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AQA PHYA4 ~ 13th June 2013 ~ A2 Physics

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Reply 80
Avatar for Xiomara
Xiomara
18
Original post by jonnyb123
22 is T=2pir/v then sub in r=mv/Be (e is Q) and cancel it down to get 2pim/Be, D

15 is Gm^2/r^2=e^2/4pi*e*r^2, (remember m and e is the same so it's just m and e squared) cancel out the r squareds, Gm^2=e^2/4pi*e, electrostatic force over gravitational force is (e^2/4pi*e)/Gm^2 which is e^2/4pi*e*Gm^2, so D


Hehe, that make sense. Thanks :smile:

Any word on Question 9?
Reply 81
Avatar for jonnyb123
jonnyb123
9
Original post by Xiomara
Hehe, that make sense. Thanks :smile:

Any word on Question 9?


you're welcome, updated the post with 9, there's a nice work through of a similar question on page 3 :smile:
(edited 10 years ago)
Original post by Jack93o
do you still add up the potential between the two masses?

so say at a random point along the line between the earth and moon, the potential due from the earth is -50MJkg-1, and the potential due from the moon was -10MJkg-1, then would the potential at that point would be -60MJkg-1?

also, slightly related to this, if I had two positive charges, how would the potential graph look like for the distance between them? like a U curve in the positive half of the Y axis?


ok, well, for the first question, thats going in to complicated stuff, and I'm pretty sure you wouldn't have to worry about that at A2, unless I am being dim and it is more simple than I thought. You could however be asked to find the distance from the earth to point X, where at the point X the resultant gravitational potential is 0, i.e. they are equal ( the gravitational potential of the earth and moon) and you can solve this algebraically. for the second question, yes, I suppose so, and likewise for two negative (attractive) charges, it would be effectively a reflection in y=0, feel free to ask anything else if you need :smile:
Reply 83
Avatar for kingm
kingm
10
Hi. Can someone explain to me Question 3 (section A) Jan 12?

Unit 4 Section A .pdf

I would have thought it was D, as T acts in the same direction as the centripetal force?
(edited 10 years ago)
Reply 84
Avatar for jonnyb123
jonnyb123
9
Original post by kingm
Hi. Can someone explain to me Question 3 (section A) Jan 12?

Unit 4 Section A .pdf

I would have thought it was D, as T acts in the same direction as the centripetal force?


T acts upwards, mg acts downwards, so the difference is the centripetal force so T-mg=mv^2/l
Reply 85
Avatar for Xiomara
Xiomara
18
Original post by jonnyb123
you're welcome, updated the post with 9, there's a nice work through of a similar question on page 3 :smile:

Just read it, and repped :smile: Thanks again!
Reply 86
Avatar for kingm
kingm
10
Original post by jonnyb123
T acts upwards, mg acts downwards, so the difference is the centripetal force so T-mg=mv^2/l


I see!! I was in As mode i think. Cheers.
Reply 87
Avatar for Jack93o
Jack93o
15
Original post by pepeeglesfield
ok, well, for the first question, thats going in to complicated stuff, and I'm pretty sure you wouldn't have to worry about that at A2, unless I am being dim and it is more simple than I thought. You could however be asked to find the distance from the earth to point X, where at the point X the resultant gravitational potential is 0, i.e. they are equal ( the gravitational potential of the earth and moon) and you can solve this algebraically. for the second question, yes, I suppose so, and likewise for two negative (attractive) charges, it would be effectively a reflection in y=0, feel free to ask anything else if you need :smile:


I thought you could never have a resultant gravitational potential of zero between two objects which are both attractive (or both repulsive) :confused:
Reply 88
Avatar for TaiGunner
TaiGunner
2
Guys anyone have any concise revision sites/notes?
Much appreciated
Reply 89
Avatar for jonnyb123
jonnyb123
9
physics question.png

Could somebody tell me how do you do this? It seems like it should be simple but I'm blanking on it...
Reply 90
Avatar for fizzbizz
fizzbizz
4
Original post by jonnyb123
physics question.png

Could somebody tell me how do you do this? It seems like it should be simple but I'm blanking on it...


This is how I'd try and do it, not sure if its correct though:

Let v = velocity of the alpha particle, and V = the velocity of the daughter nucleus (what we're trying to find)

E=12mv2 E=\frac{1}{2} mv^2

v=2Emv=\sqrt{\frac{2E}{m}}

Using the conservation of momentum law, we know the momentum going both directions cancel each other out (since the momentum to start with was zero). Also, the mass of the daughter nucleus will equal its original mass M minus the mass of the alpha particle m:

mv=(M−m)V mv=(M-m)V

Using our previous result for v:

m2Em=(M−m)V m\sqrt{\frac{2E}{m}}=(M-m)V

Bring the m inside the square root:

2Em=(M−m)V \sqrt{2Em}=(M-m)V

V=2EmM−m V = \frac{\sqrt{2Em}}{M-m}

which is option A?
(edited 10 years ago)
Original post by Jack93o
I thought you could never have a resultant gravitational potential of zero between two objects which are both attractive (or both repulsive) :confused:


well when I say a resultant of 0, I mean they are equal.. so e.g. -XJkg^-1=--YJkg^-1 hence if you were to put it on one side you would get -X-(-Y)=0 thus a resultant gravitational potential of 0. If you think about it there has to be a point along the line where the resultant Gravitational potential is 0, for instance that graph with the moon and the earth was a curve, that curve has a stationary point, if you differentiate it and make it = 0 which is true at a stationary point on a curve, the field strengh = 0, if the field strength = 0 you can effectively say the resultant potential is also 0. Of course if there wasn't a moon, the gravitational potential would have a negative value, as you are still under the effect of the earths pull, the resultant of 0 is theoretical, I think I am just confusing you and myself. But basically, the point where the resultant field strength between 2 objects is 0, their potentials are equal...sorry for over complicating that nonsense!
Reply 92
Avatar for jonnyb123
jonnyb123
9
Original post by fizzbizz


x



thanks a lot, I just didn't see that m*sqrt(2E/m)=sqrt(2Em)! (unfortunately I can't rep you twice after up-voting that pendulum explanation!)
Reply 93
Avatar for The H
The H
1
Can anyone help with Q8 please?
Thanks
http://filestore.aqa.org.uk/subjects/AQA-PHYA4-1-W-QP-JUN11.PDF
Reply 94
Avatar for jonnyb123
jonnyb123
9
Original post by The H


I asked about this a page or two ago, g=GM/r^2, p=m/v so m=pv, so g=Gp4/3pi*r^3/r^2, cancel down the rs and you'll get g=Gp4pir/3, so g is proportional to r and p. So times it by 2 and divide it by 2 and you get B

Also note how g is only proportional to r when p changes, 2 questions later you get a standard one about a planet with half the radius and a quarter the mass of earth, here g is proportional to 1/r^2, because the mass changes and the mass takes the density and volume into account. So it's just g multiplied by 4 for the radius then divided by 4 for the mass.
(edited 10 years ago)
Reply 95
Avatar for The H
The H
1
Original post by jonnyb123
I asked about this a page or two ago, g=GM/r^2, p=m/v so m=pv, so g=Gp4/3pi*r^3/r^2, cancel down the rs and you'll get g=Gp4pir/3, so g is proportional to r and p. So times it by 2 and divide it by 2 and you get B

Also note how g is only proportional to r when p changes, 2 questions later you get a standard one about a planet with half the radius and a quarter the mass of earth, here g is proportional to 1/r^2, because the mass changes and the mass takes the density and volume into account. So it's just g multiplied by 4 for the radius then divided by 4 for the mass.


Thanks a lot :smile:
Reply 96
Avatar for kingm
kingm
10
Original post by The H



Original post by jonnyb123
I asked about this a page or two ago, g=GM/r^2, p=m/v so m=pv, so g=Gp4/3pi*r^3/r^2, cancel down the rs and you'll get g=Gp4pir/3, so g is proportional to r and p. So times it by 2 and divide it by 2 and you get B


Not sure if i fluked that question or not then, but I concluded that T must be the same on the planet because the equation for a simple pendulum does not include mass, density or volume and thus T is not proportional to any of them.
Reply 97
Avatar for fizzbizz
fizzbizz
4
Original post by jonnyb123
thanks a lot, I just didn't see that m*sqrt(2E/m)=sqrt(2Em)! (unfortunately I can't rep you twice after up-voting that pendulum explanation!)


Haha no worries :smile:

Original post by The H


So we know T=2πlg T=2\pi\sqrt{\frac{l}{g}}

But g=GMr2 g=\frac{GM}{r^2}

And so

T=2πr2lGM T=2\pi \sqrt{\frac{r^2 l}{GM}}

M can be written as volume x density;

M=43πr3ρ M=\frac{4}{3}\pi r^3 \rho

So subbing this version of M into our equation for T gives us:

T=2π3l4πrρ T=2\pi \sqrt{\frac{3l}{4\pi r \rho}}

That's a confusing looking formula I know, but the main thing to pick up is that

T∝1ρr T \propto \sqrt{\frac{1}{\rho r}}

So if density is halved but radius is doubled, the overall effect will cancel;

12r×0.5ρ=1rρ \sqrt{\frac{1}{2r\times 0.5\rho}}=\sqrt{\frac{1}{r\rho}}

And so the time period remains the same at 1.0s!

There's probably an easier way but that's how I'd do it
(edited 10 years ago)
Reply 98
Avatar for jonnyb123
jonnyb123
9
Original post by kingm
Not sure if i fluked that question or not then, but I concluded that T must be the same on the planet because the equation for a simple pendulum does not include mass, density or volume and thus T is not proportional to any of them.


I don't know, that might work ha, but T is proportional to 1/sqrt(g), and g is proportional to p and r. seeing as g is the same in both cases, the effect on T is the same in both cases. We'd need an example where g wasn't the same to see who's right ha


Original post by fizzbizz


That's a confusing looking formula I know, but the main thing to pick up is that

T∝1ρr T \propto \sqrt{\frac{1}{\rho r}}

So if density is halved but radius is doubled, the overall effect will cancel;

12r×0.5ρ=1rρ \sqrt{\frac{1}{2r\times 0.5\rho}}=\sqrt{\frac{1}{r\rho}}

And so the time period remains the same at 1.0s!

There's probably an easier way but that's how I'd do it


Yeah exactly, that's what I meant only I didn't actually substitute g, I just worked out g wouldn't change
(edited 10 years ago)
Reply 99
Avatar for Pgpacker
Pgpacker
Hey, has anyone got any questions on shm or electromagnetic induction? Also does anyone know where to access the old papers before 2010?

Thanks in advance :smile:

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