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

1. Would anyone mind explaining the best method for Q12 Jan 13 paper?
2. (Original post by CD223)
It's in the book I believe as an application of SHM:

Here is a great short YouTube video that summarises what you could expect from a question.

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Ah yes!

I remember this now! I just didn't know it was called Barton's experiment.

Thanks
3. Just marking jan 13 now
can anyone explain
9,12, or 17 from the MC
http://filestore.aqa.org.uk/subjects...1-QP-JUN12.PDF
4. Hey guys!

Could someone help me with the Q2(c)(ii) in the following attachment.

I do not understand what the mark scheme did.

Why do they use 1.91 eV as their voltage value, isn't eV a unit of energy not voltage?
5. (Original post by Leonacatherine)
Just marking jan 13 now
can anyone explain
9,12, or 17 from the MC
http://filestore.aqa.org.uk/subjects...1-QP-JUN12.PDF
Hello =D

You said you're marking Jan '13 but you gave us the link to Jun '12 so which paper is it?
6. (Original post by Disney0702)
Hey guys!

Could someone help me with the Q2(c)(ii) in the following attachment.

I do not understand what the mark scheme did.

Why do they use 1.91 eV as their voltage value, isn't eV a unit of energy not voltage?
I did theway in the brackets, if you work out the field strength, V/d then you can multiply it by q to get the force because F=EQ (this is using the formulas for electric field strength and force for a uniform field which are given in the formula sheet)

Now that youve got the force, we know work = force x displacement, so displacement = work ( the amount of energy in joules needed to excite a neon atom) divided by the force created by the field and you should get the right answer
7. (Original post by Disney0702)
Hello =D

You said you're marking Jan '13 but you gave us the link to Jun '12 so which paper is it?
did I ? Whoops ! definitely jan 13
http://filestore.aqa.org.uk/subjects...W-MS-JAN13.PDF
8. (Original post by frankiejayx)
Would anyone mind explaining the best method for Q12 Jan 13 paper?
This is the way I think about it:

(I'm not that confident but the numbers seemed to drop out okay)

What I'm trying to say here is that the ratio of g for the earth to the moon is proportional the the ratio of the densities multiplied by the ratio of each radius.

As M is proportional to r cubed, and g is inversely proportional to r squared, the top and bottom ratios cancel so g is only dependent on the product of the density and radius ratio. Subbing the numbers in gives 6 divided by 5/3 which gives 3.6.

Does that clear anything up?

Again, I'm not totally sure on this so id appreciate if someone can correct me!

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9. (Original post by Disney0702)
Ah yes!

I remember this now! I just didn't know it was called Barton's experiment.

Thanks
No worries! Glad it's helped out of interest what would you put for a six marker on it?

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10. (Original post by Leonacatherine)
Just marking jan 13 now
can anyone explain
9,12, or 17 from the MC
http://filestore.aqa.org.uk/subjects...1-QP-JUN12.PDF
Question 9:
You are asked to find how long it takes before the two pendulums are in phase again. See my workings at the bottom of the post

Question 12:
http://www.thestudentroom.co.uk/show...1#post56322591

Question 17:
The answer is 360V. Here's my workings for 9 and 17:

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11. (Original post by CD223)
Question 9:
You are asked to find how long it takes before the two pendulums are in phase again. See my workings at the bottom of the post

Question 12:
http://www.thestudentroom.co.uk/show...1#post56322591

Question 17:
The answer is 360V. Here's my workings for 9 and 17:

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Just want to make sure you knew it would only be one extra before they are in phase? is it because of the ration between the time periods?
12. (Original post by Leonacatherine)
Just want to make sure you knew it would only be one extra before they are in phase? is it because of the ration between the time periods?
Essentially there's a difference of 0.05s between them. You're trying to work out how many lots of 0.05s make up the time period of P, the shorter pendulum.

This is because the soonest possible time they next become in phase is always when the shorter pendulum has completed one more oscillation. Think of it this way - it takes less time for it to complete an oscillation, so you want to work out how many times it has to oscillate before it "laps" the other pendulum and they are in phase again.

Think of two runners completing an 800m race around a race track. Runner P takes a lot less time to run a lap and so ends up running past runner Q and "lapping" him/her. In that moment that P runs past Q, they are in phase and runner P has completed one more circuit (or oscillation in the case of pendulums) in that time.

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13. (Original post by CD223)
This is the way I think about it:

(I'm not that confident but the numbers seemed to drop out okay)

What I'm trying to say here is that the ratio of g for the earth to the moon is proportional the the ratio of the densities multiplied by the ratio of each radius.

As M is proportional to r cubed, and g is inversely proportional to r squared, the top and bottom ratios cancel so g is only dependent on the product of the density and radius ratio. Subbing the numbers in gives 6 divided by 5/3 which gives 3.6.

Does that clear anything up?

Again, I'm not totally sure on this so id appreciate if someone can correct me!

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I couldnt quite follow your math My workings are attatched, they give a tenth not 3.6 so I know theyre wrong but I'm wondering if you could have a look and see where it went wrong? I always apply this method and it has always worked maybe I made a mistake somewhere I'm not sure?
14. (Original post by Leonacatherine)
I couldnt quite follow your math My workings are attatched, they give a tenth not 3.6 so I know theyre wrong but I'm wondering if you could have a look and see where it went wrong? I always apply this method and it has always worked maybe I made a mistake somewhere I'm not sure?
Hmm, is it something to do with the ratios being the inverse of 0.6?

In other words ge/gm is 6 and pe/pm is 5/3, not 3/5?

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15. (Original post by CD223)
Essentially there's a difference of 0.05s between them. You're trying to work out how many lots of 0.05s make up the time period of P, the shorter pendulum.

This is because the soonest possible time they next become in phase is always when the shorter pendulum has completed one more oscillation. Think of it this way - it takes less time for it to complete an oscillation, so you want to work out how many times it has to oscillate before it "laps" the other pendulum and they are in phase again.

Think of two runners completing an 800m race around a race track. Runner P takes a lot less time to run a lap and so ends up running past runner Q and "lapping" him/her. In that moment that P runs past Q, they are in phase and runner P has completed one more circuit (or oscillation in the case of pendulums) in that time.

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So could you just do 1.9 divided by 0.05?
16. (Original post by Leonacatherine)
So could you just do 1.9 divided by 0.05?
Yeah you could. It's just understanding why that works which is the key part.

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17. (Original post by CD223)
Yeah you could. It's just understanding why that works which is the key part.

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I do the race track analogy really helped thank you!
18. Also in Jan 13 i was unclear about the field lines for Q4b ??
19. (Original post by Leonacatherine)
did I ? Whoops ! definitely jan 13
http://filestore.aqa.org.uk/subjects...W-MS-JAN13.PDF
Here is the solutions for Q9 and Q12.
20. (Original post by CD223)
Question 9:
You are asked to find how long it takes before the two pendulums are in phase again. See my workings at the bottom of the post

Question 12:
http://www.thestudentroom.co.uk/show...1#post56322591

Question 17:
The answer is 360V. Here's my workings for 9 and 17:

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For Q17 why did you double the electric potential, I do not understand.

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