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

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Original post by gabbons
Screen Shot 2015-06-10 at 12.52.12.png

Hey guys, can anyone explain why the answer a parabola in the vertical plane? I thought it would be in the horizontal


I thought this too, I think it's to do with the fact the only force acting is now weight so that must act down, in a vertical plane


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Original post by gabbons
Screen Shot 2015-06-10 at 12.52.12.png

Hey guys, can anyone explain why the answer a parabola in the vertical plane? I thought it would be in the horizontal


It can't be a parabola to the horizontal. Once the string breaks, the mass only has a liner velocity in the horizontal direction. There is no centripetal acceleration or force towards the centre so it goes straight forward in the x component. So the answer can't be C

It can't be B because of the same reason above. For something to move in a circular orbit it needs a centripetal acceleration and force, which is lost when the string breaks.

I dont think it's A because the linear velocity of the object would be tangential to the radius of the circle at the instance when it breaks

It's D obviously. Anything you throw creates a vertical parabola
Original post by Kennethm
23 - the magnet and coil are in relative motion with each other so the coil is cutting the magnetic flux lines of the magnet therefore inducing an emf in the surrounding area which will show a reading in the ammeter as there is a current in the coil.

(you do not need to consider lenz's law here at all so don't confuse yourself with that)

24 -

ε = BANωSinωt

just apply your knowledge of maths graphs and the information they gave you to this equation.


23 - how come the ammeter stays constant after increasing - if it comes to rest, surely no magnetic flux lines are being cut so no emf can be induced

24 - I did use maths knowledge so surely the 0.5B should double all Y values and the 2w should halve all x values?
Original post by becca9621
Anybody got any predictions?


I predict I'll be needing a lot of lube for the exam tomorrow
Reply 3304
Hey, can anyone answer this question for me? I can do it just by thinking about it, but what equations could you use for this? Thanks!

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Original post by thedon96
23 - how come the ammeter stays constant after increasing - if it comes to rest, surely no magnetic flux lines are being cut so no emf can be induced

24 - I did use maths knowledge so surely the 0.5B should double all Y values and the 2w should halve all x values?


23 - they didnt say the reading was constant, answer B says that the magnetic flux linkage stays constant.

24 -

ε = BANωSinωt

B is 0.5 and ω is doubled so the first part cancels meaning the answers is out of A and B.

Sinωt is left so its Sin2ωt therefore halving the X directions ^^
Original post by bwr19
Hey, can anyone answer this question for me? I can do it just by thinking about it, but what equations could you use for this? Thanks!

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I got the answer D, if thats correct Ill explain to you :smile:
Really think the six marks will be on SHM/DAMPING/RESONANCE

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Reply 3308
Original post by Kennethm
I got the answer D, if thats correct Ill explain to you :smile:


That is indeed the correct answer. Please, do explain! :biggrin:
Original post by Kennethm
I got the answer D, if thats correct Ill explain to you :smile:


can you explain
Original post by bwr19
That is indeed the correct answer. Please, do explain! :biggrin:


literally if r is decreased so that the electric potential energy of Q is doubled then that means r has been halved.

here is the thought process:
W = QV and V has been doubled

V = KQ/r

if V is doubled then r must have halved.

referring back to the force mentioned at the start of the question F = KPQ/r^2

(forget about the constant here) then we have F is proportional to 1/r^2

so if r is halved then the force will increase by 4 times as (0.5)^2 is 0.25.

hope this helps.
(edited 8 years ago)
Reply 3311
Original post by Kennethm
literally if r is decreased so that the electric potential energy of Q is doubled then that means r has been halved.

here is the thought process:
W = QV and V has been doubled

V = (-)GM/r (dont worry about the minus)

if V is doubled then r must have halved.

referring back to the force mentioned at the start of the question F = KPQ/r^2

(forget about the constant here) then we have F is proportional to 1/r^2

so if r is halved then the force will increase by 4 times as (0.5)^2 is 0.25.

hope this helps.


Hate to be picky - but didn't it mention electrical potential energy, not gravitational?
Original post by CD223
Hate to be picky - but didn't it mention electrical potential energy, not gravitational?


ohhhh thats so embarresing! Same thing nonetheless but ill change it.
Original post by Kennethm
ohhhh thats so embarresing! Same thing nonetheless but ill change it.


not embarassing you were pretty much there.

for potential to increase by 2, radius must have been halved, as v=1/r

Force is proportional to 1/r^2.

F= 1/(1/2)^2. Therefore, 4F.

is this right?
Original post by wilson196
not embarassing you were pretty much there.

for potential to increase by 2, radius must have been halved, as v=1/r

Force is proportional to 1/r^2.

F= 1/(1/2)^2. Therefore, 4F.

is this right?


spot on my friend.
Hey guys,
So question 3 on the AQA June 2013 Physics Unit 4 paper (Section A) is:A gas molecule of mass m moving at velocity u collides at right angles with the side of a container and rebounds elastically. Which one of the following statements concerning the motion of the molecule is incorrect?
A The magnitude of the change in momentum of the molecule is zero.
B The magnitude of the change in momentum of the molecule is 2mu.
C The force exerted by the molecule on the side of the container is equal to the force exerted by the container on the molecule.
D The change in kinetic energy of the molecule is zero.

Whilst the mark scheme gives the answer to be A, the examiner's report claims:
'In the perfectly elastic collision at right angles between a molecule and the side of a container, the velocity of the molecule is exactly reversed. The change in momentum is therefore mu (−mu) = 2 mu. 70% ofthe candidates were aware of this, but almost a quarter of them thought the change in momentumwould be zero (and therefore chose distractor A).'

So there's clearly a discrepancy here. I'm inclined to agree with the examiner's report, but does anyone know the reason behind this discrepancy, and is B the correct answer?
Original post by zaxxo1
Hey guys,
So question 3 on the AQA June 2013 Physics Unit 4 paper (Section A) is:A gas molecule of mass m moving at velocity u collides at right angles with the side of a container and rebounds elastically. Which one of the following statements concerning the motion of the molecule is incorrect?
A The magnitude of the change in momentum of the molecule is zero.
B The magnitude of the change in momentum of the molecule is 2mu.
C The force exerted by the molecule on the side of the container is equal to the force exerted by the container on the molecule.
D The change in kinetic energy of the molecule is zero.

Whilst the mark scheme gives the answer to be A, the examiner's report claims:
'In the perfectly elastic collision at right angles between a molecule and the side of a container, the velocity of the molecule is exactly reversed. The change in momentum is therefore mu (−mu) = 2 mu. 70% ofthe candidates were aware of this, but almost a quarter of them thought the change in momentumwould be zero (and therefore chose distractor A).'


So there's clearly a discrepancy here. I'm inclined to agree with the examiner's report, but does anyone know the reason behind this discrepancy, and is B the correct answer?


I know this question - the answer is out of A and B since they contradict each other. A lot of students would have thought that the momentum to the right would have cancelled out completely with the momentum to the left without fully considering the basic mechanics of the CHANGE in momentum.

read the question again - it asks for the incorrect answer. B is correct so you must put A down as the answer.
(edited 8 years ago)
Original post by zaxxo1
Hey guys,
So question 3 on the AQA June 2013 Physics Unit 4 paper (Section A) is:A gas molecule of mass m moving at velocity u collides at right angles with the side of a container and rebounds elastically. Which one of the following statements concerning the motion of the molecule is incorrect?
A The magnitude of the change in momentum of the molecule is zero.
B The magnitude of the change in momentum of the molecule is 2mu.
C The force exerted by the molecule on the side of the container is equal to the force exerted by the container on the molecule.
D The change in kinetic energy of the molecule is zero.

Whilst the mark scheme gives the answer to be A, the examiner's report claims:
'In the perfectly elastic collision at right angles between a molecule and the side of a container, the velocity of the molecule is exactly reversed. The change in momentum is therefore mu (−mu) = 2 mu. 70% ofthe candidates were aware of this, but almost a quarter of them thought the change in momentumwould be zero (and therefore chose distractor A).'

So there's clearly a discrepancy here. I'm inclined to agree with the examiner's report, but does anyone know the reason behind this discrepancy, and is B the correct answer?


it asks for the incorrect answer, b, c and d are all correct. A is incorrect due to the 2mu calculation disproving this.
Not prepared for this at all :/
Original post by gabriellakhan
Not prepared for this at all :/


I was a few weeks ago. Oddly, the closer I get to the exam, and more work I do, the less prepared I feel

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