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Unit 5 Physics Edexcel A2 and Edexcel IAL

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Original post by candycake
Sorry - I'm sure this is really obvious but I still can't place the equation. Is it in the formula sheet? I've tried GPE, elastic strain energy etc.


Its not in the formula sheet however there is a long derivation to get the equation involving E tot = Ek + Ep = 1/2kA^2
Original post by XxKingSniprxX
Its not in the formula sheet however there is a long derivation to get the equation involving E tot = Ek + Ep = 1/2kA^2


Thank you so much - I'll have to learn it! Do you know why the equation for GPE can't be used in this case since E = mgh and h is decreasing by 1/2?

Original post by Pudge666
Guys , I keep making dumb mistakes with temperature calculation , how can you tell if you have to use kelvin or celsius


Always use Kelvin for calculations. The only time it doesn't matter is when its a change in temperate (e.g using mc delta theta) as the intervals between degrees are the same for Kelvin and Celcius
Original post by candycake
Thank you so much - I'll have to learn it! Do you know why the equation for GPE can't be used in this case since E = mgh and h is decreasing by 1/2?



Always use Kelvin for calculations. The only time it doesn't matter is when its a change in temperate (e.g using mc delta theta) as the intervals between degrees are the same for Kelvin and Celcius


Thank you
Anyone explain to me how standard candles are used for measuring distance to far stars?

So the SC's properties mean their distance and luminosity are known, how does that help you find non standard candle star distances?
Original post by 16characterlimit
Anyone explain to me how standard candles are used for measuring distance to far stars?

So the SC's properties mean their distance and luminosity are known, how does that help you find non standard candle star distances?


You calculate the change in brightness/variation of a star and the flux(intensity of radiation) received on earth. You then match that with stars of known Luminosity from HR-Diagram. You then use F = L/4.pi.d^2 & rearrange for Distance.

Note: DO NOT SAY 'MEASURE' AND INSTEAD WRITE 'CALCULATE'.
^ There is a clear difference between them both. :sigh:
Original post by XxKingSniprxX
You calculate the change in brightness/variation of a star and the flux(intensity of radiation) received on earth. You then match that with stars of known Luminosity from HR-Diagram. You then use F = L/4.pi.d^2 & rearrange for Distance.

Note: DO NOT SAY 'MEASURE' AND INSTEAD WRITE 'CALCULATE'.
^ There is a clear difference between them both. :sigh:


Whats that got to do with a standard candle though? You can find the distance of cepheid variables because of the period is proportional to luminosity, but how do I find the distance of a far away main sequence?
Original post by 16characterlimit
Whats that got to do with a standard candle though? You can find the distance of cepheid variables because of the period is proportional to luminosity, but how do I find the distance of a far away main sequence?


Even trignometric parallex can be used but for nearby stars. A far away star is difficult to measure accurately(from what IK) because the radiation flux that is coming is too small to be detected.
Original post by sabahshahed294
Even trignometric parallex can be used but for nearby stars. A far away star is difficult to measure accurately(from what IK) because the radiation flux that is coming is too small to be detected.


So there's no method?
Original post by 16characterlimit
So there's no method?


Well,those methods can't be applied but Hubble's Law one can work for such cases I guess.
I've just done one of the SAM papers but I can't work out what the mark scheme is driving at here. Could anyone please explain what the third marking point means?

Explain how the ultimate fate of the Universe is associated with the Hubble constant.
fate of universe depends on the density of the universe(1)
link between gravity and density(1)Hubble “constant” is changing due to gravitational forces (1)


This diagram is always used to demonstrate trig parallax, so would it be safe to draw exactly this diagram? Also I don't understand how this measures the distance from the star to the Earth because the line labeled d is going directly to the sun?
Original post by demotivated


This diagram is always used to demonstrate trig parallax, so would it be safe to draw exactly this diagram? Also I don't understand how this measures the distance from the star to the Earth because the line labeled d is going directly to the sun?


When you're comparing such vast distances it doesn't matter.
Original post by 16characterlimit
When you're comparing such vast distances it doesn't matter.


Can you also please explain why there are two angles? theta 1 and theta 2? Which would be the parsec?
Could anyone explain why the answer to question 4 is D, instead of C? in Jan 2015 IAL

http://qualifications.pearson.com/content/dam/pdf/International%20Advanced%20Level/Physics/2013/Exam%20materials/WPH05_01_que_20150122.pdf


http://qualifications.pearson.com/content/dam/pdf/International%20Advanced%20Level/Physics/2013/Exam%20materials/WPH05_01_msc_20150305.pdf
Original post by demotivated


Probably because it is 'detectable' due to its gravitational forces but is is 'invisible' because it emits no EM radiation.
Original post by demotivated


it doesnt emit or interact with EM radiation but like we know it exists and roughly how much of the universe it occupies so it is detectable
Original post by demotivated


It's D (invisible) because it emits no EM radiation, However it is detectable, otherwise we wouldn't know about it. It's typical Edexcel troll questions. You just have to really think what the words mean. In your mind you need to be thinking:

Antimatter - obviously wrong
Black dwarf stars - doesn't make any sense
Invisible - can't be seen - so emits no EM radiation
Undetectable - this question wouldn't exist if that was so
Original post by mrbeady9
It's D (invisible) because it emits no EM radiation, However it is detectable, otherwise we wouldn't know about it. It's typical Edexcel troll questions. You just have to really think what the words mean. In your mind you need to be thinking:

Antimatter - obviously wrong
Black dwarf stars - doesn't make any sense
Invisible - can't be seen - so emits no EM radiation
Undetectable - this question wouldn't exist if that was so


Original post by SuruthiG
it doesnt emit or interact with EM radiation but like we know it exists and roughly how much of the universe it occupies so it is detectable

Original post by candycake
Probably because it is 'detectable' due to its gravitational forces but is is 'invisible' because it emits no EM radiation.


Thank you guys! Another question I have about SHM.

In June 2013 (R), the answer to question 7 is D, which is understandable

http://qualifications.pearson.com/content/dam/pdf/A%20Level/Physics/2013/Exam%20materials/6PH05_01R_que_20130620.pdf

But the answer to the same question in question 8 June 2014 (IAL) is conflicting? I wouldn't pick any of the other options but I don't understand how it's D

http://qualifications.pearson.com/content/dam/pdf/International%20Advanced%20Level/Physics/2013/Exam%20materials/Question-paper-Unit-5-(WPH05)-June-2014.pdf
(edited 7 years ago)
Original post by demotivated
Thank you guys! Another question I have about SHM.

In June 2013 (R), the answer to question 7 is D, which is understandable

http://qualifications.pearson.com/content/dam/pdf/A%20Level/Physics/2013/Exam%20materials/6PH05_01R_que_20130620.pdf

But the answer to the same question in question 8 June 2014 (IAL) is conflicting? I wouldn't pick any of the other options but I don't understand how it's D

http://qualifications.pearson.com/content/dam/pdf/International%20Advanced%20Level/Physics/2013/Exam%20materials/Question-paper-Unit-5-(WPH05)-June-2014.pdf


the 2014 one is general shm not undamped like june 13 (r)
Original post by demotivated
Thank you guys! Another question I have about SHM.

In June 2013 (R), the answer to question 7 is D, which is understandable

http://qualifications.pearson.com/content/dam/pdf/A%20Level/Physics/2013/Exam%20materials/6PH05_01R_que_20130620.pdf

But the answer to the same question in question 8 June 2014 (IAL) is conflicting? I wouldn't pick any of the other options but I don't understand how it's D

http://qualifications.pearson.com/content/dam/pdf/International%20Advanced%20Level/Physics/2013/Exam%20materials/Question-paper-Unit-5-(WPH05)-June-2014.pdf


From the examiner's report for the IAL paper:
The question refers to simple harmonic motion, which may or may not be damped. If undamped, total energy would remain constant (no answer key), but this is not a condition of simple harmonic motion. In general, none of the energies stated must remain constant, and so the correct response is B.

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