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

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1. (Original post by 41jms)
look at this guys: http://www.johnbright.conwy.sch.uk/d...03JuneMSA2.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
2. (Original post by Summerdays)
Yes, they drew it (technically) the wrong way round. But you would still gain the marks
thanks! could you possible explain 10 c on the same paper for me?

heres the question, page 11: http://www.johnbright.conwy.sch.uk/d...64June2003.pdf

thanks so much!!
3. (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/d...64June2003.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.
4. (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
5. (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..
6. 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
7. (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...ideo-lectures/

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

http://www.freelance-teacher.com/videos.htm
8. Cheers, now im just going through them just trying to figure out which ones are on this exam .

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

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?
10. (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
11. (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.

(Got 68/100 on this paper, question 7 i got 1/10 )
Jesus Christ. Was panicing like hell looking through that paper then, then I realised its OCR A.
12. (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 " But yes, then realized it was OCR A
13. LOOOOOL sry about that... ive done that before when looking at a Physics B.
15. 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.
16. 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.

Thanks for all the questions on the pre-release they've been really helpful.
17. 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
18. 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

Spoiler:
Show

1.
b) 60 seconds / 70 beats. 1 beat = 0.86 seconds. Uncertainty = 1/10th so 0.086 seconds, which is 0.09s to 1 sig fig.

d) I think because the t is squared, our relative uncertainty of 0.09/2.64 is doubled, and so is 6.8% (this may seem high but he hardly worked to a great degree of accuracy?) , not 3.4, thus giving a (+-) of 0.12 but I’m not great at uncertainty so I may very well be wrong.

e) 1.72 +/- 0.12 ms-2

6.b) I would tend to argue that the earth is an oblate spheroid because the rotation produces a centrifugal force perpendicular to the spin (poles), so this force is concentrated on the equator and this is why the value for ‘g’ differs, rather than saying because ‘g’ differs, the earth is an oblate spheroid.

c) The question asks you to find the difference in radius, I think you rushed as you just gave the value of ‘r’ for the radius at the poles. :P
6374004.075m - 6368231.782m = 5772.3m difference

7.a
I would naturally choose your method also, however after discussion with a friend we were unsure if it were too simple a method, especially when the question is worth 4 marks. I have no better answer though :P

8.c) I’ve always assumed that each maxima represents a whole wavelength, as surely if you move a distance of half a wavelength back, you get a minima or destructive interference, and then if you moved it a further 0.5 lambda back, so a full wavelength, it would become a maxima again. So I got 2.3*10^-5

8.f) again, the maxima/wavelength thing means I get 1.56164*10^5 as I took 86 maxima to be 86 wavelengths….

g.) I prefer your answer as it is more distinctively different than mine to the previous wavelength haha, however using my previously calculated value I got 6.4035*10^-7

9.b) I used K.E. = 1/2mv^2, giving v = 628

c) same method, except with my answer I got 1.23*10^-12

d) I also do not like this question, I probably would have said something along the lines of ‘the Doppler effect causes a fractional change in wavelength, however this distortion because of it is comparable to the wavelength itself, so differing speeds of the sodium atoms results in noticeably different wavelengths, so the light is not monochromatic.’

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

19. (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

Spoiler:
Show

1.
b) 60 seconds / 70 beats. 1 beat = 0.86 seconds. Uncertainty = 1/10th so 0.086 seconds, which is 0.09s to 1 sig fig.

d) I think because the t is squared, our relative uncertainty of 0.09/2.64 is doubled, and so is 6.8% (this may seem high but he hardly worked to a great degree of accuracy?) , not 3.4, thus giving a (+-) of 0.12 but I’m not great at uncertainty so I may very well be wrong.

e) 1.72 +/- 0.12 ms-2

6.b) I would tend to argue that the earth is an oblate spheroid because the rotation produces a centrifugal force perpendicular to the spin (poles), so this force is concentrated on the equator and this is why the value for ‘g’ differs, rather than saying because ‘g’ differs, the earth is an oblate spheroid.

c) The question asks you to find the difference in radius, I think you rushed as you just gave the value of ‘r’ for the radius at the poles. :P
6374004.075m - 6368231.782m = 5772.3m difference

7.a
I would naturally choose your method also, however after discussion with a friend we were unsure if it were too simple a method, especially when the question is worth 4 marks. I have no better answer though :P

8.c) I’ve always assumed that each maxima represents a whole wavelength, as surely if you move a distance of half a wavelength back, you get a minima or destructive interference, and then if you moved it a further 0.5 lambda back, so a full wavelength, it would become a maxima again. So I got 2.3*10^-5

8.f) again, the maxima/wavelength thing means I get 1.56164*10^5 as I took 86 maxima to be 86 wavelengths….

g.) I prefer your answer as it is more distinctively different than mine to the previous wavelength haha, however using my previously calculated value I got 6.4035*10^-7

9.b) I used K.E. = 1/2mv^2, giving v = 628

c) same method, except with my answer I got 1.23*10^-12

d) I also do not like this question, I probably would have said something along the lines of ‘the Doppler effect causes a fractional change in wavelength, however this distortion because of it is comparable to the wavelength itself, so differing speeds of the sodium atoms results in noticeably different wavelengths, so the light is not monochromatic.’

Thank you very much I will have a look at your solution, and the ones you think I got right, to revise this section C
20. (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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