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About This Presentation
Mark scheme of Physics whole papers October November 2016
Slide Content
® IGCSE is the registered trademark of Cambridge International Examinations.
This document consists of 2 printed pages.
© UCLES 2016 [Turn ove
r
Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/11
Paper 1 Multiple Choice October/November 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 October/November 2016 series for most
Cambridge IGCSE
®
, Cambridge International A and AS Level components and some Cambridge O Level
components.
This document consists of 2 printed pages.
© UCLES 2016 [Turn ove
r
Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/11
Paper 1 Multiple Choice October/November 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 October/November 2016 series for most
Cambridge IGCSE
®
, Cambridge International A and AS Level components and some Cambridge O Level
components.
Page 2 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 11
© UCLES 2016
Question
Number
Key
Question
Number
Key
1 D 21 D
2 B 22 A
3 A 23 A
4 A 24 D
5 D 25 D
6 B 26 C
7 B 27 A
8 C 28 A
9 A 29 C
10 C 30 A
11 A 31 B
12 D 32 A
13 A 33 C
14 A 34 B
15 D 35 B
16 B 36 D
17 C 37 A
18 C 38 A
19 A 39 C
20 C 40 B
Cambridge International AS/A Level – October/November 2016 9702 11
© UCLES 2016
Question
Number
Key
Question
Number
Key
1 D 21 D
2 B 22 A
3 A 23 A
4 A 24 D
5 D 25 D
6 B 26 C
7 B 27 A
8 C 28 A
9 A 29 C
10 C 30 A
11 A 31 B
12 D 32 A
13 A 33 C
14 A 34 B
15 D 35 B
16 B 36 D
17 C 37 A
18 C 38 A
19 A 39 C
20 C 40 B
® IGCSE is the registered trademark of Cambridge International Examinations.
This document consists of 2 printed pages.
© UCLES 2016 [Turn ove
r
Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/12
Paper 1 Multiple Choice October/November 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 October/November 2016 series for most
Cambridge IGCSE
®
, Cambridge International A and AS Level components and some Cambridge O Level
components.
This document consists of 2 printed pages.
© UCLES 2016 [Turn ove
r
Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/12
Paper 1 Multiple Choice October/November 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 October/November 2016 series for most
Cambridge IGCSE
®
, Cambridge International A and AS Level components and some Cambridge O Level
components.
Page 2 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 12
© UCLES 2016
Question
Number
Key
Question
Number
Key
1 B 21 D
2 D 22 B
3 D 23 A
4 C 24 B
5 A 25 A
6 D 26 B
7 B 27 D
8 B 28 C
9 D 29 C
10 C 30 B
11 A 31 A
12 D 32 C
13 B 33 D
14 D 34 B
15 C 35 B
16 C 36 B
17 C 37 C
18 B 38 A
19 C 39 C
20 C 40 A
Cambridge International AS/A Level – October/November 2016 9702 12
© UCLES 2016
Question
Number
Key
Question
Number
Key
1 B 21 D
2 D 22 B
3 D 23 A
4 C 24 B
5 A 25 A
6 D 26 B
7 B 27 D
8 B 28 C
9 D 29 C
10 C 30 B
11 A 31 A
12 D 32 C
13 B 33 D
14 D 34 B
15 C 35 B
16 C 36 B
17 C 37 C
18 B 38 A
19 C 39 C
20 C 40 A
® IGCSE is the registered trademark of Cambridge International Examinations.
This document consists of 2 printed pages.
© UCLES 2016 [Turn ove
r
Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/13
Paper 1 Multiple Choice October/November 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 October/November 2016 series for most
Cambridge IGCSE
®
, Cambridge International A and AS Level components and some Cambridge O Level
components.
This document consists of 2 printed pages.
© UCLES 2016 [Turn ove
r
Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/13
Paper 1 Multiple Choice October/November 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 October/November 2016 series for most
Cambridge IGCSE
®
, Cambridge International A and AS Level components and some Cambridge O Level
components.
Page 2 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 13
© UCLES 2016
Question
Number
Key
Question
Number
Key
1 D 21 D
2 B 22 A
3 A 23 A
4 A 24 D
5 D 25 D
6 B 26 C
7 B 27 A
8 C 28 A
9 A 29 C
10 C 30 A
11 A 31 B
12 D 32 A
13 A 33 C
14 A 34 B
15 D 35 B
16 B 36 D
17 C 37 A
18 C 38 A
19 A 39 C
20 C 40 B
Cambridge International AS/A Level – October/November 2016 9702 13
© UCLES 2016
Question
Number
Key
Question
Number
Key
1 D 21 D
2 B 22 A
3 A 23 A
4 A 24 D
5 D 25 D
6 B 26 C
7 B 27 A
8 C 28 A
9 A 29 C
10 C 30 A
11 A 31 B
12 D 32 A
13 A 33 C
14 A 34 B
15 D 35 B
16 B 36 D
17 C 37 A
18 C 38 A
19 A 39 C
20 C 40 B
® IGCSE is the registered trademark of Cambridge International Examinations.
This document consists of 5 printed pages.
© UCLES 2016 [Turn ove
r
Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/21
Paper 2 AS Level Structured Questions October/November 2016
MARK SCHEME
Maximum Mark: 60
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 October/November 2016 series for most
Cambridge IGCSE
®
, Cambridge International A and AS Level components and some Cambridge O Level
components.
This document consists of 5 printed pages.
© UCLES 2016 [Turn ove
r
Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/21
Paper 2 AS Level Structured Questions October/November 2016
MARK SCHEME
Maximum Mark: 60
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 October/November 2016 series for most
Cambridge IGCSE
®
, Cambridge International A and AS Level components and some Cambridge O Level
components.
Page 2 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 21
© UCLES 2016
1 (a) (density =) mass / volume B1 [1]
(b) (i) d = [(6 × 7.5)
/ (π × 8100)]
1/3
= 0.12(1) m A1 [1]
(ii) percentage uncertainty = (4 + 5)
/ 3 (= 3%)
or
fractional uncertainty = (0.04 + 0.05)
/ 3 (= 0.03) C1
absolute uncertainty (= 0.03 × 0.121) = 0.0036 C1
d = 0.121 ± 0.004
m A1 [3]
2 (a) force per unit positive charge B1 [1]
(b) (i) time = 5.9 × 10
–2
/ 3.7 × 10
7
= 1.6 × 10
–9
s (1.59 × 10
–9
s) A1 [1]
(ii) E = V
/ d C1
= 2500 / 4.0 × 10
–2
= 6.3 × 10
4
N C
–1
(6.25 × 10
4
or 62500 N C
–1
) A1 [2]
(iii) a = Eq
/ m or F = ma and F = Eq C1
= (6.3 × 10
4
× 1.60 × 10
–19
) / 9.11 × 10
–31
= 1.1 × 10
16
m s
–2
A1 [2]
(iv) s = ut + ½at
2
= ½ × 1.1 × 10
16
× (1.6 × 10
–9
)
2
C1
= 1.4 × 10
–2
(m) C1
distance from plate = 2.0 – 1.4
= 0.6
cm (allow 1 or more s.f.) A1 [3]
(v) electric force ≫ gravitational force (on electron)/weight
or
acceleration due to electric field ≫ acceleration due to gravitational field B1 [1]
(vi) v
X–t graph: horizontal line at a non-zero value of v X B1
v
Y–t graph: straight line through the origin with positive gradient B1 [2]
Cambridge International AS/A Level – October/November 2016 9702 21
© UCLES 2016
1 (a) (density =) mass / volume B1 [1]
(b) (i) d = [(6 × 7.5)
/ (π × 8100)]
1/3
= 0.12(1) m A1 [1]
(ii) percentage uncertainty = (4 + 5)
/ 3 (= 3%)
or
fractional uncertainty = (0.04 + 0.05)
/ 3 (= 0.03) C1
absolute uncertainty (= 0.03 × 0.121) = 0.0036 C1
d = 0.121 ± 0.004
m A1 [3]
2 (a) force per unit positive charge B1 [1]
(b) (i) time = 5.9 × 10
–2
/ 3.7 × 10
7
= 1.6 × 10
–9
s (1.59 × 10
–9
s) A1 [1]
(ii) E = V
/ d C1
= 2500 / 4.0 × 10
–2
= 6.3 × 10
4
N C
–1
(6.25 × 10
4
or 62500 N C
–1
) A1 [2]
(iii) a = Eq
/ m or F = ma and F = Eq C1
= (6.3 × 10
4
× 1.60 × 10
–19
) / 9.11 × 10
–31
= 1.1 × 10
16
m s
–2
A1 [2]
(iv) s = ut + ½at
2
= ½ × 1.1 × 10
16
× (1.6 × 10
–9
)
2
C1
= 1.4 × 10
–2
(m) C1
distance from plate = 2.0 – 1.4
= 0.6
cm (allow 1 or more s.f.) A1 [3]
(v) electric force ≫ gravitational force (on electron)/weight
or
acceleration due to electric field ≫ acceleration due to gravitational field B1 [1]
(vi) v
X–t graph: horizontal line at a non-zero value of v X B1
v
Y–t graph: straight line through the origin with positive gradient B1 [2]
Page 3 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 21
© UCLES 2016
3 (a) force/load is proportional to extension/compression (provided proportionality limit
is not exceeded) B1 [1]
(b) (i) k = F
/ x or k = gradient C1
k = 600
N m
–1
A1 [2]
(ii) (W =) ½kx
2
or (W =) ½Fx or (W =) area under graph C1
(W =) 0.5 × 600 × (0.040)
2
= 0.48 J or (W =) 0.5 × 24 × 0.040 = 0.48 J A1 [2]
(iii) 1. (E
K =) ½mv
2
C1
= ½
× 0.025 × 6.0
2
= 0.45
J A1 [2]
2. (work done against resistive force =) 0.48 – 0.45 [= 0.03(0)
J] C1
average resistive force = 0.030
/ 0.040 C1
= 0.75
N A1 [3]
(iv) efficiency = [useful energy out
/ total energy in] (×100) C1
= [0.45
/ 0.48] (×100)
= 0.94 or 94% A1 [2]
4 (a) the number of oscillations per unit time M1
of the source/of a point on the wave/of a particle (in the medium) A1 [2]
or
the number of wavelengths/wavefronts per unit time (M1)
passing a (fixed) point (A1)
(b) T or period = 2.5 × 250 (µs) (= 625 µs) M1
frequency = 1
/ (6.25 × 10
–4
) or 1 / (2.5 × 250 × 10
–6
) = 1600 Hz A1 [2]
(c) (i) for maximum frequency: f
o = fsv / (v – v s)
1640 = (1600 × 330) / (330 – v
s) C1
v
s = 8(.0) m s
–1
(8.049 m s
–1
) A1 [2]
(ii) loudspeaker moving towards observer causes rise in/higher frequency B1
loudspeaker moving away from observer causes fall in/lower frequency B1 [2]
or
repeated rise and fall/higher and then lower frequency (M1)
caused by loudspeaker moving towards and away from observer (A1)
Cambridge International AS/A Level – October/November 2016 9702 21
© UCLES 2016
3 (a) force/load is proportional to extension/compression (provided proportionality limit
is not exceeded) B1 [1]
(b) (i) k = F
/ x or k = gradient C1
k = 600
N m
–1
A1 [2]
(ii) (W =) ½kx
2
or (W =) ½Fx or (W =) area under graph C1
(W =) 0.5 × 600 × (0.040)
2
= 0.48 J or (W =) 0.5 × 24 × 0.040 = 0.48 J A1 [2]
(iii) 1. (E
K =) ½mv
2
C1
= ½
× 0.025 × 6.0
2
= 0.45
J A1 [2]
2. (work done against resistive force =) 0.48 – 0.45 [= 0.03(0)
J] C1
average resistive force = 0.030
/ 0.040 C1
= 0.75
N A1 [3]
(iv) efficiency = [useful energy out
/ total energy in] (×100) C1
= [0.45
/ 0.48] (×100)
= 0.94 or 94% A1 [2]
4 (a) the number of oscillations per unit time M1
of the source/of a point on the wave/of a particle (in the medium) A1 [2]
or
the number of wavelengths/wavefronts per unit time (M1)
passing a (fixed) point (A1)
(b) T or period = 2.5 × 250 (µs) (= 625 µs) M1
frequency = 1
/ (6.25 × 10
–4
) or 1 / (2.5 × 250 × 10
–6
) = 1600 Hz A1 [2]
(c) (i) for maximum frequency: f
o = fsv / (v – v s)
1640 = (1600 × 330) / (330 – v
s) C1
v
s = 8(.0) m s
–1
(8.049 m s
–1
) A1 [2]
(ii) loudspeaker moving towards observer causes rise in/higher frequency B1
loudspeaker moving away from observer causes fall in/lower frequency B1 [2]
or
repeated rise and fall/higher and then lower frequency (M1)
caused by loudspeaker moving towards and away from observer (A1)
Page 4 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 21
© UCLES 2016
5 (a) wave incident on/passes by or through an aperture/edge B1
wave spreads (into geometrical shadow) B1 [2]
(b) n
λ = d sinθ C1
substitution of
θ = 90° or sin θ = 1 C1
4 × 500 × 10
–9
= d × sin 90°
line spacing = 2.0 × 10
–6
m A1 [3]
(c) wavelength of red light is longer (than 500
nm) M1
(each order/fourth order is now at a greater angle so) the fifth-order maximum
cannot be formed/not formed A1 [2]
6 (a)
charge
forms) other to electrical (from ed)(transformenergy or done work
B1 [1]
(b) (i) 1. V = IR or E = IR C1
I = 14
/ 6.0
= 2.3 (2.33)
A A1 [2]
2. total resistance of parallel resistors = 8.0
Ω C1
current = 14
/ (6.0 + 8.0)
= 1.0
A A1 [2]
(ii) P = EI (allow P = VI) or P = V
2
/ R or P = I
2
R C1
change in power = (14 × 2.33) – (14 × 1.0)
or (14
2
/ 6.0) – (14
2
/ 14)
or (2.33
2
× 6.0) – (1.0
2
× 14)
= 19
W (18 W if 2.3 A used) A1 [2]
(c) I = Anvq
ratio = (0.50n
/ n) × (1.8 A / A) or ratio = 0.50 × 1.8 C1
= 0.90 A1 [2]
Cambridge International AS/A Level – October/November 2016 9702 21
© UCLES 2016
5 (a) wave incident on/passes by or through an aperture/edge B1
wave spreads (into geometrical shadow) B1 [2]
(b) n
λ = d sinθ C1
substitution of
θ = 90° or sin θ = 1 C1
4 × 500 × 10
–9
= d × sin 90°
line spacing = 2.0 × 10
–6
m A1 [3]
(c) wavelength of red light is longer (than 500
nm) M1
(each order/fourth order is now at a greater angle so) the fifth-order maximum
cannot be formed/not formed A1 [2]
6 (a)
charge
forms) other to electrical (from ed)(transformenergy or done work
B1 [1]
(b) (i) 1. V = IR or E = IR C1
I = 14
/ 6.0
= 2.3 (2.33)
A A1 [2]
2. total resistance of parallel resistors = 8.0
Ω C1
current = 14
/ (6.0 + 8.0)
= 1.0
A A1 [2]
(ii) P = EI (allow P = VI) or P = V
2
/ R or P = I
2
R C1
change in power = (14 × 2.33) – (14 × 1.0)
or (14
2
/ 6.0) – (14
2
/ 14)
or (2.33
2
× 6.0) – (1.0
2
× 14)
= 19
W (18 W if 2.3 A used) A1 [2]
(c) I = Anvq
ratio = (0.50n
/ n) × (1.8 A / A) or ratio = 0.50 × 1.8 C1
= 0.90 A1 [2]
Page 5 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 21
© UCLES 2016
7 (a) hadron not a fundamental particle/lepton is fundamental particle
or
hadron made of quarks/lepton not made of quarks
or
strong force/interaction acts on hadrons/does not act on leptons B1 [1]
(b) (i) proton: up, up, down
/ uud B1
neutron: up, down, down
/ udd B1 [2]
(ii) composition: 2(uud) + 2(udd)
= 6 up, 6 down
/ 6u, 6d B1 [1]
(c) (i) most of the atom is empty space
or
the nucleus (volume) is (very) small compared to the atom B1 [1]
(ii) nucleus is (positively) charged B1
the mass is concentrated in (very small) nucleus/small region/small
volume/small core
or
the majority of mass in (very small) nucleus/small region/small volume/small
core B1 [2]
Cambridge International AS/A Level – October/November 2016 9702 21
© UCLES 2016
7 (a) hadron not a fundamental particle/lepton is fundamental particle
or
hadron made of quarks/lepton not made of quarks
or
strong force/interaction acts on hadrons/does not act on leptons B1 [1]
(b) (i) proton: up, up, down
/ uud B1
neutron: up, down, down
/ udd B1 [2]
(ii) composition: 2(uud) + 2(udd)
= 6 up, 6 down
/ 6u, 6d B1 [1]
(c) (i) most of the atom is empty space
or
the nucleus (volume) is (very) small compared to the atom B1 [1]
(ii) nucleus is (positively) charged B1
the mass is concentrated in (very small) nucleus/small region/small
volume/small core
or
the majority of mass in (very small) nucleus/small region/small volume/small
core B1 [2]
® IGCSE is the registered trademark of Cambridge International Examinations.
This document consists of 5 printed pages.
© UCLES 2016 [Turn ove
r
Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/22
Paper 2 AS Structured Questions October/November 2016
MARK SCHEME
Maximum Mark: 60
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 October/November 2016 series for most
Cambridge IGCSE
®
, Cambridge International A and AS Level components and some Cambridge O Level
components.
This document consists of 5 printed pages.
© UCLES 2016 [Turn ove
r
Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/22
Paper 2 AS Structured Questions October/November 2016
MARK SCHEME
Maximum Mark: 60
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 October/November 2016 series for most
Cambridge IGCSE
®
, Cambridge International A and AS Level components and some Cambridge O Level
components.
Page 2 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 22
© UCLES 2016
1 (a) (i) force / area (normal to the force) B1 [1]
(ii) (p = F
/ A so) units: kg m s
–2
/ m
2
= kg m
–1
s
–2
A1 [1]
allow use of other correct equations:
e.g. ( ∆p =
ρg∆h so) kg m
–3
m s
–2
m = kg m
–1
s
–2
e.g. ( p = W
/ ∆V so) kg m s
–2
m / m
3
= kg m
–1
s
–2
(b) units for m: kg, t: s and ρ: kg m
–3
C1
units of C: kg
/ s (kg m
–3
kg m
–1
s
–2
)
1/2
or
units of C
2
: kg
2
/ s
2
kg m
–3
kg m
–1
s
–2
C1
units of C: m
2
A1 [3]
2 (a) ∆E = mg∆h C1
= 0.030 × 9.81 × (–)0.31
= (–)0.091 J A1 [2]
(b) E = ½mv
2
C1
(initial) E = ½ × 0.030 × 1.3
2
(= 0.0254) C1
0.5 × 0.030 × v
2
= (0.5 × 0.030 × 1.3
2
) + (0.030 × 9.81 × 0.31) so v = 2.8 m s
–1
or
0.5 × 0.030 × v
2
= (0.0254) + (0.091) so v = 2.8 m s
–1
A1 [3]
(c) (i) 0.096 = 0.030
(v + 2.8) C1
v = 0.40
m s
–1
A1 [2]
(ii) F = ∆p
/ (∆)t or F = ma
= 0.096
/ 20 × 10
–3
or 0.030 (0.40 + 2.8) / 20 × 10
–3
C1
= 4.8 N A1 [2]
(d) kinetic energy (of ball and wall) decreases/changes/not conserved, so inelastic
or
(relative) speed of approach (of ball and wall) not equal to/greater than (relative)
speed of separation, so inelastic. B1 [1]
(e) force = work done
/ distance moved
= (0.091 – 0.076)
/ 0.60 C1
= 0.025
N A1 [2]
Cambridge International AS/A Level – October/November 2016 9702 22
© UCLES 2016
1 (a) (i) force / area (normal to the force) B1 [1]
(ii) (p = F
/ A so) units: kg m s
–2
/ m
2
= kg m
–1
s
–2
A1 [1]
allow use of other correct equations:
e.g. ( ∆p =
ρg∆h so) kg m
–3
m s
–2
m = kg m
–1
s
–2
e.g. ( p = W
/ ∆V so) kg m s
–2
m / m
3
= kg m
–1
s
–2
(b) units for m: kg, t: s and ρ: kg m
–3
C1
units of C: kg
/ s (kg m
–3
kg m
–1
s
–2
)
1/2
or
units of C
2
: kg
2
/ s
2
kg m
–3
kg m
–1
s
–2
C1
units of C: m
2
A1 [3]
2 (a) ∆E = mg∆h C1
= 0.030 × 9.81 × (–)0.31
= (–)0.091 J A1 [2]
(b) E = ½mv
2
C1
(initial) E = ½ × 0.030 × 1.3
2
(= 0.0254) C1
0.5 × 0.030 × v
2
= (0.5 × 0.030 × 1.3
2
) + (0.030 × 9.81 × 0.31) so v = 2.8 m s
–1
or
0.5 × 0.030 × v
2
= (0.0254) + (0.091) so v = 2.8 m s
–1
A1 [3]
(c) (i) 0.096 = 0.030
(v + 2.8) C1
v = 0.40
m s
–1
A1 [2]
(ii) F = ∆p
/ (∆)t or F = ma
= 0.096
/ 20 × 10
–3
or 0.030 (0.40 + 2.8) / 20 × 10
–3
C1
= 4.8 N A1 [2]
(d) kinetic energy (of ball and wall) decreases/changes/not conserved, so inelastic
or
(relative) speed of approach (of ball and wall) not equal to/greater than (relative)
speed of separation, so inelastic. B1 [1]
(e) force = work done
/ distance moved
= (0.091 – 0.076)
/ 0.60 C1
= 0.025
N A1 [2]
Page 3 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 22
© UCLES 2016
3 (a) resultant force (in any direction) is zero B1
resultant moment/torque (about any point) is zero B1 [2]
(b) (i) force = 33 sin 52° or 33 cos 38°
= 26
N A1 [1]
(ii) 26 × 0.30 or W × 0.20 or 12 × 0.40 C1
26 × 0.30 = (W × 0.20) + (12 × 0.40) C1
W = 15
N A1 [3]
(c) (i) E = ∆
σ / ∆ε or E = σ / ε C1
∆
σ = 2.0 × 10
11
× 7.5 × 10
–4
= 1.5 × 10
8
Pa A1 [2]
(ii) ∆
σ = ∆F / A or σ = F / A C1
A = 78
/ 1.5 × 10
8
(= 5.2 × 10
–7
m
2
) C1
5.2 × 10
–7
= πd
2
/ 4
d = 8.1 × 10
–4
m A1 [3]
4 (a) wave incident on/passes by or through an aperture/edge B1
wave spreads (into geometrical shadow) B1 [2]
(b) (i) waves (from slits) overlap (at point X) B1
path difference (from slits to X) is zero/
phase difference (between the two waves) is zero
(so constructive interference gives bright fringe) B1 [2]
(ii) difference in distances =
λ / 2 = 580 / 2
= 290
nm A1 [1]
(iii)
λ = ax / D C1
D = [0.41 × 10
–3
× (2 × 2.0 × 10
–3
)] / 580 × 10
–9
C1
= 2.8
m A1 [3]
(iv) same separation/fringe width/number of fringes
bright fringe(s)/central bright fringe/(fringe at) X less bright
dark fringe(s)/(fringe at) Y/(fringe at) Z brighter
contrast between fringes decreases
Any two of the above four points, 1 mark each B2 [2]
Cambridge International AS/A Level – October/November 2016 9702 22
© UCLES 2016
3 (a) resultant force (in any direction) is zero B1
resultant moment/torque (about any point) is zero B1 [2]
(b) (i) force = 33 sin 52° or 33 cos 38°
= 26
N A1 [1]
(ii) 26 × 0.30 or W × 0.20 or 12 × 0.40 C1
26 × 0.30 = (W × 0.20) + (12 × 0.40) C1
W = 15
N A1 [3]
(c) (i) E = ∆
σ / ∆ε or E = σ / ε C1
∆
σ = 2.0 × 10
11
× 7.5 × 10
–4
= 1.5 × 10
8
Pa A1 [2]
(ii) ∆
σ = ∆F / A or σ = F / A C1
A = 78
/ 1.5 × 10
8
(= 5.2 × 10
–7
m
2
) C1
5.2 × 10
–7
= πd
2
/ 4
d = 8.1 × 10
–4
m A1 [3]
4 (a) wave incident on/passes by or through an aperture/edge B1
wave spreads (into geometrical shadow) B1 [2]
(b) (i) waves (from slits) overlap (at point X) B1
path difference (from slits to X) is zero/
phase difference (between the two waves) is zero
(so constructive interference gives bright fringe) B1 [2]
(ii) difference in distances =
λ / 2 = 580 / 2
= 290
nm A1 [1]
(iii)
λ = ax / D C1
D = [0.41 × 10
–3
× (2 × 2.0 × 10
–3
)] / 580 × 10
–9
C1
= 2.8
m A1 [3]
(iv) same separation/fringe width/number of fringes
bright fringe(s)/central bright fringe/(fringe at) X less bright
dark fringe(s)/(fringe at) Y/(fringe at) Z brighter
contrast between fringes decreases
Any two of the above four points, 1 mark each B2 [2]
Page 4 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 22
© UCLES 2016
5 (a) total/sum of electromotive forces or e.m.f.s
= total/sum of potential differences or p.d.s M1
around a loop/(closed) circuit A1 [2]
(b) (i) (current in battery =) current in A + current in B or I
A + IB C1
( I =) 0.14 + 0.26 = 0.40
A A1 [2]
(ii) E = V + Ir
6.8 = 6.0 + 0.40 r or 6.8 = 0.40
(15 + r) C1
r = 2.0 Ω A1 [2]
(iii) R = V
/ I C1
ratio (= R
A
/ RB) = (6.0 / 0.14) / (6.0 / 0.26)
= 42.9
/ 23.1 or 0.26 / 0.14
= 1.9 (1.86) A1 [2]
(iv) 1. P = EI or VI or P = I
2
R or P = V
2
/
R C1
= 6.8 × 0.40 = 0.40
2
× 17 = 6.8
2
/ 17
= 2.7
W (2.72 W) A1 [2]
2. output power = VI
= 6.0 × 0.40 (= 2.40
W) C1
efficiency = (6.0 × 0.40)
/ (6.8 × 0.40) = 2.40 / 2.72
= 0.88 or 88% ( allow 0.89 or 89%) A1 [2]
Cambridge International AS/A Level – October/November 2016 9702 22
© UCLES 2016
5 (a) total/sum of electromotive forces or e.m.f.s
= total/sum of potential differences or p.d.s M1
around a loop/(closed) circuit A1 [2]
(b) (i) (current in battery =) current in A + current in B or I
A + IB C1
( I =) 0.14 + 0.26 = 0.40
A A1 [2]
(ii) E = V + Ir
6.8 = 6.0 + 0.40 r or 6.8 = 0.40
(15 + r) C1
r = 2.0 Ω A1 [2]
(iii) R = V
/ I C1
ratio (= R
A
/ RB) = (6.0 / 0.14) / (6.0 / 0.26)
= 42.9
/ 23.1 or 0.26 / 0.14
= 1.9 (1.86) A1 [2]
(iv) 1. P = EI or VI or P = I
2
R or P = V
2
/
R C1
= 6.8 × 0.40 = 0.40
2
× 17 = 6.8
2
/ 17
= 2.7
W (2.72 W) A1 [2]
2. output power = VI
= 6.0 × 0.40 (= 2.40
W) C1
efficiency = (6.0 × 0.40)
/ (6.8 × 0.40) = 2.40 / 2.72
= 0.88 or 88% ( allow 0.89 or 89%) A1 [2]
Page 5 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 22
© UCLES 2016
6 (a) hadron not a fundamental particle/lepton is fundamental particle
or
hadron made of quarks/lepton not made of quarks
or
strong force/interaction acts on hadrons/does not act on leptons B1 [1]
(b) (i)
e
0
1
)(+
or
)+(
β
0
1
B1
)(e
0
0
ν B1 [2]
(ii) weak (nuclear force
/ interaction) B1 [1]
(iii) • mass-energy
• momentum
• proton number
• nucleon number
• charge
Any three of the above quantities, 1 mark each B3 [3]
(c) (quark structure of proton is) up, up, down or uud B1
up/u (quark charge) is (+)⅔(e), down/d (quark charge) is –⅓(e) C1
⅔e + ⅔e – ⅓e = (+)e A1 [3]
Cambridge International AS/A Level – October/November 2016 9702 22
© UCLES 2016
6 (a) hadron not a fundamental particle/lepton is fundamental particle
or
hadron made of quarks/lepton not made of quarks
or
strong force/interaction acts on hadrons/does not act on leptons B1 [1]
(b) (i)
e
0
1
)(+
or
)+(
β
0
1
B1
)(e
0
0
ν B1 [2]
(ii) weak (nuclear force
/ interaction) B1 [1]
(iii) • mass-energy
• momentum
• proton number
• nucleon number
• charge
Any three of the above quantities, 1 mark each B3 [3]
(c) (quark structure of proton is) up, up, down or uud B1
up/u (quark charge) is (+)⅔(e), down/d (quark charge) is –⅓(e) C1
⅔e + ⅔e – ⅓e = (+)e A1 [3]
® IGCSE is the registered trademark of Cambridge International Examinations.
This document consists of 5 printed pages.
© UCLES 2016 [Turn ove
r
Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/23
Paper 2 AS Level Structured Questions October/November 2016
MARK SCHEME
Maximum Mark: 60
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 October/November 2016 series for most
Cambridge IGCSE
®
, Cambridge International A and AS Level components and some Cambridge O Level
components.
This document consists of 5 printed pages.
© UCLES 2016 [Turn ove
r
Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/23
Paper 2 AS Level Structured Questions October/November 2016
MARK SCHEME
Maximum Mark: 60
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 October/November 2016 series for most
Cambridge IGCSE
®
, Cambridge International A and AS Level components and some Cambridge O Level
components.
Page 2 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 23
© UCLES 2016
1 (a) (density =) mass / volume B1 [1]
(b) (i) d = [(6 × 7.5)
/ (π × 8100)]
1/3
= 0.12(1) m A1 [1]
(ii) percentage uncertainty = (4 + 5)
/ 3 (= 3%)
or
fractional uncertainty = (0.04 + 0.05)
/ 3 (= 0.03) C1
absolute uncertainty (= 0.03 × 0.121) = 0.0036 C1
d = 0.121 ± 0.004
m A1 [3]
2 (a) force per unit positive charge B1 [1]
(b) (i) time = 5.9 × 10
–2
/ 3.7 × 10
7
= 1.6 × 10
–9
s (1.59 × 10
–9
s) A1 [1]
(ii) E = V
/ d C1
= 2500 / 4.0 × 10
–2
= 6.3 × 10
4
N C
–1
(6.25 × 10
4
or 62500 N C
–1
) A1 [2]
(iii) a = Eq
/ m or F = ma and F = Eq C1
= (6.3 × 10
4
× 1.60 × 10
–19
) / 9.11 × 10
–31
= 1.1 × 10
16
m s
–2
A1 [2]
(iv) s = ut + ½at
2
= ½ × 1.1 × 10
16
× (1.6 × 10
–9
)
2
C1
= 1.4 × 10
–2
(m) C1
distance from plate = 2.0 – 1.4
= 0.6
cm (allow 1 or more s.f.) A1 [3]
(v) electric force ≫ gravitational force (on electron)/weight
or
acceleration due to electric field ≫ acceleration due to gravitational field B1 [1]
(vi) v
X–t graph: horizontal line at a non-zero value of v X B1
v
Y–t graph: straight line through the origin with positive gradient B1 [2]
Cambridge International AS/A Level – October/November 2016 9702 23
© UCLES 2016
1 (a) (density =) mass / volume B1 [1]
(b) (i) d = [(6 × 7.5)
/ (π × 8100)]
1/3
= 0.12(1) m A1 [1]
(ii) percentage uncertainty = (4 + 5)
/ 3 (= 3%)
or
fractional uncertainty = (0.04 + 0.05)
/ 3 (= 0.03) C1
absolute uncertainty (= 0.03 × 0.121) = 0.0036 C1
d = 0.121 ± 0.004
m A1 [3]
2 (a) force per unit positive charge B1 [1]
(b) (i) time = 5.9 × 10
–2
/ 3.7 × 10
7
= 1.6 × 10
–9
s (1.59 × 10
–9
s) A1 [1]
(ii) E = V
/ d C1
= 2500 / 4.0 × 10
–2
= 6.3 × 10
4
N C
–1
(6.25 × 10
4
or 62500 N C
–1
) A1 [2]
(iii) a = Eq
/ m or F = ma and F = Eq C1
= (6.3 × 10
4
× 1.60 × 10
–19
) / 9.11 × 10
–31
= 1.1 × 10
16
m s
–2
A1 [2]
(iv) s = ut + ½at
2
= ½ × 1.1 × 10
16
× (1.6 × 10
–9
)
2
C1
= 1.4 × 10
–2
(m) C1
distance from plate = 2.0 – 1.4
= 0.6
cm (allow 1 or more s.f.) A1 [3]
(v) electric force ≫ gravitational force (on electron)/weight
or
acceleration due to electric field ≫ acceleration due to gravitational field B1 [1]
(vi) v
X–t graph: horizontal line at a non-zero value of v X B1
v
Y–t graph: straight line through the origin with positive gradient B1 [2]
Page 3 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 23
© UCLES 2016
3 (a) force/load is proportional to extension/compression (provided proportionality limit
is not exceeded) B1 [1]
(b) (i) k = F
/ x or k = gradient C1
k = 600
N m
–1
A1 [2]
(ii) (W =) ½kx
2
or (W =) ½Fx or (W =) area under graph C1
(W =) 0.5 × 600 × (0.040)
2
= 0.48 J or (W =) 0.5 × 24 × 0.040 = 0.48 J A1 [2]
(iii) 1. (E
K =) ½mv
2
C1
= ½
× 0.025 × 6.0
2
= 0.45
J A1 [2]
2. (work done against resistive force =) 0.48 – 0.45 [= 0.03(0)
J] C1
average resistive force = 0.030
/ 0.040 C1
= 0.75
N A1 [3]
(iv) efficiency = [useful energy out
/ total energy in] (×100) C1
= [0.45
/ 0.48] (×100)
= 0.94 or 94% A1 [2]
4 (a) the number of oscillations per unit time M1
of the source/of a point on the wave/of a particle (in the medium) A1 [2]
or
the number of wavelengths/wavefronts per unit time (M1)
passing a (fixed) point (A1)
(b) T or period = 2.5 × 250 (µs) (= 625 µs) M1
frequency = 1
/ (6.25 × 10
–4
) or 1 / (2.5 × 250 × 10
–6
) = 1600 Hz A1 [2]
(c) (i) for maximum frequency: f
o = fsv / (v – v s)
1640 = (1600 × 330) / (330 – v
s) C1
v
s = 8(.0) m s
–1
(8.049 m s
–1
) A1 [2]
(ii) loudspeaker moving towards observer causes rise in/higher frequency B1
loudspeaker moving away from observer causes fall in/lower frequency B1 [2]
or
repeated rise and fall/higher and then lower frequency (M1)
caused by loudspeaker moving towards and away from observer (A1)
Cambridge International AS/A Level – October/November 2016 9702 23
© UCLES 2016
3 (a) force/load is proportional to extension/compression (provided proportionality limit
is not exceeded) B1 [1]
(b) (i) k = F
/ x or k = gradient C1
k = 600
N m
–1
A1 [2]
(ii) (W =) ½kx
2
or (W =) ½Fx or (W =) area under graph C1
(W =) 0.5 × 600 × (0.040)
2
= 0.48 J or (W =) 0.5 × 24 × 0.040 = 0.48 J A1 [2]
(iii) 1. (E
K =) ½mv
2
C1
= ½
× 0.025 × 6.0
2
= 0.45
J A1 [2]
2. (work done against resistive force =) 0.48 – 0.45 [= 0.03(0)
J] C1
average resistive force = 0.030
/ 0.040 C1
= 0.75
N A1 [3]
(iv) efficiency = [useful energy out
/ total energy in] (×100) C1
= [0.45
/ 0.48] (×100)
= 0.94 or 94% A1 [2]
4 (a) the number of oscillations per unit time M1
of the source/of a point on the wave/of a particle (in the medium) A1 [2]
or
the number of wavelengths/wavefronts per unit time (M1)
passing a (fixed) point (A1)
(b) T or period = 2.5 × 250 (µs) (= 625 µs) M1
frequency = 1
/ (6.25 × 10
–4
) or 1 / (2.5 × 250 × 10
–6
) = 1600 Hz A1 [2]
(c) (i) for maximum frequency: f
o = fsv / (v – v s)
1640 = (1600 × 330) / (330 – v
s) C1
v
s = 8(.0) m s
–1
(8.049 m s
–1
) A1 [2]
(ii) loudspeaker moving towards observer causes rise in/higher frequency B1
loudspeaker moving away from observer causes fall in/lower frequency B1 [2]
or
repeated rise and fall/higher and then lower frequency (M1)
caused by loudspeaker moving towards and away from observer (A1)
Page 4 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 23
© UCLES 2016
5 (a) wave incident on/passes by or through an aperture/edge B1
wave spreads (into geometrical shadow) B1 [2]
(b) n
λ = d sinθ C1
substitution of
θ = 90° or sin θ = 1 C1
4 × 500 × 10
–9
= d × sin 90°
line spacing = 2.0 × 10
–6
m A1 [3]
(c) wavelength of red light is longer (than 500
nm) M1
(each order/fourth order is now at a greater angle so) the fifth-order maximum
cannot be formed/not formed A1 [2]
6 (a)
charge
forms) other to electrical (from ed)(transformenergy or done work
B1 [1]
(b) (i) 1. V
= IR or E = IR C1
I
= 14 / 6.0
= 2.3 (2.33)
A A1 [2]
2.
total resistance of parallel resistors = 8.0 Ω C1
current = 14
/ (6.0 + 8.0)
= 1.0
A A1 [2]
(ii) P
= EI (allow P = VI) or P = V
2
/ R or P = I
2
R C1
change in power = (14 × 2.33) – (14 × 1.0)
or (14
2
/ 6.0) – (14
2
/ 14)
or (2.33
2
× 6.0) – (1.0
2
× 14)
= 19
W (18 W if 2.3 A used) A1 [2]
(c) I
= Anvq
ratio = (0.50n / n) × (1.8 A / A) or ratio = 0.50 × 1.8 C1
= 0.90 A1 [2]
Cambridge International AS/A Level – October/November 2016 9702 23
© UCLES 2016
5 (a) wave incident on/passes by or through an aperture/edge B1
wave spreads (into geometrical shadow) B1 [2]
(b) n
λ = d sinθ C1
substitution of
θ = 90° or sin θ = 1 C1
4 × 500 × 10
–9
= d × sin 90°
line spacing = 2.0 × 10
–6
m A1 [3]
(c) wavelength of red light is longer (than 500
nm) M1
(each order/fourth order is now at a greater angle so) the fifth-order maximum
cannot be formed/not formed A1 [2]
6 (a)
charge
forms) other to electrical (from ed)(transformenergy or done work
B1 [1]
(b) (i) 1. V
= IR or E = IR C1
I
= 14 / 6.0
= 2.3 (2.33)
A A1 [2]
2.
total resistance of parallel resistors = 8.0 Ω C1
current = 14
/ (6.0 + 8.0)
= 1.0
A A1 [2]
(ii) P
= EI (allow P = VI) or P = V
2
/ R or P = I
2
R C1
change in power = (14 × 2.33) – (14 × 1.0)
or (14
2
/ 6.0) – (14
2
/ 14)
or (2.33
2
× 6.0) – (1.0
2
× 14)
= 19
W (18 W if 2.3 A used) A1 [2]
(c) I
= Anvq
ratio = (0.50n / n) × (1.8 A / A) or ratio = 0.50 × 1.8 C1
= 0.90 A1 [2]
Page 5 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 23
© UCLES 2016
7 (a) hadron not a fundamental particle/lepton is fundamental particle
or
hadron made of quarks/lepton not made of quarks
or
strong force/interaction acts on hadrons/does not act on leptons B1 [1]
(b) (i) proton: up, up, down
/ uud B1
neutron: up, down, down
/ udd B1 [2]
(ii) composition: 2(uud) + 2(udd)
= 6 up, 6 down
/ 6u, 6d B1 [1]
(c) (i) most of the atom is empty space
or
the nucleus (volume) is (very) small compared to the atom B1 [1]
(ii) nucleus is (positively) charged B1
the mass is concentrated in (very small) nucleus/small region/small
volume/small core
or
the majority of mass in (very small) nucleus/small region/small volume/small
core B1 [2]
Cambridge International AS/A Level – October/November 2016 9702 23
© UCLES 2016
7 (a) hadron not a fundamental particle/lepton is fundamental particle
or
hadron made of quarks/lepton not made of quarks
or
strong force/interaction acts on hadrons/does not act on leptons B1 [1]
(b) (i) proton: up, up, down
/ uud B1
neutron: up, down, down
/ udd B1 [2]
(ii) composition: 2(uud) + 2(udd)
= 6 up, 6 down
/ 6u, 6d B1 [1]
(c) (i) most of the atom is empty space
or
the nucleus (volume) is (very) small compared to the atom B1 [1]
(ii) nucleus is (positively) charged B1
the mass is concentrated in (very small) nucleus/small region/small
volume/small core
or
the majority of mass in (very small) nucleus/small region/small volume/small
core B1 [2]
® IGCSE is the registered trademark of Cambridge International Examinations.
This document consists of 4 printed pages.
© UCLES 2016 [Turn ove
r
Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/31
Paper 3 Advanced Practical Skills 1 October/November 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 October/November 2016 series for most
Cambridge IGCSE
®
, Cambridge International A and AS Level components and some Cambridge O Level
components.
This document consists of 4 printed pages.
© UCLES 2016 [Turn ove
r
Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/31
Paper 3 Advanced Practical Skills 1 October/November 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 October/November 2016 series for most
Cambridge IGCSE
®
, Cambridge International A and AS Level components and some Cambridge O Level
components.
Page 2 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 31
© UCLES 2016
1 (b) Value for T in the range 0.60 s to 0.80 s with unit. [1]
Evidence of repeat timings (at least two recordings of nT where n ⩾ 5). [1]
(c) Correct calculation of k. [1]
Value of k must be given to the same number of s.f. as (or one more than) the s.f.
in the values of the raw times. [1]
(e) Six sets of readings of (h – h
1) and T (with correct trend and without help from
Supervisor) scores 5 marks, five sets scores 4 marks etc. [5]
Range: [1]
∆(h – h
1) ⩾ 30.0 cm.
Column headings: [1]
Each column heading must contain a quantity and a unit where appropriate.
The presentation of the quantity and unit must conform to accepted scientific
convention, e.g. T
/ s or T (s).
Consistency: [1]
All values of (h – h
1) must be given to the nearest mm.
(f) (i) Axes: [1]
Sensible scales must be used. Awkward scales (e.g. 3:10, fractions or non-linear)
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 must be plotted on the grid.
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 for this mark to be awarded.
All points must be within 0.01
s
on the y-axis of 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
5 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 (i.e. circled or labelled) by
the candidate. There must be at least five points left after the anomalous
point is disregarded.
Line must not be kinked or thicker than half a small square.
Cambridge International AS/A Level – October/November 2016 9702 31
© UCLES 2016
1 (b) Value for T in the range 0.60 s to 0.80 s with unit. [1]
Evidence of repeat timings (at least two recordings of nT where n ⩾ 5). [1]
(c) Correct calculation of k. [1]
Value of k must be given to the same number of s.f. as (or one more than) the s.f.
in the values of the raw times. [1]
(e) Six sets of readings of (h – h
1) and T (with correct trend and without help from
Supervisor) scores 5 marks, five sets scores 4 marks etc. [5]
Range: [1]
∆(h – h
1) ⩾ 30.0 cm.
Column headings: [1]
Each column heading must contain a quantity and a unit where appropriate.
The presentation of the quantity and unit must conform to accepted scientific
convention, e.g. T
/ s or T (s).
Consistency: [1]
All values of (h – h
1) must be given to the nearest mm.
(f) (i) Axes: [1]
Sensible scales must be used. Awkward scales (e.g. 3:10, fractions or non-linear)
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 must be plotted on the grid.
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 for this mark to be awarded.
All points must be within 0.01
s
on the y-axis of 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
5 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 (i.e. circled or labelled) by
the candidate. There must be at least five points left after the anomalous
point is disregarded.
Line must not be kinked or thicker than half a small square.
Page 3 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 31
© UCLES 2016
(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.
Both read-offs must be accurate to half a small square in both the x and y
directions.
y-intercept: [1]
Either:
Check correct read-off from a point on the line and substituted into y = mx + c.
Read-off must be accurate to half a small square in both x and y directions.
Or:
Check read-off of the intercept directly from the graph (accurate to half a
small square.
(g) Value of P = candidate’s gradient and value of Q = candidate’s intercept. [1]
Do not allow fractions.
Units for P (e.g. s
m
–1
or s cm
–1
or s mm
–1
) and Q (s) correct. [1]
2 (a) (i) Value for D with unit in the range 0.14
mm to 0.16 mm. [1]
(ii) Percentage uncertainty in D based on an absolute uncertainty of 0.01
mm.
Correct method of calculation to obtain percentage uncertainty. [1]
(c) (iii) Value of I in range 10
mA ⩽ I ⩽ 200 mA with unit (collected without help from
Supervisor). [1]
(iv) Value of V in range 0.4
V ⩽ V ⩽ 1.0 V with unit (collected without help from
Supervisor). [1]
(d) (i) Value of d > D and d < 1
mm. [1]
(ii) Correct calculation of G. [1]
(iii) Justification for s.f. in G linked to s.f. in D and d. [1]
(f) (i) Second value of V. [1]
Quality: second value of V less than first value of V. [1]
(ii) Second value of d. [1]
Cambridge International AS/A Level – October/November 2016 9702 31
© UCLES 2016
(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.
Both read-offs must be accurate to half a small square in both the x and y
directions.
y-intercept: [1]
Either:
Check correct read-off from a point on the line and substituted into y = mx + c.
Read-off must be accurate to half a small square in both x and y directions.
Or:
Check read-off of the intercept directly from the graph (accurate to half a
small square.
(g) Value of P = candidate’s gradient and value of Q = candidate’s intercept. [1]
Do not allow fractions.
Units for P (e.g. s
m
–1
or s cm
–1
or s mm
–1
) and Q (s) correct. [1]
2 (a) (i) Value for D with unit in the range 0.14
mm to 0.16 mm. [1]
(ii) Percentage uncertainty in D based on an absolute uncertainty of 0.01
mm.
Correct method of calculation to obtain percentage uncertainty. [1]
(c) (iii) Value of I in range 10
mA ⩽ I ⩽ 200 mA with unit (collected without help from
Supervisor). [1]
(iv) Value of V in range 0.4
V ⩽ V ⩽ 1.0 V with unit (collected without help from
Supervisor). [1]
(d) (i) Value of d > D and d < 1
mm. [1]
(ii) Correct calculation of G. [1]
(iii) Justification for s.f. in G linked to s.f. in D and d. [1]
(f) (i) Second value of V. [1]
Quality: second value of V less than first value of V. [1]
(ii) Second value of d. [1]
Page 4 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 31
© UCLES 2016
(g) (i) Two values of k calculated correctly, and to at least 2 significant figures. [1]
(ii) Valid comment consistent with calculated values of k, testing against a stated
numerical criterion. [1]
(h) (i) Limitations [4] (ii) Improvements [4] Do not credit
A Two readings not enough to draw a
conclusion
Take many readings (for
different diameters) and plot a
graph/
take more readings and
compare k values
Two readings not
enough for accurate
results
Repeat readings
Few readings
Take more readings and
calculate average k
B Difficult to measure diameter(s)
with reason e.g. awkward placing
micrometer round wire/
only one direction to measure
diameter
Provide separate lengths of wire
C Meter readings changed in a
particular direction over time/
repeat readings of I or V were
often different/
contact resistance varies
Use power supply/
allow reading to reach steady
value/
method of cleaning crocodile
clips or wires
D Wire is very thin introducing a large
percentage error in the diameter
Use thicker wire/
use digital micrometer
E Rheostat movement not precise
enough – overshot I reading
Method to ensure exact current
easier to produce e.g. use of
screw thread adjustment
Cambridge International AS/A Level – October/November 2016 9702 31
© UCLES 2016
(g) (i) Two values of k calculated correctly, and to at least 2 significant figures. [1]
(ii) Valid comment consistent with calculated values of k, testing against a stated
numerical criterion. [1]
(h) (i) Limitations [4] (ii) Improvements [4] Do not credit
A Two readings not enough to draw a
conclusion
Take many readings (for
different diameters) and plot a
graph/
take more readings and
compare k values
Two readings not
enough for accurate
results
Repeat readings
Few readings
Take more readings and
calculate average k
B Difficult to measure diameter(s)
with reason e.g. awkward placing
micrometer round wire/
only one direction to measure
diameter
Provide separate lengths of wire
C Meter readings changed in a
particular direction over time/
repeat readings of I or V were
often different/
contact resistance varies
Use power supply/
allow reading to reach steady
value/
method of cleaning crocodile
clips or wires
D Wire is very thin introducing a large
percentage error in the diameter
Use thicker wire/
use digital micrometer
E Rheostat movement not precise
enough – overshot I reading
Method to ensure exact current
easier to produce e.g. use of
screw thread adjustment
® IGCSE is the registered trademark of Cambridge International Examinations.
This document consists of 4 printed pages.
© UCLES 2016 [Turn ove
r
Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/33
Paper 3 Advanced Practical Skills 1 October/November 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 October/November 2016 series for most
Cambridge IGCSE
®
, Cambridge International A and AS Level components and some Cambridge O Level
components.
This document consists of 4 printed pages.
© UCLES 2016 [Turn ove
r
Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/33
Paper 3 Advanced Practical Skills 1 October/November 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 October/November 2016 series for most
Cambridge IGCSE
®
, Cambridge International A and AS Level components and some Cambridge O Level
components.
Page 2 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 33
© UCLES 2016
1 (a) (v) Value of I in range 20 mA ⩽ I ⩽ 200 mA with unit. [1]
(c) Six sets of readings of R and y (with correct trend and without help from Supervisor)
scores 5 marks, five sets scores 4 marks etc. [5]
Range: [1]
Values of R must include 33
Ω.
Column headings: [1]
Each column heading must contain a quantity and a unit where appropriate.
The presentation of the quantity and unit must conform to accepted scientific
convention, e.g. 1
/ y / m
–1
or 1 / y (m
–1
).
Consistency: [1]
All values of raw y must be given to the nearest mm.
Significant figures: [1]
Every value of 1
/ R must be given to 2 or 3 significant figures.
Calculation: [1]
1
/ y calculated correctly to the number of s.f. given by the candidate.
(d) (i) Axes: [1]
Sensible scales must be used. Awkward scales (e.g. 3:10, fractions or non-linear)
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 must be plotted on the grid.
Diameter of plotted points must be ⩽ half a small square (no “blobs”).
Plotted points must be accurate to half a small square.
Quality: [1]
All points in the table must be plotted on the grid for this mark to be awarded.
All points must be within 0.001
cm
–1
(0.1 m
–1
) (to scale) on the y-axis of 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
5 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 (i.e. circled or labelled) by
the candidate. There must be at least five points left after the anomalous
point is disregarded.
Line must not be kinked or thicker than half a small square.
Cambridge International AS/A Level – October/November 2016 9702 33
© UCLES 2016
1 (a) (v) Value of I in range 20 mA ⩽ I ⩽ 200 mA with unit. [1]
(c) Six sets of readings of R and y (with correct trend and without help from Supervisor)
scores 5 marks, five sets scores 4 marks etc. [5]
Range: [1]
Values of R must include 33
Ω.
Column headings: [1]
Each column heading must contain a quantity and a unit where appropriate.
The presentation of the quantity and unit must conform to accepted scientific
convention, e.g. 1
/ y / m
–1
or 1 / y (m
–1
).
Consistency: [1]
All values of raw y must be given to the nearest mm.
Significant figures: [1]
Every value of 1
/ R must be given to 2 or 3 significant figures.
Calculation: [1]
1
/ y calculated correctly to the number of s.f. given by the candidate.
(d) (i) Axes: [1]
Sensible scales must be used. Awkward scales (e.g. 3:10, fractions or non-linear)
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 must be plotted on the grid.
Diameter of plotted points must be ⩽ half a small square (no “blobs”).
Plotted points must be accurate to half a small square.
Quality: [1]
All points in the table must be plotted on the grid for this mark to be awarded.
All points must be within 0.001
cm
–1
(0.1 m
–1
) (to scale) on the y-axis of 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
5 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 (i.e. circled or labelled) by
the candidate. There must be at least five points left after the anomalous
point is disregarded.
Line must not be kinked or thicker than half a small square.
Page 3 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 33
© UCLES 2016
(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.
Both read-offs must be accurate to half a small square in both the x and y
directions.
y-intercept: [1]
Either:
Check correct read-off from a point on the line and substituted into y = mx + c.
Read-off must be accurate to half a small square in both x and y directions.
Or:
Check read-off of the intercept directly from the graph (accurate to half a
small square.
(e) Value of P = –
candidate’s gradient and value of Q = candidate’s intercept.
Do not allow fractions. [1]
Units for P (e.g. Ω
m
–1
) and Q (e.g. m
–1
) correct. [1]
(f) (ii) Value of X in the range 5.0
Ω to 20.0 Ω from correct calculation. [1]
2 (b) (iii) All raw readings stated to at least 0.1
s. Value for T with unit in the range
0.50
s to 0.90 s. [1]
Evidence of repeats (at least two recordings of nT where n ⩾ 5). [1]
(c) (i) Correct calculation of l. [1]
(ii) Justification of s.f. in l linked to s.f. in time readings (e.g. time, raw time, nT). [1]
(d) (ii) Value of d (l ± 0.050
m). [1]
(iii) Absolute uncertainty in d in range 2
mm–8 mm.
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]
(e) (ii) Value for t in the range 2.0
s–8.0 s. [1]
(f) Second value of T. [1]
Second value of t. [1]
Quality: Second value of t > first value of t. [1]
Cambridge International AS/A Level – October/November 2016 9702 33
© UCLES 2016
(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.
Both read-offs must be accurate to half a small square in both the x and y
directions.
y-intercept: [1]
Either:
Check correct read-off from a point on the line and substituted into y = mx + c.
Read-off must be accurate to half a small square in both x and y directions.
Or:
Check read-off of the intercept directly from the graph (accurate to half a
small square.
(e) Value of P = –
candidate’s gradient and value of Q = candidate’s intercept.
Do not allow fractions. [1]
Units for P (e.g. Ω
m
–1
) and Q (e.g. m
–1
) correct. [1]
(f) (ii) Value of X in the range 5.0
Ω to 20.0 Ω from correct calculation. [1]
2 (b) (iii) All raw readings stated to at least 0.1
s. Value for T with unit in the range
0.50
s to 0.90 s. [1]
Evidence of repeats (at least two recordings of nT where n ⩾ 5). [1]
(c) (i) Correct calculation of l. [1]
(ii) Justification of s.f. in l linked to s.f. in time readings (e.g. time, raw time, nT). [1]
(d) (ii) Value of d (l ± 0.050
m). [1]
(iii) Absolute uncertainty in d in range 2
mm–8 mm.
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]
(e) (ii) Value for t in the range 2.0
s–8.0 s. [1]
(f) Second value of T. [1]
Second value of t. [1]
Quality: Second value of t > first value of t. [1]
Page 4 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 33
© UCLES 2016
(g) (i) Two values of k calculated correctly, and to at least 2 significant figures. [1]
(ii) Valid comment consistent with calculated values of k, testing against a stated
numerical criterion. [1]
(h) (i) Limitations [4] (ii) Improvements [4] Do not credit
A Two readings not enough to draw a
conclusion
Take many readings (for
different masses) and plot a
graph/
take more readings and
compare k values
Two readings not
enough for accurate
results
Repeat readings
Few readings
Take more readings and
calculate average k
B Difficult to set d with reason e.g.
knot takes some string/knot or
weight slips
Improved method to set d e.g.
trial and error/glue knots to stop
slipping
C Difficult to measure d with reason
e.g. parallax error/cannot get ruler
close/hands move holding ruler in
place/difficult to judge centre of
mass(es)/rule not vertical
Clamped metre rule/
measure to top and bottom of
masses and find the average/
detailed method to check rule
vertical/
detailed method to reduce
parallax error
Eyes level
Eyes perpendicular
Reference to set square
without detail
D Difficult to measure (n)T with
reason e.g. difficult to judge
complete oscillation
Use fiducial marker at centre of
oscillation/
use a motion sensor placed
under masses
Reaction time error
Force on release
E Difficult to see when coil separation
is constant
or
Separation of coils constant for a
short time
Video with timer/
video and view frame by frame/
screen behind spring
Video timer
F Movement of spring along rod Method to fix spring to rod e.g.
Blu-Tack or groove in rod
Air conditioning/draughts
Unwanted modes of
oscillation
Cambridge International AS/A Level – October/November 2016 9702 33
© UCLES 2016
(g) (i) Two values of k calculated correctly, and to at least 2 significant figures. [1]
(ii) Valid comment consistent with calculated values of k, testing against a stated
numerical criterion. [1]
(h) (i) Limitations [4] (ii) Improvements [4] Do not credit
A Two readings not enough to draw a
conclusion
Take many readings (for
different masses) and plot a
graph/
take more readings and
compare k values
Two readings not
enough for accurate
results
Repeat readings
Few readings
Take more readings and
calculate average k
B Difficult to set d with reason e.g.
knot takes some string/knot or
weight slips
Improved method to set d e.g.
trial and error/glue knots to stop
slipping
C Difficult to measure d with reason
e.g. parallax error/cannot get ruler
close/hands move holding ruler in
place/difficult to judge centre of
mass(es)/rule not vertical
Clamped metre rule/
measure to top and bottom of
masses and find the average/
detailed method to check rule
vertical/
detailed method to reduce
parallax error
Eyes level
Eyes perpendicular
Reference to set square
without detail
D Difficult to measure (n)T with
reason e.g. difficult to judge
complete oscillation
Use fiducial marker at centre of
oscillation/
use a motion sensor placed
under masses
Reaction time error
Force on release
E Difficult to see when coil separation
is constant
or
Separation of coils constant for a
short time
Video with timer/
video and view frame by frame/
screen behind spring
Video timer
F Movement of spring along rod Method to fix spring to rod e.g.
Blu-Tack or groove in rod
Air conditioning/draughts
Unwanted modes of
oscillation
® IGCSE is the registered trademark of Cambridge International Examinations.
This document consists of 4 printed pages.
© UCLES 2016 [Turn ove
r
Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/34
Paper 3 Advanced Practical Skills 2 October/November 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 October/November 2016 series for most
Cambridge IGCSE
®
, Cambridge International A and AS Level components and some Cambridge O Level
components.
This document consists of 4 printed pages.
© UCLES 2016 [Turn ove
r
Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/34
Paper 3 Advanced Practical Skills 2 October/November 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 October/November 2016 series for most
Cambridge IGCSE
®
, Cambridge International A and AS Level components and some Cambridge O Level
components.
Page 2 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 34
© UCLES 2016
1 (b) (ii) Value for x in range 24.0 cm to 26.0 cm, with unit. [1]
(iv) Value for T in range 0.30
s to 1.00 s. [1]
Evidence of repeat readings (at least two recordings of nT where n ⩾ 5). [1]
(c) Six sets of values for x and T (with correct trend and without help from Supervisor)
scores 4 marks, five sets scores 3 marks etc. [4]
Range: [1]
x
min ⩽ 20.0 cm and x max ⩾ 30.0 cm.
Column headings: [1]
Each column heading must contain a quantity and an appropriate unit. The
presentation of the quantity and unit must conform to accepted scientific convention
e.g. 1
/ T
2
(s
–2
) or 1 / T
2
/ 1 / s
2
.
Consistency: [1]
All values of x must be given to the nearest mm.
Significant figures: [1]
Every value of 1/T
2
must be given to the same number of s.f. as (or one greater than)
the number of s.f. in the corresponding times.
Calculation: [1]
Values of 1/T
2
calculated correctly to the number of s.f. given by the candidate.
(d) (i) Axes: [1]
Sensible scales must be used. Awkward scales (e.g. 3:10, fractions or non-
linear) 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 must be no more than three large squares apart.
Plotting of points: [1]
All observations in the table must be plotted on the grid.
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 (at least 5) for this mark to be awarded.
All points must be within ±
1.0 cm (to scale) of a straight line in the x direction.
(ii) Line of best fit: [1]
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 if clearly indicated (i.e. circled or labelled) by the
candidate. There must be at least five points left after the anomalous point is
disregarded.
Line must not be kinked or thicker than half a small square.
Cambridge International AS/A Level – October/November 2016 9702 34
© UCLES 2016
1 (b) (ii) Value for x in range 24.0 cm to 26.0 cm, with unit. [1]
(iv) Value for T in range 0.30
s to 1.00 s. [1]
Evidence of repeat readings (at least two recordings of nT where n ⩾ 5). [1]
(c) Six sets of values for x and T (with correct trend and without help from Supervisor)
scores 4 marks, five sets scores 3 marks etc. [4]
Range: [1]
x
min ⩽ 20.0 cm and x max ⩾ 30.0 cm.
Column headings: [1]
Each column heading must contain a quantity and an appropriate unit. The
presentation of the quantity and unit must conform to accepted scientific convention
e.g. 1
/ T
2
(s
–2
) or 1 / T
2
/ 1 / s
2
.
Consistency: [1]
All values of x must be given to the nearest mm.
Significant figures: [1]
Every value of 1/T
2
must be given to the same number of s.f. as (or one greater than)
the number of s.f. in the corresponding times.
Calculation: [1]
Values of 1/T
2
calculated correctly to the number of s.f. given by the candidate.
(d) (i) Axes: [1]
Sensible scales must be used. Awkward scales (e.g. 3:10, fractions or non-
linear) 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 must be no more than three large squares apart.
Plotting of points: [1]
All observations in the table must be plotted on the grid.
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 (at least 5) for this mark to be awarded.
All points must be within ±
1.0 cm (to scale) of a straight line in the x direction.
(ii) Line of best fit: [1]
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 if clearly indicated (i.e. circled or labelled) by the
candidate. There must be at least five points left after the anomalous point is
disregarded.
Line must not be kinked or thicker than half a small square.
Page 3 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 34
© UCLES 2016
(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.
Both read-offs must be accurate to half a small square in both the x and y
directions.
y-intercept: [1]
Either:
Check correct read-off from a point on the line and substituted into y = mx + c.
Read-off must be accurate to half a small square in both x and y directions.
Or:
Check read-off of the intercept directly from the graph (accurate to half a
small square.
(e) Value of p = candidate’s gradient and value of q = candidate’s intercept.
Do not allow fractions. [1]
Units for p (e.g. cm
–1
s
–2
) and q (e.g. s
–2
) correct. [1]
2 (b) L in range 19.0
cm to 21.0 cm, with unit. [1]
(c) (iv) Values for x
1 and x 2 to nearest mm and x 2 > x1. [1]
Evidence of repeat readings of x
1 and x 2. [1]
(v) Correct calculation of X. [1]
(d) Absolute uncertainty in X in range 2
mm to 10 mm.
If repeated readings have been taken, then absolute uncertainty can be half the
range (but not zero) if working is clearly shown.
Correct method of calculation to obtain percentage uncertainty. [1]
(e) Second value for L. [1]
Second values for x
1 and x 2. [1]
Quality: X smaller for larger L. [1]
(f) (i) Two values of k calculated correctly. [1]
(ii) Justification of s.f. in k based on the s.f. in L, x
1 and x 2. [1]
(iii) Valid comment consistent with the calculated values of k, testing against a stated
numerical criterion. [1]
(g) Value for X = 50
cm. [1]
Cambridge International AS/A Level – October/November 2016 9702 34
© UCLES 2016
(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.
Both read-offs must be accurate to half a small square in both the x and y
directions.
y-intercept: [1]
Either:
Check correct read-off from a point on the line and substituted into y = mx + c.
Read-off must be accurate to half a small square in both x and y directions.
Or:
Check read-off of the intercept directly from the graph (accurate to half a
small square.
(e) Value of p = candidate’s gradient and value of q = candidate’s intercept.
Do not allow fractions. [1]
Units for p (e.g. cm
–1
s
–2
) and q (e.g. s
–2
) correct. [1]
2 (b) L in range 19.0
cm to 21.0 cm, with unit. [1]
(c) (iv) Values for x
1 and x 2 to nearest mm and x 2 > x1. [1]
Evidence of repeat readings of x
1 and x 2. [1]
(v) Correct calculation of X. [1]
(d) Absolute uncertainty in X in range 2
mm to 10 mm.
If repeated readings have been taken, then absolute uncertainty can be half the
range (but not zero) if working is clearly shown.
Correct method of calculation to obtain percentage uncertainty. [1]
(e) Second value for L. [1]
Second values for x
1 and x 2. [1]
Quality: X smaller for larger L. [1]
(f) (i) Two values of k calculated correctly. [1]
(ii) Justification of s.f. in k based on the s.f. in L, x
1 and x 2. [1]
(iii) Valid comment consistent with the calculated values of k, testing against a stated
numerical criterion. [1]
(g) Value for X = 50
cm. [1]
Page 4 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 34
© UCLES 2016
(h) (i) Limitations [4] (ii) Improvements [4] Do not credit
A Two readings not enough to draw
a conclusion
Take more readings and plot
graph/
obtain more k values and
compare
Two readings not
enough for accurate
results
Repeat readings
Few readings
Take more readings and
calculate average k
B Metre rule is not parallel to
bench/horizontal
Use a second rule and measure
at both ends/
use a (spirit) level
C Difficult to move stands with
reason e.g. friction/
bench is rough/
stands tend to stick
Guide for stands (fixed to
bench)/
mount stands on rollers/
put wheels on stands/
method to reduce friction e.g.
sand bench with sandpaper
Use a smooth(er) bench
Use lubricant
D Difficulty with rule
e.g. rule skewed/
moves sideways
Use V-shaped rods/
groove in rods/
guide for ruler with some details
Falls off
E Difficult to measure x with reason
e.g parallax error/
difficult to tell point where rod
touches ruler
Scale on vertical edge of rule/
draw a line on the rod/
use a thinner rod/
replace rods with sharp edges
e.g. prisms
Large contact area
Do not allow ‘use a computer to improve the experiment’.
Cambridge International AS/A Level – October/November 2016 9702 34
© UCLES 2016
(h) (i) Limitations [4] (ii) Improvements [4] Do not credit
A Two readings not enough to draw
a conclusion
Take more readings and plot
graph/
obtain more k values and
compare
Two readings not
enough for accurate
results
Repeat readings
Few readings
Take more readings and
calculate average k
B Metre rule is not parallel to
bench/horizontal
Use a second rule and measure
at both ends/
use a (spirit) level
C Difficult to move stands with
reason e.g. friction/
bench is rough/
stands tend to stick
Guide for stands (fixed to
bench)/
mount stands on rollers/
put wheels on stands/
method to reduce friction e.g.
sand bench with sandpaper
Use a smooth(er) bench
Use lubricant
D Difficulty with rule
e.g. rule skewed/
moves sideways
Use V-shaped rods/
groove in rods/
guide for ruler with some details
Falls off
E Difficult to measure x with reason
e.g parallax error/
difficult to tell point where rod
touches ruler
Scale on vertical edge of rule/
draw a line on the rod/
use a thinner rod/
replace rods with sharp edges
e.g. prisms
Large contact area
Do not allow ‘use a computer to improve the experiment’.
® IGCSE is the registered trademark of Cambridge International Examinations.
This document consists of 4 printed pages.
© UCLES 2016 [Turn ove
r
Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/35
Paper 3 Advanced Practical Skills 1 October/November 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 October/November 2016 series for most
Cambridge IGCSE
®
, Cambridge International A and AS Level components and some Cambridge O Level
components.
This document consists of 4 printed pages.
© UCLES 2016 [Turn ove
r
Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/35
Paper 3 Advanced Practical Skills 1 October/November 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 October/November 2016 series for most
Cambridge IGCSE
®
, Cambridge International A and AS Level components and some Cambridge O Level
components.
Page 2 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 35
© UCLES 2016
1 (a) (i) Value of V F in the range 260.0 cm
3
to 400.0 cm
3
. [1]
(c) (v) All values of raw y to the nearest mm and less than 20
cm. [1]
(d) (ii) Six sets of readings of V and y (with correct trend and without help from
Supervisor) scores 5 marks, five sets scores 4 marks etc. [5]
Range: [1]
Maximum value of V ⩾ 200
cm
3
.
Column headings: [1]
Each column heading must contain a quantity and an appropriate unit.
The presentation of the quantity and unit must conform to accepted scientific
convention, e.g. y
/ m or y (cm), V / cm
3
, 2V F
/ 3 / cm
3
.
Consistency: [1]
All values of raw V must be given to the nearest cm
3
.
(e) (i) Axes: [1]
Sensible scales must be used. Awkward scales (e.g. 3:10, fractions or non-
linear) 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 on the grid.
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 within ± 0.25
cm (to scale) in the y 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
5 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 (i.e. circled or labelled) by
the candidate.
There must be at least four points left after disregarding the anomalous point.
Line must not be kinked or thicker than half a small square.
Cambridge International AS/A Level – October/November 2016 9702 35
© UCLES 2016
1 (a) (i) Value of V F in the range 260.0 cm
3
to 400.0 cm
3
. [1]
(c) (v) All values of raw y to the nearest mm and less than 20
cm. [1]
(d) (ii) Six sets of readings of V and y (with correct trend and without help from
Supervisor) scores 5 marks, five sets scores 4 marks etc. [5]
Range: [1]
Maximum value of V ⩾ 200
cm
3
.
Column headings: [1]
Each column heading must contain a quantity and an appropriate unit.
The presentation of the quantity and unit must conform to accepted scientific
convention, e.g. y
/ m or y (cm), V / cm
3
, 2V F
/ 3 / cm
3
.
Consistency: [1]
All values of raw V must be given to the nearest cm
3
.
(e) (i) Axes: [1]
Sensible scales must be used. Awkward scales (e.g. 3:10, fractions or non-
linear) 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 on the grid.
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 within ± 0.25
cm (to scale) in the y 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
5 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 (i.e. circled or labelled) by
the candidate.
There must be at least four points left after disregarding the anomalous point.
Line must not be kinked or thicker than half a small square.
Page 3 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 35
© UCLES 2016
(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 on answer line 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:
Check correct read-off from a point on the line and substituted into y = mx + c.
Read-off must be accurate to half a small square in both x and y directions.
Or:
Check read-off of the intercept directly from the graph (accurate to half a
small square.
(f) Value of P = candidate’s gradient and value of Q = candidate’s intercept.
Do not allow fractions. [1]
Units for P (e.g. mm
–2
, cm
–2
, m
–2
but not e.g. cm / cm
3
) and Q (e.g. m) correct. [1]
(g) (i) V calculated correctly to the number of significant figures given by the candidate.
Sign of answer must be consistent with P and Q values in (f). [1]
(ii) Valid comment with a comparison of volumes e.g. V is greater than V
F. [1]
2 (a) (ii) Value of L with unit in range 0.790
m to 0.810 m. [1]
(iii) Absolute uncertainty in L either 1
mm or 2 mm.
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]
(b) (ii) All values of raw d
1 with unit to the nearest mm. [1]
(c) (ii) d
2 > d1 showing scale on rule is used correctly. [1]
(iii) Correct calculation of |d
2 – d1|. [1]
(iv) Correct calculation of M with consistent unit. Do not allow answers to 1 s.f. [1]
(d) Justification for s.f. in M linked to s.f. in m and L. [1]
Cambridge International AS/A Level – October/November 2016 9702 35
© UCLES 2016
(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 on answer line 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:
Check correct read-off from a point on the line and substituted into y = mx + c.
Read-off must be accurate to half a small square in both x and y directions.
Or:
Check read-off of the intercept directly from the graph (accurate to half a
small square.
(f) Value of P = candidate’s gradient and value of Q = candidate’s intercept.
Do not allow fractions. [1]
Units for P (e.g. mm
–2
, cm
–2
, m
–2
but not e.g. cm / cm
3
) and Q (e.g. m) correct. [1]
(g) (i) V calculated correctly to the number of significant figures given by the candidate.
Sign of answer must be consistent with P and Q values in (f). [1]
(ii) Valid comment with a comparison of volumes e.g. V is greater than V
F. [1]
2 (a) (ii) Value of L with unit in range 0.790
m to 0.810 m. [1]
(iii) Absolute uncertainty in L either 1
mm or 2 mm.
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]
(b) (ii) All values of raw d
1 with unit to the nearest mm. [1]
(c) (ii) d
2 > d1 showing scale on rule is used correctly. [1]
(iii) Correct calculation of |d
2 – d1|. [1]
(iv) Correct calculation of M with consistent unit. Do not allow answers to 1 s.f. [1]
(d) Justification for s.f. in M linked to s.f. in m and L. [1]
Page 4 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 35
© UCLES 2016
(e) Second value of d 1. [1]
Second value of d
2. [1]
Quality: Second value of |d
2 – d1| < first value of |d 2 – d1|. [1]
(f) (i) Two values of k calculated correctly. [1]
(ii) Valid comment consistent with the calculated values of k, testing against a
stated numerical criterion. [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
Two readings not
enough for accurate
results
Repeat readings
Few readings
Take more readings
and calculate average
k
B Difference between d 2 and d 1 is
small/
(d2 – d1) is small/
large % uncertainty in (d
2 – d1)
Improved method to measure
(d
2 – d1) values e.g. travelling
microscope/larger masses/clamp
vernier calipers above strip
Laser
Ultrasonic position
sensor
Longer strip
Change strip
Deflection of strip
C Difficult to distribute mass evenly/
gaps between masses vary/
masses do not lie in a straight
line/
masses have slots
Improved method of distribution
e.g. use marker or scale on
wooden strip/
use smaller masses to produce
the same m/
continuous strip of named material
e.g. modelling clay
Masses inaccurate
Stick masses to strip
Masses falling
Replace strip with
metre rule
Masses with no slots
D Difficult to measure d values with
reason e.g. rule not vertical/
difficult to hold second rule still
as a pointer/
parallax error/
rule not held stationary
Improved method to measure d
e.g. clamp metre rule/
place a set square on the floor
next to ruler (to ensure rule is
vertical)/
use of pointer (to metre rule)
Strip oscillates when
masses are put on it
Use a set square
Effects of wind
Reference to L
E Strip becomes permanently bent
or deformed
Method to overcome deformation
e.g. check d without masses
before and between readings/
lay weights on strip to flatten
Elastic properties of
wood vary
Use a new strip
Cambridge International AS/A Level – October/November 2016 9702 35
© UCLES 2016
(e) Second value of d 1. [1]
Second value of d
2. [1]
Quality: Second value of |d
2 – d1| < first value of |d 2 – d1|. [1]
(f) (i) Two values of k calculated correctly. [1]
(ii) Valid comment consistent with the calculated values of k, testing against a
stated numerical criterion. [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
Two readings not
enough for accurate
results
Repeat readings
Few readings
Take more readings
and calculate average
k
B Difference between d 2 and d 1 is
small/
(d2 – d1) is small/
large % uncertainty in (d
2 – d1)
Improved method to measure
(d
2 – d1) values e.g. travelling
microscope/larger masses/clamp
vernier calipers above strip
Laser
Ultrasonic position
sensor
Longer strip
Change strip
Deflection of strip
C Difficult to distribute mass evenly/
gaps between masses vary/
masses do not lie in a straight
line/
masses have slots
Improved method of distribution
e.g. use marker or scale on
wooden strip/
use smaller masses to produce
the same m/
continuous strip of named material
e.g. modelling clay
Masses inaccurate
Stick masses to strip
Masses falling
Replace strip with
metre rule
Masses with no slots
D Difficult to measure d values with
reason e.g. rule not vertical/
difficult to hold second rule still
as a pointer/
parallax error/
rule not held stationary
Improved method to measure d
e.g. clamp metre rule/
place a set square on the floor
next to ruler (to ensure rule is
vertical)/
use of pointer (to metre rule)
Strip oscillates when
masses are put on it
Use a set square
Effects of wind
Reference to L
E Strip becomes permanently bent
or deformed
Method to overcome deformation
e.g. check d without masses
before and between readings/
lay weights on strip to flatten
Elastic properties of
wood vary
Use a new strip
® IGCSE is the registered trademark of Cambridge International Examinations.
This document consists of 4 printed pages.
© UCLES 2016 [Turn ove
r
Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/36
Paper 3 Advanced Practical Skills 2 October/November 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 October/November 2016 series for most
Cambridge IGCSE
®
, Cambridge International A and AS Level components and some Cambridge O Level
components.
This document consists of 4 printed pages.
© UCLES 2016 [Turn ove
r
Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/36
Paper 3 Advanced Practical Skills 2 October/November 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 October/November 2016 series for most
Cambridge IGCSE
®
, Cambridge International A and AS Level components and some Cambridge O Level
components.
Page 2 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 36
© UCLES 2016
1 (b) (i) Value for x in range 45.0 cm to 55.0 cm. [1]
(iii) Value for I in range 500
µA to 1500 µA (or 0.50 mA to 1.50 mA), with unit. [1]
(c) Six sets of values for x and I (with correct trend and without help from Supervisor)
scores 5 marks, five sets scores 4 marks etc. [5]
Range: [1]
x values must include 20
cm or less and 80 cm or more.
Column headings: [1]
Each column heading must contain a quantity and an appropriate unit.
The presentation of the quantity and unit must conform to accepted scientific
convention e.g. I /
µA.
Consistency: [1]
All values of raw x must be given to the nearest mm.
(d) (i) Axes: [1]
Sensible scales must be used. Awkward scales (e.g. 3:10, fractions or non-linear)
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 must be no more than three large squares apart.
Plotting of points: [1]
All observations in the table must be plotted on the grid.
Diameter of plotted points must be ⩽ half a small square (no “blobs”).
Plotting of points must be accurate to half a small square.
Quality: [1]
All points in the table (at least 5) must be plotted on the grid.
All points must be within ±
20 µA (± 0.02 mA) of a straight line in the y
(I) direction.
(ii) Line of best fit: [1]
Judge by balance of all points on the grid about the candidate's line (at least five
points). There must be an even distribution of points either side of the line along
the full length.
One anomalous plot is allowed if clearly indicated (i.e. circled or labelled). There
must be at least five points left after disregarding the anomalous point.
Lines must not be kinked or thicker than half a small square.
Cambridge International AS/A Level – October/November 2016 9702 36
© UCLES 2016
1 (b) (i) Value for x in range 45.0 cm to 55.0 cm. [1]
(iii) Value for I in range 500
µA to 1500 µA (or 0.50 mA to 1.50 mA), with unit. [1]
(c) Six sets of values for x and I (with correct trend and without help from Supervisor)
scores 5 marks, five sets scores 4 marks etc. [5]
Range: [1]
x values must include 20
cm or less and 80 cm or more.
Column headings: [1]
Each column heading must contain a quantity and an appropriate unit.
The presentation of the quantity and unit must conform to accepted scientific
convention e.g. I /
µA.
Consistency: [1]
All values of raw x must be given to the nearest mm.
(d) (i) Axes: [1]
Sensible scales must be used. Awkward scales (e.g. 3:10, fractions or non-linear)
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 must be no more than three large squares apart.
Plotting of points: [1]
All observations in the table must be plotted on the grid.
Diameter of plotted points must be ⩽ half a small square (no “blobs”).
Plotting of points must be accurate to half a small square.
Quality: [1]
All points in the table (at least 5) must be plotted on the grid.
All points must be within ±
20 µA (± 0.02 mA) of a straight line in the y
(I) direction.
(ii) Line of best fit: [1]
Judge by balance of all points on the grid about the candidate's line (at least five
points). There must be an even distribution of points either side of the line along
the full length.
One anomalous plot is allowed if clearly indicated (i.e. circled or labelled). There
must be at least five points left after disregarding the anomalous point.
Lines must not be kinked or thicker than half a small square.
Page 3 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 36
© UCLES 2016
(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.
Both read-offs must be accurate to half a small square in both the x and y
directions.
y-intercept: [1]
Either:
Check correct read-off from a point on the line and substituted into y = mx + c.
Read-off must be accurate to half a small square in both x and y directions.
Or:
Check read-off of the intercept directly from the graph (accurate to half a
small square.
(e) Value of S = candidate’s gradient and value of T = candidate’s intercept.
Do not allow fractions. [1]
Consistent units for S (e.g. µA
cm
–1
) and T (e.g. µA). [1]
(f) Calculation: r calculated correctly to the s.f. given by the candidate. [1]
Significant figures: r given to 2 or 3 s.f. [1]
2 (b) (ii) x
1 in range 10.0 cm to 40.0 cm. [1]
(c) Value of x
2 < x1. [1]
(d) (i) Second value of x
1. [1]
(ii) Value of x
2 given to nearest mm and all other raw values of x in (b), (c) and (d)
are to the nearest mm. [1]
(e) (i) Two values of k calculated correctly. [1]
(ii) Justification of the s.f. in k based on the s.f. in x
1 and the s.f in x 2. [1]
(iii) Valid comment consistent with the calculated values of k, testing against a
stated numerical criterion. [1]
(f) (i) Raw values of D to nearest 0.001
cm and in range 1.400 cm to 2.200 cm. [1]
Evidence of repeated readings for D. [1]
(ii) Absolute uncertainty in D of 0.001
cm or 0.002 cm.
If repeated readings have been taken, then absolute uncertainty could be
half the range (but not zero) if working is clearly shown.
Correct method of calculation to obtain percentage uncertainty. [1]
Cambridge International AS/A Level – October/November 2016 9702 36
© UCLES 2016
(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.
Both read-offs must be accurate to half a small square in both the x and y
directions.
y-intercept: [1]
Either:
Check correct read-off from a point on the line and substituted into y = mx + c.
Read-off must be accurate to half a small square in both x and y directions.
Or:
Check read-off of the intercept directly from the graph (accurate to half a
small square.
(e) Value of S = candidate’s gradient and value of T = candidate’s intercept.
Do not allow fractions. [1]
Consistent units for S (e.g. µA
cm
–1
) and T (e.g. µA). [1]
(f) Calculation: r calculated correctly to the s.f. given by the candidate. [1]
Significant figures: r given to 2 or 3 s.f. [1]
2 (b) (ii) x
1 in range 10.0 cm to 40.0 cm. [1]
(c) Value of x
2 < x1. [1]
(d) (i) Second value of x
1. [1]
(ii) Value of x
2 given to nearest mm and all other raw values of x in (b), (c) and (d)
are to the nearest mm. [1]
(e) (i) Two values of k calculated correctly. [1]
(ii) Justification of the s.f. in k based on the s.f. in x
1 and the s.f in x 2. [1]
(iii) Valid comment consistent with the calculated values of k, testing against a
stated numerical criterion. [1]
(f) (i) Raw values of D to nearest 0.001
cm and in range 1.400 cm to 2.200 cm. [1]
Evidence of repeated readings for D. [1]
(ii) Absolute uncertainty in D of 0.001
cm or 0.002 cm.
If repeated readings have been taken, then absolute uncertainty could be
half the range (but not zero) if working is clearly shown.
Correct method of calculation to obtain percentage uncertainty. [1]
Page 4 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 36
© UCLES 2016
(iii) V calculated correctly. [1]
(iv) Quality: M in range 3
g to 13 g. [1]
(g) (i) Limitations [4] (ii) Improvements [4] Do not credit
A Two readings are not enough
to draw a valid conclusion
Take more readings and plot
graph/
take more readings and
compare k values
Two readings not enough
for accurate results
Repeat readings
Few readings
Take more readings and
calculate average k
B Empty beaker moves on
bench
Fix beaker with Blu-Tack/tape/
glue
C Difficult to balance rule:
rule slips on pivot/
wind disturbs balance
Make groove in rule (under
50cm mark)/
other practical method e.g.
hinge/nail through rule
Blu-tack
Tape
Switch off fans
String slips on rule
D Spheres/string/tape still wet
after immersion so mass
changes
or
string/tape adds to mass of
sphere
Use waterproof string/
use wire
Dry the spheres
Waterproof tape
E Difficult to measure x with
reason, e.g. string too thick
(so it covers graduations on
rule)
Use thin(ner) string
Parallax problems
F Marble not round Improved method of finding V
(e.g. liquid displacement)
Repeat readings and
average
Cambridge International AS/A Level – October/November 2016 9702 36
© UCLES 2016
(iii) V calculated correctly. [1]
(iv) Quality: M in range 3
g to 13 g. [1]
(g) (i) Limitations [4] (ii) Improvements [4] Do not credit
A Two readings are not enough
to draw a valid conclusion
Take more readings and plot
graph/
take more readings and
compare k values
Two readings not enough
for accurate results
Repeat readings
Few readings
Take more readings and
calculate average k
B Empty beaker moves on
bench
Fix beaker with Blu-Tack/tape/
glue
C Difficult to balance rule:
rule slips on pivot/
wind disturbs balance
Make groove in rule (under
50cm mark)/
other practical method e.g.
hinge/nail through rule
Blu-tack
Tape
Switch off fans
String slips on rule
D Spheres/string/tape still wet
after immersion so mass
changes
or
string/tape adds to mass of
sphere
Use waterproof string/
use wire
Dry the spheres
Waterproof tape
E Difficult to measure x with
reason, e.g. string too thick
(so it covers graduations on
rule)
Use thin(ner) string
Parallax problems
F Marble not round Improved method of finding V
(e.g. liquid displacement)
Repeat readings and
average
® IGCSE is the registered trademark of Cambridge International Examinations.
This document consists of 8 printed pages.
© UCLES 2016 [Turn ove
r
Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/41
Paper 4 A Level Structured Questions October/November 2016
MARK SCHEME
Maximum Mark: 100
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 October/November 2016 series for most
Cambridge IGCSE
®
, Cambridge International A and AS Level components and some Cambridge O Level
components.
This document consists of 8 printed pages.
© UCLES 2016 [Turn ove
r
Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/41
Paper 4 A Level Structured Questions October/November 2016
MARK SCHEME
Maximum Mark: 100
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 October/November 2016 series for most
Cambridge IGCSE
®
, Cambridge International A and AS Level components and some Cambridge O Level
components.
Page 2 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 41
© UCLES 2016
1 (a) gravitational force provides/is the centripetal force B1
GMm
/ r
2
= mv
2
/ r or GMm / r
2
= mrω
2
and v = 2πr
/ T or ω = 2π / T M1
with algebra to T
2
= 4π
2
r
3
/ GM A1 [3]
or
acceleration due to gravity is the centripetal acceleration (B1)
GM
/ r
2
= v
2
/ r or GM / r
2
= rω
2
and v = 2πr
/ T or ω = 2π / T (M1)
with algebra to T
2
= 4π
2
r
3
/ GM (A1)
(b) (i) equatorial orbit/orbits (directly) above the equator B1
from west to east B1 [2]
(ii) (24
× 3600)
2
= 4π
2
r
3
/ (6.67 × 10
–11
× 6.0 × 10
24
) C1
r
3
= 7.57 × 10
22
r = 4.2
× 10
7
m A1 [2]
(c) (T
/ 24)
2
= {(2.64 × 10
7
) / (4.23 × 10
7
)}
3
B1
= 0.243
T = 12 hours A1 [2]
or
k (= T
2
/ r
3
) = 24
2
/ (4.23 × 10
7
)
3
(B1)
k = 7.61
× 10
–21
T
2
(= kr
3
) = 7.61 × 10
–21
× (2.64 × 10
7
)
3
= 140
T = 12 hours (A1)
2 (a) (i) p ∝ T or pV
/ T = constant or pV = nRT C1
T (= 5
× 300 =) 1500 K A1 [2]
(ii) pV = nRT
1.0
× 10
5
× 4.0 × 10
–4
= n × 8.31 × 300
or
5.0
× 10
5
× 4.0 × 10
–4
= n × 8.31 × 1500 C1
n = 0.016 mol A1 [2]
Cambridge International AS/A Level – October/November 2016 9702 41
© UCLES 2016
1 (a) gravitational force provides/is the centripetal force B1
GMm
/ r
2
= mv
2
/ r or GMm / r
2
= mrω
2
and v = 2πr
/ T or ω = 2π / T M1
with algebra to T
2
= 4π
2
r
3
/ GM A1 [3]
or
acceleration due to gravity is the centripetal acceleration (B1)
GM
/ r
2
= v
2
/ r or GM / r
2
= rω
2
and v = 2πr
/ T or ω = 2π / T (M1)
with algebra to T
2
= 4π
2
r
3
/ GM (A1)
(b) (i) equatorial orbit/orbits (directly) above the equator B1
from west to east B1 [2]
(ii) (24
× 3600)
2
= 4π
2
r
3
/ (6.67 × 10
–11
× 6.0 × 10
24
) C1
r
3
= 7.57 × 10
22
r = 4.2
× 10
7
m A1 [2]
(c) (T
/ 24)
2
= {(2.64 × 10
7
) / (4.23 × 10
7
)}
3
B1
= 0.243
T = 12 hours A1 [2]
or
k (= T
2
/ r
3
) = 24
2
/ (4.23 × 10
7
)
3
(B1)
k = 7.61
× 10
–21
T
2
(= kr
3
) = 7.61 × 10
–21
× (2.64 × 10
7
)
3
= 140
T = 12 hours (A1)
2 (a) (i) p ∝ T or pV
/ T = constant or pV = nRT C1
T (= 5
× 300 =) 1500 K A1 [2]
(ii) pV = nRT
1.0
× 10
5
× 4.0 × 10
–4
= n × 8.31 × 300
or
5.0
× 10
5
× 4.0 × 10
–4
= n × 8.31 × 1500 C1
n = 0.016 mol A1 [2]
Page 3 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 41
© UCLES 2016
(b) (i) 1. heating/thermal energy supplied B1
2. work done on/to system B1 [2]
(ii) 1. 240
J A1
2. same value as given in 1. (= 240
J) and zero given for 3. A1
3. zero A1 [3]
3 (a) 2k/m =
ω
2
M1
ω = 2πf M1
(2
× 64 / 0.810) = (2π × f)
2
leading to f = 2.0 Hz A1 [3]
(b) v
0 = ωx0 or v0 = 2πfx 0
or
v =
ω(x0
2 – x
2
)
1/2
and x = 0 C1
v
0 = 2π × 2.0 × 1.6 × 10
–2
= 0.20
m s
–1
A1 [2]
(c) frequency: reduced/decreased B1
maximum speed: reduced/decreased B1 [2]
4 (a) (i) noise/distortion is removed (from the signal) B1
the (original) signal is reformed/reproduced/recovered/restored B1 [2]
or
signal detected above/below a threshold creates new signal (B1)
of 1s and 0s (B1)
(ii) noise is superposed on the (displacement of the) signal/cannot be
distinguished
or
analogue/signal is continuous (so cannot be regenerated)
or
analogue/signal is not discrete (so cannot be regenerated) B1
noise is amplified with the signal B1 [2]
Cambridge International AS/A Level – October/November 2016 9702 41
© UCLES 2016
(b) (i) 1. heating/thermal energy supplied B1
2. work done on/to system B1 [2]
(ii) 1. 240
J A1
2. same value as given in 1. (= 240
J) and zero given for 3. A1
3. zero A1 [3]
3 (a) 2k/m =
ω
2
M1
ω = 2πf M1
(2
× 64 / 0.810) = (2π × f)
2
leading to f = 2.0 Hz A1 [3]
(b) v
0 = ωx0 or v0 = 2πfx 0
or
v =
ω(x0
2 – x
2
)
1/2
and x = 0 C1
v
0 = 2π × 2.0 × 1.6 × 10
–2
= 0.20
m s
–1
A1 [2]
(c) frequency: reduced/decreased B1
maximum speed: reduced/decreased B1 [2]
4 (a) (i) noise/distortion is removed (from the signal) B1
the (original) signal is reformed/reproduced/recovered/restored B1 [2]
or
signal detected above/below a threshold creates new signal (B1)
of 1s and 0s (B1)
(ii) noise is superposed on the (displacement of the) signal/cannot be
distinguished
or
analogue/signal is continuous (so cannot be regenerated)
or
analogue/signal is not discrete (so cannot be regenerated) B1
noise is amplified with the signal B1 [2]
Page 4 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 41
© UCLES 2016
(b) (i) gain/dB = 10 lg (P2
/ P1)
32 = 10
lg [PMIN
/ (0.38 × 10
–6
)]
or
–32 = 10
lg (0.38 × 10
–6
/ PMIN) C1
P
MIN = 6.0 × 10
–4
W A1 [2]
(ii) attenuation = 10
lg [(9.5 × 10
–3
) / (6.02 × 10
–4
)] C1
= 12
dB
attenuation per unit length (= 12/58) = 0.21
dB km
–1
A1 [2]
5 (a) in an electric field, charges (in a conductor) would move B1
no movement of charge so zero field strength
or
charge moves until F = 0 / E = 0 B1 [2]
or
charges in metal do not move (B1)
no (resultant) force on charges so no (electric) field (B1)
(b) at P, E
A = (3.0 × 10
–12
) / [4πε0(5.0 × 10
–2
)
2
] (= 10.79 N C
–1
) M1
at P, E
B = (12 × 10
–12
) / [4πε0(10 × 10
–2
)
2
] (= 10.79 N C
–1
) M1
or
(3.0
× 10
–12
) / [4πε0(5.0 × 10
–2
)
2
] – (12 × 10
–12
) / [4πε0(10 × 10
–2
)
2
] = 0
or
(3.0
× 10
–12
) / [4πε0(5.0 × 10
–2
)
2
] = (12 × 10
–12
) / [4πε0(10 × 10
–2
)
2
]
(M2)
fields due to charged spheres are (equal and) opposite in direction, so E = 0 A1 [3]
(c) potential = 8.99
× 10
9
{(3.0 × 10
–12
) / (5.0 × 10
–2
) + (12 × 10
–12
) / (10 × 10
–2
)} C1
= 1.62
V A1 [2]
(d) ½mv
2
= qV
E
K = ½ × 107 × 1.66 × 10
–27
× v
2
C1
qV = 47
× 1.60 × 10
–19
× 1.62 C1
v
2
= 1.37 × 10
8
v = 1.2
× 10
4
m s
–1
A1 [3]
Cambridge International AS/A Level – October/November 2016 9702 41
© UCLES 2016
(b) (i) gain/dB = 10 lg (P2
/ P1)
32 = 10
lg [PMIN
/ (0.38 × 10
–6
)]
or
–32 = 10
lg (0.38 × 10
–6
/ PMIN) C1
P
MIN = 6.0 × 10
–4
W A1 [2]
(ii) attenuation = 10
lg [(9.5 × 10
–3
) / (6.02 × 10
–4
)] C1
= 12
dB
attenuation per unit length (= 12/58) = 0.21
dB km
–1
A1 [2]
5 (a) in an electric field, charges (in a conductor) would move B1
no movement of charge so zero field strength
or
charge moves until F = 0 / E = 0 B1 [2]
or
charges in metal do not move (B1)
no (resultant) force on charges so no (electric) field (B1)
(b) at P, E
A = (3.0 × 10
–12
) / [4πε0(5.0 × 10
–2
)
2
] (= 10.79 N C
–1
) M1
at P, E
B = (12 × 10
–12
) / [4πε0(10 × 10
–2
)
2
] (= 10.79 N C
–1
) M1
or
(3.0
× 10
–12
) / [4πε0(5.0 × 10
–2
)
2
] – (12 × 10
–12
) / [4πε0(10 × 10
–2
)
2
] = 0
or
(3.0
× 10
–12
) / [4πε0(5.0 × 10
–2
)
2
] = (12 × 10
–12
) / [4πε0(10 × 10
–2
)
2
]
(M2)
fields due to charged spheres are (equal and) opposite in direction, so E = 0 A1 [3]
(c) potential = 8.99
× 10
9
{(3.0 × 10
–12
) / (5.0 × 10
–2
) + (12 × 10
–12
) / (10 × 10
–2
)} C1
= 1.62
V A1 [2]
(d) ½mv
2
= qV
E
K = ½ × 107 × 1.66 × 10
–27
× v
2
C1
qV = 47
× 1.60 × 10
–19
× 1.62 C1
v
2
= 1.37 × 10
8
v = 1.2
× 10
4
m s
–1
A1 [3]
Page 5 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 41
© UCLES 2016
6 (a) reference to input (voltage) and output (voltage) B1
there is no time delay between change in input and change in output B1 [2]
or
reference to rate at which output voltage changes (B1)
infinite rate of change (of output voltage) (B1)
(b) (i) 2.00
/ 3.00 = 1.50 / R C1
or
V
+ = (3.00 × 4.5) / (2.00 + 3.00) = 2.7
2.7 = 4.5
× R / (R + 1.50) (C1)
resistance = 2.25
kΩ A1 [2]
(ii) 1. correct symbol for LED M1
two LEDs connected with opposite polarities between V
OUT and earth A1 [2]
2. below 24
°C, R T > 1.5 kΩ or resistance of thermistor increases/high B1
V
– < V + or V – decreases/low (must not contradict initial statement) M1
V
OUT is positive/+5 (V) and LED labelled as ‘pointing’ from V OUT to earth A1 [3]
7 (a) region (of space) where a force is experienced by a particle B1 [1]
(b) (i) gravitational B1
(ii) gravitational and electric B1
(iii) gravitational, electric and magnetic B1 [3]
(c) (i) force (always) normal to direction of motion M1
(magnitude of) force constant
or
speed is constant/kinetic energy is constant M1
magnetic force provides/is the centripetal force A1 [3]
(ii) mv
2
/ r = Bqv B1
momentum or p or mv = Bqr B1 [2]
Cambridge International AS/A Level – October/November 2016 9702 41
© UCLES 2016
6 (a) reference to input (voltage) and output (voltage) B1
there is no time delay between change in input and change in output B1 [2]
or
reference to rate at which output voltage changes (B1)
infinite rate of change (of output voltage) (B1)
(b) (i) 2.00
/ 3.00 = 1.50 / R C1
or
V
+ = (3.00 × 4.5) / (2.00 + 3.00) = 2.7
2.7 = 4.5
× R / (R + 1.50) (C1)
resistance = 2.25
kΩ A1 [2]
(ii) 1. correct symbol for LED M1
two LEDs connected with opposite polarities between V
OUT and earth A1 [2]
2. below 24
°C, R T > 1.5 kΩ or resistance of thermistor increases/high B1
V
– < V + or V – decreases/low (must not contradict initial statement) M1
V
OUT is positive/+5 (V) and LED labelled as ‘pointing’ from V OUT to earth A1 [3]
7 (a) region (of space) where a force is experienced by a particle B1 [1]
(b) (i) gravitational B1
(ii) gravitational and electric B1
(iii) gravitational, electric and magnetic B1 [3]
(c) (i) force (always) normal to direction of motion M1
(magnitude of) force constant
or
speed is constant/kinetic energy is constant M1
magnetic force provides/is the centripetal force A1 [3]
(ii) mv
2
/ r = Bqv B1
momentum or p or mv = Bqr B1 [2]
Page 6 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 41
© UCLES 2016
8 strong uniform magnetic field B1
nuclei precess/rotate about field (direction) (1)
radio-frequency pulse (applied) B1
R.F. or pulse is at Larmor frequency/frequency of precession (1)
causes resonance/excitation (of nuclei)/nuclei absorb energy B1
on relaxation/de-excitation, nuclei emit r.f./pulse B1
(emitted) r.f./pulse detected and processed (1)
non-uniform magnetic field B1
allows position of nuclei to be located B1
allows for location of detection to be changed/different slices to be studied (1)
any two of the points marked (1) B2 [8]
9 (a) (induced) e.m.f. proportional to rate M1
of change of (magnetic) flux (linkage) A1 [2]
(b) flux linkage = BAN
= π
× 10
–3
× 2.8 × π × (1.6 × 10
–2
)
2
× 85 = 6.0 × 10
–4
Wb B1 [1]
(c) e.m.f. = ∆N
Φ / ∆t
e.m.f. = (6.0
× 10
–4
× 2) / 0.30 C1
e.m.f. = 4.0
mV A1 [2]
(d) sketch: E = 0 for t = 0 → 0.3
s, 0.6 s → 1.0 s, 1.6 s → 2.0 s B1
E = 4 mV for t = 0.3
s → 0.6 s (either polarity) B1
E = 2 mV for t = 1.0
s → 1.6 s B1
with opposite polarity B1 [4]
Cambridge International AS/A Level – October/November 2016 9702 41
© UCLES 2016
8 strong uniform magnetic field B1
nuclei precess/rotate about field (direction) (1)
radio-frequency pulse (applied) B1
R.F. or pulse is at Larmor frequency/frequency of precession (1)
causes resonance/excitation (of nuclei)/nuclei absorb energy B1
on relaxation/de-excitation, nuclei emit r.f./pulse B1
(emitted) r.f./pulse detected and processed (1)
non-uniform magnetic field B1
allows position of nuclei to be located B1
allows for location of detection to be changed/different slices to be studied (1)
any two of the points marked (1) B2 [8]
9 (a) (induced) e.m.f. proportional to rate M1
of change of (magnetic) flux (linkage) A1 [2]
(b) flux linkage = BAN
= π
× 10
–3
× 2.8 × π × (1.6 × 10
–2
)
2
× 85 = 6.0 × 10
–4
Wb B1 [1]
(c) e.m.f. = ∆N
Φ / ∆t
e.m.f. = (6.0
× 10
–4
× 2) / 0.30 C1
e.m.f. = 4.0
mV A1 [2]
(d) sketch: E = 0 for t = 0 → 0.3
s, 0.6 s → 1.0 s, 1.6 s → 2.0 s B1
E = 4 mV for t = 0.3
s → 0.6 s (either polarity) B1
E = 2 mV for t = 1.0
s → 1.6 s B1
with opposite polarity B1 [4]
Page 7 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 41
© UCLES 2016
10 (a) electromagnetic radiation/photons incident on a surface B1
causes emission of electrons (from the surface) B1 [2]
(b) E = hc /
λ
= (6.63
× 10
–34
× 3.00 × 10
8
) / (436 × 10
–9
) C1
= 4.56
× 10
–19
J (4.6 × 10
–19
J) A1 [2]
(c) (i)
Φ = hc / λ0
λ0 = (6.63 × 10
–34
× 3.00 × 10
8
) / (1.4 × 1.60 × 10
–19
) C1
= 890
nm A1 [2]
(ii)
λ0 = (6.63 × 10
–34
× 3.00 × 10
8
) / (4.5 × 1.60 × 10
–19
)
= 280
nm A1 [1]
(d) caesium:
wavelength of photon less than threshold wavelength (or v.v.)
or
λ0 = 890 nm > 436 nm
so yes A1
tungsten:
wavelength of photon greater than threshold wavelength (or v.v.)
or
λ0 = 280 nm < 436 nm
so no A1 [2]
11 in metal, conduction band overlaps valence band/no forbidden band/no band gap B1
as temperature rises, no increase in number of free electrons/charge carriers B1
as temperature rises, lattice vibrations increase M1
(lattice) vibrations restrict movement of electrons/charge carriers M1
(current decreases) so resistance increases A1 [5]
Cambridge International AS/A Level – October/November 2016 9702 41
© UCLES 2016
10 (a) electromagnetic radiation/photons incident on a surface B1
causes emission of electrons (from the surface) B1 [2]
(b) E = hc /
λ
= (6.63
× 10
–34
× 3.00 × 10
8
) / (436 × 10
–9
) C1
= 4.56
× 10
–19
J (4.6 × 10
–19
J) A1 [2]
(c) (i)
Φ = hc / λ0
λ0 = (6.63 × 10
–34
× 3.00 × 10
8
) / (1.4 × 1.60 × 10
–19
) C1
= 890
nm A1 [2]
(ii)
λ0 = (6.63 × 10
–34
× 3.00 × 10
8
) / (4.5 × 1.60 × 10
–19
)
= 280
nm A1 [1]
(d) caesium:
wavelength of photon less than threshold wavelength (or v.v.)
or
λ0 = 890 nm > 436 nm
so yes A1
tungsten:
wavelength of photon greater than threshold wavelength (or v.v.)
or
λ0 = 280 nm < 436 nm
so no A1 [2]
11 in metal, conduction band overlaps valence band/no forbidden band/no band gap B1
as temperature rises, no increase in number of free electrons/charge carriers B1
as temperature rises, lattice vibrations increase M1
(lattice) vibrations restrict movement of electrons/charge carriers M1
(current decreases) so resistance increases A1 [5]
Page 8 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 41
© UCLES 2016
12 (a) (i) time for number of atoms/nuclei or activity to be reduced to one half M1
reference to (number of…) original nuclide/single isotope
or
reference to half of original value/initial activity A1 [2]
(ii) A = A
0 exp(–λt) and either t = t ½, A = ½A 0 or ½A 0 = A 0 exp(–λt½) M1
so ln
2 = λt½ (and ln 2 = 0.693), hence 0.693 = λt½ A1 [2]
(b) A =
λN
N = 200
/ (2.1 × 10
–6
) C1
= 9.52
× 10
7
C1
mass = (9.52
× 10
7
× 222 × 10
–3
) / (6.02 × 10
23
)
or
mass = 9.52
× 10
7
× 222 × 1.66 × 10
–27
C1
= 3.5
× 10
–17
kg A1 [4]
Cambridge International AS/A Level – October/November 2016 9702 41
© UCLES 2016
12 (a) (i) time for number of atoms/nuclei or activity to be reduced to one half M1
reference to (number of…) original nuclide/single isotope
or
reference to half of original value/initial activity A1 [2]
(ii) A = A
0 exp(–λt) and either t = t ½, A = ½A 0 or ½A 0 = A 0 exp(–λt½) M1
so ln
2 = λt½ (and ln 2 = 0.693), hence 0.693 = λt½ A1 [2]
(b) A =
λN
N = 200
/ (2.1 × 10
–6
) C1
= 9.52
× 10
7
C1
mass = (9.52
× 10
7
× 222 × 10
–3
) / (6.02 × 10
23
)
or
mass = 9.52
× 10
7
× 222 × 1.66 × 10
–27
C1
= 3.5
× 10
–17
kg A1 [4]
® IGCSE is the registered trademark of Cambridge International Examinations.
This document consists of 8 printed pages.
© UCLES 2016 [Turn ove
r
Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/42
Paper 4 A Level Structured Questions October/November 2016
MARK SCHEME
Maximum Mark: 100
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 October/November 2016 series for most
Cambridge IGCSE
®
, Cambridge International A and AS Level components and some Cambridge O Level
components.
This document consists of 8 printed pages.
© UCLES 2016 [Turn ove
r
Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/42
Paper 4 A Level Structured Questions October/November 2016
MARK SCHEME
Maximum Mark: 100
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 October/November 2016 series for most
Cambridge IGCSE
®
, Cambridge International A and AS Level components and some Cambridge O Level
components.
Page 2 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 42
© UCLES 2016
1 (a) force per unit mass B1 [1]
(b) (i) radius/diameter/size (of Proxima Centauri) ≪ /is much less than
4.0 × 10
13
km/separation (of Sun and star)
or
(because) it is a uniform sphere B1 [1]
(ii) 1. field strength = GM
/ x
2
= (6.67 × 10
–11
× 2.5 × 10
29
) / (4.0 × 10
13
× 10
3
)
2
C1
= 1.0 × 10
–14
N kg
–1
A1 [2]
2. force = field strength × mass
= 1.0 × 10
–14
× 2.0 × 10
30
C1
or
force = GMm
/ x
2
= (6.67 × 10
–11
× 2.5 × 10
29
× 2.0 × 10
30
) / (4.0 × 10
13
× 10
3
)
2
(C1)
= 2.0 × 10
16
N A1 [2]
(c) force (of 2 × 10
16
N) would have little effect on (large) mass of Sun B1
would cause an acceleration of Sun of 1.0 × 10
–14
m s
–2
/very small/negligible
acceleration B1 [2]
or
many stars all around the Sun (B1)
net effect of forces/fields is zero (B1)
2 (a) (i) number of moles/amount of substance B1 [1]
(ii) kelvin temperature/absolute temperature/thermodynamic temperature B1 [1]
(b) pV = nRT
4.9 × 10
5
× 2.4 × 10
3
× 10
–6
= n × 8.31 × 373 B1
n = 0.38 (mol) C1
number of molecules or N = 0.38 × 6.02 × 10
23
= 2.3 × 10
23
A1 [3]
Cambridge International AS/A Level – October/November 2016 9702 42
© UCLES 2016
1 (a) force per unit mass B1 [1]
(b) (i) radius/diameter/size (of Proxima Centauri) ≪ /is much less than
4.0 × 10
13
km/separation (of Sun and star)
or
(because) it is a uniform sphere B1 [1]
(ii) 1. field strength = GM
/ x
2
= (6.67 × 10
–11
× 2.5 × 10
29
) / (4.0 × 10
13
× 10
3
)
2
C1
= 1.0 × 10
–14
N kg
–1
A1 [2]
2. force = field strength × mass
= 1.0 × 10
–14
× 2.0 × 10
30
C1
or
force = GMm
/ x
2
= (6.67 × 10
–11
× 2.5 × 10
29
× 2.0 × 10
30
) / (4.0 × 10
13
× 10
3
)
2
(C1)
= 2.0 × 10
16
N A1 [2]
(c) force (of 2 × 10
16
N) would have little effect on (large) mass of Sun B1
would cause an acceleration of Sun of 1.0 × 10
–14
m s
–2
/very small/negligible
acceleration B1 [2]
or
many stars all around the Sun (B1)
net effect of forces/fields is zero (B1)
2 (a) (i) number of moles/amount of substance B1 [1]
(ii) kelvin temperature/absolute temperature/thermodynamic temperature B1 [1]
(b) pV = nRT
4.9 × 10
5
× 2.4 × 10
3
× 10
–6
= n × 8.31 × 373 B1
n = 0.38 (mol) C1
number of molecules or N = 0.38 × 6.02 × 10
23
= 2.3 × 10
23
A1 [3]
Page 3 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 42
© UCLES 2016
or
pV = NkT (C1)
4.9 × 10
5
× 2.4 × 10
3
× 10
–6
= N × 1.38 × 10
–23
× 373 (M1)
number of molecules or N = 2.3 × 10
23
(A1)
(c) volume occupied by one molecule = (2.4 × 10
3
) / (2.3 × 10
23
) C1
= 1.04 × 10
–20
cm
3
mean spacing = (1.04 × 10
–20
)
1/3
C1
= 2.2 × 10
–7
cm (allow 1 s.f.) A1 [3]
(allow other dimensionally correct methods e.g. V = (4/3)πr
3
)
3 (a) (sum of/total) potential energy and kinetic energy of (all) molecules/particles M1
reference to random (distribution) A1 [2]
(b) (i) no heat enters (gas)/leaves (gas)/no heating (of gas) B1
work done by gas (against atmosphere as it expands) M1
internal energy decreases A1 [3]
(ii) volume decreases so work done on ice/water B1
(allow work done negligible because ∆V small)
heating of ice (to break rigid forces/bonds) M1
internal energy increases A1 [3]
4 (a) (i) 0.225
s and 0.525 s A1 [1]
(ii) period or T = 0.30
s and ω = 2π / T C1
ω = 2π / 0.30
ω = 21 rad s
–1
A1 [2]
(iii) speed =
ωx0 or ω(x0
2 – x
2
)
1/2
and x = 0 C1
= 20.9 × 2.0 × 10
–2
= 0.42 m s
–1
A1 [2]
Cambridge International AS/A Level – October/November 2016 9702 42
© UCLES 2016
or
pV = NkT (C1)
4.9 × 10
5
× 2.4 × 10
3
× 10
–6
= N × 1.38 × 10
–23
× 373 (M1)
number of molecules or N = 2.3 × 10
23
(A1)
(c) volume occupied by one molecule = (2.4 × 10
3
) / (2.3 × 10
23
) C1
= 1.04 × 10
–20
cm
3
mean spacing = (1.04 × 10
–20
)
1/3
C1
= 2.2 × 10
–7
cm (allow 1 s.f.) A1 [3]
(allow other dimensionally correct methods e.g. V = (4/3)πr
3
)
3 (a) (sum of/total) potential energy and kinetic energy of (all) molecules/particles M1
reference to random (distribution) A1 [2]
(b) (i) no heat enters (gas)/leaves (gas)/no heating (of gas) B1
work done by gas (against atmosphere as it expands) M1
internal energy decreases A1 [3]
(ii) volume decreases so work done on ice/water B1
(allow work done negligible because ∆V small)
heating of ice (to break rigid forces/bonds) M1
internal energy increases A1 [3]
4 (a) (i) 0.225
s and 0.525 s A1 [1]
(ii) period or T = 0.30
s and ω = 2π / T C1
ω = 2π / 0.30
ω = 21 rad s
–1
A1 [2]
(iii) speed =
ωx0 or ω(x0
2 – x
2
)
1/2
and x = 0 C1
= 20.9 × 2.0 × 10
–2
= 0.42 m s
–1
A1 [2]
Page 4 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 42
© UCLES 2016
or
use of tangent method:
correct tangent shown on Fig. 4.2 (C1)
working e.g. ∆y
/ ∆x leading to maximum speed in range (0.38–0.46) m s
–1
(A1)
(b) sketch: reasonably shaped continuous oval/circle surrounding (0,0) B1
curve passes through (0, 0.42) and (0, –0.42) B1
curve passes through (2.0, 0) and (–2.0, 0) B1 [3]
5 (a) transducer/transmitter can be also be used as the receiver
or
transducer both transmits and receives
receives reflected pulses between the emitted pulses
(needs to be pulsed) in order to measure/determine depth(s)
(needs to be pulsed) to determine nature of boundaries
Any three of the above marking points, 1 mark each B2 [2]
(b) (i) product of speed of (ultra)sound and density (of medium) M1
reference to speed of sound in medium A1 [2]
(ii) if Z
1 and Z 2 are (nearly) equal, I T / I0 (nearly) equal to 1/unity/(very) little
reflection/mostly transmission B1
if Z
1 ≫ Z 2 or Z 1 ≪ Z 2 or the difference between Z 1 and Z 2 is (very) large,
then I
T / I0 is small/zero/mostly reflection/little transmission B1 [2]
6 (a) E = 0 or E
A = (–)E B (at x = 11 cm) B1
Q
A / x
2
= Q B
/ (20 – x)
2
= 11
2
/ 9
2
C1
Q
A
/ QB or ratio = 1.5 A1 [3]
or
E ∝ Q because r same or E = Q
/ 4πε0r
2
and r same (B1)
Q
A / QB = 48 / 32 (C1)
Q
A
/ QB or ratio = 1.5 (A1)
Cambridge International AS/A Level – October/November 2016 9702 42
© UCLES 2016
or
use of tangent method:
correct tangent shown on Fig. 4.2 (C1)
working e.g. ∆y
/ ∆x leading to maximum speed in range (0.38–0.46) m s
–1
(A1)
(b) sketch: reasonably shaped continuous oval/circle surrounding (0,0) B1
curve passes through (0, 0.42) and (0, –0.42) B1
curve passes through (2.0, 0) and (–2.0, 0) B1 [3]
5 (a) transducer/transmitter can be also be used as the receiver
or
transducer both transmits and receives
receives reflected pulses between the emitted pulses
(needs to be pulsed) in order to measure/determine depth(s)
(needs to be pulsed) to determine nature of boundaries
Any three of the above marking points, 1 mark each B2 [2]
(b) (i) product of speed of (ultra)sound and density (of medium) M1
reference to speed of sound in medium A1 [2]
(ii) if Z
1 and Z 2 are (nearly) equal, I T / I0 (nearly) equal to 1/unity/(very) little
reflection/mostly transmission B1
if Z
1 ≫ Z 2 or Z 1 ≪ Z 2 or the difference between Z 1 and Z 2 is (very) large,
then I
T / I0 is small/zero/mostly reflection/little transmission B1 [2]
6 (a) E = 0 or E
A = (–)E B (at x = 11 cm) B1
Q
A / x
2
= Q B
/ (20 – x)
2
= 11
2
/ 9
2
C1
Q
A
/ QB or ratio = 1.5 A1 [3]
or
E ∝ Q because r same or E = Q
/ 4πε0r
2
and r same (B1)
Q
A / QB = 48 / 32 (C1)
Q
A
/ QB or ratio = 1.5 (A1)
Page 5 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 42
© UCLES 2016
(b) (i) for max. speed, ∆V = (0.76 – 0.18) V or ∆V = 0.58 V C1
q ∆V = ½mv
2
2 × (1.60 × 10
–19
) × 0.58 = ½ × 4 × 1.66 × 10
–27
× v
2
C1
v
2
= 5.59 × 10
7
v = 7.5 × 10
3
m s
–1
A1 [3]
(ii) ∆V = 0.22 V C1
2 × (1.60 × 10
–19
) × 0.22 = ½ × 4 × 1.66 × 10
–27
× v
2
v
2
= 2.12 × 10
7
v = 4.6 × 10
3
m s
–1
A1 [2]
7 (a) (i) charge
/ potential (difference) or charge per (unit) potential (difference) B1 [1]
(ii) (V = Q
/ 4πε0r and C = Q / V)
for sphere, C (= Q
/ V) = 4πε 0r C1
C = 4π × 8.85 × 10
–12
× 12.5 × 10
–2
= 1.4 × 10
–11
F A1 [2]
(b) (i) 1
/ CT = 1 / 3.0 + 1 / 6.0
C
T = 2.0 µF A1 [1]
(ii) total charge = charge on 3.0
µF capacitor = 2.0 (µ) × 9.0 = 18 (µC) C1
potential difference = Q
/ C = 18 (µ)C / 3.0 (µ)F = 6.0 V A1 [2]
or
argument based on equal charges:
3.0 × V = 6.0 × (9.0 – V) (C1)
V = 6.0 V (A1)
(iii) potential difference (= 9.0 – 6.0) = 3.0
V C1
charge (= 3.0 × 2.0 (µ)) = 6.0
µC A1 [2]
Cambridge International AS/A Level – October/November 2016 9702 42
© UCLES 2016
(b) (i) for max. speed, ∆V = (0.76 – 0.18) V or ∆V = 0.58 V C1
q ∆V = ½mv
2
2 × (1.60 × 10
–19
) × 0.58 = ½ × 4 × 1.66 × 10
–27
× v
2
C1
v
2
= 5.59 × 10
7
v = 7.5 × 10
3
m s
–1
A1 [3]
(ii) ∆V = 0.22 V C1
2 × (1.60 × 10
–19
) × 0.22 = ½ × 4 × 1.66 × 10
–27
× v
2
v
2
= 2.12 × 10
7
v = 4.6 × 10
3
m s
–1
A1 [2]
7 (a) (i) charge
/ potential (difference) or charge per (unit) potential (difference) B1 [1]
(ii) (V = Q
/ 4πε0r and C = Q / V)
for sphere, C (= Q
/ V) = 4πε 0r C1
C = 4π × 8.85 × 10
–12
× 12.5 × 10
–2
= 1.4 × 10
–11
F A1 [2]
(b) (i) 1
/ CT = 1 / 3.0 + 1 / 6.0
C
T = 2.0 µF A1 [1]
(ii) total charge = charge on 3.0
µF capacitor = 2.0 (µ) × 9.0 = 18 (µC) C1
potential difference = Q
/ C = 18 (µ)C / 3.0 (µ)F = 6.0 V A1 [2]
or
argument based on equal charges:
3.0 × V = 6.0 × (9.0 – V) (C1)
V = 6.0 V (A1)
(iii) potential difference (= 9.0 – 6.0) = 3.0
V C1
charge (= 3.0 × 2.0 (µ)) = 6.0
µC A1 [2]
Page 6 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 42
© UCLES 2016
8 (a) P shown between earth symbol and voltmeter B1 [1]
(b) (i) gain = (50 × 10
3
) / 100 = 500 C1
V
IN (= 5.0 / 500) = 0.010 V A1 [2]
(ii) V
IN (= 5.0 / 5.0) = 1.0 V A1 [1]
(c) e.g. multi-range (volt)meter
c.r.o. sensitivity control
amplifier channel selector B1 [1]
9 (a) (by Newton’s third law) force on wire is up(wards) M1
by (Fleming’s) left-hand rule/right-hand slap rule to give current A1
in direction left to right shown on diagram A1 [3]
(b) force ∝ current or F = BIL or B (= 0.080
/ 6.0L) = 1 / 75L C1
maximum current = 2.5 × √2 C1
= 3.54
A
maximum force in one direction = (3.54
/ 6.0) × 0.080 C1
= 0.047
N
difference (= 2 × 0.047) = 0.094
N
or
force varies from 0.047
N upwards to 0.047 N downwards A1 [4]
10 nuclei emitting r.f. (pulse) B1
Larmor frequency/r.f. frequency emitted/detected depends on magnitude of magnetic
field B1
nuclei can be located (within a slice) B1
changing field enables position of detection (slice) to be changed B1 [4]
11 (a) (induced) e.m.f. proportional/equal to rate M1
of change of (magnetic) flux (linkage) A1 [2]
(b) (for same current) iron core gives large(r) (rates of change of) flux (linkage) B1
e.m.f induced in solenoid is greater (for same current) M1
induced e.m.f. opposes applied e.m.f. so current smaller/acts to reduce current A1 [3]
Cambridge International AS/A Level – October/November 2016 9702 42
© UCLES 2016
8 (a) P shown between earth symbol and voltmeter B1 [1]
(b) (i) gain = (50 × 10
3
) / 100 = 500 C1
V
IN (= 5.0 / 500) = 0.010 V A1 [2]
(ii) V
IN (= 5.0 / 5.0) = 1.0 V A1 [1]
(c) e.g. multi-range (volt)meter
c.r.o. sensitivity control
amplifier channel selector B1 [1]
9 (a) (by Newton’s third law) force on wire is up(wards) M1
by (Fleming’s) left-hand rule/right-hand slap rule to give current A1
in direction left to right shown on diagram A1 [3]
(b) force ∝ current or F = BIL or B (= 0.080
/ 6.0L) = 1 / 75L C1
maximum current = 2.5 × √2 C1
= 3.54
A
maximum force in one direction = (3.54
/ 6.0) × 0.080 C1
= 0.047
N
difference (= 2 × 0.047) = 0.094
N
or
force varies from 0.047
N upwards to 0.047 N downwards A1 [4]
10 nuclei emitting r.f. (pulse) B1
Larmor frequency/r.f. frequency emitted/detected depends on magnitude of magnetic
field B1
nuclei can be located (within a slice) B1
changing field enables position of detection (slice) to be changed B1 [4]
11 (a) (induced) e.m.f. proportional/equal to rate M1
of change of (magnetic) flux (linkage) A1 [2]
(b) (for same current) iron core gives large(r) (rates of change of) flux (linkage) B1
e.m.f induced in solenoid is greater (for same current) M1
induced e.m.f. opposes applied e.m.f. so current smaller/acts to reduce current A1 [3]
Page 7 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 42
© UCLES 2016
or
same supply so same induced e.m.f. balancing it (B1)
(rate of change of) flux linkage is same (M1)
smaller current for same flux when core present (A1)
(c) e.g. (heating due to) eddy currents in core
(heating due to current in) resistance of coils
hysteresis losses/losses due to changing magnetic field in core
Any two of the above marking points, 1 mark each B2 [2]
12 (a) (i) electron diffraction/electron microscope (allow other sensible suggestions) B1 [1]
(ii) photoelectric effect/Compton scattering (allow other sensible suggestions) B1 [1]
(b) (i) arrow clear from –0.54 eV to –3.40 eV B1 [1]
(ii) E = hc
/ λ or E = hf and c = f λ C1
λ = (6.63 × 10
–34
× 3.00 × 10
8
) / [(3.40 – 0.54) × 1.60 × 10
–19
] = 4.35 × 10
–7
m A1 [2]
(c) (i) wavelength associated with a particle M1
that is moving/has momentum/has speed/has velocity A1 [2]
(ii)
λ = h / mv
v = (6.63 × 10
–34
) / (9.11 × 10
–31
× 4.35 × 10
–7
) C1
= 1.67 × 10
3
m s
–1
A1 [2]
13 X-ray image of a (single) slice/cross-section (through the patient) M1
taken from different angles/rotating X-ray (beam) A1
computer is used to form/process/build up/store image B1
2D image (of the slice) B1
repeated for many/different (neighbouring) slices M1
to build up 3D image A1 [6]
Cambridge International AS/A Level – October/November 2016 9702 42
© UCLES 2016
or
same supply so same induced e.m.f. balancing it (B1)
(rate of change of) flux linkage is same (M1)
smaller current for same flux when core present (A1)
(c) e.g. (heating due to) eddy currents in core
(heating due to current in) resistance of coils
hysteresis losses/losses due to changing magnetic field in core
Any two of the above marking points, 1 mark each B2 [2]
12 (a) (i) electron diffraction/electron microscope (allow other sensible suggestions) B1 [1]
(ii) photoelectric effect/Compton scattering (allow other sensible suggestions) B1 [1]
(b) (i) arrow clear from –0.54 eV to –3.40 eV B1 [1]
(ii) E = hc
/ λ or E = hf and c = f λ C1
λ = (6.63 × 10
–34
× 3.00 × 10
8
) / [(3.40 – 0.54) × 1.60 × 10
–19
] = 4.35 × 10
–7
m A1 [2]
(c) (i) wavelength associated with a particle M1
that is moving/has momentum/has speed/has velocity A1 [2]
(ii)
λ = h / mv
v = (6.63 × 10
–34
) / (9.11 × 10
–31
× 4.35 × 10
–7
) C1
= 1.67 × 10
3
m s
–1
A1 [2]
13 X-ray image of a (single) slice/cross-section (through the patient) M1
taken from different angles/rotating X-ray (beam) A1
computer is used to form/process/build up/store image B1
2D image (of the slice) B1
repeated for many/different (neighbouring) slices M1
to build up 3D image A1 [6]
Page 8 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 42
© UCLES 2016
14 (a) (i) He
4
2
or α
4
2
B1 [1]
(ii)
n
1
0
B1 [1]
(b) (i) ∆m = (29.97830 +1.00867) – (26.98153 + 4.00260) C1
= 30.98697 – 30.98413
= 2.84 × 10
–3
u C1 [2]
(ii) E = c
2
∆m or mc
2
C1
= (3.0 × 10
8
)
2
× 2.84 × 10
–3
× 1.66 × 10
–27
= 4.2 × 10
–13
J A1 [2]
(c) mass of products is greater than mass of Al plus α
or
reaction causes (net) increase in (rest) mass (of the system) B1
α -particle must have at least this amount of kinetic energy B1 [2]
Cambridge International AS/A Level – October/November 2016 9702 42
© UCLES 2016
14 (a) (i) He
4
2
or α
4
2
B1 [1]
(ii)
n
1
0
B1 [1]
(b) (i) ∆m = (29.97830 +1.00867) – (26.98153 + 4.00260) C1
= 30.98697 – 30.98413
= 2.84 × 10
–3
u C1 [2]
(ii) E = c
2
∆m or mc
2
C1
= (3.0 × 10
8
)
2
× 2.84 × 10
–3
× 1.66 × 10
–27
= 4.2 × 10
–13
J A1 [2]
(c) mass of products is greater than mass of Al plus α
or
reaction causes (net) increase in (rest) mass (of the system) B1
α -particle must have at least this amount of kinetic energy B1 [2]
® IGCSE is the registered trademark of Cambridge International Examinations.
This document consists of 8 printed pages.
© UCLES 2016 [Turn ove
r
Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/43
Paper 4 A Level Structured Questions October/November 2016
MARK SCHEME
Maximum Mark: 100
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 October/November 2016 series for most
Cambridge IGCSE
®
, Cambridge International A and AS Level components and some Cambridge O Level
components.
This document consists of 8 printed pages.
© UCLES 2016 [Turn ove
r
Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/43
Paper 4 A Level Structured Questions October/November 2016
MARK SCHEME
Maximum Mark: 100
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 October/November 2016 series for most
Cambridge IGCSE
®
, Cambridge International A and AS Level components and some Cambridge O Level
components.
Page 2 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 43
© UCLES 2016
1 (a) gravitational force provides/is the centripetal force B1
GMm
/ r
2
= mv
2
/ r or GMm / r
2
= mrω
2
and v = 2πr
/ T or ω = 2π / T M1
with algebra to T
2
= 4π
2
r
3
/ GM A1 [3]
or
acceleration due to gravity is the centripetal acceleration (B1)
GM
/ r
2
= v
2
/ r or GM / r
2
= rω
2
and v = 2πr
/ T or ω = 2π / T (M1)
with algebra to T
2
= 4π
2
r
3
/ GM (A1)
(b) (i) equatorial orbit/orbits (directly) above the equator B1
from west to east B1 [2]
(ii) (24
× 3600)
2
= 4π
2
r
3
/ (6.67 × 10
–11
× 6.0 × 10
24
) C1
r
3
= 7.57 × 10
22
r = 4.2
× 10
7
m A1 [2]
(c) (T
/ 24)
2
= {(2.64 × 10
7
) / (4.23 × 10
7
)}
3
B1
= 0.243
T = 12 hours A1 [2]
or
k (= T
2
/ r
3
) = 24
2
/ (4.23 × 10
7
)
3
(B1)
k = 7.61
× 10
–21
T
2
(= kr
3
) = 7.61 × 10
–21
× (2.64 × 10
7
)
3
= 140
T = 12 hours (A1)
2 (a) (i) p ∝ T or pV
/ T = constant or pV = nRT C1
T (= 5
× 300 =) 1500 K A1 [2]
(ii) pV = nRT
1.0
× 10
5
× 4.0 × 10
–4
= n × 8.31 × 300
or
5.0
× 10
5
× 4.0 × 10
–4
= n × 8.31 × 1500 C1
n = 0.016 mol A1 [2]
Cambridge International AS/A Level – October/November 2016 9702 43
© UCLES 2016
1 (a) gravitational force provides/is the centripetal force B1
GMm
/ r
2
= mv
2
/ r or GMm / r
2
= mrω
2
and v = 2πr
/ T or ω = 2π / T M1
with algebra to T
2
= 4π
2
r
3
/ GM A1 [3]
or
acceleration due to gravity is the centripetal acceleration (B1)
GM
/ r
2
= v
2
/ r or GM / r
2
= rω
2
and v = 2πr
/ T or ω = 2π / T (M1)
with algebra to T
2
= 4π
2
r
3
/ GM (A1)
(b) (i) equatorial orbit/orbits (directly) above the equator B1
from west to east B1 [2]
(ii) (24
× 3600)
2
= 4π
2
r
3
/ (6.67 × 10
–11
× 6.0 × 10
24
) C1
r
3
= 7.57 × 10
22
r = 4.2
× 10
7
m A1 [2]
(c) (T
/ 24)
2
= {(2.64 × 10
7
) / (4.23 × 10
7
)}
3
B1
= 0.243
T = 12 hours A1 [2]
or
k (= T
2
/ r
3
) = 24
2
/ (4.23 × 10
7
)
3
(B1)
k = 7.61
× 10
–21
T
2
(= kr
3
) = 7.61 × 10
–21
× (2.64 × 10
7
)
3
= 140
T = 12 hours (A1)
2 (a) (i) p ∝ T or pV
/ T = constant or pV = nRT C1
T (= 5
× 300 =) 1500 K A1 [2]
(ii) pV = nRT
1.0
× 10
5
× 4.0 × 10
–4
= n × 8.31 × 300
or
5.0
× 10
5
× 4.0 × 10
–4
= n × 8.31 × 1500 C1
n = 0.016 mol A1 [2]
Page 3 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 43
© UCLES 2016
(b) (i) 1. heating/thermal energy supplied B1
2. work done on/to system B1 [2]
(ii) 1. 240
J A1
2. same value as given in 1. (= 240
J) and zero given for 3. A1
3. zero A1 [3]
3 (a) 2k/m =
ω
2
M1
ω = 2πf M1
(2
× 64 / 0.810) = (2π × f)
2
leading to f = 2.0 Hz A1 [3]
(b) v
0 = ωx0 or v0 = 2πfx 0
or
v =
ω(x0
2 – x
2
)
1/2
and x = 0 C1
v
0 = 2π × 2.0 × 1.6 × 10
–2
= 0.20
m s
–1
A1 [2]
(c) frequency: reduced/decreased B1
maximum speed: reduced/decreased B1 [2]
4 (a) (i) noise/distortion is removed (from the signal) B1
the (original) signal is reformed/reproduced/recovered/restored B1 [2]
or
signal detected above/below a threshold creates new signal (B1)
of 1s and 0s (B1)
(ii) noise is superposed on the (displacement of the) signal/cannot be
distinguished
or
analogue/signal is continuous (so cannot be regenerated)
or
analogue/signal is not discrete (so cannot be regenerated) B1
noise is amplified with the signal B1 [2]
Cambridge International AS/A Level – October/November 2016 9702 43
© UCLES 2016
(b) (i) 1. heating/thermal energy supplied B1
2. work done on/to system B1 [2]
(ii) 1. 240
J A1
2. same value as given in 1. (= 240
J) and zero given for 3. A1
3. zero A1 [3]
3 (a) 2k/m =
ω
2
M1
ω = 2πf M1
(2
× 64 / 0.810) = (2π × f)
2
leading to f = 2.0 Hz A1 [3]
(b) v
0 = ωx0 or v0 = 2πfx 0
or
v =
ω(x0
2 – x
2
)
1/2
and x = 0 C1
v
0 = 2π × 2.0 × 1.6 × 10
–2
= 0.20
m s
–1
A1 [2]
(c) frequency: reduced/decreased B1
maximum speed: reduced/decreased B1 [2]
4 (a) (i) noise/distortion is removed (from the signal) B1
the (original) signal is reformed/reproduced/recovered/restored B1 [2]
or
signal detected above/below a threshold creates new signal (B1)
of 1s and 0s (B1)
(ii) noise is superposed on the (displacement of the) signal/cannot be
distinguished
or
analogue/signal is continuous (so cannot be regenerated)
or
analogue/signal is not discrete (so cannot be regenerated) B1
noise is amplified with the signal B1 [2]
Page 4 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 43
© UCLES 2016
(b) (i) gain/dB = 10 lg (P2
/ P1)
32 = 10
lg [PMIN
/ (0.38 × 10
–6
)]
or
–32 = 10
lg (0.38 × 10
–6
/ PMIN) C1
P
MIN = 6.0 × 10
–4
W A1 [2]
(ii) attenuation = 10
lg [(9.5 × 10
–3
) / (6.02 × 10
–4
)] C1
= 12
dB
attenuation per unit length (= 12/58) = 0.21
dB km
–1
A1 [2]
5 (a) in an electric field, charges (in a conductor) would move B1
no movement of charge so zero field strength
or
charge moves until F = 0 / E = 0 B1 [2]
or
charges in metal do not move (B1)
no (resultant) force on charges so no (electric) field (B1)
(b) at P, E
A = (3.0 × 10
–12
) / [4πε0(5.0 × 10
–2
)
2
] (= 10.79 N C
–1
) M1
at P, E
B = (12 × 10
–12
) / [4πε0(10 × 10
–2
)
2
] (= 10.79 N C
–1
) M1
or
(3.0
× 10
–12
) / [4πε0(5.0 × 10
–2
)
2
] – (12 × 10
–12
) / [4πε0(10 × 10
–2
)
2
] = 0
or
(3.0
× 10
–12
) / [4πε0(5.0 × 10
–2
)
2
] = (12 × 10
–12
) / [4πε0(10 × 10
–2
)
2
]
(M2)
fields due to charged spheres are (equal and) opposite in direction, so E = 0 A1 [3]
(c) potential = 8.99
× 10
9
{(3.0 × 10
–12
) / (5.0 × 10
–2
) + (12 × 10
–12
) / (10 × 10
–2
)} C1
= 1.62
V A1 [2]
(d) ½mv
2
= qV
E
K = ½ × 107 × 1.66 × 10
–27
× v
2
C1
qV = 47
× 1.60 × 10
–19
× 1.62 C1
v
2
= 1.37 × 10
8
v = 1.2
× 10
4
m s
–1
A1 [3]
Cambridge International AS/A Level – October/November 2016 9702 43
© UCLES 2016
(b) (i) gain/dB = 10 lg (P2
/ P1)
32 = 10
lg [PMIN
/ (0.38 × 10
–6
)]
or
–32 = 10
lg (0.38 × 10
–6
/ PMIN) C1
P
MIN = 6.0 × 10
–4
W A1 [2]
(ii) attenuation = 10
lg [(9.5 × 10
–3
) / (6.02 × 10
–4
)] C1
= 12
dB
attenuation per unit length (= 12/58) = 0.21
dB km
–1
A1 [2]
5 (a) in an electric field, charges (in a conductor) would move B1
no movement of charge so zero field strength
or
charge moves until F = 0 / E = 0 B1 [2]
or
charges in metal do not move (B1)
no (resultant) force on charges so no (electric) field (B1)
(b) at P, E
A = (3.0 × 10
–12
) / [4πε0(5.0 × 10
–2
)
2
] (= 10.79 N C
–1
) M1
at P, E
B = (12 × 10
–12
) / [4πε0(10 × 10
–2
)
2
] (= 10.79 N C
–1
) M1
or
(3.0
× 10
–12
) / [4πε0(5.0 × 10
–2
)
2
] – (12 × 10
–12
) / [4πε0(10 × 10
–2
)
2
] = 0
or
(3.0
× 10
–12
) / [4πε0(5.0 × 10
–2
)
2
] = (12 × 10
–12
) / [4πε0(10 × 10
–2
)
2
]
(M2)
fields due to charged spheres are (equal and) opposite in direction, so E = 0 A1 [3]
(c) potential = 8.99
× 10
9
{(3.0 × 10
–12
) / (5.0 × 10
–2
) + (12 × 10
–12
) / (10 × 10
–2
)} C1
= 1.62
V A1 [2]
(d) ½mv
2
= qV
E
K = ½ × 107 × 1.66 × 10
–27
× v
2
C1
qV = 47
× 1.60 × 10
–19
× 1.62 C1
v
2
= 1.37 × 10
8
v = 1.2
× 10
4
m s
–1
A1 [3]
Page 5 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 43
© UCLES 2016
6 (a) reference to input (voltage) and output (voltage) B1
there is no time delay between change in input and change in output B1 [2]
or
reference to rate at which output voltage changes (B1)
infinite rate of change (of output voltage) (B1)
(b) (i) 2.00
/ 3.00 = 1.50 / R C1
or
V
+ = (3.00 × 4.5) / (2.00 + 3.00) = 2.7
2.7 = 4.5
× R / (R + 1.50) (C1)
resistance = 2.25
kΩ A1 [2]
(ii) 1. correct symbol for LED M1
two LEDs connected with opposite polarities between V
OUT and earth A1 [2]
2. below 24
°C, R T > 1.5 kΩ or resistance of thermistor increases/high B1
V
– < V + or V – decreases/low (must not contradict initial statement) M1
V
OUT is positive/+5 (V) and LED labelled as ‘pointing’ from V OUT to earth A1 [3]
7 (a) region (of space) where a force is experienced by a particle B1 [1]
(b) (i) gravitational B1
(ii) gravitational and electric B1
(iii) gravitational, electric and magnetic B1 [3]
(c) (i) force (always) normal to direction of motion M1
(magnitude of) force constant
or
speed is constant/kinetic energy is constant M1
magnetic force provides/is the centripetal force A1 [3]
(ii) mv
2
/ r = Bqv B1
momentum or p or mv = Bqr B1 [2]
Cambridge International AS/A Level – October/November 2016 9702 43
© UCLES 2016
6 (a) reference to input (voltage) and output (voltage) B1
there is no time delay between change in input and change in output B1 [2]
or
reference to rate at which output voltage changes (B1)
infinite rate of change (of output voltage) (B1)
(b) (i) 2.00
/ 3.00 = 1.50 / R C1
or
V
+ = (3.00 × 4.5) / (2.00 + 3.00) = 2.7
2.7 = 4.5
× R / (R + 1.50) (C1)
resistance = 2.25
kΩ A1 [2]
(ii) 1. correct symbol for LED M1
two LEDs connected with opposite polarities between V
OUT and earth A1 [2]
2. below 24
°C, R T > 1.5 kΩ or resistance of thermistor increases/high B1
V
– < V + or V – decreases/low (must not contradict initial statement) M1
V
OUT is positive/+5 (V) and LED labelled as ‘pointing’ from V OUT to earth A1 [3]
7 (a) region (of space) where a force is experienced by a particle B1 [1]
(b) (i) gravitational B1
(ii) gravitational and electric B1
(iii) gravitational, electric and magnetic B1 [3]
(c) (i) force (always) normal to direction of motion M1
(magnitude of) force constant
or
speed is constant/kinetic energy is constant M1
magnetic force provides/is the centripetal force A1 [3]
(ii) mv
2
/ r = Bqv B1
momentum or p or mv = Bqr B1 [2]
Page 6 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 43
© UCLES 2016
8 strong uniform magnetic field B1
nuclei precess/rotate about field (direction) (1)
radio-frequency pulse (applied) B1
R.F. or pulse is at Larmor frequency/frequency of precession (1)
causes resonance/excitation (of nuclei)/nuclei absorb energy B1
on relaxation/de-excitation, nuclei emit r.f./pulse B1
(emitted) r.f./pulse detected and processed (1)
non-uniform magnetic field B1
allows position of nuclei to be located B1
allows for location of detection to be changed/different slices to be studied (1)
any two of the points marked (1) B2 [8]
9 (a) (induced) e.m.f. proportional to rate M1
of change of (magnetic) flux (linkage) A1 [2]
(b) flux linkage = BAN
= π
× 10
–3
× 2.8 × π × (1.6 × 10
–2
)
2
× 85 = 6.0 × 10
–4
Wb B1 [1]
(c) e.m.f. = ∆N
Φ / ∆t
e.m.f. = (6.0
× 10
–4
× 2) / 0.30 C1
e.m.f. = 4.0
mV A1 [2]
(d) sketch: E = 0 for t = 0 → 0.3
s, 0.6 s → 1.0 s, 1.6 s → 2.0 s B1
E = 4 mV for t = 0.3
s → 0.6 s (either polarity) B1
E = 2 mV for t = 1.0
s → 1.6 s B1
with opposite polarity B1 [4]
Cambridge International AS/A Level – October/November 2016 9702 43
© UCLES 2016
8 strong uniform magnetic field B1
nuclei precess/rotate about field (direction) (1)
radio-frequency pulse (applied) B1
R.F. or pulse is at Larmor frequency/frequency of precession (1)
causes resonance/excitation (of nuclei)/nuclei absorb energy B1
on relaxation/de-excitation, nuclei emit r.f./pulse B1
(emitted) r.f./pulse detected and processed (1)
non-uniform magnetic field B1
allows position of nuclei to be located B1
allows for location of detection to be changed/different slices to be studied (1)
any two of the points marked (1) B2 [8]
9 (a) (induced) e.m.f. proportional to rate M1
of change of (magnetic) flux (linkage) A1 [2]
(b) flux linkage = BAN
= π
× 10
–3
× 2.8 × π × (1.6 × 10
–2
)
2
× 85 = 6.0 × 10
–4
Wb B1 [1]
(c) e.m.f. = ∆N
Φ / ∆t
e.m.f. = (6.0
× 10
–4
× 2) / 0.30 C1
e.m.f. = 4.0
mV A1 [2]
(d) sketch: E = 0 for t = 0 → 0.3
s, 0.6 s → 1.0 s, 1.6 s → 2.0 s B1
E = 4 mV for t = 0.3
s → 0.6 s (either polarity) B1
E = 2 mV for t = 1.0
s → 1.6 s B1
with opposite polarity B1 [4]
Page 7 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 43
© UCLES 2016
10 (a) electromagnetic radiation/photons incident on a surface B1
causes emission of electrons (from the surface) B1 [2]
(b) E = hc /
λ
= (6.63
× 10
–34
× 3.00 × 10
8
) / (436 × 10
–9
) C1
= 4.56
× 10
–19
J (4.6 × 10
–19
J) A1 [2]
(c) (i)
Φ = hc / λ0
λ0 = (6.63 × 10
–34
× 3.00 × 10
8
) / (1.4 × 1.60 × 10
–19
) C1
= 890
nm A1 [2]
(ii)
λ0 = (6.63 × 10
–34
× 3.00 × 10
8
) / (4.5 × 1.60 × 10
–19
)
= 280
nm A1 [1]
(d) caesium:
wavelength of photon less than threshold wavelength (or v.v.)
or
λ0 = 890 nm > 436 nm
so yes A1
tungsten:
wavelength of photon greater than threshold wavelength (or v.v.)
or
λ0 = 280 nm < 436 nm
so no A1 [2]
11 in metal, conduction band overlaps valence band/no forbidden band/no band gap B1
as temperature rises, no increase in number of free electrons/charge carriers B1
as temperature rises, lattice vibrations increase M1
(lattice) vibrations restrict movement of electrons/charge carriers M1
(current decreases) so resistance increases A1 [5]
Cambridge International AS/A Level – October/November 2016 9702 43
© UCLES 2016
10 (a) electromagnetic radiation/photons incident on a surface B1
causes emission of electrons (from the surface) B1 [2]
(b) E = hc /
λ
= (6.63
× 10
–34
× 3.00 × 10
8
) / (436 × 10
–9
) C1
= 4.56
× 10
–19
J (4.6 × 10
–19
J) A1 [2]
(c) (i)
Φ = hc / λ0
λ0 = (6.63 × 10
–34
× 3.00 × 10
8
) / (1.4 × 1.60 × 10
–19
) C1
= 890
nm A1 [2]
(ii)
λ0 = (6.63 × 10
–34
× 3.00 × 10
8
) / (4.5 × 1.60 × 10
–19
)
= 280
nm A1 [1]
(d) caesium:
wavelength of photon less than threshold wavelength (or v.v.)
or
λ0 = 890 nm > 436 nm
so yes A1
tungsten:
wavelength of photon greater than threshold wavelength (or v.v.)
or
λ0 = 280 nm < 436 nm
so no A1 [2]
11 in metal, conduction band overlaps valence band/no forbidden band/no band gap B1
as temperature rises, no increase in number of free electrons/charge carriers B1
as temperature rises, lattice vibrations increase M1
(lattice) vibrations restrict movement of electrons/charge carriers M1
(current decreases) so resistance increases A1 [5]
Page 8 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 43
© UCLES 2016
12 (a) (i) time for number of atoms/nuclei or activity to be reduced to one half M1
reference to (number of…) original nuclide/single isotope
or
reference to half of original value/initial activity A1 [2]
(ii) A = A
0 exp(–λt) and either t = t ½, A = ½A 0 or ½A 0 = A 0 exp(–λt½) M1
so ln
2 = λt½ (and ln 2 = 0.693), hence 0.693 = λt½ A1 [2]
(b) A =
λN
N = 200
/ (2.1 × 10
–6
) C1
= 9.52
× 10
7
C1
mass = (9.52
× 10
7
× 222 × 10
–3
) / (6.02 × 10
23
)
or
mass = 9.52
× 10
7
× 222 × 1.66 × 10
–27
C1
= 3.5
× 10
–17
kg A1 [4]
Cambridge International AS/A Level – October/November 2016 9702 43
© UCLES 2016
12 (a) (i) time for number of atoms/nuclei or activity to be reduced to one half M1
reference to (number of…) original nuclide/single isotope
or
reference to half of original value/initial activity A1 [2]
(ii) A = A
0 exp(–λt) and either t = t ½, A = ½A 0 or ½A 0 = A 0 exp(–λt½) M1
so ln
2 = λt½ (and ln 2 = 0.693), hence 0.693 = λt½ A1 [2]
(b) A =
λN
N = 200
/ (2.1 × 10
–6
) C1
= 9.52
× 10
7
C1
mass = (9.52
× 10
7
× 222 × 10
–3
) / (6.02 × 10
23
)
or
mass = 9.52
× 10
7
× 222 × 1.66 × 10
–27
C1
= 3.5
× 10
–17
kg A1 [4]
® IGCSE is the registered trademark of Cambridge International Examinations.
This document consists of 5 printed pages.
© UCLES 2016 [Turn ove
r
Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/51
Paper 5 Planning, Analysis and Evaluation October/November 2016
MARK SCHEME
Maximum Mark: 30
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 October/November 2016 series for most
Cambridge IGCSE
®
, Cambridge International A and AS Level components and some Cambridge O Level
components.
This document consists of 5 printed pages.
© UCLES 2016 [Turn ove
r
Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/51
Paper 5 Planning, Analysis and Evaluation October/November 2016
MARK SCHEME
Maximum Mark: 30
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 October/November 2016 series for most
Cambridge IGCSE
®
, Cambridge International A and AS Level components and some Cambridge O Level
components.
Page 2 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 51
© UCLES 2016
Question Answer Marks
1 Defining the problem
B is the independent variable and v is the dependent variable, or vary B
measure v.
1
Keep starting position of magnet constant/magnet always released from
rest.
1
Methods of data collection
Labelled diagram showing a magnet and the vertical copper tube supported. 1
Method to ensure that copper tube is vertical, e.g. set square, spirit level,
plumb line.
1
Method to determine time at bottom of tube e.g. use of light gate(s)/motion
sensor attached to timer/datalogger/computer or distance between two fixed
marks at bottom of tube and stopwatch.
Do not allow time over length of tube.
1
Method to measure B, e.g. Hall probe. 1
Method of analysis
Plot a graph of ln v against B. 1
λ = – gradient 1
v 0 = e
y-intercept
1
Cambridge International AS/A Level – October/November 2016 9702 51
© UCLES 2016
Question Answer Marks
1 Defining the problem
B is the independent variable and v is the dependent variable, or vary B
measure v.
1
Keep starting position of magnet constant/magnet always released from
rest.
1
Methods of data collection
Labelled diagram showing a magnet and the vertical copper tube supported. 1
Method to ensure that copper tube is vertical, e.g. set square, spirit level,
plumb line.
1
Method to determine time at bottom of tube e.g. use of light gate(s)/motion
sensor attached to timer/datalogger/computer or distance between two fixed
marks at bottom of tube and stopwatch.
Do not allow time over length of tube.
1
Method to measure B, e.g. Hall probe. 1
Method of analysis
Plot a graph of ln v against B. 1
λ = – gradient 1
v 0 = e
y-intercept
1
Page 3 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 51
© UCLES 2016
Question Answer Marks
Additional detail including safety considerations 6
1. Keep mass of magnet constant.
2. Measurement of an appropriate length to determine v at bottom of tube,
e.g. use ruler to measure distance between light gates/length of
magnet/between two fixed marks.
3. v = d / t for appropriate lengths (not length of tube)
4. Adjust Hall probe until maximum reading obtained/perpendicular to
field/pole
or
Use Hall probe to take readings for both poles and average.
5. Method to calibrate Hall probe using a known field.
6. Safety precaution linked to falling magnets/use sand tray/cushion to
soften fall.
7. Repeat experiment with magnets reversed and average
or
Repeat v (or t) for same B and average.
8. ln v = –
λB + ln v 0
9. Relationship is valid if the graph is a straight line.
10. Method to vary B, e.g. re-magnetise in a coil.
Cambridge International AS/A Level – October/November 2016 9702 51
© UCLES 2016
Question Answer Marks
Additional detail including safety considerations 6
1. Keep mass of magnet constant.
2. Measurement of an appropriate length to determine v at bottom of tube,
e.g. use ruler to measure distance between light gates/length of
magnet/between two fixed marks.
3. v = d / t for appropriate lengths (not length of tube)
4. Adjust Hall probe until maximum reading obtained/perpendicular to
field/pole
or
Use Hall probe to take readings for both poles and average.
5. Method to calibrate Hall probe using a known field.
6. Safety precaution linked to falling magnets/use sand tray/cushion to
soften fall.
7. Repeat experiment with magnets reversed and average
or
Repeat v (or t) for same B and average.
8. ln v = –
λB + ln v 0
9. Relationship is valid if the graph is a straight line.
10. Method to vary B, e.g. re-magnetise in a coil.
Page 4 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 51
© UCLES 2016
Question Answer Marks
2 (a)
gradient =
1
YE
y
-intercept =
1
E
1
(b)
2.7 or 2.70 0.833 or 0.8333
1.4 or 1.35 0.513 or 0.5128
0.90 or 0.900 0.426 or 0.4255
0.68 or 0.675 0.364 or 0.3636
0.54 or 0.540 0.345 or 0.3448
0.45 or 0.450 0.328 or 0.3279
All first column correct. Allow a mixture of significant figures.
All second column correct. Allow a mixture of significant figures.
Uncertainties in
X from ± 0.4 to ± 0.07 (± 0.1). Allow more than one
significant figure.
1
1
1
(c) (i) Six points plotted correctly.
Must be within half a small square. No “blobs”. 1
All error bars in X plotted correctly.
All error bars to be plotted. Length of bar must be accurate to less than half
a small square and symmetrical. 1
(ii) Line of best fit drawn.
Line must not be drawn from top point to bottom point.
The lower end of line should pass between (0.95, 0.45) and (1.1, 0.45)
and
upper end of line should pass between (2.10, 0.70) and (2.25, 0.70).
1
Worst acceptable line drawn correctly.
Steepest or shallowest possible line that passes through all the error bars.
Mark scored only if all error bars are plotted.
1
(iii) Gradient determined with a triangle that is at least half the length of the
drawn line.
Read-offs must be accurate to half a small square. 1
Method of determining absolute uncertainty.
uncertainty = gradient of line of best fit – gradient of worst acceptable line
or
uncertainty = ½(steepest worst line gradient – shallowest worst line
gradient)
1
(iv) y -intercept determined correctly by substitution into y = mx + c.
Read-offs must be accurate to half a small square. 1
Cambridge International AS/A Level – October/November 2016 9702 51
© UCLES 2016
Question Answer Marks
2 (a)
gradient =
1
YE
y
-intercept =
1
E
1
(b)
2.7 or 2.70 0.833 or 0.8333
1.4 or 1.35 0.513 or 0.5128
0.90 or 0.900 0.426 or 0.4255
0.68 or 0.675 0.364 or 0.3636
0.54 or 0.540 0.345 or 0.3448
0.45 or 0.450 0.328 or 0.3279
All first column correct. Allow a mixture of significant figures.
All second column correct. Allow a mixture of significant figures.
Uncertainties in
X from ± 0.4 to ± 0.07 (± 0.1). Allow more than one
significant figure.
1
1
1
(c) (i) Six points plotted correctly.
Must be within half a small square. No “blobs”. 1
All error bars in X plotted correctly.
All error bars to be plotted. Length of bar must be accurate to less than half
a small square and symmetrical. 1
(ii) Line of best fit drawn.
Line must not be drawn from top point to bottom point.
The lower end of line should pass between (0.95, 0.45) and (1.1, 0.45)
and
upper end of line should pass between (2.10, 0.70) and (2.25, 0.70).
1
Worst acceptable line drawn correctly.
Steepest or shallowest possible line that passes through all the error bars.
Mark scored only if all error bars are plotted.
1
(iii) Gradient determined with a triangle that is at least half the length of the
drawn line.
Read-offs must be accurate to half a small square. 1
Method of determining absolute uncertainty.
uncertainty = gradient of line of best fit – gradient of worst acceptable line
or
uncertainty = ½(steepest worst line gradient – shallowest worst line
gradient)
1
(iv) y -intercept determined correctly by substitution into y = mx + c.
Read-offs must be accurate to half a small square. 1
Page 5 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 51
© UCLES 2016
Question Answer Marks
Method of determining absolute uncertainty.
uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable
line
or
uncertainty = ½(steepest worst line y-intercept – shallowest worst line
y-intercept)
No ECF from false origin method.
1
(d) (i) E = 1/y-intercept and given to 2 or 3 s.f. 1
gradient
1
×
=
E
Y
or
gradient
intercept−y
Y in the range (0.90 to 1.20) × 10
–3
F.
Appropriate unit required.
Correct substitution of numbers must be seen.
1
(ii) Percentage uncertainty in Y
∆∆
=+×
100
mc
mc
or
∆∆
=+×
100
mE
mE
or
∆
=× 100
Y
Y
Maximum/minimum methods:
gradient min
interceptmax
gradient min min
1
max
−
=
×
=
y
E
Y
gradientmax
intercept min
gradientmax max
1
min
−
=
×
=
y
E
Y
1
Cambridge International AS/A Level – October/November 2016 9702 51
© UCLES 2016
Question Answer Marks
Method of determining absolute uncertainty.
uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable
line
or
uncertainty = ½(steepest worst line y-intercept – shallowest worst line
y-intercept)
No ECF from false origin method.
1
(d) (i) E = 1/y-intercept and given to 2 or 3 s.f. 1
gradient
1
×
=
E
Y
or
gradient
intercept−y
Y in the range (0.90 to 1.20) × 10
–3
F.
Appropriate unit required.
Correct substitution of numbers must be seen.
1
(ii) Percentage uncertainty in Y
∆∆
=+×
100
mc
mc
or
∆∆
=+×
100
mE
mE
or
∆
=× 100
Y
Y
Maximum/minimum methods:
gradient min
interceptmax
gradient min min
1
max
−
=
×
=
y
E
Y
gradientmax
intercept min
gradientmax max
1
min
−
=
×
=
y
E
Y
1
® IGCSE is the registered trademark of Cambridge International Examinations.
This document consists of 5 printed pages.
© UCLES 2016 [Turn ove
r
Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/52
Paper 5 Planning, Analysis and Evaluation October/November 2016
MARK SCHEME
Maximum Mark: 30
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 October/November 2016 series for most
Cambridge IGCSE
®
, Cambridge International A and AS Level components and some Cambridge O Level
components.
This document consists of 5 printed pages.
© UCLES 2016 [Turn ove
r
Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/52
Paper 5 Planning, Analysis and Evaluation October/November 2016
MARK SCHEME
Maximum Mark: 30
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 October/November 2016 series for most
Cambridge IGCSE
®
, Cambridge International A and AS Level components and some Cambridge O Level
components.
Page 2 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 52
© UCLES 2016
Question Answer Marks
1 Defining the problem
p is the independent variable and B is the dependent variable, or vary p and
measure B.
1
Keep the current/I (in the electromagnet) constant. 1
Methods of data collection
Labelled diagram showing Hall probe correctly positioned (along p) and
ruler correctly positioned and either Hall probe or rule supported.
1
Correct circuit diagram to include d.c. power supply in series with coil and
ammeter. Must be a workable circuit diagram to measure current through
the coil.
1
Measure p with ruler. 1
Method to determine an accurate value of p.
Examples include:
Height of P above bench – height of electromagnet
Height of P measured from ruler across the top of the electromagnet
1
Method of analysis
Plot a graph of ln B against p. 1
α = – gradient 1
IN
k
yintercept
e
−
=
1
Cambridge International AS/A Level – October/November 2016 9702 52
© UCLES 2016
Question Answer Marks
1 Defining the problem
p is the independent variable and B is the dependent variable, or vary p and
measure B.
1
Keep the current/I (in the electromagnet) constant. 1
Methods of data collection
Labelled diagram showing Hall probe correctly positioned (along p) and
ruler correctly positioned and either Hall probe or rule supported.
1
Correct circuit diagram to include d.c. power supply in series with coil and
ammeter. Must be a workable circuit diagram to measure current through
the coil.
1
Measure p with ruler. 1
Method to determine an accurate value of p.
Examples include:
Height of P above bench – height of electromagnet
Height of P measured from ruler across the top of the electromagnet
1
Method of analysis
Plot a graph of ln B against p. 1
α = – gradient 1
IN
k
yintercept
e
−
=
1
Page 3 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 52
© UCLES 2016
Question Answer Marks
Additional detail including safety considerations 6
1. Keep the number of turns/ N constant.
2. Use large number of turns/current (to increase B).
3. Avoid overheating the coil/do not touch hot coil.
4. Use of variable resistor to keep ammeter reading constant.
5. Method to ensure that Hall probe is equidistant from the poles, e.g.
determine centre of electromagnet and use of plumb line/ruler and spirit
level/set square.
6. Adjust Hall probe until maximum reading obtained/perpendicular to field.
7. Repeat each experiment for the same value of p and reverse the
current/Hall probe and average
8. ln B = – αp + ln kNI
9. Relationship is valid if the graph is a straight line.
10. Calibrate Hall probe using a known field.
Cambridge International AS/A Level – October/November 2016 9702 52
© UCLES 2016
Question Answer Marks
Additional detail including safety considerations 6
1. Keep the number of turns/ N constant.
2. Use large number of turns/current (to increase B).
3. Avoid overheating the coil/do not touch hot coil.
4. Use of variable resistor to keep ammeter reading constant.
5. Method to ensure that Hall probe is equidistant from the poles, e.g.
determine centre of electromagnet and use of plumb line/ruler and spirit
level/set square.
6. Adjust Hall probe until maximum reading obtained/perpendicular to field.
7. Repeat each experiment for the same value of p and reverse the
current/Hall probe and average
8. ln B = – αp + ln kNI
9. Relationship is valid if the graph is a straight line.
10. Calibrate Hall probe using a known field.
Page 4 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 52
© UCLES 2016
Question Answer Marks
2 (a) gradient = q
y-intercept = lg p
1
(b)
2.80 or 2.799 or 2.7993 0.28 or 0.279
2.79 or 2.792 or 2.7924 0.30 or 0.301
2.77 or 2.771 or 2.7709 0.36 or 0.362
2.72 or 2.716 or 2.7160 0.49 or 0.491
2.69 or 2.690 or 2.6902 0.57 or 0.568
2.67 or 2.672 or 2.6721 0.61 or 0.613
All first column correct – either 2 and 3 decimal places or 3 and 4 decimal
places.
All second column correct. Allow a mixture of decimal places.
Uncertainties in lg (V
/ V) from ± 0.02 to ± 0.01. Allow more than one
significant figure.
1
1
1
(c) (i) Six points plotted correctly.
Must be within half a small square. No “blobs”.
1
All error bars in lg (V / V) plotted correctly.
All error bars to be plotted. Total length of bar must be accurate to less than
half a small square and symmetrical.
1
(ii) Line of best fit drawn.
Line must not be drawn from top point to bottom point unless points are
balanced.
Upper end of line should pass between (2.694, 0.55) and (2.700, 0.55) and
lower end of line should pass between (2.770, 0.35) and (2.776, 0.35).
1
Worst acceptable line drawn correctly.
Steepest or shallowest possible line that passes through all the error bars.
Mark scored only if all error bars are plotted.
1
(iii) Gradient determined with a triangle that is at least half the length of the drawn
line.
Read-offs must be accurate to half a small square.
Gradient must be negative.
1
Method of determining absolute uncertainty.
uncertainty = gradient of line of best fit – gradient of worst acceptable line
or
uncertainty = ½(steepest worst line gradient – shallowest worst line gradient)
1
Cambridge International AS/A Level – October/November 2016 9702 52
© UCLES 2016
Question Answer Marks
2 (a) gradient = q
y-intercept = lg p
1
(b)
2.80 or 2.799 or 2.7993 0.28 or 0.279
2.79 or 2.792 or 2.7924 0.30 or 0.301
2.77 or 2.771 or 2.7709 0.36 or 0.362
2.72 or 2.716 or 2.7160 0.49 or 0.491
2.69 or 2.690 or 2.6902 0.57 or 0.568
2.67 or 2.672 or 2.6721 0.61 or 0.613
All first column correct – either 2 and 3 decimal places or 3 and 4 decimal
places.
All second column correct. Allow a mixture of decimal places.
Uncertainties in lg (V
/ V) from ± 0.02 to ± 0.01. Allow more than one
significant figure.
1
1
1
(c) (i) Six points plotted correctly.
Must be within half a small square. No “blobs”.
1
All error bars in lg (V / V) plotted correctly.
All error bars to be plotted. Total length of bar must be accurate to less than
half a small square and symmetrical.
1
(ii) Line of best fit drawn.
Line must not be drawn from top point to bottom point unless points are
balanced.
Upper end of line should pass between (2.694, 0.55) and (2.700, 0.55) and
lower end of line should pass between (2.770, 0.35) and (2.776, 0.35).
1
Worst acceptable line drawn correctly.
Steepest or shallowest possible line that passes through all the error bars.
Mark scored only if all error bars are plotted.
1
(iii) Gradient determined with a triangle that is at least half the length of the drawn
line.
Read-offs must be accurate to half a small square.
Gradient must be negative.
1
Method of determining absolute uncertainty.
uncertainty = gradient of line of best fit – gradient of worst acceptable line
or
uncertainty = ½(steepest worst line gradient – shallowest worst line gradient)
1
Page 5 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 52
© UCLES 2016
Question Answer Marks
(iv) y-intercept determined by substitution into y = mx + c.
Read-offs must be accurate to half a small square.
1
Method of determining absolute uncertainty.
uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable
line
or
uncertainty = ½(steepest worst line y-intercept – shallowest worst line
y-intercept)
No ECF from false origin method.
1
(d) Use of p = 10
answer to 2(c)(iv)
or
lg p = answer to 2(c)(iv)
1
q = gradient and in the range –2.50 to –2.70 and given to 2 or 3 s.f. 1
(e) Use of V = p × 950
q
or
lg V = q lg 950 + lg p
or
lg V = q lg 950 + y-intercept
Correct substitution of numbers must be seen to give V.
1
Cambridge International AS/A Level – October/November 2016 9702 52
© UCLES 2016
Question Answer Marks
(iv) y-intercept determined by substitution into y = mx + c.
Read-offs must be accurate to half a small square.
1
Method of determining absolute uncertainty.
uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable
line
or
uncertainty = ½(steepest worst line y-intercept – shallowest worst line
y-intercept)
No ECF from false origin method.
1
(d) Use of p = 10
answer to 2(c)(iv)
or
lg p = answer to 2(c)(iv)
1
q = gradient and in the range –2.50 to –2.70 and given to 2 or 3 s.f. 1
(e) Use of V = p × 950
q
or
lg V = q lg 950 + lg p
or
lg V = q lg 950 + y-intercept
Correct substitution of numbers must be seen to give V.
1
® IGCSE is the registered trademark of Cambridge International Examinations.
This document consists of 5 printed pages.
© UCLES 2016 [Turn ove
r
Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/53
Paper 5 Planning, Analysis and Evaluation October/November 2016
MARK SCHEME
Maximum Mark: 30
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 October/November 2016 series for most
Cambridge IGCSE
®
, Cambridge International A and AS Level components and some Cambridge O Level
components.
This document consists of 5 printed pages.
© UCLES 2016 [Turn ove
r
Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/53
Paper 5 Planning, Analysis and Evaluation October/November 2016
MARK SCHEME
Maximum Mark: 30
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 October/November 2016 series for most
Cambridge IGCSE
®
, Cambridge International A and AS Level components and some Cambridge O Level
components.
Page 2 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 53
© UCLES 2016
Question Answer Marks
1 Defining the problem
B is the independent variable and v is the dependent variable, or vary B
measure v.
1
Keep starting position of magnet constant/magnet always released from
rest.
1
Methods of data collection
Labelled diagram showing a magnet and the vertical copper tube supported. 1
Method to ensure that copper tube is vertical, e.g. set square, spirit level,
plumb line.
1
Method to determine time at bottom of tube e.g. use of light gate(s)/motion
sensor attached to timer/datalogger/computer or distance between two fixed
marks at bottom of tube and stopwatch.
Do not allow time over length of tube.
1
Method to measure B, e.g. Hall probe. 1
Method of analysis
Plot a graph of ln v against B. 1
λ = – gradient 1
v 0 = e
y-intercept
1
Cambridge International AS/A Level – October/November 2016 9702 53
© UCLES 2016
Question Answer Marks
1 Defining the problem
B is the independent variable and v is the dependent variable, or vary B
measure v.
1
Keep starting position of magnet constant/magnet always released from
rest.
1
Methods of data collection
Labelled diagram showing a magnet and the vertical copper tube supported. 1
Method to ensure that copper tube is vertical, e.g. set square, spirit level,
plumb line.
1
Method to determine time at bottom of tube e.g. use of light gate(s)/motion
sensor attached to timer/datalogger/computer or distance between two fixed
marks at bottom of tube and stopwatch.
Do not allow time over length of tube.
1
Method to measure B, e.g. Hall probe. 1
Method of analysis
Plot a graph of ln v against B. 1
λ = – gradient 1
v 0 = e
y-intercept
1
Page 3 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 53
© UCLES 2016
Question Answer Marks
Additional detail including safety considerations 6
1. Keep mass of magnet constant.
2. Measurement of an appropriate length to determine v at bottom of tube,
e.g. use ruler to measure distance between light gates/length of
magnet/between two fixed marks.
3. v = d / t for appropriate lengths (not length of tube)
4. Adjust Hall probe until maximum reading obtained/perpendicular to
field/pole
or
Use Hall probe to take readings for both poles and average.
5. Method to calibrate Hall probe using a known field.
6. Safety precaution linked to falling magnets/use sand tray/cushion to
soften fall.
7. Repeat experiment with magnets reversed and average
or
Repeat v (or t) for same B and average.
8. ln v = –
λB + ln v 0
9. Relationship is valid if the graph is a straight line.
10. Method to vary B, e.g. re-magnetise in a coil.
Cambridge International AS/A Level – October/November 2016 9702 53
© UCLES 2016
Question Answer Marks
Additional detail including safety considerations 6
1. Keep mass of magnet constant.
2. Measurement of an appropriate length to determine v at bottom of tube,
e.g. use ruler to measure distance between light gates/length of
magnet/between two fixed marks.
3. v = d / t for appropriate lengths (not length of tube)
4. Adjust Hall probe until maximum reading obtained/perpendicular to
field/pole
or
Use Hall probe to take readings for both poles and average.
5. Method to calibrate Hall probe using a known field.
6. Safety precaution linked to falling magnets/use sand tray/cushion to
soften fall.
7. Repeat experiment with magnets reversed and average
or
Repeat v (or t) for same B and average.
8. ln v = –
λB + ln v 0
9. Relationship is valid if the graph is a straight line.
10. Method to vary B, e.g. re-magnetise in a coil.
Page 4 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 53
© UCLES 2016
Question Answer Marks
2 (a)
gradient =
1
YE
y-intercept =
1
E
1
(b)
2.7 or 2.70 0.833 or 0.8333
1.4 or 1.35 0.513 or 0.5128
0.90 or 0.900 0.426 or 0.4255
0.68 or 0.675 0.364 or 0.3636
0.54 or 0.540 0.345 or 0.3448
0.45 or 0.450 0.328 or 0.3279
All first column correct. Allow a mixture of significant figures.
All second column correct. Allow a mixture of significant figures.
Uncertainties in X from ± 0.4 to ± 0.07 (± 0.1). Allow more than one
significant figure.
1
1
1
(c) (i) Six points plotted correctly.
Must be within half a small square. No “blobs”.
1
All error bars in X plotted correctly.
All error bars to be plotted. Length of bar must be accurate to less than half
a small square and symmetrical.
1
(ii) Line of best fit drawn.
Line must not be drawn from top point to bottom point.
The lower end of line should pass between (0.95, 0.45) and (1.1, 0.45) and
upper end of line should pass between (2.10, 0.70) and (2.25, 0.70).
1
Worst acceptable line drawn correctly.
Steepest or shallowest possible line that passes through all the error bars.
Mark scored only if all error bars are plotted.
1
(iii) Gradient determined with a triangle that is at least half the length of the
drawn line.
Read-offs must be accurate to half a small square.
1
Method of determining absolute uncertainty.
uncertainty = gradient of line of best fit – gradient of worst acceptable line
or
uncertainty = ½(steepest worst line gradient – shallowest worst line
gradient)
1
(iv) y-intercept determined correctly by substitution into y = mx + c.
Read-offs must be accurate to half a small square.
1
Cambridge International AS/A Level – October/November 2016 9702 53
© UCLES 2016
Question Answer Marks
2 (a)
gradient =
1
YE
y-intercept =
1
E
1
(b)
2.7 or 2.70 0.833 or 0.8333
1.4 or 1.35 0.513 or 0.5128
0.90 or 0.900 0.426 or 0.4255
0.68 or 0.675 0.364 or 0.3636
0.54 or 0.540 0.345 or 0.3448
0.45 or 0.450 0.328 or 0.3279
All first column correct. Allow a mixture of significant figures.
All second column correct. Allow a mixture of significant figures.
Uncertainties in X from ± 0.4 to ± 0.07 (± 0.1). Allow more than one
significant figure.
1
1
1
(c) (i) Six points plotted correctly.
Must be within half a small square. No “blobs”.
1
All error bars in X plotted correctly.
All error bars to be plotted. Length of bar must be accurate to less than half
a small square and symmetrical.
1
(ii) Line of best fit drawn.
Line must not be drawn from top point to bottom point.
The lower end of line should pass between (0.95, 0.45) and (1.1, 0.45) and
upper end of line should pass between (2.10, 0.70) and (2.25, 0.70).
1
Worst acceptable line drawn correctly.
Steepest or shallowest possible line that passes through all the error bars.
Mark scored only if all error bars are plotted.
1
(iii) Gradient determined with a triangle that is at least half the length of the
drawn line.
Read-offs must be accurate to half a small square.
1
Method of determining absolute uncertainty.
uncertainty = gradient of line of best fit – gradient of worst acceptable line
or
uncertainty = ½(steepest worst line gradient – shallowest worst line
gradient)
1
(iv) y-intercept determined correctly by substitution into y = mx + c.
Read-offs must be accurate to half a small square.
1
Page 5 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2016 9702 53
© UCLES 2016
Question Answer Marks
Method of determining absolute uncertainty.
uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable
line
or
uncertainty = ½(steepest worst line y-intercept – shallowest worst line
y-intercept)
No ECF from false origin method.
1
(d) (i) E = 1/y-intercept and given to 2 or 3 s.f. 1
gradient
1
×
=
E
Y
or
gradient
intercept−y
Y in the range (0.90 to 1.20) × 10
–3
F.
Appropriate unit required.
Correct substitution of numbers must be seen.
1
(ii) Percentage uncertainty in Y
∆∆
=+×
100
mc
mc
or
∆∆
=+×
100
mE
mE
or
∆
=× 100
Y
Y
Maximum/minimum methods:
gradient min
interceptmax
gradient min min
1
max
−
=
×
=y
E
Y
gradientmax
intercept min
gradientmax max
1
min
−
=
×
=y
E
Y
1
Cambridge International AS/A Level – October/November 2016 9702 53
© UCLES 2016
Question Answer Marks
Method of determining absolute uncertainty.
uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable
line
or
uncertainty = ½(steepest worst line y-intercept – shallowest worst line
y-intercept)
No ECF from false origin method.
1
(d) (i) E = 1/y-intercept and given to 2 or 3 s.f. 1
gradient
1
×
=
E
Y
or
gradient
intercept−y
Y in the range (0.90 to 1.20) × 10
–3
F.
Appropriate unit required.
Correct substitution of numbers must be seen.
1
(ii) Percentage uncertainty in Y
∆∆
=+×
100
mc
mc
or
∆∆
=+×
100
mE
mE
or
∆
=× 100
Y
Y
Maximum/minimum methods:
gradient min
interceptmax
gradient min min
1
max
−
=
×
=y
E
Y
gradientmax
intercept min
gradientmax max
1
min
−
=
×
=y
E
Y
1
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