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Cambridge International Examinations Cambridge International Advanced Subsidiary and Advanced Level

PHYSICS 9702/11

Paper 1 Multiple Choice May/June 2016

MARK SCHEME

Maximum Mark: 40

Published

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. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2016 series for most Cambridge IGCSE

®,

Cambridge International A and AS Level components and some Cambridge O Level components.

Page 2 Mark Scheme Syllabus Paper

Cambridge International AS/A Level – May/June 2016 9702 11

© Cambridge International Examinations 2016

Question Number

Key Question Number

Key

1 B 21 B

2 C 22 A

3 C 23 B

4 A 24 A

5 D 25 D

6 B 26 D

7 A 27 D

8 D 28 A

9 A 29 D

10 D 30 C

11 C 31 D

12 A 32 C

13 B 33 D

14 B 34 A

15 C 35 A

16 C 36 B

17 D 37 B

18 A 38 A

19 D 39 D

20 C 40 A





PHYSICS 9702/12


MARK SCHEME

Maximum Mark: 40

Published


®,





Question Number

Key Question Number

Key

1 C 21 D

2 C 22 A

3 A 23 B

4 B 24 C

5 D 25 C

6 A 26 D

7 A 27 D

8 A 28 C

9 C 29 B

10 C 30 C

11 D 31 C

12 D 32 A

13 A 33 B

14 D 34 B

15 C 35 D

16 B 36 B

17 A 37 D

18 C 38 C

19 B 39 B

20 D 40 D





PHYSICS 9702/13


MARK SCHEME

Maximum Mark: 40

Published


®,





Question Number

Key Question Number

Key

1 D 21 D

2 B 22 C

3 C 23 B

4 C 24 B

5 C 25 D

6 A 26 B

7 B 27 A

8 B 28 D

9 A 29 D

10 A 30 A

11 B 31 D

12 C 32 A

13 B 33 C

14 B 34 A

15 C 35 C

16 C 36 B

17 A 37 B

18 A 38 D

19 D 39 B

20 C 40 C





PHYSICS 9702/21

Paper 2 AS Level Structured Questions May/June 2016

MARK SCHEME

Maximum Mark: 60

Published


®,





1 (a) (i) (50 to 200) × 10–3 kg or (0.05 to 0.2) kg B1 [1]

(ii) (50 to 300) cm3 B1 [1]

(b) density = mass / volume or ρ = M / V C1

V = [π(0.38 × 10–3)2 × 25.0 × 10–2] / 4 (= 2.835 × 10–8 m3) C1

ρ = (0.225 × 10–3) / 2.835 × 10–8 = 7940 (kg m–3) A1

∆ρ / ρ = 2(0.01/0.38) + (0.1/25.0) + (0.001/0.225) [= 0.061] or

%ρ = 5.3% + 0.40% + 0.44% (= 6.1%) C1

∆ρ = 0.061 × 7940 = 480 (kg m–3)

density = (7.9 ± 0.5) × 103 kg m–3 or (7900 ± 500) kg m–3 A1 [5]

2 (a) (i) horizontal component (= 12 cos 50°) = 7.7 m s–1 A1 [1]

(ii) vertical component (= 12 sin 50° or 7.7 tan 50°) = 9.2 m s–1 A1 [1]

(b) v2 = u2 + 2as and v = 0 or mgh = ½mv2 or s = v2 sin2θ / 2g C1

9.22 = 2 × 9.81 × h hence h = 4.3 (4.31) m A1 [2] alternative methods using time to maximum height of 0.94 s: s = ut + ½at2 and t = 0.94 (s) (C1)

s = 9.2 × 0.94 – ½ × 9.81 × 0.942 hence s = 4.3 m (A1) or s = vt – ½at2 and t = 0.94 (s) (C1)

s = ½ × 9.81 × 0.942 hence s = 4.3 m (A1) or s = ½(u + v)t and t = 0.94 (s) (C1)

s = ½ × 9.2 × 0.94 hence s = 4.3 m (A1) (c) t (= 9.2 / 9.81) = 0.94 (0.938) s C1

horizontal distance = 0.938 × 7.7 (= 7.23 m) C1 displacement = [4.32 + 7.232]1/2 C1 = 8.4 m A1 [4]




3 (a) (i) force (= mg = 0.15 × 9.81) = 1.5 (1.47) N A1 [1] (ii) resultant force (on ball) is zero so normal contact force = weight

or the forces are in opposite directions so normal contact force = weight or normal contact force up = weight down A1 [1]

(b) (i) (resultant) force proportional/equal to rate of change of momentum B1 [1]

(ii) change in momentum = 0.15 × (6.2 + 2.5) (= 1.305 N s) C1

magnitude of force = 1.305 / 0.12 = 11 (10.9) N A1 or (average) acceleration = (6.2 + 2.5) / 0.12 (= 72.5 m s–2) (C1)

magnitude of force = 0.15 × 72.5 = 11 (10.9) N (A1) (direction of force is) upwards/up B1 [3]

(iii) there is a change/gain in momentum of the floor M1

this is equal (and opposite) to the change/loss in momentum of the ball so momentum is conserved A1 [2] or change of (total) momentum of ball and floor is zero (M1) so momentum is conserved (A1) or (total) momentum of ball and floor before is equal to the (total) momentum of ball and floor after (M1) so momentum is conserved (A1)




4 (a) the energy (stored) in a body due to its extension/compression/deformation/ change in shape/size B1 [1]

(b) (i) two values of F/x are calculated which are the same

e.g. 10.4 / 40 = 0.26 and 6.5 / 25 = 0.26 B1 or ratio of two forces and the ratio of the corresponding two extensions are calculated which are the same e.g. 5.2 / 10.4 = 0.5 and 20 / 40 = 0.5 (B1) or gradient of graph line calculated and coordinates of one point on the line used with straight line equation y = mx + c to show c = 0 (B1) (so) force is proportional to extension (and so Hooke’s law obeyed) B1 [2]

(b) (ii) 1. k = F / x or k = gradient C1

gradient or values from a single point used e.g. k = 10.4 / (40 × 10–2)

k = 26 N m–1 A1 [2]

2. work done = area under graph or ½Fx or ½(F2 + F1)(x2 – x1)

or ½kx2 or ½k(x22 – x1

2) C1

= ½ × 10.4 × 0.4 – ½ × 5.2 × 0.2 C1

or ½ × (5.2 + 10.4) × 20 × 10–2

or ½ × 26 × (0.42 − 0.22) = 1.6 J A1 [3]

(c) remove the force and the spring goes back to its original length B1 [1]

5 (a) T = 4 (ms) or 4 × 10–3 (s) C1

f = 1 / T = 1 / 0.004 = 250 Hz A1 [2]

(b) intensity ∝ (amplitude)2 and amplitude = 2.8 (2.83) (cm) B1

curve with same period and with amplitude 2.8 cm B1 curve shifted 1.0 ms to left or to right of wave X B1 [3]




(c) (i) gradient = (4.5 – 2.4) × 10–3 / (3.25 – 1.75) [= 1.4 × 10–3] B1

wavelength = 0.45 × 10–3 × 1.4 × 10–3 C1

= 6.30 × 10–7 (m) C1

= 630 nm A1 [4]

(ii) (gradient is equal to λ / a therefore) gradient of line is reduced B1

value of x will be reduced for all values of D or new line is completely below old line or intercept is less B1 [2]

6 (a) (coulomb is) ampere second B1 [1]

(b) (total) charge or Q = nAle M1

I = Q / t and l / t = v M1

I = nAle / t = nAve therefore v = I / nAe A1 [3]

(c) (i) ratio = (I / nAYe) / (I / nAZe) C1

= AZ

/ AY or 4A / A or πd2 / (πd2

/ 4) C1 = 4 A1 [3]

(ii) R = ρl / A or R = 4ρl / πd2 B1

RY = ρl / A and RZ = ρ(2l) / 4A so RY / RZ = 2 or

RY = 4ρl /πd2 and RZ = 4ρ(2l) /π4d2 or 2ρl / πd2 so RY / RZ = 2 A1 [2]

(iii) V = 12RY / (RY + RZ) or I = 12 / (RY + RZ) and V = IRY C1

V = 12 × 2/3

= 8(.0) V A1 [2]

(iv) ratio = I2RY / I2RZ or (VY

2 / RY) / (VZ

2 / RZ) or (VYI) / (VZI)

= 2 A1 [1]




7 (a) hadron: neutron/proton and

lepton: electron/(electron) neutrino B1 [1] (allow other correct particles)

(b) (i) proton: up up down or uud B1 [1] (ii) neutron: up down down or udd B1 [1]

(c) (i) neutron → proton + electron + (electron) antineutrino B1 [1] (ii) up down down (quarks) change to up up down (quarks)

or down (quark) changes to up (quark) B1 [1]





PHYSICS 9702/22


MARK SCHEME

Maximum Mark: 60

Published


®,





1 (a) acceleration = change in velocity / time (taken) or rate of change of velocity B1 [1] (b) (i) v = 0 + at or v = at C1 (a = 36 / 19 =) 1.9 (1.8947) m s–2 A1 [2] (ii) s = ½(u + v)t or s = v2

/ 2a or s = ½at2

= ½ × 36 × 19 = 362 / (2 × 1.89) = ½ × 1.89 × 192

= 340 m (342 m / 343 m / 341 m) M1 [1]

(iii) 1. (∆KE =) ½ × 95 × (36)2 C1 = 62 000 (61 560) J A1 [2]

2. (∆PE =) 95 × 9.81 × 340 sin 40° or 95 × 9.81 × 218.5 C1 = 200 000 J A1 [2]

(iv) work done (by frictional force) = ∆PE – ∆KE or work done = 200 000 – 62 000 (values from 1b(iii) 1. and 2.) C1 (frictional force = 138 000 / 340 =) 410 (406) N [420 N if full figures used] A1 [2] (v) –ma = mg sin 20° – f or ma = –mg sin 20° + f C1

–95 × 3.0 = 95 × 3.36 – f f = 600 (604) N A1 [2] 2 (a) p = F / A M1

use of m = ρV and use of V = Ah and use of F = mg M1

correct substitution to obtain p = ρgh A1 [3] (b) (i) (when h is zero the pressure is not zero due to) pressure from the air/atmosphere B1 [1]

(ii) gradient = ρg or P – 1.0 × 105 = ρgh C1

e.g. ρg = 1.0 × 105 / 0.75 (= 133333)

ρ = 133 333 / 9.81 = 14 000 (13 592) kg m–3 A1 [2]




3 (a) Young modulus = stress / strain B1 [1]

(b) (i) E = (F × l) / (A × e) or e = (F × l) / (A × E) B1

e ∝ 1 / E or

ratio eC / eS = ES / EC or (1.9 × 1011) / (1.2 × 1011) or 19 / 12 C1 (ratio =) 1.6 (1.58) A1 [3] (ii) two straight lines from (0,0) with S having the steepest gradient B1 [1] 4 (a) longitudinal: vibrations/oscillations (of the particles/wave) are parallel to the direction or in the same direction (of the propagation of energy) B1 transverse: vibrations/oscillations (of the particles/wave) are perpendicular to the direction (of the propagation of energy) B1 [2]

(b) LHS: intensity = power / area units: kg m s–2 × m × s–1 × m–2 or kg m2 s–3

× m–2 B1

RHS: units: m s–1 × kg m–3 × s–2 × m2 M1

LHS and RHS both kg s–3 A1 [3] (c) (i) change/difference in the observed/apparent frequency when the source is moving (relative to the observer) B1 [1] (ii) wavelength increases/frequency decreases/red shift B1 [1] (d) observed frequency = vfS / (v – vS) C1

550 = (340 × 510) / (340 – vS) C1 vS = 25 (24.7) m s–1 A1 [3] 5 (a) diffraction: spreading/diverging of waves/light (takes place) at (each) slit/ element/gap/aperture B1 interference: overlapping of waves (from coherent sources at each element) B1

path difference λ/phase difference of 360(°)/2π (produces the first order) B1 [3]

(b) d sinθ = nλ or sinθ = Nnλ C1

d = (2 × 486 × 10–9) / sin 29.7° (= 1.962 × 10–6) C1 number of lines = 510 (509.7) mm–1 A1 [3]




6 (a) at least six horizontal lines equally spaced and arrow to the right B1 [1] (b) charge used 2e C1

gain in KE = 15 × 1.6 × 10–19 × 103 = 2 × 1.6 × 10–19 × V (p.d.across plates) or

F (= W / d) = 15 × 1.6 × 10–19 × 103 / 16 × 10–3 C1

(hence V = 7500 V or F = 1.5 × 10–13 N) E = V / d or E = F / Q C1

E = (7500 / 16 × 10–3) or E = (1.5 × 10–13 / 3.2 × 10–19)

E = 4.7 × 105 (468 750) V m–1 A1 [4] or

KE (= ½mv2) = 15 × 103 × 1.6 × 10–19

v = [(2 × 15 × 103 × 1.6 × 10–19) / (6.68 × 10–27)]1/2 = 8.5 × 105 m s–1 (C1)

a (= v2 / 2s) = (8.5 × 105)2

/ 2 × 16 × 10–3 = 2.25 × 1013 m s–2

F (= 6.68 × 10–27 × 2.25 × 10–13) = 1.5 × 10–13 N E = F / Q (C1) Q = 2e (C1)

E = 4.7 × 105 V m–1 (A1)




7 (a) charge exists only in discrete amounts B1 [1]

(b) (i) E = I(R + r) or V = IR C1

(total resistance =) 2.7 + 0.30 + 0.25 (= 3.25 Ω) M1

I = 9.0 / (2.7 + 0.30 + 0.25) or 9.0 / 3.25 = 2.8 A A1 [3]

(ii) V = IRext C1

= 2.77 × 3.0 or 2.8 × 3.0 or

V = E – Ir (C1)

= 9.0 – 2.77 × 0.25 or 9.0 – 2.8 × 0.25 V = 8.3 (8.31) V or 8.4 V A1 [2]

(c) (i) I = nevA

v = 2.77 / (8.5 × 1029 × 1.6 × 10–19 × 2.5 × 10–6) M1

= 8.1 (8.147) × 10–6 m s–1 or 8.2 × 10–6 m s–1 A1 [2]

(ii) A reduces by a factor 4 (1/4 less) or resistance of Z goes up by 4× M1 current goes down but by less than a factor of 4 (as total resistance does not go up by a factor of 4) so drift speed goes up A1 [2] 8 (a) both electron and neutrino: lepton(s) B1 both neutron and proton: hadron(s)/baryon(s) B1 [2]

(b) (i) νβ 0

0

0

1

1

0

1

1np ++→

correct symbols for particles M1 correct numerical values (allow no values on neutrino) A1 [2]

(ii) up up down or uud → up down down or udd B1 [1] (iii) weak (nuclear) B1 [1]





PHYSICS 9702/23


MARK SCHEME

Maximum Mark: 60

Published


®,





1 (a) scalars: energy, power and time A1 vectors: momentum and weight A1 [2]

(b) (i) triangle with right angles between 120 m and 80 m, arrows in correct direction

and result displacement from start to finish arrow in correct direction and labelled R B1 [1]

(ii) 1. average speed (= 200 / 27) = 7.4 m s–1 A1 [1]

2. resultant displacement (= [1202 + 802]1/2) = 144 (m) C1 average velocity (= 144 / 27) = 5.3(3) m s–1 A1 direction (= tan–1 80 / 120) = 34° (33.7) A1 [3]

2 (a) systematic: the reading is larger or smaller than (or varying from) the true reading

by a constant amount B1

random: scatter in readings about the true reading B1 [2] (b) precision: the size of the smallest division (on the measuring instrument) or 0.01 mm for the micrometer B1 accuracy: how close (diameter) value is to the true (diameter) value B1 [2] 3 (a) (gravitational potential energy is) the energy/ability to do work of a mass that it

has or is stored due to its position/height in a gravitational field B1 kinetic energy is energy/ability to do work a object/body/mass has due to its

speed/velocity/motion/movement B1 [2] (b) (i) s = [(u + v) t] / 2 or acceleration = 9.8/9.75 (using gradient) C1

= [(7.8 + 3.9) × 0.4] / 2 or s = 3.9 × 0.4 + 1

2× 9.75 × (0.4)2 C1

s = 2.3(4) m A1 [3] (ii) a = (v – u) / t or gradient of line C1

= (7.8 – 3.9) / 0.4 = 9.8 (9.75) m s–2 (allow ± 1

2small square in readings) A1 [2]




(iii) KE = 1

2mv2 C1

change in kinetic energy = 1

2mv2 – 1

2mu2

= 1

2× 1.5 × (7.82 – 3.92) C1

= 34 (34.22) J A1 [3]

(c) work done = force × distance (moved) or Fd or Fx or mgh or mgd or mgx M1

= 1.5 × 9.8 × 2.3 = 34 (33.8) J (equals the change in KE) A1 [2] 4 (a) (resultant force = 0) (equilibrium) therefore: weight – upthrust = force from thin wire (allow tension in wire) or 5.3 (N) – upthrust = 4.8 (N) B1 [1] (b) difference in weight = upthrust or upthrust = 0.5 (N)

0.5 = ρghA or m = 0.5 / 9.81 and V = 5.0 × 13 × 10–6 (m3) C1

ρ = 0.5 / (9.81 × 5.0 × 13 × 10–6) C1

= 780 (784) kg m–3 A1 [3]

5 (a) the total momentum of a system (of colliding particles) remains constant M1

provided there is no resultant external force acting on the system/ isolated or closed system A1 [2]

(b) (i) the total kinetic energy before (the collision) is equal to the total kinetic

energy after (the collision) B1 [1]

(ii) p (= mv = 1.67 × 10–27 × 500) = 8.4 (8.35) × 10–25 N s A1 [1]

(iii) 1. mvA cos 60° + mvB cos 30° or m(vA2 + vB

2)1/2 B1

2. mvA sin 60° – mvB sin 30° B1 [2]

(iv) 8.35 × 10–25 or 500m = mvA cos 60° + mvB cos 30° and 0 = mvA sin 60° – mvB sin 30° or using a vector triangle C1 vA = 250 m s–1 A1 vB = 430 (433) m s–1 A1 [3]




6 (a) ohm is volt per ampere or volt / ampere B1 [1]

(b) (i) R = ρl / A B1

RP = 4ρ(2l) / πd2 or 8ρl / πd2 or RQ = ρl / πd2

or ratio idea e.g. length is halved hence R halved and diameter is halved hence

R is 1/4 C1

RQ (= 4ρl / π4d2) = ρl / πd2 = RP / 8

(= 12 / 8) = 1.5 Ω A1 [3]

(ii) power = I

2R or V

2 / R or VI C1

= (1.25)2 × 12 + (10)2 × 1.5 or (15)2/12 + (15)2/1.5 or 15 × 11.25 C1 = (18.75 + 150 =) 170 (168.75) W A1 [3]

(iii) IP = (15 / 12 =) 1.25 (A) and IQ = (15 / 1.5 =) 10 (A) C1

vP

/ vQ = IPnAQe / IQnAPe or (1.25 × πd

2) / (10 × πd

2/4) C1 = 0.5 A1 [3] 7 (a) (i) alter distance from vibrator to pulley alter frequency of generator (change tension in string by) changing value of the masses any two B2 [2] (ii) points on string have amplitudes varying from maximum to zero/minimum B1 [1]

(b) (i) 60° or π / 3 rad A1 [1] (ii) ratio = [3.4 / 2.2]2 C1

= 2.4 (2.39) A1 [2]




8 (a) α-particle is 2 protons and 2 neutrons; β+-particle is positive electron/positron

α-particle has charge +2e; β+-particle has +e charge

α-particle has mass 4u; β-particle has mass (1/2000)u

α-particle made up of hadrons; β+-particle a lepton any three B3 [3]

(b) νβ 0

0

0

1

1

0

1

1np ++→

all terms correct M1

all numerical values correct (ignore missing values on ν) A1 [2] (c) (i) 1. proton: up, up, down / uud B1 2. neutron: up, down, down / udd B1 [2] (ii) up quark has charge +2 / 3 (e) and down quark has charge –1 / 3 (e) total is +1(e) B1 [1]





PHYSICS 9702/31

Paper 3 Advanced Practical Skills 1 May/June 2016

MARK SCHEME

Maximum Mark: 40

Published


®,





1 (a) (ii) Value of x with consistent unit and in the range 36.0 cm to 39.0 cm. [1] (c) Value of T with unit in range 0.300 s < T < 1.00 s. [1] (d) Five sets of readings of x and time with correct trend scores 5 marks, four sets scores 4 marks etc. [5] Help from Supervisor –1. Range: [1]

xmax – xmin ≥ 30 cm. Column headings: [1] Each column heading must contain a quantity and an appropriate unit. The presentation of quantity and unit must conform to accepted scientific convention e.g. x / m or x (cm), T2

/ s2 or T2 (s2). Consistency: [1] All values of x must be given to the nearest mm. Significant figures: [1]

Every value of T2 must be given to the same number of s.f. as (or one more than) the number of s.f. in the corresponding raw values of time.

Calculation: [1]

T2 calculated correctly to the number of s.f. given by the candidate. (e) (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 occupy at least half the graph grid in both x and y directions. Scales must be labelled with the quantity that is being plotted. Scale markings should be no more than three large squares apart.

Plotting of points: [1]

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

Quality: [1]

All points in the table must be plotted on the grid (at least 5) for this mark to be awarded. All points must be no more than ±5 cm (to scale) in the x direction 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 4 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. Lines must not be kinked or thicker than half a square.

(iii) Gradient: [1]

The hypotenuse of the triangle must be greater than half the length of the drawn line.

The method of calculation must be correct. Do not allow ∆x / ∆y. Sign of gradient must match graph drawn. Both read-offs must be accurate to half a small square in both the x and y

directions. y-intercept: [1]

Either: Correct read-off from a point on the line substituted into y = mx + c or an equivalent expression. Read-offs must be accurate to half a small square in both x and y directions.

Or: Intercept read directly from the graph (accurate to half a small square).

(f) Value of P = candidate’s gradient and value of Q = candidate’s intercept. [1] Do not allow fractions. Unit for P correct (e.g. s2

m–1) and unit for Q correct (e.g. s2). [1] 2 (a) All values of w to nearest mm, with final value in range 0.200 m to 0.300 m. [1]

(b) (ii) Value(s) of θ to the nearest degree. Final value < 45° with unit. [1]

(iii) Percentage uncertainty in θ based on an absolute uncertainty in range 1° to 3°. If repeated readings have been taken, then the uncertainty can be half the range (but not zero) if the working is clearly shown. Correct method of calculation to obtain percentage uncertainty. [1] (c) (iii) Value of y in range 0.20 m to 0.50 m. [1] Evidence of repeat readings. [1] (d) (i) Correct calculation of D. [1] (ii) Correct justification for s.f. in D linked to s.f. in w and y. [1]




(e) (ii) Second value of θ. [1] Second value of y. [1]

Quality: Second value of y > first value of y (if θ2 > θ1). [1] (f) (i) Two values of k calculated correctly. [1] (ii) Sensible comment relating to the calculated values of k, testing against a criterion specified by the candidate. [1]

(g) (i) Limitations [4] (ii) Improvements [4] Do not credit

A Two readings not enough to draw a conclusion

Take more readings and plot a graph/ obtain more k values and compare

Few readings/ only one reading/ not accurate result/ “repeat readings” on its own

B Difficulty with starting position e.g. starting position of container not parallel to edge

Improved method for initial placement or release e.g. use of block with detail (aligned with side)

Change shape of container

C Small range of angles possible/cylinder slips when angle too high/cylinder moves off wrong edge on board

Workable method to increase friction e.g. sheet of paper on board/ sanding board/ use a rougher board/ roughen edge of container

“No friction” on its own/ smoother board/ use longer board

D Difficult to measure y or locate position where container moves off edge with reason e.g. moves off edge too fast/short time to observe moving off edge

Improved method for measuring y e.g. use marker or scale on board/use video and playback with scale/ paint on lid/ calibrate board

Effects of moving air/fans/ “frame by frame”

E Difficulty with set up e.g. board moves/clamp moves/board not an even height across width/board not aligned correctly

Method to improve stability of board e.g. use two clamps/ Blu-Tack to fix board to bench/ spirit level across board/ support board on long blocks

G-clamp to bench

F Difficult to release without applying force

Improved method of release e.g. card gate/block with detail of removal

Electromagnetic release





PHYSICS 9702/32


MARK SCHEME

Maximum Mark: 40

Published


®,





1 (b) (i) Value for θ in range 20° to 30° to nearest degree, with unit. [1] (iv) Value for T in range 0.50 s to 1.50 s. [1] Evidence of repeat readings. At least two measurements of nT, with n ≥ 3. [1]

(d) Six sets of values for θ and T with correct trend scores 4 marks, five sets scores

3 marks etc. No θ values over 90°. [4] Help from Supervisor –1. Range: [1]

θ values must include 30° or less and 70° or more. Column headings: [1] Each column heading must contain a quantity and an appropriate unit. The presentation of quantity and unit must conform to accepted scientific convention

e.g. θ / °, θ (°) or θ (deg) etc. θtan1/ must have no unit.

Consistency: [1] All raw values of time must be given to the nearest 0.1 s, or all to the nearest 0.01 s. Significant figures: [1]

Every value of θtan1/ must be given to 2 or 3 s.f.

Calculation: [1]

Values of θtan1/ calculated correctly to the number of s.f. given by the candidate.

(e) (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 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 must 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”). Plotted points must be accurate to half a small square. Quality: [1] All points in the table (at least 5) must be plotted for this mark to be awarded. All points must be no more than ±0.02 s (in the y (T) direction) 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 plot only if clearly indicated by the candidate. Line must not be kinked or thicker than half a small square.




(iii) Gradient: [1] Sign of gradient must match graph drawn. The hypotenuse of the triangle used must be greater than half the length of the drawn line. The method of calculation must be correct. Both read-offs must be accurate to half a small square in both x and y directions.

y-intercept: [1] Either: Correct read-off from a point on the line and substituted into y = mx + c. Read-offs must be accurate to half a small square in both x and y directions. Or: Intercept read off directly from the graph (accurate to half a small square). (f) Value of p = candidate's gradient and value of q = candidate's intercept. [1] Do not allow fractions. Correct units for p and q (both should have the unit s). [1] 2 (a) (i) h to nearest mm and in range 2.5 cm to 3.5 cm. [1] (ii) Raw values for d to nearest mm. [1] (b) Absolute uncertainty in d in range 2 mm to 5 mm. If repeated readings have been taken, then the uncertainty can be half the range (but not zero) if the working is clearly shown. Correct method of calculation to obtain percentage uncertainty. [1] (c) (i) Value for t in range 20.0 s to 90.0 s, with unit. [1] Evidence of repeat measurements of t. [1] (ii) Correct calculation of R to the s.f. used by the candidate (must be 2 or more s.f.). [1] (d) (ii) Value for x1 to nearest mm, with unit. [1] (e) Second values of h and d and t. [1] Second values of x1 and x2. [1] Quality: (x2 – x1) greater for shorter t. [1] (f) (i) Two values of k calculated correctly. [1] (ii) Sensible comment relating to the calculated values of k, testing against a criterion specified by the candidate. [1]





A Two readings are not enough to draw a conclusion

Take more readings and plot graph/ obtain more k values and compare

“Repeat readings” on its own/few readings/only one reading/take more readings and find average k

B Parallax error when measuring d

Measure on bench between two set squares/ use (vernier) calipers/ use string to find circumference then calculate d

C Bottle distorts when measuring d/ d varies along bottle/ base of bottle not flat

Collect water lost between marks and measure volume

D Difficult to judge/see/operate stopwatch when water level reaches mark

Use video with timer in view/ use frame counting/ use coloured water

Reaction time

Light gates

E Wooden strip moves continuously when water is falling on it

Use video with scale in view

F Difficult to measure height because rule not vertical/rule touches strip

Use set square on bench/ clamp rule

Only short time to measure x

G Water soaks into wooden strip/ water stays on wooden strip

Use waterproof strip Use new strip/ dry the strip

H R not constant between lines Move lines closer to top/ have lines closer together





PHYSICS 9702/33


MARK SCHEME

Maximum Mark: 40

Published


®,





1 (b) (ii) Value for y with unit in range 2.0 ≤ y ≤ 8.0 cm. [1]

(iii) Raw values of θ to the nearest degree.

Value of θ in the range 40° to 50°. [1]

(d) Six sets of readings of m, y and θ with correct trend scores 5 marks, five sets scores 4 marks etc. [5] Help from supervisor –1. Range: [1] Range of values to include m ≤ 150 g and m ≥ 400 g. 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. m sin θ / g or θ (°). Consistency: [1] All values of y must be given to the nearest mm only. Significant figures: [1]

Every value of m sin θ must be given to 2 or 3 s.f. Calculation: [1]

Values of m sin θ calculated correctly to the number of s.f. given by the candidate. (e) (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 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”). Plotted points must be accurate to half a small square. Quality: [1] All points in the table (at least 5) must be plotted on the grid for this mark to be awarded. All points must be within ±0.25 cm in the y direction 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. Lines must not be kinked or thicker than half a square.




(iii) Gradient: [1] The hypotenuse of the triangle must be greater than half of the length of the drawn line. The method of calculation must be correct. Both read-offs must be accurate to half a small square in both the x and y directions. y-intercept: [1] Either: Correct read-off from a point on the line and substituted into y = mx + c. Read-offs must be accurate to half a small square in both x and y directions. Or: Intercept read off directly from the graph (accurate to half a small square). (f) Value of P = candidate’s gradient and value of Q = candidate’s intercept. [1] Do not allow fractions. Unit for P correct (m kg–1 or cm kg–1 or mm kg–1 or m g–1 or cm g–1 or mm g–1) and consistent with value. Unit for Q correct (m or cm or mm) and consistent with value. [1] 2 (a) (ii) All raw values of d either to the nearest 0.01 or 0.001 mm with unit and in the range 0.250 mm to 0.450 mm. [1] (iii) Correct calculation of A with consistent unit and power of ten. [1] (b) (iii) Value of L with appropriate unit in range 10.0 cm ≤ L ≤ 20.0 cm. [1] (iv) Percentage uncertainty in L based on absolute uncertainty of 2 mm to 8 mm. If repeated readings have been taken, then the uncertainty can be half the range (but not zero) if the working is clearly shown. Correct method of calculation to obtain percentage uncertainty. [1] (c) (i) Correct calculation of C to the s.f. given by the candidate. [1] (ii) Correct justification for s.f. in C linked to s.f. in d and L. [1] (d) (ii) Raw values for time to the nearest 0.1 s or better. T with unit and in range 0.5 s ≤ T ≤ 2.0 s. [1] (e) (ii) Second values of d and L. [1]

Second value of T. [1]

Quality: If d1 > d2 then second value of T > first value of T. [1]




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



Take many readings and plot a graph/ obtain more k values and compare

“Repeat readings” on its own/few readings/only one reading/not enough readings for accurate value

B Difficult to judge beginning and/or end of a cycle/a complete cycle

Draw a line/mark on the mass/ (fiducial) marker at equilibrium position

C Wire not straight/kinked

Method of straightening wire e.g. use larger mass

D Difficult to measure L with reason e.g. metre rule awkward to position/parallax error

Improved method of measuring L e.g. marking L before putting into clip/ detailed method using set squares or ruler/ use a length guide (e.g. 15 cm wood)/ use string with detail/ use tape measure

Vernier calipers on its own/ set square on its own/ 30 cm ruler on its own

E Wire slips (in clip)

Better method of gripping wire e.g. wrap wire around clamp/ use two wooden blocks and wire

Any reference to attaching the mass to the wire

F Mass swings as well as rotates/ clip moves around rod/ there is a force on release

Better method of attaching clip to rod e.g. glue

G Shorter/thicker wire has too few cycles/dampens quickly/ (percentage) uncertainty greater for shorter/thicker wire

Video and timer/replay frame by frame

Repeats Longer wire





PHYSICS 9702/34


MARK SCHEME

Maximum Mark: 40

Published


®,





1 (b) (ii) 0.9VS calculated correctly and to the same number of s.f. as, or one more than, the s.f. of VS in (b)(i). [1] (c) (ii) Value for t in range 1.0 s to 9.0 s. [1]

(d) (ii) Six sets of values for VC and t with correct trend scores 5 marks, five sets scores 4 marks etc. [5]

Minor help from supervisor –1, major help from supervisor –2. Range: [1] Range of values to include VC ≤ 3.0 V and VC ≥ 8.0 V. Column headings: [1] Each column heading must contain a quantity and an appropriate unit. The presentation of quantity and unit must conform to accepted scientific convention e.g. VC / V or VC (V). Consistency: [1] All values of t must be given to the nearest 0.1 s, or all to the nearest 0.01 s. (e) (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 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 must be no more than three large squares apart. Plotting of points: [1] All observations in the table must be plotted on the grid. Diameter of plotted points must be ≤ half a small square (no “blobs”). Plotted points must be accurate to half a small square. (ii) Line of best fit: [1] Judge by balance of all points on the grid about the candidate's curve (at least 5 points). There must be an even distribution of points either side of the curve 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. (f) (ii) Tangent drawn at VC = 0.5VS. Tangent must touch curve at the candidate’s value of 0.5VS from (f)(i). [1]




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

y-intercept: [1] Either: Correct read-off from a point on the tangent is substituted into y = mx + c. Read-offs must be accurate to half a small square in both x and y directions. Or: Intercept read off directly from the graph (accurate to half a small square). (g) Value of a = candidate’s gradient and value of b = candidate’s intercept. [1] Correct units for a (e.g. V s–1) and b (s). [1] (h) Correct calculation of T. [1] Quality: T in the range 8.0 s to 14.0 s, with consistent unit. [1] 2 (a) d in the range 0.5 mm to 0.9 mm, to nearest 0.1 mm or to 0.01 mm, with unit. [1] (b) (iii) Value for x in the range 11–19 mm, with unit. [1] Evidence of repeat readings of x. [1] (c) Absolute uncertainty in x in range 2 mm to 5 mm. If repeated readings have been taken, then the uncertainty can be half the range (but not zero) if the working is clearly shown. Correct method of calculation to obtain percentage uncertainty. [1] (e) (ii) h1 recorded to nearest mm, with consistent unit. [1] (iv) Correct calculation of k to the number of s.f. given by the candidate. [1]

Value of k given to the same number of s.f. as, or one more than, the number of s.f. in (h1 – h2) or m, whichever is lower. [1]

(f) Second values of x and n. [1] Second values of h1 and h2. [1] Quality: Value of (h1 – h2) for smaller x less than the value of (h1 – h2) for larger x. [1]




(g) (i) Two values of c calculated correctly. [1] (ii) Valid comment consistent with the calculated values of c, testing against a criterion specified by the candidate. [1]

(h) (i) Limitations [4] (ii) Improvements [4] Do not credit

A Two readings are not enough to draw a conclusion

Take more readings and plot graph/ take more readings and compare c values

Repeat readings/ few readings/ only one reading /not enough readings for accurate value

B d is small/ large (percentage) uncertainty in d

Use a micrometer (to measure diameter)

Digital calipers

C n not an integer

Estimate n to the nearest ¼ turn

D Diameter not constant/ coils vary in diameter/ coils not equally spaced/ coils not circular

Method of making equally-spaced coils e.g. make small marks/grooves on wooden rod Use motor to wind spring by rotating rod

Spring not straight Use ‘factory’ spring

E Difficult to measure diameter (x) with reason e.g. calipers distort coils/end of coil gets in the way of ruler

Use thin ruler/graph paper placed between loops of spring

F h1 – h2 small, so uncertainty large

Use larger mass/larger range of masses Travelling microscope with reference to h1 – h2

Use wires of longer length to increase h1 – h2





PHYSICS 9702/35


MARK SCHEME

Maximum Mark: 40

Published


®,





1 (a) (ii) Value for L to the nearest mm with unit and in range 38.0 cm ≤ L ≤ 42.0 cm. [1]

(b) (iv) Value of I with unit in the range 25 mA to 100 mA. [1]

(c) Six sets of readings of x and I with correct trend scores 4 marks, five sets scores 3 marks etc. [4] Minor help from Supervisor –1, major help –2. Range of x: [1]

∆x ≥ 30.0 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. 1 / I / A–1 or 1 / I (A–1) or

1/I / 1/A. Do not allow 1 / I (A). Consistency: [1]

All values of I given to 0.1 mA. Significant figures: [1]

Every value of 1 / I must be given to the same number of s.f. as (or one more than)

the number of s.f. in the corresponding value of I. Calculation: [1]

Values of 1 / I calculated correctly to the number of s.f. given by the candidate. (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 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”). Plotted points 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 2.0 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. Lines must not be kinked or thicker than half a small square.




(iii) Gradient: [1] The hypotenuse of the triangle used must be greater than half of the length of the drawn line. The method of calculation must be correct.

Both read-offs must be accurate to half a small square in both the x and y directions. y-intercept: [1] Either: Correct read-off from a point on the line and substituted into y = mx + c. Read-offs must be accurate to half a small square in both x and y directions. Or: Intercept read off directly from the graph (accurate to half a small square). (e) P = –value of candidate’s gradient Q = value of candidate’s intercept. [1] Do not allow fractions. Unit for P correct (e.g. A–1

m–1, A–1 cm–1, A–1

mm–1, mA–1 m–1, mA–1

cm–1 or mA–1 mm–1). Unit for Q correct (e.g. A–1 or mA–1) and consistent with value. [1]

(f) Value of R in the range 5 Ω to 20 Ω. [1] 2 (a) (ii) Value of C in the range 35.0 cm to 40.0 cm with unit. [1] (iii) Value of d to the nearest mm with unit in range 4.0 cm to 6.0 cm. [1] (iv) Correct calculation of (C – d). [1]

(b) (ii) Value of θ to the nearest degree with unit, and θ < 90°. [1]

(iii) Absolute uncertainty in θ in range 2° to 5°. If repeated readings have been taken, then the uncertainty can be half the range (but not zero) if the working is clearly shown. Correct method of calculation to obtain percentage uncertainty. [1]

(c) Correct calculation of (tanθ – 1). Do not allow a unit. [1] Answer given to 2 s.f. or 3 s.f. [1] (d) Second value of d. [1]

Second value of θ. [1]

Quality: Second value of θ > first value of θ. [1] (e) (i) Two values of k calculated correctly. [1] (ii) Sensible comment relating to the calculated values of k, testing against a criterion specified by the candidate. [1]




(f) (i) Limitations [4] (ii) Improvements [4] Do not credit


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

Repeat readings/ few readings/ only one reading/ not enough readings for accurate value

B Difficult to measure angle with reason e.g. lack of vertical reference line/parallax/nail in the way/apparatus moves when protractor in place

Method of providing vertical reference e.g. use a plumbline/ drill hole at origin of protractor and mount on nail/ method of fixing protractor/ use a grid with angles marked/ larger protractor

C Range of θ is too small More holes further apart

D Difficult to measure C whilst balancing strip

Mark position of pivot/ add scale to wooden strip

E Difficulty in mechanical set up e.g. alignment of strip and string/horizontal string (parallel to table)

Method of improvement e.g. spirit level linked to string/axle of pulley/ add weights to stands

F Friction at nail (or axle of pulley)

Lubricate the nail/pulley axle





PHYSICS 9702/41

Paper 4 A Level Structured Questions May/June 2016

MARK SCHEME

Maximum Mark: 100

Published


®,





1 (a) (gravitational) potential at infinity defined as/is zero B1

(gravitational) force attractive so work got out/done as object moves from infinity (so potential is negative) B1 [2]

(b) (i) ∆E = m∆φ

= 180 × (14 – 10) × 108 C1

= 7.2 × 1010 J A1 increase B1 [3]

(ii) energy required = 180 × (10 – 4.4) × 108

or

energy per unit mass = (10 – 4.4) × 108 C1

½ × 180 × v2 = 180 × (10 – 4.4) × 108

or

½ × v2 = (10 – 4.4) × 108 C1

v = 3.3 × 104 m s–1 A1 [3] 2 (a) e.g. time of collisions negligible compared to time between collisions no intermolecular forces (except during collisions) random motion (of molecules) large numbers of molecules (total) volume of molecules negligible compared to volume of containing vessel or average/mean separation large compared with size of molecules any two B2 [2]

2 (b) (i) mass = 4.0 / (6.02 × 10

23) = 6.6 × 10–24 g or

mass = 4.0 × 1.66 × 10–27 × 103 = 6.6 × 10–24 g B1 [1]

(ii) 2

3kT =

2

1m <c

2> C1

2

3 × 1.38 × 10–23 × 300 =

2

1 × 6.6 × 10–27 × <c

2>

<c

2> = 1.88 × 106 (m2 s–2) C1

r.m.s. speed = 1.4 × 103 m s–1 A1 [3]




3 (a) acceleration/force proportional to displacement (from fixed point) M1 acceleration/force and displacement in opposite directions A1 [2] (b) maximum displacements/accelerations are different B1 graph is curved/not a straight line B1 [2]

(c) (i) ω = 2π / T and T = 0.8 s C1

ω = 7.9 rad s–1 A1 [2]

(ii) a = (–)ω2 x

= 7.852 × 1.5 × 10–2 C1 = 0.93 m s–2 or 0.94 m s–2 A1 [2]

(iii) ∆E = ½ mω2 (x02 – x2) C1

= ½ × 120 × 10–3 × 7.852 × (1.5 × 10–2)2 – (0.9 × 10–2)2 C1

= 5.3 × 10–4 J A1 [3]

4 (a) (i) product of speed and density M1 reference to speed in medium (and density of medium) A1 [2]

(ii) α: ratio of reflected intensity and/to incident intensity B1 Z1 and Z2: (specific) acoustic impedances of media (on each side of boundary) B1 [2]

(b) in muscle: IM = I0 e–µx

= I0 exp(–23 × 3.4 × 10–2) C1

IM

/ I0 = 0.457 C1

at boundary: α = (6.3 – 1.7)2 / (6.3 + 1.7)2

= 0.33 C1

IT

/IM = [(1 – α) =] 0.67 C1

IT

/ I0 = 0.457 × 0.67 = 0.31 A1 [5]




5 (a) (i) 1011 A1 [1] (ii)

0 0.25 0.50 0.75 1.00 1.25 1.50

1011 0110 1000 1110 0101 0011 0001

All 6 correct, 2 marks. 5 correct, 1 mark. A2 [2] (b) sketch: 6 horizontal steps of width 0.25 ms shown M1 steps at correct heights and all steps shown A1 steps shown in correct time intervals A1 [3] (c) increase sampling frequency/rate M1 so that step width/depth is reduced A1 increase number of bits (in each number) M1 so that step height is reduced A1 [4] 6 (a) sketch: from x = 0 to x = R, potential is constant at VS B1 smooth curve through (R, VS) and (2R, 0.5VS) B1 smooth curve continues to (3R, 0.33VS) B1 [3] (b) sketch: from x = 0 to x = R, field strength is zero B1 smooth curve through (R, E) and (2R, 0.25E) B1 smooth curve continues to (3R, 0.11E) B1 [3] 7 (a) line has non-zero intercept/line does not pass through origin B1

charge is/should be proportional to potential (difference) or

charge is/should be zero when p.d. is zero (therefore there is a systematic error) B1 [2]




(b) reasonable attempt at line of best fit B1

use of gradient of line of best fit clear M1

C = 2800 µF (allow ± 200 µF) A1 [3] (c) energy = ½ CV

2 or energy = ½ QV and C = Q / V C1

∆ energy = ½ × 2800 × 10–6 × (9.02 – 6.02) C1

= 6.3 × 10–2 J A1 [3] 8 (a) op-amp has infinite/(very) large gain B1

op-amp saturates if V

+ ≠ V

– M1

V+ is at earth potential so P (or V–) must be at earth A1 [3] (b) input resistance to op-amp is very large

or current in R2 = current in R1 B1

VIN (– 0) = IR2 and (0) – VOUT = IR1 M1

VOUT / VIN = –R1 / R2 A1 [3] (c) relay coil connected between VOUT and earth M1

correct diode symbol connected between VOUT and coil or between coil and earth M1 correct polarity for diode (‘clockwise’) A1 [3] 9 (a) 0.10 mm B1 [1]

(b) VH = (0.13 × 3.8) / (6.0 × 1028 × 0.10 × 10–3 × 1.60 × 10–19) C1

= 5.1 × 10–7 V A1 [2] 10 (a) (non-uniform) magnetic flux in core is changing M1

induces (different) e.m.f. in (different parts of) the core A1

(eddy) currents form in the core M1

which give rise to heating A1 [4]




(b) as magnet falls, tube cuts magnetic flux M1

e.m.f./(eddy) currents induced in metal/aluminium (tube) A1

(eddy) current heating of tube M1

with energy taken from falling magnet A1

or (eddy) currents produce magnetic field (M1)

that opposes motion of magnet (A1)

so magnet B has acceleration < g or magnet B has smaller acceleration/reaches terminal speed A1 [5]

11 (a) period = 15 ms C1

frequency (= 1 / T) = 67 Hz A1 [2] (b) zero A1 [1]

(c) Ir.m.s. = I0 / √2 C1

= 0.53 A A1 [2]

(d) energy = Ir.m.s.2 × R × t or ½ I0

2 × R × t or

power = Ir.m.s.2 × R and energy = power × t C1

energy = 0.532 × 450 × 30 × 10–3

= 3.8 J A1 [2]

12 (a) (in a solid electrons in) neighbouring atoms are close together (and influence/interact with each other) M1

this changes their electron energy levels M1

(many atoms in lattice) cause a spread of energy levels into a band A1 [3]




(b) photons of light give energy to electrons in valence band B1

electrons move into the conduction band M1

leaving holes in the valence band A1

these electrons and holes are charge carriers B1 increased number/increased current, hence reduced resistance B1 [5] 13 (a) e.g. background count (rate)/radiation multiple possible counts from each decay radiation emitted in all directions dead-time of counter (daughter) product unstable/also emits radiation self-absorption of radiation in sample or absorption in air/detector window three sensible suggestions, 1 each B3 [3]

(b) A = A0 exp(– ln 2 × t / T½)

1.21 × 102 = 3.62 × 104 exp(– ln2 × 42.0 / T½) or

1.21 × 102 = 3.62 × 104 exp(–λ × 42.0) C1 T½ = 5.1 minutes (306 s) A1 [2] (c) discrete energy levels (in nuclei) B1 [1]





PHYSICS 9702/42


MARK SCHEME

Maximum Mark: 100

Published


®,





1 (a) (i) gravitational force provides/is the centripetal force B1

same gravitational force (by Newton III) B1 [2]

(ii) ω = 2π / T

= 2π / (4.0 × 365 × 24 × 3600) C1

= 5.0 (4.98) × 10–8 rad s–1 A1 [2]

(b) (i) (centripetal force =) MAdω2 = MB(2.8 × 108 –d)ω2 or MAdA = MBdB C1

MA / MB = 3.0 = (2.8 × 108 – d) / d C1

d = 7.0 × 107 km A1 [3]

(ii) GMAMB / (2.8 × 1011)2 = MAdω2 B1

MB = (2.8 × 1011)2 × dω2 / G

= (2.8 × 1011)2 × (7.0 × 1010) × (4.98 × 10–8)2 / (6.67 × 10–11) C1

= 2.0 × 1029 kg A1 [3] 2 (a) (i) number of atoms/nuclei in 12 g of carbon-12 B1 [1] (ii) amount of substance M1

containing NA (or 6.02 × 1023) particles/molecules/atoms or which contains the same number of particles/atoms/molecules as there are atoms in 12 g of carbon-12 A1 [2]

(b) pV = nRT

2.0 × 107 × 1.8 × 104 × 10–6 = n × 8.31 × 290, so n = 149 mol or 150 mol A1 [1]

(c) (i) V and T constant and so pressure reduced by 5.0%

pressure = 0.95 × 2.0 × 107 C1

or calculation of new n (= 142.5 mol) and correct substitution into pV = nRT (C1)

pressure = 1.9 × 107 Pa A1 [2]




(ii) loss is 5 / 100 × 150 mol = 7.5 mol or

∆N = 4.52 × 1024 C1

t = (7.5 × 6.02 × 1023) / 1.5 × 1019 or

t = 4.52 × 1024 / 1.5 × 1019 C1

= 3.0 × 105 s A1 [3]

3 (a) no net energy transfer between the bodies or bodies are at the same temperature B1 [1]

(b) (i) thermocouple, platinum/metal resistance thermometer, pyrometer B1 [1] (ii) thermistor, thermocouple B1 [1] (c) (i) change = 11.5 K B1 [1] (ii) final temperature = 311.2 K B1 [1]

4 (a) (i) T = 0.60 s and ω = 2π / T C1

ω = 10 (10.47) rad s–1 A1 [2]

(ii) energy = ½mω2x02 or ½mv2 and v = ωx0 C1

= ½ × 120 × 10–3 × (10.5)2 × (2.0 × 10–2)2

= 2.6 × 10–3 J A1 [2]

(b) sketch: smooth curve in correct directions B1

peak at f M1

amplitude never zero and line extends from 0.7f to 1.3f A1 [3] (c) sketch: peaked line always below a peaked line A M1

peak not as sharp and at (or slightly less than) frequency of peak in line A A1 [2]




5 (a) amplitude of the carrier wave varies M1

in synchrony with displacement of the information/audio signal A1 [2] (b) (i) 10 kHz A1 [1] (ii) 5 kHz A1 [1]

(c) (i) 24 = 10 lg (PMIN / 5.0 × 10–13) C1

PMIN = 1.3 (1.26) × 10–10 W A1 [2]

(ii) 45 × 2 = 10 lg (500 × 10–3 / P)

P = 5.0 × 10–10 (W) M1 P > PMIN so yes A1

or maximum attenuation calculated to be 96 (dB) (M1)

96 dB > 2 × 45 dB so yes (A1)

or maximum length of wire calculated to be 48 (km) (M1) actual length 45 km < 48 km so yes (A1)

or maximum attenuation per unit length calculated to be 2.2 dB km–1 (M1) 2.2 dB km–1 > 2.0 dB km–1 so yes (A1) [2]

6 (a) lines perpendicular to surface

or lines are radial M1

lines appear to come from centre A1 [2]

(b) (i) FE = (1.6 × 10–19)2 / 4πε0x

2 C1

FG = G × (1.67 × 10–27)2 / x2 C1

FE / FG = (1.6 × 10–19)2 × (8.99 × 109) / [(1.67×10–27)2 × (6.67×10–11)]

= 1.2 (1.24) × 1036 A1 [3] (ii) FE ≫ FG B1 [1]




7 (a) e.g. storing energy blocking d.c. in oscillator circuits in tuning circuits in timing circuits any two B2 [2]

(b) (i) 1 / 6 + 1 / C + 1 / C = 1 / 4 C1

C = 24 µF A1 [2] (ii) Q = CV

= 4.0 × 10–6 × 12 C1

= 48 µC A1 [2]

(iii) 1. 48 µC A1

2. 24 µC A1 [2] 8 (a) (i) gain = voltage output / voltage input B1 [1] (ii) changes in VOUT M1

occur immediately when VIN changes A1

or changes in VIN (M1)

result in immediate changes to VOUT (A1) [2]

(b) 12 = 1 + R / (1.5 × 103) C1

R = 16.5 kΩ A1 [2]

(c) straight line from (0,0) to (0.75t1, 9.0 V) B1

horizontal line from endpoint of straight line to t1 B1

+9 V to –9 V (or v.v.) at t1 B1

correct line to t2 B1 [4]




9 (a) (i) number density of charge carriers/free electrons or number per unit volume of charge carriers/free electrons B1 [1]

(ii) PX or QY or RZ B1 [1] (b) (i) VH is inversely proportional to n B1

for semiconductors, n is (much) smaller than for metals B1 [2] (ii) magnetic field would deflect holes and electrons in same direction B1

(because) electrons are (–)ve, holes are (+)ve M1 so VH has opposite polarity/opposite sign A1 [3]

10 (a) iron rod changes flux (density)/field B1

change of flux in coil Q causes induced e.m.f. B1 [2] (b) constant reading (either polarity) from time zero to near t1 B1

spike in one direction near t1 clearly showing a larger voltage M1

of opposite polarity A1

zero reading from near t1 to t2 B1 [4] 11 (a) point P shown at ‘lower end’ of load B1 [1] (b) Vr.m.s. = 6.0 / √2 = 4.24 V C1

Ir.m.s. = 4.24 / (2.4 × 103)

= 1.8 × 10–3 A A1 [2]

(c) (i) capacitor in parallel with load B1 [1] (ii) line from peak to curve at 3.0 V for either half- or full-wave rectified M1

correct curvature on line (gradient becoming more shallow) A1 line drawn as for full-wave rectified A1 [3]




12 (a) (i) (X–ray) photon produced when electron/charged particle is stopped/accelerated (suddenly) B1

range of accelerations (in target) M1 hence distribution of wavelengths A1 [3]

(ii) electron gives all its energy to one photon B1

electron stopped in single collision B1 [2] (iii) de-excitation of (orbital) electrons in target/anode/metal B1 [1] (b) (i) aluminium sheet/filter/foil (placed in beam from tube) B1 [1] (ii) (long wavelength X-rays) do not pass through the body B1 [1] 13 (a) (photons of) electromagnetic radiation M1

emitted from nuclei A1 [2] (b) line of best fit drawn B1

recognises µ as given by the gradient of best-fit line or

ln C = ln C0 – µx B1

µ = 0.061 mm–1 (within ±0.004 mm–1, 1 mark; within ±0.002 mm–1, 2 marks) A2 [4] (c) aluminium is less absorbing (than lead)

or gradient of graph would be less M1

so µ is smaller A1 [2]





PHYSICS 9702/43


MARK SCHEME

Maximum Mark: 100

Published


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1 (a) (gravitational) potential at infinity defined as/is zero B1

(gravitational) force attractive so work got out/done as object moves from infinity (so potential is negative) B1 [2]

(b) (i) ∆E = m∆φ

= 180 × (14 – 10) × 108 C1

= 7.2 × 1010 J A1 increase B1 [3]

(ii) energy required = 180 × (10 – 4.4) × 108

or

energy per unit mass = (10 – 4.4) × 108 C1

½ × 180 × v2 = 180 × (10 – 4.4) × 108

or

½ × v2 = (10 – 4.4) × 108 C1

v = 3.3 × 104 m s–1 A1 [3] 2 (a) e.g. time of collisions negligible compared to time between collisions no intermolecular forces (except during collisions) random motion (of molecules) large numbers of molecules (total) volume of molecules negligible compared to volume of containing vessel or average/mean separation large compared with size of molecules any two B2 [2]

2 (b) (i) mass = 4.0 / (6.02 × 10

23) = 6.6 × 10–24 g or

mass = 4.0 × 1.66 × 10–27 × 103 = 6.6 × 10–24 g B1 [1]

(ii) 2

3kT =

2

1m <c

2> C1

2

3 × 1.38 × 10–23 × 300 =

2

1 × 6.6 × 10–27 × <c

2>

<c

2> = 1.88 × 106 (m2 s–2) C1

r.m.s. speed = 1.4 × 103 m s–1 A1 [3]




3 (a) acceleration/force proportional to displacement (from fixed point) M1 acceleration/force and displacement in opposite directions A1 [2] (b) maximum displacements/accelerations are different B1 graph is curved/not a straight line B1 [2]

(c) (i) ω = 2π / T and T = 0.8 s C1

ω = 7.9 rad s–1 A1 [2]

(ii) a = (–)ω2 x

= 7.852 × 1.5 × 10–2 C1 = 0.93 m s–2 or 0.94 m s–2 A1 [2]

(iii) ∆E = ½ mω2 (x02 – x2) C1

= ½ × 120 × 10–3 × 7.852 × (1.5 × 10–2)2 – (0.9 × 10–2)2 C1

= 5.3 × 10–4 J A1 [3]

4 (a) (i) product of speed and density M1 reference to speed in medium (and density of medium) A1 [2]

(ii) α: ratio of reflected intensity and/to incident intensity B1 Z1 and Z2: (specific) acoustic impedances of media (on each side of boundary) B1 [2]

(b) in muscle: IM = I0 e–µx

= I0 exp(–23 × 3.4 × 10–2) C1

IM

/ I0 = 0.457 C1

at boundary: α = (6.3 – 1.7)2 / (6.3 + 1.7)2

= 0.33 C1

IT

/IM = [(1 – α) =] 0.67 C1

IT

/ I0 = 0.457 × 0.67 = 0.31 A1 [5]




5 (a) (i) 1011 A1 [1] (ii)

0 0.25 0.50 0.75 1.00 1.25 1.50

1011 0110 1000 1110 0101 0011 0001

All 6 correct, 2 marks. 5 correct, 1 mark. A2 [2] (b) sketch: 6 horizontal steps of width 0.25 ms shown M1 steps at correct heights and all steps shown A1 steps shown in correct time intervals A1 [3] (c) increase sampling frequency/rate M1 so that step width/depth is reduced A1 increase number of bits (in each number) M1 so that step height is reduced A1 [4] 6 (a) sketch: from x = 0 to x = R, potential is constant at VS B1 smooth curve through (R, VS) and (2R, 0.5VS) B1 smooth curve continues to (3R, 0.33VS) B1 [3] (b) sketch: from x = 0 to x = R, field strength is zero B1 smooth curve through (R, E) and (2R, 0.25E) B1 smooth curve continues to (3R, 0.11E) B1 [3] 7 (a) line has non-zero intercept/line does not pass through origin B1

charge is/should be proportional to potential (difference) or

charge is/should be zero when p.d. is zero (therefore there is a systematic error) B1 [2]




(b) reasonable attempt at line of best fit B1

use of gradient of line of best fit clear M1

C = 2800 µF (allow ± 200 µF) A1 [3] (c) energy = ½ CV

2 or energy = ½ QV and C = Q / V C1

∆ energy = ½ × 2800 × 10–6 × (9.02 – 6.02) C1

= 6.3 × 10–2 J A1 [3] 8 (a) op-amp has infinite/(very) large gain B1

op-amp saturates if V

+ ≠ V

– M1

V+ is at earth potential so P (or V–) must be at earth A1 [3] (b) input resistance to op-amp is very large

or current in R2 = current in R1 B1

VIN (– 0) = IR2 and (0) – VOUT = IR1 M1

VOUT / VIN = –R1 / R2 A1 [3] (c) relay coil connected between VOUT and earth M1

correct diode symbol connected between VOUT and coil or between coil and earth M1 correct polarity for diode (‘clockwise’) A1 [3] 9 (a) 0.10 mm B1 [1]

(b) VH = (0.13 × 3.8) / (6.0 × 1028 × 0.10 × 10–3 × 1.60 × 10–19) C1

= 5.1 × 10–7 V A1 [2] 10 (a) (non-uniform) magnetic flux in core is changing M1

induces (different) e.m.f. in (different parts of) the core A1

(eddy) currents form in the core M1

which give rise to heating A1 [4]




(b) as magnet falls, tube cuts magnetic flux M1

e.m.f./(eddy) currents induced in metal/aluminium (tube) A1

(eddy) current heating of tube M1

with energy taken from falling magnet A1

or (eddy) currents produce magnetic field (M1)

that opposes motion of magnet (A1)

so magnet B has acceleration < g or magnet B has smaller acceleration/reaches terminal speed A1 [5]

11 (a) period = 15 ms C1

frequency (= 1 / T) = 67 Hz A1 [2] (b) zero A1 [1]

(c) Ir.m.s. = I0 / √2 C1

= 0.53 A A1 [2]

(d) energy = Ir.m.s.2 × R × t or ½ I0

2 × R × t or

power = Ir.m.s.2 × R and energy = power × t C1

energy = 0.532 × 450 × 30 × 10–3

= 3.8 J A1 [2]

12 (a) (in a solid electrons in) neighbouring atoms are close together (and influence/interact with each other) M1

this changes their electron energy levels M1

(many atoms in lattice) cause a spread of energy levels into a band A1 [3]




(b) photons of light give energy to electrons in valence band B1

electrons move into the conduction band M1

leaving holes in the valence band A1

these electrons and holes are charge carriers B1 increased number/increased current, hence reduced resistance B1 [5] 13 (a) e.g. background count (rate)/radiation multiple possible counts from each decay radiation emitted in all directions dead-time of counter (daughter) product unstable/also emits radiation self-absorption of radiation in sample or absorption in air/detector window three sensible suggestions, 1 each B3 [3]

(b) A = A0 exp(– ln 2 × t / T½)

1.21 × 102 = 3.62 × 104 exp(– ln2 × 42.0 / T½) or

1.21 × 102 = 3.62 × 104 exp(–λ × 42.0) C1 T½ = 5.1 minutes (306 s) A1 [2] (c) discrete energy levels (in nuclei) B1 [1]





PHYSICS 9702/51

Paper 5 Planning, Analysis and Evaluation May/June 2016

MARK SCHEME

Maximum Mark: 30

Published


®,





Question 1 Planning (15 marks) Defining the problem (2 marks)

P λ is the independent variable, or vary λ. [1] P V is the dependent variable, or measure V. [1] Methods of data collection (4 marks) M Circuit diagram showing d.c. power supply in series with diode (correct symbol needed) and

method to measure potential difference across diode. Circuit must be correct. [1]

M Instrument to change p.d. across LED e.g. variable power supply/potential divider/variable resistor. [1]

M Record wavelength of light of LED from data sheet or use Young’s slits/diffraction grating. [1]

M (Slowly) increase potential difference across LED until LED (just) emits light (or reverse procedure). [1]

Method of analysis (3 marks)

A Plot a graph of lg V against lg λ (allow natural logs). Allow lg λ against lg V. [1] A n = gradient [1] A k = 10

y-intercept [1] Additional detail (6 marks)

Relevant points might include: [6]

1 Use of a protective resistor (can be shown on the diagram). 2 Polarity of LED correct in circuit diagram. 3 Instrument to determine when LED just lights e.g. light meter/detector, LDR. 4 Method to use light detector/LDR to determine point at which LED emits light.

5 Expression that gives λ (symbols need to defined) from experimental determination of wavelength of light, e.g. Young’s slits/diffraction grating.

6 Perform experiment in a dark room/LED in tube. 7 Relationship is valid if graph is a straight line. 8 λ= +lg lg lgV n k

9 Repeat V and average for the same λ or LED. Do not allow vague computer methods.




Question 2 Analysis, conclusions and evaluation (15 marks)

Mark Expected Answer Additional Guidance

(a) A1

π

4LF

E

(b) T1 2

1

d / 106 m–2

T2

13 or 12.8

9.8 or 9.77

6.9 or 6.93

4.7 or 4.73

3.2 or 3.19

1.9 or 1.93

All values to 2 s.f. or 3 s.f. Allow a mixture of significant figures. Must be values in table.

U1 From ± 2 to ± 0.1 Allow more than one significant figure.

(c) (i) G1 Six points plotted correctly Must be within half a small square. Do not allow “blobs”. ECF allowed from table.

U2 Error bars in

2

1

d plotted correctly

All error bars to be plotted. Must be accurate to less than half a small square.

(ii) G2 Line of best fit If points are plotted correctly then lower end of line should pass between (3.2, 3.0) and (3.6, 3.0) and upper end of line should pass between (11.2, 10.0) and (11.6, 10.0).

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

Line should be clearly labelled or dashed. Examiner judgement on worst acceptable line. Lines must cross. Mark scored only if error bars are plotted.

(iii) C1 Gradient of line of best fit 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. (Should be about 9 × 10–10.)

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

(d) (i) C2 =

π×

4 60.479

gradient gradient

LF

Do not penalise POT.

(Should be about 7 × 1010.)

C3 N m–2 or Pa Allow in base units: kg m–1 s–2.

(ii) U4 Percentage uncertainty in E Must be larger than 3%.





(e) C4 e in the range 15.5 × 10–3 to

18.0 × 10–3 and given to 2 or 3 s.f.

Allow mm.

U5 Absolute uncertainty in e Note =

2

gradiente

d is possible.

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) (ii) [U4]

∆ ∆

= + + × = × +

gradient 0.01 0.5 gradientpercentage uncertainty 100 100 3.03%

gradient 2.50 19.0 gradient

× × × ×

= = =

π× π×

4 max max 4 2.51 1 9.5 62.319max

min gradient min gradient min gradient

L FE

4 min min 4 2.49 18.5 58.652

minmax gradient max gradient max gradient

L FE

× × × ×

= = =

π× π×

(e) [U5]

= + + × × + = +

0.5 0.01 0.02percentage uncertainty 2 100 % 20.4% %

19.0 2.50 0.23E E

∆

= + × ×

gradient 0.02percentage uncertainty 2 100

gradient 0.23

=2

min

max gradientmaxe

d

× ×

=

π× ×

max max

2

min min

4 max

L Fe

E d

=2

max

min gradientmine

d

× ×

=

π× ×

min min

2

max max

4 min

L Fe

E d





PHYSICS 9702/52


MARK SCHEME

Maximum Mark: 30

Published


®,





Question 1 Planning (15 marks)

Defining the problem (2 marks)

P θ is the independent variable and a is the dependent variable, or vary θ and measure a. [1]

P Keep F constant. [1] Methods of data collection (4 marks) M Diagram showing inclined plane with labelled support

(not if a ruler used as the inclined plane or as vertical support). [1] M Method to measure angle e.g. use a protractor to measure θ or use a ruler to

measure marked distances from which sin θ or θ may be determined. (Allow a labelled protractor in the correct position.) [1]

M Method to measure a time or velocity to determine a, e.g. measure the time using a

stopwatch, light gate(s) connected to a timer, motion sensor connected to a time display. [1]

M Use a balance to measure the mass of the trolley. [1] Method of analysis (3 marks)

A

Plot a graph of

a against sin θ.

or Plot a graph of

ma against sin θ.

or Plot a graph of

ma against mg sin θ.

[1]

A Relationship is valid if the graph is a straight line and does not pass through the origin [1]

A k = F – m × (y-intercept) or k = F – (y-intercept) or k = F – (y-intercept) [1]

Do not allow lg-lg graphs. Additional detail (6 marks)

Relevant points might include: [6] 1 Keep mass of trolley constant/use same trolley. 2 Correct trigonometry relationship to determine sin θ or θ using marked lengths. 3 Use ruler to measure appropriate distance to determine a, e.g. length of slope, length of

card for light gate method, position of motion sensor. 4 Equation to determine a from measurements taken appropriately with a as the subject.

5 Measurement of F for a valid method e.g. take reading from newton-meter or from stretched

elastic/spring from extension (allow falling weight e.g. F = mg). 6 Use a constant extension to produce a constant force when using stretched spring/elastic.




7 Method to ensure the inclined plane is the same height each side of the plane or spirit level

across plane or ensure force F (or string) is parallel to the plane. 8 Safety precaution linked to falling mass/trolley or spring/elastic breaking (not string).

9 Rearrangement of relationship into y = mx + c e.g. ma = –mg sin θ + (F – k) or

m

kFga

−+−= θsin or correct y-intercept (subject must be y-axis).

10 Repeat experiment for each angle θ to find average for a. Do not allow vague computer methods.






(a)

A1 ρ

π

4 L

(b) T1 2

1

d / 106

m–2

T2

1.2 or 1.21

3.2 or 3.19

4.7 or 4.73

6.9 or 6.93

9.8 or 9.77

14 or 13.7


U1 From ± 0.03 to ± 1 Allow more than one significant figure. Allow zero for first uncertainty and up to 1.2 for largest uncertainty.


U2 Error bars in2

1

d plotted correctly

All error bars to be plotted. Length of bar must be accurate to less than half a small square and symmetrical.

(ii) G2 Line of best fit Lower end of line must pass between (2.6, 4.0) and (3.0, 4.0) and upper end of line must pass between (12.4, 18.0) and (13.0, 18.0).


Line should be clearly labelled or dashed. Examiner judgement on worst acceptable line. Must be steepest/shallowest line. Mark scored only if error bars are plotted.


POT. (Should be about 1.4 –1.5 × 10–6.)

U3 Absolute uncertainty in gradient Method of determining absolute uncertainty: difference in worst gradient and gradient.





(d) (i) C2 π×

= ×

gradient0.7854 gradient

4L

Must use gradient value. Do not penalise

POT (Should be about 1 × 10–6.)

C3 Ω m Correct unit and correct power of ten.

(ii) U4 Percentage uncertainty in ρ Percentage uncertainty in gradient + 1%.

(e) C4 R in the range 25.5 to 28.4 and given to 2 or 3 s.f.

Allow 26 or 27 or 28. Allow ECF for POT error in (d)(i) e.g.

2.7 × 107.

U5 Absolute uncertainty in R Percentage uncertainty must be greater than 8.6%.





uncertainty = gradient of line of best fit – gradient of worst acceptable line

uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient) (d) (ii) [U4]

∆ ∆

= + × = × +

gradient 0.01 gradientpercentage uncertainty 100 100 1%

gradient 1.00 gradient

ρπ× π×

= =

× ×

max gradient max gradientmax

4 min 4 0.99L

ρπ× π×

= =

× ×

min gradient min gradientmin

4 max 4 1.01L

(e) [U5]

∆ ∆ = + × × = + ×

gradient 0.01 gradientpercentage uncertainty 2 100 0.086 100

gradient 0.23 gradient

ρ ρ

ρ ρ

∆ ∆ = + + × × = + ×

0.01 0.01percentage uncertainty 2 100 0.096 100

1.00 0.23

=2

min

max gradientmaxR

d

ρ× ×

=

π×

max max

2

min

4 max

LR

d

=2

max

min gradientminR

d

ρ× ×

=

π×

min min

2

max

4 min

LR

d





PHYSICS 9702/53


MARK SCHEME

Maximum Mark: 30

Published


®,





Question 1 Planning (15 marks) Defining the problem (2 marks)

P λ is the independent variable, or vary λ. [1] P V is the dependent variable, or measure V. [1] Methods of data collection (4 marks) M Circuit diagram showing d.c. power supply in series with diode (correct symbol needed) and

method to measure potential difference across diode. Circuit must be correct. [1]

M Instrument to change p.d. across LED e.g. variable power supply/potential divider/variable resistor. [1]

M Record wavelength of light of LED from data sheet or use Young’s slits/diffraction grating. [1]

M (Slowly) increase potential difference across LED until LED (just) emits light (or reverse procedure). [1]

Method of analysis (3 marks)

A Plot a graph of lg V against lg λ (allow natural logs). Allow lg λ against lg V. [1] A n = gradient [1] A k = 10

y-intercept [1] Additional detail (6 marks)

Relevant points might include: [6]

1 Use of a protective resistor (can be shown on the diagram). 2 Polarity of LED correct in circuit diagram. 3 Instrument to determine when LED just lights e.g. light meter/detector, LDR. 4 Method to use light detector/LDR to determine point at which LED emits light.

5 Expression that gives λ (symbols need to defined) from experimental determination of wavelength of light, e.g. Young’s slits/diffraction grating.

6 Perform experiment in a dark room/LED in tube. 7 Relationship is valid if graph is a straight line. 8 λ= +lg lg lgV n k

9 Repeat V and average for the same λ or LED. Do not allow vague computer methods.






(a) A1

π

4LF

E

(b) T1 2

1

d / 106 m–2

T2

13 or 12.8

9.8 or 9.77

6.9 or 6.93

4.7 or 4.73

3.2 or 3.19

1.9 or 1.93


U1 From ± 2 to ± 0.1 Allow more than one significant figure.


U2 Error bars in

2

1

d plotted correctly

All error bars to be plotted. Must be accurate to less than half a small square.

(ii) G2 Line of best fit If points are plotted correctly then lower end of line should pass between (3.2, 3.0) and (3.6, 3.0) and upper end of line should pass between (11.2, 10.0) and (11.6, 10.0).


Line should be clearly labelled or dashed. Examiner judgement on worst acceptable line. Lines must cross. Mark scored only if error bars are plotted.


POT. (Should be about 9 × 10–10.)

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

(d) (i) C2 =

π×

4 60.479

gradient gradient

LF

Do not penalise POT.

(Should be about 7 × 1010.)

C3 N m–2 or Pa Allow in base units: kg m–1 s–2.

(ii) U4 Percentage uncertainty in E Must be larger than 3%.





(e) C4 e in the range 15.5 × 10–3 to

18.0 × 10–3 and given to 2 or 3 s.f.

Allow mm.

U5 Absolute uncertainty in e Note =

2

gradiente

d is possible.


uncertainty = gradient of line of best fit – gradient of worst acceptable line uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient) (d) (ii) [U4]

∆ ∆

= + + × = × +

gradient 0.01 0.5 gradientpercentage uncertainty 100 100 3.03%

gradient 2.50 19.0 gradient

× × × ×

= = =

π× π×

4 max max 4 2.51 1 9.5 62.319max

min gradient min gradient min gradient

L FE

4 min min 4 2.49 18.5 58.652

minmax gradient max gradient max gradient

L FE

× × × ×

= = =

π× π×

(e) [U5]

= + + × × + = +

0.5 0.01 0.02percentage uncertainty 2 100 % 20.4% %

19.0 2.50 0.23E E

∆

= + × ×

gradient 0.02percentage uncertainty 2 100

gradient 0.23

=2

min

max gradientmaxe

d

× ×

=

π× ×

max max

2

min min

4 max

L Fe

E d

=2

max

min gradientmine

d

× ×

=

π× ×

min min

2

max max

4 min

L Fe

E d

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Cambridge International Examinations Cambridge ...Cambridge is publishing the mark schemes for the May/June 2016 series for most Cambridge IGCSE®, Cambridge International A and AS