HSC Physics Question Thread

[DrDusk]

Hi blyatman and Coolmate!

Thank you so much for your help.

I completely agree with your logic, however I have just checked the 'official' answer and they suggest that it is C - mass.

I understand that in accordance with the second postulate, length, time and mass are all subject to relativistic effects - however do you have any idea as to why the answer could possibly be mass instead of length?

Nope, I also agree with it being B. Official answers are wrong. Even schools teach it as that Length is what was redefined initially, as said by Blyatman.

What paper is this question taken from?

[blyatman]

Hi blyatman and Coolmate!

Thank you so much for your help.

I completely agree with your logic, however I have just checked the 'official' answer and they suggest that it is C - mass.

I understand that in accordance with the second postulate, length, time and mass are all subject to relativistic effects - however do you have any idea as to why the answer could possibly be mass instead of length?

No it's definitely not mass, the answers are wrong.

Mass dilation isn't even a thing, despite what the HSC says (but that's a topic for another day).

[DrDusk]

Mass dilation isn't even a thing, despite what the HSC says (but that's a topic for another day).

Wait what??? Doesn't mass dilate as your speed becomes higher due to E = mc^2 + (pc)^2 ?

[blyatman]

Wait what??? Doesn't mass dilate as your speed becomes higher due to E = mc^2 + (pc)^2 ?

Nah, your energy goes to infinity, but that should not be interpreted as your mass going to infinity. The HSC teaches \(E=mc^2\) where \(m=\gamma m_0\) is the dilated mass (and \(\gamma\) being the Lorentz factor), but rather it should be \(E = \gamma m_0 c^2\). In other words, rather than thinking of mass being scaled by \(\gamma\), it's more correct to think of the energy as being scaled by \(\gamma\). Likewise, momentum should be treated as \(p=\gamma m_0v\) rather than using \(p=mv\) with \(m\) being the dilated mass.

University courses in relativity often accentuate this difference to rectify what students learnt in the HSC. Recently, the physics teachers at Matrix had a long debate whether they should be teaching relativistic mass since they knew it was wrong, but ultimately decided to do so since it's what's examined in the HSC.

https://en.wikipedia.org/wiki/Mass_in_special_relativity
"The term relativistic mass tends not to be used in particle and nuclear physics and is often avoided by writers on special relativity, in favor of using the body's total energy." There's a section named "Controversy" at the bottom which goes into a bit more detail.

I was also like wtf when I first learnt this at uni. But even this doesn't take the cake as the most "wrong" thing I've learnt in HS physics.

[DrDusk]

Nah, your energy goes to infinity, but that should not be interpreted as your mass going to infinity. The HSC teaches \(E=mc^2\) where \(m=\gamma m_0\) is the dilated mass (and \(\gamma\) being the Lorentz factor), but rather it should be \(E = \gamma m_0 c^2\). In other words, rather than thinking of mass being scaled by \(\gamma\), it's more correct to think of the energy as being scaled by \(\gamma\). Likewise, momentum should be treated as \(p=\gamma m_0v\) rather than using \(p=mv\) with \(m\) being the dilated mass.

University courses in relativity often accentuate this difference to rectify what students learnt in the HSC. Recently, the physics teachers at Matrix had a long debate whether they should be teaching relativistic mass since they knew it was wrong, but ultimately decided to do so since it's what's examined in the HSC.

https://en.wikipedia.org/wiki/Mass_in_special_relativity
"The term relativistic mass tends not to be used in particle and nuclear physics and is often avoided by writers on special relativity, in favor of using the body's total energy." There's a section named "Controversy" at the bottom which goes into a bit more detail.

I was also like wtf when I first learnt this at uni. But even this doesn't take the cake as the most "wrong" thing I've learnt in HS physics.

OH YES that controversy section actually makes so much sense!

So this means essentially the formula m = gamma m_0 is actually not correct, but rather the correct definition is through the formula for momentum.

That is interesting, I'm definitely going to read up on this

[louisaaa01]

Hi,

I'm having a bit of trouble resolving forces on banked tracks. Basically, I can't seem to reconcile what I've learnt in Maths Ext 2 and Physics.

In Physics, we learn to resolve the forces parallel and perpendicular to the inclined plane. The plane is inclined at an angle θ to the horizontal. You wind up with:

Normal force (perpendicular to plane) = mgcosθ
Force parallel to plane = mgsinθ

However, in Maths, we consider the vertical and horizontal components of N (normal force) which is perpendicular to the banked track.

Neglecting friction,

Ncosθ = mg
Nsinθ = mv²/r

Yet, this essentially implies that N = mg/cosθ

But in Physics we've been taught that N = mgcosθ

Which one is correct? How do we reconcile these two equations?

Any help is very much appreciated!!

[^^^111^^^]

Wait what??? Doesn't mass dilate as your speed becomes higher due to E = mc^2 + (pc)^2 ?

Sorry I may be wrong, but shouldn't it be:

E²= (mc²)² + (pc)²

[fun_jirachi]

Hi,

I'm having a bit of trouble resolving forces on banked tracks. Basically, I can't seem to reconcile what I've learnt in Maths Ext 2 and Physics.

In Physics, we learn to resolve the forces parallel and perpendicular to the inclined plane. The plane is inclined at an angle θ to the horizontal. You wind up with:

Normal force (perpendicular to plane) = mgcosθ
Force parallel to plane = mgsinθ

However, in Maths, we consider the vertical and horizontal components of N (normal force) which is perpendicular to the banked track.

Neglecting friction,

Ncosθ = mg
Nsinθ = mv²/r

Yet, this essentially implies that N = mg/cosθ

But in Physics we've been taught that N = mgcosθ

Which one is correct? How do we reconcile these two equations?

Any help is very much appreciated!!

Correct me if I'm wrong, but I think you can't resolve these two equations, because they both describe different cases of motion on a slope. In your 'Physics' example, the motion that is experienced is down the slope (as a result of the parallel force), while in your 'Extension 2' example, the motion that is experienced is circular motion (as a result of the net force). If we were considering Circular motion in Physics on a banked track (disregarding friction), the only force acting on the object would be the horizontal net force acting towards the centre of motion ie. exact same as X2. Similarly, if we were considering an object moving up or down the slope (disregarding friction) the only component we 'see' is the parallel force up and down the slope ie. the net force, ie. same as Physics.

Hope this helps

[louisaaa01]

Correct me if I'm wrong, but I think you can't resolve these two equations, because they both describe different cases of motion on a slope. In your 'Physics' example, the motion that is experienced is down the slope (as a result of the parallel force), while in your 'Extension 2' example, the motion that is experienced is circular motion (as a result of the net force). If we were considering Circular motion in Physics on a banked track (disregarding friction), the only force acting on the object would be the horizontal net force acting towards the centre of motion ie. exact same as X2. Similarly, if we were considering an object moving up or down the slope (disregarding friction) the only component we 'see' is the parallel force up and down the slope ie. the net force, ie. same as Physics.

Hope this helps

This is really helpful, thank you so much Was just misled because the official answer to a Banked Track trial question used these inclined plane equations! Anyway, really appreciate your assistance.

[classof2019]

Can the NESA Sample Answers in the Marking Guidelines for each past HSC paper be relied upon as an indicator of what would have achieved full marks? Because I've heard that apparently not all answers included in the sample answers would have achieved full marks in a given question.

[DrDusk]

Can the NESA Sample Answers in the Marking Guidelines for each past HSC paper be relied upon as an indicator of what would have achieved full marks? Because I've heard that apparently not all answers included in the sample answers would have achieved full marks in a given question.

Usually the marking guidelines give the best possible answer, but you are right. Sometimes it cannot be relied upon.

[classof2019]

Can someone please help with this?

A power plant generates 150 MW of electricity for a town 20 km away. The transmission wires have a resistance of 0.01 W per km. The voltage drop between the plant and the town is 50 V.

(a) Calculate the power loss between the plant and the town.
(b) Calculate the voltage transmitted by the plant.

Cheers.

[louisaaa01]

Hey! Just a question, when calculating the amount of magnetic flux through a coil, do you need to multiply by the number of turns in the coil?

[fun_jirachi]

Can someone please help with this?

A power plant generates 150 MW of electricity for a town 20 km away. The transmission wires have a resistance of 0.01 W per km. The voltage drop between the plant and the town is 50 V.

(a) Calculate the power loss between the plant and the town.
(b) Calculate the voltage transmitted by the plant.

Cheers.

Hey there!

For part a), consider the fact that P=VI, and that losses are lost through resistive heating ie. P=I²R. This leads us to deduce that the power of the town is going to be (V-50)I, while the power at the plant will be VI. We are given R as equal to 0.01 ohms/km x 20km ie. R = 0.2.

Then, we can solve for I in the equation (V-50)I=VI-I²R.

For b), to find the voltage transmitted by the plant to the town is simply going to be subbing in your new value for I back into the equation in a) and solving for V. Note that the voltage that actually gets transmitted is going to be V-50, since some power is lost.

Hope this helps

Hey! Just a question, when calculating the amount of magnetic flux through a coil, do you need to multiply by the number of turns in the coil?

Correct me if I'm wrong, but I think you do!

[classof2019]

Hey there!

For part a), consider the fact that P=VI, and that losses are lost through resistive heating ie. P=I²R. This leads us to deduce that the power of the town is going to be (V-50)I, while the power at the plant will be VI. We are given R as equal to 0.01 ohms/km x 20km ie. R = 0.2.

Then, we can solve for I in the equation (V-50)I=VI-I²R.

For b), to find the voltage transmitted by the plant to the town is simply going to be subbing in your new value for I back into the equation in a) and solving for V. Note that the voltage that actually gets transmitted is going to be V-50, since some power is lost.

Hope this helps

Hey, thanks for looking into it - though, follow up q, does this rely on the assumption that current is constant? And if so, how can we assume that current doesn't decrease?