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HSC Physics Question Thread
Happy Physics Land:
--- Quote from: sire123 on February 05, 2016, 01:38:20 am ---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?
--- End quote ---
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:
--- Quote from: Happy Physics Land on February 05, 2016, 09:51:24 am ---(Im being a naughty boy going on atarnotes during my chemistry lesson hehe)
--- End quote ---
Hahahahaha you cheeky boy HPL
Happy Physics Land:
--- Quote from: brenden on February 05, 2016, 10:12:45 am ---Hahahahaha you cheeky boy HPL
--- End quote ---
Sorry Brenden I cant help with this ATARNOTES addiction
jamonwindeyer:
--- Quote from: sire123 on February 05, 2016, 01:38:20 am ---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?
--- End quote ---
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:
--- Quote from: jamonwindeyer on February 05, 2016, 06:19:59 pm ---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 ;)
--- End quote ---
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
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