## Tuesday, May 31, 2011

### Test Preparation (or not)

So, I threw the honors physics classes a bit of a curveball last week, on the day of the big (scary) test that was to cover electrostatics (static electricity, E fields, and potential), magnetism, and circuits.  The material's less familiar to students than that of mechanics, and more abstract, which is where the scary part comes in.

The test was not an "I ask you questions, you give me answers" form, but instead was more of a "show me that you understand some subset of the standards that we covered in these units."  It's very open-ended, and students couldn't get to every standard that we covered, but there are definite advantages to the "let me show you what I know" approach.  I like this type of format better as a reassessment, but I thought, given the crazy schedules of students and teachers in May and the general tone of the class, that this was a good solution for this particular day.

Ordinarily, I don't like to throw big wrenches into the process like this; if I'm going to radically change the test format for a test, I'll give advance notice, but this seemed like a better solution overall at the moment.

Before the test, however, I asked the students to fill out a short questionnaire about how they had prepared for the (normal format, as far as they knew) test.

The first part asked them to allocate their test prep time into percentages, based on the activities that they used to prepare for the test.

• Reading my old notes from class

• Reading the emailed notes from class

• Getting help from peers

• Helping peers

• Getting help from the teacher

• Reassessments

• Reworking old problems

• Working new problems

• Other: ________________

• The results were:

* The "other" on the chart is mostly "worked new problems," along with two write-ins of conceptual review and outside websites

There are some interesting trends here:
• I'd call about 52% of these activities (working problems, working with me, helping each other) active, which means that they're good things to do.
• 48% are therefore passive.  These types of work (reviewing notes, text, etc.) are particularly troublesome, because students feel like they're learning, but research shows really few returns from those hours
• The most common (by student number) were reading text and their own notes, with helping each other and reworking old problems very high as well.  Reading the notes that I send out wasn't as high, but some students really like using them, so I'll keep that up.
• I haven't done as good of a job encouraging reassessment as test prep as I should've.  Some students figure out that doing a little reassessment before the test makes the test go much better, and ultimately reduces the need for reassessment later, but most principally reassess after tests (and, really, most reassess right before the end of the term).  There's a grade focus that I'm not a fan of, but I can do a better job in the education process earlier in the year.
• Working new problems was really infrequently reported.  That's a problem.  If you have really only seen, say, conservation of energy applied in a dozen or two situations, you might start thinking that those are the only ways that it can be applied, or not have enough data to see the general relationship underlying the specific problems.  I think that there's an aspect of this  held over from chemistry, where students often get the idea that there's some finite "number of problems" that you can memorize the steps to (...and that this is learning).  Sometimes, you can get through chemistry like this, but never, ever through physics this way.  Perhaps doing more goal-less problems will help students get the idea that asking your own questions is, you know, allowed (and good!).

## Saturday, May 21, 2011

### Two Months in the Temple (Part 2)

Willie's Wheel of Fortune

The Scene: Willie is in a cage and being lowered into the fire pit by Indy (under a spell)

The Question: It seems as though when Willie is being pulled back out of the fire pit, she must be moving faster (because it takes so much less time) than when she is being lowered. Does this make sense if Indy is the only person pulling Willie (and her cage) back up?

Game Plan: We are scaling the wheel (using Ford's height and Logger Pro) and figuring out how long one rotation of the wheel takes when Willie is being lowered (to find her velocity) using kinematics. We will do the same thing while she is being pulled back up. We will compare the two velocities and see if they are reasonable speeds for Willie to be traveling at to determine whether or not movie magic or real physics was used. Kinematics will be the best route because we need to find velocity and Willie's displacement on each trip.

Indiana Jones and the Raft of Doom

The Scene: The scene begins after Indy and co. jump out of the plan on a raft and land on the side of a snowy mountain.  The group slides down the mountain on the snow, and then slides on the dirt.  They continue to slide on the dirt until they slide off the cliff.

The Question: The goal is to find the coefficient of kinetic friction for the snow and dirt as the raft flies over it. If we have time, we will find the minimum coefficient of kinetic friction needed to stop Indy and friends from falling off the cliff.

The Game Plan: From the video, we measured positions and times to get velocity of the raft, and we will use kinematics, a force diagram, and a net force equation to determine the coefficient of kinetic friction.

It's Gotta Be the Shoes!

The Scene: Mine cart scene where Indy uses his foot as a brake to stop the cart

The Question: We want to find whether or not Indiana Jones’ leather soled shoes would catch on fire if he used his shoe as a brake.

The Game Plan: We need to find the initial velocity, how fast the wheels are rotating per minute via initial velocity, the μk, the mass of Indy's cart, and the angle of Indy's shoe compared to the ground.  From the Fnets and work energy theorems, we shall find the force of friction, then using the specific heat equation, we will find the minimum energy needed for the shoes to catch fire.  Then, we will compare the energy converted to heat between Indy's shoes and the wheel to the minimum energy needed.

Don't Drag Me Down

The Scene: The raft falls from the cliff!

The Question: Would they have survived falling from that altitude and hitting the water only in a raft?

The Game Plan:

## Monday, May 9, 2011

### Two Months in the Temple (Part 1)

The Honors Physics classes are in the midst of an extended project analyzing Indiana Jones and the Temple of Doom's physics: is it real or movie magic?  There's a huge amount of physics here, both obvious (falling rafts, mine car chases, etc.) and implicit (how does that hat stay on his head?, ...just how deep was that pit, anyway?).

They formed themselves into groups, watched the movie, and picked scenes (hopefully, not in that order).

In a change from previous years, I spread out the work: a couple of days here, and a couple there, with longish (a few weeks) breaks in between.  This allows some time for groups whose initial idea didn't pan out to either get it to work or to move on to something else.  I like this model, and I'll use it in the future.

We'll do virtual posters to wrap the project up - stay tuned!

After writing proposals (best proposal gets the scene, if there's a conflict!), the groups have set out upon these quixotic quests (in their own words):

Sweating Bullets

The Scene: Mine cart chase; Indy shoots at the pursuing mine cart

The Question:  A clip in the movie shows two carts moving along the same path. The cart behind the other is rapidly firing bullets at the cart ahead. Both carts continue to move at the same speed. Our group is trying to prove the carts will not be moving the same speed after the force of firing the bullets. After firing 50 bullets, how fast will the cart who fired the bullets be moving? If we assume several of the bullets hit the cart ahead, will that cart increase speed and by how much?

The Game Plan: From the movie we are measuring the initial velocity, final velocity, and number of bullets fired. We will use conservation of momentum to calculate what the actual final velocity is and then we can determine whether or not there was movie magic or it was real phyics. We are using this method because conservation of momentum is good when there are collisions, because there are no net external forces horizontally. The firing of a bullets are like "reverse" collisions.

Bridge Over Troubled Water

The Scene: Rope bridge scene (as it falls after Indy cuts it)

The Question and Game Plan:  We are analyzing the impact received by Indiana Jones when hitting the wall.  We can measure the length of the rope and use this and height to find a final velocity.  Once we have this velocity we can use that and our estimated impact time to calculate the force he receives. We will then determine if this force received is lethal.

Short Round in Dragtown

The Scene: The plane that Indiana Jones, Willy, and Shorty are on malfunctions and is about to crash; they jump out of the crashing plane on an inflatable raft and ride the raft down to the ground.

The Questions:  What is the area of the raft (calculated from the terminal velocity)? Is this area reasonable compared to apparent area of the raft from the video?

The Game Plan: By estimating and measuring the other variables in the terminal velocity equation, we are able to calculate the area.  By finding the terminal velocity using Logger Pro and the video, we will be able to determine if the area of the raft is realistic and accurate according to the calculations.

Whip Swinger

The Scene: Indy swings from one catwalk to the other with a whip during the escape from the mine

The Questions:  What is the greatest net force exerted on the lamp by support Indiana Jones (whip's mass is omitted)?  Also, what is the coefficient of static friction between I. Jone's hand and the whip?

The Game Plan: Using Logger Pro to measure the velocity at the point where the greatest force is exerted (straight down), we can use v^2/r to determine his acceleration. Next we will substitute that into our Fnet equations to find the total force.  For the whip, we will use the same location and velocity to find the coefficient. We will research both the average grip strength of an adult male, and Harrison Ford's mass and height (estimating Indy's grip strength as 3x as strong as an average adult male), and we will use this information to determine the coefficient of static friction.

Out of Sight, Out of Mine

The Scene: During the mine cart chase, Indy's cart jumps a gap

The Questions: Was this jump realistic?  We are using LoggerPro to acquire a scaled initial velocity, height, distance, and final velocity.  Due to the fact that the camera angle changes, we must also use trigonometry to get not only the mine cart’s dimensions, but the initial and final velocities as well. Kinematics allows us to analyze both a predicted landing spot (if different from that in the movie) and the time; it is obviously movie magic if kinematics equations provide that the time elapsed was simply an impossible amount during which the cart could’ve jumped the gap. Then it can be determined whether physics supports Indiana Jones’ flawless jump, or if it was just another Hollywood hoax.

## Saturday, May 7, 2011

### What Does That Graph Tell You (Part 2)?

OK, last time we talked about the magnification and image location vs. object location graphs for plane mirror images.  These aren't too bad, because one's constant and the other's linear.  Both have nice domains - there's an image for every (non-negative) object location.

Let's go a set up the ladder, to convex mirrors.  In the hierarchy of optical complication, plane is low on the totem pole, followed by diverging (lenses or mirrors) and then converging (lenses or mirrors).  With plane mirrors, the images are all virtual, upright, the same size as the object, and the same distance behind the mirror as the object was in front of it.  With converging lenses or mirrors, you can get upright, inverted, enlarged, diminished, same-size, real, virtual, or non-existent images.

Diverging mirrors and lenses have a more varied assortment of images than plane mirrors, but are less complex than converging lenses and mirrors.

Let's start with magnification, and let's start with some ray tracing.  Students got into groups and looked at a variety of object placements, determining the locations and sizes of the images with ray diagrams.  All of the images were upright, and all were virtual - that is, all of the refracted rays appeared to come from locations that they, in fact, never came from.  We couldn't project these images on a screen.  It's nice when something's absolute!

Here's the magnification data:

That's not a super-clear trend; it's a complicated relationship, it seems.  Two things can make it easier, though.  Let's think about the vertical intercept.  Remember that the question "what's the vertical intercept" is equivalent to the question "what is the vertical quantity's value when the horizontal quantity is zero?"  In this case:

"What is the magnification when the object is right on the mirror?"

If we include that point, it can help us not only place the curve, but determine what kind of curve it is.  Here, the image is the same size as the object (magnification: 1) when the object's right on the mirror.  This means that it's not a straight inverse or inverse-square function, because those don't have vertical intercepts.  Adding this point will help, but let's think about the other side of the graph, too.

What's the long-term behavior of the graph?  The question here is: "what happens to the vertical quantity when the horizontal quantity gets very large?"  Here,

"What happens to the magnification when the object gets far away from the lens?"

Looking at the furthest objects away, the ray tracings start to show the images getting smaller and smaller, closer to zero.  There's a horizontal asymptote here, because the graph won't ever cross that axis.  If it did, that means that the image will be upside down, but all of our images are upright.

Check out the graph, now with intercept value and another point further away:

Trend looking a bit more clear?  ...it'll be something like this:
That function is approaching zero as that image gets really far away.  Not linear, but not too bad.

The image distance vs. object distance graph is a bit more complex.  In the first place, the values are negative (that's the easiest mistake to make here!), because the images are virtual.  Here's a bit of data:

The intercept?  Well, what's the image position when the object's right on the mirror?  It's not hard to try it out.  When you do it, the image is right, there, too: at position 0 when the object is.  That's a vertical intercept of zero, and it adds to the picture.

Looking at the furthest objects away, the ray tracings start to show the images getting closer and closer to the focus.  What's it like when the vertical variable keeps getting closer and closer to some value as the horizontal one keeping increasing?  That, my friends, is a horizontal asymptote, and we have one at f.   Remember that f is a negative number for diverging mirrors (and lenses)!

Put it all together, and what do you get?

Should you be able to just draw that off of the top of your head?  Probably not.  Should you be able to reason through it, step by step, and piece that together one bit at a time, thinking about each step individually and logically?

You bet.

## Wednesday, May 4, 2011

### The Ball, She Rolls...

Here's a great old AP problem for putting together some pieces:

A solid sphere of radius R and mass M is initially at rest in the position on an inclined plane, such that the lowest point of the sphere is a vertical height h above the base of the plane.  The sphere is released and rolls down the plane without slipping. The moment of inertia of the sphere about an axis through its center is 2MR2/5.   Express your answers in terms of M, R. h, g, and the angle.
a.         Determine the following for the sphere when it is at the bottom of the plane:
i.  Its translational kinetic energy
ii. Its rotational kinetic energy

Let's see, let's see...

OK, there's one connection: we can express one kinetic energy (rotational or translational) in terms of the other, through that wonderful pivot:
This only works for us when we're in that enviable "rolling without slipping" condition.  This is a cool connection, and super-useful, but it's old hat for us at this point.

What about the other kinetic energy?  Well, we could work the pivot the other way, or... think about it a little!  It started with some energy: mgh.  We just solved to see that some fraction of that (about 28%) is invested in making the sphere roll.  Where's the other five-sevenths?

It didn't turn into pumpkin pie, I'll tell you that.  This is where the kids were champs today.  Below, you see them solving for the rotational KE  - the one that they solved for first.  After that?

No prevarication, no lengthy algebra.  "Duh, it's the rest."  That's good physics thinking.

Never forget that the math is just another way to write down reasoning.

If you forget that, then you're just pushing letters.

Next part:
b.         Determine the following for the sphere when it is on the plane.
i.  Its linear acceleration
ii. The magnitude of the frictional force acting on it

There's again a connection between rotation and translation, and this time it's between the acceleration and angular acceleration:

There was a bit more algebra here, but... did there have to be?

...

Think about it for a second.  Cheat a bit: look at the answers, and think again.

We already know that the kinetic energies are proportional to the height change from top to bottom, and we also know that gravity causes the ball to move (put the rotational axis on the point of contact between the ramp and the ball if you don't believe me) and that static friction causes it to rotate.

Here are the hardest links in the chain:

- Each kinetic energy increases at a constant rate (with respect to distance), because the forces doing the work are constant and the angle between the force and displacement is constant.
- Because the kinetic energies increase at a constant rate, the energy that ends up being 2.5 times the other (5/7 mgh versus 2/7 mgh) must increase at 2.5 times the rate.  That is, the net force down the plane is 2.5 times the friction force.
- A tiny bit of algebra or (better) some thinking about the fractions that the KE split into will reveal the forces (net force and the friction force).  Each is a fraction of the weight's component down the hill: mg sin(theta).

No, it's still not done.  Yes, this is too much for a problem that you only get 15 minutes for on a stress-filled AP exam day.  Boo, College Board.

The solid sphere is replaced by a hollow sphere of identical radius R and mass M. The hollow sphere, which is released from the same location as the solid sphere, rolls down the incline without slipping.
c.         What is the total kinetic energy of the hollow sphere at the bottom of the plane?
d.               State whether the rotational kinetic energy of the hollow sphere is greater than, less than, or equal to that of the solid sphere at the bottom of the plane. Justify your answer.

I'm guessing that c. is just there to clue us in if we haven't been thinking all along (instead, being mindless algebra drones).  From a test pedagogy standpoint, I don't like this - the AP exam tends to lead students by the nose, rarely requiring them to really put together any sort of chain of reasoning.  It makes the problems easier to score consistently, though, which is (I'm sure) the purpose.

d. is a good one, though.  There are a couple of steps in the chain, which is a rare treat for the AP.

- The mass in a hollow sphere is, on average, further from the axis than in the solid sphere.
- The rotational inertia of the hollow sphere is greater than that of the solid sphere.
- If the two balls were rolling at the same speed, the hollow one would have more rotational KE, but the same translational KE.
- The two balls must, however have the same total KE at the bottom (mgh!).
- The hollow ball must therefore have a great fraction of its total KE as rotational KE (and less as translational, so it's going slower).

Try a race!  Bigger and more hollow will slow down a ball; racquetballs lose to pool balls and marbles!

## Tuesday, May 3, 2011

### Clearboards

It's almost time for the AP exam.  Our intrepid AP Physics C: Mechanics students took to the school's lobby windows with whiteboard markers for a little review.  Basically, I threw classic derivations and popular situations from the AP exam at them, and they showed (everyone) what they knew.

Those drag integrals were a bit rusty, but I think that we got them worked out - they show up frequently on the AP exam!

BTW, these are a beast to take pictures of!