Additional guidance

Section A

Question Expected Answers Marks Additional Guidance

1 a b

J s^-1 C s^-1

1 1

not W

not A

2 a 1.7(1) x 10⁸ (Pa) 1 accept 170 / 171 MPa allow inappropriate SF

b Method: evidence of using largest F smallest A e.g. 148 / (0.76 x 10^-6)

evaluation = 1.9 x 10⁸ (Pa)

1 1

accept words / numbers not any credit for F x A accept 1.95 x 10⁸ for 2/2

allow 190 / 195 scores 1/2 not 194

c area because greatest relative or % uncertainty / ± 12% / 11% / 10%

1 accept diameter as measurement not just largest uncertainty accept area because force uncertainty is only ± 0.7% / ± 1%

mark for reasoning not for choosing just area

3 a Sun’s diameter (4.5 / 8.25) x 710 pixels = 390 pixels / 1.4 x 10⁶ km / 390 pixels

= 3600 km pixel^-1

1 1

accept range 350 to 430 pixels for first mark accept range (3200 to 4000 ) km pixel^-1 for 2/2 allow 1/2 for evidence of length / number of pixels not any credit for areas

b 220 000 km 1 method not required accept in range 170 000 to 280 000 km ecf on their resolution x 60 pixels (allow 54 to 66 pixels for the ecf )

4 1 / u is negligible for Sun’s distance / 1 / f  1 / v evaluation: P = 1 / f  1 / 0.012 = 83.(3) (D)

1 1

method accept curvature of wavefronts from Sun is negligible accept 83.(3) for 2/2 allow -83.(3) for 1/2

accept full calculation with P = 1/ v – 1/ u

1

Question Expected Answers Marks Additional Guidance 5 a 1 turn on voltage is 1.2 ± 0.05V /no current below 1.2 V

2 no current for reverse voltage / bias 3 current grows at increasing rate after 1.2V

4 linear / rapid increase in current after 1.4 V or 5mA

1 1

accept any two separate features not comparison with other temperatures

accept exponentially

accept straight line after 1.4 V or 5mA or ΔI α ΔV not I α V

b turn on / threshold voltage decreases (when the temperature increases) /

above 1.3 V current at a given voltage increases (with temperature) /

voltage for same current is lower

1 accept alternative wording e.g. less voltage needed to drive a current

not sensitivity

6 a ( Q = I t = 0.29 x 5 ) = 1.5 (C) 1 accept 1.45 (C) b N = Q / e / = 1.45 / 1.6 x 10^-19

= 9.1 x 10¹⁸

1 1

method in words / numbers ecf from part a)

evaluation accept 9.06 x 10¹⁸ / 9.4 x 10¹⁸ / 9.38 x 10¹⁸ / 9.375 x 10¹⁸

allow I / e = 1.8 x 10¹⁸ for 1/2 (electrons s^-1 ) 7 (total characters) = 2 x 26 + 10 + 12 = 74

reasoning: 2 ^bits  74 / 2⁶ = 64 / 2⁷ = 128 so 7 bits

1 1 1

correct arithmetic

ecf on number of characters for 2/3 max accept bits  log 74 / log 2 = 6.2 bare 7 scores 1/3

Total section A 20

2

Section B

Question Expected Answers Marks Additional Guidance

8 a i (1 at 500 lux R_LDR = ) 570  20 () (2 at 2500 lux R_LDR = ) 130  30 ()

1 both within tolerance for the mark accept 550 to 590 (Ω) accept 100 to 160 (Ω)

ii greater confidence at 500 lux: because within data / 2500 beyond data

at 2500 lux graph has to be predicted beyond data / reading error from graph has greater  % at 2500 lux

1 1

accept interpolation

accept extrapolation, estimation, greater uncertainty accept for 2/2 greater confidence at 2500 lux because sensitivity is more at 500 lux ora

b all circuit symbols correct

R and LDR in series with 6 V battery V meter in parallel with R

1

1 1

allow any reasonable dc supply symbol

not LDRs with arrows through symbol (variable Rs) , thermistor symbol allow LDR without circle

ignore additional series ammeter, variable resistor V in series max 1/3 for symbols

c Vout = V_in x R / (R + R_LDR) / V / R ratio argument 1.6 (R + R_LDR) = 6 R / 4.4 R = 1.6 R_LDR

R = 1.6 x 570 / 4.4 = 210 () For RLDR value 550 expect R = 200

560 204 570 207 580 211

590 215 600 218(ecf)

1 1

1

correct potential divider equation for circuit drawn correctly substituted

allow full credit for correct V / R ratio argument e.g.

R / 1.6 = 570 / 4.4  R  210 

evaluation ecf on RLDR from (a)

ecf on about 1600  (2.75 x RLDR ) if V meter across LDR in circuit

allow 1/3 for finding total circuit resistance (about 780  or 2100  if V meter across LDR ) as final answer

allow methods calculating current through LDR 7.7 mA, max 1/3 for correct current only

Total question 8 9

3

Section B

Question Expected Answers Marks Additional Guidance

9 a current = 2.7  0.1 (A)

power = 2.7 x 6 = 16.2 (W)

1

1 evaluation expect in range 15.6 to 16.8 (W) ecf on their current x 6

b i 0.35 , 0.62 / 0.619 () 1 both values correct in table for the mark ii 2 points correctly plotted

reasonable line of best fit near / through plotted points, (allow ± 2 vertical graph squares by points)

1

1 tolerance  ½ graph square each way allow ecf table values expect flat at start then kinked allow continuous curve not single straight line of best fit allow ecf on plotted values iii resistance increases as current increases /

temperature increases as current rises / resistance increases as temperature increases ; conductivity decreases with increasing temperature

1

1

accept resistivity increases as current increases accept filament heats as current rises

senses of changes must be clear

allow conductivity constant for small temperature rises for 1/2

c i (I = V / R ) = 12 / 0.30 ( = 40 ) 1 requires numerical answer c ii battery has internal resistance (in series with filament) /

contact resistance in circuit reducing current / filament starts to heat up quickly and its resistance rises / response time / sampling interval of data logger means true peak current is missed

1 any valid suggestion or AW

accept resistance in connecting wires

not just internal resistance / internal resistance of lamp / resistance of circuit / internal resistance uses up current

iii 1 current levels (at 4.0 A / after 1.5 s) once filament at working resistance / temperature

2 surge current decreases as filament heats for first 1.5  0.5 second

3 surge current decreases at a decreasing rate as filament approaches working temperature for first 1.5  0.5 second

1 any correct quantitative statement with explanation for one mark

not no current for 0.05 s because battery not connected

Total question 9 10

4

Section B

Question Expected Answers Marks Additional Guidance

10 a crystalline metal: regular square / hexagonal packing of identical spheres / circles in 2-d

amorphous glass: irregular / random arrangement of one or two sizes of spheres / circles in 2-d

one appropriate technical label / annotation on either diagram (not given for contradictory labelling)

1 1 1

accept presence of dislocation(s)

accept evidence of micro-crystals / polycrystalline nature not glass fibres / wiggly polymer diagrams

accept metal: close-packed planes / free electron gas/glue / + ions / dislocation / metallic / non-directional bonds etc.

glass: random arrangement / like liquid / covalent / directional bonds

b metals: regularity of atomic planes leading to slip / plastic flow / dislocations which move through metal by a few planes / rows slipping at a time

further plausible geometric arguments (without mention of dislocations) can gain credit max 2

glasses: irregular or random arrangement means regular slip cannot occur / cannot relieve stress so stress concentrates and micro-cracks propagate leading to brittle failure max 2

structure leads to movement for metals scores 1 structure leads to lack movement in glasses scores 1 further detail for either scores 1 QoWC scores 1

3+1 4^th mark is for QoWC answer must correctly use at least one of appropriate technical terms: regular and random packing and how this leads to slip / plastic flow in metals and to brittle crack propagation / stress concentration in glasses

accept answers based on polycrystalline nature of metals with dislocations piling up on grain boundaries / isotropic properties due to random orientation of micro-crystals for full credit accept free electron gas / glue makes the non-directional metal bonding crucial to slip and crack behaviour

AW throughout but NO credit for use of brittle or ductile If QoWC awarded place tick on technical term credited c i any example of a composite material with its

components clear e.g. glass fibre reinforced plastic

1 accept concrete containing mortar and pebbles / gravel steel reinforced concrete etc. not alloys

ii initial straight line of same gradient judged by eye yield stress of 1350 MPa see overlay

plastic region linear of small slope stopping at 10%

strain

1 1

any 2 correct points but max 1 for graphs below the original allow within graph square between 1320 and 1360 MPa accept zero slope for plastic region stopping at 10% strain P.T.O.

5

Question Expected Answers Marks Additional Guidance c iii area = fracture energy / toughness /

= 14 / 0.34 x 10⁶ = 4.1(2) x 10^-5 (m²)

1 1

first mark for correct equation in words or numbers evaluation allow 41.(2) 1/2 multiplier error

Total question 10 12

6

7 Section B

Question Expected Answers Marks Additional Guidance

11 a i 16 x 44.1 x 10³ = 7.1 x 10⁵ ( > 500 kbit s^-1) 1 allow 705.6 ( kbit s^-1 ) ii resolution = 100 x 10^-3 V / (2¹⁶  1)

= 1.5(3) V

1 1

method accept 2¹⁶ / 65535 / 65536 (subd in equation) evaluation allow 1 mark for 1.8(3) V using full graph scale of 120 mV

b i 10⁶ 1

ii 1 2 000 (Hz) 2 A and F

1 1

accept 1500 to 2500 (Hz)

N.B. before marking 11 c check additional pages S12 , S13 and S14 for candidates additional answers, the marking tool is there for a reminder. Extra answers MUST be annotated to show they have been seen and credited back in the relevant question, if

appropriate.  = 1 extra mark x = wrong scores 0

^ = no added value scores 0 and NR = no further action.

c

rate = frequencies/frame x bits/frequency x frames/s = 32 x 24 x 40 = 30720 / 3.1 x 10⁴ bit s^-1 fraction = 3.1 x 10⁴ / (7.1 x 10⁵) = 0.044  1/23 < 1/20

1 1 1

QoWC method must be explicitly clear other than in numbers

evaluation

fraction shown ora 7.1 x 10⁵ / 20 = 3.5 x 10⁴ bit s^-1 > 3.1 x 10⁴ allow ecf on 500 kbit s^-1 from (ai) giving  1/16

allow ecf on other incorrect bit rates accept other valid comparisons

Total question 11 9

Total section B 40

Paper total 60

1

Qn Expected Answers

Additional guidance

(b) N kg^-1 (1)

2

2

(a) A (1);

(b) C (1);

(c) B (1) 3

3

f = c/ = 3.0 × 10⁸ m s^-1/1.8 × 10^-10 m = 1.67 × 10¹⁸ Hz (1);

E=hf = 6.6 × 10^-34 J s × 1.67 × 10¹⁸ Hz = 1.1 × 10^-15 J (1)s (1)e

3

Can use E = hc/; quoting equation (1); & then (1)s (1) e If equation wrong (e.g. E = h/c) then no marks for Q3.

No e.c.f allowed from incorrect frequency;

4 second box (phasors line up) (1);

third box (slits closer together ) (1) 2 One for each correct tick. Cancel one mark for each extra tick.

5

(a) any antinode indicated (1) (b)  = 1.2 m(1);

f = 1/0.4 s = 2.5 Hz (1)

1

2

6

(a) t = (v – u)/a = 27 m s^-1/(0.86 × 9.8 m s^-2) = 3.2 s (1)m (1)e

(b) a = (v²-u²)/2s =(- [27 m s^-1]²)/(2 × 35 m =10.4 m s^-2 F=ma=1600 kg ×10.4 m s^-2= 16 700 N= 17 000 N (1)m (1)e

2 2

As u is not specified, allow any calculation based on candidate’s own value.

Sign unimportant.

Can calc E_k/s= 583 kJ/35m = 17 kN

7

(a) W = Fs=(110kg×9.8 m s^-2) ×1.2m = 1290 J ≈ 1300 J (1)m (1)e (b) s = 30 cm – 6 cm = 24 cm (1)

F = W/ s = 1300 J/(24 × 10^-2 m) = 5420 N = 5400 N (1)

2

2

Must have evidence of calculation

1^st mark is for 24 cm / 0.24m / 24 × 10^-2 m. Accept 24 without units.

Second mark is for correct final answer

Or via suvat (u= 4.85 m s^-1, a = 49 m s^-2) & F=ma.

Allow answer which includes the weight of the barrel 6500 N.

Section A total: 21

2

Qn Expected Answers

Additional guidance

8 (a) W and AR downwards and U upwards 1 Must be vertical: ignore lengths of arrows or points of application (b) (i) Needs to be moving (relative to air) (for there be be air

resistance)(1)

(ii) F = (0.0060 kg×9.8 ms^-2) - (0.0035 kg×9.8 m s^-2) (1) = 0.0245 N (≈ 0.02 N) (1)

1

2

‘Balloon is at rest’ is enough irrespective of any qualification.

Allow 0.024 N or 0.025 N (not just 0.02 N) as evidence of evaluation (c) (i) identifying streamlining/ appropriate change in shape or

size of balloon (1)

Relating change to reduction in FAR (at constant v) (1)

(ii) Gradient dropping: resultant force drops (producing smaller acceleration) due to increased air resistance (1);

Horizontal line: FAR = U - W (1)

(iii) kv = 0.016Nsm^-1 × 1.5 ms^-1= 0.024 N (1) Comparison with answer to (b)(ii) (1)

2

2

2

QWC: maximum 1 mark if terms not correctly used or spelled.

Answer must explain in terms of the forces involved; ‘air resistance increases’ is enough

‘Resultant force = 0’ owtte is enough here. Reference to ‘terminal velocity’ is neutral.

If reverse argument is used, needs to start with F from (b)(ii) (second marking point), and then calculate k close to 0.016Nsm^-1 for the first marking point.

Total: 10

3

Qn Expected Answers

Additional guidance

 = 33° (1)m (1)e (ii) smaller (1)

 = d sin, so (for the same  d ↑ ( sin↓) ↓ (1)

2 2

ora: d sin(30°) = 650 nm  700 nm (1)m (1)e

(1) for smaller 

(1) for explanation linked to equation which may be implied.

(b) (i) 2  = d sin(90°)=1.3×10 ^-6 m × 1  = 6.50 × 10 ^-7m (1)m (1)e;

(ii) 700 nm > value calculated in (i) (1)

Would result in  > 90°, which is not possible (1)

2

2

Zero marks if n=2 not used

Calculating sin = 1.09 (1) So no solution for  possible (1)

Accept well-reasoned approach based on path differences (c) destructive interference occurs in this region (1);

for all visible wavelengths (1)/

in this region must be either infrared (first order) or ultraviolet (second order)(1);

neither is visible/present in the incident light (1)

2 ‘no visible wavelengths give solutions to n = d sin() in that region’

gets both marks

Total: 10

4

Qn Expected Answers

Additional guidance

to direction of movement of wave (1);

Longitudinal: oscillations/vibrations in direction parallel to

direction of movement of wave (1) 2

(1) for linking transverse to oscillations perpendicular (to direction of wave motion), (1) for linking longitudinal to oscillations parallel (to direction of wave motion)

Marks can be given for clear diagrams (b) (i) time of travel of P-waves is tP = D  vP / that of S-waves is

tS = D  vS (1);

 

 

   s p Ds Dp  ¹s  ¹p

V V V V

t t t D (1)

(ii) t = 50 s (1);

D = t /(1/vS-1/v_P) = 50 s/(1/3500 m s^-1 -1/8000 m s^-1) = 50 s/(1.6 × 10^-4 sm^-1) = 310 000 m (1) s (1)e

2

3

First mark is for a correct application of v/D/t to either wave

± 2 s; 48 s  299 000 m, 52 s  325 000 m For e.c.f.,D = t/(1.6 × 10^-4 sm^-1)

Correct substitution of v_s and v_p gets the (1)s This can be inferred from e.g. (1.6 × 10^-4 sm^-1)

(c) Correct amplitudes identified (1) ratio of amplitudes = 285/105 ≈ 2.67 2.67² = 7.11 ≈ 7 (1)

2

Allow amplitudes of 250 – 300 and 80 – 120 mm Allow peak-to-peak values instead of amplitudes Accept ratio in format 13:2 etc.

ecf incorrect amplitudes and rounding before squaring

Give the first mark if you can see the two amplitudes or peak-to- peak values.

Ignore units in final answer.

(d) log scale allows great range of values owtte to be registered

in a smaller range of numbers (1) 1

Total: 10

5

Qn Expected Answers

Additional guidance

hypotenuse 1000 N (1); adjacent side 900 N (1) Calculation: F_H = T cos() (1);

= 1000 N × cos(30°) (1) = 870 N ≈ 900 N (1)

3

Remember that correct answer with no working gets full marks, so 866 N = 3 marks

(b) (i) Vertical component of tension = 1000 N sin(30°) = 500N (1)

Minimum mass of kitesurfer = 500 N/ 9.8 N = 51 kg (1)

(ii) Suggesting a factor with relationship to kite size (wind speed or speed/skill/activity of surfer) (1)

Stating the direction of the effect and explaining the relationship (1)

2

2

Can be done by scale drawing, possibly from (a).

If ‘mass’ and ‘weight’ are not properly distinguished, this mark is not awarded. Can be implicit in calculation.

Need not specify direction of effect, e.g. ‘You need different sized kites in different wind speeds’ or ‘You need a different kite for speed surfing from that for acrobatics’ is OK

E.g. ‘In faster winds, you will need a smaller kite or you will go too fast/be lifted off the water’

(c) Horizontal (component of) force from kite = (horizontal) force from water on board (1);

and in opposite direction (1)

2

Ignore reference to vertical components.

Ignore attribution of horizontal force from water to e.g. resistance, friction, component of normal reaction

‘horizontal forces in equilibrium’ will gain this mark, ‘forces in equilibrium’ will not.

Total: 9 Section B total: 39

6

Qn Expected Answers

Additional guidance

Correct reference to spread/range of data excluding potential outlier (1);

Visible gap between outlier and rest of data (1)

Attempt to quantify separation of outlier from mean/minimum of remaining data in terms of spread (1)

(ii) Suggestion should make it clear that wire has been damaged in some way, e.g. narrowed by twisting (1)

2

1

accept reference to standard deviation or interquartile range, etc e.g. ‘is very much lower than the others’

‘greater than twice the spread from the mean’ gets both marks.

‘weakened’ is enough (b) Uncertainty = 1 N (1)

Rounding force to ± 1 N = 9 N (1)

2 Must be 1 s.f.

9.1 ± 1 N would get the first marking point but not the second (c) (i) %age uncertainty = 0.005mm × 100/0.38mm

= 1.3 % = 1% (1)m (1)e

(ii) %age uncertainty in d « %age uncertainty in F

2 1

Expect 1 s.f. allow 2; for > 2 s.f. maximum mark = 1 Must compare d with F (may use fractional uncertainties) (d)

mean stress = 9 N/1.1×10^-7 m² = 8.2 × 10⁷ Pa (1) m (1) e repeat with uncertainty = 1 N/1.1×10^-7 m² = 1.0 × 10⁷ Pa (1) e.c.f for both calculations from part (b)

3

Need to use candidate’s values from (b) here.

Can do via:

Max stress = (9+1)N/1.1×10^-7 m² = 9.1 × 10⁷ Pa and/or Min stress = (9-1)N/1.1×10^-7 m² = 7.3 × 10⁷ Pa

To obtain e.g uncertainty = (max – mean) = 1 × 10⁷ Pa (1) Can be found from e.g. 1/9 or 11% of mean

ignore s.f. errors in both mean and uncertainty.

Total: 11

7

Qn Expected Answers

Additional guidance

(ii) Use of gradient at start or data points for 0 & 0.5 s (1) gradient ≈ 15 m/0.5 s = 30 ms^-1/ data points give 25 ms^-1 (1)

(iii) gradient decreasing (continually) during ascent (1);

negative gradient increasing during descent (1)

1 2

2

Consistent with candidate’s curve.

Gradient triangle must have base of at least 0.2 s.

Consistent with candidate’s own tangent

Accept use of v² = u² + 2as with s = answer to (a)(i) and a = -9.8 m s^-2 to give u about 23.4 m s^-1

Must refer to both ascent and descent: can refer to both parts as a single motion if clear, e.g. ‘gradient decreases from positive to zero to negative’

(b) (i) v² = u² + 2as with v = 0 (1)

0 = 25² + 2 × (-9.8) × s s = 625/19.6 = 32 m (1)s (1)e Ignore errors in allocating – sign to g or u & s in the use of this equation: just look for 31.9 m

(ii) Air resistance/drag (1);

provided decelerating force/reduced velocity//dissipated energy (1)

3

2

Or v=u+at t = 2.55 s (1) followed by s=ut+½at² (1)m (1)e

If t = 2.75 s taken from graph, first mark is not given but can then get the remaining two marks.

Reverse argument from s = 30 m to calculate u is OK.

any mention of air resistance is enough for this first mark.

Second mark needs a clear and correct explanation.

(c) (i) Force from wind is horizontal (1);

need vertical force from wind to affect height reached (1) horizontal and vertical motion are independent (1)

(ii) Time of flight = 2×time to highest point= 2× 2.75 s (1);

speed of wind = 37m/5.5 s = 6.7 ms^-1 (1)

2

2

Any two points.

check with 13(a)(i) if different from 2.75 s

If time not doubled give zero marks unless total time of flight = 4.0 s from graph (e.g. landed on a roof).

37m/4.0 s = 9.25 ms^-1 Total: 14

8

Qn Expected Answers

Additional guidance

stopping p.d. from graph (1)

(ii) correct plotting (2);

best-fit line (1);

gradient: 3.5 V/8.5 × 10 ¹⁴ Hz (1)m = 4.1 × 10^-15 V Hz^-1 (1)e

(iii) h = e × gradient=1.6×10^-19×4.1×10^-15 = 6.6 × 10^-34 J s (1)

2

5

1

Not straight line; should be through/close to points.

Curve must reach the p.d. axis for second mark and be consistent with candidate’s graph. Expect values about 1.32 V.

All 3 points in tolerance = (2); 2 correct = (1) (overlay on scoris).

If point in blue overlay box is incorrect, check candidate’s value in(a)(i). Point on edge of overlay tolerance box is in.

Line in tolerance by eye

e.c.f. own line; can use any data points for calculation as all are close to line.

Max 1 mark for gradient calculation if 10¹⁴ not included correctly OR 4 × 10¹⁴ on x-axis read as 0.

No marks awarded if both errors made.

e.c.f. own gradient.

Must show evidence of working for answer of 6.6 × 10^-34.

Watch for incorrect gradient giving correct h by fiddling = 0 marks.

(b) (i) uncertainty in h = (0.5/100) × 6.57 × 10^-34 J s

= 3 × 10^-36 J s (1)

h_max =6.57 × 10^-34 + 3 × 10^-36 J s = 6.60 × 10^-34 J s (1) (ii) Millikan’s (maximum) value < (minimum) accepted value/

ranges do not overlap owtte (1)

suggested incorrect variable (i.e. V, f or e) (1);

suggested reason for error (can be generic, e.g. incorrect calibration of instrument involved) (1).

2

3

Or 6.57 × 10^-34 J s × 1.005 = 6.60 × 10^-34 J s (1)m (1)e

QWC is third mark, for ‘organise information clearly and coherently’.

(c) Established theories had worked well (1);

idea was too different from accepted theory(1);

insufficient experimental support for new theory (1) 2

Any two points.

Total: 15 Section C total: 40

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