HSC Physics Question Thread

[RuiAce]

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

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

Hey Jamon, whilst we're here

Since the orbits are actually elliptical, does the formula r^3/T^2=GM/(4pi^2) still hold true regardless?

[jamonwindeyer]

Hey Jamon, whilst we're here

Since the orbits are actually elliptical, does the formula r^3/T^2=GM/(4pi^2) still hold true regardless?

I believe the fact that it is elliptical means that it is more accurate to do this:



That is, the mass of the orbiting body does play a small role. Provided that we are talking about a planet orbiting a star, or a satellite orbiting a planet, that's okay. A planet orbiting a planet, it doesn't work so well

[RuiAce]

I believe the fact that it is elliptical means that it is more accurate to do this:



That is, the mass of the orbiting body does play a small role. Provided that we are talking about a planet orbiting a star, or a satellite orbiting a planet, that's okay. A planet orbiting a planet, it doesn't work so well

Kinda reminds me of the formula for binary stars in astrophysics tbh...


Edit: It IS that formula ahaha

[katnisschung]

hi again!

i did question 3 of this past paper
https://thsconline.github.io/s/?view=6521&id=0ByEFYhkkDQBKeDNxaHFsOG5GQkU&n=Hurlstone%202007%20w.%20solhttps://thsconline.github.io/s/?view=6521&id=0ByEFYhkkDQBKeDNxaHFsOG5GQkU&n=Hurlstone%202007%20w.%20sol

unfortunately there are no answers for this q.

my answer...
probably wrong...

if the scale was moved further away from the plane in which the ball moves, this adds an unknown extra
distance to the ball's projectile path. Assuming the ball is thrown with the same force,
the initial horizontal velocity would decrease as the distance is increased. Therefore, as
the initial horizontal velocity decreases, the initial velocity must decrease...

thanks ruiace or jamon or jake or anyone willing to answer another one of my questions

[katnisschung]

also quick question
into how much detail should u know about the michelson-morely
experiment... yes we just covered it at school...yes i'm concerned
lmao missed out on 2 weeks+ of lessons becos of hallelujah practice 

[FallonXay]

also quick question
into how much detail should u know about the michelson-morely
experiment... yes we just covered it at school...yes i'm concerned
lmao missed out on 2 weeks+ of lessons becos of hallelujah practice 

Relatively well. Normally, the most they'd throw at you on this question would be a 4, possibly 5 marker.
• You should be able to draw a labelled diagram of the experiment
• Identify the experimental aim (to measure the relative velocity of the Earth through the aether wind)
• Describe the experiment (splitting of beam at 90 degrees via a half-silvered mirror which were then reflected and combined to produce an interference pattern etc) and how it produced a 'null result' (The interference pattern was the same when the apparatus was rotated)
• Effects/ impact of the Michelson Morley Experiment on scientific views (i.e debates about the validity of the experiment/ doubt on the existence of the aether, use as one of the pieces of evidence supporting Einstein's Theory of Special Relativity)
• and I remember being asked in a practice trial paper once about the properties of the aether (medium through which light travelled through, permeated all matter in the universe, believed to be an absolute frame of reference, undetectable elastic material)

EDIT: Jamon explains the Michelson and Morley experiment/ aether really well in his guide here: Physics: A Complete Guide to the Course! (It's at the top of the Relativity section, you'll probably recognise the diagram)

[RuiAce]

hi again!

i did question 3 of this past paper
https://thsconline.github.io/s/?view=6521&id=0ByEFYhkkDQBKeDNxaHFsOG5GQkU&n=Hurlstone%202007%20w.%20solhttps://thsconline.github.io/s/?view=6521&id=0ByEFYhkkDQBKeDNxaHFsOG5GQkU&n=Hurlstone%202007%20w.%20sol

unfortunately there are no answers for this q.

my answer...
probably wrong...

if the scale was moved further away from the plane in which the ball moves, this adds an unknown extra
distance to the ball's projectile path. Assuming the ball is thrown with the same force,
the initial horizontal velocity would decrease as the distance is increased. Therefore, as
the initial horizontal velocity decreases, the initial velocity must decrease...

thanks ruiace or jamon or jake or anyone willing to answer another one of my questions

You seem to be answering 3b) so I'll leave 3a) alone for now.

Here's my attempt:
I wouldn't say that a fictitious distance was added in. However, if the data logger was built with a scale set for specifically, say, 3m away, then if you move the data logger back there will definitely be problems.

Rather than talking about an unknown extra distance, I would just use common sense here. In the same way that if you stand further away from a television you see things smaller, the trajectory of the ball relative to the data logger becomes smaller. Because of the fact the distance got shrunk (and obviously nothing changed to time), the velocity must supposedly decrease as well. Hence the data logger will register an underestimated velocity (keyword here is underestimated, because we're talking about how quantitative data was incorrectly predicted).

Which does, of course, mean that I agree with your final answer. Just the explanation was a bit weird...

Or I could probably use the inverse square law to argue my point but I'm a bit too scared to do that. The others would be able to give you a more confident answer.

[katnisschung]

I was doing a past paper question and it got me considering this....

according to an observer in the ship watching a ball being
dropped in the ship with constant velocity why does it
appear to fall directly down? (does the parabolic path of the ball 'compensate'
for the distance traveled by the ship thus drop directly below the observer?)

also what would happen in the situation that the ship was accelerating?

[RuiAce]

I was doing a past paper question and it got me considering this....

according to an observer in the ship watching a ball being
dropped in the ship with constant velocity why does it
appear to fall directly down? (does the parabolic path of the ball 'compensate'
for the distance traveled by the ship thus drop directly below the observer?)

also what would happen in the situation that the ship was accelerating?

If the ship was moving at a constant velocity it is its own inertial frame of reference. The ball belongs to this frame of reference. So to the observer in the ship, it will just see the ball drop vertically.

Note that the person in the ship actually thinks everything else around the ship is what's moving. Not that he is the one moving; he thinks he is stationary.


If the ship was accelerating, then we'd have a non-inertial frame of reference, to which THEN we see the parabolic path; not falling straight down.

[FallonXay]

If the ship was moving at a constant velocity it is its own inertial frame of reference. The ball belongs to this frame of reference. So to the observer in the ship, it will just see the ball drop vertically.

Note that the person in the ship actually thinks everything else around the ship is what's moving. Not that he is the one moving; he thinks he is stationary.


If the ship was accelerating, then we'd have a non-inertial frame of reference, to which THEN we see the parabolic path; not falling straight down.

Hiya Rui~

Just a quick question, 'cos this reminded me of something some people were talking about in my class earlier this year: In the non-inertial frame of reference, if the ball is dropped, how come it's a parabolic path (not diagonally linear)?

[RuiAce]

Hiya Rui~

Just a quick question, 'cos this reminded me of something some people were talking about in my class earlier this year: In the non-inertial frame of reference, if the ball is dropped, how come it's a parabolic path (not diagonally linear)?

I eventually taught myself how this works by combining both physics and maths together (although mostly still physics). You're pretty capable so I'll let you try to decipher it, but come back if you need further help (It's kinda weird to be fair - I visualised it all in my head in a matter of seconds but an explanation takes ages)

The idea is to use projectile motion principles. Pretend that you are about to throw a ball. Whilst the ball is in your hand, you have full control of its motion. However, at the instant the ball leaves your hand, it becomes a projectile.

The point of this, is that at the instant the ball leaves your hand, that's when you can guarantee that its horizontal velocity is constant. When it was in your hand, you could literally just move the ball back and forth, and that's obviously not a constant horizontal velocity.


Back to relativity.
Pretend we are in an inertial frame of reference first. Suppose you're travelling at some velocity (for visual purposes, let's consider small velocities such as 20ms^-1). Additionally, first suppose that the ball is still in your hand. Then it's obviously travelling at 20ms^-1 with you. Then suppose you release the ball. The ball is in the same inertial frame of reference as before you left it; it's also travelling at 20ms^-1. Hence because both you and the ball are travelling at 20ms^-1, you see it drop vertically.

But why is it, that when you drop the ball, it's still "moving at 20ms^-1"? (Note that I'm trying to keep away from Einsteinian physics here; I'm kinda using Newtonian physics).

Consider a stationary observer watching you travel at 20ms^-1. To him, it appears as though the ball already had an initial velocity of 20ms^-1! You were holding the ball at 20ms^-1 and that's what he saw, but the instant you let go of the ball you basically released it AT horizontal velocity 20ms^-1. So the observer can actually confirm that once the ball is dropped, you are travelling at the same speed as with the ball.


Now for the non-inertial frame of reference.
Once again, you start off by holding the ball still. Except this time you're accelerating. Acceleration is indeed what's going to play a key role here.

Suppose you start accelerating from, idk, at rest. You accelerate from being at rest (at some magnitude) all the way up to 20ms^-1, and THEN you drop the ball. This is where the analogy from earlier comes into play.

When you were accelerating, you had full control of the velocity of the ball. So if you were accelerating, the ball accelerated with you. But once you dropped the ball, it isn't with you anymore; you lost control over the ball! The ball was released when it acquired a velocity of 20ms^-1, so because it's no longer affected by anything (i.e. your hand went away) it now travels by 20ms^-1 by itself.

But whilst the ball is travelling at 20ms^-1 by itself now, what are you doing? You're still accelerating! Because you're getting faster than 20ms^-1 you start seeing the ball 'lag behind' you now.

The point, is that once you let go of the ball, it formed its own new INERTIAL frame of reference. You remain as a non-inertial frame of reference because you're accelerating, but nothing drives the ball into accelerating anymore so it's now in an inertial frame. This is also what the stationary observer would see - you're getting faster but the ball appears to stay at the same speed.

[FallonXay]

The point, is that once you let go of the ball, it formed its own new INERTIAL frame of reference. You remain as a non-inertial frame of reference because you're accelerating, but nothing drives the ball into accelerating anymore so it's now in an inertial frame. This is also what the stationary observer would see - you're getting faster but the ball appears to stay at the same speed.

Haha 'Quick Question'  (But in all seriousness, thanks for the detailed answer)

So what I don't understand is that: since you are an observer on the accelerating vehicle and in your frame you are 'stationary', wouldn't you observe the ball the be accelerating away from you (isn't the ball only travelling at a constant speed if the observer were also in the inertial frame/ at travelling at a velocity of 0 m/s)? So wouldn't this observer see a horizontal acceleration?

(Also, what effect does having two different frames of reference have? - i.e the ball in the non-inertial and the observer in the inertial)

EDIT: I mean, the ball in the inertial, and observer in the non-inertial

[RuiAce]

Haha 'Quick Question'  (But in all seriousness, thanks for the detailed answer)

So what I don't understand is that: since you are an observer on the accelerating vehicle and in your frame you are 'stationary', wouldn't you observe the ball the be accelerating away from you (isn't the ball only travelling at a constant speed if the observer were also in the inertial frame/ at travelling at a velocity of 0 m/s)? So wouldn't this observer see a horizontal acceleration?

(Also, what effect does having two different frames of reference have? - i.e the ball in the non-inertial and the observer in the inertial)

Yeah I giggled at "quick" when I reread that

Yep. So here's a twist. Normally, if you're the one accelerating you would know that you're accelerating. This is from the fact that we can't "feel" velocity as we can't "feel" momentum. Rather, we can feel acceleration because we can feel force.

Most of the analogies are given where you don't know that you're accelerating. i.e. You think everything around you is accelerating. (Just like how in physics we studied as though everything around you was moving.)
In this scenario, you're right. Somewhere in there I wrote that it appears as though the ball "lags behind" you. This is because indeed, the ball appears to carry a negative acceleration relative to you!

This is a part of why non-inertial frames of reference and general relativity isn't looked at in the course; it becomes too complex to describe. (Jake might know a thing or fifty.) To quantify the weird happening we introduce things known as "fictitious forces".

To a stationary observer, they'll probably know that some force is being exerted on you, and hence you're accelerating. As for you, these fictitious forces are a mere way to justify why the ball DOES appear to have a negative acceleration. The reason why the negative acceleration is as stated in my above post, but the outcome is basically what you determined.

Wikipedia provides this definition:

A fictitious force, also called a pseudo force, d'Alembert force or inertial force, is an apparent force that acts on all masses whose motion is described using a non-inertial frame of reference, such as a rotating reference frame.

The basic idea is that because you are actually the one accelerating, you see these weird things that go on. That's really much it! You don't realise you're accelerating, maybe, but you just are so these fictitious forces happen to 'exist'!

(As an aside, another example of a fictitious force is the 'centrifugal force', which only exists to justify Newton's third law of motion with the centripetal force.)

Having two different frames of references besides your own just means you now have three frames of references in total. It just means there's now more things to consider. It's like you're playing an MMORPG and you're the noob being overshadowed by the higher damage of the Lv20 toon, but there's a Lv60 toon behind them dealing even more explosive damage

[FallonXay]

Yeah I giggled at "quick" when I reread that

Yep. So here's a twist. Normally, if you're the one accelerating you would know that you're accelerating. This is from the fact that we can't "feel" velocity as we can't "feel" momentum. Rather, we can feel acceleration because we can feel force.

Most of the analogies are given where you don't know that you're accelerating. i.e. You think everything around you is accelerating. (Just like how in physics we studied as though everything around you was moving.)
In this scenario, you're right. Somewhere in there I wrote that it appears as though the ball "lags behind" you. This is because indeed, the ball appears to carry a negative acceleration relative to you!

This is a part of why non-inertial frames of reference and general relativity isn't looked at in the course; it becomes too complex to describe. (Jake might know a thing or fifty.) To quantify the weird happening we introduce things known as "fictitious forces".

To a stationary observer, they'll probably know that some force is being exerted on you, and hence you're accelerating. As for you, these fictitious forces are a mere way to justify why the ball DOES appear to have a negative acceleration. The reason why the negative acceleration is as stated in my above post, but the outcome is basically what you determined.

Wikipedia provides this definition:The basic idea is that because you are actually the one accelerating, you see these weird things that go on. That's really much it! You don't realise you're accelerating, maybe, but you just are so these fictitious forces happen to 'exist'!

(As an aside, another example of a fictitious force is the 'centrifugal force', which only exists to justify Newton's third law of motion with the centripetal force.)

Having two different frames of references besides your own just means you now have three frames of references in total. It just means there's now more things to consider. It's like you're playing an MMORPG and you're the noob being overshadowed by the higher damage of the Lv20 toon, but there's a Lv60 toon behind them dealing even more explosive damage

So the ball only 'appears' to be accelerating - but in reality it's travelling at a constant velocity? Would it be correct/ 'appropriate' to say that the observation of the ball as accelerating is incorrect (to scientifically make this conclusion from the observation)? I'm a little confused: I don't really understand the effect of the fictitious forces in affecting 'reality'~

[RuiAce]

So the ball only 'appears' to be accelerating - but in reality it's travelling at a constant velocity? Would it be correct/ 'appropriate' to say that the observation of the ball as accelerating is incorrect (to scientifically make this conclusion from the observation)? I'm a little confused: I don't really understand the effect of the fictitious forces in affecting 'reality'~

See this is why it's out of the HSC course. It is complicated, and what the actual correct answer is, well I have no idea.

But basically that's how I analyse the scenario.