient of friction for a single mass will be greater, less than, or the same as that for the stack of masses.  Answer:  the same, because we still have brass in contact with the same surface.  I only use a stack of masses to make the spring scale reading easier to obtain.)

So now we know everthing in the relevant equations necessary to solve for the radius r.  We find that radius to be in the neighborhood of 4-5 cm.  That's in the middle of the second mass, as measured from the center.

What does this mean experimentally?  When I turn the turntable on, the outer masses should fly off, but the first one or two should stay put.  And sure enough, that's what happens -- physics works.

Follow-up:  Next I replace the 20 g brass masses with 10 g brass masses from the same set.  Which ones should fly off now?  Answer: still, the masses more than 4-5 cm from the center will fly off, the others will stay put.  The coefficient of friction still hasn't changed, because we're using the same surfaces.  Mass cancels out when solving in variables, so the mass doesn't matter.  Sure enough, the 10 g masses fly off outside of 4-5 cm, as predicted.

GCJ

## 02 November 2010

### Grousing about Newton's Third Law

I've heard all sorts of bright ideas about teaching Newton's Third Law.  I've even tried a few myself.

One gentleman, whom I saw at a Florida AAPT meeting back in 1997, actually brought out a wooden model of a pair -- the fruity kind.  He explained that "forces come in pairs..." and opened up the pair to show, inside, two force vectors labeled F (A on B) and -F (B on A).

That's clever and laugh provoking, but not particularly useful.  It's not getting students to remember Newton's Third Law that's tough -- it's getting them to use it correctly.  Me, I tend to shy away from the videos, from the cutesy pictures of faces being punched (and delivering a punch to the fist!), and so on.  But I certainly don't have a magic formula for teaching this simultaneously easiest and toughest of laws.

As an example of the uphill battle we face teaching the third law, I'll tell a story of my own inadequacy.  I'd love some advice.

I taught the third law this year to my general physics class.  In the past I've restricted the third law to my AP sections, but I decided to give it a go... and that go turned into a full stop.

My classroom presentation was, I thought, strong and straightforward.  I wrote Newton's Third Law on the board as "the force of A on B equals the force of B on A."  In discussion with the class, we emphasized the difference between an object experiencing a force and an object applying a force.

One useful technique I have discovered is to insist that all forces be described in what I call "Newton's Third Law Language" before approaching a question.  When asked for the third law pair of the "normal force," students are up a creek; but when asked for the third law pair to the "force of the table on the block" they can handle themselves.

So in class we practiced describing forces in Newton's Third Law Language, and then finding the third law pair.  I used check-your-neighbor questions, I called on some folks randomly, I led a discussion, answered questions... all the usual techniques to make physics lecture effective.  I even specifically addressed the companion forces to an object's weight, and to a normal force.

(The one thing I didn't do is the quantitative demonstration -- that's coming shortly.  I'm attaching two force probes to two different-mass carts, and showing that the probes read the same thing when the carts collide with each other.)

On the nightly homework, I began with a conceptual question:  Can a Newton's Third Law force pair act on the same object?  Everyone got this right.  Good start, I thought.

The next homework question asked to identify two equal and opposite forces that were NOT third law pairs.  Students struggled with this one -- many said that such a thing cannot exist.  Maybe I should have asked that question later.

Finally, I showed a kid dragging a sled across a rough surface.  I asked for the Newton's Third Law companion force to the normal force, the weight, the tension in the string, and friction.  Remember, we had done two of these THE SAME DAY, IN CLASS.

Do you know, only one person out of 22 got all of these correct? Well, I know.  Their performance stunk.  They all said that the normal force's companion was the weight; that the tension's companion was friction.  BOUX!!!!!

I have confidence that, after enough iterations, most of my class will eventually be able to answer Newton's Third Law questions correctly.  I know that I have to be patient, I have to make the class confront their misconceptions by getting problems like this wrong, I have to ask the same questions in as many different ways as possible.