You are Here: Home >< Physics

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

1. I wrote my answers to Section B ( written paper) in fountain pen :0 Will it not get scanned ?
I wrote my answers to Section B ( written paper) in fountain pen :0 Will it not get scanned ?
Lol I did all my work in fountain pen too. It's fine, I did everything in fountain pen last year, no problem whatsoever
3. unofficial mark scheme for section B?
4. (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 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.
Either your 5a is wrong or the AQA text book is wrong. Because the text book says induced emf is EQUAL to rate of change of flux linkage. Not proportional. Plus if it was proportional then the equation would need a constant. But all the letters in the equation can be varied
5. (Original post by lucabrasi98)
Either your 5a is wrong or the AQA text book is wrong. Because the text book says induced emf is EQUAL to rate of change of flux linkage. Not proportional. Plus if it was proportional then the equation would need a constant. But all the letters in the equation can be varied
Both will be accepted as seen from previous mark schemes and there is a constant, that is -1.
6. (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 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.
For 5b, I put emf induced in coil by accident , I think the question was out of 3, would I get at least 1 for saying change in disk rotates, so change in flux ?
7. (Original post by DesignPredator)
Both will be accepted as seen from previous mark schemes and there is a constant, that is -1.
B,A,N and T can all be varied. A constant is something like G in the equation for gravitational firled strenth. The -1 isn't really a constant, it's just essentially demonstrating Lenz's law. It implies they're in opposite directions. And IIRC you had to mention lenz' law later in the question.

I don't recall any mark schemes where they accept both but maybe you're right
8. Im pretty sure one of the mark schemes said it was proportional
9. What if I said Faraday's law was emf is proportional to rate of change of flux, and not flux linkage? (constant of proportionality would be N, number of turns)

Will I lose any marks?
10. (Original post by mcride98)
What if I said Faraday's law was emf is proportional to rate of change of flux, and not flux linkage? (constant of proportionality would be N, number of turns)

Will I lose any marks?
Not sure, I may have said that too . To make things even worse, I said the induced emf is proportional to the rate of change of flux with time
11. (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 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.
Cheers! I can't remember all my answers now but I recall getting a lot of similar answers. Do you remember what question 4(a) asked?
12. Hey cobalt, noticed you have an offer for natsci at cambridge , which college????
13. Was the question which proceeded the energy proof asking to calculate the momentum of the alpha particle. If so, does anyone remember their working out for it?
14. (Original post by lfcrules)
Hey cobalt, noticed you have an offer for natsci at cambridge , which college????
Christ's
15. (Original post by C0balt)
Im pretty sure one of the mark schemes said it was proportional

All I know is the book said equal to in bold writing. So if it's not then I blame AQA and Nelson Thrones for ****ing me over and over again.
16. thanks a lot
17. (Original post by C0balt)
Christ's
Congrats C0balt, I've no doubt you'll make your offer!
18. (Original post by lucabrasi98)
All I know is the book said equal to in bold writing. So if it's not then I blame AQA and Nelson Thrones for ****ing me over and over again.
They are both accepted. Stop fretting.
19. (Original post by C0balt)
I initially got 8.22 but then changed to 8.0 because it gave me an exact value lmao
Yeah we always get a mark for stating it's the gradient
I got 8.0 too. Δy/Δx across the whole range returned it.
Congrats C0balt, I've no doubt you'll make your offer!
Thanks i hope so, all depends on the next 3 exams!
(Original post by micycle)
I got 8.0 too. Δy/Δx across the whole range returned it.

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: January 9, 2017
Today on TSR

### Should I ask for his number?

Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• Poll
Useful resources

Can you help? Study help unanswered threadsStudy Help rules and posting guidelinesLaTex guide for writing equations on TSR

## Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE