Category Archives: Dual Coding

Keep Calm and Draw Free Body Force Diagrams (Part 2)

You can read Part 1 which introduces the idea of free body force diagrams here.

Essentially the technique we will use is as follows:

  1. Draw a situation diagram with NO FORCE ARROWS.
  2. ‘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)
  3. Repeat step 2 for some or all of the other objects at your discretion.
  4. (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.

NOT a force diagram. (Note: whilst force arrows on situation diagrams should be discouraged, there is no equivalent argument for speed arrows)

’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:

NOT a force diagram! (Note: it is fine to draw speed/velocity/acceleration arrows on a situation diagram, but not force arrows.)

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

Situation diagram (note: no force arrows) and free body diagrams for road and tyre. Note also different style of arrow for speed and force.

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
Situation diagram for a person taking a step to the right; and free body diagrams for the person and the floor

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.

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A climbing boot

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.

Situation diagram for a person stepping off an unmoored boat; and free body force diagrams for the person and the boat. Note different style of arrow for forces and acceleration.

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 effect of stepping off an unmoored boat

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.

Next post…

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.

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Fear of Forces? Keep Calm and Draw Free Body Diagrams

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 pushing because 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

Force Diagram 1: version 1 (really bad)

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?

Force Diagram 1: version 2 (still bad)

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.

Force Diagram 1: version 3 (much better!)

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

Force Diagram 2: version 1 (very bad)

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.

Conclusion

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.

  1. Draw a situation diagram with NO FORCE ARROWS.
  2. ‘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)
  3. Repeat step 2 for some or all of the other objects at your discretion.
  4. (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.

You can read Part 2 here.

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SHM and the Top Gear Challenge

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…

Screenshot 2020-05-04 at 17.00.00.png

[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.

Screenshot 2020-05-04 at 18.12.42.png

Top Gear presenters: Assemble!

Screenshot 2020-05-10 at 11.10.37.png

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

Screenshot 2020-05-10 at 10.06.48.png

  • 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:

  1. The acceleration must always be directed towards a fixed point.
  2. The magnitude of the acceleration is directly proportional to its displacement from the fixed point.

In other words:

Screenshot 2020-05-04 at 21.33.31.png

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:

Screenshot 2020-05-10 at 10.08.29.png

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 explicitlyScreenshot 2020-05-10 at 10.17.49.png

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

Screenshot 2020-05-10 at 10.40.24.png

So we can redraw the diagram as follows:

Screenshot 2020-05-10 at 10.41.24

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.

Screenshot 2020-05-10 at 10.44.43.png

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.

Screenshot 2020-05-10 at 10.46.00.png

‘Are we there yet?’

And thence to this:

Screenshot 2020-05-10 at 10.47.18.png

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?!!’

Nearly.

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.

Screenshot 2020-05-10 at 10.48.21.png

From this, it should be easier to relate the results above to graphs of SHM:

Screenshot 2020-05-10 at 10.51.23.png

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) .

And finally…

I shall leave the final word to the estimable Top Gear team…

Screenshot 2020-05-10 at 11.08.21.png

So there you have it: JOB DONE!

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Electric Motors Without The Left Hand Rule

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’.

Screenshot 2020-02-21 at 09.48.42

(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:

Screenshot 2020-02-21 at 09.59.26.png

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:

Screenshot 2020-02-21 at 10.01.22.png

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.

Screenshot 2020-02-21 at 10.38.32.png

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.

Screenshot 2020-02-21 at 11.18.21.png

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:

Screenshot 2020-02-21 at 11.32.31.png

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.

Screenshot 2020-02-21 at 11.33.27.png

The resultant magnetic field produced by the interaction between the two magnetic fields shown above looks like this.

Screenshot 2020-02-21 at 11.37.19.png

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.

Screenshot 2020-02-21 at 11.50.55.png

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.

Screenshot 2020-02-21 at 13.42.51.png

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:

Screenshot 2020-02-21 at 14.39.36.png

And finally…

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.

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Why does kinetic energy = 1/2mv^2?

Why does kinetic energy Ek=½mv2?

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

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

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

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

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

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

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

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

Step 2:    W= F x s

But remember s = v x t so:

Step 3:    W= F x vt

And also remember that F = m x a so:

Step 4:    W= ma x vt

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

Step 5:        Wd = (v / t) x vt

The ts cancel so:

Step 6:    W= mv2

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

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

Screenshot 2019-03-09 at 15.14.17

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

But wait: Ek=mv2!?!?

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

Houston, we have a problem.

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

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

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

So Step 3 should read:

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

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

Step 3:    Wd= F x (v / 2) t

And also remember that F = m x a so:

Step 4:    Wd= ma x (v / 2) t

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

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

The ts cancel so:

Step 6:    Wd= ½mv2

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

Screenshot 2019-03-09 at 17.58.45.png

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

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

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

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

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Dual-coding SUVAT Problems

The theory of dual coding holds that the formation of mental images, in tandem with verbal processing, is often very helpful for learners. In other words, if we support verbal reasoning with visual representations, then better learning happens.

Many years ago, I was taught the dual coding technique outlined below to help with SUVAT problems. Of course, it wasn’t referred to as “dual coding” back then, but dual coding it most definitely is.

I found it a very useful technique at the time and I still find it useful to this day. And what is more, it is in my opinion a pedagogically powerful procedure. I genuinely believe that this technique helps students understand the complexities and nuances of SUVAT because it brings many things which are usually implicit out into the open and makes them explicit.

SUVAT: “Made darker by definition”?

BOSWELL. ‘He says plain things in a formal and abstract way, to be sure: but his method is good: for to have clear notions upon any subject, we must have recourse to analytick arrangement.’

JOHNSON. ‘Sir, it is what every body does, whether they will or no. But sometimes things may be made darker by definition. I see a cow, I define her, Animal quadrupes ruminans cornutum. But a goat ruminates, and a cow may have no horns. Cow is plainer.

— Boswell’s Life of Johnson (1791)

As I see it, the enduring difficulty with SUVAT problems is that such things can indeed be made darker by definition. Students are usually more than willing to accept the formal definitions of s, u, v, a and t and can apply them to straightforward and predictable problems. However, the robotic death-by-algorithm approach fails all too frequently when faced with even minor variations on a theme.

Worse still, students often treat acceleration, displacement and velocity as nearly-synonymous interchangeable quantities: they are all lumped together in that naive “intuitive physics” category called MOVEMENT.

The approach that follows attempts to make students plainly see differences between the SUVAT quantities and, hopefully, as make them as plain as a cow (to borrow Dr Johnson’s colourful phrasing).

Visual Symbols for the Dual-coding of SUVAT problems

Screenshot 2018-12-25 at 12.02.38.png

1.1 Analysing a simple SUVAT problem using dual coding

Problem: a motorcycle accelerates from rest at 0.8 m/s2 for a time of 6.0 seconds. Calculate (a) the distance travelled; and (b) the final velocity.

Screenshot 2018-12-25 at 12.09.42.png

Please note:

  1. We are using the AQA-friendly convention of substituting values before rearrangement. (Some AQA mark schemes award a mark for the correct substitution of values into an expression; however, the mark will not be awarded if the expression is incorrectly rearranged. Weaker students are strongly encouraged to substitute before rearrangement, and this is what I model.)
  2. A later time is indicated by the movement of the hands on the clock.

So far, so blindingly obvious, some might say.

But I hope the following examples will indicate the versatility of the approach.

1.2a Analysing a more complex SUVAT problem using dual coding (Up is positive convention)

Problem: A coin is dropped from rest takes 0.84 s to fall a distance of 3.5 m so that it strikes the water at the bottom of a well. With what speed must it be thrown vertically so that it takes exactly 1.5 s to hit the surface of the water?

Screenshot 2018-12-25 at 14.33.25.png

Another advantage of this method is that it makes assigning positive and negative directions to the SUVAT vectors easy as it becomes a matter of simply comparing the directions of each vector quantity (that is to say, s, u, v and a) with the arbitrarily selected positive direction arrow when we substitute values into the expression.

But what would happen if we’d selected a different positive direction arrow?

1.2b Analysing a more complex SUVAT problem using dual coding (Down is positive convention)

Problem: A well is 3.5 m deep so that a coin dropped from rest takes 0.84 s to strike the surface of the water. With what speed must it be thrown so that it takes exactly 1.5 s to hit the surface of the water?

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The answer is, of course, numerically equal to the previous answer. However, following the arbitrarily selected down is positive convention, we have a negative answer.

1.3 Analysing a projectile problem using dual coding

Let’s look at this typical problem from AQA.

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We could annotate the diagram like this:

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Guiding our students through the calculation:

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Just Show ‘Em!

Some trad-inclined teachers have embraced the motto: Just tell ’em!

It’s a good motto, to which dual coding can add the welcome corollary: Just show ’em!

The Unreasonable Effectiveness of Mathematics in the Natural Sciences

The famous phrase is, of course, from physicist Eugene Wigner (1960: 2):

My principal aim is to illuminate it from several sides. The first point is that the enormous usefulness of mathematics in the natural sciences is something bordering on the mysterious and that there is no rational explanation for it.

Further exploration of the above problem using dual coding can, I believe, give A-level students a glimpse of the truth of Wigner’s phrase.

This Is The Root You’re Looking For

In the calculation above, we found that when s = -1.8 m, v could have a value of plus or minus 6.90 m/s. Since we were interested in the velocity of the kite boarder at the end of the journey, we concluded that it was the negative root that was significant for our purposes.

But does the positive root have any physical significance? Why yes, it does. It indicates the other possible value of v when s = -1.8 m.

The displacement was -1.8 m at only one point on the real journey. However, if the kite boarder had started their projectile motion from the level of the water surface instead of from the top of the ramp, their vertical velocity at this point would have been +6.9 m/s.

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The fact that the kite boarder did not start their journey from this point is immaterial. Applying the mathematics not only tells us about their actual journey, but all other possible journeys that are consistent with the stated parameters and the subset of the laws of physics that we are considering in this problem — and that, to me, borders enough on the mysterious to bring home Wigner’s point.

And finally…

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This information allows us to annotate our final diagram as below (bearing in mind, of course, that the real journey of the kite boarder started from the top of the ramp and not from the water’s surface as shown).

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Let me end on a more cheerful note. The miracle of the appropriateness of the language of mathematics for the formulation of the laws of physics is a wonderful gift which we neither understand nor deserve. We should be grateful for it and hope that it will remain valid in future research and that it will extend, for better or for worse, to our pleasure, even though perhaps also to our bafflement, to wide branches of learning.

Wigner 1960: 9

Reference

Wigner, E. (1960). The Unreasonable Effectiveness of Mathematics in the Natural Sciences. Communications in Pure and Applied Mathematics; Vol. 13, No. 1.

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