HSC Physics Question Thread

[Fizzycyst]

Yep! But careful justwannawish, that \(\theta\) angle in the image isn't what we need. It's the other part of that 90 degree angle, so the angle formed between the coil and the magnetic field, as I've marked in green down here

(Love your work, Fizzycyst )

(Image removed from quote.)

^^ this

TY for correcting!! 

[not a mystery mark]

I have another question... please send help.

Two aeroplanes flying on a collision course at 10.0km apart.
Aeroplane A is flying at 500km/hr on a heading of North 30° East.
Aeroplane B is flying at 600km/hr on a heading of West 30° North.
If neither deviated from its original course - Calculate the time it would take before they collided.

I literally have no idea how to approach this. Thank you for checking it out. Blesu desu. <3

[jamonwindeyer]

I have another question... please send help.

Two aeroplanes flying on a collision course at 10.0km apart.
Aeroplane A is flying at 500km/hr on a heading of North 30° East.
Aeroplane B is flying at 600km/hr on a heading of West 30° North.
If neither deviated from its original course - Calculate the time it would take before they collided.

I literally have no idea how to approach this. Thank you for checking it out. Blesu desu. <3

Hey hey! I assume this is a Prelim question? If you are a current Year 12 student, you won't need to tackle something like this

That said, draw yourself a triangle showing the situation. The bottom side should be horizontal, 10km long. Then draw two lines coming up from that at the angles described in the question. So, Plane A's line should come up from the left at 30 degrees from the vertical, then Plane B's from the right at 30 degrees from the horizontal.

The two sides you have drawn both have a length that depends on the time that has passed. Plane A's will be \(500t\) and Plane B's will be \(600t\), where \(t\) is in hours!

So we have a triangle with three sides: 10, 500t and 600t. We also know the angle on the right, 30 degrees. We can tie all that together with the cosine rule, which hopefully you've seen before?



This gives you a quadratic equation that you can solve to find \(t\)! Does this help?

[itssona]

hi any help with this relativity simultaneity question please

A train conductor sets the clocks on all City-rail stations to the same time. He synchronised them. A passenger is on a train somewhere on its journey from Central to Redfern. According to this passenger, when the clock strikes noon at Central, what time is it at Redfern? Noon, before noon, or afternoon .
 

[8Dadeedo]

Hey, I was hoping to get some help with the following question

Hope this is the right place haha

Thank ya

[blasonduo]

hi any help with this relativity simultaneity question please

A train conductor sets the clocks on all City-rail stations to the same time. He synchronised them. A passenger is on a train somewhere on its journey from Central to Redfern. According to this passenger, when the clock strikes noon at Central, what time is it at Redfern? Noon, before noon, or afternoon .
 

Hey! I would say Afternoon!

As he is travelling to Redfern, by simultaneity, he should observe the clock in Redfern to be slightly ahead. This is because light from Redfern will get to the passenger before the light from central (as he is moving away from central and closer to Redfern) This change in time would be minuscule though!

[jamonwindeyer]

Hey, I was hoping to get some help with the following question

Hope this is the right place haha

Thank ya

Welcome to the forums! It is indeed

So what we are looking for is \(F_\text{moon}=F_\text{earth}\). Let's define \(r\) as the distance from the centre of the earth, making the distance from the centre of the moon \(d=3.85\times10^8-r\). We also know that \(M_e=81.4M_m\). Put all this together with Newton's Law of Gravitation, where \(m\) is the mass of the spacecraft:



After all the cancellations, just solve for \(r\) from there!!

[talitha_h]

Q: Near the end of the rocket journey to planet X the astronaut strands on the scales every hour and records the readings 82 kg, 85, 111, 128. Make an inference about the motion of the rocket during this time.

This is worth 1 mark, the answer is that craft is decelerating as it approaches the planet x. But I don't understand why this is so  ??


thanks in advance

[blasonduo]

Q: Near the end of the rocket journey to planet X the astronaut strands on the scales every hour and records the readings 82 kg, 85, 111, 128. Make an inference about the motion of the rocket during this time.

This is worth 1 mark, the answer is that craft is decelerating as it approaches the planet x. But I don't understand why this is so  ??


thanks in advance

Hello! ^~^

I think it is really useful to put yourself into the situation, like an elevator. Imagine you are currently going down to the ground floor (ie the planet in the scenario) and as you approach the floor, you'll have to slow down, and thus decelerate. When this happens, you start to feel much heavier (due to Newton's law of inertia, where our bodies want to continue with the same speed, so we apply more force to the floor and more force onto the scales)

Hope this helps!
If anything isn't clear, let me know, I'd love to explain

[jasn9776]

What is escape velocity?
Is my definition correct?:
escape velocity is the minimum initial velocity at the surface of a body required for an object to escape from the influence of the body's gravitational field and not return.

However, i am confused as there are explanations involving a tall mountain and firing horizontally.

Thanks

[RuiAce]

What is escape velocity?
Is my definition correct?:
escape velocity is the minimum initial velocity at the surface of a body required for an object to escape from the influence of the body's gravitational field and not return.

However, i am confused as there are explanations involving a tall mountain and firing horizontally.

Thanks

Whilst what you've said is true, it's missing that extra piece of information as you've stated below.

Our analysis of escape velocity is entirely dependent on firing horizontally over a tall mountain. What this means is that the projectile cannot be fired from an angle. Most of the time when throwing a ball a long distance or something, you'd fire it at an angle. If you wanted to maximise its distance, you'd fire it at 45 degrees, and if you wanted to maximise its distance, you'd fire it vertically upwards.

Here, the whole aim of firing it horizontally (i.e. 0 degrees) is to actually bring the effect of Earth's gravity into play. Normally, if you fire something horizontally (at say, chest height), it wouldn't go up, because the Earth's gravity would bring it down. However if you fired it quickly enough (as though you launched a rocket from your chest), it would probably keep going and eventually clash with some trees in front of you.

So instead we fire it from a high mountain to ensure that there's no obstacles in our way. And the idea is that the more velocity we fire it with, the more the distance it'll cover before it lands.

(That's quite important. Eventually, we'll fire it so quickly that it doesn't land, and in fact the projectile ends up orbiting around the Earth. If we keep going up, then eventually it will manage to escape. In fact, it's escaping without us firing it at an angle; what's so powerful is that it manages to escape despite the fact that we've fired it horizontally.)

[blasonduo]

What is escape velocity?
Is my definition correct?:
escape velocity is the minimum initial velocity at the surface of a body required for an object to escape from the influence of the body's gravitational field and not return.

However, I am confused as there are explanations involving a tall mountain and firing horizontally.

Thanks

EDIT just adding on because typing is slow >_<

Hello! You are absolutely right!

This throwing stuff comes from the theorization of how escape velocity came to be, kind of Newtons "Thought experiment"

His theory was a progression, such that if he were to throw an object horizontally out, it would make a parabolic shape and hit the ground.

If he threw it with a larger horizontal speed, it would still be parabolic, but it would hit the ground further away.

If he threw it fast enough so that the speed of the object and its rate of falling exactly matched the curvature of the earth, the object would be in orbit.

And this is where escape velocity comes in; At even faster speeds, he thought he could eventually escape the earth's gravitational pull entirely (11.2km/s straight up )

So this whole throwing a stone thing was Newton's way of thinking of the motions of objects due to earth's gravitational pull

Hope this helps!  ^~^

[jasn9776]

Here, the whole aim of firing it horizontally (i.e. 0 degrees) is to actually bring the effect of Earth's gravity into play. Normally, if you fire something horizontally (at say, chest height), it wouldn't go up, because the Earth's gravity would bring it down. However if you fired it quickly enough (as though you launched a rocket from your chest), it would probably keep going and eventually clash with some trees in front of you.

So instead we fire it from a high mountain to ensure that there's no obstacles in our way. And the idea is that the more velocity we fire it with, the more the distance it'll cover before it lands.

(That's quite important. Eventually, we'll fire it so quickly that it doesn't land, and in fact the projectile ends up orbiting around the Earth. If we keep going up, then eventually it will manage to escape. In fact, it's escaping without us firing it at an angle; what's so powerful is that it manages to escape despite the fact that we've fired it horizontally.)

So what your saying is that the height of the mountain doesn't matter as long as there are no obstacles in the way. But doesn't gravity decrease the further you are from the centre of the mass. So wouldn't a very tall mountain require less escape velocity than a short mountain? v=sqrt(2GM/r) so as r increases the escape velocity decreases even if minutely?
Also i don't get why firing it horizontally is the only way to 'bring the effect of Earth's gravity into play'. I would still be fully affected by gravity if i fired it vertically upwards from the surface right? So is the escape velocity the same if i fire it from the surface vertically upwards compared to horizontally

Anyways thanks for posting so fast. Its amazing what you can get free answers so quick.

[RuiAce]

So what your saying is that the height of the mountain doesn't matter as long as there are no obstacles in the way. But doesn't gravity decrease the further you are from the centre of the mass. So wouldn't a very tall mountain require less escape velocity than a short mountain? v=sqrt(2GM/r) so as r increases the escape velocity decreases even if minutely?
Also i don't get why firing it horizontally is the only way to 'bring the effect of Earth's gravity into play'. I would still be fully affected by gravity if i fired it vertically upwards from the surface right? So is the escape velocity the same if i fire it from the surface vertically upwards compared to horizontally

Anyways thanks for posting so fast. Its amazing what you can get free answers so quick.

It matters once you consider the magnitude of the velocity - I was pretty much just illustrating the concept. As per the formula, indeed theoretically speaking the higher you are on the mountain (or, tbh, just ON a higher mountain), the formula \( v = \sqrt{\frac{2GM}{r}} \) suggests that there will be a minute decrease in the velocity for some minute increase in \(r\).

(Essentially, because \(r\) actually measures the distance you are from the centre of the Earth. But again, I was really just illustrating a concept, because that's the more important bit.)

Perhaps, the word 'only' is a mistake. We should replace it with something like "fully" instead. Because a projection upwards would mean that you somewhat had a thrust in the upwards direction. Whereas when you project horizontally, the only force in the vertical direction is gravity.

[sadfd45678]

Hi guys i have a couple HSC questions where I have trouble wrapping my head around:

https://www.boardofstudies.nsw.edu.au/hsc_exams/2015/exams/2015-hsc-physics.pdf (Q19)
B was my answer because if the astronaut was in orbit, it wouldnt "fall" down, it would just go around and around the planet.

Thanks so much