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# AQA Physics PHYA5 - Thursday 18th June 2015 [Exam Discussion Thread] watch

1. (Original post by Mehrdad jafari)
L/m/t naught are from the moving reference frame "when it is stationary"

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So is L/m/t naught is the observer?

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2. I thought

And

If this is the case, how is the mass calculated in this question (June 2014 Q3 b ii) if you know the molar mass is 0.084kg?

EDIT: Never mind, just realised you're dealing with a single atom of the substance

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3. (Original post by Jimmy20002012)
So is L/m/t naught is the observer?

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t naught yeah, that is measured from the stationary observer but l/m naught are measurements from the uniformly moving object when it is stationary. So for example a uniformly moving object has a greater mass relative to its m naught mass, the same with lengh.

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4. (Original post by Fvthoms)
What is everyone's solution to the fact that there is such a small number of current-spec papers for unit 5; I was planning to do one paper every couple of days, but at this rate, I'll be finished before the end of the week! 😥
Go through all the questions in the old syllabus (from like 2001 or something). Questions are just the bog standard theory but you will gain a really good basic knowledge.
5. (Original post by gcsestuff)
Go through all the questions in the old syllabus (from like 2001 or something). Questions are just the bog standard theory but you will gain a really good basic knowledge.
How many of the older questions have you done?

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6. ive probably done them all from 2002 onwards. After we finished each topic in class I went through each set on physicsandmathstutor.
Ive just started doing each paper for both unit 5 and 4 but you have to look around a bit for the nuclear and thermal questions.
7. (Original post by gcsestuff)
ive probably done them all from 2002 onwards. After we finished each topic in class I went through each set on physicsandmathstutor.
Ive just started doing each paper for both unit 5 and 4 but you have to look around a bit for the nuclear and thermal questions.
That sounds good! I'll make sure I look at some older nuclear & thermal ones. PHYA4 I think there's enough to keep me going with the past papers 2010-2014.

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8. Things work out really strange. When I was doing PH5 last year, section A (e.g the non optional topics) seemed really really hard and I just couldn't get my head around Radioactivity + the binding energy chapter. This year i've grasped it straight away lmao

Binding energy part especially seems really simple. Last year I thought it was rocket science ffs
9. (Original post by AR_95)
Things work out really strange. When I was doing PH5 last year, section A (e.g the non optional topics) seemed really really hard and I just couldn't get my head around Radioactivity + the binding energy chapter. This year i've grasped it straight away lmao

Binding energy part especially seems really simple. Last year I thought it was rocket science ffs
Could you explain when you do and don't take into account a particle's binding energy in calculations?

I get that a standalone neutron/proton/electron won't have any binding energy, but what about a group of them? (ie: 2 neutrons)?

I only ask because I've done questions where you have to take into account the energy released including the rest mass of the neutrons, and some where you don't.

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10. (Original post by CD223)
Could you explain when you do and don't take into account a particle's binding energy in calculations?

I get that a standalone neutron/proton/electron won't have any binding energy, but what about a group of them? (ie: 2 neutrons)?

I only ask because I've done questions where you have to take into account the energy released including the rest mass of the neutrons, and some where you don't.

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I'm not really sure what you mean, you'd have to provide example of the question.

It depends on what they ask? Idk. I just did a question where the interaction was :

p+p --> 2/1 H (A= 2 Z=1) +B- + Ve

they asked for the binding energy per nucleon of the H nucleus. In that case, the energies of the protons are irrelevant because you can work out the Mass defect of the nucleus (mass of proton and neutron were given in u) and the total mass of the nucleus was also given?
11. (Original post by CD223)
Could you explain when you do and don't take into account a particle's binding energy in calculations?

I get that a standalone neutron/proton/electron won't have any binding energy, but what about a group of them? (ie: 2 neutrons)?

I only ask because I've done questions where you have to take into account the energy released including the rest mass of the neutrons, and some where you don't.

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Actually wait. I was also told once that you only need to consider the information that's given to you but that could be wrong haha

If they want you to take into consideration whatever particle which isn't the Main/daughter nucleus, most of the time they do give you the relevant information for it.
12. (Original post by AR_95)
Actually wait. I was also told once that you only need to consider the information that's given to you but that could be wrong haha

If they want you to take into consideration whatever particle which isn't the Main/daughter nucleus, most of the time they do give you the relevant information for it.
I see. Thanks for your help

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13. (Original post by CD223)
Could you explain when you do and don't take into account a particle's binding energy in calculations?

I get that a standalone neutron/proton/electron won't have any binding energy, but what about a group of them? (ie: 2 neutrons)?

I only ask because I've done questions where you have to take into account the energy released including the rest mass of the neutrons, and some where you don't.

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You wouldn't know the binding energy of 2 neutrons, for example. You would have to supply a certain amount of energy to separate them and measure their total mass after separation, the difference of which would you the mass defect and then you can calculate the binding energy of them using E=mc2.

The same argument applies for the reverse process. For example when you have 2 neutrons of known mass then after joining together a certain amount of energy is released, that is the binding energy of them. Here you don't need to take account of the mass of the neutrons if the energy released is known. If not, then the eventual mass will be given to you in which case you would have to work out the difference in their mass and then going back to E=mc2.

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14. (Original post by Mehrdad jafari)
You wouldn't know the binding energy of 2 neutrons, for example. You would have to supply a certain amount of energy to separate them and measure their total mass after separation, the difference of which would you the mass defect and then you can calculate the binding energy of them using E=mc2.

The same argument applies for the reverse process. For example when you have 2 neutrons of known mass then after joining together a certain amount of energy is released, that is the binding energy of them. Here you don't need to take account of the mass of the neutrons if the energy released is known. If not, then the eventual mass will be given to you in which case you would have to work out the difference in their mass and then going back to E=mc2.

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Thanks!

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15. No worries! Hope that helped

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16. (Original post by Mehrdad jafari)
No worries! Hope that helped

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It did! Is it worth having a go at older syllabus questions to tackle more binding energy questions?

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17. (Original post by CD223)
It did! Is it worth having a go at older syllabus questions to tackle more binding energy questions?

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I would do if I had enough time.

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18. (Original post by Mehrdad jafari)
I would do if I had enough time.

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I'll see what there is. I'm sure I'll get stuck soon enough!

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19. Hey, I was doing 1b(iii) on the 2011 astrophysics paper: http://filestore.aqa.org.uk/subjects...W-QP-JUN11.PDF
And it says the answer is 3.43*10^-6 but the CGP textbook says that formula only gives half the angle subtended so you need to multiply it by 2.

So is the book wrong? If it isn't can someone tell me when you multiply by two and when you don't when using theta = r/d to find angle subtended

Thanks
20. (Original post by JamGrip)
Hey, I was doing 1b(iii) on the 2011 astrophysics paper: http://filestore.aqa.org.uk/subjects...W-QP-JUN11.PDF
And it says the answer is 3.43*10^-6 but the CGP textbook says that formula only gives half the angle subtended so you need to multiply it by 2.

So is the book wrong? If it isn't can someone tell me when you multiply by two and when you don't when using theta = r/d to find angle subtended

Thanks

The calculation of angle subtended is denoted by:

Where h is the diameter or height of the object, and d is the distance to the object from the observer.

The data given in the AQA question effectively gives you the diameter of each of the objects, whereas CGP have carried out a two stage calculation because they are using the radius given in their question.

If you are given the radius, multiply the angle calculated by two, OR multiply the radius by two initially to get the diameter, then use this result in the calculation of the angle subtended.

Hope that clears it up.

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