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

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

    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by _Caz_)
    I think they will be much higher this year (much to my dismay)

    Posted from TSR Mobile
    It's likely they will be. Any 6 marker predictions?


    Posted from TSR Mobile
    Offline

    2
    ReputationRep:
    (Original post by CD223)
    For 17, the EPE is a maximum at P because the two particles are at the smallest distance, d, apart, and:

    

{E_{EP}}=\dfrac{Qq}{4\pi{\epsilo  n_0}d^{2}}

    The KE reduces to a minimum at P because all initial KE has transferred to EPE.

    The total kinetic energy and total potential energy constantly changes, BUT the total remains unchanged.

    The answer is C.

    For 24,

    

E=\dfrac{N\phi}{T}



\Rightarrow \phi = \dfrac{Et}{N}

    The answer is A.



    Posted from TSR Mobile
    Looks like I completely over thought 24 then. Thanks once again
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by JJBinn)
    Looks like I completely over thought 24 then. Thanks once again
    Haha no worries! It's a horrid way to phrase the question


    Posted from TSR Mobile
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by JJBinn)
    Looks like I completely over thought 24 then. Thanks once again
    Can I just clarify a mistake I made earlier?

    

{E_{EP}}\not=\dfrac{Qq}{4\pi{\ep  silon_0}d^{2}}

    

{E_{EP}}=\dfrac{Qq}{4\pi{\epsilo  n_0}d}


    Posted from TSR Mobile
    Offline

    2
    ReputationRep:
    (Original post by CD223)
    Can I just clarify a mistake I made earlier?

    

{E_{EP}}\not=\dfrac{Qq}{4\pi{\ep  silon_0}d^{2}}

    

{E_{EP}}=\dfrac{Qq}{4\pi{\epsilo  n_0}d}


    Posted from TSR Mobile
    Yeah it's cool, I only looked at the d^2 bit as that gave the answer
    Offline

    2
    ReputationRep:
    Just did the June 10 MC, I don't understand 13, 14 or number 4. On 13 I keep getting 40mm and I've no idea what I'm doing wrong. Please could somebody explain?

    http://filestore.aqa.org.uk/subjects...W-QP-JUN10.PDF
    • Thread Starter
    Offline

    3
    ReputationRep:
    Seeing as:

    V=-\dfrac{GM}{r}

    For gravitational fields, where V is always negative, how does one strictly describe the potential as a test mass moves away from the mass whose potential is being described?

    Does it "increase" or "decrease"? Because numerically it becomes a bigger number (less negative), but I've seen questions describe this as a "decrease" because the magnitude of V has decreased.


    Posted from TSR Mobile
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by JJBinn)
    Just did the June 10 MC, I don't understand 13, 14 or number 4. On 13 I keep getting 40mm and I've no idea what I'm doing wrong. Please could somebody explain?

    http://filestore.aqa.org.uk/subjects...W-QP-JUN10.PDF
    For 4, as the experiment with the rubber bung described here (http://www.nuffieldfoundation.org/pr...ircular-motion) describes, when the string breaks, it is under the influence of gravity and moves as a projectile in the vertical plane. It flies off at a tangent and falls due to the acceleration due to gravity.

    The answer is D.

    For 13 and 14, see my attached workings:
    The answer is A for both.
    Name:  ImageUploadedByStudent Room1431792346.104166.jpg
Views: 106
Size:  138.5 KB


    Posted from TSR Mobile
    Offline

    3
    ReputationRep:
    (Original post by CD223)
    Seeing as:

    V=-\dfrac{GM}{r}

    For gravitational fields, where V is always negative, how does one strictly describe the potential as a test mass moves away from the mass whose potential is being described?

    Does it "increase" or "decrease"? Because numerically it becomes a bigger number (less negative), but I've seen questions describe this as a "decrease" because the magnitude of V has decreased.


    Posted from TSR Mobile
    I always get confused on this too, it doesn't help that the mark schemes mix it up also...
    Offline

    3
    ReputationRep:
    (Original post by CD223)
    Seeing as:

    V=-\dfrac{GM}{r}

    For gravitational fields, where V is always negative, how does one strictly describe the potential as a test mass moves away from the mass whose potential is being described?

    Does it "increase" or "decrease"? Because numerically it becomes a bigger number (less negative), but I've seen questions describe this as a "decrease" because the magnitude of V has decreased.


    Posted from TSR Mobile
    The magnitude of V decreases as distance from the reference mass increases. For example, calculating V at a certain point from the reference mass gives us the energy required to move a mass per unit kg from 'that point' to infinity, that is where there is no field. But because we need a reference position to measure V, infinity is selected as our point of reference. Thats why the potential at a point in space is defined with respect to infinity, that is the energy required per unit mass to be moved from infinity to 'that point' (which is essentially the same argument as above).

    Does this answer your question?


    Posted from TSR Mobile
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by ubisoft)
    I always get confused on this too, it doesn't help that the mark schemes mix it up also...
    Lol have you seen Q4 on June 2012? It's so annoying.


    Posted from TSR Mobile
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by Mehrdad jafari)
    The magnitude of V decreases as distance from the reference mass increases. For example, calculating V at a certain point from the reference mass gives us the energy required to move a mass per unit kg from 'that point' to infinity, that is where there is no field. But because we need a reference position to measure V, infinity is selected as our point of reference. Thats why the potential at a point in space is defined with respect to infinity, that is the energy required per unit mass to be moved from infinity to 'that point' (which is essentially the same argument as above).

    Does this answer your question?


    Posted from TSR Mobile
    Thank you. I just don't like how inconsistent questions are with regards to an "increase" in potential.

    In my eyes, that could mean the number getting more negative (it's magnitude increases), or it could mean it gets less negative, as it increases numerically. Ugh.


    Posted from TSR Mobile
    Offline

    3
    ReputationRep:
    (Original post by CD223)
    Lol have you seen Q4 on June 2012? It's so annoying.


    Posted from TSR Mobile
    Nope not yet, I'm saving 2012 and 2013 for the week before the exam. What's more common in the MS, using the magnitude or the real value?
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by ubisoft)
    Nope not yet, I'm saving 2012 and 2013 for the week before the exam. What's more common in the MS, using the magnitude or the real value?
    It varies I think the real value is most common though.

    I also dislike questions like this that ask for these graphs. Name:  ImageUploadedByStudent Room1431794148.240568.jpg
Views: 70
Size:  46.7 KB


    Posted from TSR Mobile
    Offline

    3
    ReputationRep:
    (Original post by CD223)
    Thank you. I just don't like how inconsistent questions are with regards to an "increase" in potential.

    In my eyes, that could mean the number getting more negative (it's magnitude increases), or it could mean it gets less negative, as it increases numerically. Ugh.


    Posted from TSR Mobile
    No worries.
    True, but sometimes it's better not to think with equations and it's better to consider the consequences logically.
    So in the case of potential you know that more energy is required to move a mass from the surface of a planet to infinity(which is the same as the energy required to move a mass from infinity to that point), than from a distance from the surface of the planet to infinity, which means that the potential decreases as the distance increases. The negative sign is only an indication of the energy required and doesn't interfere with the value of the potential inself.


    Posted from TSR Mobile
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by Mehrdad jafari)
    No worries.
    True, but sometimes it's better not to think with equations and it's better to consider the consequences logically.
    So in the case of potential you know that more energy is required to move a mass from the surface of a planet to infinity(which is the same as the energy required to move a mass from infinity to that point), than from a distance from the surface of the planet to infinity, which means that the potential decreases as the distance increases. The negative sign is only an indication of the energy required and doesn't interfere with the value of the potential inself.


    Posted from TSR Mobile
    Oh right, so just accept that potential decreases as distance from a mass M increases, because less work is required at further distances to move a test mass from infinity to that point?


    Posted from TSR Mobile
    Offline

    3
    ReputationRep:
    (Original post by CD223)
    Oh right, so just accept that potential decreases as distance from a mass M increases, because less work is required at further distances to move a test mass from infinity to that point?


    Posted from TSR Mobile
    That is it


    Posted from TSR Mobile
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by Mehrdad jafari)
    That is it


    Posted from TSR Mobile
    Thank you

    Gonna call it a day for PHYA4 today! Now onto June 2012 PHYA5 lol!
    Offline

    3
    ReputationRep:
    (Original post by CD223)
    Thank you

    Gonna call it a day for PHYA4 today! Now onto June 2012 PHYA5 lol!
    No problem
    Have you guys finished the physics syllabus yet?


    Posted from TSR Mobile
    Offline

    3
    ReputationRep:
    (Original post by CD223)
    It varies I think the real value is most common though.

    I also dislike questions like this that ask for these graphs. Name:  ImageUploadedByStudent Room1431794148.240568.jpg
Views: 70
Size:  46.7 KB


    Posted from TSR Mobile
    Yeah I think I didn't get that when I first read it, doesn't that graph show there's no potential below the surface, but there should be as it pulls towards the centre?
 
 
 
Poll
Which pet 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.