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

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Original post by ubisoft
What?! that's only 25 marks, even if you did get 25/25 (very unlikely, I am only getting 15ish) that would still be a D max


No, of course I'm not going to ignore section B, that would be sheer insanity, but not very much far away from ignoring the six marks though


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Is the angle in the emf equation between the normal to the coil face and the field lines OR between the coil face and the field lines?


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Original post by CD223
Haha makes sense. Is it really 3 for English? That's insane. Especially given the number of points on the mark scheme that are available for answer marks.


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It's more that you have to achieve a certain level of written communication to get in each mark band (1-2, 3-4 or 5-6) than that you get marks specifically for the English if you look on th markschemes. To get 6 marks you still need enough points for that but they also have to be communicated clearly and logically and organised into paragraphs type thing. So if you did bullet points you might find yourself stuck in the 1-2 mark band, maybe 3-4, it isn't very clear on what appropriate organisation is for those ones!
Original post by CD223
Is the angle in the emf equation between the normal to the coil face and the field lines OR between the coil face and the field lines?


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You could think in both terms. You could either use the angle between the coil face and the field lines or the angle between the normal to the coil face and the field lines. But in the former case you need to use the sin(theta) and in the latter the cos(theta)


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(edited 8 years ago)
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Original post by Lau14
It's more that you have to achieve a certain level of written communication to get in each mark band (1-2, 3-4 or 5-6) than that you get marks specifically for the English if you look on th markschemes. To get 6 marks you still need enough points for that but they also have to be communicated clearly and logically and organised into paragraphs type thing. So if you did bullet points you might find yourself stuck in the 1-2 mark band, maybe 3-4, it isn't very clear on what appropriate organisation is for those ones!


Oh I see. Kinda forces people to write in prose then!


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Original post by Mehrdad jafari
You could think in both terms. You could either use the angle between the coil face and the field lines or the angle between the normal to the coil face and the field lines. But in the former case you need to use the sin(theta) and in the latter the cos(theta)


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I see. Sorry, yes, I should have specified that I was referring to the Nϕ=BANcosθN\phi=BANcos\theta form of the equation.


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(edited 8 years ago)
Original post by CD223
I see. Sorry, yes, I should have specified that I was referring to the Nϕ=BANcosθN\phi=BANcos\theta form of the equation.


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It's cool. You could still use sin(theta) in that equation but the angle has to be the angle between the coil face and the field lines so that the sin(90) would give you 1


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Original post by Mehrdad jafari
It's cool. You could still use sin(theta) in that equation but the angle has to be the angle between the coil face and the field lines so that the sin(90) would give you 1


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Cool, thanks :smile:


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Original post by CD223


No problem


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Is the equation v2=grtanθv^2=grtan\theta an equation we should be aware of for banked tracks in circular motion, or should we just be able to derive it if required?

It's in the Nelson Thornes book so I wasn't sure.


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Original post by CD223
I see. Sorry, yes, I should have specified that I was referring to the Nϕ=BANcosθN\phi=BANcos\theta form of the equation.


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The angle that the normal to the plane of the coil makes with the field lines. The plane of the coil is any plane parallel to the plane that one of its cross-sectional areas is in. So you just imagine a line perpendicular to that plane.
Original post by CD223
Is the equation v2=grtanθv^2=grtan\theta an equation we should be aware of for banked tracks in circular motion, or should we just be able to derive it if required?

It's in the Nelson Thornes book so I wasn't sure.


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I think we need to be familiar with the proof of that too as i saw in some EMPA questions demanding that. And it does not necessarily have to be a banked road for the equation to apply. The equation also applies for fair ground rides.



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Original post by CD223
Is the equation v2=grtanθv^2=grtan\theta an equation we should be aware of for banked tracks in circular motion, or should we just be able to derive it if required?

It's in the Nelson Thornes book so I wasn't sure.


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I don't have the book but it can be derived quickly.

Angle track makes with horizontal = θ\theta.
Mass of vehicle = mm
Radius of circle that vehicle moves in = rr

Resolving the reaction force R in the vertical direction:
Rcos⁡θ=mgR\cos \theta=mg
∴R=mgcos⁡θ\therefore R=\frac{mg}{\cos \theta}

Resolving R in the horizontal direction and equating to the centripetal force, (as R is causing this force):
Rsin⁡θ=mv2rR\sin\theta=\frac{mv^2}{r}
∴(mgcos⁡θ)sin⁡θ=mv2r\therefore (\frac{mg}{\cos \theta})\sin \theta=\frac{mv^2}{r}
∴v2=grtan⁡θ\therefore v^2=gr\tan \theta
Hi, I'm a bit confused about the marks and overall grade for this exam.

If an A in the multiple choice paper was 20/25 and an A in the written paper was 35/50, could I get an A in unit 4 overall by getting 18 in the multiple choice and 37 in the written paper? Or would I need to get the minimum for the grade A in each exam to get an A?

Can anyone help?:confused:
Original post by Me123456789
Hi, I'm a bit confused about the marks and overall grade for this exam.

If an A in the multiple choice paper was 20/25 and an A in the written paper was 35/50, could I get an A in unit 4 overall by getting 18 in the multiple choice and 37 in the written paper? Or would I need to get the minimum for the grade A in each exam to get an A?

Can anyone help?:confused:


You just need total A, regardless of what you get in each section.
Original post by PotterPhysics
I don't have the book but it can be derived quickly.

Angle track makes with horizontal = θ\theta.
Mass of vehicle = mm
Radius of circle that vehicle moves in = rr

Resolving the reaction force R in the vertical direction:
Rcos⁡θ=mgR\cos \theta=mg
∴R=mgcos⁡θ\therefore R=\frac{mg}{\cos \theta}

Resolving R in the horizontal direction and equating to the centripetal force, (as R is causing this force):
Rsin⁡θ=mv2rR\sin\theta=\frac{mv^2}{r}
∴(mgcos⁡θ)sin⁡θ=mv2r\therefore (\frac{mg}{\cos \theta})\sin \theta=\frac{mv^2}{r}
∴v2=grtan⁡θ\therefore v^2=gr\tan \theta


That seems too difficult to physics A-level. There were similar questions in M2.
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Original post by ubisoft
That seems too difficult to physics A-level. There were similar questions in M2.


I thought M2 was meant to be of similar difficulty to Physics A2?


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Original post by PotterPhysics
The angle that the normal to the plane of the coil makes with the field lines. The plane of the coil is any plane parallel to the plane that one of its cross-sectional areas is in. So you just imagine a line perpendicular to that plane.


Thanks!


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Original post by Mehrdad jafari
I think we need to be familiar with the proof of that too as i saw in some EMPA questions demanding that. And it does not necessarily have to be a banked road for the equation to apply. The equation also applies for fair ground rides.



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Didn't realise it also applied to fairground rides!


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Original post by ubisoft
You just need total A, regardless of what you get in each section.


Oh, thank you.:smile:
So if an A was 58/75 and I got 12/25 and 46/50 which totals to 58 I'd get an A?
And are you completely sure, it's just something my teacher said that confused me.

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