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

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    (Original post by gcsestuff)
    I agree, some questions I just look at and I have no idea how to start them, so they take a bit more thinking about. With multiple choice least you always know where to start and you can always get rid of a few answers


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    Exactly. Looking at the answers is so much easier than trying to come up with them haha.


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    (Original post by CD223)
    Well here are the boundaries:
    Attachment 414267

    Which suggest that's false, given the first section is 25 marks.


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    http://filestore.aqa.org.uk/over/sta...-JUNE-2014.PDF

    Any idea what those second grey grade boundaries suggest?
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    (Original post by muyiwaaiyenuro)
    http://filestore.aqa.org.uk/over/sta...-JUNE-2014.PDF

    Any idea what those second grey grade boundaries suggest?
    They're the individual sections' breakdown I guess. Even though your grade is actually worked out by the sum total of your marks.


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    Can someone help me with question 13 please?http://filestore.aqa.org.uk/subjects...1-QP-JUN14.PDF
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    (Original post by Adangu)
    Can someone help me with question 13 please?http://filestore.aqa.org.uk/subjects...1-QP-JUN14.PDF
    

\displaystyle F = \dfrac{Qq}{4 \pi {\epsilon_0} r^2}

    This means
    \displaystyle F \propto \dfrac{1}{r^2}.

    The original distance was d when the force was F.

    The force is halved to become 0.5 F, so the previous distance must be 20mm \div \sqrt{2} = 14mm.

    Does that help?


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    (Original post by CD223)
    

\displaystyle F = \dfrac{Qq}{4 \pi {\epsilon_0} r^2}

    This means
    \displaystyle F \propto \dfrac{1}{r^2}.

    The original distance was d when the force was F.

    The force is halved to become 0.5 F, so the previous distance must be 20mm \div \sqrt{2} = 14mm.

    Does that help?


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    Sorry to bother you more, but where does the root 2 come from?
    My brain doesn't seem to be working with me haha!
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    (Original post by Adangu)
    Sorry to bother you more, but where does the root 2 come from?
    My brain doesn't seem to be working with me haha!
    As you have:
    

\dfrac{1}{2}F

    On one side of the equation, you must have

    

\dfrac{1}{\left(\sqrt{2} d \right)^2}

    On the other side - does that make sense?


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    How would you go about mathematically proving this?

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    I got it right by guessing, because:

    

{m_{before}} = 200g

{m_{after}} = 200g + 300g = 500g

    

KE = \dfrac{1}{2}mv^2

    

\therefore \dfrac{{m_{after}}}{{m_{before}}  } = \dfrac{500}{200}

    

\therefore \dfrac{{KE_{after}}}{{KE_{before  }}} = \dfrac{5}{2}

    But this gives the inverse of the correct answer?

    I know KE depends on m as well as v, but I just didn't think enough info was given in the question about the velocity? It just says the initial velocity is U, then it coalesces with a stationary object, without saying the velocity after?

    Confused.
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    EDIT: Never mind! All sorted

    The frequency has to reduce by two thirds so
     f \rightarrow \frac{2}{3}f

    So the period would have to increase by 1.5, so
     T \rightarrow \frac{3}{2}T

    As
     T = 2 \pi \sqrt{\dfrac{m}{k}}

    This means m would have to increase by 1.5 squared so
     m \rightarrow 2.25m

     \therefore 0.40kg \rightarrow 0.90kg

    meaning 0.50kg needs to be added.
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    (Original post by Mehrdad jafari)
    Use this equation as this is quicker and less stressful in the exam
    Attachment 414673


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    Edit: oops, in the first equation its K/m and not m/k
    Aha yeah I was gonna ask about the k/m!

    Thanks

    Do you get the KE question?
    I just didn't think there was enough info on the velocity


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    http://filestore.aqa.org.uk/subjects...W-QP-JAN10.PDF Question 11 Lads, I can't seem to get my head around where to start off

    EDIT: No problem just got it by forming an equation of the earths GFS and another for the planets GFS, rearranging both so M = blah blah and dividing the pair to get X^2Y. Just wondering if there would be any quicker method or realizing some sort of ratio without having to do all of that?
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    (Original post by Mehrdad jafari)
    In case if you haven't got it

    Attachment 414675


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    Great! Thank you really should have got that one >.<


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    (Original post by lebanon95)
    http://filestore.aqa.org.uk/subjects...W-QP-JAN10.PDF Question 11 Lads, I can't seem to get my head around where to start off

    EDIT: No problem just got it by forming an equation of the earths GFS and another for the planets GFS, rearranging both so M = blah blah and dividing the pair to get X^2Y. Just wondering if there would be any quicker method or realizing some sort of ratio without having to do all of that?
    Unfortunately that method is the quickest way I know of. Form an equation for M:

    

M = \dfrac{gr^2}{G}

    Then sub in the values x and y, and divide.


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    (Original post by CD223)
    As you have:
    

\dfrac{1}{2}F

    On one side of the equation, you must have

    

\dfrac{1}{\left(\sqrt{2} d \right)^2}

    On the other side - does that make sense?


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    Sorry it still doesn't, I don't see where the root 2 comes from.
    I don't know how I don't get it!
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    (Original post by Adangu)
    Sorry it still doesn't, I don't see where the root 2 comes from.
    I don't know how I don't get it!
    F is is inversely proportional to the distance, d, squared. This means that, if you look at it another way:

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    Not sure on why the angular velocity is the same. Using w=v/r means that if the radius is smaller the speed should be greater??

    Or am I supposed to be using the g=f/m. Which shows acceleration only depends on force and mass of the object.

    Thanks again 😀😀




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    (Original post by CD223)
    F is is inversely proportional to the distance, d, squared. This means that, if you look at it another way:

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    Thank you so much! Seriously thank you!

    Looking at it like that I understand it!
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    (Original post by gcsestuff)
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    Not sure on why the angular velocity is the same. Using w=v/r means that if the radius is smaller the speed should be greater??

    Or am I supposed to be using the g=f/m. Which shows acceleration only depends on force and mass of the object.

    Thanks again 😀😀

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    Both points X and Y turn through the same angle in the same time period. (360 degrees or 2 \pi radians in 24 hours).

    As:
    

\omega = \dfrac{\Delta \theta}{\Delta t}

    This means that both points undergo the same angular displacement, theta, in the same time. They therefore have the same angular velocity.

    NB: They will have different linear velocities, however.

    With regards to the potential of both points,

    

{V_X} &gt; {V_Y}

    because
     

V = (-) \dfrac{GM}{r}

    And {r_X} &gt; {r_Y}.

    Does that make sense?




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    (Original post by Adangu)
    Thank you so much! Seriously thank you!

    Looking at it like that I understand it!
    No worries gotta love a confusing ratio question from AQA -.-


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    (Original post by CD223)
    No worries gotta love a confusing ratio question from AQA -.-


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    AQA are just the best (¬_¬)
 
 
 
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