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

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Reply 2180
Avatar for CD223
CD223
OP
10
Original post by AR_95
For the very last part, how does the N come into the pressure equation


It is like multiplying by N over N - when you square the root mean square speed to get the mean square speed, you end up with N on the bottom. To cancel this when substituting it into the u squared term on top, multiply by N on top. Does that help?


Posted from TSR Mobile
How do you work put the mass defect?
Original post by Me123456789
How do you work put the mass defect?


mass of total protons + mass of total neutrons , minus the mass of the initial nucleus
Original post by CD223
It is like multiplying by N over N - when you square the root mean square speed to get the mean square speed, you end up with N on the bottom. To cancel this when substituting it into the u squared term on top, multiply by N on top. Does that help?


Posted from TSR Mobile


yep thanks
Original post by AR_95
mass of total protons + mass of total neutrons , minus the mass of the initial nucleus


Thanks, but how would I do that for this question? (part c ii):s-smilie:

A typical fission reaction in the reactor is represented by

1 (c) (i) Calculate the number of neutrons, x.
answer = 4.neutrons
(1 mark)

1 (c) (ii) Calculate the energy released, in MeV, in the fission reaction above.

mass of neutron = 1.00867 u
mass of 233 92U nucleus = 232.98915 u
mass of 91 36Kr nucleus = 90.90368 u
mass of 139 56Ba nucleus = 138.87810 u
Original post by Leonacatherine


*3bii
Original post by Adangu
Any predictions for 6 marker on applied physics??


Finally someone else who does applied!
Jun 14 - wheel design/ moment of inertia
June 13 - comparison between real and theoretical diesel cycles
June 12 - angular momentum, moment of inertia
June 11 - design features of a flywheel
June 10- description of theoretical petrol cycle (otto)

Probably gonna swing back to engines, which i dont mind.
Hope its an easy comparison
Original post by johnh545
you dont use the masses.
You calculate the energy released using the change in binding energy.
Seeing as the neutrons are already free, they don't have any binding energy


But isn't the reason that the atom disintegrates that it has absorbed the neutron? So it wouldn't really be a free neutron, since it becomes part of the nucleus an instant before the new nucleus disintegrates...

I don't understand :tongue:
June 13 question 1cii, do the four hydrogen atoms not have one electron each? Why aren't these electrons added to the mass of the protons?

Thanks
Original post by Me123456789
Thanks, but how would I do that for this question? (part c ii):s-smilie:

A typical fission reaction in the reactor is represented by

1 (c) (i) Calculate the number of neutrons, x.
answer = 4.neutrons
(1 mark)

1 (c) (ii) Calculate the energy released, in MeV, in the fission reaction above.

mass of neutron = 1.00867 u
mass of 233 92U nucleus = 232.98915 u
mass of 91 36Kr nucleus = 90.90368 u
mass of 139 56Ba nucleus = 138.87810 u



You don't need to know any mass defect. They've given you the mass before the reaction and after the reactions.

Total mass before - Total mass after , and you'll get a small value of 0. something (u)

times that value by 931.5 to get the energy released in MeV
(edited 8 years ago)
Original post by Star Light
I'm thinking we might get a six-marker about the eye, there hasn't been one on the current spec papers so far. For example, explain for a light ray enters the eye, and how it gets received by the brain. Or maybe something about different vision defects, causes and how to fix?

Might be one about comparing some scan types, but they tend to be 2-4 markers really.
Could be one about surgery - endoscopes + why is laparoscopic surgery good, that hasn't featured at all really so far.

If only they'd give us 'Describe the events in one heart beat' again like in 2014!


Yes true, I agree with the eye one - but I don't think we need to know much about how light ray get received by the brain? Also what would you say about endoscope + laparoscopic? I am thinking more of an X-ray question!
Original post by samlyon
June 13 question 1cii, do the four hydrogen atoms not have one electron each? Why aren't these electrons added to the mass of the protons?

Thanks


Mass of an electron is insignificant compared to the mass of a proton (10^4 bigger)??

I don't think it would make much difference, Hydrogen atoms are equivalent to protons anyway

Think of it like 1000kg is the mass of blob A and 1kg is the mass of blob B, the mass of blob B doesn't really add that much to the total mass really
(edited 8 years ago)
Ok. I posted basically the same thing back on page 50 but 2b(ii) here: http://filestore.aqa.org.uk/subjects/AQA-PHYA5-2A-QP-JUN12.PDF (June 2012 astro)
http://filestore.aqa.org.uk/subjects/AQA-PHYA52A-W-MS-JUN12.PDF (markscheme)

It doesn't multiply the answer by two afterwards even though it says it's an orbital radius. Any reason why?

Thanks
Original post by DCMed96
Thanks for the quick response!!!

Anyone guess what will be on this year's six marker? (what can be the hardest thing they can throw on us) considering we are all gonna do the R place instead thanks to the lost van.....

Anyone doing med phy as well>


i'm doing med phy!! what do you think the 6marker would be on?
Reply 2194
Avatar for sykik
sykik
4
41.png

Hi I don't get why E=mc^2 gives total energy in part ii
For anyone doing Astrophysics,

what is the definition of an Airy disc?
Can anyone explain question 2 a ii

I dont get how you can answer it knowing only 1 piece of information?

http://filestore.aqa.org.uk/subjects/AQA-PHYA51-QP-JUN14.PDF
Original post by AR_95
Mass of an electron is insignificant compared to the mass of a proton (10^4 bigger)??

I don't think it would make much difference, Hydrogen atoms are equivalent to protons anyway


Thanks for the response. On the mark scheme it says you subtract the mass of the positrons produced so this is why I thought it would be significant?
Reply 2198
Avatar for sykik
sykik
4
Original post by CD223
It is like multiplying by N over N - when you square the root mean square speed to get the mean square speed, you end up with N on the bottom. To cancel this when substituting it into the u squared term on top, multiply by N on top. Does that help?


Posted from TSR Mobile


I thought N was due to N number of molecules present because if you don't have N then the equation only applies for a single molecule??
Original post by SuperMushroom
Can anyone explain question 2 a ii

I dont get how you can answer it knowing only 1 piece of information?

http://filestore.aqa.org.uk/subjects/AQA-PHYA51-QP-JUN14.PDF


Decay constant = ln2/half life

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