9702 s06 ms_all

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS

GCE Advanced/Advanced Subsidiary Level

MARK SCHEME for the May/June 2006 question paper

9702 PHYSICS

9702/01 Paper 1 – Multiple Choice

Maximum raw mark 40

This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which Examiners were initially instructed to award marks. Mark schemes must be read in conjunction with the question papers and the Report on the

Examination. The minimum marks in these components needed for various grades were previously published with these mark schemes, but are now instead included in the Report on the Examination for this session.

• CIE will not enter into discussion or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the May/June 2006 question papers for most IGCSE and GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Mark Scheme Syllabus Paper

GCE A/AS LEVEL – May/June 2006 9702 01

© University of Cambridge International Examinations 2006

Question Number

Key Question Number

Key

1 B 21 C

2 D 22 B

3 D 23 A

4 D 24 B

5 C 25 B

6 C 26 C

7 A 27 B

8 C 28 D

9 D 29 B

10 B 30 A

11 B 31 C

12 A 32 B

13 D 33 B

14 B 34 A

15 A 35 B

16 D 36 C

17 D 37 B

18 A 38 A

19 C 39 D

20 A 40 B


GCE Advanced Level


9702 PHYSICS

9702/02 Paper 2

Maximum raw mark 60

This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which Examiners were initially instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began. Any substantial changes to the mark scheme that arose from these discussions will be recorded in the published Report on the Examination. All Examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes must be read in conjunction with the question papers and the Report on the




GCE A Level – May/June 2006 9702 02


1 (a) kg m s–2 B1 [1]

(b) kg m–1 s–1 B1 [1]

(c) (i) v2 = 2gs

= 2 × 9.8 × 4.5 C1 v = 9.4 m s–1 A1 [2]

(ii) either

F (= 3.2 × 10–4 × 1.2 × 10–2 × 9.4) = 3.6 × 10–5 N M1

weight of sphere (= mg = 15 × 10–3 × 9.8) = 0.15 N M1

3.6 × 10–5 << 0.15, so justified A1 [3] or mg = crvT (M1)

terminal speed = 3.8 × 104 m s–1 (M1)

9.4 << 3.8 × 104, so justified (A1)

2 (a) (i) point at which whole weight of body M1 may be considered to act A1 [2]

(ii) sum of forces in any direction is zero B1 sum of moments about any point is zero B1 [2]

(b) either: T and W have zero moment about P M1 so F must have zero moment, i.e. pass through P A1 [2] or: if all pass through P, distance from P is zero for all forces (M1) so sum of moments about P is zero (A1)

(c) (i) Fcosα = Tcosβ B1 [1]

(ii) W = Fsinα + Tsinβ B1 [1]

(iii) 2W = 3Tsinβ B1 [1]

3 (a) sum of (random) kinetic and potential energies M1 of the atoms/molecules of the substance A1 [2]

(b) (i) potential energy unchanged as atoms remain in same positions M1 allow ‘reduced because atoms slightly closer together’ vibrational kinetic energy reduced because temperature lower M1 so internal energy less A1 [3]

(ii) potential energy increases because separation increases M1 kinetic energy unchanged because temperature unchanged M1 so internal energy increases A1 [3]

4 (a) mass per unit volume (ratio idea must be clear, not units) B1 [1]

(b) (i) pressure is same at the surface of mercury because at same horizontal level B1 [1]

(ii) hρg is same for both B1

53 × 10–2 × 1.0 × 103 × g = 71 × 10–2 × ρ × g C1

ρ = 7.5 × 102 kg m–3 A1 [3]


GCE A Level – May/June 2006 9702 02


5 (a) no hysteresis loop/no permanent deformation M1 (do not allow ‘force proportional to extension’) so elastic change A0 [1]

(b) work done = area under graph line OR average force × distance B1 = ½Fx ½(F2 + F1)(x2 – x1) A1 F = kx, so work done = = ½kx2 ½k(x2 + x1)(x2 – x1) A1 work done = ½k(x2

2 – x1

2) A0 [3] (c) gain in energy of trolley = ½k(0.0602 – 0.0452) + ½k(0.0302 – 0.0452) C1 = 0.36 J C1

kinetic energy = ½ × 0.85 × v2 = 0.36 C1 v = 0.92 m s–1 A1 [4] 6 (a) (i) correct shape drawn B1 [1] (ii) two nodes marked correctly B1 [1]

(b) ½λ = 0.324 m C1

v = fλ C1

= 512 × 2 × 0.324 = 332 m s–1 A1 [3]

(c) ¼λ = 16.2 cm C1 either antinode is 0.5 cm above top of tube or antinode is 16.2 cm above water surface A1 [2] 7 (a) lamp C M1 lamp is shorted A1 [2] (b) shorted lamp A would cause damage to the supply/lamps /blow fuse in supply B1 [1]

(c) 15 Ω B1 [1] (d) (i) V = I R C1

R = 30 Ω A1 [2] (ii) P = VI or I2R or V2 / R C1 P = 1.2 W A1 [2] (e) filament is cold when measuring with ohm-meter in (b) B1 resistance of filament rises as temperature rises B1 [2] 8 (a) nucleus emits M1

α- or β- particles and/or γ-rays A1 [2] (b) decay unaffected by environmental changes M1 such as temperature, pressure etc. (one e.g. is sufficient) A1 [2] (c) constant probability of decay (per unit time) of a nucleus B1 cannot predict which particular nucleus will decay next B1 [2]


GCE Advanced Level and GCE Advanced Subsidiary Level


9702 PHYSICS

9702/03 Paper 3

Maximum raw mark 25





GCE A/AS Level – May/June 2006 9702 03


(c) (i) First value of θ (less than 15°) [1]

(iv) Absolute uncertainty = 1 or 2 mm (1 mark). [2] Percentage uncertainty in first value of d (i.e. ratio correct) (1 mark). (v) Source of error in d: [1] meniscus effects, shape of bottle leads to parallax problems (vi) Source of error in θ : [1]

parallax effects, difficult to keep head still, difficult to move head with rule, difficult to judge point of toppling (d) Readings [6] 9 sets of readings scores 6 marks; 8 sets, 5 marks; 7 sets, 4 marks etc. Less than 4 sets scores zero. Help given by Supervisor, then –1. Excessive help then –2. Repeated readings [1]

Reasonable interval between values of d (Y 1.0 cm) [1] Quality of results [2] Judge by scatter of points about the curve (one mark) Curve for small d values steeper than curve for large d values (one mark) Column headings [1] Apply to d. The heading must contain a quantity and a unit. Consistency [1] Apply to d. Values of d must be given to the nearest millimetre. (e) Axes [1] Scales must be such that the plotted points occupy at least half the graph grid in both the x and y directions. Scales must be labelled. Do not allow awkward scales. Plotting of points [1] Check a suspect plot. Circle and tick if correct. If incorrect, show correct position with arrow, and –1. Work to half a small square. Curve of best fit [1] There must be a reasonable balance of points about the curve. (f) Determination of max value [1] (g) Second curve; all values of θ larger than before (1 mark) [2]

Similar shape (1 mark) (h) Suggestions for improvements (one mark each) [2] [Total marks: 25]




9702 PHYSICS

9702/04 Paper 4

Maximum raw mark 60







1 (a) centripetal force is provided by gravitational force B1 mv2 / r = GMm / r2 B1

hence v = √(GM / r) A0 [2] (b) (i) EK (= ½mv2) = GMm / 2r B1 [1] (ii) EP = - GMm / r B1 [1] (iii) ET = - GMm / r + GMm / 2r C1 = - GMm / 2r. A1 [2] (c) (i) if ET decreases then - GMm / 2r becomes more negative or GMm / 2r becomes larger M1 so r decreases A1 [2] (ii) EK = GMm / 2r and r decreases M1 so (EK and) v increases A1 [2] 2 (a) e.g. fixed mass/ amount of gas ideal gas (any two, 1 each) B2 [2] (b) (i) n = pV / RT C1

= (2.5 × 107 × 4.00 × 104 x 10-6) / (8.31 × 290) C1 = 415 mol A1 [3]

(ii) volume of gas at 1.85 × 105 Pa = (2.5 × 107 × 4.00 × 104) / (1.85 × 105)

= 5.41 × 106 cm3 C1

so, 5.41 × 106 = 4.00 × 104 + 7.24 × 103 N C1 N = 741 A1 [3] (answer 740 or fails to allow for gas in cylinder, max 2/3) 3 (a) gradient of graph is (a measure of) the sensitivity M1 the gradient varies with temperature A1 [2]

(b) 2040 ± 20 Ω corresponds to 15.0 ± 0.2 °C C1

T / K = T / °C + 273.15 (allow 273.2) C1 temperature is 288.2 K A1 [3] 4 (a) (i) 1.0 B1 [1] (ii) 40 Hz B1 [1]

(b) (i) speed = 2πfa C1

= 2π × 40 × 42 × 10-3 = 10.6 m s-1 A1 [2]

(ii) acceleration = 4π2f2 a C1

= (80π)2 × 42 × 10-3 = 2650 m s-2 A1 [2] (c) (i) S marked correctly (on ‘horizontal line through centre of wheel) B1 (ii) A marked correctly (on ‘vertical line’ through centre of wheel) B1 [2]




5 (a) (i) force per unit positive charge (ratio idea essential) B1 [1]

(ii) E = Q / 4πε0r2 M1

ε0 being the permittivity of free space A1 [2]

(b) (i) 2.0 × 106 = Q / (4π × 8.85 × 10-12 × 0.352) C1

Q = 2.7 × 10-5 C A1 [2]

(ii) V = (2.7 × 10-5) / (4π × 8.85 × 10-12 × 0.35) C1

= 7.0 × 105 V A1 [2] (c) electrons are stripped off the atoms B1 electrons and positive ions move in opposite directions, (giving rise to a current) B1 [2] 6 (a) (i) arrow B in correct direction (down the page) B1 (ii) arrow F in correct direction (towards Y) B1 [2] (b) (i) When two bodies interact, force on one body is equal but opposite in direction to force on the other body. B1 [1] (ii) direction opposite to that in (a)(ii) B1 [1] (c) suggested reasonable values of I and d B1 mention of expression F = BIL B1 force between wires is small M1 compared to weight of wire A1 [4] 7 (a) ‘uniform’ distribution B1 [1] (b) concentric rings B1 [1] (c) higher speed, more momentum M1

λ = h / p M1

so λ decreases and ring diameter decreases A1 [3] 8 (a) arrow labelled E pointing down the page B1 [1] (b) (i) Bqv = qE M1 forces are independent of mass and charge ‘cancels’ M1 so no deviation A1 [3] (ii) magnetic force > electric force M1 so deflects M1 ‘downwards’ A1 [3]




9702 PHYSICS

9702/05 Paper 5

Maximum mark 30





GCE A LEVEL – May/June 2006 9702 05


1 (b) (ii) Expect to see x read to nearest mm and then: Use connector with sharp edge [1] Position the wire on top of the scale on the rule to reduce parallax error (c) Readings [3] Write the number of readings as a ringed total by the results table.

6 sets of readings scores 1 mark. Check a value for lg(x/m) and a value for lg(IA). Underline checked values. Ignore small rounding errors. Tick if correct; 1 mark each. If incorrect then write in correct value. Excessive alterations in the table then –1. If minor help is given with the circuit, then –1. If excessive help is given then –2. Please indicate when help has been given to a candidate by writing SR at the top of the front page of the candidate's script. Also, please indicate the type of help that has been given by writing a brief comment by the table of results.

Column headings. Allow lg(x/m) and lg(I/A). Allow POT errors. [1]

There must be some distinguishing mark between the quantity and its unit. Please or underline each correct column heading to show that it has been seen.

Consistency of raw readings in the table of results [1]

Apply to x and I only. Expect to see I to either 0.01 A or 0.1 A. Expect to see all the values of x given to the nearest millimetre. Indicate using a vertical line down each column of raw readings to show it has been seen and a C if correct.

(d) (i) Axes [1]

Each axis must be labelled with a quantity. Allow lg(x), lgx, lg(x/m) but not lgx/m. Scales must be such that the plotted points occupy more than half the graph grid in both the x and y directions. Do not allow more than 3 large squares between scale markings. Do not allow awkward scales (e.g. 3:10, 6:10, 7:10, etc.).

Plotting of points [1]

Count the number of plots on the grid and write this value by the line and ring it. Do not allow plots in the margin area. The number of plots must correspond to the number of observations (at least 6). Do not award this mark if the number of plots is less than the number of observations. Check one suspect plot. Circle this plot. Tick if correct. If incorrect then mark the correct position with a small cross and use an arrow to indicate where the plot should have been. Allow errors less than half a small square.




(ii) Line of best fit [1] There must be a reasonable balance of points about the line of best fit. If one of the plots is a long way from the trend of the other plots then allow this plot to be ignored when the line is drawn. One mark can be awarded if the line of best fit is ‘reasonable’, but not quite right.

(iii) Measurement of gradient [1]

The hypotenuse of the triangle must be greater than half the length of the drawn line. If read-offs are inaccurate by half a small square, or more, then zero. Please indicate the vertices of the triangle used by labelling with ∆. One mark for the read-offs. The value must be negative. Value should be around -1.0. ∆x/∆y scores zero. Ignore units.

y-intercept. Expect –0.8. Ignore units. [1]

Check the read-off. Value must be within half a square. Accept correct substitution from a point on the line into y = mx + c. Allow ecf from (d)(iii) gradient for m.

(e) lg x = n lg I + lg k [1] This can be implied from the working. Value for n (from gradient). Allow ecf from (d)(iii) gradient. n = –1.0. Ignore unit. [1] Value for k (from 10y-intercept). Allow ecf from (d)(iii) y-intercept. k = 0.158. Ignore unit. [1] Working must be checked. (f) (i) Value of diameter of wire (± 0.02 mm of SV). (36 swg, diameter = 0.18 mm) [1] (ii) Cross-sectional area correct. (36 swg, Area = 2.5 x 10 –8 m2) [1] (iii) Percentage uncertainty in area [2] One mark for percentage uncertainty in r. One mark for % uncertainty x 2. (g) Correct value of ρ. No POT error allowed. Expect to see ρ = 4.7 x 10–7 Ωm [1] Check the substitution and consistency of units. (h) k = 4 x previous value. Allow ecf (g). Expect to see k = 1.9 x 10–6 Ωm [1] [Total: 20 marks]




Question 2 A1 Diagram of arrangement (light source/mesh/screen or collimator/mesh/telescope) [1] A2 Fringes or dots shown on screen. May be shown on diagram. [1] A3 Some sensible discussion of coherence [1] A4 Use of laser or single slit (and lens) in collimator [1] B1 Measurements: distance from mesh to screen and separation between fringes [1] OR measure an angle from the spectrometer table B2 n = 1; find separation between central fringe and first bright fringe [1] OR measure angle between central bright beam and first order beam using scale on table B3 Use of θλ sin=n to find d. n must be clearly identified [1]

C Any safety precaution [1] e.g. use goggles/do not look directly into laser beam/cover over sodium lamp do not touch the bulb D Any good/further detail [2] Examples of creditworthy points might be: Take readings with mesh in different positions to average d Sketch/suggestion of two-dimensional array of dots on screen Laser + mesh + screen all at same height λ = 589 nm for sodium lamp or about λ = 630 nm for He/Ne laser/semiconductor laser

D of the order of 1 m to 4 m (laser method) Measure 2θ and divide by two to reduce uncertainty in θ

Repeat experiment with 2nd order (3rd order etc.) beams Detail relating to setup/use of spectrometer Allow other valid points. [Total: 10 marks]


GCE Advanced Level


9702 PHYSICS

9702/06 Paper 6

Maximum raw mark 40





GCE A – May/June 2006 9702 06


Option A - Astrophysics and Cosmology 1 Planet: almost circular orbits B1 all in nearly the same plane B1 Comet: highly elliptical orbits B1 in many different planes B1 [4] 2 (a) (mean) density M1 of matter in the Universe A1 [2] (b) (i) symmetrical curve below given line M1 touching given line at ‘present time’ A1 [2] (ii) H0 not known with any certainty B1 mass of matter in the Universe not known B1 extent of Universe unknown B1 [3]

(allow 1 of the last 2 marks for ρ0 not known) 3 1 light-year = 0.306 pc (allow 0.3 pc) C1

1.3 × 1010 light-years = 3.98 × 103 Mpc C1 v = H0d C1

speed = 60 × 3.98 × 103 = 2.39 × 105 km s-1

ratio = (2.39 × 105 x 103)/(3 .0 × 108) = 0.8 A1 [4] 4 e.g. vast expense (M1) money could be spent on humanitarian aid (A1) observations possible that cannot be made on Earth (M1) since atmosphere limits observations (A1) technological/scientific developments on Earth (M1) greater understanding of Universe (M1) leads to ‘spin off’ benefits for individuals (A1) Any sensible comments, 1 each to max 5 B5 [5] Option F - The Physics of Fluids 5 (a) conservation of volume/mass/density or incompressible B1 [1] (b) conservation of energy B1 [1] 6 (a) air near jet is moving at speed OR water in jet is moving at speed B1 higher speed air has a lower OR high-speed water has lower pressure B1 pressure (because) air is dragged along by OR air is drawn into water jet B1 water jet air (outside pump) is not moving OR loss of air reduces pressure B1 [4] (b) (i) air/water in pump has a higher speed M1 so greater pressure difference A1 [2]


GCE A – May/June 2006 9702 06


(ii) no change in speed of air OR reference to greater ρ in Bernoulli eqn M1 so no change in pressure OR greater pressure difference A1 [2] difference (allow any logical argument based on liquid causing more/less drag on air) 7 (a) eddy currents have kinetic energy OR cause extra drag M1 eddy currents caused by movement of the car OR energy required to overcome drag A1 extra energy (of eddy currents) is derived from car’s fuel A1 [3]

(b) (i) power = force × speed B1

so power = ½CDAρv2 × v and A and ρ are constants B1 [2]

(ii) 84 × 103 = ½ × 0.34 × 1.8 × 1.1 × vmax

3 C1 vmax = 63 m s-1 A1 [2]

(iii) P = ½ × 0.34 × 1.8 × 1.1 × (63 + 9)3 C1 P = 126 kW C1 ratio = 126 / 84 = 1.5 A1 [3] Option M - Medical Physics 8 (a) alternating voltage B1 applied across (piezo-electric) crystal B1 causes crystal to vibrate B1 crystal dimensions such as to give resonance (in US range) B1 [4] (b) wavelength at 1 MHz is shorter B1

so greater detail is possible B1 [2] 9 e.g. used as a scalpel (1) further detail: causes (explosive) vaporisation of intracellular water (1) CO2 laser (1) IR radiation strongly absorbed by water (1) laser beam focused to give high power density (1) no/very little bleeding (1) accurate guidance (1) e.g. repair of retina (1) further detail: focused laser beam onto retina (1) melts tissue and forms a weld (1) (pulsed) ruby or argon laser (1) any two examples: named (1) plus further detail (2) B6 [6] (allow up to two marks for each diagnostic technique) 10 (a) minimum intensity (of sound) detected M1 where intensity = (sound) power per unit area at a stated frequency A1

value is 1 × 10-12 W m-2 B1

at 3 kHz (allow 2 kHz → 3 kHz) B1 [4]


GCE A – May/June 2006 9702 06


(b) (i) intensity = (0.14 × 10-6)/(54 × 10-6) = 2.6 × 10-3 W m-2 C1

IL = 10 lg (2.6 × 10-3)/(1 × 10-12) C1 = 94 dB A1 [3] (ii) comment e.g. would be perceived as being loud could cause tinnitus over a short period of time could cause deafness over a long period of time higher level than is acceptable in the workplace any appropriate comment, 1 mark B1 [1] Option P - Environmental Physics 11 (a) at times of low usage of electrical power B1 water pumped from low-level to high-level reservoir B1 at times of high/sudden demand for electrical power B1 water released to pass through turbines B1 [4]

(b) electrical energy generated = 78 × 106 × 4.0 × 3600 = 1.12 × 1012 J C1

energy to be stored = (1.12 × 1012)/0.75 = 1.5 × 1012 J C1

1.5 × 1012 = ρVgh C1

= 1.0 × 103 × V × 9.8 × 95

V = 1.6 × 106 m3 A1 [4] 12 (a) law: it is impossible to convert all of a given amount of thermal energy into work B1 (that is) W < QH B1 (QH – W ) is energy rejected at temperature TL B1 [3] (b) W/QH = 1 – TL/TH B1 [1] (c) efficiency = 1 – 313/393 C1 = 0.20 A1 [2] 13 (a) (i) e.g. industry setting up people preparing to go to work starting to cook breakfast (allow any two sensible suggestions, 1 each) B2 [2] (ii) e.g. change in temperature with use of heaters/air conditioning holiday or workday with more power used by industry when not on holiday (allow any two sensible suggestions, 1 each) B2 [2]


GCE A – May/June 2006 9702 06


(b) (i) sudden increase in demand (as appliances are used) B1 (ii) increased demand in the afternoon B1 [2] (allow any two sensible suggestions in (i) and (ii)) Option T - Telecommunications 14 (a) (instantaneous) displacement of information signal M1 determines the frequency of the carrier wave A1 [2] (b) (i) 12 V B1 [1] (ii) 650 kHz B1 [1] (iii) 550 kHz B1 [1] (iv) 3000 B1 [1] 15 (a) analogue-to-digital converter (do not allow ADC) B1 [1] (b) controls the time at which samples are taken B1 [1] (c) enables higher frequency components in signal to be ‘detected’ B1 [1] 16 (a) electromagnetic shielding for the inner conductor B1 the braid is earthed B1 [2] (b) increased bandwidth means more information can be carried B1 so more calls can be transmitted simultaneously B1 fewer links are required B1 [3] 17 (a) e.g. cross-talk/cross-linking interference/picking up atmospherics/picking up man-made radiation white noise associated with vibrating atoms (any two, 1 each) B2 [2] (b) (i) number of dB = 10 lg (P2/P1)

35 = 10 lg (P/7.6 × 10-6) C1 P = 0.024 W A1 [2] (ii) number of dB = 10 lg (2.6/0.024) = 20.3 C1 length = 20.3/5.8 = 3.5 km A1 [2]

9702 s06 ms_all

Education

Transcript of 9702 s06 ms_all