VCE Physics Question Thread!

[zsteve]

link doesnt seem to work for me, keen to see this

Hi, you need to copy the entire link (only part of it is hyperlinked for some reason)

[odeaa]

Hi, you need to copy the entire link (only part of it is hyperlinked for some reason)

cheers, that makes sense

surely if you wrote that in an exam you wouldnt be penalised? shows greater knowledge than just guessing that max is in the middle

[GeniDoi]

Intuitive explanation:

There was a really good PBS space time challenge episode about which particle would win the race to travel to the opposite side of the Earth (assuming it is uniform density):

One that is thrown into orbit versus one that is "dropped" down a straight hole that goes through the center of the Earth and to the other side.

Link for those who are interested (if you like physics you 100% will be): https://www.youtube.com/watch?v=MUThGpp6ze4

The solution turned out to be that they tie and meet on the opposite end of the Earth at the exact same time, and the way you solve the problem is you assume the particle that goes through the middle of the Earth behaves kinda of like a spring, so it obey's hooks law, so if you never touched the particle again and there were no frictional forces it would oscillate between the two ends of the Earth forever.

My point is, if you treat the particle as a spring (a completely valid assumption), it becomes very obvious that the maximum speed will be at the center of the oscillation (at the exact centre of the Earth) because the only type of energy it has at the center of the Earth is kinetic energy. It has no gravitational potential energy whatsoever because at the center of the Earth your height = 0. Moreover, due to the symmetry of the problem (conservation of energy + a circular earth) the particle will have no speed at the surface of the earth, so you can (somewhat naively, but still validly) say that the only logical place that it can have a maximum speed is at the midpoint, the centre of the Earth.

If this has confused anyone I'm sorry. Just trying to put in different terms.

[Floatzel98]

In the solutions for the orbiting particle how exactly do they express the period in terms of density (Equation (6))? Do they just take the volume for a sphere and substitute v = m / p to get the equation they did? Why exactly do they only want it in terms of density for?

[GeniDoi]

In the solutions for the orbiting particle how exactly do they express the period in terms of density (Equation (6))? Do they just take the volume for a sphere and substitute v = m / p to get the equation they did? Why exactly do they only want it in terms of density for?

Here are the solutions. The Newtonian one is completely accessible to a physics 3/4 understanding.

http://bit.ly/particle_challenge

[Floatzel98]

Here are the solutions. The Newtonian one is completely accessible to a physics 3/4 understanding.

http://bit.ly/particle_challenge

Yeah, that's what I was referring to. They make sense, I just don't understand why they need to express it in terms of the density? Why nothing else? Is it just because the planet is supposed to be 'uniformly dense'?

[GeniDoi]

Yeah, that's what I was referring to. They make sense, I just don't understand why they need to express it in terms of the density? Why nothing else? Is it just because the planet is supposed to be 'uniformly dense'?

Because a varying density planet changes the problem, making it significantly more complex to solve algebraically (if not downright impossible, I'm not even sure if there is an algebraic mathematical representation of the density of the earth vs depth).

As a result of that, the assumption that the falling particle only experiences a force towards the inner sphere becomes invalid and you ruin the symmetry of the problem. The point is that by making these assumptions elegant algebraic expressions can be found. There are tonnes of things which would prevent this from working in real life, from the Coriolis effect to the atmosphere to the gravitational perturbations by the sun and planets, and of course the down right impossibility of drilling a hole through the center of the Earth. If you take real life factors into account it really doesn't work anymore. Abstraction is necessary to arrive to these nice expressions.

If all that wasn't enough, it's actually wrong since it uses Newtonian mechanics but I have no idea about how the Einstein solution works.

EDIT: For anyone stumbling here worrying what this discussion is about; it's not part of the course.

[Orson]

Can someone please explain the different ways of doing Q1 VCAA 2013?

There are 2 ways, using SUVAT, and using forces.

There were two methods of approaching this question. One involved using the constant acceleration
formula
x = ut + 1/2at2. Alternatively, because the there was no friction, the net force on the trolley was
the component of the weight down the plane (5 sin10). Substituting this into Newton’s second law (5
sin10 = 0.5 × a) gave an acceleration of
1.74 m s–2.

A common error was to determine the average speed (3.5/2 = 1.75), incorrectly assume this was the
final speed at the bottom of the ramp and calculate the acceleration from v = u + at.

[GeniDoi]

You can do it the dynamics way (the 'forces' one you're talking about) by saying
F(net) = ma
=> mgsin(10) =  ma
=> gsin(10) = a
=> a = 10sin(10) = 1.74m/s^2

(note that in general for inclined planes with gravity being the only force, a = gsin(theta).
The good thing about doing it this way is you don't need to know the time or distance it's moved, which VCAA might not always tell you.

[Orson]

The good thing about doing it this way is you don't need to know the time or distance it's moved, which VCAA might not always tell you.

Cheers. I did it this way in the exam:

u = 1, t = 2, s = 3.5, a = ?
s = ut + at^2
3.5 = 0 + (1/2)*a*(2)^2
3.5 = (1/2)*a*4
3.5 = 2a
a = 1.75m/s^2

Is this correct? They did it this way for part b)

Thanks mate!

[GeniDoi]

Cheers. I did it this way in the exam:

u = 1, t = 2, s = 3.5, a = ?
s = ut + at^2
3.5 = 0 + (1/2)*a*(2)^2
3.5 = (1/2)*a*4
3.5 = 2a
a = 1.75m/s^2

Is this correct? They did it this way for part b)

Thanks mate!

Completely fine and had VCAA not given us the value of the angle, your method would be the only way to do it. However if the time/distance wasn't supplied but the angle was, the net force way would be the only way to solve it.

[Orson]

Completely fine and had VCAA not given us the value of the angle, your method would be the only way to do it. However if the time/distance wasn't supplied but the angle was, the net force way would be the only way to solve it.

Okay. Cheers heaps mate for the help! I wasn't sure whether SUVAT works on slopes...

[lzxnl]

Because a varying density planet changes the problem, making it significantly more complex to solve algebraically (if not downright impossible, I'm not even sure if there is an algebraic mathematical representation of the density of the earth vs depth).

As a result of that, the assumption that the falling particle only experiences a force towards the inner sphere becomes invalid and you ruin the symmetry of the problem. The point is that by making these assumptions elegant algebraic expressions can be found. There are tonnes of things which would prevent this from working in real life, from the Coriolis effect to the atmosphere to the gravitational perturbations by the sun and planets, and of course the down right impossibility of drilling a hole through the center of the Earth. If you take real life factors into account it really doesn't work anymore. Abstraction is necessary to arrive to these nice expressions.

If all that wasn't enough, it's actually wrong since it uses Newtonian mechanics but I have no idea about how the Einstein solution works.

EDIT: For anyone stumbling here worrying what this discussion is about; it's not part of the course.

Yeah, you can actually still represent the gravitational force if you don't have a constant density. The solution would then depend on the nature of the density. If you naturally assume spherical symmetry (i.e. the density only depends on the distance to the centre of the Earth and that rotation of the Earth doesn't change the density), then you can use Gauss's law for gravity to work out the gravitational field (and hence force) at any point within the Earth.

Essentially, Gauss's law for gravity would say that g * 4pi*r^2 = -4pi GM(enc) = -4pi GM(total) * (r/R)^3. It still does this for a spherically symmetric mass distribution; a uniform density is just a special case of this.
g = -GM(total)r/R^3
It's the same as if you had a uniform density, so GeniDoi's assumption that it behaves as a spring would be justified if and only if you dropped the spring on the surface of the Earth but that condition is met here.

Then you'd just use R^3/T^2 = GM/4pi^2
Find that T^2 = 4pi^2R^3/GM

From the spring equation above, g = a = -GMr/R^3 = -kr/m where k = GmM/R^3
Period of a spring with spring constant k is 2pi * sqrt(m/k) = 2pi * sqrt(R^3/GM) = T
If the periods are equal, so are the half periods
The two equations are equivalent
So the half periods are the same

Essentially, this proof works for any spherically symmetric body and that's the most general case possible.

[lzxnl]

Edit: my above working makes no sense as M(enc) isn't necessarily M(total)x(r/R)^3
You'd.have to replace it with an integral of the volume over the desired volume instead and that would get annoying.

[odeaa]

Hi guys,
Attempted doing a general proof but messed that up twice (misconceptions abounded )
However, here's my working for the VCAA 2014 question again. It completely avoids the assumption that max speed occurs in middle of oscillation (and is thus the mathematically rigorous method of proof). The examiners' report simply stated that max speed = middle of oscillation. This is probably a handy fact to remember then But for those who are curious or don't like assumptions:

Spoiler

https://onedrive.live.com/redir?resid=2692D5EC8060E581!6587&authkey=!AHb8GmRybJWZUUo&ithint=file%2cpdf

Does use methods knowledge which is beyond the Physics SD, but who cares. My physics teacher said it was acceptable.

yeah if anyone was wondering, I asked my teacher who used to be chief assessor and actually helped write a few of the older study designs, and he said that if you use correct physics they will never penalise you. This includes using integration for the area under graphs- although I think this is only really useful if you know the equation of the graph (eg. kinetic energy, gravity equations etc)