[madridsta]

(Original post by englishman129)
Thanks! Hows everyone scoring in the three papers and the spec paper?

I havent done em yet - just been using other practice papers as i want to save them for later.

can you please share the practice papers with me? I need some papers with answers as I only started Physics paper 4 two days ago and the exam is in 4 days. as I had a lot of other exams.

would really appreciate your help

[Vampire-Love4ever]

Edexcel Physics Jan 2011 QP & MS:
http://www.mediafire.com/?r3olp0he2px5j76

Physics Jan 2011 ER:
http://www.mediafire.com/?vnv1nmlvo8xd1h9

Unit 4 June 2010 QP & MS:
http://www.mediafire.com/?iyzyybzaodz
http://www.mediafire.com/?r92uo3wnj5ou4xb

[Vampire-Love4ever]

Jan 2010 ER:
https://eiewebvip.edexcel.org.uk/Rep...f_20100310.pdf

June 2010 ER:
http://www.scribd.com/doc/46115869/6...1-pef-20100818

[Vampire-Love4ever]

If anyone did the papers!
I have some doubts in these questions. I hope someone can help me with them. PLEASE.
I'd really appreciate any help I get.

*Jan '10:
Q15- THE WHOLE QUESTION.

*June '10:
Q3, Q4, Q13b, Q17 b, c, d(ii) and e.

*Jan '11:
Q12

[Vampire-Love4ever]

Hope this will be helpful.

[aym3n]

(Original post by Vampire-Love4ever)
If anyone did the papers!
I have some doubts in these questions. I hope someone can help me with them. PLEASE.
I'd really appreciate any help I get.

*Jan '10:
Q15- THE WHOLE QUESTION.

*June '10:
Q3, Q4, Q13b, Q17 b, c, d(ii) and e.

*Jan '11:
Q12

k im sleepy and lazy right now to actually get up and get the papers sorry about that

but but i did this one for u cuz the link to the paper was just above this post

jan '10 q15)

ai) use your ruler to measure the length of the radius above the lead plate and the thickness of the ruler and then cross multipy

so i got a length of 0.02 m so therefore

0.02 m --- 0.003m
x m --- 0.006m

so that gives u an answer of 0.04 m

aii) use the formula r= p/BQ and make p the subject formula

p= 0.04x1.5x(1.6x10^-19) = 9.6x10^-21 Ns

b) this one is easy from the markscheme

c) well im not sure of this one but the question mentions that the field is perpendicular to the plane of the photograph so it has to be into the page

but correct me if i am wrong

d) p=mv v= (9.6x10^-21)/(9.11x10^-31) = 1.05 x10^10 m/s

ii) it is greater than the speed of light so the mass increases

here you go

[aym3n]

(Original post by Vampire-Love4ever)
If anyone did the papers!
I have some doubts in these questions. I hope someone can help me with them. PLEASE.
I'd really appreciate any help I get.

*Jan '10:
Q15- THE WHOLE QUESTION.

*June '10:
Q3, Q4, Q13b, Q17 b, c, d(ii) and e.

*Jan '11:
Q12

okay so i got the papers :P

k june'10

q3 ) f= delta (momentum)/time
= 0.4-(-0.6)/ (4x10^-3)

4 ms is the time from the graph and we have to convert it cuz its in milli seconds

q4) its D

its cant be A cuz there isnt 2m and the equation doesnt show any reason to not have it

it cant be B cuz even if we take the m in the dinominator from p^2/2m and take it to the other side of the equation to get 1/2 m^2 v^2 = p^2 "2" from 2m has nowhere to be mentioned

it cant be C cuz we just cant have m^2 v^2 by just cancelling out 1/2 and 2 :P

im trying to fiqure out question 13b i got it wrong too

[inspiringsoul]

(Original post by aym3n)
k im sleepy and lazy right now to actually get up and get the papers sorry about that

but but i did this one for u cuz the link to the paper was just above this post

jan '10 q15)

ai) use your ruler to measure the length of the radius above the lead plate and the thickness of the ruler and then cross multipy

so i got a length of 0.02 m so therefore

0.02 m --- 0.003m
x m --- 0.006m

so that gives u an answer of 0.04 m

aii) use the formula r= p/BQ and make p the subject formula

p= 0.04x1.5x(1.6x10^-19) = 9.6x10^-21 Ns

b) this one is easy from the markscheme

c) well im not sure of this one but the question mentions that the field is perpendicular to the plane of the photograph so it has to be into the page

but correct me if i am wrong

d) p=mv v= (9.6x10^-21)/(9.11x10^-31) = 1.05 x10^10 m/s

ii) it is greater than the speed of light so the mass increases

here you go

Hi, there. can u please tell me, what do u cross multiply with? got the 0.02 m by measuring frm ruler, rest no.

[purplebrainz]

other than past papers and mark schemes, does anyone have any other revision material they would care to share? extra questions, revision notes, unit 4 or 5, much appreciated, i put all my energy into chemistry, and now im getting worried! ha.. also if u know any decent revision sites, quote me up!

[purplebrainz]

(Original post by Vampire-Love4ever)
To those who don't have the Edexcel A2 Physics Revision Guide.
You can download them from here:

http://www.mediafire.com/?fs5q4itt24emrxx

http://www.mediafire.com/?vvhbwhjln4lfh04 (answers)

(all thanks to M-H & saadmannan from SF)


Personally, I find it really helpful and the topics are explained briefly to the point (required according to our syllabus). It also has practice questions which are really good.

ah-may-zing. thank youuuuu.

[purplebrainz]

(Original post by Vampire-Love4ever)
Particle Physics- Edexcel Revision Guide.

http://www.studentforums.biz/referen...269/#msg407269

when i go here it says;


There are 5 attachment(s) in this post which you cannot view or download

why oh why?!

(also, in the edexcel mediafire revision guide thing, they are missing all the unit 4 answers lol )

[Vampire-Love4ever]

(Original post by purplebrainz)
when i go here it says;


There are 5 attachment(s) in this post which you cannot view or download

why oh why?!

(also, in the edexcel mediafire revision guide thing, they are missing all the unit 4 answers lol )

You need an account to see the attachments.

Really? I though it was there.

[Vampire-Love4ever]

(Original post by aym3n)
k im sleepy and lazy right now to actually get up and get the papers sorry about that

but but i did this one for u cuz the link to the paper was just above this post

jan '10 q15)

ai) use your ruler to measure the length of the radius above the lead plate and the thickness of the ruler and then cross multipy

so i got a length of 0.02 m so therefore

0.02 m --- 0.003m
x m --- 0.006m

so that gives u an answer of 0.04 m

aii) use the formula r= p/BQ and make p the subject formula

p= 0.04x1.5x(1.6x10^-19) = 9.6x10^-21 Ns

b) this one is easy from the markscheme

c) well im not sure of this one but the question mentions that the field is perpendicular to the plane of the photograph so it has to be into the page

but correct me if i am wrong

d) p=mv v= (9.6x10^-21)/(9.11x10^-31) = 1.05 x10^10 m/s

ii) it is greater than the speed of light so the mass increases

here you go

Thanks for the reply. I will go through them properly when I do the papers again.

xx

[fatima00]

can anyone help me with this one please?
In the annihilation of a proton-anti proton,the eq is

p+anti p=a neutral pion,2 +ve pions nd 2-ve pions
the incoming anti proton has an ENERGY of 1GeV.the proton is initially at rest and has a mass of 940MeV/c^2,mass of a neutral pion is 135MeV/c^2 while mass of a charged pion is 140MeV/c^2.
How much energy is available for the K.E. of the pions?

please quote me if you answer this so tht i get a notif,thanx

[aym3n]

(Original post by inspiringsoul)
Hi, there. can u please tell me, what do u cross multiply with? got the 0.02 m by measuring frm ruler, rest no.

we get the 0.02 m and the thickness of the lead plate using the measuring ruler .
this way we know that when the thickness was 0.003 m the radius was 0.02 m
so what will be the radius when the thickness is 0.006 m

if u still didnt get it...tell me ill try sum other way to explain

[aym3n]

(Original post by Vampire-Love4ever)
Thanks for the reply. I will go through them properly when I do the papers again.

xx

no problem im happy to help and im still trying to solve the question i just wokeup but i will quote u later for the rest ones

[aym3n]

(Original post by fatima00)
can anyone help me with this one please?
In the annihilation of a proton-anti proton,the eq is

p+anti p=a neutral pion,2 +ve pions nd 2-ve pions
the incoming anti proton has an ENERGY of 1GeV.the proton is initially at rest and has a mass of 940MeV/c^2,mass of a neutral pion is 135MeV/c^2 while mass of a charged pion is 140MeV/c^2.
How much energy is available for the K.E. of the pions?

please quote me if you answer this so tht i get a notif,thanx

hahah this question is in the revision guide and it sucks...cuz i tried it for an hour and still didnt get it ..and its just one mark

lol btw u shouldnt do that hard question....i wasted alot of time on these questions and so my teacher asked me to skip the really difficult ones cuz they cant give them or else the examiner council or idk what can cast a case on them

[fatima00]

(Original post by aym3n)
hahah this question is in the revision guide and it sucks...cuz i tried it for an hour and still didnt get it ..and its just one mark

lol btw u shouldnt do that hard question....i wasted alot of time on these questions and so my teacher asked me to skip the really difficult ones cuz they cant give them or else the examiner council or idk what can cast a case on them

yh its jus 1 mark:/ bt i dont get it why they've added all the masses.in annihilation qs we always subtract mass of those on the right frm mass of those on the left.tht gives us d mass defect which we substitute into E=mc^2 for energy released:/

[aym3n]

(Original post by fatima00)
yh its jus 1 mark:/ bt i dont get it why they've added all the masses.in annihilation qs we always subtract mass of those on the right frm mass of those on the left.tht gives us d mass defect which we substitute into E=mc^2 for energy released:/

hahah i dont know that either ... u c they are edexcel they can do anythhiiinngggg they want

[whizz-kid]

electricity thing is killing me