I thought that was the end, until i accidentally flicked through, with under 5 minutes left. Think i got the 4 marker right, but i didn't even read the last two questions.(Original post by Cowy97)
it felt like it was there as a troll
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OCR PHYSICS B G495~ 18th June 2015 AM ~ A2 Physics watch

Juggerman Dan
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 18062015 18:13
A friend and I have managed to remember these and *think* these are the answers. If there's anything we're missing, please do post!
SECTION A
1 Particle Physics [2]
a. (neutrino is) Lepton and Neutral
b. (proton is) Nucleon and Hadron
2 Scattering [3]
Particle deflected more
Because particle is moving slower and so force has more time to act on it
3 Graph of V vs r [3]
Field is uniform electric field
Field strength is the gradient
Gradient is constant, so E is constant
E = dV/dr = 3000/0.07 = 4.3 x 10^4 Vm1
4 Alpha Safety [1, 2]
a. Mass is NOT conserved.
b. Agree (they are dangerous inside the body)
Alpha particles have high ionising ability, because they have a 2+ charge and can take electrons away from atoms easily.
They also have a low penetrating ability, so they can’t escape the body if ingested.
(ORA: Alpha particles highly ionising, but are stopped by a few cm of air/skin, so cannot get inside the body unless ingested/swallowed)
5 Transformers [1]
C (6 V and 0.36A)
6 Transformers (again) [2, 3]
a. flux is the same everywhere
flux density is double at point Z than at the sides/top of the core [must say double/twice/etc]
b. Crack introduces air gap
Air has much lower permeability compared to iron (about 1/8000th)
So overall permeance of core decreases
Φ is proportional to Λ, so flux increases
B is proportional to Φ, so flux density increases
7 Energy Levels [2]
ΔE = 1.5 eV + 3.4 eV = 1.9 eV = 3.04x10^19 J
f = 4.6 x 10^14 Hz
λ = 6.5 x 10^7 m
Lamda or frequency?
8 ???
SECTION B
9 Electric Fields [1, 1, 1, 1, 4, 2]
a i. field strength greater at X since field lines are closer together.
a ii.draw circle around the charge passing through X.
b i. Divide by 4, so (+)405 V
b ii. Divide by 16 so 6062.5 N
c. For Fig 9.3a, resultant force is 0 N, so E = F/q = 0 NC1
(OR calculations and statement: Eleft = Eright, so no resultant field strength)
For Fig 9.3b, resultant force is 2 x kQq/r2 = 6.3 x 10^7 x q N
So E = F/q = 6.3 x 10^7 NC1
d. For +ve charge, Velec = +kQ/r
For ve charge, Velec = kQ/r
So total V = +kQ/r  kQ/r = 0
10 Motors
a. F = BIL = ???
b. If L is halved, F is halved (since F is proportional to L), so the acceleration halves, and the motor moves more slowly. QUERY: could be as F halves, but m halves, acceleration is the same, so no change in rate.
c. max NΦ = NBA = 0.0243 Wbturns
d. average e.m.f = d(NΦ)/dt = 0.12(15) V
e. When moving, the motor is experiencing a change of flux linkage, and so an emf is induced.
Current is induced too, and by Lenz's law the induced current must flow in such a way that it opposes the change creating it, so it acts in the opposite direction to the current already there.
So the resultant current decreases while the motor turns.
When the motor is jammed (not moving) there is no change in flux linkage, so no emf is induced, and so no opposing current is induced. This means the current remains as it is (higher than when rotating).
11 Particles in Magnetic Fields [1?, 1, 2, 4]
a. Force acts perpendicular to direction of motion, so motion is circular
b. r = mv/Bqc. Alpha particle is much heavier than beta particle, so you would expect the radius of the circular path it takes to be much larger, as predicted in the diagram.
Gamma straight because it is uncharged, so is unaffected by the magnetic field.
d. rα = 0.20625/B
rβ = 0.0011375/B
rβ is 180 times smaller than rα, so diagram incorrect in the sense that it only shows slight difference in radius.
12 Radiation [1, 2, 3, 2, 3]
a. LHS: Lepton number = 0 [nuclei have no lepton number]
RHS: Lepton number = 0 + 1  1 = 0 [nuclei have no lepton number, electron = +1, antineutrino = 1]
So lepton number conserved.
b. λ = 1.0 x 106 s1A0 = 5.49 x 10^5 Bq
c. A12 weeks = 394 Bq [345ish if you use their values)
d. Absorbed dose = 0.89 Gy
Assumption is all the beta particles are absorbed by the gland
e. γ rays have high penetrating ability, so can escape from body
Radiation can therefore be detected by external sensors to track movement through body.
γ rays have low ionising ability, so do not cause much damage (as long as not too much is given)
Iodine131 not used anymore, because the halflife is too long.
The total absorbed dose by the thyroid gland due to 131I was too high (so increases risk of cancer by unacceptable amounts).
SECTION C
13 Lenses [1, 2]
a. magnification = v/u = 89/5.3 = ()16.79… = ()17x
b. power = 200 D
14 Imaging [2, 2]
a. 5.2 x 107 m
b. Approx. 270 lenses in picture, so 20,000 lenses in total, but they will probably accept values in a large range
c. A part c????
15 Gases and Boltzmann Factor [3, 4]
a. ??? something here ???
b. Number of particles per cubic cm was 2.4 x 10^9
c. I is proportional to eE/kT... Boltzmann factor can be used because...
BF is proportion of electrons with enough energy to escape.
Rate of electron release is proportional to BF
Current is proportional to number of electrons flowing (since current is rate of flow of charge)
So current is proportional to BF
d. T = 2310 K
16 B Fields [2, 3]
a. N = 3800 turns [show that]
b. 380390 V (allow range as ft from a)
17 Wavelength of Electrons
a. i. E = 6.4 x 10^15 J
ii. p = 1.1 x 10^22 kgms1
b. λ = 6.1 x 10^12 m
Resolution = 3.1 x 10^12 m
c. number of electrons needed to change temperature by 1 K = 1.8 x 10^6
18 Deflecting Electrons [1, 4, 2]
a. E = 1.1 x 10^5 Vm1
b. L = 0.027m
c. E is constant everywhere (it is a uniform field) so force is the same everywhere / process in (b) is the same everywhere.
19 Relativity [3]
γ = 1.39
v = 2.1 x 10^8 ms1Last edited by Greating; 18062015 at 19:16. 
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 18062015 18:22
(Original post by Greating)
A friend and I have managed to remember these and *think* these are the answers. If there's anything we're missing, please do post!
SECTION A
1 Particle Physics [2]
a. (neutrino is) Lepton and Neutral
b. (proton is) Nucleon and Hadron
2 Scattering [3]
Particle deflected more
Because particle is moving slower and so force has more time to act on it
3 Graph of V vs r [3]
Field is uniform electric field
Field strength is the gradient
Gradient is constant, so E is constant
E = dV/dr = 3000/0.07 = 4.3 x 104 Vm1
4 Alpha Safety [1, 2]
a. Mass is NOT conserved.
b. Agree (they are dangerous inside the body)
Alpha particles have high ionising ability, because they have a 2+ charge and can take electrons away from atoms easily.
They also have a low penetrating ability, so they can’t escape the body if ingested.
(ORA: Alpha particles highly ionising, but are stopped by a few cm of air/skin, so cannot get inside the body unless ingested/swallowed)
5 Transformers [1]
C (6 V and 0.36A)
6 Transformers (again) [2, 3]
a. flux is the same everywhere
flux density is double at point Z than at the sides/top of the core [must say double/twice/etc]
b. Crack introduces air gap
Air has much lower permeability compared to iron (about 1/8000th)
So overall permeance of core decreases
Φ is proportional to Λ, so flux increases
B is proportional to Φ, so flux density increases
7 Energy Levels [2]
ΔE = 1.5 eV + 3.4 eV = 1.9 eV = 3.04x1019 J
f = 4.6 x 1014 Hz
λ = 6.5 x 107 m
Lamda or frequency?
8 ???
SECTION B
9 Electric Fields [1, 1, 4, 2]
a i. Divide by 4, so (+)405 V
a ii. Divide by 16 so 6062.5 N
b. For Fig 9.3a, resultant force is 0 N, so E = F/q = 0 NC1
(OR calculations and statement: Eleft = Eright, so no resultant field strength)
For Fig 9.3b, resultant force is 2 x kQq/r2 = 6.3 x 107 x q N
So E = F/q = 6.3 x 10^7 NC1
c. For +ve charge, Velec = +kQ/r
For ve charge, Velec = kQ/r
So total V = +kQ/r  kQ/r = 0
10 Motors
a. F = BIL = ???
b. If L is halved, F is halved (since F is proportional to L), so the acceleration halves, and the motor moves more slowly.
c. max NΦ = NBA = 0.0243 Wbturns
d. average e.m.f = d(NΦ)/dt = 0.12(15) V
e. When moving, the motor is experiencing a change of flux linkage, and so an emf is induced.
Current is induced too, and by Lenz's law the induced current must flow in such a way that it opposes the change creating it, so it acts in the opposite direction to the current already there.
So the resultant current decreases while the motor turns.
When the motor is jammed (not moving) there is no change in flux linkage, so no emf is induced, and so no opposing current is induced. This means the current remains as it is (higher than when rotating).
11 Particles in Magnetic Fields [1?, 1, 2, 4]
a. Force acts perpendicular to direction of motion, so motion is circular
b. r = mv/Bqc. Alpha particle is much heavier than beta particle, so you would expect the radius of the circular path it takes to be much larger, as predicted in the diagram.
Gamma straight because it is uncharged, so is unaffected by the magnetic field.
d. rα = 0.20625/B
rβ = 0.0011375/B
rβ is 180 times smaller than rα, so diagram incorrect in the sense that it only shows slight difference in radius.
12 Radiation [1, 2, 3, 2, 3]
a. LHS: Lepton number = 0 [nuclei have no lepton number]
RHS: Lepton number = 0 + 1  1 = 0 [nuclei have no lepton number, electron = +1, antineutrino = 1]
So lepton number conserved.
b. λ = 1.0 x 106 s1A0 = 5.49 x 105 Bq
c. A12 weeks = 394 Bq
d. Absorbed dose = 0.89 Gy
Assumption is all the beta particles are absorbed by the gland
e. γ rays have high penetrating ability, so can escape from body
Radiation can therefore be detected by external sensors to track movement through body.
γ rays have low ionising ability, so do not cause much damage (as long as not too much is given)
Iodine131 not used anymore, because the halflife is too long.
The total absorbed dose by the thyroid gland due to 131I was too high (so increases risk of cancer by unacceptable amounts).
SECTION C
13 Lenses [1, 2]
a. magnification = v/u = 89/5.3 = ()16.79… = ()17x
b. power = 200 D
14 Imaging [2, 2]a. 5.2 x 107 mb. Approx. 270 lenses in picture, so 20,000 lenses in total, but they will probably accept values in a large range
I feel like there was a question here15 Gases and Boltzmann Factor [3, 4]a. ??? something here ???b. Number of particles per cubic cm was 2.4 x 109c. I is proportional to eE/kT... Boltzmann factor can be used because...BF is proportion of electrons with enough energy to escape.Rate of electron release is proportional to BFCurrent is proportional to number of electrons flowing (since current is rate of flow of charge)So current is proportional to BFd. T = 2310 K
16 B Fields [2, 3]a. N = 3800 turns [show that]b. 380390 V (allow range as ft from a)
17 Wavelength of Electronsa. i. E = 6.4 x 1015 Jii. p = 1.1 x 1022 kgms1b. λ = 6.1 x 1012 mResolution = 3.1 x 1012 mc. number of electrons needed to change temperature by 1 K = 1.8 x 106
18 Deflecting Electrons [1, 4, 2]a. E = 1.1 x 105 Vm1b. L = 0.027mc. E is constant everywhere (it is a uniform field) so force is the same everywhere / process in (b) is the same everywhere.
19 Relativity [3]γ = 1.39v = 2.1 x 108 ms1 
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 18062015 18:36
(Original post by Cowy97)
for 10 i said when length halfed the force halfed but so did the mass so f=ma had the same acceleration
As the side length of the coil halves, the mass of the total coil would quarter, not halve (so if F halves, but m quarters, a would double, so rate of rotation would increase).
However, since the rate of rotation is determined by some hideously complicated moment of inertia (which we definitely aren't required to do in physics A Level), I just took the reasoning that let us assume the mass is still the same, so force halves and rate decreases.
You could be right, I don't know...
EDIT: this is incorrect. See QuantumHover's response below.Last edited by Greating; 18062015 at 18:49. 
megatovskimaths
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 18062015 18:38
(Original post by HennersPD)
I hope so too becauae although i liked section C, section A and B were very tough in places :/ whatd you think 80/100 on this will be in UMS and 45/60 on G494?Last edited by megatovskimaths; 18062015 at 18:47. 
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 18062015 18:38
(Original post by Greating)
Hmm... I was tempted by that argument in the exam, but there are a few points to note:
As the side length of the coil halves, the mass of the total coil would quarter, not halve (so if F halves, but m quarters, a would double, so rate of rotation would increase).
However, since the rate of rotation is determined by some hideously complicated moment of inertia (which we definitely aren't required to do in physics A Level), I just took the reasoning that let us assume the mass is still the same, so force halves and rate decreases.
You could be right, I don't know... 
QuantumHover
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 18062015 18:40
(Original post by Greating)
As the side length of the coil halves, the mass of the total coil would quarter, not halve 
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 18062015 18:46
(Original post by QuantumHover)
No, because the coil is only on the outside of the square the increase is linear not quadratic. 
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 18062015 18:50
(Original post by STATER)
Sounds good. As always the main thing holding me back is I misread questions, and do stupid calculation errors. I might have scraped an A* in maths hopefully. G495 went very well as far as I know  only around 6 marks that I lost for sure. G494 was borderline  probably an A but not an A*. 
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 18062015 18:58
I agree with a lot of those answers Cowy97 has given, so I'm pretty confident I've got an A and might be being a little optimistic but maybe an A*.... I seem to mainly get A/A* in homeworks and stuff but physics exams are a fair bit different to those homework questions, and I'd rather study in school with lessons rather than on my own haha. Let's hope all of this work pays off, and everyone elses work pays off for them!

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 18062015 19:01
(Original post by RealTime)
I agree with a lot of those answers Cowy97 has given, so I'm pretty confident I've got an A and might be being a little optimistic but maybe an A*.... I seem to mainly get A/A* in homeworks and stuff but physics exams are a fair bit different to those homework questions, and I'd rather study in school with lessons rather than on my own haha. Let's hope all of this work pays off, and everyone elses work pays off for them! 
AccountName
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 18062015 19:42
(Original post by megatovskimaths)
Personally my guess is about 48/60 for 81 ums on g494 and 82/100 for 135 ums in this paper but that may be optimistic... 😜 
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 18062015 19:44
(Original post by HennersPD)
Yeah too think i may have got an A on this or even A*! Hope so anyway! what'd you draw for the equipotential at X? 
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 18062015 19:46
(Original post by AccountName)
It was 81/100 for 135 ums last year (and the year before). Considering that (through the polls linked to this thread and last year's equivalent) this year's paper seems to have been slightly harder than last year's, I'm hoping for a slightly lower 90 percent mark, although I suppose this could be wishful thinking. 
Mutleybm1996
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 18062015 19:59
(Original post by Cowy97)
for 10 i said when length halfed the force halfed but so did the mass so f=ma had the same acceleration
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 18062015 20:10
(Original post by Mutleybm1996)
For 7 don't you do 3.41.5 or whatever it was?
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Mutleybm1996
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 18062015 20:15
(Original post by RealTime)
I'd assume that it'd be similar, the paper was around the same difficulty as others but that's just my opinion
If some people missed the last page....drops lower
Apart from that I thought it was okayish
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 18062015 20:16
(Original post by RealTime)
I just drawn a circle 
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 18062015 20:18
(Original post by Mutleybm1996)
There was about 10 marks of truly challenging stuff
If some people missed the last page....drops lower
Apart from that I thought it was okayish
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 18062015 20:18
(Original post by HennersPD)
I did around the point X, is that correct or was it through X?
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