Login

Welcome, Guest. Please login or register.

September 03, 2025, 07:58:05 pm

Author Topic: HSC Physics Question Thread  (Read 1283010 times)  Share 

0 Members and 2 Guests are viewing this topic.

RuiAce

  • ATAR Notes Lecturer
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 8814
  • "All models are wrong, but some are useful."
  • Respect: +2575
Re: Physics Question Thread
« Reply #1485 on: November 28, 2016, 06:23:28 pm »
0



katnisschung

  • Trendsetter
  • **
  • Posts: 199
  • Respect: +2
Re: Physics Question Thread
« Reply #1486 on: November 28, 2016, 06:35:32 pm »
0
GOT IT! thanks once again Ruiace
get me out of here

samuels1999

  • Adventurer
  • *
  • Posts: 21
  • "Expectation is the root of all heartache." -W.S.
  • Respect: 0
Re: Physics Question Thread
« Reply #1487 on: November 28, 2016, 07:15:16 pm »
0
Hi Everyone,

I need a bit of help with a question dealing with Kepler's Law. The question related to Mars and one of its moons. The question stated that the orbital period of the moon was 1.09x10^5 seconds and that the mass of mars was 7.1x10^23kg. It then asked me to work out the radius of the moon's orbit around Mars.

I know the question relates to Kepler's Law, but I am uncertain how to use it.

Thanks,
Samuel
"If you can't explain it simply, you do not understand it well enough" -Albert Einstein
Year 12 2017
Subjects: Adv. English, Mathematics, Mathematics Ext 1, Modern History, Physics, Design and Technology
....yeah....its a pretty odd bunch

RuiAce

  • ATAR Notes Lecturer
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 8814
  • "All models are wrong, but some are useful."
  • Respect: +2575
Re: Physics Question Thread
« Reply #1488 on: November 28, 2016, 07:23:18 pm »
+1
Hi Everyone,

I need a bit of help with a question dealing with Kepler's Law. The question related to Mars and one of its moons. The question stated that the orbital period of the moon was 1.09x10^5 seconds and that the mass of mars was 7.1x10^23kg. It then asked me to work out the radius of the moon's orbit around Mars.

I know the question relates to Kepler's Law, but I am uncertain how to use it.

Thanks,
Samuel


f_tan

  • Forum Regular
  • **
  • Posts: 69
  • Respect: 0
Re: Physics Question Thread
« Reply #1489 on: November 30, 2016, 09:18:25 pm »
0
Is angular speed just the rate of rotation?

RuiAce

  • ATAR Notes Lecturer
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 8814
  • "All models are wrong, but some are useful."
  • Respect: +2575
Re: Physics Question Thread
« Reply #1490 on: November 30, 2016, 09:19:29 pm »
0
Is angular speed just the rate of rotation?
Yes, however angular velocity is not a part of the course.

jamonwindeyer

  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 10149
  • The lurker from the north.
  • Respect: +3108
Re: Physics Question Thread
« Reply #1491 on: November 30, 2016, 10:38:34 pm »
0
Is angular speed just the rate of rotation?

It can come in just implicitly when talking about orbits, perhaps even harnessing the rotational motion of earth (the equator is the point of maximum orbital velocity on earth) ;D but yeah pretty irrelevant for the most part ;D

f_tan

  • Forum Regular
  • **
  • Posts: 69
  • Respect: 0
Re: Physics Question Thread
« Reply #1492 on: December 01, 2016, 08:26:56 pm »
0
How do I work out the answer for this question?

What g-force is experienced by a 70 kg test pilot pulling out of a downward dive in a circular arc of radius 1000 m at 200 m/s?
A) 42 g
B) 41 g
C) 5.1 g
D) 4.1 g

Thank you! :)

jakesilove

  • HSC Lecturer
  • Honorary Moderator
  • Part of the furniture
  • *******
  • Posts: 1941
  • "Synergising your ATAR potential"
  • Respect: +196
Re: Physics Question Thread
« Reply #1493 on: December 01, 2016, 09:04:42 pm »
0
How do I work out the answer for this question?

What g-force is experienced by a 70 kg test pilot pulling out of a downward dive in a circular arc of radius 1000 m at 200 m/s?
A) 42 g
B) 41 g
C) 5.1 g
D) 4.1 g

Thank you! :)

This is a seriously intense question. You need to equate net force (F=ma) with centripetal force (due to the circular arc, F=mv^2/r) which gets us



This is the 'additional' acceleration. Now, recall that the formula for working out g-forces is



Therefore, I think the answer is C. Does that all make sense?
ATAR: 99.80

Mathematics Extension 2: 93
Physics: 93
Chemistry: 93
Modern History: 94
English Advanced: 95
Mathematics: 96
Mathematics Extension 1: 98

Studying a combined Advanced Science/Law degree at UNSW

f_tan

  • Forum Regular
  • **
  • Posts: 69
  • Respect: 0
Re: Physics Question Thread
« Reply #1494 on: December 01, 2016, 09:12:14 pm »
0
Yep, that makes sense! Thanks so much!

I have another question:
A spacecraft is rising from the Earth's surface at 49 m/s. At 980 m the booster rocket tanks are jettisoned. Calculate the speed with which the booster tanks crash into the Earth.

jakesilove

  • HSC Lecturer
  • Honorary Moderator
  • Part of the furniture
  • *******
  • Posts: 1941
  • "Synergising your ATAR potential"
  • Respect: +196
Re: Physics Question Thread
« Reply #1495 on: December 01, 2016, 09:16:44 pm »
0
Yep, that makes sense! Thanks so much!

I have another question:
A spacecraft is rising from the Earth's surface at 49 m/s. At 980 m the booster rocket tanks are jettisoned. Calculate the speed with which the booster tanks crash into the Earth.

It's sort of difficult to figure out exactly what the questions asking, but let's make some educated guesses. We can imagine that the booster tank as a projectile, travelling upwards at 49 m/s, 980m above the earth's surface. We want to find out how fast it is going when it crashes into earth.

Turns out, we have a formula we can sub straight into!



Here, it's important that acceleration and displacement have the same sign, as they are both in the same direction.




Since that number pops out perfectly, I'm fairly certain that is the correct answer :)
ATAR: 99.80

Mathematics Extension 2: 93
Physics: 93
Chemistry: 93
Modern History: 94
English Advanced: 95
Mathematics: 96
Mathematics Extension 1: 98

Studying a combined Advanced Science/Law degree at UNSW

katnisschung

  • Trendsetter
  • **
  • Posts: 199
  • Respect: +2
Re: Physics Question Thread
« Reply #1496 on: December 03, 2016, 05:27:02 pm »
0
Fc=mv^2/r
(centripetal force formula)
can only be used on problems with a "small" radius correct?
get me out of here

RuiAce

  • ATAR Notes Lecturer
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 8814
  • "All models are wrong, but some are useful."
  • Respect: +2575
Re: Physics Question Thread
« Reply #1497 on: December 03, 2016, 05:29:17 pm »
0
Fc=mv^2/r
(centripetal force formula)
can only be used on problems with a "small" radius correct?
I have never heard of this

Jakeybaby

  • Trendsetter
  • **
  • Posts: 191
  • Grad Year: 2016
  • Respect: +23
Re: Physics Question Thread
« Reply #1498 on: December 03, 2016, 05:39:24 pm »
0
Fc=mv^2/r
(centripetal force formula)
can only be used on problems with a "small" radius correct?
I've only ever seen this formula used within questions regarding the spinning of an object on a string, not sure about larger radii. However, unsure about whether or not it's covered in HSC as I completed SACE this year. I can't see a large radius being used in a question though.
2016 ATAR: 98.60

2020: Bachelor of Finance @ University of Adelaide

Recipient of the 2017 University of Adelaide Principals' Scholarship

jamonwindeyer

  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 10149
  • The lurker from the north.
  • Respect: +3108
Re: Physics Question Thread
« Reply #1499 on: December 03, 2016, 05:56:40 pm »
0
Fc=mv^2/r
(centripetal force formula)
can only be used on problems with a "small" radius correct?

As long as the motion is uniformly circular, this formula works. Perhaps you are thinking of the fact that orbits are not actually uniformly circular in the real world, they are elliptical! That said, this formula is still used in the HSC course because it is still very accurate ;D

So, for the HSC course, you can use this formula for any body moving in a circular fashion ;D