What to do if your school has a batshit crazy marking policy

I’m glad to say that I don’t have to do this kind of thing any more, as my current school has a sensible marking and feedback policy.

For those of us who are less fortunate, however:

1. Don’t Panic

Refer SLT/HoD to Markopalypse Now or Adam Boxer’s Markageddon.

2. Me and My Class, Tick-boxing

Read (just read) six exercise books at random from your teaching group. Type three ebi’s and www’s and print as a label in the approved format with boxes to tick. Leave a blank space in the www and ebi section to write in the occasional bespoke comment.

Get class to hand in their exercise books held open to the page you intend to mark. If you have been able to prepare the marking labels in advance, get the class to glue them in at this point.

You will find an example label in Word format here.

Screenshot 2019-09-29 at 16.40.28.png

Huge time-saving tip: don’t copy and then cut and paste multiple copies of this into the same document file. Instead, just type “1,1,1,1” into the page range…

Screenshot 2019-09-29 at 16.43.47.png

…and then select “4 pages per sheet” for 4 crisp, identical copies on 1 one side of A4. (Thanks to Adam Boxer for this tip!)

Screenshot 2019-09-29 at 16.44.21.png

3. The Joy of Excel (and Mail Merging)

This system is especially useful when your school’s marking policy demands that students’ target grades and grade awarded for the assessed work is recorded on the feedback form.

Set up an Excel spreadsheet similar to the one below. There is a generic one available here.

Go through the student exercise books and type in comments into the spreadsheet. This sounds onerous but what you will find after marking, say, three books is that many of the comments can be directly re-used by copying and pasting directly into the relevant cell.

Screenshot 2019-09-29 at 18.11.49.png

Then, set up a marking feedback form in Word similar to this one:

Screenshot 2019-09-29 at 18.17.43.png

Next, use the Mailings menu or the Mail Merge Wizard to (1) identify the spreadsheet as the data source; and (2) transfer the data and comments from the spreadsheet to the Word document.

If you have done this correctly, it will look like this:

Screenshot 2019-09-29 at 18.25.53.png

Then Merge the data to produce a document that looks like this:

Screenshot 2019-09-29 at 18.35.26.png

Print using the 4 pages per sheet option as above or similar. Then get students to glue into their own books and complete the ebi/DIRT activities.

The other advantage of this is you can print off the Excel spreadsheet and file in your mark book.

And finally…

Hopefully, fewer and fewer teachers will need these tips as understanding of the proper role of marking and feedback permeate through the school system, but I know many schools are still demanding ‘triple marking’ when, in all truth, they shouldn’t be.

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Potential Divider Circuits and the Coulomb Train Model

A potential divider circuit is, essentially, a circuit where two or more components are arranged in series.

(a) Two resistors in series; (b) an ammeter (top) and an electric motor in series; (c) (L to R) a resistor, filament lamp and variable resistor in series

For non-physicists, these types of circuit can sometimes present problems, so in this post I am going to look in detail at the basic physics involved; and I am going to explain them using the CTM or Coulomb Train Model. (You can find the CTM model explained here.)

In the AQA GCSE Physics (and Combined Science) specifications, students are required to know that:

Extract from p.26 of the AQA spec

First, let’s look at the basics of describing electric circuits: current, potential difference and resistance.

 

1.0 Using the CTM to explain current, potential difference and resistance

 

Pupils tend to start with one concept for electricity in a direct current circuit: a concept labelled ‘current’, or ‘energy’ or ‘electricity’, all interchangeable and having the properties of movement, storability and consumption. Understanding an electrical circuit involves first differentiating the concepts of current, voltage and energy before relating them as a system, in which the energy transfer depends upon current, time and the potential difference of the battery.

The notion of current flowing in the circuit is one which pupils often meet in their introduction to a circuit and, because this relates well with their intuitive notions, this concept becomes the primary concept. (Driver 1994: 124 [italics added])


To my mind, the CTM is an excellent “bridging analogy” that helps students visualise the invisible. It is a stepping stone that provides some concrete representations of abstract quantities. In my opinion, it can help students

  1. move away from analysing circuits in terms of just current. (In my experience, even when students use terms like “potential difference”, in their eyes what they call “potential difference” behaves in a remarkably similar way to current e.g. it “flows through” components.)
  2. understand the difference between current, potential difference and resistance and how important each one is
  3. begin thinking of a circuit as a whole, interconnected system.


1.1 The CTM and electric current

Let’s begin by looking at a very simple circuit: a one ohm resistor connected across a 1 V cell.

A simple circuit

A very simple circuit

Note that it is a good teaching technique to include two ammeters on either side of the component, although the readings on both will be identical. This is to challenge the perennial misconception that electric current is “used up”. Electric charge, according to our current understanding of the universe, is a conserved quantity like energy in that it cannot be created or destroyed.

The Coulomb Train Model invites us to picture an electric circuit as a flow of positively charged coulombs carrying energy around the circuit in a clockwise fashion as shown below. The coulombs are linked together to form a continuous chain.

The CTM version of a simple circuit

The CTM applied to the very simple circuit shown above.

The name coulomb is not chosen at random: it is the SI unit of electric charge.

The current in this circuit will be given by I = V / R (equation 18 in the list on p.96 of the AQA spec, if you’re keeping track).

Using the AQA mark scheme-friendly FIFA protocol:

The otherwise inexplicable use of the letter “I” to represent electric current springs from the work André-Marie Ampère (1775–1836) and the French phrase intensité de courant (intensity of current).

From Q = I t (equation 17, p.96), current is a flow of electric charge, since I = Q / t. That is to say, if a charge of 2 coulombs passes (AQA call this a “charge flow”) in 2 seconds, the current will be …

A current of 1 amp is therefore represented on the CTM as 1 coulomb (or truck) passing by each second.

1.2 The CTM and Potential Difference

Potential difference or voltage is essentially the “energy difference” across any two parts of a circuit.

The equation used to define potential difference is not the familiar V = IR but rather the less familiar E = QV (equation 22 in the AQA list) where E is the energy transferred, Q is the charge flow (or the number of coulombs passing by in a certain time) and t is the time in seconds.

Let’s see what this would look like using the CTM:

(a) Circuit diagram showing how the measure the potential difference across a 1 V cell. (b) The same circuit represented using the CTM. (Note that the “white gloves” on the ends of the voltmeter connections are intended to be reminiscent of the white gloves of a snooker referee, indicating that the voltmeter does not disrupt the flow of the coulombs: in other words, the voltmeter has a high resistance.)

For the circuit shown, the voltmeter reading is 1 volt.

Note that on the CTM representation, one joule of energy is added to each coulomb as it passes through the cell.

If we had a 1.5 V cell then 1.5 joules would be transferred to each coulomb as it passed through, and so on.

(a) Circuit diagram showing potential difference measured across a connector with negligible resistance. (b) The same circuit represented using the CTM


If the voltmeter is moved to a different position as shown above, then the reading is 0 volts. This is because the coulombs at the points “sampled” by the voltmeter have the same amount of energy, so there is zero energy difference between them.

(a) Measuring the potential difference across a resistor. (b) The same circuit shown using the CTM.

In the position shown above, the voltmeter is measuring the potential difference across the resistor. For the circuit shown (assuming negligible resistance in all other parts of the circuit) the potential difference will be 1 V. In other words, each coulomb is losing one joule of energy as it passes through the resistance.

1.3 The CTM and Resistance

(a) Measuring the current through and the potential difference across a resistor. (b) The same circuit represented using the CTM.

In the circuit above, the potential difference across the resistor is 1 V and the current is 1 amp.

Resistance can therefore be thought of as the potential difference required to drive a current of 1 amp through that part of the circuit. It can also be thought of as the energy lost by each coulomb when a current of 1 amp flows through that part of the circuit; or, energy lost per coulomb per amp.

1.4 Summary

On the diagrams below, the coulombs are moving clockwise.

Summary of CTM

2.0 The CTM applied to a potential divider circuit

A potential divider circuit simply means that at least two resistors are in series so that the potential difference of the cell is shared across the resistors.

2.1 Two identical resistors

Because the two resistors are identical, the 3 V supply is shared equally across both resistors. That is to say, there is a potential difference of 1.5 V across each resistor. But let’s check this by applying V = IR (eq. 18). The total potential difference is 3 V and the total resistance is 1 ohm + 1 ohm = 2 ohms.

Now let’s use V = IR to check that the potential difference across each separate resistor is indeed half the total supply of 3 V. The resistance of one resistor is one ohm and the current through each one is 1.5 A. So V = 1.5 x 1 = 1.5 V.

But what would happen if we doubled the value of each resistor to 2 ohms?

Well, the current would be smaller: I = V/R = 3/4 = 0.75 amps.

The potential difference across each separate resistor would be V = I R = 0.75 x 2 = 1.5 V

So, the potential difference is always split equally when two identical resistors are placed in series (although, of course, the total resistance and the current will be different depending on the values of the resistors).

2.2a Two non-identical resistors

Let’s consider a circuit with a 2 ohm resistor in series with a 1 ohm resistor.

In this circuit, the total resistance is 1 ohm + 2 ohms = 3 ohms. The current flowing through the circuit is I = V / R = 3 / 3 = 1 amp.

So the potential difference across the 2 ohm resistor is V = IR = 1 x 2 = 2 V and the potential difference across the one ohm resistor is V = IR = 1 x 1 = 1 V.

Note that the resistor with the largest value gets the largest “share” of the potential difference.

2.2b Two non-identical resistors (different order)

Now let’s reverse the order of the resistors.

The current remains unchanged because the total resistance of the circuit is still the same.

Note that the largest resistor still gets the largest share of the potential difference, whichever way round the resistors are placed.

2.3 In Defence of the CTM and Donation Models

Many Physics teachers prefer “rope models” to so-called “donation models” like the CTM.

And it is perfectly true that rope models have some good points such as the ability to easily explain AC and a more accurate approximation of what happens when current starts to flow or stops flowing. The difficulty in their use, in my opinion, is that you are using concepts that many students barely understand (e.g. friction to model resistance) to explain how very unfamiliar concepts such as potential difference work. Also, the vagueness of some of the analogs is unhelpful: for example, when we compare potential difference to “push”, are we talking about the net resultant force on the rope or simply the force needed to balance the frictional force and keep it moving at a steady speed?

To my way of thinking, the CTM has the advantage of encouraging quantitative thinking about current, potential difference and resistance almost from the moment of first teaching. Admittedly, it cannot cope with AC — but then again, we model AC as a direct current when we use RMS values. Now admittedly, rope models are far better at picturing what happens in the initial fractions of a second when a current starts to flow after closing a switch. Be that as it may, the CTM comes into its own when we consider the “steady state” of current flow after the initial surge currents.

One of the frequent criticisms (which is usually considered quite damning) of this type of model is “How do the coulombs know how much energy to drop off at each resistor?”

For example, in the diagram above, how do the coulombs “know” to drop off 1 J at the first resistor and 2 J at the second resistor?

The answer is: they don’t. Rather, the energy loss is due to the nature of the resistor: think of a resistor as a tunnel lined with strip curtains. A coulomb loses only a small amount of its excess energy passing through a low value resistor, but a much larger amount passing through a higher value resistor, as modelled below.

Strip curtain model for CTM

A 1 ohm and 2 ohm resistor modelled as strip curtains

FWIW I therefore commend the use of the CTM to all interested parties. 



References

Driver, R., Squires, A., Rushworth, P., & Wood-Robinson, V. (1994). Making sense of secondary science: Research into children’s ideas. Routledge.

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The Acceleration Required Practical Without Light Gates (And Without Tears)

Introduction

The AQA GCSE Required Practical on Acceleration (see pp. 21-22 and pp. 55-57) has proved to be problematic for many teachers, especially those who do not have access to a working set of light gates and data logging equipment.

In version 3.8 of the Practical Handbook (pre-March 2018), AQA advised using the following equipment featuring a linear air track (LAT). The “vacuum cleaner set to blow”, (or more likely, a specialised LAT blower), creates a cushion of air that minimises friction between the glider and track.

Screenshot 2019-08-08 at 14.47.50.png

However, in version 5.0 (dated March 2018) of the handbook, AQA put forward a very different method where schools were advised to video the motion of the car using a smartphone in an effort to obtain precise timings at the 20 cm, 40 cm and other marks.

Screenshot 2019-08-08 at 13.54.07.png

It is possible that AQA published the revised version in response to a number of schools contacting them to say. “We don’t have a linear air track. Or light gates. Or a ‘vacuum cleaner set to blow’.”

The weakness of the “new” version (at least in my opinion) is that it is not quantitative: the method suggested merely records the times at which the toy car passed the lines. Many students may well be able to indirectly deduce the relationship between resultant force and acceleration from this raw timing data; but, to my mind, it would be cognitively less demanding if they were able to compare measurements of resultant force and acceleration instead.

Adapting the AQA method to make it quantitative

Screenshot 2019-08-08 at 15.39.40.png

We simplify the AQA method as above: we simply time how long the toy car takes to complete the whole journey from start to finish.

If a runway of one metre or longer is set up, then the total time for the journey of the toy car will be 20 seconds or so for the smallest accelerating weight: this makes manual timing perfectly feasible.

Important note: the length of the runway will be limited by the height of the bench. As soon as the weight stack hits the floor, the toy car will no longer experience an accelerating force and, while it may continue at a constant speed (more or less!) it will no longer be accelerating. In practice, the best way to sort this out is to pull the toy car back so that the weight stack is just below the pulley and mark this position as the start line; then slowly move the toy car forward until the weight stack is just touching the floor, and mark this position as the finish line. Measure the distance between the two lines and this is the length of your runway.

In addition, the weight stack should feature very small masses; that is to say, if you use 100 g masses then the toy car will accelerate very quickly and manual timing will prove to be impossible. In practice, we found that adding small metal washers to an improvised hook made from a paper clip worked well. We found the average mass of the washers by placing ten of them on a scale.

Then input the data into this spreadsheet (click the link to download from Google Drive) and the software should do the rest (including plotting the graph!).

The Eleventh Commandment: Thou Shalt Not Confound Thy Variables!

To confirm the straight line and directly proportional relationship between accelerating force and acceleration, bear in mind that the total mass of the accelerating system must remain constant in order for it to be a “fair test”.

The parts of our system that are accelerating are the toy car, the string and the weight stack. The total mass of the accelerating system shown below is 461 g (assuming the mass of the hook and the string are negligible).

The accelerating (or resultant) force is the weight of 0.2 g mass on the hook, which0 can be calculated using W = mg and will be equal to 0.00196 N or 1.96 mN.

In the second diagram, we have increased the mass on the weight stack to 0.4 g (and the accelerating force to 0.00392 N or 3.92 mN) but note that the total mass of the accelerating system is still the same at 461 g.

In practice, we found that using blu-tac to stick a matchbox tray to the roof of the car made managing and transferring the weight stack easier.

Personal note: as a beginning teacher, I demonstrated the linear air track version of this experiment to an A-level Physics class and ended up disconfirming Newton’s Second Law instead of confirming it; I was both embarrassed and immensely puzzled until an older, wiser colleague pointed out that the variables had been well and truly confounded by not keeping the total mass of the accelerating system constant.

It was embarrassing and that’s why I always harp on about this aspect of the experiment.

What lies beneath: the Physics underlying this method

This can be considered as “deep background” rather than necessary information, but I, for one, consider it really interesting.

Acceleration is the rate of change of a rate of change. Velocity is the rate of change of displacement with time and acceleration is the rate of change of velocity.

Interested individuals may care to delve into higher derivatives like jerk, snap, crackle and pop (I kid you not — these are the technical terms). Jerk is the rate of change of acceleration and hence can be defined as (takes a deep breath) the rate of change of a rate of change of rate of change. More can be found in the fascinating article by Eager, Pendrill and Reistad (2016) linked to above.

But on a much more prosaic level, acceleration can be defined as a = (v – u) / t where v is the final instantaneous velocity, u is the inital instantaneous velocity and t is the time taken for the change.

The instantaneous velocity is the velocity at a momentary instant of time. It is, if you like, the velocity indicated by the needle on a speedometer at a single instant of time and is different from the average velocity which is calculated from the total distance travelled divided the time taken.

This can be shown in diagram form like this:

However, our experiment is simplified because we made sure that the toy car was stationary when the timer was zero; in other words, we ensured u = 0 m/s.

This simplifies a = (v – u) / t to a = v / t.

But how can we find v, the instantaneous velocity at the end of the journey when we have no direct means of measuring it, such as a speedometer or a light gate?

No more jerks left to give

Let’s assume that, for the toy car, the jerk is zero (again, let me emphasize that jerk is a technical term defined as the rate of change of acceleration).

This means that the acceleration is constant.

This fact allows us to calculate the average velocity using a very simple formula: average velocity = (u + v) / t .

But remember that u = 0 so average velocity = v / 2 .

More pertinently for us, provided that u = 0 and jerk = 0, it allows us to calculate a value for v using v = 2 x (average velocity) .

The spreadsheet linked to above uses this formula to calculate v and then uses a = v / t.

Using this in the school laboratory

This could be done as a demonstration or, since only basic equipment is needed, a class experiment. Students may need access to computers running the spreadsheet during the experiment or soon afterwards. We found that one laptop shared between two groups was sufficient.

First experiment (relationship between force and acceleration): set up as shown in the diagram. Place washers totalling a mass of 0.8 g (or similar) and washers totalling a mass of 0.2 g on the hook or weight stack. Hold the toy car stationary at the start line. Release and start the timer. Stop the timer. Input data into the spreadsheet and repeat with different mass on the hook.

It can be useful to get students to manually “check” the value of a calculated by the spreadsheet to provide low stakes practice of using the acceleration formula.

Second experiment (relationship between mass and acceleration). Keep the accelerating force constant with (say) 0.6 g on the hook or weight stack. Hold the toy car stationary at the start line. Release and start the timer. Stop the timer. Input data into the second tab on the spreadsheet and repeat with 100 g added to the toy car (possibly blu-tac’ed into place).

Conclusion

This blog post grew in the telling. Please let me know if you try the methods outlined here and how successful you found them

References

Eager, D., Pendrill, A. M., & Reistad, N. (2016). Beyond velocity and acceleration: jerk, snap and higher derivatives. European Journal of Physics, 37(6), 065008.

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Thermal Energy and Internal Energy

The AQA GCSE Science specification calls for students to understand and apply the concepts of not only thermal energy stores but also internal energy. What follows is my understanding of the distinction between the two, which I hope will be of use to all science teachers.

My own understanding of this topic has undergone some changes thanks to some fascinating (and ongoing) discussions via EduTwitter.

What I suggest is that we look at the phenomena in question through two lenses:

  • a macroscopic lens, where we focus on things we can sense and measure directly in the laboratory
  • a microscopic lens, where we focus on using the particle model to explain phase changes such as melting and freezing.

Thermal Energy Through the Macroscopic Lens

Screenshot 2019-04-14 at 14.29.39.pngThe enojis for thermal energy stores (as suggested by the Institute of Physics) look like this (Note: ‘enoji’ = ‘energy’ + ’emoji’; and that the IoP do not use the term):Screenshot 2019-04-14 at 14.22.10.png

In many ways, they are an excellent representation. Firstly, energy is represented as a “quasi-material entity” in the form of an orange liquid which can be shifted between stores, so the enoji on the left could represent an aluminium block before it is heated, and the one on the right after it is heated. Secondly, it also attempts to make clear that the so-called forms of energy are labels added for human convenience and that energy is the same basic “stuff” whether it is in the thermal energy store or the kinetic energy store. Thirdly, it makes the link between kinetic theory and thermal energy stores explicit: the particles in a hot object are moving faster than the particles in the colder object.

However, I think the third point is not necessarily an advantage as I believe it will muddy the conceptual waters when it comes to talking about internal energy later on.

If I was a graphic designer working for the IoP these are the enojis I would present:Screenshot 2019-04-14 at 15.10.50.png

In other words, a change in the thermal energy store is always associated with a temperature change. To increase the temperature of an object, we need to shift energy into the thermal energy store. To cool an object, energy needs to be shifted out of the thermal energy store.

This has the advantage of focusing on the directly observable macroscopic properties of the system and is, I think, broadly in line with the approach suggested by the AQA specification.Screenshot 2019-04-14 at 15.32.13.png

Internal Energy Through the Microscopic Lens

Screenshot 2019-04-14 at 15.27.25.png

Internal energy is the “hidden” energy of an object.

The “visible” energies associated with an object would include its kinetic energy store if it is moving, and its gravitational potential energy store if it is lifted above ground level. But there is also a deeper, macroscopically-invisible store of energy associated with the particles of which the object is composed.

To understand internal energy, we have to look through our microscopic lens.

The Oxford Dictionary of Physics (2015) defines internal energy as:

The total of the kinetic energies of the atoms and molecules of which a system consists and the potential energies associated with their mutual interactions. It does not include the kinetic and potential energies of the system as a whole nor their nuclear energies or other intra-atomic energies.

In other words, we can equate the internal energy to the sum of the kinetic energy of each individual particle added to the sum of the potential energy due to the forces between each particle. In the simple model below, the intermolecular forces between each particle are modelled as springs, so the potential energy can be thought as stretching and squashing the “springs”. (Note: try not to talk about “bonds” in this context as it annoys the hell out of chemists, some of whom have been known to kick like a mule when provoked!)

Screenshot 2019-04-14 at 16.02.55.png

We can never measure or calculate the value of the absolute internal energy of a system in a particular state since energy will be shifting from kinetic energy stores to potential energy stores and vice versa moment-by-moment. What is a useful and significant quantity is the change in the internal energy, particularly when we are considering phase changes such as solid to liquid and so on.

This means that internal energy is not synonymous with thermal energy; rather, the thermal energy of a system can be taken as being a part (but not the whole) of the internal energy of the system.

As Rod Nave (2000) points out in his excellent web resource Hyperphysics, what we think of as the thermal energy store of a system (i.e. the sum of the translational kinetic energies of small point-like particles), is often an extremely small part of the total internal energy of the system.Screenshot 2019-04-14 at 16.31.55.png

AQA: Oops-a-doodle!

My excellent Edu-tweeting colleague @PhysicsUK has pointed out that there is indeed a discrepancy between the equations presented by AQA in their specification and on the student equation sheet.

If a change in thermal energy is always associated with a change in temperature (macroscopic lens) then we should not use the term to describe the energy change associated with a change of state when there is no temperature change (microscopic lens).

@PhysicsUK reports that AQA have ‘fessed up to the mistake and intend to correct it in the near future. Sooner would be better than later, please, AQA!

Screenshot 2019-04-14 at 16.44.38.png

References

Nave, R. (2000). HyperPhysics. Georgia State University, Department of Physics and Astronomy.

 

 

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It is F=ma, you know

“It is cheese. (Caerphilly.)”

Main research finding of the first manned Welsh mission to the Moon, as reported by Max Boyce c. 1974.

A Brief History of F=ma

When writing A Brief History Of Time, it is said that a literary agent warned Stephen Hawking that each mathematical equation that he included in the final draft would halve its eventual sales.

And one of the complex equations that Hawking wished to include? A summary of Newton’s 2nd Law: F=ma.

In other words, the force F acting on an object is equal to the mass m of the object multiplied by the acceleration experienced by the object.

(In the end, Hawking opted to include only Einstein’s E=mc2 in what turned out to be the ultra bestselling A Brief History of Time.)


F=ma is Newton’ s 2nd Law: NOT!!!

@SciByDegrees wrote an interesting post arguing that the old Physics teacher’s standby of summarising Newton’s Second Law of Motion (N2) as F=ma is wrong.

The gist of his argument (and it’s a hard argument to counter) is that F=d(mv)/dt is a far better expression of the law than the F=ma version because it covers a wider range of circumstances.

This states N2 in terms of momentum, where momentum is the product of mass m multiplied by the velocity v. More exactly, it says that force acting on an object is equal to the object’s rate of change of momentum: or, if you prefer, the change in momentum divided by the time taken for the change is equal to the force.

This is the version of N2 stated in most dictionaries of Physics. For example, the Oxford Dictionary of Physics (2015) p. 383.

I know, because on reading @SciByDegrees’ post I immediately looked up N2 in the Dictionary with the express intention of countering the argument. Imagine my consternation and horror when I found that I was wrong. (Actually, not that much consternation and horror: I am fairly inured to being wrong as it happens fairly often…)

The argument suggests that, just like V=IR is not a statement of Ohm’s Law unless R has a fixed value (like a fixed length of wire at a constant temperature), F=ma is not a sufficient statement of N2 unless the mass m is constant.

For example, if we consider a rocket capable of producing a steady 1000 N of thrust; at t=0 its mass is (say) 10 kg so its acceleration is 100 m/s2. However at t=1 s its mass has decreased by 1 kg so the acceleration is now 111 m/s2 even though the thrust is still 1000 N so obviously F is not proportional to m so F does not equal ma in this situation.


Feynman as the new Aristotle

Richard Feynman (1965) wrote along similar lines in his justly famous Lectures on Physics:

Thus at the beginning we take several things for granted. First, that the mass of an object is constant; it isn’t really, but we shall start out with the Newtonian approximation that mass is constant, the same all the time, and that, further, when we put two objects together, their masses add. These ideas were of course implied by Newton when he wrote his equation, for otherwise it is meaningless. For example, suppose the mass varied inversely as the velocity; then the momentum would never change in any circumstance, so the law means nothing unless you know how the mass changes with velocity. At first we say, it does not change.

However, I think Feynman is considerably oversimplifying what Newton said here. Dare one suppose that Feynman, who had an enviable natural facility for talking intelligently and arrestingly about nearly any subject under the Sun, had perhaps skimped a little on his background reading?

Incidentally, does anyone think that physicists (especially physics educators — myself included) are beginning to treat Feynman as the medieval scholastics are reputed to have treated Aristotle? That is to say, he is regarded as the final word on everything; or, at least, everything physics-related in the case of Feynman.


What Would Newton Do?

George Smith (2008) points out that:

The modern F=ma form of Newton’s second law nowhere occurs in any edition of the Principia [ . . . ] Instead, it has the following formulation in all three editions: A change in motion is proportional to the motive force impressed and takes place along the straight line in which that force is impressed. In the body of the Principia this law is applied both to discrete cases, in which an instantaneous impulse such as from impact is effecting the change in motion, and to continuously acting cases, such as the change in motion in the continuous deceleration of a body moving in a resisting medium. Newton thus appears to have intended his second law to be neutral between discrete forces (that is, what we now call impulses) and continuous forces.

This, I think, supports my contention that F=ma is as good a modern reformulation of Newton’s 2nd Law as any other.

If we go back to the rocket example, the instantaneous acceleration at t=0 and t=1 s can be calculated using F=ma (provided we take account of the change in m, of course). In effect, we are considering the change in motion due an instantaneous impulse here.

Please note that I would cheerfully concede that F=d(mv)/dt would yield a better and more productive analysis of rocket motion if we are considering the continuous action of the force over time rather than at isolated instants.

The analogy with V=IR is useful here. V is always equal to I times R but V is only directly proportional to I over a continuous range of values of I for a limited set of conductors we call Ohmic conductors whose resistance R is fixed over a range of physical conditions. Likewise, F is always equal to m times a but F is only directly proportional to a for a continuous range of values of a when we are considering a system whose mass is fixed.

As V=IR is neutral with respect to whether R is fixed is not, I believe that F=ma is neutral with respect to whether m is fixed or not.


Will the Real Second Law Please Stand Up?

What is Newton’s Second Law? Is it a definition of force? Is it a definition of mass? Or is it an empirical proposition linking force, mass and acceleration?

Brian Ellis (1965) argues that it partakes of all three:

Consider how Newton’s second law is actually used. In some fields it is unquestionably true that Newton’s second law is used to define a scale of force. How else, for example, can we measure interplanetary gravitational forces? But it is also unquestionably true that Newton’s second law is sometimes used to define a scale of mass. Consider, for example, the use of mass spectrography. And in yet other fields, where force, mass and acceleration are all easily and independently measurable, Newton’s second law of motion functions as an empirical correlation between these three quantities. Consider, for example, the application of Newton’s second law in ballistics and rocketry [ . . .] To suppose that Newton’s second law of motion, or any law for that matter, must have a unique role that we can describe generally and call the logical status is an unfounded and unjustifiable supposition.

In some senses, I suppose we might like those unfortunate nations in Gulliver’s Travels who fought a long and bitter war over the question of whether one should eat a boiled egg from the pointed or rounded end:

During the course of these troubles, the emperors of Blefusca … accusing us of making a schism in religion, by offending against a fundamental doctrine of our great prophet Lustrog, in the fifty-fourth chapter of the Blundecral (which is their Alcoran). This, however, is thought to be a mere strain upon the text; for the words are these: ‘that all true believers break their eggs at the convenient end.’ And which is the convenient end, seems, in my humble opinion to be left to every man’s conscience.


REFERENCES

Ellis, Brian. “The origin and nature of Newton’s laws of motion.” Beyond the edge of certainty (1965): 29-68.

Feynman, R. P., Leighton, R. B., & Sands, M. (1965). The Feynman Lectures on Physics; vol. 1 (Accessed from http://www.feynmanlectures.caltech.edu/I_09.html on 11/4/19)

Smith, George, “Newton’s Philosophiae Naturalis Principia Mathematica“, The Stanford Encyclopedia of Philosophy (Winter 2008 Edition), Edward N. Zalta (ed.), URL = <https://plato.stanford.edu/archives/win2008/entries/newton-principia/&gt;.ewto

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Ut tensio sic vis

And if two men ride of a horse, one must ride behind.

Shakespeare, Much Ado About Nothing

Sir Isaac Newton stands in popular estimation as the foremost intellect of his age; or perhaps, of any age. If a person is never truly dead while their name is spoken, then Sir Isaac stands with us still: partially overshadowed by Einstein at the dawn of the twentieth century, maybe, but never totally eclipsed.

But in the roiling intellectual cauldron of the Age of Enlightenment, even such a venerable polymath as Newton had some serious competition. As Newton himself modestly observed in a letter to a contemporary in 1676: “If I have seen a little further it is by standing on the shoulders of Giants.”

Except that one interpretation has it that the letter was not intended to be modest, but was rather a combative dig at the man to whom it was addressed: Robert Hooke, a man of but “middling” stature and, as a result of a childhood illness, also a hunchback. Not one of the “Giants” with broad philosophic shoulders to whom Newton felt indebted to, then.

Robert Hooke, as painted by Rita Greer in 2007. The painting is based on contemporary descriptions of Robert Hooke. No undisputed contemporary paintings or likenesses of Hooke have survived, possibly because of malicious intent on the part of Newton.

In popular estimation, therefore, it would appear that Hooke is fated always to sit behind Newton. At GCSE and A-level, students learn of Newton’s Laws of Motion, the eponymous unit of force, and his Law of Universal Gravitation.

And what do they learn of Hooke? They learn of his work on springs. They learn of “Hooke’s Law”: that is, the force exerted by a spring is directly proportional to its extension.

Ut tensio, sic vis.

[As extension, so is the force.]

— Robert Hooke, Lectures de Potentia Restituvia [1678]

Newton has all the laws of motion on Earth and in Heaven in the palm of his hand, and Hooke has springs. Perhaps, then, Hooke deserves to be forever second on the horse of eternal fame?

But look closer. To what objects or classes of object can we apply Hooke’s Law? The answer is: all of them.

Hooke’s Law applies to everything solid: muscle, bone, sinew, concrete, wood, ice, crystal and stone. Stretch them or squash them, and they will deform in exact proportion to the size of the force applied to them.

That is, if one power stretch or bend it one space, two will bend it two, and three will bend it three, and so forward.

The major point being that Hooke’s Law is as universal as gravity: it is baked into the very fabric of the universe: it is a direct consequence of the interactions between atoms.

Graph of interatomic force against distance between two atoms. Hooke’s Law applies in the red circle.

Now before I wax too lyrical, it must be pointed out that Hooke’s Law is a first-order linear approximation: it fails when the deforming force increases beyond a certain limit, and that limit is unique to each material. But within the limits of its domain indicated by the red circle above, it reigns supreme.

How do you calculate how much a steel beam will bow when a kitten walks across it? Hooke’s Law. How could we model the stresses on the bones of a galloping dinosaur? Hooke’s Law. How can we calculate how much Mount Everest bends when it is buffeted by wind? Hooke’s Law.

Time to re-evaluate the seating order on Shakespeare’s horse, mayhap?

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Why does kinetic energy = 1/2mv^2?

Why does kinetic energy Ek=½mv2?

Students and non-specialist teachers alike wonder: whence the half?

This post is intended to be a diagrammatic answer to this question using a Singapore Bar Model approach: so pedants, please avert your eyes.

I am indebted to Ben Rogers’ recent excellent post on showing momentum using the Bar Model approach for starting me thinking along these lines.

Part the First: How to get the *wrong* answer

Imagine pushing an object with a mass m with a constant force F so that it accelerates with a constant acceleration a so that covers a distance s in a time t. The object was initially at rest and ends up moving at velocity v.Screenshot 2019-03-09 at 14.24.59.png

(On the diagram, I’ve used the SUVAT dual coding conventions that I suggested in a previous post.)

So let’s consider the work done on the object by the force:

Step 1:    work done = force x distance moved in the direction of the force

Step 2:    W= F x s

But remember s = v x t so:

Step 3:    W= F x vt

And also remember that F = m x a so:

Step 4:    W= ma x vt

Also remember that a = change in velocity / time, so a = (v – 0) / t = v / t.

Step 5:        Wd = (v / t) x vt

The ts cancel so:

Step 6:    W= mv2

Since this is the work done on the object by the force, it is equal to the energy transferred to the kinetic energy store of the object. In other words, it is the energy the object has gained because it is moving — its kinetic energy, no less: E= mv2.

On a Singapore Bar Model diagram this can be represented as follows:

Screenshot 2019-03-09 at 15.14.17

The kinetic energy is represented by the volume of the bar.

But wait: Ek=mv2!?!?

That’s just wrong: where did the half go?

Houston, we have a problem.

Part the Second: how to get the *right* answer

The problem lies with Step 3 above. We wrongly assumed that the object has a constant velocity over the whole of the distance s.

Screenshot 2019-03-09 at 17.35.43.pngIt doesn’t because it is accelerating: it starts off moving slowly and ends up moving at the maximum, final velocity v when it has travelled the total distance s.

So Step 3 should read:

But remember that s = (average velocity) x t.

Because the object is accelerating at a constant rate, the average velocity is (v + u) / 2 and since u = 0 then average velocity is v / 2.

Step 3:    Wd= F x (v / 2) t

And also remember that F = m x a so:

Step 4:    Wd= ma x (v / 2) t

Also remember that a = change in velocity / time, so a = (v – 0) / t = v / t.

Step 5:        Wd = (v / t) x (v / 2) t

The ts cancel so:

Step 6:    Wd= ½mv2

Based on this, of course, E= ½mv2
(Phew! Houston, we no longer have a problem.)

Screenshot 2019-03-09 at 17.58.45.png

Using the Bar Model representation, the volume of the bar which is above the blue plane represents the kinetic energy of an object of mass m moving at a velocity v.

The reason it is half the volume of the bar and not the full volume (as in the incorrect Part the First analysis) is because we are considering the work done by a constant force accelerating an object which is initially at rest; the velocity of the object increases gradually from zero as the force acts upon it. It therefore takes a longer time to cover the distance s than if it was moving at a constant velocity v from the very beginning.

So there we have it, E= ½mvby a rather circuitous method.

But why go “all around the houses” in this manner? For exactly the same reason as we might choose to go by the path less travelled on some of our other journeys: quite simply, we might find that we enjoy the view.

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