UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS
GCE Advanced Subsidiary Level and GCE Advanced Level
MARK SCHEME for the May/June 2011 question paper for the guidance of teachers
9702 PHYSICS
 
9702/23
Paper 2 (AS Structured Questions), maximum raw mark 60
This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes must be read in conjunction with the question papers and the report on the examination.
 Cambridge will not enter into discussions or correspondence in connection with these mark schemes. Cambridge is publishing the mark schemes for the May/June 2011 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.
 
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Page 2 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL May/June 2011 9702 23
© University of Cambridge International Examinations 2011
1 (a)
 2nd row random, 3rd row neither, 4th row systematic all correct B2 two correct scores 1 only [2]
(b) (i) 1.
 systematic error: the average / peak is not the true value / the readings are not centred around the true value B1 [1]
2.
 random error: readings have positive and negative values around the peak value / values are scattered / wide range B1 [1]
(ii)
 
1.
 accurate: peak / average value moves towards the true value B1 [1]
2.
 precise: lines are closer together / sharper peak B1 [1]
2 (a)
resultant moment = zero / sum of clockwise moments = sum of anticlockwise moments B1 resultant force = 0 B1 [2]
(b)
 shape and orientation correct and forces labelled and arrows correct M1 angles correct / labelled A1 [2]
(c)
 
(i)
 
 cos18° =
 Scale diagram: C1
 = 520 / cos18° = 547 N ± 20 N A1 [2]
(ii)
 
 =
 sin18° = 169 N ± 20 N A1 [1]
(d)
 
θ 
 is larger hence cos
 
θ 
 is smaller,
 =
 / cos
 
θ 
 M1 hence
 is larger A0 [1]
3 (a)
 weight =
m
 ×
 = 130.5 × 9.81 = 1280 N A1 [1]
(b) (i)
 
 =
ma
 
 1280 = 130.5 × 0.57 C1
 = 1280 + 74.4 = 1350 N A1 [2]
(ii)
 1280 N A1 [1]
(c)
 1240 – 1280 = 130.5 ×
a
 C1
a
 = (–) 0.31
 
m
 
s
–2
 A1 [2]
(d) (i)
 
1.
 3.5 s A1 [1]
2.
 6.5 s A1 [1]
 
Page 3 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL May/June 2011 9702 23
© University of Cambridge International Examinations 2011
(ii)
 basic shape M1 correct points A1 [2]
4 (a)
 force is proportional to extension B1 [1]
(b) (i)
 gradient of graph determined (e.g. 50 / 40 ×10
–3
 ) = 1250 N
 
m
–1
 A1 [1]
(ii)
 
 = ½
k x 
2
 or
 = ½ final force × extension M1 = 0.5 × 1250 × (36 × 10
–3
)
2
 or 0.5 × 45 × 36 × 10
–3
 M1 = 0.81 J A0 [2]
(c) (i)
 0.81 = ½
mv 
2
 C1
 = 8.0 (8.0498) m
 
s
–1
 A1 [2]
(ii)
 4 × KE / 4 × WD or 3.24 J C1 hence twice the compression = 72 mm A1 [2]
(iii)
 Max height is when all KE or WD or elastic PE is converted to GPE C1 ratio = 1/4 or 0.25 A1 [2]
5 (a) (i)
 Start from (0,0) and smooth curve in correct direction B1 Curve correct for end section never horizontal B1 [2]
(ii)
 
 =
 /
 hence take co-ords of
 and
 from graph and calculate
 /
 B1 [1]
(b) (i)
 each lamp in parallel has a greater p.d. / greater current M1 lamp hotter M1 resistance of lamps in parallel greater A1 [3]
(ii)
 
 =
2
 /
 or
 =
 
and
 =
 C1
 = 144 / 50 = 2.88 for each lamp C1 total
 = 1.44
 A1 [3]
6 (a) (i)
 amplitude = 7.6 mm allow 7.5 mm A1 [1]
(ii)
 180° /
π
 rad A1 [1]
(iii)
 
 =
 ×
 λ
 = 15 × 0.8 C1 = 12 m
 
s
–1
 A1 [2]
(b)
 correct sketch with peak moved to the right B1 curve moved by the correct phase angle / time period of 0.25
 B1 [2]
(c) (i)
 zero (rad) A1 [1]
(ii)
 antinode maximum amplitude, node zero amplitude / displacement A1 [1]
 
Page 4 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL May/June 2011 9702 23
© University of Cambridge International Examinations 2011
(iii)
 3 A1 [1]
(iv)
 horizontal line through central section of wave B1 [1]
7 (a)
 density in solids and liquids similar M1 spacing in solids and liquids about the same A1 density in gases much less as spacing in gases much greater B1 [3]
(b)
 density = mass / volume C1 mass = 1.67 × 10
–27
 
kg and volume = 4/3
π
 
 
3
 C1 density = (1.67 × 10
–27
) / 4/3 ×
π
 × (1.0 × 10
–15
)
3
 = 3.99 × 10
17
 kg m
–3
 A1 [3]
(c)
 atoms / molecules composed of large amount of empty space / nucleus has very small volume compared to volume of atom / space between atoms in a gas is very large B1 [1]
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