HSC Physics Question Thread

[jamonwindeyer]

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'~

Tagging in yes, it would be appropriate to say that it is incorrect, it is more appropriate to say that it is an inaccuracy due to taking our measurement from a non-inertial frame of reference like, it is still a frame of reference, and so it has value, but the observation needs to be interpreted correctly.

The fictitious forces thing is best explained with an example. Imagine being in a plane. Constant speed, no turbulence, all the windows are shut. There is absolutely no way to prove that you are moving in the sky. You are in an inertial frame of reference.

Now, pretend the plane gives a burst of speed, lurches upwards, etc. This will cause things to fly around and for you to feel a force, possibly get knocked off your feet.

At this point, in your magic plane that you can't see out of, you have two options. You can accept that you are in fact flying, and have indeed accelerated. That's the logical choice. If, instead, you want to maintain the fact that you are in an inertial frame of reference and that you aren't accelerating, you have to then invent something else that caused all the commotion. This is the fictitious force, you invent it to avoid accepting the fact that you yourself are accelerating.

It's kind of like believing in the Tooth Fairy when you are young. The logical choice is the easiest, just accept that the tooth fairy doesn't exist. If you want to try and maintain the fact that she does exist, then you need to invent a lot of fiction, do a lot of storytelling, to make your view make sense

[FallonXay]

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.

Ahhh k. Cheers for the explanation/ discussion Rui, was very interesting 
(Also, I think my brain melted in the process of making sense of everything haha)

The fictitious forces thing is best explained with an example. Imagine being in a plane. Constant speed, no turbulence, all the windows are shut. There is absolutely no way to prove that you are moving in the sky. You are in an inertial frame of reference.

and very nice example, helped with the whole fictitious force thing  . Thanks Jamon~



P.s if you're interested in where this sparked from there was a similar question except it was about a pendulum hanging on a roof in which was cut (Though I'm not sure that the string would affect anything because there isn't tension anymore when the string is cut?...) - 2010 Q23b

[jamonwindeyer]

P.s if you're interested in where this sparked from there was a similar question except it was about a pendulum hanging on a roof in which was cut (Though I'm not sure that the string would affect anything because there isn't tension anymore when the string is cut?...) - 2010 Q23b

I remember this question, I had a feeling that this was the catalyst!! The string doesn't affect anything after it is cut, it is just used as a way to indicate the fact that the train is accelerating, without saying that it is

[RuiAce]

Don't worry. My brain melts all the time at uni

[katnisschung]

could anyone explain why it is (d) for q 15 (multiple choice)

i know its either a or c
(starts at 0m/s then accelerates downward at a constant rate due to acceleration)
but then why does the velocity go into the negative for when it comes back up?
pls help


http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2001exams/pdf_doc/physics_01.pdf

[jamonwindeyer]

could anyone explain why it is (d) for q 15 (multiple choice)

i know its either a or c
(starts at 0m/s then accelerates downward at a constant rate due to acceleration)
but then why does the velocity go into the negative for when it comes back up?
pls help

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2001exams/pdf_doc/physics_01.pdf

Hey! Great question; it sounds like you pretty much have it!

First of all, the graph in D has straight lines; this makes sense because the acceleration is constant. So any changes to velocity should always be linear (the gradient of those lines would be \(9.8\), to match with acceleration due to gravity).

The reason that the line goes into the negative region is because it is, after the bounce, travelling in the opposite direction to what it did initially. So say the ball is travelling \(30ms^{-1}\) downwards, after the bounce it might be travelling \(28ms^{-1}\) upwards. The difference in direction is what that shift reflects.

That said, it would be a better graph if it was negatively sloped. That is, went into the negative first, then shifted into the positive. But it asks for the best representation, so D will have to do does that help?

[katnisschung]

ahh so they have nominated upwards as negative?

[jamonwindeyer]

ahh so they have nominated upwards as negative?

Yeah, for some reason they have! Probably because it started travelling downwards, so that was made the positive direction. There is nothing wrong with it, it's just a little bit of a confusing choice

[katnisschung]

ahhh im probably overthinking it
but why is it a straight line drop in velocity?
shouldn't it be angled like a (taking into account that m= acceleration due to gravity)

[jamonwindeyer]

ahhh im probably overthinking it
but why is it a straight line drop in velocity?
shouldn't it be angled like a (taking into account that m= acceleration due to gravity)

If you were looking at position, then it would be curved for sure!! But this is velocity; the formula for velocity is this:



It's actually a straight line! It's a straight line because, yes, the gradient is the acceleration due to gravity. The acceleration due to gravity is constant. A line with a constant gradient is a straight line!

[teapancakes08]

How should I approach this question:

"A projectile fired up into the air from the top if a 75m high cliff hits the ground 500m out from the base, considering the projectile being fired at an angle and landing on a surface below that from which it was fired. Find the initial velocity, horizontal velocity, vertical velocity, range, maximum height, time taken to reach maximum height, and time of flight."

From the question I'm given the range, but am confused on where to start with everything else. It'd be easier if I could find with the maximum height though, since everything works out from there, but it doesn't seem likely to find it first. So which value should I try to find first?

[jamonwindeyer]

How should I approach this question:

"A projectile fired up into the air from the top if a 75m high cliff hits the ground 500m out from the base, considering the projectile being fired at an angle and landing on a surface below that from which it was fired. Find the initial velocity, horizontal velocity, vertical velocity, range, maximum height, time taken to reach maximum height, and time of flight."

From the question I'm given the range, but am confused on where to start with everything else. It'd be easier if I could find with the maximum height though, since everything works out from there, but it doesn't seem likely to find it first. So which value should I try to find first?

Hey!! Are you sure that you've been given all the required information here? Any missing part of the question? I ask because I don't think there is enough information to answer (happy for someone to correct me)

[Jakeybaby]

Hey!! Are you sure that you've been given all the required information here? Any missing part of the question? I ask because I don't think there is enough information to answer (happy for someone to correct me)

I do agree with you, seems that all equations will be left with 2+ unknown variables. One part seems to be missing from the question imo.

[jamonwindeyer]

I do agree with you, seems that all equations will be left with 2+ unknown variables. One part seems to be missing from the question imo.

Yeah I think so too! I can get it into an equation with only \(\theta\) and \(V\), that's best I got!

[teapancakes08]

Yeah I think so too! I can get it into an equation with only \(\theta\) and \(V\), that's best I got!

Question 9 on this sheet.