2016 Official AQA New Spec AS Level Physics Paper 1 - 24th of May 2016
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Original post by greg.h.nm
I know, Christ its like you actually had to use your brain for an exam. At least you might get a grade in your media studies exam.
What are you talking about. Can you not throw your toys out of your pram and not target someone for stating their opinion? Smh people these days.
Original post by zebra249
Would you still get the mark if you said beta plus instead of e+?
Think so at least I hope so because I put that as well.
Can anyone send me pdfs of the second set of the Physics AS papers/mark schemes?
Original post by champ_mc99
I still dunno if its nodes or antinodes and why. 
Basically the points where the chocolate was melted occured at anti-nodes I believe. So you know anti-nodes occur at maxima and minima. Hence the distance between the two melted points on the sides (not the centre) gave one wavelength?
This distance was 12 cm (hence 0.12 m) since it was a 'life-sized' chocolate bar and you work your way from there to get frequency.
I still don't see how that is 5 marks.
Original post by pippaaa99
If you use c=f*wavelength you already get frequency as in waves per second so I'm not sure you have to?
I agree with this!
Original post by champ_mc99
I forgot this bit so probably lost 2. Looks fine.
Champ, does the quote above make sense to you? It might mean you don't have to multiply by 60?
For the chocolate the in the microwave,
The calculation of the frequency is using
F=speed of light/wavelength
I put nodes for the three dots of MELTED chocolate which is wrong it is ANTINODES.
But even if you got that wrong the wavelength needed to calculate the frequency will still be the same.
Distance between a node and a node is 6cm
Distance between an antinode and antinode is 6cm. They are both half a wavelength anyway??
So 0.12m is the wavelength.
Gets you 2.5x10^9 Hz.
The calculation of the frequency is using
F=speed of light/wavelength
I put nodes for the three dots of MELTED chocolate which is wrong it is ANTINODES.
But even if you got that wrong the wavelength needed to calculate the frequency will still be the same.
Distance between a node and a node is 6cm
Distance between an antinode and antinode is 6cm. They are both half a wavelength anyway??
So 0.12m is the wavelength.
Gets you 2.5x10^9 Hz.
Original post by LightAtTheEnd
Basically the points where the chocolate was melted occured at anti-nodes I believe. So you know anti-nodes occur at maxima and minima. Hence the distance between the two melted points on the sides (not the centre) gave one wavelength?
This distance was 12 cm (hence 0.12 m) since it was a 'life-sized' chocolate bar and you work your way from there to get frequency.
I still don't see how that is 5 marks.
This distance was 12 cm (hence 0.12 m) since it was a 'life-sized' chocolate bar and you work your way from there to get frequency.
I still don't see how that is 5 marks.
Hmm? I thought antinodes were always maxima.
Original post by champ_mc99
Hmm? I thought antinodes were always maxima.
Antinodes are points which vary between the maximum and minimum displacement of a stationary wave
Original post by domy61
any thoughts on grade boundaries?
My prediction
A - 49
B - 44
C - 40
D - 36
E - 33
Original post by domy61
any thoughts on grade boundaries?
Hopefully low ahahaahahah
Original post by champ_mc99
Hmm? I thought antinodes were always maxima.
Look at a sine graph. measure from first to last for 1 full wavelength, as first to second is only a half.
Is there an unofficial mark scheme yet?
Does anyone have the unnoficial mark scheme for the AS aqa physics exam that happened today?
If you are not already aware, the markscheme is currently being edited here:
http://docs.google.com/document/d/1cjBRtaBmwyazyQyvflZgjam6dzOTlDi9quQJK1il5Po/edit?pref=2&pli=1
Anyone can leave suggestions, so feel free to leave them. Thanks. Please don't post anything stupid or unnecessary.
http://docs.google.com/document/d/1cjBRtaBmwyazyQyvflZgjam6dzOTlDi9quQJK1il5Po/edit?pref=2&pli=1
Anyone can leave suggestions, so feel free to leave them. Thanks. Please don't post anything stupid or unnecessary.
Original post by pippaaa99
Antinodes are points which vary between the maximum and minimum displacement of a stationary wave
Original post by marldor
Look at a sine graph. measure from first to last for 1 full wavelength, as first to second is only a half.
Oh right. I understand. I put antinodes but after the exam I was then convinced it was nodes.
Cheers.
for one of the last questions if you put beta instead of a positron would you get the mark
Original post by BainesyA
My prediction
A - 49
B - 44
C - 40
D - 36
E - 33
A - 49
B - 44
C - 40
D - 36
E - 33
I wish
It's based on what percentage of people they want to get an A so they put the grade boundaries so like 15% of the people that sat it get As
Grade boundaries imo will probably similar to 2015.
Around about 50 for an A.
I thought it was alright. I've certainly done better in this compared to the specimens.
Around about 50 for an A.
I thought it was alright. I've certainly done better in this compared to the specimens.
(edited 7 years ago)
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