VCE Physics Question Thread!

[schooliskool]

so friction isnt taken into account for net force if you have the acceleration?

Yeah, if you have mass and acceleration given, the net force is calculated ignoring friction. If the driving force is asked, you use net force= driving force - friction

[odeaa]

Yeah, if you have mass and acceleration given, the net force is calculated ignoring friction. If the driving force is asked, you use net force= driving force - friction

awesome, cheers

[zsteve]

Make a table with the gpe, ke and spe at the top, middle and bottom of the oscillation

Because the ke is 0 at both top and bottom, it makes sense that it would be the highest in the centre of oscillation

Hmmm... on the exam I got stuck on that q because I didn't want to assert that KE(max) was at the centre of oscillation - I know it's "intuitive" but I wasn't sure I wanted to risk marks and make that assumption.
So I turned it into a methods question, found a quadratic for KE in terms of x and got the maximum (tick)
But that was a crazy method of solving the question ... did anyone feel like the question was a bit fuzzy? After looking at the graphs and thinking about it over and over, I felt that it was possible that the maximum could occur NOT at the centre of oscillation, because SPE was quadratic and GPE was linear, thus SPE+ GPE could minimise at some weird place not at the centre ...
Hopefully I'll get better at this.

Does anyone know if you'd be penalised for using this sort of knowledge to solve the problem? Totally valid but using methods techniques.

[GeniDoi]

Hmmm... on the exam I got stuck on that q because I didn't want to assert that KE(max) was at the centre of oscillation - I know it's "intuitive" but I wasn't sure I wanted to risk marks and make that assumption.
So I turned it into a methods question, found a quadratic for KE in terms of x and got the maximum (tick)
But that was a crazy method of solving the question ... did anyone feel like the question was a bit fuzzy? After looking at the graphs and thinking about it over and over, I felt that it was possible that the maximum could occur NOT at the centre of oscillation, because SPE was quadratic and GPE was linear, thus SPE+ GPE could minimise at some weird place not at the centre ...
Hopefully I'll get better at this.

Does anyone know if you'd be penalised for using this sort of knowledge to solve the problem? Totally valid but using methods techniques.

You would probably be penalized. It's an explain style question where there are dot points you need to hit to get marks. Your solution might be valid in the methods scope but you needed to explain it in terms of physics 3/4 language and ideas.

[lzxnl]

How do you prove that maximum speed occurs at the center of oscillation?
In VCAA 2014 Q2(d), they've stated that maximum speed occurs at the centre of oscillation in the examiners report, did we have to know this as a part of our 'prerequisite' knowledge or was there a way to work this out?

Hmmm... on the exam I got stuck on that q because I didn't want to assert that KE(max) was at the centre of oscillation - I know it's "intuitive" but I wasn't sure I wanted to risk marks and make that assumption.
So I turned it into a methods question, found a quadratic for KE in terms of x and got the maximum (tick)
But that was a crazy method of solving the question ... did anyone feel like the question was a bit fuzzy? After looking at the graphs and thinking about it over and over, I felt that it was possible that the maximum could occur NOT at the centre of oscillation, because SPE was quadratic and GPE was linear, thus SPE+ GPE could minimise at some weird place not at the centre ...
Hopefully I'll get better at this.

Does anyone know if you'd be penalised for using this sort of knowledge to solve the problem? Totally valid but using methods techniques.

Let me mathematically convince you that the centre of oscillation necessarily is the point of maximum KE.
I'll do this two ways.

First way: using forces.
The centre of oscillation is the point of zero net force, so you have F = ma = 0. This means that you have zero acceleration here, so the velocity must be stationary. Now, F = -kx - mg. As F is linear in x, it changes sign at the stationary point. That means on one side of the centre of oscillation the velocity is increasing and on the other side it's decreasing.
If you increase x, F decreases so when x gets higher, the force is negative. If you decrease x, F increases so when x gets lower, the force is positive. The net effect is that as the force opposes any displacement from equilibrium, your speed will decrease either side of the centre of oscillation. So the centre of oscillation gives the maximum KE.

Second way: using energy
You have 1/2 mv^2 + 1/2 kx^2 + mgx = constant
Let's find the centre of oscillation explicitly. From above. 0 = mg + kx -> x = -mg/k. Remember this result.
Now, let's complete the square on the above.
1/2 mv^2 + 1/2 k(x^2 + 2mg/k x) = 1/2 mv^2 + k/2 ((x+mg/k)^2 - some constant) = constant
I haven't written out the constant because the exact values of these constants aren't important.
1/2 mv^2 + 1/2 k(x+mg/k)^2 = constant
So this is an ellipse. Can you see that the speed is greatest when x = -mg/k, aka the centre of oscillation?

[odeaa]

Let me mathematically convince you that the centre of oscillation necessarily is the point of maximum KE.
I'll do this two ways.

First way: using forces.
The centre of oscillation is the point of zero net force, so you have F = ma = 0. This means that you have zero acceleration here, so the velocity must be stationary. Now, F = -kx - mg. As F is linear in x, it changes sign at the stationary point. That means on one side of the centre of oscillation the velocity is increasing and on the other side it's decreasing.
If you increase x, F decreases so when x gets higher, the force is negative. If you decrease x, F increases so when x gets lower, the force is positive. The net effect is that as the force opposes any displacement from equilibrium, your speed will decrease either side of the centre of oscillation. So the centre of oscillation gives the maximum KE.

Second way: using energy
You have 1/2 mv^2 + 1/2 kx^2 + mgx = constant
Let's find the centre of oscillation explicitly. From above. 0 = mg + kx -> x = -mg/k. Remember this result.
Now, let's complete the square on the above.
1/2 mv^2 + 1/2 k(x^2 + 2mg/k x) = 1/2 mv^2 + k/2 ((x+mg/k)^2 - some constant) = constant
I haven't written out the constant because the exact values of these constants aren't important.
1/2 mv^2 + 1/2 k(x+mg/k)^2 = constant
So this is an ellipse. Can you see that the speed is greatest when x = -mg/k, aka the centre of oscillation?

You are a king, that helped heaps

[Orson]

Does anyone remember doing any questions that asked you to find the 'slowest photoelectron'? Do these even exist?

[jyce]

Does anyone remember doing any questions that asked you to find the 'slowest photoelectron'? Do these even exist?

There are definitely the 'slowest photoelectrons' - they would have the greatest ionisation energy. However, I have never come across a question asking you to find the slowest ones; only the fastest ones, using the smallest ionisation energy (i.e. the work function).

[Orson]

There are definitely the 'slowest photoelectrons' - they would have the greatest ionisation energy. However, I have never come across a question asking you to find the slowest ones; only the fastest ones, using the smallest ionisation energy (i.e. the work function).

Ohk thanks. Do you think you can make one up? I can't find any at the moment. 

[jyce]

Ohk thanks. Do you think you can make one up? I can't find any at the moment. 

I doubt you'll find one, so don't stress about it. And even if VCAA did decide, on the off chance, to do such a question, it'd be the same kind of thing as working out the greatest kinetic energy / speed.
E_{k max.} = hf - E_{ionisation min.}
E_{k min.} = hf - E_{ionisation max.}

[odeaa]

Does anyone remember doing any questions that asked you to find the 'slowest photoelectron'? Do these even exist?

they would only have just enough energy to make it past the work function, so their energy would be the same as the work function (roughly)
you could use the threshold frequency to figure this out if they dont give the WF

[jyce]

they would only have just enough energy to make it past the work function, so their energy would be the same as the work function (roughly)
you could use the threshold frequency to figure this out if they dont give the WF

Not too sure about your statement that the energy of these slowest photoelectrons "would be the same as the work function"; isn't it the case that those photoelectrons who receive the threshold frequency would have (roughly) NO kinetic energy? I thought what was instead being asked was regarding a case where some frequency, above the threshold, was selected and you would have a range of different kinetic energies - with a minimum and a maximum. In your situation, there would not be a minimum or maximum kinetic energy; rather, there would simply be no kinetic energy?

[lzxnl]

No. The 'slowest' photoelectron would have an energy almost equal to 0. Think about it. If I attached a million people to identical elastic bungee jumping ropes and threw them all off a cliff (very idealised, I know), and the elastic ropes were designed such that half the people would break the ropes (for instance they're too heavy), then the minimum kinetic energy of a 'free' person (person who broke the rope) right after breaking the rope would be 0.

More specifically, electron energies follow some kind of distribution after being excited by the light (I'm not too sure what sort of distribution tbh as you have to take into account the light absorption and interactions between electrons) and as the only requirement for a photoelectron is for its total energy to be greater than the energy required to break free, the lowest possible energy of a photoelectron is at this requirement. The kinetic energy would then be 0.

[Floatzel98]

Need some help on the attached question. My teacher wasn't able to give me an explanation why B is the correct answer. Why exactly is torque the same at the 2 points? The solutions just give numerical workings which don't explain it. I don't even really understand how Q is experiencing torque.

Thanks

[yeahm8]

Does your cheatsheet have to be taped together if its 2 A4 pages?