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

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Original post by mahmzo
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


if the answer is correct then yes, dw about it
Original post by mahmzo
He got it wrong. I got 6.6 where you had to use work out on the chart


That's right, the question had you calculate the square of L in 10E-3 m2. The value of L was 0.081m, which squares to 6.56E-3, rounded to 6.6E-3, so in the table you should write 6.6.
Original post by lanteacher
That's right, the question had you calculate the square of L in 10E-3 m2. The value of L was 0.081m, which squares to 6.56E-3, rounded to 6.6E-3, so in the table you should write 6.6.


what do you think the boundaries will be?
Reply 23
Original post by Parhomus
For question 5e how did you get the answer x 10^-4 ? With your working it gave an answer of 0.1121 ohm metres.


Typo. I'll change it.
Reply 24
Original post by mahmzo
He got it wrong. I got 6.6 where you had to use work out on the chart


Typo -- edited.
Thanks
(edited 7 years ago)
Reply 25
Original post by Parhomus
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?


There are lots of things you can write here that will score the marks
Original post by teachercol
There are lots of things you can write here that will score the marks


for the breaking stress one i wrote about finding strain at breaking point of a wire of known youngs mod and multiplying YM and strain? is that okay
Reply 27
Original post by ronnydandam
for the breaking stress one i wrote about finding strain at breaking point of a wire of known youngs mod and multiplying YM and strain? is that okay


Seems like an over complicated way of doing it to me. Have to wait for mark scheme.
Original post by teachercol
Seems like an over complicated way of doing it to me. Have to wait for mark scheme.


fair enough, hope I don't score 0 in that....
do you have the unofficial mark scheme for ocr b physics?
Reply 30
Sorry - don't teach/examine that syllabus and don't have copies of the paper.
Reply 31
does anyone have an unofficial mark scheme for the other physic paper with the multiple choice questions or at least some answers so i can see where i went wrong?
Video on the hardest question in that paper?
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; 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
(Diagram) Hang load off wire.
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







is there any chance that you would have 2017 ocr AS depth and breadth of physics QP

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