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 = 1/1.5E3 = 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 ms1 (1)
iii) v = sqrt (vv^2 + vh^2)
= sqrt ( 28.0^2 6.3^2)
= 28.7 ms1 (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 = Kgms2 . 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 = 7.5E4 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 = 100E3/5300 = 1.89E5 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 1.9E5 = 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 crosssectional 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
6.63E34 x 9.60E14 = 3.2 x 1.6E19 = Ekmax
so Ekmax = 1.25E19 J (3)
Total: 9
Q8 a) Gain in energy = 300eV = 300 x 1.6E19 J = 4.8E17J
1/2mv^2 = 4.8E17
v=1.03E7 ms1 (3)
b) lambda = h/mv = 7.07E8m (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
You are Here:
Home
> Forums
>< Study Help
>< Maths, science and technology academic help
>< Physics
>< Physics Exams

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

Four hours left to win £100 of Amazon vouchers!! Don't miss out! Take our short survey to enter  24102016 

 Follow
 1
 10062016 10:33
Last edited by teachercol; 21062016 at 09:18.Post rating:5 
 Follow
 2
 10062016 12:49
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? 
 Follow
 3
 10062016 14:15
(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? 
 Follow
 4
 10062016 14:26
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.

 Follow
 5
 10062016 19:12
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 wavelike 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?
Last edited by Parhomus; 10062016 at 19:14. 
 Follow
 6
 10062016 19:21
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.

 Follow
 7
 10062016 19:36
(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 = 1/1.5E3 = 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 ms1 (1)
iii) v = sqrt (vv^2 + vh^2)
= sqrt ( 28.0^2 6.3^2)
= 28.7 ms1 (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 = Kgms2 . 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 = 7.5E4 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 = 100E3/5300 = 1.89E5 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 1.9E5 = 1.13 E4 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 crosssectional 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
6.63E34 x 9.60E14 = 3.2 x 1.6E19 = Ekmax
so Ekmax = 1.25E19 J (3)
Total: 9
Q8 a) Gain in energy = 300eV = 300 x 1.6E19 J = 4.8E17J
1/2mv^2 = 4.8E17
v=1.03E7 ms1 (3)
b) lambda = h/mv = 7.07E8m (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 
 Follow
 8
 10062016 19:55
Yh I got like 5800 for the gradient and 0.11 for the restivity

 Follow
 9
 10062016 20:50
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.

 Follow
 10
 11062016 11:53
For 8b λ=7.3E11 m.

 Follow
 11
 11062016 17:00
(Original post by lanteacher)
For 8b λ=7.3E11 m. 
 Follow
 12
 11062016 17:43
(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=7.1E11 
 Follow
 13
 11062016 17:46
(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. 
 Follow
 14
 11062016 18:20
(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. 
 Follow
 15
 11062016 18:30
(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 
 Follow
 16
 11062016 21:08
I think 2c, the last bit should read decreasing as x is proportional to 1/a
Post rating:1 
 Follow
 17
 12062016 07:24
anyone remember what the 6.45m question was? dunno if i got that mark or not

 Follow
 18
 12062016 11:06
(Original post by ronnydandam)
anyone remember what the 6.45m question was? dunno if i got that mark or notPost rating:1 
 Follow
 19
 12062016 11:14
(Original post by mahmzo)
He got it wrong. I got 6.6 where you had to use work out on the chart 
 Follow
 20
 12062016 12:07
(Original post by ronnydandam)
oh **** yeah that, ah another mark for me then
Write a reply…
Reply
Submit reply
Register
Thanks for posting! You just need to create an account in order to submit the post Already a member? Sign in
Oops, something wasn't right
please check the following:
Sign in
Not got an account? Sign up now
Updated: September 20, 2016
Share this discussion:
Tweet
TSR Support Team
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.