HSC Physics MOD 5 (Advanced Mechanics) question - is there a mistake? (1 Viewer)

OreoMcFlurry · Aug 7, 2021

I was practicing mod 5 (advanced mechanics), and I stumbled across this question:

I put A - as it's just logical to think that as per Kepler's second law, an orbiting object covers the same area of space in the same amount of time no matter where it is in its orbit.

Hence, if an object is in it's aphelion or apogee (furthest point of orbit away from the central body) - then: Ep>Ek (energy is conserved)
and likewise, if an object is in it's perihelion or perigee (closest point of orbit from the central body) - then: Ek>Ep (energy is conserved)

It would be then logical to think that if an orbiting object were to cover a particular distance (arc length of orbit), the object at the perihelion would cover that fixed arc length in less time than the same object at the apohelion.

To prove that A is the answer, I used some HSC Maths Extension 1 techniques. If any of you are interested, I've attached it:

OreoMcFlurry · Aug 7, 2021

Hivaclibtibcharkwa said:
Is that question from Learnable?

Yes, it is

notme123 · Aug 7, 2021

OreoMcFlurry said:
I was practicing mod 5 (advanced mechanics), and I stumbled across this question:
View attachment 31425
I put A - as it's just logical to think that as per Kepler's second law, an orbiting object covers the same area of space in the same amount of time no matter where it is in its orbit.
View attachment 31426

Hence, if an object is in it's aphelion or apogee (furthest point of orbit away from the central body) - then: Ep>Ek (energy is conserved)
and likewise, if an object is in it's perihelion or perigee (closest point of orbit from the central body) - then: Ek>Ep (energy is conserved)

It would be then logical to think that if an orbiting object were to cover a particular distance (arc length of orbit), the object at the perihelion would cover that fixed arc length in less time than the same object at the apohelion.

To prove that A is the answer, I used some HSC Maths Extension 1 techniques. If any of you are interested, I've attached it:

View attachment 31428

View attachment 31429

BTW p = perihelion, a = aphelion - I'm just lazy
Qualitatively:
The comet travels faster at p than a, meaning distance covered at p in 1 sec is more than a in one sec, therefore, more seconds needed for a to cover same DISTANCE as p. Therefore, only C or D are correct.

Qualitatively:
Let areas equal and for simplicity sake, let time elapsed equal to 1 second where $\Delta l =$ distance covered in 1 second. i.e. the velocity.
$A_P = A_A$
$\frac{1}{2}{r_P}{\Delta l_P} = \frac{1}{2}{r_A}{\Delta l_A}$
${r_P}{\Delta l_P} = {r_A}{\Delta l_A}$
${\Delta l_P} = \frac{{r_A}{\Delta l_A}}{{r_P}}$
${\Delta l_P} \approx 61.2{\Delta l_A}$
Therefore, in 1 second, the comet travels more distance at p than at apheion by a factor of 61. Thus, for the comet to travel the same distance as at p in 1 second, it must travel for 61 seconds at a.

idkkdi · Aug 7, 2021

notme123 said:
BTW p = perihelion, a = aphelion - I'm just lazy
Qualitatively:
The comet travels faster at p than a, meaning distance covered at p in 1 sec is more than a in one sec, therefore, more seconds needed for a to cover same DISTANCE as p. Therefore, only C or D are correct.

Qualitatively:
Let areas equal and for simplicity sake, let time elapsed equal to 1 second where $\Delta l =$ distance covered in 1 second. i.e. the velocity.
$A_P = A_A$
$\frac{1}{2}{r_P}{\Delta l_P} = \frac{1}{2}{r_A}{\Delta l_A}$
${r_P}{\Delta l_P} = {r_A}{\Delta l_A}$
${\Delta l_P} = \frac{{r_A}{\Delta l_A}}{{r_P}}$
${\Delta l_P} \approx 61.2{\Delta l_A}$
Therefore, in 1 second, the comet travels more distance at p than at apheion by a factor of 61. Thus, for the comet to travel the same distance as at p in 1 second, it must travel for 61 seconds at a.

good.

but only d was correct after reasoning. probably a bruh moment lol

idkkdi · Aug 7, 2021

OreoMcFlurry said:
I was practicing mod 5 (advanced mechanics), and I stumbled across this question:
View attachment 31425
I put A - as it's just logical to think that as per Kepler's second law, an orbiting object covers the same area of space in the same amount of time no matter where it is in its orbit.
View attachment 31426

Hence, if an object is in it's aphelion or apogee (furthest point of orbit away from the central body) - then: Ep>Ek (energy is conserved)
and likewise, if an object is in it's perihelion or perigee (closest point of orbit from the central body) - then: Ek>Ep (energy is conserved)

It would be then logical to think that if an orbiting object were to cover a particular distance (arc length of orbit), the object at the perihelion would cover that fixed arc length in less time than the same object at the apohelion.

To prove that A is the answer, I used some HSC Maths Extension 1 techniques. If any of you are interested, I've attached it:

View attachment 31428

View attachment 31429

ur reasoning going from area at aphelion covered in one second to dA/dt is invalid. A varies with time. What you have written is actually d(Aphelion)/dt = 2.2...

2. A = 1/2 rl
A/l = 1/2 r
dA/dl = 1/2 r is invalid. You just got lucky here.

Consider A = 1/2 rl^2
dA/dl = rl
But by what you were doing,
A/l = 1/2 rl, dA/dl = 1/2 rl

OreoMcFlurry · Aug 7, 2021

Thank you everyone, I realised that I had interpreted the question incorrectly.

My calculations and reasoning was answering this question: What time does it take to cover the same distance at perihelion as during one second at aphelion?

Once I read the question again, I began to understand what they were really asking for and now understand why the answer is D. I'm sorry for the disturbance I may have caused you.
So, in short - I didn't read the question properly (again lol)

But thank you for the help though!

idkkdi · Aug 7, 2021

idkkdi said:
ur reasoning going from area at aphelion covered in one second to dA/dt is invalid. A varies with time. What you have written is actually d(Aphelion)/dt = 2.2...

2. A = 1/2 rl
A/l = 1/2 r
dA/dl = 1/2 r is invalid. You just got lucky here.

Consider A = 1/2 rl^2
dA/dl = rl
But by what you were doing,
A/l = 1/2 rl, dA/dl = 1/2 rl

on second thought,

i'm not sure if A = 1/2 rl can even be differentiated with respect to l as r varies with l.

notme123 · Aug 7, 2021

OreoMcFlurry said:
Thank you everyone, I realised that I had interpreted the question incorrectly.

My calculations and reasoning was answering this question: What time does it take to cover the same distance at perihelion as during one second at aphelion?

Once I read the question again, I began to understand what they were really asking for and now understand why the answer is D. I'm sorry for the disturbance I may have caused you.
So, in short - I didn't read the question properly (again lol)

But thank you for the help though!

It's all good I've never seen a q like this before so it's good practice.

CM_Tutor · Aug 7, 2021

idkkdi said:
on second thought,

i'm not sure if A = 1/2 rl can even be differentiated with respect to l as r varies with l.

To me, this is the key problem with the original reasoning... since $r$ and $l$ both vary with $t$ ,

$\begin{align*} A &= \cfrac{1}{2}rl \\ \cfrac{dA}{dt} &= \cfrac{1}{2}\left(l\cfrac{dr}{dt} + r\cfrac{dl}{dt}\right) \qquad \qquad \text{using the Product Rule} \end{align*}$

There is also a problem with taking the area change in one second, $\cfrac{\Delta A}{\Delta t} = 2.2854 \times 10^{15}\ \text{m}^2\ \text{s}^{-1}$ , and equating that with $\cfrac{dA}{dt}$ . The calculation is an approximation (just as we derive differentiation from first principles by approximating the gradient of a tangent with a secant) unless $\cfrac{dA}{dt}$ is constant (which I think it is, but the issue needs addressing). From a quick read about Kepler's Laws, it appears to me that $\cfrac{dA}{dt} = \cfrac{\pi ab}{T}$ where $a$ and $b$ are the lengths of the semi-major and semi-minor axes and $T$ is the orbital period. From this follows

$\begin{align*} \cfrac{dA}{dt} &= \cfrac{\pi ab}{T} \qquad \qquad \text{(?) } = 2.2854 \times 10^{15}\ \text{m}^2\ \text{s}^{-1} \\ \cfrac{2\pi ab}{T} &= l\cfrac{dr}{dt} + r\cfrac{dl}{dt} \\ \cfrac{dl}{dt} &= \cfrac{1}{r}\left(\cfrac{2\pi ab}{T} - l\cfrac{dr}{dt}\right) \\ &= \cfrac{2\pi ab - Tl\dot{r}}{rT} \end{align*}$

Since we know the values of $r$ and $l$ at the desired position in the orbit is known, the only unknown is $\dot{r}$ , but we still lack the information for an exact answer by this approach.

idkkdi · Aug 7, 2021

CM_Tutor said:
To me, this is the key problem with the original reasoning... since $r$ and $l$ both vary with $t$ ,

$\begin{align*} A &= \cfrac{1}{2}rl \\ \cfrac{dA}{dt} &= \cfrac{1}{2}\left(l\cfrac{dr}{dt} + r\cfrac{dl}{dt}\right) \qquad \qquad \text{using the Product Rule} \end{align*}$

There is also a problem with taking the area change in one second, $\cfrac{\Delta A}{\Delta t} = 2.2854 \times 10^{15}\ \text{m}^2\ \text{s}^{-1}$ , and equating that with $\cfrac{dA}{dt}$ . The calculation is an approximation (just as we derive differentiation from first principles by approximating the gradient of a tangent with a secant) unless $\cfrac{dA}{dt}$ is constant (which I think it is, but the issue needs addressing). From a quick read about Kepler's Laws, it appears to me that $\cfrac{dA}{dt} = \cfrac{\pi ab}{T}$ where $a$ and $b$ are the lengths of the semi-major and semi-minor axes and $T$ is the orbital period. From this follows

$\begin{align*} \cfrac{dA}{dt} &= \cfrac{\pi ab}{T} \qquad \qquad \text{(?) } = 2.2854 \times 10^{15}\ \text{m}^2\ \text{s}^{-1} \\ \cfrac{2\pi ab}{T} &= l\cfrac{dr}{dt} + r\cfrac{dl}{dt} \\ \cfrac{dl}{dt} &= \cfrac{1}{r}\left(\cfrac{2\pi ab}{T} - l\cfrac{dr}{dt}\right) \\ &= \cfrac{2\pi ab - Tl\dot{r}}{rT} \end{align*}$

Since we know the values of $r$ and $l$ at the desired position in the orbit is known, the only unknown is $\dot{r}$ , but we still lack the information for an exact answer by this approach.

if we had r dot,

what would you then do?

*oh, ur saying we can get r dot, but wouldnt be able to get an answer to the question?

CM_Tutor · Aug 8, 2021

idkkdi said:
if we had r dot,

what would you then do?

*oh, ur saying we can get r dot, but wouldnt be able to get an answer to the question?

I was following the idea of @OreoMcFlurry in seeking $\cfrac{dl}{dt}$ . The method that @notme123 posted is the one sought, I think.

But, if we could get $\dot{r}$ then we could get the value of $\dot{l}$ at the aphelion or perihelion. For an exact value of the time, we'd need to integrate $\dot{l}$ from 0 to $t$ for the needed distance, which would require functions for $r$ and $l$ and $\dot{r}$ as functions of $t$ ... But, for an MCQ and the period of time being at most seconds, taking $\dot{l}$ as constant and so distance as $\dot{l}t$ would give an estimate for $t$ that is sufficient to distinguish bewteen the MCQ options.

My other point was that OreoMcFlurry is illustrating a desirable characteristic of an effective learner, the ability to transfer learning between contexts - in this case, from MX1 maths to physics - but that, in doing so, the constraints come as well. Transfer of learning is a topic that has been extensively explored in the literature of educational psychology. It is essential to highly-developed problem solving skills and it is one of the characteristics that is meant to be developed in secondary and tertiary education, thought it is not really explicitly covered in any single subject.

OreoMcFlurry's idea to differentiate $A = rl / 2$ and then apply the approaches to rates of changes problems and the Chain Rule are perfectly reasonable. It illustrates that much of physics is applied mathematics and points to reasons that calculus should be brought into the HSC physics syllabus. However, differentiating needs to be done mindfully of the fact that $A$ , $r$ , and $l$ are all functions of $t$ . Differentiating a multivariable function as a function in two variables is possible - it is called partial differentiation and is a concept covered in university-level courses - but it also has rules to ensure that valid and meaningful results are produced.

I have explored the idea presented a little, but for the specific question of how to solve this MCQ, I can only point to notme123's post

idkkdi · Aug 8, 2021

CM_Tutor said:
I was following the idea of @OreoMcFlurry in seeking $\cfrac{dl}{dt}$ . The method that @notme123 posted is the one sought, I think.

But, if we could get $\dot{r}$ then we could get the value of $\dot{l}$ at the aphelion or perihelion. For an exact value of the time, we'd need to integrate $\dot{l}$ from 0 to $t$ for the needed distance, which would require functions for $r$ and $l$ and $\dot{r}$ as functions of $t$ ... But, for an MCQ and the period of time being at most seconds, taking $\dot{l}$ as constant and so distance as $\dot{l}t$ would give an estimate for $t$ that is sufficient to distinguish bewteen the MCQ options.

My other point was that OreoMcFlurry is illustrating a desirable characteristic of an effective learner, the ability to transfer learning between contexts - in this case, from MX1 maths to physics - but that, in doing so, the constraints come as well. Transfer of learning is a topic that has been extensively explored in the literature of educational psychology. It is essential to highly-developed problem solving skills and it is one of the characteristics that is meant to be developed in secondary and tertiary education, thought it is not really explicitly covered in any single subject.

OreoMcFlurry's idea to differentiate $A = rl / 2$ and then apply the approaches to rates of changes problems and the Chain Rule are perfectly reasonable. It illustrates that much of physics is applied mathematics and points to reasons that calculus should be brought into the HSC physics syllabus. However, differentiating needs to be done mindfully of the fact that $A$ , $r$ , and $l$ are all functions of $t$ . Differentiating a multivariable function as a function in two variables is possible - it is called partial differentiation and is a concept covered in university-level courses - but it also has rules to ensure that valid and meaningful results are produced.

I have explored the idea presented a little, but for the specific question of how to solve this MCQ, I can only point to notme123's post

So after dl/dt with an r dot, you would need multivariable?
If possible could you write up a quick solution?

HSC Physics MOD 5 (Advanced Mechanics) question - is there a mistake? (1 Viewer)

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