This document outlines the marking scheme for a physics exam assessing structured questions. It categorizes marks into four types: B for independent marks, M for method marks, C for compensatory method marks, and A for accuracy/answer marks which depend on M or allow C marks. It provides examples of how marks in each category are awarded, conventions like brackets and underlining, and a breakdown of the mark scheme over multiple pages assessing different exam questions.

1. NOVEMBER 2001
ADVANCED SUBSIDIARY LEVEL
MARK SCHEME
MAXIMUM MARK : 60
SYLLABUS/COMPONENT : 8702/2
PHYSICS
(STRUCTURED QUESTIONS)

2. Page 1 of 3 Mark Scheme Syllabus Paper
AS Level Examinations – November 2001 8702 2
Categorisation of marks
The marking scheme categorises marks on the MACB scheme.
B marks: These are awarded as independent marks, which do not depend on other marks. For
a B mark to be scored, the point to which it refers must be seen specifically in the
candidate’s answer.
M marks: These are method marks upon which A-marks (accuracy marks) later depend. For an
M-mark to be scored, the point to which it refers must be seen in the candidate’s
answer. If a candidate fails to score a particular M-mark, then none of the dependent
A-marks can be scored.
C marks: These are compensatory method marks which can be scored even if the points to
which they refer are not written down by the candidate, providing subsequent working
gives evidence that they must have known it. For example, if an equation carries a C-
mark and the candidate does not write down the actual equation but does correct
working which shows he/she knew the equation, then the C-mark is awarded.
A marks: These are accuracy or answer marks which either depend on an M-mark, or allow a
C-mark to be scored.
Conventions within the marking scheme
Brackets Where brackets are shown in the marking scheme, the candidate is not required to
give the bracketed information in order to earn the available marks.
Underlining In the marking scheme, underlining indicates information that is essential for marks to
be awarded.
1 mass: property of body which resists change in motion
OR quantity/amount of matter/substance
constant OR scalar quantity
weight: effect of gravitational field on a mass/body
OR force due to pull of gravity
variable OR vector quantity
(if units given, allow ½ for 2nd
and 4th
marks)
B1
B1
B1
B1 [4]
2 (a)
use of h = ut +
2
1
at2
and u = 0 (accept h =
2
1
at2
)
gradient of graph =
2
1
g
gradient = 0.7 / (0.225 – 0.082)
=4.9 ± 01
g = 9.8 m s-2
(single point solution or gradient = g, max 2/4)
B1
B1
C1
A1 [4]
(b) e.g. measurement of h OR t
repeat measurement and average
M1
A1 [2]
3 (a) sum of forces in any direction is zero
(allow vector/algebraic sum, resultant force…)
sum of moments about any point is zero
(allow algebraic sum, resultant…)
(If no mention of direction or point, allow max ½ overall)
B1
B1 [2]

3. Page 2 of 3 Mark Scheme Syllabus Paper
AS Level Examinations – November 2001 8702 2
3 (b) same force F in both springs
resultant is 2F cos60 OR scale drawing
resultant is F, so extension is x
(maximum 3
1 if uses x alone, without reference to kx)
C1
M1
A1 [3]
3 (c) (i) correct direction for W B1
(ii) correct direction for T B1
(iii) correct direction for R B1 [3]
4 (a) m = r V B1 [1]
(b) pressure in liquid depends on depth
bottom of sphere has greater pressure on it than top
so resultant force or pressure is upwards
B1
M1
A1 [3]
(c) (i)
1. kinetic energy =
2
1
mv2
=
2
1
x 2 x 10-3
x (6 x 10-2
)2
= 3.6 x 10-6
J
2. potential energy = mgh
rate of loss = mgv
= 2 x 10-3
x 9.8 x 6 x 10-2
= 1.2 x 10-3
J s-1
(no reference to time in 2., maximum 3
2 )
C1
A1
C1
C1
A1 [5]
(ii) potential energy of sphere is given/lost to the liquid
to overcome drag forces or to produce eddies or friction etc
B1
B1 [2]
5 (a) copper and aluminium (-1 for each error or omission) B2 [2]
(b) (i) A is brittle
B is ductile
C is polymeric (-1 each error or omission)
(named materials, all correct, allow 2
1 )
B2 [2]
(ii) work done is represented by area under graph
(
2
1
x 2 x 10-3
x 3) + (2 x 10-3
x 3.1)
work done = 9.2 x 10-3
J (1 s.f. –1)
C1
C1
A1 [3]
6 (a) (i) c = f λ
λ = (3.00 x 108
) / (4.8 x 1014
)
= 625 x 10-9
m
number of wavelengths = (0.1 x 10-3
) / (625 x 10-9
)
= 160
C1
C1
M1
A0 [3]
(ii) pattern seen due to diffraction (at each slit)
for large amount of diffraction, wavelength is about slit width
not so here, so very little diffraction
M1
A1
A1 [3]
(b) λ = ax / D
x = (625 x 10-9
x 2.6) / 1.5 x 10-3
= 1.1 mm
C1
C1
A1 [3]

4. Page 3 of 3 Mark Scheme Syllabus Paper
AS Level Examinations – November 2001 8702 2
(c) fringe separation is unchanged
bright fringes are brighter
dark fringes stay dark
(allow
2
1 for fringes brighter/more contrast but
2
0 for more
intense)
B1
B1
B1 [3]
7 (a) (i) P = V2
/R OR P = VI and V = IR
algebra clear
leading to R = 2.4 W
B1
B1
A0 [2]
(ii) R = r L / A
2.4 = (4.9 x 10-7
x L) / p x (0.27 x 10-3
)2
L = 1.12 m
C1
C1
A1 [3]
(b) (i) I1 + I2 = I3 B1 [1]
(ii) 1. E1 = I1R1 + I3R2 B1
2. E1 – E2 = I1R1 – I2R3 B1 [2]
8 (a) 118 B1 [1]
(b) path: correct shape
in correct position relative to nucleus
M1
A1 [2]
(c) smaller deviation B1 [1]

5. NOVEMBER 2001
ADVANCED SUBSIDIARY LEVEL
MARK SCHEME
MAXIMUM MARK : 25
SYLLABUS/COMPONENT : 8702/3
PHYSICS
(PRACTICAL)

6. Page 1 of 2 Mark Scheme Syllabus Paper
AS Level Examinations – November 2001 8702 3
Measurements
M1 Measurements
One mark for each set of readings
Intervals must be correct or –1
Check values for
a
asin
Values should be 0.0111; 0.0123; 0.0134; 0.0144; 0.0153; 0.0161
6
M2 Position of O
Measure diameter and divide by 2
Allow folding of the card ideas
1
M3 Radius value with unit
Accept 14.5 cm ± 0.2 cm
1
M4 % uncertainty in r
Accept 0.68%, 0.7%, 1%
Working must be correct
N/A 0.34% or 1.37%
1
M5 Value of y
Accept 6.3 cm ± 0.2 cm
1
M6 Quality of results
Judge by scatter of points about the line of best fit
A shallow curve gets 2
1
5 trend plots gets 2
1
2
Presentation of results
R1 Column headings
Every column heading must have a quantity
Expect to see y/cm, but ignore degrees if missing
N/A sina /degree
1
R2 Consistency of raw values of y only
Values must be given to the nearest millimetre
1
R3 SF in final value of r
Allow 2 or 3 sf only
1
Graphical work
G1 Axes
The plotted points must occupy at least half the graph grid in both the x and y
directions. The axes must be labelled.
Do not allow awkward scales.
1
G2 Plotting of points
Check one suspect plot. Work to half a small square.
1
G3 Line of best fit
At least 5 trend plots needed.
Allow a straight line to be drawn through a shallow curved trend.
1

7. Page 2 of 2 Mark Scheme Syllabus Paper
AS Level Examinations – November 2001 8702 3
G4 Determination of gradient
Check the read-offs and that ∆y/∆x has been calculated.
The length of the hypotenuse must be greater than half the length of the drawn
line.
1
Analysis of results
A1 Gradient equated with
π
120r 1
A2 Correct working for r 1
A3 r in range 14.0 cm to 15.2 cm 1
A4 Unit of r correct
Unit must be consistent with the value
1
A5 Sensible comment relating to r value
One mark for good agreement/same value/similar value/slightly different
One mark for sensible comment as to why the values are similar/not the same;
e.g. card is not a perfect circle, % uncertainty is small, theory is correct.
Statement ‘values are different’ scores zero.
Vague answers such as ‘inaccuracies’, ‘errors’ or ‘graph drawing’ are not to be
credited.
2
Special cases
S1 Something very wrong;
M1, -2; M6 = 0 (and probably A3 = 0 also)
S2 Substitution method for r;
A1 = A2 = 0
S3 Uses 2a instead of a ; calculator in radian mode; subtraction method for a ;
M1, -1; M6 = 0; A3 = 0
S4 POT error;
A3 = 0