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

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?

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
(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

For 8b λ=7.3E-11 m.

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.

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

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

oh **** yeah that, ah another mark for me then