The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

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

Scroll to see replies

for the emf question in section B was it just change in flux linkage over the time or was it rotating contiously? cant remember the question
Original post by Mehrdad jafari
Guys, let me know if you got the same answers

ImageUploadedByStudent Room1434022386.412958.jpg




Q.22 is 1/√2
I got 21 wrong (I put A)

All other answers SAME! Weeeeeeee :biggrin:
(edited 8 years ago)
Original post by QueNNch
Q.22 is 1/√2
All other answers SAME! Weeeeeeee :biggrin:

Posted from TSR Mobile


Yeah, i think i got that wrong.


Posted from TSR Mobile
On the 6 marker, I said that the string may not remain horizontal because of the frictional force of air which acts against the mass, lowering its Ke and velocity. Is that even remotely close to any mark
Original post by Alexie56
The only thing that changed was an increase in resistance. They didn't reverse the direction of the current. So why would a field be produced in the opposite direction?


I thought this, so I said that the needle lowers i.e. a lower value of current
Original post by Amanzz
Think this is incorrect. The time period of a pendulum is independent of a mass added, hence the maximum velocity is independent of the mass added. Furthermore, the only resistive force is due to the viscosity of air, an air resistive force. The ring hardly affects air resistance, and so the since the vicious force of air is dependant on velocity, which is unchanged, then the damping/viscous force is unchanged.


Posted from TSR Mobile


Definitely not incorrect, I got it incorrect I said no damping. Change in Damping is not how to air resistance varies. It his how the air resistance varies in relation to change in momentum. If you don't believe me go and do an experiment, damping has no effect on time period in the first place.
Original post by AR_95
I mentioned the velocity is constantly changing because of the centrapetal force


I did the same.

Original post by NEWT0N
Yes because a force is impressed on it. When there is no longer a force it is constant throughout its motion from release and follows a parabolic curve. The horizontal velocity now remains the same as the tangential velocity at the point of release, in accordance with Newton I.


So just to confirm; by stating the *ball it's constantly accelerating, you confirm there must be a resultant force acting on the ball while it is traveling in a circle (law 1); and relating this to F=ma is law 2?
Original post by rohan9777
for the emf question in section B was it just change in flux linkage over the time or was it rotating contiously? cant remember the question


You're correct, change in flux linkage over time, Faraday's law of induction.


Posted from TSR Mobile
Original post by AR_95
On the 6 marker, I said that the string may not remain horizontal because of the frictional force of air which acts against the mass, lowering its Ke and velocity. Is that even remotely close to any mark


I said the compinent of the weight (mg) wasnt taken into account, this component acts downwards and would bring it out of the horizontal
NEED to get an A in physics for any uni and may have just ended it today :/.
Section A I didn't think was too bad, rushed but hopefully mostly ok.
Section B though.. omg.. it was all stuff I hate and wordy. Even on the numbers that I'm good at I rushed and got a lot of it wrong.
Then the capacitor question I got completely confused by due to the 300K ohm resistor, didnt think they would want me to draw a line on the bottom of the graph so started it at the same current and made the gradient shallower with the same shape (i know I should have started below the initial current).
No gravitational questions either to get my mark back up :frown:.
Original post by rohan9777
for the emf question in section B was it just change in flux linkage over the time or was it rotating contiously? cant remember the question


It was rotating so you had to use BANwSinwt then divide by root2
Original post by Amanzz
You're correct, change in flux linkage over time, Faraday's law of induction.


Posted from TSR Mobile


I got something like 0.12?
Original post by NEWT0N
I remember getting G*m*pi*p*R/3 somewhere for the multiple choice. Anyone agree?


Yes!
Section B Q2

What did you all say for why V is negative on the graph?
Original post by NEWT0N
Why were there 50 answer boxes for the multiple choice?


I think that is a general sheet for AQA exams and not specifically for phys4 :smile:


Posted from TSR Mobile
Well I finally got home! Seemed an average paper, pretty sure I've done well enough for an A* but won't know for sure until August. I'd expect pretty normal grade boundaries, something along the lines of the last few years.
Reply 4296
Avatar for k9000
k9000
2
Original post by Disney0702
Section B Q2

What did you all say for why V is negative on the graph?


The charge was negative and so negative potentials are associated with a negative charge
Original post by NEWT0N
Why were there 50 answer boxes for the multiple choice?


There was another MC sheet, didn't you do it?
Original post by Disney0702
Section B Q2

What did you all say for why V is negative on the graph?


Q is negatively charged so work is done by the positive test charge
Do you think it's worth writing a complaint to AQA about the obvious typo in the capacitors question? Just in case it hasn't come to their attention that their proof-reading process is awful...

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you