9702 w13 ms_all

CAMBRIDGE INTERNATIONAL EXAMINATIONS

GCE Advanced Subsidiary Level and GCE Advanced Level

MARK SCHEME for the October/November 2013 series

9702 PHYSICS

9702/11 (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 October/November 2013 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.

Page 2 Mark Scheme Syllabus Paper

GCE AS/A LEVEL – October/November 2013 9702 11

© Cambridge International Examinations 2013

Question Number

Key Question Number

Key

1 B 21 A

2 C 22 C

3 B 23 D

4 B 24 A

5 C 25 D

6 C 26 C

7 C 27 A

8 C 28 D

9 D 29 B

10 B 30 A

11 C 31 B

12 D 32 C

13 A 33 C

14 D 34 B

15 B 35 B

16 B 36 A

17 A 37 D

18 C 38 C

19 B 39 C

20 D 40 A




9702 PHYSICS






Question Number

Key Question Number

Key

1 B 21 A

2 C 22 C

3 B 23 D

4 B 24 A

5 C 25 D

6 C 26 C

7 C 27 A

8 C 28 D

9 D 29 B

10 B 30 A

11 C 31 B

12 D 32 C

13 A 33 C

14 D 34 B

15 B 35 B

16 B 36 A

17 A 37 D

18 C 38 C

19 B 39 C

20 D 40 A




9702 PHYSICS






Question Number

Key Question Number

Key

1 D 21 C

2 B 22 C

3 D 23 D

4 A 24 C

5 D 25 A

6 D 26 A

7 B 27 D

8 B 28 B

9 D 29 A

10 C 30 C

11 B 31 D

12 B 32 C

13 A 33 C

14 B 34 A

15 C 35 B

16 A 36 D

17 D 37 C

18 D 38 C

19 C 39 D

20 B 40 A




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.





1 (a) kelvin / K B1 ampere / amp / A B1 [2] [allow mole / mol and candela / Cd]

(b) (i) energy OR work = force × distance [allow any energy expression] C1

units: kg m s–2 × m OR kg (m s–1)2 for ½ mv2 or mc2 M1 (ignore any numerical factor) units: = kg m2 s–2 A0 [2]

(ii) units: ρ: kg m–3 g: m s–2 A: m2 l0: m C1 C: kg m2

s–2 / kg2 m–6 m2 s–4

m2 m3 [any subject] C1

= kg–1 m s2 (allow m s2

/ kg) A1 [3]

2 (a) d = v × t C1

t = 0.2 × 4 (allow t = 0.2 × 2) C1

d = 3 × 108 × 0.8 × 10–6 OR 3 × 108 × 0.4 × 10–6 C1 d = 240 m hence distance from source to reflector = 120 m A1 [4]

(b) speed of sound 300 cf speed of light 3 × 108 OR time = 240 / 300 (= 0.8) OR time = 120 / 300 (= 0.4) C1 sound slower by factor of 106 OR time for one division 0.8 / 4 OR time for one division 0.4 / 2 C1 time base setting 0.2 s cm–1 [unit required] A1 [3]

3 (a) (work =) force × distance moved / displacement in the direction of the force OR when a force moves in the direction of the force work is done B1 [1] (b) kinetic energy = ½ mv2 C1 kinetic energy = ½ 0.4 (2.5)2 = 1.25 / 1.3 J A1 [2] (c) (i) area under graph is work done / work done = ½ Fx C1 1.25 = (14 x) / 2 C1 x = 0.18 (0.179) m [allow x = 0.19 m using kinetic energy = 1.3 J] A1 [3] (ii) smooth curve from v = 2.5 at x = 0 to v = 0 at Q M1 curve with increasing gradient A1 [2]




4 (a) torque of a couple = one of the forces / a force × distance M1 multiplied by the perpendicular distance between the forces A1 [2] (b) (i) weight at P (vertically) down B1 normal reaction OR contact force at (point of contact with the pin) P (vertically) up B1 [2]

(ii) torque = 35 × 0.25 (or 25) × 2 C1 torque = 18 (17.5) N m A1 [2] (iii) the two 35 N forces are equal and opposite and the weight and the upward /

contact / reaction force are equal and opposite B1 [1] (iv) not in equilibrium as the (resultant) torque is not zero B1 [1] 5 (a) (i) displacement is the distance the rope / particles are (above or below) from

the equilibrium / mean / rest / undisturbed position (not ‘distance moved’) B1 [1] (ii) 1. amplitude (= 80 / 4) = 20 mm B1 [1]

2. v = fλ or v = λ / T C1 f = 1 / T = 1 / 0.2 (5 Hz) C1 v = 5 × 1.5 = 7.5 m s–1 A1 [3] (b) point A of rope shown at equilibrium position B1

same wavelength, shape, peaks / wave moved ¼λ to right B1 [2] (c) (i) progressive as energy OR peaks OR troughs is/are transferred/moved

/propagated (by the waves) B1 [1] (ii) transverse as particles/rope movement is perpendicular to direction of travel

/propagation of the energy/wave velocity B1 [1] 6 (a) p.d. = work (done) / charge OR energy transferred from (electrical to other forms)

/ (unit) charge B1 [1]

(b) (i) R = ρl / A C1

ρ = 18 × 10–9 C1

R = (18 × 10–9 × 75) / 2.5 × 10–6 = 0.54 Ω A1 [3]

(ii) V = IR C1

R = 38 + (2 × 0.54) C1

I = 240 / 39.08 = 6.1 (6.14) A A1 [3]




(iii) P = I 2R or P = VI and V = IR or P = V2/R and V = IR C1

P = (6.14)2 × 2 × 0.54 C1 P = 41 (40.7) W A1 [3]

(c) area of wire is less (1/5) hence resistance greater (×5) M1

OR R is ∝ 1/A therefore R is greater p.d. across wires greater so power loss in cables increases A1 [2] 7 (a) (i) the direction of the fields is the same OR fields are uniform OR constant electric field strength OR E = V / d with symbols explained B1 [1] (ii) reduce p.d. across plates B1 increase separation of plates B1 [2]

(iii) α opposite charge to β (as deflection in opposite direction) B1

β has a range of velocities OR energies (as different deflections) and

α all have same velocity OR energy (as constant deflection) B1

α are more massive (as deflection is less for greater field strength) B1 [3] (b) W = 234 and X = 90 B1 Y = 4 and Z = 2 B1 [2] (c) A = 32 and B = 16 and C = 0 and D = –1 B1 [1]




9702 PHYSICS







1 (a) kelvin / K B1 ampere / amp / A B1 [2] [allow mole / mol and candela / Cd]

(b) (i) energy OR work = force × distance [allow any energy expression] C1

units: kg m s–2 × m OR kg (m s–1)2 for ½ mv2 or mc2 M1 (ignore any numerical factor) units: = kg m2 s–2 A0 [2]

(ii) units: ρ: kg m–3 g: m s–2 A: m2 l0: m C1 C: kg m2

s–2 / kg2 m–6 m2 s–4

m2 m3 [any subject] C1

= kg–1 m s2 (allow m s2

/ kg) A1 [3]

2 (a) d = v × t C1

t = 0.2 × 4 (allow t = 0.2 × 2) C1

d = 3 × 108 × 0.8 × 10–6 OR 3 × 108 × 0.4 × 10–6 C1 d = 240 m hence distance from source to reflector = 120 m A1 [4]

(b) speed of sound 300 cf speed of light 3 × 108 OR time = 240 / 300 (= 0.8) OR time = 120 / 300 (= 0.4) C1 sound slower by factor of 106 OR time for one division 0.8 / 4 OR time for one division 0.4 / 2 C1 time base setting 0.2 s cm–1 [unit required] A1 [3]

3 (a) (work =) force × distance moved / displacement in the direction of the force OR when a force moves in the direction of the force work is done B1 [1] (b) kinetic energy = ½ mv2 C1 kinetic energy = ½ 0.4 (2.5)2 = 1.25 / 1.3 J A1 [2] (c) (i) area under graph is work done / work done = ½ Fx C1 1.25 = (14 x) / 2 C1 x = 0.18 (0.179) m [allow x = 0.19 m using kinetic energy = 1.3 J] A1 [3] (ii) smooth curve from v = 2.5 at x = 0 to v = 0 at Q M1 curve with increasing gradient A1 [2]




4 (a) torque of a couple = one of the forces / a force × distance M1 multiplied by the perpendicular distance between the forces A1 [2] (b) (i) weight at P (vertically) down B1 normal reaction OR contact force at (point of contact with the pin) P (vertically) up B1 [2]

(ii) torque = 35 × 0.25 (or 25) × 2 C1 torque = 18 (17.5) N m A1 [2] (iii) the two 35 N forces are equal and opposite and the weight and the upward /

contact / reaction force are equal and opposite B1 [1] (iv) not in equilibrium as the (resultant) torque is not zero B1 [1] 5 (a) (i) displacement is the distance the rope / particles are (above or below) from

the equilibrium / mean / rest / undisturbed position (not ‘distance moved’) B1 [1] (ii) 1. amplitude (= 80 / 4) = 20 mm B1 [1]

2. v = fλ or v = λ / T C1 f = 1 / T = 1 / 0.2 (5 Hz) C1 v = 5 × 1.5 = 7.5 m s–1 A1 [3] (b) point A of rope shown at equilibrium position B1

same wavelength, shape, peaks / wave moved ¼λ to right B1 [2] (c) (i) progressive as energy OR peaks OR troughs is/are transferred/moved

/propagated (by the waves) B1 [1] (ii) transverse as particles/rope movement is perpendicular to direction of travel

/propagation of the energy/wave velocity B1 [1] 6 (a) p.d. = work (done) / charge OR energy transferred from (electrical to other forms)

/ (unit) charge B1 [1]

(b) (i) R = ρl / A C1

ρ = 18 × 10–9 C1

R = (18 × 10–9 × 75) / 2.5 × 10–6 = 0.54 Ω A1 [3]

(ii) V = IR C1

R = 38 + (2 × 0.54) C1

I = 240 / 39.08 = 6.1 (6.14) A A1 [3]




(iii) P = I 2R or P = VI and V = IR or P = V2/R and V = IR C1

P = (6.14)2 × 2 × 0.54 C1 P = 41 (40.7) W A1 [3]

(c) area of wire is less (1/5) hence resistance greater (×5) M1

OR R is ∝ 1/A therefore R is greater p.d. across wires greater so power loss in cables increases A1 [2] 7 (a) (i) the direction of the fields is the same OR fields are uniform OR constant electric field strength OR E = V / d with symbols explained B1 [1] (ii) reduce p.d. across plates B1 increase separation of plates B1 [2]

(iii) α opposite charge to β (as deflection in opposite direction) B1

β has a range of velocities OR energies (as different deflections) and

α all have same velocity OR energy (as constant deflection) B1

α are more massive (as deflection is less for greater field strength) B1 [3] (b) W = 234 and X = 90 B1 Y = 4 and Z = 2 B1 [2] (c) A = 32 and B = 16 and C = 0 and D = –1 B1 [1]




9702 PHYSICS







1 volume = π (14 × 10–3)2 × 12 × 10–3 (=7.389 × 10–6 m3) C1 density = mass / volume [any subject] C1

mass = 6.8 × 103 × 7.389 × 10–6 = 0.0502 weight = mg C1

weight = 0.0502 × 9.81 = 0.49 N (mark not awarded if not to two s.f.) A1 [4] 2 (a) SI units for T: s, R: m and M: kg (or seen clearly in formula) C1

K = T2 M / R3 units: s2 kg m–3 (allow s2 kg / m3 or 3

2

m

kgs) A1 [2]

(b) % uncertainty in K: 1% (for T) + 3% (for R) + 2% (for M) OR = 6% C1

K = [(86400)2 × 6 × 1024] / (4.23 × 107)3 = 5.918 × 1011 C1

6% of K = 0.355 × 1011 C1

K = (5.9 ± 0.4) × 1011 (SI units) correct power of ten required for both A1 [4] [incorrect % value then max. 1] 3 (a) (i) velocity = rate of change of displacement OR displacement change / time (taken) A1 [1] (ii) acceleration = rate of change of velocity OR change in velocity / time (taken) A1 [1] (b) (i) initial constant velocity as straight line / gradient constant B1 middle section deceleration/ speed / velocity decreases / slowing down as

gradient decreases B1 last section lower velocity (than at start) as gradient (constant and) smaller B1 [3] [special case: all three stages correct descriptions but no reasons 1/3] (ii) velocity = 45 / 1.5 = 30 m s–1 A1 [1] (iii) velocity at 4.0 s is (122 – 98) / 2.0 = 12 (m s–1) (allow 12 to 13) B1 acceleration = (12 – 30) / 2.5 = –7.2 m s–2 (if answer not this value then

comment needed to explain why, e.g. difficulty in drawing tangent) A1 [2] (iv) F = ma C1

F = (–)1500 × 7.2 = (–)11000 (10800) N A1 [2] 4 (a) gravitational PE is energy of a mass due to its position in a gravitational field B1 elastic PE energy stored (in an object) due to (a force) changing its shape / deformation / being compressed / stretched / strained B1 [2] (b) (i) 1. kinetic energy = ½ mv2 C1

kinetic energy = ½ × 0.065 × 162 = 8.3(2) J A1 [2]

2. v2 = 2gh OR PE = mgh C1

h = 162 / (2 × 9.81) = 13(.05) m A1 [2]




(ii) speed at t = ½ total time = 8 (m s–1) or total t =1.63 or t1/2 = 0.815 s C1 KE is ¼ or h at t1/2 = 9.78 (m) C1 and PE is ¾ of max ratio = 3 or ratio = 9.78 / 3.26 = 3 A1 [3] (iii) time is less because (average) acceleration is greater OR average force

is greater B1 [1] 5 (a) (i) 1. wavelength: minimum distance between two points moving in phase OR distance between neighbouring or consecutive peaks or troughs OR wavelength is the distance moved by a wavefront in time T or one

oscillation/cycle or period (of source) B1 [1] 2. frequency: number of wavefronts / (unit) time OR number of oscillations per unit time or oscillations/time B1 [1] (ii) speed = distance / time = wavelength / time period M1 speed = λ / T = λf A0 [1] (b) (i) amplitude = 4.0 mm (allow 1 s.f.) A1 [1] (ii) wavelength = 18 / 3.75 (= 4.8) C1

speed = 2.5 × 4.8 × 10–2 = 12 × 10–2 m s–1 unit consistent with numerical

answer, e.g. in cm s–1 if cm used for λ and unit changed on answer line A1 [2]

[if 18 cm = 3.5λ used giving speed 13 (12.9) cm s–1 allow max. 1].

(iii) 180º or π rad A1 [1] (c) light and screen and correct positions above and below ripple tank B1 strobe or video camera B1 [2] 6 (a) e.m.f. = total energy available (per unit charge) B1 some (of the available energy) is used/lost/wasted/given out in the internal

resistance of the battery (hence p.d. available less than e.m.f.) B1 [2]

(b) (i) V = IR C1

I = 6.9 / 5.0 = 1.4 (1.38) A A1 [2] (ii) r = lost volts / current C1

r = (9– 6.9) / 1.38 = 1.5(2) Ω A1 [2]

(c) (i) P = EI (not P = VI if only this line given or 9 V not used in second line) C1

P = 9 × 1.38 = 12 (12.4) W A1 [2] (ii) efficiency = output power / total power C1

efficiency = VI / EI = 6.9 / 9 or (9.52) / (12.4) = 0.767 / 76.7% A1 [2]




7 (a) (i) six vertical lines from plate to plate equally spaced across plates B1 [only allow if greatest to least spacing is < 1.3, condone slight curving on the

two edges. There must be no area between the plates where an additional line(s) could be added.]

arrow downwards on at least one line B1 [2] (ii) E = V / d C1

E = 1200 / 40 × 10–3 = 3.0 × 104 V m–1 (allow 1 s.f.) A1 [2] (b) (i) F = Ee C1

E = 3 × 104 × 1.6 × 10–19 = 4.8 × 10–15 N A1 [2]

(ii) couple = F × separation of charges C1

couple = 4.8 × 10–15 × 15 × 10–3 = 7.2 × 10–17 A1 unit: N m or unit consistent with unit used for the separation B1 [3]

(iii) A at top/next to +ve plate B at bottom/next to −ve plate vertically aligned M1 [could be shown on the diagram] forces are equal and opposite in same line / no resultant force and no

resultant torque A1 [2]




9702 PHYSICS

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






1 (a) (i) Value for d in the range 0.15 mm ≤ d ≤ 0.25 mm, with unit. [1]

(c) (ii) Values of V1 and V2 , and V1 ˃ V2. [1]

(d) Six sets of readings of l, V1 and V2 scores 5 marks, five sets scores 4 marks etc. [5] Major help from Supervisor –2. Minor help from Supervisor –1.

Range: ∆l ≥ 30 cm. [1] Column headings: [1]

Each column heading must contain a quantity and a unit where appropriate. The unit must conform to accepted scientific convention, e.g. l / m or 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 V1/V2 must be the same as, or one more than the least number of significant figures used in V1 and V2.

Calculation: [1] Values of V1/V2 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 point 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 on the grid for this mark to be awarded. All points must be within 0.05 (to scale) on the y-axis V1/V2 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 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: Check 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: Check read-off of the intercept directly from the graph. (f) (i) Value of P = candidate’s gradient. Value of Q = candidate’s intercept. [1]

(ii) Value of ρ in range 1.0 – 20.0 × 10–7 Ω m [1]

[Total: 20] 2 (b) Value of m to the nearest 1 g or better with consistent unit. [1]

(c) (ii) Measurement of raw θ to nearest degree with unit. [1]

Evidence of repeat readings for θ. [1]

(iii) Percentage uncertainty in θ based on absolute uncertainty of 2 to 5° (or half the range provided this is not zero), and correct method of calculation. [1]

(iv) Correct calculation of tan (θ / 2). [1] (d) (i) Second value of m > first value of m.

[1]

(ii) Second value of θ. [1]

Quality: second value of θ ˂ first value of θ. [1]

(e) Value of θ. [1] (f) (i) Two values of k calculated correctly. [1]

(ii) Justification of s.f. in k linked to significant figures in m and θ. [1] (iii) 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

A Two readings 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 / ‘few readings’ / ‘take more readings and calculate average’ / ‘only one reading’ / ‘repeat readings’ on its own

B Difficult to measure θ because hook of mass (hanger) in the way / thick band

Tie thread to centre of bottom of rubber band and hang mass from it

C Difficult to hold the protractor steady / parallax error reading angle / protractor

Improved method to measure θ e.g. project image of stretched rubber band onto a screen / mark on board /

measure lengths and calculate θ clamp protractor / take picture or video and measure angle

D Rubber band stretches over time

Take readings quickly / remove mass from rubber band between readings

E Stands moved / rods twist when loads attached to rubber band

Method of preventing movement of stands / clamp stands to bench / use nails in board

F Difficult to locate centre of band Method of locating and mark centre e.g. measure and mark centre

G Change in θ small Larger range of masses

[Total: 20]




9702 PHYSICS







1 (a) (i) Value for d in the range 0.15 mm ≤ d ≤ 0.25 mm, with unit. [1]

(c) (ii) Values of V1 and V2 , and V1 ˃ V2. [1]

(d) Six sets of readings of l, V1 and V2 scores 5 marks, five sets scores 4 marks etc. [5] Major help from Supervisor –2. Minor help from Supervisor –1.

Range: ∆l ≥ 30 cm. [1] Column headings: [1]

Each column heading must contain a quantity and a unit where appropriate. The unit must conform to accepted scientific convention, e.g. l / m or 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 V1/V2 must be the same as, or one more than the least number of significant figures used in V1 and V2.

Calculation: [1] Values of V1/V2 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 point 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 on the grid for this mark to be awarded. All points must be within 0.05 (to scale) on the y-axis V1/V2 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 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: Check 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: Check read-off of the intercept directly from the graph. (f) (i) Value of P = candidate’s gradient. Value of Q = candidate’s intercept. [1]

(ii) Value of ρ in range 1.0 – 20.0 × 10–7 Ω m [1]

[Total: 20] 2 (b) Value of m to the nearest 1 g or better with consistent unit. [1]

(c) (ii) Measurement of raw θ to nearest degree with unit. [1]

Evidence of repeat readings for θ. [1]

(iii) Percentage uncertainty in θ based on absolute uncertainty of 2 to 5° (or half the range provided this is not zero), and correct method of calculation. [1]

(iv) Correct calculation of tan (θ / 2). [1] (d) (i) Second value of m > first value of m.

[1]

(ii) Second value of θ. [1]

Quality: second value of θ ˂ first value of θ. [1]

(e) Value of θ. [1] (f) (i) Two values of k calculated correctly. [1]

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





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

A Two readings 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 / ‘few readings’ / ‘take more readings and calculate average’ / ‘only one reading’ / ‘repeat readings’ on its own

B Difficult to measure θ because hook of mass (hanger) in the way / thick band

Tie thread to centre of bottom of rubber band and hang mass from it

C Difficult to hold the protractor steady / parallax error reading angle / protractor

Improved method to measure θ e.g. project image of stretched rubber band onto a screen / mark on board /

measure lengths and calculate θ clamp protractor / take picture or video and measure angle

D Rubber band stretches over time

Take readings quickly / remove mass from rubber band between readings

E Stands moved / rods twist when loads attached to rubber band

Method of preventing movement of stands / clamp stands to bench / use nails in board

F Difficult to locate centre of band Method of locating and mark centre e.g. measure and mark centre

G Change in θ small Larger range of masses

[Total: 20]




9702 PHYSICS







1 (a) (ii) Value of h2<h1, with consistent unit. [1] (b) (iv) First values of LA and LB, with unit, and value of LA – LB in range 1.0 to 6.0 cm. [1] (c) Six sets of values for x, LA and LB scores 5 marks, five sets scores 4 marks etc. [5]

Incorrect trend –1. Help from Supervisor –1. Range: x values must include 20.0 cm or less and 80.0 cm or more. [1] Column headings: [1] Each column heading must contain a quantity and a unit where appropriate. The unit must conform to accepted scientific convention, e.g. x/cm or x (cm). Consistency of presentation of raw readings: [1] All values of x must be given to the nearest mm. Significant figures: [1] Every value of (LA–LB)/C must be given to the same s.f. as (or one more than) the least s.f. in C or in (LA–LB). Calculation: (LA–LB)/C calculated correctly. [1]

(d) (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 must be plotted.

Diameter of plotted points must be ≤ half a small square (no “blobs”). Plots must be accurate to half a small square.

Quality: [1]

All points in the table must be plotted for this mark to be scored. All points must be within ± 4 scale cm, on the x–axis, of 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 by the candidate. Line must not be kinked or thicker than half a small square.




(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 in both 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.

(e) Value of a = candidate’s gradient and value of b = candidate’s intercept. [1]

A value presented as a fraction is not allowed. Correct units for a (e.g. cm–1) and b (no unit), and gradient in range –0.050 to –0.150 cm–1 (–5.0 to –15.0 m–1). [1]

[Total: 20] 2 (a) (i) Raw value(s) for d to nearest 0.1 mm or nearest 0.01 mm. [1]

d in range 10.0 to 15.0 mm. [1] (ii) Correct calculation of R. [1] (b) Valid justification for s.f. in R based on s.f. in d and D. [1]

(d) (iv) θ in range 10o to 50o, with unit. Raw reading(s) must be integer value(s). [1]

Evidence of repeated readings of θ. [1] (e) Estimate of percentage uncertainty based on an absolute uncertainty of 2o to 10o [1]

(or half the range provided this is not zero), and correct method of calculation. (f) Second values for d and D. [1]

Second value of θ. [1]

Quality: Second θ less than first θ. [1] (g) (i) Correct calculation of two values of k. [1] (ii) Sensible comment relating to the calculated values of k, testing against a [1] criterion specified by the candidate.




(h)

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

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

Take many R values and plot a graph/ calculate more k values and compare

‘few readings’/ ‘take more readings and calculate average’/‘only one reading’

B Spool does not roll straight/ spool topples over/ jerky motion

Roll spool between guides/ use wider spool

Lubricate bench or spool

C Difficult to judge θ as it varies, equilibrium position not maintained

D Difficult to measure θ due to parallax

Method to reduce parallax (e.g. shadow method)

E Difficult to pull thread when

measuring θ / difficult to watch spool and protractor at same time

Specified thread guiding method (e.g. pulley or rod)

F Difficult to align thread with centre of protractor

Take photo or video and

measure θ on image (e.g. photo or screen)

G Difficult to make protractor vertical

Use plumb-line/detail of another sensible method (e.g. use set square on bench)

Do not allow ‘repeated readings’ Do not allow ‘use a computer to improve the experiment’

[Total: 20]


GCE Advanced Subsidiary Level


9702 PHYSICS





GCE AS LEVEL – October/November 2013 9702 35


1 (a) Value for L in the range 0.500–0.600 m. [1] (c) (iv) Value for n in the range 3–8. [1]

(d) Six sets of readings of D and n scores 5 marks, five sets scores 4 marks etc. [5] Help from Supervisor –1. Range of D: Dmin < 45 cm and Dmax > 50 cm. [1]

Column headings: [1] Each column heading must contain a quantity and a unit where appropriate.

The unit must conform to accepted scientific convention e.g. D/m. Consistency: [1] All values of D must be given to the nearest mm.

Significant figures: [1]

Every value of ((n + 1)/n)2 should be given to 2 or 3 s.f. Calculation: [1]

Values of ((n + 1)/n)2 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 point 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 on the grid for this mark to be awarded. All points must be no more than 0.04 of ((n + 1)/n)2 from a straight line. (e) (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 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: Check 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:

Check read-off of the intercept directly from the graph. (f) Value of P = –(candidate’s gradient). Value of Q = candidate’s intercept. [1]

Unit for P (e.g. m–1) consistent with value and no unit for Q. [1] [Total: 20]

2 (b) (i) Value for V0 in range 25.0 – 35.0 cm3. [1] (ii) Evidence of two volumes added together. [1]

(v) Correct calculation of V. [1] (vi) Absolute uncertainty in V in range 1 cm3–3 cm3.

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 find percentage uncertainty. [1]

(c) (iii) Value(s) of x. [1] Evidence of repeat readings of x (either here or in (d)(ii)). [1] (d) (ii) Second value of V. [1] Second value of x. [1] Quality: second value of x less than first value of x. [1]

(e) (i) Two values of k calculated correctly. [1] (ii) Justification of s.f. in k linked to significant figures in x and V. [1]

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




(f)


A Two readings not enough (to draw a conclusion)

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

‘few readings’/‘take more readings and calculate average’/‘only one reading’/ ’repeat readings’ on its own

B Difficult to remove correct amount of water because air drawn into syringe

Use syringe with longer nozzle/needle Tilt/invert ball

‘nozzle too short’ on its own

C Blu-tack not sticky enough/water leaks from ball/ syringe

Use e.g. sellotape/small cork to seal hole

D Difficult to judge lowest depth with reason e.g. parallax error/difficult to move head

Line up both sides of rubber band with ball Use mirror behind ball

Parallax measuring x Take measurements at eye level

E Difficult to judge lowest depth because ball at maximum depth for a short time

Video experiment with scale Moves too fast/too quickly Use motion sensor/high-speed cameras/slow-motion cameras Use light sensors

F Large uncertainty in value of V/scale divisions on syringe too large

Use a smaller syringe/ measure mass/weight of ball and water

G Difficult to release ball without applying force/difficult to hold the ball on the surface of the water

Method of releasing ball e.g. cut string attached to ball

Clamp ball Use card/plastic gate

Do not allow ‘use a computer to improve the experiment’.

[Total: 20]




9702 PHYSICS







1 (b) (i) Value for E in range 2.00 to 3.50 V, with unit. [1] (c) (iii) Value of x (for V = 0) with consistent unit, in range 0.250 to 0.400 m. [1] (d) Six sets of readings of x and V scores 5 marks, five sets scores 4 marks etc. [5]

Incorrect trend –1. Help from Supervisor –1. Range: [1] Range of values to include xmin ≤ 20.0 cm and xmax

≥ 80.0 cm. Column headings: [1] Each column heading must contain a quantity and a unit where appropriate. The unit must conform to accepted scientific convention, e.g. V / V or V (V). Consistency: [1] All values of x must be given to the nearest mm. Significant figures: [1] Every value of V/E must be given to the same number of s.f. (or one more than) the least s.f. in the corresponding values of V and E. Calculated values: [1] V/E calculated correctly, including sign.

(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 must be plotted. Diameter of plotted points must be ≤ half a small square (“no blobs”). Plots must be accurate to half a small square.

Quality: [1] All points in the table must be plotted on the grid for this mark to be awarded. All points must be within 2 cm (to scale) on the x-axis of 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 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 is 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 a = candidate’s gradient and value of b = candidate’s intercept.

b in range 0.25 to 0.75. [1] A value presented as a fraction is not allowed. Unit for a correct (e.g. cm–1) and consistent with value, and no unit given for b. [1]

[Total: 20] 2 (a) (i) Value for mA in range 5.0 to 30.0 g. [1] (iii) Correct calculation of R. [1] (iv) Justification for s.f. in R linked to s.f. in mA and (mA

+ mB ). [1] (b) Value for h0, with unit, to nearest mm. [1] (c) Value for hB >h0. [1]

Evidence of repeat readings for hB. [1] (d) Percentage uncertainty in hB based on absolute uncertainty of 2 to 5 mm (or half the

range provided this is not zero), and correct method of calculation. [1] (e) Second value of mB. [1]

Second value of hB. [1] Quality: hB smaller for larger mB. [1]

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





(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

‘few readings’/‘take more readings and calculate average’/ ‘only one reading’/ ‘repeat readings’ on its own

B Difficult to judge highest point/ hB

with reason (e.g. short time at highest point/doesn’t stay still for long at highest point)

Method of improved measurement of hB (e.g. video with scale/multiflash photography with scale/use marker/use pointer/trial and error method/mark track with scale/put scale on board behind/ink or chalk on ball/ motion sensors at top of track)

‘too fast’/ ball travelling too quick/high speed camera/ slow motion camera/light gates

C Error when measuring height(s) because of: parallax/ ruler not vertical/ruler not perpendicular to bench

Detailed explanation of reducing error

‘set squares’ on own/‘view perpendicular’/‘parallax error’ on own

D Bottom of ball not visible

Method of improved measurement of height(s) (e.g. measure to top of ball)

E Energy lost (e.g. as friction with track or air/sound/hitting sides/not hitting square on)

Lubricate track

F Difficult to release marble without applying a force/ difficult not to apply (sideways) velocity

Detail of a mechanical release method (e.g. card gate)

Friction on track

Do not allow ‘repeated readings’ or ‘use a computer to improve the experiment’

[Total: 20]




9702 PHYSICS

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




GCE A LEVEL – October/November 2013 9702 41


Section A 1 (a) work done in moving unit mass M1 from infinity (to the point) A1 [2] (b) (i) gravitational potential energy = GMm / x

energy = (6.67 × 10–11 × 7.35 × 1022 × 4.5) / (1.74 × 106) M1 energy = 1.27 × 107 J A0 [1]

(ii) change in grav. potential energy = change in kinetic energy B1 ½ × 4.5 × v2 = 1.27 × 107 v = 2.4 × 103 m s–1 A1 [2] (c) Earth would attract the rock / potential at Earth(’s surface) not zero / <0 / at Earth, potential due to Moon not zero M1 escape speed would be lower A1 [2] 2 (a) (i) N: (total) number of molecules B1 [1] (ii) <c2>: mean square speed/velocity B1 [1] (b) pV = ⅓Nm<c2> = NkT (mean) kinetic energy = ½ m<c2> C1 algebra clear leading to ½ m<c2> = (3/2)kT A1 [2] (c) (i) either energy required = (3/2) × 1.38 × 10–23 × 1.0 × 6.02 × 1023 C1 either energy required = 12.5 J (12J if 2 s.f.) A1 [2] or energy = (3/2) × 8.31 × 1.0 (C1) or energy = 12.5 J (A1) (ii) energy is needed to push back atmosphere/do work against

atmosphere M1 so total energy required is greater A1 [2] 3 (a) (i) any two from 0.3(0) s, 0.9(0) s, 1.50 s (allow 2.1 s etc.) B1 [1] (ii) either v = ωx and ω = 2π/T C1 either v = (2π/1.2) × 1.5 × 10–2 M1 either v = 0.079 m s–1 A0 [2] or gradient drawn clearly at a correct position (C1) working clear (M1) to give (0.08 ± 0.01) m s–1 (A0)




(b) (i) sketch: curve from (±1.5, 0) passing through (0, 25) M1 sketch: reasonable shape (curved with both intersections between

y = 12.0→13.0) A1 [2] (ii) at max. amplitude potential energy is total energy B1 total energy = 4.0 mJ B1 [2] 4 (a) (i) force proportional to product of (two) charges and inversely

proportional to square of separation M1 reference to point charges A1 [2] (ii) F = 2 × (1.6 × 10–19)2 / 4π × 8.85 × 10–12 × (20 × 10–6)2 C1 F = 1.15 × 10–18 N A1 [2] (b) (i) force per unit charge M1 on either a stationary charge or a positive charge A1 [2] (ii) 1. electric field is a vector quantity electric fields are in opposite directions charges repel Any two of the above, 1 each B2 [2] 2. graph: line always between given lines M1 crosses x-axis between 11.0 µm and 12.3 µm A1 reasonable shape for curve A1 [3] 5 (a) (i) field shown as right to left B1 [1] (ii) lines are more spaced out at ends B1 [1] (b) Hall voltage depends on angle M1 either between field and plane of probe or maximum when field normal to plane of probe or zero when field parallel to plane of probe A1 [2] (c) (i) (induced) e.m.f. proportional to rate M1 of change of (magnetic) flux (linkage) A1 [2] (allow rate of cutting of flux) (ii) e.g. move coil towards/away from solenoid e.g. rotate coil e.g. vary current in solenoid e.g. insert iron core into solenoid (any three sensible suggestions, 1 each) B3 [3]




6 (a) force due to magnetic field is constant B1 force is (always) normal to direction of motion this force provides the centripetal force A1 [3] (b) mv2 / r = Bqv M1 hence q / m = v / Br A0 [1] (c) (i) q / m = (2.0 × 107) / (2.5 × 10–3 × 4.5 × 10–2) C1 q / m = 1.8 × 1011 C kg–1 A1 [2] (ii) sketch: curved path, constant radius, in direction towards bottom of

page M1 tangent to curved path on entering and on leaving the field A1 [2] 7 (a) either if light passes through suitable film / cork dust etc. M1 either diffraction occurs and similar pattern observed A1 or concentric circles are evidence of diffraction (M1) or diffraction is a wave property (A1) [2] (b) (speed increases so) momentum increases M1 λ = h/p so λ decreases M1 hence radii decrease A1 [3] (special case: wavelength decreases so radii decreases – scores 1/3) or (speed increases so) energy increases (B1) λ = h / √(2Em) so λ decreases (M1) hence radii decrease (A1) (c) electron and proton have same (kinetic) energy C1 either E = p2 / 2m or p = √(2Em) C1 ratio = pe / pp = √(me / mp) C1 ratio = √(9.1 × 10–31) / (1.67 × 10–27) ratio = 2.3 × 10–2 A1 [4] 8 (a) energy to separate nucleons (in a nucleus) M1 separate to infinity A1 [2] (b) (i) fission B1 [1] (ii) 1. U: near right-hand end of line B1 [1] 2. Mo: to right of peak, less than 1/3 distance from peak to U B1 [1] 3. La: 0.4 → 0.6 of distance from peak to U B1 [1]




(iii) 1. right-hand side, mass = 235.922 u C1 mass change = 0.210 u A1 [2] 2. energy = mc2 C1 energy = 0.210 × 1.66 × 10–27 × (3.0 × 108)2 energy = 3.1374 × 10–11 J C1 energy = 196 MeV (need 3 s.f.) A1 [3] (use of 1 u = 934 MeV, allow 3/3; use of 1 u = 930 MeV or 932

MeV, allow 2/3) (use of 1.67 × 10–27 not 1.66 ×10–27 scores max. 2/3)

Section B 9 (a) operates on / takes signal from sensing device B1 (so that) it gives an voltage output B1 [2] (b) thermistor and resistor in series between +4 V line and earth M1 VOUT shown clearly across either thermistor or resistor A1 VOUT shown clearly across thermistor A1 [3] (c) e.g. remote switching e.g. switching large current by means of a small current e.g. isolating circuit from high voltage e.g. switching high voltage by means of a small voltage/current (any two sensible suggestions, 1 each to max. 2) B2 [2] 10 (a) pulse (of ultrasound) B1 produced by quartz / piezo-electric crystal (1) reflected from boundaries (between media) B1 reflected pulse detected B1 by the ultrasound transmitter (1) signal processed and displayed B1 intensity of reflected pulse gives information about the boundary (1) time delay gives information about depth (1) (four B marks plus any two from the four, max. 6) B2 [6] (b) shorter wavelength B1 smaller structures resolved / detected (not more sharpness) B1 [2] (c) (i) I = I0 e

–µx C1 ratio = exp(–23 × 6.4 × 10–2) C1 ratio = 0.23 A1 [3] (ii) later signal has passed through greater thickness of medium M1 so has greater attenuation / greater absorption / smaller intensity A1 [2]




11 (a) left-hand bit underlined B1 [1] (b) 1010, 1110, 1111, 1010, 1001 (5 correct scores 2, 4 correct scores 1) A2 [2] (c) significant changes in detail of V between samplings M1 so frequency too low A1 [2] 12 (a) e.g. logarithm provides a smaller number e.g. gain of amplifiers is series found by addition, (not multiplication) (any sensible suggestion) B1 [1] (b) (i) optic fibre B1 [1] (ii) attenuation/dB = 10 lg(P2/P1) C1 attenuation/dB = 10 lg(6.5 × 10–3/1.5 × 10–15) C1 attenuation/dB = 126 length = 126 / 1.8 length = 70 km A1 [3]




9702 PHYSICS







Section A 1 (a) work done in moving unit mass M1 from infinity (to the point) A1 [2] (b) (i) gravitational potential energy = GMm / x

energy = (6.67 × 10–11 × 7.35 × 1022 × 4.5) / (1.74 × 106) M1 energy = 1.27 × 107 J A0 [1]

(ii) change in grav. potential energy = change in kinetic energy B1 ½ × 4.5 × v2 = 1.27 × 107 v = 2.4 × 103 m s–1 A1 [2] (c) Earth would attract the rock / potential at Earth(’s surface) not zero / <0 / at Earth, potential due to Moon not zero M1 escape speed would be lower A1 [2] 2 (a) (i) N: (total) number of molecules B1 [1] (ii) <c2>: mean square speed/velocity B1 [1] (b) pV = ⅓Nm<c2> = NkT (mean) kinetic energy = ½ m<c2> C1 algebra clear leading to ½ m<c2> = (3/2)kT A1 [2] (c) (i) either energy required = (3/2) × 1.38 × 10–23 × 1.0 × 6.02 × 1023 C1 either energy required = 12.5 J (12J if 2 s.f.) A1 [2] or energy = (3/2) × 8.31 × 1.0 (C1) or energy = 12.5 J (A1) (ii) energy is needed to push back atmosphere/do work against

atmosphere M1 so total energy required is greater A1 [2] 3 (a) (i) any two from 0.3(0) s, 0.9(0) s, 1.50 s (allow 2.1 s etc.) B1 [1] (ii) either v = ωx and ω = 2π/T C1 either v = (2π/1.2) × 1.5 × 10–2 M1 either v = 0.079 m s–1 A0 [2] or gradient drawn clearly at a correct position (C1) working clear (M1) to give (0.08 ± 0.01) m s–1 (A0)




(b) (i) sketch: curve from (±1.5, 0) passing through (0, 25) M1 sketch: reasonable shape (curved with both intersections between

y = 12.0→13.0) A1 [2] (ii) at max. amplitude potential energy is total energy B1 total energy = 4.0 mJ B1 [2] 4 (a) (i) force proportional to product of (two) charges and inversely

proportional to square of separation M1 reference to point charges A1 [2] (ii) F = 2 × (1.6 × 10–19)2 / 4π × 8.85 × 10–12 × (20 × 10–6)2 C1 F = 1.15 × 10–18 N A1 [2] (b) (i) force per unit charge M1 on either a stationary charge or a positive charge A1 [2] (ii) 1. electric field is a vector quantity electric fields are in opposite directions charges repel Any two of the above, 1 each B2 [2] 2. graph: line always between given lines M1 crosses x-axis between 11.0 µm and 12.3 µm A1 reasonable shape for curve A1 [3] 5 (a) (i) field shown as right to left B1 [1] (ii) lines are more spaced out at ends B1 [1] (b) Hall voltage depends on angle M1 either between field and plane of probe or maximum when field normal to plane of probe or zero when field parallel to plane of probe A1 [2] (c) (i) (induced) e.m.f. proportional to rate M1 of change of (magnetic) flux (linkage) A1 [2] (allow rate of cutting of flux) (ii) e.g. move coil towards/away from solenoid e.g. rotate coil e.g. vary current in solenoid e.g. insert iron core into solenoid (any three sensible suggestions, 1 each) B3 [3]




6 (a) force due to magnetic field is constant B1 force is (always) normal to direction of motion this force provides the centripetal force A1 [3] (b) mv2 / r = Bqv M1 hence q / m = v / Br A0 [1] (c) (i) q / m = (2.0 × 107) / (2.5 × 10–3 × 4.5 × 10–2) C1 q / m = 1.8 × 1011 C kg–1 A1 [2] (ii) sketch: curved path, constant radius, in direction towards bottom of

page M1 tangent to curved path on entering and on leaving the field A1 [2] 7 (a) either if light passes through suitable film / cork dust etc. M1 either diffraction occurs and similar pattern observed A1 or concentric circles are evidence of diffraction (M1) or diffraction is a wave property (A1) [2] (b) (speed increases so) momentum increases M1 λ = h/p so λ decreases M1 hence radii decrease A1 [3] (special case: wavelength decreases so radii decreases – scores 1/3) or (speed increases so) energy increases (B1) λ = h / √(2Em) so λ decreases (M1) hence radii decrease (A1) (c) electron and proton have same (kinetic) energy C1 either E = p2 / 2m or p = √(2Em) C1 ratio = pe / pp = √(me / mp) C1 ratio = √(9.1 × 10–31) / (1.67 × 10–27) ratio = 2.3 × 10–2 A1 [4] 8 (a) energy to separate nucleons (in a nucleus) M1 separate to infinity A1 [2] (b) (i) fission B1 [1] (ii) 1. U: near right-hand end of line B1 [1] 2. Mo: to right of peak, less than 1/3 distance from peak to U B1 [1] 3. La: 0.4 → 0.6 of distance from peak to U B1 [1]




(iii) 1. right-hand side, mass = 235.922 u C1 mass change = 0.210 u A1 [2] 2. energy = mc2 C1 energy = 0.210 × 1.66 × 10–27 × (3.0 × 108)2 energy = 3.1374 × 10–11 J C1 energy = 196 MeV (need 3 s.f.) A1 [3] (use of 1 u = 934 MeV, allow 3/3; use of 1 u = 930 MeV or 932

MeV, allow 2/3) (use of 1.67 × 10–27 not 1.66 ×10–27 scores max. 2/3)

Section B 9 (a) operates on / takes signal from sensing device B1 (so that) it gives an voltage output B1 [2] (b) thermistor and resistor in series between +4 V line and earth M1 VOUT shown clearly across either thermistor or resistor A1 VOUT shown clearly across thermistor A1 [3] (c) e.g. remote switching e.g. switching large current by means of a small current e.g. isolating circuit from high voltage e.g. switching high voltage by means of a small voltage/current (any two sensible suggestions, 1 each to max. 2) B2 [2] 10 (a) pulse (of ultrasound) B1 produced by quartz / piezo-electric crystal (1) reflected from boundaries (between media) B1 reflected pulse detected B1 by the ultrasound transmitter (1) signal processed and displayed B1 intensity of reflected pulse gives information about the boundary (1) time delay gives information about depth (1) (four B marks plus any two from the four, max. 6) B2 [6] (b) shorter wavelength B1 smaller structures resolved / detected (not more sharpness) B1 [2] (c) (i) I = I0 e

–µx C1 ratio = exp(–23 × 6.4 × 10–2) C1 ratio = 0.23 A1 [3] (ii) later signal has passed through greater thickness of medium M1 so has greater attenuation / greater absorption / smaller intensity A1 [2]




11 (a) left-hand bit underlined B1 [1] (b) 1010, 1110, 1111, 1010, 1001 (5 correct scores 2, 4 correct scores 1) A2 [2] (c) significant changes in detail of V between samplings M1 so frequency too low A1 [2] 12 (a) e.g. logarithm provides a smaller number e.g. gain of amplifiers is series found by addition, (not multiplication) (any sensible suggestion) B1 [1] (b) (i) optic fibre B1 [1] (ii) attenuation/dB = 10 lg(P2/P1) C1 attenuation/dB = 10 lg(6.5 × 10–3/1.5 × 10–15) C1 attenuation/dB = 126 length = 126 / 1.8 length = 70 km A1 [3]

9701CAMBRIDGE INTERNATIONAL EXAMINATIONS



9702 PHYSICS







Section A

1 (a) force proportional to product of the two masses and inversely proportional to the square of their separation M1

either reference to point masses or separation >> ‘size’ of masses A1 [2]

(b) gravitational force provides the centripetal force B1

GMm / R2 = mRω2 M1

where m is the mass of the planet A1

GM = R3ω

2 A0 [3]

(c) ω = 2π / T C1

either Mstar / MSun = (Rstar / RSun)3 × (TSun / Tstar)

2

Mstar = 43 × (½)2 × 2.0 × 1030 C1

= 3.2 × 1031 kg A1 [3]

or Mstar = (2π)2 Rstar

3 / GT2 (C1)

= (2π)2 × (6.0 × 1011)3 / 6.67 × 10–11 × (2 × 365 × 24 × 3600)2 (C1)

= 3.2 × 1031 kg (A1)

2 (a) (i) sum of kinetic and potential energies of the molecules M1 reference to random distribution A1 [2] (ii) for ideal gas, no intermolecular forces M1 so no potential energy (only kinetic) A1 [2]

(b) (i) either change in kinetic energy = 3/2 × 1.38 × 10–23 × 1.0 × 6.02 × 1023 × 180 C1 = 2240 J A1 [2] or R = kNA

energy = 3/2 × 1.0 × 8.31 × 180 (C1) = 2240 J (A1) (ii) increase in internal energy = heat supplied + work done on system B1 2240 = energy supplied – 1500 C1 energy supplied = 3740 J A1 [3] 3 (a) work done bringing unit positive charge M1 from infinity (to the point) A1 [2]

(b) (i) either both potentials are positive / same sign M1 so same sign A1 [2] or gradients are positive & negative (so fields in opposite directions) (M1) so same sign (A1) (ii) the individual potentials are summed B1 [1] (iii) allow value of x between 10 nm and 13 nm A1 [1] (iv) V = 0.43 V (allow 0.42 V → 0.44 V) M1

energy = 2 × 1.6 × 10–19 × 0.43 A1

= 1.4 × 10–19 J A1 [3]




4 (a) e.g. store energy (do not allow ‘store charge’) in smoothing circuits blocking d.c. in oscillators any sensible suggestions, one each, max. 2 B2 [2] (b) (i) potential across each capacitor is the same and Q = CV B1 [1] (ii) total charge Q = Q1 + Q2 + Q3 M1 CV = C1V + C2V + C3V M1 (allow Q = CV here or in (i)) so C = C1 + C2 + C3 A0 [2] (c) (i) A1 [1] (ii) A1 [1] 5 (a) (i) region (of space) either where a moving charge (may) experience a force or around a magnet where another magnet experiences a force B1 [1]

(ii) (Φ =) BA sinθ A1 [1] (b) (i) plane of frame is always parallel to BV / flux linkage always zero B1 [1]

(ii) ∆Φ = 1.8 × 10–5 × 52 × 10–2 × 95 × 10–2 C1

= 8.9 × 10–6 Wb A1 [2] (c) (i) (induced) e.m.f. proportional to rate of M1 change of (magnetic) flux (linkage) A1 [2] (allow rate of cutting of flux)

(ii) e.m.f. = (8.9 × 10–6) / 0.30

= 3.0 × 10–5 V A1 [1] (iii) This question part was removed from the assessment. All candidates were

awarded 1 mark. B1 [1]




6 (a) either constant speed parallel to plate or accelerated motion / force normal to plate / in direction field B1 so not circular A0 [1] (b) (i) direction of force due to magnetic field opposite to that due to electric field B1 magnetic field into plane of page B1 [2] (ii) force due to magnetic field = force due to electric field B1 Bqv = qE B = E / v C1

= (2.8 × 104) / (4.7 × 105)

= 6.0 × 10–2 T A1 [3] (c) (i) no change / not deviated B1 [1] (ii) deviated upwards B1 [1] (iii) no change / not deviated B1 [1] 7 (a) (i) minimum photon energy B1 minimum energy to remove an electron (from the surface) B1 [2] (ii) either maximum KE is photon energy – work function energy or max KE when electron ejected from the surface B1 energies lower than max because energy required to bring electron to

the surface B1 [2]

(b) (i) threshold frequency = 1.0 × 1015 Hz (allow ±0.05 × 1015) C1

work function energy = hf0 C1

= 6.63 × 10–34 × 1.0 × 1015

= 6.63 × 10–19 J A1 [3]

(allow alternative approaches based on use of co-ordinates of points on the line)

(ii) sketch: straight line with same gradient M1 displaced to right A1 [2] (iii) intensity determines number of photons arriving per unit time B1 intensity determines number of electrons per unit time (not energy) B1 [2] 8 (a) probability of decay (of a nucleus) / fraction of number of nuclei in sample

that decay M1 per unit time A1 [2]

(allow λ =(dN / dt) / N with symbols explained – (M1), (A1) )

(b) (i) number = (1.2 × 6.02 × 1023) / 235 C1

= 3.1 × 1021 A1 [2]




(ii) N = N0 e–λt

negligible activity from the krypton B1

for barium, N = (3.1 × 1021) exp(–6.4 × 10–4 × 3600)

= 3.1 × 1020 C1

activity = λN

= 6.4 × 10–4 × 3.1 × 1020 C1

= 2.0 × 1017 Bq A1 [4]

Section B 9 (a) e.g. zero output impedance / resistance infinite input impedance / resistance infinite (open loop) gain infinite bandwidth infinite slew rate (1 each, max. 3 ) B3 [3] (b) (i) gain = 1 + (10.8 / 1.2) C1 = 10 A1 [2] (ii) graph: straight line from (0,0) towards VIN = 1.0 V, VOUT = 10 V B1 horizontal line at VOUT = 9.0 V to VIN = 2.0 V B1 correct +9.0 V → –9.0 V (and correct shape to VIN = 0) B1 [3] 10 (a) nuclei spin / precess B1 spin / precess about direction of magnetic field B1 either frequency of precession depends on magnetic field strength or large field means frequency in radio frequency range B1 [3] (b) non-uniform field means frequency of precession different in different regions

of subject B1 enables location of precessing nuclei to be determined B1 enables thickness of slice to be varied / location of slice to be changed B1 [3] 11 (a) (i) either series of ‘highs’ and ‘lows’ or two discrete values M1 with no intermediate values A1 [2] (ii) e.g. noise can be eliminated (NOT ‘no noise’) signal can be regenerated addition of extra data to check for errors larger data carrying capacity cheaper circuits more reliable circuits (any three, 1 each) B3 [3]




(b) (i) 1. amplifier B1 [1] 2. digital-to-analogue converter (allow DAC) B1 [1] (ii) output of ADC is number of digits all at one time B1 parallel-to-serial sends digits one after another B1 [2] 12 (a) e.g. no / little ionospheric reflection large information carrying capacity (any two sensible suggestions, 1 each) B2 [2] (b) prevents (very) low power signal received at satellite M1 being swamped by high-power transmitted signal A1 [2] (c) attenuation / dB = 10 lg(P2/P1) C1

185 = 10 lg(3.1 × 103/P) C1

P = 9.8 × 10–16 W A1 [3]




9702 PHYSICS

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






1 Planning (15 marks) Defining the problem (3 marks)

P f is the independent variable or vary f. [1] P h is the dependent variable or measure h. [1] P Keep current in coil constant. [1] Methods of data collection (5 marks)

M Labelled diagram of apparatus: including working circuit with a.c. supply, coil, stand and ring. [1] M Signal generator (or variable frequency power generator). [1] M Measure h with a rule/calipers. [1] M Measure f using oscilloscope/read off signal generator. [1] M Measure h from opposite sides of the ring (and determine mean)/wait for ring to stabilise. [1] Method of analysis (2 marks)

A Plot a graph of lg h against lg f. (Allow ln – ln graphs.) [1] A k = 10y-intercept and n = gradient. [1] Safety considerations (1 mark)

S Reasoned method to prevent the coil overheating – switch off when not in use. Or reasoned method to prevent injury from hot coil – do not touch hot coil, use gloves. Do not allow “small currents”. [1]

Additional detail (4 marks)

D Relevant points might include [4] 1 Use iron/steel retort stand 2 Method to keep current constant 3 Method to determine period from oscilloscope using time-base 4 When using oscilloscope, f = 1/T 5 Use coil of many turns/large current to have measurable heights 6 Logarithmic equation e.g. lg h = n lg f + lg k 7 Relationship is valid if a straight line is produced

Do not allow vague computer methods.

[Total: 15]




2 Analysis, conclusions and evaluation (15 marks)

Mark Expected Answer Additional Guidance

(a) A1 Gradient = v/4 y-intercept = –k

y-intercept must be negative.

(b) T1 1/f / 10–3 s Column heading. Allow equivalent unit. e.g. f

–1/ 10–3 Hz–1 or 1/f (10–3 s)

T2

3.13 or 3.125

2.94 or 2.941

2.65 or 2.646

2.34 or 2.336

2.08 or 2.083

1.95 or 1.953

A mixture of 3 s.f. and 4 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 d plotted correctly

Must be within half a small square. Each error bars should be five small squares

(ii) G2 Line of best fit If points are plotted correctly then lower end of line should pass between (2.06, 16) and (2.10, 16) and upper end of line should pass between (3.12, 24.5) and (3.16, 24.5).

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.

(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.

(iv) C2 y-intercept – must be negative

Check substitution in y = mx + c. Allow ecf from (c)(iii). FOX does not score. Should be about –1 (–0.35 to –1.33).

U3 Uncertainty in y-intercept Check substitution in y = mx + c. Difference in worst y-intercept and y-intercept. Allow ecf from (c)(iv) but FOX does not score.




(d) C3 k = – y-intercept cm or m

v = 4 × gradient cm s–1 or m s–1

Method required with appropriate unit. Do not penalise POT. Allow ecf from (c)(iii) and (iv).

U4 Determines absolute uncertainties in k and v

Uncertainty k = uncertainty in y-intercept.

Uncertainty in v = 4 × uncertainty in gradient.

(e) (i) C4 Between 250 and 260 given to 2 or 3 s.f.

Must be in range. Check method.

(ii) U5 Percentage uncertainty in f Absolute uncertainties in d and k must be added.

∆(d + k) = ∆d + ∆k % uncertainty in v + % uncertainty in (d + k).

[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) [U4]

Uncertainty in k = uncertainty in y-intercept

Uncertainty in v = 4 × uncertainty in gradient (e) [U5]

max f = ) min (min

gradientmax

) min 4(min

max

kdkd

v

+

=

+

min f = )max 4(max

gradient min

)max 4(max

min

kdkd

v

+

=

+

Percentage uncertainty = % uncertainty in v + % uncertainty in (d + k) Percentage uncertainty = % uncertainty in gradient + % uncertainty in (d + k)




9702 PHYSICS







1 Planning (15 marks) Defining the problem (3 marks)

P f is the independent variable or vary f. [1] P h is the dependent variable or measure h. [1] P Keep current in coil constant. [1] Methods of data collection (5 marks)

M Labelled diagram of apparatus: including working circuit with a.c. supply, coil, stand and ring. [1] M Signal generator (or variable frequency power generator). [1] M Measure h with a rule/calipers. [1] M Measure f using oscilloscope/read off signal generator. [1] M Measure h from opposite sides of the ring (and determine mean)/wait for ring to stabilise. [1] Method of analysis (2 marks)

A Plot a graph of lg h against lg f. (Allow ln – ln graphs.) [1] A k = 10y-intercept and n = gradient. [1] Safety considerations (1 mark)

S Reasoned method to prevent the coil overheating – switch off when not in use. Or reasoned method to prevent injury from hot coil – do not touch hot coil, use gloves. Do not allow “small currents”. [1]

Additional detail (4 marks)

D Relevant points might include [4] 1 Use iron/steel retort stand 2 Method to keep current constant 3 Method to determine period from oscilloscope using time-base 4 When using oscilloscope, f = 1/T 5 Use coil of many turns/large current to have measurable heights 6 Logarithmic equation e.g. lg h = n lg f + lg k 7 Relationship is valid if a straight line is produced

Do not allow vague computer methods.

[Total: 15]






(a) A1 Gradient = v/4 y-intercept = –k

y-intercept must be negative.

(b) T1 1/f / 10–3 s Column heading. Allow equivalent unit. e.g. f

–1/ 10–3 Hz–1 or 1/f (10–3 s)

T2

3.13 or 3.125

2.94 or 2.941

2.65 or 2.646

2.34 or 2.336

2.08 or 2.083

1.95 or 1.953


(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 d plotted correctly

Must be within half a small square. Each error bars should be five small squares

(ii) G2 Line of best fit If points are plotted correctly then lower end of line should pass between (2.06, 16) and (2.10, 16) and upper end of line should pass between (3.12, 24.5) and (3.16, 24.5).





(iv) C2 y-intercept – must be negative

Check substitution in y = mx + c. Allow ecf from (c)(iii). FOX does not score. Should be about –1 (–0.35 to –1.33).

U3 Uncertainty in y-intercept Check substitution in y = mx + c. Difference in worst y-intercept and y-intercept. Allow ecf from (c)(iv) but FOX does not score.




(d) C3 k = – y-intercept cm or m

v = 4 × gradient cm s–1 or m s–1

Method required with appropriate unit. Do not penalise POT. Allow ecf from (c)(iii) and (iv).

U4 Determines absolute uncertainties in k and v

Uncertainty k = uncertainty in y-intercept.

Uncertainty in v = 4 × uncertainty in gradient.

(e) (i) C4 Between 250 and 260 given to 2 or 3 s.f.

Must be in range. Check method.

(ii) U5 Percentage uncertainty in f Absolute uncertainties in d and k must be added.

∆(d + k) = ∆d + ∆k % uncertainty in v + % uncertainty in (d + k).

[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) [U4]

Uncertainty in k = uncertainty in y-intercept

Uncertainty in v = 4 × uncertainty in gradient (e) [U5]

max f = ) min (min

gradientmax

) min 4(min

max

kdkd

v

+

=

+

min f = )max 4(max

gradient min

)max 4(max

min

kdkd

v

+

=

+

Percentage uncertainty = % uncertainty in v + % uncertainty in (d + k) Percentage uncertainty = % uncertainty in gradient + % uncertainty in (d + k)




9702 PHYSICS







1 Planning (15 marks)

Defining the problem (3 marks)

P θ is the independent variable or vary θ. [1] P R is the dependent variable or measure R. [1] P Keep length of wire constant. [1] Methods of data collection (5 marks) M Labelled diagram of apparatus: wire in oil/water bath or oven or beaker with water and

source of heat. [1] M Circuit diagram to measure resistance. [1] M Use thermometer to measure the temperature of wire/water/oven. (Could be on diagram if

labelled.) [1]

M Method to determine resistance from circuit, e.g. read off ohmmeter/R = V/I [1] M Method to determine R0 e.g. use ice-water mixture. Do not allow ice (allow ice at 0 °C or melting ice). [1] Method of analysis (2 marks) Do NOT allow log-log graphs.

A R against θ R/R0 against θ θ against R [1]

A α = gradient/R0 α = gradient α = 1/ (R0 × gradient)

α = – 1/y–intercept

[1]

Safety considerations (1 mark) S Reasoned method to prevent injury from hot water/hot wire e.g. gloves (to prevent injury)

from hot water/wire; goggles to prevent splashes from hot water; do not touch hot wire/beaker. [1]

Additional detail (4 marks) D Relevant points might include [4] 1 Use insulated wire 2 Use long/thin wire to increase resistance 3 Stir liquid 4 Wait for temperature to stabilise 5 Relationship is valid if straight line, provided plotted graph is correct 6 Relationship is valid if straight line not passing through origin, provided plotted graph is

correct (any quoted expression must be correct, e.g. y-intercept = R0) 7 Use small current to minimise heating effect Do not allow vague computer methods. [Total 15]






(a) A1 Gradient = 1/2a y-intercept = t

(b) T1 d/v /s Column heading. Allow equivalent unit.

T2

1.3 or 1.30

1.6 or 1.63

2.0 or 1.98

2.3 or 2.30

2.6 or 2.63

2.9 or 2.94


U1 ±0.2 / ±0.16 to ±0.09 / ±0.1 Uncertainties in d/v.

(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 d/v plotted correctly Must be within half a small square. Ecf allowed from table.

(ii) G2 Line of best fit If points are plotted correctly then lower end of line should pass between (9.5, 1.3) and (10.5, 1.3) and upper end of line should pass between (34.0, 2.9) and (35.5, 2.9). Allow ecf from points plotted incorrectly – examiner judgement.








(iv) C2 y-intercept of best fit line Check substitution in y = mx + c. Should be about 0.6 – 0.7. FOX does not score.

U4 Uncertainty in y-intercept Check substitution in y = mx + c. Difference in worst y-intercept and y-intercept. FOX does not score. Allow ecf from (c)(iv).

(d) (i) C3 a = 1/(2 × gradient) and in the range 7.25 to 7.74 and given to 2 s.f. or 3 s.f.

Allow 7.3 to 7.7.

C4 t = y-intercept and units for a [m s–2] and t [s]

(ii) U5 Percentage uncertainties in a and t Percentage uncertainty in gradient. Percentage uncertainty in y-intercept.

[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)

(iv) [U4]

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) (ii) [U5]

Percentage uncertainty in a = 100100 ×∆

=×∆

m

m

a

a

Percentage uncertainty in t = 100100 ×∆

=×∆

c

c

t

t

Education

9702 w13 ms_all