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

Original post by Lau14

Confuse the examiners by answering 26 questions :P

Confuse the examiner by answering 42 questions, because you always wonder about the meaning of life :wink:

Reply 1881

Absent Agent

Original post by CD223

Confuse the examiner by answering 42 questions, because you always wonder about the meaning of life :wink:

What are you guys talking about? Are there more questions on the exam papers I didn't know?? :smile:

Reply 1882

Original post by Mehrdad jafari

What are you guys talking about? Are there more questions on the exam papers I didn't know?? :smile:

http://www.thestudentroom.co.uk/showthread.php?p=56207383#post56207383

Aha

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Reply 1883

AR_95

Original post by Mehrdad jafari

We both share the same misfortune lol. I'm doing that option too. Have done any question on that unit? I've heard the questions on special relativity are a bit vague

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I haven't done any questions other than the ones on the discovery of the electron. I've done a few practices on special relativity before and a tiny bit today. Hopefully I should have all of turning points practised with all 5 papers tomorrow. I think because they missed out Milikans experiment that it should be on it this year.

Tbh I find turning points quite interesting. Looking back, it seems like the easiest option out of the 4 but I haven't really looked at the other 3. Does seem like the medical physics one can be a pain

Reply 1884

Absent Agent

Original post by AR_95

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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Reply 1885

Lau14

Original post by CD223

Confuse the examiner by answering 42 questions, because you always wonder about the meaning of life :wink:

I wonder if any of them would notice!

Reply 1886

Original post by Lau14

I wonder if any of them would notice!

Probably if they had 700 other papers to mark which answered 25 :wink:

Reply 1887

AR_95

Original post by Mehrdad jafari

waves as in wave particle duality? Definitely finding it the most boring. Only thing I can get my head around easily is the photoelectric effect.

Reply 1888

Original post by Lau14

I wonder if any of them would notice!

Don't think they even look at them, they just shove it under an OMR machine I think

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Reply 1889

Absent Agent

Original post by AR_95

waves as in wave particle duality? Definitely finding it the most boring. Only thing I can get my head around easily is the photoelectric effect.

Yeah, photoelectric is cool. I cannot make sense of the duality nor did I at AS. And the electron microscopes at the end of the chapter are just horrible. I guess we are far from the truth in waves business lol

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(edited 8 years ago)

Reply 1890

Lau14

Original post by CD223

Probably if they had 700 other papers to mark which answered 25 :wink:

True, it would stand out then :tongue:

Reply 1891

PotterPhysics

Original post by CD223

Good idea! How many papers have you done? All of them? I feel unprepared tbh

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Yes, once each I think, though I'm sure I've left out a few since I tend to skip around and not work through them in linear order (june12 --> jan13, etc). I only started fairly recently because I tend to forget everything if I start too early, but I'm planning to work through them at least once more (twice would be ideal) before the exam. If you feel unprepared I suggest locating your weaker topics and working through the topic-specific papers here: http://www.physicsandmathstutor.com/physics-revision/aqa-unit-4/ This will really give you a boost for a given topic.

Reply 1892

Original post by PotterPhysics

Same here! Thanks for that.

I hope I can get through them all again once. Not sure I will :s-smilie:

Reply 1893

Can someone help me on this question. Nor lamely really good at capacitance but this one has thrown me, I know it has something to do with E is proportional with V^2 but cant seem to figure it out for some reason. Do you have to set up two simultaneous equations one with V^2 and one with (V-2)^2. Have no clue. Thanks

ImageUploadedByStudent Room1432580585.393480.jpg

ImageUploadedByStudent Room1432580585.393480.jpg

Reply 1894

huniibehi

Original post by Jimmy20002012

That's a difficult one burdittphysics on YouTube does great examples

Reply 1895

Original post by Jimmy20002012

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]

Reply 1896

PotterPhysics

Original post by Jimmy20002012

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

Reply 1897

Original post by CD223

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]

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

Reply 1898