FIFA and Really Challenging GCSE Physics Calculations

‘FIFA’ in this context has nothing to do with football; rather, it is a mnemonic that helps KS3 and KS4 students from across the attainment range engage productively with calculation questions.

FIFA stands for:

  • Formula
  • Insert values
  • Fine-tune
  • Answer

From personal experience, I can say that FIFA has worked to boost physics outcomes in the schools I have worked in. What is especially gratifying, however, is that a number of fellow teaching professionals have been kind enough to share their experience of using it:


Framing FIFA as a modular approach

Straightforward calculation questions (typically 2 or 3 marks) can be ‘unlocked’ using the original FIFA approach. More challenging questions (typically 4 or 5 marks) can often be handled using the FIFA-one-two approach.

However, what about the most challenging 5 or 6 mark questions that are targeted at Grade 8/9? Can FIFA help in solving these?

I believe it can. But before we dive into that, let’s look at a more traditional, non-FIFA, algebraic approach.


A challenging freezing question: the traditional (non-FIFA) algebraic approach

Note: this is a ‘made up’ question written in the style of the GCSE exam.

A pdf of this question is here. A traditional algebraic approach to solving this problem would look like this:

This approach would be fine for confident students with high previous attainment in physics and mathematics. I will go further and say that it should be positively encouraged for students who possess — in Edward Gibbon’s words — that ‘happy disposition’:

But the power of instruction is seldom of much efficacy, except in those happy dispositions where it is almost superfluous.

Edward Gibbon, The Decline and Fall of the Roman Empire

But what about those students who are more akin to the rest of us, and for whom the ‘power of instruction’ is not a superfluity but rather a necessity on which they depend?


A challenging freezing question: the FIFA-1-2-3 approach

Since this question involves both cooling and freezing it seems reasonable to start with the specific heat capacity formula and then use the specific latent heat formula:

FIFA-one-two isn’t enough. We must resort to FIFA-1-2-3.

What is noteworthy here is that the third FIFA formula isn’t on the formula sheet and is not on the list of formulas that need to be memorised. Instead, it is made by the student based on their understanding of physics and a close reading of the question.

Challenging? Yes, undoubtedly. But students will have unlocked some marks (up to 4 out of 6 by my estimation).

FIFA isn’t a royal road to mathematical mastery (although it certainly is a better bet than the dreaded ‘formula triangle’ that I and many other have used in the past). FIFA is the scaffolding, not the finished product.

Genuine scientific understanding is the clock tower; FIFA is simply some temporary scaffolding that helps students get there.

We complete the FIFA-1-2-3 process as follows:


Conclusion: FIFA fixes it

The FIFA-system was born of the despair engendered when you mark a set of mock exam papers and the majority of pages are blank: students had not even attempted the calculation skills.

In my experience, FIFA fixes that — students are much more willing to start a calculation question. And that means that, even when they cannot successfully navigate to a ‘full mark’ conclusion, they gain at least some marks, and and one does not have to be a particularly perceptive scholar of the human heart to understand that gaining ‘some marks‘ is more motivating than ‘no marks‘.

Update: Ed Southall makes a persuasive case against formula triangles in this 2016 article.

The Coulomb Train Model Revisited (Part 3)

In this post, we will look at explaining electrical resistance using the Coulomb Train Model.

This is part 3 of a continuing series (click to read part 1 and part 2).

The Coulomb Train Model (CTM) is a helpful model for both explaining and predicting the behaviour of real electric circuits which I think is useful for KS3 and KS4 students.

Without further ado, here is a a summary.

A summary of the Coulomb Train Model

Representing Resistance on the CTM

To measure resistance, we would set up this circuit.

We can represent this same circuit on the CTM as follows:

If we count how many ‘coulombs’ (grey trucks) pass one point in a certain time then on this animation we get 5 coulombs in 20 seconds.
This is equivalent to a current of
5 coulombs / 20 seconds = 0.2 coulombs per second = 0.2 amperes.

This way of thinking about current is consistent with the formula charge flow = current x time or Q=It which can be rearranged to give I=Q/t.

We have used identical labels on the circuit diagram and the CTM animation to encourage students to view them as different representations of a real situation. The ammeter at X would read 0.2 amps. We could place the ammeter at any other point in the circuit and still get a reading of 0.2 amps since ammeters only ‘count coulombs per second’ and don’t make any measurement of energy (represented by the orange substance in the trucks).

However, the voltmeter does make a measurement of energy: it compares the energy difference between a single coulomb at Y and a single coulomb at Z. If (say) 1.5 joules of energy is transferred from each coulomb as it passes through the bulb from Y to Z then the voltmeter will read a potential difference (or ‘voltage’ if you prefer) of 1.5 volts.

This way of thinking about potential difference is consistent with the formula energy transferred = charge flow x potential difference or E=QV which we can rearrange to give V=E/Q.

So as you can see, one volt is really equivalent to an energy change of one joule for every coulomb (!)

We can calculate the resistance of the bulb by using R=V/I so R = 1.5/0.2 = 7.5 ohms.


Resistance is not futile . . .

Students sometimes have difficulty accepting the idea of a ‘resistor’: ‘Why would anyone in their right minds deliberately design something that reduces the flow of electric current?’ It’s important to explain that it is vital to be able to control the flow of electric current and that one of the most common electronic components in your phone or games console is — the humble resistor.

One of many resistors on a circuit board. The colour codes tell us the value of each resistor.

Teachers often default to explaining electric circuits using bulbs as the active component. There is a lot to recommend this practice, not least the fact that changes in the circuit instantaneously affect the brightness of the bulb. However, it vital (especially at GCSE) to allow students to learn about circuits featuring resistors and other components rather than just the pedagogically overused (imho) filament lamp.


Calculating the resistance of a resistor

Consider this circuit where we have a resistor R1.

This can be represented as a coulomb train model like this:

The resistor does not glow with visible light as the bulb does, but it would glow pretty brightly if viewed through an infra red camera since the energy carried by the coulombs is transferred to the thermal energy store of the resistor. The only way we can observe this energy shift without such a special camera is to use a voltmeter.

Let’s begin by analysing this circuit qualitatively.

  • The coulombs are moving faster in this circuit than the previous circuit. This means that the current is larger. (Remember: current is coulombs per second.)
  • Because the current is larger, R1 must have a smaller resistance than the bulb. (Remember: resistance is a quantity that reduces the current.)
  • The energy transferred to each coulomb is the same in each example so the potential difference of the cell is the same in both circuits. (Of course, V can be altered by adding a second cell or turning up the setting on a power supply, but in many circuits V is, loosely speaking, a ‘fixed’ or ‘quasi-constant’ value.)
  • Because the ‘push’ or potential difference is the same size but the resistance of R1 is smaller, then the same cell is able to push a larger current around the circuit.

Now let’s analyse the circuit quantitatively.

  • 5 coulombs pass a single point in 13 seconds so the current is 5/13 = 0.38 coulombs per second = 0.4 amperes. (Double the current in the bulb circuit.)
  • The resistance can be calculated using R=V/I = 1.5/0.4 = 3.75 ohms. (Half the resistance of the bulb.)
  • Each coulomb is being loaded with 1.5 J of energy as it passes through the cell. Since this is happening twice as often in the resistor circuit as the bulb circuit, the cell will ‘go flat’ or ’empty its chemical energy store’ in half the time of the bulb cell.

So there we have it: more fun and high jinks with the CTM.

I hope that I have persuaded a few more teachers that the CTM is useful for getting students to think productively and, more importantly, quantitatively using correct scientific terminology about electric circuits.

In the next installment, we will look at series and parallel circuits.

The Coulomb Train Model Revisited (Part 1)

The Coulomb Train Model (CTM) is a helpful model for both explaining and predicting the behaviour of real electric circuits which I think is suitable for use with KS3 and KS4 students (that’s 11-16 year olds for non-UK educators).

I have written about it before (see here and here) but I have recently been experimenting with animated versions of the original diagrams.

Essentially, the CTM is a donation model akin to the famous ‘bread and bakery van’ or even the ‘penguins and ski lift’ models, but to my mind it has some major advantages over these:

  • The trucks (‘coulombs’) in the CTM are linked in a continuous chain. If one ‘coulomb’ stops then they all stop. This helps students grasp why a break anywhere in a circuit will stop all current.
  • The CTM presents a simplified but still quantitatively accurate picture of otherwise abstract entities such as coulombs and energy rather than the more whimsical ‘bread van’ = ‘charge carrier’ and ‘bread’ = ‘energy’ (or ‘penguin’ = ‘charge carrier’ and ‘gpe of penguin’ = ‘energy of charge carrier’) for the other models.
  • Explanations and predictions scripted using the CTM use direct but substantially correct terminiology such as ‘One ampere is one coulomb per second’ rather than the woolier ‘current is proportional to the number of bread vans passing in one second’ or similar.

Modelling current flow using the CTM

The coulombs are the ‘trucks’ travelling clockwise in this animation. This models conventional current.

Charge flow (in coulombs) = current (in amps) x time (in seconds)

So a current of one ampere is one coulomb passing in one second. On the animation, 5 coulombs pass through the ammeter in 25 seconds so this is a current of 0.20 amps.

We have shown two ammeters to emphasise that current is conserved. That is to say, the coulombs are not ‘used up’ as they pass through the bulb.

The ammeters are shown as semi-transparent as a reminder that an ammeter is a ‘low resistance’ instrument.


Modelling ‘a source of potential difference is needed to make current flow’ using the CTM

Energy transferred (in joules) = potential difference (in volts) x charge flow (in coulombs) 

So the potential difference = energy transferred divided by the number of coulombs.

The source of potential difference is the number of joules transferred into each coulomb as it passes through the cell. If it was a 1.5 V cell then 1.5 joules of energy would be transferred into each coulomb.

This is represented as the orange stuff in the coulombs on the animation.

What is this energy? Well, it’s not ‘electrical energy’ for certain as that is not included on the IoP Energy Stores and Pathways model. In a metallic conductor, it would be the sum of the kinetic energy stores and electrical potential energy stores of 6.25 billion billion electrons that make up one coulomb of charge. The sum would be a constant value (assuming zero resistance) but it would be interchanged randomly between the kinetic and potential energy stores.

For the CTM, we can be a good deal less specific: it’s just ‘energy’ and the CTM provides a simplified, concrete picture that allows us to apply the potential difference equation in a way that is consistent with reality.

Or at least, that would be my argument.

The voltmeter is shown connected in parallel and the ‘gloves’ hint at it being a ‘high resistance’ instrument.

More will follow in part 2 (including why I decided to have the bulb flash between bright and dim in the animations).

You can read part 2 here.

Teaching Newton’s Third Law

Newton’s First and Second Laws of Motion are universal: they tell us how any set of forces will affect any object.

The “Push-Me-Pull-You” made famous by Dr Doolittle

If the forces are ‘balanced’ (dread word! — saying ‘total force is zero’ is better, I think) then the object will not accelerate: that is the essence of the First Law. If the sum of the forces is anything other than zero, then the object will accelerate; and what is more, it will accelerate at a rate that is directly proportional to the total force and inversely proportional to the mass of the object; and let’s not forget that it will also accelerate in the direction in which the total force acts. Acceleration is, after all, a vector quantity.

So far, so good. But what about the Third Law? It goes without saying, I hope, that Newton’s Third Law is also universal, but it tells us something different from the first two.

The first two tell us how forces affect objects; the third tells us how objects affect objects: in other words, how objects interact with each other.

The word ‘interact’ can be defined as ‘to act in such a way so as to affect each other’; in other words, how an action produces a reaction. However, the word ‘reaction’ has some unhelpful baggage. For example, you tap my knee (lightly!) with a hammer and my leg jerks. This is a reaction in the biological sense but not in the Newtonian sense; this type of reaction (although involuntary) requires the involvement of an active nervous system and an active muscle system. Because of this, there is a short but unavoidable time delay between the stimulus and the response.

The same is not true of a Newton Third Law reaction: the action and reaction happen simultaneously with zero time delay. The reaction is also entirely passive as the force is generated by the mere fact of the interaction and requires no active ‘participation’ from the ‘acted upon’ object.

I try to avoid the words ‘action’ and ‘reaction’ in statements of Newton’s Third Law for this reason.

If body A exerts a force on body B, then body B exerts an equal and opposite force on body A.

The best version of Newton’s Third Law (imho)

In our universe, body B simply cannot help but affect body A when body A acts on it. Newton’s Third Law is the first step towards understanding that of necessity we exist in an interconnected universe.


Getting the Third Law wrong…

Let’s consider a stationary teapot. (Why not?)

This is NOT a good illustation of Newton’s Third Law…

We can reject this as an appropriate example of Newton’s Third Law for two reasons:

  • Reason 1: Force X and Force Y are acting on a single object. Newton’s Third Law is about the forces produced by an interaction between objects and so cannot be illustrated by a single object.
  • Reason 2: Force X and Force Y are ‘equal’ only in the parochial and limited sense of being merely ‘equal in magnitude’ (8.2 N). They are very different types of force: X is an action-at-distance gravitational force and Y is an electromagnetic contact force. (‘Electromagnetic’ because contact forces are produced by electrons in atoms repelling the electrons in other atoms.) The word ‘equal’ in Newton’s Third Law does some seriously heavy lifting…

Getting the Third Law right…

The Third Law deals with the forces produced by interactions and so cannot be shown using a single diagram. Free body diagrams are the answer here (as they are in a vast range of mechanics problems).

This is a good illustration of Newton’s Third Law

The Earth (body A) pulls the teapot (body B) downwards with the force X so the teapot (body B) pulls the Earth (body A) upwards with the equal but opposite force W. They are both gravitational forces and so are both colour-coded black on the diagram because they are a ‘Newton 3 pair’.

It is worth noting that, applying Newton’s Second Law (F=ma), the downward 8.2 N would produce an acceleration of 9.8 metres per second per second on the teapot if it was allowed to fall. However, the upward 8.2 N would produce an acceleration of only 0.0000000000000000000000014 metres per second per second on the rather more massive planet Earth. Remember that the acceleration produced by the resultant force is inversely proportional to the mass of the object being accelerated.

Similarly, the Earth’s surface pushes upward on the teapot with the force Y and the teapot pushes downward on the Earth’s surface with the force Z. These two forces form a Newton 3 pair and so are colour-coded red on the diagram.

We can summarise this in the form of a table:


Testing understanding

One the best exam questions to test students’ understanding of Newton’s Third Law (at least in my opinion) can be found here. It is a really clever question from the legacy Edexcel specificiation which changed the way I thought about Newton’s Third Law because I was suddenly struck by the thought that the only force that we, as humans, have direct control over is force D on the diagram below. Yes, if D increases then B increases in tandem, but without the weighty presence of the Earth we wouldn’t be able to leap upwards…

The Force Is Zero With This One

There is nothing absurd about perpetual motion; everywhere we look—at the planets wheeling around the Sun, or the electrons circling the heart of the atom—we see examples of it. Where there is no friction, as in airless space, an object can keep moving forever.

Arthur C. Clarke, Things That Can Never Be Done (1972)

So, according to Arthur C. Clarke, famous author and originator of the idea of the geosynchronous communications satellite, perpetual motion is no big deal. What is a big deal, of course, is arranging for such perpetual motion to occur on Earth. This is a near impossibilty, although we can get close: Clarke suggests magnetically suspending a heavy flywheel in a vacuum. (He also goes on to make the point that perpetual motion machines that can do useful external work while they run are, in fact, utterly impossible.)

The lack of visible examples of perpetual motion on Earth is, I think, why getting students to accept Newton’s First Law of Motion is such a perpetual struggle.

Newton’s First Law of Motion

Every body continues in its state of rest or uniform motion in a straight line, except insofar as it doesn’t.

A. S. Eddington, The Nature Of The Physical World (1958)

I think Sir Arthur Eddington was only half-kidding when he mischieviously rewrote the First Law as above. There is a sense where the First Law is superfluous.

It is superfluous because, technically, it is subsumed by the more famous Second Law which can be stated as F=ma where F is the resultant (or total) force acting on an object. Where the acceleration a is zero, as it would be for ‘uniform motion in a straight line’, then the resultant force is zero.

However, the point of the First Law is to act as the foundation stone of Newtonian dynamics: any deviation of an object from a straight line path is taken as implying the existence of a resultant force; if there is no deviation there no force and vice versa.

The First Law is an attempt to reset our Earth-centric intuition that a resultant force is needed to keep things moving, rather than change the perpetual motion of an object. In other words, our everyday, lived experience that to make (say) a box of books move with uniform motion in a straight line we need to keep pushing is wrong

Newton’s First Law, for real this time

A more modern formulation of Newton’s First Law might read:

An object experiencing zero resultant force will either: (a) remain stationary; or, (b) keep moving a constant velocity.

In my experience, students generally have no difficulty accepting clause (a) as long as they understand what we mean by a ‘zero resultant force’. As in so many things, example is possibly the best teacher:

The meaning of zero and non-zero resultant force

Clause (b) uses the concept of ‘constant velocity’ to avoid the circuitous ‘uniform motion in a straight line’. It is this second clause which gives many students difficulty because they hold the misconception that force is needed to sustain motion rather than change it.

And, truth be told, bearing in mind what students’ lived experience of motion on Earth is, it’s easy to see why they find clause (b) so uncongenial.

Is there any way of making clause (b) more palatable to your typical GCSE student?

Galileo, Galileo — magnifico!

Galileo, Galileo,
Galileo, Galileo,
Galileo Figaro - magnifico!

Queen, Bohemian Rhapsody

Galileo Galilei was one of the giants on whose shoulders Newton stood. His principle of inertia anticipated Newton’s First Law by nearly a century.

What follows is a variation of a ‘thought experiment’ that Galileo advanced in support of the principle of inertia; that is to say, that objects will continue moving at a constant velocity unless they are acted on by a resultant force. (A similar version from the IoP can be found here.)

Galileo’s U-shaped Track for the principle of inertia

Picture a ball placed at point A on the track and released.

Galieo’s U-shaped track (1/4): what we see in the real world…

What we see is the ball oscillating back and forth along the track. However, what we also observe is that the height reached by the ball gradually decreases. This is because of resistive forces that slow down the ball (e.g. friction between the track and the ball and air resistance).

What would happen if we stretched out one side of the curve to make a flat line?

Galieo’s U-shaped track (2/4): an extrapolation from what we see in the real world…

We surmise that the ball would come to a stop at some distant point B because of the same resistive forces we observed above.

Next, we return to the U-shaped track and think about what would happen if we lived in a world without any resistive forces.

Galieo’s U-shaped track (3/4): an extrapolation of what we would see in a world with no resistive forces…

The ball would oscillate back-and-forth between A and B. The height would not decrease as there would be no resistive forces.

Finally, what would happen in our imaginary, perfectly frictionless world if we stretched out one side of the ‘U’?

Galieo’s U-shaped track (4/4): a further extrapolation of what we might see in a world with no resistive force…

The ball would keep moving at a constant velocity because there would be no resistive forces to make it slow down.

This, then, is the way things move when no forces are acting on them: when the (resultant) force is zero, in other words.

Conclusion

Galileo framed the argument above (although he used a V-shaped track rather than a U-shaped one) to persuade a ‘tough crowd’ of Aristotleans of the plausibility of the principle of inertia.

In my experience, it can be a helpful argument to persuade even a ‘tough crowd’ of GCSE students to look at the world anew through a Newtonian lens…

Postscript

My excellent edu-Twitter colleague Matt Perks (@dodiscimus) points out that you can model Galileo’s U-shaped track using the PhET Energy Skate Park simulation and that you can even set the value of friction to zero and other values.

Click on the link above, select Playground, build a U-shaped track, set the friction slider to a certain value and away you go!

Using the PhET Energy Skate Park to model Galileo’s U-shaped track
You can even model the decrease in speed because of frictional forces!

This could be a real boon to helping students visualise the thought experiment.

Photosynthesis and Energy Stores

Getting a group of British physics teachers to agree to a new consensus is like herding cats: much easier in principle than in practice.

Herding cats is much easier with CGI… (Screenshot taken from https://www.youtube.com/watch?v=qola8nvoZm4)

However, it seems to be me that, generally speaking, the IoP (Institute of Physics) has persuaded a critical mass of physics teachers that their ‘Energy Stores and Pathways’ model is indeed a Good Thing.

It very much helps, of course, that all the examination boards have committed to using the language of the Energy Stores and Pathways model. This means that the vast majority of physics education resources (textbooks and revision guides and so on) now use it as well — or at least, the physics sections do.

Energy Stores and Pathways: a very brief overview

There’s a bit more to the new model than adding the word ‘store’ to energy labels so that ‘kinetic energy’ becomes ‘kinetic energy store’; although, truth be told, that’s not a bad start.

I have banged on about this model many times before (see the link here) so I won’t go into detail now. For now, I suggest that we stick with the First Rule of the IoP Energy Club….

With apologies to Brad Pitt and Chuck Palahniuk, Image from ‘Fight Club’ (1999)

You can also read the IoP’s own introduction to the Energy Stores and Pathways model (see link here).

The Problem with Photosynthesis

The problem with photosynthesis is that it is often described in terms of ‘light energy’. The IoP Energy Stores and Pathways model does not recognise ‘light’ as an energy store because it does not persist over a significant period of time in a single well-defined location. Rather, light is classified as an ‘energy carrier’ or pathway (see also this link)

A ‘bad’ description of photosynthesis which doesn’t use the Energy Stores and Pathways language

It is possible that the problem is simply one of resource authors using familiar but outdated language. It would seem that exam board specifications are punctilious in avoiding the term ‘light energy’; for example, see below.

From the AQA Combined Science (Trilogy) specification, page 39

How to describe Photosynthesis using the Energy Stores and Pathways model

It’s very simple: just say ‘plants absorb the energy carried by light’ rather than ‘plants absorb light energy’.

In diagram form, the difference can represented as follows:

Conclusion

Over time, I think that the vast majority of physics teachers (in at least in the UK) have come to see the value of the ESP (Energy Stores and Pathways) approach.

I this that I speak for most physics teachers when we hope that biology and chemistry teachers will come to the same conclusion.

H’mmm…if navigating physics teachers towards a consensus is like herding cats, then to what can we liken doing the same for a combined group of physics, biology and chemistry teachers? Perhaps herding a conglomeration of cats, dogs and gerbils across the boundless, storm-wracked prairies of Tornado Alley. In the dark. With both hands tied behind your back.

Wish us luck.

The Acceleration Required Practical Without Light Gates PART DEUX

I have written about completing the acceleration practical without light gates before but I thought I’d share a slight variation on the original method that I have found to work well with my teaching groups. Links to some digital resources (spreadsheet, powerpoint and worksheet) will be included.

The method does not require light gates or a data logger. In fact, the only measuring instruments needed are a metre rule and a stop clock. The other items are standard laboratory equipment (dynamic trolley, bench pulley, string. 4 x 10 g masses on a hanger, 4 x 100 g masses on a hanger, and wooden runway). If your class can access IT then a rather clever spreadsheet is included, but this is not essential.

We use small 10 g masses to accelerate the trolley so the time it takes to travel a certain distance (between 0.50 to 0.90 m) can be timed manually with a stop clock (typical time for the 10 g mass is between 3 and 5 seconds).

This works well as a class practical, especially if you follow Adam Boxer’s excellent ‘Slow Practical’ method.

The Powerpoint that I use to run this practical can be downloaded here.

Set up a friction-compensated slope

The F in Newton’s Second Law stands for the resultant force (or total force) so ideally we should eliminate any frictional force tending to slow down the trolley. This can be done by tilting the runway slightly as shown.

Using one or two 100 g slotted masses propped under one end of the runway provides enough of a slope so that the trolley continues moving at a steady speed when given a short, gentle push. Use trial and error to find the precise angle of the slope needed.

Students should mark START and STOP lines on the runway and measure the distance s between them and record it on the worksheet (or in the spreadsheet).

Make sure the weight stack does not hit the ground before the trolley crosses the stop line, otherwise the results will be unreliable as the trolley will not be accelerating over the full distance.

Use a system of constant mass

Increase the mass of the trolley, but keep F fixed

Calculate the acceleration

The force of the weight stack on the trolley can be calculated using W=mg where m is the mass in kilograms and g is the gravitational field strength of 9.81 N/kg, although the approximation 10 g = 0.10 N can be useful if students are performing the calculations and plotting the graph manually.

Students can use the formula a = 2s/t2 to calculate the acceleration manually. Note that the units of this expression are m/s2 as we would expect for a valid equation for acceleration.

A derivation of this expression suitable for GCSE students is outlined on Slide 5 of the Powerpoint.

If students have access to tablets or computers, they can use this spreadsheet to automatically calculate the results and plot the graph. Students can print the graph if they click on the relevant tab. (The line of best fit is not included as all students generally benefit from practicing this skill!)

Evaluate the results

Students can evaluate the results using Slide 7 of the Powerpoint.

Note that in the graph shown, although there is a convincing straight line of best fit, there is also a noticeable systematic error: the acceleration is slightly too small for the indicated force. This would suggest that the runway was not tilted steeply enough to eliminate all frictional forces.

If you find this blog and resources useful, please leave a comment and/or share it on Twitter 🙂

Magnetism? THERE IS NO MAGNETISM!!!!

Has a school physics experiment or demonstration ever changed the course of human history?

On 21 April 1820, one such demonstration most definitely did. According to physics lore, Hans Christian Øersted was attempting to demonstrate to his students that, according to the scientific understanding of the day, there was in fact no connection between magnetism and electricity.

To this laudable end, he placed a compass needle near to a wire to show that when the current was switched on, the needle would not be affected.

Except that it was affected. Frequently. Each and every time Øersted switched on the electric current, the needle was deflected from pointing North.

Everybody has heard that wise old saw that ‘If it doesn’t work, it’s physics…” except that in this case ‘It did actually work as it was supposed to but in an unexpected way due to a hitherto-unknown-completely-new-branch-of-physics.’

Øersted, to his eternal credit, did not let it lie there and was a pioneer of the new science of electromagnetism.

Push-me-pull-you: or, two current-carrying conductors

One curious consequence of Øersted’s new science was the realisation that, since electric currents create magnetic fields, two wires carrying electric currents will exert a force on each other.

Let’s consider two long, straight conductors placed parallel to each other as shown.

Screenshot 2020-03-22 at 14.39.37.png

In the diagram above, the magnetic field produced by the current in A is shown by the green lines. Applying Fleming’s Left Hand Rule* to conductor B, we find that a force is produced on B which acts towards conductor A. We could go through a similar process to find the force acting on B, but it’s far easier to apply Newton’s Third Law instead: if body A exerts a force on body B, then body B exerts an equal and opposite force on body A. Hence, conductor A experiences a force which pulls it towards conductor B.

So, two long, straight conductors carrying currents in the same direction will be attracted to each other. By a similar analysis, we find that two long, straight conductors carrying currents in opposite directions will be repelled from each other.

In the past, this phenomenon was used to define the ampere as the unit of current: ‘The ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed 1 m apart in vacuum, would produce between these conductors a force equal to 2×10−7 newton per metre of length.‘ However, the 2019 redefinition of the SI system has ditched this and adopted a new definition in terms of the transfer of the elementary charge, e.

Enter Albert Einstein, pursuing an enigma

What is the connection between magnetism and electricity? It was precisely this puzzle that started Albert Einstein on the road to special relativity. It is one of the unsung triumphs of this theory that it lays bare the connection between magnetism and electricity.

In what follows, we’re going to apply Einstein’s analysis to the situation of two long, straight current-carrying conductors. Acknowledgment: I’m going to following a line of argument laid out in Beiser 1988: 19-22.

It’s gotta be perfect (or ‘idealised’, if you prefer)

Let’s consider two idealised conductors A and B both at rest in the inertial reference frame of the laboratory. The flow of charge in both conductors is made up of positive and negative charge carriers moving in opposite directions with a speed v.

None of the charges in A interact with the other charges in A because we are considering an idealised conductor. However, the charges in A will interact with the charges in B.

Screenshot 2020-03-22 at 16.08.06.png
Two conductors viewed from the inertial frame of the laboratory

Flip the inertial reference frame

Now let’s look at the situation from the inertial reference frame of one of the positive charges in A. For simplicity, we can focus on a single positive charge in A since it does not interact with any of the other charges in A.

With reference to this inertial frame, the positive charge in A is stationary and the positive charges in B are also stationary.

However, the inertial frame of the laboratory is moving right-to-left with a speed v and the negative charges are moving right-to-left with a speed of 2v.

Screenshot 2020-03-22 at 16.13.28.png
The same two conductors viewed from the inertial frame of one of the positive charges in conductor A. Note that all the positive charges are now stationary; the laboratory is moving with speed v right to left, and the negative charges are moving with speed 2v right to left

Since the positive charges in B are stationary with respect to the positive charge in A, the distance between them is the same as it was in the laboratory inertial frame. However, since the negative charges in B are moving with speed 2v with respect to positive charge in A, the spacing between is contracted due to relativistic length contraction (see Lottie and Lorentzian Length Contraction).

Because of this, the negative charge density of B increases since they are closer together. However, the positive charge density of B remains the same since they are stationary relative to the positive charge in A so there is no length contraction.

This means that, as far as the positive charge in A is concerned, conductor B has a net negative charge which means the positive charge experiences an attractive Coulomb’s Law electrical force towards B.

A similar analysis applied to electric currents in opposite directions would show that the positive charge in A would experience a repulsive Coulomb’s Law electrical force. The spacing between the positive charges in B would be contracted but the spacing between the negative charges remains unchanged, so conductor B has a net positive charge because the positive charge density has increased but the negative charge density is unchanged.

Magnetism? THERE IS NO MAGNETISM!!!!

So what we normally think of as a ‘magnetic’ force in the inertial frame of the laboratory can be explained as a consequence of special relativity altering the charge densities in conductors. Although we have just considered a special case, all magnetic phenomena can be interpreted on the basis of Coulomb’s Law, charge invariance** and special relativity.

For the interested reader, Duffin (1980: 388-390) offers a quantitative analysis where he uses a similar argument to derive the expression for the magnetic field due to a long straight conductor.

Update: I’m also indebted to @sbdugdale who points out the there’s a good treatment of this in the Feynman Lectures on Physics, section 13.6.

Notes and references

* Although you could use a non-FLHR catapult field analysis, of course

** ‘A current-carrying conductor that is electrically neutral in one frame of reference might not be neutral in another frame. How can this observation be reconciled with charge invariance? The answer is that we must consider the entire circuit of which the conductor is a part. Because a circuit must be closed for a current to occur in it, for every current element in one direction that a moving observer find to have, say, a positive charge, there must be another current element in the opposite direction which the same observer finds to have a negative charge. Hence, magnetic forces always act between different parts of the same circuit, even though the circuit as a whole appears electrically neutral to all observers.’ Beiser 1988: 21

Beiser, A. (1988). Concepts of modern physics. Tata McGraw-Hill Education

Duffin, W. J. (1980). Electricity and magnetism. McGraw-Hill.

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.

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.

Postscript

You can read Part Deux of this blogpost, which details an adaptation of this experiment to work with dynamics trolleys and other standard laboratory equipment.