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

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    (Original post by Mehrdad jafari)
    Yeah sorry i meant starts at the top my mind is switched off lol.
    Yeah you should be right


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    Thanks, that's a relief.

    See my last post: How many times do you do each past paper? (And how do you get good at 6 markers... pls.)
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    (Original post by Paul Dirac)
    How many times do you guys do each past paper?
    Spoiler:
    Show
    Edit: Also why am I so bad at 6 mark questions? Do people just memorize the mark schemes for those? Because it seems everyone is better at it than I am (and I got A*A* in English GCSE )
    I usually do each past paper once though it might be useful if you think doing them once again will help you.
    I don't really have insight into six markers but i will start doing them from tomorrow


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    (Original post by JJBinn)
    Oh man I was setting M to zero in that equation because I was so fixated on the fact it was weightless.

    On a side note, for this equation, is the mass representative of both objects? Ie gravitational attraction occurs between 2 masses, do you add the masses together for example if it was a satellite orbiting the earth, in using that equation do you add the mass of the satellite to the earth's mass?
    I see. The mass M is representative of the mass of the object creating the field. In this case m is around 5.9\times10^{24} as that is the mass of the earth only. The satellite is the small test mass orbiting the mass creating the field. In equations, capital M is the mass creating the field, lower case m is the small test mass (eg: satellite) in that field.

    (Original post by Mehrdad jafari)
    me too. To be honest i cannot remember studying unit 5 lool.
    In theory definitely
    Hahaha. Need to start revising it properly for sure. Thanks for that - really hoping i didn't do badly

    (Original post by Kiki05)
    Ay yo
    Any1 out there plz help me with this
    Q 14 June 10

    I dont know y the answer is A
    can some1 plz explain y its A, it would b much appreciated

    Thanks
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    (Original post by Paul Dirac)
    Thanks

    For June 2014 Section B question 3bii would this be correct? (see attached)

    Attachment 406017
    It would be correct. They're usually quite vague when it comes to diagram mark schemes but as long as the general shape is there (positive or negative) they usually give the mark.



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    Cn anyone help me in these momentum questions. For the hose pipe one I have found the mass by using the density equation but haven't idea on how to get velocity from the given information. For the truck momentum one, seem really easy but can't seem to get it I know the initial momentum is 20,000 now what? Thanks


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    (Original post by Mehrdad jafari)
    For the first question i would suggest you to work out with units to figure out the velocity because that can be something of a challenge if you don't.
    Attachment 406515


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    Oh how did I not get that momentum question, thinking about it they were pretty easy. Thanks, the units really helped on the hosepipe question quite a lot.


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    (Original post by Jimmy20002012)
    Oh how did I not get that momentum question, thinking about it they were pretty easy. Thanks, the units really helped on the hosepipe question quite a lot.


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    It's cool . Yeah, the wording can catch you out as well


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    looks simple but i cant get to the answer
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    (Original post by Jimmy20002012)
    Oh how did I not get that momentum question, thinking about it they were pretty easy. Thanks, the units really helped on the hosepipe question quite a lot.


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    I remember doing those questions thinking it was odd - that hose pipe question has appeared twice now on papers. The exact same question and answer!


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    and this one
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    (Original post by CD223)
    I remember doing those questions thinking it was odd - that hose pipe question has appeared twice now on papers. The exact same question and answer!


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    Yeah I know think they rebank at least 3 questions each year. I have a feeling that they could do put that in the written section as well. As in June 2013 they put the phase difference question of two pendulums in section B and it appeared in section A the previous year.


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    (Original post by AR_95)


    looks simple but i cant get to the answer
    

v=\omega r

    

v = \dfrac{108 \times 10^{3}}{3600}

    

\omega = \dfrac{v}{r}

    

\therefore \omega = \dfrac{30}{0.2}

    

\therefore \omega = 150\ rads^{-1}


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    What I don't get is why is the 108000 divided by 3600 and not multiplied. Or is it different in this case because the units stated ^-1?
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    (Original post by AR_95)
    What I don't get is why is the 108000 divided by 3600 and not multiplied. Or is it different in this case because the units stated ^-1?
    It's because 108 kilometres per hour is equal to 108,000m in 3600 seconds. Dividing one by the other gives a speed in metres per second.


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    (Original post by CD223)
    It's because 108 kilometres per hour is equal to 108,000m in 3600 seconds. Dividing one by the other gives a speed in metres per second.


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    lmao that's just silly from me. Forgot the meaning that it was km PER HOÚR not km x hoúr in seconds haha
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    (Original post by AR_95)


    and this one
    The answer is D. On the right I've explained what the formula breaks down into.

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    (Original post by AR_95)
    lmao that's just silly from me. Forgot the meaning that it was km PER HOÚR not km x hoúr in seconds haha
    No problem! The best of us make that mistake a thousand times over. Comes back to the classic "read the question" mistake we all make


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    would anyone mind helping me with Q2biii in June 2012? I dont understand why the proton's acceleration increases whereas the electron's decreases?
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    Done for the day. Hoping someone can help me with these two




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    (Original post by frankiejayx)
    would anyone mind helping me with Q2biii in June 2012? I dont understand why the proton's acceleration increases whereas the electron's decreases?
    The direction of the magnetic field lines is the direction a free positive charge would move along.

    Hence, they point towards a negative charge in this case.

    The proton accelerates to the right as it is attracted to the negative charge. It's acceleration increases as the distance between it and the negative charge decreases, and due to the electric field strength equalling:

    

\dfrac{Qq}{4\pi {\epsilon_0} r^2}

    This means that the proton's acceleration increases as it gets closer to the negative charge.

    Conversely, the electron accelerates to the left as it is repelled by the negative charge. It's acceleration decreases as the distance between it and the negative charge increases, and due to the electric field strength equalling:

    

\dfrac{Qq}{4\pi {\epsilon_0} r^2}

    This means that the electron's acceleration decreases as it gets further away from the negative charge.
 
 
 
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