HSC Physics Question Thread

[jakesilove]

HSC PHYSICS Q&A THREAD

To go straight to posts for the new syllabus, click here.

What is this thread for?
If you have general questions about the HSC Physics course or how to improve in certain areas, this is the place to ask! 👌

Who can/will answer questions?
Everyone is welcome to contribute; even if you're unsure of yourself, providing different perspectives is incredibly valuable.

Please don't be dissuaded by the fact that you haven't finished Year 12, or didn't score as highly as others, or your advice contradicts something else you've seen on this thread, or whatever; none of this disqualifies you from helping others. And if you're worried you do have some sort of misconception, put it out there and someone else can clarify and modify your understanding! 

There'll be a whole bunch of other high-scoring students with their own wealths of wisdom to share with you. So you may even get multiple answers from different people offering their insights - very cool.


To ask a question or make a post, you will first need an ATAR Notes account. You probably already have one, but if you don't, it takes about four seconds to sign up - and completely free!

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* Physics MEGA THREAD
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* Foolproof Method for Maths Questions in HSC Physics
* Physics: 8 Mistakes Every Student Makes!
* A Quick Guide to Motion and Kinematics
* A Guide to Understanding Special Relativity
* A Guide to Motors and the Motor Effect
* A Guide to Generators and Induction
* A Guide to Transformers and AC Power
* A Guide to Cathode Rays
* A Guide to The Photoelectric Effect
* 60 Hours Until Physics
* Quantum Physics in 10 Minutes
* How to Go About Studying for Any Science

Original post.

Before you can ask a question, 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!

Hey everyone!

A lot of you will have met me at the HSC Head Start lectures, where I lectured in 2U and 3U Maths, Physics and Chemistry.
My role on these forums is to help you. The HSC syllabus is tricky, nuanced and pretty damn huge. To help you out, I thought it would be a great idea to have a forum where you can just post questions, and myself or other forum members can post answers!

This is a community, so we want you to feel like you can post any type of Physics question, no matter how "basic" you might think it is. Remember, IF YOU'RE HAVING TROUBLE WITH A TOPIC, THERE ARE THOUSANDS OF OTHERS HAVING THE SAME ISSUE. The best way to learn Physics is by looking through practice questions, and their associated answers. I honestly think a forum like this, and a place where I could always go to have difficult questions answered would have helped me in my HSC year.

Remember that Physics can be a difficult course. There will be lots of answers to the same questions, and I'll try give you the best or easiest to remember ones.

I got an ATAR of 99.80, and a mark of 93 in the the Physics course. There are similar forums for a bunch of other subjects, so make sure to take a look at them as well!

[chloe9756]

A truck is travelling along a straight highway at a speed of 30 ms-1. Ahead of it is a car travelling at 28 ms-1.
The truck driver wishes to pass the car but notices an oncoming car which he estimates to be 1.0 km away.
The truck driver assumes that the oncoming car is also travelling at 30 ms-1. If the car is 3.0 m long, the truck 15
m long and two car lengths are allowed as clearance before and after passing:
(a) Assuming that the truck driver’s assumptions are correct, does the truck make it? Give reasons for your
answer.
(b) The truck driver’s distance estimate is correct but the oncoming car is travelling at a speed greater than 30
ms-1. What is the maximum speed that the oncoming car can travel and still avoid a collision?

ans:

a) yes, by 100m
b) 36.7 ms-1

[jamonwindeyer]

A truck is travelling along a straight highway at a speed of 30 ms-1. Ahead of it is a car travelling at 28 ms-1.
The truck driver wishes to pass the car but notices an oncoming car which he estimates to be 1.0 km away.
The truck driver assumes that the oncoming car is also travelling at 30 ms-1. If the car is 3.0 m long, the truck 15
m long and two car lengths are allowed as clearance before and after passing:
(a) Assuming that the truck driver’s assumptions are correct, does the truck make it? Give reasons for your
answer.
(b) The truck driver’s distance estimate is correct but the oncoming car is travelling at a speed greater than 30
ms-1. What is the maximum speed that the oncoming car can travel and still avoid a collision?

ans:

a) yes, by 100m
b) 36.7 ms-1

Hi Chloe! I'm not Jake (obviously  ) but let me field this one for you! I've written up a solution below. The quality is a little blurry, I'm not quite sure why. I'd normally write up the solution in the response itself, but I'm having some issues with my LaTex equation formatting, so this will hopefully be a good substitute!

Have a read, maybe twice, it's a little tricky to wrap your head around, and of course let me know if anything doesn't quite make sense. This is a question on relative velocities, which can be difficult, but if you are struggling, picture driving in a car on the motorway. When a car overtakes you, they are moving at say, 120 kilometres an hour. But you don't see them whiz past, you see them slowly edge in front of you. Relative to you (your frame of reference, if you remember from Space topics), they are only moving 10 kilometres an hour, since you are travelling at 110km/h. If a police officer was sitting on the side of the freeway, in their frame of reference, they see the overtaker moving at 120km/h (and would likely go and fine them for speeding).

Thanks so much for the question! I guarantee a whole bunch of people will benefit from you asking, be sure to keep sending questions to Jake and myself! We are happy to assist  enjoy your weekend!

[chloe9756]

1. a sailor hoists a heavy weight up by pulling down a rope slung over a pulley:

a) is the tension in the rope greater when the mass is at rest or moving upwards with constant speed? justify your answer.
b) if the mass is travelling upwards, is the tension in the rope greatest when the crate is speeding up or when it is slowing down?

these are my answers, but i have no idea if they're correct or not.
a) the tension is greater when the mass is moving upwards. An acceleration of the crate would require more tension to prevent it from falling.
b) the tension is greater as the crate speeds up, as there is more force required to pull a mass up as it accelerates.

2. a small car (500kg) pulls on a caravan (200kg). the car's engine supplies a driving force of 18 000N. The car and caravan travel along with equal velocities and speed up at the same rate as they are connected with a rope.

while the car starts off from a red light, is the force it moves itself with greater, equal or less than the force it uses to move the caravan? give a qualitative and quantitative explanation.

thank you so much

[Happy Physics Land]

1. a sailor hoists a heavy weight up by pulling down a rope slung over a pulley:

a) is the tension in the rope greater when the mass is at rest or moving upwards with constant speed? justify your answer.
b) if the mass is travelling upwards, is the tension in the rope greatest when the crate is speeding up or when it is slowing down?

these are my answers, but i have no idea if they're correct or not.
a) the tension is greater when the mass is moving upwards. An acceleration of the crate would require more tension to prevent it from falling.
b) the tension is greater as the crate speeds up, as there is more force required to pull a mass up as it accelerates.

thank you so much

Hey Chloe:

I am a year 12 student currently undertaking physics, and in regards to question 1, I believe that I have a valid explanation to these questions however I cannot be 100% certain. I will post my solutions and explanations here and if jake happens to scroll through this chat he can confirm whether or not my reasonings are correct. Tension questions as such as perhaps the most difficult part of Moving About, and to be honest some teachers simply skip this section of the textbook because its not frequently asked in HSC exams, but more so in engineering studied exams.

a) The tension in the rope when the mass is at rest should be the same as the tension of the rope when the mass is moving upwards with constant velocity. This would seem very counter-intuitive at first because in real life when we pull up an object away from the ground, we would always feel more exhausted than simply holding the object statically in one position. However, keep in mind that we are dealing with FORCE here. In your response, chloe, you have made a very valid point that the upwards acceleration of the cradle would require more tensile force. However your answer is a little inaccurate in saying that the point of a higher tensile force is to prevent the object from falling. The correct explanation to your answer of having a greater tensile force should be to accelerated the object to our desired "constant speed" value, because as we all know, according to Newton's second law, when we have an acceleration, we have a force (F=ma).

Ok, you have truly state that a greater tensile force is required to accelerate the mass. But when the mass begins to travel at a constant velocity, as phrased in the question, it would no longer be accelerating. Therefore, the value of acceleration would be 0, and if we substitute a=0 into F=ma, then we would discover that the extra tensile force required for a constant velocity is indeed 0N.  The weight force of the mass is still equal but opposite to the tension on the rope (Newton's 3rd Law) and there are no other forces acting in the system.Hence the tensile force in the rope should be the same in both scenarios.

The reason why a lot of students would be confused in this case is that in real life, when we pull an object up at constant velocity, we feel exhausted after a while not because of the extra force we apply, but because as the object is pulled up, we are required to exert more kinetic energy to the mass as the object increases in its gravitational potential energy.

b) I feel like there is a very sneaky trap in the second question, your answer seems very logical and it does make sense. Let's first look at how you would have structured your answer better. Something you should mention is as the mass speeds up, it possesses an acceleration and hence according to Newton's 2nd law, F=ma, a tensile force would be required to speed up the mass. This reference to Newton's law would make your response stronger and there would be a higher chance for you to score full marks if you refer to an established scientific law.

But heres the problem. Lets say the crate is accelerating at 10ms^-2, then F = ma will give a net force of 10m Newtons. However, consider the case where the crate is decelerating at 10ms^-2, meaning that a= -10ms^-2, F=ma will give a net force of -10m Newtons. If we compare their magnitude, the force in both scenarios would be the same, since the negative sign is only indicative of a direction. So personally I would say that when the crate is speeding up at a larger rate than its slowing down, then your answer would be correct. If the rate of speeding up is the same as slowing down, then the tension in the rope would be the same.

I hope my answers make sense to you and if you remain dubious about question b), perhaps Jake can confirm my argument when he comes around. But these are very good mechanical questions to think about.

Best Regards
Happy Physics Land

[jamonwindeyer]

1. a sailor hoists a heavy weight up by pulling down a rope slung over a pulley:

a) is the tension in the rope greater when the mass is at rest or moving upwards with constant speed? justify your answer.
b) if the mass is travelling upwards, is the tension in the rope greatest when the crate is speeding up or when it is slowing down?

these are my answers, but i have no idea if they're correct or not.
a) the tension is greater when the mass is moving upwards. An acceleration of the crate would require more tension to prevent it from falling.
b) the tension is greater as the crate speeds up, as there is more force required to pull a mass up as it accelerates.

2. a small car (500kg) pulls on a caravan (200kg). the car's engine supplies a driving force of 18 000N. The car and caravan travel along with equal velocities and speed up at the same rate as they are connected with a rope.

while the car starts off from a red light, is the force it moves itself with greater, equal or less than the force it uses to move the caravan? give a qualitative and quantitative explanation.

thank you so much

Hey Chloe! Happy Physics Land, you did an awesome job answering the question, spot on!

The important thing to remember is that the tension in the rope is a force which corresponds directly to the force that the rope is applying to the crate. In this way, it is a reactive force.

To confirm the answer, Chloe, your reasoning that an acceleration of the crate would result in greater tension in the rope. But Happy has correctly identified that, when moving at a constant speed, there is no acceleration in the crate. Therefore, the tension is identical to if it was stationary. This is assuming the crate isn't on the ground, in which case, there would be no tension in the rope.
For the second part however, I think my interpretation is different. Consider the crate travelling upward at a constant speed, with some level of tension. To accelerate it upwards would require an additional upwards force, provided by the rope, and thus tension would increase. If the crate then accelerated downward (still moving upwards, remember), this downwards acceleration would in fact be caused by gravity, no tension involved. I would therefore say that tension is higher in the rope when accelerating upwards.

Your second question is very interesting indeed. The key is in this phrase:

The car and caravan travel along with equal velocities and speed up at the same rate   as they are connected with a rope. 

What this tells us is that acceleration is identical for the car and trailer. So, if acceleration is equal, but the masses are different, then the forces must also be different. The resultant calculation is an application of Newton's 2nd Law:

Total Force = Car Force + Trailer Force
Car Force = 500a
Trailer Force = 200a

So Total Force = 700a
         18000 = 700a
            a = 25.7...

So Car Force = 12850, Trailer Force = 5140 (there is a rounding error there), and so the car uses more force pulling itself than the trailer! 

I hope this explanation helps Chloe, and Happy Physics Land, thanks heaps for tackling that doozy of a first question! I hope it is also helping you on your HSC journey, it is certainly helping a lot of other people 

[Happy Physics Land]

Hey Jamon:

Thank you very much for your kind words, I believe that I have over-complicated 1b) after seeing your solution, but yes thank you so much for your approval and your encouragement. It means a lot to me!

Best Regards
Happy Physics Land

[chloe9756]

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

 

[brenden]

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

Well, I know nothing about Physics, but I know that people asking questions is the foundation for everything we do - so don't be sorry for that! Keep asking them! Thanks Chloe

[cajama]

I have a question from motors and gens (from the Surfing book) 

A metal bar, mass 0.04kg, resistance 4 ohms, and length 0.5m, is placed on two frictionless conducting rails in a magnetic field of 0.3T. The bar is connected to a 36 V power source.

a) Calculate the current in the bar (ans: 9A)
b) Calculate the magnitude of the force which acts on the bar (ans: 1.35 N)
c) If the bar is free to move, and the force is in the appropriate direction, calculate how far it will move along the rails in 0.25 s. (ans: 1.05m)

Thanks in advance 
P.s. I can't say for sure if the answers are correct on the book because some of the answers we get from questions can be really dodgy.

[jakesilove]

I have a question from motors and gens (from the Surfing book) 

A metal bar, mass 0.04kg, resistance 4 ohms, and length 0.5m, is placed on two frictionless conducting rails in a magnetic field of 0.3T. The bar is connected to a 36 V power source.

a) Calculate the current in the bar (ans: 9A)
b) Calculate the magnitude of the force which acts on the bar (ans: 1.35 N)
c) If the bar is free to move, and the force is in the appropriate direction, calculate how far it will move along the rails in 0.25 s. (ans: 1.05m)

Thanks in advance 
P.s. I can't say for sure if the answers are correct on the book because some of the answers we get from questions can be really dodgy.

Hey cajama!

Really really great questions, and definitely one of the most difficult of its type. It requires a good knowledge of multiple sections of the course, which is where the most difficult questions in your HSC will lie. I hope you can follow my explanation; once you've covered even one of these questions, and gone over it a few times, you'll be totally fine with anything even similar to it. My biggest piece of advice for M&G questions where you're just like WHAT THE HELL DO I EVEN DO??!??!?!! is write down the information you have, and just try sub them into a formula.

[Happy Physics Land]

I have a question from motors and gens (from the Surfing book) 

A metal bar, mass 0.04kg, resistance 4 ohms, and length 0.5m, is placed on two frictionless conducting rails in a magnetic field of 0.3T. The bar is connected to a 36 V power source.

a) Calculate the current in the bar (ans: 9A)
b) Calculate the magnitude of the force which acts on the bar (ans: 1.35 N)
c) If the bar is free to move, and the force is in the appropriate direction, calculate how far it will move along the rails in 0.25 s. (ans: 1.05m)

Thanks in advance 
P.s. I can't say for sure if the answers are correct on the book because some of the answers we get from questions can be really dodgy.

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.



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]

I have a question from motors and gens (from the Surfing book) 

A metal bar, mass 0.04kg, resistance 4 ohms, and length 0.5m, is placed on two frictionless conducting rails in a magnetic field of 0.3T. The bar is connected to a 36 V power source.

a) Calculate the current in the bar (ans: 9A)
b) Calculate the magnitude of the force which acts on the bar (ans: 1.35 N)
c) If the bar is free to move, and the force is in the appropriate direction, calculate how far it will move along the rails in 0.25 s. (ans: 1.05m)

Thanks in advance 
P.s. I can't say for sure if the answers are correct on the book because some of the answers we get from questions can be really dodgy.

I am really sorry I cant upload my 2D diagram because of some technical issues that state that my image is "unwritable" but I hope my explanation was still helpful and clear to you!

[brenden]

I am really sorry I cant upload my 2D diagram because of some technical issues that state that my image is "unwritable" but I hope my explanation was still helpful and clear to you!

That issue should be fixed tomorrow. For now, you can go to Imgur and upload the image, then once the image is uploaded, right click it, "open image in new tab", then take the URL of that and put it in between the code [img ][/img]

[Happy Physics Land]

That issue should be fixed tomorrow. For now, you can go to Imgur and upload the image, then once the image is uploaded, right click it, "open image in new tab", then take the URL of that and put it in between the code [img ][/img]

Noice Brenden!!! Thank you so much you are a lifesaver!!!!!