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
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.
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.
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
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?
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