OCR Physics A G485 FP and FoP 28/6/16
Usual disclaimers. These are just my quick answers . They may and probably will contain errors and typos.
My students were pretty happy when they came out of the exam room yesterday.
First impression is that this was pretty straight down the middle.
Not much in the way of long extended answers
Here we go
Q1 a) 1T = 1N / 1A x 1m = 1 kg ms2 / 1 Cs1 x 1m = kg s1 C1 (2)
b) i) Bqv = mv^2 / r so B = mv / qr = (9.11E31 x 7.0E6)/(1.6E19 x 2.5E2) = 1.59E3 (3)
ii) T = 2 pi r / v = 2 pi x 2.5E2 / 7.0E6 =2.2E8s (1)
iii) If v doubles and m, B, and q are same then r is prop to v, so r doubles
T is prop to r/v so stays constant. (2)
Total: 8
Q2 a) Direction should be + to 
Field lines must meet plates at 90 (2)
b) i) Grad = dy/dx = 1.0E21/8.0E27 = 1.25E7
E = Q / 4 pi eo r^2 so Grad = Q / 4 pi eo so Q = 4 pi eo x grad
= 1.19E17 C (2)
ii) Charge decreases by 2e (so now Q = 1.358E17C)
Grad is prop to Q so will decrease (2)
c) i) mg = Eq = Vq/d
4/3 pi r^3 rho = Vq/d so q = 1.12E18C (3)
ii) N=Q/e = 7 (1)
Total: 10
Q3 a) i) Mag flux linkage = product of no of turns on coil and magnetic flux through 1 turn of coil (1)
ii) 1) L = N x 2 pi r so N = ? / 2 pi r (1)
2) N phi = NBA = L / 2 pi r x B x pi r^2 = BrL/2 (1)
b) i) Faraday's law says induced emf  rate of change of magnetic flux linkage (1)
ii) emf =  gradient > cos graph (2)
c) Ac supply creates alternating current in primary coil which creates alternating flux in primary
Core 'conducts' flux through secondary coil.
Alternating flux in secondary induces an alternating emf in secondary (2)
Total: 8
Q4 a) In series so same charge
v=Q/C so V is prop to 1/C
so if B has twice the capacitance it has half the PD. (2)
b) 1/C = 1/200 + 1/200 so C = 100uF
R = 18 + 9 = 27k so tau = 27E3 x 100E6 = 2.7s (3)
c) i) Initial I = 1.5E4A from graph
so Vo = IR = 6.0v
so Qo = CV = 7.2E3C (2)
ii) 2 x 40k in parallel means resistance halves so time constant halves
so graph starts at twice the current (3.0) and has quarter value at each 20s
Total: 9
Q5 a) i) 2 x 1n0 (1)
ii) 0e1 + antineutrino (2)
b) i) No of decay/s = 2000/9.0E3 = 2.22E15
so Activity = 2.22E15
lambda = ln2/half life = 2.293E10 s1 (convert y to s!)
N = A / lambda = 8.906E24
n = N/NA = 14.79 moles so m = 0.238x 14.79 = 3.52 kg (4)
ii) W = Pt = 120/1000 x 24 = 2.88 kW.h (2)
Total: 9
Q6 a) Hadrons experience strong force (1)
b) u +2/3 ; d 1/3 (1)
c) proton = uud (1)
d) uud + udd = uud + uud + pi
so pi = d +ubar(3)
e) i) E = mc^2 where E = energy m = change in mass c = speed of light in a vacuum (1)
ii) Min KE = 1.4E8eV = 2.24E11 J
so m = E/c^2 = 2.29E28 kg (2)
Total: 9
Q7 a) 1) fission occurs with heavy nuclei; fusion with light
2) fission: 1 nucleus splits into 2 smaller ones; fusion two small nuclei join to form 1 larger one (2)
b) Nuclei need to get close enough for strong force to pull them together
so need enough energy to overcome coulomb repulsion
KE = 3/2 kT so need high temperatures
Need a reasonable chance that collision will occur so need lots of nuclei ie high pressures (4)
c) i) Bombard nucleus with thermal neutrons. Neutron is absorbed by nucleus.
Unstable nucleus splits ( releasing more neutrons) (2)
ii) 3/2 kT = 1/2 mv^2
T = 300+273 = 573K so v = 3736 ms1 (3)
Total: 11
Q8 a) Photoelectric effect
Photon absorbed; electron released
Compton scattering
Photon scattered with lower energy at an angle and electron emitted
Pair production
Photon absorbed and electron/positron pair created.
1st and 3rd involve photon being absorbed so transmitted intensity decreases. (4)
b) i) E = hf = hc/ lambda = .141E14J (2)
ii) Gradient = mu so mu =  gradient = 0.20 (2)
Total: 8
Q9 a) Frequency of radio waves = angular frequency / 2 pi = 6.37E6 Hz
so lambda = 3.0E8/6.37E6 = 4.7m (3)
b) relaxation time = time between proton being excited to higher energy state and dropping back to ground state (1)
Total:4
Q10 a) piezoelectric effect: when apply PD to a crystal, it expands or contracts. (1)
b) i) z = product of density and speed of sound in material (1)
ii) In patient, ultrasound will reflect off boundaries between different tissue types (media with different z) (1)
c) z muscle = 1.38E6; z fat = 1.69E6
so Ir/I = (z2z1)^2 / (z2+z1)^2 = 0.0961/9.42 = 1.0%
so 99% is transmitted (3)
Total: 6
Q11 a) tan(theta) = 1.3E20/2.4E22 so theta = 0.31 degrees (1)
b) dlambda / lanbda = v/c so d lambda = 656.3 x 2.5E5 /3.0E8 = 0.547 nm (2)
c) GMm/R^2 = mv^/r so M  v^r/G = 6.09E40 kg
so N = M/m = 3.03E10 stars (4)
Total: 7
Q12 a) Initially all 4 forces unified
Grav force splits first; primordial quark soup
Strong force separates; leptons form from photons; more matter than antimatter
weak and EM force separate; protons and neutrons formed ; p:n = 4:1; annihilation results in matter only
Fusion forms He and Li; matter is plasma
Atoms form; recombination; universe is transparent; first light CMB formed (5)
b) Hubble: for distant galaxies, velocity of recession is prop to distance (1)
c) i) rho = 8 x 1.673E27
= 8 H0^2 / 8 pi G so Ho = 2.73E18 s1 (3)
ii) t = 1 / Ho = 3.6E17 s = 1.14E10 y (2)
Total: 11
Hopefully that adds up to 100.
Not bad
I wouldn't expect grade boundaries to be below what they were last year.
Good exam paper. Well done examiners.
OCR Physics A G485 FPFoP Unofficial mark scheme
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 1
 29062016 14:58
Last edited by teachercol; 29062016 at 16:49.Post rating:1 
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 29062016 15:36
Yes I expect so.

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 3
 29062016 15:37
(Original post by teachercol)
Yes I expect so. 
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 29062016 15:48
(Original post by Jm098)
Thank you and also sorry but could yiu have that fusion realeases energy as electromagnetic and kinetic whereas fission is just kinetic? 
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 5
 29062016 15:55
For question 1bi and ii, was 5cm the diameter and not the radius? And for 4ci would the graph not start and twice the original current as the capacitor was charged to the same voltage but resistance was halved so current increased? Or is my memory playing tricks on me?

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 6
 29062016 16:02
FOr Q1, 5cm was the diameter not radius.

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 29062016 16:03
(Original post by boxcirclebox)
For question 1bi and ii, was 5cm the diameter and not the radius? And for 4ci would the graph not start and twice the original current as the capacitor was charged to the same voltage but resistance was halved so current increased? Or is my memory playing tricks on me? 
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 8
 29062016 16:05
would I get the mark for saying hadrons contain 3 quarks? Also does anyone have any idea of boundaries, I think I'm on about 85 will this be a*?

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 9
 29062016 16:06
(Original post by jtt1998)
would I get the mark for saying hadrons contain 3 quarks? Also does anyone have any idea of boundaries, I think I'm on about 85 will this be a*? 
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 10
 29062016 16:20
(Original post by teachercol)
OCR Physics A G485 FP and FoP 28/6/16
Usual disclaimers. These are just my quick answers . They may and probably will contain errors and typos.
My students were pretty happy when they came out of the exam room yesterday.
First impression is that this was pretty straight down the middle.
Not much in the way of long extended answers
Here we go
Q1 a) 1T = 1N / 1A x 1m = 1 kg ms2 / 1 Cs1 x 1m = kg s1 C1 (2)
b) i) Bqv = mv^2 / r so B = mv / qr = (9.11E31 x 7.0E6)/(1.6E19 x 5.0E2) = 7.97E4T (3)
ii) T = 2 pi r / v = 2 pi x 5.0E2 / 7.0E6 = 4.49E8s (1)
iii) If v doubles and m, B, and q are same then r is prop to v, so r doubles
T is prop to r/v so stays constant. (2)
Total: 8
Q2 a) Direction should be + to 
Field lines must meet plates at 90 (2)
b) i) Grad = dy/dx = 1.0E21/8.0E27 = 1.25E7
E = Q / 4 pi eo r^2 so Grad = Q / 4 pi eo so Q = 4 pi eo x grad
= 1.19E17 C (2)
ii) Charge decreases by 2e (so now Q = 1.358E17C)
Grad is prop to Q so will decrease (2)
c) i) mg = Eq = Vq/d
4/3 pi r^3 rho = Vq/d so q = 1.12E18C (3)
ii) N=Q/e = 7 (1)
Total: 10
Q3 a) i) Mag flux linkage = product of no of turns on coil and magnetic flux through 1 turn of coil (1)
ii) 1) L = N x 2 pi r so N = ? / 2 pi r (1)
2) N phi = NBA = L / 2 pi r x B x pi r^2 = BrL/2 (1)
b) i) Faraday's law says induced emf  rate of change of magnetic flux linkage (1)
ii) emf =  gradient > cos graph (2)
c) Ac supply creates alternating current in primary coil which creates alternating flux in primary
Core 'conducts' flux through secondary coil.
Alternating flux in secondary induces an alternating emf in secondary (2)
Total: 8
Q4 a) In series so same charge
v=Q/C so V is prop to 1/C
so if B has twice the capacitance it has half the PD. (2)
b) 1/C = 1/200 + 1/200 so C = 100uF
R = 18 + 9 = 27k so tau = 27E3 x 100E6 = 2.7s (3)
c) i) Initial I = 1.5E4A from graph
so Vo = IR = 6.0v
so Qo = CV = 7.2E3C (2)
ii) 2 x 40k in parallel means resistance halves so time constant halves
so graph starts at same current and has quarter value at each 20s
Total: 9
Q5 a) i) 2 x 1n0 (1)
ii) 0e1 + antineutrino (2)
b) i) No of decay/s = 2000/9.0E3 = 2.22E15
so Activity = 2.22E15
lambda = ln2/half life = 2.293E10 s1 (convert y to s!)
N = A / lambda = 8.906E24
n = N/NA = 14.79 moles so m = 0.238x 14.79 = 3.52 kg (4)
ii) W = Pt = 120/1000 x 24 = 2.88 kW.h (2)
Total: 9
Q6 a) Hadrons experience strong force (1)
b) u +2/3 ; d 1/3 (1)
c) proton = uud (1)
d) uud + udd = uud + uud + pi
so pi = d +ubar(3)
e) i) E = mc^2 where E = energy m = change in mass c = speed of light in a vacuum (1)
ii) Min KE = 1.4E8eV = 2.24E11 J
so m = E/c^2 = 2.29E28 kg (2)
Total: 9
Q7 a) 1) fission occurs with heavy nuclei; fusion with light
2) fission: 1 nucleus splits into 2 smaller ones; fusion two small nuclei join to form 1 larger one (2)
b) Nuclei need to get close enough for strong force to pull them together
so need enough energy to overcome coulomb repulsion
KE = 3/2 kT so need high temperatures
Need a reasonable chance that collision will occur so need lots of nuclei ie high pressures (4)
c) i) Bombard nucleus with thermal neutrons. Neutron is absorbed by nucleus.
Unstable nucleus splits ( releasing more neutrons) (2)
ii) 3/2 kT = 1/2 mv^2
T = 300+273 = 573K so v = 3736 ms1 (3)
Total: 11
Q8 a) Photoelectric effect
Photon absorbed; electron released
Compton scattering
Photon scattered with lower energy at an angle and electron emitted
Pair production
Photon absorbed and electron/positron pair created.
1st and 3rd involve photon being absorbed so transmitted intensity decreases. (4)
b) i) E = hf = hc/ lambda = .141E14J (2)
ii) Gradient = mu so mu =  gradient = 0.20 (2)
Total: 8
Q9 a) Frequency of radio waves = angular frequency / 2 pi = 6.37E6 Hz
so lambda = 3.0E8/6.37E6 = 4.7m (3)
b) relaxation time = time between proton being excited to higher energy state and dropping back to ground state (1)
Total:4
Q10 a) piezoelectric effect: when apply PD to a crystal, it expands or contracts. (1)
b) i) z = product of density and speed of sound in material (1)
ii) In patient, ultrasound will reflect off boundaries between different tissue types (media with different z) (1)
c) z muscle = 1.38E6; z fat = 1.69E6
so Ir/I = (z2z1)^2 / (z2+z1)^2 = 0.0961/9.42 = 1.0%
so 99% is transmitted (3)
Total: 6
Q11 a) tan(theta) = 1.3E20/2.4E22 so theta = 0.31 degrees (1)
b) dlambda / lanbda = v/c so d lambda = 656.3 x 2.5E5 /3.0E8 = 0.547 nm (2)
c) GMm/R^2 = mv^/r so M  v^r/G = 6.09E40 kg
so N = M/m = 3.03E10 stars (4)
Total: 7
Q12 a) Initially all 4 forces unified
Grav force splits first; primordial quark soup
Strong force separates; leptons form from photons; more matter than antimatter
weak and EM force separate; protons and neutrons formed ; p:n = 4:1; annihilation results in matter only
Fusion forms He and Li; matter is plasma
Atoms form; recombination; universe is transparent; first light CMB formed (5)
b) Hubble: for distant galaxies, velocity of recession is prop to distance (1)
c) i) rho = 8 x 1.673E27
= 8 H0^2 / 8 pi G so Ho = 2.73E18 s1 (3)
ii) t = 1 / Ho = 3.6E17 s = 1.14E10 y (2)
Total: 11
Hopefully that adds up to 100.
Not bad
I wouldn't expect grade boundaries to be below what they were last year.
Good exam paper. Well done examiners.
Hey, for 4c)ii) if the voltage across the capacitor is the same, with half the resistance, surely the initial current is twice that in the first graph. 
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 11
 29062016 16:28
(Original post by teachercol)
I wouldn't give that a mark. 
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 12
 29062016 16:43
(Original post by boxcirclebox)
For question 1bi and ii, was 5cm the diameter and not the radius? And for 4ci would the graph not start and twice the original current as the capacitor was charged to the same voltage but resistance was halved so current increased? Or is my memory playing tricks on me?Last edited by teachercol; 29062016 at 16:49. 
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 13
 29062016 16:46
(Original post by jtt1998)
would I get the mark for saying hadrons contain 3 quarks? Also does anyone have any idea of boundaries, I think I'm on about 85 will this be a*? 
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 14
 29062016 16:47
(Original post by TelevisedTadpole)
I talked about how fusion requires much higher temperatures than fission to occur. What do you think about that point? 
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 15
 29062016 16:48
What does rho and uber mean?

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 16
 29062016 16:52
(Original post by Ramil Ahmadov)
What does rho and uber mean?
ubar means anti u 
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 17
 29062016 16:53

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 18
 29062016 16:54
(Original post by Ramil Ahmadov)
Thank you for the answer. I thought pi is dddPost rating:2 
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 29062016 17:00
any predication about grade boundary?
Posted from TSR Mobile 
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 20
 29062016 18:57
(Original post by teachercol)
Yes quite right for both  misread question. I'll update.
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Updated: July 11, 2016
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