9702 s09 ms_all

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS
GCE Advanced Subsidiary Level and GCE Advanced Level
MARK SCHEME for the May/June 2009 question paper
for the guidance of teachers
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 May/June 2009 question papers for most IGCSE, GCE
Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Mark Scheme: Teachers’ version Syllabus Paper
GCE A/AS LEVEL – May/June 2009 9702 01
© UCLES 2009
Question
Number
Key
Question
Number
Key
1 D 21 A
2 C 22 B
3 B 23 C
4 B 24 B
5 D 25 A
6 C 26 B
7 A 27 A
8 C 28 B
9 B 29 D
10 C 30 A
11 A 31 D
12 D 32 A
13 B 33 B
14 C 34 D
15 B 35 D
16 C 36 B
17 B 37 B
18 D 38 C
19 A 39 C
20 D 40 D

Location Entry Codes
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Question Paper Mark Scheme Principal Examiner’s Report
Introduction Introduction Introduction
First variant Question Paper First variant Mark Scheme First variant Principal
Examiner’s Report
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Examiner’s Report
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9702 PHYSICS
9702/21 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,
which would have considered the acceptability of alternative answers.
examination.
First variant Mark Scheme

© UCLES 2009
1 (a) (i) micrometer (screw gauge) / travelling microscope ...................... B1 [1]
(ii) either ohm-meter or voltmeter and ammeter
or multimeter/avo on ohm setting ................................................ B1 [1]
(iii) either (calibrated) c.r.o. or a.c. voltmeter and × √2 ...................... B1 [1]
(b) density = mass / volume .................................................................... C1
= 580 / 63
= 2.685 g cm-3
…(allow 2.68, 2.69, 2.7) ............. A1
% uncertainty in mass = (10 / 580) × 100 = 1.7% .............................. C1
% uncertainty in volume = 3 × (0.1 / 6) × 100 = 5.0% ........................ C1
uncertainty in density = 0.18 g cm-3
density = 2.7 ± 0.2 g cm-3
................................................................... A1 [5]
(answer 2.69 ± 0.09 g cm-3
scores 4 marks)
2 (a) ball moving in opposite direction (after collision) ................................ B1 [1]
(b) (i) change in momentum = 1.2 (4.0 + 0.8) ....................................... C2
(correct values, 1 mark; correct sign {values added}, 1 mark )
= 5.76 N s …(allow 5.8) ......................... A1 [3]
(ii) force = ∆p / ∆t or m∆v / ∆t ................................................... C1
= 5.76 / 0.08 or 1.2 × 4.8 / 0.08 ........................................ C1
= 72 N ............................................................................... A1 [3]
(c) 5.76 = 3.6 × V ..................................................................................... C1
V = 1.6 m s-1
....................................................................................... A1 [2]
(d) either speed of approach = 4.0 m s-1
and
speed of separation = 2.4 m s-1
.............................................. M1
not equal and so inelastic ....................................................... A1
or kinetic energy before = 9.6 J and
kinetic energy after collision = 4.99 J ...................................... M1
kinetic energy after is less / not conserved so inelastic .......... A1 [2]
3 (a) product of (magnitude of one) force and distance between forces ..... M1
reference to either perpendicular distance between forces
or line of action of forces and perpendicular distance .... A1 [2]
(b) (i) 90° ............................................................................................... B1 [1]
(ii) 130 = F × 0.45 (allow e.c.f. for angle in (i)) ................................ C1
F = 290 N ..................................................................................... A1 [2]
(allow 1 mark only if angle stated in (i) is not used in (ii))

© UCLES 2009
4 (a) (i) change of shape / size / length / dimension ................................ C1
when (deforming) force is removed, returns to original shape / size A1 [2]
(ii) L = ke ........................................................................................... B1 [1]
(b) 2e ....................................................................................................... B1
½k …(allow e.c.f. from extension) .................................................... B1
½e and 2k ........................................................................................... B1
2
3
e …(allow e.c.f. from extension in part 2) ...................................... B1
3
2
k …(allow e.c.f. from extension) .................................................... B1 [5]
5 (a) either phase difference is π rad / 180°
or path difference (between waves from S1 and S2) is ½λ / (n + ½)λ . B1
either same amplitude / intensity at M
or ratio of amplitudes is 1.28 / ratio of intensities is 1.282
.................. B1 [2]
(b) path difference between waves from S1 and S2 = 28 cm ................. B1
wavelength changes from 33 cm to 8.25 cm ...................................... B1
minimum when λ = (56 cm,) 18.7 cm, 11.2 cm, (8.0 cm) ................ B1
so two minima .................................................................................... B1 [4]
6 (a) (i) E = V / d ....................................................................................... C1
= 350 / (2.5 × 10-2
)
= 1.4 × 104
N C-1
.......................................................................... A1 [2]
(ii) force = Eq .................................................................................... C1
= 1.4 × 104
× 1.6 × 10-19
............................................................... M1
= 2.24 × 10-15
............................................................................... A0 [2]
(b) (i) F = ma ......................................................................................... C1
a = (2.24 × 10-15
) / (9.1 × 10-31
)
= 2.46 × 1015
m s-2
…(allow 2.5 × 105
) ...................................... A1 [2]
(ii) s = ½at2
....................................................................................... C1
2.5 × 10-2
= ½ × 2.46 × 1015
× t2
t = 4.5 × 10-9
s ............................................................................. A1 [2]
(c) either gravitational force is normal to electric force
or electric force horizontal, gravitational force vertical ................ B2 [2]
special case: force/acceleration due to electric field >> force/acceleration
due to gravitational field, allow 1 mark

© UCLES 2009
7 (a) (i) R .................................................................................................. B1 [1]
(ii) 0.5R ............................................................................................. B1 [1]
(iii) 2.5R …(allow e.c.f. from (ii)) ..................................................... B1 [1]
(b) (i) I1 + I2 = I3 .................................................................................... B1 [1]
(ii) E2 = I3R + I2R .............................................................................. B1 [1]
(iii) E1 – E2 = 2I1R – I2R .................................................................... B1 [1]
8 (a) rate of decay / activity / decay (of nucleus) is not affected by external factors / environment /
surroundings ....................................................................................... B2 [2]
(If states specific factor(s), rather than giving general statement above, then give 2 marks for
two stated factors, but 1 mark only if one factor stated)
(b) (i) gamma / γ .................................................................................... B1 [1]
(ii) alpha / α ...................................................................................... B1 [1]
(iii) gamma / γ .................................................................................... B1 [1]
(iv) beta / β ......................................................................................... B1 [1]

9702 PHYSICS
9702/22 Paper 2 (AS Structured Questions), maximum raw mark 60
examination.
Second variant Mark Scheme

© UCLES 2009
1 (a) e.g. time (s), current (A), temperature (K), amount of substance (mol),
luminous intensity (cdl)
1 each, max 3 …………………………………………………………………………. B3 [3]
(b) density = mass / volume ……………………………………………………………… C1
unit of density: kg m–3
…………………………………………….………… C1
unit of acceleration: m s–2
………………………………………………………… C1
unit of pressure: kg m–3
m s–2
m ………………………………………..……… B1
kg m–1
s–2
……………………………………………………… B1 [5]
(allow 4/5 for solution in terms of only dimensions)
2 (a) 2.4s ……………………………………………………………………………………. A1 [1]
(b) in (b) and (c), allow answers as (+) or (–)
recognises distance travelled as area under graph line ………………………….. C1
height = (½ × 2.4 × 9.0) – (½ × 1.6 × 6.0) ……………………………………….. C1
= 6.0m (allow 6m) …………………………………………………………… A1 [3]
(answer 15.6 scores 2 marks
answer 10.8 or 4.8 scores 1 mark)
alternative solution: s = ut – ½at2
= (9 × 4) – ½ × (9 / 2.4) × 42
= 6.0m
(answer 66 scores 2 marks
answer 36 or 30 scores 1 mark)
(c) (i) change in momentum = 0.78 (9.0 + 4.2) (allow 4.2 ± 0.2) ………………. C1
= 10.3N s (allow 10 N s) …………………………. A1 [2]
(ii) force = ∆p / ∆t or m∆v / ∆t ………………………………………. C1
= 10.3 / 3.5 / 0.08
= 2.9N ……………………………………………………………………… A1 [2]
(d) (i) 2.9N ……………………………………………………………………………….. A1 [1]
(ii) g = weight / mass ………………………………………………………………. C1
= 2.9 / 0.78
= 3.7m s–2
…………………………………………………………………….. A1 [2]
3 (a) product of (magnitude of one) force and distance between forces ………………. M1
reference to either perpendicular distance between forces
or line of action of forces & perpendicular distance ………………… A1 [2]
(b) (i) 90° ………………………………………………………………………………….. B1 [1]
(ii) 130 = F × 0.45 (allow e.c.f. for angle in (i)) ……………………………….. C1
F = 290N …………………………………………………………………………. A1 [2]
(allow 1 mark only if angle stated in (i) is not used in (ii))

© UCLES 2009
4 (a) (i) change of shape / size / length / dimension ……………………………………. C1
when (deforming) force is removed, returns to original shape / size ………… A1 [2]
(ii) L = ke ……………………………………………………………………………… B1 [1]
(b) 2e ………………………………………………………………………………………… B1
½k (allow e.c.f. from extension) ……………………………………………………… B1
½e and 2k ……………………………………………………………………………… B1
2
3
e (allow e.c.f. from extension in part 2) ………………………………………….. B1
3
2
k (allow e.c.f. from extension) ……………………………………………………… B1 [5]
5 (a) constant phase difference …………………………………………………………….. B1 [1]
(b) allow wavelength estimate 750nm → 550nm …………………………………….. C1
separation = λD / x ……………………………………………………………………. C1
= (650 × 10–9
× 2.4) / (0.86 × 10–3
)
= 1.8mm ………………………………………………………………….. A1 [3]
(allow 2 marks from inappropriate estimate if answer is in range 10cm → 0.1mm)
(c) no longer complete destructive interference /
amplitudes no longer completely cancel …………………………………………….. M1
so dark fringes are lighter …………………………………………………………….. A1 [2]
6 (a) (i) E = V / d …………………………………………………………………………… C1
= 350 / (2.5 × 10–2
)
= 1.4 × 104
N C–1
……………………………………………………………… A1 [2]
(ii) force = Eq …………………………………………………………………………. C1
= 1.4 × 104
× 1.6 × 10–19
…………………………………………………… M1
= 2.24 × 10–15
………………………………………………………………. A0 [2]
(b) (i) F = ma ……………………………………………………………………………. C1
a = (2.24 × 10–15
) / (9.1 × 10–31
)
= 2.46 × 1015
m s–2
(allow 2.5 × 105
) ……………………………………… A1 [2]
(ii) s = ½at2
…………………………………………………………………………… C1
2.5 × 10–2
= ½ × 2.46 × 1015
× t2
t = 4.5 × 10–9
s ……………………………………………………………………. A1 [2]
(c) either gravitational force is normal to electric force
or electric force horizontal, gravitational force vertical ………………………. B2 [2]
special case: force/acceleration due to electric field >> force/acceleration
due to gravitational field, allow 1 mark

© UCLES 2009
7 (a) ∞…………………………………………………………………………………………… A1
2R ………………………………………………………………………………………… A1
R ………………………………………………………………………………………….. A1 [3]
(b) (i) I1 + I 3 = I 2 + I 4 ………………………………………………………………… A1 [1]
(ii) E2 – E1 = I 3R ……………………………………………………………………. A1 [1]
(iii) E2 = I 3R + 2I 4R …………………………………………………………………. A1 [1]
8 (a) rate of decay / activity / decay (of nucleus) is not affected by external
factors / environment / surroundings B2 [2]
(If states specific factor(s), rather than giving general statement above,
then give 2 marks for two stated factors, but 1 mark only if one factor stated)
(b) (i) gamma / γ …………………………………………………………………………. B1 [1]
(ii) alpha / α …………………………………………………………………………… B1 [1]
(iii) gamma / γ …………………………………………………………………………. B1 [1]
(iv) beta / β …………………………………………………………………………….. B1 [1]

9702 PHYSICS
9702/31 Paper 31 (Advanced Practical Skills 1),
maximum raw mark 40
examination.

© UCLES 2009
1 (b) Measurements [6]
One mark for each set of readings for different Rtotal 47Ω.
Incorrect trend –1 (wrong trend is R ↑ I ↑ / negative gradient).
1 or more incorrect values of R –1.
Apparatus setup correctly without help from supervisor. [1]
Range of R: to include (12 / 16 Ω) and (71 / 94 Ω) and (141 / 188 Ω). [1]
Column headings (R/Ω, I/A, 1/I/A–1
). Must have R and 1/I columns. [1]
Each column heading must contain a quantity and a unit where appropriate.
Ignore units in the body of the table. Do not accept 1/I/ A or 1/I (A).
There must be some distinguishing mark between the quantity and the unit
(i.e. solidus is expected I/A, but accept, for example, I (A)).
Consistency of presentation of raw readings. [1]
All values of raw I must be given to the same number of decimal places.
Ignore converted current columns. If trailing zeros consistency = 0.
If current same consistency = 0.
Significant figures [1]
Apply to 1/I.
If raw I is given to 2 sf, then accept 1/I to 2 or 3 sf.
Values of 1/I correct. Underline and check a value for 1/I at R = lowest value. [1]
If incorrect, write in the correct value.
(c) (i) (Graph)
Axes. Sensible scales must be used. Awkward scales (e.g. 3:10) are not allowed. [1]
Scales must be chosen so that the plotted points must occupy at least half
the graph grid in both x and y directions. Indicate false origin with FO.
Scales must be labelled with the quantity which is being plotted. Ignore units.
All observations must be plotted. Do not accept blobs (points > 0.5 square). [1]
Ring and check a suspect plot. Tick if correct. Re-plot if incorrect.
Work to an accuracy of half a small square.
Line of best fit. Judge by scatter of points about the candidate's line. [1]
There must be a fair scatter of points either side of the line.
5 trend points. No kinked lines.
Quality. Judge by scatter of all points. All table values need to be plotted.
Min 6 needed
If wrong trend Q = 0. If any plot out by 10 Ω from examiners line Q = 0. [1]
(ii) Gradient [1]
The hypotenuse of the ∆ must be equal to or greater than half the length of the
drawn line. Read-offs must be accurate to half a small square.
Check for ∆y/∆x (i.e. do not allow ∆x/∆y).
y-intercept from graph or substitute correct read-offs into y = mx + c. [1]
Penalise for incorrect algebra. Label FO.

© UCLES 2009
(d) Correct method for finding P and Q. m = 1/P. c = Q/P [1]
Correct method needed. Value for P and Q. Ignore negative sign. [1]
P = 1.0 – 5.0 V (or AΩ). Q = 50 – 150 Ω (or V/A) (Resistor X). Unit required.
Penalise AE.
[Total: 20]
Special case: If I same, Measurements = 5 max, Consistency = 0, Axes = 0, Q = 0, Gradient = 0.
If no I, consistency = 0, 1/I calculation = 0, SF = 0. Allow CH mark on columns present.
2 (a) (ii) Allow reference to measuring cylinder and consistent number of significant figures.
Reference to precision of measuring cylinder. Consistent with SF in their vol. [1]
(b) (i) All raw heights to nearest mm. (heights < 30.0 cm) [1]
(ii) θ < 90° [1]
(iii) Percentage uncertainty in θ. ∆θ = 2 – 5°. [1]
If repeated readings have been done then the uncertainty can be half the range.
Correct ratio idea required. ∆θ⁄θ (×100%) (×100% can be implied)
(c) Measurement of 2nd
height less than first height. [1]
Measurement of 2nd
raw θ (any value) to nearest degree or half a degree [1]
Measurement of 2nd
volume [1]
(c)/(b)(ii) Evidence of repeats in angle measurement [1]
θ(b)(ii) > θ(c) [1]
(c), (b)(iv) Volume in (c) half of volume in (b)(iv). 0.4 Y Vc/ Vb Y 0.6. [1]
(d) Correct calculation to check inverse proportionality. h × cosθ = k [1]
One numerical check: check 2nd
value if available.
Conclusion. Sensible comments relating to calculations to within 20% or their own
value and suggested relation. Allow ecf in conclusion if arithmetical error in calculation.
If incorrect ideas or no ratio then conclusion = 0. [1]
Special case: If 2nd
Volume ¾ and not ¼ full, then 2nd
Vol = 0 and allow for 2nd
height and 2nd
angle
greater than the first height and first angle respectively.

© UCLES 2009
(e)(i) and (ii)
Sources of error or limitation.
[4]
Improvements. Use of other apparatus or
different procedures. [4]
Ap Two readings are not enough (to draw a
valid conclusion).
As Take many (sets of) readings and plot a
graph of the results. Be clear NOT just
repeat readings.
Bp Parallax error in measuring h/θ . Bs Get eye level/‘eye level’ perpendicular (to
protractor lines, ruler scale or meniscus).
Put scale onto bottle.
Cp Difficult to measure height owing to
refraction/shape of bottle/thickness of
bottom not taken into account/ruler does not
start at zero/cannot see meniscus clearly.
Cs Add dye/use ruler with a zero at the start.
Dp Difficulty in deciding the toppling point. Ds Move by increments/hold with newtonmeter
and tilt until F = 0/bottle on tilting ramp idea.
Ep Difficulty in measuring θ owing to container
not perfectly right angled (curved) at the
bottom/difficult to line up protractor/
horizontal line of protractor not on table/
difficult to manipulate bottle and measure
angle/flexible container/shape of bottle.
Es Make bottom square with plasticine/use
protractor with horizontal line flush to table
top/freestanding or clamped protractor.
[Total: 20]
No reference to light gates, motion sensors, video, reaction time, volume measurements, pointers,
changing bottle, repeat readings, calipers or movement of container.

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

© UCLES 2009
1 Table
(b) Measurements. One mark for each set of readings for l and t. [6]
If incorrect trend then –1 (incorrect trend is l ↑ t ↓).
For values of l in specified range, if any value of time <1s then –1.
Help from supervisor then –1.
Repeated values of t for each length. [1]
Range. l min ≤ 12cm and l max ≥ 48cm [1]
Column headings – each must include a quantity and a unit where appropriate. [1]
Ignore units in the body of the table.
There must be some distinguishing mark between the quantity and the unit
(solidus is expected but accept, for example, t (s) or t in s or t in sec).
Consistency of raw readings – all values of l must be given to the nearest mm. [1]
Significant figures. Apply to √l. [1]
If l is given to 2 sf, then accept √l to 2 or 3 sf.
Check the value of √l for largest l. If incorrect, write in the correct value. [1]
Graph
(c) (i) Axes – scales must be chosen so that the plotted points must occupy at least half the
grid in both x and y directions. [1]
Sensible scales must be used (not 3:10 etc). Indicate false origin with FO.
Scales must be labelled with the quantity which is being plotted. Ignore units.
Scale value labels must be no further apart than three large squares.
Plots – all observations must be plotted (write a ringed total on the graph). [1]
Ring and check the ‘worst’ plot. Tick if correct. Re-plot if incorrect.
Work to an accuracy of half a small square. Do not allow ‘blobs’ > half a small square.
Line of best fit. There must be at least five trend plots after allowing one ‘rogue’ point. If
trend curved then allow curve but not straight line. [1]
Indicate best line if candidate's line is not the best line.
Quality of results – judge by scatter of points about a straight line.
Allow up to ±0.25cm½
. [1]
All points must be plotted for this mark to be scored (minimum 6 points).

© UCLES 2009
Analysis
(c) (ii) Calculation of gradient – the hypotenuse of the ∆ must be at least half the length of the
drawn line, and read-offs must be accurate to half a small square. Method of calculation
must be correct. [1]
y-intercept correctly read from graph or calculated using correct read-offs. [1]
Conclusions
(d) Method: p = gradient and k = intercept [1]
Valid units (s cm–½
or s m–½
for p, and s for k). [1]
Check for and penalise power-of-ten error in unit for p.
[Total: 20]
2 First readings
(b) (i) No help from SV with setting up the apparatus. [1]
(ii) Sensible practical detail such as ‘position scale close to mass holder’, ‘view horizontally’,
‘view at eye level’, ’view perpendicular to scale’, ‘allow for zero error’, ‘measure at
several positions’, ‘use setsquare on bench to make ruler vertical’. (Do not allow
‘repeated readings’). [1]
(c) (ii) New height, with unit, to nearest mm. [1]
New height < previous height. [1]
Uncertainty
(c) (iv) Percentage uncertainty in x, using ∆x = 1 or 2 mm or half the range of repeated
readings. Correct ratio idea required. [1]
Second readings
(d) (i) 2nd
measurement of height with no current. [1]
(ii) Value of second I < first I (but don’t allow zero for second I). [1]
(iii) Measurement of height with new current ≤ hd(i) [1]
Calculation
(d) (iii) Correct calculation of second deflection x [1]
Quality of data
(d) (iii) Larger I produces larger deflection (check from raw values). [1]

© UCLES 2009
Analysis and conclusions
(e) Correct calculation to check proportionality (e.g. two values of k). [1]
Sensible comment relating to calculations and suggested relationship. [1]
Use 50% permitted variation in k if candidate does not suggest a value.
(f) (i) (ii) Limitations and improvements
Limitation (4 max) Improvement (4 max)
A Two readings not enough A Take more readings and plot graph
B Change in height very small B1 Use longer wire / larger current / higher
voltage
B2 Use travelling microscope / vernier
calipers (if method described)
C Parallax error in height
measurement
C Use setsquare from rule to mass* / use
mirror
D Rule not vertical D Use setsquare on bench* / use plumbline /
clamp rule
E Could not achieve 1.2A / contact
Resistance / current fluctuating
E Use higher voltage supply / clean contacts
/ use continuously variable supply
F Hard to measure h because mass
moves
F Turn off fans / method of checking mass
hasn’t moved
G Hard to measure 45cm length
because wire not straight / croc clips
move / wire slips in clamps
G Use smaller croc clips / reduce load on
clips / solder connections / tighten
clamps / measure the 45cm with the wire
straight
* do not credit here if already credited in (b)(ii) [8]
[Total: 20]

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

© UCLES 2009
Section A
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM / R2
C1
8.6 × (0.6 × 107
)2
= M × 6.67 × 10–11
C1
M = 4.6 × 1024
kg A1 [3]
(c) (i) either potential decreases as distance from planet decreases
or potential zero at infinity and X is closer to zero
or potential α –1/r and Y more negative M1
so point Y is closer to planet. A1 [2]
(ii) idea of ∆φ = ½v2
C1
(6.8 – 5.3) × 107
= ½v2
v = 5.5 × 103
ms–1
A1 [2]
2 (a) either the half-life of the source is very long
or decay constant is very small
or half-life >> 40 days
or decay constant << 0.02 day–1
B1 [1]
(b) number of helium atoms = 3.5 × 106
× 40 × 24 × 3600 C1
= 1.21 × 1013
either pV = NkT or pV = nRT and n = N / NA C1
1.5 × 105
× V = 1.21 × 1013
× 1.38 × 10–23
× 290
V = 3.2 × 10–13
m3
A1 [3]
(if uses T/°C or n = 1 or n = 4, then 1 mark max for calculation of number of
atoms)
3 (a) increasing separation of molecules / breaking bonds between molecules B1
(allow atoms/molecules, overcome forces)
doing work against atmosphere (during expansion) B1 [2]
(b) (i) 1 either bubbles produced at a constant rate / mass evaporates/lost at
constant rate
or find mass loss more than once and this rate should be constant
or temperature of liquid remains constant B1 [1]
2 to allow/cancel out/eliminate/compensate for heat losses (to atmosphere) B1 [1]
(do not allow ‘prevent’/‘stop’)
(ii) use of power × time = mass × specific latent heat C1
(70 – 50) × 5 × 60 = (13.6 – 6.5) × L C1
L = 845 J g–1
A1 [3]

© UCLES 2009
4 (a) (i) (θ =) ω t (allow any subject if all terms given) B1 [1]
(ii) (SQ =) r sinωt (allow any subject if all terms given) B1 [1]
(b) this is the solution of the equation a = –ω2
x M1
a = –ω2
x is the (defining) equation of s.h.m. A1 [2]
(c) (i) f = ω / 2π C1
= 4.7 / 2π
= 0.75 Hz A1 [2]
(ii) v = rω (r must be identified) C1
= 4.7 × 12
= 56 cm s–1
A1 [2]
5 (a) (i) ratio of charge (on body) and its potential B1 [1]
(do not allow reference to plates of a capacitor)
(ii) (potential at surface of sphere =) V = Q / 4πε0r M1
C = Q / V = 4πε0r A0 [1]
(b) (i) C = 4 × π × 8.85 × 10–12
× 0.36
= 4.0 × 10–11
F (allow 1 s.f.) A1 [1]
(ii) Q = CV
= 4.0 × 10–11
× 7.0 × 105
= 2.8 × 10–5
C A1 [1]
(c) plastic is an insulator / not a conductor / has no free electrons B1
charges do not move (on an insulator) B1
either so no single value for the potential
or charge cannot be considered to be at centre B1 [3]
(d) either energy = ½CV2
or energy = ½QV and C = Q/V C1
energy = ½ × 4 × 10–11
× {(7.0 × 105
)2
– (2.5 × 105
)2
)} C1
= 8.6 J A1 [3]

© UCLES 2009
6 (a) unit of magnetic flux density / magnetic field strength B1
(uniform) field normal to wire carrying current of 1 A M1
giving force (per unit length) of 1 N m–1
A1 [3]
(b) (i) force on magnet / balance is downwards (so by Newton’s third law) B1
force on wire is upwards M1
pole P is a north pole A1 [3]
(ii) F = BIL and F = mg (g missing, then 0/3 in (ii)) C1
2.3 × 10–3
× 9.8 = B × 2.6 × 4.4 × 10–2
(g = 10, loses this mark) C1
B = 0.20 T A1 [3]
(c) reading for maximum current = 2.3 × √2 C1
total variation = 2 × 2.3 × √2
= 6.5 g A1 [2]
7 coil in series with meter (do not allow inclusion of a cell) B1
push known pole into coil B1
observe current direction (not reading) B1
(induced) field / field from coil repels magnet B1
either states rule to determine direction of magnetic field in coil
or reversing magnet direction gives opposite deflection on meter B1
direction of induced current such as to oppose the change producing it B1 [6]
8 (a) wave theory predicts any frequency would give rise to emission of electron M1
if exposure time is sufficiently long A1
photon has (specific value of) energy dependent on frequency M1
emission if energy greater than threshold / work function / energy to remove
electron from surface A1 [4]
(b) photon is packet/quantum of energy M1
of electromagnetic radiation A1
(photon) energy = h × frequency B1 [3]
every particle has an (associated) wavelength B1
wavelength = h / p M1
where p is the momentum (of the particle) A1 [3]
9 (a) (i) ∆N / ∆t (ignore any sign) B1 [1]
(ii) ∆N / N (ignore any sign) B1 [1]
(b) source must decay by 8% C1
A = A0 exp(–ln2 t / T½) or A/ A0 = 1 / (2t/T
) C1
0.92 = exp(–ln2 × t / 5.27) or 0.92 = 1 / (2t/5.27
) C1
t = 0.634 years
= 230 days A1 [4]
(allow 2 marks for A/ A0 = 0.08, answer 7010 days
allow 1 mark for A/ A0 = 0.12, answer 5880 days)

© UCLES 2009
Section B
10 (a) (part of) the output is added to /returned to / mixed with the input B1
and is out of phase with the input / fed to inverting input B1 [2]
(b) 25 = 1 + (120 / R) C1
R = 5 kΩ A1 [2]
(c) (i) –2 V A1 [1]
(ii) 9 V A1 [1]
11 (a) pulse of ultrasound (1)
reflected at boundaries / boundary (1)
received / detected (at surface) by transducer (1)
signal processed and displayed (1)
time between transmission and receipt of pulse gives
(information about) depth of boundary (1)
reflected intensity gives information as to nature of boundary (1)
(any four points, 1 each, max 4) B4 [4]
(b) (i) coefficient = (Z2 – Z1)2
/ (Z2 + Z1)2
= (6.3 – 1.7)2
/ (6.3 + 1.7)2
C1
= 0.33 (unit quoted, then –1) A1 [2]
(ii) fraction = exp(–µx) C1
= exp(–23 × 4.1 × 10–2
)
= 0.39 A1 [2]
(iii) intensity = 0.33 × 0.392
× I C1
= 0.050 I A1 [2]
(do not allow e.c.f. from (i) and (ii) if these answers are greater than 1)
12 (a) loss / reduction in power / energy / voltage/ amplitude (of the signal) B1 [1]
(b) (i) attenuation = 125 × 7 = 875 dB A1 [1]
(ii) 20 amplifiers
gain = 20 × 43 = 860 dB A1 [1]
(c) gain = 10 lg(P1/P2) C1
overall gain = –15 dB / attenuation is 15 dB C1
–15 = 10 lg(P / 450)
P = 14 mW A1 [3]

© UCLES 2009
13 (a) switch; tuning cct; (r.f.) amplifier; demodulator;
serial-to-parallel converter; DAC; (a.f.) amplifier
mark as 2 sets of 2 marks each
5 blocks identified correctly B2
(each error or omission, deduct 1 mark)
5 blocks in correct order B2 [4]
(4 or 3 blocks in correct order, allow 1 mark)
(b) phone transmits signal (to identify itself) (1)
signal received by (several) base stations (1)
transferred to cellular exchange (1)
computer selects base station with strongest signal (1)
assigns a (carrier) frequency (1)
(any four, 1 each, max 4) B4 [4]

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

© UCLES 2009
1 Planning (15 marks)
Defining the problem (3 marks)
P1 Vary l or l is the independent variable [Allow M] [1]
P2 Determine the period T (for each l [M]) or T is the dependent variable [1]
P3 M is kept constant [l is kept constant] [1]
Methods of data collection (5 marks)
M1 Diagram showing the cantilever is fixed e.g. g-clamp & bench, retort stand & clamp [1]
M2 Many oscillations repeated to determine average T (n [ 10 or t [ 10 s for stopwatch) [1]
M3 Weigh M using balance [1]
M4 Measure w and d and measure/record l [1]
M5 Use of vernier caliper/micrometer to measure d and/or w [1]
Method of analysis (2 marks)
A1 Appropriate graph plotted i.e. T 2
against l 3
; [T 2
against M] or lg T against lg l or lg M [1]
A2
gradient
k
gradientwd
M
E =
×
= 3
2
16π
gradientwd
E
×
= 3
2
16 3
lπ
Allow logarithmic solutions e.g. intercept-intercept-2
10010 yxy
kk
E == [1]
Safety considerations (1 mark)
S1 Relevant safety precaution related to the use of loads [1]
e.g. cushion/sand in case load falls, keep feet away, keep distance from experiment.
Additional detail (4 marks)
D Relevant points might include [4]
1. Use same rule or keep w and/or d constant.
2. Repeat measurements of d and/or w along rule and average.
3. Discussion of use of motion sensor e.g. orientation or light gates with detail.
4. Use small amplitude or small angle oscillations (to ensure equation is valid).
5. Method of securing load to rule e.g. with tape/glue.
6. Discussion of magnitude of load: large enough to make T large enough.
7. Use of fiducial marker to help to time.
8. Start timing after oscillations have settled.
Do not allow vague use of computers/light gates, video cameras, dataloggers.
[Total: 15]

© UCLES 2009
2 Analysis, conclusions and evaluation (15 marks)
Part Mark Expected Answer Additional Guidance
(a) A1 gradient = q
y-intercept = lg p or log p
(b) T1
T2
1.176 or 1.18 0.415 or 0.41
1.279 or 1.28 0.544 or 0.54
1.362 or 1.36 0.643 or 0.64
1.431 or 1.43 0.732 or 0.73
1.491 or 1.49 0.806 or 0.81
T1 is awarded for correct values lg d
T2 is awarded for correct values lg I
A mixture of 2dp and 3dp is allowed within
each column
E1 ± 0.016 or ± 0.017 or ± 0.02
decreasing to ± 0.006 or ± 0.007
or ± 0.01
Allow more than one significant figures.
(c) (i) G1 Five points plotted correctly Must be within half a small square. Use
transparency. Ecf allowed from table.
E2 Error bars in lg I plotted correctly. Check first and last point. Must be accurate
within half a small square. Ecf allowed from
table.
(ii) G2 Line of best fit If points are plotted correctly then lower end
of line should pass between (1.15, 0.370) and
(1.15, 0.385) and upper end of line should
pass between (1.50, 0.815) and (1.50, 0.825).
Allow ecf 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.
(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.
E3 Error in gradient Method of determining absolute error
Difference in worst gradient and gradient.
(iv) C2 y-intercept Must be negative and the gradient must be
used. Check substitution into c = y – mx.
Allow ecf from (c)(iii).
If gradient within range given, then y-intercept
should be about –1.1
E4 Method of determining error in
y-intercept
Determines worst y-intercept using worst
gradient and finds difference.
Check substitution but do not check
calculation.
Do not allow ecf from false origin read-off.

© UCLES 2009
(d) C3 p = 10candidate’s y-intercept
p should be about 0.08. Allow ecf from
(c)(iv).
If FO used then p should be about 2.34 to
2.43.
C4 q = in the range 1.20–1.30 and
given to 2 or 3 sf.
Candidate’s gradient must be used.
E5 Method for determining errors in
values of p and q.
Determines worst p using worst y-intercept
and finds difference. Allow ecf from (c)(iv).
q error must be same as error in gradient.
[Total: 15]
Uncertainties in Question 2
(c) (iii) Uncertainty in 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)
(c) (iv) Uncertainty in the y-intercept [E4]
1. Uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line
2. Uncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)
N.B. Must use gradient from worst acceptable line and a point on the same worst acceptable line
to determine y-intercept of worst acceptable line.
(d) Uncertainty in p [E5]
1. Uncertainty = p from y-intercept of BFL – p from y-intercept of WAL
2. Uncertainty = ½ (p from y-intercept of shallowest WAL – p from y-intercept of steepest
WAL)

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