Cambridge Physics Exam Mark Schemes

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS
GCE Advanced Subsidiary Level and GCE Advanced Level
MARK SCHEME for the May/June 2010 question paper
for the guidance of teachers
9702 PHYSICS
9702/11 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 2010 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 2010 9702 11
© UCLES 2010
Question
Number
Key
Question
Number
Key
1 D 21 A
2 A 22 C
3 A 23 B
4 B 24 A
5 C 25 C
6 B 26 C
7 B 27 A
8 C 28 D
9 B 29 D
10 A 30 A
11 C 31 D
12 D 32 D
13 A 33 A
14 C 34 A
15 B 35 A
16 D 36 D
17 A 37 C
18 B 38 A
19 D 39 C
20 D 40 C

9702 PHYSICS
examination.

© UCLES 2010
Question
Number
Key
Question
Number
Key
1 B 21 D
2 D 22 C
3 B 23 A
4 A 24 C
5 B 25 B
6 C 26 D
7 A 27 D
8 B 28 A
9 C 29 C
10 D 30 D
11 A 31 A
12 A 32 D
13 C 33 A
14 C 34 A
15 D 35 A
16 B 36 D
17 B 37 A
18 A 38 C
19 D 39 C
20 A 40 C

9702 PHYSICS
examination.

© UCLES 2010
Question
Number
Key
Question
Number
Key
1 A 21 D
2 C 22 B
3 B 23 C
4 B 24 C
5 B 25 A
6 D 26 D
7 A 27 D
8 C 28 C
9 D 29 A
10 C 30 D
11 A 31 A
12 A 32 A
13 B 33 D
14 C 34 D
15 A 35 A
16 B 36 A
17 D 37 C
18 B 38 C
19 A 39 C
20 D 40 A

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.

GCE AS/A LEVEL – May/June 2010 9702 21
© UCLES 2010
1 10–9
…………………………………………….…………..………….…………………........ B1
c …………………………………………….…………..………….………………………….. B1
mega ….……………………………………….…………..………….………………………. B1
tera …….……………………………………….…………..………….…………………….... B1 [4]
2 (a) scalar …………………………………………………………..………………………… B1
scalar …………………………………………………………..………………………… B1
vector …………………………………………………………..………………………… B1 [3]
(b) (i) 1 gradient (of graph) is the speed/velocity (can be scored here or in 2)………. B1
initial gradient is zero …………………………………………………………… B1 [2]
2 gradient (of line/graph) becomes constant ……….……..…………………… B1 [1]
(ii) speed = (2.8 ± 0.1) m s–1
……… ………………………………………………… A2 [2]
(if answer > ±0.1 but ≤ ±0.2, then award 1 mark)
(iii) curved line never below given line and starts from zero …..………………….. B1
continuous curve with increasing gradient …………………..…………………. B1
line never vertical or straight ………………………………..……………………. B1 [3]
3 (a) either energy (stored)/work done represented by area under graph
or energy = average force × extension ………………………………………… B1
energy = ½ × 180 × 4.0 × 10–2
……………………………………..………………… C1
= 3.6 J …………………………………………………………………………. A1 [3]
(b) (i) either momentum before release is zero ………………………………………. M1
so sum of momenta (of trolleys) after release is zero …..……………. A1
or force = rate of change of momentum (M1)
force on trolleys equal and opposite (A1)
or impulse = change in momentum (M1)
impulse on each equal and opposite (A1) [2]
(ii) 1 M1V1 = M2V2 ……………..……………………………………..………………. B1 [1]
2 E = ½ M1V1
2
+ ½ M2V2
2
………………………………………………………… B1 [1]
(iii) 1 EK = ½mv2
and p = mv combined to give …………………………………… M1
EK = p2
/ 2m …………………………………………………………………….. A0 [1]
2 m smaller, EK is larger because p is the same/constant …………………… M1
so trolley B …..………………………………………………………………….. A0 [1]

© UCLES 2010
4 (a) when a wave (front) passes by/incident on an edge/slit ….…..…………………… M1
wave bends/spreads (into the geometrical shadow) …………..…………………… A1 [2]
(b) tan θ =
165
38
θ = 13° …………….………………………………..…………………………………… C1
d sin θ = nλ …………….………………………………..……….……………………… C1
d = 2.82 × 10–6
…………….……………………………….……………………………. C1
number = (1/d =) 3.6 × 105
……………….……………………………………………. A1 [4]
(c) P remains in same position …………………………………………………………… B1
X and Y rotate through 90° ……………………………………....……………………. B1 [2]
(d) either screen not parallel to grating
or grating not normal to (incident) light …………………………………………. B1 [1]
5 (a) region/area where a charge experiences a force ……………….………………….. B1 [1]
(b) (i) left-hand sphere (+), right-hand sphere (–) ……………………..………………. B1 [1]
(ii) 1 correct region labelled C within 10 mm of central part of plate
otherwise within 5 mm of plate ………….…………………………………….. B1 [1]
2 correct region labelled D area of field not included for (b)(ii)1 …….………. B1 [1]
(c) (i) arrows through P and N in correct directions …………………………………… B1 [1]
(ii) torque = force × perpendicular distance (between forces) ….………………… C1
= 1.6 × 10–19
× 5.0 × 104
× 2.8 × 10–10
× sin30
= 1.1 × 10–24
N m …….…………….……………………………………… A1 [2]
6 (a) (i) P = VI …………………………..……………………………….…..……………… C1
60 = 12 × I
I = 5.(0) A …………………………………………….…………………………… A1 [2]
(ii) either V = IR or P = I 2
R or P = V2
/ R ….………..………………. C1
either 12 = 5 × R or 60 = 52
× R or 60 = 122
/R ….……….…………… M1
R = 2.4 Ω …………………………………………………………………………. A0 [2]
(b) R = ρL/A …………………………..…………………………………………………….. C1
A = π × (0.4 × 10–3
)2
(= 5.03 × 10–7
) .…………..……………………………………… C1
L = (2.4 × 5.03 × 10–7
)/(1.0 × 10–6
)
= 1.2 m …………..……………….……………………………………………………. A1 [3]
(c) resistance is halved ……………………………….…………………………………… M1
either current is doubled or power ∝ 1/R ….……… ……………………………… M1
power is doubled …………………….……..…………………………………………… A1 [3]

© UCLES 2010
7 (a) nuclei/atoms with same proton number/atomic number …...………………………. B1
nuclei/atoms contain different numbers of neutrons/different atomic mass ..……. B1 [2]
(b) (i) 92 …………………………………………………………………………………… A1 [1]
(ii) 146 ………………………………..………………………………………………… A1 [1]
(c) (i) mass = 238 × 1.66 × 10–27
…..……………………….…………………………… C1
= 3.95 × 10–25
kg ………………….………………………………………… A1 [2]
(ii) volume =
3
4
π × (8.9 × 10–15
)3
(= 2.95 × 10–42
) ……..………………………… C1
density = (3.95 × 10–25
)/(2.95 × 10–42
)
= 1.3 × 1017
kg m–3
……………………………………………................ A1 [2]
(d) nucleus contains most of mass of atom ……………………………………………… B1
either nuclear diameter/volume very much less than that of atom
or atom is mostly (empty) space .......................................................…………… B1 [2]

9702 PHYSICS
9702/22 Paper 2 (AS Structured Questions)
examination.

© UCLES 2010
1 (a) micrometer/screw gauge/digital callipers ………………………………………. B1 [1]
(b) (i) look/check for zero error ……………………………………………………. B1 [1]
(ii) take several readings ……………………………………………………….. M1
around the circumference/along the wire …………………………………. A1 [2]
2 (a) e.g. initial speed is zero
constant acceleration
straight line motion
(any two, one mark each) ……………………………………………………………….B2 [2]
(b) (i) s = ½a t 2
0.79 = ½ × 9.8 × t 2
………………………………………………………….. C1
t = 0.40 s allow 1 SF or greater ……………………………………………. A1
2 or 3 SF answer ……………………………………………………….. A1 [3]
(ii) distance travelled by end of time interval = 90 cm ………………………. C1
0.90 = ½ × 9.8 × t 2
t = 0.43 s allow 2 SF or greater ……………………………………………. C1
time interval = 0.03 s ………………………………………………………... A1 [3]
(c) (air resistance) means ball’s speed/acceleration is less ……………………… M1
length of image is shorter ………………………………………………………… A1 [2]
3 (a) (i) force is rate of change of momentum ………………………………………… B1 [1]
(ii) force on body A is equal in magnitude to force on body B (from A) …………M1
forces are in opposite directions ……………………………………………… A1
forces are of the same kind ………………………………………………………A1 [3]
(b) (i) 1 FA = – FB ……………………………………………………………………. B1 [1]
2 t A = t B ……………………………………………………………………… B1 [1]
(ii) ∆p = FA t A = – FB t B ………………………………………………………….. B1 [1]
(c) graph: momentum change occurs at same times for both spheres …………. B1
final momentum of sphere B is to the right …………………………………….. M1
and of magnitude 5 N s …………………………………………………………… A1 [3]
4 (a) e.g. no energy transfer
amplitude varies along its length/nodes and antinodes
neighbouring points (in inter-nodal loop) vibrate in phase, etc.
(any two, 1 mark each to max 2 ………………………………………………………..B2 [2]

© UCLES 2010
(b) (i) λ = (330 × 102
)/550 ………………………………………………………….. M1
λ = 60cm ……………………………………………………………………… A0 [1]
(ii) node labelled at piston ………………………………………………………. B1
antinode labelled at open end of tube ……………………………………… B1
additional node and antinode in correct positions along tube …………… B1 [3]
(c) at lowest frequency, length = λ/4 ………………………………………………... C1
λ = 1.8 m
frequency = 330/1.8 ………………………………………………………………. C1
= 180Hz ………………………………………………………………………….... A1 [3]
5 (a) (i) Young modulus = stress/strain ……………………………………………… C1
data chosen using point in linear region of graph ………………………… M1
Young modulus = (2.1 × 108
)/(1.9 × 10–3
)
= 1.1 × 1011
Pa ……………………………………………………………….. A1 [3]
(ii) This mark was removed from the assessment, owing to a power-of-ten
inconsistency in the printed question paper.
(b) area between lines represents energy/area under curve represents energy .. M1
when rubber is stretched and then released/two areas are different ……...... A1
this energy seen as thermal energy/heating/difference represents energy
released as heat …………………………………………………………………… A1 [3]
6 (a) either P ∝ V 2
or P = V 2
/R …………………………………………………………. C1
reduction = (2302
– 2202
)/2302
= 8.5 % ………………………………………………………………….. A1 [2]
(b) (i) zero ……………………………………………………………………………. A1 [1]
(ii) 0.3(0)A ……………………………………………………………………….. A1 [1]
(c) (i) correct plots to within ± 1mm ………………………………………………. B1 [1]
(ii) reasonable line/curve through points giving current as 0.12A
allow ± 0.005A) ………………………………………………………………. B1 [1]
(iii) V = IR …………………………………………………………………………. C1
V = 0.12 × 5.0
= 0.6(0)V …………………………………………………………………... A1 [2]
(d) circuit acts as a potential divider/current divides/current in AC not the same as
current in BC ………………………………………………………………………. B1
resistance between A and C not equal to resistance between C and B ……. B1
or current in wire AC × R is not equal to current in wire BC × R B1 [2]
any 2 statements

© UCLES 2010
7 (a) (i) either helium nucleus
or contains 2 protons and 2 neutrons ………………………………… B1 [1]
(ii) e.g. range is a few cm in air/sheet of thin paper
speed up to 0.1 c
causes dense ionisation in air
positively charged or deflected in magnetic or electric fields
(any two, 1 each to max 2) ………………………………………………….. B2 [2]
(b) (i) α4
2 ……………………………………………………………………………… B1
either p1
1 or H1
1 ……………………………………………………………….. B1 [2]
(ii) 1 initially, α-particle must have some kinetic energy ………………….. B1 [1]
(ii) 2 1.1 MeV = 1.1 × 1.6 × 10–13
= 1.76 × 10–13
J …………………………. C1
EK = ½mv 2
……………………………………………………………….. C1
1.76 × 10–13
= ½ × 4 × 1.66 × 10–27
× v 2
……………………………… C1
v = 7.3 × 106
m s–1
……………………………………………………..... A1 [4]
use of 1.67 × 10–27
kg for mass is a maximum of 3/4

9702 PHYSICS
9702/23 Paper 2 (AS Structured Questions), maximum raw mark 60
examination.

© UCLES 2010
1 (a) (i) 1% of ±2.05 is ±0.02 A1 [1]
(ii) max. value is 2.08 V A1 [1]
(b) there may be a zero error/calibration error/systematic error M1
which makes all readings either higher or lower than true value A1 [2]
2 (a) no resultant force/sum of forces zero B1
no resultant moment/torque/sum of moments/torques zero B1 [2]
(b) (i) each force is represented by the side of a triangle/by an arrow M1
in magnitude and direction A1
arrows joined, head to tail B1 [3]
(could be shown on a sketch diagram)
(ii) if the triangle is ‘closed’ (then the forces are in equilibrium) B1 [1]
(c) triangle drawn with correct shape (incorrect arrows loses this mark) B1
T1 = 5.4 ± 0.2N B1
T2 = 4.0 ± 0.2N B1 [3]
(d) forces in strings would be horizontal B1
(so) no vertical force to support the weight B1 [2]
3 (a) evidence of use of area below the line B1
distance = 39 m (allow ±0.5m) A2 [3]
(if > ±0.5m but ≤ 1.0m, then allow 1 mark)
(b) (i) 1 EK = ½mv 2
C1
∆ EK = ½ × 92 × (62
– 32
)
= 1240 J A1 [2]
2 EP = mgh C1
∆ EP = 92 × 9.8 × 1.3
= 1170J A1 [2]
(ii) E = Pt C1
E = 75 × 8
= 600J A1 [2]
(c) (i) energy = (1240 + 600) – 1170 M1
= 670J A0 [1]
(ii) force = 670/39 = 17N A1 [1]
(d) frictional forces include air resistance B1
air resistance decreases with decrease of speed B1 [2]

© UCLES 2010
4 (a) (i) solid has fixed volume and fixed shape/incompressible B1 [1]
(ii) gas fills any space into which it is put B1 [1]
(b) atoms/molecules have (elastic) collisions with the walls (of the vessel) B1
momentum of atom/molecule changes B1
so impulse (on wall)/force on wall B1
random motion/many collisions (per unit time) gives rise to
(constant) force/pressure B1 [4]
(c) spacing (much) greater in gases than in liquids/about ten times C1
either spacing depends on 1/3
√ρ
or ratio of spacings is about 8.8 A1 [2]
5 (a) (i) 1 number of oscillations per unit time (not per second) B1 [1]
2 nλ A1 [1]
(ii) v = distance / time = nλ/t M1
n/t = f hence v= fλ A1
or f oscillations per unit time so fλ is distance per unit time M1
distance per unit time is v so v = fλ A1 [2]
(b) (i) 1.0 period is 3 × 2 = 6.0 ms C1
frequency = 1/(6 × 10–3
) = 170Hz A1 [2]
(ii) wave (with approx. same amplitude and) with correct phase difference B1 [1]
6 (a) (i) movement/flow of charged particles B1 [1]
(ii) work done per unit charge (transferred) B1 [1]
(b) straight line through origin B1
resistance = V/I , with values for V and I shown M1
= 20 Ω A0 [2]
(using the gradient loses the last mark)
(c) (i) 0.5A A1 [1]
(ii) either resistance of each resistor is 20Ω or total current = 0.8A C1
either combined resistance = 10 Ω or R = E/I = 10Ω A1 [2]
(d) (i) 10V A1 [1]
(ii) power = EI C1
= 10 × 0.2 = 2.0W A1 [2]

© UCLES 2010
7 (a) (i) either helium nucleus
or particle containing two protons and two neutrons B1 [1]
(ii) allow any value between 1cm and 10cm B1 [1]
(b) (i) energy = (8.5 × 10–13
)/(1.6 × 10–13
) M1
= 5.3MeV A0 [1]
(ii) number = (5.3 × 106
)/31 C1
= 1.7 × 105
(allow 2 s.f. only) A1 [2]
(iii) number per unit length = (1.7 × 105
) /(a)(ii)
correct numerical value A1
correct unit B1 [2]

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

© UCLES 2010
1 (a) Ring e.m.f. value
(c) Six sets of values for V and I scores 5 marks, five sets scores 4 marks, etc. [5]
Indicate the number of sets of readings.
Incorrect trend –1 (wrong trend N increases, I increases).
Apparatus correctly set up without help from supervisor. [2]
Major help –2, minor help –1
Range of N in table to include 1 or 2 and 11 or 12. [1]
Column headings (N (no unit), V/V, I/A, R/Ω, (1/R)/Ω–1
) [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
(solidus is expected but accept for example, V(V)).
Consistency of presentation of raw readings of I and V. [1]
All values of I must be given to the same number of decimal places.
All values of V must be given to the same number of decimal places.
Significant figures. [1]
S.f. for 1/R must be the same as, or one more than, the least number of s.f. used in I or V.
Check each row.
Values of 1/R correct. Underline and check the specified value of 1/R. [1]
If incorrect, write in the correct value.
(d) Graph
(i) Axes [1]
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. Indicate false origin with FO.
Scales must be labelled with the quantity that is being plotted. Ignore units.
Allow inverted axes but do not allow the wrong graph.
Scale markings should be no more than three large squares apart.
All observations must be plotted. [1]
Write a ringed total of plotted points.
Do not accept blobs (points > 0.5 small square).
Ring and check a suspect plot. Tick if correct. Re-plot if incorrect.
Work to an accuracy of half a small square.
(ii) Line of best fit [1]
Judge by balance of at least 5 trend points about the candidate’s line.
There must be an even distribution of points either side of the line along the whole
length. Indicate best line if candidate’s line is not the best line.
Lines must not be kinked.
Quality
Judge by scatter of all points about a straight line. [1]
All plots from table (minimum 5) must be within 1 mA of a straight line.
Do not award if wrong graph or wrong trend.

© UCLES 2010
(iii) Gradient [1]
The hypotenuse of the triangle must be at least half the length of the drawn line.
Both 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).
y-intercept from graph or substitute correct read-offs into y = mx + c [1]
(Expect close to 0). Label FO.
(e) M = gradient value. L = y–intercept value. No substitution method. [1]
If inverted axes not corrected for –1
Value of M = value from part (a) ± 0.5V. [1]
Value of L = 0 ± 1 mA.
Appropriate units
[Total: 20]
2 (b) Total time over which swings are measured > 10 s. [1]
Correct calculation of T = Tn /n. [1]
(c) (i) Value of l = 5 cm ± 1 cm [1]
Evidence of repeats in length value (here or in d(iii)). [1]
(ii) Measure in two different places/check zero error. [1]
(iii) Percentage uncertainty in length. Consistent units. ∆l = 0.1 mm [1]
If repeated readings have been taken, then the uncertainty can be half the range.
Correct ratio idea required (0.1/length × 100%).
(d) (ii) Measurement of time for longer tube. [1]
t longer tube < t shorter tube [1]
(iii) Measurement of length for longer tube to the nearest 1 mm. [1]
Consistent unit
(iv) Add two lengths together correctly. Allow rounding. [1]
(e) Correct calculation of two values of k = T2
/l. [1]
Valid conclusion based on the calculated values of k. [1]
Candidate must test against a specified criterion.

© UCLES 2010
Limitations (4) Improvements (4) Ignore
A Ap Two readings not enough
(to support conclusion)/too few
readings.
As Take many readings and plot a
graph/compare values of k. Do not
allow average k.
Repeat readings
B Bp (l inaccurate because) gap
between long and short tube/
ends of tubes uneven. Tubes
not straight/kinked/disjointed.
Bs Get one long tube without a
break/stick two tubes together/use
longer tube on its own. Method of
smoothing ends.
Parallax error
C Cp Tube(s) not vertical when
stationary/ not aligned with
string.
Cs Smaller diameter tube/thicker
walled tube/suitable method of
alignment.
Thicker string
D Dp Not swinging in one plane
only/idea of non-uniform
oscillation.
Ds Method of reducing draught e.g.
close windows, turn off fans, screen
experiment.
E Ep Time difficult to measure
because difficult to know when
oscillation returns to original
position/maximum height.
Es A marker to time as passes
centre/reaches maximum
displacement. Light gate at centre
with timer/motion sensor at end with
data logger/video with timer
(playback) in slow motion.
Difficult to release from
same point each time/
human error/reaction
time/unqualified use of
light gates/sensors
Xp/Xs Other valid suggestions (e.g. knot slipping) with valid method.
[Total: 20]

9702 PHYSICS
examination.

© UCLES 2010
1 (b) Six sets of values for N and I scores 5 marks, five sets scores 4 marks, etc. [5]
Incorrect trend –1.
Apparatus set up correctly without help from supervisor. Minor help –1, [2]
major help –2
Range – [1]
To include N = 1 or 2 and N = 11 or 12.
Column headings – [1]
(solidus is expected, but accept, for example, I (A))
Consistency of presentation of raw readings of I – [1]
All raw values of I must be given to the same number of decimal places.
Significant figures – [1]
S.f. for 1/I must be the same as, or one more than, the s.f. for I.
Check each row.
Values of 1/I correct – [1]
Underline and check the specified value of 1/I. If incorrect, write in the correct
value.
(c) (i) Graph
Axes – [1]
Plots – [1]
All observations must be plotted. Write a ringed total of plotted points.
Do not accept blobs (points > half a small square).
(ii) Line of best fit – [1]
Judge by the balance of at least 5 trend plots about the candidate’s line.
Line must not be kinked or thicker than 1 mm.
Quality – [1]
Judge by scatter of all points about a straight line.
All plots in the table must be within 10 Ω of a straight line.

© UCLES 2010
(iii) Gradient – [1]
Both read-offs must be accurate to half a small square. If incorrect, write in the correct
value.
y-intercept – [1]
Either from graph or by substitution of correct read-offs into y = mx + c.
Check for and label false origin.
(d) G = gradient value and H = intercept value. [1]
Do not credit if a substitution method is used.
Range of values (–70Ω Y H Y –30 Ω and 3.5 V Y G Y 5.5 V) with appropriate units. [1]
Do not credit if a substitution method is used.
[Total: 20]
2 (b) (i) Value of maximum force to 1 d.p. in raw data and greater than 0N. [1]
Evidence of repeated measurements of F in (b)(i) or (d). [1]
(ii) Reaches maximum force suddenly (short time); no notice given when releases. [1]
(iii) Percentage uncertainty in maximum force. [1]
0.1N Y ∆F Y 0.4 N. If repeated readings have been done then the uncertainty could be
half the range. Correct ratio idea required (e.g. 0.2 / F × 100%).
(c) (i) Measurement of raw t to the nearest 0.01 mm. [1]
(ii) Take repeats in different places / (account for) zero errors. [1]
(iii) Maximum force with three slides. Unit required. [1]
(d) Measurement of thickness of one slide. [1]
Measurement of maximum force with one slide. [1]
Quality: F(b)(i) > F(d) > F(c)(iii) [1]
(e) Calculation of two values of k. [1]
Valid conclusion based on the calculated values of k. [1]
Candidates must test against a specified criterion.

© UCLES 2010
(f)(i),(ii) Identify limitations and improvements
Limitations (4) Improvements (4) Do not credit
A Two readings are not enough
(to support conclusion
Take more (sets of) readings and
plot a graph
Repeat readings.
B Maximum force reached
without warning (if not already
credited in (b)(ii))
Bs Practical method of recording
maximum value e.g. video with
playback in slow motion / max-min
newton metre / force sensor with
data logger / masses with pulley.
Parallax error.
Solution for parallax
error.
‘Use of computer’ to
measure maximum
force.
C t changes due to compression
force of magnets / slide
thickness non uniform (if not
already credited) / thread
thickness adds to separation.
Method of attaching newton meter
without thread / measure and add
thread thickness.
D Zero error on newton meter
when used horizontally.
Adjust zero / practical vertical
arrangement.
Condition of newton
meter.
E Glass may affect magnetic
force / effect of surrounding
magnetic materials
(e.g. G clamp).
Use a variety of materials to separate
magnets and test if material affects
results / use a non magnetic clamp /
glue first magnet to bench.
Reference to Earth’s
field.
F Friction with bench. Method of reducing friction.
G Difficulties with alignment of
force with magnets.
Method of raising magnets / longer
loop.
X Difficult to measure force due
to weak magnets / small force
(if validated by SR)
More sensitive newton meter.
[Total: 20]

9702 PHYSICS
examination.

© UCLES 2010
1 (c) Six sets of readings of I and V scores 5 marks, five sets scores 4 marks, etc. [5]
Incorrect trend –1 (wrong trend is I increases, V10
decreases).
Apparatus correctly setup without help from supervisor. [2]
Range of I: Imin Y 10 mA and Imax [ 35 mA. Ignore POT errors. [1]
Column headings (e.g. V/V, I/A, V10
/V10
). [1]
Must have V and I columns.
There must be some distinguishing mark between the quantity and the unit.
(solidus is expected but accept, for example, V(V))
Consistency of presentation of raw readings. [1]
All values of V must be given to the same number of decimal places (must have dp).
All values of I must be given to the same number of decimal places.
Significant figures.
Sf for V10
must be the same as or one more than the sf used in V. Check each row. [1]
Values of V10
correct. Underline and check the specified value of V10
. [1]
(d) Graph
(i) Axes
Sensible scales must be used. Awkward scales (e.g. 3:10) are not allowed. [1]
Plots
All observations must be plotted. [1]
Write a ringed total of plotted points.
Quality
All points in table (minimum 5) must be within 2 mA of a straight line. [1]

© UCLES 2010
(iii) Gradient
The hypotenuse of the triangle must be at least half the length of the drawn line. [1]
Label FO.
(e) a = gradient value and b = y–intercept value. [1]
If inverted axes not corrected for –1
Range of values (0.1AV–10
Y a Y 0.9AV–10
, b = 0 ± 0.01A) and appropriate units [1]
[Total: 20]
2 (a) Raw value(s) of x: 25.0 cm Y x Y 35.0 cm with unit to nearest mm. [1]
(b) (i) Evidence of repeated measurements of d in (b)(i) or (e) [1]
Value of d = 3.0 mm ± 1.0 mm or SV ± 1.0 mm [1]
Raw values of d to at least 0.1 mm
(ii) Value of t in range 1 s to 10 s unless SV indicates otherwise. Allow SV ± 5 s [1]
(c) Absolute uncertainty in t1 in the range 0.1 to 0.6 s [1]
If repeated readings have been taken, then the uncertainty could be half the range.
Correct calculation to get % uncertainty.
(d) v calculated correctly with consistent units. [1]
(e) Second value for d. [1]
Second value for t. [1]
Quality: t2 less than t1. (d increases, t decreases) [1]
(f) (i) Calculation of two values of k. [1]
(ii) Valid conclusion based on the calculated values. [1]
(iii) Relate raw values of x, t and d. Any decimal place arguments score zero. [1]

© UCLES 2010
(to support conclusion)/too few
readings.
As Take more (sets of) readings and
plot a graph/compare values of k.
Repeat readings.
B Bp Time too short/reaction
time large compared to
measured time/parallax error
in judging start/stop.
Bs Increase x/lengthen tube/smaller
balls/video with timer (playback) in
slow motion.
Light gates, motion
sensors, data loggers,
computers, helpers,
solution for parallax
error.
Set squares, rulers, etc.
C Cp Difficult to see glass balls. Cs Use coloured balls/shine light
through.
Use ball bearings (type
of ball and oil stays
fixed).
D Dp Terminal velocity not
reached (by the first marker).
Ds A valid method to check reached
TV, e.g. time constant over three
markers/video with timer (playback)
in slow motion, multi-flash
photography/stroboscope.
References to starting
point.
Do not accept ‘move x
down’ on its own.
Change viscosity of oil
(oil and glass must
remain fixed).
E Ep Balls not all the same
diameter/size/shape/mass
Es Use micrometer screwgauge/top
pan balance
X Xp Balls had a hole in/air
bubbles on ball or oil.
Xs Clean balls/immerse in oil
[Total: 20]

9702 PHYSICS
examination.

© UCLES 2010
1 (a) Apparatus set up without help from Supervisor. [1]
Value of L to nearest mm. [1]
(d) Table –
Six sets of readings of d and h scores 5 marks, five sets scores 4 marks, etc.
Incorrect trend –1. [5]
Range –
Range of values of d [ 15 cm. [1]
Column headings –
Each column heading must contain a quantity and a unit. Ignore units in the body of the
table.
There must be some distinguishing mark between the quantity and the unit e.g. 1/d / m–1
or
1/d (m–1
). [1]
Consistency –
All raw values of h must be given to the nearest mm. [1]
Significant figures –
S.f. for 1/d must be the same as, or one more than, the s.f. given for raw d. Check each row.
[1]
Calculated values –
Check the specified value of 1/d. If wrong, write in the correct value. [1]
(e) (i) Graph
Axes –
both x and y directions. Indicate a false origin with FO.
Scale markings should not be more than three large squares apart. [1]
Plotting of points –
All observations must be plotted.
Do not accept ‘blobs’ (points > half a small square).
Ring and check a suspect point. Tick if correct. Re-plot if incorrect.
Work to an accuracy of half a small square. [1]
(ii) Line of best fit –
Judge by the balance of at least 5 trend points about the candidate’s line. There must be
an even distribution of points either side of the line along the whole length.
Indicate best line if candidate’s line is not the best line.
Line must not be kinked or thicker than 1 mm. [1]
Quality –
Judge by scatter of all points about a best line. All plots from table (minimum 5) must be
within 0.1 m–1
of a straight line (in 1/d direction).
Do not credit if it is the wrong graph or if the trend is wrong. [1]

© UCLES 2010
(iii) Gradient –
Read-offs must be accurate to half a small square – if wrong write in the correct value(s).
Check for ∆y/∆x (i.e. do not allow ∆x/∆y). [1]
y-intercept –
Value must be read from graph to nearest half small square (after checking for false
origin) or calculated using ratios or y = mx + c. [1]
(f) Correct calculation of z (gradient value must be used).
Ignore sign. [1]
Value of z given with unit of length (gradient value must be used). [1]
[Total: 20]
2 (a) Measurement of I in range 1.5 A–2.5 A and to 0.1A or better. [1]
(c) Measurement of x to the nearest mm. [1]
(d) Measurement of θ (less than 45°). Raw values to no more than nearest degree or half
degree. [1]
(e) Percentage uncertainty in θ : Correct method, using ∆θ = half the range, or ∆θ = 2° to 10°. [1]
(f) (i) Evidence of repeated measurements either here or in (d). [1]
(ii) Correct average value of θ. [1]
(g) Second measurement of x. [1]
Second measurement of I. [1]
Quality: I decreases as x decreases. [1]
(h) (i) Correct calculation of two values of k. [1]
(ii) Valid conclusion based on the calculated values of k. Candidate must test against a
specified criterion. [1]
(iii) Statement that the s.f. for k depend on the s.f. for I and x. Ignore any reference to d.p. [1]

© UCLES 2010
(i) Identifying limitations and suggesting improvements
A Two readings (of x and I) are
not enough (to draw a valid
conclusion).
Take more readings and plot a
graph.
Repeat readings.
B Difficult to measure x / difficult
to keep x constant / difficult to
keep distance between wire
and magnet constant / difficult
to keep distance between wire
and stick constant.
Use a clamped ruler / method of
fixing the string
Parallax error in
measuring x.
C Magnet does not come to rest. Practical method of damping / shield
from draughts / turn off fans.
Magnet swings too fast.
D Measured angles are very
small
Use larger currents / use bigger
protractor
Use stronger / larger
magnet.
E Parallax error in measuring θ /
reading protractor / reading
deflection.
Method of bringing protractor closer
to wire / shine light from above
Increase x / use mirror.
F Difficult to alter rheostat while
holding string.
Method of fixing the string (unless
already credited in B) / method of
fixing rheostat to bench / use
assistant.
G (θ affected by) magnetic
materials nearby / stray
magnetic fields.
Use wooden / non-magnetic stands. Move object further
away.
H Fluctuating current. Method of improving contact with
wire (e.g. cleaning contacts, soldered
connections).
Do NOT credit: Use sensors / use lightgates / use video.
[Total: 20]

9702 PHYSICS
examination.

© UCLES 2010
1 (c) Six sets of readings of I and V scores 5 marks, five sets scores 4 marks, etc. [5]
Incorrect trend then –1 (wrong trend P increases, R4
decreases).
Apparatus correctly set up without help from supervisor. [2]
Major help –2, minor help –1
Range of V: Vmin Y 2 V and Vmax [ 10 V. [1]
Column headings (V/V, I/A, P/W, R/Ω, R 4
/Ω4
) [1]
Must have V and I columns.
(solidus is expected but accept, for example, V(V)).
Consistency of presentation of raw readings. [1]
All raw values of V must be given to the same number of decimal places
and this must be 0.1 V.
All raw values of I must be given to the same number of decimal places
Significant figures. [1]
S.F. for P must be the same as, or one more than, the least number of S.F. used
in V or I. Check each row.
Values of R4
correct. Underline and check the specified value of R4
. [1]
(d) (i) Graph
Axes: Sensible scales must be used, no awkward scales (e.g. 3:10). [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.
Allow inverted axes but do not allow wrong graph.
Plots
All observations must be plotted.
Write a ringed total of plotted points. [1]
Quality [1]
All points in the table (minimum 5) must be within 50 mW of a straight line.

© UCLES 2010
(iii) Gradient [1]
Check for ∆y / ∆x (i.e. do not allow ∆x / ∆y).
Label FO.
(e) a = gradient value and b = y–intercept value. [1]
Units for a and b are correct (expect WΩ–4
for a and W for b). [1]
Range: a = 3 × 10–9
± 1 × 10–9
or SV ± 33%
[Total: 20]
2 (a) (ii) Value of d, with consistent unit. Range of d: 5 ± 1 cm [1]
d to nearest mm. [1]
(c) (ii) Evidence of repeated measurements of t either in (c)(ii) or (e)(ii). [1]
Value of t in range 5 to 30 s. [1]
(d) Absolute uncertainty in t in the range 0.5 to 1.0 s. [1]
If repeated readings have been taken, then the uncertainty can be half the range.
Correct calculation to get % uncertainty. [1]
(e) (ii) Second value for d. [1]
Second value for t. [1]
Quality: t2 less than t1. [1]
(f) (i) Correct calculation of two values of k or equivalent. [1]
(ii) Valid conclusion based on the calculated values of k. [1]
(iii) Justification with reference to the significant figures in t and d. [1]

© UCLES 2010
(g)
(to support conclusion) / too
few readings.
As Take more (sets of) readings and
plot a graph / compare values of k.
Repeat readings
B Bp Marker never exactly on
2 cm or 0.5 cm: either above
or below / increments in
changes in amplitude too large
/ difficult to judge 2 cm and 0.5
cm.
Bs Video with timer (playback) in
slow motion / position sensor above
with data logger / measure the
amplitudes over time.
Use computer to
improve the experiment.
Multi-flash photography?
Light gates.
C Cp Straw not vertical (straight)
/ straw bumping into sides/
non-vertical oscillation.
Cs Wider container / glue straw /
method of alignment.
No ref to changing oil
D Dp Difficult to measure ‘d’
because of lining up meniscus
/ refraction of curved container.
Ds Mark straw/ mark container / use
travelling microscope / vernier
calliper?
E Ep Difficult to measure time
because moves past the
marker quickly / small
distances involved.
Es Video with timer (playback) in
slow motion / position sensor above
with data logger. Credit once only.
[Total: 20]

9702 PHYSICS
9702/41 Paper 4 (A2 Structured Questions), maximum raw mark 100
examination.

© UCLES 2010
Section A
1 (a) angle (subtended) at centre of circle B1
(by) arc equal in length to radius B1 [2]
(b) (i) point S shown below C B1 [1]
(ii) (max) force / tension = weight + centripetal force C1
centripetal force = mrω2
C1
15 = 3.0/9.8 × 0.85 × ω2
C1
ω = 7.6 rad s–1
A1 [4]
2 (a) (i) 27.2 + 273.15 or 27.2 + 273.2 C1
300.4 K A1 [2]
(ii) 11.6 K A1 [1]
(b) (i) (<c2
> is the) mean / average square speed B1 [1]
(ii) ρ = Nm/V with N explained B1
so, pV = 1/3 Nm<c2
> B1
and pV = NkT with k explained B1
so mean kinetic energy / <EK> = ½m<c2
> = 3/2 kT B1 [4]
(c) (i) pV = nRT
2.1 × 107
× 7.8 × 10–3
= n × 8.3 × 290 C1
n = 68 mol A1 [2]
(ii) mean kinetic energy = 3/2 kT
= 3/2 × 1.38 × 10–23
× 290 C1
= 6.0 × 10–21
J A1 [2]
(iii) realisation that total internal energy is the total kinetic energy C1
energy = 6.0 × 10–21
× 68 × 6.02 × 1023
C1
= 2.46 × 105
J A1 [3]
3 (a) (i) to-and-fro / backward and forward motion (between two limits) B1 [1]
(ii) no energy loss or gain / no external force acting / constant energy / constant amplitude
B1 [1]
(iii) acceleration directed towards a fixed point B1
acceleration proportional to distance from the fixed point / displacement B1 [2]
(b) acceleration is constant (magnitude) M1
so cannot be s.h.m. A1 [2]

© UCLES 2010
4 (a) ability to do work B1
as a result of the position/shape, etc. of an object B1 [2]
(b) (i) 1 ∆Egpe = GMm / r C1
= (6.67 × 10–11
× {2 × 1.66 × 10–27
}2
) / (3.8 × 10–15
) C1
= 1.93 × 10–49
J A1 [3]
2 ∆Eepe = Qq / 4πε0r C1
= (1.6 × 10–19
)2
/ (4π × 8.85 × 10–12
× 3.8 × 10–15
) C1
= 6.06 × 10–14
J A1 [3]
(ii) idea that 2EK = ∆Eepe – ∆Egpe B1
EK = 3.03 × 10–14
J
= (3.03 × 10–14
) / 1.6 × 10–13
M1
= 0.19 MeV A0 [2]
(iii) fusion may occur / may break into sub-nuclear particles B1 [1]
5 (a) (i) VH depends on angle between (plane of) probe and B-field B1
either VH max when plane and B-field are normal to each other
or VH zero when plane and B-field are parallel
or VH depends on sine of angle between plane and B-field B1 [2]
(ii) 1 calculates VHr at least three times M1
to 1 s.f. constant so valid or approx constant so valid
or to 2 s.f., not constant so invalid A1 [2]
2 straight line passes through origin B1 [1]
(b) (i) e.m.f. induced is proportional / equal to M1
rate of change of (magnetic) flux (linkage) A1
constant field in coil / flux (linkage) of coil does not change B1 [3]
(ii) e.g. vary current (in wire) / switch current on or off / use a.c. current
rotate coil
move coil towards / away from wire (1 mark each, max 3) B3 [3]
6 (a) all four diodes correct to give output, regardless of polarity M1
connected for correct polarity A1 [2]
(b) NS / NP = VS / VP C1
V0 = √2 × Vrms C1
ratio = 9.0 / (√2 × 240)
= 1/38 or 1/37 or 0.027 A1 [3]

© UCLES 2010
7 (a) arrow pointing up the page B1 [1]
(b) (i) Eq = Bqv C1
v = (12 × 103
) / (930 × 10–6
) C1
= 1.3 × 107
m s–1
A1 [3]
(ii) Bqv = mv 2
/ r C1
q/m = (1.3 × 107
) / (7.9 × 10–2
× 930 × 10–6
) C1
= 1.8 × 1011
C kg–1
A1 [3]
8 (a) momentum conservation hence momenta of photons are equal (but opposite) M1
same momentum so same energy A1 [2]
(b) (i) (∆)E = (∆)mc2
C1
= 1.2 × 10–28
× (3.0 × 108
)2
= 1.08 × 10–11
J A1 [2]
(ii) E = hc / λ
λ = (6.63 × 10–34
× 3.0 × 108
) / (1.08 × 10–11
) C1
= 1.84 × 10–14
m A1 [2]
(iii) λ = h / p
p = (6.63 × 10–34
) / (1.84 × 10–14
) C1
= 3.6 × 10–20
N s A1 [2]
Section B
9 (a) (i) point X shown correctly B1 [1]
(ii) op-amp has very large / infinite gain M1
non-inverting input is at earth (potential) / earthed / at 0 V M1
if amplifier is not to saturate, inverting input must be (almost)
at earth potential / 0 (V) same potential as inverting input A1 [3]
(b) (i) total input resistance = 1.2 kΩ C1
(amplifier) gain (= –4.2 / 1.2) = –3.5 C1
(voltmeter) reading = –3.5 × –1.5
= 5.25 V A1 [3]
(total disregard of signs or incorrect sign in answer, max 2 marks)
(ii) (less bright so) resistance of LDR increases M1
(amplifier) gain decreases M1
(voltmeter) reading decreases A1 [3]

© UCLES 2010
10 (a) X-ray taken of slice / plane / section B1
repeated at different angles B1
images / data is processed B1
combined / added to give (2-D) image of slice B1
repeated for successive slices B1
to build up a 3-D image B1
image can be viewed from different angles / rotated B1
max 6 [6]
(b) (i) 16 A1 [1]
(ii) evidence of deducting 16 then dividing by 3 C1
to give A1 [2]
3 2
6 5
11 (a) frequency of carrier wave varies (in synchrony) with signal M1
(in synchrony) with displacement of signal A1 [2]
(b) advantages e.g. less noise / less interference
greater bandwidth / better quality
(1 each, max 2)
disadvantages e.g. short range / more transmitters / line of sight
more complex circuitry
greater expense
(1 each, max 2) B4 [4]
12 (a) gain / loss/dB = 10 lg(P1/P2) C1
190 = 10 lg(18 × 103
/ P2)
or –190 = 10 lg P2 / 18 × 103
) C1
power = 1.8 × 10–15
W A1 [3]
(b) (i) 11 GHz / 12 GHz B1 [1]
(ii) e.g. so that input signal to satellite will not be ‘swamped’
to avoid interference of uplink with / by downlink B1 [1]

9702 PHYSICS
examination.

© UCLES 2010
Section A
1 (a) work done moving unit mass M1
from infinity to the point A1 [2]
(b) (i) at R, φ = 6.3 × 107
J kg–1
(allow ± 0.1 × 107
) B1
φ = GM / R
6.3 × 107
= (6.67 × 10–11
× M) / (6.4 × 106
) C1
M = 6.0 × 1024
kg (allow 5.95 → 6.14) A1 [3]
Maximum of 2/3 for any value chosen for φ not at R
(ii) change in potential = 2.1 × 107
J kg–1
(allow ± 0.1 × 107
) C1
loss in potential energy = gain in kinetic energy B1
½ mv 2
= φ m or ½ mv 2
= GM / 3R C1
½ v 2
= 2.1 × 107
v = 6.5 × 103
m s–1
………..……(allow 6.3 → 6.6) A1 [4]
(answer 7.9 × 103
m s–1
, based on x = 2R, allow max 3 marks)
(iii) e.g. speed / velocity / acceleration would be greater B1
deviates / bends from straight path B1 [2]
(any sensible ideas, 1 each, max 2)
2 (a) (i) reduction in energy (of the oscillations) (B1)
reduction in amplitude / energy of oscillations (B1)
due to force (always) opposing motion / resistive forces (B1) [2]
any two of the above, max 2
(ii) amplitude is decreasing (very) gradually / oscillations would
continue (for a long time) /many oscillations M1
light damping A1 [2]
(b) (i) frequency = 1 / 0.3
= 3.3 Hz A1 [1]
allow points taken from time axis giving f = 3.45 Hz
(ii) energy = ½ mv 2
and v = ωa C1
= ½ × 0.065 × (2π/0.3)2
× (1.5 × 10–2
)2
M1
= 3.2 mJ A0 [2]
(c) amplitude reduces exponentially / does not decrease linearly M1
so will be not be 0.7 cm A1 [2]

© UCLES 2010
3 (a) (i) 1 deg C corresponds to (3840 – 190) / 100 Ω C1
for resistance 2300 Ω, temperature is 100 × (2300 – 3840) / (190 – 3840)
temperature is 42°C A1 [2]
(ii) either 286 K ≡ 13°C or 42°C ≡ 315 K B1
thermodynamic scale does not depend on the property of a substance M1
so change in resistance (of thermistor) with temperature is non-linear A1 [3]
(b) heat gained by ice in melting = 0.012 × 3.3 × 105
J C1
= 3960 J
heat lost by water = 0.095 × 4.2 × 103
× (28 – θ) C1
3960 + (0.012 × 4.2 × 103
× θ) = 0.095 × 4.2 × 103
× (28 – θ) C1
θ = 16°C A1 [4]
(answer 18°C – melted ice omitted – allow max 2 marks)
(use of (θ – T) then allow max 1 mark)
4 (a) force = q1q2 / 4πε0x2
C1
= (6.4 × 10–19
)2
/ (4π × 8.85 × 10–12
× {12 × 10–6
}2
) C1
= 2.56 × 10–17
N A1 [3]
(b) potential at P is same as potential at Q B1
work done = q∆V M1
∆V = 0 so zero work done A0 [2]
(c) at midpoint, potential is 2 × (6.4 × 10–19
) / (4πε0 × 6 × 10–6
) C1
at P, potential is (6.4 × 10–19
) / (4πε0 × 3 × 10–6
) + (6.4 × 10–19
) / (4πε0 × 9 × 10–6
) C1
change in potential = (6.4 × 10–19
) / (4πε0 × 9 × 10–6
)
energy = 1.6 × 10–19
× (6.4 × 10–19
) / (4πε0 × 9 × 10–6
) C1
= 1.0 × 10–22
J A1 [4]
5 (a) e.g. ‘storage of charge’ / storage of energy
blocking of direct current
producing of electrical oscillations
smoothing
(any two, 1 mark each) B2 [2]
(b) (i) capacitance of parallel combination = 60 µF C1
total capacitance = 20 µF A1 [2]
(ii) p.d. across parallel combination = ½ × p.d. across single capacitor C1
maximum is 9V A1 [2]
(c) either energy = ½CV 2
or energy = ½QV and Q = CV C1
energy = ½ × 4700 × 10–6
× (182
– 122
) C1
= 0.42 J A1 [3]

© UCLES 2010
6 (a) (i) straight line with positive gradient M1
through origin A1 [2]
(ii) maximum force shown at θ = 90° M1
zero force shown at θ = 0° M1
reasonable curve with F about ½ max at 30° A1 [3]
(b) (i) force on electron due to magnetic field B1
force on electron normal to magnetic field and direction of electron B1 [2]
(ii) quote / mention of (Fleming’s) left hand rule M1
electron moves towards QR A1 [2]
7 (a) either the value of steady / constant voltage M1
that produces same power (in a resistor) as the alternating voltage A1 [2]
or if alternating voltage is squared and averaged (M1)
the r.m.s. value is the square root of this averaged value (A1)
(b) (i) 220 V A1 [1]
(ii) 156 V A1 [1]
(iii) 60 Hz A1 [1]
(c) power = Vrms
2
/ R C1
R = 1562
/ 1500
= 16 Ω A1 [2]
8 (a) (i) number = (5.1 × 10–6
× 6.02 × 1023
) / 241 C1
= 1.27 × 1016
A1 [2]
(ii) A = λN C1
5.9 × 105
= λ × 1.27 × 1016
λ = 4.65 × 10–11
s–1
A1 [2]
(iii) 4.65 × 10–11
× t½ = ln2 C1
t½ = 1.49 × 1010
s
= 470 years A1 [2]
(b) sample / activity would decay appreciably whilst measurements are being made B1 [1]

© UCLES 2010
Section B
9 (a) (i) fraction of the output (signal) is added to the input (signal) M1
out of phase by 180° / π rad / to inverting input A1 [2]
(ii) e.g. reduces gain
increases bandwidth
greater stability
reduces distortion
(b) (i) gain = 4.4 / 0.062
= 71 A1 [1]
(ii) 71 = 1 + 120/R C1
R = 1.7 × 103
Ω A1 [2]
(c) for the amplifier not to saturate B1
maximum output is (71 × 95 × 10–3
=) approximately 6.7 V M1
supply should be +/– 9 V A1 [3]
10 (a) (i) strain gauge B1 [1]
(ii) piezo-electric / quartz crystal / transducer B1 [1]
(b) circuit: coil of relay connected between sensing circuit output and earth B1
switch across terminals of external circuit B1
diode in series with coil with correct polarity for diode B1
second diode with correct polarity B1 [4]
11 either quartz or piezo-electric crystal B1
opposite faces /two sides coated (with silver) to act as electrodes B1
either molecular structure indicated
or centres of (+) and (–) charge not coincident B1
potential difference across crystal causes crystal to change shape B1
alternating voltage (in US frequency range) applied across crystal B1
causes crystal to oscillate / vibrate B1
(crystal cut) so that it vibrates at resonant frequency B1 [6]
(max 6)

© UCLES 2010
12 (a) signal becomes distorted / noisy B1
signal loses power / energy / intensity / is attenuated B1 [2]
(b) (i) either numbers involved are smaller / more manageable / cover wider range
or calculations involve addition & subtraction rather than multiplication and division
B1 [1]
(ii) 25 = 10 lg(Pmin / (6.1 × 10–19
)) C1
minimum signal power = 1.93 × 10–16
W C1
signal loss = 10 lg(6.5 × 10–3
)/(1.93 × 10–16
)
= 135 dB C1
maximum cable length = 135 / 1.6 C1
= 85 km so no repeaters necessary A1 [5]

9702 PHYSICS
examination.

© UCLES 2010
Section A
1 (a) work done moving unit mass M1
from infinity to the point A1 [2]
(b) (i) at R, φ = 6.3 × 107
J kg–1
(allow ± 0.1 × 107
) B1
φ = GM / R
6.3 × 107
= (6.67 × 10–11
× M) / (6.4 × 106
) C1
M = 6.0 × 1024
kg (allow 5.95 → 6.14) A1 [3]
Maximum of 2/3 for any value chosen for φ not at R
(ii) change in potential = 2.1 × 107
J kg–1
(allow ± 0.1 × 107
) C1
loss in potential energy = gain in kinetic energy B1
½ mv 2
= φ m or ½ mv 2
= GM / 3R C1
½ v 2
= 2.1 × 107
v = 6.5 × 103
m s–1
………..……(allow 6.3 → 6.6) A1 [4]
(answer 7.9 × 103
m s–1
, based on x = 2R, allow max 3 marks)
(iii) e.g. speed / velocity / acceleration would be greater B1
deviates / bends from straight path B1 [2]
(any sensible ideas, 1 each, max 2)
2 (a) (i) reduction in energy (of the oscillations) (B1)
reduction in amplitude / energy of oscillations (B1)
due to force (always) opposing motion / resistive forces (B1) [2]
any two of the above, max 2
(ii) amplitude is decreasing (very) gradually / oscillations would
continue (for a long time) /many oscillations M1
light damping A1 [2]
(b) (i) frequency = 1 / 0.3
= 3.3 Hz A1 [1]
allow points taken from time axis giving f = 3.45 Hz
(ii) energy = ½ mv 2
and v = ωa C1
= ½ × 0.065 × (2π/0.3)2
× (1.5 × 10–2
)2
M1
= 3.2 mJ A0 [2]
(c) amplitude reduces exponentially / does not decrease linearly M1
so will be not be 0.7 cm A1 [2]

© UCLES 2010
3 (a) (i) 1 deg C corresponds to (3840 – 190) / 100 Ω C1
for resistance 2300 Ω, temperature is 100 × (2300 – 3840) / (190 – 3840)
temperature is 42°C A1 [2]
(ii) either 286 K ≡ 13°C or 42°C ≡ 315 K B1
thermodynamic scale does not depend on the property of a substance M1
so change in resistance (of thermistor) with temperature is non-linear A1 [3]
(b) heat gained by ice in melting = 0.012 × 3.3 × 105
J C1
= 3960 J
heat lost by water = 0.095 × 4.2 × 103
× (28 – θ) C1
3960 + (0.012 × 4.2 × 103
× θ) = 0.095 × 4.2 × 103
× (28 – θ) C1
θ = 16°C A1 [4]
(answer 18°C – melted ice omitted – allow max 2 marks)
(use of (θ – T) then allow max 1 mark)
4 (a) force = q1q2 / 4πε0x2
C1
= (6.4 × 10–19
)2
/ (4π × 8.85 × 10–12
× {12 × 10–6
}2
) C1
= 2.56 × 10–17
N A1 [3]
(b) potential at P is same as potential at Q B1
work done = q∆V M1
∆V = 0 so zero work done A0 [2]
(c) at midpoint, potential is 2 × (6.4 × 10–19
) / (4πε0 × 6 × 10–6
) C1
at P, potential is (6.4 × 10–19
) / (4πε0 × 3 × 10–6
) + (6.4 × 10–19
) / (4πε0 × 9 × 10–6
) C1
change in potential = (6.4 × 10–19
) / (4πε0 × 9 × 10–6
)
energy = 1.6 × 10–19
× (6.4 × 10–19
) / (4πε0 × 9 × 10–6
) C1
= 1.0 × 10–22
J A1 [4]
5 (a) e.g. ‘storage of charge’ / storage of energy
blocking of direct current
producing of electrical oscillations
smoothing
(b) (i) capacitance of parallel combination = 60 µF C1
total capacitance = 20 µF A1 [2]
(ii) p.d. across parallel combination = ½ × p.d. across single capacitor C1
maximum is 9V A1 [2]
(c) either energy = ½CV 2
or energy = ½QV and Q = CV C1
energy = ½ × 4700 × 10–6
× (182
– 122
) C1
= 0.42 J A1 [3]

© UCLES 2010
6 (a) (i) straight line with positive gradient M1
through origin A1 [2]
(ii) maximum force shown at θ = 90° M1
zero force shown at θ = 0° M1
reasonable curve with F about ½ max at 30° A1 [3]
(b) (i) force on electron due to magnetic field B1
force on electron normal to magnetic field and direction of electron B1 [2]
(ii) quote / mention of (Fleming’s) left hand rule M1
electron moves towards QR A1 [2]
7 (a) either the value of steady / constant voltage M1
that produces same power (in a resistor) as the alternating voltage A1 [2]
or if alternating voltage is squared and averaged (M1)
the r.m.s. value is the square root of this averaged value (A1)
(b) (i) 220 V A1 [1]
(ii) 156 V A1 [1]
(iii) 60 Hz A1 [1]
(c) power = Vrms
2
/ R C1
R = 1562
/ 1500
= 16 Ω A1 [2]
8 (a) (i) number = (5.1 × 10–6
× 6.02 × 1023
) / 241 C1
= 1.27 × 1016
A1 [2]
(ii) A = λN C1
5.9 × 105
= λ × 1.27 × 1016
λ = 4.65 × 10–11
s–1
A1 [2]
(iii) 4.65 × 10–11
× t½ = ln2 C1
t½ = 1.49 × 1010
s
= 470 years A1 [2]
(b) sample / activity would decay appreciably whilst measurements are being made B1 [1]

© UCLES 2010
Section B
9 (a) (i) fraction of the output (signal) is added to the input (signal) M1
out of phase by 180° / π rad / to inverting input A1 [2]
(ii) e.g. reduces gain
increases bandwidth
greater stability
reduces distortion
(b) (i) gain = 4.4 / 0.062
= 71 A1 [1]
(ii) 71 = 1 + 120/R C1
R = 1.7 × 103
Ω A1 [2]
(c) for the amplifier not to saturate B1
maximum output is (71 × 95 × 10–3
=) approximately 6.7 V M1
supply should be +/– 9 V A1 [3]
10 (a) (i) strain gauge B1 [1]
(ii) piezo-electric / quartz crystal / transducer B1 [1]
(b) circuit: coil of relay connected between sensing circuit output and earth B1
switch across terminals of external circuit B1
diode in series with coil with correct polarity for diode B1
second diode with correct polarity B1 [4]
11 either quartz or piezo-electric crystal B1
opposite faces /two sides coated (with silver) to act as electrodes B1
either molecular structure indicated
or centres of (+) and (–) charge not coincident B1
potential difference across crystal causes crystal to change shape B1
alternating voltage (in US frequency range) applied across crystal B1
causes crystal to oscillate / vibrate B1
(crystal cut) so that it vibrates at resonant frequency B1 [6]
(max 6)

© UCLES 2010
12 (a) signal becomes distorted / noisy B1
signal loses power / energy / intensity / is attenuated B1 [2]
(b) (i) either numbers involved are smaller / more manageable / cover wider range
or calculations involve addition & subtraction rather than multiplication and division
B1 [1]
(ii) 25 = 10 lg(Pmin / (6.1 × 10–19
)) C1
minimum signal power = 1.93 × 10–16
W C1
signal loss = 10 lg(6.5 × 10–3
)/(1.93 × 10–16
)
= 135 dB C1
maximum cable length = 135 / 1.6 C1
= 85 km so no repeaters necessary A1 [5]

9702 PHYSICS
9702/51 Paper 5 (Planning, Analysis and Evaluation),
maximum raw mark 30
examination.

© UCLES 2010
1 Planning (15 marks)
Defining the problem (3 marks)
P1 Vary v and measure d, or v is the independent variable and d is the dependent
variable [1]
P2 Keep mass constant [1]
P3 Keep the wood constant/keep same type of nails [1]
Methods of data collection (5 marks)
M1 Diagram of apparatus showing mass falling onto centre of nail [1]
M2 Change height of falling mass (to change v) [1]
M3 Measurement(s) from which v can be determined, e.g. measure height fallen; light
gate(s) connected to timer/data-logger measuring time, and ticker tape/motion
sensor. Do not award stopwatch methods. [1]
M4 Appropriate equation to determine v (the velocity of the mass at the instant it hits the
nail) [1]
M5 Detail on measuring d ; subtract, needle, mark nail, depth gauge [1]
Method of analysis (2 marks)
A1 Plot a graph of log d against log v [1]
A2 n = gradient [1]
Safety considerations (1 mark)
S1 Precaution linked to falling masses, e.g. keep well away/sand trays [1]
Additional detail (4 marks)
D 1/2/3/4 Relevant points might include [4]
1. Method to create a large d, e.g. large mass, thin nails, soft wood
2. Use of a guide for falling mass/guide for nail
3. Use of vernier scale to measure d
4. Repeat experiment and determine an average
5. Use different part of wood for each test
6. Method to make nail vertical e.g. set square
7. Discussion / preliminary experiment about thin nails going totally into wood
8. lg d = n lg v + lg k
[Total: 15]

© UCLES 2010
2 Analysis, conclusions and evaluation (15 marks)
Part Mark Expected Answer Additional Guidance
(a) A1
Cπ2
1
Allow
CC
0.159
6.28
1
=
(b) T1
T2
4.55 330 or
333
4.00 290 or
294
3.33 240 or
238
2.86 210 or
208
2.50 180 or
179
2.22 160 or
161
T1 awarded for 1/f column; ignore rounding and sf
e.g. allow 4.54 or 4.544 or 4.545
T2 awarded for Xc column – must be values in table
U1 ± 20 (allow ± 17 or 18 or
19), decreasing to ± 10
(allow ± 7)
Allow one significant figure.
Do not allow ± 10 for 1st
row.
(c) (i) G1 Six points plotted
correctly
Must be within half a small square. Allow ecf from
table.
U2 Error bars in Xc plotted
correctly
Check first and last point. Must be accurate within half
a small square. All plots must have error bars.
(ii) G2 Line of best fit If points are plotted correctly then lower end of line
should pass between (2.0, 142) and (2.0, 148) and
upper end of line should pass between (4.85, 360) and
(4.95, 360). 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 at least half the length of
the drawn line. Check the read-offs. Work to half a
small square. Do not penalise POT.
U3 Error in gradient Method of determining absolute error.
Difference in worst gradient and gradient.

© UCLES 2010
(d) C2 C = 1/(2π × gradient)
= 0.159/gradient
Gradient must be used correctly.
Allow ecf from (c)(iii). Do not penalise POT.
If gradient within range given, then C in range
(2.08 – 2.21) × 10–6
U4 Method of determining
error in C
Uses worst gradient and finds difference.
Allow fractional error methods.
Do not check calculation.
C3 Consistent unit of C : F Penalise POT; allow s Ω–1
or Ω–1
Hz–1
. Should be
about 10–6
F.
Unit must be consistent with working.
(e) (i) C4 0.455 – 0.490 given to 3 sf
or 0.46 – 0.49 given to 2 sf
Answer must be in ranges given.
(ii) U5 Percentage uncertainty in
gradient + 10%
Expect to see similar calculation to above.
Allow using largest or smallest value 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) C [E4]
1. Uncertainty = C from gradient – C from worst acceptable line
2.
gradient
gradient∆
=
∆
C
C
(e) τ [E5]
1. Substitution method to find worst acceptable τ using either
largest C × 242 × 103
or smallest C × 198 × 103
Percentage uncertainty = 100×
∆
τ
τ
2. Percentage uncertainty = 10100
gradient
gradient
+×
∆
= 10100 +×
∆
C
C

9702 PHYSICS
maximum raw mark 30
examination.

© UCLES 2010
P1 x is the independent variable, B is the dependent variable or vary x and measure B [1]
P2 Keep the number of turns on the coil/radius of the coil constant [1]
Do not accept same coil – ‘coil’ is not a variable
P3 Keep the current (in the coil) constant [1]
M1 Diagram showing coil and Hall probe with a means of read out appropriately
positioned along axis [1]
M2 Coil connected to a power supply [1]
M3 Measure x with a ruler [1]
M4 Hall probe at right angles to direction of magnetic field or gives maximum output for
each reading [1]
M5 Method to determine axis of coil or to find x = 0 [1]
A1 Plot a graph of ln B against x [1]
A2 Relationship valid if a straight line is produced (ignore reference to y-intercept) [1]
S1 Precaution linked to (large) current in coil/heating, e.g. switch off when not in use to
avoid overheating coil; do not touch because it is hot [1]
1. Method to create a large magnetic field, e.g. use large current or large number
of turns.
2. Reasoned method to keep current constant.
3. Reasoned method to keep Hall probe in same orientation (e.g. use of set
square, fix to rule, optical bench or equivalent).
4. B is proportional to voltage across Hall probe/calibrate Hall probe (in a known
magnetic field).
5. Repeat experiment with Hall probe reversed or equivalent.
6. Identifies logarithmic equation i.e. ln B = –p x + ln B0
7. Avoid external magnetic fields.
8. Method to keep Hall probe along axis.
Do not allow vague computer methods.
[Total: 15]

© UCLES 2010
(a) A1
g
2
4π Allow
g
39.5
(b) T1 Column headings:
T / s and T 2
/ s2
There must be a dividing mark between the quantity
and the unit, i.e. “in”; “/”; (unit) e.g. T (s).
T2
3.57 or 3.572
3.20 or 3.204
2.79 or 2.789
2.40 or 2.403
1.99 or 1.988
1.59 or 1.588
Must be values in the table.
U1 From ± 0.04 to ± 0.02 or ±
0.03
Allow more than one significant figure, e.g. ± 0.038.
(c) (i) G1 Six points plotted correctly Must be within half a small square. Ecf allowed
from table.
U2 Error bars in T 2
plotted
correctly.
Check first and last point. Must be accurate within
half a small square. All plots must have error bars.
should pass between (37, 1.5) and (38, 1.5) and
upper end of line should pass between (92, 3.7)
and (94, 3.7). Allow ecf from points plotted
incorrectly – examiner judgement.
G3 Worst acceptable straight
line.
(iii) C1 Gradient of best fit line The triangle used should be at least half the length
of the drawn line. Check the read-offs. Work to
half a small square. Do not penalise POT.
(d) C2 g = 4π2
/gradient =
39.5/gradient
Allow ecf from (c)(iii).
U4 Determines uncertainty in g Uses worst calculated g value or fractional method.
C3 Consistent unit: cm s–2
or
m s–2
Penalise POT. Allow equivalent cm/s2
and m/s2

© UCLES 2010
(e) (i) C4 24.6 to 25.9 given to 3 sf
or 25 or 26 given to 2 sf.
Allow m, etc.
(ii) U5 Determines percentage
uncertainty in l
Check method; allow with or without consideration
of ∆T.
[Total: 15]
(d) [E4]
1. Uncertainty = g from gradient – g from worst acceptable line
2.
gradient
gradient∆
=
∆
g
g
(e) [E5]
1. Works out worst l then finds difference then uses 2
2. 100100
gradient
gradient
100 ×
∆
=×
∆
=×
∆
l
l
g
g
3. 100100
2
gradient
gradient
100
2
×
∆
=×




 ∆
+
∆
=×




 ∆
+
∆
l
l
T
T
T
T
g
g

9702 PHYSICS
maximum raw mark 30
examination.

© UCLES 2010
P1 x is the independent variable, B is the dependent variable or vary x and measure B [1]
P2 Keep the number of turns on the coil/radius of the coil constant [1]
Do not accept same coil – ‘coil’ is not a variable
P3 Keep the current (in the coil) constant [1]
M1 Diagram showing coil and Hall probe with a means of read out appropriately
positioned along axis [1]
M2 Coil connected to a power supply [1]
M3 Measure x with a ruler [1]
M4 Hall probe at right angles to direction of magnetic field or gives maximum output for
each reading [1]
M5 Method to determine axis of coil or to find x = 0 [1]
A1 Plot a graph of ln B against x [1]
A2 Relationship valid if a straight line is produced (ignore reference to y-intercept) [1]
S1 Precaution linked to (large) current in coil/heating, e.g. switch off when not in use to
avoid overheating coil; do not touch because it is hot [1]
1. Method to create a large magnetic field, e.g. use large current or large number
of turns.
2. Reasoned method to keep current constant.
3. Reasoned method to keep Hall probe in same orientation (e.g. use of set
square, fix to rule, optical bench or equivalent).
4. B is proportional to voltage across Hall probe/calibrate Hall probe (in a known
magnetic field).
5. Repeat experiment with Hall probe reversed or equivalent.
6. Identifies logarithmic equation i.e. ln B = –p x + ln B0
7. Avoid external magnetic fields.
8. Method to keep Hall probe along axis.
Do not allow vague computer methods.
[Total: 15]

© UCLES 2010
(a) A1
g
2
4π Allow
g
39.5
(b) T1 Column headings:
T / s and T 2
/ s2
There must be a dividing mark between the quantity
and the unit, i.e. “in”; “/”; (unit) e.g. T (s).
T2
3.57 or 3.572
3.20 or 3.204
2.79 or 2.789
2.40 or 2.403
1.99 or 1.988
1.59 or 1.588
Must be values in the table.
U1 From ± 0.04 to ± 0.02 or ±
0.03
Allow more than one significant figure, e.g. ± 0.038.
(c) (i) G1 Six points plotted correctly Must be within half a small square. Ecf allowed
from table.
U2 Error bars in T 2
plotted
correctly.
Check first and last point. Must be accurate within
half a small square. All plots must have error bars.
should pass between (37, 1.5) and (38, 1.5) and
upper end of line should pass between (92, 3.7)
and (94, 3.7). Allow ecf from points plotted
incorrectly – examiner judgement.
G3 Worst acceptable straight
line.
(iii) C1 Gradient of best fit line The triangle used should be at least half the length
of the drawn line. Check the read-offs. Work to
half a small square. Do not penalise POT.
(d) C2 g = 4π2
/gradient =
39.5/gradient
Allow ecf from (c)(iii).
U4 Determines uncertainty in g Uses worst calculated g value or fractional method.
C3 Consistent unit: cm s–2
or
m s–2
Penalise POT. Allow equivalent cm/s2
and m/s2

© UCLES 2010
(e) (i) C4 24.6 to 25.9 given to 3 sf
or 25 or 26 given to 2 sf.
Allow m, etc.
(ii) U5 Determines percentage
uncertainty in l
Check method; allow with or without consideration
of ∆T.
[Total: 15]
(d) [E4]
1. Uncertainty = g from gradient – g from worst acceptable line
2.
gradient
gradient∆
=
∆
g
g
(e) [E5]
1. Works out worst l then finds difference then uses 2
2. 100100
gradient
gradient
100 ×
∆
=×
∆
=×
∆
l
l
g
g
3. 100100
2
gradient
gradient
100
2
×
∆
=×




 ∆
+
∆
=×




 ∆
+
∆
l
l
T
T
T
T
g
g

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