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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 JellyTeapot)
How were you supposed to draw the graph of the second ball in the container that wasn't full with oil?
Constant positive gradient until it hits the oil, then it slopes off downwards sharply, and levels off to a constant speed before hitting the bottom
2. (Original post by Raimonduo)
Uhh, it did refract though. The angle of incidence was below that critical angle. Well, if you calculated the angles correctly that is.
I didn't round anything up and I got the angle of incidence slightly over the critical angle.
3. For the question about drawing on the graph for what red light would look like is the central maxima slightly bigger and is the intensity of the fringes slightly less and more spaced out ?
4. (Original post by Raimonduo)
Uhh, it did refract though. The angle of incidence was below that critical angle. Well, if you calculated the angles correctly that is.
30+22 is 52 the critical angle was 51 hence TIR

Posted from TSR Mobile
5. (Original post by Davelittle)
For prove the angle is 30 degrees I got 31.9, as did 3 other people.

Did anyone else?
Yeah, it said to show 'about', so if you kept it at 3.s.f you should be fine.
6. (Original post by Raimonduo)
Uhh, it did refract though. The angle of incidence was below that critical angle. Well, if you calculated the angles correctly that is.
Critical angle was 51 degrees

Incidence was 53 or so :/
7. Hi everyone! Today's paper was actually harder than normal paper. I recon you will need around 44-46/70 to get an A. However, there were some stupid questions in which I made silly mistakes. Overall, it was an ok paper so I am definitely expecting an A. EDIT_____----- Now, who the heck voted me down? I know, anyone who is jealous with my efforts is pissed off.
8. (Original post by Goods)
30+22 is 52 the critical angle was 51 hence TIR

Posted from TSR Mobile
lol... I forgot the normal line wasn't where I thought it was.. yeah it does TIR.
9. (Original post by Raimonduo)
Yeah, it said to show 'about', so if you kept it at 3.s.f you should be fine.
It should have been show its about 32 degrees, I got worried as 1.9 degrees is quite a lot
10. (Original post by Raimonduo)
lol... I forgot the normal line wasn't where I thought it was.. yeah it does TIR.
You might get ecf marks
11. I think I got about 57 marks in unit 2 (under estimated) and about 65 marks in unit 1( under estimated). What do people think what you got in unit 1 and 2?
12. I did worse in unit 1 than unit 2 which paper is worth more ums ?
13. (Original post by StalkeR47)
x
(Original post by x-Sophie-x)
x
See if we hadn't argued about it yesterday I may have forgotten the criteria to TIR.
Thanks to you all I gained two marks. Cheers
14. (Original post by Bookaky)
I did worse in unit 1 than unit 2 which paper is worth more ums ?
Equal amount.
15. (Original post by Bookaky)
I did worse in unit 1 than unit 2 which paper is worth more ums ?
Depends on the country's performance.
16. (Original post by StalkeR47)
I think I got about 57 marks in unit 2 (under estimated) and about 65 marks in unit 1( under estimated). What do people think what you got in unit 1 and 2?
57 in this sounds like it could be 112-115, cant speak for unit 1 as I didn't sit it, but 65 is often 120ums
17. PHYA2 (Provisional, and very shaky) Mark scheme;

Question 1 – Simple matter of working out forces and that stuff, efficiency question may cause problems though.

1) 6082N, which is 6100 to 2sf (2)
b) 9.5 degrees using arctan of 1000/6000 (2)
c) Acceleration was 9.2ms^-2 (2)
d) Can’t remember what this one was asking but I got 3.9x10^6 (not sure) (2)
e) I got 20.5% but I’m not sure as I’ve heard people give various answers (3)

Question 2 – Fairly straightforward question about projectiles.

2) Time taken was 0.55s (2)
b) The horizontal distance was 248m I think (2)
c) As both bullets have the same mass, S=ut will be smaller for the bullet with the lower horizontal velocity (3?) An addition as suggested by ThomT94; 'Also, bullet b's distance is not affected by its mass. It's because both accelerate towards the ground at the same rate from the same height, therefore, same time of travel, thus lower horizontal speed causes bullet to travel shorter distance in the given time.'
-- Can’t remember if there was anything more to this question

Question 3 – Perhaps the best 6 marker they’ve ever given, very straightforward.

3) The ball begins to accelerate with a large acceleration as its weight force is (one of) the only forces acting as it begins to move. The gradient then begins to slowly decrease as the resistive forces begin to balance out the weight force. The ball finally reaches terminal velocity as the resistive forces balance out the weight force. Appropriate reference to Newton’s first law needed. Then the ball remains at that velocity, as there is no external force, give reference to Second Law. (6)
b) Initially a straight line indicating uniform acceleration, then a curve of decreasing gradient ending with the ball at terminal velocity. (3)

Question 4 – A decent question, I guess.

4) Weight of the ball was 0.44N (2sf) using volume of a sphere which is given in the data sheet!!! (4)
b)Hooke’s law states that force applied is proportional to the extension up to the limit of proportionality (2)
c) Gradient was roughly 1150/1100 Nm^-1needed (3)
d) Using the area of a triangle and then counting (roughly) 9 small boxes each of area 0.025 gave 1.08J (3?)
e) Straight line parallel to the first one but not finishing at zero as plastic deformation has occurred (2)
f) Plastic deformation means the object won’t return to its original shape/length once the load is removed (1)
g) Work is less as area under graph is less energy absorbed as heat and in permanent extension (1)

Question 5 – Wasn’t too bad..

5) Two conditions for TIR were an angle greater than the critical angle and moving from a medium of higher refractive index to lower (2)
b) Defining the frequency of a progressive wave. Just explain the equation? (1)
c) Speed of light in the object thing was 1.7x10^8 (2)
d) Proof that the angle was about 31.8 degrees using Snell’s Law
e) Critical angle was 51.1 degrees using 1/sinc (2)
f) Undergoes TIR as (20+31.9) > 51 (2)

Question 6 – Don’t even know, seemed cool..

6) First phase difference was 90 degrees or pi/2. Second was 270 degrees or 3pi/2 (2)
b) Oscillation between maximum and minimum amplitude; i.e moves up and down (2)
c) Polarisation only occurs in transverse waves; mention something about plane of polarisation (2)
d) Define frequency (1)
e) 750m (2)

Question 7 – I may have flopped

7) All of the light has the same wavelength (1)
b) Can’t really remember – According to Frogs491 its ‘Fringes larger so intensity equal’ -- If this is the graph of intensity, then Yes, this was the graph of intensity showing the maxima at (roughly) the same height as the central. (2)
c) Safety precautions that need to be taken when dealing with laser light can include; wearing reflective glasses; taking care not to shine the light in your eyes (or other people's eyes for that matter); any other sensible answer (2)
e) Last part; white at central maxima and very bright as all of the wavelengths converge at the centre; blue tinge closer to central maxima and red tinge further away; not sure about anything else. (3)

That’s all I can remember guys. I’ve probably omitted quite a few things so if anyone can spot any mistakes, let me know and I’ll make amendments

EDIT: I'm going to start adding in the marks for each question slowly so that should be half done soon. If anyone can remember what the marks are for each question please let me know! Thanks. Done

Okay its all done now. Mucho gracias to ThomT94 and .raiden. for helping clear a few things up and to anyone else that contributed!

Still missing some stuff over here guys. Although this post is likely long forgotten now..

Not missing anything now, and I sure hope it doesn't get forgotten. Hopefully this finds it way into the OP..
18. (Original post by FLLF)
57 in this sounds like it could be 112-115, cant speak for unit 1 as I didn't sit it, but 65 is often 120ums
The amount of raw marks needed for 120 UMS:
62 Jan 13
60 Jun
58 Jan 12
59 Jun
62 Jan 11
57 Jun
60 Jan 10
70 Jun
61 Jan 09

If we count June 09 as anomalous, not been higher than 62 and as low as 57.
19. (Original post by FLLF)
57 in this sounds like it could be 112-115, cant speak for unit 1 as I didn't sit it, but 65 is often 120ums
No idea m8. We will see what happens.
20. (Original post by Davelittle)
You might get ecf marks
I'm seriously relying on ecf marks, half the time I got the method right for the follow on question but the initial one was off the chart of wrongness.

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