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AQA Physics PHYA4 - 20th June 2016 [Exam Discussion Thread]

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Original post by kingaaran
A positive ion will have weight going down and so if it is in equilibrium the electric force on it must be acting upwards.

The direction of the electric force is the direction a positive particle would go and thus it is upwards


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So to hold an electron in equilibrium, field would be downwards?


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Original post by Yo12345
The one with the thumb being the current and the fingers being the field direction?


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No, check page 122 of the textbook, the dynamo rule.
Original post by lucabrasi98
1. It's not about what you get the marks for. I still got the question right, I just hate seeing them write 2.6 as and an answer when you the accurate solution is 2.666. That was obviously an accident by them. They pull stuff like that every year.

2. Honestly I didn't notice the atomic mass unit. But I still fail to see why they're using that when an alpha particle is just 2 protons and 2 neutrons. What would I do in an exam then? And what's the point of specifying proton and neutron masses if sometimes they just use atomic mass instead?

3. The whole issue with whether to use rounded answers for the next question or accurate answers still gets on my nerves. I wouldn't mind if they just settled on one way but they genuinely change their minds as they see fit. Which is a problem as sometimes using the one they don't want gets you "incorrect" rounding. E.g them wanting 2700 but you getting 2763 and rounding to 2800 because you either used the accurate or rounded answer from the last question and they wanted you to use the opposite


1. 2.666 is not accurate because your speed is only to two significant figures, meaning any other significant figures after the second you cannot be certain about.

2. The atomic mass unit is defined such that both the proton and neutron have mass 1amu. You can use the separate ones for accuracy in greater significant figures or if you only want to know the mass of some protons, but as far as the required accuracy of the questions go, amu is fine to use.

3. Well if you use an answer correct to two significant figures and they use the full answer correct to 199283737126271891 significant figures, you will both still be correct to two significant figures. So it doesn't matter. The best bet obviously is to use the most accurate one, because you never can be wrong.

Don't look too much into this :tongue:


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Original post by Yo12345
So to hold an electron in equilibrium, field would be downwards?


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Yes, because a downwards field will create an upwards force an electron - which is exactly what you need to balance it in equilibrium!


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Reply 564
Can someone help me with question 19 on june 2014, https://60abffc9b401b1c0936e01291c15112cab0778c4.googledrive.com/host/0B1ZiqBksUHNYVFI1YlJBOWhwRW8/June%202014%20QP%20-%20Unit%204A%20AQA%20Physics.pdf
I dont understand how you are meant to get 200, I got 600 when I did:

c = Q/V
1600/x = 400/(x-2)
1600x - 3200 = 400x
1200x = 3200
x = 8/3
c = 1600/(8/3)
c=600
Hi all,


I posted about having some A2 paper packs I created myself before




Here is another one ... this one is multiple choice and I go through in the mark scheme I created how you should be thinking through each question.

Sorry there is only 7 questions in this one, but the mark scheme is quite good and took a bit longer to make than the paper did.


Again if you would like more please email me at; [email protected]


Cheers,


Will
Original post by cmrol
Can someone help me with question 19 on june 2014, https://60abffc9b401b1c0936e01291c15112cab0778c4.googledrive.com/host/0B1ZiqBksUHNYVFI1YlJBOWhwRW8/June%202014%20QP%20-%20Unit%204A%20AQA%20Physics.pdfI dont understand how you are meant to get 200, I got 600 when I did:c = Q/V1600/x = 400/(x-2)1600x - 3200 = 400x1200x = 3200x = 8/3c = 1600/(8/3)c=600


using E = (1/2)cV2

(1) 1600 = (1/2)cV2
(2) 400 = (1/2)c(V-2)2

4 = [cV2]/[c(V-2)2]
4(V-2)2 = V2
4V2 - 16V + 16 = V2
3V2 - 16V + 16 = 0
V = 4, V = 4/3

sub V = 4 into (1), 1600 = (1/2)16c
sub V = 4 into (2), 400 = (1/2)4c

1600 = 8c

c = 200
(edited 7 years ago)
Hello Again,



In case the other unit 4 paper I have finished got lost in this thread for yo, I have attached the other paper and its mark scheme.



Enjoy !!,


Will
Reply 568
Original post by alevelstresss
using E = (1/2)cV2

(1) 1600 = (1/2)cV2
(2) 400 = (1/2)c(V-2)2

4 = [cV2]/[c(V-2)2]
4(V-2)2 = V2
4V2 - 16V + 16 = V2
3V2 - 16V + 16 = 0
V = 4, V = 4/3

sub V = 4 into (1), 1600 = (1/2)16c
sub V = 4 into (2), 400 = (1/2)4c

1600 = 8c

c = 200


Thank you
Original post by alevelstresss
using E = (1/2)cV2

(1) 1600 = (1/2)cV2
(2) 400 = (1/2)c(V-2)2

4 = [cV2]/[c(V-2)2]
4(V-2)2 = V2
4V2 - 16V + 16 = V2
3V2 - 16V + 16 = 0
V = 4, V = 4/3

sub V = 4 into (1), 1600 = (1/2)16c
sub V = 4 into (2), 400 = (1/2)4c

1600 = 8c

c = 200


thats a daft amount of working for a 1 marker imo
Original post by WillRose
thats a daft amount of working for a 1 marker imo


yeah there must be a quicker way, but i did what i could
ImageUploadedByStudent Room1466330148.602982.jpg

Does this always apply for kinetic energy?


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Original post by alevelstresss
yeah there must be a quicker way, but i did what i could


not bashing you at all,just AQA
Original post by alevelstresss
yeah there must be a quicker way, but i did what i could


You have 45 ish minutes, so it's fine. I always spend longer on the MC anyway


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Original post by cmrol
Can someone help me with question 19 on june 2014, https://60abffc9b401b1c0936e01291c15112cab0778c4.googledrive.com/host/0B1ZiqBksUHNYVFI1YlJBOWhwRW8/June%202014%20QP%20-%20Unit%204A%20AQA%20Physics.pdf
I dont understand how you are meant to get 200, I got 600 when I did:

c = Q/V
1600/x = 400/(x-2)
1600x - 3200 = 400x
1200x = 3200
x = 8/3
c = 1600/(8/3)
c=600


Here's what I did, it's super long winded but it got there, there must be an easier way though! IMG_20160619_105723[1].jpg
Original post by cmrol
Can someone help me with question 19 on june 2014, https://60abffc9b401b1c0936e01291c15112cab0778c4.googledrive.com/host/0B1ZiqBksUHNYVFI1YlJBOWhwRW8/June%202014%20QP%20-%20Unit%204A%20AQA%20Physics.pdf
I dont understand how you are meant to get 200, I got 600 when I did:

c = Q/V
1600/x = 400/(x-2)
1600x - 3200 = 400x
1200x = 3200
x = 8/3
c = 1600/(8/3)
c=600


You know e is proportional to v^2, so if you quarter energy, you must half the pd, so the original pd must be 4, as that's the only value that works in the situation. Just sub that into E=1/2CV^2.
do people do section a or b first?
What could a potential 6 marker about Induction be about?
Original post by liverpool2044
do people do section a or b first?


section b first because i can usually guess some of the MC if i am stretched for time
Original post by Seclusion
You know e is proportional to v^2, so if you quarter energy, you must half the pd, so the original pd must be 4, as that's the only value that works in the situation. Just sub that into E=1/2CV^2.


ah, clever

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