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AQA A Physics Unit 4 24th Jan 2012 Tweet

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1. Re: AQA A Physics Unit 4 24th Jan 2012
Could someone explain gravitational potential to me please? I can do the questions via formal but I don't really get it...
Can anyone think of anything strange that could come up?. Also If I cover everything on the spec that should be enough right?
2. Re: AQA A Physics Unit 4 24th Jan 2012
(Original post by kimmey)
thanks for this but can you please tell me how volume/area = v ? thanks
Well 2.0*10^-4 is how much volume of water is flowing past one point each second, so it is cross-sectional area * velocity, and you want just velocity, you divide it by the cross-sectional area to get velocity.

Plus you can just look at the units, to get the unit of velocity (ms^-1) using m^3s^-1 (meters cubed per second) and m^2 (meters squared), simply divide the first by the second and you're left with ms^-1 (meters per second).
3. Re: AQA A Physics Unit 4 24th Jan 2012
After being rejected by Cambridge, I'm even more pumped to own this exam.
4. Re: AQA A Physics Unit 4 24th Jan 2012
(Original post by Femto)
After being rejected by Cambridge, I'm even more pumped to own this exam.
Lol same here, although I got rejected from oxford :P
5. Re: AQA A Physics Unit 4 24th Jan 2012
(Original post by maths134)
Could someone explain gravitational potential to me please? I can do the questions via formal but I don't really get it...
Can anyone think of anything strange that could come up?. Also If I cover everything on the spec that should be enough right?
Nothing strange should come up, but if it does we cant really prepare for it.
6. Re: AQA A Physics Unit 4 24th Jan 2012
Does anyone have a graph for kinetic energy against velocity for SHM, if that exists?
7. Re: AQA A Physics Unit 4 24th Jan 2012
(Original post by --NWzD9--)
Does anyone have a graph for kinetic energy against velocity for SHM, if that exists?
Surely it would take the shape of a parabola?

EDIT: Like this:

http://img.sparknotes.com/content/te...8/KEgraphs.gif

But it would have to have a negative velocity as well surely as, with any SHM system, the velocity changes direction.
Last edited by Femto; 05-01-2012 at 18:17.
8. Re: AQA A Physics Unit 4 24th Jan 2012
(Original post by Femto)
Surely it would take the shape of a parabola?

EDIT: Like this:

http://img.sparknotes.com/content/te...8/KEgraphs.gif

But it would have to have a negative velocity as well surely as, with any SHM system, the velocity changes direction.
Oops, sorry, what i wrote was wrong. I was meant to say kinetic energy against time, starting from maximum displacement?
9. Re: AQA A Physics Unit 4 24th Jan 2012
(Original post by --NWzD9--)
Oops, sorry, what i wrote was wrong. I was meant to say kinetic energy against time, starting from maximum displacement?
You could draw a displacement time graph then find the gradient and that would be v against t and square it? Might seem too long though.
10. Re: AQA A Physics Unit 4 24th Jan 2012
(Original post by --NWzD9--)
Oops, sorry, what i wrote was wrong. I was meant to say kinetic energy against time, starting from maximum displacement?
Hmm, I don't think you'd be asked that. Although I'm not 100% sure about what I've just said - would somebody be able to clarify this?
11. Re: AQA A Physics Unit 4 24th Jan 2012
http://www.has.vcu.edu/new-phy/WSG/C...mages/fig1.jpg

I was told it would be shaped like this, instead with the y-axis as KE
12. Re: AQA A Physics Unit 4 24th Jan 2012
How do you work out the answers to Q.13 & 24 from the Multiple Choice section of the June 2010 paper? Mark scheme says answers are A & C respectively

13. Re: AQA A Physics Unit 4 24th Jan 2012
(Original post by --NWzD9--)
How do you work out the answers to Q.13 & 24 from the Multiple Choice section of the June 2010 paper? Mark scheme says answers are A & C respectively

for 13 us v=1/4pi epsilon nought*(q/r)
then use v4=v-16
then you have 4/x=16/(120-x)
then working out you get x =24
or you could do it as a sort of ratio so 4+16=20
and 4 in proportion of that is 5
then 120/5=24

then for 24 it rotates twice as fast so cuts twice as much flux so there has got to be 2 waves in the graph space, and then as it is double omega then amplitude will double so its c

hope that helps
14. Re: AQA A Physics Unit 4 24th Jan 2012
(Original post by nexl)
The 6 mark questions in past papers are:

Specimen - SHM
Jan 10 - Transformers
Jun 10 - Momentum
Jan 11 - Capacitors
Jun 11 - Electricity

There could be a question on fields this time, since it hasn't come up before.
is there anything else that could show up apart from fields???
15. Re: AQA A Physics Unit 4 24th Jan 2012
Does anyone have the mark scheme for June 2011 cos I just did it as a mock and wanna I know how I did lol.
16. Re: AQA A Physics Unit 4 24th Jan 2012
can any1 explain how to do this Question

19. A particle of mass m moves in a circle of radius r at uniform speed, taking time T for each revolution. What is the kinetic energy of the particle?
17. Re: AQA A Physics Unit 4 24th Jan 2012
(Original post by mk207)
can any1 explain how to do this Question

19. A particle of mass m moves in a circle of radius r at uniform speed, taking time T for each revolution. What is the kinetic energy of the particle?
Ok. W=2pie/T. V=wr so v=2rpie/T, square it and sub into Ek=1/2mv2
18. Re: AQA A Physics Unit 4 24th Jan 2012
and another 1

23. Protons, each of mass m and charge e, follow a circular path when travelling perpendicular to a magnetic field of uniform flux density B. What is the time taken for one complete orbit?
19. Re: AQA A Physics Unit 4 24th Jan 2012
(Original post by mk207)
and another 1

23. Protons, each of mass m and charge e, follow a circular path when travelling perpendicular to a magnetic field of uniform flux density B. What is the time taken for one complete orbit?
20. Re: AQA A Physics Unit 4 24th Jan 2012
(Original post by mk207)
and another 1

23. Protons, each of mass m and charge e, follow a circular path when travelling perpendicular to a magnetic field of uniform flux density B. What is the time taken for one complete orbit?
do mw^2r=BQV dont forget w=2pi/T solve for T
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