See that equation abovet? It's the equation governing the period of a mass-on-a-spring. At least, in conceptual physics, it is.
Heresy, you shout. You just ignored the 2π term! You can't do that!
I remember being utterly shocked by my sophomore quantum physics professor exclaiming wildly that "factors of 2 don't matter!" We were a bit miffed that some homework problem predicted an answer that was off from a book value by about a factor of 2. What we, silly newbies that we were, didn't understand was that the homework problem was showing an alternate means of approximating a well-known value. We should have been amazed that we were that close; instead, we complained of inaccuracy. It took a couple more years before the "order of magnitude estimate" became a part of our collective soul. I want to begin the incorcism process a bit earlier.
In a laboratory in which we are trying to measure to a precision of 5 or 10 percent, a factor of 6 does indeed matter. So why do I write the period of a mass on a spring like this?
I'm teaching CONCEPTUAL physics to 9th graders. Most are finishing their first year of algebra. The idea of a square root is still new to them. And they react to π about as well as a bunch of trainee umpires.
Most of the questions I ask in conceptual physics are of the form, "The mass attached to the spring is doubled. Does the period increase or decrease? By a factor of 2? Greater than a factor of 2? Less?"
Well, the heretical equation above demonstrates the dependence of the period on mass and spring constant just fine. I've eliminated the 2π and I've broken up the square root signs above and below the fraction bar for clarity's sake. I'm not in the business of teaching mathematical notation. I'm pleased if my students can recognize that doubling the mass increases the period, but doesn't double the period. And if they can sketch a graph of period vs. mass for a constant k.
My hope is that when these students attempt AP Physics 1 in their senior year, they have a head start toward the necessary conceptual understanding. They can learn how to linearize graphs, to tease out the 2π factor, in future courses. For now, my class is understanding well the RELATIONSHIPS among period, mass, and spring constant. That's my goal, here in 9th grade, even if I burn at the stake the next time I see professional physicists.
27 April 2013
22 April 2013
Mail Time: Does light have mass?
Michael Herrin writes in:
A student asked me, “Does light have mass? If it has mass, this implies it is matter. If that is so, what state of matter is it?”
My answer, though I freely admit that any area of physics in which experiments cannot be set up in the classroom is outside my expertise:
Light does NOT have mass. Light carries momentum, but the equation for momentum of a photon is p=h/λ. Light carries energy, but the equation for energy of a photon is hc/λ. No experiment or theory that I'm aware of has ever shown or postulated massive photons.
Michael follows up:
That’s what I thought, as well but wanted to make sure I got it right. I explained that yes, we can take mass and convert it to energy (mass defect, e=mc^2, etc) but to my knowledge, we haven’t done the reverse, although Einstein showed that it is theoretically possible.
Oh, but "we" have, or at least the universe has. It's called "pair production": Some energy, say after a huge collision between particles, is converted to mass of a positron-electron pair, or some other pair that conserves charge, mass-energy, etc. In fact, the energy-time uncertainty principle insists that such pairs be produced all the time, they just re-annihilate back into energy before we have, um, time to see them.
I always appreciate questions via email... Thanks, Michael!
GCJ
21 April 2013
Precise language: lose the "potential" in "potential energy"?
Although I hate traveling during the school year, I thoroughly enjoy being around good physics teachers. Yesterday's AP Physics 1 Consultant Training in Chicago provoked interesting discussion, powerful insight, plenty of arguments, and other features of effective collaboration between teachers.
I'll distill the what insight I have about the new exams eventually. The AP Physics 1 and 2 curriculum framework is 150 pages worth of dense eduspeak. It's going to take a while to parse the text and expose the intent of the authors, especially when the College Board has released so few exam items. All I can say for now: The exams are beyond excellent, and you should teach to them.
The new exams will require significant amounts of quality writing. Gone are the days when "justify your answer" meant give one sentence with an equation. Students will have 90 minutes to answer about five free response questions; so the readers will expect adroit use of verbiage in those responses.
The Committee has spent years parsing their own language, so that every statement in the curriculum framework is self-consistent, and consistent with the language on the exam, and consistent with their perception of best physics teaching practice. Precision of language is, I agree, important in teaching first-year physics; I've written much on the subject. That said, sometimes pedantic consistency conflicts with common sense... while I'm more than willing to take a stand with students on the difference between "work is done by the gas" and "work increases", I'm not willing to call out someone who says "a 1 kg mass is thrown off of a cliff.* "
*Objects and systems have a property called "mass." Thus, it is imprecise to refer to a "mass." No, we should say "an object with a mass of 1 kg is thrown off of a cliff." While that's entirely correct, and I am glad the exams will be stated precisely, I have no intention of having an argument about such a subtle difference with a first-year student.
Even in the midst of what sounds like heinous baloney, useful nuggets emerge. On the tail end of the discussion about objects with mass, Jeff Funkhouser pointed out to me that he doesn't use the term "potential energy." Part of his reasoning I'd categorize as correct but pedantic -- since potential energy only can arise as a result of a conservative force, and since a force requires an interaction between multiple objects, it's not entirely correct to say that an object "has" potential energy. Rather, a ball at the top of a cliff has energy stored via the interaction between the ball and the earth (or between the ball and the earth's gravitational field); a mass* attached to a spring has energy stored via its interaction with the stretched spring.
*sorry
The pedagogical point Jeff is making -- the point which I think is entirely correct -- is that this omnibus term "potential energy" gets in the way of understanding. Focus on the source or storage of the energy, not on the catch-all term "potential."
I'm in the midst of a 9th grade conceptual physics unit on energy. In any problem, we begin by writing in words the energy conversion: for example, "work done by a rope is converted to kinetic energy." That allows us to write the equation "Fd = (1/2)mv2." Finally, we can use semiquantitative reasoning to determine whether doubling the force of the rope doubles, more than doubles, or less than doubles the object's speed.
So consider a ball dropped from 2 m high onto a vertical spring. What's the energy conversion? I've trained my students to write "gravitational potential energy is converted to spring potential energy." But having read through stacks of papers, I see that Jeff is right. How often have I read "potential energy is converted to potential energy?" Or even with a correct statement of energy conversion, how many times have I seen the resulting equation written as "mgh = mgh"?
Enough already. The term "potential energy" is hereby banishéd from my classroom. We will instead write "gravitational energy is converted to spring energy."
14 April 2013
AP Physics 1 and 2: What questions do you have for consultants?
This coming weekend (April 19-21), I'll be attending a College Board Physics Consultant meeting / training session in Chicago. The topic: The new courses, AP Physics 1 and AP Physics 2.
For the 2013 AP Summer Institutes, the College Board is asking us to spend about 1 day out of five discussing the new exams rather than AP Physics B. In summer 2014, the algebra-based institutes will, of course, be exclusively AP Physics 1 and 2, because Physics B will no longer be offered after 2013-14.
You can find out most of what you need to know about AP Physics 1 and 2 at the official College Board physics redesign page.
The best place to look for understanding the new courses is starting on page 131 of the curriculum framework. I can't stand the majority of this document -- it's as dense as the watergate tape transcripts, with unfortunately fewer swear words -- but you can ferret out the general format and thrust of the exam questions by reading the sample problems on page 131 and beyond.
What other questions do you have about these new exams? I will have a chance this weekend to talk to people who know. What do you want me to ask them? What sort of information that's NOT on the webpage and NOT in the curriculum framework do you want to know about? Please email me, or post a comment, before Saturday; I'll try to investigate and share what I can.
GCJ
For the 2013 AP Summer Institutes, the College Board is asking us to spend about 1 day out of five discussing the new exams rather than AP Physics B. In summer 2014, the algebra-based institutes will, of course, be exclusively AP Physics 1 and 2, because Physics B will no longer be offered after 2013-14.
You can find out most of what you need to know about AP Physics 1 and 2 at the official College Board physics redesign page.
The best place to look for understanding the new courses is starting on page 131 of the curriculum framework. I can't stand the majority of this document -- it's as dense as the watergate tape transcripts, with unfortunately fewer swear words -- but you can ferret out the general format and thrust of the exam questions by reading the sample problems on page 131 and beyond.
What other questions do you have about these new exams? I will have a chance this weekend to talk to people who know. What do you want me to ask them? What sort of information that's NOT on the webpage and NOT in the curriculum framework do you want to know about? Please email me, or post a comment, before Saturday; I'll try to investigate and share what I can.
GCJ
13 April 2013
Experiment: photogate to measure the speed of a dropped cylinder
Photo credit: Robert Edwards |
I'm lucky to work at a school with four physics teachers. I'm good at a lot of stuff, but when I need help or ideas, three other experts are available in the other third-floor classrooms.
I knew I wanted to have my students measure the speed of a ball-on-a-string when dropped from various vertical heights. A motion detector can make the measurement, but it's difficult to set the detector at the right height, and even more difficult to train a ninth grader to filter out the meaningful speed from the 50 m/s reading the detector spits out when the mass rises above the detector's range.
My colleague Curtis told me to use a photogate. "That's difficult," I argued. "The ball is nearly as wide as the gate. If they don't set the photogate beam precisely at the center of the ball, they get a skewed reading, because the ball will have a different diameter intersecting the beam. And, incidentally, do you have any boy scout tricks for tying a string to a ball?"
"Nah," he said. "Use a cylindrical mass." Then no matter where the mass hits the photogate beam, it will always travel the same distance through the photogate. Woo-hoo! Solved!
It was a bit tricky to get the labquest programmed properly, but I figured it out -- use "motion" mode for the photogate with a distance equal to the diameter of the cylinder. (That's 2 cm for the cylinder pictured above.)
So, in class, I have students predict: If I double the vertical height from which the mass is dropped, what does that do to the mass's speed at the bottom? Do I double the speed? Quadruple the speed? Increase the speed by less than a factor of 2?
It takes a bit of work to determine that the speed should increase, but not double; good mathematicians see that the speed should be multiplied by the square root of 2.
The picture shows Logi measuring the vertical height from which he's going to drop the cylinder. The beauty of Curtis's setup is, the labquest just SPITS OUT a measurement of speed each time the cylinder goes through the photogate. There's no interpretation or calculation necessary, just a reading of a device. Woo-hoo!
And, boy, is this generally accurate. The students who are careful to measure vertical heights from the bottom of the cylinder consistently find that doubling the height of the drop increases the speed at the bottom... by a factor of about 1.4. And even those who are less careful still find a new speed that's greater than the original speed, but not doubled.
GCJ
10 April 2013
Collisions -- in-class lab exercises
I've had considerable success this year with 9th graders replacing quantitative demonstrations with in-class laboratory exercises. Everyone individually solves a problem at his desk; when I approve the solution, each student in turn goes to the back of the room where the solved problem serves as a prediction to an actual laboratory setup.
We've gone through this process with position-time graphs; velocity-time graphs; direction of force and motion; friction forces; forces in 2-d; collisions; impulse; and energy.
Eventually I will proofread all of these, compile them, make sure they work in multiple years, and publish them all. But for now, I offer this set of collision problems. These are particularly easy to set up if you have PASCO carts and tracks.
Posts I intend to write soon: The baseball schedule is pretty crazy right now. I'm umpiring or broadcasting five games a week for the next month. Yet I am still here and willing to write about physics! Soon, perhaps, I will get to:
* Why the in-class lab exercises have worked well in ninth grade, but I have no intention of every trying them in 12th grade classes
* a fishing line experiment with impulse
* Our overall, full-year conceptual physics outline -- would you like to take our exam?
* Any and all of your AP physics 1 questions, after I get back from the April 20-21 consultant training.
Please send any other requests via email, and I'll see what I can do.
GCJ
02 April 2013
Avoiding the negative sign in impulse = change in momentum
Classic question: I have two identical balls. One is a "happy ball" which rebounds upon impact with the floor, reaching almost the same height from which it was dropped; one is a "sad ball" which essentially stops upon hitting the floor. It's a reasonable assumption that upon hitting the floor, both remain in contact with the floor for about the same amount of time.
Which ball applies a larger force to the floor?
The correct approach, of course, is to use the impulse-momentum formula. The time of collision is the same for each ball, so whichever ball experiences a larger change in momentum applies a larger force.
So which ball changes its momentum by more? It's often useful to make up values here... assume 1 Ns of momentum for both balls right before they collide.
The classic approach is to define up as the positive direction, and to define the change in momentum as pfinal - pinitial. The sad ball has a momentum change equal to (0 - (-1 Ns)) = 1 Ns; the happy ball has a momentum change equal to (1 Ns - (-1 Ns) = 2 Ns. The happy ball changes its momentum by more, thus applies a larger force to the ground.
I don't believe in negative signs where they can be avoided. So how have my freshmen answered this same question without the negative sign?
They recognize that the sad ball goes from 1 Ns to 0 Ns, so changes its momentum by 1 Ns. That's easy.
The happy ball also starts the collision by coming to rest, losing 1 Ns of momentum. But then, the happy ball has to rebound, too. It has to not only lose its original 1 Ns to stop, but it also has to gain 1 Ns in order to get back up to speed. So its total momentum change is 2 Ns: 1 Ns to stop, 1 Ns to rebound.
Explained this way (and then practiced a few times), the students are entirely comfortable with this conceptual approach. The follow up question is to make sure they understand why the happy ball doesn't have zero momentum change -- after all, the happy ball had the same amount of momentum before and after collision! But they quickly recognize that "zero momentum change" means the object never collided with anything. Rebounding takes more momentum change than just stopping.
Which ball applies a larger force to the floor?
The correct approach, of course, is to use the impulse-momentum formula. The time of collision is the same for each ball, so whichever ball experiences a larger change in momentum applies a larger force.
So which ball changes its momentum by more? It's often useful to make up values here... assume 1 Ns of momentum for both balls right before they collide.
The classic approach is to define up as the positive direction, and to define the change in momentum as pfinal - pinitial. The sad ball has a momentum change equal to (0 - (-1 Ns)) = 1 Ns; the happy ball has a momentum change equal to (1 Ns - (-1 Ns) = 2 Ns. The happy ball changes its momentum by more, thus applies a larger force to the ground.
I don't believe in negative signs where they can be avoided. So how have my freshmen answered this same question without the negative sign?
They recognize that the sad ball goes from 1 Ns to 0 Ns, so changes its momentum by 1 Ns. That's easy.
The happy ball also starts the collision by coming to rest, losing 1 Ns of momentum. But then, the happy ball has to rebound, too. It has to not only lose its original 1 Ns to stop, but it also has to gain 1 Ns in order to get back up to speed. So its total momentum change is 2 Ns: 1 Ns to stop, 1 Ns to rebound.
Explained this way (and then practiced a few times), the students are entirely comfortable with this conceptual approach. The follow up question is to make sure they understand why the happy ball doesn't have zero momentum change -- after all, the happy ball had the same amount of momentum before and after collision! But they quickly recognize that "zero momentum change" means the object never collided with anything. Rebounding takes more momentum change than just stopping.