9702 w08 ms_all

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS

GCE Advanced Subsidiary Level and GCE Advanced Level

MARK SCHEME for the October/November 2008 question paper

9702 PHYSICS

9702/01 Paper 1 (Multiple Choice), maximum raw mark 40

Mark schemes must be read in conjunction with the question papers and the report on the examination.

• CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the October/November 2008 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Mark Scheme Syllabus Paper

GCE A/AS LEVEL – October/November 2008 9702 01

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Question Number

Key Question Number

Key

1 C 21 C

2 B 22 C

3 A 23 C

4 D 24 D

5 D 25 B

6 D 26 A

7 A 27 C

8 A 28 D

9 B 29 B

10 D 30 A

11 A 31 D

12 D 32 B

13 A 33 B

14 A 34 D

15 A 35 D

16 A 36 B

17 C 37 B

18 D 38 D

19 D 39 A

20 C 40 A




9702 PHYSICS

9702/02 Paper 2 (AS Structured Questions), maximum raw mark 60

This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began.

All Examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated.





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1 (a) (i) Q = It (allow any subject for the equation) B1 [1] (ii) I B1 t B1 [2] (allow 1 mark only if all three quoted) (b) (i) base unit of I is A base unit of n is m–3 (not /m–3) base unit of S is m2 base unit of q is A s (not C) base unit of v is m s–1 (–1 for each error or omission) B3 [3] (ii) A = m–3 m2 A s (m s–1)k M1 e.g. for m: 0 = –3 + 2 + k k = 1 A1 [2]

2 (a) (i) v2 = 2as

v2 = 2 × 0.85 × 9.8 × 12.8 C1 v = 14.6 m s-1 A1 [2] (ii) time = 29.3 / 14.6 C1 = 2.0 s A1 [2] (any acceleration scores 0 marks; allow 1 s.f.)

(b) either 60 km h–1 = 16.7 m s–1 or 14.6 m s–1 = 53 km h–1 or 22.1 m s–1 = 79.6 km h–1 M1 so driving within speed limit A1 but reaction time is too long / too slow B1 [3]

3 (a) moment: force × perpendicular distance M1 of force from pivot / axis / point A1

couple: (magnitude of) one force × perpendicular distance M1 between the two forces A1 [4] (penalise the ‘perpendicular’ omission once only)

(b) (i) W × 4.8 = (12 × 84) + (2.5 × 72) C1 W = 250 N (248 N) A1 [2] (ii) either friction at the pivot or small movement of weights B1 [1]

4 (a) (i) either force = e × (V / d) or E = V/d C1

= 1.6 × 10–19 × (250 / 7.6 × 10–3) C1

= 5.3 × 10–15 N A1 [3]

(ii) either ∆EK = eV or ∆EK = Fd C1

= 1.6 × 10–19 × 250 = 5.3 × 10–15 × 7.6 × 10–3 M1

= 4.0 × 10–17 J A0 [2] (allow full credit for correct working via calculation of a and v)



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(iii) either ∆EK = ½mv2

4.0 × 10–17 = ½ × 9.1 × 10–31 × v2 C1

v = 9.4 × 106 m s–1 A1 [2] or v2 = 2as and a = F/m

v2 = (2 × 5.3 × 10–15 × 7.6 × 10–3)/(9.11 × 10–31) (C1)

v = 9.4 × 106 m s–1 (A1) (b) speed depends on (electric) potential difference M2

(If states ∆EK does not depend on uniformity of field, then award 1 mark, treated as an M mark) so speed always the same A1 [3] 5 (a) haphazard / random / erratic / zig-zag movement M1 of (smoke) particles (do not allow molecules / atoms) A1 [2]

(b) motion is due to unequal / unbalanced collision rates (on different faces) B1 (unequal collision rate due to) random motion of (gas) molecules / atoms B1 [2] (c) either collisions with air molecules average out M1 this prevents haphazard motion A1 [2] or particle is more massive / heavier / has large inertia (M1) collisions cause only small movements / accelerations (A1) 6 (a) wave incident at an edge / aperture / slit /(edge of) obstacle M1 bending / spreading of wave (into geometrical shadow) A1 [2] (award 0/2 for bending at a boundary) (b) (i) apparatus e.g. laser & slit / point source & slit / lamp and slit & slit microwave source & slit water / ripple tank, source & barrier B1 detector e.g. screen aerial / microwave probe strobe / lamp B1 what is observed B1 [3] (ii) apparatus e.g. loudspeaker, and slit / edge B1 detector e.g. microphone & c.r.o. / ear B1 what is observed B1 [3] 7 (a) either V = IP B1 current in circuit = E / (P + Q) B1 hence V = EP / (P + Q) A0 [2] or current is the same throughout the circuit (M1) V / P = E / (P + Q) (A1) hence V = EP / (P + Q) (A0)



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(b) (i) (as temperature rises), resistance of (thermistor) decreases M1 either resistance of parallel combination decreases

or p.d. across 5 kΩ resistor / thermistor decreases M1

p.d. across 2000 Ω resistor / voltmeter reading increases A1 [3] (ii) if R is the resistance of the parallel combination,

either 3.6 = (2 × 6) / (2 + R) or current in 2 kΩ resistor = 1.8 mA C1

R = 1.33 kΩ current in 5 kΩ resistor = 0.48 mA C1

1.33

1 =

5

1 +

T

1 current in thermistor = 1.32 mA C1

T = 1.82 kΩ T = 2.4 / 1.32 = 1.82 kΩ A1 [4] 8 (a) nucleus has constant probability of decay M1 per unit time / in a given time A1 [2] (allow 1 mark for ‘cannot predict which nucleus will decay next’) (b) (i) count rate / activity decreases B1 [1] (ii) count rate fluctuates / is not smooth B1 [1] (c) either the (decay) curves are similar / same or curves indicate same half-life B1 [1]




9702 PHYSICS

9702/32 Paper 32 (Advanced Practical Skills 2), maximum raw mark 40





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1 (a) (ii) Measurement of θ. 5 Y θ Y 10° Ignore d.p. [1]

(b) Six sets of readings scores 6 marks, five sets scores 5 marks, etc. Help given, –1 (e.g. putting plumbline into position). Generally wrong trend, –1. Allow n = 0. [6]

Range. Maximum angle θ max [ 45°. [1]

Table headings. θ /° θ (°) No unit for 1/cosθ. [1]

Consistency in raw data – all values of θ given to the nearest 1° or 0.5°. [1]

Calculated quantities. Allow small rounding errors.

– check the specified value of 1/cosθ and tick if correct.

Specified value is the largest value of θ. [1] Significant figures. [1]

– all values of 1/cosθ should be to the same s.f. as (or one more than) the raw value of θ. Quality of data. 5 points close to Examiner’s straight line. Wrong trend/curved trend – no mark. [1]

(c) Points should occupy at least half the grid in both directions and scales should be sensible (not 3, 6, 9 or other awkward) and labelled with a quantity. Do not penalise reversed axes. Label FO. Ignore units. [1]

Check that one point is correctly plotted (error Y half a small square). All tabulated results to be plotted on graph grid.

Do not allow blobs (points [ half a small square). If plot incorrect indicate correct position. [1]

Line of best fit. At least 5 trend plots. Allow curved trend.

No hairy or thick lines ([ half a small square). No kinks. [1]

(d) Gradient. Triangle chosen for gradient as a hypotenuse at least half the length of the drawn line. Read-offs are on the line correct to within half a small square and correct substitution.

Gradient mark = 0 if curve used. If wrong write in correct read-off. Correct sub into ∆y/∆x. [1]

Intercept calculated by a correct method or using the graph. Allow for extrapolation for curve at n = 0 (i.e. do not allow algebraic errors with y = mx + c). [1]

(e) Correct method and substitution. k equal to

m2

gradient. [1]

Method and value of M within 50 % of Supervisor’s value. M = intercept / k. Allow e.c.f. for k. Write in Supervisor’s value for M underneath. [1]

[Total: 20]



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2 (b) (i) Measurement of l 19.0 Y l Y 21.0 cm. Ignore d.p. Supervisor’s help –1. [1] (ii) Correct method of estimation of percentage uncertainty.

∆l = 1 mm or 2 mm or half the range. [1] (iii) Correct calculation of first value of l3 (20³ = 8000). If incorrect write in correct value. Accept small rounding errors. [1] (iv) Justification for s.f. for l3. Same or one more than the raw value of l. Consistent with their own data. [1]

(c) Measurement of T. 0.2 Y T Y 2.0 s [1] (c) or (d) Measurement of raw t to the nearest 0.1 s or 0.01 s. [1] Evidence of repeat readings of t. [1]

Evidence of n [ 10 oscillations. [1] (d) Measurement of second l to nearest mm. [1]

Measurement of second T(d) I T(c) . Penalise wrong trend. [1] (e) Correct method and calculation of k values. [1] Valid comment on whether equation applies to results. [1] Allow e.c.f. on arithmetic errors of k values. Evidence of correct ratio for one value of k is

necessary to access this mark. k values within 10% to support relationship. Allow up to 20% if candidate stated a value.

(f) (i) Problems [4] (f) (ii) Improvements [4]

Ap Not enough readings (to draw a conclusion). As More readings and plot a graph.

Bp Time too fast/moves too fast/error in timing large compared to time measured.

Bs 1 Video recorder, playback frame by frame/ slow motion with timer/stroboscope with scale.

Bs 2 Longer hacksaw blade/heavier mass (to increase time of oscillation)/more oscillations than already used (larger n).

Cp Judging beginning/end of oscillation/complete oscillation.

Cs Motion/position sensor placed at side of mass/fiducial marker/(stationary) reference marker and stated purpose.

Dp Length error e.g. parallax error in reading the ruler/difficulty in establishing centre of mass/ ends of blocks.

Ds Find the mid-point of the mass by finding the distance to both ends and taking an average/ thinner rule with reason/scale starts at 0 cm with reason/scale on blade/corrections for parallax error.

Ep Difficulty in setting up the apparatus horizontally/difficulty in assembly with detail.

Es Use spirit level/measure up from bench/ partner to help with set up.

[Total: 20]




9702 PHYSICS

9702/04 Paper 4 (A2 Structured Questions), maximum raw mark 100







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Section A 1 (a) (i) F = GMm / R2 B1 [1]

(ii) F = mRω2 B1 [1]

(iii) reaction force = GMm / R2 – mRω2 (allow e.c.f.) B1 [1]

(b) (i) either value of R in expression Rω2 varies

or mRω2 no longer parallel to GMm / R2 / normal to surface B1

becomes smaller as object approaches a pole / is zero at pole B1 [2]

(ii) 1. acceleration = 6.4 × 106 × (2π / 8.6 × 104)2 C1 = 0.034 m s–2 A1 [2] 2. acceleration = 0 A1 [1] (c) e.g. ‘radius’ of planet varies density of planet not constant planet spinning nearby planets / stars (any sensible comments, 1 mark each, maximum 2) B2 [2] 2 (a) (Thermal) energy / heat required to convert unit mass of solid to liquid M1 at its normal melting point / without any change in temperature A1 [2]

(reference to 1 kg or to ice → water scores max 1 mark) (b) (i) To make allowance for heat gains from the atmosphere B1 [1] (ii) e.g. constant rate of production of droplets from funnel constant mass of water collected per minute in beaker (any sensible suggestion, 1 mark) B1 [1]

(iii) mass melted by heater in 5 minutes = 64.7 – ½ × 16.6 = 56.4 g C1

56.4 × 10–3 × L = 18 C1 L = 320 kJ kg–1 A1 [3] (Use of m = 64.7, giving L = 278 kJ kg–1, scores max 1 mark use of m = 48.1, giving L = 374 kJ kg–1, scores max 2 marks) 3 (a) acceleration / force (directly) proportional to displacement M1 and either directed towards fixed point or acceleration & displacement in opposite directions A1 [2] (b) (i) maximum / minimum height / 8 mm above cloth / 14 mm below cloth B1 [1] (ii) 1. a = 11 mm A1 [1]

2. ω = 2πf C1

= 2π × 4.5 = 28.3 rad s–1 (do not allow 1 s.f.) A1 [2]



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(c) (i) v = ωa C1

= 28.3 × 11 × 10–3 = 0.31 m s–1 (do not allow 1 s.f.) A1 [2]

(ii) v = ω √(a2 – y2) y = 3 mm C1

= 28.3 × 10–3 √(112 – 32) C1 = 0.30 m s–1 (allow 1 s.f.) A1 [3]

4 (a) ∆U = q + w (allow correct word equation) B1 [1] (b) either kinetic energy constant because temperature constant M1 potential energy constant because no intermolecular forces M1 so no change in internal energy A1 [3] or kinetic energy and potential energy both constant (M1) so no change in internal energy (A1) reason for either constant k.e. or constant p.e. given (A1) 5 (a) change/loss in kinetic energy = change/gain in electric potential energy B1

2 × ½mv2 = q2 / 4πε0r C1

2 × ½ × 2 × 1.67 × 10–27 × v2

= (1.6 × 10–19)2 / (4π × 8.85 × 10–12 × 1.1 × 10–14) M1

v = 2.5 × 106 m s–1 A0 [3] (b) pV = ½Nm<c2> and pV = NkT C1

½ m<c2> = 23

kT (award 1 mark of first two if <c2> not used) C1

½ × 2 × 1.67 × 10–27 × (2.5 × 106)2 = 23

× 1.38 × 10–23 × T C1

T = 5 × 108 K A1 [4]

(c) e.g. this is very high temperature temperature found in stars (any sensible comment, 1 mark) (if T < 106 K, should comment that too low for fusion to occur) B1 [1] 6 (a) (i) either prevent loss of magnetic flux or improves flux linkage with secondary B1 [1] (ii) reduces eddy current (losses) B1 reduces losses of energy (in core) B1 [2] (b) (i) (induced) e.m.f. proportional to / equal to M1 rate of change of (magnetic) flux (linkage) A1 [2] (ii) changing current in primary gives rise to (1) changing flux in core (1) flux links with the secondary coil (1) changing flux in secondary coil, inducing e.m.f. (1)



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(any three, 1 each to max 3) B3 [3] (c) e.g. can change voltage easily / efficiently high voltage transmission reduces power losses (any two sensible suggestions, 1 each) B2 [2] 7 (a) e.g. ‘instantaneous’ emission (of electrons) threshold frequency below which no emission (max) electron energy dependent on frequency (max) electron energy not dependent on intensity rate of emission (of electrons) depends on intensity (any three sensible suggestions, 1 each) B3 [3] (b) (i) ‘packet’ / quantum of energy M1 of electromagnetic energy / radiation A1 [2] (ii) discrete wavelengths mean photons have particular energies M1 energy of photon determined by energy change of (orbital) electron M1 so discrete energy levels A0 [2] (c) (i) three energy changes shown correctly B1 arrows ‘pointing’ in correct direction B1 wavelengths correctly identified B1 [3]

(ii) chooses λ = 486 nm C1

∆E = hc / λ C1

= (6.63 × 10–34 × 3.0 × 108) / (4.86 × 10–9)

= 4.09 × 10–19 J (allow 2 s.f.) A1 [3] 8 (a) region (of space) / area where B1 a force is experienced by M1 current-carrying conductor / moving charge / permanent magnet A1 [3] (b) (i) electric B1 [1] (ii) gravitational B1 [1] (iii) magnetic B1 [1] (iv) magnetic B1 [1]



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Section B 9 (a) IR has less attenuation (per unit length) B1 fewer (repeater) amplifiers / longer uninterrupted length B1 [2] (b) either limited range B1 (so) cells do not overlap (appreciably) B1 [2] or short wavelength (B1) so convenient length aerial (on mobile phone) (B1) (c) large bandwidth / large information carrying capacity B1 different so that uplink signal not swamped by downlink B1 [2] 10 (a) (i) 1. inverting (amplifier) B1 [1] 2. gain of op-amp is very large / infinite B1 non-inverting input is at earth / 0 V B1 for amplifier not to saturate, P must be at about earth / 0 V B1 [3] (ii) input resistance is very large B1 (so) current in R1 = current in R2 B1 I = VIN / R1 B1 I = – VOUT / R2 (minus sign can be in either of the equations) B1 hence gain = VOUT / VIN = –R2 / R1 A0 [4]

(b) (i) 1. feedback resistance = 33.3 kΩ C1 gain (= 33.3 / 5) = 6.66 C1

VOUT (= 6.66 × 1.2) = 8.0 V (+ or – acceptable, allow 1 s.f.) A1 [3]

2. feedback resistance = 8.33 kΩ C1

VOUT (= 6.66 × 1.2 / 5) = 2.0 V (+ or – acceptable, allow 1 s.f.) A1 [2] (ii) (Increase in lamp-LDR distance gives) decrease in intensity M1 Feedback / LDR resistance increases M1 voltmeter reading increases / becomes more negative A1 [3] 11 (a) CT image: (thin) slice (through structure) B1 any further detail e.g. built up from many ‘slices’ / 3-D image B1 X-ray image: ‘shadow’ image (of whole structure) / 2-D image B1 [3] (b) X-ray image of slice taken from many different angles (1) these images are combined (and processed) (1) repeated for many different slices (1) to build up a 3-D image (1) 3-D image can be rotated (1) computer required to store and process huge quantity of data (1) (any five, 1 each to max 5) B5 [5]




9702 PHYSICS

9702/05 Paper 5 (Planning, Analysis and Evaluation), maximum raw mark 30







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Question 1

Planning (15 marks)

Defining the problem (3 marks)

P1 d is the independent variable or vary d (allow in table if numbers given) [1]

P2 R is the dependent variable or measure R as d varied (allow in table) [1]

P3 Keep output of light source constant (allow constant current / e.m.f. / voltage / power) [1] Methods of data collection (5 marks)

M1 Diagram showing an LDR in a circuit and an independent lamp. [1]

M2 Diagram showing ruler measuring appropriate distance or d labelled correctly. [1]

M3 Correct circuit diagram for LDR using conventional symbols; allow labelled diagram [1] Ammeter and voltmeter with power supply, or potential divider methods ohmmeter without power supply, or bridge methods.

M4 Method of determining R. [1] Ohmmeter. R = V/I justified. Potential divider equation Description of balancing bridge with correct equation.

M5 Perform experiment in a dark room/tube [1] Method of analysis (2 marks)

A1 Plot a graph of log R against log d [1]

A2 Relationship is correct if log R against log d graph is a straight line [1] Safety considerations (1 mark)

S1 Do not look directly at bright light source / do not touch hot light source. [1] Allow safety glasses with reference to light source.

Additional detail (4 marks)

D1/2/3/4 Relevant points might include [4] Detail on measuring the distance Keep orientation of LDR with respect to the light source constant Reasoned method for keeping light and LDR in correct orientation. (E.g. use of set square, fix to rule, optical bench or equivalent) Determination of a typical current Range of ammeter / ohmmeter Control (or monitoring) of an additional variable e.g. temperature Reason for performing experiment in a dark room related to the LDR Method for checking the output of the light source is constant. Identifies gradient = n and/or y-intercept = log k for log R against log d graph

Do not allow parallax when reading ruler, or reflectors. [Total: 15]



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Question 2 Analysis, conclusions and evaluation (15 marks)

Part Mark Expected Answer Additional Guidance

(a) A1

2

8

eB

m Allow

2

8

Bgradient

m

e

=

(b) T1

4.4 or 4.41

7.8 or 7.84

12 or 11.6 (or 11.56)

15 or 15.2 (or 15.21)

18 or 18.5 (or 18.49)

22 or 22.1 (or 22.09)

Ignore significant figures

T2 All values given to two or three significant figures.

Must be to two or three significant figures. A mixture of 2s.f. and 3s.f. is allowed.

E1 ± 0.4 (allow ± 0.5), ± 0.6, ± 0.7,

± 0.8, ± 0.9, ± 0.9 or ± 1.0

Allow more than one significant figure.

(c) (i) G1 Six points plotted correctly. Must be within half a small square. Use transparency. E.c.f. allowed from table.

E2 Error bars in d

2 plotted correctly. Check first and last point. Must be accurate within half a small square.

(c) (ii) G2 Line of best fit. If points are plotted correctly then lower end of line should pass between (150, 2) and (200, 2) and upper end of line should pass between (3200, 24) and (3250, 23.7). Allow e.c.f. from points plotted incorrectly – examiner judgement.

G3 Worst acceptable straight line. Steepest or shallowest possible line that passes through all the error bars.

Line should be clearly labelled or dashed. Should pass from top of top error bar to bottom of bottom error bar or bottom of top error bar to top of bottom error bar. Mark scored only if error bars are plotted.

(c) (iii) C1 Gradient of best fit line. The triangle used should be greater than half the length of the drawn line. Check the read offs. Work to half a small square. Do not penalise POT. If points and BFL correct then gradient should be in numerical range (7.00 – 7.35) (× 10–7).

E3 Error in gradient Method of determining absolute error. Difference in worst gradient and gradient.

(d) C2 e/m = 8/(gradient × B2) = 1.28 × 105/gradient = 1.8 × 1011

Gradient must be used. Allow e.c.f. from (c) (iii) but penalise POT.

If gradient within range given, then e/m in range (1.74 – 1.83) × 1011.

E4 Method of determining error in e/m. Uses worst gradient and finds difference. Allow fractional error methods. Do not check calculation.

C3 Unit of e/m: C kg–1. Accept V m–2 T–2.



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(e) C4 3.80 – 4.00 × 10-3

[If POT in (d) allow 0.38 – 0.40] Check method.

22 )10(3.8

5008−

××

×

=

m

e

B

Answer must be in range given.

E5 Method for determining largest error in correct value of B.

This mark can only be scored if B is in range. Expect to see similar calculation to above with largest e/m x (3.9 × 10–2)2 or smallest e/m x (3.7 × 10–2)2. Allow fractional error methods.

[Total: 15] Uncertainties in Question 2 (c) (iii) Gradient [E3]

1. Uncertainty = gradient of line of best fit – gradient of worst acceptable line

2. Uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient) (d) e/m [E4]

1. Uncertainty = e/m from gradient – e/m from worst acceptable line

2. gradient

gradient

m

e

m

e

∆∆=

(e) B [E5]

1. Substitution method to find worst acceptable B using either largest e/m × (3.9 × 10–2)2 or smallest e/m × (3.7 × 10–2)2.

2.

+=

+=

d

d

d

d

B

B

m

e

m

e

m

e

m

e

∆

2

∆2∆∆

2

1∆



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Summary of shorthand notation which may be used in annotating scripts:

XEX Wrong experiment

SFP Significant figure penalty

ECF Error carried forward

AE Arithmetical error

POT Power of ten error

NV Not valid

NR Not relevant

NBL Not best line

NWL Not worst line

FO False origin

NE Not enough

NGE Not good enough

BOD Benefit of the doubt

NA Not allowed

SV Supervisor's value

SR Supervisor’s report

OOR Candidate's value is out of range

CON Contradictory physics not to be credited

∆ Used to show that the size of a triangle is appropriate

M3 Used to show the type of mark awarded for a particular piece of work

C Used to show that the raw readings are consistent

SF Used to show calculated quantities have been given to an appropriate number of significant

figures

^ Piece of work missing (one mark penalty)

^^ Several pieces of work missing (more than one mark penalty)

↔ Scale can be doubled in the x-direction

b Scale can be doubled in the y-direction

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