OCR Physic A G485 FPFoP Unoffiial mark Scheme
OCR Physics AG485 FP and FoP June 18 2015
Usual disclaimers. These are just my answers and are in no sense official. They may contain errros and typos.
The view of my folks coming out was that it was pretty easy. i don't often hear that. A few lamentations that there was nothing on EM induction, Fission, fusion or MRI. I could have save a few weeks teaching .... The questions seem mostly things we've seen before. Lots of calculations - not much descriptive / memory stuff which will suit some but not others.
Q1 a) A tricky one to word straight away
An electric field is in a region of space where there is a force on a test charge placed at a point. (1)
b) difference - E can be attractive or repulsive; g always attractive
similarity - both inverse square laws 1/r^2 (2)
c) Uniform deceleration until reaches a stop 2/3 way across gap (4ev electron in 6v gap), then accelerates uniformly back through gap. Same energy/speed going out as had when went in. (3)
d) i) E = V/d = 60E3/25E-2 = 2.4E5 Vm-1
F=Eq = 2.4E5 x 1.5E-13 = 3.6E-8N (2)
ii) Time = d/v = 1.8/1/2 = 1/5s
Horiz acc = F/m = 3.6E-8/8.0E-7 = 0.045 ms-2
s=1/2at^2 = 1/2 x 0.045 x15^2 = 0.051m (3)
Total : 11
Q2 a) i) F=BIL so I = 6.8E-5/(0.070 x 1.0E-2) = 0.097A (2)
ii) Force would alternate up and down (1)
(Earths field too weak / cable too heavy ) probably not accepted.
b) i) F=ma so Bqv = mv^2/r so r = mv/Bq (2)
ii) r is prop to v so KE is prop to r^2 so ratio of KE = ratio of r^2 = 16 (2)
Total :7
Q3 a) P and n are NOT fundamental because made from quarks (1)
b) leptons include (two from) electrons, muons tau , neutrinos (x3) (1)
c) i) 40K19 -> 40Ca20 + 0e-1 + vbar (2)
ii) Total mass after reaction is less than mass before + some is converted to energy / BE per nucleon of Ca40 is more than BE/nuc of K-40 so release of energy as nucleus becomes more stable (2)
iii) N = n x NA = 4.5E-4 x 0.012/100 x 6.0E23/0.040 = 8.1E17
lambda = ln2/t1/2 = 1.65E-17
A = lambda x N = 13.4 Bq (3)
Total : 9
Q4 a) i) Exponential graph (1)
ii) ntc so when temp increases, R decreases so time constant is less so decays faster.
exponential graph that starts at same PD but decays faster (1)
- both asymptotic to x axis
iii) Read off values of V at equal intervals of t (=V0, V1, V2 ...). Ratios should be constant V0/V1 = V1/V2 = V2/V3 etc (1)
b) i) Cp = C1 + C2 = 2.2 + 4.7 = 6.9uF
T = CR = 6.9E-6 x 240 = 1.66E-3s (1)
ii) This used to be used as an experiment to measure C - look up red switch!
I = total Q per unit time = CV x 1/f = 6.9E-6 x 1.4 x 120 = 1.16E-3 A (3)
iii) If double f then T = 4.2E-3 = 2.5 ish time constants
so capacitor wont have time to completely discharge (2)
Total : 9
Q5 a) Standard bit of book work for 3 marks
Fire alphas at gold foil and count how many deflected at different angles
Most go straight through but 1 in 800 deflected more than 90
Must be positive to repel alpha
Must be < 1E-13m so get force big enough to bounce alpha back
etc (3)
b) i) Been on a previous paper
Al nucleus is moving to the right at lower velocity than alpha initially
Momentum is conserved so momentum of Al = initial momentum of alpha (2)
i) Convert 8Mev to J and equate to KE
1/2 x 6.6E-27 x v^2 = 8.0 E6 x 1.6E-19
so v = 2.0E7 ms-1 (2)
iii) F = Q1Q2/4pieor^2 Q1 = 2e and Q2 = 13e
270 = (2 x 1.6E-19 x 13 x 16E-19) / ( 4 x pi x 8.85E-12 x r^2)
so r = 4.7E-15m (3)
iv) Range of strong force is about 4fm. if get within that there is an attractive force. Will be much bigger than coulomb repulsion so net attractive force. (2)
Total 12
Q6 a) BE = energy needed to separate a nucleus into individual protons and neutrons (1)
b) BE / nuc = 4.53E-12 / 4 = 1.13E-12 J (1)
c) He has higher charge on nucleus so stronger repulsive coulomb forces
so need more KE to get within 4fm so need higher temp (2)
d) 1/2 mv^2 = 3/2 kT
1/2 x 6.6E-27 x v^2 = 3/2 x 1.38E-23 x 1E8 so v = 7.9E5 ms-1 (2)
Total = 6
Q7 a) a photon is a quantum / packet of light energy/
travel at 3.0E8 in a vacuum (2)
b) i) I = ne
so n = 4.8E-3/1.6E-19 = 3.0E16 QED (1)
ii) Total KE of electrons = n x QV = 3.0E16 x 150E3 x 1.6E-19 = 720 J
99% converted to heat
so heat = 0.99 x 720 = mcdT = 8.6E-3 x 140 x dT so dT = 592 C s-1
(NB no 273 factor for changes in temp!) (3)
iii) energy of 1 electron = QV = hc/lambda
so lambda = (6.6E-34 x 3.0E8) / (150E3 x 1.6E-19) = 8.25E-12 m (2)
c) Inject a contrast medium which is a high Z material (high attenuation coefficient)
so absorbs Xrays eg iodine (not barium!)
Subtractive X ray (2)
Total 10
Q8 a) Gammas are more penetrating so leave patient rather than being absorbed.
Means lower exposure of patient to ionizing radiation and can pinpoint location because not absorbed / deflected on way to detector. (2)
b) CT / CAT scan - more book work
Xrays emitted by tungsten target bombarded by high energy electrons
Bremsstrahlung / line emission
Emitted in a fan shaped beam
Absorbed more by high Z matter such as bone
Ring of 720 detectors + rotating source
Patient moved through about 1cm for each rotation
Get a series of slices that computer uses to construct 3D image
etc (5)
Total 7
Q9 a) Doppler effect is change of frequency or wavelength when we have a moving source and/or a moving observer of waves. (1)
b) Place detector / receiver at an angle to artery
Ultrasound reflects off moving blood cells
Frequency change
df/f = 2v/c cos theta
change if f is prop to velocity of blood cells
f increases if moving toward and decreases if moving away (3)
c) i) z = rho x c so rho = 1.66E6 / 1.57E3 = 1057 kgm-3 (1)
ii) lambda = c/f = 1.57E3 / 2.4E6 = 6.54 E-4
d) Ir = (formula) = (4-1)^2 / (4+1)^2 = 9/25 = 36% so 64% is transmitted (3)
e) Big difference between z for air and z for skin means that 99.99% would be reflected.
Need impedance match by using silica gel with same impedance as skin
in air gap between Tx and skin so max transmitted. (2)
Total 11
Q10 a) white dwarf : when red giant runs out of fuel to fuse, it collapses - throws off outer layers as planetary nebula and forms white dwarf. Small (earth size). Electron degeneracy pressure stops further grav collapse. Hot but no fusion. Gradually cools . (1)
b) dist in pc / 1/ angle = 1/0.0059 = 169.5 pc
dist in ly = 169.5 x 3.26 = 552ly (2)
c) i) I = power per unit cross sectional area (1)
ii) density = m /v so ratio = 12 / (1.1E5)63 = 9.0E-15 (2)
power = I x area so ratio = (4300^4 x 1.1E5^2) / 25000^4 = 1.06E7 (3)
Total 9
Q11 a) Hubble ; speed of recession of distance galaxies is prop to distance (1)
b) i) v = zc = 3.38E-3 x 3.0E5 = 1014 kms-1
v is pop to d so v1/v2 = d1/d2 so d = 4.49E23m (2)
ii) Draw straight line (through origin )
gradient = Ho = 678E3/3.0E23
so t = 1/Ho = 4.42E17s = 1.4E10 y (3)
c) Cosmic microwave background - continuous spectrum at 2.7K / isotropic and homogeneous
Primordial abundances H/He = 75/25 or p/n = 4:1 observations from intergalalctic gas clouds. (3)
Total 9
Hopefully that adds ap to 100.
Not too bad compared to some other papers this year.
Last years boundary was 69A and 39E. I don't think this is going to be too far from that. A little bit easier imho so lets go for
A 72
B 64
C 56
D 48
E 40
which makes the 100% level 72+8+8 = 88
Good Luck folks. I'll be marking this paper soon.
Col
Usual disclaimers. These are just my answers and are in no sense official. They may contain errros and typos.
The view of my folks coming out was that it was pretty easy. i don't often hear that. A few lamentations that there was nothing on EM induction, Fission, fusion or MRI. I could have save a few weeks teaching .... The questions seem mostly things we've seen before. Lots of calculations - not much descriptive / memory stuff which will suit some but not others.
Q1 a) A tricky one to word straight away
An electric field is in a region of space where there is a force on a test charge placed at a point. (1)
b) difference - E can be attractive or repulsive; g always attractive
similarity - both inverse square laws 1/r^2 (2)
c) Uniform deceleration until reaches a stop 2/3 way across gap (4ev electron in 6v gap), then accelerates uniformly back through gap. Same energy/speed going out as had when went in. (3)
d) i) E = V/d = 60E3/25E-2 = 2.4E5 Vm-1
F=Eq = 2.4E5 x 1.5E-13 = 3.6E-8N (2)
ii) Time = d/v = 1.8/1/2 = 1/5s
Horiz acc = F/m = 3.6E-8/8.0E-7 = 0.045 ms-2
s=1/2at^2 = 1/2 x 0.045 x15^2 = 0.051m (3)
Total : 11
Q2 a) i) F=BIL so I = 6.8E-5/(0.070 x 1.0E-2) = 0.097A (2)
ii) Force would alternate up and down (1)
(Earths field too weak / cable too heavy ) probably not accepted.
b) i) F=ma so Bqv = mv^2/r so r = mv/Bq (2)
ii) r is prop to v so KE is prop to r^2 so ratio of KE = ratio of r^2 = 16 (2)
Total :7
Q3 a) P and n are NOT fundamental because made from quarks (1)
b) leptons include (two from) electrons, muons tau , neutrinos (x3) (1)
c) i) 40K19 -> 40Ca20 + 0e-1 + vbar (2)
ii) Total mass after reaction is less than mass before + some is converted to energy / BE per nucleon of Ca40 is more than BE/nuc of K-40 so release of energy as nucleus becomes more stable (2)
iii) N = n x NA = 4.5E-4 x 0.012/100 x 6.0E23/0.040 = 8.1E17
lambda = ln2/t1/2 = 1.65E-17
A = lambda x N = 13.4 Bq (3)
Total : 9
Q4 a) i) Exponential graph (1)
ii) ntc so when temp increases, R decreases so time constant is less so decays faster.
exponential graph that starts at same PD but decays faster (1)
- both asymptotic to x axis
iii) Read off values of V at equal intervals of t (=V0, V1, V2 ...). Ratios should be constant V0/V1 = V1/V2 = V2/V3 etc (1)
b) i) Cp = C1 + C2 = 2.2 + 4.7 = 6.9uF
T = CR = 6.9E-6 x 240 = 1.66E-3s (1)
ii) This used to be used as an experiment to measure C - look up red switch!
I = total Q per unit time = CV x 1/f = 6.9E-6 x 1.4 x 120 = 1.16E-3 A (3)
iii) If double f then T = 4.2E-3 = 2.5 ish time constants
so capacitor wont have time to completely discharge (2)
Total : 9
Q5 a) Standard bit of book work for 3 marks
Fire alphas at gold foil and count how many deflected at different angles
Most go straight through but 1 in 800 deflected more than 90
Must be positive to repel alpha
Must be < 1E-13m so get force big enough to bounce alpha back
etc (3)
b) i) Been on a previous paper
Al nucleus is moving to the right at lower velocity than alpha initially
Momentum is conserved so momentum of Al = initial momentum of alpha (2)
i) Convert 8Mev to J and equate to KE
1/2 x 6.6E-27 x v^2 = 8.0 E6 x 1.6E-19
so v = 2.0E7 ms-1 (2)
iii) F = Q1Q2/4pieor^2 Q1 = 2e and Q2 = 13e
270 = (2 x 1.6E-19 x 13 x 16E-19) / ( 4 x pi x 8.85E-12 x r^2)
so r = 4.7E-15m (3)
iv) Range of strong force is about 4fm. if get within that there is an attractive force. Will be much bigger than coulomb repulsion so net attractive force. (2)
Total 12
Q6 a) BE = energy needed to separate a nucleus into individual protons and neutrons (1)
b) BE / nuc = 4.53E-12 / 4 = 1.13E-12 J (1)
c) He has higher charge on nucleus so stronger repulsive coulomb forces
so need more KE to get within 4fm so need higher temp (2)
d) 1/2 mv^2 = 3/2 kT
1/2 x 6.6E-27 x v^2 = 3/2 x 1.38E-23 x 1E8 so v = 7.9E5 ms-1 (2)
Total = 6
Q7 a) a photon is a quantum / packet of light energy/
travel at 3.0E8 in a vacuum (2)
b) i) I = ne
so n = 4.8E-3/1.6E-19 = 3.0E16 QED (1)
ii) Total KE of electrons = n x QV = 3.0E16 x 150E3 x 1.6E-19 = 720 J
99% converted to heat
so heat = 0.99 x 720 = mcdT = 8.6E-3 x 140 x dT so dT = 592 C s-1
(NB no 273 factor for changes in temp!) (3)
iii) energy of 1 electron = QV = hc/lambda
so lambda = (6.6E-34 x 3.0E8) / (150E3 x 1.6E-19) = 8.25E-12 m (2)
c) Inject a contrast medium which is a high Z material (high attenuation coefficient)
so absorbs Xrays eg iodine (not barium!)
Subtractive X ray (2)
Total 10
Q8 a) Gammas are more penetrating so leave patient rather than being absorbed.
Means lower exposure of patient to ionizing radiation and can pinpoint location because not absorbed / deflected on way to detector. (2)
b) CT / CAT scan - more book work
Xrays emitted by tungsten target bombarded by high energy electrons
Bremsstrahlung / line emission
Emitted in a fan shaped beam
Absorbed more by high Z matter such as bone
Ring of 720 detectors + rotating source
Patient moved through about 1cm for each rotation
Get a series of slices that computer uses to construct 3D image
etc (5)
Total 7
Q9 a) Doppler effect is change of frequency or wavelength when we have a moving source and/or a moving observer of waves. (1)
b) Place detector / receiver at an angle to artery
Ultrasound reflects off moving blood cells
Frequency change
df/f = 2v/c cos theta
change if f is prop to velocity of blood cells
f increases if moving toward and decreases if moving away (3)
c) i) z = rho x c so rho = 1.66E6 / 1.57E3 = 1057 kgm-3 (1)
ii) lambda = c/f = 1.57E3 / 2.4E6 = 6.54 E-4
d) Ir = (formula) = (4-1)^2 / (4+1)^2 = 9/25 = 36% so 64% is transmitted (3)
e) Big difference between z for air and z for skin means that 99.99% would be reflected.
Need impedance match by using silica gel with same impedance as skin
in air gap between Tx and skin so max transmitted. (2)
Total 11
Q10 a) white dwarf : when red giant runs out of fuel to fuse, it collapses - throws off outer layers as planetary nebula and forms white dwarf. Small (earth size). Electron degeneracy pressure stops further grav collapse. Hot but no fusion. Gradually cools . (1)
b) dist in pc / 1/ angle = 1/0.0059 = 169.5 pc
dist in ly = 169.5 x 3.26 = 552ly (2)
c) i) I = power per unit cross sectional area (1)
ii) density = m /v so ratio = 12 / (1.1E5)63 = 9.0E-15 (2)
power = I x area so ratio = (4300^4 x 1.1E5^2) / 25000^4 = 1.06E7 (3)
Total 9
Q11 a) Hubble ; speed of recession of distance galaxies is prop to distance (1)
b) i) v = zc = 3.38E-3 x 3.0E5 = 1014 kms-1
v is pop to d so v1/v2 = d1/d2 so d = 4.49E23m (2)
ii) Draw straight line (through origin )
gradient = Ho = 678E3/3.0E23
so t = 1/Ho = 4.42E17s = 1.4E10 y (3)
c) Cosmic microwave background - continuous spectrum at 2.7K / isotropic and homogeneous
Primordial abundances H/He = 75/25 or p/n = 4:1 observations from intergalalctic gas clouds. (3)
Total 9
Hopefully that adds ap to 100.
Not too bad compared to some other papers this year.
Last years boundary was 69A and 39E. I don't think this is going to be too far from that. A little bit easier imho so lets go for
A 72
B 64
C 56
D 48
E 40
which makes the 100% level 72+8+8 = 88
Good Luck folks. I'll be marking this paper soon.
Col
(edited 8 years ago)
Scroll to see replies
brilliant thank you!
I think 90
for the F=BIL
I chucked the numbers in to find F which obviously didn't work, dropped most marks on first parts
for the F=BIL I chucked the numbers in to find F which obviously didn't work, dropped most marks on first parts
Original post by teachercol
Good Luck folks.
Col
Good Luck folks.
Col
Great help.
Original post by Raizel
Might wanna fix the grade boundaries --> You put C twice instead of E
Thanks
Fixed
Original post by teachercol
.
Original post by Raizel
Might wanna fix the grade boundaries --> You put C twice instead of E
And also made a typo on 'unofficial'. Doesn't take away the helpfulness though...
Original post by SH0405
And also made a typo on 'unofficial'. Doesn't take away the helpfulness though...
I know
Original post by fatart123
85 min and about 90 max. Talked about the disadvantages of beta instead of advantages of gamma for 8a... Will that lose both marks or will the ms accept both approaches?
Might be Ok - but the question definitely asks for advantages of gamma.
Original post by Elcor
Thank you once again!
What do you mean by 'not barium' in the contrast medium question?
What do you mean by 'not barium' in the contrast medium question?
Several of my students said barium which is used for stomach / intestinal x rays but not angiograms.
Original post by teachercol
Might be Ok - but the question definitely asks for advantages of gamma.
Mhm, I don't really expect to get the marks tbh... Only needed to get a low C today tho so I don't suppose it matters too much
How accurate are your grade boundary predictions for G485?
As in, are you normally close?
As in, are you normally close?
Original post by teachercol
Several of my students said barium which is used for stomach / intestinal x rays but not angiograms.
Is it iodine 131? I hadn't learnt a number so wrote 131 because it sounded vaguely familiar
Thank you teachercol!!!!!
I'm defo not getting into manc. It's all banter I suppose.
Posted from TSR Mobile
I'm defo not getting into manc. It's all banter I suppose.
Posted from TSR Mobile
Original post by Genghis_Khan
How accurate are your grade boundary predictions for G485?
As in, are you normally close?
As in, are you normally close?
I'm usually too harsh. Generally my students are very good and so they tend to do a lot better than the average student. This time they thought the paper was pretty easy so I use their feelings and my own judgement.
Ok - to be honest - its a pure guess.
The biggest unknown is the mark scheme. last year G484was easy but the mark scheme was a nit picking nightmare and we all put the boundaries way too high - but then students didn't get the marks they expected so it evens out.
The grade boundaries aren't set until nearly all the papers have been marked. and the chief examiner looks at the stats.
Original post by teachercol
Several of my students said barium which is used for stomach / intestinal x rays but not angiograms.
Fair enough but I don't believe this is on the spec; it's been barium every time in every past paper question on contrast media
Do you think I'd lose a mark for saying 'power per unit area' rather than 'power per unit cross-sectional area'?
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