HSC Physics Question Thread

[RuiAce]

I'm going to blow in here and just give my two cents lol    Of course HSC physics is not anywhere near uni level physics, they can't expect people to do uni level stuff in year 12 and excel at it.. how much actual physics and chemistry could possibly be squeezed into the curriculium? not to mention that a lot of the stuff everyone gets taught, regardless of the maths involved, is wrong
Obviously not doing high level maths could be a drawback, if you commit to doing HSC physics then you can commit to putting in the extra effort for the maths involved, but don't go blaming us general kids for all the wordy crap in exams, we don't like it either   
_{please don't hate me}

Lol I never said it had to be like "uni" physics.

Look at the exams before the year 2001. That's some real physics and chemistry.


I wouldn't say it's the general maths' students "fault" that the exam got dragged down in difficulty. That was a decision by BOS - not petitioned. But I doubt at the time they realised how many cons they'd introduce and that they potentially outweigh the pros.

[brontem]

Look at the exams before the year 2001. That's some real physics and chemistry.

I'm good mate, I'll just stick to the 2016 fantasy physics _{ba dum tsh} 

[jamonwindeyer]

I'm going to blow in here and just give my two cents lol    Of course HSC physics is not anywhere near uni level physics, they can't expect people to do uni level stuff in year 12 and excel at it.. how much actual physics and chemistry could possibly be squeezed into the curriculium? not to mention that a lot of the stuff everyone gets taught, regardless of the maths involved, is wrong
Obviously not doing high level maths could be a drawback, if you commit to doing HSC physics then you can commit to putting in the extra effort for the maths involved, but don't go blaming us general kids for all the wordy crap in exams, we don't like it either   
_{please don't hate me}

Love the way you put this. You are absolutely right, and tbh, if it were up to me I'd maintain the current level of mathematics in Physics and introduce a new subject for those who want to move on and do tertiary physics, that is a bit more mathematical 

PS - No one will ever hate you for your opinion!

Lol I never said it had to be like "uni" physics.

Look at the exams before the year 2001. That's some real physics and chemistry.

I wouldn't say it's the general maths' students "fault" that the exam got dragged down in difficulty. That was a decision by BOS - not petitioned. But I doubt at the time they realised how many cons they'd introduce and that they potentially outweigh the pros.

I don't think the difficulty was changed too much: It is just the difficulty shifted from the mathematical end of the spectrum to the analytic side of the spectrum that is more similar to the HSIE subjects  of course many students find the math stuff harder than the HSIE stuff,

Very interesting discussion guys, but probably let's keep it back to Physics Q+A now, I might do some research regarding the Physics Extension subject and start a discussion thread if I find more! Or feel free to start one now, don't let me interrupt the conversation 

I'm good mate, I'll just stick to the 2016 fantasy physics _{ba dum tsh}

 

[brontem]

Since everyone seems to be up
Why is it D? It's probably really obvious but idk? I'm pretty good with M&G but again I don't recall going through emf in any depth besides defining it basically
Thanks 

[Swagadaktal]

Since everyone seems to be up
Why is it D? It's probably really obvious but idk? I'm pretty good with M&G but again I don't recall going through emf in any depth besides defining it basically
Thanks 

Yo I think it's faraday's law (or some other physics hoe)( yo jamon or something will probs do this 100x better but if you need an answer in the mean time hehe )
The equation goes: Induced EMF = -dflux/dt 
So as you can see, there's a sin wave there for fluf. The dy/dx function of a sin wave is a cos wave f(x) = sin(x), f'(x) = cos(x)

Now, with induced current, the sign is reversed (positive becomes negative) - so you'd have a -cos(x) wave for induced emf. Which matches up with D -- are you expected to know diff of functions in HSC physics? I'm a vce nomad

[brontem]

Yo I think it's faraday's law (or some other physics hoe)

100% going to be starting every answer to any Faraday question with this
But thanks  What we're meant to know I'm not sure (hence me asking the question hahaha), graphs like this I just force into my memory 

[jamonwindeyer]

100% going to be starting every answer to any Faraday question with this
But thanks  What we're meant to know I'm not sure (hence me asking the question hahaha), graphs like this I just force into my memory 

Swag's got you covered, it's definitely Faraday, my favourite Physics hoe 

Faraday's Law goes like this:



In words, the induced EMF/voltage is given by the rate of change of magnetic flux with respect to time (there are more complex versions of this formula out there, but this is the principle).

Note that electromotive force (EMF) is interchangeable with the term "voltage" at this level, they are really one in the same. So, the voltage we generate is given by how quickly we are changing the magnetic field. The fact that the negative sign is there reflects Lenz's Law, the induced voltage should OPPOSE the changing field that created it, so essentially:



The two cancel each other out (as we'd expect), if they didn't that would violate Lenz's Law.

Anyway, in terms of the question, the formula itself is enough to answer mathematically. A sine curve is shown, so the negative derivative is a negative cosine curve, answer D. However, a Physics student isn't expected to know how to do this, so let's do it the longer way.

The rate of change of the magnetic field can be considered properly by taking the derivative of the sine curve, or just simply notice that the EMF should be at a peak when the magnetic field is changing rapidly, and when the magnetic field is at a peak (not changing), the induced EMF should be zero. Remember we only have induced EMF when magnetic flux is changing. That eliminates options A and C, because they don't have peaks/troughs in the correct places. We want either B or D.

Now, the next one is where we bring in Faraday's Law for ease of use. The negative sign makes selecting D easier to see. However, Faraday's Law isn't explicitly in the syllabus in the mathematical form, so instead, you could use a Lenz's Law explanation. The EMF must be opposite in sign to the rate of change of magnetic field, because the induced EMF must oppose the change that created it. So, as the rate of change of magnetic flux is positive (EG - at the start of the top graph), we expect a negative peak (trough) for the EMF. D matches this response 

Does this make sense? Definitely a tough question this one, a tad beyond the syllabus IMO, but definitely within realms of reason, where did you find it? 

[brontem]

Does this make sense? Definitely a tough question this one, a tad beyond the syllabus IMO, but definitely within realms of reason, where did you find it? 

Ahh yep that definitley makes sense - got it now, thank you!!  its from the 2012 HSC exam 

[jamonwindeyer]

Ahh yep that definitley makes sense - got it now, thank you!!  its from the 2012 HSC exam 

Oh lol, okay never mind BOSTES wanted to push with that question then!! 

To clarify, you definitely need to know Faraday's, but the mathematical version that makes that question easier is not compulsory as to my knowledge. Great to know about though if you like it!

[MysteryMarker]

An external observer on a planet witnesses a space probe to undergo the slingshot effect (on the same planet). Would an observer on the planet agree that the space probe's speed has increased? discuss your reasons.

Need help with this question, thanks guys.

[jakesilove]

An external observer on a planet witnesses a space probe to undergo the slingshot effect (on the same planet). Would an observer on the planet agree that the space probe's speed has increased? discuss your reasons.

Need help with this question, thanks guys.

I've blatantly stolen this answer from another source, but here ya go! Hope it makes sense

No, the observer will argue that the probe's speed has not increased.

Relative to the observer, the probe's speed will appear to be unchanged. As the probe approaches the assisting planet, its GPE will decrease and its kinetic energy will increase but as the probe moves away from the assisting planet, its GPE will increase and kinetic energy will decrease relative to the observer. This means that the speed at which the probe approaches will be the same as the speed at which the probe moves away from the assisting planet.

In actual fact, the velocity of the probe relative to the sun has changed in both magnitude and direction because the probe's output velocity is equal to the input velocity added to the velocity component of the assisting planet. This makes an observer from the sun argue that the probe's velocity (speed and magnitude) has increased.

Jake

[MysteryMarker]

I've blatantly stolen this answer from another source, but here ya go! Hope it makes sense

No, the observer will argue that the probe's speed has not increased.

Relative to the observer, the probe's speed will appear to be unchanged. As the probe approaches the assisting planet, its GPE will decrease and its kinetic energy will increase but as the probe moves away from the assisting planet, its GPE will increase and kinetic energy will decrease relative to the observer. This means that the speed at which the probe approaches will be the same as the speed at which the probe moves away from the assisting planet.

In actual fact, the velocity of the probe relative to the sun has changed in both magnitude and direction because the probe's output velocity is equal to the input velocity added to the velocity component of the assisting planet. This makes an observer from the sun argue that the probe's velocity (speed and magnitude) has increased.

Jake

Oh, because what i was thinking is that for the maximum effect of the slingshot effect, the probe would gain a significant amount of energy, almost double the planets orbital speed. When this occurs there is a compromise. The planet itself will lose a bit of KE to compensate for the probe's gain, but due to the planets large mass, its velocity change is minimal. SO therefore, an observer on the planet will see an increase in speed.

Is this wrong or is the answer for the above question subjective?

[Happy Physics Land]

Oh, because what i was thinking is that for the maximum effect of the slingshot effect, the probe would gain a significant amount of energy, almost double the planets orbital speed. When this occurs there is a compromise. The planet itself will lose a bit of KE to compensate for the probe's gain, but due to the planets large mass, its velocity change is minimal. SO therefore, an observer on the planet will see an increase in speed.

Is this wrong or is the answer for the above question subjective?

Hey Mysterymarker!

To be honest l dont like the way the question was written. You can think about Jake's answer in another way: suppose you are an observer standing on the assisting planet and you are rotating along with the planet's rotation about its central axis. The space probe undergoing slingshot effect will "steal" the planet's angular momentum, meaning that it will then be accelerating in the same direction of rotation as the assisting planet. This means that for you, there will be little change in relative velocity of the space probe since you are both rotating in the same direction with approximately the same angular velocity. So I can only say here that it is POSSIBLE that the observer wouldnt see a change in relative velocity. But for me, the best answer would be there would be a decrease in relative velocity observed compared to when you are an observer from another planet and therefore the increase in relative velocity would be too insignificant to be noticed.

Hope it helps!

Best Regards
Happy Physics Land

[RuiAce]

Can I have some help please

A weight lifter Can lift 140kg on earth he has been sent to Venus to improve his. Given the acceleration due to gravity on Venus is 8.93m/s/s, what should he be able to lift on Venus with the same effort?

What is he improving?

Anyway, this question treats force as constant.

On Earth:
F = m_Eg_E = 140kg * 9.8 m s^-2 = 1372 N

On Venus:
m_V = F/a_V = 1372 N / 8.93 m s^-2 = 153.63941...kg = 150 kg (2 s. f.)

[conic curve]

What is he improving?

Anyway, this question treats force as constant.

On Earth:
F = m_Eg_E = 140kg * 9.8 m s^-2 = 1372 N

On Venus:
m_V = F/a_V = 1372 N / 8.93 m s^-2 = 153.63941...N = 15 N (2 s. f.)

Just curious but why is "g" (or whatever value it was, I forgot) always 9.8 ms^-2?

Thanks