9702 w09 ms_all

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS

GCE Advanced Subsidiary Level and GCE Advanced Level

MARK SCHEME for the October/November 2009 question paper

for the guidance of teachers

9702 PHYSICS

9702/11 Paper 11 (Multiple Choice), maximum raw mark 40

Mark schemes must be read in conjunction with the question papers and the report on the examination.

• CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the October/November 2009 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Page 2 Mark Scheme: Teachers’ version Syllabus Paper

GCE A/AS LEVEL – October/November 2009 9702 11

© UCLES 2009

Question Number

Key Question Number

Key

1 C 21 C

2 A 22 B

3 B 23 B

4 A 24 B

5 A 25 D

6 B 26 D

7 C 27 D

8 D 28 B

9 A 29 A

10 C 30 C

11 C 31 C

12 B 32 D

13 D 33 A

14 A 34 A

15 D 35 B

16 C 36 C

17 C 37 C

18 D 38 D

19 B 39 C

20 D 40 B





9702 PHYSICS

9702/12 Paper 12 (Multiple Choice), maximum raw mark 40





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Question Number

Key Question Number

Key

1 C 21 B

2 B 22 B

3 A 23 B

4 A 24 D

5 B 25 D

6 C 26 D

7 D 27 B

8 A 28 A

9 C 29 C

10 C 30 C

11 B 31 D

12 D 32 A

13 A 33 A

14 D 34 B

15 C 35 C

16 C 36 C

17 D 37 D

18 B 38 C

19 D 39 B

20 C 40 A





9702 PHYSICS

9702/21 Paper 21 (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.





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1 (a) (i) car uses 210 / 14 = 15 litres of fuel ....................................................................C1 volume reading = 45 litres . ................................................................................. A1 [2] (ii) from ‘full’ to ‘3/4’ mark ......................................................................................... B1 [1] (b) (i) line/graph does not pass through (‘empty, 0) / there is an intercept ................... B1 [1] (do not allow ‘non-linear’) (ii) (meter shows zero fuel when there is some left in the tank so) acts as a ‘reserve’ ............................................................................................... B1 [1]

[Total: 5] 2 (a) (i) (air) resistance increases with speed .................................................................M1 resultant / accelerating force decreases ............................................................. A1 [2] (ii) either (air) resistance is zero or weight / gravitational force is only force ................................................... B1 [1] (b) use of gradient of a tangent .......................................................................................M1 acceleration = 1.9 ± 0.2 m s-2 .................................................................................. A2 [3] (for values > ± 0.2 but ≤ 0.4, allow 1 mark) (answer 3.3 m s-2 scores no marks)

(c) (i) 1 weight = 90 × 9.8 = 880 N ........................................................................... A1 [1] (use of g = 10 m s-2 then deduct mark but once only in the Paper)

2 accelerating force = 90 × 1.9 = 170 N …(allow ecf) ................................. A1 [1] (ii) resistive force = 880 – 170 = 710 N ................................................................ A1 [1] (allow ecf but only if resistive force remains positive)

[Total: 9] 3 (a) (i) either sum / total momentum (of system of bodies) is constant or total momentum before = total momentum after ......................................M1 for an isolated system / no (external) force acts on system ............................... A1 [2] (ii) zero momentum before / after decay ..................................................................M1

so α-particle and nucleus D must have momenta in opposite directions ........... A1 [2] (b) (i) kinetic energy = ½ mv2 .. ...................................................................................C1

1.0 × 10-12 = ½ × 4 × 1.66 × 10-27 × v2 ..............................................................M1

v = 1.7 × 107 m s-1 ............................................................................................. A0 [2]

(ii) 1.7 × 107 × 4u = 216u × V ................................................................................C1

V = 3.1 × 105 m s-1 ............................................................................................ A1 [2]

(accept 3.2 × 105 m s-1, do not accept 220 rather than 216)



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(c) (1.7 × 107)2 = 2 × deceleration × 4.5 × 10-2 ..............................................................C1

deceleration / a = 3.2 × 1015 m s-2 ........................................................................... A1 [2]

(accept calculation based on calculating F = 2.22 × 10-11 N and then use of F = ma)

[Total: 10] 4 (a) (i) returns to original shape / size / length etc. ........................................................ B1 when load / distorting forces / weight / strain is removed ................................... B1 [2]

(ii) 1 R = ρL / A ....................................................................................................... B1 [1] 2 E = WL / Ae ................................................................................................... B1 [1]

(b) E = WR / eρ ............................................................................................................C1

= (34 × 0.44) / (7.7 × 10-4 × 9.2 × 10-8) ....................................................................C1

= 2.1 × 1011 Pa ......................................................................................................... A1 [3]

[Total: 7] 5 (a) transfer / propagation of energy ................................................................................M1 as a result of oscillations / vibrations ......................................................................... A1 [2] (b) (i) displacement / velocity / acceleration (of particles in the wave) ......................... B1 [1] (ii) displacement etc. is normal to direction of energy transfer / travel of wave / propagation of wave ……(not ‘wave motion’) ............................ B1 [1] (iii) displacement etc. along / same direction of energy transfer / travel of wave / propagation of wave ……(not ‘wave motion’) ............................ B1 [1] (c) diffraction: suitable object, means of observation ......................................................M1 either laser or lamp and aperture or distant source ........................................................................................................M1 light region where darkness expected ....................................................................... A1 interference: suitable object, means of observation and illumination ........................ B1 light and dark fringes observed ................................................................................. B1 appropriate reference to a dimension for diffraction or for interference .......................................................................................................... B1 [6]

[Total: 11] 6 (a) energy transferred from source / changed from some form to electrical ...................M1 per unit charge (to drive charge round a complete circuit) ........................................ A1 [2] (b) and power in R = I 2X ............................................................................................M1 E = I (X + r) ................................................................................................................M1 power in cell = EI and algebra clear leading to ratio = X / (X + r) ............................. A1 [3]



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(c) (i) 1.4 W .................................................................................................................. A1

0.40 Ω ………(allow ±0.05 Ω) ......................................................................... A1 [2]

(ii) current in circuit = 1.4/0.4 = 1.87 A .............................................................C1

1.5 = 1.87 (r + 0.40) ..........................................................................................C1

r = 0.40 Ω …...................................................................................................... A1 [3] (d) either less power lost / energy wasted / lost or greater efficiency (of energy transfer) .......................................................... B1 [1]

[Total: 11] 7 (a) deviation shown correctly .......................................................................................... B1 [1] (b) smaller deviation (not zero deviation) ........................................................................M1 acceptable path wrt position of N .............................................................................. A1 [2] (c) the nucleus is (very) small .........................................................................................M1 in comparison to the atom ......................................................................................... A1 [2] (special case: ‘atom is mostly empty space’ scores 1 mark) (d) deviation depends on charge on the nucleus / N / electrostatic repulsion ................. B1 same charge so no change in deviation .................................................................... B1 [2]

[Total: 7]





9702 PHYSICS

9702/22 Paper 22 (AS Structured Questions), maximum raw mark 60






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1 (a) (i) either 1.55% or 1.6% …(not 1.5 or 2) ............................................ A1 [1] (ii) either 1.09% or 1.1% …(not 1.0 or 1) ............................................ A1 [1] (b) answer of (ii) + 2 × (i) to any number of sig. fig. either 4.2% or 4.3% .................................................................................... A1 [1] (c) (i) either the value has more significant figures than the data or uncertainty of ±0.4 renders more than 2 s.f. meaningless) ......................... B1 [1]

(ii) uncertainty in g = ±0.41 / ±0.42 to any number of s.f. ....................................C1

g = (9.8 ± 0.4) m s-2 ........................................................................................ A1 [2]

[Total: 6] 2 (a) (i) e.g. (phase) change from liquid to gas / vapour thermal energy required to maintain constant temperature ......................... B1 [1] (do not allow ‘convert water to steam’) (ii) e.g. evaporation takes place at surface .............................................................. B1 boiling takes place in body of the liquid ....................................................... B1 e.g. evaporation occurs at all temperatures ....................................................... B1 boiling occurs at one temperature ............................................................... B1 [4]

(b) (i) volume = (4.5

48=) 10.7 cm3 .............................................................................. A1 [1]

(ii) 1 volume = 10.7 / (6.0 × 1023) = 1.8 × 10-23 cm3 ................................................................................................ A1 [1] 2 separation = 3√(1.8 × 10-23) = 2.6 × 10-8 cm ................................................................................................... A1 [1]

[Total: 8] 3 (a) (i) speed = 4.0 m s-1 …(allow 1 s.f.) ................................................................... A1 [1] (ii) v2 = 2gh

= 2 × 9.8 × 1.96 .............................................................................................M1 v = 6.2 m s-1 ..................................................................................................... A0 [1] (use of g = 10 m s-2 loses the mark) (b) correct basic shape with correct directions for vectors ..............................................M1 speed = (7.4 ± 0.2) m s-1 ......................................................................................... A1 at (33 ± 2)° to the vertical .......................................................................................... A1 [3] (for credit to be awarded, speed and angle must be correct on the diagram – not calculated)



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(c) (i) either v2 = 2 × 9.8 × 0.98 or v = 6.2 / √2 ............................................C1 speed = 4.4 m s-1 .............................................................................................. A1 [2] (allow calculation of t = 0.447 s, then v = 4.4 m s-1) (ii) 1 momentum = mv ...........................................................................................C1 change in momentum = 0.034 (6.2 + 4.4) ........................................................C1 = 0.36 kg m s-1 .............................................................. A1 [3] (use of 0.034 (6.2 - 4.4) loses last two marks) 2 force = ∆p / ∆t …….(however expressed) ...................................................C1

= 0.12

0.36

= 3.0 N ……(allow 1 s.f.) ................................................................... A1 [2]

[Total: 12] 4 (a) ability to do work ........................................................................................................ B1 as a result of a change of shape of an object/stretched etc ...................................... B1 [2] (b) work = average force ×distance moved (in direction of the force) ........................... B1 either work = ½ × F × x or work is area under F/x graph which is ½Fx ................................................. B1 F = kx ........................................................................................................................ B1 so work / energy = ½kx2 .......................................................................................... A0 [3]

(c) (i) spring constant = 2.1

3.8 .....................................................................................M1

= 1.8 N cm-1 ........................................................................... A0 [1] (ii) 1 ∆EP = mg∆h or W∆h ...................................................................................C1 = 3.8 × 1.5 × 10-2 = 0.057 J ................................................................................................ A1 [2] 2 ∆ES = ½ × 1.8 × 102 (0.0362 – 0.0212) ...........................................................M1 = 0.077 J ................................................................................................ A0 [1] 3 work done = 0.077 – 0.057 = 0.020 J ....................................................................................... A1 [1] (allow e.c.f. if ∆ES > ∆EP)

[Total: 10]



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5 (a) (i) frequency f ......................................................................................................... B1 [1] (ii) amplitude A ....................................................................................................... B1 [1]

(b) π rad or 180° ………(unit necessary) .................................................................... B1 [1] (c) (i) speed = f × L ..................................................................................................... B1 [1] (ii) wave is reflected at end / at P ............................................................................ B1 either incident and reflected waves interfere or two waves travelling in opposite directions interfere ................................M1 speed is the speed of incident or reflected wave / one of these waves .............. A1 [3]

[Total: 7] 6 (a) total resistance in series = 2R total resistance in parallel = ½R ................................................................................M1 ratio is 2R / ½R = 4 ......(allow mark if clear numbers in the ratio) ........................ A0 [1] (b) at 1.5 V, current is 0.10 A ..........................................................................................C1

resistance = V/I = 0.1

1.5

= 15 Ω .................................................................................................... A1 [2] (use of tangent or any other current scores no marks) (c)

p.d. across each lamp / V

resistance of each lamp / Ω

combined resistance / Ω

series parallel

1.5 3.0

15 20

30 10

column 1 .................................................................................................................... A1 columns 2 and 3: max 3 marks with -1 mark for each error or omission .................. A3 [4] (d) (i) ratio is 3 ...............(allow e.c.f.) ......................................................................... A1 [1] (ii) resistance increases as potential difference increases ...................................... B1 increasing p.d. increases current ........................................................................ B1 current increases non-linearly so resistance increases ..................................... B1 [3]

[Total: 11]



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7 (a) either forms of same element or atoms / nuclei with same number of protons ................................................M1 atoms / nuclei contain different numbers of neutrons ................................................ A1 [2] (use of ‘element’ rather than atoms / nuclei scores max 1 mark) (b) (i) decay is not affected by environmental factors .................................................. B1 [1] (allow two named factors) (ii) either time of decay (of a nucleus) cannot be predicted or nucleus has constant probability in a given time .................................. B1 [1] (c) Re

185

75 ......................................................................................................................... B1

either e0

1− or 0

1−β ......................................................................................... B1 [2]

[Total: 6]





9702 PHYSICS

9702/31 Paper 31 (Advanced Practical Skills 1), maximum raw mark 40






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1 (a) First values for h and z, to the nearest mm. [1] (b) Measurements – Add up the number of sets of values of z and h and put a ringed total [4] by the table. Four marks for six sets of readings of z and h, three for five sets, etc.

(–1 if help given by supervisor, -1 if wrong trend i.e. h↑ z↓) Maximum value for z - h greater than 6.0 cm [1] Column headings [1] Each column heading must contain a quantity and a unit where appropriate. Ignore units in the body of the table. There must be some distinguishing mark between the quantity and the unit (i.e. solidus is expected, but accept, for example, h (mm)). Consistency of presentation of raw readings [1] All raw values of h and z must be given to the same number of decimal places. (c) (i) (Graph) Axes [1] Sensible scales must be used. Awkward scales (e.g. 3:10) are not allowed. Scale markings should be no more than 3 large squares apart. Scales must be chosen so that the plotted points occupy at least half the graph grid in both x and y directions. Scales must be labelled with the quantity being plotted. Ignore units. Allow reversed axes but do not allow the wrong graph. (Graph) Plotting All observations must be plotted. Put a ringed total of plotted points. [1] Ring and check a suspect plot. Tick if correct. Re-plot if incorrect. Work to an accuracy of half a small square. Penalise ‘blobs’ – dia. of plots must be < ½ a small square. (Graph) Line of best fit [1] Judge by scatter of at least 5 trend points about the candidate's line. There must be a fair scatter of points either side of the line. Indicate best line if candidate's line is not the best line. (Graph) Quality of results Judge by scatter of points about the best fit line. [1] All points in the table (of which there must be at least 5) must be within + 0.3 cm (to scale) on the h axis. (ii) Gradient The hypotenuse must be at least half the length of the drawn line. [1] Read-offs must be accurate to half a small square. If incorrect write in the correct [1] value. Check for ∆y/∆x (i.e. do not allow ∆x/∆y).



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(d) (i) Raw value(s) for d to nearest 0.1 mm or 0.01 mm [1] 18.00 mm < d < 27.00 mm. Unit required. Repeated readings for d. [1] (ii) A calculated correctly. Allow ecf. Check value. If incorrect, write in the correct value. [1] Significant figures for A must be the same as, or one more than, the sig. figs. of the [1] raw values of d. (e) Method – value from (c) (ii) equated to k/ρAg + 1 [1] Substitution methods lose both (e) marks Calculation - value for k in range 4 to 6 Nm-1.(allow 3.50 < k < 6.49). [1] (or refer to supervisor’s value). Unit required. Ignore SF. This mark is conditional on achieving the previous mark.

[Total: 20]



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2 (a) (i) First value of l, with unit, to nearest mm. (40 ≤ l ≤ 60 cm) [1] (-1 if help given by supervisor) (b) (i) First value of d ( 18 cm < d < 22 cm) with consistent unit. [1] (ii) Method of measuring d accurately – two details of procedure e.g: [2]

• Method of consistent release of marble

• Use of named item(s) as marker(s)

• Refining position of marker

• Place ruler underneath and view vertically from above. Do not allow ‘repeats’ (iii) Percentage uncertainty in d. [1] Range of absolute uncertainty: 2 mm < ∆d < 10 mm. If repeated readings have been done then the uncertainty can be half the range. Correct ratio idea required. x 100% implied. (c) (i) First value of k, substitution correct. Consistent unit. [1] (ii) Justification for s.f. in value of k. [1] Either: k must be given to same no. of SF, or one more than, l and d. Or: k must be given to same no. of SF, or one more than, l or d, whichever has the least no. of SF. (d) Second values of l and d. [1] Evidence of repeat readings for first or second value of d. [1] Second d less than first d. [1] (e) Percentage difference (or fractional difference) in k values calculated. [1] Sensible conclusion consistent with uncertainty of 20% [1] in values of k, or candidate’s stated uncertainty.



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(f) Identifying limitations and suggesting improvements:

(f) (i) Limitations/ sources of error

(max 4 marks)

(f) (ii) Improvements (max 4 marks))

Ignore:

A Only two readings/two readings are not enough (to draw a valid conclusion)

Take more readings and plot a graph/calculate more k values

repeat readings

B Hard to measure d because ball moves too quickly/ too fast/only stationary for short time …

Use video and play back slowly/ frame by frame Use slow motion camera Use position sensor/motion sensor Allow light gates, adjusting position until beam interrupted

Use a high-speed camera/computer/data logger

C Difficulty in releasing marble consistently/ from rest/without applying a force

Description of a mechanism to release marble e.g. slot in tube + card

Change angle

D Parallax error in measurement of d

Description of method of reducing parallax error requiring additional equipment e.g. shadow projection

view at eye level view from above use a marker

E Incorrect alignment/ inconsistent collisions/ different paths down tube

Use narrower tube

F Motion of ball affected by air movement/ ball swings around

Turn off fans/air con. Shield from draughts

Use a closed room/vacuum refs to air resistance heavier ball

G Difficult to measure l because it is hard to judge the position of the centre of the ball

Measure diameter of ball using vernier calipers Measure l to top and bottom of ball and average.

[Total:20]





9702 PHYSICS







© UCLES 2009

1 (a) First value for h to nearest mm [1] (b) Measurements table [4] Four marks for six sets of readings for m and h, three for five sets, etc. (–1 if trend is positive, –1 if help from supervisor) Table – range [1]

Values of m must be [ 10 g and Y 100 g. Values must include 10 or 20 g and 90 or 100 g with no interval greater than 20 g.

Table – column headings [1] Each column heading must contain a quantity and a unit where appropriate. Ignore units in the body of the table. There must be some distinguishing mark between the quantity and the unit (i.e. solidus is

expected, but accept, for example, m (g)). Table – consistency of presentation of raw readings [1] All values of h must be given to the same number of decimal places. (c) (i) (Graph) Axes – [1] Sensible scales must be used. Awkward scales (e.g. 3:10) are not allowed. Scales must be chosen so that the plotted points occupy at least half the graph grid in

both x and y directions. Scales must be labelled with the quantity which is being plotted. Ignore units. Allow reversed axes but do not allow wrong graph. Gaps between labels must not be greater than three large squares. (Graph) Plotting – [1] All observations must be plotted. Ring and check a suspect plot. Tick if correct. Re-plot if incorrect (i.e. if plot is more than

half a small square from the correct position). Do not allow plots with diameter greater than half a small square. (Graph) Line of best fit – [1] Judge by scatter of at least 5 trend plots about the candidate's line. There must be a fair scatter of points either side of the line. Indicate best line if candidate's line is not the best line. (Graph) Quality of results – [1] Judge by scatter of points about a best fit line All points in the table (which must be at least 5) must be within 0.5 ‘h-scale cm’ of a

straight line. Do not award if wrong trend. (ii) Gradient –

The hypotenuse must be at least half the length of the drawn line. [1] Read-offs must be no more than half a small square from the line (if incorrect, write in correct value). [1] Check for ∆y/∆x. Check value is consistent with trend.



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(d) (i) Measurement – value for raw d in range 18.00 to 27.00 mm (or SV ±2.00 mm), and

given to nearest 0.1 mm or nearest 0.01 mm. Unit must be given. [1] Measurement – repeated readings for d. [1] (ii) A calculated correctly. Allow ecf. Check value. Penalise power of ten error. If incorrect, write in the correct value. [1] S.f. in A the same as or one more than the s.f. in raw d. [1] (e) Gradient value from (c)(ii) equated to – (k+ρAg). Allow ecf. Penalise sign error. [1] Value for k in range 3.50 to 6.49 Nm–1 (or SV ±30%). [1] Ignore sign. Unit required. Do not award this mark if the gradient has not been used. [Total: 20]



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2 (a) Value of l, with unit, to nearest mm. [1] If SV help, then –1.

(b) (i) First value of a (Y 25 cm) [1] (ii) First value of b (less than a) [1] (c) (i) Use named item as marker for rebound distance/ [1] Place ruler under path and view vertically from above/ Use second brick as releasing point. (ii) Percentage uncertainty in b [1] If repeated readings have been done then the uncertainty could be half the range,

otherwise absolute uncertainty must be at least 2 mm and no more than 10 mm. Correct ratio idea required. (d) First value of k substitution correct and value <1. There must be no unit. [1] S.f. in value of k – must be 2 or 3 s.f. (but allow 4 s.f. if all raw data is to 3 s.f.) [1] (e) Second values of a and b. [1] Evidence of repeat readings for first or second value of b [1] Second value of b shows correct trend. [1] (f) Calculation of % difference (or equivalent) in k values. [1] Valid conclusion based on the two values of k (e.g. k is constant because values close),

consistent with 20% difference as border between ‘close’ and ‘not close’ unless candidate has defined his own % difference. [1]



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(g) Identifying limitations and improvements

(g) (i) Difficulties (one from each box – max. 4)

(g) (ii) Improvements (one from each box – max. 4)

But not

A Two sets of readings not enough.

Take more readings and plot a graph / calculate more k values.

Repeated readings.

B Difficult to judge rebound point/distance because of movement / short static time.

Use video with slow playback / use position sensor to measure rebound / use sound of ball striking a block to judge rebound / use lightgate and refine its position.

Use computer or data logger / attach pointer to ball / change length of string / time rebound instead of measuring.

C Difficult to release without exerting a force/movement.

Named, realistic method of release without a force (e.g. remote-controlled clamp).

D Parallax error in measuring rebound distance.

Observe shadow on screen. View at eye level.

E Inconsistent bounce / ball bounces at an angle.

Use smoother brick. Use heavier ball.

F Motion affected by air movement / ball swings around.

Turn off fans or air con / shield from draughts.

Air resistance / carry out in vacuum / constraining guides.

G When measuring l it is difficult to judge centre of ball.

Suitable method for measuring diameter of ball.

[8] [Total: 20]





9702 PHYSICS







© UCLES 2009

1 (b) (i) Value for l between 0.010 and 0.080 m (1.0–8.0 cm), or ± 2.0 cm of supervisor’s value. Raw value(s) to nearest mm. [1]

(c) Two values of height given. [1] Check calculation. Ignore POT error. If method incorrect to work out v, final (f) mark not available. [1]

(d) No help from supervisor. [1]

Six sets of values scores 3 marks, five sets scores 2 marks etc. [3] Add up number of sets of readings for M and l and put a ringed total by the table.

Wrong trend –1 (Correct trend M increases, l increases).

Range of M includes 100 g or 150 g and 400 g or 450 g. [1]

Each column heading must contain a quantity and a unit where appropriate. [1] Ignore units in the body of the table. There must be a distinguishing mark between the quantity and the unit. (solidus is expected, accept brackets e.g. M/kg, l/m, v/m, M/v / kg m–1)

Consistency of presentation of raw readings. All values of raw l are given to the same number of decimal places. [1]

Significant figures for M/v must be the same as, or one more than the least number of [1] significant figures used in M or v. Check each row. If v = constant, quality mark not available AND final (f) mark not available.

Check the specified value of M/v correct. (Expect around 1–3 kg m–1 or 10–30 g cm–1) [1] Ignore POT. If incorrect write in correct value. Allow small rounding errors.

Graph (e) (i) Axes [1] Sensible scales must be used. Awkward scales (e.g. 3:10) are not allowed. Scale markings should be no more than three large squares apart. Scales must be chosen so that the plotted points must occupy at least half the graph grid in both x and y directions. Allow inverted axes. Do not allow wrong graph. Scales must be labelled with the quantity which is being plotted. Ignore units.

All observations must be plotted. Put a ringed total of plotted points. [1] Ring and check a suspect plot. Tick if correct. Re-plot if incorrect. Work to an accuracy of not greater than half a small square.

Do not allow blobs (i.e. diameter > half a small square). (e) (ii) Line of best fit [1] Judge by scatter of points about the candidate's line. There must be a fair scatter of points either side of the line. At least 5 trend plots required.

Quality. This mark is not available for the wrong graph or wrong trend. [1] Judge by scatter of all the points about a best fit line. All points in the table (of which there must be at least 5 plots) must be plotted. Allow ± 0.3 cm to scale on the x-axis. (If v = constant, quality mark not available AND

final (f) mark not available.)



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(e) (iii) Gradient. Check dy/dx [1] The hypotenuse must be at least half the length of the drawn line on the graph grid. Read-offs must be read to at least half a small square. If read-off incorrect write in correct value. Be prepared to check both read-offs. If both

incorrect do not allow ecf in the y-intercept if using one of the read-offs from the gradient. Intercept. Check substitution only. Check both read-offs to half a small square. [1] or read from graph to half a small square as long as no false origin. (f) Look for values of y-intercept and gradient used correctly to find C. [1] i.e. grad = qk AND y-intercept = qC or y-intercept = (grad/k) × C. Value of C in range 0 to ± 1 N, consistent with unit or refer to supervisor’s results. [1] Correct method needed. If method of working out v incorrect or if v = constant in table, this mark is not available. [Total: 20] 2 (a) Evidence of repeat measurements of d. [1] Value of raw d(s) given to nearest 0.1 mm or 0.01 mm (–1 if help given by supervisor). [1] (b) Percentage uncertainty in d. [1] If repeated readings have been done then the uncertainty could be half the range, otherwise absolute uncertainty must be 0.1 mm or 0.01 mm consistent with above. Correct ratio idea required. (d) Method of calculation of l correct. 1.5πd [1] Significant figures in l same or one more than the raw values of d. Ignore units. [1] (e) Value of m1 in range 60 to 300 g, consistent with unit. [1] If Supervisor notes that hanger moved at 50 g allow m1 = 50 g. (f) Evidence of repeat readings for first or second value of m. [1] Second value of m. [1] Second value of l greater than first l. [1] Second value of m

≥ 2 × m1. [1] (g) Calculation of the two values of m2

/ l3 or equivalent.

Check one value and correct substitutions. [1] Conclusion consistent with candidate’s k values. Use 20% permitted variation in k if candidate does not suggest a value. [1]



© UCLES 2009

(h) (i) and (ii)

Sources of error or limitation. [4]

Improvements. Use of other apparatus or different procedures. [4]

Ap Only two readings/Two readings are not enough (to draw a valid conclusion).

As Take many (sets of) readings and plot a graph/find more values of k’s.

Be clear NOT just repeat readings.

Bp Circumference/l imprecise because helical/coiled/slanted/spiral/thickness of thread/non-uniform diameter of rod.

Bs Mark string and measure length/wrap so coils are closer/allow for thickness of thread/ diameter to be taken at different places along/diameter taken at different angles (at same position).

Cp Use of (10 g) increments imprecise.

Cs Use smaller mass increments/use newtonmeter/other valid method (water or sand).

Dp Difficulty to judge/tell when the string starts to slip/gradual movement.

Ds Practical method of detecting movement: fixed marker or scale/motion sensor/(travelling) microscope/measure height from table.

Ep Large scatter in repeated readings of mass/non-uniform surface bar/varying friction.

Es

Fp Difficult to add masses without swinging/pushing the hanger/masses do not fit hanger.

Fs Lower masses slowly/support underneath and remove hand slowly/scissor jack.

Ignore reference to light gates, video, reaction time, repeat readings, micrometer, fans, parallax or sanding. [Total: 20]





9702 PHYSICS







© UCLES 2009

1 (b) Value for V0 in range 1.3 to 1.7V, with unit [1] (c) (ii) First value of V less than V0 [1] (d) No help from Supervisor (-1 for minor help, -2 for major help) [2] (d) Measurements table [3] Six sets of readings of R and V scores 3 marks, five sets scores 2 marks etc. Wrong trend in table then –1. (d) Table - range [1] Values of R must include one of 100/220Ω and one of 3300/4700Ω. (d) Table - column headings [1] Each column heading must contain a quantity and a unit where appropriate. There must be some distinguishing mark between the quantity and the unit. Ignore units in the body of the table. R/(1000+R) has no unit. (d) Table - consistency of presentation of raw readings. [1] All values of raw V must be given to the same number of decimal places. (d) Table – calculated values [1] Check the specified value of R/(1000+R) is calculated correctly. If incorrect, write in the correct value. Ignore rounding errors. (d) Table - significant figures [1] S.f. for 1/V must be the same as, or one more than, s.f. for raw V. Check each row in the table. (e) (i) (Graph) Axes – [1] Sensible scales must be used. Awkward scales (e.g. 3:10) are not allowed. Scales must be chosen so that the plotted points occupy at least half the graph grid in both x and y directions. Scales must be labelled with the quantity

plotted. Ignore units. Allow inverted axes, –1 wrong quantities plotted. No more than 3 large squares between scale markings. (Graph) Plotting – [1] All observations must be plotted. Ring and check a suspect plot, tick if correct. Re-plot if incorrect. Plots should be no more than ½ a small square from

correct position in x or y direction. Diameter must be less than or equal to ½ small square.

(e) (ii) (Graph) Line of best fit – [1] At least 5 trend plots are needed. Judge by scatter of points about the candidate's line. There must be a fair scatter of points either side of the line. Indicate best line if candidate’s line is not the best line. If trend curved allow a smooth drawn curve not straight line. (Graph) Quality of results – [1] All table points must be plotted (minimum of 5 needed). Judge by scatter of all plots which must be within ± 0.02 V –1 of assessors line.



© UCLES 2009

(e) (iii) Gradient – [1] The hypotenuse of the ∆ must be at least half the length of the drawn line. Both read-offs must be accurate to half a small square. Check for ∆y /∆x. Check sign is consistent with trend. (e) (iii) Intercept – [1] Correctly read-off from graph (indicate a false origin) or the method of calculation is correct (check substitution of point on line). (f) Method of calculation of P is correct with gradient and intercept values used. [1] (f) Value for P in range 630 to 730Ω, with unit. [1] Substitution loses both marks.

[Total: 20] 2 (c) (i) Value of l < 25cm, with unit. [1] (c) (i) l to nearest mm. [1] (c) (iii) Evidence of repeated measurements of hfinal [1] (c) (iii) Value of hfinal in range 5.0 to 50.0 cm. [1] (d) Percentage uncertainty in hfinal. [1] If repeated readings have been done then the uncertainty could be half the range, otherwise

absolute uncertainty must be in range 2 mm to 20 mm. Correct ratio idea required. (e) Ep to no more than 3 s.f. [1] (e) Value for Ep consistent with unit. [1] (f) Second value of l greater than first value. [1] (f) Second value of hfinal [1] (f) Second value of hfinal shows correct trend (i.e. l ↑ h ↑ or l ↓ h ↓). [1] (g) Check calculation of the two values of Ep/√l or equivalent. [1] (g) Valid conclusion based on the calculated values. Consistent with 20% or with candidate’s

stated criterion. [1]



© UCLES 2009

(h)

Limitation (4 max) Improvement (4 max)

A Two sets of readings are not enough / only two sets

Take more readings and plot a graph

B Difficult to take measurements (h/l) because the ruler moves / is not vertical

Clamp rule / ensure rule is vertical using a set square on the bench

C Change in properties / deterioration of the thread due to repeated drops

Use a new thread each time

D Poor accuracy due to size of increment / only note measured hfinal values not the values between.

Use smaller increments

E Obtaining constant loop length for repeats at one value of loop length / variation in hfinal values for repeats at one loop length

Sensible method to ensure constant loop length for repeats

F Tangling cotton

Do not allow ‘repeated readings’, centres of mass, or nail, knots, time ideas. Do not allow use of video, ‘use a computer to improve experiment’, sensors. Do not allow amount of tape/plasticine/glue, thinner/thicker thread, fans. Do not allow ‘eye level’.

[Total: 20]





9702 PHYSICS

9702/41 Paper 41 (A2 Structured Questions), maximum raw mark 100






© UCLES 2009

Section A

1 (a) F ∝ Mm / R2 …..…(words or explained symbols) ................................................M1 either M and m are point masses or R >> diameter of masses …(do not allow ‘size’) ....................................... A1 [2] (b) (i) equatorial orbit .................................................................................................... B1 period 24 hours / same angular speed ............................................................... B1 from west to east / same direction of rotation ..................................................... B1 [3] (allow one of the last two marks for ‘always overhead’ if 2nd or 3rd marks not scored) (ii) gravitational force provides centripetal force / gives rise to centripetal acceleration ….(in ‘words’) ........................................ B1

GM / x2 = xω2 ....................................................................................................M1 g = GM / R2 .......................................................................................................M1

to give gR2 = x3ω2 ............................................................................................ A0 [3]

(iii) ω = 2π / (24 × 3600) = 7.27 × 10-5 rad s-1 ........................................................C1

9.81 × (6.4 × 106)2 = x3 × (7.27 × 10-5)2 .............................................................C1

x3 = 7.6 × 1022

x = 4.2 × 107 m ................................................................................................. A1 [3] (use of g = 10 m s-2, loses 1 mark but once only in the Paper)

[Total: 11] 2 (a) either pV = NkT or pV = nRT and n = N / NA .....................................................C1 clear correct substitution e.g.

2.5 × 105 × 4.5 × 103 × 10-6 = N × 1.38 × 10-23 × 290 ...............................................M1

N = 2.8 × 1023 .......................................................................................................... A0 [2] (allow 1 mark for calculation of n = 0.467 mol)

(b) (i) volume = (1.2 × 10-10)3 × 2.8 × 1023 or 3

4 πr3 × 2.8 × 1023 ..............................C1

= 4.8 × 10-7 m3 2.53 × 10-7 m3 ..................................... A1 [2]

(ii) either 4.5 × 103 cm3 >> 0.48 cm3 or ratio of volumes is about 10-4 ................ B1 justified because volume of molecules is negligible ........................................... B1 [2]

[Total: 6]



© UCLES 2009

3 (a) e.g. two objects of different masses at same temperature (M1) same material would have different amount of heat (A1) e.g. temperature shows direction of heat transfer (M1) from high to low regardless of objects (A1) e.g. when substance melts/boils (M1) heat input but no temperature change (A1) any two, M1 + A1 each, max 4 ………………………………..…………........................... [4] (b) (i) energy losses (to the surroundings) .................................................................M1 either increase as the temperature rises or rise is zero when heat loss = heat input ............................................... A1 [2] (ii) idea of input power = maximum rate of heat loss .............................................C1

power = m × c × ∆θ / ∆t

54 = 0.96 × c × 3.7 / 60 .....................................................................................C1 c = 910 J kg-1 K-1 ............................................................................................... A1 [3]

[Total: 9] 4 (a) (i) amplitude = 0.2 mm .......................................................................................... A1 [1] (ii) period = 1.2 ms .................................................................................................C1 frequency = 830 Hz .......................................................................................... A1 [2] (b) (i) any two of zero, 0.6 ms and 1.2 ms .................................................................... A1 [1] (ii) any two of 0.3 ms, 0.9 ms, 1.5 ms ...................................................................... A1 [1]

(c) either v = ωx0 = 2πfx0

= 2π × 830 × 0.2 × 10-3 = 1.05 m s-1 or slope of graph = 1.0 m s-1 ……(allow ± 0.1 m s-1) .......................................C1 EK = ½mv2

= ½ × 2.5 × 10-3 × 1.052 .....................................................................................C1

= 1.4 × 10-3 J ...................................................................................................... A1 [3] (d) (i) large / maximum amplitude of vibration .............................................................. B1 when impressed frequency equals natural frequency of vibration ...................... B1 [2] (ii) e.g. metal panels on machinery vibrate / oscillate ........................................... (M1) motor in machine impresses frequency on panel ......................................(A1) e.g. car suspension system vibrates / oscillates................................................. (M1) going over bumps would give large amplitude vibrations .............................(A1) any feasible example, M1 + A1 ............................................................................... [2]

[Total: 12]



© UCLES 2009

5 (a) work done per / on unit positive charge .....................................................................M1 moving charge from infinity to the point ..................................................................... A1 [2]

(b) (i) α-particle and gold nucleus repel each other ..................................................... B1

all kinetic energy of α-particle converted into electric potential energy .............. B1 [2]

(ii) 1 potential energy = (79 × 2 × 1.6 × 10-192) / (4π × 8.85 × 10-12 × d) ..............C1

kinetic energy = 4.8 × 1.6 × 10-13 = 7.68 × 10-13 J ...........................................C1

equating to give d = 4.7 × 10-14 m ..................................................................... A1 [3]

(ii) 2 F = Qq / 4πε0d × 1 / d = 7.68 × 10-13 × 1 / (4.7 × 10-14) ..............................C1 = 16 N ....................................................................................................... A1 [2]

[Total: 9] 6 (a) concentric circles …(at least three lines) ................................................................M1 with increasing separation ......................................................................................... A1 correct direction clear ................................................................................................ B1 [3] (b) (i) correct position to left of wire .............................................................................. B1 [1]

(ii) B = (4π × 10-7 × 1.7) / (2π × 1.9 × 10-2) .............................................................C1

= 1.8 × 10-5 T ................................................................................................. A1 [2]

(c) distance ∝ current ...................................................................................................C1

current = (2.8 / 1.9) × 1.7 = 2.5 A ........................................................................................................ A1 [2]

[Total: 8] 7 (a) e.g. more (output) power available e.g. less ripple for same smoothing capacitor any sensible suggestion ............................................................................................ B1 [1] (b) (i) curve showing half-wave rectification ................................................................. B1 [1]

(ii) similar to (i) but phase shift of 180° .................................................................... B1 [1] (c) (i) correct symbol, connected in parallel with R ...................................................... B1 [1] (ii) 1 larger capacitor / second capacitor in parallel with R ..................................... B1 [1] (not increase R) 2 same peak values ........................................................................................... B1 correct shape giving less ripple .......................................................................... B1 [2]

[Total: 7]



© UCLES 2009

8 (a) neutron is a single nucleon / particle ......................................................................... B1 [1]

(b) binding energy = 4 × 7.07 × 1.6 × 10-13 ....................................................................C1

= 4.52 × 10-12 J

binding energy = c2 ∆m ........................................................................................... C1

4.52 × 10-12 = (3.0 × 108)2 × ∆m

∆m = 5.03 × 10-29 kg ................................................................................................ A1 [3] (c) (i) fusion ……(do not allow fussion) .................................................................... B1 [1]

(ii) (2 × 1.12) + 3x = 28.28 ...................................................................................C1 …... –17.7 ..........................................................................................................C1 x = 2.78 MeV per nucleon ................................................................................ A1 [3] (use of +17.7 gives x = 14.6 MeV, allow 1 mark only)

[Total: 8]



© UCLES 2009

Section B

9 (a) resistance of wire = ρL / A ....................................................................................... B1 as crack widens, L increases ....................................................................................M1 and A decreases ............................................................................M1 so resistance increases ............................................................................................. A0 [3]

(b) ∆L / L = ∆R / R ........................................................................................................ B1

= (146.2 – 143.0) / 143.0 × 100 ..................................................................C1

∆L / L = 2.24% ......................................................................................................... A1 [3]

[Total: 6]

10 at 16 °C, V+ = 1.00 V and V

– = 0.98 V or V+ > V

– ........................................................ B1

at 16 °C, output is positive ................................................................................................M1 diode R is ‘on’ and diode G is ‘off’ .................................................................................... A1 as temperature rises, diode R goes ‘off’ and diode G goes ‘on’ ....................................... B1 [4] (allow e.c.f. from 2nd to 3rd marks and also 3rd to 4th marks)

[Total: 4] 11 large / 1 T magnetic field applied along body (allow ‘across’) (1) r.f. pulse applied ............................................................................................................... (1) causes hydrogen nuclei / protons ..................................................................................... (1) to resonate ....................................................................................................................... (1) (nuclei) return to equilibrium state / after relaxation time ................................................. (1) r.f. (pulse) emitted ............................................................................................................ (1) pulses detected, processed and displayed ...................................................................... (1) resonant frequency depends on magnetic field strength .................................................. (1) calibrated non-uniform field enables nuclei to be located ................................................ (1) any six points, one mark each .......................................................................................... B6 [6]

[Total: 6]



© UCLES 2009

12 (a) e.g. signal can be regenerated .................................................................................M1 so that there is minimal noise .................................................................................... A1 e.g. extra data can be added ....................................................................................M1 so that signal can be checked for errors .................................................................... A1 [4] (any two, sensible suggestions, M1 + A1, max 4) (b) (i) 1101 ................................................................................................................... B1 [1] (ii) 5 ......................................................................................................................... B1 [1] (c) (i) block X: serial-to parallel ................................................................................... B1 block Y: DAC / digital-to-analogue (converter) .................................................. B1 [2] (ii) takes the simultaneous / all bits of a number ....................................................M1 and transmits them one after another / down a single line ................................ A1 [2] (d) increase number of bits in digital number at each sampling ......................................M1 so that step height is reduced ................................................................................... A1 increase sampling frequency / reduce time between samples ..................................M1 so that depth / width of step is reduced ..................................................................... A1 [4] (do not allow ‘smoother output’)

[Total: 14]





9702 PHYSICS

9702/42 Paper 42 (A2 Structured Questions),






© UCLES 2009

Section A 1 (a) (i) force per (unit) mass ……(ratio idea essential) ................................................. B1 [1] (ii) g = GM / R2 .......................................................................................................C1

9.81 = (6.67 × 10-11 × M) / (6.38 × 106)2 ……(all 3 s.f) ......................................M1

M = 5.99 × 1024 kg ........................................................................................... A0 [2]

(b) (i) either GM = ω2r3 or gR2 = ω2r3 ..................................................................C1

either 6.67 × 10-11 x 5.99 × 1024 = ω2 × (2.86 × 107)3

or 9.81 × (6.38 × 106)2 = ω2 × (2.86 × 107)3 ...............................................C1

ω = 1.3 × 10-4 rad s-1 ......................................................................................... A1 [3] (use of r = 2.22 × 107m scores max 2 marks)

(ii) period of orbit = 2π / ω .......................................................................................C1

= 4.8 × 104 s (= 13.4 hours) ....................................................... A1

period for geostationary satellite is 24 hours (= 8.6 × 104 s) ............................. A1 so no ................................................................................................................... A0 [3] (c) satellite can then provide cover at Poles ................................................................... B1 [1]

[Total: 10] 2 (a) sum of kinetic and potential energies of molecules / particles / atoms ......................M1 random (distribution) ................................................................................................. A1 [2]

(b) +∆U: increase in internal energy ............................................................................... B1 +q: heating of / heat supplied to system ................................................................. B1 +w: work done on system ....................................................................................... B1 [3]

(c) (i) work done = p∆V ..............................................................................................C1

= 1.0 × 105 × (2.1 – 1.8) × 10-3 = 30 J ..............................................................................................M1

w = 30 J, q = 0 so ∆U = 30 J .............................................................................. A1 [3] (ii) these three marks were removed, as insufficient data was given in the question.

[Total: 8]



© UCLES 2009

3 (a) straight line through origin ......................................................................................... B1 negative gradient ....................................................................................................... B1 [2]

(b) a = -ω2x and ω = 2πf ..........................................................................................C1

750 = (2πf)2 × 0.3 × 10-3 ...........................................................................................C1 f = 250 Hz ................................................................................................................ A1 [3] (c) straight line between(-0.3,+190) and (+0.3,-190) ...................................................... A2 [2] (allow 1 mark for end of line incorrect by one grid square or line does not extend to +/- 0.3 mm)

[Total: 7] 4 (a) charge / potential ………(ratio must be clear) ........................................................ B1 [1]

(b) potential (at surface of sphere) = Q / 4πε0R ............................................................M1

C = Q / V = 4πε0R ................................................................................................. A0 [1]

(c) (i) C = 4π × 8.85 × 10-12 × 0.63 ..............................................................................C1

= 7.0 × 10-11 .................................................................................................. A1 farad / F ............................................................................................................. B1 [3] (ii) energy = ½CV

2 .................................................................................................C1

0.25 × ½C × (1.2 × 106)2 = ½CV

2 .....................................................................C1

V = 6.0 × 105 V ................................................................................................. A1 [3] (use of 0.75 rather than 0.25, allow max 2 marks)

[Total: 8] 5 (a) (i) concentric circles, anticlockwise ……(minimum 3 circles) ..............................M1 separation of lines increases with distance from wire ........................................ A1 [2] (ii) direction from Y towards X ................................................................................. A1 [1]

(b) (i) flux density at wire Y = (4π × 10-7 × 5.0) / (2π × 2.5 × 10-2) ...............................C1

= 4.0 × 10-5 T ..................................................................C1

force per unit length = BI

= 4.0 × 10-5 × 7.0 .............................................................C1

= 2.8 × 10-4 N .................................................................. A1 [4] (ii) either force depends on product of the currents in the two wires .....................M1 so equal .................................................................................................. A1 or (isolated system so) Newton’s 3rd law applies ..................................... (M1) so equal ................................................................................................(A1) [2]

[Total: 9]



© UCLES 2009

6 (a) (i) e.m.f. induced proportional / equal to ................................................................M1 rate of change of (magnetic) flux (linkage) ......................................................... A1 [2]

(ii) e.m.f. (induced) only when flux is changing / cut ................................................ B1 direct current gives constant flux ........................................................................ B1 [2]

(b) (i) (induced) e.m.f. / current acts in such a direction to produce effects ................. B1 to oppose the change causing it ......................................................................... B1 [2]

(ii) (induced) current in secondary produces magnetic field ....................................M1 opposes (changing) field produced in primary ....................................................M1 so not in phase .................................................................................................. A0 [2]

(c) (i) alternating means that voltage / current is easy to change ................................ B1 [1]

(ii) high voltage means less power / energy loss (during transmission) .................. B1 [1]

[Total: 10]

7 (a) each line corresponds to a (specific) photon energy ................................................. B1 photon emitted when electron changes its energy level ............................................ B1 discrete energy changes so discrete levels ............................................................... B1 [3]

(b) (i) E = hc / λ …(allow ratio ideas) ......................................................................C1

= (6.63 × 10-34 × 3.0 × 108) / (486 × 10-9)

= 4.09 × 10-19 J .............................................................................................. A1 [2]

(ii) four transitions to/from – 5.45 × 10-19 J level ....................................................... B1 all transitions shown from higher to lower energy (level) .................................... B1 [2]

[Total: 7]

8 (a) (constant) probability of decay ..................................................................................M1 per unit time ............................................................................................................... A1 [2] (reference to decay of isotope / mass / sample / nuclide, allow max 1 mark)

(b) either when time = t½, N = ½N0

or ½N0 = N exp(-λt½) t½ ..........................................................................M1

either 2 = exp(λt½)

or ½ = exp(-λt½) ..................................................................................M1

(taking logs), ln2 = 0.693 = λt½ ............................................................................ A1 [3]

(c) A = λN

1.8 × 105 = N × (0.693 / 1.66 × 108) .....................................................................C1

N = 4.3 × 1013

mass = 60 × (N / NA) or 60 × N × u ...................................................................C1

= (60 × 4.3 × 1011) / (6.02 × 1023)

= 4.3 × 10-9 g ................................................................................................. A1 [3]

[Total: 8]



© UCLES 2009

Section B 9 (a) e.g. reduces gain increases bandwidth less distortion greater stability ………(1 each, max 2) ......................................................... B2 [2] (b) gain = – RF / RI = – 8.0 / 4.0 ......................................................................................................M1 numerical value is 2 ................................................................................................... A0 [1] (c) (i) 2, 6 and 7 ........................................................................................................... A1 [1] (ii) e.g. digital-to-analogue converter (allow DAC) adding / mixing signals with ‘weighting’ ....................................................... B1 [1]

[Total: 5] 10 (a) (i) e.m. radiation / photons is produced whenever a charged particle is accelerated .....................................................................................................M1 wavelength depens on magnitude of acceleration ............................................. A1 electrons have a distribution of accelerations ..................................................... A1 so continuous spectrum ...................................................................................... A0 [3] (ii) either when electron loses all its energy in one collision or when energy of electron produces a single photon ................................ B1 [1] (b) (i) parallel beam (in matter) ..................................................................................... B1

I = I0 exp(-µx) ....................................................................................................M1

I, I0, (µ) and x explained ..................................................................................... A1 [3] (ii) either low-energy photons absorbed (much) more readily or low-energy photons (far) less penetrating ............................................... B1 low-energy photons do not contribute to X-ray image ........................................ B1 low energy photons could cause tissue damage ................................................ B1 [3]

[Total: 10]



© UCLES 2009

11 (a) amplitude modulation ……(allow AM) ..................................................................... B1 [1] (b) (i) frequency = 1 / period ......................................................................................C1 = 100 kHz ........................................................................................ A1 [2] (ii) frequency = 10 kHz .......................................................................................... A1 [1] (c) (i) vertical line at 100 kHz ....................................................................................... B1 vertical lines at 90 kHz and 110 kHz .................................................................. B1 lines at 90 kHz and 110 kHz same length and shorter than at 100 kHz ............. B1 [3] (ii) 20 kHz ................................................................................................................ B1 [1]

[Total: 8] 12 (a) (i) base stations ...................................................................................................... B1 [1] (ii) cellular exchange ................................................................................................ B1 [1] (b) base station / X sends / receives signal to / from handset ...................... .................. B1 call relayed to cellular exchange / Y (and on to PSTN) ............................................. B1 computer at cellular exchange monitors signal from base stations ........................... B1 selects base station with strongest signal ................................................................. B1 allocates a (carrier) frequency / time slot for the call ................................................ B1 [5]

[Total: 7]





9702 PHYSICS

9702/51 Paper 51 (Planning, Analysis and Evaluation), maximum raw mark 30






© UCLES 2009

Question 1 Planning (15 marks) Defining the problem (3 marks) P1 Vary V or f [1] P2 Measure f for different V or measure V for different f [1] P3 Keep temperature constant [1] Methods of data collection (5 marks) M1 labelled diagram including source of sound adjacent to the opening e.g. loudspeaker/tuning fork [1] M2 Method of producing sound of different frequencies e.g. several tuning forks or signal generator [1] M3 Method of measuring volume of air – volume of container - volume of water or find total volume of

each different container [1] M4 Method of determining resonant frequency e.g. largest sound heard or displayed [1] M5 Perform experiment in quiet room or avoid other noise [1] Method of analysis (2 marks) A1 Plot a graph of f

2 against 1/V or lg f against lg V or or lg f against lg 1/V [1] A2 Relationship is correct if graph is a straight line through the origin or straight line for log-log graph [1] Safety considerations (1 mark) S Switch off power supply when not in use/ ear defenders for loudspeaker method [1] Additional detail (4 marks) D Relevant points might include [4]

1. Detail on measuring volume – use of measuring cylinder/burette 2. Determination of frequency using oscilloscope/read off tuning fork or signal generator 3. Detailed timebase calculation 4. Detail determining resonance e.g. adding/subtracting small amounts of water/changing signal

generator to create resonance 5. Discussion of container e.g. end correction/shape of mouth of bottle 6. Gradient = k or lg f = -0.5 lg V + 0.5 lg k or lg f = 0.5 lg 1/V + 0.5 lg k 7. Constant amplitude/intensity of source of sound 8. Method to check fundamental frequency.

15 marks can be scored in total.



© UCLES 2009

Question 2 Analysis, conclusions and evaluation (15 marks)

Part Mark Expected Answer Additional Guidance

(a) A1 Gradient =h

y-intercept = lgg

1 or – lg g

Allow log and/or ln

(b) T1 T2

2.467 or 2.4669 3.00 or 2.996 2.481 or 2.4814 2.93 or 2.934 2.496 or 2.4955 2.88 or 2.881 2.509 or 2.5092 2.83 or 2.833 2.522 or 2.5224 2.79 or 2.785

T1 for lg T T2 for lg R Allow mixture of dp.

U1 ± 0.004 to ± 0.007 Allow more than one significant figure.

(c) (i) G1 Five points plotted correctly Must be within half a small square. Use transparency. Ecf allowed from table.

U2 Error bars in lg R plotted correctly Check first and last point. Must be accurate within half a small square. Allow ecf from (b)

(c) (ii) G2 Line of best fit There must at least four trend plots with a reasonable balance of points about the line. Allow ecf from points plotted incorrectly. Examiner judgement.

G3 Worst acceptable straight line. Steepest or shallowest possible line.

Line should be clearly labelled or dashed. Should pass from top of top error bar to bottom of bottom error bar or bottom of top error bar to top of bottom error bar. Mark scored only if error bars are plotted. Allow ecf from (b) and (c) (i)

(c) (iii) C1 Gradient of best fit line The triangle used should be greater than half the length of the drawn line. Check the read offs. Work to half a small square. Do not penalise POT or sign of gradient.

U3 Uncertainty in gradient Method of determining absolute uncertainty Difference in worst gradient and gradient.

(c) (iv) C2 y-intercept Gradient must be used. Check substitution into c = y – mx. Allow ecf from (c) (iii). If gradient negative then y-intercept should be about 11-13. If gradient positive then y-intercept should be about -4 or -5.

U4 Uncertainty in y-intercept Method of determining absolute uncertainty Difference in worst y-intercept and y-intercept. Do not allow ecf from false origin read-off. Allow ecf from (c) (iv)



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(d) C3 g = 1/10y-intercept = 10 – y-intercept

y-intercept must be used. g should be about 10-13 Allow ecf from (c) (iv). If FO or positive gradient used then g should be about 10-4.

C4 h = candidate’s gradient value Answer must be negative and given to 2 or 3 sf.

U5 Method for uncertainty in g and uncertainty in h.

Expect to see difference in values for g. Uncertainty in h must be the same as the gradient.

[Total: 15]

Uncertainties in Question 2 (c) (iii) Gradient [U3]

1. Uncertainty = gradient of line of best fit – gradient of worst acceptable line 2. Uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)

(c) (iv) y-intercept [U4]

1. Uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line 2. Uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)

(d) [U5]

1. Uncertainty = 10 – best y-intercept - 10 – worst y-intercept





9702 PHYSICS

9702/52 Paper 52 (Planning, Analysis and Evaluation), maximum raw mark 30






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Question 1 Planning (15 marks) Defining the problem (3 marks) P1 Vary d and measure y or d is the independent variable and y is the dependent variable [1] P2 Keep current constant [1] P3 Keep length of wire constant [1] Methods of data collection (5 marks) M1 Diagram showing ruler positioned and power supply connected to wire or diagram showing initial

and final marks on screen and power supply connected to wire [1] M2 Use of ammeter to check current – penalise incorrect circuit diagrams [1] M3 Measurement of d using micrometer [1] M4 Allow time for displacement of wire to stabilise [1] M5 Detail on measuring y; final reading - initial reading [1] Method of analysis (2 marks) A1 Plot a graph of log y against log d [1] A2 q = gradient [1] Safety considerations (1 mark) S Safety related to hot wire – use of gloves, wait to cool down/switch off before changing wire, do

not touch hot wire [1] Additional detail (4 marks) D Relevant points might include [4]

1. Use of vernier scale to measure y /well described optical method/use of set square 2. Method for keeping current constant e.g. use of rheostat 3. Check starting position for y for same wire 4. lg y = q lg d + lg p 5. Repeat measurements of d at different points along the wire and determine average 6. Control of additional variables e.g. separation between supports, room temperature 7. Use of protective resistor (either labelled or explained).

15 marks can be scored in total.



© UCLES 2009

Question 2 Analysis, conclusions and evaluation (15 marks)

Part Mark Expected Answer Additional Guidance

(a) A1

g

2

(b) T1 t2 / s2 Column heading: allow t2 (s2) or t2in s2 Do not allow (t / s)2

T2 0.12 or 0.123 0.15 or 0.152 0.18 or 0.185 0.20 or 0.203 0.24 or 0.240 0.27 0r 0.270

Must be to two or three significant figures. A mixture of 2sf and 3sf is allowed.

U1 ± 0.007 to ± 0.010

(allow ± 0.011)

Allow more than one significant figure.

(c) (i) G1 Six points plotted correctly. Must be within half a small square. Use transparency. Ecf allowed from table.

U2 Error bars in t2 plotted correctly. Check first and last point. Must be accurate within half a small square.

(c) (ii) G2 Line of best fit. If points are plotted correctly then lower end of line should pass between (0.60, 0.116) and (0.60, 0.123) and upper end of line should pass between (1.30, 0.268) and (1.30, 0.272). Allow ecf from points plotted incorrectly – examiner judgement. Five good trend plots needed.

G3 Worst acceptable straight line. Steepest or shallowest possible line that passes through all the error bars.

Line should be clearly labelled or dashed. Should pass from top of top error bar to bottom of bottom error bar or bottom of top error bar to top of bottom error bar. Mark scored only if error bars are plotted.

(c) (iii) C1 Gradient of best fit line. The triangle used should be greater than half the length of the drawn line. Check the read offs. If incorrect circle and write in correct value. Work to half a small square. Do not penalise POT.

U3 Uncertainty in gradient. Method of determining absolute error Difference in worst gradient and gradient.

(d) C2 g = 2/gradient

Gradient must be used. Allow ecf from (c) (iii)

U4 Method of determining uncertainty in g.

Uses worst gradient and finds difference. Allow fractional error methods. Do not check calculation.



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C3 Unit of g: m s-2 Accept N kg-1

(e) (i) C4 21.9 – 23.5 Answer must be in range given to 2 or 3sf. Allow 22 or 23.

(e) (ii) U5 Method for percentage uncertainty in b.

Calculates percentage uncertainty in t2 and adds to percentage uncertainty in gradient or g. Allow ecf from (c) (iii) and/or (d).

[Total: 15]

Uncertainties in Question 2 (c) (iii) Gradient [U3]

1. Uncertainty = gradient of line of best fit – gradient of worst acceptable line 2. Uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)

(d) g [U4]

1. Uncertainty = g from gradient - g from worst acceptable line

2. gradient

gradient

g

g ∆∆=

(e) b [U5]

1. Substitution method to find worst acceptable g using either largest g × 2.222 or smallest g × 2.202 then determines percentage uncertainty

2. 0.9% + percentage uncertainty in gradient or percentage uncertainty in g

3. 1002100 ×

∆+

∆=×

∆

t

t

g

g

b

b

Education

9702 w09 ms_all