You can read Part 1 which introduces the idea of free body force diagrams here.
Essentially the technique we will use is as follows:
Draw a situation diagram with NO FORCE ARROWS.
‘Now let’s look at the forces acting on just object 1’ and draw a separate free body diagram (i.e. a diagram showing just object 1 and the forces acting on it)
Repeat step 2 for some or all of the other objects at your discretion.
(Optional) Link all the diagrams with dotted lines to emphasise that they are facets of a more complex, nuanced whole
The Wheel Thing
Let’s consider a car travelling at a constant velocity of 20 miles per hour.
’20 m.p.h.’ is such an uncivilised unit so let’s use the FIFA system to change it into more civilised scientific S.I. units:
Note that point A on the car tyre is moving at 8.9 m/s due to the rotation of the wheel, as well as moving at 8.9 m/s with the rest of the car. This means that point A is moving at 8.9 + 8.9 = 17.8 m/s relative to the ground.
More strangely, point B on the car tyre is moving backwards at speed of 8.9 m/s due to the rotation of the wheel, as well as moving forwards at 8.9 m/s with the rest of the car. Point B is therefore momentarily stationary with respect to the ground.
The tyres can therefore ‘grip’ the road surface because the contact points on each tyre are stationary with respect to the road surface for the moment that they are in position B. If this was not the case, then the car would be difficult to control as it would be in a skid.
(Apologies for emphasising this point — I personally find it incredibly counterintuitive! Who says wheels are not technologically advanced!)
Forces on a tyre
Assuming the car in the diagram is a four wheel drive, the total force driving it forward would be 4 x 330 N = 1320 N. Since it is travelling at a constant speed, this means that there is zero resultant force (or total force). We can therefore infer that the total resistive force acting on the car is 1320 N.
It is can also be slightly disconcerting that the force driving the car forward is a frictional force because we usually speak of frictional forces having a tendency to ‘oppose motion’.
And so they are in this case also. The movement they are opposing is the relative motion between the tyre surface and the road. Reduce the frictional force between the road with oil or mud, and the tyre would not ‘lock’ on the surface and instead would ‘spin’ in place. It’s worth bearing in mind (and communicating to students) that the tread pattern on the tyre is designed to maximise the frictional force between the tyre surface and the road
And then a step to the right…
It’s just a jump to the left
And then a step to the right
The Time Warp, Rocky Horror Picture Show
We can see how important friction is for taking a step forward in the above diagrams. Again, it is worth pointing out to students how much effort goes into designing the ‘tread’ on certain types of footwear so as to maximise the frictional force. On climbing boots, the ‘tread’ extends on to the upper surface of the boot for that very reason.
One step beyond
Let’s apply a similar analysis to the case of a person stepping off a boat that happens not be tied to the mooring.
The person pushes back on the boat (gripping the boat with friction as above). By Newton’s Third Law, this generates an equal an opposite force on the boat. There is no horizontal force to the right due to the tension in the rope, since there is no rope(!) This means that there is a resultant force on the boat to the left so the boat accelerates to the left.
The forces on the person and the boat will be equal in magnitude, but the acceleration will depend on the mass of each object from F = ma.
Since the boat (e.g. a rowing boat) is likely to have a smaller mass than the person, its acceleration to the left will be higher in magnitude than the acceleration of the person to the right — which will lead to the unfortunate consequence shown below.
The acceleration of the person and the boat happens only when the person and boat are in contact with each other, since this is the only time when there will be a resultant force in the horizontal direction.
Note that although force arrows on a situation diagram should be discouraged for the sake of clarity, there is an argument for drawing velocity and acceleration arrows on the situation diagram as a form of dual coding. Further details can be found here, and an explanation of why acceleration is shown as a double headed arrow.
The velocity to the left built up by the boat in this short instant will be greater than the velocity to the right built up by the person, because the acceleration of the boat is greater, as argued above.
The outcome, of course, is that the person falls in the water, which has been the subject of countless You’ve Been Framed clips.
In the next post, I will try to move beyond horizontal forces and take account of the normal reaction force when an object rests on both horizontal surfaces and inclined surfaces.
Why do so many students hold pernicious and persistent misconceptions about forces?
Partly, I think, because of the apparent clash between our intuitive, gut-level knowledge of real world physics. For example, a typical student might find the statement ‘If I push this box, it will stop moving shortly after I stop pushingbecause force is needed to move things‘ entirely unobjectionable; whilst in the theoretical, rarefied world of the physicist the statement ‘The box will keep moving at a constant velocity after I stop pushing it, unless it is acted on by a resultant force such as friction‘ would get a tick whereas the former would get a big angry X and and a darkly muttered comment about ‘bloody Aristotleans.’
After all, ‘pernicious’ is in the eye of the beholder. Physics teachers have to remember that they suffer mightily under the ‘curse of knowledge’ and have forgotten what it’s like to look at the world through anything than the lens of Newtonian mechanics.
We learn about the world through the power of example. Human beings are ‘inference engines’: we strive to make sense of the world by constructing general rules based on the examples presented to us.
Many of the examples of forces in action presented to students are in the form of force diagrams; and in my experience, all too many force diagrams add to students’ confusion.
A bad force diagram
Over the years, I have seen many versions of this diagram. To my own chagrin, I must admit that I, personally, have drawn versions of this diagram in the past. But I now recognise it has one major, irredeemable flaw: the arrows are drawn hanging in mid-air.
OK, let’s address this. Is this better?
No, it isn’t because it is still unclear which forces are acting on which object. Is the blue 75 N arrow the person pushing the cart forward or the cart pulling the person forward? Is the red 75 N arrow the cart pushing back on the person or the person pulling back on the cart?
From both versions of this diagram shown above: we simply cannot tell.
As a consequence, I think the explanatory value of this diagram is limited.
Free Body Diagrams to the Rescue!
A free body diagram is simply one where we consider the forces on each object in the situation in turn.
We begin with a situation diagram. This shows the relationship between the objects we are considering. Next, we draw a free body diagram for each object; that is, we draw each object involved and consider the forces acting on it.
From version 3 of Force Diagram 1, we can see that it was an attempt to illustrate Newton’s Third Law i.e. that if body A exerts a force on body B then body B exerts an equal and opposite force on body A.
Another bad force diagram
This is a bad force diagram because it is unclear which forces are acting on the cart and which are acting on the person. Apart from a very general ‘Well, 50 N minus 50 N means zero resultant force so zero acceleration’, there is not a lot of information that can be extracted from this diagram.
Also, the most likely mechanism to produce the red retarding force of 50 N is friction between the wheels of the cart and the ground (and note that since the cart is being pushed by an external body and the wheels are not powered like those of a car, the frictional force opposes the motion). Showing this force acting on the handle of the cart is not helpful, in my opinion.
Free body diagrams to the rescue (again)!
The Newton 3 pairs are colour coded. For example, the orange 50 N forward force on the person (object A) is produced as a direct result of Newton’s 3rd Law because the person’s foot is using friction to grip the floor surface (object B) and push backwards on it (the orange arrow in the bottom diagram).
This diagram shows a complete free body diagram body analysis for all three objects (cart, person, floor) involved in this simple interaction.
I’m not suggesting that all three free body diagrams always need to be discussed. For example, at KS3 the discussion might be limited at the teacher’s discretion to the top ‘Forces on Cart’ diagram as an example of Newton’s First Law in action. Or equally, the teacher may wish to extend the analysis to include the second and third diagrams, depending on their own judgement of their students’ understanding. The Key Stage ticks and crosses on the diagram are indicative suggestions only.
At KS3 and KS4, there is not a pressing need to explicitly label this technique as ‘free body force diagrams’. Instead, what I suggest (perhaps after drawing the situation diagram without any force arrows on it) is the simple statement that ‘OK, let’s look at the forces acting on just the cart’ before drawing the top diagram. Further diagrams can be introduced with a similar statements such as ‘Next, let’s look at the forces acting on just the person’ and so on. Linking the diagrams with dotted lines as shown is, I think, useful in not losing sight of the fact that we are dealing piecemeal with a complex and nuanced whole.
The free body force diagram technique (whether or not the teacher decides to explicitly call it that) offers a useful tool that will allow us all to (fingers crossed!) draw better force diagrams.
Draw a situation diagram with NO FORCE ARROWS.
‘Now let’s look at the forces acting on just object 1’ and draw a separate free body diagram (i.e. a diagram showing just object 1 and the forces acting on it)
Repeat step 2 for some or all of the other objects at your discretion.
(Optional) Link all the diagrams with dotted lines to emphasise that they are facets of a more complex, nuanced whole
In the next post, I hope to show how the technique can be used to explain common problems such as how a car tyre interacts with the ground to drive a car forward.
There are three things that everyone should know about simple harmonic motion (SHM).
Firstly, it is simple;
Secondly, it is harmonic;
Thirdly, it is a type of motion.
There, my work here is done. H’mmm — it looks like this physics teaching lark is much easier than is generally acknowledged…
[The above joke courtesy of the excellent Blackadder 2 (1986), of course.]
Misconceptions to the left of us, misconceptions to the right of us…
In my opinion, the misconceptions which hamper students’ attempts to understand simple harmonic motion are:
A shallow understanding of dynamics which does not differentiate between ‘displacement’ ‘velocity’ and ‘acceleration’ but lumps them together as interchangeable flavours of ‘movement’
The idea that ‘acceleration’ invariably leads to an increase in the magnitude of velocity and that only the materially different ‘deceleration’ (which is exclusively produced by resistive forces such as friction or drag) can result in a decrease.
Not understanding the positive and negative direction conventions when analysing motion.
All of these misconceptions can, I believe, be helpfully addressed by using a form of dual coding which I outlined in a previous post.
Top Gear presenters: Assemble!
The discussion context which I present is that of a rather strange episode of the motoring programme Top Gear. You have been given the opportunity to win the car of your dreams if — and only if — you can drive it so that it performs SHM (simple harmonic motion) with a period of 30 seconds and an amplitude of 120 m.
This is a fairly reasonable challenge as it would lead to a maximum acceleration of 5.3 m s-2. For reference, a typical production car can go 0-27 m/s in 4.0 s (a = 6.8 m s-2)) but a Tesla Model S can go 0-27 m/s in a scorching 2.28 s (a = 11.8 m s-2). BTW ‘0-27 m/s’ is the SI civilised way of saying 0-60 mph. It can also be an excellent extension activity for students to check the plausibility of this challenge(!)
Timing and the Top Gear SHM Challenge
At what time should the car reach E on its outward journey to ensure we meet the Top Gear SHM Challenge? (15 s since A to E is half of a full oscillation and T should be 30 seconds according to the challenge)
At what time should the car reach C? (7.5 s since this is a quarter of a full oscillation.)
All physics teachers, to a greater or lesser degree, labour under the ‘curse of knowledge’. What we think is ‘obvious’ is not always so obvious to the learner. There is an egregiously underappreciated value in making our implicit assumptions and thinking explicit, and I think diagrams like the above are invaluable in this process.
But what is this SHM (of which you speak of so knowledgeably) anyway?
Simple harmonic motion must fulfil two conditions:
The acceleration must always be directed towards a fixed point.
The magnitude of the acceleration is directly proportional to its displacement from the fixed point.
In other words:
Let’s look at this definition in terms of our fanciful Top Gear challenge. More to the point, let’s look at the situation when t = 0 s:
Questions that could be discussed here:
Why is the displacement at A labelled as ‘+120 m’? (Displacement is a vector and at A it is in the same direction as the [arbitrary] positive direction we have selected and show as the grey arrow labelled +ve.)
The equation suggests that the value of a should be negative when x is positive. Is the diagram consistent with this? (Yes. The acceleration arrow is directed towards the fixed point C and is in the opposite direction to the positive direction indicated by the grey arrow.)
What is the value of a indicated on the diagram? Is this consistent with the terms of the challenge? (Zero. Yes, since 100 m is the required amplitude or maximum displacement so if v was greater than zero at this point the car would go beyond 100 m.)
How could you operate the car controls so as to achieve this part of simple harmonic motion? (You should be depressing the gas pedal to the floor, or ‘pedal to the metal’, to achieve maximum acceleration.)
Model the thinking explicitly
Hands up who thinks the time on the second clock on the diagram above should read 3.75 seconds? It makes sense, doesn’t it? It takes 7.5 s to reach C (one quarter of an oscillation) so the temptation to ‘split the difference’ is nigh on irresistible — except that it would be wrong — and I must confess, it took several revisions of this post before I spotted this error myself (!).
The vehicle is accelerating, so it does not cover equal distances in equal times. It takes longer to travel from A to B than B to C on this part of the journey because the vehicle is gaining speed.
So what is the time when x = 60 m
So we can redraw the diagram as follows:
Some further questions that could be asked are:
Is the acceleration arrow at B smaller or larger than the acceleration arrow at A? Is this consistent with what we know about SHM? (Smaller. Yes, because for SHM, acceleration is proportional to displacement. The displacement at B is +60 m; the acceleration at B is half the value of the acceleration at A because of this. Note that the magnitude of the acceleration is reduced but the direction of a is still negative since the displacement is positive.)
Is the velocity at B positive or negative? (Negative, since it is opposite to the positive direction selected on the diagram and shown by the grey ‘+ve’ arrow.)
Is the magnitude of the velocity at B smaller or larger than at A, and is this consistent with a negative acceleration? (Larger. Yes, since both acceleration and velocity are in the same direction. Note that this is an important point to highlight since many students hold the misconception that a negative acceleration is always a ‘deceleration’.)
How could you operate the car controls so as to achieve this part of simple harmonic motion? (You should have eased off the gas pedal at this point to achieve half the acceleration obtained at A.)
Next, we move on to this diagram and ask students to use their knowledge of SHM to decide the values of the question marks on the diagram.
Which hopefully should lead to a diagram like the one below, and realisation that at this point, the driver’s foot should be entirely off the gas pedal.
‘Are we there yet?’
And thence to this:
One of the most salient points to highlight in the above diagram is the question: how could you operate the car controls at this point? The answer is of course, that you would be pressing the foot brake pedal to achieve a medium magnitude deceleration. This is often a point of confusion for students: how can a positive acceleration produce a decrease in the magnitude of the velocity? Hopefully, the dual coding convention suggested in this blog post will make this clearer to students.
‘No, really, ARE WE THERE YET?!!’
Over time, we can build up a picture of a complete cycle of SHM, such as the one show below. This shows the car reversing backwards at t = 25 s while the driver gradually increases the pressure on the brake.
From this, it should be easier to relate the results above to graphs of SHM:
A quick check reveals that the displacement is positive and half its maximum value; the acceleration is negative and half of its maximum magnitude; and the velocity is positive and just below its maximum value (since the average deceleration is smaller between C and B than it will be between B and A) .
I shall leave the final word to the estimable Top Gear team…
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.
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.
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.
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.
** ‘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.
There is little doubt that students find understanding how an electric motor works hard.
What follows is an approach that neatly sidesteps the need for applying Fleming’s Left Hand Rule (FLHR) by using the idea of the catapult field.
The catapult field is a neat bit of Physics pedagogy that appears to have fallen out of favour in recent years for some unknown reason. I hope to rehabilitate and publicise this valuable approach so that more teachers may try out this electromagnetic ‘road less travelled’.
(Incidentally, if you are teaching FLHR, the mnemonic shown above is not the best way to remember it: try using this approach instead.)
The magnetic field produced by a long straight conductor
Moving electric charges produce magnetic fields. When a current flows through a conductor, it produces a magnetic field in the form of a series of cylinders centred on the wire. This is usually shown on a diagram like this:
If we imagine looking down from a point directly above the centre of the conductor (as indicated by the disembodied eye), we would see a plan view like this:
We are using the ‘dot and cross‘ convention (where an X represents an arrow heading away from us and a dot represents an arrow heading towards us) to easily render a 3D situation as a 2D diagram.
The direction of the magnetic field lines is found by using the right hand grip rule.
The thumb is pointed in the direction of the current. The field lines ‘point’ in the same direction as the fingers on the right hand curl.
3D to 2D
Now let’s think about the interaction between the magnetic field of a current carrying conductor and the uniform magnetic field produced by a pair of magnets.
In the diagrams below, I have tried to make the transition between a 3D and a 2D representation explicit, something that as science teachers I think we skip over too quickly — another example of the ‘curse of knowledge’, I believe.
Magnetic Field on Magnetic Field
If we place the current carrying conductor inside the magnetic field produced by the permanent magnets, we can show the magnetic fields like this:
Note that, in the area shaded green, the both sets of magnetic field lines are in the same direction. This leads a to stronger magnetic field here. However, the opposite is true in the region shaded pink, which leads to a weaker magnetic field in this region.
The resultant magnetic field produced by the interaction between the two magnetic fields shown above looks like this.
Note that the regions where the magnetic field is strong have the magnetic field lines close together, and the regions where it is weak have the field lines far apart.
The Catapult Field
This arrangement of magnetic field lines shown above is unstable and is called a catapult field.
Essentially, the bunched up field lines will push the conductor out of the permanent magnetic field.
If I may wax poetic for a moment: as an oyster will form a opalescent pearl around an irritant, the permanent magnets form a catapult field to expel the symmetry-destroying current-carrying conductor.
The conductor is pushed in the direction of the weakened magnetic field. In a highly non-rigorous sense, we can think of the conductor being pushed out of the enfeebled ‘crack’ produced in the magnetic field of the permanent magnets by the magnetic field of the current carrying conductor…
Also, the force shown by the green arrow above is in exactly the same direction as the force predicted by Fleming’s Left Hand Rule, but we have established its direction using only the right hand grip rule and a consideration of the interaction between two magnetic field.
The Catapult Field for an electric motor
First, let’s make sure that students can relate the 3D arrangement for an electric motor to a 2D diagram.
The pink highlighted regions show where the field lines due to the current in the conductor (red) are in the opposite direction to the field line produced by the permanent magnet (purple). These regions are where the purple field lines will be weakened, and the clear inference is that the left hand side of the coil will experience an upward force and the right hand side of the coil will experience a downward force. As suggested (perhaps a little fancifully) above, the conductors are being forced into the weakened ‘cracks’ produced in the purple field lines.
The catapult field for the electric motor would look, perhaps, like this:
On a practical teaching note, I wouldn’t advise dispensing with Fleming’s Left Hand Rule altogether, but hopefully the idea of a catapult field adds another string to your pedagogical bow as far as teaching electric motors is concerned (!)
I have certainly found it useful when teaching students who struggle with applying Fleming’s Left Hand Rule, and it is also useful when introducing the Rule to supply an understandable justification why a force is generated by a current in a magnetic field in the first place.
The catapult field is a ‘road less travelled’ in terms of teaching electromagnetism, but I would urge you to try it nonetheless. It may — just may — make all the difference.
Student: Did you know FIFA is also the name of a video game, Sir?
Student: Yeah. It’s part of a series. I just got FIFA 20. It’s one of my favourite games ever.
Me: Goodness me. I had no idea. I just chose the letters ‘FIFA’ completely and utterly at random!
The FIFA method is an AQA mark scheme-friendly* way of approaching GCSE Physics calculation questions. (It is also useful for some Y12 Physics students.)
I mentioned it in a previous blog and @PedagogueSci was kind enough to give it a boost here, so I thought I’d explain the method in a separate blog post.
The FIFA method:
Avoids the use of formula triangles
Minimises the cognitive load on students when approaching calculations.
Why we shouldn’t use formula triangles
Formula triangles are bad news. They are a cognitive dead end.
During a university admissions interview for veterinary medicine, I asked a prospective student to explain how they would make up a solution for infusion into a dog. Part of the answer required them to work out the volume required for a given amount and concentration. The candidate started off by drawing a triangle, then hesitated, eventually giving up in despair. […]
They are a trick that hides the maths: students don’t apply the skills they have previously learned. This means students don’t realise how important maths is for science.
I’m also concerned that if students can’t rearrange simple equations like the one above, they really can’t manage when equations become more complex.
I believe the use of formula triangle also increases (rather than decreases) the cognitive load on students when carrying out calculations. For example, if the concentration c is 0.5 mol dm-3 and the number of moles n required is 0.01 mol, then in order to calculate the volume V they need to:
recall the relevant equation and what each symbol means and hold it in working memory
recall the layout of symbols within the formula triangle and either (a) write it down or (b) hold it in working memory
recall that n and c are known values and that V is the unknown value and hold this information in working memory when applying the formula triangle to the problem
The FIFA method in use (part 1)
The FIFA acronym stands for:
FINE TUNE (this often, but not always, equates to rearranging the formula)
Lets look at applying it for a typical higher level GCSE Physics calculation question
We add the FIFA rubric:
Students have to recall the relevant equation as it is not given on the Data and Formula Sheet. They write it down. This is an important step as once it is written down they no longer have to hold it in their working memory.
Note that this is less cognitively demanding on the student’s working memory as they only have to recall the formula on its own; they do not have to recall the formula triangle associated with it.
Students find it encouraging that on many mark schemes, the selection of the correct equation may gain a mark, even if no further steps are taken.
Next, we insert the values. I find it useful to provide a framework for this such as:
We can ask general questions such as: “What data are in the question?” or more focused questions such as “Yes or no: are we told what the kinetic energy store is?” and follow up questions such as “What is the kinetic energy? What units do we use for that?” and so on.
Note that since we are considering each item of data individually and in a sequence determined by the written formula, this is much less cognitively demanding in terms of what needs to be held in the student’s working memory than the formula triangle method.
Note also that on many mark schemes, a mark is available for the correct substitution of values. Even if they were not able to proceed any further, they would still gain 2/5 marks. For many students, the notion of incremental gain in calculation questions needs to be pushed really hard otherwise they will not attempt these “scary” calculation questions.
Next we are going to “fine tune” what we have written down in order to calculate the final answer. In this instance, the “fine tuning” process equates to a simple algebraic rearrangement. However, it is useful to leave room for some “creative ambiguity” here as we can also use the “fine tuning” process to resolve difficulties with units. Tempting though it may seem, DON’T change FIFA to FIRA.
We fine tune in three distinct steps (see addendum):
Finally, we input the values on a calculator to give a final answer. Note that since AQA have declined to provide a unit on the final answer line, a mark is available for writing “kg” in the relevant space — a fact which students find surprising but strangely encouraging.
The key idea here is to be as positive and encouraging as possible. Even if all they can do is recall the formula and remember that mass is measured in kg, there is an incremental gain. A mark or two here is always better than zero marks.
The FIFA method in use (part 2)
In this example, we are using the creative ambiguity inherent in the term “fine tune” rather than “rearrange” to resolve a possible difficulty with unit conversion.
In this example, we resolve another potential difficulty with unit conversion during the our creatively ambiguous “fine tune” stage:
The emphasis, as always, is to resolve issues sequentially and individually in order to minimise cognitive overload.
The FIFA method and low demand Foundation tier calculation questions
I teach the FIFA method to all students, but it’s essential to show how the method can be adapted for low demand Foundation tier questions. (Note: improving student performance on these questions is probably a more significant and quicker and easier win than working on their “extended answer” skills).
For the treatment below, the assumption is that students have already been taught the FIFA method in a number of contexts and that we are teaching them how to apply it to the calculation questions on the foundation tier paper, perhaps as part of an examination skills session.
For the majority of low demand questions, the required formula will be supplied so students will not need to recall it. What they will need, however, is support in inserting the values correctly. Providing a framework as shown below can be very helpful:
Also, clearly indicating where the data came from is useful.
The fine tune stage is not needed, so we can move straight to the answer.
Note also that the FIFA method can be applied to all calculation questions, not just the ones that could be answered using formula triangle methods, as in part (c) of the question above.
I believe that using FIFA helps to make our thinking and methods in Physics calculations more explicit and clearer for students.
My hope is that science teachers reading this will give it a go.
*Disclaimer: AQA has not endorsed the FIFA method. I describe it as “AQA mark scheme-friendly” using my professional own judgment and interpretation of published AQA mark schemes.
I am embarrassed to admit that this was the original version published. Somehow I missed the more straightforward way of “fine tuning” by squaring the 30 and multiplying by 0.5 and somehow moved straight to the cross multiplication — D’oh!
My thanks to @BenyohaiPhysics and @AdamWteach for pointing it out to me.
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.
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…
…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!)
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.
Then, set up a marking feedback form in Word similar to this one:
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:
Then Merge the data to produce a document that looks like this:
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.
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.
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
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.)
understand the difference between current, potential difference and resistance and how important each one is
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 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 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.
On the diagrams below, the coulombs are moving clockwise.
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.
A 1 ohm and 2 ohm resistor modelled as strip curtains
FWIW I therefore commend the use of the CTM to all interested parties.
Driver, R., Squires, A., Rushworth, P., & Wood-Robinson, V. (1994). Making sense of secondary science: Research into children’s ideas. Routledge.
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.
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.
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
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.
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).
This blog post grew in the telling. Please let me know if you try the methods outlined here and how successful you found them
Eager, D., Pendrill, A. M., & Reistad, N. (2016). Beyond velocity and acceleration: jerk, snap and higher derivatives. European Journal of Physics, 37(6), 065008.
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
The 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):
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:
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.
Internal Energy Through the Microscopic Lens
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!)
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.
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!
Nave, R. (2000). HyperPhysics. Georgia State University, Department of Physics and Astronomy.