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AQA Physics PHYA5 - 28th June 2016 [Exam Discussion Thread]

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Original post by allofthestars
Im almost sure it was 2 marks like haha


Can confirm it was 2 was the specific charge 1.8x10^11 Ckg^-1
Original post by FireBLue97
Can confirm it was 2 was the specific charge 1.8x10^11 Ckg^-1


I got 1.9x10^11 Ckg^-1

Did you use em=2VB2r2\dfrac{e}{m} = \dfrac{2V}{B^2r^2}?
For the question asking about the population experiencing a higher background radiation and the possible causes for it - would the answer of "Radioactive building materials used in construction" got the mark?

It is an answer in the text book however I don't know if it's relevant to the "in the past 100 years" part
Original post by js.int
I got 1.9x10^11 Ckg^-1

Did you use em=2VB2r2\dfrac{e}{m} = \dfrac{2V}{B^2r^2}?


Do you remember what the numbers were for V, r and B? Isn't the two supposed to be on the bottom for that formula?
(edited 7 years ago)
Original post by -jordan-
Do you remember what the numbers were for V, r and B?


Nope :biggrin: but I remember it the specific charge to be something like 1.87 or 1.89 x 10^11 Ckg^-1
Original post by -jordan-
... Isn't the two supposed to be on the bottom for that formula?


Nope:

[br]Ek=eV=12mv2,v=Berm[br][br]∴eV=12m(Berm)2[br][br]∴eV=12(B2e2r2m)[br][br]∴V=12(B2er2m)[br][br]∴2V=B2er2m[br][br]∴em=2VB2r2[br][br]E_k = eV = \dfrac{1}{2}mv^2, v = \dfrac{Ber}{m}[br][br]\therefore eV = \dfrac{1}{2}m(\dfrac{Ber}{m})^2[br][br]\therefore eV = \dfrac{1}{2}(\dfrac{B^2e^2r^2}{m})[br][br]\therefore V = \dfrac{1}{2}(\dfrac{B^2er^2}{m})[br][br]\therefore 2V = \dfrac{B^2er^2}{m}[br][br]\therefore \dfrac{e}{m} = \dfrac{2V}{B^2r^2}[br]
Original post by js.int
Nope:

[br]Ek=eV=12mv2,v=Berm[br][br]∴eV=12m(Berm)2[br][br]∴eV=12(B2e2r2m)[br][br]∴V=12(B2er2m)[br][br]∴2V=B2er2m[br][br]∴em=2VB2r2[br][br]E_k = eV = \dfrac{1}{2}mv^2, v = \dfrac{Ber}{m}[br][br]\therefore eV = \dfrac{1}{2}m(\dfrac{Ber}{m})^2[br][br]\therefore eV = \dfrac{1}{2}(\dfrac{B^2e^2r^2}{m})[br][br]\therefore V = \dfrac{1}{2}(\dfrac{B^2er^2}{m})[br][br]\therefore 2V = \dfrac{B^2er^2}{m}[br][br]\therefore \dfrac{e}{m} = \dfrac{2V}{B^2r^2}[br]


Hm that's odd. Antonine has the two on the bottom and I got a two by using Bev=eV/d and replacing d by 2r...
Original post by -jordan-
Hm that's odd. Antonine has the two on the bottom and I got a two by using Bev=eV/d and replacing d by 2r...


Isn't eVd\dfrac{eV}{d} for uniform electric fields?... not magnetic fields?
Original post by js.int
Isn't eVd\dfrac{eV}{d} for uniform electric fields?... not magnetic fields?


The question had both an electric field and a magnetic field, you were told the potential at which it's accelerated.

1/2mv^2 = eV is for a uniform electric field, using Work Done = QV. I used the force instead rather than work done. There is some reason why my method doesn't work I just can't think why.
(edited 7 years ago)
Original post by -jordan-
The question had both an electric field and a magnetic field, you were told the potential at which it's accelerated.

1/2mv^2 = eV is for a uniform electric field, using Work Done = QV. I used the force instead rather than work done


But surely once an electron has been accelerated and has left the anode, it isn't in the electric field any more? So there is no electric force on the electron...

Say, if in fact, there is still an electric force on the electrons, wouldn't it cause the electron to accelerate? (Tangental to the circle it follows) ... and hence cause the radius of the circular beam to increase?
Original post by js.int
But surely once an electron has been accelerated and has left the anode, it isn't in the electric field any more? So there is no electric force on the electron...

Say, if in fact, there is still an electric force on the electrons, wouldn't it cause the electron to accelerate? (Tangental to the circle it follows) ... and hence cause the radius of the circular beam to increase?


Yes, you are correct. I still got V/B^2r^2 anyway, so I have half the expected value, so hopefully only dropped a mark. Was it two?
(edited 7 years ago)
Original post by -jordan-
Yes, you are correct. I still got V/B^2r^2 anyway, so I have half the expected value, so hopefully only dropped a mark. Was it two?


I think it was 4 marks for that question because it was meant to be quite a long derivation :s-smilie:
Original post by js.int
I think it was 4 marks for that question because it was meant to be quite a long derivation :s-smilie:


Marks were surely not for derivation though, it's given in the books and one you can just learn.
Original post by -jordan-
Marks were surely not for derivation though, it's given in the books and one you can just learn.


I don't know tbh... I remembered that the question had quite a large blank space and I thought it was a 4 marker. I could be wrong though, what other parts to that question was there? I might be thinking of the wrong question
Original post by js.int
I don't know tbh... I remembered that the question had quite a large blank space and I thought it was a 4 marker. I could be wrong though, what other parts to that question was there? I might be thinking of the wrong question


The question just asked to find the specific charge. There likely was a large space for the derivations as they were expecting them but they'll also surely be expecting people to just have learnt the formula... and account for mistakes that people may make in remembering it. I'm hoping anyway :P

I just knew something was wrong, all the questions I've done in the past used figures which gave you the expected charge of 1.8x10^11. Was surprised when it wasn't but didn't get a chance to have a look
(edited 7 years ago)
Original post by -jordan-
The question just asked to find the specific charge. There likely was a large space for the derivations as they were expecting them but they'll also surely be expecting people to just have learnt the formula... and account for mistakes that people may make in remembering it. I'm hoping anyway :P


Tbh they may put something in the additional guidance if a lot of people are making the mistake. But I think you would still get some marks out of 4.
Original post by js.int
I don't know tbh... I remembered that the question had quite a large blank space and I thought it was a 4 marker. I could be wrong though, what other parts to that question was there? I might be thinking of the wrong question


It was 4 marks because the equation isn't given, so you either have to know the equation or the derivation for it. It asked you to calculate the specific charge given some values of B, r and V - it never asked anything about deriving it. They will probably give marks for derivation, but also the mark scheme will likely say "If 1.9x10^11 Ckg^-1 seen, 4 marks". They can't take marks away because you know the equation MORE than other people, that'd be completely nonsensical.
Can anyone remmeber how many marks was in question 5 and how they were split up for each ? I`m a bit panicy as i got the first part wrong and carried my answer onto the rest, will there be marks awarded for error carried forward?

Thanks
Original post by Jay1421
Could you possibly tell me which question this was? I really don't remember it and I'm worried I've missed one :frown:


It was something along the lines of: Give one reason why background radiation has increased within the last 100 years.
Original post by Aprkrheiqkk
It was something along the lines of: Give one reason why background radiation has increased within the last 100 years.


Does anybody know if Radon levels have increased due to mining? I put that :frown:

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