9702 w01 ms_all

NOVEMBER 2001

ADVANCED SUBSIDIARY LEVEL

MARK SCHEME

MAXIMUM MARK : 60

SYLLABUS/COMPONENT : 8702/2

PHYSICS(STRUCTURED QUESTIONS)

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. Fora B mark to be scored, the point to which it refers must be seen specifically in thecandidate’s answer.

M marks: These are method marks upon which A-marks (accuracy marks) later depend. For anM-mark to be scored, the point to which it refers must be seen in the candidate’sanswer. If a candidate fails to score a particular M-mark, then none of the dependentA-marks can be scored.

C marks: These are compensatory method marks which can be scored even if the points towhich they refer are not written down by the candidate, providing subsequent workinggives 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 correctworking 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 aC-mark to be scored.

Conventions within the marking scheme

Brackets Where brackets are shown in the marking scheme, the candidate is not required togive the bracketed information in order to earn the available marks.

Underlining In the marking scheme, underlining indicates information that is essential for marks tobe 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

1at

2 and u = 0 (accept h =

2

1at

2)

gradient of graph = 2

1g

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 (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 = � 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

1mv

2

= 2

1x 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

1x 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 10

14)

= 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]



(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�

B1

B1

A0 [2]

(ii) R = � L / A

2.4 = (4.9 x 10-7

x L) / � 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]

NOVEMBER 2001

ADVANCED SUBSIDIARY LEVEL

MARK SCHEME

MAXIMUM MARK : 25

SYLLABUS/COMPONENT : 8702/3

PHYSICS(PRACTICAL)



Measurements

M1 Measurements

One mark for each set of readings

Intervals must be correct or –1

Check values for �

�sin

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 sin� /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 ydirections. 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



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 drawnline.

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 becredited.

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 2� instead of � ; calculator in radian mode; subtraction method for � ;

M1, -1; M6 = 0; A3 = 0

S4 POT error;

A3 = 0

9702 w01 ms_all

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