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‘.

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 Revisited (Part 2)

In this post, we will look at understanding potential difference (or voltage) using the Coulomb Train Model.

This is part 2 of a continuing series. You can read part 1 here.

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).

To summarise what has been discussed so far:


Modelling potential difference using the CTM

Potential difference is the ‘push’ needed to make electric charge move around a closed circuit. On the CTM, we can represent the ‘push’ as a gain in the energy of the coulomb. (This is consistent with the actual definition of the volt V = E/Q, where one volt is a change in energy of one joule per coulomb.)

How can we observe this gain in energy? Simple, we use a voltmeter.

Kudos to https://www.circuit-diagram.org/editor/ for the lovely circuit diagrams

On the CTM, this would look like this:

What the voltmeter does is compare the energy contained by two coulombs: one at A and the other at B. The coulombs at B, having passed through the 1.5 V cell, each have 1.5 joules of energy more than than the coulombs at A. This means that the voltmeter in this position reads 1.5 volts. We would say that the potential difference across the cell is 1.5 V. (Try and avoid talking about the potential difference ‘through’ or ‘of’ any part of the circuit.)


More potential difference measurements using the CTM

Let’s move the voltmeter to a different position.

On the CTM, this would look like this:

Let’s make the very reasonable assumption that the connecting wires have zero resistance. This would mean that the coulombs at C have 1.5 joules of energy and that the coulombs at D have 1.5 joules of energy. They have not lost any energy since they have not passed through any part of the circuit that actually has a resistance. The voltmeter would therefore read 0 volts since it cannot detect any energy difference.

Now let’s move the voltmeter one last time.

On the CTM, this would look like this:

Notice that the coulombs at F have 1.5 fewer joules than the coulombs at E. The coulombs transfer 1.5 joules of energy to the bulb because the bulb has a resistance.

Any part of the circuit that has non-zero resistance will ‘rob’ coulombs of their energy. On this very simple model, we assume that only the bulb has a resistance and so only the bulb will ‘push back’ against the movement of the coulombs and cost them energy.

Also on this simple model, the potential difference across the bulb is identical to the potential difference across the cell — but this is not always the case. For example, if the wires had a small but non-negligible resistance and if the cell had an internal resistance, but these would only come into play at A-level.

The bulb is shown as ‘flashing’ on the CTM to provide a visual cue to help students mentally model the transfer of energy from the coulombs to the bulb. In reality, instead of just one coulomb transferring a largish ‘chunk’ of energy, there would be approximately 1.25 billion billion electrons continuously transferring a tiny fraction of this energy over the course of one second (assuming a d.c. current of 0.2 amps) so we wouldn’t see the bulb ‘flash’ in reality.


How do the coulombs ‘know’ how much energy to drop off?

This section is probably more of interest to specialist physics teachers, but all are welcome.

One frequent criticism of donation models like the CTM is how do the coulombs ‘know’ to drop off all their energy at the bulb?

The response to this, of course, is that they don’t. This criticism is an artefact of an (arguably) over-simplified model whereby we assume that only the bulb has resistance. The energy carried by the coulombs according to this model could be shown as a sketch graph, and let’s be honest it does look a little dodgy…

But, more accurately, of course, the energy loss is a process rather than an event. And the connecting wires actually have a small resistance. This leads to this graph:

Realistically speaking, the coulombs don’t lose all their energy passing through the bulb: they merely lose most of their energy here due to the process of passing through a high resistance part of the circuit.

In part 3 of this series, we’ll look at how resistance can be modelled using the CTM.

You can read part 3 here.

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…