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# AQA Physics PHYA4 - Thursday 11th June 2015 [Exam Discussion Thread] watch

1. (Original post by ubisoft)
I'm getting close to 90%, but as I flopped the ISA, I need 112 on each paper so yeah not happening. The good thing is I only need a B to get an A overall

How much are you getting on mc ? I'm only getting 15/25
I'm getting about 20 on the MC but I always flop the written part more lol.

I need 207 UMS for an A. Which is essentially a B, although technically a very high C.

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2. (Original post by ubisoft)
I'm getting close to 90%, but as I flopped the ISA, I need 112 on each paper so yeah not happening. The good thing is I only need a B to get an A overall

How much are you getting on mc ? I'm only getting 15/25
I usually make stupid mistakes on MC too. Got 15/25 in my mock and in the past papers I've done I've averaged about 15-20, always seem to make stupid mistakes on it. When I go through after there's only usually 1 or 2 that I think ahh right I really didn't know that.
3. (Original post by JJBinn)
I usually make stupid mistakes on MC too. Got 15/25 in my mock and in the past papers I've done I've averaged about 15-20, always seem to make stupid mistakes on it. When I go through after there's only usually 1 or 2 that I think ahh right I really didn't know that.
They're not usually too difficult when you take it out of exam conditions but I can totally understand dropping 5-10 when you're in the actual thing!

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4. June 2011 MC Question 9:

Why is the damping force always in the opposite direction to the velocity?
I always thought it was always towards the centre of the oscillations, like the acceleration, which is always hence the minus sign.

Is it because the force and velocity are in phase?

Also, Question 23:
Is it the rate of change of flux linkage that becomes zero but the magnetic flux linkage that remains constant when the magnet comes to rest? I got confused. I think I get it now but I wanna make sure.

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5. (Original post by CD223)
June 2011 MC Question 9:

Why is the damping force always in the opposite direction to the velocity?
I always thought it was always towards the centre of the oscillations, like the acceleration, which is always hence the minus sign.

Is it because the force and velocity are in phase?

Also, Question 23:
Is it the rate of change of flux linkage that becomes zero but the magnetic flux linkage that remains constant when the magnet comes to rest? I got confused. I think I get it now but I wanna make sure.

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What are the options? If you think about it, the damping force is logically against velocity as the damping force is acting upon the oscillating object such that it will reduce its amplitude. As it's amplitude is proportional to its max speed, reducing its max speed will reduce its amplitude and therefore the system has been damped (so, to reduce its max speed the damping force opposes the direction of speed)

Number 23, I am struggling to understand it. My instinct would be B because I think the magnetic flux linkage is the area exposed to the field or something, so as it goes inside it will increase and then become constant. If this is wrong I sound stupid but oh well.
6. (Original post by JJBinn)
What are the options? If you think about it, the damping force is logically against velocity as the damping force is acting upon the oscillating object such that it will reduce its amplitude. As it's amplitude is proportional to its max speed, reducing its max speed will reduce its amplitude and therefore the system has been damped (so, to reduce its max speed the damping force opposes the direction of speed)

Number 23, I am struggling to understand it. My instinct would be B because I think the magnetic flux linkage is the area exposed to the field or something, so as it goes inside it will increase and then become constant. If this is wrong I sound stupid but oh well.
That makes sense, thank you!

Not having a good day today. Accidentally cancelled and on both sides of the equation:

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7. (Original post by CD223)
That makes sense, thank you!

Not having a good day today. Accidentally cancelled and on both sides of the equation:

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Haha that's an easy thing to do by accident. I've been doing maths today really, I don't like differential equations at all
8. (Original post by JJBinn)
Haha that's an easy thing to do by accident. I've been doing maths today really, I don't like differential equations at all
Ah differential equations from C4 or M2? I suppose it's the same topic either way but for some reason I find M2 ones easier haha.

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9. (Original post by CD223)
Ah differential equations from C4 or M2? I suppose it's the same topic either way but for some reason I find M2 ones easier haha.

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It's question 8 on C4 June 2005 that I'm struggling with. Thankfully I found out the grade boundaries were 54 for 80 ums
10. (Original post by JJBinn)
It's question 8 on C4 June 2005 that I'm struggling with. Thankfully I found out the grade boundaries were 54 for 80 ums
Oh I see! Do you understand it with the MS or would you like an explanation?

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11. Does anyone know why they omitted question 15 of the MC from January 2011 of the PHYA4 paper? Is it because it was ambiguous as to whether it was a uniform or radial electric field?

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12. (Original post by CD223)
Oh I see! Do you understand it with the MS or would you like an explanation?

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On part B I can get to the part where you have t and c but I don't understand how to get the answer in the form V=A+Be^-kt

I tried to sub in t and c to -1/kln(20-kV)=t+c but I wasn't able to simplify that to get the answer, not sure if I just couldn't simplify it or if this is the wrong thing to do?
13. (Original post by JJBinn)
On part B I can get to the part where you have t and c but I don't understand how to get the answer in the form V=A+Be^-kt

I tried to sub in t and c to -1/kln(20-kV)=t+c but I wasn't able to simplify that to get the answer, not sure if I just couldn't simplify it or if this is the wrong thing to do?
I've not done this question as I do AQA unfortunately so I don't have a written solution to it :/
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14. (Original post by CD223)
I've not done this question as I do AQA unfortunately so I don't have a written solution to it :/
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Oh yeah I didn't consider exam boards haha, got it explained by someone now anyway, mark scheme just over complicated it. Still pretty hard though.
15. (Original post by JJBinn)
Oh yeah I didn't consider exam boards haha, got it explained by someone now anyway, mark scheme just over complicated it. Still pretty hard though.
Ah sorry I couldn't help aha.

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16. (Original post by CD223)
Does anyone know why they omitted question 15 of the MC from January 2011 of the PHYA4 paper? Is it because it was ambiguous as to whether it was a uniform or radial electric field?

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How do you know if it was omitted?

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17. (Original post by Mehrdad jafari)
How do you know if it was omitted?

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There was an erratum notice with the paper telling invigilators to instruct candidates to cross out the question.

Only reason I can think of is maybe because it could refer to uniform or radial fields?

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18. (Original post by CD223)
There was an erratum notice with the paper telling invigilators to instruct candidates to cross out the question.

Only reason I can think of is maybe because it could refer to uniform or radial fields?

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The question refers to a fixed charge. I think you cannot have a uniform electric field with a fixed charge, could you?

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19. I realise this is a Unit 1 question, but I thought this was the best place to ask.
Could someone please explain how to do 2b?
https://www.dropbox.com/s/z534tiu1hk...04516.jpg?dl=0
Thanks
20. (Original post by l1lvink)
I realise this is a Unit 1 question, but I thought this was the best place to ask.
Could someone please explain how to do 2b?
https://www.dropbox.com/s/z534tiu1hk...04516.jpg?dl=0
Thanks
I did OCR at AS and i didn't do particle physics last year so my approach might be a bit different but i hope it's understandable.

For a photon to create an electron and a positron the energy of the photon must be converted into the mass of the positron and the mass of the electron. Since the mass of the positron and electron are the same, the equivalent energy required to create that mass can be calculated using E=(2)mc2.( 2 because rest mass of positron+rest mass of electron). Then that energy is the energy required to create the pair particles. Note that it is assumed the pair particles, electron and positron, will have no kinetic energy after formation and hence that energy would be the minimum energy to create them.

Let me know if anything is not ok.

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