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Physic A Level 9702 w15 Markscheme 51

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® IGCSE is the registered trademark of Cambridge International Examinations.
CAMBRIDGE INTERNATIONAL EXAMINATIONS
Cambridge International Advanced Subsidiary and Advanced Level
MARK SCHEME for the October/November 2015 series
9702 PHYSICS
9702/51
Paper 5 (Planning, Analysis and Evaluation), maximum raw mark 30
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 2015 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 2015
9702 51
© Cambridge International Examinations 2015
1 Planning (15 marks) Defining the problem (3 marks)
P
m
is the independent variable, or vary
m
.
[1] P (tan)
φ
is the dependent variable, or measure (tan)
φ
. [1] P Keep the temperature of the oil constant. [1]
Methods of data collection (5 marks)
M Labelled diagram showing labelled protractor positioned to determine
φ
for tilted cylinder. Allow distances marked to determine
φ
and use of a rule.
[1] M Use of balance/scales to measure the mass of the oil/cylinder. [1] M Mass of oil = mass of (oil + cylinder) – mass of cylinder. [1] M Use of vernier calipers/micrometer/rule to measure
d
. [1] M Repeat each experiment for the same value of
m
and average
φ
. [1]
Method of analysis (2 marks)
A Plot a graph of
φ
tan1
against
m
. (Allow
3
d m
or
3
d m
ρ
or
ρ
m
. Do not allow log-log graphs.)
[1] A
a
= gradient
×
ρ
d
3
and
b
=
y
-intercept; must be consistent with suggested graph. [1]
Safety considerations (1 mark)
S Precaution linked to preventing spilling oil, e.g. use a tray/lid/cloth to absorb oil (do not allow just wiping or mopping)
or
precaution linked to preventing glass cylinder breaking, e.g. padding/cushion
or
use of gloves to prevent skin irritation (do not allow “because oil is slippery”). [1]
Additional detail (4 marks)
D Relevant points might include [4] 1 Repeat measurements of
d
in different directions and average 2 Use of video with slow motion/frame by frame playback to determine
φ
3 Use large protractor to reduce percentage uncertainty
or
trigonometry relationship related to measurements to be taken 4 Use the same (diameter) cylinder (not “same size” but allow “same size and shape”) 5 Slowly/gently/gradually tilt cylinder of oil/use of rough surface (to prevent sliding) 6 Experimental method to determine density of oil and
ρ
=
m
/
V
7 Relationship is valid if the graph is a straight line that does NOT pass through the srcin
/
has an intercept; must be consistent with suggested graph Do not allow vague computer methods.
Page 3 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2015
9702 51
© Cambridge International Examinations 2015
2 Analysis, conclusions and evaluation (15 marks)
Mark Expected Answer Additional Guidance
(a)
A1 gradient =
2
Ed
π
ρ
4
y
-intercept =
E r
(b)
T1
I
1
/
A
–1
Allow
I
1
(A
–1
) or
A
11
I
. T2
4.2 or 4.17 5.0 or 5.00 5.9 or 5.88 6.7 or 6.67 7.7 or 7.69 8.3 or 8.33 Allow a mixture of significant figures. Must be table values. U1
±
0.2 to
±
0.6 or
±
0.7 or
±
0.8 Allow more than one significant figure.
(c) (i)
G1 Six points plotted correctly Must be within half a small square. Do not allow “blobs”. ECF allowed from table. U2 Error bars in 1
/
I
plotted correctly All error bars to be plotted. Must be accurate to less than half a small square. Length of bar must be accurate to less than half a small square. Do not allow less than 0.05.
(ii)
G2 Line of best fit If points are plotted correctly then lower end of line should pass between (41, 4.5) and (44, 4.5)
and
upper end of line should pass between (83, 8.0) and (88, 8.0). Line should not go from bottom to top points. G3 Worst acceptable straight line.Steepest or shallowest possible line that passes through all the error bars. Line should be clearly labelled or dashed. Examiner judgement on worst acceptable line. Lines must cross. Mark scored only if error bars are plotted.
(iii)
C1 Gradient of line of best fit The triangle used should be at least half the length of the drawn line. Check the read-offs. Work to half a small square. Do not penalise POT. (Should be about 8.) U3 Absolute uncertainty in gradient Method of determining absolute uncertainty: difference in worst gradient and gradient.
(iv)
C2
y
-intercept Check substitution into
y
=
mx
+
c
. Allow ECF from
(c)(iii)
. (Should be about 0.7
–
1.5.)
Page 4 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2015
9702 51
© Cambridge International Examinations 2015
U4 Absolute uncertainty in
y
-intercept Uses worst gradient and point on WAL. Do not check calculation.
(d) (i)
C3
ρ
= 2.415
×
10
–7
×
gradient Must be in the range 1.80
×
10
–6
to 2.10
×
10
–6
and given to 2 or 3 s.f.
Must use gradient.
ρ
=
4
2
Ed
π
×
gradient [2
×
10
–6
Ω
m = 2
×
10
–4
Ω
cm = 2
×
10
–3
Ω
mm] C4
r
=
E
×
y
-intercept = 3.2
×
y
-intercept and
Ω
m and
Ω
given Must include units for
ρ
and
r
. Allow V
A
–1
or kg
m
2
A
–2
s
–3
for
Ω
.
(ii)
U5 Percentage uncertainty in
ρ
Must be greater than 9.6%.
Uncertainties in Question 2 (c) (iii)
Gradient [U3]
uncertainty = gradient of line of best fit – gradient of worst acceptable line uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)
(iv)
[U4]
uncertainty =
y
-intercept of line of best fit –
y
-intercept of worst acceptable line uncertainty = ½ (steepest worst line
y
-intercept – shallowest worst line
y
-intercept)
(d) (ii)
[U5]
percentage uncertainty =
×++∆
0.310.0123.20.1
mm
×
100 =
×∆
100
mm
+ 3.125 + 2
×
3.226
max.
p
=
( )
4100.323.3
23
−
×××π
×
max. gradient min.
p
=
( )
4100.303.1
23
−
×××π
×
min. gradient

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