Category Archives: physics

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

Introduction

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

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

Screenshot 2019-08-08 at 14.47.50.png

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

Screenshot 2019-08-08 at 13.54.07.png

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

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

Adapting the AQA method to make it quantitative

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

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

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

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

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

The Eleventh Commandment: Thou Shalt Not Confound Thy Variables!

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

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

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

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

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

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

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

What lies beneath: the Physics underlying this method

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

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

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

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

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

This can be shown in diagram form like this:

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

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

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

No more jerks left to give

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

This means that the acceleration is constant.

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

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

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

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

Using this in the school laboratory

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

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

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

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

Conclusion

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

References

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

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Filed under physics, Required Practicals, Science

Kinetic Energy Using The Singapore Bar Model

I think the Singapore Bar Model is a neat bit of pedagogy that has great potential in Science education.

Essentially, the Singapore Bar Model uses pictorial representations (often in the form of a bar or line) to help students bridge the gap between concrete and abstract reasoning. I wrote about one possible application here.

A recent discussion on Twitter started me thinking about if it could be applied to kinetic energy.

For example, how would you explain what happens to the kinetic energy of an object if its velocity is halved?

Many students assume that the KE would halve as well, instead of reducing to a quarter of its original value.

How can we help students grasp this slippery concept without using algebra? Algebra would work fine with your higher sets, of course, but not necessarily for other groups.

This gives a clear visual representation of the fact that the KE quarters when the velocity halves. In other words, 0.5 x 0.5 = 0.25.

(Note that I have purposefully used decimals as we know that many students struggle with fractions(!))

Many students found the following question on an AQA paper extremely challenging:

The correct answer is that the power output drops to one eighth of its original value.

Could the Singapore Bar Model helps students to see why this is the case?

I think it could:

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Teaching Electric Circuits? Climb On Board The Coulomb Train!

I’ve said it before and I’ll say it again: teaching electric circuits is hard.

Providing your students with a conceptual model can, in my opinion, be immensely helpful. In recent years, I have used what I call the Coulomb Train Model (CTM). It is essentially a variation on the “stiff chain” analogies that some researchers have argued as being particularly useful in developing students’ understanding.

One reason why I like the CTM is that it provides a physical picture to aid students’ comprehension of many of the electrical equations needed at GCSE.

Of course, any analogy or model will have its flaws, but on the whole I think the CTM has fewer than many of its rivals!

Essentially, the CTM pictures an electric circuit as a continuously moving chain of postively-charged “trucks” called coulombs that carry energy from the cell to (say) the bulb. In the diagram below, they should be pictured as moving clockwise.

The coulomb is, of course, the S.I. unit of electric charge, so rest assured that there is method in the apparent madness of naming our “trucks” with a word that would be unfamiliar to most of our students.

Charge flow = current x time

Charge flow = number of coulombs that pass a given point in time.

Current = number of coulombs that pass by in one second (i.e. current = charge flow / time).

In other words, an ammeter counts the coulombs passing by in one second. The ammeter only “sees” the coulombs and does not register how much (or how little) energy each one contains. Therefore current I1 and current I2 are equal.

The ammeters are shown as being semi-transparent in order to provide a visual cue that they are low resistance devices.

Energy transferred = charge flow x potential difference

On the CTM, potential difference can be pictured as energy being added to, or removed from, each coulomb.

For example, if one joule is removed from each coulomb as they pass through the bulb, the potential difference across the bulb is one volt. If one joule is added to each coulomb as they pass through the cell, then the potential difference (or e.m.f.) across the cell is one volt.

The total energy transferred from (say) ten coulombs passing through the bulb would be charge flow (10 coulombs) x potential difference (1 volt) = 10 joules.

The white gloves on the voltmeter are intended to be reminiscent of the white gloves of a snooker referee.

The intention is to disrupt the flow of the coulombs as little as possible and so this is a visual reminder that a voltmeter is a high resistance instrument.

To emphasise the fact that potential difference is an “energy difference”, challenge students to predict the reading on this voltmeter.

The potential difference V3 is, of course, zero since there is no transfer of energy to or from the coulombs.

Current in Series and Parallel Circuits

I think the CTM can be really effective in allowing students to a see a comprehensible physical analogue of the circuits.

For example, I3 = I4 = I5 = 0.5 amps; I6 = I11 = 2 amps; and I7 = I8 = I9 = I10 = 1 amp.

Potential difference in series and parallel circuit

Equally, I think the CTM can give a comprehensible physical picture of the situation.

In this case (assuming the the cell has a p.d. of 1 V and the bulbs are identical), V4 = V5 = 0.5 V.

In the parallel circuit, each bulb tranfers one joule of energy from each bulb, and so the potential difference across each bulb is one volt.

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Ohm From Ohm

Amongst the myriad inconveniences and troubles of a Physics teacher’s life, the choice of the symbols commonly used to represent voltage, current and resistance, must surely rank in the top ten.

V is for voltage in volts, V

Well, OK, that’s sensible enough. On a good day, I may even remember to call it “potential difference”. The sage advice of Never use two words when one will do is commonly accepted by everyone; however, Physics teachers have, as a profession, decided to go it alone and completely ignore this tired old saw. Thus, voltage is become potential difference because of — erm, reasons (?)

One can only hope that everyone got the memo . . .

R is for resistance in ohms, Ω

R for resistance? That’s fairly sensible too.

“But what’s that weird squiggly thing, Sir?”

“Ah, you mean the Greek letter omega? Because Physics is soooo enormous that the measly 26 letters of the Latin alphabet ain’t big enough for it…”

I is for current in amps, A

“WTφ? Are you taking the πΣΣ, Sir?”

“I know, I know! Look, if it helps, think of it as short for intensité du courant . . . Wait, don’t leave! Stop, I have many more fun Physics facts to teach you! Look, here’s a picture of Richard Feynman playing his bongo drums — nooooooooo!”

Ohm’s Law: or is it more a sort of guideline?

Let’s start with a brief statement of Ohm’s Law:

V = I R

Except, that’s not Ohm’s Law; it’s actually the definition of resistance:

R = V / I

There is not a single instance where it is not true by definition. The value of resistance will always be equal to the ratio of the potential difference and the current.

Think of it like this. At room temperature, 1 V of potential difference can push (say) 0.5 A of current through the wire in a filament bulb. (I just love that retro 1890s tech, don’t you?)

This means it has a resistance of 1/0.5 = 2 ohms. However, bump up the potential difference to 6 V and the current is (say) 0.75 A. This means that is has a resistance of 6/0.75 = 8 ohms. Its resistance has changed because it has become hotter. In other words, its resistance is not constant.

Ohm’s Law is perhaps most simply stated as:

The potential difference is directly proportional to the current over a range of physical conditions (including temperature).

Using standard symbols:

V α I

or, taking R’ as a constant of proportionality:

V = I R’

You do see the difference, don’t you?

In the first example, R is not a constant value for a given range of physical conditions: for example it can get higher as the temperature increases.

In the second, R’ is constant over a range of temperatures and other physical conditions.

And so there we have it: V=IR can be a perfectly valid statement of Ohm’s Law, provided it is specified that R is constant. If one does not do that, then all bets are off…

In the meantime, here’s another picture of Richard Feynman playing the bongo drums. Enjoy!

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Lottie and Lorentzian Length Contraction

@_youhadonejob tweeted this textbook picture with the amusing and sardonic comment “Little girl in this textbook is 5 m tall”.

I liked @jim_henderson60’s take on this when he tweeted: “You see. Physics helps us all grow tall.”

But then I started thinking, what if the 5 m measuring stick was in an inertial frame moving past Lottie’s inertial frame at a substantial fraction of light speed? (In my head, I named the girl “Lottie”, although “Alice” would be more in the more usual tradition of SR* pedagogy, I guess.)

The illustration could represent that single instant at which both ends of the 5 m ruler were precisely opposite Lottie’s head and feet as its inertial frame whizzed by hers…

A quick calculation indicated that Lorentz length contraction could indeed account for the relative measurements on the illustration if v = 0.97c

Of course, Lorentz length contraction is a two way street. From the 5 m ruler’s inertial frame, length contraction would make Lottie appear even shorter than her compact 1.2 m. Given that v = 0.97c, I calculate that she would appear only 0.29 m tall.

Correction: not appear. She would genuinely be only 0.29 m tall when viewed from that inertial frame, just as the 5 m rule would genuinely be only 1.2 m long when viewed from Lottie’s inertial frame.

We live in an universe where everything is indeed relative. However, for most of us that takes a fair amount of getting used to…

*SR = special relativity. My brain is currently too small to handle GR (general relativity).

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