NCEA Level 2, 2019 Physics exam
91171 Demonstrate understanding of mechanics
One of my students in 2023 was given this question from 2019 as a practice question. It has four parts, but part (d) is an unrelated torque question. The first three parts follow on from each other. Let's see how these questions stack up.
QUESTION TWO: HALF TIME AT THE HOCKEY MATCH
setup and part (a) | part (b) | part (c) | how can part (c) be solved | conclusion
Setup and part (a)
This question is confusing straight away. Any object spun in a horizontal circle on the end of a string is not going to have a horizontal string. Sadly, a sloping string is not at all referenced in the wording of the question, including in subsequent parts, but it appears not to matter for part (a) because the plan diagram – a view looking straight down – apparently shows the horizontal radius is 0.500 m, not that the string length is 0.500 m. The string itself will be longer, but it doesn't matter... for the moment.
The whistle's direction of travel is continually changing, so speed is the appropriate word, not velocity; v is also used for speed in formulas, so the diagram is quite helpful in clarifying what's required. The calculation is not hard.
`speed = (dist ance) / (time)`
`speed = (ci rcumference) / (period)`
`speed = c / T`
`speed = (2pir) / T`
`speed = (2pi times 0.5) / 1.4`
`speed = pi / 1.4`
`speed = 2.24399...`
`speed approx 2.24 m*s^-1`
Part (b)
This part is specifically excluding vertical forces, so again, the vertical forces won't matter for the moment.
`F = ma`
`a = v^2 / r`
`F = (mv^2) / r`
These are the centripetal acceleration and centripetal force, the acceleration and force on the whistle, acting towards the centre of the circle (NOT the force acting towards the inner/upper end of the string, which would be larger, and at whatever vertical angle the string was at). Don't forget to convert grams to kilograms, the SI base unit for mass. Three significant figures is appropriate in the answer because all the stated values have 3 sig figs. Don't forget a unit.
`F = (0.04 times 2.24399...^2) / 0.5`
`F = 0.403 N`
Using the rounded speed value of 2.24 m/s is probably fine, and gives a centripetal force of 0.401 N. Horizontal forces in other directions are all zero.
The string exerts a centripetal force on the whistle, continually accelerating it into a circular path. There are no forces in the direction of travel, so no acceleration/deceleration in the direction of travel, so the speed is constant.
Still not too much to worry about.
Part (c)
Oh wow. NCEA, you've done it again.
A student who just blindly (or resignedly, in the face of ongoing NCEA stupidity) uses the original horizontal radius of 0.5 m will get a completely wrong value.
`v = 1 m*s^-1 larr` given
`r = 0.5 m larr` WRONG!
`a = v^2 / r`
`a = 1^2 / 0.5`
`a = 2 m*s^-2 larr` WRONG!
`F = (mv^2) / r`
`F = (0.04 times 1^2) / 0.5`
`F = (0.04 times 1^2) / 0.5`
`F = 0.08 N larr` WRONG!
But this is the value stated in the assessment schedule! Why is it wrong?
The centripetal acceleration and the force are both proportional to the square of the speed, remember. This means reducing the speed of the whistle will change the centripetal acceleration required to keep the whistle rotating in a circular path.
But if the horizontal acceleration changes and the vertical acceleration (gravity) doesn't change, the angle of depression of the string will change. The string length hasn't changed (this is an assumption, but a really good one), so the whistle's rotation radius changes. The horizontal acceleration is inversely proportional to the radius, so the acceleration changes again, which changes the string angle, which changes the radius, which changes the acceleration. It goes on and on until the whistle gradually settles into a new stable orbit. (Or fails to do that and falls out of orbit, hitting the referee in the face.)
We weren't given any information about how the change in speed was made, just that it was changed. We know it cannot have been instant, because that would require an infinite acceleration, so it is completely unreasonable for a student to assume that. However, that is apparently what the assessment schedule assumes and requires. That is the only way the new horizontal force on the whistle could be momentarily 0.08 N. It's just completely physically impossible, that's all.
What is a reasonable assumption a student could make? Should the student assume that some unspecified amount of time has passed and the change has already been made? "The speed ... is reduced" implies it is done in a physically possible way over a short period of time. But that would put the whistle and string in a new and different arrangement, and not an easy one to figure out at NCEA Level 2. That problem is probably best solved by successive approximation using a recursive formula, but that technique is not taught at NCEA Level 2.
Even if it was taught, this question would be very hard to answer recursively in an exam. With the extremely poor framing of the overall question (that may be deliberate!), just understanding what's actually going on is not at all easy. Throw in the time needed to form a complicated recursive equation then do a rather large number of iterations – it does not converge quickly – and it could easily take up more time than the whole exam slot. (But how long is that? The front page of the exam does not state any time considerations or limits!)
But it's worse than that, because like the 2019 NCEA rectangle having a component with negative height, it may be only the brightest students who realise the radius would change as soon as the speed started to change, or that the question appears to be trying to use an infinite acceleration. They would be the only ones who would be lured into wasting a huge amount of time in the process of trying to solve the problem.
How are students supposed to answer this question with a correct answer? Is the intent for them to calculate the new horizontal force using the incorrect old value of the radius?
Yes. The intent of this 2019 NCEA Level 2 Physics exam: "We want you to give a wrong answer!"
The assessment schedule says the force would be 0.08 N and expects the student to comment along the lines of "Whistle would either fall out of circular motion or the radius would have to diminish." But for the speed to suddenly and instantly drop to 1 m·s-1 would require an infinite deceleration. Not physically possible. As the whistle decelerated, its radius of motion would change until its speed stablised again. The situation with the stated force of 0.08 N and having the assumed radius? Not physically possible.
There's that word "assumed" again. Students should NOT have to make physically impossible assumptions to answer physics questions.
The question is completely inappropriate to present to students in an exam, although it might make an interesting assignment for older students (perhaps at university level) who have been taught the necessary techniques.
How can part (c) be solved correctly?
First, we need to be clear on what's happening.
The whistle experiences a constant centripetal acceleration in order to keep it travelling in a circular path and a constant downward acceleration due to gravity. The total of these horizontal and vertical components is provided by the string, which will be taut, and at a downward angle from the hand powering the rotation. The faster the whistle rotates the greater the centripetal acceleration and the more horizontal the string is.
Pro tip: I store calculated values like the ones below in my calculator's variable memories (A through F, and X and Y; or A through Z, depending on the calculator), noting what's been stored where, so I can use the full, unrounded values later. This avoids the problem of rounding errors compounding when doing successive calculations.
The initial centripetal acceleration in part (a) (using unrounded values):
`a = v^2 / r`
`a = (2.24399...^2) / 0.5`
`a = 10.0710...`
`a = 10.1 m*s^-2`
Gravitational acceleration is normally taken as 9.81 m·s-2 straight down, to 3 significant figures, but the given value in the resource sheet for the exam is 9.8 m·s-2. (Which is not too bad, because around Auckland it's actually around 9.7994 m·s-2 to 9.7997 m·s-2 – it varies by location.)
From F = ma, force is directly proportional to acceleration, and for our situation the mass (the whistle) is constant and the same for both components, so we can use the acceleration components and trigonometry to calculate the initial angle of depression (the angle below horizontal) of the string in part (a):
`tan theta = (opp) / (adj)`
`theta = tan^-1((opp) / (adj))`
`theta = tan^-1 (-9.81 / (10.0710...))`
`theta = -44.2477...°`
`theta = -44.2°`
How 'bout that. This hockey referee is not at all doing what's implied in the question – spinning in a flat circle with the string essentially horizontal. Realising that there is a vertical acceleration and force that need to be taken into account is essential for being able to answer part (c) correctly.
But as I explained earlier, the request in part (c) for the size of the new horizontal force is a nearly impossible question for an NCEA 2 student. I can only conclude that the examiner and everyone who checked the exam didn't realise what's actually going on, or simply didn't care – a sort of "Meh, the wrong value is fine" attitude.
Because the speed of the whistle has changed, and the centripetal acceleration is proportional to the square of the speed, the horizontal acceleration changes. This in turn means the angle of depression of the string is also going to change from the -44.2° we previously calculated it had in part (a). That changes the radius of the whistle's rotation, which in turn changes the acceleration again – which is inversely proportional to the radius – which changes the string's angle, and so on.
How can we calculate the string length?
Knowing the string length is an essential part of creating a recursive formula. We can use trigonometry again for this, and plug in the horizontal radius and the angle of depression (which because of alternate angles on parallel lines works out nicely).
`cos theta = (adjacent) / (hypoten use)`
`hypoten use = (adjacent) / (cos theta)`
`stri ng l eng th = 0.5 / (cos 44.2477...°)`
`stri ng l eng th = 0.698003...`
`stri ng l eng th = 0.698 m`
Alternatively, Pythagoras' Theorem can be applied by scaling the vertical acceleration component (but don't try to use Pythagoras for a recursive formula, because the proportions of the triangle change as the radius changes).
`c^2 = a^2 + b^2`
`c = sqrt(a^2 + b^2)`
`stri ng l eng th = sqrt(r^2 + h^2)`
`stri ng l eng th = sqrt(0.5^2 + (-9.81(0.5/(10.0710...)))^2)`
`stri ng l eng th = 0.698003...`
`stri ng l eng th = 0.698 m`
How can we calculate the new force?
Basically, we first need to calculate the new horizontal acceleration, then use it to calculate the new force. Calculating the horizontal acceleration is the tricky part, which requires a recursive formula.
The recursive formula I came up with for the horizontal acceleration took a long time to converge (67 iterations for just 3dp – nuts in an exam context; it's a full half minute of mashing the equals button, assuming you get the formula right). Basically, the formula uses the vertical and horizontal acceleration components to get the string angle, then uses that angle to get the new radius, the reciprocal of which is the next iteration of the acceleration. (Curiously, plugging the formula into Desmos, using x instead of a, gives a vertical line at 3.888, but without any marked intersection points with intersecting lines.)
Results are rounded to 2 significant figures because the new speed has 2 sig figs.
`v = 1.0 m*s^-1 larr` given
`theta = -68°`
`r = 0.26 m`
`a_(n+1) = 1 / (0.698 times cos(tan^-1(9.81 / a_n))`
`a = 3.9 m*s^-2`
`F = 0.16 N`
The circle made by the whistle is much smaller than it was in part (a) because the horizontal and vertical forces are no longer almost the same magnitude as each other, so the string is now much more vertical.
The new horizontal acceleration, and hence the new horizontal force, are both much larger than calculated using the original horizontal radius.
Oh yeah – because it's NCEA, don't forget to clearly state the result.
The size of the new horizontal force on the whistle is 0.16 N.
Conclusion
This is a very bad exam question.
It requires the student to choose between two assumptions – one which is physically impossible, the other requiring maths not taught at any NCEA level. The student should NOT have to make ANY assumptions unless directed to.
What is wrong with NCEA exam setting and checking that questions like this are presented to students? How does deliberately leading students toward incorrect answers encourage them to correctly "demonstrate understanding of mechanics"?
The only simple way for students to avoid going down a very time-consuming rabbit hole would be for them to make a completely unjustified assumption such as the referee lengthens the string when she slows down the whistle rotation so that it ends up providing the exact same horizontal radius as originally. If the examiner had included this change in the setup in part (c) it would have been very contrived, but at the same time made answering the question correctly actually possible for an NCEA Level 2 Physics student.
Flawed questions like this happen far too often in NCEA maths – so often that it is expected that maths students will have questions they will have to make wild guesses about just to be able to start answering them.
It's sad that it happens in physics as well.
Such questions undermine the numeracy of New Zealand students because they subtly teach and emphasise that maths cannot be relied on to give definite answers. These students are New Zealand's future, but NCEA and the Ministry of Education is destroying their faith in mathematics. This will only make finding maths teachers and engineers harder in the future.
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