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

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Original post by CD223
I thought M2 was meant to be of similar difficulty to Physics A2?


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No way near. You don't even get a formula book, you have to remember all the formulae
Original post by Me123456789
Oh, thank you.:smile:
So if an A was 58/75 and I got 12/25 and 46/50 which totals to 58 I'd get an A?
And are you completely sure, it's just something my teacher said that confused me.


How do you know what an A is on each section? You just get the total grade boundary for both sections, so obviously it's the total marks that matter.
Original post by ubisoft
How do you know what an A is on each section? You just get the total grade boundary for both sections, so obviously it's the total marks that matter.


The grade boundaries put marks for both of the papers:

http://filestore.aqa.org.uk/over/stat_pdf/AQA-A-LEVEL-GDE-BOUND-JAN13.PDF
Original post by Me123456789
The grade boundaries put marks for both of the papers:

http://filestore.aqa.org.uk/over/stat_pdf/AQA-A-LEVEL-GDE-BOUND-JAN13.PDF


It is the total
Original post by CD223
Didn't realise it also applied to fairground rides!


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Yeah it does. If you work out the horizontal and vertical components of the the body rotating at an angle and then combining them together then you arrive at the same equation.


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Original post by Mehrdad jafari
Yeah it does. If you work out the horizontal and vertical components of the the body rotating at an angle and then combining them together then you arrive at the same equation.


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I think it applies to any body in circular motion only under its reaction/levitation force and without friction. An aeroplane should give the same formula.
Original post by PotterPhysics
I think it applies to any body in circular motion only under its reaction/levitation force and without friction. An aeroplane should give the same formula.


Yeah, that's true.


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Original post by ubisoft
No way near. You don't even get a formula book, you have to remember all the formulae


Oh right! I find M2 easier than physics though :L

I think it's just the amount of theory and wordy questions that make it harder.


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Original post by Mehrdad jafari
Yeah it does. If you work out the horizontal and vertical components of the the body rotating at an angle and then combining them together then you arrive at the same equation.


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Ah I take it they're horizontal fairground rides and not things like Ferris wheels?


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Original post by PotterPhysics
I think it applies to any body in circular motion only under its reaction/levitation force and without friction. An aeroplane should give the same formula.


Does it not have to be in horizontal circular motion for this equation to apply?


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Original post by CD223
Ah I take it they're horizontal fairground rides and not things like Ferris wheels?


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Yeah, no ferris wheels, because in ferris wheels the centripetal force varies around the circle, and even if the centripetal force didn't vary, that's fair wheels in the absence of gravitational fields, there would be no component of the weight and with using that equation it would be assumed the body is in circular motion at and angle of 90 or 0 degrees


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Original post by Mehrdad jafari
Yeah, no ferris wheels, because in ferris wheels the centripetal force varies around the circle, and even if the centripetal force didn't vary, that's fair wheels in the absence of gravitational fields, there would be no component of the weight and with using that equation it would be assumed the body is in circular motion at and angle of 90 or 0 degrees


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I see. Thanks.

Ferris wheels would only have R, the reaction force, and mg, the weight, acting if all other resistance forces are ignored, so in this case the centripetal force would vary between R+mg at the top and R-mg at the bottom, right?


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One month today guys! Let's make it count!


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Original post by CD223
I see. Thanks.

Ferris wheels would only have R, the reaction force, and mg, the weight, acting if all other resistance forces are ignored, so in this case the centripetal force would vary between R+mg at the top and R-mg at the bottom, right?


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Yeah, but R+mg at the bottom and R-mg at the top because at the top the body has overcome it's weight. It's like you throw a ball into the air and try to measure it's weight. While it's moving upwards the weight is 'negative' and at the the turning point the weight would be zero


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Original post by Mehrdad jafari
Yeah, but R+mg at the bottom and R-mg at the top because at the top the body has overcome it's weight. It's like you throw a ball into the air and try to measure it's weight. While it's moving upwards the weight is 'negative' and at the the turning point the weight would be zero


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I'm talking about centripetal motion - the resultant force is always towards the centre so whichever force points towards the centre is greatest.


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Original post by CD223
I'm talking about centripetal motion - the resultant force is always towards the centre so whichever force points towards the centre is greatest.


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That's what I meant as well :smile:, but i think my explanation lacked details.
The point at the bottom of the fair wheel would experience the greatest resultant force.


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Original post by Mehrdad jafari
That's what I meant as well :smile:, but i think my explanation lacked details.
The point at the bottom of the fair wheel would experience the greatest resultant force.


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Surely the magnitude of the resultant force would remain constant around the circle if the speed was constant?


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Original post by CD223
Surely the magnitude of the resultant force would remain constant around the circle if the speed was constant?


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Yeah, definitely, but not if it was circular motion in vertical plane in a gravitational field


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Original post by CD223
Does it not have to be in horizontal circular motion for this equation to apply?


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Ah of course, forgot to mention that
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Original post by Mehrdad jafari
Yeah, definitely, but not if it was circular motion in vertical plane in a gravitational field


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Ah right :smile: do you think they would make us comment on gravitational fields in the circular motion topic though?

Only think I can envisage is them using the gravitational formula F=GMmr2F=\frac{GMm}{r^2} and equating it to the centripetal force for a satellite?


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