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CAMBRIDGE INTERNATIONAL EXAMINATIONS
GCE Advanced Subsidiary Level and GCE Advanced Level
MARK SCHEME for the May/June 2013 series
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 should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2013 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.
w w w . X t r e m e P a p e r s . c o m
Page 2 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9702 23
© Cambridge International Examinations 2013
1 (a)
force: kg m
s
–2
A1 [1]
(b) (i)
I
2
: A
2
l
: m
x
: m C1
K
: kg m
s
–2
A
–2
A1 [2]
(ii)
curve of the correct shape (for inverse proportionality) M1
clearly approaching each axis but never touching the axis
A1 [2]
(iii)
curving upwards and through srcin A1 [1]
2 (a) (i) 1.
distance of path / along line AB B1 [1]
2.
shortest distance between AB / distance in straight line between AB or displacement from A to B B1 [1]
(ii)
acceleration = rate of change of velocity A1 [1]
(b) (i)
distance = area under line or (
v
/2)
t
or
s
= (8.8)
2
/ (2 × 9.81) C1
= 8.8 / 2 × 0.90 = 3.96
m or
s
= 3.95
m = 4(.0)
m A1 [2]
(ii)
acceleration = (– 4.4 – 8.8) / 0.50 C1 = (–) 26(.4)
m
s
–2
A1 [2]
(c) (i)
the accelerations are constant as straight lines B1
the accelerations are the same as same gradient or no air resistance as acceleration is constant or change of speed in opposite directions (one speeds up one slows down) B1 [2]
(ii)
area under the lines represents height or KE at trampoline equals PE at maximum height B1
second area is smaller / velocity after rebound smaller hence KE less B1
hence less height means loss in potential energy A0 [2]
3 (a) (i)
the total momentum of a system (of interacting bodies) remains constant M1
provided there are no resultant external forces / isolated system A1 [2]
(ii)
elastic: total kinetic energy is conserved, inelastic: loss of kinetic energy B1 [1] [allow elastic: relative speed of approach equals relative speed of separation]
Page 3 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9702 23
© Cambridge International Examinations 2013
(b) (i)
initial mom: 4.2 × 3.6 – 1.2 × 1.5 (= 15.12 – 1.8 = 13.3) C1
final mom: 4.2 ×
v
+ 1.5 × 3 C1
v
= (13.3 – 4.5) / 4.2 = 2.1
m
s
–1
A1 [3]
(ii)
initial kinetic energy = ½
m
A
(
v
A
)
2
+ ½
m
B
(
v
B
)
2
= 27.21 + 1.08 = 28(.28) M1
final kinetic energy = 9.26 + 6.75 = 16 M1
initial KE is not the same as final KE hence inelastic A1 [3]
provided final KE less than initial KE
[allow in terms of relative speeds of approach and separation]
4 (a) (i)
stress = force / cross-sectional area B1 [1]
(ii)
strain = extension / srcinal length B1 [1]
(b) (i)
E
= stress / strain C1
E
= 0.17 × 10
12
C1
stress = 0.17 × 10
12
× 0.095 / 100 C1
= 1.6(2) × 10
8
Pa A1 [4]
(ii)
force = (stress × area) = 1.615 × 10
8
× 0.18 × 10
–6
C1
= 29(.1)
N A1 [2]
5 (a)
when waves overlap / meet B1
the resultant displacement is the sum of the individual displacements of the waves B1 [2]
(b) (i) 1.
phase difference = 180
º / (
n
+ ½) 360
º (allow in rad) B1 [1]
2.
phase difference = 0 / 360
º / (
n
360
º) (allow in rad) B1 [1]
(ii)
v
=
f
λ
C1
λ
= 320 / 400 = 0.80
m A1 [2]
(iii)
path difference = 7 – 5 = 2 (m) = 2.5
λ
M1
hence minimum or maximum if phase change at P is suggested A1 [2]
6 (a)
p.d. = work done / energy transformed (from electrical to other forms) charge B1 [1]
(b) (i)
maximum 20
V
A1 [1]
(ii)
minimum = (600 / 1000) × 20 C1
= 12
V A1 [2]
Page 4 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9702 23
© Cambridge International Examinations 2013
(c) (i)
use of 1.2
k
Ω
M1
1/1200 + 1/600 = 1/
R
,
R
= 400
Ω
A1 [2]
(ii)
total parallel resistance (R
2
+ LDR) is less than R
2
M1
(minimum) p.d. is reduced A1 [2]
7 (a) (i)
nucleus contains 92 protons B1
nucleus contains 143 neutrons (missing
‘nucleus’ 1/2)
B1
outside / around nucleus 92 electrons (B1) most of atom is empty space / mass concentrated in nucleus (B1) total charge is zero (B1) diameter of atom ~ 10
–10
m or size of nucleus ~ 10
–15
m (B1) any two of (B1)
marks [4]
(ii)
nucleus has same number / 92 protons B1
nuclei have 143 and 146 neutrons (missing
‘nucleus’ 1/2) B1 [2]
(b) (i)
Y
= 35 A1
Z
= 85 A1 [2]
(ii)
mass-energy is conserved in the reaction B1 mass on rhs of reaction is less so energy is released explained in terms of
E
=
mc
2
B1 [2]

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