9702 s13 ms_all

CAMBRIDGE INTERNATIONAL EXAMINATIONS

GCE Advanced Subsidiary Level and GCE Advanced Level

MARK SCHEME for the May/June 2013 series

9702 PHYSICS

9702/11 Paper 1 (Multiple Choice), maximum raw mark 40

Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2013 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.

Mark Scheme Syllabus Paper

GCE A LEVEL – May/June 2013 9702 11

© Cambridge International Examinations 2013

Question Number

Key Question Number

Key

1 D 21 D

2 C 22 A

3 C 23 A

4 D 24 C

5 D 25 A

6 D 26 C

7 B 27 A

8 D 28 C

9 A 29 B

10 D 30 B

11 D 31 D

12 A 32 A

13 C 33 A

14 B 34 A

15 A 35 B

16 C 36 D

17 B 37 D

18 B 38 C

19 B 39 A

20 B 40 B




9702 PHYSICS






Question Number

Key Question Number

Key

1 B 21 C

2 B 22 B

3 A 23 B

4 D 24 D

5 C 25 D

6 B 26 B

7 D 27 C

8 B 28 B

9 D 29 C

10 B 30 D

11 D 31 A

12 A 32 C

13 B 33 A

14 D 34 A

15 B 35 D

16 B 36 D

17 C 37 C

18 A 38 C

19 C 39 B

20 D 40 B




9702 PHYSICS






Question Number

Key Question Number

Key

1 B 21 D

2 A 22 A

3 C 23 C

4 A 24 C

5 D 25 C

6 B 26 B

7 B 27 D

8 C 28 D

9 B 29 C

10 B 30 D

11 A 31 D

12 B 32 C

13 D 33 C

14 A 34 B

15 A 35 C

16 A 36 B

17 C 37 C

18 A 38 C

19 C 39 D

20 A 40 B




9702 PHYSICS

9702/21 Paper 2 (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.



GCE AS/A LEVEL – May/June 2013 9702 21


1 (a) the wire returns to its original length (not ‘shape’) M1 when the load is removed A1 [2] (b) energy: N m / kg m2

s–2 and volume m3 C1 energy / volume: kg m2

s–2 / m3 M1 energy / volume: kg m–1

s–2 A0 [2]

(c) ε has no units B1 E: kg m s–2 m–2 M1 units of RHS: kg m–1

s–2 = LHS units / satisfactory conclusion to show C has no units A1 [3] 2 (a) mass is the property of a body resisting changes in motion / quantity of matter in a body / measure of inertia to changes in motion B1 weight is the force due to the gravitational field/force due to gravity or gravitational force B1 [2] Allow 1/2 for ‘mass is scalar weight is vector’ (b) (i) arrow vertically down through O B1

tension forces in correct direction on rope B1 [2]

(ii) 1. weight = mg = 4.9 × 9.81 (= 48.07) C1

69 sin θ = mg C1

θ = 44.(1)° scale drawing allow ± 2° A1 [3] use of cos or tan 1/3 only

2. T = 69 cos θ C1 = 49.6 / 50 N scale drawing 50 ±2 (2/2) 50 ±4 (1/2) A1 [2]

correct answers obtained using scale diagram or triangle of forces will score full marks cos in 1. then sin in 2. (2/2)

3 (a) loss in potential energy due to decrease in height (as P.E. = mgh) (B1) gain in kinetic energy due to increase in speed (as K.E. = ½ mv2) (B1) special case ‘as PE decreases KE increases’ (1/2) increase in thermal energy due to work done against air resistance (B1) loss in P.E. equals gain in K.E. and thermal energy (B1) max. 3 [3]




(b) (i) kinetic energy = ½ mv2 C1

= ½ × 0.150 × (25)2 C1 = 46.875 = 47 J A1 [3]

(ii) 1. potential energy (= mgh) = 0.150 × 9.81 × 21 C1 loss = KE – mgh = 46.875 – (30.9) C1 = 15.97 = 16 J A1 [3] 2. work done = 16 J

work done = force × distance C1 F = 16 / 21 = 0.76 N A1 [2] 4 (a) pressure = force / area (normal to force) A1 [1] (b) molecules/atoms/particles in (constant) random/haphazard motion B1 molecules have a change in momentum when they collide with the walls M1 (force exerted on molecules) therefore force on the walls A1 reference to average force from many molecules/many collisions A1 [4] (c) elastic collision when kinetic energy conserved B1 temperature constant for gas B1 [2] 5 (a) waves overlap / meet / superpose (B1)

coherence / constant phase difference (not constant λ or frequency) (B1)

path difference = 0, λ, 2λ or phase difference = 0, 2π, 4π (B1) same direction of polarisation/unpolarised (B1) max. 3 [3]

(b) λ = v / f C1

f = 12 × 109 Hz C1

λ = 3 × 108 / 12 × 109 (any subject) M1 = 0.025 m A0 [3] (c) maximum at P B1 several minima or maxima between O and P B1 5 maxima / 6 minima between O and P or 7 maxima / 6 minima including O and P B1 [3] (d) slits made narrower B1 slits put closer together B1 [2] (not just ‘make slits smaller’) Allow tilting the slits M1 and explanation of axes of rotation A1




6 (a) (i) chemical to electrical B1 [1] (ii) electrical to thermal / heat or heat and light B1 [1]

(b) (i) (PB =) EI or I2(R1 + R2) A1 [1]

(ii) (PR =) I2R1 A1 [1]

(c) R = ρl / A or clear from the following equation B1

ratio = I2R1 / I2R2 =

( ) ( )212

2

of resistance 8 has or 22

RRd

d×

π/

π/

l

l

ρ

ρ C1

= 8 or 8:1 A1 [3] (d) P = V2 / R or E2 / R C1 (V or E the same) hence ratio is 1/8 or 1:8 = 0.125 (allow ecf from (c)) A1 [2] 7 (a) the majority/most went straight through or were deviated by small angles B1 a very small proportion/a few were deviated by large angles B1

small angles described as < 10° and large angles described as >90° B1 [3] (b) most of the atom is empty space/nucleus very small compared with atom B1 mass and charge concentrated in (very small) nucleus B1 correct links made with statements in (a) B1 [3]




9702 PHYSICS







1 (a) power = energy / time C1 = (force × distance / time) = kg m2 s–2 / s C1 = kg m2 s–3 A1 [3]

(b) (i) units of L2: m2 and units of ρ : kg m–3 and units of v3: m3 s–3 C1

(C = P / L2 ρ v3) hence units of C: kg m2 s–3 m–2 kg–1 m3 m–3 s3 or any correct statement of component units M1 argument /discussion / cancelling leading to C having no units A1 [3]

(ii) power available from wind = 3.5 × 105 × 100 / 55 (= 6.36 × 105) C1 v3 = 3.5 × 105 × 100 / (55 × 0.931 × (25)2 × 1.3) C1 v = 9.4 m s–1 A1 [3] (iii) not all kinetic energy of wind converted to kinetic energy of blades B1 generator / conversion to electrical energy not 100% efficient / heat produced in generator / bearings etc B1 [2] (there must be cause of loss and where located) 2 (a) force = rate of change of momentum A1 [1] (b) (i) horizontal line on graph from t = 0 to t about 2.0 s ± ½ square, a > 0 M1 horizontal line at 3.5 on graph from 0 to 2 s A1 vertical line at t = 2.0 s to a = 0 or sharp step without a line B1 horizontal line from t = 2 s to t = 4 s with a = 0 B1 [4] (ii) straight line and positive gradient M1 starting at (0,0) A1

finishing at (2,16.8) A1 horizontal line from 16.8 M1 from 2.0 to 4.0 A1 [5] 3 (a) the point where (all) the weight (of the body) M1

is considered / seems to act A1 [2]

(b) (i) vertical component of T (= 30 cos 40°) = 23 N A1 [1] (ii) the sum of the clockwise moments about a point equals the sum of the anticlockwise moments (about the same point) B1 [1] (iii) (moments about A): 23 × 1.2 (27.58) M1

= 8.5 × 0.60 + 1.2 × W M1 working to show W = 19 or answer of 18.73 (N) A1 [3]

(iv) (M = W / g = 18.73 / 9.81 =) 1.9(09) kg A1 [1]




(c) (for equilibrium) resultant force (and moment) = 0 B1 upward force does not equal downward force / horizontal component of T not balanced by forces shown B1 [2]

4 (a) apparatus: cell with particles e.g. smoke (container must be closed) B1

diagram showing suitable arrangement with light illumination and microscope B1 [2]

(b) specks / flashes of light M1 in random motion A1 [2] (c) cannot see what is causing smoke to move hence molecules smaller than smoke particles (B1)

continuous motion of smoke particles implies continuous motion of molecules (B1) random motion of particles implies random motion of molecules (B1) max. 2 [2]

5 (a) (i) v = fλ C1

λ = 40 / 50 = 0.8(0) m A1 [2] (ii) waves (travel along string and) reflect at Q / wall / fixed end B1 incident and reflected waves interfere / superpose B1 [2] (b) (i) nodes labelled at P, Q and the two points at zero displacement B1 antinodes labelled at the three points of maximum displacement B1 [2]

(ii) (1.5λ for PQ hence PQ = 0.8 × 1.5) = 1.2 m A1 [1] (iii) T = 1 / f = 1/50 = 20 ms C1 5 ms is ¼ of cycle A1

horizontal line through PQ drawn on Fig. 5.2 B1 [3] 6 (a) charge = current × time B1 [1] (b) (i) P = V

2 / R C1 = (240)2 / 18 = 3200 W A1 [2]

(ii) I = V / R = 240 / 18 = 13.3 A A1 [1]

(iii) charge = It = 13.3 × 2.6 × 106 C1 = 3.47 × 107 C A1 [2] (iv) number of electrons = 3.47 × 107 / 1.6 × 10–19 (= 2.17 × 1026) C1 number of electrons per second = 2.17 × 1026 / 2.6 × 106 = 8.35 × 1019 A1 [2]




7 (a) (i) W = 206 and X = 82 A1 Y = 4 and Z = 2 A1 [2] (ii) mass-energy is conserved B1 mass on rhs is less because energy is released B1 [2] (b) not affected by external conditions/factors/environment B1 [1] or two examples temperature and pressure




9702 PHYSICS







1 (a) force: kg m s–2 A1 [1]

(b) (i) I2: A2 l: m x: m C1

K: kg m s–2 A–2 A1 [2] (ii) curve of the correct shape (for inverse proportionality) M1 clearly approaching each axis but never touching the axis A1 [2] (iii) curving upwards and through origin A1 [1] 2 (a) (i) 1. distance of path / along line AB B1 [1] 2. shortest distance between AB / distance in straight line between AB or displacement from A to B B1 [1] (ii) acceleration = rate of change of velocity A1 [1] (b) (i) distance = area under line or (v/2)t or s = (8.8)2 / (2 × 9.81) C1 = 8.8 / 2 × 0.90 = 3.96 m or s = 3.95 m = 4(.0) m A1 [2] (ii) acceleration = (– 4.4 – 8.8) / 0.50 C1 = (–) 26(.4) m s–2 A1 [2] (c) (i) the accelerations are constant as straight lines B1 the accelerations are the same as same gradient or no air resistance as acceleration is constant or

change of speed in opposite directions (one speeds up one slows down) B1 [2]

(ii) area under the lines represents height or KE at trampoline equals PE at maximum height B1 second area is smaller / velocity after rebound smaller hence KE less B1

hence less height means loss in potential energy A0 [2]

3 (a) (i) the total momentum of a system (of interacting bodies) remains constant M1

provided there are no resultant external forces / isolated system A1 [2]

(ii) elastic: total kinetic energy is conserved, inelastic: loss of kinetic energy B1 [1] [allow elastic: relative speed of approach equals relative speed of separation]




(b) (i) initial mom: 4.2 × 3.6 – 1.2 × 1.5 (= 15.12 – 1.8 = 13.3) C1 final mom: 4.2 × v + 1.5 × 3 C1 v = (13.3 – 4.5) / 4.2 = 2.1 m s–1 A1 [3] (ii) initial kinetic energy = ½ mA(vA)2 + ½ mB(vB)2 = 27.21 + 1.08 = 28(.28) M1 final kinetic energy = 9.26 + 6.75 = 16 M1 initial KE is not the same as final KE hence inelastic A1 [3] provided final KE less than initial KE [allow in terms of relative speeds of approach and separation] 4 (a) (i) stress = force / cross-sectional area B1 [1] (ii) strain = extension / original length B1 [1] (b) (i) E = stress / strain C1 E = 0.17 × 1012 C1 stress = 0.17 × 1012 × 0.095 / 100 C1 = 1.6(2) × 108

Pa A1 [4] (ii) force = (stress × area) = 1.615 × 108 × 0.18 × 10–6 C1 = 29(.1) N A1 [2] 5 (a) when waves overlap / meet B1 the resultant displacement is the sum of the individual displacements of the waves B1 [2] (b) (i) 1. phase difference = 180 º / (n + ½) 360 º (allow in rad) B1 [1] 2. phase difference = 0 / 360 º / (n360 º) (allow in rad) B1 [1]

(ii) v = f λ C1

λ = 320 / 400 = 0.80 m A1 [2]

(iii) path difference = 7 – 5 = 2 (m)

= 2.5 λ M1 hence minimum or maximum if phase change at P is suggested A1 [2] 6 (a) p.d. = work done / energy transformed (from electrical to other forms) charge B1 [1] (b) (i) maximum 20 V A1 [1] (ii) minimum = (600 / 1000) × 20 C1 = 12 V A1 [2]




(c) (i) use of 1.2 kΩ M1

1/1200 + 1/600 = 1/R, R = 400 Ω A1 [2] (ii) total parallel resistance (R2 + LDR) is less than R2 M1 (minimum) p.d. is reduced A1 [2] 7 (a) (i) nucleus contains 92 protons B1

nucleus contains 143 neutrons (missing ‘nucleus’ 1/2) B1 outside / around nucleus 92 electrons (B1) most of atom is empty space / mass concentrated in nucleus (B1) total charge is zero (B1) diameter of atom ~ 10–10

m or size of nucleus ~ 10–15 m (B1)

any two of (B1) marks [4] (ii) nucleus has same number / 92 protons B1 nuclei have 143 and 146 neutrons (missing ‘nucleus’ 1/2) B1 [2] (b) (i) Y = 35 A1 Z = 85 A1 [2] (ii) mass-energy is conserved in the reaction B1 mass on rhs of reaction is less so energy is released explained in terms of E = mc2 B1 [2]




9702 PHYSICS

9702/31 Paper 3 (Advanced Practical Skills 1), maximum raw mark 40






1 (a) Value of L in the range 0.790–0.810 m. [1] (c) (ii) Value of d to the nearest mm and d < 0.600 m. [1] (d) Six sets of readings of m and d scores 5 marks, five sets scores 4 marks etc. [5] Correct trend is d decreases as m increases. Help from Supervisor –1.

Range of m: mmin = 0 g or 10 g; mmax > 100 g. [1] Column headings: [1] Each column heading must contain a quantity and a unit. The presentation of quantity and unit must conform to accepted scientific convention e.g. 1/d /m–1. Consistency: [1] All values of d must be given to the nearest mm. Significant figures: [1] Significant figures for every row of values of 1/d same as, or one greater than, d as recorded in table. Calculation: [1] Values of 1/d calculated correctly. (e) (i) Axes: [1] Sensible scales must be used, no awkward scales (e.g. 3:10). 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 that is being plotted. Scale markings should be no more than three large squares apart. Plotting of points: [1] All observations in the table must be plotted. Diameter of plotted points must be ≤ half a small square (no “blobs”). Work to an accuracy of half a small square. Quality: [1] All points in the table must be plotted (at least 5) for this mark to be awarded. Scatter of points must be less than ± 0.05 m–1 of 1/d from a straight line. (ii) Line of best fit: [1] Judge by balance of all points on the grid about the candidate’s line (at least 5 points). There must be an even distribution of points either side of the line along the full length. Allow one anomalous point only if clearly indicated (i.e. circled or labelled) by the

candidate. Line must not be kinked or thicker than half a small square.




(iii) Gradient: [1] The hypotenuse of the triangle must be at least half the length of the drawn line. Both read-offs must be accurate to half a small square in both the x and y directions. The method of calculation must be correct. y–intercept: [1] Either: Correct read-off from a point on the line and substituted into y = mx + c. Read-off must be accurate to half a small square in both x and y directions. Or: Correct read-off of the intercept directly from the graph. (f) Value of P = candidate’s gradient. Value of Q = candidate’s intercept. Unit for P (e.g. kg–1 m–1) and Q (m–1). [1] (g) Value of k in range 1.0–2.0. [1] [Total: 20]

2 (a) (ii) Value of θ with unit. Help from Supervisor –1. [1]

θ in range 72°–92°. [1]

(iii) Absolute uncertainty in θ in range 2º–10º. If repeated readings have been taken, then the uncertainty can be half the range (but NOT zero if values are equal). Correct method of calculation to obtain percentage uncertainty. [1]

(iv) Correct calculation of sin θ. Ignore unit. Do not allow sin θ = O/H ideas as triangle not a right-angled triangle. [1]

(b) Value of T with unit in range 1.0 ≤ T ≤ 2.0 s. [1] Evidence of repeats here or in (c)(ii). [1]

(c) (ii) Second value of θ. [1] Second value of T. [1] Second value of T < first value of T. [1] (d) (i) Two values of k calculated correctly. [1]

(ii) Justification of s.f. in k linked to significant figures in T (or t) and θ. [1] (iii) Sensible comment relating to the calculated values of k, testing against a criterion specified by the candidate. [1]




(e)

(i) Limitations max. 4 (ii) Improvements max. 4 Do not credit

A two readings not enough (to draw a conclusion)

take more readings and plot a graph/ calculate more k values and compare

“repeat readings” on its own /few readings/only one reading /take more readings and (calculate) average k

B end of nail slips in bracket/bracket moves/is not stable

use something with a sharper point e.g. cocktail stick/dent in bracket (to seat head of nail) valid method to fix bracket e.g. use blu-tack/glue/use bigger/heavier bracket/fix bracket/ clamp to bench

method of fixing nail

C difficult to measure T with reason e.g. heavily damped/oscillations die away quickly

‘too few oscillations’ on its own/ T small

D difficult to judge start of/end of/complete oscillation

use a fixed/fiducial marker /improved timing method e.g. video with timer/video and view frame-by-frame multiflash photography with strobe rate

human error/ reaction time /record time for more oscillations marker fixed to rod /marker placed at extreme of oscillation use light gate

E difficult to read θ/angle/protractor with reason e.g. difficult to hold steady in the air

clamp protractor parallax error use a larger protractor

F fans/air conditioning affect oscillations

[Total: 20]




9702 PHYSICS







1 (b) (ii) Value of t in the range 10.0 s ≤ t ≤ 20.0 s. [1]

Evidence of repeat measurements of t.

[1]

(c) Six sets of readings of S and t scores 5 marks, five sets scores 4 marks etc. If trend wrong or no S or t column –1. Major help from Supervisor –2 (setting up circuit). Minor help from Supervisor –1.

[5]

Range:

Values of S must include 22 (kΩ) or 10 (kΩ) and 1.2 (kΩ) or 1.0 (kΩ).

[1]

Column headings: Each column heading must contain a quantity and a unit. The presentation of quantity and unit must conform to accepted scientific

convention e.g. 1/t / s–1 1/S / kΩ –1 1/S (kΩ –1) t / s t (s).

( 1/t(s) 1/S 1/kΩ 1/S (kΩ) –1 are not allowed.)

[1]

Consistency: All values of raw t must be given to the same precision (either 0.1 s or 0.01 s). Significant figures: Significant figures for every row of values of 1/S must be the same as or one greater than the s.f. in S as recorded in table.

[1]

[1]

Calculation:

Values of 1/t calculated correctly.

[1]

(d) (i) Axes: Sensible scales must be used, no awkward scales (e.g. 3:10). 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 that is being plotted. Scale markings should be no more than three large squares apart.

[1]

Plotting of points: All observations in the table must be plotted. Points must be plotted to an accuracy of half a small square. Diameter of points must be ≤ half a small square (no “blobs”).

[1]

Quality: All points in the table must be plotted (at least 5) for this mark to be awarded. Judge by the scatter of all the points about the straight line. Points must be

less than 0.05 kΩ 1 from a straight line on the 1/S axis.

[1]

(ii) Line of best fit: Judge by balance of all points on the grid about the candidate’s line (at least 5 points). There must be an even distribution of points either side of the line along the full length. Allow one anomalous point only if clearly indicated (i.e. circled or labelled) by the candidate. Line must not be kinked or thicker than half a small square.

[1]




(iii) Gradient:

The hypotenuse of the triangle must be at least half the length of the drawn line. Both read-offs must be accurate to half a small square in both the x and y directions. The method of calculation must be correct.

[1]

y-intercept: Either: Correct read-off from a point on the line substituted into y = mx + c. Read-off must be accurate to half a small square in both x and y directions. Or: Intercept read off directly from the graph.

[1]

(e) Value of a = candidate’s gradient. Value of b = candidate’s intercept / candidate’s gradient

= candidate’s intercept / a

[1]

Unit for a correct and consistent with value e.g. kΩ s–1, Ω s–1. [1]

[Total: 20] 2 (a) (i) Measurement of d with unit in range 0.5 mm – 2.5 mm.

[1]

Evidence of repeated readings of d. [1]

(ii) Absolute uncertainty in d in the range 0.2 – 0.5 mm. If repeated readings have been taken, then the absolute uncertainty can be half the range (but not zero if values are equal). Correct method of calculation to get percentage uncertainty.

[1]

(c) (i) Measurement of r1 recorded to nearest 0.1 cm3, and in range 1 to 5 cm3. [1]

(ii) (iii)

Value for n. Correct calculation of V.

[1]

[1]

(d)

Justification of s.f. in V linked to significant figures in (r1 – r2) and in n.

[1]

(e) (ii) (iii)

Second value of d.

Second value of n. Quality: V larger for larger d.

[1]

[1]

[1]

(f) (i) Two values of k calculated correctly.

[1]

(ii) Sensible comment relating to the calculated values of k, testing against a criterion specified by the candidate. [1]




(g)

(i) Limitations 4 max. (ii) Improvements 4 max. Do not credit


take many readings and plot a graph/take many readings and calculate more k values and compare

repeat readings/few readings/take more readings and (calculate) average k/only one reading

B1 large uncertainty in d because d is small

improved method to measure d e.g. measure OD and wall thickness/image of cross-section with scale

flexible tube/rigid tube/wire in tube/micrometer/d is small

B2 difficult to measure d with reason e.g. tube distorts/difficult to insert both prongs inside tube/difficult to judge when jaws are in line with edges of d

use travelling microscope/measure volume and calculate d/use a magnifying glass with scale

C difficult to count bubbles with reason e.g. plunger moves unsteadily/plunger sticks/bubbles unexpected/bubbles emerge too quickly

method to improve control e.g. use a G-clamp or screw to move plunger/use a narrower diameter syringe with reason (e.g.smaller force needed so better control)

bubbles difficult to count/lubrication

D difficult to watch the syringe scale and the bubbles at the same time

description of method to allow simultaneous measurement e.g. use video with playback/use video to allow slow motion/use video to see bubbles

parallax/use assistant/change syringe scale/high speed camera/slow motion camera/light gates

E difficult to keep end of tube at 2 cm depth

method to keep tube at 2 cm depth e.g. use of clamp/Blu-Tack/tape

stick tube/attach tube/water in tube/comments about Blu-Tack seal

[Total: 20]




9702 PHYSICS







1 (a) (i) Value of raw d in the range 0.15 mm ≤ d ≤ 0.44 mm. [1]

(b) (v) Value of l in range 0.1 m < l < 1 m. Value of V in range 0.1 V ≤ V ≤ 2.0 V. [1] (d) Six sets of readings of l and V scores 5 marks; five sets scores 4 marks etc. [5] Major help from Supervisor –2 (setting up apparatus). Minor help from Supervisor –1. Range of l :∆l ≥ 60 cm. [1] Column headings: [1] Each column heading must contain a quantity and a unit. The presentation of quantity and unit must conform to accepted scientific convention. e.g. 1/l /m–1, V/l / V m–1. Do not allow 1/l (m), V(V) / l (m). Consistency: [1] All values of raw l must be given to the nearest mm. Significant figures: [1] Significant figures for every row of values of 1/l same as or one greater than l as recorded in table. Calculation: [1] Values of V/l calculated correctly (e) (i) Axes: [1] Sensible scales must be used, no awkward scales (e.g. 3:10). 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 that is being plotted. Scale markings should be no more than three large squares apart. Plotting of points: [1] All observations in the table must be plotted. Diameter of points must be ≤ half a small square (no “blobs”). Check that the points are plotted correctly. Work to an accuracy of half a small square. Quality: [1] All points in the table must be plotted (at least 5) for this mark to be awarded. Scatter of points must be less than 0.1 m–1 from a straight line on the 1/l axis. (ii) Line of best fit: [1] Judge by balance of all points on the grid about the candidate’s line (at least 5 points) There must be an even distribution of points either side of the line along the full length. Allow one anomalous point only if clearly indicated (i.e. circled or labelled) by the candidate. Line must not be kinked or thicker than half a small square.




(iii) Gradient: [1] The hypotenuse of the triangle must be at least half the length of the drawn line. Both read-offs must be accurate to half a small square in both the x and y directions. The method of calculation must be correct. y-intercept: [1] Either: Correct read-off from a point on the line and substituted into y = mx + c. Read-off must be accurate to half a small square in both x and y directions. Or: Correct read-off of the intercept directly from the graph. (f) (i) Value of M = candidate’s gradient. Value of N = –(candidate’s intercept). [1]

(ii) Answer in range ρ: 2.0 ≤ ρ ≤ 20.0 × 10–7 Ω m. Consistent with units. [1]

[Total: 20]

2 (a) (ii) Measurement of raw H in range 10.0 cm < H < 20.0 cm consistent with unit. [1] (b) (ii) Measurement of raw h1 to nearest mm with unit. [1] (iii) Absolute uncertainty in h1 in the range 2–5 mm. If repeated readings have been taken,

then the absolute uncertainty can be half the range. Correct method of calculation to get percentage uncertainty. [1]

(c) (iii) Measurement of h2 less than h1. [1] Evidence of repeat readings here or in (e). [1] (d) Correct calculation of F with no units. [1] (e) Second value of h1. [1] Second value of h2. [1] Second value of h2 < first value of h2. [1] (f) (i) Two values of k calculated correctly. [1] (ii) Justification of s.f. in k linked to significant figures in h1 and (h1 – h2). [1] (iii) Sensible comment relating to the calculated values of k, testing against a criterion specified by the candidate. [1]




(g)



take many readings and plot a graph/calculate more k values and compare

repeat readings /few readings /take more readings and (calculate) average k /only one reading

B discontinuous movement at bottom

method of providing continuous ramp e.g. tape join

alignment /stick /fix

C parallax error (or wtte) in h1 or h2 or heights

ruler and set square with detail e.g. set square from ruler to track or ball

ruler perpendicular to bench /parallax error in height

D difficult to measure h1/h2 with reason e.g. cannot see bottom of marble/bottom of track not at bottom of marble/thickness of track not taken into account

measure to top of marble. /measure diameter of marble and subtract it from height to top of marble

H /clear ramps

E difficult to release marble without applying a force

description of mechanical method of releasing marble e.g. card gate

string method /use of helpers

F difficult to measure h2 with reason related to time e.g. short time interval/doesn’t stay still at h2 for long

method of improved measurement of h2 e.g. video with (clamped) rule/multiflash photography with (clamped) rule/trial and improvement method/position sensore at top of ramp/grid behind runway/scale on runway

too fast/ball travelling too quick, etc. /high speed camera or slow motion camera

[Total: 20]




9702 PHYSICS







1 (b) (i) Value of d in the range 0.480 – 0.500 m, with unit.

[1]

(c) Six sets of readings of d and F scores 6 marks, five sets scores 5 marks etc. Incorrect trend or no d data –1. Minor help from Supervisor –1, major help –2.

[6]

Range: dmax – dmin ≥ 30 cm.

[1]

Column headings: Each column heading must contain a quantity and a unit. The presentation of quantity and unit must conform to accepted scientific convention e.g. 1/d / m–1.

[1]

Consistency: All values of d must be given to the nearest mm and all values of F must be given to the nearest 0.1 N.

[1]

Significant figures: Significant figures for every row of values of 1 / d same as, or one greater than, d as recorded in table.

[1]

Calculation: Values of 1 / d calculated correctly.

[1]

(d) (i) Axes: Sensible scales must be used, no awkward scales (e.g. 3:10). 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 that is being plotted. Scale markings should be no more than three large squares apart.

[1]

Plotting of points: All observations in the table must be plotted on the grid. Points must be plotted to an accuracy of half a small square. Diameter of plotted points must be ≤ half a small square (no blobs).

[1]

Quality:

All points in the table must be plotted on the grid (at least 5) for this mark to be awarded. Judge by the scatter of all the points about a straight line. Scatter of points must be less than ± 0.001 cm–1 from a straight line in the 1/d direction.

[1]

(ii)

Line of best fit: Judge by balance of all the points on the grid about the candidate’s line (at least 5 points). There must be an even distribution of points either side of the line along the full length. One anomalous point is allowed only if clearly indicated (i.e. circled or labelled) by the candidate. Line must not be kinked or thicker than half a square.

[1]




(iii) Gradient: The hypotenuse of the triangle must be at least half the length of the drawn line. Both read-offs must be accurate to half a small square in both the x and y directions. The method of calculation must be correct.

[1]

y-intercept: Either: Correct read-off from a point on the line substituted into y = mx + c or an equivalent expression, with read-off accurate to half a small square in both x and y directions. Or: Intercept read directly from the graph, with read-off accurate to half a small squarein both x and y directions.

[1]

(e) Value of a = candidate’s gradient. Value of b = candidate’s intercept. Unit for a correct and consistent with value, e.g. N cm. Unit for b is correct and consistent with value, e.g. N.

[1]

[1] [Total: 20] 2 (a) (i) All values of D to nearest 0.01 cm or all to nearest 0.001 cm, and in

range 3.0 to 5.0 mm. Evidence of repeat readings of D.

[1]

[1]

(ii) Absolute uncertainty in D in range 0.02 to 0.05 cm and correct method of calculation to obtain percentage uncertainty. If repeated readings have been taken, then the absolute uncertainty can be half the range (but not zero if values are equal).

[1]

(c) (i)

l in range 19.0 to 21.0 cm, with unit, to nearest mm.

[1]

(iii) t in range 2.0 to 10.0 s and value(s) to nearest 0.1s or 0.01s.

[1]

(iv) Correct calculation of v.

[1]

(d) Justification for s.f. in v linked to s.f. in D and t.

[1]

(e) (ii) Second value of l. Second value of t. Second value of t > first value of t.

[1] [1] [1]

(f) (i) Two values of k calculated correctly.

[1]

(ii) Sensible comment relating to the calculated values of k, testing against a criterion specified by the candidate.

[1]




(g)


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

take more readings and plot a graph / take more readings and calculate more k values and compare

“repeat readings” on its own / few readings /

only one reading / take more readings and (calculate) average k

B1 large uncertainty in D because D is small

measure outside diameter and wall thickness / measure an image showing the cross-section and a scale

use micrometer

B2 tube distorts when measuring D

use travelling microscope / measure volume and calculate D

C tube not straight so difficult to make tube vertical / tube not straight so difficult to measure length

tape to a straight rod / increase attached mass

use stiffer tube

D difficult to judge moment (or operate stopwatch) when level reaches syringe graduations

use video with timer / view video frame by frame / collect water for a timed interval and measure volume / use light gates and timer with practical detail / use different diameter syringe with reason / use position sensor above water surface

‘reaction time’ on its own / human error / ‘light gates’ on its own /slow motion (or high speed) camera

E difficult to see water level

use coloured water (or dye) / use clear syringe / view against black background

F clay stretches (or squashes) tube

measure length after attaching clay

References to parallax error are ignored for this experiment.

[Total: 20]




9702 PHYSICS







1 (a) (ii) Value of T in range 0.4 ≤ T ≤ 1.4 s. [1] Evidence of repeats. [1] (b) Six sets of readings of m and t (or T) scores 5 marks, four sets scores 4 marks etc. [5] Help from Supervisor –1.

Range of m : ∆m ≥ 0.600 kg [1] Column headings: [1]

Each column heading must contain a quantity and a unit. The presentation of quantity and unit must conform to accepted scientific convention e.g. 1/T2/ s–2. Do not allow 1/T2(s)2

Consistency: [1] All values of raw t must all be given to the nearest 0.1 s or 0.01 s. Significant figures: [1] Significant figures for every row of values of 1/T2 same as or one greater than t (or T) as recorded in table. Calculation: [1] Values of 1/T2 calculated correctly (c) (i) Axes: [1]

Sensible scales must be used, no awkward scales (e.g. 3:10). 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 that is being plotted. Scale markings should be no more than three large squares apart.

Plotting of points: [1]

All observations in the table must be plotted. Diameter of points must be ≤ half a small square (no “blobs”). Work to an accuracy of half a small square.

Quality: [1]

All points in the table must be plotted (at least 5) for this mark to be awarded. Scatter of points must be less than 0.1 s–2 of 1/T2 from a straight line.

(ii) Line of best fit: [1]

Judge by balance of all points on the grid about the candidate’s line (at least 5 points) There must be an even distribution of points either side of the line along the full length. Allow one anomalous point only if clearly indicated (i.e. circled or labelled) by the candidate. Line must not be kinked or thicker than half a small square.

(iii) Gradient: [1] The hypotenuse of the triangle must be at least half the length of the drawn line. Both read-offs must be accurate to half a small square in both the x and y directions.




y-intercept: [1] Either: Correct-read off from a point on the line and substituted into y = mx + c. Read-off must be accurate to half a small square in both x and y directions. Or: Correct read-off of the intercept directly from the graph.

(d) Value of P = candidate’s gradient. Value of Q = candidate’s intercept. [1] Unit for P (kg–1 s–2) correct and consistent with value and Q (s–2) [1] [Total: 20]

2 (a) (i) Value of L in range 8.0 ≤ L ≤ 12.0 cm with consistent unit to the nearest mm. [1]

(ii) Absolute uncertainty 1 ≤ ∆L ≤ 3 mm. If repeated readings have been taken, then the uncertainty can be half the range. Correct method of calculation to get percentage uncertainty. [1]

(b) (iii) Value of raw N1 an integer. [1]

(c) (iii) Value of N2 ≥ N1. [1]

Evidence of repeats for N1 or N2 either here or in (b)(iii). [1]

(d) Correct calculation of F. [1]

(e) Second value of L. [1] Second values of N2 and N1. [1] Second (average) value of N1 > first (average) value of N1. [1]

(f) (i) Two values of k calculated correctly. [1] (ii) Justification of s.f. in k linked to significant figures in L and (N1 – N2) and m. [1] (iii) Sensible comment relating to the calculated values of k, testing against a criterion

specified by the candidate. [1]




(g)



take more readings and plot a graph/calculate more k values and compare

repeat readings/few readings/only one reading/take more readings and average k

B friction at pulley

method of reducing friction of pulley with location

C wet string added to force/ mass of string not accounted for

waterproof/nylon/wire

D can only measure to nearest 0.4g / paperclip

use smaller masses e.g. half paperclips, riders, graph paper

newton meter

E change in N are very small

reasoned explanation for changing length of wire

helpers parallax errors

F copper wire is not flat/straight/exit not parallel to water level

circular wire shape change liquid/wire

[Total: 20]




9702 PHYSICS

9702/41 Paper 4 (A2 Structured Questions), maximum raw mark 100






Section A 1 (a) region of space area / volume B1 where a mass experiences a force B1 [2] (b) (i) force proportional to product of two masses M1 force inversely proportional to the square of their separation M1 either reference to point masses or separation >> ‘size’ of masses A1 [3]

(ii) field strength = GM / x2 or field strength ∝ 1 / x2 C1

ratio = (7.78 × 108)2 / (1.5 × 108)2 C1 = 27 A1 [3] (c) (i) either centripetal force = mRω

2 and ω = 2π / T

or centripetal force = mv2 / R and v = 2πR /T B1 gravitational force provides the centripetal force B1 either GMm / R2 = mRω

2 or GMm / R2 = mv2 / R M1 M = 4π2R3 / GT2 A0 [3] (allow working to be given in terms of acceleration) (ii) M = 4π2 × (1.5 × 1011)3 / 6.67 × 10–11 × (3.16 × 107)2 C1 = 2.0 × 1030

kg A1 [2] 2 (a) obeys the equation pV = constant × T or pV = nRT M1 p, V and T explained A1 at all values of p, V and T/fixed mass/n is constant A1 [3] (b) (i) 3.4 × 105 × 2.5 × 103 × 10–6 = n × 8.31 × 300 M1 n = 0.34 mol A0 [1] (ii) for total mass/amount of gas 3.9 × 105 × (2.5 + 1.6) × 103 × 10–6 = (0.34 + 0.20) × 8.31 × T C1 T = 360 K A1 [2] (c) when tap opened gas passed (from cylinder B) to cylinder A B1 work done on gas in cylinder A (and no heating) M1 so internal energy and hence temperature increase A1 [3]




3 (a) (i) 1. amplitude = 1.7 cm A1 [1] 2. period = 0.36 cm C1 frequency = 1/0.36 frequency = 2.8 Hz A1 [2] (ii) a = (–)ω2x and ω = 2π/T C1 acceleration = (2π/0.36)2 × 1.7 × 10–2 M1 = 5.2 m s–2 A0 [2] (b) graph: straight line, through origin, with negative gradient M1 from (–1.7 × 10–2, 5.2) to (1.7 × 10–2, –5.2) A1 [2] (if scale not reasonable, do not allow second mark) (c) either kinetic energy = ½mω

2(x02 – x2)

or potential energy = ½mω2x2 and potential energy = kinetic energy B1

½mω2(x0 – x2) = ½ × ½mω

2x02 or ½mω

2x2 = ½ × ½mω2x0

2 C1 x0

2 = 2x2 x = x0 / √2 = 1.7 / √2 = 1.2 cm A1 [3] 4 (a) work done moving unit positive charge M1 from infinity (to the point) A1 [2] (b) (gain in) kinetic energy = change in potential energy B1 ½mv2 = qV leading to v = (2Vq/m)½ B1 [2] (c) either (2.5 × 105)2 = 2 × V × 9.58 × 107 C1 V = 330 V M1 this is less than 470 V and so ‘no’ A1 [3]

or v = (2 × 470 × 9.58 × 107) (C1) v = 3.0 × 105

m s–1 (M1) this is greater than 2.5 × 105 m s–1 and so ‘no’ (A1) or (2.5 × 105)2 = 2 × 470 × (q/m) (C1) (q/m) = 6.6 × 107

C kg–1 (M1) this is less than 9.58 × 107

C kg–1 and so ‘no’ (A1)




5 (a) (uniform magnetic) flux normal to long (straight) wire carrying a current of 1 A M1 (creates) force per unit length of 1 N m–1 A1 [2] (b) (i) flux density = 4π × 10–7 × 1.5 × 103 × 3.5 C1 = 6.6 × 10–3

T A1 [2] (ii) flux linkage = 6.6 × 10–3 × 28 × 10–4 × 160 C1 = 3.0 × 10–3

Wb A1 [2] (c) (i) (induced) e.m.f. proportional to rate of M1 change of (magnetic) flux (linkage) A1 [2]

(ii) e.m.f. = (2 × 3.0 × 10–3) / 0.80 C1 = 7.4 × 10–3

V A1 [2] 6 (a) (i) to reduce power loss in the core B1 due to eddy currents/induced currents B1 [2] (ii) either no power loss in transformer or input power = output power B1 [1] (b) either r.m.s. voltage across load = 9.0 × (8100 / 300) C1 peak voltage across load = √2 × 243 = 340 V A1 [2] or peak voltage across primary coil = 9.0 × √2 (C1) peak voltage across load = 12.7 × (8100/300) = 340 V (A1) 7 (a) (i) lowest frequency of e.m. radiation M1 giving rise to emission of electrons (from the surface) A1 [2] (ii) E = hf C1 threshold frequency = (9.0 × 10–19) / (6.63 × 10–34) = 1.4 × 1015

Hz A1 [2] (b) either 300 nm ≡ 10 × 1015

Hz (and 600 nm ≡ 5.0 × 1014 Hz)

or 300 nm ≡ 6.6 × 10–19 J (and 600 nm ≡ 3.3 × 10–19

J) or zinc λ0 = 340 nm, platinum λ0 = 220 nm (and sodium λ0 = 520 nm) M1 emission from sodium and zinc A1 [2] (c) each photon has larger energy M1 fewer photons per unit time M1 fewer electrons emitted per unit time A1 [3]




8 (a) two (light) nuclei combine M1 to form a more massive nucleus A1 [2] (b) (i) ∆m = (2.01410 u + 1.00728 u) – 3.01605 u = 5.33 × 10–3 u C1 energy = c2 × ∆m C1 = 5.33 × 10–3 × 1.66 × 10–27 × (3.00 × 108)2 = 8.0 × 10–13

J A1 [3] (ii) speed/kinetic energy of proton and deuterium must be very large B1 so that the nuclei can overcome electrostatic repulsion B1 [2]

Section B

9 (a) (i) light-dependent resistor/LDR B1 [1] (ii) strain gauge B1 [1] (iii) quartz/piezo-electric crystal B1 [1] (b) (i) resistance of thermistor decreases as temperature increses M1 etiher VOUT = V × R / (R + RT) or current increases and VOUT = I R A1 VOUT increases A1 [3] (ii) either change in RT with temperature is non-linear or VOUT is not proportional to RT/ change in VOUT with RT is non-linear M1 so change is non-linear A1 [2] 10 (a) sharpness: how well the edges (of structures) are defined B1 contrast: difference in (degree of) blackening between structures B1 [2] (b) e.g. scattering of photos in tissue/no use of a collimator/no use of lead grid large penumbra on shadow/large area anode/wide beam large pixel size (any two sensible suggestions, 1 each) B2 [2] (c) (i) I = I0e

–µx C1 ratio = exp(–2.85 × 3.5) / exp(–0.95 × 8.0) C1 = (4.65 × 10–5) / (5.00 × 10–4) = 0.093 A1 [3] (ii) either large difference (in intensities) or ratio much less than 1.0 M1 so good contrast A1 [2] (answer given in (c)(ii) must be consistent with ratio given in (c)(i))




11 (a) (i) amplitude of the carrier wave varies M1 (in synchrony) with the displacement of the information signal A1 [2] (ii) e.g. more than one radio station can operate in same region/less interference enables shorter aerial increased range/less power required/less attenuation less distortion (any two sensible answers, 1 each) B2 [2] (b) (i) frequency = 909 kHz C1 wavelength = (3.0 × 108) / (909 × 103) = 330 m A1 [2] (ii) bandwidth = 18 kHz A1 [1] (iii) frequency = 9000 Hz A1 [1] 12 (a) for received signal, 28 = 10 lg(P / 0.36 × 10–6) C1 P = 2.3 × 10–4

W A1 [2] (b) loss in fibre = 10 lg(9.8 × 10–3 / 2.27 × 10–4) C1 = 16 dB A1 [2] (c) attenuation per unit length = 16 / 85 = 0.19 dB km–1 A1 [1]




9702 PHYSICS







Section A 1 (a) equatorial orbit / above equator B1

satellite moves from west to east / same direction as Earth spins B1 period is 24 hours / same period as spinning of Earth B1 [3] (allow 1 mark for ‘appears to be stationary/overhead’ if none of above marks scored)

(b) gravitational force provides/is the centripetal force B1 GMm/R2 = mRω

2 or GMm/R2 = mv2/R M1 ω = 2π /T or v = 2πR / T or clear substitution M1 clear working to give R3 = (GMT2 / 4π

2) A1 [4]

(c) R3 = 6.67 × 10–11 × 6.0 × 1024 × (24 × 3600)2 / 4π2 C1

= 7.57 × 1022 C1 R = 4.2 × 107 m A1 [3] (missing out 3600 gives 1.8 × 105 m and scores 2/3 marks)

2 (a) (i) 1. pV = nRT

1.80 × 10–3 × 2.60 × 105 = n × 8.31 × 297 C1 n = 0.19 mol A1 [2]

2. ∆q = mc∆T

95.0 = 0.190 × 12.5 × ∆T B1 ∆T = 40 K A1 [2] (allow 2 marks for correct answer with clear logic shown)

(ii) p/T = constant

(2.6 × 105) / 297 = p / (297 + 40) M1 p = 2.95 × 105 Pa A0 [1]

(b) change in internal energy is 120 J / 25 J B1

internal energy decreases / ∆U is negative / kinetic energy of molecules decreases M1 so temperature lower A1 [3]




3 (a) (i) ω = 2π / T = 2π / 0.69 C1 = 9.1 rad s–1 A1 [2] (allow use of f = 1.5 Hz to give ω = 9.4 rad s–1)

(ii) 1. x = 2.1 cos 9.1t

2.1 and 9.1 numerical values B1 use of cos B1 [2]

2. v0 = 2.1 × 10–2 × 9.1 (allow ecf of value of x0 from (ii)1.)

v0 = 0.19 m s–1 B1 v = v0 sin 9.1t (allow cos 9.1t if sin used in (ii)1.) B1 [2]

(b) energy = either ½ mv0

2 or ½ mω2x0

2 = either ½ × 0.078 × 0.192 or ½ × 0.078 × 9.12 × (2.1 × 10–2)2 C1 = 1.4 × 10–3 J A1 [2]

4 (a) (i) V = q / 4πε0R B1 [1]

(ii) (capacitance is) ratio of charge and potential or q/V M1 C = q/V = 4πε0R A0 [1]

(b) (i) C = 4π × 8.85 × 10–12 × 0.45 C1 = 50 pF A1 [2]

(ii) either energy = ½ CV2 or energy = ½ QV and Q = CV C1

energy of spark = ½ × 50 × 10–12 (9.0 × 105)2 – (3.6 × 105)2 C1 = 17 J A1 [3]

5 (a) (uniform magnetic) flux normal to long (straight) wire carrying a current of 1 A M1

(creates) force per unit length of 1 N m–1 A1 [2]

(b) (i) sketch: concentric circles M1 increasing separation (must show more than 3 circles) A1 correct direction (anticlockwise, looking down) B1 [3]

(ii) B = (4π × 10–7 × 6.3) / (2π × 4.5 × 10–2) C1

= 2.8 × 10–5 T A1 [2]

(iii) F = BIL (sinθ) C1 = 2.8 × 10–5 × 9.3 × 1

F/L = 2.6 × 10–4 N m–1 A1 [2]

(c) force per unit length depends on product IXIY / by Newton’s third law / action and reaction are equal and opposite M1 so same for both A1 [2]




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

(b) (i) positive terminal identified (upper connection to load) B1 [1]

(ii) VP = √2 × VRMS C1 ratio = 240 √2 / 9 C1

ratio = 38 A1 [3] (VP = VRMS / √2 gives ratio = 18.9 and scores 1/3) (ratio = 240 / 9 = 26.7 scores 1/3) (ratio = 9 / (240 / √2) = 0.0265 is inverted ratio and scores 1/3)

(c) (i) e.g. (output) p.d. / voltage / current does not fall to zero e.g. range of (output) p.d. / voltage / current is reduced (any sensible answer) B1 [1]

(ii) sketch: same peak value at start of discharge M1

correct shape between one peak and the next A1 [2] 7 (a) each wavelength is associated with a discrete change in energy M1

discrete energy change / difference implies discrete levels A1 [2]

(b) (i) 1. arrow from –0.54 eV to –0.85 eV, labelled L B1 [1]

2. arrow from –0.54 eV to –3.4 eV , labelled S B1 [1] (two correct arrows, but only one label – allow 2 marks) (two correct arrows, but no labels – allow 1 mark)

(ii) E = hc / λ C1

(3.4 – 0.54) × 1.6 × 10–19 = (6.63 × 10–34 × 3.0 × 108) / λ C1 λ = 4.35 × 10–7 m A1 [3]

(c) –1.50 → –3.4 = 1.9 eV

–0.85 → –3.4 = 2.55 eV (allow 2.6 eV) –0.54 → –3.4 = 2.86 eV (allow 2.9 eV) 3 correct, 2 marks with –1 mark for each additional energy 2 correct, 1 mark but no marks if any additional energy differences B2 [2]




8 (a) energy is given out / released on formation of the α-particle (or reverse argument) M1 either E = mc2 so mass is less or reference to mass-energy equivalence A1 [2]

(b) (i) mass change = 18.00567 u – 18.00641 u C1 = 7.4 × 10–4 u (sign not required) A1 [2]

(ii) energy = c2

∆m = (3.0 × 108)2 × 7.4 × 10–4 × 1.66 × 10–27 C1 = 1.1 × 10–13 J A1 [2]

(allow use of u = 1.67 × 10–27 kg) (allow method based on 1u equivalent to 930 MeV to 933 MeV)

(iii) either mass of products greater than mass of reactants M1

this mass/energy provided as kinetic energy of the helium-4 nucleus A1 or both nuclei positively charged (M1)

energy required to overcome electrostatic repulsion (A1) [2]




Section B 9 (a) 30 litres → 54 litres (allow ± 4 litres on both limits) A1 [1]

(b) (i) only 0.1 V change in reading for 10 litre consumption (or similar numbers) B1 above about 60 litres gradient is small compared to the gradient at about 40 litres B1 [2]

(ii) voltmeter reading (nearly) zero when fuel is left C1

voltmeter reads only about 0.1 V when 10 litres of fuel left in tank A1 [2] (“voltmeter reads zero when about 4 litres of fuel left in tank” scores 2 marks)

10 (a) product of density and speed of sound / wave M1

(density of medium and) speed of sound / wave in medium A1 [2]

(b) if (Z1 – Z2) is small, mostly transmission M1 if (Z1 – Z2) is large, mostly reflection M1 (if ‘mostly’ not stated allow 1/2 marks for these first two marks) either reflection / transmission also depends on (Z1 + Z2) or intensity reflection coefficient = (Z1 – Z2)

2 / (Z1 + Z2)2 A1 [3]

(c) e.g. smaller structures can be distinguished B1 because better resolution at shorter wavelength / higher frequency B1 [2]

11 (a) changing voltage changes energy / speed of electrons M1

changing electron energy changes maximum X-ray photon energy A1 [2]

(b) (i) 1. loss of power / energy / intensity B1 [1]

2. intensity changes when beam not parallel C1 decreases when beam is divergent A1 [2]

(ii) ratio = (exp –2.9 × 2.5) / (exp –0.95 × 6.0) C1

= 0.21 (min. 2 sig. fig.) A1 [2] (values of both lengths incorrect by factor of 10–2 to give ratio of 0.985 scores 1 mark)




12 (a) takes all the simultaneous digits for one number B1 and ‘sends’ them one after another (along the transmission line) B1 [2]

(b) (i) 0111 A1 [1]

(ii) 0110 A1 [1]

(c) levels shown

t 0 0.2 0.4 0.6 0.8 1.0 1.2

0 8 7 15 6 5 8

(–1 for each error or omission) A2 correct basic shape of graph i.e. series of steps M1 with levels staying constant during correct time intervals A1 [4] (vertical lines in steps do not need to be shown)

(d) increasing number of bits reduces step height M1 increasing sampling frequency reduces step depth / width M1 reproduction of signal is more exact A1 [3]




9702 PHYSICS







Section A 1 (a) region of space area / volume B1 where a mass experiences a force B1 [2] (b) (i) force proportional to product of two masses M1 force inversely proportional to the square of their separation M1 either reference to point masses or separation >> ‘size’ of masses A1 [3]

(ii) field strength = GM / x2 or field strength ∝ 1 / x2 C1

ratio = (7.78 × 108)2 / (1.5 × 108)2 C1 = 27 A1 [3] (c) (i) either centripetal force = mRω

2 and ω = 2π / T

or centripetal force = mv2 / R and v = 2πR /T B1 gravitational force provides the centripetal force B1 either GMm / R2 = mRω

2 or GMm / R2 = mv2 / R M1 M = 4π2R3 / GT2 A0 [3] (allow working to be given in terms of acceleration) (ii) M = 4π2 × (1.5 × 1011)3 / 6.67 × 10–11 × (3.16 × 107)2 C1 = 2.0 × 1030

kg A1 [2] 2 (a) obeys the equation pV = constant × T or pV = nRT M1 p, V and T explained A1 at all values of p, V and T/fixed mass/n is constant A1 [3] (b) (i) 3.4 × 105 × 2.5 × 103 × 10–6 = n × 8.31 × 300 M1 n = 0.34 mol A0 [1] (ii) for total mass/amount of gas 3.9 × 105 × (2.5 + 1.6) × 103 × 10–6 = (0.34 + 0.20) × 8.31 × T C1 T = 360 K A1 [2] (c) when tap opened gas passed (from cylinder B) to cylinder A B1 work done on gas in cylinder A (and no heating) M1 so internal energy and hence temperature increase A1 [3]




3 (a) (i) 1. amplitude = 1.7 cm A1 [1] 2. period = 0.36 cm C1 frequency = 1/0.36 frequency = 2.8 Hz A1 [2] (ii) a = (–)ω2x and ω = 2π/T C1 acceleration = (2π/0.36)2 × 1.7 × 10–2 M1 = 5.2 m s–2 A0 [2] (b) graph: straight line, through origin, with negative gradient M1 from (–1.7 × 10–2, 5.2) to (1.7 × 10–2, –5.2) A1 [2] (if scale not reasonable, do not allow second mark) (c) either kinetic energy = ½mω

2(x02 – x2)

or potential energy = ½mω2x2 and potential energy = kinetic energy B1

½mω2(x0 – x2) = ½ × ½mω

2x02 or ½mω

2x2 = ½ × ½mω2x0

2 C1 x0

2 = 2x2 x = x0 / √2 = 1.7 / √2 = 1.2 cm A1 [3] 4 (a) work done moving unit positive charge M1 from infinity (to the point) A1 [2] (b) (gain in) kinetic energy = change in potential energy B1 ½mv2 = qV leading to v = (2Vq/m)½ B1 [2] (c) either (2.5 × 105)2 = 2 × V × 9.58 × 107 C1 V = 330 V M1 this is less than 470 V and so ‘no’ A1 [3]

or v = (2 × 470 × 9.58 × 107) (C1) v = 3.0 × 105

m s–1 (M1) this is greater than 2.5 × 105 m s–1 and so ‘no’ (A1) or (2.5 × 105)2 = 2 × 470 × (q/m) (C1) (q/m) = 6.6 × 107

C kg–1 (M1) this is less than 9.58 × 107

C kg–1 and so ‘no’ (A1)




5 (a) (uniform magnetic) flux normal to long (straight) wire carrying a current of 1 A M1 (creates) force per unit length of 1 N m–1 A1 [2] (b) (i) flux density = 4π × 10–7 × 1.5 × 103 × 3.5 C1 = 6.6 × 10–3

T A1 [2] (ii) flux linkage = 6.6 × 10–3 × 28 × 10–4 × 160 C1 = 3.0 × 10–3

Wb A1 [2] (c) (i) (induced) e.m.f. proportional to rate of M1 change of (magnetic) flux (linkage) A1 [2]

(ii) e.m.f. = (2 × 3.0 × 10–3) / 0.80 C1 = 7.4 × 10–3

V A1 [2] 6 (a) (i) to reduce power loss in the core B1 due to eddy currents/induced currents B1 [2] (ii) either no power loss in transformer or input power = output power B1 [1] (b) either r.m.s. voltage across load = 9.0 × (8100 / 300) C1 peak voltage across load = √2 × 243 = 340 V A1 [2] or peak voltage across primary coil = 9.0 × √2 (C1) peak voltage across load = 12.7 × (8100/300) = 340 V (A1) 7 (a) (i) lowest frequency of e.m. radiation M1 giving rise to emission of electrons (from the surface) A1 [2] (ii) E = hf C1 threshold frequency = (9.0 × 10–19) / (6.63 × 10–34) = 1.4 × 1015

Hz A1 [2] (b) either 300 nm ≡ 10 × 1015

Hz (and 600 nm ≡ 5.0 × 1014 Hz)

or 300 nm ≡ 6.6 × 10–19 J (and 600 nm ≡ 3.3 × 10–19

J) or zinc λ0 = 340 nm, platinum λ0 = 220 nm (and sodium λ0 = 520 nm) M1 emission from sodium and zinc A1 [2] (c) each photon has larger energy M1 fewer photons per unit time M1 fewer electrons emitted per unit time A1 [3]




8 (a) two (light) nuclei combine M1 to form a more massive nucleus A1 [2] (b) (i) ∆m = (2.01410 u + 1.00728 u) – 3.01605 u = 5.33 × 10–3 u C1 energy = c2 × ∆m C1 = 5.33 × 10–3 × 1.66 × 10–27 × (3.00 × 108)2 = 8.0 × 10–13

J A1 [3] (ii) speed/kinetic energy of proton and deuterium must be very large B1 so that the nuclei can overcome electrostatic repulsion B1 [2]

Section B

9 (a) (i) light-dependent resistor/LDR B1 [1] (ii) strain gauge B1 [1] (iii) quartz/piezo-electric crystal B1 [1] (b) (i) resistance of thermistor decreases as temperature increses M1 etiher VOUT = V × R / (R + RT) or current increases and VOUT = I R A1 VOUT increases A1 [3] (ii) either change in RT with temperature is non-linear or VOUT is not proportional to RT/ change in VOUT with RT is non-linear M1 so change is non-linear A1 [2] 10 (a) sharpness: how well the edges (of structures) are defined B1 contrast: difference in (degree of) blackening between structures B1 [2] (b) e.g. scattering of photos in tissue/no use of a collimator/no use of lead grid large penumbra on shadow/large area anode/wide beam large pixel size (any two sensible suggestions, 1 each) B2 [2] (c) (i) I = I0e

–µx C1 ratio = exp(–2.85 × 3.5) / exp(–0.95 × 8.0) C1 = (4.65 × 10–5) / (5.00 × 10–4) = 0.093 A1 [3] (ii) either large difference (in intensities) or ratio much less than 1.0 M1 so good contrast A1 [2] (answer given in (c)(ii) must be consistent with ratio given in (c)(i))




11 (a) (i) amplitude of the carrier wave varies M1 (in synchrony) with the displacement of the information signal A1 [2] (ii) e.g. more than one radio station can operate in same region/less interference enables shorter aerial increased range/less power required/less attenuation less distortion (any two sensible answers, 1 each) B2 [2] (b) (i) frequency = 909 kHz C1 wavelength = (3.0 × 108) / (909 × 103) = 330 m A1 [2] (ii) bandwidth = 18 kHz A1 [1] (iii) frequency = 9000 Hz A1 [1] 12 (a) for received signal, 28 = 10 lg(P / 0.36 × 10–6) C1 P = 2.3 × 10–4

W A1 [2] (b) loss in fibre = 10 lg(9.8 × 10–3 / 2.27 × 10–4) C1 = 16 dB A1 [2] (c) attenuation per unit length = 16 / 85 = 0.19 dB km–1 A1 [1]




9702 PHYSICS

9702/51 Paper 5 (Planning, Analysis and Evaluation), maximum raw mark 30






1 Planning (15 marks)

Defining the problem (3 marks)

P h is the independent variable or vary h. [1] P Q is the dependent variable or measure Q (allow t). [1] P Keep l constant. [1] Methods of data collection (5 marks)

M Labelled diagram of apparatus: including labelled measuring cylinder/calibrated beaker to receive water. (Measurement may be credited in the text.) [1]

M Vary position of vertical/larger tube. [1] M Measure h and l with a rule/caliper. [1] M Measure d with a travelling microscope or vernier calipers. [1] M Measure t with stopwatch. [1] Method of analysis (2 marks)

A Plot a graph of Q against h. [Allow lg Q against lg h] [1]

A gradient

2 4

×

=

l

gd πρη

Must include gradient and η must be subject of formula. [1]

Safety considerations (1 mark)

S Reasoned method to prevent spills, e.g. use tray/sink/cloths on floor. Reasoned method to prevent injury when adjusting metal/glass tubes by wearing protective gloves. Mark may not be scored from the diagram. [1]

Additional detail (4 marks) D Relevant points might include [4] 1 Repeat experiment for same h and average 2 Method to determine the density of water including method to measure mass and volume

and equation 3 Take many readings of d and average 4 Relationship is valid if straight line passing through origin [if lg-lg graph allow straight line with

gradient = 1] 5 Method to check that tube is horizontal 6 Detail on measuring h to the centre of the horizontal tube e.g. add radius of tube 7 Keep temperature of water constant Do not allow vague computer methods. [Total: 15]




2 Analysis, conclusions and evaluation (15 marks)

Mark Expected Answer Additional Guidance

(a) A1 Gradient = 1 / EW y-intercept = 1 / E

(b) T1

V

1 / V–1

Column heading. Allow equivalent unit. e.g. V–1 / V–1 or 1/V / 1/V or 1/V (V–1)

T2

0.20 or 0.196

0.22 or 0.222

0.25 or 0.250

0.30 or 0.303

0.33 or 0.333

0.37 or 0.370

A mixture of 2 s.f. and 3 s.f. is allowed.

(c) (i) G1 Six points plotted correctly Must be less than half a small square. Ecf allowed from table. Penalise ‘blobs’.

U1 All error bars in C plotted correctly

Must be within half a small square. Ecf allowed from table. Horizontal.

(c) (ii) G2 Line of best fit If points are plotted correctly then lower end of line should pass between (0.75, 0.200) and (0.75, 0.205) and upper end of line should pass between (3.0, 0.352) and (3.0, 0.358). Allow ecf from points plotted incorrectly – examiner judgement.

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 at least half the length of the drawn line. Check the read offs. Work to half a small square. Do not penalise POT.

U2 Uncertainty in gradient Method of determining absolute uncertainty Difference in worst gradient and gradient.

(c) (iv) C2 y-intercept Expect to see point substituted into y = mx + c FOX does not score. Do not penalise POT. Should be about 0.15.

U3 Uncertainty in y-intercept Difference in worst y-intercept and y-intercept. FOX does not score. Allow ecf from (c)(iv).




(d) (i) C3 E = 1/ y-intercept and V Method required. Do not check calculation. Allow ecf from (c)(iv).

U4 Absolute uncertainty in E

(d) (ii) C4 Between 2.00 × 10–3 F and

2.40 × 10–3 F and given to 2 or 3 s.f.

Must be in range. Allow use of mF.

(d) (iii) U5 Percentage uncertainty in W %uncertainty in E + %uncertainty in gradient

[Total: 15] Uncertainties in Question 2

(c) (iii) Gradient [U2]

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

(c) (iv) y-intercept [U3]

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

(d) (i) [U4]

Absolute uncertainty = max E – E = E – min E = 2

minmax EE −

Absolute uncertainty = Ec

c×

∆

(d) (iii) [U5]

Percentage uncertainty = 100×

W

W∆

∆W = max W – W = W – min W = 2

minmax WW −

max W = mE minmin

1

×

min W= mE maxmax

1

×


+

E

E

m

m ∆∆= 100×

+

c

c

m

m ∆∆




9702 PHYSICS









P f is the independent variable or vary f. [1]

P I0 is the dependent variable or measure I0. [1] P Keep V0 constant. [1] Methods of data collection (5 marks)

M Labelled/ workable circuit diagram of apparatus: including coil connected to power supply/signal generator. [1]

M Variable frequency ac power supply or signal generator (method of varying frequency). [1] M Measure current using ammeter or p.d. across resistor. [1] M Measure V across power supply or across coil e.g. voltmeter or c.r.o. explained. [1] M Measure f using oscilloscope/read off signal generator. [1] Method of analysis (2 marks)

A Plot a graph of 1/I0

2 against f

2 or V0

2/I0

2 against f

2 or equivalent. Do not allow log-log graph. [1]

A gradient24

gradient0

2

2

0

π

=

π

=

VVL or

24

gradient

π

=L [1]


S Reasoned method to prevent overheating of coil or burns from coil. [1] Additional detail (4 marks)

D Relevant points might include [4] 1 Use lower frequencies to produce larger currents 2 Keep resistance of circuit/coil constant 3 Additional detail on measuring T using timebase 4 f = 1 / period 5 Additional detail on measuring V0 using y-gain 6 Relationship is valid if straight line, provided plotted graph is correct 7 Relationship is valid if straight line not passing through origin, provided plotted graph is

correct (any quoted expression must be correct) 8 Detail on changing r.m.s. to peak Do not allow vague computer methods. Ignore reference to iron core / other magnetic fields.

[Total: 15]






(a) A1 Gradient =

E

b4

(b) T1 a2 / 10–4 m2 Column heading. Allow equivalent unit. Do not award if 10–4 omitted.

T2

44 or 43.6

52 or 51.8

61 or 60.8

71 or 70.6

79 or 79.2

88 or 88.4


U1 1.3 increasing to 1.9 Allow 1 increasing to 2.

(c) (i) G1 Six points plotted correctly Must be within half a small square. Ecf allowed from table. Penalise ‘blobs’.

U2 All error bars in a2 plotted correctly

Must be within half a small square. Ecf allowed from table.

(c) (ii) G2 Line of best fit If points are plotted correctly then lower end of line should pass between (1000, 43) and (1000,44.5) and upper end of line should pass between (2000, 87) and (2000, 89). Allow ecf from points plotted incorrectly – examiner judgement.








(d) (i) C2 E = 4b/gradient Method required. Do not check calculation. Should be about 36 000. Do not penalise POT error.

C3 V m–1 or N C–1 Penalise POT error.

(d) (ii) U4 Percentage uncertainty in E Must be greater than 2.5%.

(e) C4 V = 540 to 580 V and given to 2 or 3 s.f.

Working must be correct; check for inconsistent units. Must be in stated range.

U5 Absolute uncertainty in V




(d) [U4]

Percentage uncertainty = 100×∆

E

E

max E = m

b

min

max

min E = m

b

max

min

Percentage uncertainty = 2.5100100 +×

∆=×

∆+

∆

m

m

b

b

m

m

(e) [U5]

Absolute uncertainty = max V – V = V – min V = 2

minmax VV −

max V = b

aE

min4

max max 2

×

= m

a

min

max2

min V = b

aE

max 4

min min2

×

= m

a

max

min2

Absolute uncertainty = Vb

b

E

E

a

aV

m

m

a

a

∆+

∆+

∆=

∆+

∆ 22




9702 PHYSICS



Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. 12 Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2013 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.






P h is the independent variable or vary h. [1] P Q is the dependent variable or measure Q (allow t). [1] P Keep l constant. [1] Methods of data collection (5 marks)

M Labelled diagram of apparatus: including labelled measuring cylinder/calibrated beaker to receive water. (Measurement may be credited in the text.) [1]

M Vary position of vertical/larger tube. [1] M Measure h and l with a rule/caliper. [1] M Measure d with a travelling microscope or vernier calipers. [1] M Measure t with stopwatch. [1] Method of analysis (2 marks)

A Plot a graph of Q against h. [Allow lg Q against lg h] [1]

A gradient

2 4

×

=

l

gd πρη

Must include gradient and η must be subject of formula. [1]


S Reasoned method to prevent spills, e.g. use tray/sink/cloths on floor. Reasoned method to prevent injury when adjusting metal/glass tubes by wearing protective gloves. Mark may not be scored from the diagram. [1]

Additional detail (4 marks) D Relevant points might include [4] 1 Repeat experiment for same h and average 2 Method to determine the density of water including method to measure mass and volume

and equation 3 Take many readings of d and average 4 Relationship is valid if straight line passing through origin [if lg-lg graph allow straight line with

gradient = 1] 5 Method to check that tube is horizontal 6 Detail on measuring h to the centre of the horizontal tube e.g. add radius of tube 7 Keep temperature of water constant Do not allow vague computer methods. [Total: 15]






(a) A1 Gradient = 1 / EW y-intercept = 1 / E

(b) T1

V

1 / V–1

Column heading. Allow equivalent unit. e.g. V–1 / V–1 or 1/V / 1/V or 1/V (V–1)

T2

0.20 or 0.196

0.22 or 0.222

0.25 or 0.250

0.30 or 0.303

0.33 or 0.333

0.37 or 0.370


(c) (i) G1 Six points plotted correctly Must be less than half a small square. Ecf allowed from table. Penalise ‘blobs’.

U1 All error bars in C plotted correctly

Must be within half a small square. Ecf allowed from table. Horizontal.

(c) (ii) G2 Line of best fit If points are plotted correctly then lower end of line should pass between (0.75, 0.200) and (0.75, 0.205) and upper end of line should pass between (3.0, 0.352) and (3.0, 0.358). Allow ecf from points plotted incorrectly – examiner judgement.





(c) (iv) C2 y-intercept Expect to see point substituted into y = mx + c FOX does not score. Do not penalise POT. Should be about 0.15.

U3 Uncertainty in y-intercept Difference in worst y-intercept and y-intercept. FOX does not score. Allow ecf from (c)(iv).




(d) (i) C3 E = 1/ y-intercept and V Method required. Do not check calculation. Allow ecf from (c)(iv).

U4 Absolute uncertainty in E

(d) (ii) C4 Between 2.00 × 10–3 F and

2.40 × 10–3 F and given to 2 or 3 s.f.

Must be in range. Allow use of mF.

(d) (iii) U5 Percentage uncertainty in W %uncertainty in E + %uncertainty in gradient




(c) (iv) y-intercept [U3]

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

(d) (i) [U4]

Absolute uncertainty = max E – E = E – min E = 2

minmax EE −

Absolute uncertainty = Ec

c×

∆

(d) (iii) [U5]


W

W∆

∆W = max W – W = W – min W = 2

minmax WW −

max W = mE minmin

1

×

min W= mE maxmax

1

×


+

E

E

m

m ∆∆= 100×

+

c

c

m

m ∆∆

9702 s13 ms_all

Education

gce advanced level mark

level mayjune

mark schemes

mark scheme syllabus

ordinary level components

question paper

maximum raw mark

multiple choice