Edexcel Physics Unit 4 - 21st June 2011

OP

PHYSICS UNIT 4 JAN 2010.pdf542.1KB

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

(edited 12 years ago)

PHYSICS UNIT 4 MS JAN 2010.pdf280.1KB

Reply 62

OP

Jan 2010 ER:
https://eiewebvip.edexcel.org.uk/Reports/Confidential%20Documents/1001/6ph04_01_pef_20100310.pdf

June 2010 ER:
http://www.scribd.com/doc/46115869/6PH04-01-pef-20100818

Reply 63

OP

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. :h:

*Jan '10:
Q15- THE WHOLE QUESTION. :eek:

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

*Jan '11:
Q12

Reply 64

OP

Particles, Radiation & Quantum Phenomena.ppt387.1KB

Hope this will be helpful. :smile:

Reply 65

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. :h:

*Jan '10:
Q15- THE WHOLE QUESTION. :eek:

*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 :colondollar:

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

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 :smile:

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 :biggrin:

Reply 66

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. :h:

*Jan '10:
Q15- THE WHOLE QUESTION. :eek:

*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 :tongue:

i got it wrong too :tongue:

Reply 67

inspiringsoul

Original post by aym3n

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

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

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 :smile:

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 :biggrin:

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

Reply 68

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!

Reply 69

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. :h:

ah-may-zing. thank youuuuu.

Reply 70

purplebrainz

Original post by Vampire-Love4ever

Particle Physics- Edexcel Revision Guide.

http://www.studentforums.biz/reference-material-83/phy-unit-4-particle-physics/msg407269/#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 :frown:

)

Reply 71

OP

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 :frown:

)

You need an account to see the attachments.

Really? I though it was there. :s-smilie:

Reply 72

OP

Original post by aym3n

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

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

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 :smile:

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 :biggrin:

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

xx

Reply 73

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 :smile:

Reply 74

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 :smile:

Reply 75

Original post by Vampire-Love4ever

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

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 :smile:

Reply 76

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 :smile:

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 :tongue:

Reply 77

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 :tongue:

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:/

Reply 78