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# AQA Physics PHYA4 - 20th June 2016 [Exam Discussion Thread] watch

1. doyou get a mark for saying limitation of t=2pirootl/g is it dosnt take into consideration air resistance?????
2. (Original post by BioStudentx)
Well done! 65 is solid and will be near 100% imo.

I got around 60-64 so hopefully (fingers crossed) anything below 60 is an A*. Ofc the grade doesn't matter and ums does which is something everyone needs to remember.
Ah thank you! I hope so, I really need at least one A*. An A is much easier for me to get this year because of spare UMS from last year but UCL is asking for A*AA.
Fingers crossed everything goes well for us both, good luck with PHYA5 also!
3. im so dumb for the graph of earth and venus g values
i drew it from 9.81 going down ( i knew it would go to 0 at some point but i wasnt thinking clear and drew like a u shape) then it went up to 7.71 or whatever the value waas ffs

i drew it like a sideways smiley face ( u shaped) with the left being higher than the right
4. (Original post by james2mid)
No, they were both positive so considering the charge on the left it would move to the left (Y to X)
They were both positive but X was 4nc and Y was 1nc. So a charge inbetween would move towards Y?
5. (Original post by Yo12345)
Got around 38-40 on Section B. How many would I need on the multiple choice to get a mid-A?

Posted from TSR Mobile
I'd guess around 15. Assuming grade boundaries for A* is around 60.
6. anyone got rough grade boundaries. i feel it was hrder than last years paper
7. i think this paper will be ~56 for A*, the boundaries cannot increase from 54 (2015) to 60 for an A* the paper wasnt that easy.
8. (Original post by IMZL)
Someting about limitations of the equation for time period, section B, 4a.
nit out there but put thermal expansion of pendulum length e.q in colder climates length is shorter so temperature must remain constant,think i read it in a book somewhere
9. for the limitation i wrote it doesnt work for compound pendulums where mass is spread out?
10. (Original post by shuu00)
I think you're right. When the gradient is 1/C, the energy stored is the area between the line, the limits, and the x-axis.

I just used the equation E=(1/2)(C)(V^2) and then did the difference so deltaE=(1/2)(C)(12^2-9^2)
I used the equation too. Do you remember what you got for your capacitance?
11. For the braking system thing, I said the advantage is that the energy used in braking can be recovered and used to start the vehicle again (should have said through the use of a flywheel but I didn't), and the disadvantage is that the braking system is complex and heavy so it increases the inertia of the vehicle, making it 'harder to accelerate'. Is this okay?
12. Here are my answers to Written, the question parts might not be correct.

1ai) 8.26 (will be a range)
ii) 2.6x10^-4

bi)28
ii)5.9x10^4
iii) bottom box ticked

2ai) Derivation
ii) 6.06x10^6 (3SF only)

b) Downward Curve staring at 9.81
Minimum at g=0 at a point slightly offset from halfway closer to Venus
Rises up to 8.87

3a) No external forces

bi) V=mv/N

bii) Derivation

ci) 216
ii) 1.15x10^-19 kgms-1 OR Ns
iii) Anti electron neutrino released also so Beta minus particles have range of Kinetic Energies upon release.

4a) Small amplitude oscillation as only valid for small angular amplitudes.

b) -Set up light string with Bob on the end,
attach top of string to clamp stand boss.
-Measure length from top to centre of Bob.
-Release the Bob so it oscillates with small amplitude.
-Use stop clock to measure time taken for 20 oscillations.
-Divide this by 20 for mean T
-Plot T^2 against L
-Repeat procedure

c) -Student value 4x true value
-Time period half so all values for T^2 will be 1/4 the true
-State equation or show that g is inversely proportional to gradient.

5a) Lenz's Law: Direction of induced EMF/current is always in a direction that opposes the change that caused it.
Faraday's Law: Induced EMF is proportional to rate of change of flux linkage.

b) -Change in flux linkage when current flows.
-EMF induced in wheel (or whatever it was called)
-Current induced in the wheel as good conductor.
-Current carrying conductor (wheel) opposes the field due to coil, therefore there is a force against the direction of motion.

c) Wheel not in contact with electromagnet so no wear, but in brake pads there is wear due to friction.

More energy used or less effective or heating.
13. Was the capacitance in the first question 8.2 microfarads????
14. (Original post by rxns_00)
Ah thank you! I hope so, I really need at least one A*. An A is much easier for me to get this year because of spare UMS from last year but UCL is asking for A*AA.
Fingers crossed everything goes well for us both, good luck with PHYA5 also!
What course did you apply for?
15. (Original post by CourtlyCanter)
I suspect there'll be a backup power supply for that ie when your car just shuts down its system or runs out of battery. You may be right though.
I think both answers for increased power drain and electrical failure will be valid. Although my view is that it would be unreasonable to use a powered braking system, because if you park your car on a hill your battery will eventually go flat and then your car will have f***Ed off down the hill by the time you come back with your shopping.
16. (Original post by Jordenwilder1998)
doyou get a mark for saying limitation of t=2pirootl/g is it dosnt take into consideration air resistance?????
That's exactly what I put but i'm not too sure really, it probably depends how many other people put that!
17. (Original post by DesignPredator)
Here are my answers to Written, the question parts might not be correct.

1ai) 8.26 (will be a range)
ii) 2.6x10^-4
bi)28
ii)5.9x10^4
iii) bottom box ticked
2ai) Derivation
ii) 6.06x10^6 (3SF only)
b) Downward Curve staring at 9.81
Minimum at zero at a point slightly offset from halfway towards Venus
Rises up to 8.87
3a) V=mv/N
b) Derivation
ci) 1.15x10^-19 kgms-1 OR Ns
cii) Anti electron neutrino released also so Beta minus particles have range of Kinetic Energies upon release.
4a) Small amplitude oscillation as only valid for small angular amplitudes.
b) -Set up light string with Bob on the end, attach top of string to clamp stand boss.
-Measure length from top to centre of Bob.-Release the Bob so it oscillates with small amplitude.
-Use stop clock to measure time taken for 20 oscillations.
-Divide this by 20 for mean T
-Plot T^2 against L
-Repeat procedure
c) -Student value 4x true value
-Time period half so all values for T^2 will be 1/4 the true
-State equation or show that g is inversely proportional to gradient.

5a) Lenz's Law: Direction of induced EMF/current is always in a direction that opposes the change that caused it.
Faraday's Law: Induced EMF is proportional to rate of change of flux linkage.

b) -Change in flux linkage when current flows.
-EMF induced in wheel (or whatever it was called)
-Current induced in the wheel as good conductor.
-Current carrying conductor (wheel) opposes the field due to coil, therefore there is a force against the direction of motion.

c) Wheel not in contact with electromagnet so no wear, but in brake pads there is wear due to friction.

More energy used or less effective or heating.
Wasn't it V = -mv/N?
18. (Original post by Mango Milkshake)
Daughter
(Original post by Yo12345)
No, daughter one
(Original post by shuu00)
It couldn't have been b/c I couldn't do the proof. People that got N as the daughter nucleus ended up proving it . . .
Okay assuming I'd completed the question thinking N was that of the original, would I only lose marks on the 4-marker proof question or any of the ones following? Do you guys remember what the follow-up ones were?
19. (Original post by johnapplebottom)
for the limitation i wrote it doesnt work for compound pendulums where mass is spread out?
I wrote it doesn't work for pendulums located as places at a significance distance as the value 'g' will vary?
20. (Original post by Questioner1234)
Was the capacitance in the first question 8.2 microcoulombs????
That's what I got (rounded of course).

And it would be in farads, not coulombs

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