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

I think they will be much higher this year (much to my dismay)

Posted from TSR Mobile

For 17, the EPE is a maximum at P because the two particles are at the smallest distance, d, apart, and:

$[br]{E_{EP}}=\dfrac{Qq}{4\pi{\epsilon_0}d^{2}}[br]$

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,

$[br]E=\dfrac{N\phi}{T}[br][br]\Rightarrow \phi = \dfrac{Et}{N}[br]$

The answer is A.



Posted from TSR Mobile

Looks like I completely over thought 24 then. Thanks once again

Looks like I completely over thought 24 then. Thanks once again

Can I just clarify a mistake I made earlier?

$[br]{E_{EP}}\not=\dfrac{Qq}{4\pi{\epsilon_0}d^{2}}[br]$

$[br]{E_{EP}}=\dfrac{Qq}{4\pi{\epsilon_0}d}[br]$


Posted from TSR Mobile

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/AQA-PHYA4-1-W-QP-JUN10.PDF

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

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...

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

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?

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

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

Thank you

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

It varies I think the real value is most common though.

I also dislike questions like this that ask for these graphs.


Posted from TSR Mobile