HSC Physics Question Thread

[RuiAce]

I think Syndicate meant each individual question wouldn't take 2 minutes, meaning MC wouldn't take 40 minutes to complete. I agree that 15 minutes minimum is required to properly attempt the questions.

That'd make more sense.

[RuiAce]

Aite thanks guys! And btw could anyone say a list of commonly chucked 6 markers? Ik the impacts of transformers come out a lot, and the MM experiment. Anything else?

Those two are the main popular ones. Note that transformers can be asked as AC generators as well.

And then, you have a variety of occasionally appearing ones.

Applications of superconductors
Applications of semiconductors
Launching a rocket into space, slingshot and reentry
Hertz experiment all the way down to Einstein with regards to photoelectric effect

Edison VS Westinghouse appears rarely

And that's as far as my memory goes.

[jakesilove]

Aite thanks guys! And btw could anyone say a list of commonly chucked 6 markers? Ik the impacts of transformers come out a lot, and the MM experiment. Anything else?

My advice would be to look through past HSC papers from something like 2005-2015, and keep a list of any questions over 6 marks. Write out answers, get your teachers to mark them, write out answers again until they are perfect. Then, put them into your Physics notes, because without a doubt you will be asked the exact same questions in your HSC (Not necessarily all of them, but I promise you that you'll see at least a few). That way, when you go to answer questions you know precisely what information you need to include to get full marks!

Glad to see everyone so active on the forums and helping each other out!

Jake

[Charli33]

Hi guys
I am pretty lost in regards to the Motors and Generators course and was wondering if someone could explain the concept of back EMF?
It would be much appreciated!

[Happy Physics Land]

Hi guys
I am pretty lost in regards to the Motors and Generators course and was wondering if someone could explain the concept of back EMF?
It would be much appreciated!

Hey Charlie:

Awesome! Back emf is actually a concept that my classmates all struggled with for two lessons before they eventually understood what it is, simply because you just feel like you have current and emf produced everywhere and you have your motor effect and all these other fancy physics principles and you would kinda feel lost. So don't worry, you are in the same boat as majority of the other students and I will definitely help you to understand this concept better.

So lets think back to out first principles - the motor effect. We know that when we supply a current to a coil inside a magnetic field, the coil experiences a force (motor effect) and it rotates. (N.B. if you would like to know why it experiences a force, it is because the current passing through the coil actually produces a magnetic field that interacts with the external magnetic field caused by a pair of magnets. This exerts a force onto the coil, making it spin). We call this current that is supplied to the rotor the "supply current". Since a current requires voltage(which pushes the electrons through the conductor) to flow through the coil, we call this voltage "supply voltage" or "supply emf".

We have also learnt however, that whenever there is a change in magnetic flux, an emf is induced and hence a current is induced (Faraday's law). Essentially there will be a change in magnetic flux whenever there is a relative motion which cuts magnetic field lines. The magnitude of this current/emf produced will be proportional to the rate of change of flux. This induced emf will be in such a direction that it opposes the change in magnetic flux (Lenz's law).

Ok now, we need to combine the two parts I talked about above to explain the back emf. According to the motor effect, the rotor coil would be rotating in the external magnetic field. As it rotates, there is relative motion between the rotor coil and the magnetic field. As a result, it constantly cuts different amount of magnetic field lines, causing change in magnetic flux and induces an emf　(Faraday's law). But according to lenz's law, this induced emf will be in a direction to oppose the change in magnetic flux, hence this emf will be acting against the supply emf to try to maintain the previous magnetic flux condition. Because it opposes the supply emf, we call this INDUCED EMF "back emf".

When the armature (rotor coil) is parallel to the magnetic field, the back emf induced will be the greatest because torque is the greatest when armature is parallel to magnetic field lines (T = nBIAcostheta, theta=0, costheta = 1, hence T is maximum when theta=0 i.e. parallel). This maximum torque means that the armature will be cutting the magnetic field lines at a greater rate, and according to faraday's law, more emf will be produced because there is a faster change in magnetic flux. When the armature coil is perpendicular to the magnetic field, i.e. theta=90 degrees, the torque will be minimum, meaning that change in magnetic flux is minimum and hence back emf produced in minimum.

Ok so I hope my explanation made sense to you and if you didnt understand any of those, just remember two things at least:
1. back emf acts against the supply emf
2. back emf is induced due to the rotor's rotation which causes change in magnetic flux

Best Regards
Happy Physics Land

[Charli33]

Thank you so much Happy Physics Land, that really helps! 

[Maz]

Hey
i have a relativity question i was wandering if you could please help me in?
two protons in an accelerator are moving towards each other at 0.75c. At what speed are the protons approaching each other at relative to a stationary observer in the laboratory and relative to each other?

one more thing please...how would you do it if instead of then particles moving towards each other, they were moving in the same direction?

thankyou soo much 

[RuiAce]

Hey
i have a relativity question i was wandering if you could please help me in?
two protons in an accelerator are moving towards each other at 0.75c. At what speed are the protons approaching each other at relative to a stationary observer in the laboratory and relative to each other?

one more thing please...how would you do it if instead of then particles moving towards each other, they were moving in the same direction?

thankyou soo much 

Admittedly, I'm beginning to forget this part of physics so don't rush to quote me.

When the two protons are moving towards each other, they will see each other approaching at the speed of light 'c'. This is because the speed of light is constant in all frames of references.

Note that the 'light' that the proton gives off to the other proton, is what the other proton, effectively, 'sees'. (Analogy: When you look at your friend, you're just seeing light reflected off their body)

If they were moving in the same direction however, at the same speed, then no experiment can be done to distinguish those two frames of references. They are the same frame of reference and they will appear stationary to each other.
On the initial question, the stationary observer himself is in a rest frame. He will see them at 0.75c.

[jamonwindeyer]

Admittedly, I'm beginning to forget this part of physics so don't rush to quote me.

When the two protons are moving towards each other, they will see each other approaching at the speed of light 'c'. This is because the speed of light is constant in all frames of references.

Note that the 'light' that the proton gives off to the other proton, is what the other proton, effectively, 'sees'. (Analogy: When you look at your friend, you're just seeing light reflected off their body)

If they were moving in the same direction however, at the same speed, then no experiment can be done to distinguish those two frames of references. They are the same frame of reference and they will appear stationary to each other.
On the initial question, the stationary observer himself is in a rest frame. He will see them at 0.75c.

I agree on all counts except the "approach each other at the speed of light" aspect. If I see a car approaching me, and it was travelling at 100 kilometres per hour, while I'm moving towards it at 100 kilometres an hour, then I would say it was travelling 200 kilometres an hour. Relative velocity. So I don't think we can automatically say that the answer is 'c' based on the fact that the reflected light travels at this speed.

That being said, the method of analysis above doesn't sit right with me either. It would mean that each proton sees the other approaching at 1.5c, which violates the Special Theory of Relativity. So, the answer could very well be c, I am not sure, but I don't think we can conclude it based on this reasoning.

So yeah, everything else I totally agree with, just that one bit I don't think the explanation is quite correct. The correct answer eludes me for the moment, give me a bit! Unless someone else wants to weigh in 

[Maz]

i think i still don't quite get the idea...
heres another question...
A pion is an unstable particle that decays into 2 photons. A particular pion travelling at 0.95c, with respect to an observer at rest, decays into 2 photons, X and Y, which then travel in opposite directions. photons travel at the speed of ight

so if this is the 'Pion' travelling at 0.95c in ---> direction                                   <----- 'X'       'Y'---->
 and there is a stationary observer
when calculating the velocity of photon Y with respect to the stationary observer...
wouldn't it be:
Formula:  u+v/1+((uv)/c²)
so the answer is saying (c-0.95c)/((1-0.95c x c)/c²)=c (i know how the speed of light is constant...but idk how they got to it)
i don't really get how they got to that as they are both travelling in the same direction...so why is there a '-0.95'?
i don't think i really get how to do these...
thankyou in advance

[jamonwindeyer]

i think i still don't quite get the idea...
heres another question...
A pion is an unstable particle that decays into 2 photons. A particular pion travelling at 0.95c, with respect to an observer at rest, decays into 2 photons, X and Y, which then travel in opposite directions. photons travel at the speed of ight

so if this is the 'Pion' travelling at 0.95c in ---> direction                                   <----- 'X'       'Y'---->
 and there is a stationary observer
when calculating the velocity of photon Y with respect to the stationary observer...
wouldn't it be:
Formula:  u+v/1+((uv)/c²)
so the answer is saying (c-0.95c)/((1-0.95c x c)/c²)=c (i know how the speed of light is constant...but idk how they got to it)
i don't really get how they got to that as they are both travelling in the same direction...so why is there a '-0.95'?
i don't think i really get how to do these...
thankyou in advance

Hmm, I think this is one of the areas that you guys do that HSC students don't cover. I've never seen that formula before. I just did some asking around and this is only covered at the highest level of First Year University Physics (UNSW). Aka, way beyond my capabilities. However, with regard to that question, I am quite certain the -0.95c comes from the motion of the pion. When it decays, the two photons are emitted at the speed of light with respect to the pion's reference frame. I know the speed of light is constant, but I think relative velocity still applies here, which is where the 0.95c would come from (the pion's speed).

Sorry I can't be of more help, but you have officially exceeded the limits of my knowledge! Let me do some more sleuthing and see if I can get you a better answer 

[jakesilove]

i think i still don't quite get the idea...
heres another question...
A pion is an unstable particle that decays into 2 photons. A particular pion travelling at 0.95c, with respect to an observer at rest, decays into 2 photons, X and Y, which then travel in opposite directions. photons travel at the speed of ight

so if this is the 'Pion' travelling at 0.95c in ---> direction                                   <----- 'X'       'Y'---->
 and there is a stationary observer
when calculating the velocity of photon Y with respect to the stationary observer...
wouldn't it be:
Formula:  u+v/1+((uv)/c²)
so the answer is saying (c-0.95c)/((1-0.95c x c)/c²)=c (i know how the speed of light is constant...but idk how they got to it)
i don't really get how they got to that as they are both travelling in the same direction...so why is there a '-0.95'?
i don't think i really get how to do these...
thankyou in advance

Hey MQ!

Unfortunately, as Jamon said, this is an area that falls outside of the HSC curriculum. As such, as far as Relativity questions go, I'm afraid we won't be of much help. Whilst I've touched on this briefly, I don't feel like I could adequately explain relative velocity in terms of Relativistic effect. However, if you click on this link, hopefully you'll get a good idea of how the formulas work.

Sorry we couldn't be of more help!

Jake

[Maz]

Hey MQ!

Unfortunately, as Jamon said, this is an area that falls outside of the HSC curriculum. As such, as far as Relativity questions go, I'm afraid we won't be of much help. Whilst I've touched on this briefly, I don't feel like I could adequately explain relative velocity in terms of Relativistic effect. However, if you click on this link, hopefully you'll get a good idea of how the formulas work.

Sorry we couldn't be of more help!

Jake

hey
no problem...but thank you for trying...and the link isn't working properly i think...not unless it was supposed to go to my Facebook page 

[jakesilove]

hey
no problem...but thank you for trying...and the link isn't working properly i think...not unless it was supposed to go to my Facebook page 

Try now!

[jamonwindeyer]

hey
no problem...but thank you for trying...and the link isn't working properly i think...not unless it was supposed to go to my Facebook page 

It was Jake's way of saying the answer is inside you, you just have to find it...

Metaphors are strong