9702 s12 ms_all

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS

GCE Advanced Subsidiary Level and GCE Advanced Level

MARK SCHEME for the May/June 2012 question paper

for the guidance of teachers

9702 PHYSICS

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

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

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

Cambridge is publishing the mark schemes for the May/June 2012 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Mark Scheme: Teachers’ version Syllabus Paper

GCE AS/A LEVEL – May/June 2012 9702 11

© University of Cambridge International Examinations 2012

Question Number

Key Question Number

Key

1 A 21 D

2 D 22 B

3 C 23 C

4 D 24 B

5 C 25 B

6 C 26 C

7 D 27 B

8 B 28 B

9 D 29 C

10 D 30 C

11 B 31 D

12 A 32 C

13 C 33 C

14 D 34 A

15 D 35 C

16 B 36 B

17 C 37 A

18 B 38 C

19 D 39 A

20 D 40 B





9702 PHYSICS








Question Number

Key Question Number

Key

1 C 21 C

2 D 22 C

3 D 23 C

4 B 24 B

5 B 25 C

6 A 26 D

7 A 27 C

8 D 28 D

9 A 29 C

10 A 30 B

11 B 31 C

12 D 32 C

13 A 33 D

14 B 34 B

15 D 35 D

16 D 36 D

17 B 37 D

18 D 38 C

19 B 39 B

20 D 40 B





9702 PHYSICS








Question Number

Key Question Number

Key

1 C 21 B

2 A 22 C

3 D 23 B

4 C 24 D

5 D 25 D

6 D 26 B

7 C 27 C

8 D 28 C

9 B 29 B

10 A 30 C

11 D 31 D

12 B 32 A

13 D 33 C

14 C 34 C

15 D 35 B

16 C 36 C

17 B 37 C

18 D 38 A

19 B 39 B

20 B 40 A





9702 PHYSICS

9702/21 Paper 2 (AS Structured Questions), maximum raw mark 60

This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers.


• Cambridge will not enter into discussions or correspondence in connection with these mark schemes. Cambridge is publishing the mark schemes for the May/June 2012 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.




1 (a) (i) V units: m3 (allow metres cubed or cubic metres) A1 [1]

(ii) Pressure units: kg m s–2 / m2 (allow use of P = ρgh) M1 Units: kg m–1

s–2 A0 [1] (b) V / t units: m3 s–1 B1 Clear substitution of units for P, r4 and l M1

msm

msmkg

8 1

3

4

2

1

1

−

−−

−

==

l tV

rPC

4π

Units: kg m–1

s–1 A1 [3]

(8 or π in final answer –1. Use of dimensions max 2/3) 2 (a) (i) v = u + at C1

= 4.23 + 9.81 × 1.51 M1 = 19.0(4) m s–1 (Allow 2 s.f.) A0 [2] (Use of –g max 1/2. Use of g = 10 max 1/2. Allow use of 9.8. Allow 19 m s–1) (ii) either s = ut + ½ at2 (or v2 = u2 + 2as etc.) = 4.23 × 1.51 + 0.5 × 9.81 × (1.51)2 C1 = 17.6 m (or 17.5 m) A1 [2] (Use of –g here wrong physics (0/2))

(b) (i) F = ∆P / ∆t need idea of change in momentum C1 = [0.0465 × (18.6 + 19)] / 12.5 × 10–3 C1 = 140 N A1 (Use of – sign max 2/4. Ignore –ve sign in answer) Direction: upwards B1 [4]

(ii) h = ½ × (18.6)2 / 9.81 C1 = 17.6 m (2 s.f. –1) A1 [2] (Use of 19 m s–1, 0/2 wrong physics) (c) either kinetic energy of the ball is not conserved on impact or speed before impact is not equal to speed after hence inelastic B1 [1] 3 (a) Resultant force (and resultant torque) is zero B1 Weight (down) = force from/due to spring (up) B1 [2] (b) (i) 0.2, 0.6, 1.0 s (one of these) A1 [1] (ii) 0, 0.8 s (one of these) A1 [1] (iii) 0.2, 0.6, 1.0 s (one of these) A1 [1]




(c) (i) Hooke’s law: extension is proportional to the force (not mass) B1 Linear/straight line graph hence obeys Hooke’s law B1 [2]

(ii) Use of the gradient (not just F = kx) C1 K = (0.4 × 9.8) / 15 × 10–2 M1 = 26(.1) N m–1 A0 [2] (iii) either energy = area to left of line or energy = ½ ke2 C1 = ½ × [(0.4 × 9.8) / 15 × 10–2] × (15 × 10–2)2 C1 = 0.294 J (allow 2 s.f.) A1 [3]

4 (a) (i) R = V2 / P or P = IV and V = IR C1 = (220)2 / 2500

= 19.4 Ω (allow 2 s.f.) A1 [2]

(ii) R = ρl / A C1 l = [19.4 × 2.0 × 10–7] / 1.1 × 10–6 C1 = 3.53 m (allow 2 s.f.) A1 [3] (b) (i) P = 625, 620 or 630 W A1 [1] (ii) R needs to be reduced C1

Either length ¼ of original length or area 4× greater or diameter 2× greater A1 [2]

5 (a) (i) sum of e.m.f.’s = sum of p.d.’s around a loop/circuit B1 [1] (ii) energy B1 [1]

(b) (i) 2.0 = I × (4.0 + 2.5 + 0.5) C1

I = 0.286 A (allow 2 s.f.) A1 [2]

(If total resistance is not 7 Ω, 0/2 marks) (ii) R = [0.90 / 1.0] × 4 (= 3.6) C1

V = I R = 0.286 × 3.6 = 1.03 V A1 [2] (If factor of 0.9 not used, then 0/2 marks) (iii) E = 1.03 V A1 [1] (iv) either no current through cell B or p.d. across r is zero B1 [1] 6 (a) (i) coherence: constant phase difference M1 between (two) waves A1 [2]

(ii) path difference is either λ or nλ

or phase difference is 360° or n × 360° or n2π rad B1 [1]




(iii) path difference is either λ/2 or (n + ½) λ

or phase difference is odd multiple of either 180° or π rad B1 [1]

(iv) w = λD / a C1

= [630 × 10–9 × 1.5] / 0.45 × 10–3 C1

= 2.1 × 10–3 m A1 [3]

(b) no change to dark fringes B1

no change to separation/fringe width B1 bright fringes are brighter/lighter/more intense B1 [3]

7 (a) (i) 2 protons and 2 neutrons B1 [1] (ii) e.g. positively charged 2e mass 4u constant energy absorbed by thin paper or few cm of air (3 cm → 8 cm) (not low penetration) highly ionizing deflected in electric/magnetic fields (One mark for each property, max 2) B2 [2] (b) mass-energy is conserved B1

difference in mass ‘changed’ into a form of energy B1

energy in the form of kinetic energy of the products / γ-radiation photons / e.m. radiation B1 [3]





9702 PHYSICS








1 (a) l 8

4

C

r P

t

V π

=

C = [π × 2.5 × 103 × (0.75 × 10–3)4] / (8 × 1.2 × 10–6 × 0.25) C1 = 1.04 × 10–3 N s m–2 A1 [2] (b) 4 × %r C1 %C = %P + 4 × %r + %V/t + %l = 2% + 5.3% + 0.83% + 0.4% (= 8.6%) A1

∆C = ± 0.089 × 10–3 N s m–2 A1 [3] (c) C = (1.04 ± 0.09) × 10–3 N s m–2 A1 [1] 2 (a) (i) v2 = u2 + 2as = (8.4)2 + 2 × 9.81 × 5 C1 = 12.99 m s–1 (allow 13 to 2 s.f. but not 12.9) A1 [2] (ii) t = (v – u) / a or s = ut + ½at2 = (12.99 – 8.4) / 9.81 or 5 = 8.4t + ½ × 9.81t2 M1 t = 0.468 s A0 [1] (b) reasonable shape M1 suitable scale A1 correctly plotted 1st and last points at (0,8.4) and (0.88 – 0.96,0) with non-vertical line at 0.47 s A1 [3]

(c) (i) 1. kinetic energy at end is zero so ∆KE = ½ mv2 or ∆KE = ½ mu2 – ½ mv2 C1 = ½ × 0.05 × (8.4)2

= (–) 1.8 J A1 [2] 2. final maximum height = (4.2)2 / (2 × 9.8) = (0.9 (m)) change in PE = mgh2 – mgh1 C1 = 0.05 × 9.8 × (0.9 – 5) C1 = (–) 2.0 J A1 [3] (ii) change is – 3.8 (J) B1

energy lost to ground (on impact) / energy of deformation of the ball / thermal energy in ball B1 [2]

3 (a) A body continues at rest or constant velocity unless acted on by a resultant (external) force B1 [1] (b) (i) constant velocity/zero acceleration and therefore no resultant force M1

no resultant force (and no resultant torque) hence in equilibrium A1 [2]

(ii) component of weight = 450 × 9.81 × sin 12° (= 917.8) C1 tension = 650 + 450 g sin12° = (650 + 917.8) C1 = 1600 (1570) N A1 [3]




(iii) work done against frictional force or friction between log and slope M1 output power greater than the gain in PE / s A1 [2]

4 (a) total resistance = 20 (kΩ) C1 current = 12 / 20 (mA) or potential divider formula C1 p.d. = [12 / 20] × 12 = 7.2 V A1 [3]

(b) parallel resistance = 3 (kΩ) C1

total resistance 8 + 3 = 11 (kΩ) C1 current = 12 / 11 × 103 = 1.09 × 10–3 or 1.1 × 10–3

A A1 [3] (c) (i) LDR resistance decreases M1

total resistance (of circuit) is less hence current increases A1 [2] (ii) resistance across XY is less M1

less proportion of 12 V across XY hence p.d. is less A1 [2] 5 (a) E = stress / strain B1 [1] (b) (i) 1. diameter / cross sectional area / radius

2. original length B1 [1] (ii) measure original length with a metre ruler / tape B1 measure the diameter with micrometer (screw gauge) B1 [2] allow digital vernier calipers (iii) energy = ½ Fe or area under graph or ½ kx2 C1

= ½ × 0.25 × 10–3 × 3 = 3.8 × 10–4 J A1 [2]

(c) straight line through origin below original line M1

line through (0.25, 1.5) A1 [2] 6 (a) two waves travelling (along the same line) in opposite directions overlap/meet M1 same frequency / wavelength A1 resultant displacement is the sum of displacements of each wave / produces nodes and antinodes B1 [3] (b) apparatus: source of sound + detector + reflection system B1 adjustment to apparatus to set up standing waves – how recognised B1 measurements made to obtain wavelength B1 [3]

(c) (i) at least two nodes and two antinodes A1 [1]

(ii) node to node = λ / 2 = 34 cm (allow 33 to 35 cm) C1

c = fλ C1 f = 340 / 0.68 = 500 (490 to 520) Hz A1 [3]




7 (a) W = 1 and X = 0 A1 [1] Y = 2 A1 [1] Z = 55 A1 [1] (b) explanation in terms of mass – energy conservation B1 energy released as gamma or photons or kinetic energy of products or em radiation B1 [2]





9702 PHYSICS








1 (a) displacement is a vector, distance is a scalar B1 displacement is straight line between two points / distance is sum of lengths moved / example showing difference B1 [2] (either one of the definitions for the second mark)

(b) a body continues at rest or at constant velocity unless acted on by a resultant (external) force B1 [1]

(c) (i) sum of T1 and T2 equals frictional force B1 these two forces are in opposite directions B1 [2] (allow for 1/2 for travelling in straight line hence no rotation / no resultant torque)

(ii) 1. scale vector triangle with correct orientation / vector triangle with correct orientation both with arrows B1

scale given or mathematical analysis for tensions B1 [2] 2. T1 = 10.1 × 103 (± 0.5 × 103) N A1 T2 = 16.4 × 103 (± 0.5 × 103) N A1 [2]

2 (a) weight = 452 × 9.81

component down the slope = 452 × 9.81 × sin 14° M1 = 1072.7 = 1070 N A0 [1]

(b) (i) F = ma C1 T – (1070 + 525) = 452 × 0.13 C1 T = 1650 (1653.76) N any forces missing 1/3 A1 [3]

(ii) 1. s = ut + ½at2 hence 10 = 0 + ½ × 0.13t2 C1 t = [(2 × 10) / 0.13]1/2 = 12.4 or 12 s A1 [2] 2. v = (0 + 2 × 0.13 × 10)1/2 = 1.61 or 1.6 m s–1 A1 [1]

(c) straight line from the origin B1 line down to zero velocity in short time compared to stage 1 B1 line less steep negative gradient B1 final velocity larger than final velocity in the first part – at least 2× B1 [4]

3 (a) V = h × A

m = V × ρ B1

W = h × A × ρ × g B1 P = F / A B1

P = hρg

P is proportional to h if ρ is constant (and g) B1 [4]

(b) density changes with height B1 hence density is not constant with link to formula B1 [2]




4 (a) electric field strength is the force per unit positive charge (acting on a stationary charge) B1 [1]

(b) (i) E = V / d C1 = 1200 / 14 × 10–3 = 8.57 × 104

V m–1 A1 [2]

(ii) W = QV or W = F × d and therefore W = E × Q × d C1 = 3.2 × 10–19 × 1200 = 3.84 × 10–16

J A1 [2]

(iii) ∆U = mgh C1 = 6.6 × 10–27 × 9.8 × 14 × 10–3 = 9.06 × 10–28

J A1 [2]

(iv) ∆K = 3.84 × 10–16 – ∆U = 3.84 × 10–16

J A1 [1]

(v) K = ½mv2 C1

v = [(2 × 3.8 × 10–16) / 6.6 × 10–27]1/2 = 3.4 × 105

m s–1 A1 [2]

5 (a) (i) sum of currents into a junction = sum of currents out of junction B1 [1]

(ii) charge B1 [1]

(b) (i) ΣE = ΣIR 20 – 12 = 2.0(0.6 + R) (not used 3 resistors 0/2) C1

R = 3.4 Ω A1 [2]

(ii) P = EI C1 = 20 × 2 = 40 W A1 [2]

(iii) P = I2R C1 P = (2)2 × (0.1 + 0.5 + 3.4) = 16 W A1 [2]

(iv) efficiency = useful power / output power C1 24 / 40 = 0.6 or 12 × 2 / 20 × 2 or 60% A1 [2]




6 (a) (i) diffraction bending/spreading of light at edge/slit B1 this occurs at each slit B1 [2]

(ii) constant phase difference between each of the waves B1 [1]

(iii) (when the waves meet) the resultant displacement is the sum of the displacements of each wave B1 [1]

(b) d sinθ = nλ

n = d / λ = 1 / 450 × 103 × 630 × 10–9 C1

n = 3.52 M1 hence number of orders = 3 A1 [3]

(c) λ blue is less than λ red M1 more orders seen A1 each order is at a smaller angle than for the equivalent red A1 [3]

7 (a) thin paper reduces count rate hence α B1

addition of 1 cm of aluminium causes little more count rate reduction hence only

other radiation is γ B1 [2]

(b) magnetic field perpendicular to direction of radiation B1 look for a count rate in expected direction / area if there were negatively charged radiation present. If no count rate recorded then β not present. B1 [2]





9702 PHYSICS

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







1 (a) (iii) Value of x in the range 0.50 – 0.60 m. [1] (b) (ii) Value of T with unit: 0.9 s < T < 1.3 s. [1] Evidence of repeats. [1] (c) Six sets of readings of x and T scores 4 marks, five sets scores 3 marks etc.

Incorrect trend –1. Minor help from Supervisor –1; major help –2. [4] Range of x at least 25 cm. [1] Column headings:

Each column heading must contain a quantity and a unit where appropriate. [1] The unit must conform to accepted scientific convention e.g. x/m or x(m) or x in m.

Consistency of presentation of raw readings: [1]

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

Significant figures: [1] Significant figures for √x should be the same as, or one more than, s.f. for x.

Calculation: √x calculated correctly. [1] (d) (i) Axes: [1]

Sensible scales must be used. Awkward scales (e.g. 3:10) are not allowed. Scales must be chosen so that the plotted points on the grid occupy at least half the graph grid in both x and y directions. Scales must be labelled with the quantity that is being plotted. Scale markings should not be greater than three large squares apart.

Plotting of points: [1] All the observations in the table must be plotted. Check the points are plotted correctly. Work to an accuracy of half a small square. Do not accept ‘blobs’ (points with diameter greater than half a small square).

Quality: [1] All points in the table must be plotted (at least 5) for this mark to be scored. Judge by the scatter of all the points about a straight line. All points must be within 0.04 m½ (0.4 cm½) on the √x axis from a straight line.

(ii) Line of best fit: [1] Judge by the balance of all the points on the grid (at least 5) about the candidate’s line. There must be an even distribution of points either side of the line along the full length. Allow one anomalous point if clearly indicated (e.g. circled or labelled) by the candidate. Line must not be kinked or thicker than half a small square.




(iii) Gradient: [1] The hypotenuse of the triangle must be at least half the length of the drawn line. Both read-offs must be accurate to half a small square in both the x and y directions. Do not allow ∆x/∆y.

y-intercept: [1] Either: Check correct read-off from a point on the line, and substitution into y = mx + c. Read-off must be accurate to half a small square in both the x and y directions. Allow ecf of gradient value.

Or: Check the read-off of the intercept directly from the graph.

(e) Value of P = candidate’s gradient and Q = value of candidate’s intercept. Do not allow fractions. [1]

Unit for P (s m–½ or s cm–½ or s mm–½) consistent with value, and Q (s). [1]

[Total: 20]

2 (a) (iii) Value of F0 with unit. [1] Evidence of repeats. [1] (iv) Absolute uncertainty in F0 in range 0.4 – 1 N.

If repeated readings have been taken, then the uncertainty can be half the range. Correct method of calculation of percentage uncertainty. [1]

(v) Value of µ given to 2 or 3 s.f. [1] (b) (ii) Value of θ with unit to the nearest degree. [1] (iii) Correct calculation of (sin θ + µ cos θ). [1] (c) (ii) Value of F. [1] (d) Second value of θ. [1]

Second value of θ < first value of θ. [1] Second value of F < first value of F. [1] Allow F2 > F1 if θ2 > θ1.

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

criterion. [1]




(f)

(i) Limitations 4 max. (ii) Improvements 4 max. No credit/not enough

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

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

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

B some parts of board rougher than others/ surface of board is uneven/ board not flat

method to ensure same section of board used in each experiment (e.g. mark one section)

board is rough/ there is friction between the block and the board/ use a smoother surface/ references to oil/lubricants

C large (percentage) uncertainty in F

use larger/heavier masses values of F very similar

D difficulty in arranging newton-meter parallel to board/pulling in line with board

use (long) piece of string to connect the newton-meter to the block

newton-meter touching board when attached

E block moves suddenly/without warning (so difficult to read newton-meter at the instant the block starts to move) value of F changes when block moves

use system of pulley and weights/ sand to measure F/ use a newton-meter with a max hold facility/ use video and playback/ use force sensor and datalogger/computer

F board tends to slip/ board not stable/ supporting block can topple

method described to secure board/block/support e.g. clamp the board, fix the supporting block to the bench with tape/blu-tack

G cannot zero newton-meter when used horizontally

use system of pulley and weights/ sand to measure F/ use force sensor and datalogger/computer

zero error in newton-meter

Ignore ‘parallax problems’, ‘use assistant’ or references to draughts, fans, a.c. [Total: 20]





9702 PHYSICS








1 (a) Value of L in range 0.80 m > L > 0.60 m. Consistent with unit. [1] (b) (iii) Value of h0, less than 50 cm, to the nearest mm. [1] (c) Six sets of readings of d and h scores 5 marks, five sets scores 4 marks etc.

Help from Supervisor –1. [5] Range of d: [1] To include 25.0 cm (0.250 m) or more and 10.0 cm (0.100 m) or less Column headings: [1] Each column heading must contain a quantity and a unit The unit must conform to accepted scientific convention e.g. d / m, d(m) or d in m, (h – h0)/m, (L/2 – d)2/m2 Consistency: [1] All values of d and h must be given to the nearest mm. Significant figures: [1] All values of (L/2 – d)2 to 2 or 3 s.f. Calculation: [1] Values of (L/2 – d)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 which is being plotted. Scale markings must be no more than 3 large squares apart. Plotting of points: [1] All observations in the table must be plotted. Diameter of plots must be < half a small square (no blobs). Plots must be accurate to half a small square. Quality: [1] All points in the table must be plotted (at least 5) for this mark to be awarded. Scatter of points must be less than 0.5 cm (0.005 m) of (h – h0) of a straight line.

(ii) Line of best fit: [1]

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




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

Do not allow ∆x / ∆y. 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) Value of a = candidate’s gradient. Value of b = candidate’s intercept. [1]

Unit for a (e.g. m) and b (e.g. m2) consistent with values. [1]

[Total: 20]

2 (b) (i) Value of ball diameter or d to the nearest 0.1 mm (or 0.01 mm). [1]

Values of ball diameter and d in range 5 mm < d < 25 mm. [1] (ii) Absolute uncertainty is between 2 mm and 5 mm. [1]

If repeated readings have been taken, then the absolute uncertainty can be half the range. Correct method shown to find the percentage uncertainty.

(iii) Correct calculation of A with consistent unit. [1] (c) (ii) Value of F, with unit. [1]

Evidence of repeat measurements of F here or in (d)(ii). [1] (d) (ii) Second value of d. [1]

Second value of A is given to the same number of s.f. (or one more s.f.) than d2. [1] Second value of F. [1] Quality: When d increases (second d value is larger than first d value) F also increases (second F value is larger than first F value) and vice versa. [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)


A two results not enough take more readings and plot a graph/ calculate more k values and compare

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

B difficult to form a perfect sphere or disc/diameter of sphere or disc varied

method to make uniform spheres/discs e.g. moulds

pre-sized spheres/ repeat diameter measurement and average

C reason for difficulty in measuring d e.g. viewed through ruler/parallax error in d

method to improve measurement of d e.g. travelling microscope

eyes in line

D difficult to pull newton-meter parallel to ruler/bench

method to ensure force is parallel to ruler e.g. use a long string/pulley and weights*

E difficult to judge reading on newton-meter when detaches with reason e.g. ruler moves suddenly/without warning (so difficult to read newton-meter at the instant the ruler starts to move)/force drops to zero immediately after detachment

method to read force at detachment e.g. newton meter with a ‘max hold’ facility/video and playback or freeze frame/ use system of pulley and weights or sand to measure F*/ use force sensor and datalogger or computer*

video to take reading/ digital (electronic) newton meter/ parallax related to newton meter/ difficult to measure force/ issue of viewing ruler and meter simultaneously

F contact area less than calculated disc area/bulging disc

G difficult to zero newton-meter when used horizontally

improved method to measure F: e.g. use system of pulley and weights or sand*/use force sensor with datalogger or computer*

zero error in newton-meter/ just a pulley

Do not allow: reaction time/human error/using vernier caliper/helpers/use of micrometer screw gauge/effect of temperature/change in stickiness of Blu-Tack. *This answer can be credited as D, E or G (but not more than once).

[Total: 20]





9702 PHYSICS








1 (a) (ii) Value of h0 in range 0.70 m > h0 > 0.50 m. Consistent with unit. [1] (b) (iii) Value of h, less than h0 in (a)(ii), with unit. [1] (c) Five sets of readings of h and m scores 5 marks, four sets scores 4 marks etc.

Major help from Supervisor –2 (setting up apparatus). Minor help from Supervisor –1. [5] Range of m: [1] To include 0.350 kg. Column headings: [1]

Each column heading must contain a quantity and a unit. The unit must conform to accepted scientific convention e.g. m / kg, m(kg) or m in kg, (h0 – h)/m / m kg–1, 1/m / kg–1

Consistency: [1]

All values of h must be given to the nearest mm.

Significant figures: [1] Significant figures for every row of values of 1/m same as or one greater than m as recorded in the table.

Calculation: [1]

Values of (h0 – h)

/m 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 of points: [1]

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

Quality: [1]

All points in the table must be plotted (at least 4) for this mark to be awarded. Scatter of points must be less than 0.5 kg–1 (0.0005 g–1) of 1/m 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 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.

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.

Do not allow ∆x / ∆y.




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) Value of P = candidate’s gradient. Value of Q = candidate’s intercept. [1] Unit for P (e.g. m) consistent with value, and Q (m kg–1) [1] [Total: 20]

2 (b) (ii) Value of θ0 to the nearest degree or 0.5° in range 70° # θ # 80° [1]

(iii) Value of θ with unit, θ < θ0 [1]

(iv) Correct calculation of (θ0 – θ) [1]

(c) (i) Value of raw d with unit to nearest mm. [1]

(ii) Absolute uncertainty in 2 mm < d < 5 mm. [1] If repeated readings have been taken, then the absolute uncertainty can be half the range. Correct method shown to find the percentage uncertainty.

(d) Second value of θ0 within 1 ºC of first value of θ0. [1] Second value of θ. [1]

Second value of ∆θ > first value of ∆θ (check second value of d > first value of d). [1] Evidence of repeat readings of d here or in (c)(i). [1]

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

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

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




(f)


A two results not enough

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


B heat lost through sides and /or bottom

method to reduce heat loss/ lag/ insulate/ polystyrene container

use of lid/ heat loss in warming bowl/cup/ draughts/ heat loss to surroundings

C temperature change is small/

∆θ values too close

time for longer/ higher starting temperature/ greater range of surface areas

D large (percentage) uncertainty

in ∆θ

use thermometer with greater sensitivity or precision/ use thermometer that can read to 0.1 ºC

use more accurate thermometer/ thermometer not precise enough/ not just ‘digital thermometer’

E water in bowl barely covers (bulb of) thermometer

use larger volume of water/ use of thermocouple/ other small temperature sensor (e.g. probe)

not just ‘digital thermometer’ any reference to stirrer/ non-uniform temperature/ thermometer touching base

F parallax error in measuring d / reason for difficulty in access in measuring d

use dividers/calipers string measurements to measure d

G difficult to mark level with reason

method of making mark stay e.g. depth gauge/ calibrated marks/ marker on outside

Do not allow: use of coloured ink/reaction time/fans/draughts/water left behind/beakers not

accurate/ helpers. [Total: 20]





9702 PHYSICS








1 (a) (iii) Value for I0 in range 2.0 to 4.0 mA, with unit. [1]

(b) (ii) First value of I (greater than I0). [1]

(c) Six sets of readings of R and I scores 5 marks, five sets scores 4 marks etc. [5] Major help from Supervisor –2. Minor help from Supervisor –1. Incorrect trend then –1. Range: [1]

Values of R must include 0.22 kΩ or 0.33 kΩ and 3.3 kΩ or 4.7 kΩ. Column headings: [1] Each column heading must contain a quantity and a unit. There must be some distinguishing mark between the quantity and the unit. Consistency: [1]

Values of I must be given either all to the nearest 0.1 mA or all to the nearest 0.01 mA. Significant figures: [1] Every value of 1/R must be given to either 2 or 3 significant figures. Calculated values: [1] 1/R calculated correctly.

(d) (i) Axes: [1] Sensible scales must be used (no awkward scales such as 3:10). Scales must be chosen so that the plotted points must occupy at least half the graph grid in both x and y directions. Scales must be labelled with the quantity which is being plotted. Scale markings must be no more than 3 large squares apart. Plotting of points: [1] All observations in the table must be plotted. Diameter of plots must be < half a small square (no blobs). Plotting must be accurate to half a small square.

Quality: [1]

Range of I must be at least 2 mA, and all points must be within 0.5 mA of a straight line. All points in the table must be plotted (at least 5) for this mark to be scored.

(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. One anomalous point is allowed only if clearly indicated (i.e. circled or labelled) by the candidate. Line must not be kinked or thicker than half a small square.




(d) (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.

Do not allow ∆x / ∆y. y-intercept: [1] Either: Correct read off from a point on the line is substituted into y = mx + c. Or: Check read-off of the intercept directly from the graph.

(e) Calculation of b is correct, [1] i.e. b = (candidate’s gradient value)/(candidate’s intercept value).

Value for b in range 0.8 kΩ to 1.2 kΩ, with unit. [1] [Total: 20] 2 (b) Value of d in range 0.80 to 0.99 mm, to nearest 0.01 mm, with unit. [1]

Evidence of repeated measurements for d. [1] (c) Percentage uncertainty in d based on absolute uncertainty of 0.01 mm. [1]

Correct calculation to get percentage uncertainty.

(d) (iv) Value of θ in range 91° to 180° to nearest degree, with unit. [1]

Evidence of repeated measurements for θ. [1]

(v) Correct calculation of sin(180° – θ ). [1]

sin(180° – θ ) given to 2 or 3 s.f. [1] (e) Second value of d. [1]

Second value of θ. [1]

Quality: θ larger for smaller d. [1] (f) (i) Correct calculation of two values of k. [1] (ii) Valid conclusion based on the calculated values of k. Candidate must test correctly

against a stated criterion. [1]




(g)


A two results not enough take more readings and plot a graph/ calculate more k values and compare


B θ (or angle, or scale reading, or protractor reading, or pointer reading) is difficult to measure, with reason linked to rapid motion or short time

video and view playback/ slow motion camera/ video to read angle/ add a ‘max hold’ pointer/ angle sensor with data logger (or computer)

just ‘use a computer’/ ‘reading’ difficult to measure

C parallax error in θ measurement

use mirror scale/ description of method to reduce error

view at right angles/ trial and improvement

D θ (or reading) is difficult (or inaccurate, or imprecise) because pointer is thick

– use thinner pointer/ use larger scale

E pointer attachment moves description of secure method of attachment

F – description of method of fixing block to bench

[Total: 20]





9702 PHYSICS








1 (b) (ii) Ammeter reading with unit, in range 1 mA < I < 1 A. Must see n = 3. [1]

(c) Six sets of readings of I and n scores 5 marks, five sets scores 4 marks etc.

Incorrect trend then –1. Correct trend is I decreases as n increases. Major help from Supervisor –2. Minor help from Supervisor –1. [5]

Range of 6 or 7. [1]

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

The unit must conform to accepted scientific convention e.g. I / A, I (A), I in A, n + 1 /I / A–1. Consistency: [1]

All values of I must be given to the nearest 0.1 mA or better. Significant figures: [1]

Significant figures for every row of values of (n + 1) / I same as or one greater than s.f. in I, as recorded in the table.

Calculation: [1]

Values of (n + 1) / I 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 must 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 of points: [1]

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

Quality: [1]

Judge by scatter of all points about best fit line. All points in the table must be plotted for this mark to be scored. At least 5 plots needed.

All points must be within 0.2 of n from a best 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 x and y directions.

Do not allow ∆x / ∆y.




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 intercept directly from the graph.

(e) Value of P = candidate’s gradient. Value of Q = candidate’s intercept. [1]

Do not allow fractions.

(f) Value of V in range 1V ≤ V ≤ 2V. [1]

(g) R with appropriate unit Ω or V A–1. Expect 50 Ω or 0.05 V mA–1 or 0.05 kΩ [1] [Total: 20]

2 (b) (ii) Value of x with unit to the nearest mm in range: 40.0 cm ≤ x ≤ 60.0 cm. [1] (c) (ii) Value of x1 with consistent unit. [1] (iii) Correct calculation of d1 with unit. [1] (iv) Absolute uncertainty in d1 in range 2 – 5 mm. [1]

If repeated readings have been taken, then the absolute uncertainty can be half the range. Correct method shown to find the percentage uncertainty

(d) (ii) Value of x2. [1]

(e) (iii) Value of 1 s < T < 4 s. [1]

Evidence of repeats. [1] (f) Second value of T. [1]

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

(g) (i) Two values of k calculated correctly. [1] (ii) Justification of sf in k linked to significant figures in d and T. [1] (iii) Sensible comment relating to the calculated values of k, testing against a criterion

specified by the candidate. [1]




(h)


A two results not enough

take more readings with discs of other materials / mass and plot a graph/ calculate more k values and compare

repeat readings few readings

B reason why difficult to record/ measure x2/x1 directly

use a taller /narrower shape take measurement to each end and average/ hole in middle to see x1/x2/ hang masses with string

C difficult to get circular shape/flat top/ same shape/ two shapes not the same because of groove in 100 g mass

use a mould/ use a plane surface to press down on plasticine

use rubber masses

D pivot/100 g mass moved while x2 being determined

method of securing 100 g mass to rule/ rubber pivot

fix pivot and ruler

E oscillation not in one plane only

F difficult to determine end/start of oscillation/ difficult to turn through 90° each time

use of (fiducial) marker(s)/ video with timer

use a protractor

[Total: 20]





9702 PHYSICS

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







Section A 1 (a) work done in bringing unit mass from infinity (to the point) B1 [1] (b) gravitational force is (always) attractive B1 either as r decreases, object/mass/body does work or work is done by masses as they come together B1 [2] (c) either force on mass = mg (where g is the acceleration of free fall /gravitational field strength) B1 g = GM/r2 B1 if r @ h, g is constant B1 ∆EP = force × distance moved M1 = mgh A0 or ∆EP = m∆φ (C1) = GMm(1/r1 – 1/r2) = GMm(r2 – r1)/r1r2 (B1) if r2 ≈ r1, then (r2 – r1) = h and r1r2 = r2 (B1) g = GM/r2 (B1) ∆EP = mgh (A0) [4] (d) ½mv2 = m∆φ v2 = 2 × GM/r C1 = (2 × 4.3 × 1013) / (3.4 × 106) C1 v = 5.0 × 103

m s–1 A1 [3] (Use of diameter instead of radius to give v = 3.6 × 103 m s–1 scores 2 marks) 2 (a) (i) either random motion or constant velocity until hits wall/other molecule B1 [1] (ii) (total) volume of molecules is negligible M1 compared to volume of containing vessel A1 or radius/diameter of a molecule is negligible (M1) compared to the average intermolecular distance (A1) [2] (b) either molecule has component of velocity in three directions or c2 = cX

2 + cY2 + cZ

2 M1 random motion and averaging, so <cX

2> = <cY2> = <cZ

2> M1 <c2> = 3<cX

2> A1 so, pV = ⅓Nm<c2> A0 [3]

(c) <c2> ∝ T or crms ∝ T C1

temperatures are 300 K and 373 K C1 crms = 580 m s–1 A1 [3] (Do not allow any marks for use of temperature in units of ºC instead of K)




3 (a) (numerically equal to) quantity of (thermal) energy required to change the state of unit mass of a substance M1 without any change of temperature A1 [2] (Allow 1 mark for definition of specific latent heat of fusion/vaporisation) (b) either energy supplied = 2400 × 2 × 60 = 288000 J C1 energy required for evaporation = 106 × 2260 = 240000 J C1 difference = 48000 J rate of loss = 48000 / 120 = 400 W A1 or energy required for evaporation = 106 × 2260 = 240000 J (C1) power required for evaporation = 240000 / (2 × 60) = 2000 W (C1) rate of loss = 2400 – 2000 = 400 W (A1) [3] 4 (a) a = (–)ω2x and ω = 2π/T C1 T = 0.60 s C1 a = (4π2 × 2.0 × 10–2) / (0.6)2 = 2.2 m s–2 A1 [3] (b) sinusoidal wave with all values positive B1 all values positive, all peaks at EK and energy = 0 at t = 0 B1 period = 0.30 s B1 [3] 5 (a) force per unit positive charge acting on a stationary charge B1 [1] (b) (i) E = Q / 4πε0r

2 C1 Q = 1.8 × 104 × 102 × 4π × 8.85 × 10–12 × (25 × 10–2)2 M1 Q = 1.25 × 10–5

C = 12.5 µC A0 [2] (ii) V = Q / 4πε0r = (1.25 × 10–5) / (4π × 8.85 × 10–12 × 25 × 10–2) C1 = 4.5 × 105

V A1 [2] (Do not allow use of V = Er unless explained)




6 (a) (i) peak voltage = 4.0 V A1 [1] (ii) r.m.s. voltage (= 4.0/√2) = 2.8 V A1 [1] (iii) period T = 20 ms M1 frequency = 1 / (20 × 10–3) M1 frequency = 50 Hz A0 [2] (b) (i) change = 4.0 – 2.4 = 1.6 V A1 [1] (ii) ∆Q = C∆V or Q = CV C1 = 5.0 × 10–6 × 1.6 = 8.0 × 10–6

C A1 [2] (iii) discharge time = 7 ms C1 current = (8.0 × 10–6) / (7.0 × 10–3) M1 = 1.1(4) × 10–3

A A0 [2] (c) average p.d. = 3.2 V C1 resistance = 3.2 / (1.1 × 10–3) = 2900 Ω (allow 2800 Ω) A1 [2] 7 (a) sketch: concentric circles (minimum of 3 circles) M1 separation increasing with distance from wire A1 correct direction B1 [3] (b) (i) arrow direction from wire B towards wire A B1 [1] (ii) either reference to Newton’s third law or force on each wire proportional to product of the two currents M1 so forces are equal A1 [2] (c) force always towards wire A/always in same direction B1 varies from zero (to a maximum value) (1) variation is sinusoidal / sin2 (1) (at) twice frequency of current (1) (any two, one each) B2 [3] 8 (a) packet/quantum/discrete amount of energy M1 of electromagnetic radiation A1 (allow 1 mark for ‘packet of electromagnetic radiation’) energy = Planck constant × frequency (seen here or in b) B1 [3] (b) each (coloured) line corresponds to one wavelength/frequency B1 energy = Planck constant × frequency implies specific energy change between energy levels B1 so discrete levels A0 [2]




9 (a) (i) either probability of decay (of a nucleus) M1 per unit time A1 [2] or λ = (–)(dN/dt) / N (M1) (–)dN/dt and N explained (A1) (ii) in time t½, number of nuclei changes from N0 to ½N0 B1 ½ = exp(–λ t½) or 2 = exp (λ t½) B1 ln (½) = –λ t½ and ln (½) = –0.693 or ln 2 = λ t½ and ln 2 = 0.693 B1 0.693 = λ t½ A0 [3] (b) 228 = 538 exp(–8λ) C1 λ = 0.107 (hours–1) C1 t½ = 6.5 hours (do not allow 3 or more SF) A1 [3] (c) e.g. random nature of decay background radiation daughter product is radioactive (any two sensible suggestions, 1 each) B2 [2]




Section B 10 (a) light-dependent resistor (allow LDR) B1 [1] (b) (i) two resistors in series between +5 V line and earth M1 midpoint connected to inverting input of op-amp A1 [2] (ii) relay coil between diode and earth M1 switch between lamp and earth A1 [2] (c) (i) switch on/off mains supply using a low voltage/current output B1 [1] (allow ‘isolates circuit from mains supply’) (ii) relay will switch on for one polarity of output (voltage) C1 switches on when output (voltage) is negative A1 [2] 11 (a) (i) e.m. radiation produced whenever charged particle is accelerated M1 electrons hitting target have distribution of accelerations A1 [2] (ii) either wavelength shorter/shortest for greater/greatest acceleration or λmin = hc/ Emax or minimum wavelength for maximum energy B1 all electron energy given up in one collision/converted to single photon B1 [2] (b) (i) hardness measures the penetration of the beam C1 greater hardness, greater penetration A1 [2] (ii) controlled by changing the anode voltage C1 higher anode voltage, greater penetration/hardness A1 [2] (c) (i) long-wavelength radiation more likely to be absorbed in the body/less likely to penetrate through body B1 [1] (ii) (aluminium) filter/metal foil placed in the X-ray beam B1 [1] 12 (a) strong uniform (magnetic) field M1 either aligns nuclei or gives rise to Larmor/resonant frequency in r.f. region A1 non-uniform (magnetic) field M1 either enables nuclei to be located or changes the Larmor/resonant frequency A1 [4] (b) (i) difference in flux density = 2.0 × 10–2 × 3.0 × 10–3 = 6.0 × 10–5

T A1 [1] (ii) ∆f = 2 × c × ∆B C1 = 2 × 1.34 × 108 × 6.0 × 10–5 = 1.6 × 104 Hz A1 [2]




13 (a) (i) no interference (between signals) near boundaries (of cells) B1 [1]

(ii) for large area, signal strength would have to be greater and this could be hazardous to health B1 [1]

(b) mobile phone is sending out an (identifying) signal M1 computer/cellular exchange continuously selects cell/base station with strongest signal A1 computer/cellular exchange allocates (carrier) frequency (and slot) A1 [3]





9702 PHYSICS








Section A 1 (a) force proportional to product of masses and inversely proportional to square of separation (do not allow square of distance/radius) M1 either point masses or separation @ size of masses A1 [2] (b) (i) ω = 2π / (27.3 × 24 × 3600) or 2π / (2.36 x 106) M1 = 2.66 × 10–6

rad s–1 A0 [1] (ii) GM = r3

ω2 or GM = v2r C1

M = (3.84 × 105 × 103)3 × (2.66 × 10–6)2 / (6.67 × 10–11) M1 = 6.0 × 1024

kg A0 [2] (special case: uses g = GM/r2 with g = 9.81, r = 6.4 × 106 scores max 1 mark) (c) (i) grav. force = (6.0 × 1024) × (7.4 × 1022) × (6.67 × 10–11)/(3.84 × 108)2 C1 = 2.0 × 1020

N (allow 1 SF) A1 [2] (ii) either ∆EP = Fx because F constant as x ! radius of orbit B1 ∆EP = 2.0 × 1020 × 4.0 × 10–2 C1 = 8.0 × 1018

J (allow 1 SF) A1 [3]

or ∆EP = GMm/r1 – GMm/r2 C1 Correct substitution B1 8.0 × 1018

J A1 (∆EP = GMm/r1 + GMm/r2 is incorrect physics so 0/3) 2 (a) energy = ½mω

2a2 and ω = 2πf C1 = ½ × 37 × 10–3 × (2π × 3.5)2 × (2.8 × 10–2)2 M1 = 7.0 × 10–3

J A0 [2] (allow 2π × 3.5 shown as 7π) Energy = ½ mv2 and v = rω (C1) Correct substitution (M1) Energy = 7.0 × 10–3

J (A0) (b) EK = EP ½mω

2 (a2 – x2) = ½mω2x2 or EK or EP = 3.5 mJ C1

x = a/√2 = 2.8 /√2 or EK = ½mω2 (a2 – x2) or EP = ½mω

2x2 C1 = 2.0 cm A1 [3] (EK or EP = 7.0 mJ scores 0/3) Allow: k = 17.9 (C1) E = ½ kx2 (C1) x = 2.0 cm (A1)




(c) (i) graph: horizontal line, y-intercept = 7.0 mJ with end-points of line at +2.8 cm and –2.8 cm B1 [1] (ii) graph: reasonable curve B1

with maximum at (0,7.0) end-points of line at (–2.8, 0) and (+2.8, 0) B1 [2]

(iii) graph: inverted version of (ii) M1 with intersections at (–2.0, 3.5) and (+2.0, 3.5) A1 [2]

(Allow marks in (iii), but not in (ii), if graphs K & P are not labelled) (d) gravitational potential energy B1 [1] 3 (a) sum of potential energy and kinetic energy of atoms/molecules/particles M1 reference to random (distribution) A1 [2] (b) (i) as lattice structure is ‘broken’/bonds broken/forces between molecules reduced (not molecules separate) B1 no change in kinetic energy, potential energy increases M1 internal energy increases A1 [3] (ii) either molecules/atoms/particles move faster/ <c2> is increasing

or kinetic energy increases with temperature (increases) B1 no change in potential energy, kinetic energy increases M1 internal energy increases A1 [3] 4 (a) (i) as r decreases, energy decreases/work got out (due to) M1 attraction so point mass is negatively charged A1 [2] (ii) electric potential energy = charge × electric potential B1 electric field strength is potential gradient B1 field strength = gradient of potential energy graph/charge A0 [2] (b) tangent drawn at (4.0, 14.5) B1 gradient = 3.6 × 10–24 A2 (for < ±0.3 allow 2 marks, for < ±0.6 allow 1 mark) field strength = (3.6 × 10–24) / (1.6 × 10–19) = 2.3 × 10–5

V m–1 (allow ecf from gradient value) A1 [4] (one point solution for gradient leading to 2.3 × 10–5

Vm–1 scores 1 mark only)




5 (a) (long) straight conductor carrying current of 1 A M1 current/wire normal to magnetic field M1 (for flux density 1 T,) force per unit length is 1 N m–1 A1 [3] (b) (i) (originally) downward force on magnet (due to current) B1 by Newton’s third law (allow “N3”) M1 upward force on wire A1 [3] (ii) F = BIL 2.4 × 10–3 × 9.8 = B × 5.6 × 6.4 × 10–2 C1 B = 0.066 T (need 2 SF) A1 [2] (g missing scores 0/2, but g = 10 leading to 0.067T scores 1/2) (c) new reading is 2.4√2 g C1 either changes between +3.4 g and –3.4 g or total change is 6.8 g A1 [2] 6 (a) oil drop charged by friction/beta source B1 between parallel metal plates B1 plates are horizontal (1) adjustable potential difference/field between plates B1 until oil drop is stationary B1 mg = q × V/d B1 symbols explained (1) oil drop viewed through microscope (1) m determined from terminal speed of drop (when p.d. is zero) (1) (any two extras, 1 each) B2 [7] (b) 3.2 × 10–19

C A1 [1] 7 (a) minimum energy to remove an electron from the metal/surface B1 [1] (b) gradient = 4.17 × 10–15 (allow 4.1 → 4.3) C1 h = 4.15 × 10–15 × 1.6 × 10–19 or h = 4.1 to 4.3 × 10–15 eV s A1 = 6.6 × 10–34

J s A0 [2] (c) graph: straight line parallel to given line with intercept at any higher frequency B1 intercept at between 6.9 × 1014

Hz and 7.1 × 1014 Hz B1 [3]




8 (a) nuclei having same number of protons/proton (atomic) number B1 different numbers of neutrons/neutron number B1 [2] (allow second mark for nucleons/nucleon number/mass number/atomic mass if made clear that same number of protons/proton number) (b) probability of decay per unit time is the decay constant C1 λ = ln 2 / t½ = 0.693 / (52 × 24 × 3600) C1 = 1.54 × 10–7

s–1 A1 [3] (c) (i) A = A0 exp(–λt) 7.4 × 106 = A0 exp(–1.54 × 10–7 × 21 × 24 × 3600) C1 A0 = 9.8 × 106

Bq A1 [2] (alternative method uses 21 days as 0.404 half-lives) (ii) A = λN and mass = N × 89 / NA C1 mass = (9.8 × 106 × 89) / (1.54 × 10–7 × 6.02 × 1023) = 9.4 × 10–9

g A1 [2]




Section B 9 (a) e.g. infinite input impedance/resistance zero output impedance/resistance infinite (open loop) gain infinite bandwidth infinite slew rate (any four, one mark each) B4 [4] (b) graph: square wave M1 180° phase change A1 amplitude 5.0 V A1 [3] (c) correct symbol for LED M1 diodes connected correctly between VOUT and earth A1 diodes identified correctly A1 [3] (special case: if diode symbol, not LED symbol, allow 2nd and 3rd marks to be scored) 10 (a) e.g. beam is divergent/obeys inverse square law absorption (in block) scattering (of beam in block) reflection (at boundaries) (any two sensible suggestions, 1 each) B2 [2]

(b) (i) I = I0

exp(–µx) C1 I0/I

= exp(0.27 × 2.4) = 1.9 A1 [2] (ii) I0/I = exp(0.27 × 1.3) × exp(3.0 × 1.1) C1 = 1.42 × 27.1 = 38.5 A1 [2] (c) either much greater absorption in bone than in soft tissue or Io / I much greater for bone than soft tissue B1 [1] 11 (a) (i) loss of (signal) power B1 [1] (ii) unwanted power (on signal) M1 that is random A1 [2] (b) for digital, only the ‘high’ and the ‘low’ / 1 and 0 are necessary M1 variation between ‘highs’ and ‘lows’ caused by noise not required A1 [2] (c) attenuation = 10 lg(P2 / P1) C1 either 195 = 10 lg(2.4 × 103 / P) or –195 = 10 lg(P / 2.4 × 103) C1 P = 7.6 × 10–17

W A1 [3]




12 (a) (i) modulator B1 [1] (ii) serial-to-parallel converter (accept series-to-parallel converter) B1 [1] (b) (i) enables one aerial to be used for transmission and receipt of signals A1 [1] (ii) all bits for one number arrive at one time B1 bits are sent out one after another B1 [2]





9702 PHYSICS








Section A 1 (a) work done in bringing unit mass from infinity (to the point) B1 [1] (b) gravitational force is (always) attractive B1 either as r decreases, object/mass/body does work or work is done by masses as they come together B1 [2] (c) either force on mass = mg (where g is the acceleration of free fall /gravitational field strength) B1 g = GM/r2 B1 if r @ h, g is constant B1 ∆EP = force × distance moved M1 = mgh A0 or ∆EP = m∆φ (C1) = GMm(1/r1 – 1/r2) = GMm(r2 – r1)/r1r2 (B1) if r2 ≈ r1, then (r2 – r1) = h and r1r2 = r2 (B1) g = GM/r2 (B1) ∆EP = mgh (A0) [4] (d) ½mv2 = m∆φ v2 = 2 × GM/r C1 = (2 × 4.3 × 1013) / (3.4 × 106) C1 v = 5.0 × 103

m s–1 A1 [3] (Use of diameter instead of radius to give v = 3.6 × 103 m s–1 scores 2 marks) 2 (a) (i) either random motion or constant velocity until hits wall/other molecule B1 [1] (ii) (total) volume of molecules is negligible M1 compared to volume of containing vessel A1 or radius/diameter of a molecule is negligible (M1) compared to the average intermolecular distance (A1) [2] (b) either molecule has component of velocity in three directions or c2 = cX

2 + cY2 + cZ

2 M1 random motion and averaging, so <cX

2> = <cY2> = <cZ

2> M1 <c2> = 3<cX

2> A1 so, pV = ⅓Nm<c2> A0 [3]

(c) <c2> ∝ T or crms ∝ T C1

temperatures are 300 K and 373 K C1 crms = 580 m s–1 A1 [3] (Do not allow any marks for use of temperature in units of ºC instead of K)




3 (a) (numerically equal to) quantity of (thermal) energy required to change the state of unit mass of a substance M1 without any change of temperature A1 [2] (Allow 1 mark for definition of specific latent heat of fusion/vaporisation) (b) either energy supplied = 2400 × 2 × 60 = 288000 J C1 energy required for evaporation = 106 × 2260 = 240000 J C1 difference = 48000 J rate of loss = 48000 / 120 = 400 W A1 or energy required for evaporation = 106 × 2260 = 240000 J (C1) power required for evaporation = 240000 / (2 × 60) = 2000 W (C1) rate of loss = 2400 – 2000 = 400 W (A1) [3] 4 (a) a = (–)ω2x and ω = 2π/T C1 T = 0.60 s C1 a = (4π2 × 2.0 × 10–2) / (0.6)2 = 2.2 m s–2 A1 [3] (b) sinusoidal wave with all values positive B1 all values positive, all peaks at EK and energy = 0 at t = 0 B1 period = 0.30 s B1 [3] 5 (a) force per unit positive charge acting on a stationary charge B1 [1] (b) (i) E = Q / 4πε0r

2 C1 Q = 1.8 × 104 × 102 × 4π × 8.85 × 10–12 × (25 × 10–2)2 M1 Q = 1.25 × 10–5

C = 12.5 µC A0 [2] (ii) V = Q / 4πε0r = (1.25 × 10–5) / (4π × 8.85 × 10–12 × 25 × 10–2) C1 = 4.5 × 105

V A1 [2] (Do not allow use of V = Er unless explained)




6 (a) (i) peak voltage = 4.0 V A1 [1] (ii) r.m.s. voltage (= 4.0/√2) = 2.8 V A1 [1] (iii) period T = 20 ms M1 frequency = 1 / (20 × 10–3) M1 frequency = 50 Hz A0 [2] (b) (i) change = 4.0 – 2.4 = 1.6 V A1 [1] (ii) ∆Q = C∆V or Q = CV C1 = 5.0 × 10–6 × 1.6 = 8.0 × 10–6

C A1 [2] (iii) discharge time = 7 ms C1 current = (8.0 × 10–6) / (7.0 × 10–3) M1 = 1.1(4) × 10–3

A A0 [2] (c) average p.d. = 3.2 V C1 resistance = 3.2 / (1.1 × 10–3) = 2900 Ω (allow 2800 Ω) A1 [2] 7 (a) sketch: concentric circles (minimum of 3 circles) M1 separation increasing with distance from wire A1 correct direction B1 [3] (b) (i) arrow direction from wire B towards wire A B1 [1] (ii) either reference to Newton’s third law or force on each wire proportional to product of the two currents M1 so forces are equal A1 [2] (c) force always towards wire A/always in same direction B1 varies from zero (to a maximum value) (1) variation is sinusoidal / sin2 (1) (at) twice frequency of current (1) (any two, one each) B2 [3] 8 (a) packet/quantum/discrete amount of energy M1 of electromagnetic radiation A1 (allow 1 mark for ‘packet of electromagnetic radiation’) energy = Planck constant × frequency (seen here or in b) B1 [3] (b) each (coloured) line corresponds to one wavelength/frequency B1 energy = Planck constant × frequency implies specific energy change between energy levels B1 so discrete levels A0 [2]




9 (a) (i) either probability of decay (of a nucleus) M1 per unit time A1 [2] or λ = (–)(dN/dt) / N (M1) (–)dN/dt and N explained (A1) (ii) in time t½, number of nuclei changes from N0 to ½N0 B1 ½ = exp(–λ t½) or 2 = exp (λ t½) B1 ln (½) = –λ t½ and ln (½) = –0.693 or ln 2 = λ t½ and ln 2 = 0.693 B1 0.693 = λ t½ A0 [3] (b) 228 = 538 exp(–8λ) C1 λ = 0.107 (hours–1) C1 t½ = 6.5 hours (do not allow 3 or more SF) A1 [3] (c) e.g. random nature of decay background radiation daughter product is radioactive (any two sensible suggestions, 1 each) B2 [2]




Section B 10 (a) light-dependent resistor (allow LDR) B1 [1] (b) (i) two resistors in series between +5 V line and earth M1 midpoint connected to inverting input of op-amp A1 [2] (ii) relay coil between diode and earth M1 switch between lamp and earth A1 [2] (c) (i) switch on/off mains supply using a low voltage/current output B1 [1] (allow ‘isolates circuit from mains supply’) (ii) relay will switch on for one polarity of output (voltage) C1 switches on when output (voltage) is negative A1 [2] 11 (a) (i) e.m. radiation produced whenever charged particle is accelerated M1 electrons hitting target have distribution of accelerations A1 [2] (ii) either wavelength shorter/shortest for greater/greatest acceleration or λmin = hc/ Emax or minimum wavelength for maximum energy B1 all electron energy given up in one collision/converted to single photon B1 [2] (b) (i) hardness measures the penetration of the beam C1 greater hardness, greater penetration A1 [2] (ii) controlled by changing the anode voltage C1 higher anode voltage, greater penetration/hardness A1 [2] (c) (i) long-wavelength radiation more likely to be absorbed in the body/less likely to penetrate through body B1 [1] (ii) (aluminium) filter/metal foil placed in the X-ray beam B1 [1] 12 (a) strong uniform (magnetic) field M1 either aligns nuclei or gives rise to Larmor/resonant frequency in r.f. region A1 non-uniform (magnetic) field M1 either enables nuclei to be located or changes the Larmor/resonant frequency A1 [4] (b) (i) difference in flux density = 2.0 × 10–2 × 3.0 × 10–3 = 6.0 × 10–5

T A1 [1] (ii) ∆f = 2 × c × ∆B C1 = 2 × 1.34 × 108 × 6.0 × 10–5 = 1.6 × 104 Hz A1 [2]




13 (a) (i) no interference (between signals) near boundaries (of cells) B1 [1]

(ii) for large area, signal strength would have to be greater and this could be hazardous to health B1 [1]

(b) mobile phone is sending out an (identifying) signal M1 computer/cellular exchange continuously selects cell/base station with strongest signal A1 computer/cellular exchange allocates (carrier) frequency (and slot) A1 [3]





9702 PHYSICS

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







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

P1 Frequency or period of rotation or ω is the independent variable and θ is the dependent variable

or vary f or T or ω and measure θ. [1]

P2 ω = 2πf = 2π/T [1] P3 Keep the length of the rigid rod constant; ignore reference to mass. [1] Methods of data collection (5 marks) M1 Labelled diagram of apparatus: small object, pole attached to a rotating device

(motor, turntable). [1] M2 Method to change the speed of the rotating device. [1] M3 Method to determine frequency or time period (e.g. stop watch to time a number of rotations,

rev counter/tachometer, light gates connected to a timer/frequency meter). [1] M4 Use fiducial mark or light gates perpendicular to motion of object. [1] M5 Method to measure angle – use protractor or rule for measurements for trigonometry methods.

This must be shown correctly on diagram or explained in text. [1] Method of analysis (2 marks)

A1 Plot a graph of cos θ against 1/ω 2. [1] A2 Relationship is valid if straight line through the origin [1] Safety considerations (1 mark) S1 Use a protective screen in case mass detaches from the pole. Do not use goggles. [1] Additional detail (4 marks) Relevant points might include [4]

1 Large motor speed to produce measurable θ. 2 Additional detail on measuring angle e.g. large protractor fixed to pole. 3 Projection method, slow motion freeze frame video, camera with detail.

4 cos θ = h/l or equivalent. 5 Method of checking pole is vertical – use a set square. 6 Additional detail on measuring angular velocity, e.g. time at least 10 rotations. 7 Wait for motion to become stable. Do not allow vague computer methods. [Total: 15]




2 Analysis, conclusions and evaluation (15 marks)

Part Mark Expected Answer Additional Guidance

(a) A1 Gradient = r y-intercept = lg s

Allow log or ln

(b) T1 T2

1.70 or 1.699 0.41 or 0.415 1.78 or 1.778 0.53 or 0.531 1.85 or 1.845 0.64 or 0.643 1.90 or 1.903 0.73 or 0.732 1.95 or 1.954 0.82 or 0.820 1.98 or 1.978 0.86 or 0.857

Ignore significant figures. A mixture is allowed.

U1 From ± 0.03 or ± 0.04, to

± 0.01 (±0.012)

Allow more than one significant figure.

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

U2 Error bars in lg (y / mm) plotted correctly.

Must be accurate within half a small square.

(ii) G2 Line of best fit If points are plotted correctly then lower end of line should pass between (1.655, 0.35) and (1.665, 0.35) and upper end of line should pass between (2.00, 0.89) and (2.00, 0.90). Allow ecf from points plotted incorrectly – examiner judgement.

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

Line should be clearly labelled or dashed. Should pass from top of top error bar to bottom of bottom error bar or bottom of top error bar to top of bottom error bar. Mark scored only if error bars are plotted.

(iii) C1 Gradient of best fit line

The triangle used should be at least half the length of the drawn line. Check the read offs. Work to half a small square. Do not penalise POT. (Should be about 1.6)

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

(iv) C2 Negative y-intercept Must be negative. FOX does not score. Expect to see point substituted into y = mx + c Allow ecf from (c)(iii)

U4 Uncertainty in y-intercept Uses worst gradient and point on WAL. Do not check calculation. FOX does not score.




(d) C3 r = gradient and is given to 2 or 3 s.f. and in the range 1.57 to 1.64

Allow 1.6 to 2 s.f. Penalise 1 s.f. or >3 s.f.

C4 s = 10y-intercept y-intercept must be used.

(Should be about 0.005 or 5 × 10–3) Allow ecf for method from (c)(iv).

U5 Absolute uncertainty in r and s

Uncertainty in r should be the same as the uncertainty in the gradient. Difference in worst s and s.

[Total: 15] Uncertainties in Question 2 (c) (iii) Gradient [U3] Uncertainty = gradient of line of best fit – gradient of worst acceptable line Uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient) (iv) [U4] Uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line Uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)

(d) [U5] Uncertainty = best s –worst s





9702 PHYSICS









P1 v is the independent variable and θ is the dependent variable or vary v and measure θ. [1]

P2 Keep the (shape and) size/volume/surface area/mass of balloon/helium constant [1]

Do not credit ‘same balloon’. P3 Keep the temperature (air/helium/balloon) constant. [1] Methods of data collection (5 marks) M1 Labelled diagram of apparatus: balloon, string fixed and method of producing wind.

Method of producing wind to be approximately horizontal to balloon. [1] M2 Suspend mass from balloon. [1] M3 Method to change wind speed, e.g. change setting, variable power supply/resistor/change

distance from fan. [1] M4 Method to measure wind speed, e.g. wind speed indicator/detector, anemometer [1] M5 Method to measure angle – use protractor or rule for measurements for trigonometry methods.


A1 Plot a graph of tan θ against 1/v2. [1]

A2 Relationship valid if straight line through origin [1] Safety considerations (1 mark) S1 Avoid the moving blades of the fan (safety screen, switch off when changing experiment);

goggles to avoid air stream into eye. [1] Additional detail (4 marks) D1/2/3/4 Relevant points might include [4] 1 Large wind speed to produce measurable deflection/large cross-sectional area of balloon. 2 Additional detail on measuring angle e.g. use a large protractor, projection method.

3 tan θ = h/l.

4 Measuring air speed at point where balloon is positioned. 5 Adjust height of fan so that air flow is horizontally aligned to the balloon. 6 Reason for adding mass to increase stability/deflection. 7 Keep windows shut/air conditioning switched off/use of wind tunnel to avoid draughts. 8 Wait for the balloon to become stable. Do not allow vague computer methods. [Total: 15]






(a) A1 Gradient =

µ

T

2

1 Allow equivalent, e.g.

µ4

T

(b) T1 1/L / m–1 or (1/L) / m–1 Allow 1/L (m–1), 1/L / 1/m, 1/L (1/m)

T2 1.83 or 1.835 2.08 or 2.083 2.35 or 2.353 2.50 or 2.500 2.82 or 2.817 3.13 or 3.125

Values must correspond to table. A mixture of 3 s.f. and 4 s.f. is allowed

U1 From ± 0.01 or ± 0.02, to ± 0.05 Allow more than one significant figure.


U2 All Error bars in 1/L/m–1 plotted correctly.

Check second and last point for accuracy. Must be accurate within half a small square.

(ii) G2 Line of best fit If points are plotted correctly then lower end of line should pass between (1.8, 250) and (1.8, 254) and upper end of line should pass between (3.18, 450) and (3.2, 448). Allow ecf from points plotted incorrectly – examiner judgement.


Line should be clearly labelled or dashed. Should pass from left of top error bar to right of bottom error bar or right of top error bar to left of bottom error bar. Mark scored only if all error bars are plotted.


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 140)

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

(d) (i) C2 Value of µ using gradient µ = 7.5/gradient2

Gradient must be used. (Should be about 0.00037 or 3.7 × 10–4)

C3 kg m–1 or N Hz–2 m–2 Allow other correct units e.g. N s2

m–2 or Pa s2 or N (Hz m)–2

(ii) U4 10% + 2 × percentage uncertainty in gradient

Check working. Must be larger than 10%.




(e) C4 r given to 2 or 3 s.f. and in the range 1.15 × 10–4 to 1.18 × 10–4


U5 (d)(ii) / 2 Check working if not (d)(ii) / 2

[Total: 15] Uncertainties in Question 2 (c) (iii) Gradient [U3] Uncertainty = gradient of line of best fit – gradient of worst acceptable line Uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient) (d) (ii) [U4]

Percentage uncertainty = 10 + 2 m

m∆

Percentage uncertainty = 100 min4

max

2

×

−

×

µ

µ

m

T

Percentage uncertainty = 100max 4

min

2

×

−

×

µ

µ

m

T

(e) (ii) [U5]

Percentage uncertainty = 100 max ×

−

r

rr

Percentage uncertainty = 100 min×

−

r

rr





9702 PHYSICS









P1 Frequency or period of rotation or ω is the independent variable and θ is the dependent variable

or vary f or T or ω and measure θ. [1]

P2 ω = 2πf = 2π/T [1] P3 Keep the length of the rigid rod constant; ignore reference to mass. [1] Methods of data collection (5 marks) M1 Labelled diagram of apparatus: small object, pole attached to a rotating device

(motor, turntable). [1] M2 Method to change the speed of the rotating device. [1] M3 Method to determine frequency or time period (e.g. stop watch to time a number of rotations,

rev counter/tachometer, light gates connected to a timer/frequency meter). [1] M4 Use fiducial mark or light gates perpendicular to motion of object. [1] M5 Method to measure angle – use protractor or rule for measurements for trigonometry methods.


A1 Plot a graph of cos θ against 1/ω 2. [1] A2 Relationship is valid if straight line through the origin [1] Safety considerations (1 mark) S1 Use a protective screen in case mass detaches from the pole. Do not use goggles. [1] Additional detail (4 marks) Relevant points might include [4]

1 Large motor speed to produce measurable θ. 2 Additional detail on measuring angle e.g. large protractor fixed to pole. 3 Projection method, slow motion freeze frame video, camera with detail.

4 cos θ = h/l or equivalent. 5 Method of checking pole is vertical – use a set square. 6 Additional detail on measuring angular velocity, e.g. time at least 10 rotations. 7 Wait for motion to become stable. Do not allow vague computer methods. [Total: 15]






(a) A1 Gradient = r y-intercept = lg s

Allow log or ln

(b) T1 T2

1.70 or 1.699 0.41 or 0.415 1.78 or 1.778 0.53 or 0.531 1.85 or 1.845 0.64 or 0.643 1.90 or 1.903 0.73 or 0.732 1.95 or 1.954 0.82 or 0.820 1.98 or 1.978 0.86 or 0.857

Ignore significant figures. A mixture is allowed.

U1 From ± 0.03 or ± 0.04, to

± 0.01 (±0.012)

Allow more than one significant figure.


U2 Error bars in lg (y / mm) plotted correctly.

Must be accurate within half a small square.

(ii) G2 Line of best fit If points are plotted correctly then lower end of line should pass between (1.655, 0.35) and (1.665, 0.35) and upper end of line should pass between (2.00, 0.89) and (2.00, 0.90). Allow ecf from points plotted incorrectly – examiner judgement.


Line should be clearly labelled or dashed. Should pass from top of top error bar to bottom of bottom error bar or bottom of top error bar to top of bottom error bar. Mark scored only if error bars are plotted.


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 1.6)

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

(iv) C2 Negative y-intercept Must be negative. FOX does not score. Expect to see point substituted into y = mx + c Allow ecf from (c)(iii)

U4 Uncertainty in y-intercept Uses worst gradient and point on WAL. Do not check calculation. FOX does not score.




(d) C3 r = gradient and is given to 2 or 3 s.f. and in the range 1.57 to 1.64


C4 s = 10y-intercept y-intercept must be used.

(Should be about 0.005 or 5 × 10–3) Allow ecf for method from (c)(iv).

U5 Absolute uncertainty in r and s

Uncertainty in r should be the same as the uncertainty in the gradient. Difference in worst s and s.

[Total: 15] Uncertainties in Question 2 (c) (iii) Gradient [U3] Uncertainty = gradient of line of best fit – gradient of worst acceptable line Uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient) (iv) [U4] Uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line Uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)

(d) [U5] Uncertainty = best s –worst s

9702 s12 ms_all

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Transcript of 9702 s12 ms_all