9702 w07 ms_all

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS
GCE Advanced Subsidiary Level and GCE Advanced Level
MARK SCHEME for the October/November 2007 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 2007 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 2007 9702 01
© UCLES 2007
Question
Number
Key
Question
Number
Key
1 C 21 B
2 A 22 B
3 B 23 B
4 D 24 D
5 C 25 B
6 A 26 C
7 C 27 D
8 D 28 B
9 B 29 D
10 D 30 D
11 B 31 A
12 A 32 A
13 C 33 D
14 D 34 D
15 B 35 A
16 C 36 C
17 C 37 C
18 A 38 C
19 B 39 B
20 A 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.
examination.
syllabuses.

© UCLES 2007
1 (a) systematic: e.g. constant error (in all readings)
cannot be eliminated by averaging
error in measuring instrument B1
random: e.g. readings scattered (equally) about true value
error due to observer
can be eliminated by averaging
(only if averaging not included for systematic) B1 [2]
(b) 15 = π × R2
× 20
R = 0.4886 cm (accept any number of s.f.) C1
% uncertainty in V = 3.3 % (or 0.5/15) C1
% uncertainty in L = 0.5 % (or 0.1/20) C1
% uncertainty in R = 1.9 % (i.e. one half of the sum) C1
R = 0.489 ± 0.009 cm A1 [5]
2 (a) 3.5 T B1 [1]
(b) (i) distance= average speed × time (however expressed) C1
= 14 m A1 [2]
(ii) distance= 5.6 × (T – 5) (or 3.5T – 14) A1 [1]
(c) 3.5T = 14 + 5.6(T – 5) C1
T = 6.7 s A1 [2]
(d) (i) acceleration = (5.6 / 5 =) 1.12 m s–2
C1
force = ma C1
= 75 N A1 [3]
(ii) power = (force × speed =) {75 + 23} × 4.5 C1
= 440 W A1 [2]
(allow 1/2 for 234 W, 0/2 for 338 W or 104 W)
3 (a) (i) potential energy: stored energy available to do work B1 [1]
(ii) gravitational:due to height/position of mass OR distance from mass
OR moving mass from one point to another B1
elastic: due to deformation/stretching/compressing B1 [2]
(b) (i) height raised = (61 – {61 cos18} =) 3.0 cm C1
energy = (mgh = 0.051 × 9.8 × 0.030 =) 1.5 × 10–2
J A1 [2]
(ii) moment = force × perpendicular distance
= 0.051 × 9.8 × 0.61 × sin18 C1
= 0.094 N m A1 [2]

© UCLES 2007
4 (a) brittle B1 [1]
(b) Young modulus= stress / strain C1
= (9.5 × 108
) / 0.013
= 7.3 × 1010
Pa (allow ± 0.1 × 1010
Pa) A1 [2]
(c) stress = force / area C1
(minimum) area = (1.9 × 103
) / (9.5 × 108
)
= 2.0 × 10–6
m2
C1
(max) area of cross-section= (3.2 – 2.0) × 10–6
= 1.2 × 10–6
m2
A1 [3]
(d) when bent, ‘top’ and ‘bottom’ edges have different extensions M1
with thick rod, difference is greater (than with a thin rod) A1
so breaks with less bending A0 [2]
5 (a) amplitude between 6.5 squares and 7.5 squares on 3 peaks B2
(allow 1 mark if outside this range but between 6.0 and 8.0 squares)
correct phase (ignore lead/lag, look at x-axis only and allow ±½ square B1 [3]
(b) λ = ax / D C1
540 × 10–9
= (0.700 × 10–3
x) / 2.75 C1
x = 2.12 mm A1 [3]
(c) (i) same separation B1
bright areas brighter (1)
dark areas, no change (1)
(allow ‘contrast greater’ for 1 mark if dark/light areas not discussed)
fewer fringes observed (1) any two, 1 each B2 [3]
(ii) smaller separation of fringes B1
no change in brightness B1 [2]
6 (a) power = VI C1
current = 10.5 × 103 / 230 M1
= 45.7 A A0 [2]
(b) (i) p.d. across cable = 5.0 V C1
R = 5.0 / 46 C1
= 0.11 Ω A1 [3]
(ii) R = ρL / A C1
0.11 = (1.8 × 10–8
× 16 × 2) / A C1
A = 5.3 × 10–6
m2
A1 [3]
(wires in parallel, not series, allow max 1/3 marks)

© UCLES 2007
(c) (i) either power = V 2
/ R or power ∝ V 2
C1
ratio = (210 / 230)2
= 0.83 A1 [2]
(ii) resistance of cable is greater M1
greater power loss/fire hazard/insulation may melt
wire may melt/cable gets hot A1 [2]
7 (a) most α-particles deviated through small angles B1
(accept ‘undeviated’)
few α-particles deviated through angles greater than 90° B1 [2]
(b) (i) allow 10–9
m → 10–11
m B1 [1]
(ii) allow 10–13
m → 10–15
m B1 [1]
(if (i) and (ii) out of range but (ii) = 10–4
(i), then allow 1 mark)
(if no units or wrong units but (ii) = 10–4
(i), then allow 1 mark)

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

© UCLES 2007
Question 1
Manipulation, measurement and observation
Successful collection of data
(a) (i) Diameter of wire. 2 d.p. (mm) in raw data. Allow 0.195 or 0.190 mm.
(0.19 mm ± 0.02 mm or SV ± 0.02 mm). Consistent unit. Unit needed. [1]
(i) Repeat measurement [1]
(c) 6 measurements in table
Five marks for six sets of readings for V and l
Four marks for five sets
Three marks for four sets, etc.
Major or unspecified help –2 (e.g. setting up circuit)
Minor help: –1 (e.g. minor changes with circuit) AND –1 (help with reading
micrometer)
Unreasonable values of V −1
(e.g. Voltage values are the same (Vmax – Vmin < 0.5V ), wrong trend
(l↑, V↓) or if any one value of V < 0.5 V.) [5]
Range and distribution of values
(c) (lmax – lmin) must be greater than or equal to 70 cm. Ignore POT error. [1]
Presentation of data and observations
Table: layout
(c) Column headings: V/V; l/cm; l/A/m–1
. [1]
Each column heading must contain a quantity and a unit where appropriate.
Ignore units in the body of the table. Ignore POT errors.
There must be some distinguishing mark between the quantity and the unit
(i.e. solidus is expected, but accept, for example, l (cm)).
Table: raw data
(c) Consistency of presentation of raw readings
All values of l must be given to the same number of decimal places.
l read to 1 mm or 1 cm [1]
Table: calculated quantities
(c) Significant figures. Apply to l/A.
l/A should be given to the same number or one more than the lowest number of
significant figures from l or raw values of d. [1]
(c) Values of l/A correct using candidate’s figures. Allow small rounding errors.
Check a value. If incorrect, write in the correct value. Ignore POT error. [1]

© UCLES 2007
Graph: layout
(Graph) Axes. If wrong graph plotted (e.g. l against l/A) do not award mark.
Sensible scales must be used. Awkward scales (e.g. 3:10) are not allowed.
Scales must be chosen so that the plotted points occupy at least half
the graph grid in both x and y directions. Allow inverted axes.
Scales must be labelled with the quantity which is being plotted. Ignore units. [1]
Graph: plotting of points
(Graph) All observations must be plotted. No blobs (points [ half a small square φ).
Ring and check a suspect plot. Tick if correct. Re-plot if incorrect.
Work to an accuracy of half a small square. [1]
Graph: trend line
(Graph) Line of best fit. Allow 1 point off. At least 5 trend plots needed.
Do not award mark if large scatter.
Judge by scatter of points about the candidate's line.
There must be a fair scatter of points either side of the line.
Indicate best line if candidate's line is not the best line. [1]
Quality of data
(Graph) Judge by scatter of points about the best fit line. Allow up to ± 0.05V.
All plotted points are assessed for this mark.
At least 5 plots needed. If V constant do not award mark. [1]
Analysis, conclusions and evaluation
Interpretation of graph
(d) (iii) Gradient
The hypotenuse of the ∆ must be at least half the length of the drawn line.
Read-offs must be accurate to half a small square. Do not allow table values
unless on the line of best fit. Write in correct read off.
Check for ∆y/∆x (i.e. do not allow ∆x/∆y). [1]
(d) (iii) y-intercept
The value must be read to the nearest half square.
The value can be calculated using ratios or y = mx + c. Incorrect algebra –1.
If a false origin has been used then label FO. [1]
Drawing conclusions
(e) Value for k, 0.5 V Y k Y 2.5 V
Should be y-intercept. Unit required. 2 or 3 SF. [1]
(e) Value for I, 0.05 A Y I Y 0.20 A
Must come from gradient. Working must be checked. Unit required A (VΩ–1
).
2 or 3SF.
Sensible answer checked using candidate’s figures into correct substitution. m=ρI [1]
[Total: 20]

© UCLES 2007
Question 2
(a) (iii) Position of end of rule at equilibrium. Nearest cm or mm. < 1m
Consistent unit. [1]
(b) First value of d between 1 and 5 cm. If lowest position given write in correct value of d. [1]
(b) First value of highest position within 5 cm of the equilibrium position. [1]
(d) Second value of d. Different value to the first. Allow out of range. [1]
(d) Second value of highest position. [1]
(d) Repeated measurements for highest position (evidence from (b) or (d)) [1]
Quality of data
(d) Bigger d gives bigger x. Check with corrected values of d and x.
If d < x in either case or if d = x in both cases, this loses the mark. [1]
Display of calculation and reasoning
(b) First value of x calculated correctly
Calculation must be checked. Write down the correct value if answer wrong. [1]
(d) Second value of x calculated correctly
Calculation must be checked. Write down the correct value if answer wrong. [1]
(e) Correct calculation to check proportionality ecf if candidates value of d is the lowest position.
Possibilities include: two calculations of x/d
ratio of x values and ratio of d values both calculated [1]
Drawing conclusions
(e) Conclusion based on calculation. Consistent argument.
Incorrect ideas score zero. [1]
Estimating uncertainties
(c) Percentage uncertainty in x
Allow uncertainty in x; 2 mm Y ∆x Y 10 mm.
If repeated readings have been done then the uncertainty could be half the range.
Correct ratio idea required x 100 stated/implied. [1]

© UCLES 2007
Identifying limitations
(f) (i) Relevant points must be underlined and ticked.
Some of these might be:
A Only two readings (are not enough to draw a valid conclusion).
B Hard to judge/see (when/where) highest position with reference to movement.
Do not accept reaction time ideas.
C Parallax (error) or good diagram demonstrating this.
D Difficulty in release/keeping rule still prior to release (reference to force).
E Equilibrium position changes with evidence shown in measurements.
X Other additional source of error. [4]
Suggesting improvements
(f) (ii) Relevant points must be underlined and ticked.4
A Take more readings and plot a graph/calculate k values.
B High speed (camera to take) photographs/film the motion and play back frame by
frame/ slow motion/ use pause OR
motion/position sensor above/ below mass OR
trial and error with light gate/ horizontal marker.
C Measure at eye level/repeat to get eye in right place/ place rule as close as possible
to vertical rule/use helper to release or measure/use mounted pin at end of rule (to
help locate position on scale).
D Use a named method to release the rule e.g. cotton and candle or
scissors/electromagnet/end stop or clamp.
X cm rule – use a mm rule. Need to see evidence in their previous measurements that
their readings are taken to the nearest cm or 0.5 cm.
Y Other additional solution, well explained.
Do not allow ‘repeated readings, vacuum, draft free room’
Do not allow ‘use a computer to improve the experiment’
Do not allow ‘increase range/change load on ruler/change length of ruler/changing
quality of ruler’ [4]
[Total: 20]

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

© UCLES 2007
Question 1
(b) Measurements [5]
Five marks for six sets of readings for l and R2,
four for five sets, etc.
(–1 for unreasonable values of R2, e.g. R2>40 or R2<2.5, e.g. impossible R2)
(b) Circuit set up without help from Supervisor (minor help –1, major help –2) [2]
Range and distribution of values
(b) R2 values must include 40 Ω and one value Y 5 Ω [1]
Table: layout
(b) Column headings [1]
Each column heading must contain a quantity and a unit where appropriate.
Ignore units in the body of the table.
There must be some distinguishing mark between the quantity and the unit
(i.e. solidus is expected, but accept, for example, l (cm)).
Table: raw data
(b) Consistency of presentation of raw readings [1]
All values of l must be given to the same number of decimal places.
(l can be to nearest 1 mm or 1 cm).
Table: calculated quantities
(b) Significant figures [1]
Apply to 1/l only.
If l is given to 2 sf, then accept 1/l to 2 or 3 sf.
If l is given to 3 sf, then accept 1/l to 3 or 4 sf.
Values of 1/l given as fractions lose this mark.
(b) Values of 1/l correct. [1]
Check a value. If incorrect, write in the correct value.
Allow values of 1/l given as fractions for this mark.
Graph: layout
(Graph) Axes [1]
Sensible scales must be used. Awkward scales (e.g. 3:10) are not allowed.
There should not be more than three large squares between axis labels.
Scales must be chosen so that the plotted points must occupy at least half
the graph grid in both x and y directions.
Scales must be labelled with the quantity which is being plotted (do not accept R
instead of R2). Ignore units.
Do not penalise reversed axes but penalise if the wrong graph has been plotted.

© UCLES 2007
Graph: plotting of points
(Graph) All observations must be plotted. [1]
Ring and check a suspect plot, tick if correct. Re-plot if incorrect (and re-check
quality mark).
Work to an accuracy of half a small square.
Penalise blobs [ half a small square diameter.
Graph: trend line
(Graph) Line of best fit (must be 5 or more plots, do not allow if scatter is large). [1]
Judge by scatter of points about the candidate's line.
There must be a fair scatter of points either side of the line.
Indicate best line if candidate's line is not the best line.
Quality of data
(Graph) Judge by scatter of points about the best fit line (all points ±1 Ω) [1]
Trend must be correct. At least 5 plots are needed for this mark to be scored.
Interpretation of graph
(c) (iii) Gradient [1]
The hypotenuse must be at least half the length of the drawn line.
Read-offs must be accurate to half a small square (if incorrect, write in correct
value).
Check for ∆y/∆x (i.e. do not allow ∆x/∆y). Ignore POTE.
(c) (iii) y-intercept [1]
The value must be read to the nearest half square.
The value can be calculated using ratios or y = mx + c (if algebra is not
obviously wrong).
If a false origin has been used then label FO.
Drawing conclusions
(d) Value for R1 [1]
There should be evidence that it is obtained from 1/(100cm x gradient).
Must be in range 5 to 15 Ω.
2 or 3 sf. Unit required.
(d) Value for k [1]
Should be candidate’s intercept. 2 or 3 sf.
Unit required.
Should be in range 0.0050 to 0.0150 cm–1
.
[Total for Question 1: 20]

© UCLES 2007
Question 2
(b) (ii) First value of d to nearest cm or mm. [1]
(c) (ii) First value of t (must be between 0.1 and 10 s). [1]
(f) (ii) Second value of d (must be less than first value) [1]
(f) (ii) Second value of t. [1]
(f) (ii) Two values of h in range 0 to 130 cm. (both values could be the same) [1]
(f) (ii) Repeated measurements for t (first or second reading) [1]
Quality of data
(f) (ii) Smaller d gives greater v (use corrected values of v). [1]
Display of calculation and reasoning
(e) First value of v calculated correctly. Calculations must be checked (if wrong, write in correct
value). [1]
(f) (ii) Second value of v calculated correctly. Calculations must be checked (if wrong, write in
correct value). [1]
(g) Correct calculation to check proportionality [1]
Possibilities include: Two calculations of vd.
Ratio of v values and inverse ratio of d values both calculated.

© UCLES 2007
Drawing conclusions
(g) Conclusion [1]
Sensible comments relating to proportionality calculations and to the suggested relation.
Incorrect ideas score zero.
Estimating uncertainties
(d) Percentage uncertainty in t. [1]
Absolute uncertainty must be 0.1 to 0.5 s, or if repeated readings have been done then the
uncertainty could be half the range.
Correct ratio idea and x100 required.
Identifying limitations
(h) (i) Relevant points must be underlined and ticked. [4]
A Two sets of readings not enough (to draw valid conclusion).
B Cone may have not reached terminal velocity.
C Hard to see when cone strikes floor.
D Cone falls at an angle (due to draughts/imbalance of cone).
E Human error in timing/reaction time.
F Difficult to measure diameter because cone flexible.
G Parallax error (at reading positions).
X Other source of error
Suggesting improvements
(h) (ii) Relevant points must be underlined and ticked. [4]
A Take more readings and plot a graph/calculate ratios.
B Ensure terminal velocity by increasing release height/measure velocity at two
intervals to check terminal velocity reached.
C Use pressure/other sensor (on floor) to stop timer/use assistant to judge when it
reaches the floor.
D Turn off fans/balance the cone e.g. extra strip of tape.
E1 Use light gate to trigger stopwatch/use video camera with slow motion replay/use
multiflash photography/use high speed camera with known time intervals.
E2 Time over greater distance.
F Measure diameter of cone in two directions and average.
G Drop in front of rule/read at eye level.
Y Another improvement, well explained.
Do not allow ‘repeated readings’ (unless qualified by ‘plot a graph’).
Do not allow ‘use a computer to improve the experiment’
[Total for Question 2: 20]

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

© UCLES 2007
Section A
1 (a) (i) angle subtended at centre of circle .......................................................................B1
arc equal in length to the radius ...........................................................................B1 [2]
(ii) arc = rθ and for one revolution, arc = 2πr .............................................................M1
so, θ = 2πr/r = 2π ..................................................................................................A0 [1]
(b) (i) either weight provides/equals the centripetal force
or acceleration of free fall is centripetal acceleration ....................................B1
9.8 = 0.13 × ω2
......................................................................................................M1
ω = 8.7 rad s-1
.......................................................................................................A0 [2]
(ii) force in cord = weight + centripetal force (can be an equation) ..........................C1
force in cord = (L – 13) × 5/1.8 or force constant = 5.0/1.8 ................................C1
(L – 13) × 5/1.8 = 5.0 + 5/9.8 × L × 10-2
× 8.72
..................................................C1
L = 17.2 cm ...........................................................................................................A1 [4]
(constant centripetal force of 5.0 N gives L = 16.6 cm allow 2/4)
2 (a) (i) pV = nRT
V = (8.31 × 300)/(1.02 × 105
) ...............................................................................C1
= 0.0244 m3
(if uses Celsius, then 0/2) ..........................................................A1 [2]
(ii) volume occupied by one atom = 0.0244 / (6.02 × 1023
) = 4.06 × 10-26
m3
............M1
separation ≈ 3
√(4.06 × 10-26
) ................................................................................A1
= 3.44 × 10-9
m ...................................................................................A0 [2]
(b) (i) F = GMm / r2
.......................................................................................................C1
= (6.67 × 10-11
× {4 × 1.66 × 10-27
}2
) / (3.44 × 10-9
)2
..........................................C1
= 2.49 × 10-46
N ................................................................................................A1 [3]
(ii) ratio = (4 × 1.66 × 10-27
× 9.8) / 2.49 × 10-46
........................................................C1
= 2.6 × 1020
..........................................................................................................A1 [2]
(c) assumption that forces between atoms are negligible .................................................B1
comment e.g. ratio shows gravitational force to be very small
e.g. force is very much less than weight
e.g. if there are forces, they are not gravitational .......................................B1 [2]

© UCLES 2007
3 (a) (i) 0.8 cm ...................................................................................................................B1 [1]
(ii) (max.) kinetic energy = 2.56 mJ ...........................................................................C1
v(MAX) = ωa ............................................................................................................C1
(max.) kinetic energy = ½mω2
a2
or ½mω2
(a2
– x2
) ............................................C1
2.56 × 10-3
= ½ × 0.130 × ω2
× (0.8 × 10-2
)2
..........................................................M1
ω = 24.8 rad s-1
.....................................................................................................C1
f = ω/2π ................................................................................................................M1
= 4.0 Hz (3.95 Hz) ..............................................................................................A0 [6]
(b) (i) line parallel to x-axis at 2.56 mJ ...........................................................................B1 [1]
(ii) 1 4.0 Hz ................................................................................................................B1
2 0.50 cm (allow ±0.03 cm) ................................................................................B1 [2]
4 (a) (i) either lines directed away from sphere
or lines go from positive to negative
or line shows direction of force on positive charge .......................................M1
so positively charged ............................................................................................A1 [2]
(ii) either all lines (appear to) radiate from centre
or all lines are normal to surface of sphere ...................................................B1 [1]
(b) tangent to curve ...........................................................................................................B1
in correct position and direction ...................................................................................B1 [2]
(c) (i) V = (0.76 × 10-9
) / (4π × 8.85 × 10-12
× 0.024) .....................................................C1
= 285 V ...........................................................................................................A1 [2]
(ii) negative charge is induced on (inside of) box ......................................................M1
formula applies to isolated (point) charge
OR less work done moving test charge from infinity ..........................................A1
so potential is lower ..............................................................................................A1 [3]
(d) either gravitational field is always attractive
or field lines must be directed towards both box and sphere ..............................B1 [1]

© UCLES 2007
5 (a) e.g. separate charges, store energy, smoothing circuit. etc. .......................................B1 [1]
(allow ‘stores charge’)
(b) (i) charge = current × time ........................................................................................B1 [1]
(ii) area is 21.2 cm2
(allow ±0.5 cm2
) .......................................................................C2
(allow 1 mark if outside ±0.5 cm2
but within ±1.0 cm2
)
1.0 cm2
represents (0.125 × 10-3
× 1.25 =) 156 µC ..............................................C1
charge = 3300 µC .................................................................................................A1 [4]
(iii) capacitance = Q/V ..............................................................................................C1
= (3300 × 10-6
) / 15
= 220 µF ..............................................................................................................A1 [2]
(c) either energy = ½CV2
or energy = ½QV and C = Q/V .................................................C1
½ × C × 152
= 2 × ½ × C × V2
......................................................................................C1
V = 10.6 V ...................................................................................................................A1 [3]
6 (a) (i) BI sinθ ..................................................................................................................B1 [1]
(ii) (downwards) into (the plane of) the paper ............................................................B1 [1]
(b) (i) magnetic field (due to current) in one loop OR each loop acts as a coil ...............B1
cuts/is normal to current in second loop OR produces magnetic field ..............B1
causing force on second loop OR fields in same direction ...............M1
either Newton’s 3rd discussed
or vice versa clear gives rise to attraction OR so attracts ...................................A1 [4]
(ii) B = 2 × 10-7
I/0.75 × 10-2
(= 2.67 × 10-5
I) .............................................................C1
force = 0.26 × 10-3
× 9.81 (= 2.55 × 10-3
N) .........................................................C1
F = BIL
2.55 × 10-3
= 2.67 × 10-5
× I2
× 2π × 4.7 × 10-2
....................................................C1
I = 18 A .................................................................................................................A1 [4]

© UCLES 2007
7 (a) energy required to (completely) separate the nucleons (in a nucleus) ........................B1 [1]
(b) (i) U labelled near right-hand end of line ...................................................................B1
Ba and Kr in approximately correct positions .......................................................B1 [2]
(ii) binding energy is A × EB .......................................................................................B1
either binding energy of U < binding energy of (Ba + Kr)
or EB of U < EB of (Ba + Kr) ...........................................................................B1 [2]
(c) Krypton-92 reduced to 1/8 in 9 s .................................................................................M1
in 9 s, very little decay of Barium-141 ..........................................................................M1
so, approximately 9 s ..................................................................................................A1 [3]
OR
λKr = 0.231 or λBa = 6.42 × 10-4
(M1)
8 = e-λB × t
/e-λK × t
(C1)
t = 9.0 s (A1)

© UCLES 2007
Section B
8 (a) (i) - 9 V
(ii) + 9 V (both (i) and (ii) correct for the mark) ........................................................B1 [1]
(b) ..........................................................................................................................B1
..........................................................................................................................B1
..........................................................................................................................B1 [3]
(no e.c.f. from (a))
(c) (i) cct: thermistor and resistor in series ………………………………………………...M1
output connections across thermistor ...................................................................A1 [2]
(ii) as temperature decreases, thermistor resistance increases ................................B1
p.d. across thermistor = RT / (R + RT) × V ...........................................................M1
as RT increases, output increases ........................................................................A1 [3]
9 (a) product of density (of medium) and speed of sound (in medium) ...............................B1 [1]
(b) difference in acoustic impedance ................................................................................M1
determines fraction of incident intensity
that is reflected/amount of reflection ............................................................................A1 [2]
(c) pulse of ultrasound (directed into body) ......................................................................B1
reflected at boundary (between tissues) ......................................................................B1
(reflected pulse is) detected and processed ................................................................B1
time for return of echo gives (information on) depth ....................................................B1
amount of reflection gives information on tissue structures .........................................B1 [5]
10 (a) (i) amplitude (modulated) (allow ‘AM’) .....................................................................B1 [1]
(ii) carrier (frequency / wave) .....................................................................................B1 [1]
(iii) sideband (frequency) ............................................................................................B1 [1]
(b) 10 kHz .........................................................................................................................B1 [1]
(c) sketch: general shape i.e. any wave that is amplitude modulated ..............................M1
correct period for modulating waveform (200 µs) .......................................................A1
correct period for carrier waveform (20 µs) ................................................................A1 [3]

© UCLES 2007
11 (a) carrier frequencies can be re-used (simultaneously without interference) ...................B1
so that number of handsets possible is increased .......................................................B1
OR anything sensible e.g.UHF used (B1)
so ‘line of sight’ (B1) [2]
(b) handset sends out an (identifying) signal ....................................................................M1
communicated by base stations to (computer at) exchange .......................................A1
computer selects base station with strongest signal ...................................................B1
and allocates a (carrier) frequency ..............................................................................B1 [4]

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

© UCLES 2007
Question 1
Planning (15 marks)
Defining the problem (3 marks)
P1 p is the independent variable or vary p. [1]
P2 A is the dependent variable or determine A for different p. [1]
P3 Keep the incident amplitude constant (allow volume of sound or power/intensity output).
[1]
Methods of data collection (5 marks)
M1 Labelled diagram of a workable arrangement including source of sound – glass window
– detector of sound. [1]
Allowed sources: loudspeaker, bell, buzzer, siren but not musical instruments.
Allowed detectors: microphone, sound meter, sound detector.
M2 Method of measuring pressure. Use bourdon gauge/manometer or pressure gauge [1]
Do not allow barometer or pressure meter.
M3 Method of reducing pressure. Use (vacuum) pump to withdraw air from glass
window. [1]
M4 Method of measuring amplitude. Measure amplitude from oscilloscope. [1]
M5 Perform experiment in quiet room. [1]
Method of analysis (2 marks)
A1 Appropriate quantities plotted i.e. A2
against p; or lg A against lg p or A against √p. [1]
A2 Relationship to be correct i.e. graph is a straight line through the origin. [1]
Gradient should equal 0.5 if lg A against lg p plotted.
An explicit statement is required.
Safety considerations (1 mark)
S1 Relevant safety precaution related to either the use of glass or intensity of sound, [1]
e.g. wear ear defenders, switch on sound source for short period of time.
Protection (goggles/safety screen) in case glass breaks.
Awareness of low pressures in glass.
Do not allow gloves because glass is sharp.

© UCLES 2007
Additional detail (4 marks).
D1/2/3/4 Relevant points might include: [4]
Method of ensuring that output from speaker is constant.
Method of reducing sound reflections from e.g. foam/speaker & microphone
close to glass.
Window perpendicular to sound source.
Detail explaining use of oscilloscope to measure amplitude.
Difficulty in measuring amplitude at small pressures/use a loud incident source.
Control (or monitoring) of one additional variable e.g. temperature, frequency,
distances.
Allow time for the temperature/pressure between the glass to stabilise.
Discussion of attenuation in air/frame/glass.
[Total: 15]

© UCLES 2007
Question 2
Analysis, conclusions and evaluation (15 marks)
Approach to data analysis (1 mark)
(a) lg I = n lg V + lg k
Relationship confirmed by straight line. [1]
Table of results (2 marks)
(b) Correct column of lg (V/V). [1]
Correct column of lg (I/A) [1]
lg (V/V) lg (I/A)
0.301 or 0.30 0.079 or 0.08
0.602 or 0.60 0.230 or 0.23
0.778 or 0.78 0.322 or 0.32
0.903 or 0.90 0.380 or 0.38
1.000 or 1.00 0.431 or 0.43
1.079 or 1.08 0.477 or 0.48
2 or 3 dp needed (except allow 4 dp for last two rows in lg V).
Allow small rounding errors in the 3 dp. Must be exact for 2 dp.
Graph (3 marks)
(c) (i) Points plotted correctly. All six required for this mark. Circle error(s). [1]
(ii) Line of best-fit. Must be within tolerances. [1]
Worst acceptable straight line. Line should be clearly labelled. [1]
Conclusion (4 marks)
(iii) Gradient of best-fit line (range 0.500 to 0.525 or check substitution). [1]
If points incorrect, then check size of ∆ and substitution within half a small
square.
(iv) y-intercept. Must be negative. Check substitution from point on best-fit line. [1]
(d) Method of determining k (= 10candidate’s y-intercept
). [1]
Value of n = candidate’s gradient value and in the range 0.500 to 0.525. [1]
Answers to k and n should be to 2 or 3 sf.

© UCLES 2007
Treatment of errors (5 marks)
(b) Errors in lg I allow 2 or 3 dp. [1]
Values should descend from 0.04 to 0.01.
(c) (i) Error bars in lg I plotted correctly [1]
(iii) Error in gradient [1]
Method of determining absolute error
(iv) Error in y-intercept [1]
Method of determining absolute error
(d) Method for error in k and error in n (same as error in gradient) [1]
[Total: 15]

9702 w07 ms_all

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