9702 s11 ms_all

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS

GCE Advanced Subsidiary Level and GCE Advanced Level

MARK SCHEME for the May/June 2011 question paper

for the guidance of teachers

9702 PHYSICS

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

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

• Cambridge will not enter into discussions or correspondence in connection with these mark schemes.

Cambridge is publishing the mark schemes for the May/June 2011 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Mark Scheme: Teachers’ version Syllabus Paper

GCE AS/A LEVEL – May/June 2011 9702 11

© University of Cambridge International Examinations 2011

Question Number

Key Question Number

Key

1 C 21 A

2 C 22 D

3 C 23 B

4 C 24 A

5 A 25 D

6 A 26 B

7 B 27 D

8 B 28 A

9 A 29 C

10 D 30 D

11 B 31 C

12 D 32 A

13 C 33 A

14 C 34 C

15 A 35 C

16 B 36 C

17 C 37 D

18 C 38 D

19 D 39 B

20 B 40 D





9702 PHYSICS








Question Number

Key Question Number

Key

1 C 21 C

2 B 22 B

3 A 23 C

4 D 24 B

5 D 25 C

6 D 26 B

7 D 27 D

8 A 28 C

9 D 29 B

10 B 30 A

11 A 31 B

12 B 32 C

13 D 33 D

14 A 34 D

15 B 35 A

16 C 36 C

17 C 37 B

18 D 38 C

19 D 39 C

20 A 40 B





9702 PHYSICS








Question Number

Key Question Number

Key

1 C 21 B

2 C 22 A

3 C 23 D

4 A 24 B

5 C 25 B

6 B 26 D

7 A 27 D

8 B 28 D

9 D 29 A

10 A 30 C

11 D 31 A

12 C 32 C

13 B 33 C

14 C 34 A

15 A 35 C

16 C 36 D

17 B 37 C

18 C 38 D

19 D 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.


• Cambridge will not enter into discussions or correspondence in connection with these mark schemes. Cambridge is publishing the mark schemes for the May/June 2011 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.




1 (a) (i) metre rule / tape (not ‘rule’) B1 [1] (ii) micrometer (screw gauge) / digital caliper B1 [1] (iii) ammeter and voltmeter / ohmmeter / multimeter on ‘ohm’ setting B1 [1] (b) (i) resistivity = RA / L C1

= [7.5 × π × (0.38 × 10–3)2 / 4] / 1.75 M1

= 4.86 × 10–7 Ω m A0 [2] (ii) (uncertainty in R =) [0.2 / 7.5] × 100 = 2.7% and (uncertainty in L =) [3 / 1750] × 100 = 0.17% C1 (uncertainty in A =) 2 × (0.01 / 0.38) × 100 = 5.3 % C1 total = 8.13% C1

uncertainty = 0.395 × 10–7 (Ω m) A1 [4] (missing 2 factor in uncertainty in A, then allow max 3/4)

(c) resistivity = (4.9 × 10–7 ± 0.4 × 10–7) Ω m A1 [1]

2 (a) work done is the force × the distance moved / displacement in the direction of the force

or work is done when a force moves in the direction of the force B1 [1] (b) component of weight = 850 × 9.81 × sin 7.5° C1 = 1090 N A1 [2] (use of incorrect trigonometric function, 0/2)

(c) (i) Σ F = 4600 – 1090 = (3510) M1 deceleration = 3510 / 850 A1 = 4.1 m s–2 A0 [2] (ii) v2 = u2 + 2as 0 = 252 + 2 × – 4.1 × s C1 s = 625 / 8.2 = 76 m A1 [2] (allow full credit for calculation of time (6.05 s) & then s) (iii) 1. kinetic energy = ½ mv2 C1 = 0.5 × 850 × 252 = 2.7 × 105 J A1 [2] 2. work done = 4600 × 75.7 = 3.5 × 105 J A1 [1] (iv) difference is the loss in potential energy (owtte) B1 [1]




3 (a) point where the weight of an object / gravitational force M1 may be considered to act A1 [2] (b) product of the force and the perpendicular distance (to the pivot) B1 [1] (c) (i) 1. sum / net / resultant force is zero B1 2. net / resultant moment is zero sum of clockwise moments = sum of anticlockwise moments B1 [2] (ii) W × 0.2 = 80 × 0.5 + 70 × 1.3 C1 = 40 + 91 C1 W = 655 N A1 [3] (allow 2/3 for one error in distance but 0/3 if two errors) (iii) move pivot to left (M1) gives greater clockwise moment / smaller anticlockwise moment (A1) or move W to right (M1) gives smaller anticlockwise moment (A1) [2] 4 (a) (i) stress is force / area B1 [1] (ii) strain is extension / original length B1 [1]

(b) (i) E = [F / A] ÷ [e / l] C1 e = (25 × 1.7) / (5.74 × 10–8 × 1.6 × 1011) C1 e = 4.6 × 10–3 m A1 [3] (ii) A becomes A/2 or stress is doubled B1

e ∝ l / A or substitution into full formula B1 total extension increase is 4e A1 [3]

5 (a) (i) I = 12 / (6 + 12) C1 minimum current = 0.67 A A1 [2] (ii) correct start and finish points M1 correct shape for curve with decreasing gradient A1 [2] (b) maximum current = 2.0 A A1 minimum current = 0 A1 [2] (c) (i) smooth curve starting at (0,0) with decreasing gradient M1 end section not horizontal A1 [2] (ii) full range of current / p.d. possible or currents / p.d. down to zero or brightness ranging from off to full brightness B1 [1]




6 (a) any two of: large number of molecules / atoms / particles molecules in random motion no intermolecular forces elastic collisions time of collisions much less than time between collisions volume of molecules much less than volume of containing vessel B1 + B1 [2] (b) molecules collide with the walls change in momentum of molecules implies force (on molecules) molecules exert equal and opposite force on wall pressure is averaging effect of many collisions (any three statements, 1 each) B3 [3]

7 (a) when waves overlap / meet, (resultant) displacement is the sum of the individual displacements B1 [1]

(b) (i) two (ball-type) dippers (M1) connected to the same vibrating source /motor (A1) or one wave source described (M1) with two slits (A1) [2] (ii) lamp with viewing screen on opposite side of tank B1 means of freezing picture e.g. strobe B1 [2] (c) (i) two correct lines labelled X B1 [1] (ii) correct line labelled N B1 [1]





9702 PHYSICS








1 (a) scalar has only magnitude B1 vector has magnitude and direction B1 [2] (b) kinetic energy, mass, power all three underlined B1 [1] (c) (i) s = ut + ½ at2

15 = 0.5 × 9.81 × t2 C1 T = 1.7 s A1 [2] if g = 10 is used then –1 but only once on paper (ii) vertical component vv: vv

2 = u2 + 2as = 0 + 2 × 9.81 × 15 or vv = u + at = 9.81 × 1.7(5) vv =17.16 C1 resultant velocity: v2 = (17.16)2 + (20)2 C1 v = 26 m s–1 A1 [3] If u = 20 is used instead of u = 0 then 0/3 Allow the solution using: initial (potential energy + kinetic energy) = final kinetic energy (iii) distance is the actual path travelled B1 displacement is the straight line distance between start and finish points (in

that direction) / minimum distance B1 [2] 2 (a) (i) base units of D: force: kg m s–2 B1 radius: m velocity: m s–1 B1 base units of D: [F / (R × v)] kg m s–2 / (m × m s–1) M1 = kg m–1

s–1 A0 [3]

(ii) 1. F = 6π × D × R × v = [6π × 6.6 × 10–4 × 1.5 × 10–3 × 3.7] = 6.9 × 10–5 N A1 [1] 2. mg – F = ma hence a = g – [F / m]

m = ρ × V = ρ × 4/3 π R3 = (1.4 × 10–5) C1

a = 9.81 – [6.9 × 10–5] / ρ × 4/3 π × (1.5 × 10–3)3 (9.81 – 4.88) M1 a = 4.9(3) m s–2 A1 [3] (b) (i) a = g at time t = 0 B1 a decreases (as time increases) B1 a goes to zero B1 [3] (ii) Correct shape below original line M1 sketch goes to terminal velocity earlier A1 [2]




3 (a) (i) work done equals force × distance moved / displacement in the direction of the force B1 [1]

(ii) power is the rate of doing work / work done per unit time B1 [1] (b) (i) kinetic energy = ½ mv2 C1 = 0.5 × 600 (9.5)2 C1 = 27075 (J) = 27 kJ A1 [3] (ii) potential energy = mgh = 600 × 9.81 × 4.1 M1 = 24132 (J) A1 = 24 kJ A0 [2] (iii) work done = 27 – 24 = 3.0 kJ A1 [1] (iv) resistive force = 3000 / 8.2 (distance along slope = 4.1 / sin 30°) C1 = 366 N A1 [2]

4 (a) clamped horizontal wire over pulley or vertical wire attached to ceiling with mass attached B1

details: reference mark on wire with fixed scale alongside B1 [2] (b) measure original length of wire to reference mark with metre ruler / tape (B1) measure diameter with micrometer / digital calipers (B1) measure initial and final reading (for extension) with metre ruler or other suitable

scale (B1) measure / record mass or weight used for the extension (B1) good physics method: measure diameter in several places / remove load and check wire returns to

original length / take several readings with different loads (B1) MAX of 4 points B4 [4] (c) determine extension from final and initial readings (B1) plot a graph of force against extension (B1) determine gradient of graph for F / e (B1)

calculate area from πd2 / 4 (B1) calculate E from E = F l / e A or gradient × l / A (B1) MAX of 4 points B4 [4]




5 (a) (i) energy converted from chemical to electrical when charge flows through cell or round complete circuit B1

(ii) (resistance of the cell) causing loss of voltage or energy loss in cell B1 [2]

(b) (i) EB – EA = I (R + rB + rA)

12 – 3 = I (3.3 + 0.1 + 0.2) C1

I = 2.5 A A1 [2]

(ii) Power = E × I = 12 × 2.5 C1 = 30 W A1 [2]

(iii) P = I2 × R or P = V2 / R or P = VI = (2.5)2 × 3 = 92 / 3.6 = 9 × 2.5 C1 = 22.5 J s–1 A1 [2] (c) power supplied from cell B is greater than energy lost per second in circuit B1 [1]

6 (a) (i) to produce coherent sources or constant phase difference B1 [1]

(ii) 1. 360° / 2π rad allow n × 360° or n × 2π (unit missing –1) B1 [1]

2. 180° / π rad allow (n × 360°) – 180° or (n × 2π) – π B1 [1] (iii) 1. waves overlap / meet B1 (resultant) displacement is sum of displacements of each wave B1 [2] 2. at P crest on trough (OWTTE) B1 [1]

(b) λ = ax / D C1 = 2 × 2.3 × 10–3 × 0.25 ×10–3 / 1.8 C1 = 639 nm A1 [3]





9702 PHYSICS








1 (a) 2nd row random, 3rd row neither, 4th row systematic all correct B2 two correct scores 1 only [2] (b) (i) 1. systematic error: the average / peak is not the true value / the readings

are not centred around the true value B1 [1] 2. random error: readings have positive and negative values around the

peak value / values are scattered / wide range B1 [1] (ii) 1. accurate: peak / average value moves towards the true value B1 [1] 2. precise: lines are closer together / sharper peak B1 [1] 2 (a) resultant moment = zero / sum of clockwise moments = sum of anticlockwise

moments B1 resultant force = 0 B1 [2] (b) shape and orientation correct and forces labelled and arrows correct M1 angles correct / labelled A1 [2] (c) (i) T cos18° = W Scale diagram: C1 T = 520 / cos18° = 547 N ± 20 N A1 [2] (ii) R = T sin18° = 169 N ± 20 N A1 [1]

(d) θ is larger hence cos θ is smaller, T = W / cos θ M1 hence T is larger A0 [1] 3 (a) weight = m × g = 130.5 × 9.81 = 1280 N A1 [1] (b) (i) F = ma T – 1280 = 130.5 × 0.57 C1 T = 1280 + 74.4 = 1350 N A1 [2] (ii) 1280 N A1 [1] (c) 1240 – 1280 = 130.5 × a C1 a = (–) 0.31 m s–2 A1 [2] (d) (i) 1. 3.5 s A1 [1] 2. 6.5 s A1 [1]




(ii) basic shape M1 correct points A1 [2] 4 (a) force is proportional to extension B1 [1]

(b) (i) gradient of graph determined (e.g. 50 / 40 ×10–3 ) = 1250 N m–1 A1 [1]

(ii) W = ½ k x2 or W = ½ final force × extension M1 = 0.5 × 1250 × (36 × 10–3)2 or 0.5 × 45 × 36 × 10–3 M1 = 0.81 J A0 [2]

(c) (i) 0.81 = ½ mv2 C1 v = 8.0 (8.0498) m s–1 A1 [2]

(ii) 4 × KE / 4 × WD or 3.24 J C1 hence twice the compression = 72 mm A1 [2]

(iii) Max height is when all KE or WD or elastic PE is converted to GPE C1 ratio = 1/4 or 0.25 A1 [2]

5 (a) (i) Start from (0,0) and smooth curve in correct direction B1 Curve correct for end section never horizontal B1 [2]

(ii) R = V / I hence take co-ords of V and I from graph and calculate V / I B1 [1]

(b) (i) each lamp in parallel has a greater p.d. / greater current M1 lamp hotter M1 resistance of lamps in parallel greater A1 [3]

(ii) P = V2 / R or P = VI and V = IR C1 R = 144 / 50 = 2.88 for each lamp C1

total R = 1.44 Ω A1 [3]

6 (a) (i) amplitude = 7.6 mm allow 7.5 mm A1 [1]

(ii) 180° / π rad A1 [1]

(iii) v = f × λ = 15 × 0.8 C1 = 12 m s–1 A1 [2]

(b) correct sketch with peak moved to the right B1 curve moved by the correct phase angle / time period of 0.25 T B1 [2]

(c) (i) zero (rad) A1 [1]

(ii) antinode maximum amplitude, node zero amplitude / displacement A1 [1]




(iii) 3 A1 [1]

(iv) horizontal line through central section of wave B1 [1] 7 (a) density in solids and liquids similar M1 spacing in solids and liquids about the same A1 density in gases much less as spacing in gases much greater B1 [3] (b) density = mass / volume C1 mass = 1.67 × 10–27

kg and volume = 4/3 π r 3 C1 density = (1.67 × 10–27) / 4/3 × π × (1.0 × 10–15)3 = 3.99 × 1017 kg m–3 A1 [3] (c) atoms / molecules composed of large amount of empty space / nucleus has very

small volume compared to volume of atom / space between atoms in a gas is very large B1 [1]


GCE Advanced Level



9702 PHYSICS

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





GCE A LEVEL – May/June 2011 9702 31


1 (b) (iii) Value of I non-zero value and 10 mA with unit. [1]

Value of V 0.5 V 1.5 V with unit. [1]

(c) Six sets of readings of I and V scores 5 marks, five sets scores 4 marks etc. [5] Incorrect trend then –1. Minor help from supervisor –1; major help –2.

Range: Range of I at least 0.3 mA. [1] 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 e.g. V / V. Consistency of presentation of raw readings: [1]

All values of I must be given to the same number of decimal places. All values of V must be given to the same number of decimal places.

Significant figures: [1] Significant figures for 1/V the same as, or one more than, that for V.

Calculation: Check the values of 1/V and 1/I. [1] (d) (i) 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 on the grid occupy at least half the graph grid in both x and y directions. Scales must be labelled with the quantity that is being plotted. Ignore units.

Scale markings should not be greater than three large squares apart. Plotting of points: [1] All the observations in the table must be plotted. Check the points are plotted correctly.

Work to an accuracy of half a small square. Do not accept blobs (points with diameter greater than half a small square). Quality: [1] All points in the table must be plotted (at least 5) for this mark to be scored. Judge by

the scatter of all points about a straight line. All points must be within 0.01 V–1 on the 1/V axis from a straight line.

(ii) Line of best fit: [1] Judge by the balance of all the points (at least 5) about the candidate’s line. There must

be an even distribution of points either side of the line along the full length. (iii) Gradient: [1] The hypotenuse of the triangle must be at least half the length of the drawn line. Read-

offs must be accurate to half a small square. Intercept: [1] Either: Check correct read-off from a point on the line, and substitution into y = mx + c. Read-

off must be accurate to half a small square. Allow ecf of gradient value. Or: Check the read-off of the intercept directly from the graph.




(e) Correct method to find R. [1]

Answer in range 40 – 60 Ω with unit. [1] [Total: 20] 2 (a) Measurement of t in the range 0.20 cm – 1.00 cm to 0.1 mm or 0.01 mm with unit. [1] (b) (i) Measurement of d in the range 3 cm – 9 cm with unit. [1] (ii) Correct calculation of w. [1] (c) (ii) Value of T in the range 3 s – 5 s. [1] Evidence of repeat readings. [1] (d) Absolute uncertainty in T in range 0.1 s – 0.6 s. [1] If repeated readings have been taken, then the uncertainty can be half the range. Correct method of calculation of percentage uncertainty. (e) Second value of d in the range 14 cm – 31 cm . [1] Correct calculation of second value of w. [1] Second value of T. [1] Second value of T < first value of T. [1] (f) (i) Correct calculation of two values of k. [1] (ii) Sensible comment relating to the calculated values of k, testing against a specified

criterion. [1]




(g)

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

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

Take more readings and plot a graph/ calculate more k values (and compare) Allow ‘repeat readings and plot a graph’

Few readings Take more readings and calculate average k Only one reading

B Rule hits bench

Method of preventing rule hitting bench e.g. project end of cylinder over bench or elevate apparatus

Ignore amplitude changes/difficult to start at the same amplitude each time

C rule used for wider diameter/ couldn’t use calipers

Method to improve measurement of larger diameter e.g. use set squares held against ruler/wrap string or paper around and measure circumference/use calipers and hold against ruler/travelling microscope

Use larger Vernier calipers

D Difficult to judge when oscillation is complete

1.Use video (+ playback) + timer/use clock on video / use position or motion sensor placed above/below rule (not above centre) / use of light gate with detailed method. 2.Use (fiducial) marker/pointer at centre (of oscillation)

Difficult to measure the time/human error/ references to reaction times

E Oscillations die away quickly/too few oscillations/ damped

F Use same surface/material (for cylinders)

Ignore ‘parallax problems’, ‘use assistant’ or references to draughts, fans, air conditioning.

[Total: 20]


GCE Advanced Level



9702 PHYSICS








1 (b) Raw reading for nail height H, to nearest mm. [1] (d) (i) Reading for string height h, less than H. [1] (e) No help from supervisor. [1]

Six sets of readings scores 4 marks, five sets scores 3 marks etc. [4] Incorrect trend then –1. Range: m values must include 180 g or more. [1] 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. Consistency of presentation of raw readings: [1] All values of h must be given to the nearest mm. All values of m must be given to the nearest g. Significant figures: [1] S.f. for 1/(H–h)2 must be the same as, or one more than, the s.f. given for (H–h). Calculation: [1] 1/(H–h)2 calculated correctly.

(f) (i) Axes: [1] Sensible scales must be used, no awkward scales (e.g. 3:10). Scales must be chosen so that the plotted points must 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. Scale markings must be no more than 3 large squares apart. Plotting of points: [1] All observations in the table must be plotted. Check that the points are correctly plotted. Work to an accuracy of half a small square. Do not accept blobs (points with diameter greater than half a small square).

Quality: [1] All points in the table must be plotted (at least 5) for this mark to be scored. Scatter of points must be less than ± 2000 g2 on the m2 axis from a straight line.

(ii) Line of best fit: [1]

Judge by balance of all the points (at least 5) about the candidate's line. There must be an even distribution of points either side of the line along the full length.




(f) (iii) Gradient: [1] The hypotenuse must be at least half the length of the drawn line. Both read-offs must be accurate to half a small square. If incorrect, write in the correct value(s). Do not allow use of points from the table unless they are on the

line. Do not allow ∆x/∆y. Intercept: [1] Either: Check correct read-off from a point on the line, and substitution into y = mx + c. Read-off must be accurate to half a small square. Allow ecf of gradient value. Or: Check the read-off of the intercept directly from the graph.

(g) Correct method used to find a and b. [1] Correct unit for a and correct unit for b. [1] [Total: 20] 2 (a) (ii) y in range 65 to 75 cm. [1] (iii) Value for h to nearest mm and in range 1 to 20 cm, with unit. [1] (b) (ii) First value of x in range 8 to 11 cm. [1] (iii) First value of h1. [1] (c) (i) First value of d calculated correctly. [1] (ii) Percentage uncertainty in d calculated using correct method and using absolute

uncertainty of 1 or 2 mm (or half the range if repeated readings have been taken). [1] (e) (ii) Second value of x. [1]

Second value of h1. [1] Repeats: Any evidence of repeats for height values or x values. [1] Quality: Second value of d less than first value. [1]

(f) (i) Two values of k calculated correctly. [1] (ii) Sensible comment relating to the calculated values of k, testing against a specified

criterion. [1]




(g)

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

A Two readings are not enough (to draw a conclusion).

Take more readings and plot a graph/calculate more k values (and compare). Allow ‘repeat readings and plot a graph’

Few readings/take more readings and calculate average k/only one reading

B d is very small. 1. Use larger mass/use larger x value.

2. Use thinner rule.

Parallax error.

C Difficult to measure h (with reason).

Use vernier caliper/travelling microscope/dial gauge/position sensor above rule.

D Difficult to measure x (with reason)/difficult to judge position of mass.

Method of improving measurement of x (e.g. hang masses below rule).

X Other specific relevant problem with apparatus.

Relevant solution. Apparatus slips.

Do not accept ‘repeated readings’ or ‘light gates’. [Total: 20]


GCE Advanced Level



9702 PHYSICS








1 (a) (i) Value of x in the range 1 cm – 3 cm. [1] (b) (ii) Value of T in range 1.8 s T 4.5 s with consistent unit. [1] If outside this range allow SV ± 40% (write in SV if used). Evidence of repeat times. [1] (c) Six sets of readings of x and T scores 4 marks, five sets scores 3 marks etc. [4] Incorrect trend then –1. Help from supervisor –1. Range of x : To include 1 cm and 6 cm. [1] Column headings: [1] Each column heading must contain a quantity and a unit. There must be some distinguishing mark between the quantity and the unit e.g. T / s. Ignore POT errors. Ignore units in body of table. Consistency of presentation of raw readings: [1] All values of x must be given to the nearest mm. Significant figures: Significant figures for every row of 1/x same as, or one more than, raw x. [1] Calculation: 1/x calculated correctly. [1] (d) (i) 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 on the grid occupy at least half the graph grid in both x and y directions. Indicate false origin with FOX.

Scales must be labelled with the quantity which is being plotted. Ignore units. Scale markings should not be more than three large squares apart. Plotting of points: [1] All observations in table must be plotted. Write a ringed total of plotted points ignoring any point off the grid. Check points plotted correctly. Tick if correct. Re-plot if incorrect. Work to an accuracy of half a small square. Do not accept ‘blobs’ (points with diameter greater than half a small square). Quality: [1] All points in the table must be plotted (at least five) for this mark to be scored. Judge by

scatter of all points about straight line. All points must be within 0.05 cm–1 of 1/x from a straight line.

(ii) Line of best fit: [1] Judge by the balance of all the points (at least five) about candidate’s line. There must

be an even distribution of points either side of the line along the whole length. If mark is not awarded indicate rotation or direction of best fit line. Lines must not be kinked.




(iii) Gradient: [1] The hypotenuse of the triangle must be at least half the length of the drawn line. Read-

offs must be accurate to half a small square. Check for ∆y/∆x (i.e. do not allow ∆x/∆y). If incorrect, write in the correct value(s).

y-intercept: [1] Either: check correct read off from a point on the line and substitute into y = mx + c. Read off

must be accurate to half a small square. Allow ecf of gradient value. Or: check read-off of intercept directly from graph. (e) a is the value of candidate’s gradient with consistent unit (s (c)m or (c)m s). [1] b is the value of candidate’s y-intercept with consistent unit (s). (f) Either: Strip too wide for clips. [1] Or: time too small (to measure). [Total: 20] 2 (a) (ii) Measurement of raw l to nearest mm in the range 90 cm – 100 cm. [1] (iii) Value of h0 with unit. [1] (b) (ii) Value of h < h0. [1] (iii) Check correct calculation of d. [1] (c) Absolute uncertainty in d in the range 1 mm – 2 mm or half the range of repeated readings,

unless zero. Correct method of calculation to get percentage uncertainty. [1] (d) Second value of l in range 55 cm l 65 cm. [1]

Second value of h0. [1] Second value of h < h

0

. [1] Quality : second value of d < first value of d. [1] (e) (i) Correct calculation of two values of k. [1] (ii) Sensible comment relating to the calculated values of k, testing against a specified

criterion. [1] (iii) Justification of sf in k linked to l and d. [1]




(f)

(i) Limitations 4 max (ii) Improvements 4 max Do not credit Ap Two readings (of d and l) not enough/

only two readings/ too few readings As Take more readings and plot a

graph/ more values of k (and compare).

Take more readings and calculate average k / only one reading

Bp Difficult to measure h with reason/ parallax error in h

Bs Detailed use of set square or pointer to improve parallax/ method for easier access/ method of reducing parallax

Mass gets in the way.

Cp d is small Cs1

Cs2

Larger mass Method to measure d directly e.g. using a travelling microscope or position sensor

Dp Rule may not be vertical (when measuring h)

Ds Detailed use of set square (table level)

Xp Specific problem candidate encountered e.g. ruler slips on support/supports slip on block

Xs e.g. glue support to block Ignore reference to computers, using assistance, draughts

[Total: 20]


GCE Advanced Level



9702 PHYSICS








1 (c) Angle x, with unit. [1] (d) (iii) Angle y, greater than x. [1] (e) Six sets of readings scores 4 marks, five sets scores 3 marks etc. [4]

Incorrect trend then –1. Help from supervisor then –1. Range: m values must include 190 g or greater. [1] 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, e.g. m / g. Consistency of presentation of raw readings: [1] All values of y must be given to the nearest degree or half degree. All values of m must be given to the nearest gram (e.g. 190 g or 0.190 kg). Significant figures: [1]

S.f. for sinθ must be the same as, or one more than, the s.f. given for θ .

Calculation: Values of sinθ calculated correctly. [1]

(f) (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. Ignore units. Scale markings must be no more than 3 large squares apart. Plotting of points: [1] All observations in the table must be plotted. Check that the points are correctly plotted. Work to an accuracy of half a small square. Do not accept blobs (points with diameter greater than half a small square).

Quality: [1] All points in the table must be plotted (at least 5) for this mark to be scored.

Scatter of points must be less than ± 0.02 on the sinθ axis from a straight line.

(ii) Line of best fit: [1] Judge by balance of all the points (at least 5) about the candidate's line. There must be an even distribution of points either side of the line along the full length.




(iii) Gradient: [1] The hypotenuse of the triangle used must be at least half the length of the drawn line. Both read-offs must be accurate to half a small square. The method of calculation must be correct. Intercept: [1] Either: Check correct read-off from a point on the line, and substitution into y = mx + c. Read-off must be accurate to half a small square. Allow ecf of gradient value. Or: Check the read-off of the intercept directly from the graph.

(g) (ii) Raw value(s) of r in range 30 to 50 mm (or SV diameter/2 ± 10 mm) and given to

nearest mm, with unit. [1] (h) Method of calculation of a is correct and uses the gradient value. [1] Unit for a has dimensions mass × length (e.g. g cm). [1] [Total: 20] 2 (a) (ii) Value for R, with unit, in range 10 to 50 mm. [1] Diameter is measured to determine R (either here or in (d)). [1] (iii) Percentage uncertainty in R calculated by correct method, with absolute

uncertainty of 0.5 mm or 1 mm or half the range of any repeats. [1] (b) (ii) First measurement of T, with unit, in range 0.5 s to 10.0 s. [1] Evidence of repeat measurements of T. [1] (c) First value of C calculated correctly, with correct unit (e.g. kg mm2). [1] (d) (ii) Second value for R. [1]

Second value for T. [1] Quality: Second T < first T. [1] Second value of C calculated correctly. [1]

(e) (i) Both values of k calculated correctly. [1] (ii) Sensible comment relating to the calculated values of k, testing against a specified

criterion. [1]




(f)

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

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

Take more readings and plot a graph/calculate more k values (and compare). Allow ‘repeat readings and plot a graph’

Few readings/take more readings and calculate average k/only one reading.

B Difficult to judge the end of an oscillation.

1. Use video (+ playback) + timer/use clock on video

2. Use (fiducial) marker/ pointer, with reference point on mass hanger

Difficult to measure the time/human error/references to reaction times/difficult to release from the same point each time. Data logging/light gates motion sensor/“release when marks line up”.

C Diameter/radius of a mass hanger not constant.

Comparison of diameters of 50 g and 100 g mass hangers.

D Mass tends to swing as well as rotate.

Switch off fans.

E T affected when rubber band extends.

F Method of measuring diameter. Use more precisely (e.g. vernier calipers).

G Method of increasing T (e.g. use larger mass/diameter or longer/thinner rubber band).

H Labelled values of mass may not be accurate.

Method of finding mass (e.g. top pan balance).

Do not allow “parallax error”. [Total: 20]


GCE Advanced Level



9702 PHYSICS








1 (a) Value of l0 with unit in range 1.5 cm l0 3.0 cm. [1] (b) (iv) Value of T with unit 20 s T 5 s. [1] Evidence of repeat times. (c) Six sets of readings of l and T scores 4 marks, five sets scores 3 marks etc. [4] Incorrect trend then –1 Help from supervisor –1. Range : ∆x ≥ 7 cm. [1] 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 e.g. T / s, x / cm. Precision of x from raw values of l and l0. [1] Check values of x the same as the least precision in l or l0. [1] (d) (i) 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 on the grid 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. Scale markings should not be more than three large squares apart.

Plotting of points: [1] All observations in table must be plotted. Write a ringed total of plotted points ignoring any point off the grid. Check points plotted

correctly. Tick if correct. Re-plot if incorrect. Work to an accuracy of half a small square. Do not accept ‘blobs’ (points with diameter greater than half a small square).

Quality: [1] All points in the table must be plotted (at least five) for this mark to be scored. Judge by

scatter of all points about straight line. All points must be within 4 mm to scale from a straight line. Indicate tolerance on graph. Indicate reason if mark is not awarded.

(ii) Line of best fit: [1] Judge by the balance of all the points (at least five) about candidates’ line. There must

be an even distribution of points either side of the line along the whole length. If mark is not awarded indicate rotation or direction of best fit line. Lines must not be kinked.

(iii) Gradient: [1] The hypotenuse of the triangle must be at least half the length of the drawn line. Read-offs must be accurate to half a small square. Check for ∆y/∆x (i.e. do not allow ∆x/∆y). If incorrect, write in the correct value(s). y-intercept: [1] Either: check correct read off from a point on the line and substitute into y = mx + c. Read off must be accurate to half a small square. Allow ecf of gradient value.




Or: check read-off of intercept directly from graph. (e) p is the value of the candidate’s gradient in s m–1 , s cm–1 , s mm–1 , mm–1

s. [1] q is the value of the candidate’s y-intercept in s. (f) Value of x (10 – 100 cm) with consistent unit when T = 75 s. [1] Correct method seen. [Total: 20] 2 (a) Measurement of d to nearest 0.01 mm with consistent unit. [1] Evidence of repeat readings. (c) (ii) Value of h in the range 9 cm – 11 cm with unit. [1] (d) (ii) Value of x in the range 1 cm – 5 cm to the nearest mm with unit. [1] (iii) Value of y = x – (10 ± 2) mm. [1] (e) Absolute uncertainty in y in range 2 – 5 mm (or half the range of repeated readings). [1] Correct calculation to get percentage uncertainty. (f) Second value of d < (a). [1]

Second value of x. [1] Quality : second value of y > first value of y. [1] (g) (i) Values of k calculated correctly. [1] (ii) Sensible comment relating to the calculated values of k, testing against a specified

criterion. [1] (iii) Justification of s.f. in k linked to least s.f. in d and y or x. [1]




(h)

(i) Limitations 4 max (ii) Improvements 4 max Do not credit Ap Two readings (of d and l) not enough/

only two readings/too few readings. As Take more readings and plot a

graph/more values of k (and compare).

Take more readings and calculate average k / only one reading.

Bp Maintaining h constant.

Bs Clamp mass hanger/specified release mechanism/hold against fixed pointer.

Cp Explain difficulty in getting measurement of x/depth accurately with finger/position of finger and line may not be in line.

Cs

Put mark on rod/use a clip/ measure rod out of sand with scale or ruler/scale marked on ruler/draw mark all the way round.

Dp Rod falls sideways/not entering sand vertically.

Ds Practical method to keep rod vertical e.g. guide for rod.

Ep Cannot see if mass is directly above rod.

Es Practical method to ensure centralisation of mass e.g. guide for mass.

Do not credit use of computers, assistants, dataloggers.

Fp Depth/x very small Fs Increase height/mass Xp Specific problem candidate

encountered e.g. uniformity of sand. Xs e.g. solution to specific problem

candidate encountered. Ignore uneven surface.

[Total: 20]





9702 PHYSICS

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







Section A 1 (a) (i) force proportional to product of masses B1 force inversely proportional to square of separation B1 [2] (ii) separation much greater than radius / diameter of Sun / planet B1 [1]

(b) (i) e.g. force or field strength ∝ 1 / r 2

potential ∝ 1 / r B1 [1] (ii) e.g. gravitational force (always) attractive B1 electric force attractive or repulsive B1 [2] 2 (a) number of atoms of carbon-12 M1 in 0.012 kg of carbon-12 A1 [2] (b) pV = NkT or pV = nRT C1 substitutes temperature as 298 K C1 either 1.1 × 105 × 6.5 × 10–2 = N × 1.38 × 10–23 × 298 or 1.1 × 105 × 6.5 × 10–2 = n × 8.31 × 298 and n = N / 6.02 × 1023 C1 N = 1.7 × 1024 A1 [4] 3 (a) acceleration / force proportional to displacement from a fixed point M1 acceleration / force (always) directed towards that fixed point / in opposite direction to displacement A1 [2]

(b) (i) Aρg / m is a constant and so acceleration proportional to x B1 negative sign shows acceleration towards a fixed point / in opposite direction to displacement B1 [2]

(ii) ω

2 = (Aρg / m) C1

ω = 2πf C1

(2 × π × 1.5)2 = (4.5 × 10–4 × 1.0 × 103 × 9.81 / m) C1 m = 50 g A1 [4] 4 (a) work done in bringing unit positive charge M1 from infinity (to that point) A1 [2] (b) (i) field strength is potential gradient B1 [1] (ii) field strength proportional to force (on particle Q) B1 potential gradient proportional to gradient of (potential energy) graph B1 so force is proportional to the gradient of the graph A0 [2]




(c) energy = 5.1 × 1.6 × 10–19 (J) C1

potential energy = Q1Q2 / 4πε0r C1

5.1 × 1.6 × 10–19 = (1.6 × 10–19)2 / 4π × 8.85 × 10–12 × r C1 r = 2.8 × 10–10

m A1 [4] (d) (i) work is got out as x decreases M1 so opposite sign A1 [2] (ii) energy would be doubled B1 gradient would be increased B1 [2] 5 (a) region (of space) where there is a force M1 either on / produced by magnetic pole or on / produced by current carrying conductor / moving charge A1 [2] (b) (i) force on particle is (always) normal to velocity / direction of travel B1 speed of particle is constant B1 [2] (ii) magnetic force provides the centripetal force B1 mv2 / r = Bqv M1 r = mv / Bq A0 [2] (c) (i) direction from ‘bottom to top’ of diagram B1 [1] (ii) radius proportional to momentum C1 ratio = 5.7 / 7.4 = 0.77 A1 [2] (answer must be consistent with direction given in (c)(i)) 6 (a) (i) to concentrate the (magnetic) flux / reduce flux losses B1 [1] (ii) changing flux (in core) induces current in core M1 currents in core give rise to a heating effect A1 [2] (b) (i) e.m.f. induced proportional to M1 rate of change of (magnetic) flux (linkage) A1 [2] (ii) magnetic flux in phase with / proportional to e.m.f. / current in primary coil M1 e.m.f. / p.d. across secondary proportional to rate of change of flux M1 so e.m.f. of supply not in phase with p.d. across secondary A0 [2] (c) (i) for same power (transmission), high voltage with low current B1 with low current, less energy losses in transmission cables B1 [2] (ii) voltage is easily / efficiently changed B1 [1]




7 (a) for a wave, electron can ‘collect’ energy continuously B1 for a wave, electron will always be emitted / electron will be emitted at all frequencies….. M1 after a sufficiently long delay A1 [3] (b) (i) either wavelength is longer than threshold wavelength or frequency is below the threshold frequency or photon energy is less than work function B1 [1]

(ii) hc / λ = φ + EMAX C1

(6.63 × 10–34 × 3.0 × 108) / (240 × 10–9) = φ + 4.44 × 10–19 C1

φ = 3.8 × 10–19 J (allow 3.9 × 10–19

J) A1 [3] (c) (i) photon energy larger M1 so (maximum) kinetic energy is larger A1 [2] (ii) fewer photons (per unit time) M1 so (maximum) current is smaller A1 [2] 8 (a) (i) Fe shown near peak A1 [1] (ii) Zr shown about half-way along plateau A1 [1] (iii) H shown at less than 0.4 of maximum height A1 [1] (b) (i) heavy / large nucleus breaks up / splits M1 into two nuclei / fragments of approximately equal mass A1 [2] (ii) binding energy of nucleus = BE × A B1 binding energy of parent nucleus is less than sum of binding energies of fragments B1 [2]




Section B

9 (a) to compare two potentials / voltages M1 output depends upon which is greater A1 [2]

(b) (i) resistance of thermistor = 2.5 kΩ C1

resistance of X = 2.5 kΩ A1 [2]

(ii) at 5 ˚C / at < 10 ˚C, V – > V + M1 so VOUT is –9 V A1 at 20 ˚C / at > 10 ˚C, V – < V + and VOUT is +9 V B1 VOUT switches between negative and positive at 10 ˚C B1 [4] (allow similar scheme if 20 ˚C treated first)

10 (a) product of density (of medium) and speed of sound (in the medium) B1 [1]

(b) α would be nearly equal to 1 M1 either reflected intensity would be nearly equal to incident intensity

or coefficient for transmitted intensity = (1 – α) M1 transmitted intensity would be small A1 [3]

(c) (i) α = (1.7 – 1.3)2 / (1.7 + 1.3)2 C1 = 0.018 A1 [2]

(ii) attenuation in fat = exp(–48 × 2x × 10–2) C1 0.012 = 0.018 exp(–48 × 2x × 10–2) C1 x = 0.42 cm A1 [3]

11 (a) frequency of carrier wave varies M1 (in synchrony) with the displacement of the information signal A1 [2]

(b) (i) 5.0 V A1 [1]

(ii) 640 kHz A1 [1]

(iii) 560 kHz A1 [1]

(iv) 7000 (condone unit) A1 [1]

12 (a) e.g. acts as ‘return’ for the signal shields inner core from noise / interference / cross-talk (any two sensible answers, 1 each, max 2) B2 [2]

(b) e.g. greater bandwidth less attenuation (per unit length) less noise / interference (any two sensible answers, 1 each, max 2) B2 [2]

(c) attenuation is 2.4 dB C1 attenuation = 10 lg(P1 / P2) C1 ratio = 1.7 A1 [3]





9702 PHYSICS








Section A 1 (a) region (of space) where a particle / body experiences a force B1 [1]

(b) similarity: e.g. force ∝ 1 / r

2 potential ∝ 1 / r B1 [1] difference: e.g. gravitation force (always) attractive B1 electric force attractive or repulsive B1 [2] (c) either ratio is Q1Q2 / 4πε0m1m2G C1 = (1.6 × 10–19)2 / 4π × 8.85 × 10–12 × (1.67 × 10–27)2 × 6.67 × 10–11 C1 = 1.2 × 1036 A1 [3] or FE = 2.30 × 10–28 × R –2 (C1) FG = 1.86 × 10–64 × R –2 (C1) FE / FG = 1.2 × 1036 (A1) 2 (a) amount of substance M1 containing same number of particles as in 0.012 kg of carbon-12 A1 [2] (b) pV = nRT C1 amount = (2.3 × 105 × 3.1 × 10–3) / (8.31 × 290) + (2.3 × 105 × 4.6 × 10–3) / (8.31 × 303) C1 = 0.296 + 0.420 C1 = 0.716 mol A1 [4] (give full credit for starting equation pV = NkT and N = nNA) 3 (a) charges on plates are equal and opposite M1 so no resultant charge A1 energy stored because there is charge separation B1 [3] (b) (i) capacitance = Q / V C1 = (18 × 10–3) / 10 = 1800 µF A1 [2] (ii) use of area under graph or energy = ½CV2 C1 energy = 2.5 × 15.7 × 10–3 or energy = ½ × 1800 × 10–6 × (102 – 7.52) = 39 mJ A1 [2] (c) combined capacitance of Y & Z = 20 µF or total capacitance = 6.67 µF C1 p.d. across capacitor X = 8 V or p.d. across combination = 12 V C1 charge = 10 × 10–6 × 8 or 6.67 × 10–6 × 12 = 80 µC A1 [3]




4 (a) +∆U: increase in internal energy B1 +q: thermal energy / heat supplied to the system B1 +w: work done on the system B1 [3] (b) (i) (thermal) energy required to change the state of a substance M1 per unit mass A1 without any change of temperature A1 [3] (ii) when evaporating greater change in separation of atoms/molecules M1 greater change in volume M1 identifies each difference correctly with ∆U and w A1 [3] 5 (a) (i) (induced) e.m.f. proportional to M1 rate of change of (magnetic) flux (linkage) / rate of flux cutting A1 [2] (ii) 1. moving magnet causes change of flux linkage B1 [1] 2. speed of magnet varies so varying rate of change of flux B1 [1] 3. magnet changes direction of motion (so current changes direction) B1 [1] (b) period = 0.75 s C1 frequency = 1.33 Hz A1 [2] (c) graph: smooth correctly shaped curve with peak at f0 M1 A never zero A1 [2] (d) (i) resonance B1 [1] (ii) e.g. quartz crystal for timing / production of ultrasound A1 [1] 6 (a) (i) 2πf = 380 C1 frequency = 60 Hz A1 [2]

(ii) IRMS × √ 2 = I0 C1 IRMS = 9.9 / √ 2 = 7.0 A A1 [2]

(b) power = I2R C1 R = 400 / 7.02 = 8.2 Ω A1 [2]




7 (a) wavelength of wave associated with a particle M1 that is moving A1 [2] (b) (i) energy of electron = 850 × 1.6 × 10–19 M1 = 1.36 × 10–16

J energy = p2 / 2m or p = mv and EK = ½mv 2 momentum = √ (1.36 × 10–16 × 2 × 9.11 × 10–31) M1 = 1.6 × 10–23

N s A0 [2] (ii) λ = h / p C1 wavelength = (6.63 × 10–34) / (1.6 × 10–23) = 4.1 × 10–11

m A1 [2] (c) diagram or description showing: electron beam in a vacuum B1 incident on thin metal target / carbon film B1 fluorescent screen B1 pattern of concentric rings observed M1 pattern similar to diffraction pattern observed with visible light A1 [5] 8 (a) energy required to separate nucleons in a nucleus M1 to infinity A1 [2] (b) 1u = 1.66 × 10–27

kg E = mc2 C1 = 1.66 × 10–27 × (3.0 × 108)2 M1 = 1.49 × 10–10 J = (1.49 × 10–10) / (1.6 × 10–13) M1 = 930 MeV A0 [3] (c) (i) ∆m = 2.0141u – (1.0073 + 1.0087)u = –1.9 × 10–3

u C1 binding energy = 1.9 × 10–3 × 930 =1.8 MeV A1 [2] (ii) ∆m = (57 × 1.0087u) + (40 × 1.0073u) – 97.0980u C1 = (–)0.69 u binding energy per nucleon = (0.69 × 930) / 97 C1 = 6.61 MeV A1 [3]




Section B 9 (a) thin / fine metal wire B1 lay-out shown as a grid B1 encased in plastic B1 [3] (b) (i) gain (of amplifier) B1 [1] (ii) for VOUT = 0, then V + = V – or V1 = V2 C1 V1 = (1000/1125) × 4.5 C1 V1 = 4.0 V A1 [3] (iii) V2 = (1000 / 1128) × 4.5 = 3.99 V C1 VOUT = 12 × (3.99 – 4.00) = (–) 0.12 V A1 [2] 10 strong / large (uniform) magnetic field B1 nuclei precess / rotate about field direction (1) radio frequency pulse B1 at Larmor frequency (1) causes resonance / nuclei absorb energy B1 on relaxation / de-excitation, nuclei emit r.f. pulse B1 pulse detected and processed (1) non-uniform field superposed on uniform field B1 allows position of resonating nuclei to be determined B1 allows for location of detection to be changed (1) (six points, 1 each plus any two extra – max 8) [8] 11 (a) e.g. unreliable communication (M1) because ion layers vary in height / density (A1) e.g. cannot carry all information required (M1) bandwidth too narrow (A1) e.g. coverage limited (M1) reception poor in hilly areas (A1) (any two sensible suggestions, M1 & A1 for each, max 4) [4] (b) signal must be amplified (greatly) before transmission back to Earth B1 uplink signal would be swamped by downlink signal B1 [2]




12 (a) (i) ratio / dB = 10 lg(P1 / P2) C1 24 = 10 lg(P1 / 5.6 × 10–19) C1 P1 = 1.4 × 10–16

W A1 [3] (ii) attenuation per unit length = 1 / L × 10 lg(P1 / P2) C1 1.9 = 1 / L × 10 lg(3.5 × 10–3/1.4 × 10–16) C1 L = 1 km A1 [3] or attenuation = 10 lg(3.5 × 10–3/5.6 × 10–19) (C1) = 158 dB attenuation along fibre = (158 – 24) (C1) L = (158 – 24) / 1.9 = 71 km (A1) (b) less attenuation (per unit length) / longer uninterrupted length of fibre B1 [1]





9702 PHYSICS








Section A 1 (a) region (of space) where a particle / body experiences a force B1 [1]

(b) similarity: e.g. force ∝ 1 / r

2 potential ∝ 1 / r B1 [1] difference: e.g. gravitation force (always) attractive B1 electric force attractive or repulsive B1 [2] (c) either ratio is Q1Q2 / 4πε0m1m2G C1 = (1.6 × 10–19)2 / 4π × 8.85 × 10–12 × (1.67 × 10–27)2 × 6.67 × 10–11 C1 = 1.2 × 1036 A1 [3] or FE = 2.30 × 10–28 × R –2 (C1) FG = 1.86 × 10–64 × R –2 (C1) FE / FG = 1.2 × 1036 (A1) 2 (a) amount of substance M1 containing same number of particles as in 0.012 kg of carbon-12 A1 [2] (b) pV = nRT C1 amount = (2.3 × 105 × 3.1 × 10–3) / (8.31 × 290) + (2.3 × 105 × 4.6 × 10–3) / (8.31 × 303) C1 = 0.296 + 0.420 C1 = 0.716 mol A1 [4] (give full credit for starting equation pV = NkT and N = nNA) 3 (a) charges on plates are equal and opposite M1 so no resultant charge A1 energy stored because there is charge separation B1 [3] (b) (i) capacitance = Q / V C1 = (18 × 10–3) / 10 = 1800 µF A1 [2] (ii) use of area under graph or energy = ½CV2 C1 energy = 2.5 × 15.7 × 10–3 or energy = ½ × 1800 × 10–6 × (102 – 7.52) = 39 mJ A1 [2] (c) combined capacitance of Y & Z = 20 µF or total capacitance = 6.67 µF C1 p.d. across capacitor X = 8 V or p.d. across combination = 12 V C1 charge = 10 × 10–6 × 8 or 6.67 × 10–6 × 12 = 80 µC A1 [3]




4 (a) +∆U: increase in internal energy B1 +q: thermal energy / heat supplied to the system B1 +w: work done on the system B1 [3] (b) (i) (thermal) energy required to change the state of a substance M1 per unit mass A1 without any change of temperature A1 [3] (ii) when evaporating greater change in separation of atoms/molecules M1 greater change in volume M1 identifies each difference correctly with ∆U and w A1 [3] 5 (a) (i) (induced) e.m.f. proportional to M1 rate of change of (magnetic) flux (linkage) / rate of flux cutting A1 [2] (ii) 1. moving magnet causes change of flux linkage B1 [1] 2. speed of magnet varies so varying rate of change of flux B1 [1] 3. magnet changes direction of motion (so current changes direction) B1 [1] (b) period = 0.75 s C1 frequency = 1.33 Hz A1 [2] (c) graph: smooth correctly shaped curve with peak at f0 M1 A never zero A1 [2] (d) (i) resonance B1 [1] (ii) e.g. quartz crystal for timing / production of ultrasound A1 [1] 6 (a) (i) 2πf = 380 C1 frequency = 60 Hz A1 [2]

(ii) IRMS × √ 2 = I0 C1 IRMS = 9.9 / √ 2 = 7.0 A A1 [2]

(b) power = I2R C1 R = 400 / 7.02 = 8.2 Ω A1 [2]




7 (a) wavelength of wave associated with a particle M1 that is moving A1 [2] (b) (i) energy of electron = 850 × 1.6 × 10–19 M1 = 1.36 × 10–16

J energy = p2 / 2m or p = mv and EK = ½mv 2 momentum = √ (1.36 × 10–16 × 2 × 9.11 × 10–31) M1 = 1.6 × 10–23

N s A0 [2] (ii) λ = h / p C1 wavelength = (6.63 × 10–34) / (1.6 × 10–23) = 4.1 × 10–11

m A1 [2] (c) diagram or description showing: electron beam in a vacuum B1 incident on thin metal target / carbon film B1 fluorescent screen B1 pattern of concentric rings observed M1 pattern similar to diffraction pattern observed with visible light A1 [5] 8 (a) energy required to separate nucleons in a nucleus M1 to infinity A1 [2] (b) 1u = 1.66 × 10–27

kg E = mc2 C1 = 1.66 × 10–27 × (3.0 × 108)2 M1 = 1.49 × 10–10 J = (1.49 × 10–10) / (1.6 × 10–13) M1 = 930 MeV A0 [3] (c) (i) ∆m = 2.0141u – (1.0073 + 1.0087)u = –1.9 × 10–3

u C1 binding energy = 1.9 × 10–3 × 930 =1.8 MeV A1 [2] (ii) ∆m = (57 × 1.0087u) + (40 × 1.0073u) – 97.0980u C1 = (–)0.69 u binding energy per nucleon = (0.69 × 930) / 97 C1 = 6.61 MeV A1 [3]




Section B 9 (a) thin / fine metal wire B1 lay-out shown as a grid B1 encased in plastic B1 [3] (b) (i) gain (of amplifier) B1 [1] (ii) for VOUT = 0, then V + = V – or V1 = V2 C1 V1 = (1000/1125) × 4.5 C1 V1 = 4.0 V A1 [3] (iii) V2 = (1000 / 1128) × 4.5 = 3.99 V C1 VOUT = 12 × (3.99 – 4.00) = (–) 0.12 V A1 [2] 10 strong / large (uniform) magnetic field B1 nuclei precess / rotate about field direction (1) radio frequency pulse B1 at Larmor frequency (1) causes resonance / nuclei absorb energy B1 on relaxation / de-excitation, nuclei emit r.f. pulse B1 pulse detected and processed (1) non-uniform field superposed on uniform field B1 allows position of resonating nuclei to be determined B1 allows for location of detection to be changed (1) (six points, 1 each plus any two extra – max 8) [8] 11 (a) e.g. unreliable communication (M1) because ion layers vary in height / density (A1) e.g. cannot carry all information required (M1) bandwidth too narrow (A1) e.g. coverage limited (M1) reception poor in hilly areas (A1) (any two sensible suggestions, M1 & A1 for each, max 4) [4] (b) signal must be amplified (greatly) before transmission back to Earth B1 uplink signal would be swamped by downlink signal B1 [2]




12 (a) (i) ratio / dB = 10 lg(P1 / P2) C1 24 = 10 lg(P1 / 5.6 × 10–19) C1 P1 = 1.4 × 10–16

W A1 [3] (ii) attenuation per unit length = 1 / L × 10 lg(P1 / P2) C1 1.9 = 1 / L × 10 lg(3.5 × 10–3/1.4 × 10–16) C1 L = 1 km A1 [3] or attenuation = 10 lg(3.5 × 10–3/5.6 × 10–19) (C1) = 158 dB attenuation along fibre = (158 – 24) (C1) L = (158 – 24) / 1.9 = 71 km (A1) (b) less attenuation (per unit length) / longer uninterrupted length of fibre B1 [1]





9702 PHYSICS

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







1 Planning (15 marks) Defining the problem (3 marks) P1 n is the independent variable and V is the dependent variable or vary n and measure V [1] P2 Keep distance from light to photocell constant [1] P3 Keep intensity of light constant. Allow constant voltage across lamp/current through

lamp/brightness. Do not allow ‘same lamp/output’. [1] Methods of data collection (5 marks) M1 Labelled diagram of apparatus: lamp, glass sheet and photocell in line. [1] M2 Voltmeter connected to photocell. Penalise unworkable photocell circuit. [1] M3 Use micrometer (screw gauge) to measure thickness of glass sheet. [1] M4 Take many readings of thickness and average. [1] M5 Perform experiment in a dark room or shield apparatus. [1] Method of analysis (2 marks) A1 Plot a graph of ln V against n. Allow ln V against nt [1]

A2 α = (–)gradient / t. (ln V against nt then α = (–)gradient) [1]

Safety considerations (1 mark) S Reasoned method to prevent burns from hot source, e.g. use gloves Reasoned method to prevent eye damage from bright/intense source, e.g. shield lamp/

dark glasses/do not look at source directly Reasoned method to prevent cuts from glass e.g. use gloves. [1]

Additional detail (4 marks) D Relevant points might include [4] 1 Use small distance/high intensity to gain large reading. 2 Method to check output of lamp is constant e.g. measure current through/p.d. across

lamp/regularly check V0 with no glass. 3 Reasoned method to ensure output of lamp is constant e.g. workable circuit diagram

with variable resistor or variable power supply. 4 Clean sheets of glass before use. 5 Direction of light is perpendicular to glass sheets/constant orientation.

6 ln V = –αnt + ln V0. 7 Further safety consideration. Do not allow vague computer methods.

[Total: 15]




2 Analysis, conclusions and evaluation (15 marks)

Part Mark Expected Answer Additional Guidance

(a) A1 NkT/A 290Nk/A

T1

h

1/ m–1

Column heading. Allow equivalent unit. e.g. h–1 / m–1

T2 2.5 or 2.50 2.8 or 2.78 3.1 or 3.13 3.6 or 3.57 4.2 or 4.17 4.8 or 4.76

A mixture of 2sf and 3sf is allowed.

(b)

U1 From ± 0.03 to ± 0.1, ± 0.11 or ± 0.12 Allow more than one significant figure.

G1 Six points plotted correctly Check second and fifth plots and other anomalous plots. Must be less than half a small square. Ecf allowed from table.

(c) (i)

U2 All error bars in

h

1 plotted correctly

Half square or greater loses the mark. Ecf allowed from table.

G2 Line of best fit If points are plotted correctly then lower end of line should pass between (2.20, 1.0) and (2.30, 1.0) and upper end of line should pass between (4.75, 2.1) and (4.85, 2.1). Allow ecf from points plotted incorrectly – examiner judgement.

(ii)

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.

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.

(iii)

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

C2 Value of N =

kT

A×gradient

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

(d)

U4 Determines uncertainty in N Method required. Do not check calculation.

C3 Method to determine h N

pA

NkTh ××==

−20101.111 ; T = 278 K

Must use answer from (d).

(e) (i)

C4 Between 0.361 and 0.391 given to 2 or 3 sf

Must be in range. Allow 0.36, 0.37, 0.38 or 0.39. Assume metres unless otherwise specified.

(ii) U5 Percentage uncertainty % uncertainty in N + % uncertainty in T)

[Allow ∆T to be 0.5 or 1]

[Total: 15]




Uncertainties in Question 2 (c) (iii) Gradient [U3] Uncertainty = gradient of line of best fit – gradient of worst acceptable line Uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)

(d) [U4] Uncertainty = worst N – N

Nm

mN ×

∆=∆

kT

AmN ×∆=∆

(e) [U5]

Percentage uncertainty = 100×∆

h

h

Percentage uncertainty = percentage uncertainty in N + percentage uncertainty in T





9702 PHYSICS








1 Planning (15 marks) Defining the problem (3 marks) P1 p is the independent variable or vary p [1] P2 q is the dependent variable or measure q [1] P3 Keep (horizontal) velocity (v) constant [1] Methods of data collection (5 marks) M1 Labelled diagram of apparatus including method to vary p. [1] M2 Method to determine position of ball on surface e.g. carbon paper/dye/video/sand. [1] M3 Use ruler/caliper to measure p and/or q. [1] M4 Method to ensure velocity is constant e.g. releasing ball from same height on a track/

spring loaded device or impulse device set to a constant value. [1] M5 Method to ensure that the moved surface remains horizontal, e.g. spirit level/check

height at different places. [1] Method of analysis (2 marks)

A1 q2 against p q against p p against q2 p against q [1]

A2 2

gradient×

=

gv

2gradient

gv ×=

gradient2×=

gv

2gradient

1 gv ×= [1]

Allow valid logarithmic graph e.g. lg q against lg p, lg 2p, lg 2p/g and valid calculation of v from y-intercept. Safety considerations (1 mark) S Reasoned method to prevent ball rolling on floor e.g. box below/storage box for balls/

sand box. Reasoned method to prevent ball causing injury e.g. goggles/safety screen [1] Additional detail (4 marks) D Relevant points might include [4] 1 Method to ensure that velocity of ball is horizontal only when it reaches table, e.g.

curved track. 2 Ensure that the ball leaves the table at 90°, e.g. set square/protractor on upper surface. 3 Detail on measuring q – location of landing position e.g. centre of crater/start of track. 4 Detail on determining location of zero position for p and q e.g. set square, plumb line. 5 Detail on method of determining position of ball e.g. slow motion playback including

scale. 6 Take many readings of q for each p and average. 7 Straight line through the origin shows that p is proportional to q2/relationship is valid –

this mark may only be awarded when A1 is given. 8 Use of high density ball to minimise the effects of air resistance. Do not allow vague computer methods.

[Total: 15]






(a) A1 F

T1 T2

6.7 or 6.67 9.0 or 9.00 4.5 or 4.55 6.5 or 6.50 3.0 or 3.03 4.6 or 4.63 2.1 or 2.13 3.5 or 3.50 1.5 or 1.47 2.8 or 2.75 1.2 or 1.16 2.4 or 2.38

T1 for 1/R. T2 for V/E. Must be 2sf or 3sf; a mixture is allowed

(b)

U1 From ± 0.6 or ± 0.7, to ± 0.2 Allow more than one significant figure.


(c) (i)

U2 All error bars in V/E plotted correctly Half square or greater loses the mark. Ecf allowed from table.

G2 Line of best fit If points are plotted correctly then lower end of line should pass between (0, 0.9) and (0, 1.1) and upper end of line should pass between (7, 9.3) and (7, 9.5). Allow ecf from points plotted incorrectly – examiner judgement.

(ii)




(iii)

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

C2 (Gradient value) Ω Gradient must be used correctly. Expect

about 1200 Ω Allow ecf from (c)(iii) but penalise POT. Do not penalise sf or rounding errors.

(d)

U4 Determines uncertainty in F Allow ecf from POT.

C3 Determines V/E correctly. Answer should be approximately 11; F from (d) must be used.

1+R

F

(e) (i)

U5 Determines absolute uncertainty Should be approximately 15% of RF/R1 (about 1.5). Several possible methods.

(ii) C4 In the range 17.2 to 18.0 given to 2 or 3sf

Allow 17 or 18 to 2sf. Must be (e)(i) × 1.6

[Total: 15]




Uncertainties in Question 2 (c) (iii) Gradient [U3] Uncertainty = gradient of line of best fit – gradient of worst acceptable line Uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient) (d) [U4] Uncertainty = uncertainty in gradient (e) [U5] Uncertainty = worst V/E – V/E Note worst V/E is calculated either by max V/min E or by min V/max E Or max gradient of WAL/114 or min gradient of WAL/126

Uncertainty = R

F

m

m×

∆+0.05 [Allow V/E instead of F/R]

Uncertainty = R

F

F

F×


Uncertainty = R

F

m

m×

∆+

120

6 [Allow V/E instead of F/R]

Uncertainty = R

F

F

F×

∆+

120






9702 PHYSICS








1 Planning (15 marks) Defining the problem (3 marks) P1 p is the independent variable or vary p [1] P2 q is the dependent variable or measure q [1] P3 Keep (horizontal) velocity (v) constant [1] Methods of data collection (5 marks) M1 Labelled diagram of apparatus including method to vary p. [1] M2 Method to determine position of ball on surface e.g. carbon paper/dye/video/sand. [1] M3 Use ruler/caliper to measure p and/or q. [1] M4 Method to ensure velocity is constant e.g. releasing ball from same height on a track/

spring loaded device or impulse device set to a constant value. [1] M5 Method to ensure that the moved surface remains horizontal, e.g. spirit level/check

height at different places. [1] Method of analysis (2 marks)

A1 q2 against p q against p p against q2 p against q [1]

A2 2

gradient×

=

gv

2gradient

gv ×=

gradient2×=

gv

2gradient

1 gv ×= [1]

Allow valid logarithmic graph e.g. lg q against lg p, lg 2p, lg 2p/g and valid calculation of v from y-intercept. Safety considerations (1 mark) S Reasoned method to prevent ball rolling on floor e.g. box below/storage box for balls/

sand box. Reasoned method to prevent ball causing injury e.g. goggles/safety screen [1] Additional detail (4 marks) D Relevant points might include [4] 1 Method to ensure that velocity of ball is horizontal only when it reaches table, e.g.

curved track. 2 Ensure that the ball leaves the table at 90°, e.g. set square/protractor on upper surface. 3 Detail on measuring q – location of landing position e.g. centre of crater/start of track. 4 Detail on determining location of zero position for p and q e.g. set square, plumb line. 5 Detail on method of determining position of ball e.g. slow motion playback including

scale. 6 Take many readings of q for each p and average. 7 Straight line through the origin shows that p is proportional to q2/relationship is valid –

this mark may only be awarded when A1 is given. 8 Use of high density ball to minimise the effects of air resistance. Do not allow vague computer methods.

[Total: 15]






(a) A1 F

T1 T2

6.7 or 6.67 9.0 or 9.00 4.5 or 4.55 6.5 or 6.50 3.0 or 3.03 4.6 or 4.63 2.1 or 2.13 3.5 or 3.50 1.5 or 1.47 2.8 or 2.75 1.2 or 1.16 2.4 or 2.38

T1 for 1/R. T2 for V/E. Must be 2sf or 3sf; a mixture is allowed

(b)

U1 From ± 0.6 or ± 0.7, to ± 0.2 Allow more than one significant figure.


(c) (i)

U2 All error bars in V/E plotted correctly Half square or greater loses the mark. Ecf allowed from table.

G2 Line of best fit If points are plotted correctly then lower end of line should pass between (0, 0.9) and (0, 1.1) and upper end of line should pass between (7, 9.3) and (7, 9.5). Allow ecf from points plotted incorrectly – examiner judgement.

(ii)




(iii)

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

C2 (Gradient value) Ω Gradient must be used correctly. Expect

about 1200 Ω Allow ecf from (c)(iii) but penalise POT. Do not penalise sf or rounding errors.

(d)

U4 Determines uncertainty in F Allow ecf from POT.

C3 Determines V/E correctly. Answer should be approximately 11; F from (d) must be used.

1+R

F

(e) (i)

U5 Determines absolute uncertainty Should be approximately 15% of RF/R1 (about 1.5). Several possible methods.

(ii) C4 In the range 17.2 to 18.0 given to 2 or 3sf

Allow 17 or 18 to 2sf. Must be (e)(i) × 1.6

[Total: 15]




Uncertainties in Question 2 (c) (iii) Gradient [U3] Uncertainty = gradient of line of best fit – gradient of worst acceptable line Uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient) (d) [U4] Uncertainty = uncertainty in gradient (e) [U5] Uncertainty = worst V/E – V/E Note worst V/E is calculated either by max V/min E or by min V/max E Or max gradient of WAL/114 or min gradient of WAL/126

Uncertainty = R

F

m

m×


Uncertainty = R

F

F

F×


Uncertainty = R

F

m

m×

∆+

120


Uncertainty = R

F

F

F×

∆+

120


9702 s11 ms_all

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