HSC Physics Question Thread

[jakesilove]

Hello Cajama:

I am a year 12 student currently undertaking physics and I am more than happy to help you out here. In regards to questions like these, a lot of people get stuck on it because of how abstract the question is (after all, we cant really see the magnetic field and whoever made this question is so time-poor that they wouldnt even bother giving us a diagram). A lot of my friends are in the same situation and what I recommend to do is to draw a simple 2D diagram just to make life easier. In a HSC exam, questions like these would often be accompanied by a 3D diagram, and similarly what I would recommend for you to do is to draw a 2D diagram which allows you to observe whats happening much more clearly.

Ok so recommendations aside, lets get into the question.

(Image removed from quote.)

N.B. In the 2D diagram I have attached, I have assumed the direction of the magnetic field vectors and the negative and positive terminals on the voltage supplier. Now, you will soon discover that even if the magnetic field direction reverses, or if the negative and positive terminals reverse, the current will still be travelling in a direction that is perpendicular to the direction of the magnetic field. Hence in this case I am allowed to assume these directions. The importance of this will be discussed later on in part B.

a) This question is rather simple. After you manage to draw out your 2D diagram, you will see thats it's simply an application of Ohm's Law, something we learnt back in year 11 in the "Electricity at homes" module.
Ohm's law states that R = V/I. In this question, our R = 4 ohms, V = 36V and hence when we substitute these values into the formula, we get an answer of 9A currents (R = V/I, I = V/R, I = 36/4, I = 9A)

b) Usually in a question that is divided into several parts, the outcome of the previous part will usually relate to solving the second part. In part a), we were demanded to calculate the current flowing through the metal bar. Recall all the formulae you have learnt in Motors and Generators module, there are two formulae that heavily relate to the concept of force: F = BILsin(theta) and F=qvBsin(theta). If we think about F=qvBsin(theta), we will soon discover that we dont exactly have a value for q, and nor do we have a value for v. Hence F=qvB would be unsuitable for solving part b). Let's consider F =BIL sin(theta). We calculated the current flowing through the bar in part a), which we found out to be 9A. We are also provided with the length of the bar which is 0.5m and the magnitude of the magnetic field which is 0.3T.

But, what about theta? Clearly we are NOT provided with the value of theta. BUT THROUGH DRAWING A 2D DIAGRAM, we can find out what theta is. Lets go back to the first principles and define what this theta is, in the equation F = BILsin(theta). This theta is defined as the angle between the direction of the current and the magnetic field vector. So, when you consider the 2D diagram, it becomes apparent that the current is travelling in a direction that is perpendicular to the magnetic field direction. Therefore, the value of theta is 90 degrees.

Now let's substitute everything into the equation F= BILsin(theta). B = 0.3T, I = 9A, L = 0.5m and theta = 90 degrees. Hence
F = 0.3 x 0.5 x 9 x sin(90) = 1.35 N  which is the magnitude of the force. The direction of the force here cannot be calculated because we are not provided with the direction of the magnetic field, we only know that the direction of the magnetic field, whether to the right or to the left, will be perpendicular to the current flow within the metal bar.

c) Ok so this is perhaps the most tricky part of the question because it combined the knowledge from TWO MODULES!! The way I figured out this question is through observing which part of the question I havent used yet. So far, we have used resistance, voltage, length and magnetic field values provided by the question. So now we are left with 0.04kg mass, and like what I said beforehand, the previous part in a question will usually relate to its subsequent part. In part b) we found out the magnitude of force which is 1.35N, hence using Newton's 2nd law F=ma, we discover that 1.35 = 0.04(a), hence a=33.75 ms^-2. Sweet, now we have acceleration, but we were asked to find the displacement of the bar. A convenient way to figure out what to do is to just skim through the formula sheet, finding out the equation that relate acceleration and displacement together. Hence in this case it would be appropriate to use the formula y = ut + 1/2 at^2, because we have all the necessary details to figure out displacement (y). Initially, velocity of the bar is 0, hence u= 0 ms^-1. The acceleration we have worked out is 33.75 ms^-2 and the time in this case is provided (t = 0.25s). Everything after this step now is easy, substitute in all the values, we will obtain that y = 0(0.25) + 1/2 (33.75) (0.25)^2 and throw this into the calculator will will obtain an answer of 1.05m (2.d.p.).

Happy Physics Land, this is a great answer to Cajama's questions! Again, I'd recommend anyone who wants to ask a question, or answer a question, please does so.

Remember that before you can ask a question, or answer one, you'll have to make an ATAR Notes account here. Once you've done that, a little 'reply' button will come up when you're viewing threads, and you'll be able to post whatever you want!

I think that my answer essentially covers most of what Happy Physics Land has included, with one key area lacking: I didn't draw a diagram. I think that's extremely, extremely important and so thank you HPL for doing so. When you get a complicated question like this, ALWAYS draw a diagram trying to figure out what's actually going on. Great one!

Jake

[Happy Physics Land]

Happy Physics Land, this is a great answer to Cajama's questions! Again, I'd recommend anyone who wants to ask a question, or answer a question, please does so.

Remember that before you can ask a question, or answer one, you'll have to make an ATAR Notes account here. Once you've done that, a little 'reply' button will come up when you're viewing threads, and you'll be able to post whatever you want!

Thank you jake for your recognition, I do believe that your answer was much more succinctly stated than mine and its a lot more easier to understand actually. I hope I can soon get on your level jake!!!

I think that my answer essentially covers most of what Happy Physics Land has included, with one key area lacking: I didn't draw a diagram. I think that's extremely, extremely important and so thank you HPL for doing so. When you get a complicated question like this, ALWAYS draw a diagram trying to figure out what's actually going on. Great one!

Jake

Thank you very much jake!!! I  think that your response was much clearer and succint than mine actually and I really understood every part of your explanation very easily. I HOPE I CAN GET ON YOUR LEVEL SOON JAKE!    (I posted this comment half an hour ago ldk why it wasnt there damn o.O)

[cajama]

You guys are absolute legends! Massive thanks for the help.

[jamonwindeyer]

sorry for all the questions but i have no idea how to do some of these

1. a car of mass 2.0 x 10^3kg cruises North down the high way at 100 km/h, with a driving force of 1.2 x 10^4N.

a) calculate the retarding force of friction acting on the car.
b) the car speeds up to 110m/s in 5 seconds. Calculate the acceleration of the car, and thus determine the new driving force of the car.

2. a dog pulls a 80kg sled along the ground with 500N, which encounters a friction force of 150N. On the sled is a 20kg box.

a) calculate the acceleration of the sled.
b) calculate the friction force of the sled on the box which moves the box forwards along with the sled.
c) calculate the net force on the sled.

these are some of my answers but i have no idea whether they're correct:
2. a) a = (500-150)/(80+20) = 3.5 m/s/s forwards
    b) F = 3.5 x 20 = 70N forwards

Hello again Chloe! Don't be sorry, it's what we are here for! Let's begin with your first question (I apologise in advance that I am not using formatted formulae, this is a brief technical issue which will be fixed tomorrow!).

The car is cruising at 100 km/h. What this means (though perhaps not stated explicitly) is that the car is travelling at a constant speed. Thus, it has an acceleration of 0! What does this mean? Well, consider Newton's 2nd Law:

Net Force = Mass x Acceleration

If the acceleration is zero, then the net force must also be zero. But how can this be? The answer is friction. When a car is cruising at constant speed, the driving force of the motor is perfectly balanced by the retarding force of friction, caused by the friction of the road on the tyres, and the friction caused by air resistance.

Therefore, the answer to your first question is 12 000N, the retarding frictional force (retarding simply means de-accelerating, or acting against) is equal to the driving force, since speed is constant.

For your second question, remember that acceleration is simply the rate of change of velocity. Ie - velocity increasing by 1 metre per second, in 1 second, corresponds to an acceleration of 1ms^-2. In this question, the velocity increases by 10 metres per second, in 5 seconds. This corresponds to an acceleration of:

Acceleration = Change in Velocity/Time = 2 ms^-2

So, that's the acceleration. To get the force, we need to be a bit tricky. We will calculate the additional force, which causes this acceleration, by plugging this value and the mass into Newtons 2nd Law.

F = Mass x Acceleration = 2 x 2000 = 4000N additional force to speed the car up.

We then add this to the original driving force, to obtain an answer of 16 000  Newtons . Note, this question does require the assumption that the retarding force does not change based on speed, which in real life, it certainly does. But Physics at this level is about keeping things simple 

Right, second question:

For part A, you are spot on. Acceleration is just obtained by plugging in to Newton's 2nd Law, and you've made all the correct inclusions (box + sled weight, and net force). Awesome!

For part B, again, spot on. The force of friction "drags" the box along with the sled, and this friction must provide the same acceleration as in Part A. Therefore, your answer is correct again, well done!

Part C requires a little clever thinking. Let's think about all the forces on the sled, we have:

- The 500N driving force (or dog force, if you will) FORWARD
- The 150N of friction from the ground BACKWARD
- 70 N from the box BACKWARD

But where does this third force come from? Just as the sled is dragging along the box (due to friction), the box is dragging the sled backwards. This is a consequence of Newton's 3rd Law. Therefore, the answer is:

F = 500 - 150 - 70 = 280 Newtons in the direction of motion.

You could also consider the 350N net force on the sled and box together, and that since 70N of this must be spent accelerating the box, the rest must be spent accelerating the sled. Both are valid interpretations which lead to the same answer.

I hope these answers helped Chloe! And never apologise for asking a question, ask as many as you can, as much as you want, we and your peers want you to improve and succeed (the way Physics scales is entirely dependent on how well Physics students do in the HSC, so it is in everyones best interests to help each other out  ) .

[sire123]

Yo guys, does anyone have a summary of just the formulas in Physics? The core topics I mean (not option) cos I feel like the one I wrote doesn't cover all of them. It'll be really great if u can send it via email, or just pics of anything. Cheers!

[Happy Physics Land]

Yo guys, does anyone have a summary of just the formulas in Physics? The core topics I mean (not option) cos I feel like the one I wrote doesn't cover all of them. It'll be really great if u can send it via email, or just pics of anything. Cheers!

Hello Sire 123:

Here are ALL the formulae you will need to know for the Topic of Space in HSC Physics. Everything that I have underlined using a red colour is compulsory for you to know, everything l underlined using a yellow-ish colour is everything that I would recommend you to know because you can refer to them in your questions. Everything else that are not underlined are just beneficial for you to know but you wouldnt need to worry too much about them. I will post another one for motors and generators soon and I will organise it in such a way that it would be easier for you to see. Sorry about the layout of this formula sheet which wouldnt be the most user-friendly.



If you have any queries dont hesitate to pop up and ask!!!

Best Regards
Happy Physics Land

[Happy Physics Land]

Yo guys, does anyone have a summary of just the formulas in Physics? The core topics I mean (not option) cos I feel like the one I wrote doesn't cover all of them. It'll be really great if u can send it via email, or just pics of anything. Cheers!

Hello Sire 123:

Here is the complete set of notes on all the formulae you are required to know for the topic Motors and Generators, again, if you have any queries on what the letters represent or how to derive any formula, please dont hesitate to pop up and ask, because others may have the same question as you and you are actually helping everyone by proposing questions. Likewise to my previous formulae for the space topic, the red ticks suggest those formulae that you must know by heart, the yellow ticks indicate those formulaw that is strongly recommended for you to know. Those formulae without a tick would be beneficial for you to know however they are not too necessary for HSC exams (you can still refer to them). Just a sidenote, it is really important when you are answering motors and generators questions to refer to one of these formulae or refer to Faraday or Lenz's law wherever appropriate. These will enhance your reasoning and your argument. Also, by doing so, the markers can easily recognise that you have a solid understanding of these knowledge.



In regards to the formulae on Ideas into Implementation, Jake will be able to help you out, since I have not done the topic yet. But anyways, I hope you a successful HSC year and all the best in Physics!

Best Regards
Happy Physics Land

[jakesilove]

Yo guys, does anyone have a summary of just the formulas in Physics? The core topics I mean (not option) cos I feel like the one I wrote doesn't cover all of them. It'll be really great if u can send it via email, or just pics of anything. Cheers!

Hey Sire!

Firstly, let's just give HPL a massive round of applause for the massive effort he's put into answering this question. A stellar summary, and honestly that is definitely something we will be adapting into a resource. Utter legend.

As for Ideas to Implementation: It really isn't a formula-based topic, so having a glace through the dotpoint you actually don't need to know ANY formulas that aren't on your formula sheet!

Jake

[sire123]

Orritey thanks Jake, and Happy Physics Land, you are an absolute lege Thanks man!

[sire123]

And btw, I'm not sure with some of the ones that aren't underlined.
Like with the rocket launch formula T - (Fweight-Fair), isn't air resistance disregarded in the whole topic?
Also, I'm assuming the formulas of thrust force and rocket velocity are outside of the syllabus? Could you explain how you derive it, and which kind of Qs you'll use it in (or can you actually use it if we technically haven't "learnt" it?)
Lastly, why is EMF produced by rotation of conductor negative value?

[Happy Physics Land]

And btw, I'm not sure with some of the ones that aren't underlined.
Like with the rocket launch formula T - (Fweight-Fair), isn't air resistance disregarded in the whole topic?
Also, I'm assuming the formulas of thrust force and rocket velocity are outside of the syllabus? Could you explain how you derive it, and which kind of Qs you'll use it in (or can you actually use it if we technically haven't "learnt" it?)
Lastly, why is EMF produced by rotation of conductor negative value?

Hello Sire 123:

Honestly man, these problems that you have put forward are such good questions and this will not only help you, but also other students in NSW as well. Honestly heaps good questions. Ok, those formulae that I have not underlined are definitely not directly related to the syllabus, however they are beneficial for you to know because it will help you understand the theory better and you can definitely refer to them in some questions.

I understand that in projectile motion, air resistance is completely ignored for the sake of simplicity of calculation. However, when we draw the free body diagram for rocket launch, we know that there will be three forces acting upon the rocket: thrust force, weight force and air resistance. You will not be required to perform a calculation based on this formula but in questions where they ask you the forces experienced by the rocket, these three forces MUST be mentioned.

And yes the thrust force and rocket velocity are not directly related to the syllabus HOWEVER IT IS RELEVANT for answering questions in HSC EXAMS, they enable you to see these quantities are derived and you can still refer to them in HSC exam questions. Sorry I can give you an example of a HSC question right now and I will show you the derivation process for this formula when I get back home this afternoon (Im being a naughty boy going on atarnotes during my chemistry lesson hehe).

And the last question is my absolute favourite. The reason why the value of EMF is negative is because of Lenz's law, which states that the EMF (and hence the current) induced is such that it will oppose the change in magnetic flux. Hence, we need to use the negative sign to indicate that this EMF induce is a resistive force that acts against the direction of change in magnetic flux. And if this happens to be a question in your exam, remember to refer to Faraday's law for the induction of EMF and Lenz's law for the negative value of EMF.

So yeah this afternoon when I get the time I will reply your questions on:
- HSC question for thrust force and rocket velocity
- Derivation of rocket velocity from law of conservation of momentum

Best Regards
Happy Physics Land

[brenden]

(Im being a naughty boy going on atarnotes during my chemistry lesson hehe)

Hahahahaha you cheeky boy HPL

[Happy Physics Land]

Hahahahaha you cheeky boy HPL

Sorry Brenden I cant help with this ATARNOTES addiction

[jamonwindeyer]

And btw, I'm not sure with some of the ones that aren't underlined.
Like with the rocket launch formula T - (Fweight-Fair), isn't air resistance disregarded in the whole topic?
Also, I'm assuming the formulas of thrust force and rocket velocity are outside of the syllabus? Could you explain how you derive it, and which kind of Qs you'll use it in (or can you actually use it if we technically haven't "learnt" it?)
Lastly, why is EMF produced by rotation of conductor negative value?

HPL, you are a legend in the making. However, I can't resist a derivation, plus I'm dying to use LaTex in the forums now that it is back, so I'll jump in here, if yours is different post it, I want to see how you do it! Sire123, let's start with a derivation of rocket thrust force.

We learned in Prelim that force is equal to the rate of change of momentum, that is:



In this case, F is our thrust force. Now, lets blend that with a formula for momentum p = mv, also from Year 11:



However, this thrust force is referring to the exhaust from the rocket engine. The exhaust is expelled at a constant velocity (this is just a property of rocket engines, to do with the conservation of chemical energy, but totally irrelevant). So, if velocity is constant, we know that it must be the mass that is changing in the top part of the fraction. This leaves us with the formula:



The derivation of rocket velocity is more mathematical. We know from the law of conservation of momentum that the momentum in an isolated system must be constant. That is, the momentum of the rocket and its fuel must stay the same. If the rocket is initially at rest, it obviously has zero momentum, so interestingly, the momentum of the rocket and its fuel must remain zero throughout launch . How can this be?

It is because the momentum of the fuel (i.e. - exhaust) is opposite in direction to the momentum of the rocket. One goes up, one goes down, and they 'cancel' to give a total momentum of zero. With this in mind, the proof is below (using the formula for momentum from above):





And there is the second formula! Note that I used fuel, HPL used Gas, we're both referring to the speed/mass of the exhaust from the rocket engines 

[Happy Physics Land]

HPL, you are a legend in the making. However, I can't resist a derivation, plus I'm dying to use LaTex in the forums now that it is back, so I'll jump in here, if yours is different post it, I want to see how you do it! Sire123, let's start with a derivation of rocket thrust force.

We learned in Prelim that force is equal to the rate of change of momentum, that is:



In this case, F is our thrust force. Now, lets blend that with a formula for momentum p = mv, also from Year 11:



However, this thrust force is referring to the exhaust from the rocket engine. The exhaust is expelled at a constant velocity (this is just a property of rocket engines, to do with the conservation of chemical energy, but totally irrelevant). So, if velocity is constant, we know that it must be the mass that is changing in the top part of the fraction. This leaves us with the formula:



The derivation of rocket velocity is more mathematical. We know from the law of conservation of momentum that the momentum in an isolated system must be constant. That is, the momentum of the rocket and its fuel must stay the same. If the rocket is initially at rest, it obviously has zero momentum, so interestingly, the momentum of the rocket and its fuel must remain zero throughout launch . How can this be?

It is because the momentum of the fuel (i.e. - exhaust) is opposite in direction to the momentum of the rocket. One goes up, one goes down, and they 'cancel' to give a total momentum of zero. With this in mind, the proof is below (using the formula for momentum from above):





And there is the second formula! Note that I used fuel, HPL used Gas, we're both referring to the speed/mass of the exhaust from the rocket engines 

Oh crap I totally forgot that I still had the job to derive this for you sire 123!! Im so sorry and thank you so much Jamon for remembering to derive the two formulae for sire 123!!! You are such a legend!

But yeah I dont feel like I have done my job properly so l will take the rocket propulsion proof a step further, way back to the first principles so that you can see how rocket propulsion is related to the Law of Conversation of Momentum, keeping in mind that when a rocket launches, the force with which gas acts on the rocket (F_{gr} = Force of gas on rocket) is equal to the force with which the rocket acts onto the gas (F_{rg} = Force of rocket on the gas).



So yeah anyways sorry sire for such a late reply, hope this will help you to understand how the law of conservation of momentum is involved with rocket launch, great question!

Best Regards
Happy Physics Land