HSC Physics Question Thread

[jakesilove]

Thanks so much for your reply Jake, greatly appreciated. Great minds must think alike because i did use my ipod to film the drops then counted how many frames the falling body was in the air for and worked out the time in seconds. I'm able to work out average velocity since i know what vertical height I dropped the metal sheets from and this gives quite a nice graph showing a trend that indicates that as the surface area was decreased the average velocity of the falling body increased! I suppose one limitation that I could talk about in my discussion is that fact that I can only calculate average velocity as opposed to instantaneous velocity which would better show when the object approached terminal velocity etc. Maybe using a data logger would have done the trick, or a background erected behind for the fall which had 10cm increments lined on it, which I could use in conjunction with a frame by frame analysis to see when uniform motion was reached. My teacher is always going on about trying to find a straight line relationship when you have a graph so that's what i'm now attempting to do. I was able to solve a differential equation, (Sum of forces acting) = m (dv/dt) = mg-F(drag), (with surface area proportional to the drag force), but it doesn't really give a straight line relation, more logarithmic. Reckon it's worth putting in, or is that over analysing the problem? Anyway, very interesting stuff, thanks again for your help!

Glad that you had such success with your experiment!

I'd definitely say that you can NEVER over analyse a Physics question. As long as you're doing the correct maths (which it seems like you are), any further calculations are beneficial. If you want to knock the socks off your teacher, I would definitely try to cater to what he wants! Straight line relationships are standard in higher levels of Physics, and so if you can work something out you absolutely should. That being said, if you don't get a straight line (but seem to have some other nice function) I would still put that in, with maybe an attempted explanation as to why the graph looks as it does. Just make sure to explain each step, label everything, and increase a "discussion" of each point (ie. this graph proves that.....).

Loving your enthusiasm! Keep it up.

Jake

[mijomo]

Sorry one last question. My iPod shoots video in 30 fps. That means each frame is 1/30th of a second (0.03s roughly). Does that mean when converting back into time in seconds I can have times with 3 sig figs? Something like 1.11s or only 2 sig figs (1.1s), or only 1 sig fig (1s). I feel like 3 is the way to go but i'm not really sure on the 'rules' of sig figs haha. Thanks again!

[jakesilove]

Sorry one last question. My iPod shoots video in 30 fps. That means each frame is 1/30th of a second (0.03s roughly). Does that mean when converting back into time in seconds I can have times with 3 sig figs? Something like 1.11s or only 2 sig figs (1.1s), or only 1 sig fig (1s). I feel like 3 is the way to go but i'm not really sure on the 'rules' of sig figs haha. Thanks again!

Don't be sorry! Honestly these questions are all fantastic ones, and shows that you have a very in depth understanding of the concepts.

I would stick with two sig figs. Remember that you always select the LOWEST number of sig figs you used in the calculation. In this case, 30 fps could be 1 or 2 sig figs, but I think it's fair to assume that it's 2. So I would write something like "assuming 30 fps is correct to 2 sig figs, t = 1.1s".

Jake

[Happy Physics Land]

[jamonwindeyer]

And also, another question (sorry!), what are the benefits of the glass disc insulators' shape (still talking about the transmission tower oops) and without it, would the current just travel through the tower and be earthed? Like if the live wires were not separated from their supporting pylons, would all the electricity just be earthed? Thank you!

Neutron

Hey Neutron! Before I begin, once again I'll give props to HPL for an awesome answer to both your questions. His first answer was based on a definition of magnetic flux and, essentially, that the rate of change of magnetic flux cannot be infinite (instantaneous change is not possible). His second answer is also an awesome explanation of exactly why insulators are disc shaped. However, I do want to add my two cents here.

In terms of transformers, flux leakage refers to the fact that not all of the magnetic flux from the primary winding will pass through the secondary winding, inducing a new EMF. It "leaks" through the surrounding oil or winding insulation. How do Eddy Currents contribute to Flux Leakage (not energy loss, the more general notion focused on in the HSC syllabus).

Eddy Currents are produced by a changing magnetic flux in the iron core of a transformer. We know that Lenz's Law/Faraday's Law (Lenz's Law is just a common way of explaining/understanding the Conservation of Energy in Faraday's Law) says that any induced EMF (and thus, currents) will be induced in such a direction that it opposes that process which created it. But how do Eddy Currents oppose the changing flux from the primary winding?

Eddy Currents have their own magnetic field. The spinning currents induce new magnetic fields (magnetic flux) which act in the opposite direction to the flux from the primary winding. We know the new magnetic fields act in the opposite direction as an extension of Lenz's Law from above; opposing current, opposing magnetic field. So, to put it very colloquially, the flux from the Eddy Currents cancels some of the flux from the primary winding out!

Now, for your second question, HPL's answer is more than sufficient (and actually contains stuff I didn't think of, nice!). To answer your extra little question at the end, if the live power lines were not insulated, it is very likely they would be earthed. At voltages of 25000+, you need lots of insulation to prevent sparking, and since the power poles are earthed, I would say that yes, that is a very logical conclusion. 

[dhungelsajal123]

Hey Jake! I attended the ATAR notes lectures, do you know where I can access the powerpoints?

[Syndicate]

Hey Jake! I attended the ATAR notes lectures, do you know where I can access the powerpoints?

Hey,

You will find them over here

[sire123]

Can anyone explain the slingshot effect? And how in depth we would needa know for a possible hsc q on it, cheers

[jakesilove]

Can anyone explain the slingshot effect? And how in depth we would needa know for a possible hsc q on it, cheers

Hey Sire!

You really don't need to know the slingshot effect to any great depth. You know that the slingshot effect is basically when a space shuttle uses a planet's gravitational field to "turn"; let's examine the benefits.

The first benefit is that usually, when accelerating (ie. turning), the shuttle would have to use thrusters to create a turning force. This would waste fuel, and quite a lot of fuel. Using a planet's gravitational field gives a "free" turning force: the object will be attracted towards the planet, and as it moves forward will "turn" towards to planet. Thrusters and still used to get into and out of the planet's gravitational field, but to a much lower extent.

Another benefit is that the spacecraft will not LOSE speed, and it can even GAIN speed. But how that that make sense? Conservation of momentum. Basically, the spacecraft will "steal" some of the planet's momentum to increase its own. Do you need to understand that in any more depth? Definitely not.

You would never get an in depth question about the slingshot effect. If you have a general understanding of the above answer, you will seriously be totally fine. It's cool stuff!

Go watch the Martian if you want some more details, and a great understanding from a great movie.

Hope this helps!

Jake

[Happy Physics Land]

Can anyone explain the slingshot effect? And how in depth we would needa know for a possible hsc q on it, cheers

Hello Sire 123,

Seeing that Jake has provided so many valid points and the explanation is so great, I cannot help but to give in my 2 cents in terms of the conversation of momentum. As the space probe passes tangentially to the large planet (usually Jupiter or Saturn), it gains a significant increase in velocity whilst the large planet loses a tiny amount of rotation velocity. This is because the planet is thousands of times larger than the space probe, and since momentum of the system has to be conserved, the momentum of space probe increases and the momentum of the planet descreases in its momentum. Since the mass of the space probe and planet remains constant, then velocity must change for both objects to allow the momentum of each object to increase/decrease.

Let's have a look at this example: suppose the space probe is 10kg and the planet is 1x10^8 kg, the space probe is travelling at 100m/s and the planet is rotating with a velocity of 100m/s as well.

momentum of planet = 10^8 x 100 = 10^10 Ns
momentum of space probe = 10 x 100 = 1000 Ns

when the space probe passes tangentially near the planet, it "steals" 200 m/s off the planet

momentum of space probe = 10 x 300 = 3000 Ns
momentum of planet = 10^10 - (3000-1000) = 9.99998 x 10^9 Ns (since the momentum of the system has to be conserved, then it follows that if the space probe gains 2000 Ns momentum, then planet must lose 2000 Ns momentum, which is not a significant amount for the planet)
Velocity of planet = 9.999998 x 10^9 / 10^8 = 99.98 m/s, which is virtually the same as the 100m/s prior to slingshot effect (i.e. a trivial decrease in the velocity of planet)

Through this example (definitely not a real circumstance, but just to let you see whats going on), we can see that the planet only needs to lose a little bit of velocity for the space probe to gain a significant change in velocity, and this is because of the law of conservation of momentum.

I hope that was a good explanation on how the law of conservation of momentum relates to the slingshot effect. If you have any questions, please dont hesitate to ask!

Best Regards
Happy Physics Land

[Happy Physics Land]

This is a post regarding a very valuable question that JellyBeanz, one of our members sent me. Im posting the solution to his question here because it is one of the most typical form of question you can ever get in your preliminary yearly exams. Extremely good question which covers essentially all the calculation you would need for the topic of momentum. If anyone ever needs to look at samples of momentum/impulse questions to prepare for exams, this would be a rather good one to have a look at. So yeah honourary credits to Jelly Beanz, really appreciate it.

The question is as follows,
A car and driver of total mass 850 kg are travelling east along a straight road at a constant speed of 75.0 km/h. The car collides with a rubbish bin of mass 120 kg, which had been left on the road. The bin becomes wedged under the car within 0.350 s. The driver removed his foot from the accelerator and did not apply the brakes during collision.

1) Calculate initial momentum of the car
2) what is the final momentum of car-bin system
3) determine the speed of the car immediately after collision.
4)What is the change in momentum of the rubbish bin during the collision?
5) determine the magnitude of the average force exerted on the bin by the car.
6) What is the direction of the average force exerted by the bin on the car?

Here is the solution:


Thank you Jelly Beanz, very valuable stuff for our community, the question is certainly going to do all preliminary students a big favour.

N.B. THANK YOU to adequace for pointing out my mistake in question 4. In question four I subtracted initial momentum from final momentum, however I divided by time which is an unnecessary step in question 4. All you would have to do is to subtract initial momentum from final momentum for question 4, and in question five, you use the value from question 4 and divide that by 0.35 seconds because the formula for impulse (or change in momentum) is I = F x t. Much apologies for this mistake and any inconveniences caused

[Neutron]

Hey guys! I need help with this question "A square has a magnetic flux of intensity B through it. What is the new intensity of the area of the square is doubled?" The answer is 0.25 B and I have no idea why (I thought it was 0.5 ) And could you please also clarify magnetic flux and magnetic flux density, thank you 😅

[jakesilove]

Hey guys! I need help with this question "A square has a magnetic flux of intensity B through it. What is the new intensity of the area of the square is doubled?" The answer is 0.25 B and I have no idea why (I thought it was 0.5 ) And could you please also clarify magnetic flux and magnetic flux density, thank you 😅

Hey Neutron!

Firstly, to define Magnetic Flux and Magnetic Flux Density:

Magnetic Flux:
The total flux measured across a surface. This value is a Scalar.
Density per volume, essentially.

Magnetic Flux Density:
The flux measured at a specific point. This value is a Vector.
Density per area, essentially.
Same as a Magnetic Field at a point.

As for your actual question: unless I am reading the formulas wrong, I think that 0.5B is the correct answer (and maybe your textbook has got it wrong?).

Check with your teacher, but if you have x magnetic field lines across a square, causing a flux density of B, then doubling the area of the square (maintaining x magnetic field lines) should result in a flux density half that of the original.

Unless the question states that the side length doubles (in which case the area is multiplied by 4), I think your textbook may have just got this wrong.

Jake

[Maz]

hello
Can someone please help me with this?
Alpha Centauri is a star 4.367light year from earth. A time traveller taking this journey at near light speed in a spaceship was confused by the fact that he made the journey in less than 4 years based on his clock. how would you explain this situation to the traveller? How would this dilemma be viewed by an observer on Earth.
As an answer i put: To the observer on Earth, time dilation has affected their view of time taken by the the ship to be more than time taken relative to the traveler. Time dilation is more noticeable as speed increases.
i was wandering if that would be enough as an answer- or if i was even right and how you would calculate that?
Thankyou so much in advance 

[jakesilove]

hello
Can someone please help me with this?
Alpha Centauri is a star 4.367light year from earth. A time traveller taking this journey at near light speed in a spaceship was confused by the fact that he made the journey in less than 4 years based on his clock. how would you explain this situation to the traveller? How would this dilemma be viewed by an observer on Earth.
As an answer i put: To the observer on Earth, time dilation has affected their view of time taken by the the ship to be more than time taken relative to the traveler. Time dilation is more noticeable as speed increases.
i was wandering if that would be enough as an answer- or if i was even right and how you would calculate that?
Thankyou so much in advance 

Hey!

I think that you would definitely need to be a little bit clearer with your answer, just to make sure you fully get across to the marker that you know what you're talking about.

How would you explain this situation to the traveler?

Firstly, I would start by explaining that his situation is impossible. If something is 4.367 light years from earth, then it takes light 4.367 years to get from Alpha Centauri to Earth. Therefore, it is impossible to get there in less than 4 years as this would be faster than the speed of light.

Maybe his clock reads more than 4.367 years (let's just say 5 years). In that case, the fact that he was traveling near the speed of light would explain why he got there so quickly. Length contraction occurs at high speeds, meaning that, to the traveler, the distance between himself and Alpha Centauri has decreased. This allows him to get there quicker. Remember that to the traveler, time operates exactly as normal. To the traveler, it is the outside world that appears to be moving more slowly. So time dilation should not actually make a difference to a traveler viewing a clock on the space ship: to him, the clock moves as normal.

Observer on Earth

To an observer on earth, time dilation would mean that the time experienced by a traveler appears to be slower. To the person on Earth, the clock on the spaceship will move more slowly. However, time will move forward as normal on Earth, and so it will appear that the traveler takes longer than 5 years to get there. The observer on earth will also notice the mass of the spaceship increase (due to mass dilation) and the length of the spaceship decrease (based on length contraction).

I hope this helps! Take a look at the formulas in order to know exactly what's going on when something moves very quickly. Remember that a person moving near to the speed of light will still experience everything (time, length etc.) to be the same; it is only the outside world that appears to be different.

Jake