Happy Thanksgiving from a different sort of family...

 

Although I thoroughly enjoy hunting for hand turkeys on AP exams with my friend Jen, I don't actually like turkey.  And the other two members of my household are both vegetarian.  So the main course of our Thanksgiving dinner is shown above: Shari's macaroni and cheese, based on a recipe I discovered on a sports blog.

The best part of a trimester system is giving a major exam right before Thanksgiving break.  Invariably, my students outperform what I might have predicted a month ago - because they take this major exam quite seriously, because they have matured both physically and mentally in the early part of the year, because we spend significant time in and out of class reviewing facts and basic recall items.  And I discover these predominantly positive exam results right before the national day of thanks. 

Keep the faith, folks.  I know that most of us teaching physics don't have a major exam to see direct evidence of the great progress our classes have been making.  But I know it's there.  Maybe you have to wait until a December or January exam; maybe you'll have to manufacture a few high-stakes testing situations to assure yourself that the progress you're seeking is, in fact, there.  

This is what a "summative assessment" is for.  No, it's not torture intended to shame and expose our students failings; it's like a regular season football game, a chance for our students to see where their strengths and weaknesses lie.  And, if we've been practicing well and building strength all season, we expect we'll be seeing far more strength than weakness.

The strengths are there, and will continue to get, well, stronger throughout the year.  Eventually, you'll hear your students say "remember when these kinds of problems were hard?"  Not yet.  When we get back from break, they'll say, "wow, I did way better on this exam than I thought I would!"

Right NOW, though, they're eating and watching both kinds of football, enjoying the break from rigorous academic effort.  As am I.  Happy Thanksgiving.

Modified Atwood machine that *slows down*

The problem above is from 5 Steps to a 5: AP Physics 1, the "Elite Student Edition."  This situation provides an opportunity to find out how well students are doing with understanding the meaning of acceleration.

Ask: immediately after the carts are pushed or released, in which experiment is the magnitude of the carts' acceleration greater?  And, are the directions of the top cart's acceleration the same, or different, in the two experiments?

The correct answer refers to free body diagrams of both carts in each experiment.  The hanging block experiences forces from the rope and the earth; the only unbalanced force on the top block is that of the rope.*

*Yes, the top block also experiences balanced forces from earth and the surface.  Those aren't relevant to the acceleration, because they are balanced. 

Those free body diagrams are the same in both experiments!  By Newton's second law, the same unbalanced force acting on the same mass causes the same amount of acceleration.  And similarly, because the unbalanced force is in the same direction in each experiment, the accelerations have the same direction.  Done.

But what about the initial push?  All forces other than gravity** require contact.  While I was pushing the cart, the cart experienced a force from my hand.  After that - and the question explicitly is asking about AFTER the cart has been pushed - my hand isn't touching the cart, and so cannot apply a force.

** in mechanics, anyway

Okay, but isn't the cart on the surface moving to the right in Experiment B?  So its acceleration is in a different direction than in Experiment A, when the cart moves to the left.

Um, no - here is precisely the misconception this question is designed to test!  Motion is emphatically NOT the same thing as acceleration.  

When an object speeds up, its acceleration is the same as the direction of motion.  When an object slows down, its acceleration is opposite the direction of motion.  

In Experiment A, the cart is moving left and speeding up, giving it a leftward acceleration.  In Experiment B, the cart is moving right and slowing down.  Thus, its acceleration is opposite the direction of motion... so the acceleration is also leftward.

Mail Time: how do you rank the forces on a block dropped onto a spring?

I was asked about a ranking task in which block 1 is dropped onto block 2, which is attached to a vertical spring.  The blocks compress the spring, and come briefly to rest in the position shown in the right-hand diagram:

We're asked to rank the (magnitude of) three forces at this instant of maximum spring compression:

• Fa is the upward force of block 2 on block 1
• Fb is the downward force of block 1 on block 2
• Fc is the downward gravitational force of the earth on block 1

 Student 1's reasoning was:

We know that Fa is equal to Fb because they are a Newton's 3rd law force pair.  Because the blocks are at rest when the spring is maximally compressed, they are in equilibrium and thus forces on each block are balanced.  The forces acting on the top block are the force of block 2 upward (here called Fa), the force of the spring upward, and weight downward (here called Fc).  Setting up = down, we see that Fc has to be greater than Fa.  Final ranking: Fc > (Fa = Fb).

Which parts of this reasoning are right?  Which are wrong?

Correct: Fa and Fb are in fact a Newton's 3rd law force pair.  The force of block 2 on block 1 is equal to the force of block 1 on block 2.

Incorrect: The blocks are not in equilibrium at the instant of maximum compression.  Equilibrium means no acceleration, so no change in speed.  Here, the blocks may be at rest momentarily, but their speed is always changing!  For the blocks to be in equilibrium, they must be at rest and not changing their speed from zero.  

Incorrect: The force of the spring does not act on block 1!  All forces (in physics 1) other than gravity require contact.  The spring is not touching block 1, and so does not apply a force to block 1.

So how to solve, then?

Draw a correct free body diagram for the top block.  This has the weight downward (called Fc), and the upward contact force of block 2 on block 1 (called Fa).  The top block has an upward acceleration: when an object speeds up, its acceleration is in the direction of motion.  This block is starting to speed up and move upward, so its acceleration is also upward.*  Acceleration is in the direction of the unbalanced force - so the forces on the top block are unbalanced upward, meaning the upward Fa is greater than the downward Fc.  This, paired with the Newton's 3rd law pair, leads to the correct answer (Fa = Fb) > Fc.

*What about right before the block came to rest?  When an object slows down, its acceleration is opposite the direction of motion.  Just before the position of maximum compression, the block was slowing down and moving downwards... also giving an upward acceleration!