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A2 (Edexcel) Physics Unit 4 'Physics on the Move' - 24th January 2012

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Original post by Alpha-Omega


To get a discussion for Unit 4 revision started, why is the answer here C?


You are looking to calculate the flux linkage for coil Y, where Flux linkage=NBA

Where N= number of turns in coil, B=magnetic flux density and A=cross-sectional area of wire

Here, you know the magnetic flux density affecting coil Y is B=0.002T\mathrm{B=0.002T} as the current is only flowing through coil X.

From the question, you know that for coil Y, N=50\mathrm{N=50} and A=4cm2A=4cm^2.

But to calculate the flux linkage, you need the cross-sectional area to be in metres squared. To convert cm to m, you would divide by 100, but here, to convert squared cms to squared metres, you have to divide by 100x100, giving you A=0.0004m2A=0.0004m^2.

Another way to think of this, is:
4cm2=2cm×2cm[br]=0.02m×0.02m[br]=0.0004m24cm^2=2cm \times 2cm[br]=0.02m \times 0.02m[br]=0.0004m^2

And so to find the flux linkage, you multiply N, B and A.

Fluxlinkage=NBA[br]=50×0.002T×0.0004m2[br]=0.00004Wb[br]=4x105Wb\mathrm{Flux linkage}=NBA[br]=50 \times 0.002T \times 0.0004m^2[br]=0.00004 Wb[br]=4x10^{-5} Wb
(edited 12 years ago)
1. A body is falling freely under gravity. The rate at which the body's momentum is changing is equal to its
A acceleration
B kinetic energy
C potential energy
D weight

Why is the answer to this D?
Original post by whooshpaddy
Why is the answer to this D?


Thanks for the answer. Very well explained.

Resultant force=ma=m(v-u)/t = (mv - mu)/t = change in momentum / time = rate of change of momentum

So rate of change of momentum is equal to resultant force.

The only force acting on this free-falling object is weight.

So, weight is the rate of change of momentum.
Reply 43
is it me or is there too many content to remember??? how are people revising? and what's the best way to revise for this paper??

are there only 4 past papers???
Original post by Numan786
is it me or is there too many content to remember??? how are people revising? and what's the best way to revise for this paper??

are there only 4 past papers???


You should see A2 Chemistry....

Try the past paper posted in this thread. Just go back a page or two and you should see it there.
Original post by whooshpaddy
Not that I know of, although someone (really sorry but I can't remember who - please don't murder me!) uploaded this mock paper that their teacher had made for them in the thread for this exam in June 2011, although there is no mark scheme. I have attached it to this post. (Apologies apologies apologies to the original poster of this!)

I am assuming you have done the past papers from the edexcel website? Jan 2010, June 2010 and Jan 2011. Do you have the June 2011 past paper? If not I am sure could try and find it for you.


Do you still have your answers with you on the mock paper with the corrections? Would you
be able to scan and post it here? I got stuck in some questions.
(edited 12 years ago)
Reply 46
say I'm curious how does one predict questions from a examiers report or the word escapes me but it begins with an s??
Original post by Alpha-Omega
Do you still have your answers with you on the mock paper with the corrections? Would you
be able to scan and post it here? I got stuck in some questions.


As I said before, wasn't originally my paper - but am in process of making a mark scheme, for most of section B at least - I am finding that most of the Qs are from old OCR papers. Hopefully will be able to finish and upload it soon !
Original post by whooshpaddy
As I said before, wasn't originally my paper - but am in process of making a mark scheme, for most of section B at least - I am finding that most of the Qs are from old OCR papers. Hopefully will be able to finish and upload it soon !


That's what I meant. I thought you've done it already. Looking forward to the sample mark scheme.
Original post by Alpha-Omega
That's what I meant. I thought you've done it already. Looking forward to the sample mark scheme.


Haven't finished it - looks like we are going to have to decide on answers for question 3 though - I know what paper etc the question is from, but all the mark schemes I have found are blank for that section?! :s-smilie:

What did you write for the answers for Part B Q3?

Nearly done section B - 3Qs left to do
Mark Scheme attached for the paper I posted earlier - as good as I can get for the time being.

Section A and the last question are from a book and I'm not sure which one, so can't find the answers. When I have done them I could check with you guys on here and then if agreed, add to mark scheme.
Original post by whooshpaddy
As I said before, wasn't originally my paper - but am in process of making a mark scheme, for most of section B at least - I am finding that most of the Qs are from old OCR papers. Hopefully will be able to finish and upload it soon !


Section A

1. B ---- V = Q / C
2. A ---- Along the equipotential line, potential difference is constant; hence, no energy transfer.
3. B ---- F = BQV
4. C ---- Energy = VQ ---> V = E/Q = 150/30 = 5 (the x10-6 cancels out)

Section B

3ai) "If they come at rest at the same time, the total momentum would be zero. Initially, the total momentum is not
zero (similar speed in opposite directions, but different mass); therefore, this scenario can't happen as total momentum
has to be conserved."

3aii) (In this question, we're using the relative units for the value of mass)

Initial total momentum = 3u - 2u = u
Total momentum when they're closest together = (3 + 2) v, where v is the velocity (as they have the same velocity at this point)

so...

u = (3 + 2)v = 5v

v = u/5

3bi) Total initial kinetic energy = Total kinetic energy (when closest to each other) + total electric potential energy

3bii) Total initial kinetic energy = (1/2)(1.67 x 10-27)(3u2 + 2u2) = 4.175 x 10-27 u2 4.18 x 10-27 u2

3biii) E = (k(3e)(2e)) / (1.5 x 10-15) = 9.20576 x 10-13

This energy need to be overcome so total initial kinetic energy should be equal or greater than this.

Using value from 3bii:

9.20576 x 10-13 = 4.175 x 10-27 u2

Solve for u and you should get something like 1.48 x 107.
(edited 12 years ago)
Section A

Q1 I discounted B and D as there's just not enough info to work these out... Not C, as the time constant isn't time to halve, but time to fall to 37%.
So it must be A, as using 0.1ms=RC gives the capacitance as 1E-7F, which is 10μF.

Q2 ? I have no idea, brain is feeling fried...

Q3 B? as F=BQV, F=2 x 1.6E-19 x 1.6E7, F=5.12E-12 N
So we disagree on this one...

Q4 C? as E=F/Q and E=V/d,
so F/Q=V/d
so Fd=VQ, as work done=Forcexdistance,
150E-6 = 30E-6 x V
so V= 150/30 = 5V

I will do the last question later too, having a break now though... Back in 1 hour!
(edited 12 years ago)
Original post by Alpha-Omega
Section A

3. A ---- F = BQV


I'm till getting 5.12E-12 as an answer for this one...
Original post by whooshpaddy
I'm till getting 5.12E-12 as an answer for this one...


I got that on my calculation, but for some reason, I wrote A instead of B.
Reply 55
anyone make any predictions of what they may ask from looking at the specification.
Reply 56
guyz the problem with june 2011 paper is that theyve started asking so much irrelevant stuff.. I wish i was born 2 years earlier lol..
Reply 57
3 days to go :/
Anybody know off any great websites/pdf or whatever explaining the electric field & capacitors topic? They screw me over in all papers
Original post by nevetstreblig
Anybody know off any great websites/pdf or whatever explaining the electric field & capacitors topic? They screw me over in all papers


Have a look at these:

http://www.youtube.com/user/Steve4Physics/featured
http://www.youtube.com/user/ibphysicshelp/featured

I'm not sure if those topics you mentioned are here, but they're generally very good for A-Level Physics.

Which is it specifically on Electric Fields and Capacitors that you're finding the most difficult?

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