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# OCR Physics A G485 FPFoP Unofficial mark scheme

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1. 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 ms-2 / 1 Cs-1 x 1m = kg s-1 C-1 (2)
b) i) Bqv = mv^2 / r so B = mv / qr = (9.11E-31 x 7.0E6)/(1.6E-19 x 2.5E-2) = 1.59E-3 (3)
ii) T = 2 pi r / v = 2 pi x 2.5E-2 / 7.0E6 =2.2E-8s (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 = 1.25E-7
E = Q / 4 pi eo r^2 so Grad = Q / 4 pi eo so Q = 4 pi eo x grad
= 1.19E-17 C (2)
ii) Charge decreases by 2e (so now Q = 1.358E-17C)
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 = 1.12E-18C (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 100E-6 = 2.7s (3)
c) i) Initial I = 1.5E-4A from graph
so Vo = IR = 6.0v
so Qo = CV = 7.2E-3C (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) 0e-1 + antineutrino (2)
b) i) No of decay/s = 2000/9.0E-3 = 2.22E15
so Activity = 2.22E15
lambda = ln2/half life = 2.293E-10 s-1 (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 = 2.24E-11 J
so m = E/c^2 = 2.29E-28 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 ms-1 (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 = (z2-z1)^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 1.673E-27
= 8 H0^2 / 8 pi G so Ho = 2.73E-18 s-1 (3)
ii) t = 1 / Ho = 3.6E17 s = 1.14E10 y (2)
Total: 11

Hopefully that adds up to 100.

I wouldn't expect grade boundaries to be below what they were last year.
Good exam paper. Well done examiners.
2. Yes I expect so.
3. (Original post by teachercol)
Yes I expect so.
Thank you and also sorry but could yiu have that fusion realeases energy as electromagnetic and kinetic whereas fission is just kinetic?
4. (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?
I wouldn't give that a mark.
5. 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?
6. FOr Q1, 5cm was the diameter not radius.
7. (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?
Omg thank god!!
8. 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*?
9. (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*?
Me too.. Should be an a* if not more
10. (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 ms-2 / 1 Cs-1 x 1m = kg s-1 C-1 (2)
b) i) Bqv = mv^2 / r so B = mv / qr = (9.11E-31 x 7.0E6)/(1.6E-19 x 5.0E-2) = 7.97E-4T (3)
ii) T = 2 pi r / v = 2 pi x 5.0E-2 / 7.0E6 = 4.49E-8s (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 = 1.25E-7
E = Q / 4 pi eo r^2 so Grad = Q / 4 pi eo so Q = 4 pi eo x grad
= 1.19E-17 C (2)
ii) Charge decreases by 2e (so now Q = 1.358E-17C)
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 = 1.12E-18C (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 100E-6 = 2.7s (3)
c) i) Initial I = 1.5E-4A from graph
so Vo = IR = 6.0v
so Qo = CV = 7.2E-3C (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) 0e-1 + antineutrino (2)
b) i) No of decay/s = 2000/9.0E-3 = 2.22E15
so Activity = 2.22E15
lambda = ln2/half life = 2.293E-10 s-1 (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 = 2.24E-11 J
so m = E/c^2 = 2.29E-28 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 ms-1 (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 = (z2-z1)^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 1.673E-27
= 8 H0^2 / 8 pi G so Ho = 2.73E-18 s-1 (3)
ii) t = 1 / Ho = 3.6E17 s = 1.14E10 y (2)
Total: 11

Hopefully that adds up to 100.

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.
11. (Original post by teachercol)
I wouldn't give that a mark.
I talked about how fusion requires much higher temperatures than fission to occur. What do you think about that point?
12. (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?
Yes -quite right for both - misread question. I'll update.
13. (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*?
No not all hadrons have 3 quarks. Mesons are quark + antiquark.
I talked about how fusion requires much higher temperatures than fission to occur. What do you think about that point?
I wouldn't give a mark for that.
15. What does rho and uber mean?
16. (Original post by Ramil Ahmadov)
What does rho and uber mean?
rho is the greek letter for density.

ubar means anti u
17. (Original post by teachercol)
rho is the greek letter for density.

ubar means anti u
Thank you for the answer. I thought pi- is ddd
18. (Original post by Ramil Ahmadov)
Thank you for the answer. I thought pi- is ddd
Nope - a pion is a meson. Quark + antiquark.

Posted from TSR Mobile
20. (Original post by teachercol)
Yes -quite right for both - misread question. I'll update.

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