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# OCR AS Physics A H156 Depth in Physics Thursday 9th June 2016 Unofficial mark scheme Watch

1. OCR AS Physics A H156 Depth in Physics Thursday 9th June 2016

Usual disclaimers - these are just my answers done at great speed.
They may contain errors, typos and omissions.

My students were generally smiling when they came out feeling that this was a more accessible paper than the breadth one had been.
My impression is that i was a pretty straight down the middle paper with few tricks and catches. Well done examiners.

Q1 a) L: oscillates parallel to direction of travel of wave
T: oscillates perpendicular to direction of travel of wave (2)
b) i) Amplitude = 4 squares = 40mV (1)
ii) Period = 3 squares = 1.5ms
Frequency + 1/ Period = 1/1.5E-3 = 667 Hz (2)
iii) V = f x lambda so lambda = 330/667 = 0.49m (1)
c) Intensity is prop to amplitude squared
so if intensity is 1/4 then amp is 1/2, freq doesnt change
so half height; same period. (2)
Total;: 8

Q2 a) i) Principle of superposition: when 2 or more waves overlap, the resultant displacement at a point is equal to the
vector sum of the individual displacements at that point (1)
ii) coherent = constant phase difference (1)
b) 1 lambda
2) 1.5 lambda (2)
c) lambda = ax/D so x = lambda D/ a; lambda and D are constant so
so if a is increased separation of max increase (2)
Total : 6

Q3 a) i) suvat v=u+at so t = 6.3 / 9.81 = 0.642s (1)
ii) vh x t = dx so vh = 18/0.642 = 28.0 ms-1 (1)
iii) v = sqrt (vv^2 + vh^2)
= sqrt ( 28.0^2 6.3^2)
= 28.7 ms-1 (2)
b) i) Ek = 1/2 mv^2 = 65.9J (1)
ii) dEp = mg dh = 3.14J (1)
ii) Ek at top = 65.9 - 3.14 = 62.8J (1)
c) with air resistance; reaches lower max height; will reach max eight before hits wall; same initial path (2)
Total: 9

Q4 a) V = W / Q so units are J/C = N.m/A.s = Kgms-2 . m / As = kg m^2 A^-1 s^-3 (3)
b) i) PD across 1.2K = 0.9v
so Rldr / 1.2k = 5.1/0.9 so Rldr = 6.8k = 6800 ohms (2)
ii) I = V/R = 0.9 / 1200 = 7.5E-4 A (1)
c) Rldr will decrease so total resistance in circuit is less so current increases.
so PD across 1.2k will increase (=IR) so PD across RDR must decrease (PDs add up to 6v) (4)
Total: 10

Q5 a) pho = m/V so V = m/rho = 100E-3/5300 = 1.89E-5 m^3 (1)
b) i) so that current flows through all cross sectional area of putty (1)
ii) use micrometer / vernier calipers at different points and different orientations
use a template with hole of correct diameter and extrude (??) (2)
c) i) 6.6 (1)
ii) %uncert in L = 0.001/0.049 x 100 = 2.0%
so %uncert in L^2 = 2 x 2.0 = 4.0% (1)
d) i) Plot point and draw line of best fit (NOT through false origin!) (2)
ii) Gradient = dy/dx = 5.7E3 (2)
e) Gradient = rho / V so rho = 5.7E3 x 1.9E-5 = 0.108 ohm.meter (3)
Total: 13

Q6 a) Safety - wear safety glasses incase wire whiplashes into eyes
place cushion under load so that doesnt break anything when it drops
Measure average diameter of wire with micrometer - several places and orientations
Calc cross-sectional area = pi x (d/2)^2
Add weights one at a time until breaks Record breaking mass.
Breaking stress = mass x 9.81 / cross sectional area (6)
b) glass = straight line - steep
rubber - curves - lower line when unload (Hysteresis) (2)
Total: 8

Q7 a) Photoelectric effect : one photon is absorbed by and may release one electron
if frequency of light is above threshold frequency of metal then electrons are emitted
UV has higher frequency than visible
E = hf so UV photons have more energy
if energy of photons > work function energy of metal, electrons emitted.
UV has enough energy, visible doesnt.
when electrons emitted charge on electroscope drops and leafs fall (no longer repelling each other)
when close, plate absorbs more photons per unit time so more electrons emitted and falls faster (6)
b) hf = phi + Ekmax
6.63E-34 x 9.60E14 = 3.2 x 1.6E-19 = Ekmax
so Ekmax = 1.25E-19 J (3)
Total: 9

Q8 a) Gain in energy = 300eV = 300 x 1.6E-19 J = 4.8E-17J
1/2mv^2 = 4.8E-17
v=1.03E7 ms-1 (3)
b) lambda = h/mv = 7.07E-8m (2)
c) Higher PD = more energy = greater v = smaller wavelength
lambda = ax/D so x = lambda D/a so x reduces
Rings will be close together (and brighter) (2)
Total: 7

Hopefully that all adds up to 70.

Let me know if you spot any errors.

I dont know what the grade boundaries will be.
I dont even know if the two papers will have different grade boundaries and are converted to UMS before adding or if the raw marks are added than converted.
OCR havent told us. The first method is fairer but who knows.

Good Luck
Col
2. Thanks for this, really appreciate it!

Do you know how lenient they are with error carried forward marks? I stupidly misread the period as 4 squares on the first question so got the wrong frequency which obviously got me the wrong wavelength, would get any marks for the wavelength one or not?
3. (Original post by Polapod)
Thanks for this, really appreciate it!

Do you know how lenient they are with error carried forward marks? I stupidly misread the period as 4 squares on the first question so got the wrong frequency which obviously got me the wrong wavelength, would get any marks for the wavelength one or not?
I'd expect an ecf there so you would only lose one mark
4. Hey there was some speculation about the suvat question. Some people used different values. I got an answer of 30.6. I found the angle of theta by equating the horizontal and vertically components together then, put that value back to find v. apparently some people used their value in the previous parts to calculate, however some people used the values given.
5. For part 7a do you think you were supposed to link the one to one ratio with the increase in intensity; when it was above threshold frequency, that allowed more photoelectrons to be emitted per second. Also could you have shown why the observations proved why it wasn't only wave-like in nature such as saying the fact that increasing the intensity for em radiation below the threshold frequency would speed up the process assuming the wavelike nature assumes a continuous flow of energy; and since it didn't happen it could only be explained by the particle nature of em radiation?
6. Also for the question about ensuring the diameter was constant; I thought it meant that it wouldn't change shape rather than being the same throughout the putty; so I ended up explaining it with R=pl/A and saying if the value on the ohm meter changes then so must the area >>diameter. I think the wording for the question was quite confusing for me.
7. (Original post by teachercol)
OCR AS Physics A H156 Depth in Physics Thursday 9th June 2016

Usual disclaimers - these are just my answers done at great speed.
They may contain errors, typos and omissions.

My students were generally smiling when they came out feeling that this was a more accessible paper than the breadth one had been.
My impression is that i was a pretty straight down the middle paper with few tricks and catches. Well done examiners.

Q1 a) L: oscillates parallel to direction of travel of wave
T: oscillates perpendicular to direction of travel of wave (2)
b) i) Amplitude = 4 squares = 40mV (1)
ii) Period = 3 squares = 1.5ms
Frequency + 1/ Period = 1/1.5E-3 = 667 Hz (2)
iii) V = f x lambda so lambda = 330/667 = 0.49m (1)
c) Intensity is prop to amplitude squared
so if intensity is 1/4 then amp is 1/2, freq doesnt change
so half height; same period. (2)
Total;: 8

Q2 a) i) Principle of superposition: when 2 or more waves overlap, the resultant displacement at a point is equal to the
vector sum of the individual displacements at that point (1)
ii) coherent = constant phase difference (1)
b) 1 lambda
2) 1.5 lambda (2)
c) lambda = ax/D so x = lambda D/ a
so if a is increased separation of max increase (2)
Total : 6

Q3 a) i) suvat v=u+at so t = 6.3 / 9.81 = 0.642s (1)
ii) vh x t = dx so vh = 18/0.642 = 28.0 ms-1 (1)
iii) v = sqrt (vv^2 + vh^2)
= sqrt ( 28.0^2 6.3^2)
= 28.7 ms-1 (2)
b) i) Ek = 1/2 mv^2 = 65.9J (1)
ii) dEp = mg dh = 3.14J (1)
ii) Ek at top = 65.9 - 3.14 = 62.8J (1)
c) with air resistance; reaches lower max height; will reach max eight before hits wall; same initial path (2)
Total: 9

Q4 a) V = W / Q so units are J/C = N.m/A.s = Kgms-2 . m / As = kg m^2 A^-1 s^-3 (3)
b) i) PD across 1.2K = 0.9v
so Rldr / 1.2k = 5.1/0.9 so Rldr = 6.8k = 6800 ohms (2)
ii) I = V/R = 0.9 / 1200 = 7.5E-4 A (1)
c) Rldr will decrease so total resistance in circuit is less so current increases.
so PD across 1.2k will increase (=IR) so PD across RDR must decrease (PDs add up to 6v) (4)
Total: 10

Q5 a) pho = m/V so V = m/rho = 100E-3/5300 = 1.89E-5 m^3 (1)
b) i) so that current flows through all cross sectional area of putty (1)
ii) use micrometer / vernier calipers at different points and different orientations
use a template with hole of correct diameter and extrude (??) (2)
c) i) 6.45 (1)
ii) %uncert in L = 0.001/0.049 x 100 = 2.0%
so %uncert in L^2 = 2 x 2.0 = 4.0% (1)
d) i) Plot point and draw line of best fit (NOT through false origin!) (2)
ii) Gradient = dy/dx = (58 -10.5)/ (10 -2) x 10-^-3 = 5.9E3 (2)
e) Gradient = rho / V so rho = 5.9E3 x 1.9E-5 = 1.13 E-4 ohm.meter (3)
Total: 13

Q6 a) Safety - wear safety glasses incase wire whiplashes into eyes
place cushion under load so that doesnt break anything when it drops
Measure average diameter of wire with micrometer - several places and orientations
Calc cross-sectional area = pi x (d/2)^2
Add weights one at a time until breaks Record breaking mass.
Breaking stress = mass x 9.81 / cross sectional area (6)
b) glass = straight line - steep
rubber - curves - lower line when unload (Hysteresis) (2)
Total: 8

Q7 a) Photoelectric effect : one photon is absorbed by and may release one electron
if frequency of light is above threshold frequency of metal then electrons are emitted
UV has higher frequency than visible
E = hf so UV photons have more energy
if energy of photons > work function energy of metal, electrons emitted.
UV has enough energy, visible doesnt.
when electrons emitted charge on electroscope drops and leafs fall (no longer repelling each other)
when close, plate absorbs more photons per unit time so more electrons emitted and falls faster (6)
b) hf = phi + Ekmax
6.63E-34 x 9.60E14 = 3.2 x 1.6E-19 = Ekmax
so Ekmax = 1.25E-19 J (3)
Total: 9

Q8 a) Gain in energy = 300eV = 300 x 1.6E-19 J = 4.8E-17J
1/2mv^2 = 4.8E-17
v=1.03E7 ms-1 (3)
b) lambda = h/mv = 7.07E-8m (2)
c) Higher PD = more energy = greater v = smaller wavelength
lambda = ax/D so x = lambda D/a so x reduces
Rings will be close together (and brighter) (2)
Total: 7

Hopefully that all adds up to 70.

Let me know if you spot any errors.

I dont know what the grade boundaries will be.
I dont even know if the two papers will have different grade boundaries and are converted to UMS before adding or if the raw marks are added than converted.
OCR havent told us. The first method is fairer but who knows.

Good Luck
Col
For question 5e how did you get the answer x 10^-4 ? With your working it gave an answer of 0.1121 ohm metres.
8. Yh I got like 5800 for the gradient and 0.11 for the restivity
9. I really hope that the grade boundary for an A isn't higher than 57. I think I can still get an A and make up for the first paper if that's the case.
10. For 8b λ=7.3E-11 m.
11. (Original post by lanteacher)
For 8b λ=7.3E-11 m.
If you use v=1.0E7 I used the value i got in the working from the one before so i ended up with lambda=7.1E-11
12. (Original post by Parhomus;[url="tel:65676235")
65676235[/url]]If you use v=1.0E7 I used the value i got in the working from the one before so i ended up with lambda=7.1E-11
Both would be correct and should be accepted by the mark scheme, so not to worry. I was just pointing out the power of 10 was -11, not -8.
13. (Original post by lanteacher)
Both would be correct and should be accepted by the mark scheme, so not to worry. I was just pointing out the power of 10 was -11, not -8.
Oh, right.
14. (Original post by lanteacher)
Both would be correct and should be accepted by the mark scheme, so not to worry. I was just pointing out the power of 10 was -11, not -8.
So will there be answers for using both the values they give. Plus the values you work out? I did that on the projectile question. I got 30.6 He said on the mark scheme its 20.7
15. (Original post by mahmzo)
So will there be answers for using both the values they give. Plus the values you work out? I did that on the projectile question. I got 30.6 He said on the mark scheme its 20.7
From previous mark schemes over the years, in the "show that the value is X" questions they accept the worked out (technically more correct) AND the approximated value they have given you in the question for the next questions. I hope that this will be the case in the new papers as well (and there's no reason why it wouldn't be).
16. I think 2c, the last bit should read decreasing as x is proportional to 1/a
17. anyone remember what the 6.45m question was? dunno if i got that mark or not
18. (Original post by ronnydandam)
anyone remember what the 6.45m question was? dunno if i got that mark or not
He got it wrong. I got 6.6 where you had to use work out on the chart
19. (Original post by mahmzo)
He got it wrong. I got 6.6 where you had to use work out on the chart
oh **** yeah that, ah another mark for me then
20. (Original post by ronnydandam)
oh **** yeah that, ah another mark for me then
Do you think ill get full marks, on the velocity projectile question. I got the correct answer however my working out was everywhere, like I had no room so had to do it in a cramped space :/ I am scared they will mark me wrong

Updated: April 8, 2017
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