9702 w12 ms_all

CAMBRIDGE INTERNATIONAL EXAMINATIONS

GCE Advanced Subsidiary Level and GCE Advanced Level

MARK SCHEME for the October/November 2012 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 October/November 2012 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.

Page 2 Mark Scheme Syllabus Paper

GCE AS/A LEVEL – October/November 2012 9702 11

© Cambridge International Examinations 2012

Question Number

Key Question Number

Key

1 B 21 D

2 C 22 D

3 D 23 A

4 B 24 B

5 B 25 B

6 C 26 D

7 A 27 A

8 D 28 B

9 A 29 C

10 C 30 B

11 C 31 B

12 A 32 A

13 A 33 C

14 A 34 D

15 B 35 C

16 B 36 C

17 C 37 B

18 A 38 C

19 A 39 A

20 C 40 B




9702 PHYSICS







Question Number

Key Question Number

Key

1 B 21 D

2 A 22 A

3 B 23 A

4 C 24 D

5 B 25 D

6 A 26 D

7 D 27 B

8 A 28 C

9 C 29 A

10 D 30 D

11 A 31 A

12 D 32 B

13 D 33 C

14 D 34 C

15 B 35 C

16 D 36 C

17 C 37 C

18 C 38 B

19 A 39 C

20 D 40 D




9702 PHYSICS







Question Number

Key Question Number

Key

1 B 21 B

2 D 22 A

3 B 23 B

4 C 24 B

5 B 25 C

6 A 26 B

7 C 27 B

8 B 28 A

9 D 29 B

10 A 30 A

11 B 31 A

12 C 32 C

13 B 33 D

14 B 34 B

15 D 35 D

16 D 36 B

17 A 37 D

18 B 38 A

19 A 39 C

20 D 40 B




9702 PHYSICS

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

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

Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the October/November 2012 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.




1 (a) (i) acceleration = change in velocity / time (taken) or acceleration = rate of change of velocity B1 [1] (ii) a body continues at constant velocity unless acted on by a resultant force B1 [1] (b) (i) distance is represented by the area under graph C1 distance = ½ × 29.5 × 3 = 44.3 m (accept 43.5 m for 29 to 45 m for 30) A1 [2] (ii) resultant force = weight – frictional force B1 frictional force increases with speed B1 at start frictional force = 0 / at end weight = frictional force B1 [3] (iii) 1. frictional force increases B1 [1] 2. frictional force (constant) and then decreases B1 [1] (iv) 1. acceleration = (v2 – v1) / t = (20 – 50) / (17 – 15) C1 = (–) 15 m s–2 A1 [2] 2. W – F = ma C1 W = 95 × 9.81 (= 932) C1 F = (95 × 15) + 932 = 2400 (2360) (2357) N A1 [3] 2 (a) resistance = potential difference / current B1 [1] (b) (i) metal wire in series with power supply and ammeter B1 voltmeter in parallel with metal wire B1 rheostat in series with power supply or potential divider arrangement or variable power supply B1 [3] (ii) 1. intercept on graph B1 [1] 2. scatter of readings about the best fit line B1 [1] (iii) correction for zero error explained B1

use of V and corrected І values from graph C1

resistance = V / І = 22.(2) Ω [e.g. 4.0 / 0.18] A1 [3] (c) R = 6.8 / 0.64 = 10.625 C1

%R = %V + %І = (0.1 / 6.8) × 100 + (0.01 / 0.64) × 100 C1 = 1.47% + 1.56%

∆R = 0.0303 × 10.625 = 0.32 Ω

R = 10.6 ± 0.3 Ω A1 [3]




3 (a) pressure = force / area B1 [1] (b) molecules collide with object / surface and rebound B1 molecules have change in momentum hence force acts B1 fewer molecules per unit volume on top of mountain / temperature is less hence lower speed of molecules B1 hence less pressure A0 [3]

(c) (i) ρ = m / V C1

W = Vρg = 0.25 × 0.45 × 9.81 × 13600 C1 = 15000 (15009) N A1 [3]

(ii) p = W / A (or using p = ρgh) = 15009 / 0.45 = 3.3 × 104 Pa A1 [1]

(iii) pressure will be greater due to the air pressure (acting on the surface of the liquid) B1 [1]

4 (a) waves pass through the elements / gaps / slits in the grating M1 spread into geometric shadow A1 [2] (b) (i) 1. displacements add to give resultant displacement B1 each wavelength travels the same path difference or are in phase B1 hence produce a maximum A0 [2]

2. to obtain a maximum the path difference must be λ or phase difference

360° / 2π rad B1

λ of red and blue are different B1 hence maxima at different angles / positions A0 [2]

(ii) nλ = d sin θ C1 N = sin 61° / (2 × 625 × 10–9) = 7.0 × 105 A1 [2]

(iii) nλ = 2 × 625 is a constant (1250) C1

n = 1 → λ = 1250 outside visible

n = 3 → λ = 417 in visible

n = 4 → λ = 312.5 outside visible

λ = 420 nm A1 [2] 5 (a) when the load is removed then the wire / body object does not return to its original shape /

length B1 [1]

(b) (i) stress = force / area C1 F = 220 × 106 × 1.54 × 10–6 = 340 (338.8) N A1 [2]

(ii) E = (F × l) / (A × e) C1 e = (90 × 106) × 1.75 / (1.2 × 1011) = 1.31 × 10–3

m A1 [2]

(c) the stress is no longer proportional to the extension B1 [1]




6 (a) 92 protons in the nucleus and 92 electrons around nucleus B1 143 neutrons (in the nucleus) B1 [2]

(b) (i) α-particle travels short distance in air B1 [1] (ii) very small proportion in backwards direction / large angles B1 majority pass through with no /small deflections B1 either most of mass is in very small volume (nucleus) and is charged or most of atom is

empty space B1 [3]

(c) I = Q / t C1 n / t = (1.5 × 10–12) /( 2 × 1.6 × 10–19) C1 n / t = 4.7 × 106

s–1 A1 [3]




9702 PHYSICS







1 (a) units for D identified as kg m s–2 M1

all other units shown: units for A: m2 units for v2: m2 s–2 units for ρ: kg m–3

22

23

2

smmmkg

smkg−−

−

=C with cancelling / simplification to give C no units A1 [2]

(b) (i) straight line from (0,0) to (1,9.8) ± half a square B1 [1] (ii) ½ mv2 = mgh or using v2 = 2 as C1 v = (2 × 9.81 × 1000)1/2 = 140 m s–1 A1 [2] (c) (i) weight = drag (D) ( + upthrust) B1 [1] Allow mg or W for weight and D or expression for D for drag (ii) 1. mg = 1.4 ×10–5 × 9.81 C1 1.4 × 10–5 × 9.81 = 0.5 × 0.6 × 1.2 × 7.1 × 10–6 × v2 M1 v = 7.33 m s–1 A0 [2] 2. line from (0,0) correct curvature to a horizontal line at velocity of 7 m s–1 M1 line reaches 7 m s–1 between 1.5 s and 3.5 s A1 [2] 2 (a) (resultant) force = rate of change of momentum / allow proportional to or change in momentum / time (taken) B1 [1]

(b) (i) ∆p = (–) 65 × 10–3 (5.2 + 3.7) C1 = (–) 0.58 N s A1 [2] (ii) F = 0.58 / 7.5 × 10–3 = 77(.3) N A1 [1] (c) (i) 1. force on the wall from the ball is equal to the force on ball from the wall M1 but in the opposite direction A1 [2] (statement of Newton’s third law can score one mark) 2. momentum change of ball is equal and opposite to momentum change of the wall / change of momentum of ball and wall is zero B1 [1] (ii) kinetic energy (of ball and wall) is reduced / not conserved so inelastic B1 [1] (Allow relative speed of approach does not equal relative speed of separation.)




3 (a) metal: regular / repeated / ordered arrangement / pattern / lattice or long range order (of atoms / molecules / ions) B1 polymer: tangled chains (of atoms / molecules) or long chains (of atoms / molecules / ions) B1 amorphous: disordered / irregular arrangement or short range order (of atoms / molecules / ions) B1 [3] (b) metal: straight line or straight line then curving with less positive gradient B1 polymer: curve with decreasing gradient with steep increasing gradient at end B1 [2] 4 (a) waves (travels along tube) reflect at closed end / end of tube B1 incident and reflected waves or these two waves are in opposite directions M1 interfere or stationary wave formed if tube length equivalent to

λ / 4, 3λ / 4, etc. A1 [3] (b) (i) 1. no motion (as node) / zero amplitude B1 [1] 2. vibration backwards and forwards / maximum amplitude along length B1 [1]

(ii) λ = 330 / 880 (= 0.375 m) C1

L = 3λ / 4 C1 L = 3 / 4 × (0.375) = 0.28 (0.281) m A1 [3]

5 (a) (i) І1 = І2 + І3 B1 [1]

(ii) І = V / R or І2 = 12 / 10 (= 1.2 A) C1

R = [1/6 + 1 / 10]–1 [total R = 3.75 Ω] or І3 = 12 / 6 (= 2.0 A) C1

І1 = 12 / 3.75 = 3.2 A or І1 = 1.2 + 2.0 = 3.2 A A1 [3]

(iii) power = VІ or І2R or V2 / R C1

s2

w2

3

2

s

2

3

w

2

2

/

/ or or

resistorsseriesinpower

wireinpower

RV

RV

R

Rx

I

I

I

I

V

V== C1

x = 12 × 1.2 / 12 × 2.0 = 0.6(0) allow 3 / 5 or 3:5 A1 [3] (b) p.d. BC: 12 – 12 × 0.4 = 7.2 (V) / p.d. AC = 4.8 (V) C1 p.d. BD: 12 – 12 × 4 / 6 = 4.0 (V) / p.d. AD = 8.0 (V) C1 p.d. = 3.2 V A1 [3] 6 (a) extension is proportional to force / load B1 [1] (b) F = mg C1 x = (mg / k ) = 0.41 × 9.81 / 25 = (4.02 / 25) M1 x = 0.16 m A0 [2]




(c) (i) weight and (reaction) force from spring (which is equal to tension in spring) B1 [1] (ii) F – weight or 0.06 × 25 = ma C1 F = 0.2209 × 25 = 5.52 (N) or 0.22 × 25 = 5.5 a = (5.52 – 0.41× 9.81) / 0.41 or 1.5 / 0.41 and (5.5 – 4.02) C1 a = 3.7 (3.66) m s–2 gives 3.6 m s–2 A1 [3] (d) elastic potential energy / strain energy to kinetic energy and gravitational potential energy B1 stretching / extension reduces and velocity increases / height increases B1 [2]

7 (a) He3

2 + He

3

2 → He

4

2 + 2 p

1

1 + Q

A numbers correct (4 and 1) B1 Z numbers correct (2 and 1) B1 [2] (b) both nuclei have 2 protons B1 the two isotopes have 1 neutron and two neutrons B1 [2] [allow 1 for ‘same number of protons but different number of neutrons’] (c) proton number and neutron number B1 energy – mass B1 momentum B1 [2]

(d) (i) γ radiation B1 [1] (ii) product(s) must have kinetic energy B1 [1] (e) 13.8 MeV = 13.8 × 1.6 × 10–19 × 106 (= 2.208 × 10–12) C1 60 = n × 13.8 × 1.6 × 10–13 n = 2.7(2) × 1013 s–1 A1 [2]




9702 PHYSICS







1 (a) spacing = 380 or 3.8 × 102 pm B1 [1]

(b) time = 24 × 3600 time = 0.086 (0.0864) Ms B1 [1]

(c) time = distance / speed = 8

11

103

101.5

×

×

C1

= 500 (s) = 8.3 min A1 [2] (d) momentum and weight B1 [1] (e) (i) arrow to the right of plane direction (about 4° to 24°) B1 [1]

(ii) scale diagram drawn or use of cosine formula v2 = 2502 + 362 – 2 × 250 × 36 × cos 45°

or resolving v = [(36 cos 45°)2 + (250 – 36 sin 45°)2]1/2 C1 resultant velocity = 226 (220 – 240 for scale diagram) m s–1

allow one mark for values 210 to 219 or 241 to 250 m s–1

or use of formula (v2 = 51068) v = 230 (226) m s–1 A1 [2] 2 (a) (i) accelerations (A to B and B to C) are same magnitude B1 accelerations (A to B and B to C) are opposite directions or both accelerations are toward B B1 (A to B and B to C) the component of the weight down the slope provides the acceleration B1 [3] (ii) acceleration = g sin15 ° C1 s = 0 + ½ at2 s = 0.26 / sin 15 ° = 1.0 C1

°×

×=

sin1589

2012

.

.t t = 0.89 s A1 [3]

(iii) v = 0 + g sin15t or v2 = 0 + 2g sin15 × 1.0 C1 v = 2.26 m s–1 A1 [2] (using loss of GPE = gain KE can score full marks) (b) loss of GPE at A = gain in GPE at C or loss of KE at B = gain in GPE at C B1 h1 = h2 = 0.26 m or ½ mv2 = mgh h2 = 0.5 × (2.26)2 / 9.81 = 0.26 m x = 0.26 / sin 30° = 0.52 m A1 [2] 3 (a) power is the rate of doing work or power = work done / time (taken) or power = energy transferred / time (taken) B1 [1] (b) (i) as the speed increases drag / air resistance increases B1 resultant force reduces hence acceleration is less B1 constant speed when resultant force is zero B1 [3] (allow one mark for speed increases and acceleration decreases)




(ii) force from cyclist = drag force / resistive force B1 P = 12 × 48 M1 P = 576 W A0 [2] (iii) tangent drawn at speed = 8.0 m s–1 M1 gradient values that show acceleration between 0.44 to 0.48 m s–2 A1 [2] (iv) F – R = ma C1

600 / 8 – R = 80 × 0.5 [using P = 576] 576 / 8 – R = 80 × 0.5 C1 R = 75 – 40 = 35 N R = 72 – 40 = 32 N A1 [3] (v) at 12 m s–1 drag is 48 N, at 8 m s–1 drag is 35 or 32 N R / v calculated as 4 and 4 or 4.4 and consistent response for whether R is proportional to v or not B1 [1] 4 (a) e.m.f. = chemical energy to electrical energy M1 p.d. = electrical energy to thermal energy M1

idea of per unit charge A1 [3]

(b) E = I (R +r) or I = E / (R +r) (any subject) B1 [1] (c) (i) E = 5.8 V B1 [1] (ii) evidence of gradient calculation or calculation with values from graph e.g. 5.8 = 4 + 1.0 × r C1

r = 1.8 Ω A1 [2] (d) (i) P = VI C1 P = 2.9 × 1.6 = 4.6 (4.64) W A1 [2] (ii) power from battery = 1.6 × 5.8 = 9.28 or efficiency = VI / EI C1 efficiency = (4.64 / 9.28) × 100 = 50 % or (2.9 / 5.8) × 100 = 50% A1 [2] 5 (a) travel through a vacuum / free space B1 [1]

(b) (i) B : name: microwaves wavelength: 10– 4 to 10–1

m B1 C : name: ultra-violet / UV wavelength: 10–7 to 10–9

m B1 F : name: X –rays wavelength: 10–9 to 10–12

m B1 [3]

(ii) f = 9

8

10 500

103

−

×

×

C1

f = 6(.0) × 1014 Hz A1 [2]




(c) vibrations are in one direction M1 perpendicular to direction of propagation / energy transfer or good sketch showing this A1 [2] 6 (a) (i) electron B1 [1]

(ii) any two: can be deflected by electric and magnetic fields or negatively charged /

absorbed by few (1 – 4) mm of aluminum / 0.5 to 2 m or metres for range in air / speed up to 0.99c / range of speeds / energies B2 [2]

(iii) decay occurs and cannot be affected by external / environmental factors or two stated factors such as chemical / pressure / temperature / humidity B1 [1] (b) 3 and 0 for superscript numbers B1 2 and –1 for subscript numbers B1 [2] (c) energy = 5.7 × 103 × 1.6 × 10–19 (= 9.12 × 10–16 J) C1

v2 = 31

16

10 × 9.11

10×9.12×2

−

−

C1

v = 4.5 × 107 m s–1 A1 [3] (d) both have 1 proton and 1 electron B1 1 neutron in hydrogen-2 and 2 neutrons in hydrogen-3 B1 [2] (special case: for one mark ‘same number of protons / atomic number different number of neutrons’)




9702 PHYSICS

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






1 (b) (ii) Values of raw L in range 2.0 cm Y L Y 8.0 cm consistent with unit. [1]

(iii) Value of θ < 90 ° with unit. No raw value greater than 0.5 ° precision. [1]

(c) Five sets of readings of L, m and θ scores 5 marks, four sets scores 4 marks etc. [5] Incorrect trend then –1. Major help from Supervisor –2. Minor help from Supervisor –1.

Range: mmin Y 0.100 kg, mmax [ 0.350 kg. [1] Column headings: [1] Each column heading must contain a quantity and a unit where appropriate.

The unit must conform to accepted scientific convention e.g. m / kg, m sin θ / kg, θ / °. Consistency: [1] All values of L must be given to the nearest mm. Significant figures: [1]

All values of m sin θ must have the same number of significant figures as, or one more

than, the least number of significant figures in m and θ. Calculation: [1]

Values of m sin θ calculated correctly. (d) (i) Axes: [1] Sensible scales must be used. Awkward scales (e.g. 3:10) are not allowed. Scales must be chosen so that the plotted points occupy at least half the graph grid in both x and y directions. Scales must be labelled with the quantity that is being plotted. Scale markings must be no more than three large squares apart. Plotting of points: [1] All observations in the table must be plotted on the graph grid.

Diameter of plots must be Y half a small square (no blobs). Check that the points are plotted correctly. Work to an accuracy of half a small square in

both the x and y directions. Quality: [1] All points in the table must be plotted (at least 4) for this mark to be scored. Judge by the scatter of all the points about a straight line.

All points must be within ± 0.01 kg in the m sin θ direction of a straight line. (ii) Line of best fit: [1] Judge by balance of all the points on the grid (at least 4) 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 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 sign of the gradient must match the graph. The hypotenuse of the triangle used

must be at least half the length of the drawn line. Both read-offs must be accurate to half a small square 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 substitution into y = mx + c. Read-off must be accurate to half a small square in both the x and y directions. Or: Check the read-off of the intercept directly from the graph. (e) Value of P = candidate’s gradient. Value of Q = candidate’s intercept. [1] Do not allow a value presented as a fraction. Unit for P (m kg–1 or cm kg–1 or mm kg–1 or m g–1 or cm g–1 or mm g–1) and Q (m or cm or mm)

correct and consistent with value. [1] [Total: 20] 2 (a) (ii) Value of circumference in range 30.0 – 50.0 cm to the nearest mm with unit. [1] (iii) Absolute uncertainty in circumference is between 2 mm – 6 mm. [1] If repeated readings have been taken, then the absolute uncertainty can be half the

range. Correct method used to calculate the percentage uncertainty. (iv) Value of circumference within 2 cm of first value. [1] (b) (ii) Raw time values to at least 0.1s or 0.01 s, value of 0.5 s < T < 2.0 s. [1] Evidence of repeats. [1] (c) (i) Second value of T. [1] Second value of T > first value of T. [1] (ii) Third value of T. [1] (d) (ii) Correct calculation of two values of k. [1] Correct calculation of third value of k. [1] (iii) Justification of significant figures in k linked to significant figures in time and m (not just

“raw readings”) [1] (iv) Sensible comment relating to the calculated values of k, testing against a criterion

specified by the candidate. [1]




(e)

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

A three results not enough /not enough results

take more readings and plot a graph

two results not enough /repeat readings /few readings

B string too wide for markings on rule

use thinner string

C rules have different thicknesses so effective length of loop changes/ /different lengths so not a fair test

use rulers of similar thicknesses/ readings/method to take thickness into account /use rulers of the same length

D times are small /large uncertainty in time

use longer strings/improved method of timing

E difficult to judge start/ end of/complete oscillation

Position/motion sensor facing the rule /video with timer

position sensor at end or in middle

F swings of 30 cm ruler highly damped

G difficult to make two loops of the same circumference

method by which this can be achieved

H large uncertainty in mass method of measuring mass more precisely

accurate balance

[Total: 20]




9702 PHYSICS







1 (b) (i) Value of h in range 0.085 m Y h Y 0.095 m consistent with unit. [1]

(c) Value of T in range 0.6 s Y T Y 1.5 s consistent with unit. [1] Evidence of repeats. [1] (d) Six sets of readings of h and T or raw times scores 4 marks, five sets scores 3 marks etc. Help from Supervisor –1. [4]

Range: hmax – hmin [ 15.5 cm [1] Column headings: Each column heading must contain a quantity and a unit where appropriate. The unit must conform to accepted scientific convention e.g. T2h / s2m (or m s2) and h2/ m2. [1] Consistency: All raw values of h must be given to the nearest mm. [1] Significant figures: All values of h2 must have the same number of significant figures as, or one more than, the number of significant figures in h. [1] Calculation: Values of T2h calculated correctly. [1] (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 must be no more than three large squares apart. Plotting of points: [1] All observations in the table must be plotted on the graph grid.

Diameter of plots must be Y half a small square (no “blobs”). Check that the points are plotted correctly. Work to an accuracy of half a small square in

both the x and y directions. 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 ± 0.0025 m2 (25 cm2) in the h2 direction of a straight line. (ii) Line of best fit: [1] Judge by 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 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 sign of the gradient must match the graph. The hypotenuse of the triangle should be greater than half the length of the drawn line. Both read-offs must be accurate to half a small square in both the x and y directions. The method of calculation must be correct. y intercept: [1] Either: Correct read-off from a point on the line and substitution into y = mx + c. Read-off must be accurate to half a small square in both the x and y directions. Or: Correct read-off of the intercept directly from the graph. (f) Value of P = candidate’s gradient. Value of Q = candidate’s intercept. [1] Do not allow a value presented as a fraction. Unit for P (s2 m–1 or s2

cm–1 or s2 mm–1) and Q (s2

m or s2 cm or s2

mm) correct and [1] consistent with value. [Total: 20]

2 (a) (ii) Value of L in range: 5.0 cm Y L Y 15.0 cm with unit to nearest mm. [1]

(b) (ii) Value of s in range: 50.0 cm Y s Y 70.0 cm with unit. [1] Supervisor’s help –1. Evidence of repeat measurements. [1] (iii) Absolute uncertainty in s is between 2 cm – 10 cm. [1] If repeated readings have been taken, then the absolute uncertainty can be half the

range. Correct method used to calculate the percentage uncertainty. (iv) Correct calculation of x. [1] (c) Raw value(s) of t greater than 1 s to a precision of 0.1 or 0.01 s with unit. [1] (d) (i) Correct calculation of v using either value of x with consistent unit. [1] (ii) Justification of significant figures in v linked to significant figures in t and x or (s – L) (not just “raw readings”). [1] (e) (iii) Second value of t. [1] Second value of s. [1] Quality: correct trend; If s increases, t increases. [1] (f) Sensible comment relating to the calculated values of v, testing against a criterion specified

by the candidate. [1]




(g)


A two readings not enough (to draw a conclusion)

take many readings (for different masses) and plot a graph /calculate more v values and compare

‘repeat readings’ /few readings /take more readings and calculate average v

B the car does not travel in a straight line

method of determining the distance e.g. video + scale/method of marking a path /method of guiding trolley in straight line

C times are short /large uncertainty in t

use a longer slope /use a steeper slope

trolley too fast

D difficult to judge when trolley stopped/ difficult to start the stopwatch when all wheels on bench/when trolley at B/when trolley horizontal

improved method of timing eg video with timer or frame by frame/motion sensor placed at end of path/ticker tape timer

light gate(s) /reaction time /human error

E there is a drop when the trolley reaches the end of the board/at B there is a loss of velocity/kinetic energy

method to smooth transition e.g. thinner board/bevelled edge/thin card placed at transition

F difficult to release without applying a force/ velocity /difficult to position head at B after releasing trolley A

method of releasing trolley e.g. card/barrier or electromagnet

air resistance

G calculation of x doesn’t take back of trolley into account

detailed method of measuring from wheel to the back of the trolley

measuring l

[Total: 20]




9702 PHYSICS







1 (a) (iv) Value for I1 < 200 mA, with consistent unit. [1]

(v) Value for I2 with unit of current. [1]

I2 > I1 [1]

(b) Six sets of readings of I1, I2 and x scores 4 marks, five sets scores 3 marks etc. [4] Incorrect trend –1. Major help from Supervisor –2. Minor help from Supervisor –1.

Range: xmax – xmin [ 0.500 m. [1] Column headings: [1] Each column heading must contain a quantity and a unit where appropriate. The unit must conform to accepted scientific convention

e.g. I / A or I(A), 1/x (m–1), I1/I2 Consistency: [1] All values of x must be given to the nearest mm. Significant figures: [1]

All values of I2 / I1 must have the same significant figures as, or one more than, the

least number of significant figures in raw I1 and I2 Calculation: [1]

Values of I2 / I1 calculated correctly. (c) (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 of points: [1] All observations in the table must be plotted on the graph grid.

Diameter of plots must be Y half a small square. Check that the points are plotted correctly. Work to an accuracy of half a small square in

both the x and y directions. 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 ± 0.25 m–1 in the 1/x direction of a straight line. (ii) Line of best fit: [1] Judge by 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 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 sign of the gradient must match the graph. The hypotenuse of the triangle used must be greater than half the length of the drawn

line. Both read-offs must be accurate to half a small square in both the x and y directions. The method of calculation must be correct. y intercept: [1] Either: Correct read-off from a point on the line and substitution into y = mx + c. Read-off must be accurate to half a small square in both the x and y directions. Or: Correct read-off of the intercept directly from the graph. (d) Value of P = candidate’s gradient and value of Q = candidate’s intercept. [1] Do not allow a value presented as a fraction. Unit for P (m or cm or mm, consistent with value) and Q (no unit) correct. [1] [Total: 20] 2 (a) (i) Value for D in range 10 to 20 mm to the nearest mm, with unit. [1] (ii) Percentage uncertainty in D based on an absolute uncertainty of 0.5, 1, 2 or 3 mm. [1] If repeated readings have been taken, then the absolute uncertainty can be half the

range. Correct method used to calculate the percentage uncertainty. (b) (ii) Value of x to the nearest mm, in range 1.3 – 1.7 cm, with unit. [1] (iii) Correct calculation of V with consistent unit. [1] (c) (iv) Raw time values to 0.1 s or 0.01 s. Value of T in range 0.1 – 1.0 s. [1]

Evidence of repeat measurements. [1] (d) (iv) Second value of x. [1] (e) Second value of T. [1]

Second value of T < first value of T. [1] (f) (i) Correct calculation of two values of k. [1] (ii) Justification of significant figures in k linked to significant figures in D, x and time (not just “raw readings”). [1] (iii) Sensible comment relating to the calculated values of k, testing against a specified

criterion. [1]




(g)

* Credit in Bs or Cs, but not both. ** Credit in Es or Fs, but not both.

[Total: 20]


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/ only one reading

take more readings and (calculate) average k

B parallax error in D/ difficult to measure D because loop is in the way

use Vernier calipers/micrometer/travelling microscope to measure D*

use string

C V not accurate because D not internal diameter

measure thickness/diameter of wire using micrometer

use travelling microscope/Vernier calipers to measure D*

D mass swings side-to-side/ horizontal movement/ moves in more than one plane/non-uniform oscillation

E times are small/large uncertainty in T

use bigger mass

improved timing method e.g. motion/position sensor below weight/video with timer/video and view frame-by-frame**

light gates/ human error/reaction time/ time more cycles/ high frequency oscillations

F

difficult to judge start of/end of/complete oscillation

fixed/fiducial marker

improved timing method

e.g. motion/position sensor below weight/video with timer/video and view frame-by-frame**

marker fixed to spring/ marker placed at extreme(s) of oscillation light gates

G metal strip bends/ not horizontal

use stiffer strip/ thicker strip/support strip at both ends.

strip not straight/ move spring/use stronger strip




9702 PHYSICS

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






Section A 1 (a) force is proportional to the product of the masses and inversely proportional to the square of the separation M1 either point masses or separation >> size of masses A1 [2] (b) (i) gravitational force provides the centripetal force B1 mv2/r = GMm/r2 and EK = ½mv2 M1 hence EK = GMm/2r A0 [2]

(ii) 1. ∆EK = ½ × 4.00 × 1014 × 620 × (7.30 × 106–1 – 7.34 × 106–1) C1 = 9.26 × 107

J (ignore any sign in answer) A1 [2] (allow 1.0 × 108

J if evidence that EK evaluated separately for each r)

2. ∆EP = 4.00 × 1014 × 620 × (7.30 × 106–1 – 7.34 × 106–1) C1 = 1.85 × 108

J (ignore any sign in answer) A1 [2] (allow 1.8 or 1.9 × 108

J)

(iii) either (7.30 × 106)–1 – (7.34 × 106)–1 or ∆EK is positive / EK increased M1 speed has increased A1 [2] 2 (a) (i) sum of potential energy and kinetic energy of atoms / molecules / particles M1 reference to random A1 [2] (ii) no intermolecular forces B1 no potential energy B1 internal energy is kinetic energy (of random motion) of molecules B1 [3] (reference to random motion here then allow back credit to (i) if M1 scored)

(b) kinetic energy ∝ thermodynamic temperature B1 either temperature in Celsius, not kelvin so incorrect or temperature in kelvin is not doubled B1 [2] 3 (a) temperature of the spheres is the same B1 no (net) transfer of energy between the spheres B1 [2]

(b) (i) power = m × c × ∆θ where m is mass per second C1 3800 = m × 4.2 × (42 – 18) C1 m = 38 g s–1 A1 [3] (ii) some thermal energy is lost to the surroundings M1 so rate is an overestimate A1 [2] 4 (a) straight line through origin M1 shows acceleration proportional to displacement A1 negative gradient M1

shows acceleration and displacement in opposite directions A1 [4]




(b) (i) 2.8 cm A1 [1]

(ii) either gradient = ω2 and ω = 2πf or a = –ω2x and ω = 2πf C1

gradient = 13.5 / (2.8 × 10–2) = 482

ω = 22 rad s–1 C1

frequency = (22/2π =) 3.5 Hz A1 [3] (c) e.g. lower spring may not be extended e.g. upper spring may exceed limit of proportionality / elastic limit (any sensible suggestion) B1 [1] 5 (a) (i) ratio of charge and potential (difference) / voltage (ratio must be clear) B1 [1] (ii) capacitor has equal magnitudes of (+)ve and (-)ve charge B1 total charge on capacitor is zero (so does not store charge) B1 (+)ve and (-)ve charges to be separated M1 work done to achieve this so stores energy A1 [4] (b) (i) capacitance of Y and Z together is 24 µF C1 1 / C = 1 / 24 + 1 / 12 C = 8.0 µF (allow 1 s.f.) A1 [2] (ii) some discussion as to why all charge of one sign on one plate of X B1 Q = (CV =) 8.0 × 10–6 × 9.0 M1 = 72 µC A0 [2] (iii) 1. V = (72 × 10–6) / (12 × 10–6) = 6.0 V (allow 1 s.f.) (allow 72/12) A1 [1] 2. either Q = 12 × 10–6 × 3.0 or charge is shared between Y and Z C1 charge = 36 µC A1 [2] Must have correct voltage in (iii)1 if just quote of 36 µC in (iii)2. 6 (a) (i) particle must be moving M1 with component of velocity normal to magnetic field A1 [2]

(ii) F = Bqv sin θ M1

q, v and θ explained A1 [2] (b) (i) face BCGF shaded A1 [1] (ii) between face BCGF and face ADHE A1 [1] (c) potential difference gives rise to an electric field M1 either FE = qE (no need to explain symbols) or electric field gives rise to force (on an electron) A1 [2]




7 (a) induced e.m.f./current produces effects / acts in such a direction / tends M1 to oppose the change causing it A1 [2] (b) (i) 1. to reduce flux losses / increase flux linkage / easily magnetised and

demagnetised B1 [1] 2. to reduce energy / heat losses (do not allow ‘to prevent energy losses’) M1 caused by eddy currents A1 [2] (allow 1 mark for ‘reduce eddy currents’) (ii) alternating current / voltage B1 gives rise to (changing) flux in core B1 flux links the secondary coil M1 (by Faraday’s law) changing flux induces e.m.f. (in secondary coil) A1 [4] 8 (a) discrete quantity / packet / quantum of energy of electromagnetic radiation B1 energy of photon = Planck constant × frequency B1 [2] (b) threshold frequency (1) rate of emission is proportional to intensity (1) max. kinetic energy of electron dependent on frequency (1) max. kinetic energy independent of intensity (1) (any three, 1 each, max 3) B3 [3]

(c) either E = hc/λ or hc/λ = eV C1

λ = 450 nm to give work function of 3.5 eV

energy = 4.4 × 10–19 or 2.8 eV to give λ = 355 nm M1 2.8 eV < 3.5 eV so no emission 355 nm < 450 nm so no A1 [3] or work function = 3.5 eV threshold frequency = 8.45×1014

Hz C1 450 nm = 6.67×1014

Hz M1 6.67 × 1014

Hz < 8.45 × 1014 Hz A1




Section B 9 (a) e.g. zero output impedance / resistance infinite input impedance / resistance infinite (open loop) gain infinite bandwidth infinite slew rate 1 each, max. 3 B3 [3] (b) (i) graph: square wave M1 correct cross-over points where V2 = V1 A1 amplitude 5 V A1 correct polarity (positive at t = 0) A1 [4] (ii) correct symbol for LED M1 diodes connected correctly between VOUT and earth A1 correct polarity consistent with graph in (i) A1 [3] (R points ‘down’ if (i) correct) 10 X-ray images taken from different angles / X-rays directed from different angles B1 of one section / slice (1) all images in the same plane (1) images combined to give image of section / slice B1 images of successive sections / slices combined B1 image formed using a computer B1 image formed is 3D image (1) that can be rotated / viewed from different angles (1) (four B-marks plus any two additional marks) B2 [6] 11 (a) e.g. noise can be eliminated / filtered / signal can be regenerated extra bits can be added to check for errors multiplexing possible digital circuits are more reliable / cheaper data can be encrypted for security any sensible advantages, 1 each, max. 3 B3 [3] (b) (i) 1. higher frequencies can be reproduced B1 [1] 2. smaller changes in loudness / amplitude can be detected B1 [1] (ii) bit rate = 44.1 × 103 × 16 C1 = 7.06 × 105 s–1 number = 7.06 × 106 × 340 = 2.4 × 108 A1 [2] 12 (a) (i) signal in one wire (pair) is picked up by a neighbouring wire (pair) B1 [1] (ii) outer of coaxial cable is earthed B1 outer shields the core from noise / external signals B1 [2]




(b) attenuation per unit length = 1/L × 10 lg(P2/P1) C1 signal power at receiver = 102.5 × 3.8 × 10–8 = 1.2 × 10–5

W C1 attenuation in wire pair = 10 lg(3.0 × 10–3 / 1.2 × 10–5) = 24 dB C1 attenuation per unit length = 24 / 1.4 = 17 dB km–1 A1 [4] (other correct methods of calculation are possible)




9702 PHYSICS







Section A 1 (a) force is proportional to the product of the masses and inversely proportional to the square of the separation M1 either point masses or separation >> size of masses A1 [2] (b) (i) gravitational force provides the centripetal force B1 mv2/r = GMm/r2 and EK = ½mv2 M1 hence EK = GMm/2r A0 [2]

(ii) 1. ∆EK = ½ × 4.00 × 1014 × 620 × (7.30 × 106–1 – 7.34 × 106–1) C1 = 9.26 × 107

J (ignore any sign in answer) A1 [2] (allow 1.0 × 108

J if evidence that EK evaluated separately for each r)

2. ∆EP = 4.00 × 1014 × 620 × (7.30 × 106–1 – 7.34 × 106–1) C1 = 1.85 × 108

J (ignore any sign in answer) A1 [2] (allow 1.8 or 1.9 × 108

J)

(iii) either (7.30 × 106)–1 – (7.34 × 106)–1 or ∆EK is positive / EK increased M1 speed has increased A1 [2] 2 (a) (i) sum of potential energy and kinetic energy of atoms / molecules / particles M1 reference to random A1 [2] (ii) no intermolecular forces B1 no potential energy B1 internal energy is kinetic energy (of random motion) of molecules B1 [3] (reference to random motion here then allow back credit to (i) if M1 scored)

(b) kinetic energy ∝ thermodynamic temperature B1 either temperature in Celsius, not kelvin so incorrect or temperature in kelvin is not doubled B1 [2] 3 (a) temperature of the spheres is the same B1 no (net) transfer of energy between the spheres B1 [2]

(b) (i) power = m × c × ∆θ where m is mass per second C1 3800 = m × 4.2 × (42 – 18) C1 m = 38 g s–1 A1 [3] (ii) some thermal energy is lost to the surroundings M1 so rate is an overestimate A1 [2] 4 (a) straight line through origin M1 shows acceleration proportional to displacement A1 negative gradient M1

shows acceleration and displacement in opposite directions A1 [4]




(b) (i) 2.8 cm A1 [1]

(ii) either gradient = ω2 and ω = 2πf or a = –ω2x and ω = 2πf C1

gradient = 13.5 / (2.8 × 10–2) = 482

ω = 22 rad s–1 C1

frequency = (22/2π =) 3.5 Hz A1 [3] (c) e.g. lower spring may not be extended e.g. upper spring may exceed limit of proportionality / elastic limit (any sensible suggestion) B1 [1] 5 (a) (i) ratio of charge and potential (difference) / voltage (ratio must be clear) B1 [1] (ii) capacitor has equal magnitudes of (+)ve and (-)ve charge B1 total charge on capacitor is zero (so does not store charge) B1 (+)ve and (-)ve charges to be separated M1 work done to achieve this so stores energy A1 [4] (b) (i) capacitance of Y and Z together is 24 µF C1 1 / C = 1 / 24 + 1 / 12 C = 8.0 µF (allow 1 s.f.) A1 [2] (ii) some discussion as to why all charge of one sign on one plate of X B1 Q = (CV =) 8.0 × 10–6 × 9.0 M1 = 72 µC A0 [2] (iii) 1. V = (72 × 10–6) / (12 × 10–6) = 6.0 V (allow 1 s.f.) (allow 72/12) A1 [1] 2. either Q = 12 × 10–6 × 3.0 or charge is shared between Y and Z C1 charge = 36 µC A1 [2] Must have correct voltage in (iii)1 if just quote of 36 µC in (iii)2. 6 (a) (i) particle must be moving M1 with component of velocity normal to magnetic field A1 [2]

(ii) F = Bqv sin θ M1

q, v and θ explained A1 [2] (b) (i) face BCGF shaded A1 [1] (ii) between face BCGF and face ADHE A1 [1] (c) potential difference gives rise to an electric field M1 either FE = qE (no need to explain symbols) or electric field gives rise to force (on an electron) A1 [2]




7 (a) induced e.m.f./current produces effects / acts in such a direction / tends M1 to oppose the change causing it A1 [2] (b) (i) 1. to reduce flux losses / increase flux linkage / easily magnetised and

demagnetised B1 [1] 2. to reduce energy / heat losses (do not allow ‘to prevent energy losses’) M1 caused by eddy currents A1 [2] (allow 1 mark for ‘reduce eddy currents’) (ii) alternating current / voltage B1 gives rise to (changing) flux in core B1 flux links the secondary coil M1 (by Faraday’s law) changing flux induces e.m.f. (in secondary coil) A1 [4] 8 (a) discrete quantity / packet / quantum of energy of electromagnetic radiation B1 energy of photon = Planck constant × frequency B1 [2] (b) threshold frequency (1) rate of emission is proportional to intensity (1) max. kinetic energy of electron dependent on frequency (1) max. kinetic energy independent of intensity (1) (any three, 1 each, max 3) B3 [3]

(c) either E = hc/λ or hc/λ = eV C1

λ = 450 nm to give work function of 3.5 eV

energy = 4.4 × 10–19 or 2.8 eV to give λ = 355 nm M1 2.8 eV < 3.5 eV so no emission 355 nm < 450 nm so no A1 [3] or work function = 3.5 eV threshold frequency = 8.45×1014

Hz C1 450 nm = 6.67×1014

Hz M1 6.67 × 1014

Hz < 8.45 × 1014 Hz A1




Section B 9 (a) e.g. zero output impedance / resistance infinite input impedance / resistance infinite (open loop) gain infinite bandwidth infinite slew rate 1 each, max. 3 B3 [3] (b) (i) graph: square wave M1 correct cross-over points where V2 = V1 A1 amplitude 5 V A1 correct polarity (positive at t = 0) A1 [4] (ii) correct symbol for LED M1 diodes connected correctly between VOUT and earth A1 correct polarity consistent with graph in (i) A1 [3] (R points ‘down’ if (i) correct) 10 X-ray images taken from different angles / X-rays directed from different angles B1 of one section / slice (1) all images in the same plane (1) images combined to give image of section / slice B1 images of successive sections / slices combined B1 image formed using a computer B1 image formed is 3D image (1) that can be rotated / viewed from different angles (1) (four B-marks plus any two additional marks) B2 [6] 11 (a) e.g. noise can be eliminated / filtered / signal can be regenerated extra bits can be added to check for errors multiplexing possible digital circuits are more reliable / cheaper data can be encrypted for security any sensible advantages, 1 each, max. 3 B3 [3] (b) (i) 1. higher frequencies can be reproduced B1 [1] 2. smaller changes in loudness / amplitude can be detected B1 [1] (ii) bit rate = 44.1 × 103 × 16 C1 = 7.06 × 105 s–1 number = 7.06 × 106 × 340 = 2.4 × 108 A1 [2] 12 (a) (i) signal in one wire (pair) is picked up by a neighbouring wire (pair) B1 [1] (ii) outer of coaxial cable is earthed B1 outer shields the core from noise / external signals B1 [2]




(b) attenuation per unit length = 1/L × 10 lg(P2/P1) C1 signal power at receiver = 102.5 × 3.8 × 10–8 = 1.2 × 10–5

W C1 attenuation in wire pair = 10 lg(3.0 × 10–3 / 1.2 × 10–5) = 24 dB C1 attenuation per unit length = 24 / 1.4 = 17 dB km–1 A1 [4] (other correct methods of calculation are possible)




9702 PHYSICS







Section A 1 (a) (i) number of molecules B1 [1] (ii) mean square speed B1 [1] (b) (i) 1. pV = nRT C1 n = (6.1 × 105 × 2.1 × 104 × 10–6) / (8.31 × 285) C1 n = 5.4 mol A1 [3] 2. either N = nNA = 5.4 × 6.02 × 1023 C1 = 3.26 × 1024 A1 or pV = NkT N = (6.1 × 105 × 2.1 × 104 × 10–6) / (1.38 × 10–23 × 285) (C1) N = 3.26 × 1024 (A1) [2] (ii) either 6.1 × 105 × 2.1 × 10–2 = 1/3 × 3.25 × 1024 × 4 × 1.66 × 10–27× <c2> C1 <c2> = 1.78 × 106 C1 cRMS = 1.33 × 103 m s–1 A1 or 1/2 × 4 × 1.66 × 10–27 × <c2> = 3/2 × 1.38 × 10–23 × 285 (C1) <c2> = 1.78 × 106 (C1) cRMS = 1.33 × 103 m s–1 (A1) [3] 2 (a) (i) 1. 0.1 s, 0.3 s, 0.5 s, etc (any two) A1 [1] 2. either 0, 0.4 s, 0.8 s, 1.2 s or 0.2 s, 0.6 s, 1.0 s (any two) A1 [1] (ii) period = 0.4 s C1 frequency = (1/0.4 =) 2.5 Hz A1 [2] (iii) phase difference = 90 ° or ½ π rad B1 [1] (b) frequency = 2.4 – 2.5 Hz B1 [1] (c) e.g. attach sheet of card to trolley M1 increases damping / frictional force A1 e.g. reduce oscillator amplitude (M1) reduces power/energy input to system (A1) [2]




3 (a) (i) (tangent to line gives) direction of force on a (small test) mass B1 [1] (ii) (tangent to line gives) direction of force on a (small test) charge M1 charge is positive A1 [2] (b) similarity: e.g. radial fields lines normal to surface greater separation of lines with increased distance from sphere field strength ∝ 1 / (distance to centre of sphere)2

(allow any sensible answer) B1 difference: e.g. gravitational force (always) towards sphere B1 electric force direction depends on sign of charge on sphere / towards or away from sphere B1 e.g. gravitational field/force is attractive (B1) electric field/force is attractive or repulsive (B1) (allow any sensible comparison) [3] (c) gravitational force = 1.67 × 10–27 × 9.81 = 1.6 × 10–26

N A1 electric force = 1.6 × 10–19 × 270 / (1.8 × 10–2) C1 = 2.4 × 10–15

N A1 electric force very much greater than gravitational force B1 [4] 4 (a) force on proton is normal to velocity and field M1 provides centripetal force (for circular motion) A1 [2] (b) magnetic force = Bqv B1 centripetal force = mrω2 or mv2/r B1 v = rω B1

Bqv = Bqrω = mrω2 ω = Bq/m A1 [4] 5 (a) either φ = BA sinθ M1 where A is the area (through which flux passes) θ is the angle between B and (plane of) A A1 or

φ = BA (M1) where A is area normal to B (A1) [2] (b) graph: VH constant and non zero between the poles and zero outside M1 sharp increase/decrease at ends of magnet A1 [2]




(c) (i) (induced) e.m.f. proportional to M1 rate of change of (magnetic) flux (linkage) A1 [2] (ii) short pulse on entering and on leaving region between poles M1 pulses approximately the same shape but opposite polarities A1 e.m.f. zero between poles and outside A1 [3] 6 (a) (i) connection to ‘top’ of resistor labelled as positive B1 [1] (ii) diode B and diode D B1 [1] (b) (i) VP = 4.0 V C1 mean power = VP

2/2R C1 = 42 / (2 × 2700) = 2.96 × 10–3

W A1 [3] (ii) capacitor, correct symbol, connected in parallel with R B1 [1] (c) graph: half-wave rectification M1 same period and same peak value A1 [2] 7 (a) wavelength associated with a particle M1 that is moving A1 [2] (b) (i) kinetic energy = 1.6 × 10–19 × 4700 C1 = 7.52 × 10–16

J either energy = p2/2m or EK = ½mv2 and p = mv C1 p = √(7.52 × 10–16 × 2 × 9.1 × 10–31) C1 = 3.7 × 10–23

N s λ = h/p C1 = (6.63 × 10–34) / (3.7 × 10–23) = 1.8 × 10–11

m A1 [5] (ii) wavelength is about separation of atoms B1 can be used in (electron) diffraction B1 [2] 8 (a) (i) x = 2 A1 [1] (ii) either beta particle or electron B1 [1] (b) (i) mass of separate nucleons = (92 × 1.007) + (143 × 1.009) u C1 = 236.931 u C1 binding energy = 236.931 u – 235.123 u = 1.808 u A1 [3]




(ii) E = mc2 C1 energy = 1.808 × 1.66 × 10–27 × (3.0 × 108)2 = 2.7 × 10–10

J C1 binding energy per nucleon = (2.7 × 10–10) / (235 × 1.6 × 10–13) M1 = 7.18 MeV A0 [3] (c) energy released = (95 × 8.09) + (139 × 7.92) – (235 × 7.18) C1 = 1869.43 – 1687.3 = 182 MeV A1 [2] (allow calculation using mass difference between products and reactants)

Section B 9 (a) light-emitting diode (allow LED) B1 [1] (b) gives a high or a low output / +5 V or –5 V output M1 dependent on which of the inputs is at a higher potential A1 [2] (c) (i) provides a reference/constant potential B1 [1] (ii) determines temperature of ‘switch-over’ B1 [1] (d) (i) relay A1 [1] (ii) relay connected correctly for op-amp output and high-voltage circuit B1 diode with correct polarity in output from op-amp B1 [2] 10 (a) background reading = 19 B1 [1] (b) A = 2 A1 B = 5 A1 C = 9 A1 D = 3 A1 [4] (Allow 1 mark if only subtracts background reading) (c) (i) either 5, 14 or 14, 5 (A+D, B+C or v.v.) B1 [1] (ii) Three numbers and ‘inside’ number is 8 (B+D) B1 Three numbers and ‘outside’ numbers are either 2,9 or 9,2 (A,C or v.v.) B1 [2] 11 (a) high frequency wave B1 the amplitude or the frequency is varied M1 the variation represents the information signal / in synchrony with (the displacement of) the information signal. A1 [3]




(b) e.g. shorter aerial required longer transmission range / lower transmitter power / less attenuation allows more than one station in a region less distortion (allow any three sensible suggestions, 1 mark each) B3 [3] 12 (a) (i) e.g. linking a (land) telephone to the (local) exchange B1 [1] (ii) e.g. connecting an aerial to a television B1 [1] (iii) e.g. linking a ground station to a satellite B1 [1] (b) (i) attenuation = 10 lg (P2 / P1) C1 total attenuation = 2.1 × 40 (= 84 dB) C1 84 = 10 lg (450 × 10–3 / P) P = 1.8 × 10–9

W A1 [3] (answer 1.1 ×108

W scores 1 mark only) (ii) maximum attenuation = 10 lg (450 × 10–3 / 7.2 × 10–11) = 98 dB C1 maximum length = 98/2.1 = 47 km A1 [2]




9702 PHYSICS

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






1 Planning (15 marks)

Defining the problem (3 marks)

P v is the independent variable or vary v. [1] P E is the dependent variable or measure E. [1] P Keep the number of turns on the coil constant. [1]

Methods of data collection (5 marks)

M1 Labelled diagram showing magnet falling vertically through coil. [1] M2 Voltmeter or c.r.o. connected to the coil. Allow voltage sensor connected to datalogger. [1] M3 Method to change speed e.g. change height. [1] M4 Measurements to determine v. Use metre rule to measure distance magnet falls to the

bottom of the coil or metre rule/ruler to measure length of coil or ruler to measure length of the magnet. [Allow timing instrument to measure the time of the fall from the start to the bottom of the coil.] [1]

M5 Method of determining v corresponding to appropriate distance e.g. v = √2gh or v=2h/t (for height method) or v = L/t for length of magnet or coil and by stopwatch, timer or lightgate(s) connected to datalogger. [Allow v = gt for timing fall to bottom of coil.] [1]

Method of analysis (2 marks)

A Plot a graph of E against v. [Allow lg E against lg v] [1] A Relationship valid if straight line through origin. [1]

[If lg-lg then straight line with gradient = (+)1 (ignore reference to y-intercept)] Safety considerations (1 mark)

S Keep away from falling magnet/use sand tray/cushion to catch magnet. [1] Additional detail (4 marks)

D1/2/3/4 Relevant points might include [4] Use coil with large number of turns/drop magnet from large heights/strong magnet 1 Detailed use of datalogger/storage oscilloscope to determine maximum E; allow video

camera including slow motion play back 2 Use same magnet or magnet of same strength. 3 Use of short magnet so that v is (nearly) constant 4 Use short/thin coil so that v is (nearly) constant 5 Use a non-metallic vertical guide/tube 6 Method to support vertical coil or guide/tube 7 Repeat experiment for each v and average

Do not allow vague computer methods.

[Total: 15]




2 Analysis, conclusions and evaluation (15 marks)

Part Mark Expected Answer Additional Guidance

(a) A1 Gradient = hc/e y-intercept = – B/e

Note y-intercept must be negative

(b) T1 1/λ / 106 m–1 Appropriate column heading

T2

1.05 or 1.053

1.14 or 1.143

1.53 or 1.527

1.79 or 1.786

1.98 or 1.980

2.33 or 2.326

Must be values in table. A mixture of 3 s.f. and 4 s.f. is allowed.

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

U1 All error bars in V / V plotted correctly.

Do not allow near misses

(c) (ii) G2 Line of best fit If points are plotted correctly then lower end of line should pass between (1.12, 0.7) and (1.16,0.7) and upper end of line should pass between (2.32, 2.25) and (2.34, 2.25). 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.

(c) (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 1.3 × 10

–6.

U2 Uncertainty in gradient Method of determining absolute uncertainty Difference in worst gradient

and gradient. [± 0.08]

(c) (iv) C2 y-intercept Must be negative Expect to see point substituted into y = mx + c FOX does not score. Do not penalise POT. Should be between –0.72 and –0.86

U3 Method of determining uncertainty in y-intercept

Difference in worst y-intercept and y-intercept.

[Should be about ± 0.14]. FOX does not score. Allow ecf from (c)(iv).




(d) (i) C3 h in the range 6.77 × 10–34 to 7.14 × 10–34 and given to 2 or 3 significant figures

Gradient must be used. Penalise 1 s.f. or >3 s.f.

h = gradient × e/c = gradient × 5.33 × 10–28

Allow 6.8 × 10–34 to 7.1 × 10–34 to 2 s.f.

(d) (ii) U4 Percentage uncertainty in h 100×

m

m∆ or 100×

h

h∆

[should be about 6%]

(e) C4 B = –e × y-intercept and J or C V or V C

Ignore ‘–’ signs. y-intercept must be used but allow ecf from FOX. Should be between 1.16 × 10–19 J and 1.37 × 10–19 J. If FOX 8.3 × 10–20 J

U5 Absolute uncertainty in B Uncertainty = best B – worst B

= ∆y-intercept × e

[Total: 15] Uncertainties in Question 2 (c) (iii) Gradient [U2]

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

Uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient) (iv) [U3]

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

Uncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept) (d) (ii) [U4]

Percentage uncertainty = 100×

m

m∆

Percentage uncertainty = 100) min(max

1002

1×

−

=×

h

hh

h

h∆

(e) [U5]

Absolute uncertainty = best B – worst B

Absolute uncertainty = ∆y-intercept × e

Absolute uncertainty = Bc

c×

∆




9702 PHYSICS









P v is the independent variable or vary v. [1] P E is the dependent variable or measure E. [1] P Keep the number of turns on the coil constant. [1]

Methods of data collection (5 marks)

M1 Labelled diagram showing magnet falling vertically through coil. [1] M2 Voltmeter or c.r.o. connected to the coil. Allow voltage sensor connected to datalogger. [1] M3 Method to change speed e.g. change height. [1] M4 Measurements to determine v. Use metre rule to measure distance magnet falls to the

bottom of the coil or metre rule/ruler to measure length of coil or ruler to measure length of the magnet. [Allow timing instrument to measure the time of the fall from the start to the bottom of the coil.] [1]

M5 Method of determining v corresponding to appropriate distance e.g. v = √2gh or v=2h/t (for height method) or v = L/t for length of magnet or coil and by stopwatch, timer or lightgate(s) connected to datalogger. [Allow v = gt for timing fall to bottom of coil.] [1]

Method of analysis (2 marks)

A Plot a graph of E against v. [Allow lg E against lg v] [1] A Relationship valid if straight line through origin. [1]

[If lg-lg then straight line with gradient = (+)1 (ignore reference to y-intercept)] Safety considerations (1 mark)

S Keep away from falling magnet/use sand tray/cushion to catch magnet. [1] Additional detail (4 marks)

D1/2/3/4 Relevant points might include [4] Use coil with large number of turns/drop magnet from large heights/strong magnet 1 Detailed use of datalogger/storage oscilloscope to determine maximum E; allow video

camera including slow motion play back 2 Use same magnet or magnet of same strength. 3 Use of short magnet so that v is (nearly) constant 4 Use short/thin coil so that v is (nearly) constant 5 Use a non-metallic vertical guide/tube 6 Method to support vertical coil or guide/tube 7 Repeat experiment for each v and average

Do not allow vague computer methods.

[Total: 15]






(a) A1 Gradient = hc/e y-intercept = – B/e

Note y-intercept must be negative

(b) T1 1/λ / 106 m–1 Appropriate column heading

T2

1.05 or 1.053

1.14 or 1.143

1.53 or 1.527

1.79 or 1.786

1.98 or 1.980

2.33 or 2.326

Must be values in table. A mixture of 3 s.f. and 4 s.f. is allowed.


U1 All error bars in V / V plotted correctly.

Do not allow near misses

(c) (ii) G2 Line of best fit If points are plotted correctly then lower end of line should pass between (1.12, 0.7) and (1.16,0.7) and upper end of line should pass between (2.32, 2.25) and (2.34, 2.25). Allow ecf from points plotted incorrectly – examiner judgement.




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 1.3 × 10

–6.

U2 Uncertainty in gradient Method of determining absolute uncertainty Difference in worst gradient

and gradient. [± 0.08]

(c) (iv) C2 y-intercept Must be negative Expect to see point substituted into y = mx + c FOX does not score. Do not penalise POT. Should be between –0.72 and –0.86

U3 Method of determining uncertainty in y-intercept

Difference in worst y-intercept and y-intercept.

[Should be about ± 0.14]. FOX does not score. Allow ecf from (c)(iv).




(d) (i) C3 h in the range 6.77 × 10–34 to 7.14 × 10–34 and given to 2 or 3 significant figures

Gradient must be used. Penalise 1 s.f. or >3 s.f.

h = gradient × e/c = gradient × 5.33 × 10–28

Allow 6.8 × 10–34 to 7.1 × 10–34 to 2 s.f.

(d) (ii) U4 Percentage uncertainty in h 100×

m

m∆ or 100×

h

h∆

[should be about 6%]

(e) C4 B = –e × y-intercept and J or C V or V C

Ignore ‘–’ signs. y-intercept must be used but allow ecf from FOX. Should be between 1.16 × 10–19 J and 1.37 × 10–19 J. If FOX 8.3 × 10–20 J

U5 Absolute uncertainty in B Uncertainty = best B – worst B

= ∆y-intercept × e


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

Uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient) (iv) [U3]

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

Uncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept) (d) (ii) [U4]

Percentage uncertainty = 100×

m

m∆


1002

1×

−

=×

h

hh

h

h∆

(e) [U5]

Absolute uncertainty = best B – worst B

Absolute uncertainty = ∆y-intercept × e

Absolute uncertainty = Bc

c×

∆




9702 PHYSICS









P λ is the independent variable or vary λ. [1]

P θ is the dependent variable or measure θ (for each λ). [1] P Light sources to be of similar intensity/brightness. [1]

Methods of data collection (5 marks) M1 Labelled diagram showing observer, light sources with method of producing monochromatic

light e.g. filter/coloured LED. [1] M2 Method to measure wavelength: record from filter/LED or Young’s slit/diffraction grating

method. [1] M3 Use a rule to measure the distances. [1]

M4 Method to determine θ, e.g. θ (or sin θ or tan θ) = separation/distance or

( )distance2

separation

2tan

×

=θ

Do not allow protractor methods. [1] M5 Carry out the experiment in a dark room. [1] Method of analysis (2 marks)

A Plot a graph of θ against λ. [Allow lg θ against lg λ]. [1] A Relationship valid if straight line through origin. [1]

[If lg-lg then straight line with gradient = (+)1 (ignore reference to y-intercept)]

Safety considerations (1 mark) S Lamp becomes hot, therefore do not touch/switch off when not in use or use gloves when

moving hot lamp. OR Light may damage eyes, therefore wear dark glasses or do not look at unprotected lamps. [1]

Additional detail (4 marks) D1/2/3/4 Relevant points might include [4] 1 Use vertical filament lamps. Allow vertical slits.

2 Additional detail on measuring λ e.g. use of equation for Young’s slit/diffraction grating method.

3 Use of vernier calipers to measure the separation of light sources. 4 Use large distances/separations.

5 θ = sin θ = tan θ for small angles. 6 View with the same eye. 7 Method to ensure distances are perpendicular or observer equidistant from pair of lamps. 8 Repeat experiment for each λ and average. Do not allow vague computer methods.

[Total: 15]






(a) A1 Gradient = kA2

(b) T1 T2

1.3 or 1.33 1.2

0.8(0)(0)(0) 0.74

0.571 or 0.5714

0.54 or 0.55

0.444 or 0.4444

0.41 or 0.411 or 0.410

0.364 or 0.3636

0.34

0.308 or 0.3077

0.29 or 0.30

T1 must be values in 1/M. Ignore row 2.

T2 must be to 2 s.f. or 3 s.f.

U1 From ± 0.2 or ± 0.15 to ± 0.02 or

± 0.03

Allow more than one significant figure.

Do not allow ± 0.1 for row 1.


U2 All error bars in v2 plotted correctly

Must be accurate within half a small square.

(c) (ii) G2 Line of best fit There must be a balance of points about the line of best fit – examiner judgement. Allow ecf from points plotted incorrectly.




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 0.9.

U3 Uncertainty in gradient Method of determining absolute uncertainty.

Difference in worst gradient and gradient.

(d) (i) C2 k = gradient / A2

= gradient / 0.04

Should be about 22.

C3 N m–1 Allow kg s–2

(d) (ii) U4 Percentage uncertainty in k 5%1001002100 +×=××+×

m

m

A

A

m

m ∆∆∆




(e) C4 v in the range 0.534 to 0.559 and given to 2 or 3 s.f.

For 2 s.f. 0.53 to 0.56

U5 Uncertainty in v


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 = 5%1001002100 +×=××+×

m

m

A

A

m

m ∆∆∆

Maximum k = ) (min

max 2

A

m

Minimum k = )(max

min2

A

m


1002

1×

−

=×

k

kk

k

k∆

(e) [U5]

Percentage uncertainty = 1001002

1××+×

k

k

A

A ∆∆

Absolute uncertainty = v × percentage uncertainty/100

Maximum v = 0.75

max max

kA×

Minimum v = 0.75

min min

kA×

Absolute uncertainty = max v – v or v – min v or )min(max vv −

2

1

Education

9702 w12 ms_all