9702 s14 ms_all

CAMBRIDGE INTERNATIONAL EXAMINATIONS

GCE Advanced Subsidiary Level and GCE Advanced Level

MARK SCHEME for the May/June 2014 series

9702 PHYSICS

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

Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2014 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 – May/June 2014 9702 11

© Cambridge International Examinations 2014

Question Number

Key Question Number

Key

1 A 21 C

2 B 22 D

3 A 23 B

4 D 24 D

5 B 25 B

6 D 26 B

7 D 27 D

8 A 28 D

9 B 29 A

10 A 30 A

11 B 31 D

12 C 32 B

13 B 33 A

14 D 34 D

15 C 35 B

16 B 36 A

17 D 37 C

18 D 38 B

19 B 39 B

20 B 40 A




9702 PHYSICS






Question Number

Key Question Number

Key

1 C 21 C

2 B 22 C

3 B 23 C

4 C 24 C

5 D 25 B

6 B 26 D

7 C 27 D

8 C 28 A

9 C 29 D

10 B 30 B

11 A 31 A

12 D 32 A

13 A 33 A

14 C 34 D

15 B 35 B

16 B 36 B

17 C 37 B

18 B 38 C

19 D 39 C

20 A 40 C




9702 PHYSICS






Question Number

Key Question Number

Key

1 C 21 D

2 D 22 D

3 A 23 A

4 B 24 D

5 C 25 C

6 C 26 C

7 C 27 B

8 D 28 A

9 C 29 B

10 C 30 B

11 B 31 A

12 C 32 C

13 C 33 D

14 A 34 D

15 D 35 A

16 D 36 A

17 C 37 A

18 B 38 C

19 C 39 C

20 C 40 D




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) (i) either rate of change of displacement or (change in) displacement / time (taken) B1 [1] (ii) speed has magnitude only B1 velocity has magnitude and direction B1 [2]

(b) (i) idea of area under graph / use of tvu

s ×+

=

2

)( C1

2.52

32)(18×

+=s C1

= 62.5 m A1 [3] (ii) a = (18 – 32) / 2.5 (= –5.6) C1 F = ma C1

F = 1500 × (–) 5.6 = (–) 8400 N A1 [3] (c) arrow labelled A and arrow labelled F both to the left B1 [1] 2 (a) (i) work (done) / time (taken) B1 [1]

(ii) work = force × displacement (in direction of force) B1

power = force × displacement / time (taken) = force × velocity B1 [2]

(b) (i) weight = mg C1

P = Fv = 2500 × 9.81 × sin 9° × 8.5 (or use cos 81°) C1 = 33 (32.6) kW A1 [3]

(ii) no gain or loss of KE B1 no work (done) against air resistance B1 [2]

3 (a) (i) resultant force is zero B1

weight of plank + weight of man = FA + FB or 200 (N) + 880 (N) or 1080 = FA + FB B1 [2]

(ii) principle of moments used C1

(anticlockwise moments) FB × 5.0 C1

(clockwise moments) 880 × 0.5 + 200 × 2.5 C1 FB = (440 + 500) / 5.0 = 188 N A1 [4]

(b) straight line with positive gradient (allow freehand) M1

start point (0, 100) A1 finish point (5, 980) A1 [3]




4 (a) kinetic energy = ½ mv2 C1

= ½ × 0.040 × (2.8)2 = 0.157 J or 0.16 J A1 [2] (b) (i) k = F / x or F = kx C1

XB = 14 / 800 = 0.0175 m A1 [2]

(ii) area under graph = elastic potential energy stored C1

or ½ kx2 or ½ Fx (energy stored =) 0.1225 J less than KE (of 0.16 J) A1 [2]

5 (a) (i) displacement is the distance from the equilibrium position / undisturbed position / midpoint / rest position B1

amplitude is the maximum displacement B1 [2]

(ii) frequency is the number of wavefronts / crests passing a point per unit time / number of oscillations per unit time B1

time period is the time between adjacent wavefronts or time for one oscillation B1 [2]

(b) (i) 1. amplitude = 1.5 mm A1 [1] 2. wavelength = 25 / 6 C1

= 4.2 cm or 4.2 × 10–2 m A1 [2]

(ii) v = λ / T or v = f λ and T= 1 / f C1 T = 4.2 / 7.5 = 0.56 s A1 [2] (c) (i) progressive M0

wavefront / crests moving / energy is transferred by the waves A1 [1] (ii) transverse M0

the vibration is perpendicular to the direction of energy transfer / wave velocity or travel of the wave / wavefronts A1 [1]

6 (a) e.m.f.: energy converted from chemical / other forms to electrical

per unit charge B1 p.d.: energy converted from electrical to other forms per unit charge B1 [2]

(b) (i) the p.d. across the lamp is less than 12 V

or there are lost volts / power / energy in the battery / internal resistance B1 [1]

(ii) R = V2 / P (or V = RI and P = VI) C1

= 144 / 48

= 3.0 Ω A1 [2]




(iii) I = E / (RT + r) C1 = 12 / 2.0 = 6.0 A A1 [2]

(iv) power of each lamp = I

2R

= (3.0)2 × 3.0 C1 = 27 W A1 [2] (c) less resistance (in circuit) / more current M1 more lost volts / less p.d. across battery A1 [2]

7 (a) α: helium nucleus

β: electron

γ: electromagnetic radiation / wave / ray or photon three correct 2 / 2, two correct 1 / 2 B2 [2] (b) (i) atomic number / proton number / Z –2, nucleon / mass number / A –4 B1 [1] (ii) atomic number / proton number / Z +1 nucleon / mass number / A no change B1 [1] (iii) no change in proton or mass number or “no change” B1 [1]




9702 PHYSICS







1 (a) power = energy / time or work done / time B1 force: kg m s–2 (including from mg in mgh or Fv)

or kinetic energy (2

1mv2): kg (m s–1)2 B1

(distance: m and (time) –1: s–1) and hence power: kg m s–2 m s–1 = kg m2 s–3 B1 [3]

(b) Q / t : kg m2

s–3 C1 A: m2 and x: m and T: K C1 correct substitution into C = (Qx) / tAT or equivalent, or with cancellation C1 units of C : kg m s–3

K–1 A1 [4]

2 (a) ρ = m / V C1

V = (π d

2 / 4) × t = 7.67 × 10–7

m3

ρ = (9.6 × 10–3) / [π(22.1/2 × 10–3)2 × 2.00 × 10–3] C1

ρ = 12513 kg m–3 (allow 2 or more s.f.) A1 [3]

(b) (i) ∆ρ / ρ = ∆m / m + ∆t / t + 2∆d / d C1 = 5.21% + 0.50% + 0.905% [or correct fractional uncertainties] C1 = 6.6% (6.61%) A1 [3]

(ii) ρ = 12 500 ± 800 kg m–3 A1 [1] 3 (a) a body / mass / object continues (at rest or) at constant / uniform velocity unless

acted on by a resultant force B1 [1] (b) (i) weight vertically down B1 normal / reaction / contact (force) perpendicular / normal to the slope B1 [2]

(ii) 1. acceleration = gradient or (v – u) / t or ∆v / t C1 = (6.0 – 0.8) / (2.0 – 0.0) = 2.6 m s–2 M1 [2] 2. F = ma

= 65 × 2.6 = 169 N (allow to 2 or 3 s.f.) A1 [1]

3. weight component seen: mg sinθ (218 N) C1 218 – R = 169 C1 R = 49 N (require 2 s.f.) A1 [3]




4 (a) GPE: energy of a mass due to its position in a gravitational field B1 KE: energy (a mass has) due to its motion / speed / velocity B1 [2]

(b) (i) 1. KE = 2

1 mv2 C1

= 2

1 × 0.4 × (30)2 C1

= 180 J A1 [3]

2. s = 0 + 2

1 × 9.81 × (2.16)2 or s = (30 sin 45°)2

/ (2 × 9.81) C1

= 22.88 (22.9) m = 22.94 (22.9) m A1 [2] 3. GPE = mgh C1

= 0.4 × 9.81 × 22.88 = 89.8 (90) J A1 [2] (ii) 1. KE = initial KE – GPE = 180 – 90 = 90 J A1 [1] 2. (horizontal) velocity is not zero / (object) is still moving / answer explained

in terms of conservation of energy B1 [1] 5 (a) (Young modulus / E =) stress / strain B1 [1] (b) (i) stress = F / A

or = F / (π d2/4)

or = F / (π d2) M1 ratio = 4 (or 4:1) A1 [2] (ii) E is the same for both wires (as same material) [e.g. EP = EQ] M1 strain = stress / E ratio = 4 (or 4:1) [must be same as (i)] A1 [2] 6 (a) there are no lost volts / energy lost in the battery or there are no lost volts / energy lost in the internal resistance B1 [1]

(b) the current / I decreases (as R increases) M1 p.d. decreases (as R increases) A1 or the parallel resistance (of X and R) increases M1 p.d. across parallel resistors increases, so p.d. (across Y) decreases A1 [2]




(c) (i) current = 2.4 (A) C1

p.d. across AB = 24 – 2.4 × 6 = 9.6 V M1 or

total resistance = 10 Ω (= 24 V / 2.4 A) C1

(parallel resistance = 4 Ω), p.d. = 24 × (4 / 10) = 9.6 V M1 [2]

(ii) R (AB) = 9.6 / 2.4 = 4.0 Ω C1 1 / 6 + 1 / X = 1 / 4 [must correctly substitute for R] C1

X = 12 Ω A1 or

IR = 9.6 / 6.0 = 1.6 (A) (C1)

IX = 2.4 – 1.6 = 0.8 (A) (C1)

X (= 9.6 / 0.8) = 12 Ω (A1) [3]

(iii) power = VI or EI or V

2 / R or E

2 / R or І

2R C1

= 24 × 2.4 or (24)2 / 10 or (2.4)2 × 10

= 57.6 W (allow 2 or more s.f.) A1 [2] (d) power decreases M0

e.m.f. constant or power = 24 × current, and current decreases or e.m.f. constant or power = 242

/ resistance, and resistance increases A1 [1] 7 (a) waves from the double slit are coherent / constant phase difference B1 waves (from each slit) overlap / superpose / meet (not interfere) B1

maximum / bright fringe where path difference is nλ

or phase difference is n360U / 2πn rad

or minimum / dark fringe where path difference is (n + 2

1)λ

or phase difference is (2n + 1) 180U / (2n + 1)π rad B1 [3]

(b) v = fλ C1

λ = (3 × 108) / 670 × 1012 = 448 (or 450) (nm) M1 [2] (c) w = 12 / 9 C1

a (= Dλ / w) = (2.8 × 450 × 10–9) / (12 / 9 × 10–3) [allow nm, mm] C1

= 9.5 × 10–4 m [9.4 × 10–4

m using λ = 448 nm] A1 [3] (d) (red light has) larger / higher / longer wavelength (must be comparison) M1 fringes further apart / larger separation A1 [2]




9702 PHYSICS







1 (a) current, mass and temperature two correct 2/2, one omission or error 1/2 A2 [2]

(b) σ : no units, V: m3 C1 EP: kg m2

s–2 C1 C: kg m2 s–2 × m–3 = kg m–1

s–2 A1 [3] 2 (a) scalar has magnitude only B1 vector has magnitude and direction B1 [2]

(b) (i) v2 = 0 + 2 × 9.81 × 25 (or using 2

1 m v2 = mgh) C1

v = 22(.1) m s–1 A1 [2]

(ii) 22.1 = 0 + 9.81 × t (or 25 = 2

1 × 9.81 × t

2) M1

t (=22.1 / 9.81) = 2.26 s or t [=(5.097)1/2] = 2.26 s A0 [1] (iii) horizontal distance = 15 × t = 15 × 2.257 = 33.86 (allow 15 × 2.3 = 34.5) C1 (displacement)2 = (horizontal distance)2 + (vertical distance)2 C1 = (25)2 + (33.86)2 C1 displacement = 42 (42.08) m (allow 43 (42.6) m, allow 2 or more s.f.) A1 [4] (iv) distance is the actual (curved) path followed by ball B1 displacement is the straight line / minimum distance P to Q B1 [2] 3 (a) work done is the product of force and the distance moved in the direction of the

force or product of force and displacement in the direction of the force B1 [1]




(b) (i) work done equals the decrease in GPE – gain in KE B1 [1] (ii) 1. distance = area under line C1 = (7.4 × 2.5) / 2 = 9.3 m (9.25 m) M1 [2] or acceleration from graph a = 7.4 / 2.5 (= 2.96) (C1) and equation of motion (7.4)2 = 2 × 2.96 × s gives s = 9.3 (9.25) m (A1)

2. kinetic energy = 2

1 m v

2 C1

= 2

1 × 75 × (7.4)2 C1

= 2100 J A1 [3] 3. potential energy = mgh C1 h = 9.3 sin 30 ° C1 PE = 75 × 9.81 × 9.3 sin 30 ° = 3400 J A1 [3] 4. work done = energy loss C1 R = (3421 – 2054) / 9.3 C1 = 150 (147) N A1 [3] 4 (a) add small mass to cause extension then remove mass to see if spring returns to original length M1 repeat for larger masses and note maximum mass for which, when load is

removed, the spring does return to original length A1 [2] (b) Hooke’s law requires force proportional to extension B1 graph shows a straight line, hence obeys Hooke’s law M1 [2] (c) k = force / extension C1 = (0.42 × 9.81) / [(30 – 21.2) × 10–2] C1 = 47 (46.8) N m–1 A1 [3] 5 (a) lost volts / energy used within the cell / internal resistance B1 when cell supplies a current B1 [2]




(b) (i) E = І(R + r) C1 4.5 = 0.65 (6.0 + r)

r = 0.92 Ω A1 [2]

(ii) І = 0.65 (A) and V = ІR C1 V = 0.65 × 6 = 3.9 V A1 [2]

(iii) P = V

2 / R or P = І2R and P = ІV C1

= (3.9)2 / 6 = 2.5 W A1 [2]

(iv) efficiency = power out / power in C1

= І

2R / І

2(R + r) = R / (R + r) = 6.0 / ( 6.0 + 0.92 ) = 0.87 A1 [2] (c) (circuit) resistance decreases B1 current increases M1 more heating effect A1 [3] 6 (a) (i) progressive wave transfers energy, stationary wave no transfer of energy / keeps energy within wave B1 [1] (ii) (progressive) wave / wave from loudspeaker reflects at end of tube B1 reflected wave overlaps (another) progressive wave B1 same frequency and speed hence stationary wave formed B1 [3] (iii) (side to side) along length of tube / along axis of tube B1 [1] (b) all three nodes clearly marked with N / clearly labelled at cross-over points B1 [1] (c) phase difference = 0 A1 [1]

(d) (i) v = fλ C1

λ = 330 / 440 = 0.75 m A1 [2]

(ii) L = 5/4 λ C1 = 5/4 × 0.75 = 0.94 m A1 [2]




9702 PHYSICS

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






1 (a) Value of L0 in range 0.045 m–0.070 m (4.5 to 7.0 cm). [1]

If out of range, compare to Supervisor’s value ± 20%. (b) (iii) Value of L > L0. [1] (c) Six sets of readings of m and L scores 5 marks, five sets scores 4 marks, etc. [5]

Incorrect trend then –1. Correct trend is L decreases as m increases for all m values. Major help from Supervisor –2. Minor help from Supervisor –1.

Range: at least one value of m less than 200 g and one value more than 200 g. [1] 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 /

o, L / m, m2 / kg2, e / m, e2(m2).

Consistency: [1] All values of L must be given to the nearest mm only. Significant figures: Significant figures for every row of m2 same as (or one greater than) the s.f. in m as recorded in table. [1] Calculation: [1] Values of e2 calculated correctly to the number of significant figures given by the candidate.

(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 plots must be Y half a small square (no “blobs”). Work to an accuracy of half a small square.

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

Scatter of points must be less than ± 0.0005 m2 of e2 from a straight line. (ii) Line of best fit: [1]

Judge by balance of all points on the grid about the candidates’ 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.

(e) P = – value of the gradient and Q = value of the y-intercept. [1]

Do not allow fractions. Do not allow substitution methods. (f) Check substitution and value of M in range 0.100–0.500 kg with unit. [1] [Total: 20] 2 (b) (ii) Value of x in the range 25.0–35.0 cm with unit. [1]

x to the nearest mm [1] (iii) Absolute uncertainty in x in range 2 mm–5 mm. [1]

If repeated readings have been taken, then the uncertainty can be half the range (not zero) only if working shown. Correct method of calculation to obtain percentage uncertainty.

(c) (ii) Value of T with unit. [1]

Evidence of repeats. [1] (iii) Correct calculation of f. [1] (d) (ii) Second value of x. [1]

Second value of T. [1] Second value of T < first value of T. [1]

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

T / time. (Do not allow “raw readings”.) [1] (iii) Valid comment relating to the calculated values of k, testing against a

criterion specified by the candidate. [1]




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

A Two readings not enough to draw a conclusion.

Take more readings for different lengths and plot a graph or take more readings and compare k values.

Not enough repeat readings. Few readings. Idea of repeats. “Too few readings / two readings” on its own.

B Difficulty linked to timing with reason, e.g. time small / short / vibrates fast / high frequency / oscillates fast / swings fast. or Large uncertainty in time with reason. or Human reaction time with reason, e.g. short time.

Improved method of timing, e.g. video with timer / video and view frame by frame / light gate placed at the centre / motion sensor at side of blade (to timer / datalogger display).

“Human errors / reaction time” on their own. “Light gate” on its own. Moves fast. Video and playback. Fans. Longer blade. Simultaneous release of blade and start timer. Amplitude. High speed camera or slow motion cameras.

C Unevenness of oscillation or G-clamp moving.

Improved method of smoother oscillation, e.g. use of wooden block either side of hacksaw blade. Method of fixing G-clamp, e.g. clamp G-clamp (to table).

D Difficulty judging centre of masses (due to slots).

Use masses without slots / measure to the top and bottom and average.

Blu Tack. Masses different heights.

E Difficult to measure x with reason, e.g. difficult to know where to start x as jaws of clamp are rounded / blade may not be vertical (due to clamp).

Improved method to measure x, e.g. description of use of set square

Uncertainty in metre rule. Parallax error.

F Small range for x for measurable times

[Total: 20]


GCE Advanced Level


9702 PHYSICS





GCE A LEVEL – May/June 2014 9702 32


1 (a) (i) Value of l 0 in range 4.0 cm Y l 0 Y 8.0 cm. [1]

(b) (iii) Value of h to nearest mm, in the range 40.0 cm Y h Y 50.0 cm. [1] (c) Six sets of values for h and l scores 5 marks, five sets scores 4 marks, etc. [5] Incorrect trend –1. Help from Supervisor –1. Range: [1] h values must include 20 cm or less. 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. 1/h2/cm–2 or 1/h2 (1/cm2) but not 1/h2(cm2). Consistency: [1] All values of h and l must be given to the nearest mm only. Significant figures: [1] Every value of 1/h2 must be given to the same s.f. as (or one greater than) the

s.f. in the corresponding h. Calculation: [1] Values of (l –l 0)

2 calculated correctly. (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 must be no more than 3 large squares apart. Plotting: [1] All observations in the table must be plotted.

Diameter of plotted points must be Y half a small square (no “blobs”). Work to an accuracy of half a small square. Quality: [1] All points must be plotted (at least 5) for this mark to be scored. Scatter of

points must be within ±10 cm2 of (l –l 0)2 from a straight line.

(ii) Line of best fit: [1] Judge by balance of all points 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 (i.e. circled or labelled) by the candidate.

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




(iii) Gradient: [1] The hypotenuse 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: 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: Check read-off of the intercept directly from the graph. (e) p = value of the gradient, and q = value of the intercept. [1] Dimensionally correct units for p and q. [1] [Total: 20]

2 (a) (i) Value for d to nearest mm, in range 1.0 cm Y d Y 2.0 cm. [1] (ii) Correct calculation of l. [1]

(b) (ii) Value for t in range 4.0s Y t Y 10.0 s, with unit. [1] Evidence of repeat readings of t. [1] (c) (iii) Value for A to nearest mm, with unit. [1] (d) Absolute uncertainty in A in range 2 to 5 mm. [1] If repeated readings have been taken, then absolute uncertainty could be half the

range (but not zero) only if working shown. Correct method of calculation to obtain percentage uncertainty.

(e) Second value of t. [1] Second value of A. [1] (f) (i) Two values of k calculated correctly. [1] kl in range 0.20 to 0.30 cm s–2. [1] (ii) Justification based on the number of s.f. in d, n, t and A (not just “raw readings”). [1] (iii) Valid 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 are not enough to draw a valid conclusion

Take more readings (for different masses) and plot a graph, or take more readings and compare k values

Not enough repeat readings Few readings Idea of repeats “Too few readings/two readings” on its own

B Difficult to measure d, with reason, e.g. parallax error/ measuring outside diameter/loop gets in the way

Use vernier calipers/micrometer Measure inside and outside diameter and find average

“Parallax error” on its own “Calipers” on its own

C Difficult to judge exactly when 10 oscillations completed

Video + timer/video and view frame-by-frame. Use distance sensor in stated and correct position Use a (fiducial) marker at centre of oscillation/equilibrium position Light gate at equilibrium position/centre of oscillation

Human/reaction time error High speed cameras/slow motion cameras Oscillations too fast

D Mass swings as it oscillates/non-uniform oscillation/spring moves along bolt

Use tube to act as a guide Use deeper groove on bolt Fix top of spring to bolt with, e.g. Blu-tack/Sellotape

E Difficult to judge when contact is lost/hard to see gap

Use pressure sensor on bolt Use video close-up/zoom lens/ magnifying glass Video + slow-motion, linked to observing gap Better/more sensitive method of adjusting height of bolt, e.g. lab jack

“Video + slow motion” on its own

F n is not a whole number Measure n to nearest ¼ turn

Do not credit use of an assistant, fans, air conditioning, or use of computers/data loggers on its own.

[Total: 20]




9702 PHYSICS







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

Scatter of points must be less than ± 0.25 A–1 of 1/I from a straight line.

[1]

1 (a) Value of L with unit in range 90.0 cm ≤ L ≤ 110.0 cm. [1]

(d) (ii) Value of I with unit in the range 50 mA ≤ I ≤ 150 mA. Allow Supervisor’s value ± 20%.

[1]

(e)

Six sets of readings of x and I scores 5 marks, five sets scores 4 marks etc.

Incorrect trend –1 (correct trend is I increases as x increases for all values of x). Major help from Supervisor –2. Minor help from Supervisor –1.

[5]

Range: ∆x ≥ 70 cm. [1]

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

convention e.g. x/cm or x(cm), x2/(x+L)/m but not x2/(x+L)/m2/m, 1/I (A–1), 1/I

(1/A) but not 1/I (A) or 1/I(A)–1.

[1]

Consistency: All values of x must be given to the nearest mm only.

[1]

Significant figures: Significant figures for every row of x2/(x+L) same as (or one greater than) the s.f. in x. Calculation: Values of x2/(x + L) calculated correctly to the number of significant figures given by the candidate.

[1] [1]

(f) (i) Axes: Sensible scales must be used, no awkward scales (e.g. 3:10). Scales must be chosen so that the plotted points occupy at least half the graph grid in both x and y directions. Scales must be labelled with the quantity that is being plotted. Scale markings should be no more than three large squares apart.

[1]

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

[1]




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

[1]

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

[1]

y-intercept: Either: 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.

[1]

(g) P = – the value of the gradient and Q = the value of the y-intercept. Do not allow fractions. Do not allow a substitution method. Unit for P (A–1m–1) and Q (A–1) consistent with value.

[1]

[1]

[Total: 20]

2 (b) (i) Value of y in range 8.0 cm ≤ y ≤ 12.0 cm with unit. [1]

(ii) Absolute uncertainty in y in range 2 mm – 5 mm. If repeated readings have been taken, then allow uncertainty to be half the range (but not zero) only if working shown. Correct method of calculation to obtain percentage uncertainty.

[1]

(iii) Value of x to the nearest mm with unit. [1]

(iv) Correct calculation of R. [1]

(c) (iii) Value of T with unit in range 1.5 s ≤ T ≤ 3.0 s. Evidence of repeats.

[1] [1]

(d) Second value of R. Second value of T. Correct trend for T with respect to R (T decreases as R decreases).

[1] [1] [1]

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




(ii) Correct justification of significant figures in k linked to significant figures in x and y and time (do not allow “raw readings”).

[1]

(iii) Valid comment consistent with the calculated values of k, testing against a criterion specified by the candidate.

[1]


A Two readings not enough to draw a conclusion.

Take more readings (for different radii) and plot a graph or take more readings and compare k values.

Not enough repeat readings. Few readings. Idea of repeats. “Too few readings/two readings” on its own.

B Only a few cycles possible/(heavily) damped/ball does not return to the original height.

Use a longer track/heavier ball. Smoother or lubricated track.

C Difficult to judge exactly when an oscillation is complete.

Improved method of timing e.g. video with timer/video and view frame by frame/measure time from centre of oscillation/put marker at the centre of oscillation/light gate at centre.

“Ball too fast” on its own. Human reaction time. Video and play back. High speed cameras or slow motion cameras. Use of motion sensor.

D Difficult to measure y with reason e.g. not sure where horizontal is at the top/ruler not vertical/allowing for the thickness/lips of the track/can’t hold ruler steady.

Improved method of measuring y e.g. use a ruler across the top/tie a string across the top/ set square on bench/use of plumbline/measure two heights and find difference/use of micrometer for thickness of track/clamp ruler to measure y/clamp ruler vertically.

Parallax error. Not clear where the lowest point of track is.

E Limitation of track e.g. is not entirely circular/portion of track in clamps is straight/too stiff to bend/track twisted/skewed.

Improved method of supporting track e.g. attach track to board/clamp in more places.

Changing track e.g. more flexible or rubber track. Difficult to clamp. “Friction on track” on its own.

F Difficulty in placing ball at the same point each time/difficulty in releasing the ball with no force.

Method of marking the same point on the track/method of releasing the ball e.g. use card as a gate/electromagnet release and steel ball

If electromagnetic release do not allow ‘metal’ ball. Clamp or robotic arm. Burning through string.

Do not credit measurement of x, use of an assistant, fans, air conditioning, use of computers/dataloggers on its own. [Total: 20]




9702 PHYSICS







1 (a) (i) Value for θ in range 80° to 100°, with unit. [1] (b) (ii) Value for t in range 10 to 40 s, with unit.. [1]

Evidence of repeat readings of t. [1]

(c) Five sets of values for θ and t scores 4 marks, four sets scores 3 marks etc. [4] Incorrect trend –1. Help from Supervisor –1. Range: [1]

θ values must include 75° or less and 105° 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. t2 / s2 or t2 (s2), θ (°) or θ (deg) etc. sin2(θ / 2) must have no unit. Consistency: [1] All values of t must be given to the nearest 0.1 s, or all to the nearest 0.01 s. Significant figures: [1] Every value of t2 must be given to the same s.f. as (or one greater than) the s.f. in the corresponding t. Calculation: [1]

Values of sin2(θ / 2) calculated correctly. (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 must be no more than three large squares apart.

Plotting: [1] All observations in the table must be plotted on the grid. Diameter of plotted points must be ≤ half a small square (no “blobs”). Plotting must be accurate to half a small square. Quality: [1] All points in the table must be plotted (at least 5) for this mark to be awarded.

Scatter of points must be within ± 20 s2 of a straight line in the y (t2) direction.




(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 plot 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 x and y directions.

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

(e) q = candidate's gradient and p = candidate's intercept. [1]

Correct units for q and p (s2 for q and s2 for p). [1] [Total: 20]




2 (a) r in range 5.0 mm to 12.0 mm and to nearest 0.1 mm or better. [1] (b) (i) Value for l in range 61.0 mm to 65.0 mm. [1] (ii) Value for e in range 6.0 mm to 8.0 mm. [1] (c) (ii) Value for x. [1]

Evidence of repeat readings of x. [1] (iii) Absolute uncertainty in x in range 2 to 9 mm. [1]

If repeated readings have been taken, then absolute uncertainty could be half the range (but not zero) only if working is shown. Correct method of calculation to obtain percentage uncertainty.

(iv) Calculated value of θ correct. [1]

θ given to 2 or 3 significant figures. [1] (d) Second values of e and x. [1] (e) (i) Two values of k calculated correctly. [1]

Both values for k in range 0.80 to 1.20. [1] (ii) Valid comment consistent with the calculated values of k, testing against a





(f) Limitations (4 max) Improvements (4 max) Do not credit

A Two readings are not enough to draw a valid conclusion

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

Repeat readings / too few readings / two readings

B Difficult to align groove with line / difficult to estimate centre of groove

Mark centre line of groove / method of aligning groove (e.g. use lines on paper / use graph paper / transparent ramp)

C e is small so uncertainty in e is large / small change in e gives large

change in θ

Use larger spheres to enable larger e

Just “use larger spheres”

D Parallax error when measuring x

Use set square (with detail of workable method)

E Difficult to locate centre of sphere when measuring x

Measure to edge of sphere and add r

Difficult to locate centre of sphere when measuring r

F Sphere rolls slightly after hitting tape / sphere does not stick

Use video with scale (in view) / Description of workable improvement (e.g. powder on strip / plasticine surface)

High speed camera / slow motion camera / video camera / stickier surface / magnets

[Total: 20]




9702 PHYSICS







1 (b) (iii) Values of a in range 53.0 cm – 57.0 cm and b < a with unit. [1] (iv) Value of L in the range 12.0 cm – 16.0 cm with unit. [1] (d) Six sets of readings of a and b scores 5 marks, five sets scores 4 marks, etc. [5]

Incorrect trend –1 (correct trend is b increases as a increases). Help from Supervisor –1.

Range: ∆a [ 39 cm. [1] Column headings: [1]

Each column heading must contain a quantity and unit. The presentation of quantity and unit must conform to accepted scientific convention, e.g. (1 / b) / m–1.

Consistency: [1]

All values of a and b must be given to the nearest mm. Significant figures: [1]

Significant figures for every row of values of 1 / b same as (or one greater than) b as recorded in table.

Calculation: [1]

Values of a / b 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 must be plotted.

Diameter of plotted points must be Y half a small square (no “blobs”). Work to an accuracy of half a small square. Quality: [1] All points in the table must be plotted (at least 5) for this mark to be awarded.

Scatter of points must be less than ± 0.001 cm–1 (0.1 m–1) of 1 / b 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) Value of P = –gradient and value of Q = intercept. [1] (g) Value of M in range 40–200 g with unit. No POT error allowed. [1] [Total: 20] 2 (a) (ii) Values of d to the nearest 0.01 mm with unit. [1]

6.00 mm Y d Y 10.00 mm. If out of range allow Supervisor’s value ± 2.00 mm. [1] Evidence of repeat readings. [1]

(iii) Absolute uncertainty in d in range 0.05 mm–2.00 mm. [1]

If repeated readings have been taken, then the uncertainty could be half the range (but not zero) only if the working is shown. Correct method of calculation to get percentage uncertainty.

(c) (iv) Value of x1 > x with unit to the nearest mm. [1] (v) Correct calculation of e. [1] (d) (ii) Second value of d. [1]

Second value of x1. [1] Second value of e < first value of e. [1]

(e) (i) Two values of k calculated correctly. [1] (ii) Justification based on the number of significant figures in e (or x1 – x) and d. [1] (iii) Valid comment relating to the calculated values of k, testing against a






A Two readings not enough to draw a conclusion

Take many readings for different diameters and plot a graph

Repeat readings Few readings Too few readings / only two readings

B Large uncertainty in extension because extension small

Use longer / thinner cylinders or thinner central portion / time for hanging longer / greater mass to give greater extension

C Difficult to roll uniform cylinder / cylinder not symmetrical / not uniform diameter / density or consistency not the same

Viable suggestion for improvement, e.g. spacers, mould, force through hole

D Reading of x1 is imprecise because marks have widened

Improved method of marking without a dent

E Difficulty with clamping the plasticine, e.g. breaks prematurely / twists in clamp

Improved method to attach weight to plasticine, e.g string loop through handles / place clamp lengthways

F Micrometer digs into plasticine and may weaken it / gives incorrect diameter reading

Improved method to measure d, e.g. travelling microscope / work out diameter from volume or circumference

G Difficulties relating to the properties of the plasticine over time, e.g. high temperature making too soft / picking up impurities / re-breaking at fractured points / as roll temperature increases and affects

Use new piece of plasticine each time / roll and leave until reaches room temperature

[Total: 20]




9702 PHYSICS

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






Section A

1 (a) work done bringing unit mass M1 from infinity (to the point) A1 [2]

(b) EP = –mφ B1 [1]

(c) φ ∝ 1/x C1

either at 6R from centre, potential is (6.3 × 107)/6 (= 1.05 × 107 J kg–1)

and at 5R from centre, potential is (6.3 × 107)/5 (= 1.26 × 107 J kg–1) C1

change in energy = (1.26 – 1.05) × 107 × 1.3 C1

= 2.7 × 106 J A1

or change in potential = (1/5 – 1/6) × (6.3 × 107) (C1)

change in energy = (1/5 – 1/6) × (6.3 × 107) × 1.3 (C1)

= 2.7 × 106 J (A1) [4]

2 (a) the number of atoms M1 in 12 g of carbon-12 A1 [2] (b) (i) amount = 3.2/40 = 0.080 mol A1 [1] (ii) pV = nRT

p × 210 × 10–6 = 0.080 × 8.31 × 310 C1

p = 9.8 × 105 Pa A1 [2] (do not credit if T in °C not K)

(iii) either pV = 1/3 × Nm <c2>

N = 0.080 × 6.02 × 1023 (= 4.82 × 1022)

and m = 40 × 1.66 × 10–27 (= 6.64 × 10–26) C1

9.8 × 105 × 210 × 10–6 = 1/3 × 4.82 × 1022 × 6.64 × 10–26 × <c2> C1

<c2> = 1.93 × 105 cRMS = 440 m s–1 A1 [3]

or Nm = 3.2 × 10–3 (C1)

9.8 × 105 × 210 × 10–6 = 1/3 × 3.2 × 10–3 × <c2> (C1)

<c2> = 1.93 × 105 cRMS = 440 m s–1 (A1) or 1/2 m<c2> = 3/2 kT (C1)

1/2 × 40 × 1.66 × 10–27 <c2> = 3/2 × 1.38 × 10–23 × 310 (C1)

<c2> = 1.93 × 105 cRMS = 440 m s–1 (A1)

(if T in °C not K award max 1/3, unless already penalised in (b)(ii))




3 (a) either change in volume = (1.69 – 1.00 × 10–3) or liquid volume << volume of vapour M1

work done = 1.01 × 105 × 1.69 = 1.71 × 105 (J) A1 [2] (b) (i) 1. heating of system/thermal energy supplied to the system B1 [1] 2. work done on the system B1 [1]

(ii) ∆U = (2.26 × 106) – (1.71 × 105) C1

= 2.09 × 106 J (3 s.f. needed) A1 [2] 4 (a) kinetic (energy)/KE/EK B1 [1] (b) either change in energy = 0.60 mJ

or max E proportional to (amplitude)2/equivalent numerical working B1 new amplitude is 1.3 cm B1 change in amplitude = 0.2 cm B1 [3] 5 (a) graph: straight line at constant potential = V0 from x = 0 to x = r B1

curve with decreasing gradient M1 passing through (2r, 0.50V0) and (4r, 0.25V0) A1 [3] (b) graph: straight line at E = 0 from x = 0 to x = r B1 curve with decreasing gradient from (r, E0) M1 passing through (2r, ¼E0) A1 [3] (for 3rd mark line must be drawn to x = 4r and must not touch x-axis) 6 (a) (i) energy = EQ C1

= 9.0 × 22 × 10–3 = 0.20 J A1 [2] (ii) 1. C = Q / V

V = (22 × 10–3)/(4700 × 10–6) C1 = 4.7 V A1 [2] 2. either E = ½CV

2 C1

= ½ × 4700 × 10–6 × 4.72

= 5.1 × 10–2 J A1 [2] or E = ½QV (C1)

= ½ × 22 × 10–3 × 4.7

= 5.1 × 10–2 J (A1) or E = ½Q2/C (C1)

= ½ × (22 × 10–3)2/4700 ×10–6

= 5.1 × 10–2 J (A1)




(b) energy lost (as thermal energy) in resistance/wires/battery/resistor B1 [1] (award only if answer in (a)(i) > answer in (a)(ii)2) 7 (a) graph: VH increases from zero when current switched on B1 VH then non-zero constant B1 VH returns to zero when current switched off B1 [3] (b) (i) (induced) e.m.f. proportional to rate M1 of change of (magnetic) flux (linkage) A1 [2] (ii) pulse as current is being switched on B1 zero e.m.f. when current in coil B1 pulse in opposite direction when switching off B1 [3] 8 (a) discrete and equal amounts (of charge) B1 [1]

allow: discrete amounts of 1.6 × 10–19C/elementary charge/e

integral multiples of 1.6 × 10–19C/elementary charge/e (b) weight = qV / d

4.8 × 10–14 = (q × 680)/(7.0 × 10–3) C1

q = 4.9 × 10–19 C A1 [2]

(c) elementary charge = 1.6 × 10–19 C (allow 1.6 × 10–19 C to 1.7 × 10–19 C ) M0 either the values are (approximately) multiples of this or it is a common factor C1 it is the highest common factor A1 [2] 9 (a) e.g. no time delay between illumination and emission max. (kinetic) energy of electron dependent on frequency max. (kinetic) energy of electron independent of intensity rate of emission of electrons dependent on/proportional to intensity

(any three separate statements, one mark each, maximum 3) B3 [3] (b) (i) (photon) interaction with electron may be below surface B1 energy required to bring electron to surface B1 [2]




(ii) 1. threshold frequency = 5.8 × 1014 Hz A1 [1] 2. Φ = hf0 C1

= 6.63 × 10–34 × 5.8 × 1014

= 3.84 × 10–19 (J) C1

= (3.84 × 10–19)/(1.6 × 10–19) = 2.4 eV A1 [3]

or

hf = Φ + EMAX (C1) chooses point on line and substitutes values EMAX, f and h into equation with the units of the hf term converted from J to eV (C1) Φ = 2.4 eV (A1) 10 (a) energy required to separate the nucleons (in a nucleus) M1 to infinity A1 [2] (allow reverse statement)

(b) (i) ∆m = (2 × 1.00867) + 1.00728 – 3.01551 C1

= 9.11 × 10–3 u C1

binding energy = 9.11 × 10–3 × 930 = 8.47 MeV A1 [3]

(allow 930 to 934 MeV so answer could be in range 8.47 to 8.51 MeV) (allow 2 s.f.)

(ii) ∆m = 211.70394 – 209.93722 = 1.76672 u C1

binding energy per nucleon = (1.76672 × 930)/210 C1 = 7.82 MeV A1 [3] (allow 930 to 934 MeV so answer could be in range 7.82 to 7.86 MeV) (allow 2 s.f.) (c) total binding energy of barium and krypton M1 is greater than binding energy of uranium A1 [2]

Section B 11 (a) (i) inverting amplifier B1 [1] (ii) gain is very large/infinite B1 V+ is earthed/zero B1 for amplifier not to saturate, P must be (almost) earth/zero B1 [3]

(b) (i) RA = 100 kΩ A1

RB = 10 kΩ A1 VIN = 1000 mV A1 [3] (ii) variable range meter B1 [1]




12 (a) series of X-ray images (for one section/slice) M1 taken from different angles M1 to give image of the section/slice A1 repeated for many slices M1 to build up three-dimensional image (of whole object) A1 [5] (b) deduction of background from readings C1 division by three C1 P = 5 Q = 9 R = 7 S =13 (four correct 2/2, three correct 1/2) A2 [4] 13 (a) e.g. noise can be eliminated/waveform can be regenerated extra bits of data can be added to check for errors cheaper/more reliable greater rate of transfer of data (1 each, max 2) B2 [2] (b) receives bits all at one time B1 transmits the bits one after another B1 [2] (c) sampling frequency must be higher than/(at least) twice frequency to be sampled M1 either higher (range of) frequencies reproduced on the disc or lower (range of) frequencies on phone A1 either higher quality (of sound) on disc

or high quality (of sound) not required for phone B1 [3] 14 (a) reduction in power (allow intensity/amplitude) B1 [1]

(b) (i) attenuation = 2.4 × 30 = 72 dB A1 [1] (ii) gain/attenuation/dB = 10 lg(P2/P1) C1 72 = 10 lg(PIN/POUT) or –72 = 10 lg(POUT /PIN) C1

ratio = 1.6 × 107 A1 [3] (c) e.g. enables smaller/more manageable numbers to be used e.g. gains in dB for series amplifiers are added, not multiplied B1 [1]


GCE Advanced Level


9702 PHYSICS







Section A 1 (a) gravitational force provides/is the centripetal force B1 GMm / r2 R mv2

/ r M1 v R √(GM / r) A0 [2] allow gravitational field strength provides/is the centripetal acceleration (B1)

GM / r2 R v2 / r (M1)

(b) (i) kinetic energy increase/change R loss / change in (gravitational) potential

energy B1 ½mV0

2 R GMm / x C1 V0

2 R 2GM / x

V0 R √(2GM / x) A1 [3] (max. 2 for use of r not x) (ii) V0 is (always) greater than v (for x = r) M1 so stone could not enter into orbit A1 [2] (expressions in (a) and (b)(i) must be dimensionally correct) 2 (a) use of kelvin temperatures B1 both values of (V / T) correct (11.87), V / T is constant so pressure is constant M1 [2] (allow use of n R1. Do not allow other values of n.) (b) (i) work done R p∆V R 4.2 × 105 × (3.87 – 3.49) × 103 × 10–6 C1 = = = = = R 160 J A1 [2] (do not allow use of V instead of ∆V) (ii) increase / change in internal energy R heating of system N work done on system C1 R 565 – 160 = = = = R 405 J A1 [2] (c) internal energy R sum of kinetic energy and potential energy / EK N EP B1 no intermolecular forces M1 no potential energy (so ∆U R ∆EK) A1 [3] 3 (a) resonance B1 [1] (b) Pt R mc ∆θ C1 750 × 2 × 60 R 0.28 × c × (98 – 25) C1 c R 4400 J kg–1 K–1 A1 [3] (use of ∆θ R 73 N 273 max. 1 / 3) (use of t R 2 s not 120 s max. 2 / 3)




(c) e.g. some microwave leakage from the cooker e.g. container for the water is also heated (any sensible suggestion) B1 [1] 4 (a) (i) FE R Q1Q2

/ 4πε0r2 C1

= = = = R 8.99 × 109 × (1.6 × 10–19)2 / (2.0 × 10–15)2

R 58 N A1 [2] (ii) FG R Gm1m2

/ r2 C1 = = = R 6.67 × 10–11 × (1.67 × 10–27)2

/ (2.0 × 10–15)2 = = = R 4.7 × 10–35 N A1 [2] (b) (i) force of repulsion (much) greater than force of attraction B1 must be some other force of attraction M1 to hold nucleus together A1 [3] (Do not allow if FG > FE in (a) or one of the forces not calculated in (a)) (ii) outside nucleus there is repulsion between protons B1 either attractive force must act only in nucleus or if not short range, all nuclei would stick together B1 [2] 5 (a) only curve with decreasing gradient M1 acceptable value near xR 0 and does not reach zero A1 [2] (if graph line less than 4.0 cm do not allow A1 mark) (no credit if graph line has positive and negative values of VH) (b) graph: from 0 to 2T, two cycles of a sinusoidal wave M1 all peaks above 3.5 mV C1 peaks at 4.95 / 5.0 mV (allow 4.8 mV to 5.2 mV) A1 [3] (c) e.m.f. induced in coil when magnetic field / flux is changing / cutting B1 either at each position, magnetic field does not vary so no e.m.f. is induced in the coil / no reading on the millivoltmeter or at each position, switch off current and take millivoltmeter reading or at each position, rapidly remove coil from field and take meter reading B1 [2] 6 (a) electric and magnetic fields normal to each other B1 either charged particle enters region normal to both fields or correct B direction w.r.t. E for zero deflection B1 for no deflection, v R E / B B1 [3] (no credit if magnetic field region clearly not overlapping with electric field region)




(b) (i) m R Bqr / v C1 = = = R=(640 × 10–3 × 1.6 × 10–19 × 6.2 × 10–2) / (9.6 × 104) C1 = = = R=6.61 × 10–26

kg C1 = = = R=(6.61 × 10–26) / (1.66 × 10–27) u = = = R=40 u A1 [4] (ii) q / m ∝ 1 / r or m constant and q ∝ 1 / r B1 q / m for A is twice that for B B1 ions in path A have (same mass but) twice the charge (of ions in path B) B1 [3] 7 (a) angle subtended at the centre of a circle B1 by an arc equal in length to the radius B1 [2] (b) (i) arc R distance × angle C1

diameter R 3.8 × 105 × 9.7 × 10–6 = = = R 3.7 km A1 [2] (ii) Mars is (much) further from Earth / away (answer must be comparative) B1 angle (at telescope is much) smaller B1 [2] 8 (a) photon energy R hc / λ

R (6.63 × 10–34 × 3.0 × 108) / (590 × 10–9) C1 R 3.37 × 10–19

J C1 number R (3.2 × 10–3) / (3.37 × 10–19) = = = R 9.5 × 1015 (allow 9.4 × 1015) A1 [3] (b) (i) p R h / λ C1 = = = R (6.63 × 10–34) / (590 × 10–9) = = = R 1.12 × 10–27

kg m s–1 C1 total momentum R 9.5 × 1015 × 1.12 × 10–27 = = = R 1.06 × 10–11

kg m s–1 A1 [3] (ii) force R 1.06 × 10–11 N A1 [1] 9 (a) time for number of atoms / nuclei / activity (of the isotope) M1 to be reduced to one half (of its initial value) A1 [2] (b) (i) A R λN C1 460 R N × ln 2 / (8.1 × 24 × 60 × 60) C1 N R 4.6 × 108 A1 [3] (ii) number of water molecules in 1.0 kg R (6.02 × 1023) / (18 × 10–3) C1

R 3.3 × 1025

ratio R (3.3 × 1025) / (4.6 × 108) R 7.2 (7.3) × 1016 A1 [2]




(c) A R A0 e–λt and λt½ R ln 2 C1

170 R 460 exp (–ln 2 t / 8.1) C1 t R 11.6 days (allow 2 s.f.) A1 [3]

Section B 10 (a) compares the potentials/voltages at the (inverting and non-inverting) inputs B1 either output (potential) dependent on which input is the larger or V+ > V–, then VOUT is positive B1 states the other condition B1 [3] (b) (i) ring drawn around both the LEDs (and series resistors) B1 [1] (ii) V– R (1.5 × 2.4) / (1.2 N 2.4) R 1.0 V B1 [1] (allow 1.5 × 2.4 / 3.6 R 1.0 V) (iii) 1. VOUT switches at N1.0 V B1 maximum VOUT is 5.0 V B1 when curve is above N1.0 V, VOUT is negative (or v.v.) B1 [3] 2. at time t1, diode R is emitting light, diode G is not emitting B1 at time t2, diode R is not emitting, diode G is emitting B1 [2] (must be consistent with graph line. If no graph line then 0 / 2) 11 (a) X-ray: flat / shadow / 2D image B1 regardless of depth of object / depth not indicated B1 CT scan: built up from (many) images at different angles B1 image is three-dimensional B1 image can be rotated / viewed at different angles B1 [5] (b) (i) I R I0 e

–µx C1

0.25 R e–0.69x x R2.0 mm (allow 1 s.f.) A1 [2] (ii) for aluminium, I / I0 R e–0.46 × 2.4 R 0.33 C1 fraction R 0.33 × 0.25 = = = R 0.083 A1 [2] (iii) gain / dB R 10 lg(I / I0) C1 R 10 lg(0.083) = = = R (–) 10.8 dB (allow 2 s.f.) A1 with negative sign B1 [3] 12 (a) (i) satellite is in equatorial orbit B1 travelling from west to east B1 period of 24 hours / 1 day B1 [3]




(ii) either uplink signal is highly attenuated or signal is highly amplified (before transmission) as downlink signal B1

prevents downlink signal swamping the uplink signal B1 [2] (b) speed of signal is same order of magnitude in both systems B1 optic fibre link (much) shorter than via satellite M1 time delay using optic fibre is less A1 [3]




9702 PHYSICS







Section A

1 (a) work done bringing unit mass M1 from infinity (to the point) A1 [2]

(b) EP = –mφ B1 [1]

(c) φ ∝ 1/x C1

either at 6R from centre, potential is (6.3 × 107)/6 (= 1.05 × 107 J kg–1)

and at 5R from centre, potential is (6.3 × 107)/5 (= 1.26 × 107 J kg–1) C1

change in energy = (1.26 – 1.05) × 107 × 1.3 C1

= 2.7 × 106 J A1

or change in potential = (1/5 – 1/6) × (6.3 × 107) (C1)

change in energy = (1/5 – 1/6) × (6.3 × 107) × 1.3 (C1)

= 2.7 × 106 J (A1) [4]

2 (a) the number of atoms M1 in 12 g of carbon-12 A1 [2] (b) (i) amount = 3.2/40 = 0.080 mol A1 [1] (ii) pV = nRT

p × 210 × 10–6 = 0.080 × 8.31 × 310 C1

p = 9.8 × 105 Pa A1 [2] (do not credit if T in °C not K)

(iii) either pV = 1/3 × Nm <c2>

N = 0.080 × 6.02 × 1023 (= 4.82 × 1022)

and m = 40 × 1.66 × 10–27 (= 6.64 × 10–26) C1

9.8 × 105 × 210 × 10–6 = 1/3 × 4.82 × 1022 × 6.64 × 10–26 × <c2> C1

<c2> = 1.93 × 105 cRMS = 440 m s–1 A1 [3]

or Nm = 3.2 × 10–3 (C1)

9.8 × 105 × 210 × 10–6 = 1/3 × 3.2 × 10–3 × <c2> (C1)

<c2> = 1.93 × 105 cRMS = 440 m s–1 (A1) or 1/2 m<c2> = 3/2 kT (C1)

1/2 × 40 × 1.66 × 10–27 <c2> = 3/2 × 1.38 × 10–23 × 310 (C1)

<c2> = 1.93 × 105 cRMS = 440 m s–1 (A1)

(if T in °C not K award max 1/3, unless already penalised in (b)(ii))




3 (a) either change in volume = (1.69 – 1.00 × 10–3) or liquid volume << volume of vapour M1

work done = 1.01 × 105 × 1.69 = 1.71 × 105 (J) A1 [2] (b) (i) 1. heating of system/thermal energy supplied to the system B1 [1] 2. work done on the system B1 [1]

(ii) ∆U = (2.26 × 106) – (1.71 × 105) C1

= 2.09 × 106 J (3 s.f. needed) A1 [2] 4 (a) kinetic (energy)/KE/EK B1 [1] (b) either change in energy = 0.60 mJ

or max E proportional to (amplitude)2/equivalent numerical working B1 new amplitude is 1.3 cm B1 change in amplitude = 0.2 cm B1 [3] 5 (a) graph: straight line at constant potential = V0 from x = 0 to x = r B1

curve with decreasing gradient M1 passing through (2r, 0.50V0) and (4r, 0.25V0) A1 [3] (b) graph: straight line at E = 0 from x = 0 to x = r B1 curve with decreasing gradient from (r, E0) M1 passing through (2r, ¼E0) A1 [3] (for 3rd mark line must be drawn to x = 4r and must not touch x-axis) 6 (a) (i) energy = EQ C1

= 9.0 × 22 × 10–3 = 0.20 J A1 [2] (ii) 1. C = Q / V

V = (22 × 10–3)/(4700 × 10–6) C1 = 4.7 V A1 [2] 2. either E = ½CV

2 C1

= ½ × 4700 × 10–6 × 4.72

= 5.1 × 10–2 J A1 [2] or E = ½QV (C1)

= ½ × 22 × 10–3 × 4.7

= 5.1 × 10–2 J (A1) or E = ½Q2/C (C1)

= ½ × (22 × 10–3)2/4700 ×10–6

= 5.1 × 10–2 J (A1)




(b) energy lost (as thermal energy) in resistance/wires/battery/resistor B1 [1] (award only if answer in (a)(i) > answer in (a)(ii)2) 7 (a) graph: VH increases from zero when current switched on B1 VH then non-zero constant B1 VH returns to zero when current switched off B1 [3] (b) (i) (induced) e.m.f. proportional to rate M1 of change of (magnetic) flux (linkage) A1 [2] (ii) pulse as current is being switched on B1 zero e.m.f. when current in coil B1 pulse in opposite direction when switching off B1 [3] 8 (a) discrete and equal amounts (of charge) B1 [1]

allow: discrete amounts of 1.6 × 10–19C/elementary charge/e

integral multiples of 1.6 × 10–19C/elementary charge/e (b) weight = qV / d

4.8 × 10–14 = (q × 680)/(7.0 × 10–3) C1

q = 4.9 × 10–19 C A1 [2]

(c) elementary charge = 1.6 × 10–19 C (allow 1.6 × 10–19 C to 1.7 × 10–19 C ) M0 either the values are (approximately) multiples of this or it is a common factor C1 it is the highest common factor A1 [2] 9 (a) e.g. no time delay between illumination and emission max. (kinetic) energy of electron dependent on frequency max. (kinetic) energy of electron independent of intensity rate of emission of electrons dependent on/proportional to intensity

(any three separate statements, one mark each, maximum 3) B3 [3] (b) (i) (photon) interaction with electron may be below surface B1 energy required to bring electron to surface B1 [2]




(ii) 1. threshold frequency = 5.8 × 1014 Hz A1 [1] 2. Φ = hf0 C1

= 6.63 × 10–34 × 5.8 × 1014

= 3.84 × 10–19 (J) C1

= (3.84 × 10–19)/(1.6 × 10–19) = 2.4 eV A1 [3]

or

hf = Φ + EMAX (C1) chooses point on line and substitutes values EMAX, f and h into equation with the units of the hf term converted from J to eV (C1) Φ = 2.4 eV (A1) 10 (a) energy required to separate the nucleons (in a nucleus) M1 to infinity A1 [2] (allow reverse statement)

(b) (i) ∆m = (2 × 1.00867) + 1.00728 – 3.01551 C1

= 9.11 × 10–3 u C1

binding energy = 9.11 × 10–3 × 930 = 8.47 MeV A1 [3]

(allow 930 to 934 MeV so answer could be in range 8.47 to 8.51 MeV) (allow 2 s.f.)

(ii) ∆m = 211.70394 – 209.93722 = 1.76672 u C1

binding energy per nucleon = (1.76672 × 930)/210 C1 = 7.82 MeV A1 [3] (allow 930 to 934 MeV so answer could be in range 7.82 to 7.86 MeV) (allow 2 s.f.) (c) total binding energy of barium and krypton M1 is greater than binding energy of uranium A1 [2]

Section B 11 (a) (i) inverting amplifier B1 [1] (ii) gain is very large/infinite B1 V+ is earthed/zero B1 for amplifier not to saturate, P must be (almost) earth/zero B1 [3]

(b) (i) RA = 100 kΩ A1

RB = 10 kΩ A1 VIN = 1000 mV A1 [3] (ii) variable range meter B1 [1]




12 (a) series of X-ray images (for one section/slice) M1 taken from different angles M1 to give image of the section/slice A1 repeated for many slices M1 to build up three-dimensional image (of whole object) A1 [5] (b) deduction of background from readings C1 division by three C1 P = 5 Q = 9 R = 7 S =13 (four correct 2/2, three correct 1/2) A2 [4] 13 (a) e.g. noise can be eliminated/waveform can be regenerated extra bits of data can be added to check for errors cheaper/more reliable greater rate of transfer of data (1 each, max 2) B2 [2] (b) receives bits all at one time B1 transmits the bits one after another B1 [2] (c) sampling frequency must be higher than/(at least) twice frequency to be sampled M1 either higher (range of) frequencies reproduced on the disc or lower (range of) frequencies on phone A1 either higher quality (of sound) on disc

or high quality (of sound) not required for phone B1 [3] 14 (a) reduction in power (allow intensity/amplitude) B1 [1]

(b) (i) attenuation = 2.4 × 30 = 72 dB A1 [1] (ii) gain/attenuation/dB = 10 lg(P2/P1) C1 72 = 10 lg(PIN/POUT) or –72 = 10 lg(POUT /PIN) C1

ratio = 1.6 × 107 A1 [3] (c) e.g. enables smaller/more manageable numbers to be used e.g. gains in dB for series amplifiers are added, not multiplied B1 [1]




9702 PHYSICS

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






1 Planning (15 marks)

Defining the problem (3 marks)

P r is the independent variable or vary r. [1] P T (or t) is the dependent variable or measure T (or t). [1] P Keep the radius of curvature (of the track) or C constant (or radius of track

constant). Do not allow “use same track”. [1] Methods of data collection (5 marks)

M Diagram showing ball in a (curved) track with supports for track, e.g. retort stands. Minimum of two labels (from ball, track, supports; not stopwatch, bench, micrometer). Supports making contact with track higher than ball / at least half way up. [1]

M Measure time using stopwatch or light gates and timer or datalogger with motion

sensor. Detail needed for video camera. [1]

M Use many oscillations (at least 10 or at least 10 s of timing) and determine

T = t / n. [1] M Measure diameter (radius) of ball with a micrometer / vernier calipers. Do not allow travelling microscope. [1] M radius = diameter / 2. [1] Method of analysis (2 marks)

A Plot a graph of T

2 against r (or r against T

2) Do not allow log graphs. [1]

A C = y-intercept × gradient

intercepty

28

5

2

−

=

π

g (or for r against T

2, C = y-intercept) [1]

Safety considerations (1 mark) S Precaution linked to ball escaping on to floor, e.g. use barrier / safety

screen / sand tray to prevent balls rolling on to floor. [1]




Additional detail (4 marks) D Relevant points might include [4] 1 Add weights to / G-clamp retort stands 2 Keep the material / density of the ball constant 3 Use of fiducial marker near centre of track / mark on the track 4 Clean track / balls. Do not allow oil the track. 5 Repeat measurements of t (for each ball) and average 6 Repeat measurement for d (or r) and average 7 Relationship is valid if straight line, provided plotted graph is correct 8 Relationship is valid if straight line not passing through origin or has an intercept,

provided plotted graph is correct (any quoted expression must be correct, e.g.

y-intercept = g

C

5

282

π)

Do not allow vague computer methods. [Total: 15]




2 Analysis, conclusions and evaluation (15 marks)

Mark Expected Answer Additional Guidance

(a) A1 gradient = –R y-intercept = –R/P

Must be negative.

(b) T1 T2

6.7 or 6.67 –5.1 or –5.13

4.5 or 4.55 –3.8 or –3.75

3.0 or 3.03 –2.8 or –2.75

2.0 or 2.00 –2.1 or –2.06

1.5 or 1.52 –1.8 or –1.75

1.1 or 1.11 –1.5 or –1.50

Allow a mixture of significant figures. T1 and T2 must be table values. Ignore “–” omissions.

U1 From ± 0.3 or ± 0.4 to ± 0.1 or ± 0.2 Allow more than one significant figure.

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

U2 Error bars in V / E 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 (6.4,–5.0) and (6.6,–5.0) and upper end of line should pass between (1.0,–1.5) and (1.2,–1.5).

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 all error bars are plotted.

(iii) C1 Gradient of best fit line Must be negative. 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 –650.)

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




(iv) C2 Negative y-intercept Must be negative. Check substitution into y = mx + c Allow ecf from (c)(iii). (Should be about –0.8.) FOX does not score.

U4 Uncertainty in y-intercept Uses worst gradient and point on WAL. Check method. FOX does not score.

(d) (i) C3 P = R / y-intercept = –gradient / y-intercept

Include unit [Ω] for P and R. Do not penalise POT.

C4 R = gradient in the range 620 to 680 and given to 2 or 3 s.f.

(ii) U5 Percentage uncertainty in P 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)

(c) (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) (i) [U5]

Percentage uncertainty in gradient + percentage uncertainty in y-intercept

max P = intercept ymin

gradientmax

intercept ymin

max

−

=

−

R

min f = interceptmax y

gradient min

interceptmax y

min

−

=

−

R




9702 PHYSICS







1 Planning (15 marks) Defining the problem (3 marks) P r is the independent variable, B is the dependent variable or vary r and measure B. [1] P Keep the number of turns on the coil(s) constant. [1]

Do not accept “same coil”. P Keep the current in the coil constant. [1] Methods of data collection (5 marks) M Diagram showing flat coils and labelled Hall probe positioned at X. Minimum two labels

needed. Solenoids will not be credited. [1] M Workable circuit diagram for coil connected to a (d.c.) power supply and ammeter.

Do not allow a.c. power supply or incorrect circuit diagrams. [1] M Connect Hall probe to voltmeter / c.r.o.

Allow galvanometer but do not allow ammeter. [1] M Measure diameter (radius) with a ruler / vernier calipers. Do not allow micrometer. [1] M Calibrate Hall probe with a known magnetic flux density. [1] Method of analysis (2 marks) A Plot a graph of B against 1 / r [allow lg B against lg r or other valid graph] [1]

A µ 0

IN0.72

gradient= [1]

Safety considerations (1 mark) S Precaution linked to (large) heating of coil, e.g. switch off when not in use to avoid

overheating coil; do not touch coil because it is hot. [1]




Additional detail (4 marks) D Relevant points might include [4]

1 Use large current / large number of turns to create a large magnetic field 2 Use rheostat (to adjust current in circuit) (with ammeter) to keep the current constant 3 Hall probe at right angles to direction of magnetic field / parallel to coils. Allow adjust to obtain

maximum reading 4 Reasoned method to keep Hall probe perpendicular to direction of magnetic field or at X (e.g.

use of set square, fix to rule, optical bench or equivalent) 5 Method to check coils are correctly aligned in parallel 6 Repeat experiment with Hall probe reversed and average 7 Repeat measurement for d (or r) and average 8 Relationship is valid if the graph is a straight line passing through the origin for appropriate

graph [if lg–lg then straight line with gradient = –1 (ignore reference to y-intercept)]

Do not allow vague computer methods. [Total: 15] 2 Analysis, conclusions and evaluation (15 marks)


(a) A1 gradient =

g

24π−

y-intercept = kg

24π

Gradient must be negative.

Allow y-intercept = –gradient × k

(b) T1 (mean) t / s, T / s and T

2 / s2 All column headings to be correct.

T2

31.8 or 31.81

30.8 or 30.80

29.6 or 29.59

28.7 or 28.73

27.8 or 27.77

26.8 or 26.83

Check all values of T

2. Allow a mixture of significant figures.

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

U1 Error bars in d plotted correctly

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

(c) (ii) G2 Line of best fit Lower end of line should pass between (1.60, 27.0) and (1.64,27.0) and upper end of line should pass between (0.44,31.8) and (0.48,31.8).





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

(c) (iii) C1 Gradient of best fit line Must be negative. 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 –4.)

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

(c) (iv) C2 y-intercept FOX does not score. Check substitution into y = mx + c Allow ecf from (c)(iii). (Should be about 33.7.)

U3 Uncertainty in y-intercept Uses worst gradient and point on WAL. Do not check calculation. FOX does not score.

(d) (i) C3 g between 9.20 and 9.90 given to 2 or 3 s.f. and correct unit (m s–2) having used gradient.

mg

24π

−= ; allow N kg–1

C4 k determined correctly with correct unit (m) m

cgck

−

==

24π

(k must be positive.)

(d) (ii) U4 Percentage uncertainty in g

U5 Percentage uncertainty in k Percentage uncertainty in k must be larger than the percentage uncertainty in g.

[Total: 15]

Uncertainties in Question 2 (c) (iii) Gradient [U2] Uncertainty = gradient of line of best fit – gradient of worst acceptable line

Uncertainty = 2

1 (steepest worst line gradient – shallowest worst line gradient)

(c) (iv) [U3]

Uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line

Uncertainty = 2

1 (steepest y-intercept – shallowest y-intercept)




(d) (ii) [U4]

Percentage uncertainty in g = 100100 ×∆

=×∆

g

g

m

m

[U5]

Percentage uncertainty in k = 100100100 ×∆

+×∆

=×∆

cc

g

g

kk

max k = 24

interceptmax max

π

× -yg =

gradient min

interceptmax −y

min k = 24

interceptmin min

π

× -g =

gradientmax

intercept min −y




9702 PHYSICS







1 Planning (15 marks)

Defining the problem (3 marks)

P r is the independent variable or vary r. [1] P T (or t) is the dependent variable or measure T (or t). [1] P Keep the radius of curvature (of the track) or C constant (or radius of track

constant). Do not allow “use same track”. [1] Methods of data collection (5 marks)

M Diagram showing ball in a (curved) track with supports for track, e.g. retort stands. Minimum of two labels (from ball, track, supports; not stopwatch, bench, micrometer). Supports making contact with track higher than ball / at least half way up. [1]

M Measure time using stopwatch or light gates and timer or datalogger with motion

sensor. Detail needed for video camera. [1]

M Use many oscillations (at least 10 or at least 10 s of timing) and determine

T = t / n. [1] M Measure diameter (radius) of ball with a micrometer / vernier calipers. Do not allow travelling microscope. [1] M radius = diameter / 2. [1] Method of analysis (2 marks)

A Plot a graph of T

2 against r (or r against T

2) Do not allow log graphs. [1]

A C = y-intercept × gradient

intercepty

28

5

2

−

=

π

g (or for r against T

2, C = y-intercept) [1]

Safety considerations (1 mark) S Precaution linked to ball escaping on to floor, e.g. use barrier / safety

screen / sand tray to prevent balls rolling on to floor. [1]




Additional detail (4 marks) D Relevant points might include [4] 1 Add weights to / G-clamp retort stands 2 Keep the material / density of the ball constant 3 Use of fiducial marker near centre of track / mark on the track 4 Clean track / balls. Do not allow oil the track. 5 Repeat measurements of t (for each ball) and average 6 Repeat measurement for d (or r) and average 7 Relationship is valid if straight line, provided plotted graph is correct 8 Relationship is valid if straight line not passing through origin or has an intercept,

provided plotted graph is correct (any quoted expression must be correct, e.g.

y-intercept = g

C

5

282

π)

Do not allow vague computer methods. [Total: 15]




2 Analysis, conclusions and evaluation (15 marks)


(a) A1 gradient = –R y-intercept = –R/P

Must be negative.

(b) T1 T2

6.7 or 6.67 –5.1 or –5.13

4.5 or 4.55 –3.8 or –3.75

3.0 or 3.03 –2.8 or –2.75

2.0 or 2.00 –2.1 or –2.06

1.5 or 1.52 –1.8 or –1.75

1.1 or 1.11 –1.5 or –1.50

Allow a mixture of significant figures. T1 and T2 must be table values. Ignore “–” omissions.

U1 From ± 0.3 or ± 0.4 to ± 0.1 or ± 0.2 Allow more than one significant figure.

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

U2 Error bars in V / E 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 (6.4,–5.0) and (6.6,–5.0) and upper end of line should pass between (1.0,–1.5) and (1.2,–1.5).


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 all error bars are plotted.

(iii) C1 Gradient of best fit line Must be negative. 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 –650.)

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




(iv) C2 Negative y-intercept Must be negative. Check substitution into y = mx + c Allow ecf from (c)(iii). (Should be about –0.8.) FOX does not score.

U4 Uncertainty in y-intercept Uses worst gradient and point on WAL. Check method. FOX does not score.

(d) (i) C3 P = R / y-intercept = –gradient / y-intercept

Include unit [Ω] for P and R. Do not penalise POT.

C4 R = gradient in the range 620 to 680 and given to 2 or 3 s.f.

(ii) U5 Percentage uncertainty in P 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)

(c) (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) (i) [U5]

Percentage uncertainty in gradient + percentage uncertainty in y-intercept

max P = intercept ymin

gradientmax

intercept ymin

max

−

=

−

R

min f = interceptmax y

gradient min

interceptmax y

min

−

=

−

R

Education

9702 s14 ms_all