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OCR Physics B G495 Field and Particle Pictures June 21st 2011 Exam Thread

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Original post by 41jms
look at this guys: http://www.johnbright.conwy.sch.uk/depts/science/Content/Past_Papers/Y13/Physics/2003JuneMSA2.pdf

as it doesnt have page number type in the find (Ctrl + F) "cosine curve" it'll take you to the bit im on about.. erm the faint line on the graph is flux linkage. it asks you to draw on the graph how the voltage across the coil changes with time.

Can someone please explain how that markscheme is right?

I thought voltage was proportional to negative gradient. therefore the graph would look a bit different, like stationary points would still be the same but the maxima and minima the other way round?


Yes, they drew it (technically) the wrong way round. But you would still gain the marks :smile:
Reply 101
Original post by Summerdays
Yes, they drew it (technically) the wrong way round. But you would still gain the marks :smile:


thanks! could you possible explain 10 c on the same paper for me?

heres the question, page 11: http://www.johnbright.conwy.sch.uk/depts/science/Content/Past_Papers/Y13/Physics/2864June2003.pdf

thanks so much!!
Original post by 41jms
thanks! could you possible explain 10 c on the same paper for me?

heres the question, page 11: http://www.johnbright.conwy.sch.uk/depts/science/Content/Past_Papers/Y13/Physics/2864June2003.pdf

thanks so much!!


The fraction will decrease because the protons now have a greater speed (due to having a greater kinetic energy), which means that the time the lead nucleus has in trying to repel the protons (with a force 1.6x10^-19N) is less. Thus, the impulse acting on each proton is less and thus, less protons are scattered by 90 degrees.
Reply 103
Original post by Summerdays
The fraction will decrease because the protons now have a greater speed (due to having a greater kinetic energy), which means that the time the lead nucleus has in trying to repel the protons (with a force 1.6x10^-19N) is less. Thus, the impulse acting on each proton is less and thus, less protons are scattered by 90 degrees.


ahh thanks! please stay on this thread for the next few days :tongue:
Reply 104
Original post by Summerdays
The fraction will decrease because the protons now have a greater speed (due to having a greater kinetic energy), which means that the time the lead nucleus has in trying to repel the protons (with a force 1.6x10^-19N) is less. Thus, the impulse acting on each proton is less and thus, less protons are scattered by 90 degrees.


tried to rep you, but i have ran out..
Reply 105
Guys I've done most of the past papers I have and Ive gone through both the specification and the advancing physics textbook and tbh Im not really sure what else I should to to prepare... Can I have some interesting ideas rather than just go through the specification again because one of the problems is that i find it all really boring and so find it hard to actually work, and I have today and tommorow scheduled to prepare for the exam plus monday afternoon and evening. any websites that I could work through that are interesting... I just feel like if i just go through the textbook more It will be so boring I barely pay attention to the notes I make
Original post by Sereni
Guys I've done most of the past papers I have and Ive gone through both the specification and the advancing physics textbook and tbh Im not really sure what else I should to to prepare... Can I have some interesting ideas rather than just go through the specification again because one of the problems is that i find it all really boring and so find it hard to actually work, and I have today and tommorow scheduled to prepare for the exam plus monday afternoon and evening. any websites that I could work through that are interesting... I just feel like if i just go through the textbook more It will be so boring I barely pay attention to the notes I make


http://ocw.mit.edu/courses/physics/8-02-electricity-and-magnetism-spring-2002/video-lectures/

http://freevideolectures.com/Course/2876/Fundamentals-of-Physics-III

http://www.khanacademy.org/#physics

http://www.freelance-teacher.com/videos.htm
Reply 107


Cheers, now im just going through them just trying to figure out which ones are on this exam :smile:.

Also is it just me or is anyone else suspisious over how easy the advance notice material is...
Reply 108
Original post by Sereni
Cheers, now im just going through them just trying to figure out which ones are on this exam :smile:.

Also is it just me or is anyone else suspisious over how easy the advance notice material is...


Yeah I am, but maybe it just will be easy?

My Chemistry exam had a 'synoptic element' and the exam was basically all AS stuff so was really quite easy, easier then the past papers at least. I'm hoping for the same with physics haha.

I still hate magnets.
Is this a morning exam?
Reply 109
Original post by Ollie901
Yeah I am, but maybe it just will be easy?

My Chemistry exam had a 'synoptic element' and the exam was basically all AS stuff so was really quite easy, easier then the past papers at least. I'm hoping for the same with physics haha.

I still hate magnets.
Is this a morning exam?


its morning for me (england). mm I need an A in physics to meet my med offer
Reply 110
Original post by firstevermember
Question 4)B)ii) 4marks for drawing this graph... I dont get why it should be a 'square pulse’ shaped graph. The mark squeme is at bottom of pdf.

http://www.ocr.org.uk/download/assess_mat/ocr_7977_sam_gce_unit_g485.pdf

(Got 68/100 on this paper, question 7 i got 1/10 :biggrin: )


Jesus Christ. Was panicing like hell looking through that paper then, then I realised its OCR A.
Original post by Ollie901
Jesus Christ. Was panicing like hell looking through that paper then, then I realised its OCR A.


Yer me too, I read cosmic background radiation and I was like "When did we ever learn that :confused:" But yes, then realized it was OCR A
LOOOOOL sry about that... ive done that before when looking at a Physics B.
Reply 113
Can someone please explain magnets/motors to me please pleasee?! :frown: :frown:
Reply 114
Really nervous for this exam, its my toughest exam and the most important one. Out of 150UMS I only need 96 to get into Uni so long as I pass my other exams, which is only about 64/100 marks but still, im not good at Physics at all and some questions really bum me over. Still cant tackle any special relativity stuff and long answer questions are baaaddd.

Does anyone know any equations with voltage in them that you need to know, excluding the obvious like V=IR, P=V^2/R and P = I^2R.... would really be useful, thanks.
Reply 115
Does anyone have any resources for revision notes??? I find the textbook is really unhelpful and the CGP book doesn't go into much depth. :s-smilie:

Thanks for all the questions on the pre-release they've been really helpful. :smile:
Reply 116
can someone explain question 3bii) from the Jan 06 paper to me? It's a one mark question and the answer is 1.8ev, i have no idea how they got that
Reply 117
After comparing my answers with Summerdays, here are the ones where I differed, or used a different method, all the others I got the same answer. The vast majority seem right to me :smile:


Spoiler



ALL OBSOLETE NOW WITH ANSWERS GIVEN ON PAGE 7!!!

I'll add more tomorrow hopefully. :biggrin:
(edited 12 years ago)
Original post by JoeCarr
After comparing my answers with Summerdays, here are the ones where I differed, or used a different method, all the others I got the same answer. The vast majority seem right to me :smile:


Spoiler



I'll add more tomorrow hopefully. :biggrin:



Thank you very much :biggrin: I will have a look at your solution, and the ones you think I got right, to revise this section C :tongue:
Original post by bear54
can someone explain question 3bii) from the Jan 06 paper to me? It's a one mark question and the answer is 1.8ev, i have no idea how they got that


Due to a 12eV electron colliding inelastically with the hydrogen atom, the answer is based on the assumption that the electron, that is bound to the nucleus of the hydrogen atom, gains as much energy as possible. We also assume that the electron was in the ground state, before the collision took place. So -13.6eV + 12eV = -1.6eV. But this means that the electron (bound to the hydrogen nucleus) has moved to the energy level that is equal to -3.4eV (there wasn't enough energy for the electron to move to the energy level "-1.5eV", because this energy level is less negative than -1.6eV)
This means that 3.4eV-1.6eV = 1.8eV is left as kinetic energy, for the electron.

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