Join TSR now and get all your revision questions answeredSign up now

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

    Offline

    2
    ReputationRep:
    (Original post by Alexie56)
    Surely if they were charged with the same voltage the initial current should be the same. They were juat discharged through different resistors. So I drew a shallower graph?😕
    No V=IR so I=V/R is proportional to R since V is constant. R increases so 1/R decreases, hence I is smaller.
    Offline

    3
    ReputationRep:
    (Original post by kevincarreira)
    First one is weight
    June 11 paper
    Name:  ImageUploadedByStudent Room1434020322.932684.jpg
Views: 336
Size:  134.7 KB


    Posted from TSR Mobile
    Offline

    3
    ReputationRep:
    (Original post by Alexie56)
    Surely if they were charged with the same voltage the initial current should be the same. They were juat discharged through different resistors. So I drew a shallower graph?😕
    thought of V=IR, so initial current would be less
    Offline

    2
    ReputationRep:
    (Original post by NEWT0N)
    There were 3? Hm, maybe they would want you to extend the graph to cover the whole length of the x-axis or something. Or they would not want it to intercept the x-axis but only tend asymptotically to it. I don't know
    Yeah I thought so?

    Erm yeah that would be likely, if they have really made a typo they might just omit the question though...

    How was the rest of the paper for you?


    Posted from TSR Mobile
    Offline

    2
    ReputationRep:
    (Original post by Disney0702)
    75
    Thanks, are the grade boundaries usually that low? (I don't check them)
    Offline

    3
    ReputationRep:
    Why did I get around 31nC for the charge?? I found the gradient and multiplied by the constant
    Offline

    1
    ReputationRep:
    (Original post by QueNNch)
    Thanks for putting them!

    I think you question 17 is incorrect because the question asked for the number on electrons on the NEGATIVE plate ONLY. Therefore the answer is half of what you got. (something like 2.2 x 10^10)

    Posted from TSR Mobile
    Q 22 i got 1/root 2
    Offline

    9
    ReputationRep:
    (Original post by Fvthoms)
    55 for A*
    50 for A
    etc...
    Hmmm just out of curiosity why'd ya 55 for A*?

    Last it was 61/75 do you think this exam was significantly harder than last year.

    Coz i dont know if i found harder than last year after reading peoples comments
    Offline

    2
    ReputationRep:
    I thought the same, and also, the capacitance is the same. I started mine at the same current level and ended it above the other graph.

    (Original post by Alexie56)
    Surely if they were charged with the same voltage the initial current should be the same. They were juat discharged through different resistors. So I drew a shallower graph?😕
    Offline

    3
    ReputationRep:
    (Original post by NEWT0N)
    No time period isn't affected because it only depends on the length of the spring and a simple pendulum is isochronus.

    I put the time period stays the same and there is a larger restoring force because the weight is greater at maximum amplitude so the horizontal compoent of weight (restoring force) is larger. But then I went on to say something silly like how this causes a smaller amplitude, but I think it should be a bigger amplitude? My thinking is still a bit cloudy after the exam so I'm hoping someone can clarify this for me!
    I also wrote about a larger restoring force, hence a larger energy loss due to resistive forces, which would result in heavier damping. Don't know if it's correct or not...
    Offline

    1
    ReputationRep:
    (Original post by NEWT0N)
    No V=IR so I=V/R is proportional to R since V is constant. R increases so 1/R decreases, hence I is smaller.
    They were charged through the same circuit in the exact same way though I think so the initial current is the same. When it discharges it goes through a different circuit with the resistor. Not entirely sure though
    Offline

    9
    ReputationRep:
    (Original post by DannySmith420)
    Q 22 i got 1/root 2
    Same!
    Offline

    2
    ReputationRep:
    (Original post by Mehrdad jafari)
    Here are my answers. The ones I'm not sure with i have mentioned next to them.

    I have failed to take the answer of one of the MC questions. Let me know if you found which one.
    I would be very happy if you could correct the ones that are not right
    Attachment 425411


    Posted from TSR Mobile
    1. Is newtons as rate of change of momentum is force (newtons 2nd law)
    21. I put A as it had the longest side perpendicular to the magnetic field. Its a couple and not looking for flux (so largest area is irrelevant) force on couple is F=BIl so biggest couple was A as it had largest length perpendicular to field
    22. I think is 1/root 2 as kinetic energy is proportional to the square of velocity
    Offline

    1
    ReputationRep:
    anyone else get like DDDAAA near the end?
    Offline

    2
    ReputationRep:
    (Original post by Will177)
    They were charged through the same circuit in the exact same way though I think so the initial current is the same. When it discharges it goes through a different circuit with the resistor. Not entirely sure though
    The graph was for discharing, so my analysis applies there

    V is the same as before
    R for the discharing circuit is higher

    Therefore V=IR for the discharing circuit gives I proportional to 1/R, so I for the discharing circuit is smaller as claimed
    Offline

    3
    ReputationRep:
    Past grade boundaries:Name:  1434020555233.jpg
Views: 377
Size:  38.8 KB

    I reckon
    51 for A
    57 for A*
    Offline

    0
    ReputationRep:
    On 2nd law points I talked about how ma is mv^2/r due to v^2/r being the centripetal acceleration
    Offline

    2
    ReputationRep:
    (Original post by NEWT0N)
    It went well but I was disappointed at the sheer number of repeats from previous papers which probably gave people who did lots of past papers (including me) a slight advantage on time taken to complete the paper.
    Yeah the multiple choice certainly felt quite familiar! Ah well long time to wait to find out how we acc did, good luck!


    Posted from TSR Mobile
    Offline

    2
    ReputationRep:
    (Original post by QueNNch)
    Alright guys I drew my graph like this

    Attachment 425409

    I think in probably wrong because the initial current (y axis) would be lower.

    Thoughts?
    Totally wrong as gradient should also be shallower due to higher RC. Don't think they'll include the question though as it was worded incorrectly should have been 300ohm not 300kohm
    Offline

    9
    ReputationRep:
    (Original post by NEWT0N)
    Thanks, are the grade boundaries usually that low? (I don't check them)
    At the beginning of the spec the grade boundaries were pretty low. You could lose 12 marks and get full UMS.

    But last year its gone up a bit.

    It was 61/75 for an A* and 69/75 to get full UMS.
 
 
 
Poll
Which Fantasy Franchise is the best?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.