9702 w15 ms_all

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CAMBRIDGE INTERNATIONAL EXAMINATIONS

Cambridge International Advanced Subsidiary and Advanced Level

MARK SCHEME for the October/November 2015 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 2015 series for most Cambridge IGCSE

®, Cambridge International A and AS Level components and some

Cambridge O Level components.

Page 2 Mark Scheme Syllabus Paper

Cambridge International AS/A Level – October/November 2015 9702 11

© Cambridge International Examinations 2015

Question Number

Key Question Number

Key

1 B 21 B

2 B 22 A

3 A 23 B

4 A 24 C

5 B 25 A

6 C 26 A

7 C 27 D

8 B 28 A

9 D 29 B

10 D 30 A

11 A 31 C

12 D 32 A

13 C 33 B

14 A 34 C

15 B 35 A

16 D 36 A

17 B 37 D

18 D 38 D

19 D 39 D

20 B 40 B





9702 PHYSICS








Question Number

Key Question Number

Key

1 C 21 D

2 B 22 D

3 C 23 C

4 B 24 B

5 D 25 D

6 A 26 D

7 B 27 A

8 C 28 D

9 A 29 C

10 A 30 C

11 C 31 A

12 A 32 A

13 C 33 A

14 C 34 C

15 A 35 A

16 B 36 B

17 C 37 A

18 C 38 D

19 B 39 C

20 B 40 D





9702 PHYSICS








Question Number

Key Question Number

Key

1 A 21 C

2 D 22 C

3 A 23 C

4 B 24 D

5 B 25 B

6 B 26 B

7 A 27 B

8 B 28 C

9 C 29 B

10 C 30 A

11 C 31 A

12 D 32 A

13 D 33 D

14 A 34 A

15 B 35 A

16 D 36 A

17 B 37 D

18 C 38 C

19 A 39 C

20 B 40 D





9702 PHYSICS

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

This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. 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 2015 series for most Cambridge IGCSE






1 (a) temperature B1 current B1 [2]

(allow amount of substance, luminous intensity) (b) (i) 1. E = (stress / strain =) [force / area] / [extension / original length]

units of stress: kg m s–2

/ m2 and no units for strain B1

units of E: kg m

–1 s

–2 A0 [1]

2. units for T: s, l: m and M: kg

K

2 = T

2 E / M l

3 hence units: s

2 kg m

–1 s

–2 / kg

3 (= m

–4) C1

units of K: m–2

A1 [2]

(ii) % uncertainty in E = 4% (for T

2) + 0.6% (for l

3) + 0.1% (for M) + 3% (for K

2)

= 7.7% B1

E = [(1.48 × 105)2 × 0.2068 × (0.892)

3] / (0.45)

2

= 1.588 × 1010 C1

7.7% of E = 1.22 × 109 C1

E = (1.6 ± 0.1) × 1010

kg m–1

s–2 A1 [4]

2 (a) ps = 10–12

(s) or T = 4 × 50 × 10–12

(s) B1

v = fλ or v = λ / T C1

λ = 3.0 × 108 × 4 × 50 × 10

–12 C1

= 0.06(0) m A1 [4]

(b) 1500 = 3.0 × 108 × 4 × time-base setting or T = 5 × 10

–6 s C1

time-base setting = 1.3 (1.25) µs cm–1

A1 [2]

3 (a) work done is force × distance moved in direction of force

or no work done along PQ as no displacement/distance moved in direction of force B1 work done is same in vertical direction as same distance moved in direction of

force B1 [2]




(b) (i) at maximum height t = 1.5 (s) or s = ½(u + v) t, s = 11 m and t = 1.5 s C1

Vv = 0 + 9.81 × 1.5 Vv = (11 × 2) / 1.5

= 15 (14.7) m s

–1 A1 [2]

(ii) straight line from (0,0) to (3.00, 25.5) B1 [1]

(iii) at maximum height Vh = 25.5 / 3 (= 8.5 m s–1

) B1 ratio = mgh / ½ mv2 C1

= (2 × 9.81 × 11.0) / (8.5)2

= 3.0 (2.99) A1 [3]

(iv) deceleration is greater/resultant force (weight and friction force) is greater M1

time is less A1 [2] 4 (a) density = mass / volume C1

mass = 7900 × 4.5 × 24 × 10–6

= 0.85 (0.853) kg M1 [2]

(b) pressure = force / area C1

force = W cos 40° C1

pressure = (0.85 × 9.81 cos 40°) / 24 × 10–4

= 2.7 (2.66) × 103

Pa A1 [3]

(c) F = ma C1 W sin 40° – f = ma C1

0.85 × 9.81 × sin 40° – f = 0.85 × 3.8

f (= 5.36 – 3.23) = 2.1 N [5.38 – 3.242 if 0.8532 kg is used for the mass] A1 [3]




5 (a) progressive: all particles have same amplitude

stationary: no nodes or antinodes or maximum to minimum/zero amplitude B1

progressive: adjacent particles are not in phase

stationary: waves particles are in phase (between adjacent nodes) B1 [2]

(b) (i) wavelength 1.2 m (zero displacement at 0.0, 0.60 m, 1.2 m, 1.8 m, 2.4 m)

either peaks at 0.30 m and 1.5 m and troughs at 0.90 m and 2.1 m

or vice versa (but not both) B1

maximum amplitude 5.0 mm B1 [2]

(ii) 180° or π rad A1 [1]

(iii) at t = 0 particle has kinetic energy as particle is moving B1 at t = 5.0 ms no kinetic energy as particle is stationary

so decrease in kinetic energy (between t = 0 and t = 5.0 ms) B1 [2]

6 (a) energy converted from chemical to electrical per unit charge B1 [1]

(b) (i) current = E / (R + r) C1 = 6.0 / (16 + 0.5)

= 0.36 (0.364) A A1 [2]

(ii) terminal p.d. = (0.36 × 16) = 5.8 V or (6 – 0.36 × 0.5)

= 5.8 V A1 [1]

(c) (i) use of R = ρl / A or proportionality with length and inverse

proportionality with area or d

2 C1

d / 2 and l / 2 gives resistance of Z = 2RY = 24 (Ω) C1

R = resistance of parallel combination = [1/24 + 1/12]–1

= 8(.0) (Ω) A1 [3]

(ii) resistance of circuit less therefore current larger B1

lost volts greater therefore terminal p.d. less B1 [2]

(d) power = I

2 R or VI or V

2 / R C1

current in second circuit (= 6.0 / 12.5) = 0.48 (A) B1

ratio = [(0.36)2 × 16] / [(0.48)

2 × 12] = 0.75 [0.77 if full s.f. used] B1 [3]




7 (a) (i) curved path towards negative (–) plate (right-hand side) B1 [1]

(ii) range of α-particle is only few cm in air/loss of energy of the α-particles due

to collision with air molecules/ionisation of the air molecules B1 [1]

(iii) V = E × d C1

= 140 × 106 × 12 × 10

–3 = 1.7 (1.68) MV A1 [2]

(b) β have opposite charge to α therefore deflection in opposite direction B1

β has a range of velocities/energies hence number of different deflections B1

β have less mass or q / m is larger hence deflection is greater

or

β with (very) high speed (may) have less deflection B1 [3]

(c)

emitted particle change in Z change in A

α-particle –2 –4

β-particle +1 0

A1 [1]





9702 PHYSICS








1 (a) v = fλ C1

λ = (3.0 × 108) / (4.6 × 10

20) C1

( = 6.52 × 10–13

=) 0.65(2) pm A1 [3]

(b) t = (8.5 × 1016

) / (3.0 × 108) C1

( = 2.83 × 108 =) 0.28(3) Gs A1 [2]

(c) mass, power and temperature all underlined and no others B1 [1]

(d) (i) arrow in the direction 30° to 40° south of east B1 [1]

(ii) triangle of velocities completed (i.e. correct scale diagram) or correct working

given C1

e.g. [142 + 8.0

2 – 2(14)(8.0) cos 60°]

1/2

or [(14 – 8.0 cos 60°)2 + (8.0 sin 60°)

2]1/2

resultant velocity = 12(.2) (or 12.0 to 12.4 from scale diagram) m s–1 A1 [2]

2 (a) (i) v = u + at C1

0 = 3.6 – 3.0t t (= 3.6 / 3.0) = 1.2 s A1 [2]

(ii) (distance to rest from P = (3.6 × 1.2) / 2 =) 2.2 (2.16) m A1 [1]

or

[0 – (3.6)2] / [2 × (–3.0)] = 2.2 (2.16) m

or

3.6 × 1.2 – ½ × 3.0 × (1.2)2 = 2.2 (2.16) m

or

0 + ½ × 3.0 × (1.2)2 = 2.2 (2.16) m

(b) distance = 6.0 – 2.16 (= 3.84) C1

v2 = u2

+ 2as = 2 × 3.0 × 3.84 (= 23.04) M1

or

x + 2 × 2.16 = 6.0 gives x = 1.68 (m) (C1)

v2 = 3.6

2 + 2 × 1.68 × 3.0 (= 23.04) (M1)

or correct method with intermediate time calculated (t = 1.6 s from Q to R)

v = 4.8 m s

–1 A0 [2]




(c) straight line from v = 3.6 m s

–1 to v = 0 at t = 1.2 s B1

straight line continues with the same gradient as v changes sign B1

straight line from v = 0 intercept to v = –4.8 m s–1

B1 [3]

(d) difference in KE = ½m(v2

– u2)

= 0.5 × 0.45 (4.82 – 3.6

2) [= 5.184 – 2.916] C1

= 2.3 (2.27) J A1 [2]

3 (a) (i) k = F / x or 1 / gradient C1

(k = 4.4 / (5.4 × 10–2

) =) 81 (81.48) N m–1

A1 [2]

(ii) work done = area under line or ½Fx or ½kx2 C1

(= 0.5 × 4.4 × 5.4 × 10–2

=) 0.12 (0.119) J A1 [2]

(b) (i) kinetic energy/Ek of trolley/T (and block) changes to EPE/strain

energy/elastic energy of spring B1

EPE changes to KE of trolley/T and KE of block or to give lower KE to trolley B1 [2]

(ii) change in momentum = m(v + u) C1

= 0.25 (0.75 + 1.2) = 0.49 (0.488) N s A1 [2] 4 (a) product of the force and the perpendicular distance to/from a point/pivot B1 [1]

(b) (i) 4000 × 2.8 × sin 30° or 500 × 1.4 × sin 30° or T × 2.8

or 4000 × 1.4 or 500 × 0.7 B1

4000 × 2.8 × sin 30° + 500 × 1.4 × sin 30° = T × 2.8 M1

hence T = 2100 (2125) N A0 [2] (ii) (Tv = 2100 cos 60° =) 1100 (1050) N A1 [1]

(iii) there is an upward (vertical component of) force at A B1

upward force at A + Tv = sum of downward forces/weight+load/4500 N B1 [2]




5 (a) (i) I = V / R C1 ( = 240 / 1500 =) 0.16 A A1 [2]

(ii) I2 = 0.40 – 0.16 (= 0.24) C1

0.24 (350 + R) = 240

R = 650 Ω A1 [2]

(iii) power = IV or I2R or V2/R C1

ratio = (84 × 0.24) / (88 × 0.16)

or [(0.24)2 × 350] / [(0.16)

2 × 550]

or (842/350) / (88

2 / 550)

or 20.16 / 14.08

= 1.4(3) A1 [2]

(b) (i) p.d. across 350 Ω resistor = 0.24 × 350

or p.d. across 550 Ω resistor = 0.16 × 550 C1

V350 = 84 (V) and V550 = 88 (V) gives VAB = 4.0 V

or V950 = 152 (V) and VR = 156 V gives VAB = 4.0 V A1 [2]

(ii) p.d. across R increases or potential at B increases or V350 decreases hence

VAB increases B1 [1] 6 (a) internal resistance causes lost volts B1

p.d. across lamp is less than 12 V, power is less than 48 W B1 [2]

(b) (i) greater lost volts or p.d. across cell/lamp reduced, less current in lamp B1 [1]

(ii) p.d. across lamp/current in lamp decreases, hence resistance decreases B1 [1] 7 (a) (i) 3.2 mm A1 [1] (ii) 20 mm A1 [1] (b) (i) energy is transferred/propagated (through the water) or wave

profile/wavefronts move (outwards from dipper) so progressive B1 [1] (ii) to produce waves with constant/zero phase difference/coherent waves B1 [1]




(c) (i) path difference is λ B1

water vibrates/oscillates with amplitude about 2 × 3.2 mm B1 [2]

(ii) path difference is λ / 2 so little/no motion/displacement/amplitude B1 [1]

8 (a) result: majority / most (of the α-particles) went straight through/were deviated by

small angles M1

conclusion: most of the atom is (empty) space or size/volume of nucleus very

small compared with atom A1

result: a small proportion were deflected through large angles or >90° or came

straight back M1 conclusion: the mass or majority of mass is in a (very) small charged

volume/region/nucleus A1 [4]

(b) ρ = m / V C1

mass of atom and mass of nucleus (approx.) equal stated or cancelled or values

given e.g. 63 u or 63 × 1.66 × 10–27 C1

ratio = (rA)3

/ (rN)3 = (1.15 × 10

–10)3

/ (1.4 × 10–14

)3

or

ratio = (dA)3

/ (dN)3 = (2.3 × 10

–10)3

/ (2.8 × 10–14

)3

= 5.5 × 1011

A1 [3]





9702 PHYSICS








1 (a) energy or W: kg m2

s–2

or power or P: kg m

2 s

–3 M1

intensity or I: kg m2

s–2

m–2

s–1

(from use of energy expression)

or kg m

2 s

–3 m

–2 (from use of power expression)

indication of simplification to kg s–3 A1 [2]

(b) (i) ρ: kg m–3

, c: m s–1

, f: s–1, x0: m M1

substitution of terms in an appropriate equation and simplification to show K

has no units A1 [2]

(ii) I = 20 × 1.2 × 330 × (260)2

× (0.24 × 10–9

)2 C1

= 3.1 × 10–11

(W m–2

) C1

= 31 (30.8) pW m

–2 A1 [3]

2 (a) (i) (the loudspeakers) are connected to the same signal generator B1 [1]

(ii) 1. the waves (that overlap) have phase difference of zero or path difference

of zero and so

either constructive interference

or displacement larger B1 [1]

2. the waves (that overlap) have phase difference of (n + ½) × 360° or

(n + ½) × 2π rad or path difference of (n + ½)λ and so

either destructive interference

or displacements cancel/smaller B1 [1]

3. the waves (that overlap) are in phase or have phase difference of n360°

or 2πn rad or path difference of nλ and so

either constructive interference

or displacement larger B1 [1] (b) time period = 0.002 s or 2 ms C1

wave drawn is half time period B1

amplitude 1.0 cm (same as Fig. 2.2) B1 [3]




3 (a) (i) 1. s = ut + ½ at2

192 = ½ × 9.81 × t2 C1 t = 6.3 (6.26) s A1 [2]

2. max Ek (= mgh) = 0.27 × 9.81 × 192 C1

or

calculation of v (= 61.4) and use of EK (= ½ mv2) = ½ × 0.27 × (61.4)

2 (C1)

max Ek = 510 (509) J A1 [2]

(ii) velocity is proportional to time or velocity increases at a constant rate

as acceleration is constant or resultant force is constant B1 [1]

(iii) use of v = at or v2

= 2as or E = ½ mv2 to give v = 61(.4) m s

–1 B1 [1]

(b) (i) R increases with velocity B1

resultant force is mg – R or resultant force decreases B1

acceleration decreases B1 [3]

(ii) at v = 40 m s

–1, R = 0.6 (N) C1

0.27 × 9.8 – 0.6 = 0.27 × a

a = 7.6 (7.58) m s–2

A1 [2]

(iii) R = weight for terminal velocity B1

either weight requires velocity to be about 80 m s

–1

or at 60 m s–1

, R is less than weight

so does not reach terminal velocity B1 [2]

4 (a) (i) reaction/vertical force = weight – P cos 60° C1

= 180 – 35 cos 60°

= 160 (163) N A1 [2]

(ii) work done = 35 sin 60° × 20 C1

= 610 (606) J A1 [2]




(b) (i) work done by force P = work done against frictional force B1 [1]

(ii) horizontal component of P is equal and opposite to frictional force B1

vertical component of P + normal reaction force equal and opposite to weight B1 [2]

5 (a) (i) resistance = V / I B1

very high/infinite resistance at low voltages B1

resistance decreases as V increases B1 [3]

(ii) p.d. from graph 0.50 (V) C1

resistance = 0.5 / (4.4 × 10–3

)

= 110 (114) Ω A1 [2]

(b) (i) current (= 1.2 / 375) = 3.2 × 10–3

A A1 [1]

(ii) current in diode = 4.4 × 10–3

(A)

total resistance = 1.2 / 4.4 × 10–3

= 272.7 (Ω) C1

resistance of R1 = 272.7 – 113.6 = 160 (159) Ω A1

or

p.d. across diode = 0.5 V and p.d. across R1 = 0.7 V (C1)

resistance of R1 = 0.7 / 4.4 × 10–3

= 160 (159) Ω (A1) [2]

(iii) power = IV or I2R or V2/ R C1

ratio = (4.4 × 0.5) / (3.2 × 1.2)

or [(4.4)2 × 114] / [(3.2)

2 × 375]

or [(0.5)2 × 375] / [114 × (1.2)

2]

= 0.57 A1 [2]

6 (a) waves from loudspeaker (travel down tube and) are reflected at closed end B1

two waves (travelling) in opposite directions with same frequency/wavelength

overlap B1 [2]

(b) (i) 0.51 m A1

0.85 m A1 [2]

(ii) A at open end, N at closed end, with an N and A in between, equally spaced

(by eye) B1 [1]




7 (a) stress or σ = F / A C1

max. tension = UTS × A = 4.5 × 108 × 15 × 10

–6 = 6800 (6750) N A1 [2]

(b) ρ = m / V C1

weight = mg = ρVg = ρALg

6750 = 7.8 × 103 × 15 × 10

–6 × L × 9.81 C1

L = 5.9 (5.88) × 103

m A1

or

maximum mass = 6750 / 9.81 = 688 kg (C1)

mass per unit length = ρA = 0.117 kg m–1

(C1)

L = 688 / 0.117 = 5.9 × 103

m (A1)

or maximum mass = 6750 / 9.81 = 688 kg (C1)

volume = m / ρ = 0.0882 m3 = LA (C1)

L = 0.0882 / 15 × 10–6

= 5.9 × 103

m (A1) [3]

8 (a) mass-energy

proton number or charge

nucleon number B2 [2]

(b) (i) Ek = ½ mv2

and p = mv with working leading to

[via Ek = ½ m2v2 / m or ½ m (p / m)

2]

to Ek = m

p

2

2

B1 [1]

(ii) p = (2Ekm)½ hence (2[Ekm]α)

½ = (2[Ekm]Th)

½ C1

2 × [Ek]Th × 234 = 2 × 6.69 × 10–13

× 4 C1

[Ek]Th = 1.14 × 10–14

J = 71(.5) keV A1

or

calculation of speed of α-particle = 1.42 × 107

m s–1

calculation of momentum of α-particle/nucleus = 9.43 × 10–20

N s (C1)

[Ek]Th = 1.14 × 10–14

J (C1)

= 71(.5) keV (A1) [3]





9702 PHYSICS

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







1 (a) (iv) Value of d in range 19.5 cm to 20.5 cm with unit. [1]

(c) (ii) Value of N with evidence of repeat readings. [1]

(e) Six sets of readings of d and N scores 5 marks, five sets scores 4 marks etc. [5]

Incorrect trend –1. Help from Supervisor –1.

Range: [1]

Smallest value of d < 9.5 cm.

Column headings: [1]

Each column heading must contain a quantity and a unit where appropriate. The

presentation of quantity and unit must conform to accepted scientific convention

e.g. d –1 / m

–1.

Consistency: [1]

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

Significant figures: [1]

Every value of 1 / d must be given to the same number of significant figures as (or

one more than) the number of significant figures in the corresponding value of d.

Calculation: [1]

√N calculated correctly to the number of significant figures given by the candidate.

(f) (i) Axes: [1]

Sensible scales must be used. Awkward scales (e.g. 3:10) are not allowed.

Scales must be chosen so that the plotted points occupy at least half the

graph grid in both x and y directions.

Scales must be labelled with the quantity that is being plotted.

Scale markings should be no more than three large squares apart.

Plotting: [1]

All observations in the table must be plotted on the grid.

Diameter of plotted points must be ⩽ half a small square (no “blobs”).

Points must be plotted to an accuracy of half a small square.

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

Scatter of points must be no more than ±0.5 in the √N direction from a

straight line.

(ii) Line of best fit: [1]

Judge by balance of all points on the grid about the candidate’s line (at least

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 (i.e. circled or labelled) by

the candidate.

Lines must not be kinked or thicker than half a square.




(iii) Gradient: [1]

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

line. Do not allow ∆x / ∆y. Sign of gradient must match graph drawn.

Both read-offs must be accurate to half a small square in both the x and y directions.

y-intercept: [1]

Either:

Correct read-offs from a point on the line substituted into y = mx + c or an

equivalent expression. Read-offs must be accurate to half a small square in

both x and y directions.

Or:

Intercept read directly from the graph, with read-off accurate to half a small

square.

(g) Value of A = candidate’s gradient and value of B = candidate’s intercept. [1]

Unit for A correct (e.g. m or cm or mm) and consistent with value.

No unit given for B. [1]

2 (a) (ii) Value for x to the nearest mm. [1]

x in the range 0.155 m to 0.165 m. [1]

(b) (i) Correct calculation of C in m (correct to 2 s.f.). [1]

(ii) Justification for significant figures in C linked to significant figures in x and h. [1]

(c) (ii) Value for R with unit. [1]

Evidence of repeat readings. [1]

(iii) Absolute uncertainty in R in range 5 mm to 20 mm and correct method of

calculation to obtain percentage uncertainty. If repeated readings have been

taken, then the uncertainty can be half the range (but not zero) if the working

is clearly shown. [1]

(d) Second value of x. [1]

Second value of R. [1]

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

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

(ii) Valid comment consistent with the calculated values of k, testing against a

criterion specified by the candidate. [1]




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

A Not enough readings to draw a

conclusion Take many readings for different

holes and plot a graph/

obtain more k values and

compare

Few readings/

only one reading/

not enough readings for

an accurate result/

“repeat readings”

on its own/

take more readings

and (calculate)

average k

B Ball rolls off block/ball does not

move along straight line from the

wood/rod does not hit marble

square on each time/rod hits

marble at an angle

Small groove in the wood to

place the marble

C Difficult to measure distance rod is

pulled back/difficult to hold rod still

before release

Use another stand or stop

D Difficult to measure R with reason

e.g. marble skids in sand leaving

elongated hole/can’t fit ruler in

sand tray/parallax error

Improved method for measuring

R e.g. video with scale/use

carbon paper/ink on marble/put

scale on the sand

E Difficult to flatten sand/know when

sand is horizontal

Use a straight edge/

use a spirit level

F Difficult to measure x with reason

e.g. wooden rod moves

Method of measuring x/

clamp rule close by/

draw scale on rod





9702 PHYSICS



®,

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




1 (b) (i) Value for L to the nearest mm, with unit. [1]

(c) Second value of h > first value of h. [1]

(d) (ii) Six sets of readings of m, h and θ scores 5 marks, five sets scores 4 marks etc. [5]

Help from Supervisor –1.

Incorrect trend –1. Correct trend is h increases as m increases.

Range: [1]

Range of values to include mmin < 60 g and mmax > 80 g.

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

The presentation of quantity and unit must conform to accepted scientific

convention e.g. h / cos θ (cm).

Consistency: [1] All values of h must be given to the nearest mm.


Every value of h / cos θ must be given to 2 or 3 significant figures only.

Calculation: [1]

Values of h / cosθ calculated correctly to the number of significant figures

given by the candidate.

(e) (i) Axes: [1]






Plotting: [1]




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

Scatter of points must be no more than 10 g in the m direction of a straight

line.





Allow one anomalous point only if clearly indicated (i.e. circled or labelled) by the

candidate.





(iii) Gradient: [1] The hypotenuse of the triangle must be greater than half the length of the

drawn line.

Do not allow ∆x / ∆y. Sign of gradient must match graph drawn.


y-intercept: [1] Either:


equivalent expression.

Read-offs must be accurate to half a small square in both x and y directions.

Or:

Intercept read directly from the graph, with read-off accurate to half a small square.

(f) Value of A = candidate’s gradient and value of B = candidate’s intercept. [1]

Unit for A correct (e.g. m kg–1

or cm g–1

) and unit for B correct (m or cm or mm). [1]

2 (a) (iii) Values of x, y and z to the nearest mm with unit. [1]

Value of z > value of x. [1] (iv) Absolute uncertainty in y of 1 mm to 4 mm and correct method of calculation to

obtain percentage uncertainty. If repeated readings have been taken, then the

uncertainty can be half the range (but not zero) if working is clearly shown. [1]

(b) Correct calculation of C with consistent unit. [1]

(c) (ii) Value for T with unit in range 5.0 s > T > 0.5 s. [1] Evidence of repeat readings for T. [1]

(iv) Justification for significant figures in T2 linked to significant figures in the (raw) times. [1]

(d) Second values of x, y and z. Value of y within 5 mm of value in (a)(iii). [1]

Second value of T. [1]

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

(e) (i) Two values of k calculated correctly. [1] (ii) Valid comment consistent with the calculated values of k, testing against

a criterion specified by the candidate. [1]






conclusion

Take many readings (for different

masses) and plot a graph/

obtain more k values and compare

“Repeat readings” on its

own/

few readings/

only one reading/

not enough readings for

an accurate result/

take more readings and

(calculate) average k

B Rod is bent when loaded Use smaller masses/

rigid/stiff/thick rod

Just “rod is bent”/

shorter rod

C Difficult to get horizontal Use a spirit level or named

instrument.

D Difficult to measure distances with

reason e.g. rod unstable/awkward

with metre rule/rod moves/holding

ruler mid-air

Add a scale on rod/

use travelling microscope/

clamp ruler

Parallax

Do not award if reason

given is bent rod.

E y not constant with a reason e.g.

spring/loop moves around during

oscillations

Cut groove or drill hole in wooden

rod/

tape to wooden rod

F Difficult to judge the start/end of an

oscillation

or Difficult to judge when to start/stop

the stopwatch

(Fiducial) marker at centre/

video and timer/view frame by

frame/

motion sensor placed below/above

More oscillations/

high speed camera/

reaction time/

human error

G Oscillation in more than one

plane/irregular oscillations

Wind/draughts/

switch off air

conditioning/

close doors and

windows





9702 PHYSICS








1 (b) (i) Value of θ to the nearest degree and in the range 135° to 165°. [1]

(ii) Value of L in range 5.0 to 10.0 cm, with unit. [1]

(d) Six sets of readings of θ and L scores 5 marks, five sets scores 4 marks etc. [5]


Range: [1]

θmax ⩾ 160° and θmin ⩽ 140°.


Each column heading must contain a quantity and a unit. The presentation of

quantity and unit must conform to accepted scientific convention e.g. θ / °.

1/sin(θ – 90°) must have no unit.

Consistency: [1]

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


Every value of 1/sin(θ – 90°) must be given to 2 or 3 significant figures only.

Calculation: [1]

Values of 1/sin(θ – 90°) calculated correctly to the number of significant figures

given by the candidate.

(e) (i) Axes: [1]






Plotting: [1]




Quality: [1]

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

Scatter of points must be no more than ±0.3 (to scale) cm in the L direction

from a straight line.






the candidate.





(iii) Gradient: [1]

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

Do not allow ∆x / ∆y. Sign of gradient must match graph drawn.


y-intercept: [1]

Either:




Or:

Intercept read directly from the graph, with read-off accurate to half a small square.

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

Unit for a is correct (e.g. cm–1

) and no unit for b. [1] 2 (a) (i) All values of d to nearest mm, with unit, in range 5 to 30 mm. [1]

Value of l greater than value of d. [1]

(ii) Correct calculation of V with consistent unit. [1]

(b) Justification for significant figures in V linked to significant figures in d and l. [1]

(c) (iii) t in range 5.00 s to 30.00 s, with unit. [1]

Evidence of repeat readings of t. [1]

(d) Absolute uncertainty in t in range 0.5 s to 5.0 s and correct method of calculation to


absolute uncertainty can be half the range (but not zero) if working is clearly shown. [1]

(e) Second values of d and l. [1]

Second value of t. [1]

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

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






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


conclusion

Take more readings and plot a

graph/


compare

Few readings/

only one reading/

not enough readings

for an accurate result/


on its own/

take more readings

and (calculate)

average k

B d is small/large uncertainty in d Improved method of measuring

d e.g. micrometer/vernier

calipers/digital calipers/travelling

microscope

Difficult to measure d/ parallax error/

“calipers” on its own/

use bigger/larger

components

C Volume of component not accurate,

with reason e.g. component not

cylindrical/has groove.

Method to find volume of

component e.g. use liquid

displacement method

D Difficult to judge/know/see when

LED goes out.

Use dark(ened) room/

light meter/

light sensor/

cardboard tube over LED/

voltmeter to measure time for

p.d. to fall below specific value

Use video

E Poor/dirty/loose contacts Method of cleaning contacts e.g.

iron wool





9702 PHYSICS








1 (b) (i) Value of a with consistent unit and in the range 0 to 50.0 cm. [1]

(v) Value of V with unit to nearest 0.001 V and in range 0 to 2 V. [1]

(c) Six sets of readings of a, b and V scores 5 marks, five sets scores 4 marks etc. [5]

Incorrect trend –1. Minor help from Supervisor –1. Major help from Supervisor –2.

Range: [1] amax – amin ⩾ 30.0 cm.


Each column heading must contain a quantity and a unit. The presentation of

quantity and unit must conform to accepted scientific convention. e.g. a / m or

a (m), 1/V / V–1

.

Consistency: [1]

All values of a and b must be given to the nearest mm.


Every value of 1 / V must be given to the same number of significant figures as (or

one more than) the number of significant figures in the corresponding value of V.

Calculation: [1]

Values of 1 / V calculated correctly to the number of significant figures given by

the candidate.

(d) (i) Axes: [1]






Plotting: [1]




Quality: [1]


Scatter of points must be no more than ± 0.050 m (5.0 cm) to scale in the b direction from a straight line.






the candidate.





(iii) Gradient: [1]

The hypotenuse of the triangle must be greater than half the length of the

drawn line. Do not allow ∆x / ∆y. Sign of gradient must match graph drawn.


y-intercept: [1]

Either:

Correct read-off from a point on the line substituted into y = mx + c or an



Or:

Intercept read directly from the graph, with read-off accurate to half a small

square.

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

Unit for P is correct (e.g. m–1

V–1

) and unit for Q is correct (e.g. V–1

). [1]

2 (b) (iv) All values of x with unit to nearest mm. Average x ⩽ 20.0 cm. [1]

(c) (i) Value of θ to the nearest degree in the range 25° to 35°. [1]

(ii) Absolute uncertainty in θ in range 2° to 5° and correct method of calculation

to obtain percentage uncertainty. If repeated readings have been taken, then

the absolute uncertainty can be half the range (but not zero) if working is

clearly shown. [1]

(iii) Correct calculation of cos2(θ / 2). No unit. [1]

(iv) Justification for significant figures in cos2(θ / 2) linked to significant figures in θ. [1]

(d) (ii) T in range 0.5 s to 2.5 s, with unit. [1]

Evidence of repeat readings of T. [1]

(f) Second value of θ. [1]

Second values of T1 and T2. [1]

Second value of T1 / T2 < first value of T1 / T2 when rounded to 2 s.f. [1]

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






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


conclusion


graph/


compare

Few readings/

only one reading/

not enough readings

for an accurate result/


on its own/

take more readings

and (calculate)

average k

B Difficult to measure θ or read

protractor with reason e.g. rod is

above protractor/rod obscures view

Improved method of measuring

θ e.g. shadow projection with

light above/thinner rod/rod with

smaller diameter/plumb-lines

hung from rod/larger

protractor/360° protractor

Just “difficult to

measure θ”/ smaller rod/

“protractor reads to

1°”

C Oscillations or T affected by...

e.g. air movement

force on release

different forces

angle of release

unwanted modes of oscillation

Improved method of release e.g.

card gate

or switch off air conditioning/close

windows/closed room

Release by

electromagnet/

cutting string/

damping/

air resistance/

friction

D Large percentage uncertainty in

time/period T is short

Valid method to improve timing

e.g. use video with timer/frame

by frame/motion sensor and

position at side of cradle

or Increase T by... e.g. heavier

nail/longer nail/string

Oscillations too fast/

high/low speed

camera/

video on its own/

human reaction time/

just “difficult to

determine time”/

fiducial marker

E Valid problem linked to magnetism

e.g. nail weakly magnetised/metal

stand attracts nail/interference by

Earth’s magnetic field.

Valid method to overcome

problem linked with magnetism

e.g. stroke nail more times/use

of coil

Nail loses magnetism

F Method of fixing

paper/protractor/magnets e.g.

tape/Blu-tack/draw protractor on

paper

Just “stick paper”

without method





9702 PHYSICS








1 (a) Value of H in the range 13.0 cm to 17.0 cm. [1]

(c) (iii) Value of F to nearest 0.1 N. [1]

(iv) Value of x correct and in range 3.5 cm to 4.5 cm. [1]

(d) Six sets of readings of hw, hb and F scores 4 marks, five sets scores 3 marks etc. [4]


Range: [1] xmax – xmin ⩾ 6.0 cm.


Each column heading must contain a quantity and a unit.

The presentation of quantity and unit must conform to accepted scientific convention,

e.g. (H – x)3

/ cm3.

Consistency: [1]

All values of hw and hb must be given to the nearest mm.


Significant figures for (H – x)3 must be the same as, or one greater than, the

number of s.f. for (H – x).

Calculation: [1]

Values of (H – x)3 calculated correctly.

(e) (i) Axes: [1]


Scales must be chosen so that the plotted points occupy at least half the graph

grid in both x and y directions.



Plotting: [1]



Points must be plotted to an accuracy of half a small square. Quality: [1]


Scatter of points must be less than ± 0.2 N from a straight line in the F direction.


Judged 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 (i.e. circled or labelled) by the

candidate.





(iii) Gradient: [1]

The hypotenuse of the triangle used must be greater than half the length of the

drawn line.

The method of calculation must be correct.


y-intercept: [1]

Either:




Or:

Intercept read directly from the graph, with read-off accurate to half a small square

in the y direction.

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

Unit for a is correct (e.g. N cm–3

) and unit for b is correct (e.g. N). [1]

2 (a) (i) All values of d to nearest 0.001 cm and in range 0.100 cm to 0.500 cm. [1]

(ii) All values of D to nearest 0.1 cm. [1]

(b) Values of l, h and s present.

Value of h to nearest 0.1 cm and in range 7.5 cm to 8.5 cm, with unit. [1]

(c) (ii) t on answer line in range 1.00 s to 20.00 s, with unit. [1]

Evidence of repeated readings for t. [1] (iii) Absolute uncertainty in t in range 0.2 s to 0.5 s and correct method of calculation to


absolute uncertainty can be half the range (but not zero) if the working is clearly

shown. [1]

(iv) Calculation of g correct to the second s.f., with consistent unit (e.g. m s

–2). [1]

(d) (ii) Second values of d and t. [1]

t greater for smaller d. [1]

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

(ii) Justification for significant figures in k linked to significant figures in t and d. [1]

(iii) Valid comment consistent with the calculated values of k, tested against a







conclusion


graph/


compare

“Repeat readings”

on its own/

few readings/

only one reading/

take more readings

and (calculate)

average k/

two readings not

enough for accurate

results

B Parallax error when measuring

h or D

Use calipers/

use set square as pointer

Rule not vertical

when measuring h

C Plastic deforms when measuring

larger d/diameter with tubing

Use larger diameter axle instead

of tubing

D Push force to start flywheel may

vary/push force may affect time

Release mechanism, with detail/

use steeper ramp

“Force may be too

large”/

start before top line

E Flywheel doesn’t travel straight/

flywheel hits (sides of) track

Sensible method of preventing

collision, e.g. level the track

sideways/widen the track

Reduce friction/

mark both sides of

track

F Use video with timer/

view frame by frame/

use light gates at start and end/

use longer track

Reaction time





9702 PHYSICS

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







Section A

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

GMmS / x2 = mSv2

/ x (allow x or r; allow m or mS) M1

EK = ½mSv2 and clear algebra leading to EK = GMmS / 2x A1 [3]

(ii) EP = – GMmS / x (sign essential) B1 [1]

(iii) ET = EK + EP

= GMmS / 2x – GMmS / x C1

= – GMmS / 2x (allow ECF from (a)(ii)) A1 [2]

(b) (i) decreases B1 [1]

(ii) decreases B1 [1]

(iii) decreases B1 [1]

(iv) increases B1 [1]

(for answers in (b) allow ECF from (a)(iii))

2 (a) obeys the equation pV = nRT or pV / T = constant M1

all symbols explained; T in kelvin/thermodynamic temperature A1 [2]

(b) (i) temperature rise = 48 K A1 [1]

(ii) <c2> ∝ T or equivalent C1

<c2> = (353 / 305) × 1.9 × 10

6 C1

cr.m.s. = 1480 m s–1

A1 [3]

3 (a) heat/thermal energy gained by system or energy transferred to system by heating B1

plus work done on the system or minus work done by the system B1 [2]

(b) (i) either volume decreases so work done on the system

or small volume change so work done on system negligible M1

(thermal) energy absorbed to break lattice structure M1

internal energy increases A1 [3]

(ii) gas expands so work done by gas (against atmosphere) M1

no time for thermal energy to enter or leave the gas M1

internal energy decreases A1 [3]

4 (a) free: (body oscillates) without any loss of energy/no resistive forces/no external

forces applied B1

forced: continuous energy input (required)/body is made to vibrate by an

(external) periodic force/driving oscillator B1 [2]




(b) (i) idea of resonance B1

maximum amplitude at natural frequency B1

frequency = 2.1 Hz (allow 2.08 to 2.12 Hz) B1 [3]

(ii) peak not very sharp/amplitude not infinite so frictional forces are present B1 [1]

(c) v = ωx0

= 2π × 2.1 × 4.7 × 10–2

(allow ECF from (b)(i)) C1 = 0.62 m s

–1 A1 [2]

5 (a) (i) force proportional to the product of the two/point charges B1

and inversely proportional to the square of their separation B1 [2]

(ii) 1. force radially away from sphere/to right/to east B1 [1]

2. (maximum) at/on surface of sphere or x = r B1 [1]

3. F ∝ 1 / x2 or F = q1q2

/ (4πε 0x2) C1

ratio = 16 A1 [2]

(b) E = q / (4πε0x2) or E ∝ q C1

maximum charge = (2.0 / 1.5) × 6.0 × 10–7

C1

= 8.0 × 10–7

C

additional charge = 2.0 × 10–7

C A1 [3]

6 (a) (i) force = mg M1

along the direction of the field/of the motion A1 [2]

(ii) no force B1 [1]

(b) (i) force due to E-field downwards so force due to B-field upwards B1

into the plane of the paper B1 [2]

(ii) force due to magnetic field = Bqv B1

force due to electric field = Eq B1

(use of FB and FE not explained, allow 1/2)

forces are equal (and opposite) so Bv = E or Eq = Bqv so E = Bv B1 [3]

(c) sketch: smooth curved path M1

in ‘upward’ direction A1 [2]

7 (a) minimum frequency of e.m. radiation/a photon (not “light”) M1

for emission of electrons from a surface A1 [2]

(reference to light/UV rather than e.m. radiation, allow 1/2)




(b) EMAX corresponds to electron emitted from surface B1

electron (below surface) requires energy to bring it to surface, so less than EMAX B1 [2]

(c) (i) 1/λ0 = 1.85 × 106 (allow 1.82 to 1.88) C1

f0 = c / λ0

= 3.00 × 108 × 1.85 × 10

6

= 5.55 × 1014

Hz A1 [2]

(ii) Φ = hf0

= 6.63 × 10–34

× 5.55 × 1014

(allow ECF from (c)(i)) C1

= 3.68 × 10–19

J A1 [2]

(d) sketch: straight line with same gradient M1

intercept between 1.0 and 1.5 A1 [2]

8 (a) nucleus: small central part/core of an atom B1

nucleon: proton or a neutron B1

particle contained within a nucleus B1 [3]

(b) (i) 1. decay constant = ln 2 / (3.8 × 24 × 3600) C1

= 2.1 × 10–6

s–1

A1 [2]

2. A = λN

97 = 2.1 × 10–6

× N C1

N = 4.6 × 107 A1 [2]

(ii) 1.0 m3 contains (6.02 × 10

23) / (2.5 × 10

–2) air molecules C1

ratio = (4.6 × 107 × 2.5 × 10

–2) / (6.02 × 10

23)

= 1.9 × 10–18

A1 [2]




Section B 9 (a) (i) (+) 3.0 V B1 [1]

(ii) potential = 6.0 × 2.0 / (2.0 + 2.8) C1

= 2.5 V A1 [2]

(iii) potential = 6.0 × 2.0 / (2.0 + 1.8)

= 3.2 V A1 [1]

(b) at 10 °C, VA > VB M1

VOUT is –9.0 V (allow “negative saturation”) A1

at 20 °C, VOUT is +9.0 V B1

(if 20 °C considered initially, mark as M1,A1,B1) sudden switch (from –9 V to +9 V) when VA = VB B1 [4]

10 (a) sharpness: clarity of edges/resolution (of image) B1

contrast: difference in degree of blackening (of structures) B1 [2]

(b) (i) X-rays produced when (high speed) electrons hit target/anode B1

either electrons have been accelerated through 80 kV

or electrons have (kinetic) energy of 80 keV B1 [2]

(ii) IT / I = e–3.0

×

1.4 C1

= 0.015 A1 [2]

(c) for good contrast, µx or eµx or e

–µx must be very different B1

µx or eµx or e

–µx for bone and muscle will be different than that for muscle M1

so good contrast A1 [3]

11 (a) frequency of carrier wave varies M1

in synchrony with the displacement of the signal/information wave A1 [2]

(b) (i) 5.0 V A1 [1]

(ii) 720 kHz A1 [1]

(iii) 780 kHz A1 [1]

(iv) 7500 A1 [1]




12 (a) (i) (gradual) loss of power/intensity/amplitude (not “signal”) B1 [1]

(ii) e.g. noise can be eliminated (not “there is no noise”) M1

because pulses can be regenerated A1

e.g. much greater data handling/carrying capacity M1

because many messages can be carried at the same time/greater

bandwidth A1

e.g. more secure (M1)

because it can be encrypted (A1)

e.g. error checking (M1)

because extra information/parity bit can be added (A1) [4]

(allow any two sensible suggestions with ‘state’ M1 and ‘explain’ A1)

(b) attenuation = 10 lg (145 / 29) (= 7.0) C1

attenuation per unit length = 7.0 / 36

= 0.19 dB km–1

A1 [2]





9702 PHYSICS








Section A

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

GMmS / x2 = mSv2

/ x (allow x or r; allow m or mS) M1

EK = ½mSv2 and clear algebra leading to EK = GMmS / 2x A1 [3]

(ii) EP = – GMmS / x (sign essential) B1 [1]

(iii) ET = EK + EP

= GMmS / 2x – GMmS / x C1

= – GMmS / 2x (allow ECF from (a)(ii)) A1 [2]

(b) (i) decreases B1 [1]

(ii) decreases B1 [1]

(iii) decreases B1 [1]

(iv) increases B1 [1]

(for answers in (b) allow ECF from (a)(iii))

2 (a) obeys the equation pV = nRT or pV / T = constant M1

all symbols explained; T in kelvin/thermodynamic temperature A1 [2]

(b) (i) temperature rise = 48 K A1 [1]

(ii) <c2> ∝ T or equivalent C1

<c2> = (353 / 305) × 1.9 × 10

6 C1

cr.m.s. = 1480 m s–1

A1 [3]

3 (a) heat/thermal energy gained by system or energy transferred to system by heating B1

plus work done on the system or minus work done by the system B1 [2]

(b) (i) either volume decreases so work done on the system

or small volume change so work done on system negligible M1

(thermal) energy absorbed to break lattice structure M1

internal energy increases A1 [3]

(ii) gas expands so work done by gas (against atmosphere) M1

no time for thermal energy to enter or leave the gas M1

internal energy decreases A1 [3]

4 (a) free: (body oscillates) without any loss of energy/no resistive forces/no external

forces applied B1

forced: continuous energy input (required)/body is made to vibrate by an

(external) periodic force/driving oscillator B1 [2]




(b) (i) idea of resonance B1

maximum amplitude at natural frequency B1

frequency = 2.1 Hz (allow 2.08 to 2.12 Hz) B1 [3]

(ii) peak not very sharp/amplitude not infinite so frictional forces are present B1 [1]

(c) v = ωx0

= 2π × 2.1 × 4.7 × 10–2

(allow ECF from (b)(i)) C1 = 0.62 m s

–1 A1 [2]

5 (a) (i) force proportional to the product of the two/point charges B1

and inversely proportional to the square of their separation B1 [2]

(ii) 1. force radially away from sphere/to right/to east B1 [1]

2. (maximum) at/on surface of sphere or x = r B1 [1]

3. F ∝ 1 / x2 or F = q1q2

/ (4πε 0x2) C1

ratio = 16 A1 [2]

(b) E = q / (4πε0x2) or E ∝ q C1

maximum charge = (2.0 / 1.5) × 6.0 × 10–7

C1

= 8.0 × 10–7

C

additional charge = 2.0 × 10–7

C A1 [3]

6 (a) (i) force = mg M1

along the direction of the field/of the motion A1 [2]

(ii) no force B1 [1]

(b) (i) force due to E-field downwards so force due to B-field upwards B1

into the plane of the paper B1 [2]

(ii) force due to magnetic field = Bqv B1

force due to electric field = Eq B1

(use of FB and FE not explained, allow 1/2)

forces are equal (and opposite) so Bv = E or Eq = Bqv so E = Bv B1 [3]

(c) sketch: smooth curved path M1

in ‘upward’ direction A1 [2]

7 (a) minimum frequency of e.m. radiation/a photon (not “light”) M1

for emission of electrons from a surface A1 [2]

(reference to light/UV rather than e.m. radiation, allow 1/2)




(b) EMAX corresponds to electron emitted from surface B1

electron (below surface) requires energy to bring it to surface, so less than EMAX B1 [2]

(c) (i) 1/λ0 = 1.85 × 106 (allow 1.82 to 1.88) C1

f0 = c / λ0

= 3.00 × 108 × 1.85 × 10

6

= 5.55 × 1014

Hz A1 [2]

(ii) Φ = hf0

= 6.63 × 10–34

× 5.55 × 1014

(allow ECF from (c)(i)) C1

= 3.68 × 10–19

J A1 [2]

(d) sketch: straight line with same gradient M1

intercept between 1.0 and 1.5 A1 [2]

8 (a) nucleus: small central part/core of an atom B1

nucleon: proton or a neutron B1

particle contained within a nucleus B1 [3]

(b) (i) 1. decay constant = ln 2 / (3.8 × 24 × 3600) C1

= 2.1 × 10–6

s–1

A1 [2]

2. A = λN

97 = 2.1 × 10–6

× N C1

N = 4.6 × 107 A1 [2]

(ii) 1.0 m3 contains (6.02 × 10

23) / (2.5 × 10

–2) air molecules C1

ratio = (4.6 × 107 × 2.5 × 10

–2) / (6.02 × 10

23)

= 1.9 × 10–18

A1 [2]




Section B 9 (a) (i) (+) 3.0 V B1 [1]

(ii) potential = 6.0 × 2.0 / (2.0 + 2.8) C1

= 2.5 V A1 [2]

(iii) potential = 6.0 × 2.0 / (2.0 + 1.8)

= 3.2 V A1 [1]

(b) at 10 °C, VA > VB M1

VOUT is –9.0 V (allow “negative saturation”) A1

at 20 °C, VOUT is +9.0 V B1

(if 20 °C considered initially, mark as M1,A1,B1) sudden switch (from –9 V to +9 V) when VA = VB B1 [4]

10 (a) sharpness: clarity of edges/resolution (of image) B1

contrast: difference in degree of blackening (of structures) B1 [2]

(b) (i) X-rays produced when (high speed) electrons hit target/anode B1

either electrons have been accelerated through 80 kV

or electrons have (kinetic) energy of 80 keV B1 [2]

(ii) IT / I = e–3.0

×

1.4 C1

= 0.015 A1 [2]

(c) for good contrast, µx or eµx or e

–µx must be very different B1

µx or eµx or e

–µx for bone and muscle will be different than that for muscle M1

so good contrast A1 [3]

11 (a) frequency of carrier wave varies M1

in synchrony with the displacement of the signal/information wave A1 [2]

(b) (i) 5.0 V A1 [1]

(ii) 720 kHz A1 [1]

(iii) 780 kHz A1 [1]

(iv) 7500 A1 [1]




12 (a) (i) (gradual) loss of power/intensity/amplitude (not “signal”) B1 [1]

(ii) e.g. noise can be eliminated (not “there is no noise”) M1

because pulses can be regenerated A1

e.g. much greater data handling/carrying capacity M1

because many messages can be carried at the same time/greater

bandwidth A1

e.g. more secure (M1)

because it can be encrypted (A1)

e.g. error checking (M1)

because extra information/parity bit can be added (A1) [4]

(allow any two sensible suggestions with ‘state’ M1 and ‘explain’ A1)

(b) attenuation = 10 lg (145 / 29) (= 7.0) C1

attenuation per unit length = 7.0 / 36

= 0.19 dB km–1

A1 [2]





9702 PHYSICS








1 (a) (gravitational) force proportional to product of masses and inversely proportional to square of separation M1 either point masses or particles or ‘size’ ≪ separation A1 [2] (b) gravitational force provides the centripetal force B1

either GMm / x2 = mxω2 or mv2 / x M1

either ω = 2π / T or v = 2πx / T and working to GM = 4π2x 3 / T

2 A1 [3] (c) either use of gradient of graph or line through origin so can use single point or line shown extrapolated to origin B1

gradient = (4.5 × 1014) / 0.35

6.67 × 10–11 × M = 4π2 × (4.5 × 1014 × 109) / (0.35 × 24 × 36002) correct conversion for km3 and power of 10 C1 correct conversion for day2 C1

M = 1.02 × 1026 kg A1 [4] 2 (a) total volume of molecules negligible compared to that of containing vessel

no intermolecular forces molecules in random motion time of collision small compared with the time between collisions large number of molecules any two B2 [2]

(b) in a real gas there is a range of velocities or must take the average of v2 B1 [1]

(c) (i) either p = 3

1ρ <c2>

or 1.0 × 105 = 3

1× 1.2 × <c2> C1

<c2> = 2.5 × 105 C1 cr.m.s. = 500 m s–1 A1 [3]

(ii) T ∝ <c2> C1

<c2> = 2.5 × 105 × 480 / 300

= 4.0 × 105 m2

s–2 (allow ECF from (c)(i)) A1 [2] 3 (a) same temperature B1 no (net) transfer of thermal energy (between the bodies) B1 [2] (b) (i) 41.3 K B1 [1] (ii) 330.4 K B1 [1]




(c) ∆EK = 2

3 × 1.9 × 60

= 171 J C1 work done = p∆V

= 1.2 × 105 × 950 × 10–6 C1 = 114 J C1 thermal energy = 114 + 171 = 285 (290) J A1 [4] 4 (a) acceleration/force proportional to distance from a fixed point or displacement M1 either acceleration/force and displacement in opposite directions or acceleration/force (always) directed towards a fixed point/mean

position/equilibrium position A1 [2]

(b) hρ g = Mg / A B1

h × 790 × 4.9 × 10–4 = 70 × 10–3 leading to h = 0.18 m or 18 cm A1 [2]

(c) (i) 1. ω2 = (790 × 4.9 × 10–4 × 9.81) / (70 × 10–3) C1 = 54.25

ω = 7.37 (rad s–1)

period (= 2π / ω) = 0.85 s C1 t1 = 0.43 s A1 [3] 2. t3 = 1.28 s (allow 2 s.f.) A1 [1]

(ii) energy of peak = ½Mω2x0

2 B1

change = ½ × 70 × 10–3 × 54.25 (2.2 × 10–2)2 – (1.0 × 10–2)2 C1

= 7.3 × 10–4 J A1 [3]




5 (a) charges in metal do not move B1 no (resultant) force on charges so no (electric) field B1 [2] (allow 1/2 for “no field inside sphere”) (b) either average field strength = ½ (28 + 54) N C–1 C1

average force = 8.5 × 10–9 × ½ (28 + 54) C1

= 3.49 × 10–7 N

change in potential energy = 3.49 × 10–7 × 2.0 × 10–2

= 7.0 × 10–9 J (allow 1 s.f.) A1 (allow range 54 ± 1) or (for a point charge) V = Ex (C1)

∆V = (54 × 5.0 × 10–2) – (28 × 7.0 × 10–2) (C1)

change in potential energy = 8.5 × 10–9 × (2.70 – 1.96)

= 6.3 × 10–9 J (allow 1 s.f.) (A1) (allow range 54 ± 1) or ∆V is area under curve (C1) ∆V = 0.74 V (C1)

change in potential energy = 8.5 × 10–9 × 0.74

= 6.3 × 10–9 J (allow 1 s.f.) (A1) [3] (allow range 0.70 to 0.84) 6 (a) magnetic fields are equal in magnitude/strength/flux density M1 magnetic fields are opposite in direction M1 fields superpose/add/cancel to give zero/negligible resultant field A1 [3] (b) core causes increase in magnetic flux in the solenoid/induced poles in core or field induced in core B1 changing flux threads/cuts the turns on the solenoid M1 (by Faraday’s law) an e.m.f. is induced in the solenoid A1 by Lenz’s law, this e.m.f. opposes the battery e.m.f. A1 [4]

7 (a) (i) V0 (= 14 2 ) = 19.8 (20) V A1 [1]

(ii) ω (= 2π × 750) = 4700 rad s–1 A1 [1] (b) large amount of charge required to charge capacitor M1 capacitor would charge and discharge rapidly/in a very short time or capacitor would charge and discharge 750/1500 times per second M1

I = Q / t, so large current A1 [3]




8 (a) hc / λ = Φ + EMAX M1 h = Planck constant, c = speed of light / e.m. radiation A1 [2] (b) (i) gradient of line is hc M1 h and c are both constants A1 [2]

(ii) Φ = 2.28 × 1.6 × 10–19 C1

= 3.65 × 10–19 (J)

hc / λ0 = 3.65 × 10–19

λ0 = (6.63 × 10–34 × 3.0 × 108) / (3.65 × 10–19) C1

= 5.45 × 10–7 m A1 [3]

9 (a) energy required to separate the nucleons (in a nucleus) or energy required to separate the protons and neutrons in a nucleus M1 (or energy released when nucleons combine (to form a nucleus)/energy released

when protons and neutrons combine to form a nucleus) either completely or to infinity A1 [2] (either free protons and neutrons or from infinity) (b) (i) either different forms of same element or nuclei having same number of M1 protons with different numbers of neutrons A1 [2] (ii) 1784 MeV (accept min. 3 s.f.) A1 7.57 MeV A1 [2]

(c) (i) λ = ln 2 / (7.1 × 108 × 365 × 24 × 3600) = 3.1 × 10–17 s–1 B1 [1]

(ii) A = λN

5000 = 3.1 × 10–17 × N C1

N = 1.61 × 1020

mass = 235 × (1.61 × 1020) / (6.02 × 1023) C1 = 0.063 g (accept min. 2 s.f.) A1 [3]




Section B

10 (a) correct LED symbol B1 separately connected between VOUT and earth with opposite polarities M1 diode B ‘pointing’ from VOUT to earth A1 [3] (ignore protective resistors) (b) diode in VOUT line M1 diode ‘pointing’ towards VOUT from earth A1 relay coil connected between VOUT and earth M1 switch connected across lamp A1 [4] (if a diode is placed across the relay it must point down otherwise max. 2/4; one diode but wrong direction max. 3/4) 11 (a) e.g. scattering (in metal) non-parallel beam (not just “A closer than B”) reflection (from metal) diffraction in the metal/lattice any two B2 [2]

(b) (i) 1. ratio = eµx

= exp(0.27 × 4.0) C1 = 2.94 (2.9) A1 [2]

2. ratio = exp(0.27 × 2.5) × exp(3.0 × 1.5) C1

= 1.96 × 90 = 177 (180) A1 [2] (do not penalise unit error more than once) (ii) each ratio gives measure of transmission B1 ratios (in (i)) very different so good contrast B1 [2] 12 (a) (i) serial-to-parallel converter B1 [1] (ii) digital-to-analogue converter or DAC B1 [1] (iii) (audio) amplifier or AF amplifier B1 [1] (b) (i) 4 A1 [1] (ii) 1011 A1 [1] (c) correct levels at 0.25 ms intervals 0, 8, 11, 10, 15 A1 and 7, 4 A1 series of steps, each of depth 0.25 ms M1 voltage levels shown in correct intervals A1 [4]




13 (a) advantage: e.g. shorter time delay greater coverage over a long time B1 disadvantage: e.g. satellite needs to be tracked more satellites for (continuous) coverage/communication (any sensible suggestions) B1 [2] (b) (i) frequencies linking Earth with satellite B1 6 GHz is uplink frequency 4 GHz is downlink frequency (allow vice versa) B1 [2] (ii) either signal from Earth to satellite is attenuated greatly or downlink must be amplified greatly before transmission B1 downlink would swamp uplink unless frequencies are different B1 [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 m is the independent variable, or vary m. [1]

P (tan)φ is the dependent variable, or measure (tan) φ. [1]

P Keep the temperature of the oil constant. [1]

Methods of data collection (5 marks)

M Labelled diagram showing labelled protractor positioned to determine φ for tilted

cylinder.

Allow distances marked to determine φ and use of a rule. [1]

M Use of balance/scales to measure the mass of the oil/cylinder. [1]

M Mass of oil = mass of (oil + cylinder) – mass of cylinder. [1]

M Use of vernier calipers/micrometer/rule to measure d. [1]

M Repeat each experiment for the same value of m and average φ. [1]

Method of analysis (2 marks)

A Plot a graph of φtan

1 against m.

(Allow 3d

m or 3d

m

ρ or ρ

m

. Do not allow log-log graphs.) [1]

A a = gradient ×ρd3 and b = y-intercept; must be consistent with suggested graph. [1]

Safety considerations (1 mark)

S Precaution linked to preventing spilling oil, e.g. use a tray/lid/cloth to absorb oil (do not

allow just wiping or mopping)

or precaution linked to preventing glass cylinder breaking, e.g. padding/cushion

or use of gloves to prevent skin irritation (do not allow “because oil is slippery”). [1]

Additional detail (4 marks)

D Relevant points might include [4]

1 Repeat measurements of d in different directions and average

2 Use of video with slow motion/frame by frame playback to determine φ

3 Use large protractor to reduce percentage uncertainty or trigonometry

relationship related to measurements to be taken

4 Use the same (diameter) cylinder (not “same size” but allow “same size and shape”)

5 Slowly/gently/gradually tilt cylinder of oil/use of rough surface (to prevent sliding)

6 Experimental method to determine density of oil and ρ = m / V

7 Relationship is valid if the graph is a straight line that does NOT pass through the

origin / has an intercept; must be consistent with suggested graph

Do not allow vague computer methods.




2 Analysis, conclusions and evaluation (15 marks)

Mark Expected Answer Additional Guidance

(a) A1 gradient =

2Edπ

ρ4

y-intercept = E

r

(b) T1

I

1 / A

–1 Allow

I

1 (A

–1) or

A

11

I.

T2

4.2 or 4.17

5.0 or 5.00

5.9 or 5.88

6.7 or 6.67

7.7 or 7.69

8.3 or 8.33

Allow a mixture of significant figures.

Must be table values.

U1 ± 0.2 to ± 0.6 or ± 0.7 or ± 0.8 Allow more than one significant figure.

(c) (i) G1 Six points plotted correctly Must be within half a small square.

Do not allow “blobs”.

ECF allowed from table.

U2 Error bars in 1 / I plotted

correctly

All error bars to be plotted. Must be accurate to

less than half a small square. Length of bar

must be accurate to less than half a small

square. Do not allow less than 0.05.

(ii) G2 Line of best fit If points are plotted correctly then lower end of

line should pass between (41, 4.5) and (44, 4.5)

and upper end of line should pass between

(83, 8.0) and (88, 8.0).

Line should not go from bottom to top points.

G3 Worst acceptable straight line.

Steepest or shallowest

possible line that passes

through all the error bars.

Line should be clearly labelled or dashed.

Examiner judgement on worst acceptable line.

Lines must cross. Mark scored only if error bars

are plotted.

(iii) C1 Gradient of line of best fit The triangle used should be at least half the

length of the drawn line. Check the read-offs.

Work to half a small square. Do not penalise

POT. (Should be about 8.)

U3 Absolute uncertainty in

gradient

Method of determining absolute uncertainty:

difference in worst gradient and gradient.

(iv) C2 y-intercept Check substitution into y = mx + c.

Allow ECF from (c)(iii). (Should be about 0.7–1.5.)




U4 Absolute uncertainty in y-

intercept

Uses worst gradient and point on WAL.

Do not check calculation.

(d) (i) C3 ρ = 2.415 × 10–7

× gradient

Must be in the range

1.80 × 10–6

to 2.10 × 10–6

and

given to 2 or 3 s.f.

Must use gradient.

ρ = 4

2Edπ

× gradient

[2 × 10–6

Ω m = 2 × 10–4

Ω cm = 2 × 10–3

Ω mm]

C4 r = E × y-intercept

= 3.2 × y-intercept

and Ω m and Ω given

Must include units for ρ and r.

Allow V A–1

or kg m2

A–2

s–3

for Ω.

(ii) U5 Percentage uncertainty in ρ Must be greater than 9.6%.

Uncertainties in Question 2

(c) (iii) Gradient [U3]

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

uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)

(iv) [U4]

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

uncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)

(d) (ii) [U5]

percentage uncertainty =

×++

∆

0.31

0.012

3.2

0.1

m

m × 100

=

×

∆100

m

m + 3.125 + 2 × 3.226

max. p = ( )4

100.323.32

3−×××π

× max. gradient

min. p = ( )4

100.303.12

3−×××π

× min. gradient





9702 PHYSICS










P m is the independent variable, or vary m. [1]

P (tan)φ is the dependent variable, or measure (tan) φ. [1]

P Keep the temperature of the oil constant. [1]


M Labelled diagram showing labelled protractor positioned to determine φ for tilted

cylinder.

Allow distances marked to determine φ and use of a rule. [1]

M Use of balance/scales to measure the mass of the oil/cylinder. [1]

M Mass of oil = mass of (oil + cylinder) – mass of cylinder. [1]

M Use of vernier calipers/micrometer/rule to measure d. [1]

M Repeat each experiment for the same value of m and average φ. [1]


A Plot a graph of φtan

1 against m.

(Allow 3d

m or 3d

m

ρ or ρ

m

. Do not allow log-log graphs.) [1]

A a = gradient ×ρd3 and b = y-intercept; must be consistent with suggested graph. [1]


S Precaution linked to preventing spilling oil, e.g. use a tray/lid/cloth to absorb oil (do not

allow just wiping or mopping)

or precaution linked to preventing glass cylinder breaking, e.g. padding/cushion

or use of gloves to prevent skin irritation (do not allow “because oil is slippery”). [1]



1 Repeat measurements of d in different directions and average

2 Use of video with slow motion/frame by frame playback to determine φ

3 Use large protractor to reduce percentage uncertainty or trigonometry

relationship related to measurements to be taken

4 Use the same (diameter) cylinder (not “same size” but allow “same size and shape”)

5 Slowly/gently/gradually tilt cylinder of oil/use of rough surface (to prevent sliding)

6 Experimental method to determine density of oil and ρ = m / V

7 Relationship is valid if the graph is a straight line that does NOT pass through the

origin / has an intercept; must be consistent with suggested graph







(a) A1 gradient =

2Edπ

ρ4

y-intercept = E

r

(b) T1

I

1 / A

–1 Allow

I

1 (A

–1) or

A

11

I.

T2

4.2 or 4.17

5.0 or 5.00

5.9 or 5.88

6.7 or 6.67

7.7 or 7.69

8.3 or 8.33


Must be table values.

U1 ± 0.2 to ± 0.6 or ± 0.7 or ± 0.8 Allow more than one significant figure.




U2 Error bars in 1 / I plotted

correctly




square. Do not allow less than 0.05.


line should pass between (41, 4.5) and (44, 4.5)

and upper end of line should pass between

(83, 8.0) and (88, 8.0).

Line should not go from bottom to top points.








are plotted.






gradient



(iv) C2 y-intercept Check substitution into y = mx + c.

Allow ECF from (c)(iii). (Should be about 0.7–1.5.)




U4 Absolute uncertainty in y-

intercept

Uses worst gradient and point on WAL.

Do not check calculation.

(d) (i) C3 ρ = 2.415 × 10–7

× gradient

Must be in the range

1.80 × 10–6

to 2.10 × 10–6

and


Must use gradient.

ρ = 4

2Edπ

× gradient

[2 × 10–6

Ω m = 2 × 10–4

Ω cm = 2 × 10–3

Ω mm]

C4 r = E × y-intercept

= 3.2 × y-intercept

and Ω m and Ω given

Must include units for ρ and r.

Allow V A–1

or kg m2

A–2

s–3

for Ω.

(ii) U5 Percentage uncertainty in ρ Must be greater than 9.6%.





(iv) [U4]

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

uncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)

(d) (ii) [U5]

percentage uncertainty =

×++

∆

0.31

0.012

3.2

0.1

m

m × 100

=

×

∆100

m

m + 3.125 + 2 × 3.226

max. p = ( )4

100.323.32

3−×××π

× max. gradient

min. p = ( )4

100.303.12

3−×××π

× min. gradient





9702 PHYSICS










P m is the independent variable and E is the dependent variable or vary m and measure E.

Do not allow time. [1]

P Keep the temperature change of water constant. Allow two specified temperatures.

Do not allow “keep temperature constant”. [1]

P Keep the mass or volume of water constant. [1]


M Labelled diagram including labelled thermometer with bulb in water and at least one

other label. [1]

M Workable circuit diagram to determine E: power supply, heater and ammeter and

voltmeter, or joulemeter or wattmeter. [1]

M Method to determine change in temperature: measure initial temperature, measure final

temperature and subtract, or measure initial temperature and specific temperature

change. [1]

M Use balance/scales to measure mass of blocks. [1]

M Stir water (so that metal is in thermal equilibrium). [1]


A Plot a graph of E against m.

Do not allow log–log graphs. [1]

A a = gradient and b = y-intercept; must be consistent with suggested graph. [1]


S Precaution linked to hot heater/water, e.g. use gloves or use tongs for hot blocks.

Do not allow goggles. [1]



1 Method to ensure that e.m.f. of the power supply is constant/current in heater is constant,

e.g. adjust variable power supply/variable resistor to ensure p.d./current is constant

2 Keep the starting temperature of water/metal constant

3 Wait for water and metal temperatures to equalise

4 Add insulation to sides of beaker/lid (to prevent energy losses)

5 Use of timer and equation, e.g. E = Pt = ItV for candidate’s method

6 Use large temperature change to reduce percentage uncertainty

7 Relationship is valid if the graph is a straight line that does not pass through the origin







(a) A1 gradient =

m

Pg

(b) T1 T / s, v / m s–1

and v2 / m

2 s

–2 Allow T (s), v (m s

–1) and v2

(m2

s–2

).

T2

8.7 or 8.74

19 or 19.3

27 or 27.4

37 or 36.6

45 or 44.9

52 or 52.3

Must be values of v2 in table (if v not rounded).

All values of v2 must be 2 s.f. or 3 s.f.


U1 From ± 0.9 or ± 1 to ± 3 Allow more than one significant figure.




U2 Error bars in v2 plotted

correctly




square.


line should pass between (0.16, 10) and

(0.18, 10) and upper end of line should pass

between (0.70, 50) and (0.72, 50).

Line should not go from top to bottom points.








are plotted.






gradient



(d) (i) C2 P =

g

m × gradient

= 2.55 × 10–3

× gradient

Must use gradient. Should be about 0.19.

C3 kg

(ii) U4 Percentage uncertainty in P Must be greater than 4%.




(e) (i) C4 v in the range 4.70 to 4.90 and


(ii) U5 Percentage uncertainty in v Allow credit if absolute uncertainty in mass used

correctly.





(d) (ii) [U4]

percentage uncertainty = 10004.0gradient

gradient 100

0.025

0.001

gradient

gradient ×

+

∆=×

+

∆

max. P = 9.81

0.026 × max. gradient

max. P = 9.81

0.024 × min. gradient

(e) (ii) [U5]

percentage uncertainty = 1000.5

0.005

2

1×

+

∆×

P

P

max. v = 0.040

0.5059.81max. ××P

min. v = 0.040

0.4959.81min. ××P

Education

9702 w15 ms_all