9702 s16 ms_all

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Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/11
Paper 1 Multiple Choice May/June 2016
MARK SCHEME
Maximum Mark: 40
Published
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.
Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for
Teachers.
Cambridge will not enter into discussions about these mark schemes.
Cambridge is publishing the mark schemes for the May/June 2016 series for most Cambridge IGCSE®
,
Cambridge International A and AS Level components and some Cambridge O Level components.

Mark Scheme Syllabus Paper
Cambridge International AS/A Level – May/June 2016 9702 11
© Cambridge International Examinations 2016
Question
Number
Key
Question
Number
Key
1 B 21 B
2 C 22 A
3 C 23 B
4 A 24 A
5 D 25 D
6 B 26 D
7 A 27 D
8 D 28 A
9 A 29 D
10 D 30 C
11 C 31 D
12 A 32 C
13 B 33 D
14 B 34 A
15 C 35 A
16 C 36 B
17 D 37 B
18 A 38 A
19 D 39 D
20 C 40 A

PHYSICS 9702/12
MARK SCHEME
Maximum Mark: 40
Published
Teachers.
,

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

PHYSICS 9702/13
MARK SCHEME
Maximum Mark: 40
Published
Teachers.
,

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

PHYSICS 9702/21
Paper 2 AS Level Structured Questions May/June 2016
MARK SCHEME
Maximum Mark: 60
Published
Teachers.
,

1 (a) (i) (50 to 200) × 10–3
kg or (0.05 to 0.2)kg B1 [1]
(ii) (50 to 300)cm3
B1 [1]
(b) density = mass/volume or ρ = M/V C1
V = [π(0.38 × 10–3
)2
× 25.0 × 10–2
]/4 (= 2.835 × 10–8
m3
) C1
ρ =(0.225 × 10–3
)/2.835 × 10–8
=7940(kgm–3
) A1
∆ρ/ρ = 2(0.01/0.38) + (0.1/25.0) + (0.001/0.225) [= 0.061]
or
%ρ = 5.3% + 0.40% + 0.44% (= 6.1%) C1
∆ρ = 0.061 × 7940 = 480 (kgm–3
)
density = (7.9 ± 0.5) × 103
kgm–3
or (7900 ± 500)kgm–3
A1 [5]
2 (a) (i) horizontal component (= 12cos50°) = 7.7ms–1
A1 [1]
(ii) vertical component (= 12sin50° or 7.7tan50°) = 9.2ms–1
A1 [1]
(b) v2
= u2
+ 2as and v = 0 or mgh = ½mv2
or s = v2
sin2
θ/2g C1
9.22
= 2 × 9.81 × h hence h = 4.3 (4.31) m A1 [2]
alternative methods using time to maximum height of 0.94s:
s = ut + ½at2
and t = 0.94(s) (C1)
s = 9.2 × 0.94 – ½ × 9.81 × 0.942
hence s = 4.3m (A1)
or
s = vt – ½at2
and t = 0.94(s) (C1)
s = ½ × 9.81 × 0.942
hence s = 4.3m (A1)
or
s = ½(u + v)t and t = 0.94(s) (C1)
s = ½ × 9.2 × 0.94 hence s = 4.3m (A1)
(c) t (= 9.2/9.81)= 0.94 (0.938)s C1
horizontal distance = 0.938 × 7.7 (= 7.23m) C1
displacement = [4.32
+ 7.232
]1/2
C1
= 8.4m A1 [4]

3 (a) (i) force (= mg = 0.15 × 9.81) = 1.5 (1.47) N A1 [1]
(ii) resultant force (on ball) is zero so normal contact force = weight
or
the forces are in opposite directions so normal contact force = weight
or
normal contact force up = weight down A1 [1]
(b) (i) (resultant) force proportional/equal to rate of change of momentum B1 [1]
(ii) change in momentum = 0.15 × (6.2 + 2.5) (= 1.305 Ns) C1
magnitude of force = 1.305/0.12
= 11 (10.9) N A1
or
(average) acceleration = (6.2 + 2.5) / 0.12 (= 72.5 ms–2
) (C1)
magnitude of force = 0.15 × 72.5
=11 (10.9)N (A1)
(direction of force is) upwards/up B1 [3]
(iii) there is a change/gain in momentum of the floor M1
this is equal (and opposite) to the change/loss in momentum of the ball so
momentum is conserved A1 [2]
or
change of (total) momentum of ball and floor is zero (M1)
so momentum is conserved (A1)
or
(total) momentum of ball and floor before is equal to the (total) momentum
of ball and floor after (M1)
so momentum is conserved (A1)

4 (a) the energy (stored) in a body due to its extension/compression/deformation/
change in shape/size B1 [1]
(b) (i) two values of F/x are calculated which are the same
e.g. 10.4/40 = 0.26 and 6.5/25 = 0.26 B1
or
ratio of two forces and the ratio of the corresponding two extensions are
calculated which are the same
e.g. 5.2/10.4 = 0.5 and 20/40 = 0.5 (B1)
or
gradient of graph line calculated and coordinates of one point on the
line used with straight line equation y = mx + c to show c = 0 (B1)
(so) force is proportional to extension (and so Hooke’s law obeyed) B1 [2]
(b) (ii) 1. k = F/x or k = gradient C1
gradient or values from a single point used e.g. k = 10.4/(40 × 10–2
)
k = 26Nm–1
A1 [2]
2. work done = area under graph
or ½Fx or ½(F2 + F1)(x2 – x1)
or ½kx2
or ½k(x2
2
– x1
2
) C1
= ½ × 10.4 × 0.4 – ½ × 5.2 × 0.2 C1
or ½ × (5.2 + 10.4) × 20 × 10–2
or ½ × 26 × (0.42
− 0.22
)
= 1.6 J A1 [3]
(c) remove the force and the spring goes back to its original length B1 [1]
5 (a) T = 4(ms) or 4 × 10–3
(s) C1
f = 1/T = 1/0.004
= 250Hz A1 [2]
(b) intensity ∝ (amplitude)2
and amplitude = 2.8 (2.83)(cm) B1
curve with same period and with amplitude 2.8cm B1
curve shifted 1.0ms to left or to right of wave X B1 [3]

(c) (i) gradient = (4.5 – 2.4) × 10–3
/(3.25 – 1.75) [= 1.4 × 10–3
] B1
wavelength = 0.45 × 10–3
× 1.4 × 10–3
C1
= 6.30 × 10–7
(m) C1
= 630nm A1 [4]
(ii) (gradient is equal to λ/a therefore) gradient of line is reduced B1
value of x will be reduced for all values of D
or new line is completely below old line
or intercept is less B1 [2]
6 (a) (coulomb is) ampere second B1 [1]
(b) (total) charge or Q = nAle M1
I = Q/t and l /t = v M1
I = nAle/t = nAve therefore v = I/nAe A1 [3]
(c) (i) ratio = (I/nAYe)/(I /nAZe) C1
= AZ/AY or 4A/A or πd2
/(πd2
/4) C1
= 4 A1 [3]
(ii) R = ρl/A or R = 4ρl/πd2
B1
RY = ρl/A and RZ = ρ(2l)/4A so RY /RZ = 2
or
RY = 4ρl /πd2
and RZ =4ρ(2l) /π4d2
or 2ρl /πd2
so RY /RZ = 2 A1 [2]
(iii) V = 12RY /(RY + RZ) or I = 12/(RY + RZ) and V = IRY C1
V = 12 × 2/3
= 8(.0)V A1 [2]
(iv) ratio = I2
RY /I2
RZ or (VY
2
/RY)/(VZ
2
/RZ) or (VYI)/(VZI)
= 2 A1 [1]

7 (a) hadron: neutron/proton
and
lepton: electron/(electron) neutrino B1 [1]
(allow other correct particles)
(b) (i) proton: up up down or uud B1 [1]
(ii) neutron: up down down or udd B1 [1]
(c) (i) neutron → proton + electron + (electron) antineutrino B1 [1]
(ii) up down down (quarks) change to up up down (quarks)
or
down (quark) changes to up (quark) B1 [1]

PHYSICS 9702/22
MARK SCHEME
Maximum Mark: 60
Published
Teachers.
,

1 (a) acceleration = change in velocity / time (taken) or rate of change of velocity B1 [1]
(b) (i) v = 0 + at or v = at C1
(a = 36/19 =) 1.9 (1.8947) ms–2
A1 [2]
(ii) s = ½(u + v)t or s = v2
/2a or s = ½at2
= ½ × 36 × 19 = 362
/(2 × 1.89) = ½ × 1.89 × 192
= 340 m (342 m/343 m/341 m) M1 [1]
(iii) 1. (∆KE =) ½ × 95 × (36)2
C1
= 62000 (61560) J A1 [2]
2. (∆PE =) 95 × 9.81 × 340 sin 40° or 95 × 9.81 × 218.5 C1
= 200000 J A1 [2]
(iv) work done (by frictional force) = ∆PE – ∆KE
or
work done = 200000 – 62000 (values from 1b(iii) 1. and 2.) C1
(frictional force = 138000/340 =) 410 (406) N [420 N if full figures used] A1 [2]
(v) –ma = mg sin 20° – f or ma = –mg sin 20° + f C1
–95 × 3.0 = 95 × 3.36 – f
f = 600 (604) N A1 [2]
2 (a) p = F/A M1
use of m = ρV and use of V = Ah and use of F = mg M1
correct substitution to obtain p = ρgh A1 [3]
(b) (i) (when h is zero the pressure is not zero due to) pressure from the
air/atmosphere B1 [1]
(ii) gradient = ρg or P – 1.0 × 105
= ρgh C1
e.g. ρg = 1.0 × 105
/0.75 (= 133333)
ρ = 133333/9.81
= 14000 (13592) kgm–3
A1 [2]

3 (a) Young modulus = stress/strain B1 [1]
(b) (i) E = (F × l)/(A × e) or e = (F × l)/(A × E) B1
e ∝ 1/E
or
ratio eC /eS = ES /EC or (1.9 × 1011
)/(1.2 × 1011
) or 19/12 C1
(ratio =) 1.6 (1.58) A1 [3]
(ii) two straight lines from (0,0) with S having the steepest gradient B1 [1]
4 (a) longitudinal: vibrations/oscillations (of the particles/wave) are parallel to the
direction or in the same direction (of the propagation of energy) B1
transverse: vibrations/oscillations (of the particles/wave) are perpendicular to
the direction (of the propagation of energy) B1 [2]
(b) LHS: intensity = power/area units: kgms–2
× m × s–1
× m–2
or kgm2
s–3
× m–2
B1
RHS: units: ms–1
× kgm–3
× s–2
× m2
M1
LHS and RHS both kgs–3
A1 [3]
(c) (i) change/difference in the observed/apparent frequency when the source is
moving (relative to the observer) B1 [1]
(ii) wavelength increases/frequency decreases/red shift B1 [1]
(d) observed frequency = vfS /(v – vS) C1
550 = (340 × 510)/(340 – vS) C1
vS = 25 (24.7) ms–1
A1 [3]
5 (a) diffraction: spreading/diverging of waves/light (takes place) at (each) slit/
element/gap/aperture B1
interference: overlapping of waves (from coherent sources at each element) B1
path difference λ/phase difference of 360(°)/2π (produces the first order) B1 [3]
(b) d sinθ = nλ or sinθ = Nnλ C1
d = (2 × 486 × 10–9
)/sin 29.7° (= 1.962 × 10–6
) C1
number of lines = 510 (509.7) mm–1
A1 [3]

6 (a) at least six horizontal lines equally spaced and arrow to the right B1 [1]
(b) charge used 2e C1
gain in KE = 15 × 1.6 × 10–19
× 103
= 2 × 1.6 × 10–19
× V (p.d.across plates)
or
F (= W/d) = 15 × 1.6 × 10–19
× 103
/16 × 10–3
C1
(hence V = 7500 V or F = 1.5 × 10–13
N)
E = V/d or E = F/Q C1
E = (7500/16 × 10–3
) or E = (1.5 × 10–13
/3.2 × 10–19
)
E = 4.7 × 105
(468750) Vm–1
A1 [4]
or
KE (= ½mv2
) = 15 × 103
× 1.6 × 10–19
v = [(2 × 15 × 103
× 1.6 × 10–19
)/(6.68 × 10–27
)]1/2
= 8.5 × 105
ms–1
(C1)
a (= v2
/2s) = (8.5 × 105
)2
/2 × 16 × 10–3
= 2.25 × 1013
ms–2
F (= 6.68 × 10–27
× 2.25 × 10–13
) = 1.5 × 10–13
N
E = F/Q (C1)
Q = 2e (C1)
E = 4.7 × 105
Vm–1
(A1)

7 (a) charge exists only in discrete amounts B1 [1]
(b) (i) E = I(R + r) or V = IR C1
(total resistance =) 2.7 + 0.30 + 0.25 (= 3.25 Ω) M1
I = 9.0/(2.7 + 0.30 + 0.25) or 9.0/3.25 = 2.8 A A1 [3]
(ii) V = IRext C1
= 2.77 × 3.0 or 2.8 × 3.0
or
V = E – Ir (C1)
= 9.0 – 2.77 × 0.25 or 9.0 – 2.8 × 0.25
V = 8.3 (8.31) V or 8.4 V A1 [2]
(c) (i) I = nevA
v = 2.77/(8.5 × 1029
× 1.6 × 10–19
× 2.5 × 10–6
) M1
= 8.1 (8.147) × 10–6
ms–1
or 8.2 × 10–6
ms–1
A1 [2]
(ii) A reduces by a factor 4 (1/4 less) or resistance of Z goes up by 4× M1
current goes down but by less than a factor of 4 (as total resistance
does not go up by a factor of 4) so drift speed goes up A1 [2]
8 (a) both electron and neutrino: lepton(s) B1
both neutron and proton: hadron(s)/baryon(s) B1 [2]
(b) (i) νβ 0
0
0
1
1
0
1
1 np ++→
correct symbols for particles M1
correct numerical values (allow no values on neutrino) A1 [2]
(ii) up up down or uud → up down down or udd B1 [1]
(iii) weak (nuclear) B1 [1]

PHYSICS 9702/23
MARK SCHEME
Maximum Mark: 60
Published
Teachers.
,

1 (a) scalars: energy, power and time A1
vectors: momentum and weight A1 [2]
(b) (i) triangle with right angles between 120m and 80m, arrows in correct direction
and result displacement from start to finish arrow in correct direction and
labelled R B1 [1]
(ii) 1. average speed (= 200/27) = 7.4ms–1
A1 [1]
2. resultant displacement (= [1202
+ 802
]1/2
) = 144 (m) C1
average velocity (= 144/27) = 5.3(3)ms–1
A1
direction (= tan–1
80/120) = 34° (33.7) A1 [3]
2 (a) systematic: the reading is larger or smaller than (or varying from) the true reading
by a constant amount B1
random: scatter in readings about the true reading B1 [2]
(b) precision: the size of the smallest division (on the measuring instrument)
or
0.01mm for the micrometer B1
accuracy: how close (diameter) value is to the true (diameter) value B1 [2]
3 (a) (gravitational potential energy is) the energy/ability to do work of a mass that it
has or is stored due to its position/height in a gravitational field B1
kinetic energy is energy/ability to do work a object/body/mass has due to its
speed/velocity/motion/movement B1 [2]
(b) (i) s = [(u + v)t]/2 or acceleration = 9.8/9.75 (using gradient) C1
= [(7.8 + 3.9) × 0.4]/2 or s = 3.9 × 0.4 + 1
2 × 9.75 × (0.4)2
C1
s = 2.3(4)m A1 [3]
(ii) a = (v – u)/t or gradient of line C1
= (7.8 – 3.9)/0.4 = 9.8 (9.75) ms–2
(allow ± 1
2
small square in readings) A1 [2]

(iii) KE = 1
2
mv2
C1
change in kinetic energy = 1
2
mv2
– 1
2
mu2
= 1
2
× 1.5 × (7.82
– 3.92
) C1
= 34 (34.22) J A1 [3]
(c) work done = force × distance (moved) or Fd or Fx or mgh or mgd or mgx M1
= 1.5 × 9.8 × 2.3 = 34 (33.8) J (equals the change in KE) A1 [2]
4 (a) (resultant force = 0) (equilibrium)
therefore: weight – upthrust = force from thin wire (allow tension in wire)
or
5.3 (N) – upthrust = 4.8 (N) B1 [1]
(b) difference in weight = upthrust or upthrust = 0.5 (N)
0.5 = ρghA or m = 0.5/9.81 and V = 5.0 × 13 × 10–6
(m3
) C1
ρ = 0.5/(9.81 × 5.0 × 13 × 10–6
) C1
= 780 (784) kgm–3
A1 [3]
5 (a) the total momentum of a system (of colliding particles) remains constant M1
provided there is no resultant external force acting on the system/isolated or
closed system A1 [2]
(b) (i) the total kinetic energy before (the collision) is equal to the total kinetic
energy after (the collision) B1 [1]
(ii) p (= mv = 1.67 × 10–27
× 500) = 8.4 (8.35) × 10–25
Ns A1 [1]
(iii) 1. mvA cos60° + mvB cos30° or m(vA
2
+ vB
2
)1/2
B1
2. mvA sin60° + mvB sin30° B1 [2]
(iv) 8.35 × 10–25
or 500m = mvA cos60° + mvB cos30°
and
0 = mvA sin60° + mvB sin30°
or using a vector triangle C1
vA = 250ms–1
A1
vB = 430 (433)ms–1
A1 [3]

6 (a) ohm is volt per ampere or volt/ampere B1 [1]
(b) (i) R = ρl/A B1
RP = 4ρ(2l)/πd2
or 8ρl/πd2
or RQ = ρl/πd2
or
ratio idea e.g. length is halved hence R halved and diameter is halved hence
R is 1/4 C1
RQ (= 4ρl/π4d2
) = ρl/πd2
= RP /8
(= 12/8) = 1.5 Ω A1 [3]
(ii) power = I2
R or V2
/R or VI C1
= (1.25)2
× 12 + (10)2
× 1.5 or (15)2
/12 + (15)2
/1.5 or 15 × 11.25 C1
= (18.75 + 150 =) 170 (168.75) W A1 [3]
(iii) IP = (15/12 =) 1.25 (A) and IQ = (15/1.5 =) 10 (A) C1
vP/vQ = IPnAQe/IQnAPe or (1.25 × πd2
)/(10 × πd2
/4) C1
= 0.5 A1 [3]
7 (a) (i) alter distance from vibrator to pulley
alter frequency of generator
(change tension in string by) changing value of the masses
any two B2 [2]
(ii) points on string have amplitudes varying from maximum to zero/minimum B1 [1]
(b) (i) 60° or π/3 rad A1 [1]
(ii) ratio = [3.4/2.2]2
C1
= 2.4 (2.39) A1 [2]

8 (a) α-particle is 2 protons and 2 neutrons; β+
-particle is positive electron/positron
α-particle has charge +2e; β+
-particle has +e charge
α-particle has mass 4u; β-particle has mass (1/2000)u
α-particle made up of hadrons; β+
-particle a lepton
any three B3 [3]
(b) νβ 0
0
0
1
1
0
1
1 np ++→
all terms correct M1
all numerical values correct (ignore missing values on ν) A1 [2]
(c) (i) 1. proton: up, up, down/uud B1
2. neutron: up, down, down/udd B1 [2]
(ii) up quark has charge +2/3 (e) and down quark has charge –1/3 (e)
total is +1(e) B1 [1]

PHYSICS 9702/31
Paper 3 Advanced Practical Skills 1 May/June 2016
MARK SCHEME
Maximum Mark: 40
Published
Teachers.
,

1 (a) (ii) Value of x with consistent unit and in the range 36.0cm to 39.0cm. [1]
(c) Value of T with unit in range 0.300s < T < 1.00s. [1]
(d) Five sets of readings of x and time with correct trend scores 5 marks, four sets scores
4 marks etc. [5]
Help from Supervisor –1.
Range: [1]
xmax – xmin ≥ 30cm.
Column headings: [1]
Each column heading must contain a quantity and an appropriate unit.
The presentation of quantity and unit must conform to accepted scientific convention
e.g. x/m or x (cm), T2
/s2
or T2
(s2
).
Consistency: [1]
All values of x must be given to the nearest mm.
Significant figures: [1]
Every value of T2
must be given to the same number of s.f. as (or one more than)
the number of s.f. in the corresponding raw values of time.
Calculation: [1]
T2
calculated correctly to the number of s.f. given by the candidate.
(e) (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.
Scales must be labelled with the quantity that is being plotted.
Scale markings should be no more than three large squares apart.
Plotting of points: [1]
All observations in the table must be plotted.
Diameter of plotted points must be ≤ half a small square (no “blobs”).
Points must be plotted to an accuracy of half a small square.
Quality: [1]
All points in the table must be plotted on the grid (at least 5) for this mark to be
awarded.
All points must be no more than ±5cm (to scale) in the x direction from a
straight line.

(ii) Line of best fit: [1]
Judge by balance of all points on the grid about the candidate’s line (at least 4
points). There must be an even distribution of points either side of the line along
the full length.
Allow one anomalous point only if clearly indicated by the candidate.
Lines must not be kinked or thicker than half a square.
(iii) Gradient: [1]
The hypotenuse of the triangle must be greater than half the length of the
drawn line.
The method of calculation must be correct. Do not allow ∆x/∆y.
Sign of gradient must match graph drawn.
Both read-offs must be accurate to half a small square in both the x and y
directions.
y-intercept: [1]
Either:
Correct read-off from a point on the line substituted into y = mx + c or an
equivalent expression.
Read-offs must be accurate to half a small square in both x and y directions.
Or:
Intercept read directly from the graph (accurate to half a small square).
(f) Value of P = candidate’s gradient and value of Q = candidate’s intercept. [1]
Do not allow fractions.
Unit for P correct (e.g. s2
m–1
) and unit for Q correct (e.g. s2
). [1]
2 (a) All values of w to nearest mm, with final value in range 0.200m to 0.300m. [1]
(b) (ii) Value(s) of θ to the nearest degree. Final value < 45° with unit. [1]
(iii) Percentage uncertainty in θ based on an absolute uncertainty in range 1° to 3°.
If repeated readings have been taken, then the uncertainty can be half the range
(but not zero) if the working is clearly shown.
Correct method of calculation to obtain percentage uncertainty. [1]
(c) (iii) Value of y in range 0.20m to 0.50m. [1]
Evidence of repeat readings. [1]
(d) (i) Correct calculation of D. [1]
(ii) Correct justification for s.f. in D linked to s.f. in w and y. [1]

(e) (ii) Second value of θ. [1]
Second value of y. [1]
Quality: Second value of y > first value of y (if θ2 > θ1). [1]
(f) (i) Two values of k calculated correctly. [1]
(ii) Sensible comment relating to the calculated values of k, testing against a criterion
specified by the candidate. [1]
(g) (i) Limitations [4] (ii) Improvements [4] Do not credit
A Two readings not enough to draw
a conclusion
Take more readings and plot a
graph/
obtain more k values and
compare
Few readings/
only one reading/
not accurate result/
“repeat readings”
on its own
B Difficulty with starting position e.g.
starting position of container not
parallel to edge
Improved method for initial
placement or release
e.g. use of block with detail
(aligned with side)
Change shape of
container
C Small range of angles
possible/cylinder slips when angle
too high/cylinder moves off wrong
edge on board
Workable method to increase
friction e.g. sheet of paper on
board/
sanding board/
use a rougher board/
roughen edge of container
“No friction” on its
own/
smoother board/
use longer board
D Difficult to measure y or locate
position where container moves
off edge with reason e.g. moves
off edge too fast/short time to
observe moving off edge
Improved method for measuring
y e.g. use marker or scale on
board/use video and playback
with scale/
paint on lid/
calibrate board
Effects of moving
air/fans/
“frame by frame”
E Difficulty with set up e.g. board
moves/clamp moves/board not an
even height across width/board
not aligned correctly
Method to improve stability of
board e.g. use two clamps/
Blu-Tack to fix board to bench/
spirit level across board/
support board on long blocks
G-clamp to bench
F Difficult to release without
applying force
Improved method of release e.g.
card gate/block with detail of
removal
Electromagnetic
release

PHYSICS 9702/32
MARK SCHEME
Maximum Mark: 40
Published
Teachers.
,

1 (b) (i) Value for θ in range 20° to 30° to nearest degree, with unit. [1]
(iv) Value for T in range 0.50s to 1.50s. [1]
Evidence of repeat readings. At least two measurements of nT, with n ≥ 3. [1]
(d) Six sets of values for θ and T with correct trend scores 4 marks, five sets scores
3 marks etc. No θ values over 90°. [4]
Help from Supervisor –1.
Range: [1]
θ values must include 30° or less and 70° or more.
e.g. θ/°, θ (°) or θ (deg) etc. θtan1/ must have no unit.
Consistency: [1]
All raw values of time must be given to the nearest 0.1s, or all to the nearest 0.01s.
Every value of θtan1/ must be given to 2 or 3 s.f.
Calculation: [1]
Values of θtan1/ calculated correctly to the number of s.f. given by the candidate.
(e) (i) Axes: [1]
Scales must be chosen so that the plotted points occupy at least half the graph grid
in both x and y directions.
Scale markings must be no more than three large squares apart.
All observations must be plotted.
Plotted points must be accurate to half a small square.
Quality: [1]
All points in the table (at least 5) must be plotted for this mark to be awarded.
All points must be no more than ±0.02s (in the y (T) direction) of a straight line.
Judge by balance of all points on the grid about the candidate's line (at least 5
points). There must be an even distribution of points either side of the line along the
full length.
Allow one anomalous plot only if clearly indicated by the candidate.
Line must not be kinked or thicker than half a small square.

(iii) Gradient: [1]
Sign of gradient must match graph drawn.
The hypotenuse of the triangle used must be greater than half the length of the
drawn line.
The method of calculation must be correct.
Both read-offs must be accurate to half a small square in both x and y directions.
y-intercept: [1]
Either:
Correct read-off from a point on the line and substituted into y = mx + c.
Or:
Intercept read off directly from the graph (accurate to half a small square).
(f) Value of p = candidate's gradient and value of q = candidate's intercept. [1]
Correct units for p and q (both should have the unit s). [1]
2 (a) (i) h to nearest mm and in range 2.5cm to 3.5cm. [1]
(ii) Raw values for d to nearest mm. [1]
(b) Absolute uncertainty in d in range 2mm to 5mm.
(c) (i) Value for t in range 20.0s to 90.0s, with unit. [1]
Evidence of repeat measurements of t. [1]
(ii) Correct calculation of R to the s.f. used by the candidate (must be 2 or more s.f.). [1]
(d) (ii) Value for x1 to nearest mm, with unit. [1]
(e) Second values of h and d and t. [1]
Second values of x1 and x2. [1]
Quality: (x2 – x1) greater for shorter t. [1]

A Two readings are not enough
to draw a conclusion
Take more readings and plot
graph/
compare
“Repeat readings” on its
own/few readings/only
one reading/take more
readings and find
average k
B Parallax error when
measuring d
Measure on bench between
two set squares/
use (vernier) calipers/
use string to find
circumference then calculate d
C Bottle distorts when
measuring d/
d varies along bottle/
base of bottle not flat
Collect water lost between
marks and measure volume
D Difficult to judge/see/operate
stopwatch when water level
reaches mark
Use video with timer in
view/
use frame counting/
use coloured water
Reaction time
Light gates
E Wooden strip moves
continuously when water is
falling on it
Use video with scale in view
F Difficult to measure height
because rule not vertical/rule
touches strip
Use set square on bench/
clamp rule
Only short time to
measure x
G Water soaks into wooden
strip/
water stays on wooden strip
Use waterproof strip Use new strip/
dry the strip
H R not constant between lines Move lines closer to top/
have lines closer together

PHYSICS 9702/33
MARK SCHEME
Maximum Mark: 40
Published
Teachers.
,

1 (b) (ii) Value for y with unit in range 2.0 ≤ y ≤ 8.0cm. [1]
(iii) Raw values of θ to the nearest degree.
Value of θ in the range 40° to 50°. [1]
(d) Six sets of readings of m, y and θ with correct trend scores 5 marks, five sets scores
4 marks etc. [5]
Help from supervisor –1.
Range: [1]
Range of values to include m ≤ 150g and m ≥ 400g.
Each column heading must contain a quantity and a unit where appropriate.
The unit must conform to accepted scientific convention, e.g. msinθ /g or θ (°).
Consistency: [1]
All values of y must be given to the nearest mm only.
Every value of msinθ must be given to 2 or 3 s.f.
Calculation: [1]
Values of msinθ calculated correctly to the number of s.f. given by the candidate.
(e) (i) Axes: [1]
Quality: [1]
All points in the table (at least 5) must be plotted on the grid for this mark to be
awarded.
All points must be within ±0.25cm in the y direction of a straight line.
points). There must be an even distribution of points either side of the line along the
full length.
Lines must not be kinked or thicker than half a square.

(iii) Gradient: [1]
The hypotenuse of the triangle must be greater than half of the length of the
drawn line.
Both read-offs must be accurate to half a small square in both the x and y directions.
y-intercept: [1]
Either:
Or:
(f) Value of P = candidate’s gradient and value of Q = candidate’s intercept. [1]
Unit for P correct (mkg–1
or cmkg–1
or mmkg–1
or mg–1
or cmg–1
or mmg–1
)
and consistent with value.
Unit for Q correct (m or cm or mm) and consistent with value. [1]
2 (a) (ii) All raw values of d either to the nearest 0.01 or 0.001mm with unit and in the
range 0.250mm to 0.450mm. [1]
(iii) Correct calculation of A with consistent unit and power of ten. [1]
(b) (iii) Value of L with appropriate unit in range 10.0 cm ≤ L ≤ 20.0cm. [1]
(iv) Percentage uncertainty in L based on absolute uncertainty of 2mm to 8mm.
(c) (i) Correct calculation of C to the s.f. given by the candidate. [1]
(ii) Correct justification for s.f. in C linked to s.f. in d and L. [1]
(d) (ii) Raw values for time to the nearest 0.1s or better.
T with unit and in range 0.5s ≤ T ≤ 2.0s. [1]
(e) (ii) Second values of d and L. [1]
Second value of T. [1]
Quality: If d1 > d2 then second value of T > first value of T. [1]

A Two readings not enough to
draw a conclusion
Take many readings and plot
a graph/
compare
“Repeat readings” on its
own/few readings/only one
reading/not enough
readings for accurate
value
B Difficult to judge beginning
and/or end of a cycle/a complete
cycle
Draw a line/mark on the
mass/
(fiducial) marker at
equilibrium position
C Wire not straight/kinked Method of straightening wire
e.g. use larger mass
D Difficult to measure L with
reason
e.g. metre rule awkward to
position/parallax error
Improved method of
measuring L
e.g. marking L before putting
into clip/
detailed method using set
squares or ruler/
use a length guide (e.g.
15cm wood)/
use string with detail/
use tape measure
Vernier calipers on its
own/
set square on its own/
30cm ruler on its own
E Wire slips (in clip) Better method of gripping
wire
e.g. wrap wire around clamp/
use two wooden blocks and
wire
Any reference to attaching
the mass to the wire
F Mass swings as well as rotates/
clip moves around rod/
there is a force on release
Better method of attaching
clip to rod e.g. glue
G Shorter/thicker wire has too few
cycles/dampens quickly/
(percentage) uncertainty greater
for shorter/thicker wire
Video and timer/replay frame
by frame
Repeats
Longer wire

PHYSICS 9702/34
MARK SCHEME
Maximum Mark: 40
Published
Teachers.
,

1 (b) (ii) 0.9VS calculated correctly and to the same number of s.f. as, or one more than,
the s.f. of VS in (b)(i). [1]
(c) (ii) Value for t in range 1.0s to 9.0s. [1]
(d) (ii) Six sets of values for VC and t with correct trend scores 5 marks, five sets scores
4 marks etc. [5]
Minor help from supervisor –1, major help from supervisor –2.
Range: [1]
Range of values to include VC ≤ 3.0 V and VC ≥ 8.0 V.
e.g. VC /V or VC (V).
Consistency: [1]
All values of t must be given to the nearest 0.1s, or all to the nearest 0.01s.
(e) (i) Axes: [1]
Scale markings must be no more than three large squares apart.
All observations in the table must be plotted on the grid.
Judge by balance of all points on the grid about the candidate's curve (at least 5
points). There must be an even distribution of points either side of the curve along
the full length.
Line must not be kinked or thicker than half a small square.
(f) (ii) Tangent drawn at VC = 0.5VS.
Tangent must touch curve at the candidate’s value of 0.5VS from (f)(i). [1]

(iii) Gradient: [1]
The hypotenuse of the triangle used must be greater than half the length of the
drawn line.
Both read-offs must be accurate to half a small square in both x and y directions.
y-intercept: [1]
Either:
Correct read-off from a point on the tangent is substituted into y = mx + c.
Or:
(g) Value of a = candidate’s gradient and value of b = candidate’s intercept. [1]
Correct units for a (e.g. Vs–1
) and b (s). [1]
(h) Correct calculation of T. [1]
Quality: T in the range 8.0s to 14.0s, with consistent unit. [1]
2 (a) d in the range 0.5mm to 0.9mm, to nearest 0.1mm or to 0.01mm, with unit. [1]
(b) (iii) Value for x in the range 11–19mm, with unit. [1]
Evidence of repeat readings of x. [1]
(c) Absolute uncertainty in x in range 2mm to 5mm.
(e) (ii) h1 recorded to nearest mm, with consistent unit. [1]
(iv) Correct calculation of k to the number of s.f. given by the candidate. [1]
Value of k given to the same number of s.f. as, or one more than, the number of
s.f. in (h1 – h2) or m, whichever is lower. [1]
(f) Second values of x and n. [1]
Second values of h1 and h2. [1]
Quality: Value of (h1 – h2) for smaller x less than the value of (h1 – h2) for larger x. [1]

(g) (i) Two values of c calculated correctly. [1]
(ii) Valid comment consistent with the calculated values of c, testing against a criterion
(h) (i) Limitations [4] (ii) Improvements [4] Do not credit
A Two readings are not enough to
draw a conclusion
Take more readings and plot
graph/
take more readings and
compare c values
Repeat readings/
few readings/
only one reading/not
enough readings for
accurate value
B d is small/
large (percentage) uncertainty in
d
Use a micrometer (to measure
diameter)
Digital calipers
C n not an integer Estimate n to the nearest ¼
turn
D Diameter not constant/
coils vary in diameter/
coils not equally spaced/
coils not circular
Method of making equally-
spaced coils e.g. make small
marks/grooves on wooden rod
Use motor to wind spring by
rotating rod
Spring not straight
Use ‘factory’ spring
E Difficult to measure diameter (x)
with reason e.g. calipers distort
coils/end of coil gets in the way
of ruler
Use thin ruler/graph paper
placed between loops of spring
F h1 – h2 small, so uncertainty
large
Use larger mass/larger range of
masses
Travelling microscope with
reference to h1 – h2
Use wires of longer length to
increase h1 – h2

PHYSICS 9702/35
MARK SCHEME
Maximum Mark: 40
Published
Teachers.
,

1 (a) (ii) Value for L to the nearest mm with unit and in range 38.0cm ≤ L ≤ 42.0cm. [1]
(b) (iv) Value of I with unit in the range 25mA to 100mA. [1]
(c) Six sets of readings of x and I with correct trend scores 4 marks, five sets scores
3 marks etc. [4]
Minor help from Supervisor –1, major help –2.
Range of x: [1]
∆x ≥ 30.0cm.
Each column heading must contain a quantity and a unit where appropriate.
The unit must conform to accepted scientific convention, e.g. 1/I/A–1
or 1/I(A–1
) or
1/I / 1/A. Do not allow 1/I(A).
Consistency: [1]
All values of I given to 0.1mA.
Every value of 1/I must be given to the same number of s.f. as (or one more than)
the number of s.f. in the corresponding value of I.
Calculation: [1]
Values of 1/I calculated correctly to the number of s.f. given by the candidate.
(d) (i) Axes: [1]
Quality: [1]
All points in the table must be plotted on the grid for this mark to be awarded.
All points must be within 2.0cm (to scale) on the x-axis of a straight line.
points). There must be an even distribution of points either side of the line along
the full length.
Lines must not be kinked or thicker than half a small square.

(iii) Gradient: [1]
The hypotenuse of the triangle used must be greater than half of the length of the
drawn line.
Both read-offs must be accurate to half a small square in both the x and y directions.
y-intercept: [1]
Either:
Or:
(e) P = –value of candidate’s gradient Q = value of candidate’s intercept. [1]
Unit for P correct (e.g. A–1
m–1
, A–1
cm–1
, A–1
mm–1
, mA–1
m–1
, mA–1
cm–1
or mA–1
mm–1
).
Unit for Q correct (e.g. A–1
or mA–1
) and consistent with value. [1]
(f) Value of R in the range 5 Ω to 20 Ω. [1]
2 (a) (ii) Value of C in the range 35.0cm to 40.0cm with unit. [1]
(iii) Value of d to the nearest mm with unit in range 4.0cm to 6.0cm. [1]
(iv) Correct calculation of (C – d). [1]
(b) (ii) Value of θ to the nearest degree with unit, and θ < 90°. [1]
(iii) Absolute uncertainty in θ in range 2° to 5°.
(c) Correct calculation of (tanθ – 1). Do not allow a unit. [1]
Answer given to 2 s.f. or 3 s.f. [1]
(d) Second value of d. [1]
Second value of θ. [1]
Quality: Second value of θ > first value of θ. [1]
(e) (i) Two values of k calculated correctly. [1]

(f) (i) Limitations [4] (ii) Improvements [4] Do not credit
A Two readings not enough to
draw a conclusion
Take more readings and plot a
graph/
take more readings and
compare k values
Repeat readings/
few readings/
only one reading/
not enough readings for
accurate value
B Difficult to measure angle
with reason e.g. lack of
vertical reference
line/parallax/nail in the
way/apparatus moves when
protractor in place
Method of providing vertical
reference e.g.
use a plumbline/
drill hole at origin of protractor
and mount on nail/
method of fixing protractor/
use a grid with angles marked/
larger protractor
C Range of θ is too small More holes further apart
D Difficult to measure C whilst
balancing strip
Mark position of pivot/
add scale to wooden strip
E Difficulty in mechanical set
up e.g. alignment of strip
and string/horizontal string
(parallel to table)
Method of improvement e.g.
spirit level linked to string/axle
of pulley/
add weights to stands
F Friction at nail (or axle of
pulley)
Lubricate the nail/pulley axle

PHYSICS 9702/41
Paper 4 A Level Structured Questions May/June 2016
MARK SCHEME
Maximum Mark: 100
Published
Teachers.
,

1 (a) (gravitational) potential at infinity defined as/is zero B1
(gravitational) force attractive so work got out/done as object moves from infinity
(so potential is negative) B1 [2]
(b) (i) ∆E = m∆φ
= 180 × (14 – 10) × 108
C1
= 7.2 × 1010
J A1
increase B1 [3]
(ii) energy required = 180 × (10 – 4.4) × 108
or
energy per unit mass = (10 – 4.4) × 108
C1
½ × 180 × v2
= 180 × (10 – 4.4) × 108
or
½ × v2
= (10 – 4.4) × 108
C1
v = 3.3 × 104
ms–1
A1 [3]
2 (a) e.g. time of collisions negligible compared to time between collisions
no intermolecular forces (except during collisions)
random motion (of molecules)
large numbers of molecules
(total) volume of molecules negligible compared to volume of containing vessel
or
average/mean separation large compared with size of molecules
any two B2 [2]
2 (b) (i) mass = 4.0 / (6.02 × 1023
) = 6.6 × 10–24
g
or
mass = 4.0 × 1.66 × 10–27
× 103
= 6.6 × 10–24
g B1 [1]
(ii)
2
3
kT =
2
1
m <c2
> C1
2
3
× 1.38 × 10–23
× 300 =
2
1
× 6.6 × 10–27
× <c2
>
<c2
> = 1.88 × 106
(m2
s–2
) C1
r.m.s. speed = 1.4 × 103
ms–1
A1 [3]

3 (a) acceleration/force proportional to displacement (from fixed point) M1
acceleration/force and displacement in opposite directions A1 [2]
(b) maximum displacements/accelerations are different B1
graph is curved/not a straight line B1 [2]
(c) (i) ω = 2π / T and T = 0.8s C1
ω = 7.9 rads–1
A1 [2]
(ii) a = (–)ω2
x
= 7.852
× 1.5 × 10–2
C1
= 0.93 ms–2
or 0.94 ms–2
A1 [2]
(iii) ∆E = ½ mω2
(x0
2
– x2
) C1
= ½ × 120 × 10–3
× 7.852
× {(1.5 × 10–2
)2
– (0.9 × 10–2
)2
} C1
= 5.3 × 10–4
J A1 [3]
4 (a) (i) product of speed and density M1
reference to speed in medium (and density of medium) A1 [2]
(ii) α: ratio of reflected intensity and/to incident intensity B1
Z1 and Z2: (specific) acoustic impedances of media (on each side of boundary) B1 [2]
(b) in muscle: IM = I0 e–µx
= I0 exp(–23 × 3.4 × 10–2
) C1
IM / I0 = 0.457 C1
at boundary: α = (6.3 – 1.7)2
/ (6.3 + 1.7)2
= 0.33 C1
IT /IM = [(1 – α) =] 0.67 C1
IT / I0 = 0.457 × 0.67
= 0.31 A1 [5]

5 (a) (i) 1011 A1 [1]
(ii)
0 0.25 0.50 0.75 1.00 1.25 1.50
1011 0110 1000 1110 0101 0011 0001
All 6 correct, 2 marks. 5 correct, 1 mark. A2 [2]
(b) sketch: 6 horizontal steps of width 0.25ms shown M1
steps at correct heights and all steps shown A1
steps shown in correct time intervals A1 [3]
(c) increase sampling frequency/rate M1
so that step width/depth is reduced A1
increase number of bits (in each number) M1
so that step height is reduced A1 [4]
6 (a) sketch: from x = 0 to x = R, potential is constant at VS B1
smooth curve through (R, VS) and (2R, 0.5VS) B1
smooth curve continues to (3R, 0.33VS) B1 [3]
(b) sketch: from x = 0 to x = R, field strength is zero B1
smooth curve through (R, E) and (2R, 0.25E) B1
smooth curve continues to (3R, 0.11E) B1 [3]
7 (a) line has non-zero intercept/line does not pass through origin B1
charge is/should be proportional to potential (difference)
or
charge is/should be zero when p.d. is zero
(therefore there is a systematic error) B1 [2]

(b) reasonable attempt at line of best fit B1
use of gradient of line of best fit clear M1
C = 2800 µF (allow ± 200 µF) A1 [3]
(c) energy = ½ CV2
or energy = ½ QV and C = Q / V C1
∆ energy = ½ × 2800 × 10–6
× (9.02
– 6.02
) C1
= 6.3 × 10–2
J A1 [3]
8 (a) op-amp has infinite/(very) large gain B1
op-amp saturates if V+
≠ V–
M1
V+
is at earth potential so P (or V–
) must be at earth A1 [3]
(b) input resistance to op-amp is very large
or
current in R2 = current in R1 B1
VIN (– 0) = IR2 and (0) – VOUT = IR1 M1
VOUT / VIN = –R1 / R2 A1 [3]
(c) relay coil connected between VOUT and earth M1
correct diode symbol connected between VOUT and coil or between coil and earth M1
correct polarity for diode (‘clockwise’) A1 [3]
9 (a) 0.10 mm B1 [1]
(b) VH = (0.13 × 3.8) / (6.0 × 1028
× 0.10 × 10–3
× 1.60 × 10–19
) C1
= 5.1 × 10–7
V A1 [2]
10 (a) (non-uniform) magnetic flux in core is changing M1
induces (different) e.m.f. in (different parts of) the core A1
(eddy) currents form in the core M1
which give rise to heating A1 [4]

(b) as magnet falls, tube cuts magnetic flux M1
e.m.f./(eddy) currents induced in metal/aluminium (tube) A1
(eddy) current heating of tube M1
with energy taken from falling magnet A1
or
(eddy) currents produce magnetic field (M1)
that opposes motion of magnet (A1)
so magnet B has acceleration < g
or
magnet B has smaller acceleration/reaches terminal speed A1 [5]
11 (a) period = 15 ms C1
frequency (= 1 / T) = 67 Hz A1 [2]
(b) zero A1 [1]
(c) Ir.m.s. = I0 / √2 C1
= 0.53 A A1 [2]
(d) energy = Ir.m.s.
2
× R × t or ½ I0
2
× R × t
or
power = Ir.m.s.
2
× R and energy = power × t C1
energy = 0.532
× 450 × 30 × 10–3
= 3.8 J A1 [2]
12 (a) (in a solid electrons in) neighbouring atoms are close together
(and influence/interact with each other) M1
this changes their electron energy levels M1
(many atoms in lattice) cause a spread of energy levels into a band A1 [3]

(b) photons of light give energy to electrons in valence band B1
electrons move into the conduction band M1
leaving holes in the valence band A1
these electrons and holes are charge carriers B1
increased number/increased current, hence reduced resistance B1 [5]
13 (a) e.g. background count (rate)/radiation
multiple possible counts from each decay
radiation emitted in all directions
dead-time of counter
(daughter) product unstable/also emits radiation
self-absorption of radiation in sample or absorption in air/detector window
three sensible suggestions, 1 each B3 [3]
(b) A = A0 exp(–ln2 × t / T½)
1.21 × 102
= 3.62 × 104
exp(–ln2 × 42.0/T½)
or
1.21 × 102
= 3.62 × 104
exp(–λ × 42.0) C1
T½ = 5.1 minutes (306 s) A1 [2]
(c) discrete energy levels (in nuclei) B1 [1]

PHYSICS 9702/42
MARK SCHEME
Maximum Mark: 100
Published
Teachers.
,

1 (a) (i) gravitational force provides/is the centripetal force B1
same gravitational force (by Newton III) B1 [2]
(ii) ω = 2π/T
= 2π/(4.0 × 365 × 24 × 3600) C1
= 5.0 (4.98) × 10–8
rads–1
A1 [2]
(b) (i) (centripetal force =) MAdω2
= MB(2.8 × 108
–d)ω2
or
MAdA = MBdB C1
MA /MB = 3.0 = (2.8 × 108
– d)/d C1
d = 7.0 × 107
km A1 [3]
(ii) GMAMB /(2.8 × 1011
)2
= MAdω2
B1
MB = (2.8 × 1011
)2
× dω2
/G
= (2.8 × 1011
)2
× (7.0 × 1010
) × (4.98 × 10–8
)2
/(6.67 × 10–11
) C1
= 2.0 × 1029
kg A1 [3]
2 (a) (i) number of atoms/nuclei in 12 g of carbon-12 B1 [1]
(ii) amount of substance M1
containing NA (or 6.02 × 1023
) particles/molecules/atoms
or
which contains the same number of particles/atoms/molecules as there
are atoms in 12g of carbon-12 A1 [2]
(b) pV = nRT
2.0 × 107
× 1.8 × 104
× 10–6
= n × 8.31 × 290, so n = 149mol or 150mol A1 [1]
(c) (i) V and T constant and so pressure reduced by 5.0%
pressure = 0.95 × 2.0 × 107
C1
or
calculation of new n (= 142.5mol) and correct substitution into pV = nRT (C1)
pressure = 1.9 × 107
Pa A1 [2]

(ii) loss is 5/100 × 150mol = 7.5mol
or
∆N = 4.52 × 1024
C1
t = (7.5 × 6.02 × 1023
)/1.5 × 1019
or
t = 4.52 × 1024
/1.5 × 1019
C1
= 3.0 × 105
s A1 [3]
3 (a) no net energy transfer between the bodies
or
bodies are at the same temperature B1 [1]
(b) (i) thermocouple, platinum/metal resistance thermometer, pyrometer B1 [1]
(ii) thermistor, thermocouple B1 [1]
(c) (i) change = 11.5K B1 [1]
(ii) final temperature = 311.2K B1 [1]
4 (a) (i) T = 0.60s and ω = 2π/T C1
ω = 10(10.47) rads–1
A1 [2]
(ii) energy = ½mω2
x0
2
or ½mv2
and v = ωx0 C1
= ½ × 120 × 10–3
× (10.5)2
× (2.0 × 10–2
)2
= 2.6 × 10–3
J A1 [2]
(b) sketch: smooth curve in correct directions B1
peak at f M1
amplitude never zero and line extends from 0.7f to 1.3f A1 [3]
(c) sketch: peaked line always below a peaked line A M1
peak not as sharp and at (or slightly less than) frequency of peak in line A A1 [2]

5 (a) amplitude of the carrier wave varies M1
in synchrony with displacement of the information/audio signal A1 [2]
(b) (i) 10kHz A1 [1]
(ii) 5kHz A1 [1]
(c) (i) 24 = 10 lg (PMIN /{5.0 × 10–13
}) C1
PMIN = 1.3 (1.26) × 10–10
W A1 [2]
(ii) 45 × 2 = 10 lg ({500 × 10–3
}/P)
P = 5.0 × 10–10
(W) M1
P > PMIN so yes A1
or
maximum attenuation calculated to be 96(dB) (M1)
96dB > 2 × 45dB so yes (A1)
or
maximum length of wire calculated to be 48 (km) (M1)
actual length 45km < 48km so yes (A1)
or
maximum attenuation per unit length calculated to be 2.2dBkm–1
(M1)
2.2dBkm–1
> 2.0dBkm–1
so yes (A1) [2]
6 (a) lines perpendicular to surface
or
lines are radial M1
lines appear to come from centre A1 [2]
(b) (i) FE = (1.6 × 10–19
)2
/4πε0x2
C1
FG = G × (1.67 × 10–27
)2
/x2
C1
FE /FG = (1.6 × 10–19
)2
× (8.99 × 109
)/[(1.67×10–27
)2
× (6.67×10–11
)]
= 1.2 (1.24) × 1036
A1 [3]
(ii) FE ≫ FG B1 [1]

7 (a) e.g. storing energy
blocking d.c.
in oscillator circuits
in tuning circuits
in timing circuits
any two B2 [2]
(b) (i) 1/6 + 1/C + 1/C = 1/4 C1
C = 24µF A1 [2]
(ii) Q = CV
= 4.0 × 10–6
× 12 C1
= 48µC A1 [2]
(iii) 1. 48µC A1
2. 24µC A1 [2]
8 (a) (i) gain = voltage output/voltage input B1 [1]
(ii) changes in VOUT M1
occur immediately when VIN changes A1
or
changes in VIN (M1)
result in immediate changes to VOUT (A1) [2]
(b) 12 = 1 + R/(1.5 × 103
) C1
R = 16.5 kΩ A1 [2]
(c) straight line from (0,0) to (0.75t1, 9.0V) B1
horizontal line from endpoint of straight line to t1 B1
+9V to –9V (or v.v.) at t1 B1
correct line to t2 B1 [4]

9 (a) (i) number density of charge carriers/free electrons
or
number per unit volume of charge carriers/free electrons B1 [1]
(ii) PX or QY or RZ B1 [1]
(b) (i) VH is inversely proportional to n B1
for semiconductors, n is (much) smaller than for metals B1 [2]
(ii) magnetic field would deflect holes and electrons in same direction B1
(because) electrons are (–)ve, holes are (+)ve M1
so VH has opposite polarity/opposite sign A1 [3]
10 (a) iron rod changes flux (density)/field B1
change of flux in coil Q causes induced e.m.f. B1 [2]
(b) constant reading (either polarity) from time zero to near t1 B1
spike in one direction near t1 clearly showing a larger voltage M1
of opposite polarity A1
zero reading from near t1 to t2 B1 [4]
11 (a) point P shown at ‘lower end’ of load B1 [1]
(b) Vr.m.s. = 6.0/√2 = 4.24V C1
Ir.m.s. = 4.24/(2.4 × 103
)
= 1.8 × 10–3
A A1 [2]
(c) (i) capacitor in parallel with load B1 [1]
(ii) line from peak to curve at 3.0V for either half- or full-wave rectified M1
correct curvature on line (gradient becoming more shallow) A1
line drawn as for full-wave rectified A1 [3]

12 (a) (i) (X–ray) photon produced when electron/charged particle is
stopped/accelerated (suddenly) B1
range of accelerations (in target) M1
hence distribution of wavelengths A1 [3]
(ii) electron gives all its energy to one photon B1
electron stopped in single collision B1 [2]
(iii) de-excitation of (orbital) electrons in target/anode/metal B1 [1]
(b) (i) aluminium sheet/filter/foil (placed in beam from tube) B1 [1]
(ii) (long wavelength X-rays) do not pass through the body B1 [1]
13 (a) (photons of) electromagnetic radiation M1
emitted from nuclei A1 [2]
(b) line of best fit drawn B1
recognises µ as given by the gradient of best-fit line
or
ln C = ln C0 – µx B1
µ = 0.061mm–1
(within ±0.004mm–1
, 1 mark; within ±0.002mm–1
, 2 marks) A2 [4]
(c) aluminium is less absorbing (than lead)
or
gradient of graph would be less M1
so µ is smaller A1 [2]

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1 (a) (gravitational) potential at infinity defined as/is zero B1
(gravitational) force attractive so work got out/done as object moves from infinity
(so potential is negative) B1 [2]
(b) (i) ∆E = m∆φ
= 180 × (14 – 10) × 108
C1
= 7.2 × 1010
J A1
increase B1 [3]
(ii) energy required = 180 × (10 – 4.4) × 108
or
energy per unit mass = (10 – 4.4) × 108
C1
½ × 180 × v2
= 180 × (10 – 4.4) × 108
or
½ × v2
= (10 – 4.4) × 108
C1
v = 3.3 × 104
ms–1
A1 [3]
2 (a) e.g. time of collisions negligible compared to time between collisions
no intermolecular forces (except during collisions)
random motion (of molecules)
large numbers of molecules
(total) volume of molecules negligible compared to volume of containing vessel
or
average/mean separation large compared with size of molecules
any two B2 [2]
2 (b) (i) mass = 4.0 / (6.02 × 1023
) = 6.6 × 10–24
g
or
mass = 4.0 × 1.66 × 10–27
× 103
= 6.6 × 10–24
g B1 [1]
(ii)
2
3
kT =
2
1
m <c2
> C1
2
3
× 1.38 × 10–23
× 300 =
2
1
× 6.6 × 10–27
× <c2
>
<c2
> = 1.88 × 106
(m2
s–2
) C1
r.m.s. speed = 1.4 × 103
ms–1
A1 [3]

3 (a) acceleration/force proportional to displacement (from fixed point) M1
acceleration/force and displacement in opposite directions A1 [2]
(b) maximum displacements/accelerations are different B1
graph is curved/not a straight line B1 [2]
(c) (i) ω = 2π / T and T = 0.8s C1
ω = 7.9 rads–1
A1 [2]
(ii) a = (–)ω2
x
= 7.852
× 1.5 × 10–2
C1
= 0.93 ms–2
or 0.94 ms–2
A1 [2]
(iii) ∆E = ½ mω2
(x0
2
– x2
) C1
= ½ × 120 × 10–3
× 7.852
× {(1.5 × 10–2
)2
– (0.9 × 10–2
)2
} C1
= 5.3 × 10–4
J A1 [3]
4 (a) (i) product of speed and density M1
reference to speed in medium (and density of medium) A1 [2]
(ii) α: ratio of reflected intensity and/to incident intensity B1
Z1 and Z2: (specific) acoustic impedances of media (on each side of boundary) B1 [2]
(b) in muscle: IM = I0 e–µx
= I0 exp(–23 × 3.4 × 10–2
) C1
IM / I0 = 0.457 C1
at boundary: α = (6.3 – 1.7)2
/ (6.3 + 1.7)2
= 0.33 C1
IT /IM = [(1 – α) =] 0.67 C1
IT / I0 = 0.457 × 0.67
= 0.31 A1 [5]

5 (a) (i) 1011 A1 [1]
(ii)
0 0.25 0.50 0.75 1.00 1.25 1.50
1011 0110 1000 1110 0101 0011 0001
All 6 correct, 2 marks. 5 correct, 1 mark. A2 [2]
(b) sketch: 6 horizontal steps of width 0.25ms shown M1
steps at correct heights and all steps shown A1
steps shown in correct time intervals A1 [3]
(c) increase sampling frequency/rate M1
so that step width/depth is reduced A1
increase number of bits (in each number) M1
so that step height is reduced A1 [4]
6 (a) sketch: from x = 0 to x = R, potential is constant at VS B1
smooth curve through (R, VS) and (2R, 0.5VS) B1
smooth curve continues to (3R, 0.33VS) B1 [3]
(b) sketch: from x = 0 to x = R, field strength is zero B1
smooth curve through (R, E) and (2R, 0.25E) B1
smooth curve continues to (3R, 0.11E) B1 [3]
7 (a) line has non-zero intercept/line does not pass through origin B1
charge is/should be proportional to potential (difference)
or
charge is/should be zero when p.d. is zero
(therefore there is a systematic error) B1 [2]

(b) reasonable attempt at line of best fit B1
use of gradient of line of best fit clear M1
C = 2800 µF (allow ± 200 µF) A1 [3]
(c) energy = ½ CV2
or energy = ½ QV and C = Q / V C1
∆ energy = ½ × 2800 × 10–6
× (9.02
– 6.02
) C1
= 6.3 × 10–2
J A1 [3]
8 (a) op-amp has infinite/(very) large gain B1
op-amp saturates if V+
≠ V–
M1
V+
is at earth potential so P (or V–
) must be at earth A1 [3]
(b) input resistance to op-amp is very large
or
current in R2 = current in R1 B1
VIN (– 0) = IR2 and (0) – VOUT = IR1 M1
VOUT / VIN = –R1 / R2 A1 [3]
(c) relay coil connected between VOUT and earth M1
correct diode symbol connected between VOUT and coil or between coil and earth M1
correct polarity for diode (‘clockwise’) A1 [3]
9 (a) 0.10 mm B1 [1]
(b) VH = (0.13 × 3.8) / (6.0 × 1028
× 0.10 × 10–3
× 1.60 × 10–19
) C1
= 5.1 × 10–7
V A1 [2]
10 (a) (non-uniform) magnetic flux in core is changing M1
induces (different) e.m.f. in (different parts of) the core A1
(eddy) currents form in the core M1
which give rise to heating A1 [4]

(b) as magnet falls, tube cuts magnetic flux M1
e.m.f./(eddy) currents induced in metal/aluminium (tube) A1
(eddy) current heating of tube M1
with energy taken from falling magnet A1
or
(eddy) currents produce magnetic field (M1)
that opposes motion of magnet (A1)
so magnet B has acceleration < g
or
magnet B has smaller acceleration/reaches terminal speed A1 [5]
11 (a) period = 15 ms C1
frequency (= 1 / T) = 67 Hz A1 [2]
(b) zero A1 [1]
(c) Ir.m.s. = I0 / √2 C1
= 0.53 A A1 [2]
(d) energy = Ir.m.s.
2
× R × t or ½ I0
2
× R × t
or
power = Ir.m.s.
2
× R and energy = power × t C1
energy = 0.532
× 450 × 30 × 10–3
= 3.8 J A1 [2]
12 (a) (in a solid electrons in) neighbouring atoms are close together
(and influence/interact with each other) M1
this changes their electron energy levels M1
(many atoms in lattice) cause a spread of energy levels into a band A1 [3]

(b) photons of light give energy to electrons in valence band B1
electrons move into the conduction band M1
leaving holes in the valence band A1
these electrons and holes are charge carriers B1
increased number/increased current, hence reduced resistance B1 [5]
13 (a) e.g. background count (rate)/radiation
multiple possible counts from each decay
radiation emitted in all directions
dead-time of counter
(daughter) product unstable/also emits radiation
self-absorption of radiation in sample or absorption in air/detector window
three sensible suggestions, 1 each B3 [3]
(b) A = A0 exp(–ln2 × t / T½)
1.21 × 102
= 3.62 × 104
exp(–ln2 × 42.0/T½)
or
1.21 × 102
= 3.62 × 104
exp(–λ × 42.0) C1
T½ = 5.1 minutes (306 s) A1 [2]
(c) discrete energy levels (in nuclei) B1 [1]

PHYSICS 9702/51
Paper 5 Planning, Analysis and Evaluation May/June 2016
MARK SCHEME
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Question 1 Planning (15 marks)
Defining the problem (2 marks)
P λ is the independent variable, or vary λ. [1]
P V is the dependent variable, or measure V. [1]
Methods of data collection (4 marks)
M Circuit diagram showing d.c. power supply in series with diode (correct symbol needed) and
method to measure potential difference across diode. Circuit must be correct. [1]
M Instrument to change p.d. across LED e.g. variable power supply/potential divider/variable
resistor. [1]
M Record wavelength of light of LED from data sheet or use Young’s slits/diffraction grating. [1]
M (Slowly) increase potential difference across LED until LED (just) emits light (or reverse
procedure). [1]
Method of analysis (3 marks)
A Plot a graph of lg V against lg λ (allow natural logs). Allow lg λ against lg V. [1]
A n = gradient [1]
A k = 10y-intercept
[1]
Additional detail (6 marks)
Relevant points might include: [6]
1 Use of a protective resistor (can be shown on the diagram).
2 Polarity of LED correct in circuit diagram.
3 Instrument to determine when LED just lights e.g. light meter/detector, LDR.
4 Method to use light detector/LDR to determine point at which LED emits light.
5 Expression that gives λ (symbols need to defined) from experimental determination of
wavelength of light, e.g. Young’s slits/diffraction grating.
6 Perform experiment in a dark room/LED in tube.
7 Relationship is valid if graph is a straight line.
8 λ= +lg lg lgV n k
9 Repeat V and average for the same λ or LED.
Do not allow vague computer methods.

Question 2 Analysis, conclusions and evaluation (15 marks)
Mark Expected Answer Additional Guidance
(a) A1
π
4LF
E
(b) T1
2
1
d
/106
m–2
T2 13 or 12.8
9.8 or 9.77
6.9 or 6.93
4.7 or 4.73
3.2 or 3.19
1.9 or 1.93
All values to 2 s.f. or 3 s.f. Allow a mixture of
significant figures. Must be values in table.
U1 From ± 2 to ± 0.1 Allow more than one significant figure.
(c) (i) G1 Six points plotted correctly Must be within half a small square.
Do not allow “blobs”.
ECF allowed from table.
U2
Error bars in 2
1
d
plotted correctly
All error bars to be plotted. Must be accurate to
less than half a small square.
(ii) G2 Line of best fit If points are plotted correctly then lower end of
line should pass between (3.2, 3.0) and (3.6,
3.0) and upper end of line should pass between
(11.2, 10.0) and (11.6, 10.0).
G3 Worst acceptable straight line.
Steepest or shallowest possible
line that passes through all the
error bars.
Line should be clearly labelled or dashed.
Examiner judgement on worst acceptable line.
Lines must cross. Mark scored only if error bars
are plotted.
(iii) C1 Gradient of line of best fit 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. (Should be about 9 × 10–10
.)
U3 Absolute uncertainty in gradient Method of determining absolute uncertainty
Difference in worst gradient and gradient.
(d) (i) C2
=
π×
4 60.479
gradient gradient
LF Do not penalise POT.
(Should be about 7 × 1010
.)
C3 Nm–2
or Pa Allow in base units: kgm–1
s–2
.
(ii) U4 Percentage uncertainty in E Must be larger than 3%.

(e) C4 e in the range 15.5 × 10–3
to
18.0 × 10–3
and given to 2 or 3 s.f.
Allow mm.
U5 Absolute uncertainty in e
Note = 2
gradient
e
d
is possible.
Uncertainties in Question 2
(c) (iii) Gradient [U3]
uncertainty = gradient of line of best fit – gradient of worst acceptable line
uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)
(d) (ii) [U4]
   ∆ ∆
= + + × = × +   
   
gradient 0.01 0.5 gradient
percentage uncertainty 100 100 3.03%
× × × ×
= = =
π× π×
4 max max 4 2.51 19.5 62.319
max
min gradient min gradient min gradient
L F
E
4 min min 4 2.49 18.5 58.652
min
max gradient max gradient max gradient
L F
E
× × × ×
= = =
π× π×
(e) [U5]
  
= + + × × + = +  
  
0.5 0.01 0.02
percentage uncertainty 2 100 % 20.4% %
19.0 2.50 0.23
E E
 ∆  
= + × ×  
  
gradient 0.02
percentage uncertainty 2 100
gradient 0.23
= 2
min
max gradient
max e
d
× ×
=
π× ×
max max
2
min min
4
max
L F
e
E d
= 2
max
min gradient
mine
d
× ×
=
π× ×
min min
2
max max
4
min
L F
e
E d

PHYSICS 9702/52
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P θ is the independent variable and a is the dependent variable, or vary θ and measure a. [1]
P Keep F constant. [1]
M Diagram showing inclined plane with labelled support
(not if a ruler used as the inclined plane or as vertical support). [1]
M Method to measure angle e.g. use a protractor to measure θ or use a ruler to
measure marked distances from which sinθ or θ may be determined.
(Allow a labelled protractor in the correct position.) [1]
M Method to measure a time or velocity to determine a, e.g. measure the time using a
stopwatch, light gate(s) connected to a timer, motion sensor connected to a time
display. [1]
M Use a balance to measure the mass of the trolley. [1]
A Plot a graph of
a againstsin θ.
or Plot a graph of
maagainstsin θ.
or Plot a graph of
maagainstmgsin θ. [1]
A Relationship is valid if the graph is a straight line and does not pass through the origin [1]
A k = F – m × (y-intercept) or k = F – (y-intercept) or k = F – (y-intercept) [1]
Do not allow lg-lg graphs.
1 Keep mass of trolley constant/use same trolley.
2 Correct trigonometry relationship to determine sin θ or θ using marked lengths.
3 Use ruler to measure appropriate distance to determine a, e.g. length of slope, length of
card for light gate method, position of motion sensor.
4 Equation to determine a from measurements taken appropriately with a as the subject.
5 Measurement of F for a valid method e.g. take reading from newton-meter or from stretched
elastic/spring from extension (allow falling weight e.g. F = mg).
6 Use a constant extension to produce a constant force when using stretched spring/elastic.

7 Method to ensure the inclined plane is the same height each side of the plane or spirit level
across plane or ensure force F (or string) is parallel to the plane.
8 Safety precaution linked to falling mass/trolley or spring/elastic breaking (not string).
9 Rearrangement of relationship into y = mx + c e.g. ma = –mg sin θ + (F – k) or
m
kF
ga
−
+−= θsin or correct y-intercept (subject must be y-axis).
10 Repeat experiment for each angle θ to find average for a.

(a) A1
ρ
π
4 L
(b) T1
2
1
d
/106
m–2
T2
1.2 or 1.21
3.2 or 3.19
4.7 or 4.73
6.9 or 6.93
9.8 or 9.77
14 or 13.7
U1 From ± 0.03 to ± 1 Allow more than one significant figure.
Allow zero for first uncertainty and up to 1.2 for
largest uncertainty.
U2 Error bars in 2
1
d
plotted correctly
All error bars to be plotted. Length of bar must
be accurate to less than half a small square
and symmetrical.
(ii) G2 Line of best fit Lower end of line must pass between
(2.6, 4.0) and (3.0, 4.0) and upper end of line
must pass between (12.4, 18.0) and
(13.0, 18.0).
error bars.
Must be steepest/shallowest line. Mark
scored only if error bars are plotted.
POT. (Should be about 1.4 –1.5 × 10–6
.)
U3 Absolute uncertainty in gradient Method of determining absolute uncertainty:
difference in worst gradient and gradient.

(d) (i) C2 π×
= ×
gradient
0.7854 gradient
4L
Must use gradient value. Do not penalise
POT (Should be about 1 × 10–6
.)
C3 Ωm Correct unit and correct power of ten.
(ii) U4 Percentage uncertainty in ρ Percentage uncertainty in gradient + 1%.
(e) C4 R in the range 25.5 to 28.4 and
given to 2 or 3 s.f.
Allow 26 or 27 or 28.
Allow ECF for POT error in (d)(i) e.g.
2.7 × 107
.
U5 Absolute uncertainty in R Percentage uncertainty must be greater than
8.6%.

(d) (ii) [U4]
   ∆ ∆
= + × = × +   
   
gradient 0.01 gradient
percentage uncertainty 100 100 1%
ρ
π× π×
= =
× ×
max gradient max gradient
max
4 min 4 0.99L
ρ
π× π×
= =
× ×
min gradient min gradient
min
4 max 4 1.01L
(e) [U5]
   ∆ ∆ 
= + × × = + ×    
    
percentage uncertainty 2 100 0.086 100
ρ ρ
ρ ρ
   ∆ ∆ 
= + + × × = + ×    
    
0.01 0.01
percentage uncertainty 2 100 0.096 100
1.00 0.23
= 2
min
max gradient
maxR
d
ρ× ×
=
π×
max max
2
min
4
max
L
R
d
= 2
max
min gradient
minR
d
ρ× ×
=
π×
min min
2
max
4
min
L
R
d

PHYSICS 9702/53
MARK SCHEME
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Published
Teachers.
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P λ is the independent variable, or vary λ. [1]
P V is the dependent variable, or measure V. [1]
M Circuit diagram showing d.c. power supply in series with diode (correct symbol needed) and
method to measure potential difference across diode. Circuit must be correct. [1]
M Instrument to change p.d. across LED e.g. variable power supply/potential divider/variable
resistor. [1]
M Record wavelength of light of LED from data sheet or use Young’s slits/diffraction grating. [1]
M (Slowly) increase potential difference across LED until LED (just) emits light (or reverse
procedure). [1]
A Plot a graph of lg V against lg λ (allow natural logs). Allow lg λ against lg V. [1]
A n = gradient [1]
A k = 10y-intercept
[1]
1 Use of a protective resistor (can be shown on the diagram).
2 Polarity of LED correct in circuit diagram.
3 Instrument to determine when LED just lights e.g. light meter/detector, LDR.
4 Method to use light detector/LDR to determine point at which LED emits light.
5 Expression that gives λ (symbols need to defined) from experimental determination of
wavelength of light, e.g. Young’s slits/diffraction grating.
6 Perform experiment in a dark room/LED in tube.
7 Relationship is valid if graph is a straight line.
8 λ= +lg lg lgV n k
9 Repeat V and average for the same λ or LED.

(a) A1
π
4LF
E
(b) T1
2
1
d
/106
m–2
T2 13 or 12.8
9.8 or 9.77
6.9 or 6.93
4.7 or 4.73
3.2 or 3.19
1.9 or 1.93
U1 From ± 2 to ± 0.1 Allow more than one significant figure.
U2
Error bars in 2
1
d
plotted correctly
All error bars to be plotted. Must be accurate to
less than half a small square.
(ii) G2 Line of best fit If points are plotted correctly then lower end of
line should pass between (3.2, 3.0) and (3.6,
3.0) and upper end of line should pass between
(11.2, 10.0) and (11.6, 10.0).
error bars.
Lines must cross. Mark scored only if error bars
are plotted.
POT. (Should be about 9 × 10–10
.)
U3 Absolute uncertainty in gradient Method of determining absolute uncertainty
Difference in worst gradient and gradient.
(d) (i) C2
=
π×
4 60.479
gradient gradient
LF Do not penalise POT.
(Should be about 7 × 1010
.)
C3 Nm–2
or Pa Allow in base units: kgm–1
s–2
.
(ii) U4 Percentage uncertainty in E Must be larger than 3%.

(e) C4 e in the range 15.5 × 10–3
to
18.0 × 10–3
and given to 2 or 3 s.f.
Allow mm.
U5 Absolute uncertainty in e
Note = 2
gradient
e
d
is possible.
(d) (ii) [U4]
   ∆ ∆
= + + × = × +   
   
percentage uncertainty 100 100 3.03%
× × × ×
= = =
π× π×
4 max max 4 2.51 19.5 62.319
max
min gradient min gradient min gradient
L F
E
4 min min 4 2.49 18.5 58.652
min
max gradient max gradient max gradient
L F
E
× × × ×
= = =
π× π×
(e) [U5]
  
= + + × × + = +  
  
0.5 0.01 0.02
percentage uncertainty 2 100 % 20.4% %
19.0 2.50 0.23
E E
 ∆  
= + × ×  
  
gradient 0.02
percentage uncertainty 2 100
gradient 0.23
= 2
min
max gradient
max e
d
× ×
=
π× ×
max max
2
min min
4
max
L F
e
E d
= 2
max
min gradient
mine
d
× ×
=
π× ×
min min
2
max max
4
min
L F
e
E d

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