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# The Physics PHYA2 thread! 5th June 2013 Watch

• View Poll Results: What mark do you think you got out of 70?
0-20
6
3.00%
21-40
12
6.00%
41-50
29
14.50%
51-60
79
39.50%
61-70
74
37.00%

1. (Original post by navyspaces)
I did this also, for good reason. Having seen countless mark schemes now, I know that questions where you are asked to 'show that something equals something' allow two answers, generally - one coming from the exact, calculated value from the 'show that' question, or the rough value given in the question. I worked it out both ways, and I believe I got 9.6o or 10.4o, and went with 10.4.

See similar mark schemes if you'd like to be reassured!
yep I knew that It seems safer, in case the previous value you calculated was wrong for some reason (even though it was a prove that one)
2. Just to add to the Unoficial MS, (not the UMS ) wasn't the second wave 180 degrees, not 270? Or have I just been a stupid idiot? (as opposed to a clever idiot)
3. (Original post by g.k.galloway)
What UMS do you think 62 will be?
Firstly, be careful estimating. I'm doing this is a retake, and I assure you I'd not be retaking had I got what I thought I had the first time...

That said, this will help you. I compiled the marks for the grades I need, where the marks are in chronological order, with the average at the end. m stands for full UMS, so on average 62 would give you that.

m 61 70 60 57 62 59 58 60 62 avg. 61.0
A* 55 63 54 51 56 54 53 55 56 avg. 55.2
A 49 56 48 45 50 49 48 50 50 avg. 49.4

4. (Original post by OliverG)
Just to add to the Unoficial MS, (not the UMS ) wasn't the second wave 180 degrees, not 270? Or have I just been a stupid idiot? (as opposed to a clever idiot)
It was definitely 270, or 3/2 Pi, as it was a wavelength and a half away from A. A, B and C were involved, and the distance between A and B was Pi/2, the distance between A and C was 3/2 Pi, and the distance between B and C was Pi (180 degrees). Did you get A and B mixed up?
5. (Original post by .raiden.)
This is what I did, it was probably wrong lol
I'm pretty sure it looks like this

6. [QUOTE=g.k.galloway;42989607]
(Original post by gandanmo)

I know this wasn't directed at me but on a past paper I did --

- 1 mark for a straight line ending behind the loading line
- 1 mark for it being parallel to the loading line.

Whether it will be the same again I don't know though
FEEL FREE BRO! thanks so much! so id get one mark? also you know the last last question like 7c or whatever it was, i wrote the white light would be less intense, maximas would be wider (or larger) and minimas would be shorter, and also colours would be observed in white light, how many marks out of 3 would i get? THANKS!
7. (Original post by HenryD)
Hello, hmm:
2C) You'll get marks for that yes, you may even get full marks if you explained why t will be the same? If not then 1 or 2 marks (probably 2 if you explained).
5A) Depending on how you wrote it you may get the mark for that, sounds a little ambigious but if the examiner understands you you'll get the mark. 1-2/2
6A) Someone else on here did the same, it's possible that they'll give you that if they're lenient. Probably a 50 50 toss up between whether you get the marks or not.
ONE LAST THING BRO !
you know the last last question like 7c or whatever it was, i wrote the white light would be less intense, maximas would be wider (or larger) and minimas would be shorter, and also colours would be observed in white light, how many marks out of 3 would i get?
8. (Original post by michelfish)
I'm pretty sure it looks like this

There is an almost identical question regarding a parachutist on June 2009 (I think...)

Straight line of constant gradient showing constant accelleration
Deceleration due to entering the oil, down to new terminal velocity
Horizontal straight line showing terminal velocity, then vertical line showing the ball coming instantaneously to rest.
9. So what did everyone conclude about the white light question?

I said about a central white maximum, then each subsidiary maxima is a spectrum with red inner and white outer, and also that the maxima are wider.

Apparently it's not that because it was single slit or something?
10. (Original post by michelfish)
I'm pretty sure it looks like this

I hope so. That's what I did.

I was thinking it could be one of two graphs, depending if the ball reaches a velocity higher than the terminal velocity in the oil, before it hits the oil. I went with the assumption that it does.
11. Guys I'm pretty sure it was refraction and not TIR. The angle was like 31 degrees which falls below the critical angle. Hence refraction?
12. Forgive moi if I'm wrong but for the Phase difference question wouldn't In Phase & Anti phase be enough ???
13. (Original post by navyspaces)
There is an almost identical question regarding a parachutist on June 2009 (I think...)

Straight line of constant gradient showing constant accelleration
Deceleration due to entering the oil, down to new terminal velocity
Horizontal straight line showing terminal velocity, then vertical line showing the ball coming instantaneously to rest.
Ah, all good then
I don't do old past papers, but clearly they must help
14. (Original post by Davelittle)
It would accelerate in the oil and the acceleration would decrease (leading to a curve on this portion of the graph similar to the start of the question) until it reached 0 where the graph would be a straight line at a constant maximum velocity.
Disagree about the acceleration, it's not necessary if you think the ball is already travelling faster than it would in the oil. If you look at the height it was dropped from then it would've been, so it would've decelerated into a curve. But they may allow both as technically there isn't a way to really know this.

(Original post by gandanmo)

oh bro one other thing you know for the unloading graph in Q4 i drew a curve coming back down but ending towards the right of the original(loading) but because its a curve do i lose both marks or would i still get one mark for the line ending at the right point on the x-axis> THANKS!!
(Original post by g.k.galloway)

I know this wasn't directed at me but on a past paper I did --

- 1 mark for a straight line ending behind the loading line
- 1 mark for it being parallel to the loading line.

Whether it will be the same again I don't know though
Yeah this is what I'm expecting it to be too.

(Original post by gandanmo)
ONE LAST THING BRO !
you know the last last question like 7c or whatever it was, i wrote the white light would be less intense, maximas would be wider (or larger) and minimas would be shorter, and also colours would be observed in white light, how many marks out of 3 would i get?
I think probably two Possibly three but more likely 2.

(Original post by SamuelJ)
So what did everyone conclude about the white light question?

I said about a central white maximum, then each subsidiary maxima is a spectrum with red inner and white outer, and also that the maxima are wider.

Apparently it's not that because it was single slit or something?
That is a single slit diffraction graph. But it would be blue on the inside, red on the outside.
15. (Original post by navyspaces)
Firstly, be careful estimating. I'm doing this is a retake, and I assure you I'd not be retaking had I got what I thought I had the first time...

That said, this will help you. I compiled the marks for the grades I need, where the marks are in chronological order, with the average at the end. m stands for full UMS, so on average 62 would give you that.

m 61 70 60 57 62 59 58 60 62 avg. 61.0
A* 55 63 54 51 56 54 53 55 56 avg. 55.2
A 49 56 48 45 50 49 48 50 50 avg. 49.4

That's fantastic thank you, I just hoped for something near full ums, so even if I lose another mark or two I should get that

Thanks again
16. (Original post by hassanen9)
Guys I'm pretty sure it was refraction and not TIR. The angle was like 31 degrees which falls below the critical angle. Hence refraction?
This was the trick here The angle was indeed around 30° to the normal, but that didn't include the angle of incidence, which was around 20°. Adding these together gave around 52°, which was over the critical angle.
17. (Original post by g.k.galloway)
That's fantastic thank you, I just hoped for something near full ums, so even if I lose another mark or two I should get that

Thanks again
Me too; my university place is effectively resting on this retake, as if I don't ace it I have to make up for it in Units Four and Five, which would be a lot more effort... Very pleased to have just realised that I didn't get all the TIR stuff wrong. In a burst of post exam madness, my friend convinced me that TIR worked the opposite way round...
18. (Original post by michelfish)
Ah, all good then
I don't do old past papers, but clearly they must help
In my opinion (as an A2 student predicted ridiculously over the top grades, hehe) the only real way to prepare for exams like Physics is to do past papers... I don't know how else you'd manage it!
19. (Original post by HenryD)
Disagree about the acceleration, it's not necessary if you think the ball is already travelling faster than it would in the oil. If you look at the height it was dropped from then it would've been, so it would've decelerated into a curve. But they may allow both as technically there isn't a way to really know this.

Yeah this is what I'm expecting it to be too.

I think probably two Possibly three but more likely 2.

That is a single slit diffraction graph. But it would be blue on the inside, red on the outside.
sorry I misread your post!
20. Have to start work on the PAT soon, and start my extended project on something in physics. So much work.

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