HSC Physics Question Thread

[Maz]

It was Jake's way of saying the answer is inside you, you just have to find it...

Metaphors are strong 

haha 
and thank you both guys

[RuiAce]

I agree on all counts except the "approach each other at the speed of light" aspect. If I see a car approaching me, and it was travelling at 100 kilometres per hour, while I'm moving towards it at 100 kilometres an hour, then I would say it was travelling 200 kilometres an hour. Relative velocity. So I don't think we can automatically say that the answer is 'c' based on the fact that the reflected light travels at this speed.

That being said, the method of analysis above doesn't sit right with me either. It would mean that each proton sees the other approaching at 1.5c, which violates the Special Theory of Relativity. So, the answer could very well be c, I am not sure, but I don't think we can conclude it based on this reasoning.

So yeah, everything else I totally agree with, just that one bit I don't think the explanation is quite correct. The correct answer eludes me for the moment, give me a bit! Unless someone else wants to weigh in 

I'll try to explain a bit of my thought

Model A uses classical physics
Model B uses relativistic physics

The relative velocity of 200km/hr seems reasonable enough at those speeds you gave, however I reckon this is actually why. At low velocities, the principles of classic physics can be regarded as true.

This occurs, because there is no need for the Lorentz Factor of √(1-v^2/c^2). Note that by plugging v=200/3.6 ms^-1 this value is negligible.

At relativistic speeds, we need to specifically consult relativistic physics in order to understand what is going on. Recall that Einstein postulated that speed was an absolute quantity. Length and time are what become relative - this is why at relativistic speeds we have length contraction and time dilation taking place.

(We do not see length contraction or time dilation so clearly at speeds that we can associate with. Yet it's there - It's just so negligible. But we will observe this happening quite clearly at the relativistic speeds. )

[FallonXay]

Question 19) in the 2012 HSC Physics exam asks: Which set of conditions would result in the most rapid heating of the base of the cooking pot in an induction cooking system?

The answers say High AC voltage frequency and Low Electrical resistant of the pot base.

I don't understand why the base of the cooking base pot would need to have low resistance, as wouldn't you want the induced eddy currents to face resistance to convert the energy into heat?

Thanks.

[Happy Physics Land]

Question 19) in the 2012 HSC Physics exam asks: Which set of conditions would result in the most rapid heating of the base of the cooking pot in an induction cooking system?

The answers say High AC voltage frequency and Low Electrical resistant of the pot base.

I don't understand why the base of the cooking base pot would need to have low resistance, as wouldn't you want the induced eddy currents to face resistance to convert the energy into heat?

Thanks.

Hey Fallon Xay:

Interesting! This is actually one of the problems that I had an argument on with my teacher. So the way I understood the question is exactly the same as what you are thinking now, i.e. "the resistance of the metallic base of the pot required to heat it up as rapidly as possible". The question in itself is quite vague and one lesson l learn from this is to never assume anything.

So lets just start analysing this question. Of course we would need a high AC voltage frequency in order to cause constant change in magnetic flux in order to induce increasing eddy currents flowing through the base of the pot. Tick, thats the right answer. Now we are up to the electrical resistance of the base of the pot. The base of the pot would have to be metallic in order for eddy current to flow through and any conductive metal would have to have relatively low electrical resistance compared to things like plastic or rubber. This is why the answer would be low electrical resistance not high.

But yes I beyond all doubts agree with what you are saying. If we just assume that they are talking about a metallic pot base, then high electrical resistance would be the correct answer, because with more resistance in the metal, more heat is produced as the eddy current experiences more impedance. I.e. copper pot would be unsuitable because it has an extremely low electrical resistance and therefore a lot of eddy current may flow through but heat may not necessarily be produced. A steel pot on the other hand, because it has carbon atoms interstitially alloyed to iron, the lattice structure is more solid and its electrical resistance is higher than copper, hence able to produce a lot of heat but is still able to conduct electricity. In this question however, I think the key idea is not to assume that the question is only talking about metal, but just generally the electrical resistance of the material used for pot base. So, generally the material of pot base would need to have low electrical resistance to allow the flow of eddy current.

Best Regards
Happy Physics Land

[jamonwindeyer]

Hey Fallon Xay:

Interesting! This is actually one of the problems that I had an argument on with my teacher. So the way I understood the question is exactly the same as what you are thinking now, i.e. "the resistance of the metallic base of the pot required to heat it up as rapidly as possible". The question in itself is quite vague and one lesson l learn from this is to never assume anything.

So lets just start analysing this question. Of course we would need a high AC voltage frequency in order to cause constant change in magnetic flux in order to induce increasing eddy currents flowing through the base of the pot. Tick, thats the right answer. Now we are up to the electrical resistance of the base of the pot. The base of the pot would have to be metallic in order for eddy current to flow through and any conductive metal would have to have relatively low electrical resistance compared to things like plastic or rubber. This is why the answer would be low electrical resistance not high.

But yes I beyond all doubts agree with what you are saying. If we just assume that they are talking about a metallic pot base, then high electrical resistance would be the correct answer, because with more resistance in the metal, more heat is produced as the eddy current experiences more impedance. I.e. copper pot would be unsuitable because it has an extremely low electrical resistance and therefore a lot of eddy current may flow through but heat may not necessarily be produced. A steel pot on the other hand, because it has carbon atoms interstitially alloyed to iron, the lattice structure is more solid and its electrical resistance is higher than copper, hence able to produce a lot of heat but is still able to conduct electricity. In this question however, I think the key idea is not to assume that the question is only talking about metal, but just generally the electrical resistance of the material used for pot base. So, generally the material of pot base would need to have low electrical resistance to allow the flow of eddy current.

Best Regards
Happy Physics Land

Spot on, a little bit interpretive there. Here is another thought.

The effect we are discussing here is called Joule Heating (or Resistive Heating), and essentially, the formula for the heating of an object is:



This heating is, as suggested by the formula, caused primarily by the passage of an electrical current through resistance. A phenomenon known as hysteresis also plays a role, but most engineers argue that it isn't as important as Joule heating.

So, this formula suggests that, given a constant current, that increasing the resistance will increase the heating effect, and this is true. However, it is important to remember that our resistance impacts the size of the induced eddy currents.

Let me explain. The size of the induced EMF/voltage in the pot base is proportional to the rate of change of magnetic flux. This is Faraday's Law. Let's assume we have a constant voltage. We know that:



So, given our voltage is constant, if we want a higher current, we need a smaller resistance. Now, going back to the heating formula above, doubling the current will cause the heating effect to be quadrupled (since the current is squared), whereas doubling the resistance only results in the heating effect to be doubled.

Based on this interpretation, it is definitely more effective to have a lower resistance, and thus, larger eddy currents in the pot base, to achieve the maximum heating effect. This becomes more obvious if we use ohms law to re-write the heat dissipation law:



So, in fact, having a lower resistance means more current will flow, and thus, more heat will be generated than it would be by having a high resistance. This is why a low valued resistor, plugged into a huge voltage, blows up (for lack of a better term).

This is a very simplified model and by no means totally accurate, but it is an interesting way to look at the problem.

Now, this does not get rid of the idea that high resistance = more heat. It does. But we need a sizeable voltage drop across the said resistance for it to matter. This is quite a complex topic, definitely not worth investigating in depth. In actuality, there are many things at play here.

Just a different way of looking at the question. At its core, I am saying the same as Happy Physics Land. If the resistance goes too high, no current flows, and so heating does not occur. This is just a more detailed/mathematical explanation of that intuition 

[RuiAce]

Spot on, a little bit interpretive there. Here is another thought.

The effect we are discussing here is called Joule Heating (or Resistive Heating), and essentially, the formula for the heating of an object is:

Keep in mind that

is actually taught in the HSC when mentioning resistive heating. This can be easily combined with the definition of power:


Where W = work done, measured in joules

[jamonwindeyer]

Keep in mind that

is actually taught in the HSC when mentioning resistive heating. This can be easily combined with the definition of power:


Where W = work done, measured in joules

Good call, this is a much better way to explain it in terms of the HSC syllabus.

[Neutron]

Hey guys! Neutron here, I've got a few basic physics problems that I got kinda stuck with D: I was wondering whether you guys could help!

1. A student was standing on bathroom scales placed on a plank of wood and was recording the reading of weight from the scales. The plank was slowly raised at one end, so that the angle of the inline to the horizontal increased. What would the student observed about the weight reading?

2. The Earth's radius is 638 x 10^6 m and mass 5.98 x 10^24 kg. A 670 kg satellite is placed into circular orbit where teh acceleration due to gravity is 0.225 ms^-2. What is the altitude of this orbit?

For question 1, I was wondering how you can explain it through physics principles? Like I know that the weight would decrease but how would you describe it? Also, for question 2, I kinda forgot which equation to use. And also, how do you derive the equations for orbital radius and centripedal acceleration? Thank you so much!

[jamonwindeyer]

Hey guys! Neutron here, I've got a few basic physics problems that I got kinda stuck with D: I was wondering whether you guys could help!

1. A student was standing on bathroom scales placed on a plank of wood and was recording the reading of weight from the scales. The plank was slowly raised at one end, so that the angle of the inline to the horizontal increased. What would the student observed about the weight reading?

2. The Earth's radius is 638 x 10^6 m and mass 5.98 x 10^24 kg. A 670 kg satellite is placed into circular orbit where teh acceleration due to gravity is 0.225 ms^-2. What is the altitude of this orbit?

For question 1, I was wondering how you can explain it through physics principles? Like I know that the weight would decrease but how would you describe it? Also, for question 2, I kinda forgot which equation to use. And also, how do you derive the equations for orbital radius and centripedal acceleration? Thank you so much!

Hey Neutron! I'll tackle Question 2 first, then I'll work on a diagram for Question 1. For Question 2, we have a satellite placed into orbit. This means that the force acting on the satellite is caused by gravity. We equate Newton's Formula for Gravitational Force to Nerwtons 2nd Law to get the answer. Note that we use r+h, where r is the earth's radius and h is the altitude. Centripetal acceleration is taken from the centre of earth.





To derive formulae for centripetal acceleration,orbital radius, etc etc, you will be looking at combinations of Newton's 2nd Law, Newton's Law of Universal Gravitation, and the Formula for Centripetal Acceleration. Totally depends on what you want the value with respect to. Did you have any specific derivations in mind?

[jamonwindeyer]

Hey guys! Neutron here, I've got a few basic physics problems that I got kinda stuck with D: I was wondering whether you guys could help!

1. A student was standing on bathroom scales placed on a plank of wood and was recording the reading of weight from the scales. The plank was slowly raised at one end, so that the angle of the inline to the horizontal increased. What would the student observed about the weight reading?

2. The Earth's radius is 638 x 10^6 m and mass 5.98 x 10^24 kg. A 670 kg satellite is placed into circular orbit where teh acceleration due to gravity is 0.225 ms^-2. What is the altitude of this orbit?

For question 1, I was wondering how you can explain it through physics principles? Like I know that the weight would decrease but how would you describe it? Also, for question 2, I kinda forgot which equation to use. And also, how do you derive the equations for orbital radius and centripedal acceleration? Thank you so much!

Okay, so I put together a diagram in Word, and then it decided to crash. Lesson: Save often when running a Microsoft Program on an Apple Machine.  

I'll have another crack at a diagram tomorrow, but a quick explanation in the mean time. What is happening is best explained using vectors and forces.

We assume that the scale measures the weight force as directed downwards onto its surface. That is, it measures the magnitude of the equivalent force pushing directly down on it. When the plank and scale are flat, all the weight force is directed downwards and so 100% of the weight force is registered. As the plank is raised, the scale becomes inclined. However, the students weight force continues to be directed downwards. If we break this into two vector-components, one pointing perpendicular to the scales surface and one pointing parallel to it, we can see that the component of the weight force directed in the correct direction (such that the scale registers the weight) is lessened. In English, the weight force is redistributed so that less and less acts directly downwards on the scale. The rest is (and this is irrelevant to the question), balanced by frictional forces.

This explanation makes much more sense with a diagram, I'll get on it for you tomorrow as soon as I can! 

[Happy Physics Land]

Hey guys! Neutron here, I've got a few basic physics problems that I got kinda stuck with D: I was wondering whether you guys could help!

1. A student was standing on bathroom scales placed on a plank of wood and was recording the reading of weight from the scales. The plank was slowly raised at one end, so that the angle of the inline to the horizontal increased. What would the student observed about the weight reading?

2. The Earth's radius is 638 x 10^6 m and mass 5.98 x 10^24 kg. A 670 kg satellite is placed into circular orbit where teh acceleration due to gravity is 0.225 ms^-2. What is the altitude of this orbit?

For question 1, I was wondering how you can explain it through physics principles? Like I know that the weight would decrease but how would you describe it? Also, for question 2, I kinda forgot which equation to use. And also, how do you derive the equations for orbital radius and centripedal acceleration? Thank you so much!

Hey Neutron:

Wow there are a whole heap of questions right there haha! But thats ok, I thoroughly enjoy them all. I will start off with the easy works of derivation first then I will move on to answering your questions yeah?

For centripetal acceleration:

F = ma (Newton's 2nd law)
Hence F(c) = ma(c) (where F(c) means centripetal force and a(c) is centripetal acceleration)
But using the centripetal force formula we also know that F(c) = mv^2/r
Hence mv^2/r = ma(c)
Cancelling m from both sides, we get a(c) = v^2/r
So the formula for centripetal acceleration is v^2/r

For Orbital Velocity:

*Theres a lot of ways to work out orbital radius, and theres no set formula for it, the most common equation we use for working out orbital radius would be using Kepler's third law, i.e. r^3/T^2 = GM/4pi^2. So Im just assuming that you are asking for a derivation of orbital velocity instead of orbital radius here?

When an object is in uniform circular motion (orbit) around a planet, centripetal force = gravitational force of attraction
Hence applying the formula for both, we get:
mv^2/r = GMm/r^2
Cancelling m from the numerator of both sides, we get:
v^2/r = GM/r^2
Cancelling r from the denominator of both sides, we get:
v^2 = GM/r
Hence v(orb) = squareroot(GM/r)

Right, I hope you can understand my derivation (I havent really Learnt LaTex yet, sorry for not being able to present this in normal words and format)

Answer to question 1:

So when the plank of wood is elevated (i.e. angle of inclination to the horizontal ground rises), the weight force you exert onto the scale is actually broken into components. I have included a diagram below to aid my explanation. Essentially, when we just weigh ourselves normally, the scale shows our weight as a measure of our normal force, which is the reaction force to your weight force. However when you are elevated to an angle, (let the weight force you exert be W), your normal force no longer equals to your weight force, but are broken into the vertical component of Wsin(theta) and horizontal force of Wcos(theta). So now, the normal force that is measured by the scale becomes the vertical component Wsin(theta), and since sine of any acute angle would be less than 1, hence Wsin(theta) is less compared to W, and hence the reading on the scale also decreases.



Answer to question 2:

So here I would be using the formula for gravitational acceleration: g = GM/r^2
So what we WANT TO figure out here is r, because once we find r, we get to know what the altitude is by doing r - 638 x 10^6
We know what , we know what M is (5.98 x 10^24), and we also know what g is (0.225ms^-2).
Now, you might ask, whats the purpose of the question giving us the mass of satellite then? Well there is absolutely no purpose, its just a red herring. The question wants to trick you into thinking that you need to use m somewhere, so some students might end up applying the formula GMm/r^2. But thats not the formula for gravitational acceleration! The formula for gravitational acceleration would be GM/r^2 and is independent from the mass of the orbiting object!
Now, substitute in every value:
0.225 = 5.98x10^24 x 6.67x10^-11 / r^2
Hence r^2 = 1.77 x 10^15
Hence r = 4.21 x 10^7
So now, we do r - 6.38x10^6, we get 3.572x10^7 m altitude

I might have made a mistake in a place or two, just double check with your answer. And double check also that the Earth's radius is 6.38 x 10^6m not 638 x 10^6 m?

But overall great questions! Thanks for posting and if you have any further questions, dont hesitate to ask!

Best Regards
Happy Physics Land

[Happy Physics Land]

Okay, so I put together a diagram in Word, and then it decided to crash. Lesson: Save often when running a Microsoft Program on an Apple Machine.  

I'll have another crack at a diagram tomorrow, but a quick explanation in the mean time. What is happening is best explained using vectors and forces.

We assume that the scale measures the weight force as directed downwards onto its surface. That is, it measures the magnitude of the equivalent force pushing directly down on it. When the plank and scale are flat, all the weight force is directed downwards and so 100% of the weight force is registered. As the plank is raised, the scale becomes inclined. However, the students weight force continues to be directed downwards. If we break this into two vector-components, one pointing perpendicular to the scales surface and one pointing parallel to it, we can see that the component of the weight force directed in the correct direction (such that the scale registers the weight) is lessened. In English, the weight force is redistributed so that less and less acts directly downwards on the scale. The rest is (and this is irrelevant to the question), balanced by frictional forces.

This explanation makes much more sense with a diagram, I'll get on it for you tomorrow as soon as I can! 

Jamon fam I got your back covered!

[Neutron]

Thank you so much to you both for the comprehensive responses! Although HPL, where did you get the equation you used for Q2 from? I don't seem to be able to find it on the data sheet D: And also, I was revising some past papers and I forgot the relationship between the voltage in DC motors and the resulting speed current. The question I"m having trouble with is:

A 12V DC motor has the input voltage increased from 0V to 12V. What happens to the speed and current?
a) The rotor speed is constant but the current in the motor increases
b) The rotor speed increases and the current in the motor increases
c) The rotor speed is constant but the current in the motor decreases
d) The rotor speed increases and the current in the motor decreases

The answer is b and I was wondering why? Thank you!

(Sorry for the amount of questions I have D:)

Neutron

[jamonwindeyer]

Jamon fam I got your back covered!

Thanks HPL, glad to know I've got backup remind me I need to teach you LaTex! It is super easy once you know what is going on 

[jamonwindeyer]

Thank you so much to you both for the comprehensive responses! Although HPL, where did you get the equation you used for Q2 from? I don't seem to be able to find it on the data sheet D: And also, I was revising some past papers and I forgot the relationship between the voltage in DC motors and the resulting speed current. The question I"m having trouble with is:

A 12V DC motor has the input voltage increased from 0V to 12V. What happens to the speed and current?
a) The rotor speed is constant but the current in the motor increases
b) The rotor speed increases and the current in the motor increases
c) The rotor speed is constant but the current in the motor decreases
d) The rotor speed increases and the current in the motor decreases

The answer is b and I was wondering why? Thank you!

(Sorry for the amount of questions I have D:)

Neutron

The formula he used was from equating Newton's 2nd Law with Newton's Law for Universal Gravitation, the exact same as my response. You end up with:



As for your question, consider the motor as just a general resistor in a circuit. As we increase the voltage, the current will also increase by Ohm's Law (Back EMF will play a part, but the current will still increase). So the current increases for sure.

Now, the rotational force, or torque, is what causes a rotor to spin. The larger this force, the greater the rotational acceleration, and the rotor will end up spinning faster.

The formula for torque is:



So, since current is increasing, torque is increasing. Therefore, the motor speed will also increase. Hence the Answer of B 

Was that overview good or did you need some help with Ohm's Law and the ideas behind it? Just let me know