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

No problem! This is exactly how I would do it in an exam because it's the first approach that came to mind and takes at most a few seconds to carry out if you know what you're doing, but CD's one is definitely nicer. (What I did would best be described as a "bash".)

Yes it is June 2014, has some pretty tough multi choice questions in there. Thanks for that makes much more sense.


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Lol mine was more logic than maths (I chose the dumb way ). I tend to prefer logical thinking but I do appreciate its impressive if you can manage the maths. I certainly couldn't under pressure haha.


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Nah, mathematicians would definitely prefer your method as being more "elegant". (Side note: That was nothing compared to STEP, where you have to crunch through much harder calculations without a calculator.)

That's true. For me studying it makes unit 4 acceptable.
I've heard the astro option is the hardest alongside medical one. Though i found the waves chapter a bit abstract and hard to tolerate


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It's the hardest Capacitance question I've seen. June 2014, right?

Essentially, as $E \propto V^2$, this means that, if we set $E = 1600\mu J$, then when V has decreased by 2.0V, E has become $\frac{E}{4} = 400\mu J$.

As E has decreased by a factor of 4 and $E \propto V^2$, V must have decreased by a factor of 2 $\left(\sqrt{\frac{V^2}{4}}\right)$.

As V has halved by decreasing by 2.0V, the initial pd must have been 4.0V.

This follows that, as $E = \frac{1}{2}CV^2$,

$C = \frac{2 \times 1600 \times 10^{-6}}{4^2}$

$[br]\therefore C = 200 \times 10^{-6}F[br]$


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Thats weird, I think astro is the easiest option. Much easier than turning points definitely. Its just a load of new stuff to learn, thats all. Not to mention its quite an interesting topic (stars and all) :P


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Alternative (and longer!)

We are given information about V and we need to find C, so the equation to use is $E=\frac{1}{2}CV^2$. Initially, $E_0=\frac{1}{2}CV^2=1600\mu$ implying that $CV^2=3200\mu$. Then the voltage decreases by 2V and the energy stored becomes $400\mu$, which we can express as the equation $E_1=\frac{1}{2}C(V-2)^2=400\mu$, or $C(V-2)^2=800\mu$.

Now square rooting both sides of each equation we obtain $V\sqrt{C}=\sqrt{3200\mu}$ and $(V-2)\sqrt{C}=\sqrt{800\mu}$. Thus $V=\frac{\sqrt{3200\mu}}{\sqrt{C}}=\frac{\sqrt{800\mu}}{\sqrt{C}}+2$, or equivalently $\sqrt{3200\mu}-\sqrt{800\mu}=2\sqrt{C}$. Now we just square both sides and divide by 4 to get $C=\frac{(\sqrt{3200\mu}-\sqrt{800\mu})^2}{4}=200\mu F$

Hmmmmmmmmmmmmmmmmmmmm. That sounds like essentially what I've done. Here's what I did.

It seems we won't see the EMPA mark scheme until 10 says after the actual exam
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Oh no, is that 10 days after PHYA4?



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I'm still not sure if that's the case because the exam papers taken in the last 10 days have been released



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It's just it's been more than 10 days (11 days actually) since the last date the EMPA was taken on?


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https://60abffc9b401b1c0936e01291c15112cab0778c4.googledrive.com/host/0B1ZiqBksUHNYVFI1YlJBOWhwRW8/June%202010%20QP%20-%20Unit%204A%20AQA%20Physics.pdf

Can anyone help with question 14? thanks a lot

Yeah, true, I really don't know what's happening. If the files are released electronically then we would have the papers by now which I don't think is the case. And if the files are released manually then they will be released by the end of today or maximum tomorrow

https://60abffc9b401b1c0936e01291c15112cab0778c4.googledrive.com/host/0B1ZiqBksUHNYVFI1YlJBOWhwRW8/June%202010%20QP%20-%20Unit%204A%20AQA%20Physics.pdf

Can anyone help with question 14? thanks a lot

Hope so! Wanna see how badly I did 😖
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can anyone help with jan 2010 - q21?? I keep getting C ;(

Thanks