HSC Physics Question Thread

[Kle123]

I am referring to question 10 of a past HY paper of my school. I thought the answer was D, however the marking guideline says C. It may be wrong though because question 4 was a complete dud (as seen in marking guideline). Could someone explain to me why the answer is C.

[RuiAce]

I am referring to question 10 of a past HY paper of my school. I thought the answer was D, however the marking guideline says C. It may be wrong though because question 4 was a complete dud (as seen in marking guideline). Could someone explain to me why the answer is C.

Where's the question?

[Kle123]

whoops should have previewed my post.

[RuiAce]

whoops should have previewed my post.

I actually don't see how it could possibly be C...

[Kle123]

I actually don't see how it could possibly be C...

wow, my school is so dodgy. Thanks RuiAce for helping and responding so fast to me today.

[jamonwindeyer]

I actually don't see how it could possibly be C...

Just as extra assurance, definitely agree with Rui - And I agree with your answer of D too Kle123

[Bubbly_bluey]

Hey guys! i need a little help with this past hsc question. From what i hopefully understand so far, BC and the current of the wire are running in parallel in the same direction so they would attract. AD runs in the opposite direction as the wire so they would repel. I'm not so sure what happens to AB and CD: do they not experience a force? And so what would happen to the overall square wire because it can't rotate unless there is a magnetic field...
I feel like this response is no enough for the 3 marks... :/

btw thanks jamon and elyse (if you're here) for your positive feedback for english, literally made my night

[jamonwindeyer]

Hey guys! i need a little help with this past hsc question. From what i hopefully understand so far, BC and the current of the wire are running in parallel in the same direction so they would attract. AD runs in the opposite direction as the wire so they would repel. I'm not so sure what happens to AB and CD: do they not experience a force? And so what would happen to the overall square wire because it can't rotate unless there is a magnetic field...
I feel like this response is no enough for the 3 marks... :/

btw thanks jamon and elyse (if you're here) for your positive feedback for english, literally made my night

Hey Bubbly! You are welcome let me see if I can help here, a few points:

- You're correct on all counts; AB and CD definitely do not experience the force! This is because the currents are perpendicular; you don't get a force in that circumstance
- You have identified all the major points, except you should specify that the attractive force due to BC is larger than the repulsive force due to AD. Therefore, the net force on the loop is attractive. There is no rotation, the loop just moves towards the straight wire (or, as a better answer, just say the loop experiences a net force upwards!

[katnisschung]

Hi there,

so for motors and generators, when calculating the force on a current
carrying conductor within a magnetic field, is the angle measured
from the line parallel to that of the magnetic field or from an
'imaginary' line perpendicular to the magnetic field?

thanks... i hope this makes sense

[Kle123]

Hi there,

so for motors and generators, when calculating the force on a current
carrying conductor within a magnetic field, is the angle measured
from the line parallel to that of the magnetic field or from an
'imaginary' line perpendicular to the magnetic field?

thanks... i hope this makes sense

you would use the angle between the current carrying wire and the magnetic field. So the answer to your question is parallel (as all magnetic field lines i've seen in force questions i imagine are from areas really close to a magnet's pole or from uniform magnetic plates). Also, It wouldn't matter if you use the obtuse or acute angle as sine would give the same answer

[Kle123]

In this question I had some idea of how to answer it, however in marking it afterwards i gave myself a zero as i didnt talk about kepler's law of periods. I still can't relate kepler's law into my answer. Please help ataranote's leaders . Thanks alot!

[jamonwindeyer]

In this question I had some idea of how to answer it, however in marking it afterwards i gave myself a zero as i didnt talk about kepler's law of periods. I still can't relate kepler's law into my answer. Please help ataranote's leaders . Thanks alot!

Hey Kle! So this question first requires a knowledge of WHAT the required manoeuvre is - What was your previous answer (even roughly)? Just want to see whether you've got that down before I explain how Kepler's can be linked to it - Want to make sure I give you the right answer for where you are at

Ps - Thanks for your awesome answer above!

[Kle123]

Thanks Jamon,
I wrote in my answer, assuming that the altitude is high enough that that there is negligible orbital decay, to most efficiently dock the craft, only the thrust is to be applied in the direction of the space station for only a minimal time, so that a small velocity is required which eventually moves the spacecraft to the station. As approaching the station, the thrust is to be forced in the opposite direction for the exact same amount of time so that the ship would decelerate back to being stationary relative to the station. As the craft is now hovering on top of the station, the thrust is to be forced upwards relative to the space station floor for a minimal amount of time so that it can slowly move into docking mode.

I didn't know whether the 5000m is important and also don't know how docking looks like.

[jamonwindeyer]

Thanks Jamon,
I wrote in my answer, assuming that the altitude is high enough that that there is negligible orbital decay, to most efficiently dock the craft, only the thrust is to be applied in the direction of the space station for only a minimal time, so that a small velocity is required which eventually moves the spacecraft to the station. As approaching the station, the thrust is to be forced in the opposite direction for the exact same amount of time so that the ship would decelerate back to being stationary relative to the station. As the craft is now hovering on top of the station, the thrust is to be forced upwards relative to the space station floor for a minimal amount of time so that it can slowly move into docking mode.

I didn't know whether the 5000m is important and also don't know how docking looks like.

Great! Solid answer - I love that you notice that orbital decay could play a role, I like that. One issue with your answer - If you speed up, will you be in the same orbit as the Space Station? No! Speeding up would mean you break away from the earth's orbit, so your answer doesn't quite work. Instead, what we need to do is apply forces such that we increase/decrease our orbital radius. To start, we apply a force to decrease our orbital radius. According to Kepler's 3rd Law, that will also reduce our period, and thus, increase our orbital velocity! We start moving faster, kind of like we are moving to overtake the station on the inside. At the precisely correct moment, we apply another force to return to our initial orbit and dock with the station, having used our reduction in radius to speed ourselves up

Does that make sense? It's a little tricky to comprehend, but if you run it through your head a few times hopefully something clicks!! Once you understand the manoeuvre, it becomes a little easier to tie Kepler's Law into it (it is the reason it works)

[Kle123]

Great! Solid answer - I love that you notice that orbital decay could play a role, I like that. One issue with your answer - If you speed up, will you be in the same orbit as the Space Station? No! Speeding up would mean you break away from the earth's orbit, so your answer doesn't quite work. Instead, what we need to do is apply forces such that we increase/decrease our orbital radius.

OMG that makes so much sense thanks!
One thing though... how would we decrease our orbital radius? Do we provide thrust in front of direction of motion to slow down? But wouldn't that cause us to follow an elliptical orbit/or slowly fall into the Earth's atmosphere / or does slowing down automatically drop us into a lower orbit?

So to reduce orbital radius do we slow down through providing forward thrust then accelerate to the required speed a lower orbit (which is what i'm thinking)?