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Cambridge International Examinations Cambridge International Advanced Subsidiary and Advanced Level

PHYSICS 9702/11 Paper 1 Multiple Choice May/June 2017

MARK SCHEME

Maximum Mark: 40

Published

This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2017 series for most Cambridge IGCSE®, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components.

9702/11 Cambridge International AS/A Level – Mark Scheme PUBLISHED

May/June 2017

© UCLES 2017 of 3

Question Answer Marks

1 D 1

2 D 1

3 A 1

4 D 1

5 C 1

6 B 1

7 B 1

8 C 1

9 D 1

10 A 1

11 A 1

12 D 1

13 C 1

14 A 1

15 A 1

16 C 1

17 C 1

18 D 1

19 A 1

20 D 1

21 A 1

22 B 1

23 B 1

24 B 1

25 C 1

26 B 1

27 C 1

28 C 1


May/June 2017

© UCLES 2017 of 3


29 A 1

30 A 1

31 B 1

32 A 1

33 C 1

34 A 1

35 C 1

36 B 1

37 C 1

38 A 1

39 B 1

40 C 1






MARK SCHEME

Maximum Mark: 40

Published



May/June 2017

© UCLES 2017 Page 2 of 3


1 B 1

2 A 1

3 D 1

4 C 1

5 B 1

6 A 1

7 C 1

8 B 1

9 C 1

10 A 1

11 C 1

12 D 1

13 D 1

14 C 1

15 B 1

16 C 1

17 C 1

18 A 1

19 A 1

20 D 1

21 B 1

22 C 1

23 B 1

24 A 1

25 D 1

26 B 1

27 D 1

28 B 1


May/June 2017



29 B 1

30 C 1

31 C 1

32 A 1

33 A 1

34 C 1

35 A 1

36 B 1

37 B 1

38 A 1

39 B 1

40 A 1






MARK SCHEME

Maximum Mark: 40

Published



May/June 2017



1 B 1

2 B 1

3 D 1

4 C 1

5 B 1

6 A 1

7 D 1

8 C 1

9 D 1

10 A 1

11 D 1

12 A 1

13 A 1

14 A 1

15 D 1

16 C 1

17 B 1

18 D 1

19 B 1

20 A 1

21 B 1

22 A 1

23 C 1

24 D 1

25 D 1

26 A 1

27 B 1

28 C 1


May/June 2017



29 D 1

30 D 1

31 B 1

32 B 1

33 C 1

34 B 1

35 C 1

36 A 1

37 A 1

38 D 1

39 A 1

40 A 1





PHYSICS 9702/21 Paper 2 AS Level Structured Questions May/June 2017

MARK SCHEME

Maximum Mark: 60

Published

This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2017 series for most Cambridge IGCSE®, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components.


May/June 2017



1(a) (stress =) force / area or kg m s–2 / m2 B1

= kg m–1 s–2 A1

1(b)(i) 0.58 = 2π × [(4 × 0.500 × 0.6003 ) / (E × 0.0300 × 0.005003)]0.5 C1

E = [4π2 × 4 × 0.500 × (0.600)3] / [(0.58)2 × 0.0300 × (0.00500)3] = 1.35 × 1010 (Pa)

C1

= 14 (13.5) GPa A1

1(b)(ii)1. (accuracy determined by) the closeness of the value(s)/measurement(s) to the true value B1

(precision determined by) the range of the values/measurements B1

1(b)(ii)2. l is (cubed so) 3 × (percentage/fractional) uncertainty and T is (squared so) 2 × (percentage / fractional) uncertainty and (so) l contributes more

B1


May/June 2017



2(a) resultant force (in any direction) is zero B1

resultant torque/moment (about any point) is zero B1

2(b)(i) a = (v − u) / t or gradient or ∆v / (∆)t C1

e.g. a = (8.8 − 4.6) / (7.0 – 4.0) = 1.4 m s–2 A1

2(b)(ii) s = 4.6 × 4 + [(8.8 + 4.6) / 2] × 3 C1

= 18.4 + 20.1 = 39 (38.5) m

A1

2(b)(iii) ∆E = ½ × 95 [(8.8)2 − (4.6)2] C1

= 3678 – 1005 = 2700 (2673) J

A1

2(b)(iv)1. weight = 95 × 9.81 (= 932 N) C1

vertical tension force = 280 sin 25° or 280 cos 65° (=118.3 N) C1

F = 932 + 118 = 1100 (1050) N

A1

2(b)(iv)2. horizontal tension force = 280 cos 25° or 280 sin 65° (= 253.8 N) C1

resultant force = 95 × 1.4 (= 133 N) C1

133 = 253.8 – R R = 120 (120.8) N

A1


May/June 2017



3(a) ρ = m / V C1

V = πd2L / 4 or πr2L C1

weight = 2.7 × 103 × π (1.2 × 10–2)2 × 5.0 × 10–2 × 9.81 = 0.60 N A1

3(b)(i) the point from where (all) the weight (of a body) seems to act B1

3(b)(ii) W × 12 C1

(0.25 × 8) + (0.6 × 38) C1

W = (2 + 22.8) / 12 = 2.1 (2.07) N

A1

3(c)(i) pressure changes with depth (in water) or pressure on bottom (of cylinder) different from pressure on top

B1

pressure on bottom of cylinder greater than pressure on top or force (up) on bottom of cylinder greater than force (down) on top

B1

3(c)(ii) anticlockwise moment reduced and reducing the weight of X reduces clockwise moment or anticlockwise moment reduced so clockwise moment now greater than (total) anticlockwise moment

B1


May/June 2017



4(a) (two) waves travelling (at same speed) in opposite directions overlap B1

waves (are same type and) have same frequency/wavelength B1

4(b)(i) λ = 12 / 250 (= 0.048 m)

C1

distance = 1.5 × 0.048 = 0.072 m

A1

4(b)(ii) T = 1 / 250 = 0.004 (s) or 4 (ms)

C1

1. curve drawn is mirror image of that in Fig. 4.2 and labelled P A1

2. horizontal line drawn between A and B and labelled Q A1


5(a) observed frequency is different to source frequency when source moves relative to observer B1

5(b) 360 = (400 × 340) / (340 ± v) C1

v = 38 (37.8) m s–1 A1

away (from the observer) B1


May/June 2017



6(a) volt / ampere B1

6(b)(i) RT = [1 / 3.0 + 1 / 6.0]–1 + 4.0 (= 6.0 Ω) C1

Ι = 1.5 / 6.0 C1

= 0.25 A A1

6(b)(ii) VB = 0.5 V I = 0.5 / 3.0 = 0.17 (0.167) A

A1

6(b)(iii) P = I 2R or VI or V 2 / R C1

ratio = (0.1672 × 3.0) / (0.252 × 4.0) = 0.33

A1

6(c)(i) vary/change/different radius/diameter/cross-sectional area (of wire) B1

6(c)(ii) v = I / Ane

( )( )

= B

C

/ratio

/AA

B

C

II

or × C

B

AA

B

C

II

C1

(R ∝ 1 / A so) ratio = × B

C

RR

B

C

II

= ×0.167 3.00.25 4.0

= 0.50

A1

6(d)(i) 0.25 A to 0.13 (0.125) A or halved A1

6(d)(ii) no change A1


May/June 2017



7(a)(i) (proton is uud so) (2 / 3)e + (2 / 3)e – (1 / 3)e = e B1

7(a)(ii) (neutron is udd so) (2 / 3)e – (1 / 3)e –(1 / 3)e = 0 B1

7(b)(i)

β– β+

nucleon number 90 64

proton number 39 28 all correct

B1

7(b)(ii) weak (nuclear force/interaction) B1

7(b)(iii) β– decay: electron and (electron) antineutrino β+ decay: positron and (electron) neutrino all correct

B1





PHYSICS 9702/22 Paper 2 AS Level Structured Questions May/June 2017

MARK SCHEME

Maximum Mark: 60

Published

This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2017 series for most Cambridge IGCSE®, Cambridge International A and AS Level and Cambridge Pre–U components, and some Cambridge O Level components.


May/June 2017



1(a) kelvin, mole, ampere, candela any two

B1

1(b) use of resistivity = RA / l and V = IR (to give ρ = VA / Il) C1

units of V: (work done / charge) kg m2 s–2 (A s)–1 C1

units of resistivity: (kg m2 s–3 A–1 A–1 m) = kg m3 s–3 A–2

A1

or

use of R = ρL / A and P = I2R (gives ρ = PA / I2L) (C1)

units of P: kg m2 s–3 (C1)

units of resistivity: (kg m2 s–3 × m2) / (A2 × m) = kg m3 s–3 A–2

(A1)

1(c)(i) ρ = (RA/l) C1

= (0.03 × 1.5 × 10–6) / 2.5 (= 1.8 × 10–8) C1

= 18 nΩ m A1

1(c)(ii) 1. precision is determined by the range in the measurements/values/readings/data/results B1

2. metre rule measures to ± 1 mm and micrometer to ± 0.01 mm (so there is less (percentage) uncertainty/random error) B1


May/June 2017



2(a) rate of change of displacement or change in displacement/time taken B1

2(b)(i) s = ut + ½at2 C1

t = [(2 × 1.25) / 9.81]1/2 (= 0.5048 s) C1

or

v2 = u2 + 2as vvert = (2 × 9.81 × 1.25)1/2 (= 4.95)

(C1)

t = [2s / (u + v)] = 2 × 1.25 / 4.95 (= 0.5048 s) (C1)

v = d / t = 1.5 / 0.50(48) = 3.0 (2.97) m s–1

A1

2(b)(ii) vertical velocity = at = 9.81 × 0.5048 (= 4.95) [using t = 0.50 gives 4.9]

C1

velocity = [(vh)2 + (vv)2]1/2 C1

= [(2.97)2 + (4.95)2]1/2 = 5.8 (5.79) [using t = 0.50 leads to 5.7]

A1

direction (= tan–1 4.95/2.97) = 59° A1

2(b)(iii) kinetic energy = ½mv2 C1

= ½ × 0.45 × (5.8)2 = 7.6 (7.57) J [using t = 0.50 leads to 7.3 J]

A1


May/June 2017



2(b)(iv) potential energy = mgh C1

= (0.45 × 9.81 × 1.25) = 5.5 (5.52) J

A1

2(c) there is KE of the ball at the start/leaving table or the ball has an initial/constant horizontal velocity or the ball has velocity at start/leaving table

B1


3(a) E = stress / strain or (F / A) / (e / l) C1

= [gradient × 3.5] / [π × (0.19 × 10–3)2] e.g. E = [(40 – 5) / ([11.6 – 3.2] × 10–3) × 3.5] / [π × (0.19 × 10–3)2] or [4170 × 3.5] / [π × (0.19 × 10–3)2]

C1

E (= 1.3 × 1011) = 0.13 TPa (allow answers in range 0.120–0.136 TPa) A1

3(b) a larger range of F required or range greater than 35 N B1


May/June 2017



4(a) a body/mass/object continues (at rest or) at constant/uniform velocity unless acted on by a resultant force B1

4(b)(i) initial momentum = final momentum m1u1 + m2u2 = m1v1 + m2v2

C1

0.60 × 100 − 0.80 × 200 = −0.40 × 100 + v × 200 v = (−) 0.3(0) m s–1

A1

4(b)(ii) kinetic energy is not conserved/is lost (but) total energy is conserved/constant or some of the (initial) kinetic energy is transformed into other forms of energy

B1


5(a) frequency is the number of vibrations/oscillations per unit time or the number of wavefronts passing a point per unit time B1

5(b) vibrations/oscillation of the air particles are parallel to the direction of it (the direction of travel of the sound wave) B1

5(c)(i) T = 2(.0) (ms) C1

f = 500 Hz A1

5(c)(ii) 1. amplitude increases (time) period decreases 2. amplitude decreases (time) period increases any 3 points

B3


May/June 2017



6(a)(i) waves at (each) slit/aperture spread B1

(into the geometric shadow) wave(s) overlap/superpose/sum/meet/intersect B1

6(a)(ii) there is not a constant phase difference/coherence (for two separate light source(s)) or waves/light from the double slit are coherent/have a constant phase difference

B1

6(b) x = λD / a C1

λ = (36 × 10–3 × 0.48 × 10–3) / (16 × 2.4) C1

= 4.5 × 10–7 m A1

6(c)(i) no movement of the water/water is flat/no ripples/disturbance B1

the path difference is 2.5λ or the phase difference is 900° or 5π rad B1

6(c)(ii) 1. surface/water/P vibrates/ripples and as (waves from the two dippers) arrive in phase

B1

2. surface/water/P vibrates/ripples and as amplitudes/displacements are no longer equal/do not cancel

B1


May/June 2017



7(a) energy transformed from chemical to electrical / unit charge (driven around a complete circuit) B1

7(b)(i) the current decreases (as resistance of Y increases) M1

lost volts go down (as resistance of Y increases) M1

p.d. AB increases (as resistance of Y increases) A1

7(b)(ii)1. 1.50 = 0.180 × (6.00 + 0.200 + RX) C1

RX = 2.1(3) Ω A1

7(b)(ii)2. p.d. AB = 1.5 − (0.180 × 0.200) or 0.18 × (2.13 + 6.00) C1

= 1.46(4) V A1

7(b)(ii)3. efficiency = (useful) power output / (total) power input or IV / IE C1

( = 1.46 / 1.5) = 0.97 [0.98 if full figures used] A1


8(a) β– emission: neutron changes to proton (+ beta–/electron) and β+ emission: proton changes to neutron (+ beta+/positron)

B1

β– emission: (electron) antineutrino also emitted and β+ emission: (electron) neutrino also emitted

B1

8(b) proton: up up down (and zero strange) neutron: up down down (and zero strange)

B1





PHYSICS 9702/31 Paper 3 Advanced Practical Skills 1 May/June 2017

MARK SCHEME

Maximum Mark: 40

Published



May/June 2017


Question Answer Mark

1(b)(iii) Value(s) of I1 with unit in the range 30–200 mA. 1

1(c)(iii) I2 > I1 (ECF unit from 1(b)(iii)). 1

1(d) Six sets of readings of x (different values), I1, I2 with correct trend and without help from Supervisor scores 5 marks, five sets scores 4 marks etc.

5

Range: xmax ⩾ 90 cm. 1

Column headings: 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. x / m, I2 / mA, I2 / I1 no unit.

1

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

1

Significant figures: All values of I2 / I1 must be given to the same number of s.f. as (or one more than) the s.f. in raw I1 and I2 (using the s.f. of the I value with the lowest number of s.f.).

1

Calculation: Values of I2 / I1 are correct.

1

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

1

Plotting of points: All observations must be plotted 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.

1


May/June 2017



Quality: All values of I2 / I1 must be greater than 1. All points in the table must be plotted for this mark to be awarded. It must be possible to draw a straight line that is within ± 0.05 on the I2 / I1 axis (normally y axis) of all plotted points.

1

1(e)(ii) Line of best fit: Judge by balance of all points on the grid about the candidate’s line (at least 5). 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. There must be at least five points left after the anomalous point is disregarded. Lines must not be kinked or thicker than half a small square.

1

1(e)(iii) Gradient: Gradient sign on answer line matches graph drawn. The hypotenuse of the triangle used must be greater than half the length of the drawn line. Method of calculation must be correct. Do not allow ∆x / ∆y. Both read-offs must be accurate to half a small square in both the x and y directions.

1

y-intercept: Correct read-off from a point on the line substituted correctly into y = mx + c or an equivalent expression. Read-off accurate to half a small square in both x and y directions. or Intercept read directly from the graph, with read-off at x = 0, accurate to half a small square in y direction.

1

1(f) Value of P = candidate’s gradient and value of Q = candidate’s intercept. The values must not be fractions.

1

Unit for P is correct (e.g. m–1) and Q stated without a unit. 1


May/June 2017



2(b)(ii) Values of time in the range 3–7 s and within 0.4 s of each other. All raw readings stated to same precision and to at least 0.1 s.

1

2(c)(ii) x on the answer line in the range 35.0–40.0 cm and all value(s) of raw x to nearest mm with unit. 1

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

1

2(e)(i) Value of n on answer line in the range 10 to 20. No unit. 1

Evidence of repeat readings. 1

2(e)(ii) Correct calculation of (n + 1)2 / n2. 1

2(f) Second value of x. 1

Second value of n. 1

Quality: second value of n < first value of n (if x2 < x1 allow n2 > n1). 1

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

2(g)(ii) Valid comment consistent with calculated values of k, testing against a criterion specified by the candidate. 1

2(h) Correct calculation of L written to 3 s.f. and correctly rounded to 3 s.f. 1


May/June 2017



2(i)(i) A Two readings are not enough to draw a valid conclusion (not “not enough for accurate results”, “few readings”).

B Difficult to judge when the balls are back in phase.

C Balls hit each other/blown by moving air/balls move sideways/irregular movement.

D Difficult to ensure both lengths 35 cm or difficult to measure x with a reason e.g. string kinked/parallax error.

E Damping/oscillations die out quickly.

F Difficulty linked to fixing block/wood with reason e.g. longer block requires more tape (therefore an unfair test)/block comes unstuck/blocks attached at an angle (x not constant).

G For two/both balls the release point is different or release instant is different. 1 mark for each point up to a maximum of 4.

4

2(i)(ii) A Take more readings and plot a graph/take more readings and compare k values (not “repeat readings” on its own).

B Video/film/record and perpendicular to plane of oscillation/from the side.

C Turn off air conditioning/close windows/use windshield/use a longer rod.

D Improved method of measuring 35 cm or x e.g. use clamps to fix loop and ball. F Improved method of attachment e.g. Blu-Tack/glue/Velcro/stickier tape/stronger adhesive on the tape.

G Improved method of release e.g. bar to hold both balls then drop away like a gate. 1 mark for each point up to a maximum of 4.

4






MARK SCHEME

Maximum Mark: 40

Published



May/June 2017



1(a)(v) Value of p in the range 20.0–30.0 cm, with unit. 1

Value of q less than p. 1

1(b) Six sets of readings of p, q and V (with correct trend and without help from Supervisor) scores 5 marks, five sets scores 4 marks etc.

5

Range: pmax ⩾ 40.0 cm and pmin ⩽ 10.0 cm.

1

Column headings: 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. 1 / p / cm–1.

1

Consistency: All values of p and q must be given to the nearest mm.

1

Significant figures: Significant figures for every value of 1 / q must be same as, or one greater than, the s.f. of q as recorded in table.

1

Calculation: Values of 1 / q calculated correctly.

1

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

1

Plotting of points: All observations must be plotted on the grid. Diameter of plotted points must be ⩽ half a small square (no “blobs”). Plots must be accurate to within half a small square in both x and y directions.

1


May/June 2017



Quality: All points in the table must be plotted (at least 5) for this mark to be awarded. It must be possible to draw a straight line that is within ± 1.0 m–1 (± 0.01 cm–1) of all the plotted points in the 1 / q direction.

1

1(c)(ii) Line of best fit: Judge by balance of all 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. Lines must not be kinked or thicker than half a small square.

1

1(c)(iii) Gradient: The hypotenuse of the triangle used must be greater than half the length of the drawn line. Method of calculation must be correct. Do not allow ∆x / ∆y. Both read-offs must be accurate to half a small square in both the x and y directions.

1

y-intercept: Correct read-off from a point on the line substituted into y = mx + c or an equivalent expression. Read-off accurate to half a small square in both x and y directions. or Intercept read directly from the graph, with read-off at x = 0 accurate to half a small square in y direction.

1

1(d) Value of a = candidate’s gradient and value of b = candidate’s intercept. Values must not be fractions.

1

There must be no unit for a. Unit for b correct e.g. m–1 or cm–1.

1


May/June 2017



2(a)(ii) Value of h1 to nearest 0.1 cm. 1

h2 < h1. 1

2(a)(iv) Absolute uncertainty in h2 of 2–5 mm 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 the working is clearly shown.

1

2(b)(i) Correct calculation of k. 1

2(b)(ii) Justification for s.f. in k linked to s.f. in m, g and (h1–h2) or m, g, h1 and h2. 1

2(c)(iii) Value of T in range 1.00–3.00 s. 1

Evidence of repeat readings of nT with n ⩾ 5. 1

2(d) Second values of h1 and h2. 1

Second value of T. 1

Quality: T for three springs > T for two springs. 1

2(e)(i) Two values of C calculated correctly. 1

2(e)(ii) Valid comment consistent with the calculated values of C, testing against a criterion specified by the candidate. 1


May/June 2017



2(f)(i) A Two readings are not enough to draw a (valid) conclusion (not “not enough for accurate results”, “few readings”).

B (Difficulty) measuring h because of parallax/metre rule not vertical.

C Oscillation dies away quickly/oscillation damped.

D Difficult to judge/tell end of oscillation/decide when to operate stopwatch.

E Other modes of oscillation occur.

F Some movement at joints still occurs.

1 mark for each point up to a maximum of 4.

4

2(f)(ii) A Take more readings and plot a graph/take more readings and compare C values (not “repeat readings” on its own).

B Use set square against rule (with detail)/use pointer attached to bottom of mass/use set square from rule to bottom of mass.

C Twist through larger angle to get more rotations.

D View (video) recording/film of motion with timer in view or put mark on mass to make it easier to see motion.

E Workable method of restricting vertical/swinging movement (e.g. enclose in transparent tube).

F Better method of fixing joints/use two complete springs. 1 mark for each point up to a maximum of 4.

4






MARK SCHEME

Maximum Mark: 40

Published



May/June 2017



1(a) Value of L with unit to the nearest mm in the range 2.5–3.5 cm. 1

1(c) Value of T with unit in the range 0.7 s to 1.5 s. 1

Evidence of repeated timings. Must see nT repeated where n ⩾ 5. 1

1(d) Six sets of readings of n (different values) and time with correct trend and without help from Supervisor scores 4 marks, five sets scores 3 marks etc.

4

Range of n ⩾ 9. 1

Column headings: 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. T / s. No unit for n or √n.

1

Consistency: All values of raw time must be given to either 0.1 s or 0.01 s.

1

Significant figures: All values of √n must be given to 3 significant figures.

1

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

1


1

Quality: All points in the table must be plotted for this mark to be awarded. It must be possible to draw a straight line that is within ± 0.10 on the √n axis of all plotted points.

1


May/June 2017



1(e)(ii) Line of best fit: Judge by balance of all points on the grid about the candidate’s line (at least 5). 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. There must be at least five points left after the anomalous point is disregarded. Lines must not be kinked or thicker than half a small square.

1

1(e)(iii) Gradient: The hypotenuse of the triangle used must be greater than half the length of the drawn line. Method of calculation must be correct. Do not allow ∆x / ∆y. Both read-offs must be accurate to half a small square in both x and y directions.

1

y-intercept: Correct read-off from a point on the line substituted correctly into y = mx + c or an equivalent expression. Read-off accurate to half a small square in both x and y directions. or Intercept read directly from the graph, with read-off at x = 0, accurate to half a small square in y direction.

1

1(f) Value of P = candidate’s gradient and value of Q = candidate’s intercept. The values must not be fractions.

1

Units for P and Q both s. 1

1(g) Correct calculation of g = Lπ2 / P2 with consistent unit. 1


May/June 2017



2(a) Value(s) of A with unit in the range 97.5–99.5 cm. 1

2(b)(ii) Values of all raw x to nearest mm. 1

2(c)(iii) Value of y ˃ x. 1

2(c)(iv) Correct calculation of (y – x). 1

2(c)(v) Percentage uncertainty in (y – x) based on absolute uncertainty of 2–5 mm. If repeated readings have been taken, then the uncertainty can be half the range (but not zero) if the working is clearly shown. Correct method of calculation to obtain percentage uncertainty.

1

2(d)(i) Correct calculation of m(A – 2y). 1

2(d)(ii) Justification for s.f. in m(A – 2y) linked to s.f. in A, y and m or (A – 2y) and m. 1

2(e)(i) Second value of y. 1

Quality: second value of y > first value of y. 1

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

2(f)(ii) Valid comment consistent with calculated values of k, testing against a criterion specified by the candidate. 1

2(g) Correct calculation of M = 1 / k – B. 1


May/June 2017



2(h)(i) A Two readings are not enough to draw a (valid) conclusion (not “not enough for accurate results”, “few readings”).

B Difficult to measure x or y with reason e.g. markings on rule obscured because of thickness of string/twist in string/ruler oscillating/ruler swinging/rule slides in loop/string not vertical.

C Large (%) uncertainty in (y – x) or (y – x) is small. D Little difference in y values. E Difficult to balance rule. 1 mark for each point up to a maximum of 4.

4

2(h)(ii) A Take more readings and plot a graph/take more readings and compare k values (not “repeat readings” on its own).

B Use thinner string/tie a knot in loop/use two loops/use a hook/use longer loop/string.

C Heavier (added/slotted) masses (not heavier hangers).

D Use wider range of (added) masses.

E Suspend rule from its top edge/balance on fulcrum/use thicker string with reason such as to make easier to balance/prevent rule from slipping.

1 mark for each point up to a maximum of 4.

4






MARK SCHEME

Maximum Mark: 40

Published



May/June 2017



1(a)(v) Value of T in the range 0.10–0.90 s. 1

Evidence of repeated readings. Must see nT repeated where n ⩾ 5. 1

1(b) Six sets of readings of M, h and T showing the correct trend and without help from the Supervisor scores 5 marks, five sets scores 4 marks etc.

5

Range: Mmax ⩾ 450 g and Mmin ⩽ 200 g. 1

Column headings: 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. T 3 / s3.

1

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

1

Significant figures: Significant figures for every value of T 3 must be the same as (or one greater than) the s.f. of raw times as recorded in table.

1

Calculation: Values of T 3 calculated correctly.

1

1(c)(i)

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

1


1

Quality: All points in the table must be plotted (at least 5) for this mark to be awarded. It must be possible to draw a straight line that is within ± 1.0 cm (to scale) of all the plotted points in the h direction.

1


May/June 2017



1(c)(ii) Line of best fit: Judge by balance of all 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. There must be at least five points left after the anomalous point is disregarded. Lines must not be kinked or thicker than half a square.

1

1(c)(iii) Gradient: The hypotenuse of the triangle used must be greater than half the length of the drawn line. Method of calculation must be correct. Do not allow ∆x / ∆y. Both read-offs must be accurate to half a small square in both the x and y directions.

1

y-intercept: Correct read-off from a point on the line substituted into y = mx + c or an equivalent expression. Read-off accurate to half a small square in both x and y directions. or Intercept read directly from the graph, with read-off at h = 0, accurate to half a small square in y direction.

1

1(d)

Value of a = candidate’s gradient and value of b = candidate’s intercept. The values must not be fractions.

1

Unit for a correct (e.g. s3 cm–1).

Unit for b is s3. 1


May/June 2017



2(a) Value for x0 to nearest 0.1 cm. 1

2(b)(ii) x greater than x0. 1

2(b)(iii) Raw θ in range 91–110° and recorded to nearest degree. 1

2(b)(v) Absolute uncertainty in (φ – 90°) of 2–5° and correct method of calculation to obtain percentage uncertainty. If repeated readings of φ have been taken, then the absolute uncertainty can be half the range (but not zero) only if working shown clearly.

1

2(c)(ii) Second value of x. 1

Second value of φ. 1

Quality: second value of φ ⩾ first value of φ. 1

2(d)(i) Two values of k calculated correctly. 1

2(d)(ii) Justification for s.f. in k linked to s.f. in θ, x and x0. 1

2(d)(iii) Valid comment consistent with calculated values of k, testing against a criterion specified by the candidate. 1

2(e)(ii) Raw value(s) of D and d recorded to the nearest 0.1 cm. 1

2(e)(iv) Value of ρ calculated correctly. 1


May/June 2017



2(f)(i) A Two readings are not enough to draw a (valid) conclusion (not “not enough for accurate results”, “few readings”). B Difficult to get/adjust wooden strip so that it is horizontal/parallel to bench. C Difficulty when measuring x (or x0), with reason e.g. parallax error/thickness of string/difficult to judge centre of nail. D Difficult to measure angle with reason, e.g. parallax error/hard to hold protractor steady/hard not to touch wooden

strip/difficult to align protractor correctly. (Do not credit parallax error twice for both C and D.)

E Large (percentage) uncertainty in (φ – 90°) or (φ – 90°) is small.

F Difficult to measure D (or d) with reason linked to use of ruler. 1 mark for each point up to a maximum of 4.

4

2(f)(ii) A Take more readings and plot a graph/take more readings and compare k values (not “repeat readings” on its own). B Use thinner nail/fulcrum/prism. C Mark a scale on the wooden strip/use a thinner string. D Hold protractor in a clamp/

take photograph and measure angle on photo/ trigonometric method with detail e.g. measure height(s) of end(s) of wooden strip

E Method to increase (φ – 90°) e.g. more paper clips/larger values of x/move pipe and pivot closer.

F Use vernier/digital calipers/travelling microscope. 1 mark for each point up to a maximum of 4.

4






MARK SCHEME

Maximum Mark: 40

Published



May/June 2017



1(a)(iii) Value of x with unit to the nearest mm in the range 10.0–15.0 cm. 1

1(b)(ii) Evidence of repeated timings. Must see nT repeated where n ⩾ 5. 1

1(c) Six sets of readings of x (different values) and time with correct trend and without help from Supervisor scores 5 marks, five sets scores 4 marks etc.

5

Range: xmin ⩽ 5.0 cm and xmax ⩾ 20.0 cm. 1

Column headings: 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. x2 / m2.

1

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

1

Significant figures: All values of x2 must be given to the same number of s.f. as (or one more than) the s.f. in raw x.

1

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

1


1

Quality: All points in the table must be plotted for this mark to be awarded. It must be possible to draw a straight line that is within ± 0.025 s on the T axis of all plotted points.

1


May/June 2017



1(d)(ii) Line of best fit: Judge by balance of all points on the grid about the candidate’s line (at least 5). 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. There must be at least five points left after the anomalous point is disregarded. Lines must not be kinked or thicker than half a small square.

1

1(d)(iii) Gradient: The hypotenuse of the triangle used must be greater than half the length of the drawn line. Gradient sign must match graph. Method of calculation must be correct. Do not allow ∆x / ∆y. Both read-offs must be accurate to half a small square in both the x and y directions.

1

y-intercept: Correct read-off from a point on the line substituted correctly into y = mx + c or an equivalent expression. Read-off accurate to half a small square in both x and y directions. or Intercept read directly from the graph, with read-off at x2 = 0, accurate to half a small square in the y direction.

1

1(e) Value of P = candidate’s gradient and value of Q = candidate’s intercept. The values must not be fractions.

1

Unit for P dimensionally correct (e.g. s m–2 or s cm–2 or s mm–2). Unit for Q correct (s).

1

1(f)(iii) Correct calculation of x. 1


May/June 2017



2(a) Value of raw L with unit in range 1.5–2.5 cm. 1

2(c)(ii) Value of y ⩾ L and to the nearest mm. 1

2(c)(iii) Value of raw θ to the nearest degree. 1

2(d) Percentage uncertainty in θ based on absolute uncertainty of 2–10°. If repeated readings have been taken, then the uncertainty can be half the range (but not zero) if the working is clearly shown. Correct method of calculation to obtain percentage uncertainty.

1

2(e)(i) Correct calculation of (y – L). 1

2(e)(ii) Correct calculation of cos (θ / 2). 1

2(e)(iii) Justification for s.f. in cos (θ / 2) linked to s.f. in θ. 1

2(f) Second value of y. 1

Second value of θ. 1

Quality: second value of θ > first value of θ. 1

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

2(g)(ii) Valid comment consistent with calculated values of k, testing against a criterion specified by the candidate. 1


May/June 2017



2(h)(i) A Two readings are not enough to draw a (valid) conclusion (not “not enough for accurate results”, “few readings”). B Moving hands affecting keeping 15 cm the same or measuring y or θ.

C Parallax error affecting the measurement of y or θ. D Reason for d not remaining constant e.g. loops slip on stand. E Difficult to know where to measure from for the angle or y. 1 mark for each point up to a maximum of 4.

4

2(h)(ii) A Take more readings and plot a graph/take more readings and compare k values (not “repeat readings” on its own). B Use another stand for valid purpose (e.g. instead of hand to hold spring). C Photo or still from video to measure θ or including scale in frame to measure y. D Improved method of fixing springs onto stand e.g. use Blu-Tack/tape/vertically clamped hacksaw blade/bulldog clips or

use sandpaper to make stand rough. E Use a grid behind the springs/markers on spring. 1 mark for each point up to a maximum of 4.

4





PHYSICS 9702/41 Paper 4 A Level Structured Questions May/June 2017

MARK SCHEME

Maximum Mark: 100

Published



May/June 2017



1(a) gravitational force (of attraction between satellite and planet) B1

provides / is centripetal force (on satellite about the planet) B1

1(b) M = (4/3) × πR3ρ B1

ω = 2π / T or v = 2πnR / T B1

GM / (nR)2 = nRω2 or v 2 / nR M1

substitution clear to give ρ = 3πn3 / GT 2 A1

1(c) n = (3.84 × 105) / (6.38 × 103) = 60.19 or 60.2 C1

ρ = 3π × 60.193 / [(6.67 × 10–11) × (27.3 × 24 × 3600)2] C1

ρ = 5.54 × 103 kg m–3 A1


May/June 2017



2(a) e.g. period = 3 / 2.5 C1

frequency = 0.83 Hz A1

2(b) light (damping) B1

2(c) at 2.7 s, A0 = 1.5 (cm) B1

energy = ½ m × 4π2f 2A02 B1

= ½ × 0.18 × 4π2 × 0.832 × (1.5 × 10–2)2

= 5.51 × 10–4 (J)

C1

at 7.5 s, A0 = 0.75 (cm) B1

energy = ¼ × 5.51 × 10–4 or energy = ½ × 0.18 × 4π2 × 0.832 × (0.75 × 10–2)2

C1

energy = 1.38 × 10–4 (J) change = (5.51 × 10–4 – 1.38 × 10–4) = 4.13 J

A1


May/June 2017



3(a)(i) signal consists of (a series of) 1s and 0s or offs and ons or highs and lows B1

3(a)(ii) component X: parallel-to-serial converter B1

component Y: DAC/digital-to-analogue converter B1

3(a)(iii) sample the (analogue) signal M1

at regular intervals and converts the analogue number to a digital number A1

3(b)(i) attenuation in fibre = 84 × 0.19 (= 16 dB) C1

ratio = 16 + 28 = 44 dB

A1

3(b)(ii) ratio / dB = 10 lg (P2 / P1) C1

44 = 10 lg (9.7 × 10–3 / P) or –44 = 10 lg (P / 9.7 × 10–3)

C1

power = 3.9 × 10–7 W A1


May/June 2017



4(a) random/haphazard B1

constant velocity or speed in a straight line between collisions or distribution of speeds/different directions

B1

4(b) (small) specks of light/bright specks/pollen grains/dust particles/smoke particles M1

moving haphazardly/randomly/jerky/in a zigzag fashion A1

4(c)(i) pV = ⅓ Nm⟨c2⟩ 1.05 × 105 × 0.0240 = ⅓ × 4.00 × 10–3 × ⟨c2⟩

C1

⟨c2⟩ = 1.89 × 106 C1

or

½ m⟨c2⟩ = (3 / 2) kT 0.5 × (4.00 × 10–3 / 6.02 × 1023) × ⟨c2⟩ = 1.5 × 1.38 × 10–23 × 300

(C1)

⟨c2⟩ = 1.87 × 106 (C1)

or

nRT = ⅓ Nm⟨c2⟩ 1.00 × 8.31 × 300 = ⅓ × 4.00 × 10–3 × ⟨c2⟩

(C1)

⟨c2⟩ = 1.87 × 106 (C1)

cr.m.s. = 1.37 × 103 m s–1 A1


May/June 2017



4(c)(ii) ⟨c2⟩ ∝ T C1

⟨c2⟩ at 177 °C = 1.89 × 106 × (450 / 300) C1

cr.m.s. at 177 °C = 1.68 × 103 m s–1 A1


5(a) (loss in) kinetic energy of α-particle = Qq / 4πε0r or 7.7 × 10–13 = Qq / 4πε0r

C1

7.7 × 10–13 = 8.99 × 109 × 79 × 2 × (1.60 × 10–19)2 / r M1

r = 4.7 × 10–14 m r is closest distance of approach so radius less than this

A1

5(b) force = Qq / 4πε0r 2 = 4u × a C1

8.99 × 109 × 79 × 2 × (1.60 × 10–19)2 / (4.7 × 10–14)2 = 4 × 1.66 × 10–27 × a C1

a = 2.5 × 1027 m s–2 A1

5(c) so that single interactions between nucleus and α-particle can be studied or so that multiple deflections with nucleus do not occur

B1


May/June 2017



6(a)(i) lamp needs ‘high’ power/‘large’ current/‘large’ voltage B1

op-amp can deliver only a small current/small voltage B1

6(a)(ii) correct symbol for relay coil connected between output and earth B1

switch between mains supply and lamp B1

6(b)(i) vary light intensity at which lamp is switched on/off B1

6(b)(ii) so that relay operates for only one current/voltage direction or so that relay/lamp operates for either dark or light conditions

B1

6(c) when light level increases, LDR resistance decreases B1

(RLDR low,) so V – > V+, so VOUT negative/–5 V (must be consistent with B1 mark) M1

or

when light level decreases, LDR resistance increases (B1)

(RLDR high,) so V – < V+, so VOUT is positive/+5 V (must be consistent with B1 mark) (M1)

lamp comes on as light level decreases or lamp goes off as light level increases

A1


May/June 2017



7(a) (magnetic) force (always) normal to velocity/direction of motion M1

(magnitude of magnetic) force constant or speed is constant/kinetic energy is constant

M1

so provides the centripetal force A1

7(b) increase in KE = loss in PE or ½ mv 2 = qV M1

p = mv with algebra leading to p = √(2mqV) A1

7(c) Bqv = mv2 / r mv = Bqr or p = Bqr

C1

(2 × 9.11 × 10–31 × 1.60 × 10–19 × 120)1/2 = B × 1.60 × 10–19 × 0.074 C1

B = 5.0 × 10–4 T A1

7(d) greater momentum M1

(p = Bqr and) so r increased A1


May/June 2017



8 strong (uniform) magnetic field B1

* nuclei precess/rotate about field (direction)

radio frequency pulse/RF pulse (applied) B1

* RF or pulse is at Larmor frequency / frequency of precession

causes resonance / excitation (of nuclei)/nuclei to absorb energy B1

on relaxation/de-excitation, nuclei emit RF/pulse B1

* (emitted) RF/pulse detected and processed

non-uniform field (superposed on uniform field) B1

allows positions of (resonating) nuclei to be determined B1

* allows for position of detection to be changed/different slices to be studied

max. 2 of additional detail points marked * B2


May/June 2017



9(a)(i) core reduces loss of (magnetic) flux linkage/improves flux linkage B1

9(a)(ii) reduces (size of eddy) currents in core B1

(so that) heating of core is reduced B1

9(b) alternating voltage gives rise to changing magnetic flux in core M1

(changing) flux links the secondary coil A1

induced e.m.f. (in secondary) only when flux is changing/cut B1


10(a)(i) penetration of beam M1

greater hardness means greater penetration/shorter wavelength/higher frequency/higher photon energy A1

10(a)(ii) greater accelerating potential difference or greater p.d. between anode and cathode

B1

10(b) I = I0 exp(–µx) ratio = (exp –1.5 × 2.9) / (exp –4.0 × 0.95) (= exp –0.55)

C1

= 0.58 A1


May/June 2017



11(a) electrons (in gas atoms/molecules) interact with photons B1

photon energy causes electron to move to higher energy level/to be excited B1

photon energy = difference in energy of (electron) energy levels B1

when electrons de-excite, photons emitted in all directions (so dark line) B1

11(b)(i) photon energy ∝ 1 / λ C1

energy = 1.68 eV A1

or

E = hc / λ E = 6.63 × 10–34 × 3.0 × 108 / (740 × 10–9) = 2.688 × 10–19 J

(C1)

energy = 1.68 eV (A1)

11(b)(ii) 3.4 eV → 1.5 eV 3.4 eV → 0.85 eV 3.4 eV → 0.54 eV all correct and none incorrect 2/2 2 correct and 1 incorrect or only 2 correctly drawn 1/2

B2


May/June 2017



12(a) x = 7 A1

12(b)(i) E = mc2 C1

= 1.66 × 10–27 × (3.0 × 108)2

= 1.494 × 10–10 J

C1

division by 1.6 × 10–13 clear to give 934 MeV A1

12(b)(ii) ∆m = (235.123 + 1.00863) – (94.945 + 138.955 + 2 × 1.00863 + 7 × 5.49 × 10–4) or ∆m = 235.123 – (94.945 + 138.955 + 1 × 1.00863 + 7 × 5.49 × 10–4)

C1

= 0.21053 u C1

energy = 0.21053 × 934 = 197 MeV

A1

12(c) kinetic energy of nuclei/particles/products/fragments B1

γ–ray photon energy B1






MARK SCHEME

Maximum Mark: 100

Published



May/June 2017



1(a) force per unit mass B1

1(b)(i) g = GM / r 2 = (6.67 ×10–11 × 1.0 × 1013) / (3.6 × 103)2

C1

= 5.1 × 10–5 N kg–1 A1

1(b)(ii) mass = (960 / 9.81) kg weight on comet = (960 / 9.81) × 5.1 ×10–5

C1

= 5.0 × 10–3 N A1

1(c) similarity: e.g. both attractive/pointed towards the comet e.g. same order of magnitude

B1

difference: e.g. radial/non-radial e.g. same (over surface)/varies (over surface)

B1


May/June 2017



2(a)(i) mean/average square speed/velocity B1

2(a)(ii) pV = NkT or pV = nRT B1

ρ = Nm / V or ρ = nNAm / V and k = nR / N

B1

EK = ½ m⟨c2⟩ with algebra to (3 / 2)kT B1

2(b)(i) no (external) work done or ∆U = q or w = 0 B1

q = NA × (3 / 2)k × 1.0 M1

NAk = R so q = (3 / 2)R A1

2(b)(ii) specific heat capacity = (3 / 2) × R / 0.028 C1

= 450 J kg–1 K–1 A1


May/June 2017



3(a)(i) e.g. period = 6 / 2.5 C1


3(a)(ii) energy = ½ m × 4π2f 2y02 C1

= ½ × 0.25 × 4π2 × 0.422 × (1.5 × 10–2)2 C1

= 2.0 × 10–4 J A1

3(b)(i) (induced) e.m.f. proportional to rate of M1

change of magnetic flux (linkage)or cutting of magnetic flux

A1

3(b)(ii) coil cuts flux/field (of moving magnet) inducing e.m.f. in coil B1

(induced) current in resistor causes heating (effect) M1

thermal energy/heat derived from energy of oscillations (of magnet) A1


May/June 2017



4(a) pulse (of ultrasound) B1

* produced by quartz crystal/piezo-electric crystal

* gel/coupling medium (on skin) used to reduce reflection at skin

reflected from boundaries (between media) B1

reflected pulse/wave detected by (ultrasound) transmitter B1

reflected wave processed and displayed B1

* intensity of reflected pulse/wave gives information about boundary

* time delay gives information about depth of boundary


4(b) IT = I0 exp (–µx) C1

2.9 = exp (4.6µ) C1

µ = 0.23 cm–1 A1


May/June 2017



5(a) any two reasonable suggestions e.g. • signal can be regenerated/noise removed (not “no noise”) • circuits more reliable • circuits cheaper to produce • multiplexing (is possible) • error correction/checking • easier encryption/better security

B2

5(b)(i) samples the analogue signal M1

at regular intervals and converts it (to a digital number) A1

5(b)(ii) 1. smaller step depth B1

2. smaller step height B1


May/June 2017



6(a) force proportional to product of charges and inversely proportional to the square of the separation M1

reference to point charges A1

6(b)(i) (near to each sphere,) fields are in opposite directionsor point (between spheres) where fields are equal and opposite or point (between spheres) where field strength is zero

M1

so same (sign of charge) A1

6(b)(ii) (at x = 5.0 cm,) E = 3.0 × 103 V m–1 and a = qE / m C1

E = (1.60 × 10–19 × 3.0 × 103) / (1.67 × 10–27) C1

= 2.9 × 1011 m s–2 A1

6(c) field strength or E is potential gradient or field strength is rate of change of (electric) potential

M1

(field strength) maximum at x = 6 cm A1


May/June 2017



7(a) equal and opposite charges on the plates so no resultant charge B1

+ve and –ve charges separated so energy stored B1

7(b) charge / potential difference M1

reference to charge on one plate and p.d. between plates A1

7(c) energy = ½ CV2 or energy = ½ QV and C = Q / V

C1

(1 / 16) × ½ CV02 = ½ CV2

V = ¼ V0

A1


May/June 2017



8(a)(i) circle around both diodes B1

8(a)(ii) indicates (whether) temperature M1

(is) above or below a set value A1

8(b)(i) (when resistance of C > RV,) V– > V+ or V+ < 3 Vor p.d. across RV < p.d. across R/Y/3 V or p.d. across C > p.d. across R/ X/3 V

M1

op-amp output is negative M1

(only) green A1

8(b)(ii) resistance of C becomes less than RV or V– < V+

B1

green (LED) goes out A1

blue (LED) comes on A1

8(c) changes/determines temperature at which LEDs switch B1


May/June 2017



9(a)(i) Hall voltage depends on thickness of slice C1

thinner slice, larger Hall voltage A1

9(a)(ii) Hall voltage depends on current in slice B1

9(b) sinusoidal wave, one cycle B1

at θ = 0 and at θ = 360°, VH = VMAX B1

at θ = 180°, VH = –VMAX B1


May/June 2017



10(a) two from: • frequency below which electrons not ejected • maximum energy of electron depends on frequency • maximum energy of electrons does not depend on intensity • instantaneous emission of electrons

B2

10(b)(i) (λ0 is the) threshold wavelength or wavelength corresponding to threshold frequency or maximum wavelength for emission of electrons

B1

10(b)(ii)1. intercept = 1 / λ0 = 2.2 × 106 m–1 λ0 = 4.5 × 10–7 m or 450 nm

A1

10(b)(ii)2. gradient = hc C1

gradient = 2.0 × 10–25 or correct substitution into gradient formula C1

h = (2.0 × 10–25) / (3.0 × 108) = 6.7 × 10–34 J s A1

10(c) line: same gradient B1

straight line, positive gradient, intercept at greater than 2.2 × 106 when candidate’s line extrapolated B1


May/June 2017



11(a) loss of (electric) potential energy = gain in kinetic energy or qV = ½ mv2 or EK = p2 / 2m = qV

B1

p = mv with algebra leading to p = √(2mqV) B1

11(b)(i) particle/electron has a wavelength (associated with it) B1

dependent on its momentum or when/because particle is moving

B1

11(b)(ii) p = (2 × 9.11 × 10–31 × 1.60 × 10–19 × 120)1/2 C1

λ = (6.63 × 10–34) / (5.91 × 10–24) C1

= 1.12 × 10–10 m A1

11(c) wavelength is similar to separation of atoms M1

so diffraction observed A1


May/June 2017



12(a) 7 e01− A1

12(b)(i) E = mc2 C1

= 1.66 × 10–27 × (3.00 × 108)2 M1

= 1.494 × 10–10 J division by 1.60 × 10–13 clear to give 934 MeV

A1

12(b)(ii) ∆m = (82 × 1.00863u) + (57 × 1.00728u) – 138.955u = (–) 1.16762 (u)

C1

energy = 1.16762 × 934 C1

energy per nucleon = (1.16762 × 934) / 139 = 7.85 MeV

A1

12(c) above A = 56, binding energy per nucleon decreases as A increases B1

U-235 has larger nucleon number M1

so less (binding energy per nucleon) A1

or

fission takes place with uranium (B1)

fission reaction releases energy (M1)

binding energy per nucleon less (for uranium than for products) (A1)






MARK SCHEME

Maximum Mark: 100

Published



May/June 2017



1(a) gravitational force (of attraction between satellite and planet) B1

provides / is centripetal force (on satellite about the planet) B1

1(b) M = (4/3) × πR3ρ B1

ω = 2π / T or v = 2πnR / T B1

GM / (nR)2 = nRω2 or v 2 / nR M1

substitution clear to give ρ = 3πn3 / GT 2 A1

1(c) n = (3.84 × 105) / (6.38 × 103) = 60.19 or 60.2 C1

ρ = 3π × 60.193 / [(6.67 × 10–11) × (27.3 × 24 × 3600)2] C1

ρ = 5.54 × 103 kg m–3 A1


May/June 2017



2(a) e.g. period = 3 / 2.5 C1


2(b) light (damping) B1

2(c) at 2.7 s, A0 = 1.5 (cm) B1

energy = ½ m × 4π2f 2A02 B1

= ½ × 0.18 × 4π2 × 0.832 × (1.5 × 10–2)2

= 5.51 × 10–4 (J)

C1

at 7.5 s, A0 = 0.75 (cm) B1

energy = ¼ × 5.51 × 10–4 or energy = ½ × 0.18 × 4π2 × 0.832 × (0.75 × 10–2)2

C1

energy = 1.38 × 10–4 (J) change = (5.51 × 10–4 – 1.38 × 10–4) = 4.13 J

A1


May/June 2017



3(a)(i) signal consists of (a series of) 1s and 0s or offs and ons or highs and lows B1

3(a)(ii) component X: parallel-to-serial converter B1

component Y: DAC/digital-to-analogue converter B1

3(a)(iii) sample the (analogue) signal M1

at regular intervals and converts the analogue number to a digital number A1

3(b)(i) attenuation in fibre = 84 × 0.19 (= 16 dB) C1

ratio = 16 + 28 = 44 dB

A1

3(b)(ii) ratio / dB = 10 lg (P2 / P1) C1

44 = 10 lg (9.7 × 10–3 / P) or –44 = 10 lg (P / 9.7 × 10–3)

C1

power = 3.9 × 10–7 W A1


May/June 2017



4(a) random/haphazard B1

constant velocity or speed in a straight line between collisions or distribution of speeds/different directions

B1

4(b) (small) specks of light/bright specks/pollen grains/dust particles/smoke particles M1

moving haphazardly/randomly/jerky/in a zigzag fashion A1

4(c)(i) pV = ⅓ Nm⟨c2⟩ 1.05 × 105 × 0.0240 = ⅓ × 4.00 × 10–3 × ⟨c2⟩

C1

⟨c2⟩ = 1.89 × 106 C1

or

½ m⟨c2⟩ = (3 / 2) kT 0.5 × (4.00 × 10–3 / 6.02 × 1023) × ⟨c2⟩ = 1.5 × 1.38 × 10–23 × 300

(C1)

⟨c2⟩ = 1.87 × 106 (C1)

or

nRT = ⅓ Nm⟨c2⟩ 1.00 × 8.31 × 300 = ⅓ × 4.00 × 10–3 × ⟨c2⟩

(C1)

⟨c2⟩ = 1.87 × 106 (C1)

cr.m.s. = 1.37 × 103 m s–1 A1


May/June 2017



4(c)(ii) ⟨c2⟩ ∝ T C1

⟨c2⟩ at 177 °C = 1.89 × 106 × (450 / 300) C1

cr.m.s. at 177 °C = 1.68 × 103 m s–1 A1


5(a) (loss in) kinetic energy of α-particle = Qq / 4πε0r or 7.7 × 10–13 = Qq / 4πε0r

C1

7.7 × 10–13 = 8.99 × 109 × 79 × 2 × (1.60 × 10–19)2 / r M1

r = 4.7 × 10–14 m r is closest distance of approach so radius less than this

A1

5(b) force = Qq / 4πε0r 2 = 4u × a C1

8.99 × 109 × 79 × 2 × (1.60 × 10–19)2 / (4.7 × 10–14)2 = 4 × 1.66 × 10–27 × a C1

a = 2.5 × 1027 m s–2 A1

5(c) so that single interactions between nucleus and α-particle can be studied or so that multiple deflections with nucleus do not occur

B1


May/June 2017



6(a)(i) lamp needs ‘high’ power/‘large’ current/‘large’ voltage B1

op-amp can deliver only a small current/small voltage B1

6(a)(ii) correct symbol for relay coil connected between output and earth B1

switch between mains supply and lamp B1

6(b)(i) vary light intensity at which lamp is switched on/off B1

6(b)(ii) so that relay operates for only one current/voltage direction or so that relay/lamp operates for either dark or light conditions

B1

6(c) when light level increases, LDR resistance decreases B1

(RLDR low,) so V – > V+, so VOUT negative/–5 V (must be consistent with B1 mark) M1

or

when light level decreases, LDR resistance increases (B1)

(RLDR high,) so V – < V+, so VOUT is positive/+5 V (must be consistent with B1 mark) (M1)

lamp comes on as light level decreases or lamp goes off as light level increases

A1


May/June 2017



7(a) (magnetic) force (always) normal to velocity/direction of motion M1

(magnitude of magnetic) force constant or speed is constant/kinetic energy is constant

M1

so provides the centripetal force A1

7(b) increase in KE = loss in PE or ½ mv 2 = qV M1

p = mv with algebra leading to p = √(2mqV) A1

7(c) Bqv = mv2 / r mv = Bqr or p = Bqr

C1

(2 × 9.11 × 10–31 × 1.60 × 10–19 × 120)1/2 = B × 1.60 × 10–19 × 0.074 C1

B = 5.0 × 10–4 T A1

7(d) greater momentum M1

(p = Bqr and) so r increased A1


May/June 2017



8 strong (uniform) magnetic field B1

* nuclei precess/rotate about field (direction)

radio frequency pulse/RF pulse (applied) B1

* RF or pulse is at Larmor frequency / frequency of precession

causes resonance / excitation (of nuclei)/nuclei to absorb energy B1

on relaxation/de-excitation, nuclei emit RF/pulse B1

* (emitted) RF/pulse detected and processed

non-uniform field (superposed on uniform field) B1

allows positions of (resonating) nuclei to be determined B1

* allows for position of detection to be changed/different slices to be studied



May/June 2017



9(a)(i) core reduces loss of (magnetic) flux linkage/improves flux linkage B1

9(a)(ii) reduces (size of eddy) currents in core B1

(so that) heating of core is reduced B1

9(b) alternating voltage gives rise to changing magnetic flux in core M1

(changing) flux links the secondary coil A1

induced e.m.f. (in secondary) only when flux is changing/cut B1


10(a)(i) penetration of beam M1

greater hardness means greater penetration/shorter wavelength/higher frequency/higher photon energy A1

10(a)(ii) greater accelerating potential difference or greater p.d. between anode and cathode

B1

10(b) I = I0 exp(–µx) ratio = (exp –1.5 × 2.9) / (exp –4.0 × 0.95) (= exp –0.55)

C1

= 0.58 A1


May/June 2017



11(a) electrons (in gas atoms/molecules) interact with photons B1

photon energy causes electron to move to higher energy level/to be excited B1

photon energy = difference in energy of (electron) energy levels B1

when electrons de-excite, photons emitted in all directions (so dark line) B1

11(b)(i) photon energy ∝ 1 / λ C1

energy = 1.68 eV A1

or

E = hc / λ E = 6.63 × 10–34 × 3.0 × 108 / (740 × 10–9) = 2.688 × 10–19 J

(C1)

energy = 1.68 eV (A1)

11(b)(ii) 3.4 eV → 1.5 eV 3.4 eV → 0.85 eV 3.4 eV → 0.54 eV all correct and none incorrect 2/2 2 correct and 1 incorrect or only 2 correctly drawn 1/2

B2


May/June 2017



12(a) x = 7 A1

12(b)(i) E = mc2 C1

= 1.66 × 10–27 × (3.0 × 108)2

= 1.494 × 10–10 J

C1

division by 1.6 × 10–13 clear to give 934 MeV A1

12(b)(ii) ∆m = (235.123 + 1.00863) – (94.945 + 138.955 + 2 × 1.00863 + 7 × 5.49 × 10–4) or ∆m = 235.123 – (94.945 + 138.955 + 1 × 1.00863 + 7 × 5.49 × 10–4)

C1

= 0.21053 u C1

energy = 0.21053 × 934 = 197 MeV

A1

12(c) kinetic energy of nuclei/particles/products/fragments B1

γ–ray photon energy B1





PHYSICS 9702/51 Paper 5 Planning, Analysis and Evaluation May/June 2017

MARK SCHEME

Maximum Mark: 30

Published



May/June 2017



1 Defining the problem

(sin) θ is the independent variable and v is the dependent variable or vary (sin) θ and measure v 1

keep s (PQ) constant 1

Methods of data collection

labelled diagram showing inclined plane with labelled support and P and Q marked 1

method to measure angle e.g. use a protractor to measure θ or use a ruler to measure marked distances from which sin θ or θ may be determined

1

method of timing for an appropriate distance to determine v (at Q) e.g. use a stopwatch/timer or correctly positioned light gate(s) connected to a timer/data-logger or correctly positioned motion sensor connected to data-logger

1

measurement of an appropriate distance to determine v (at Q) e.g. rule to measure an appropriate length or length of a card to interrupt light beam or distance from motion sensor to Q

1

Method of analysis

plot a graph of v2 against sin θ 1

relationship valid if a straight line produced (not passing through the origin) 1

+= − ×gradient

2B mg

ms

or

=g +×-intercept

2B my

Bs

1


May/June 2017



Additional detail including safety considerations Max. 6

D1 use cushion/foam/sandbox for falling body (B)

D2 (sin) θ determined using trigonometry relationship using marked lengths

D3 appropriate equation to determine v (at Q) e.g. =

2svt

D4 repeat experiment for each θ and average v or t

D5 use of balance to measure mass of wooden block m and falling body B and rule to measure s

D6 =

+2-intercept .BsgyB m

D7 clean surfaces of blocks/inclined plane/ensure surface of the plane is smooth

D8 keep B and m constant or keep mass of block and mass of falling body constant

D9 method to ensure that wooden block starts at the same position P, e.g. put a mark on the block or align front or back of block

D10 method to prevent plane slipping so that angle being measured remains the same, e.g. a mass as a stop


May/June 2017



2(a) gradient = − 2

v

y-intercept = 2Lv

1

2(b) 0.80 ± 0.01

0.77 ± 0.01

0.73 ± 0.01

0.70 ± 0.01

0.66 ± 0.01

0.62 ± 0.01

First mark for all values of t correct. Second mark for uncertainties correct.

2

2(c)(i) Six points plotted correctly. Must be accurate to less than half a small square. No “blobs”. Diameter of points must be less than half a small square.

1

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

1


May/June 2017



2(c)(ii) Line of best fit drawn. If points are plotted correctly then upper end of line should pass between (4.8, 0.76) and (5.6, 0.76) and lower end of line should pass between (17.6, 0.64) and (18.8, 0.64). Line should not be from first to last plot.

1

Worst acceptable line drawn (steepest or shallowest possible line). All error bars must be plotted.

1

2(c)(iii) Gradient determined with a triangle that is at least half the length of the drawn line. Gradient must be negative.

1

uncertainty = gradient of line of best fit – gradient of worst acceptable line or uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)

1

2(c)(iv) y-intercept read-off y-axis to less than half small square or determined by substitution into y = mx + c. 1

uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line or uncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)

1

2(d)(i) v determined from gradient and units for v and L correct with correct power of ten.

= − = −2 2

gradient 2(c)(iii)v

1

L determined from y-intercept and v and L given to 2 or 3 significant figures. Correct substitution of numbers must be seen.

= × = × = − = −-intercept (c)(iv)-intercept (c)(iv)

2 2 gradient (c)(iii)v v yL y

1


May/June 2017



2(d)(ii) % uncertainty in v = % uncertainty in gradient 1

% uncertainty in L = % uncertainty in y-intercept + % uncertainty in gradient or % uncertainty in L = % uncertainty in y-intercept + % uncertainty in v Correct substitution of numbers must be seen. Maximum/minimum methods:

= ×max -interceptMax max -intercept max or

mingradientyL y v

= ×min -interceptMin min -intercept min or max gradient

yL y v

1






MARK SCHEME

Maximum Mark: 30

Published



May/June 2017




r is the independent variable and f (frequency of turntable) is the dependent variable or vary r and measure f (frequency of turntable)

1

keep m constant 1


labelled diagram showing power supply connected to motor (two leads) within turntable; circuits must be workable 1

method to change frequency of rotation of the turntable, e.g. adjust output of (variable) power supply or adjust variable resistor

1

increase frequency until the cube moves (relative to the turntable) 1

method to determine period of rotation of the turntable, e.g. stopwatch, light gate attached to a timer/data-logger or stroboscope

1

Method of analysis

plots a graph of f against 1 / r (allow log f against log r) 1

relationship valid if a straight line produced passing through the origin (for lg f vs. lg r straight line of gradient of –1)

1

K = gradient × 4π2m (for lg f vs. lg r, K = 10y-intercept × 4π2m)

1


May/June 2017




D1 use safety screen

D2 time at least 10 rotations of turntable or detailed use of stroboscope

D3 f = 1 / T for correct determination of period of rotation of turntable

D4 repeat experiment for each r and average f

D5 use balance to measure mass of cube

D6 wait for turntable to rotate steadily before increasing frequency or gradual/incremental/slowly increase in frequency

D7 use a spirit level to check that turntable is horizontal or clean cube/surface

D8 use a rule to measure r

D9 method to ensure r is measured to the centre of the cube, e.g. put a mark on the cube or align front or back of cube by a set distance

D10 method to determine centre of the turntable e.g. measure two or more diameters/maximum distance ideas


May/June 2017



2(a) gradient = 1

E

y-intercept = QE

1

2(b) P / Ω 1

I / A–1

± 9 29 or 29.4

± 11 36 or 35.7

± 16.5 53 or 52.6

± 23.5 71 or 71.4

± 28 83 or 83.3

± 34 100 First mark for uncertainties correct. Allow 1 s.f. e.g. 10, 10, 20, 20, 30, 30. Second mark for all second column correct. Allow a mixture of significant figures.

2


1

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

1


May/June 2017



2(c)(ii) Line of best fit drawn. If points are plotted correctly then lower end of line should pass between (200, 32) and (200, 34) and upper end of line should pass between (600, 88) and (600, 91).

1


1

2(c)(iii) Gradient determined with a triangle that is at least half the length of the drawn line. 1


1

2(c)(iv) y-intercept determined by substitution of correct point into y = mx + c. 1


1


May/June 2017



2(d)(i) E determined using gradient and units for E and Q with correct power of ten.

= =1 1

gradient 2(c)(iii)E

1

Q determined using y-intercept and E and Q given to 2 or 3 significant figures. Correct substitution of numbers must be seen.

= × = × = =-intercept 2(c)(iv)-intercept 2(c)(iv)gradient 2(c)(iii)

yQ E y E

1

2(d)(ii) % uncertainty in E = % uncertainty in gradient 1

% uncertainty in Q = % uncertainty in E + % uncertainty in y-intercept or % uncertainty in Q = % uncertainty in gradient + % uncertainty in y-intercept. Correct substitution of numbers must be seen. Maximum/minimum methods:


mingradientyQ y E


yQ y E

1






MARK SCHEME

Maximum Mark: 30

Published



May/June 2017




(sin) θ is the independent variable and v is the dependent variable or vary (sin) θ and measure v 1

keep s (PQ) constant 1


labelled diagram showing inclined plane with labelled support and P and Q marked 1

method to measure angle e.g. use a protractor to measure θ or use a ruler to measure marked distances from which sin θ or θ may be determined

1

method of timing for an appropriate distance to determine v (at Q) e.g. use a stopwatch/timer or correctly positioned light gate(s) connected to a timer/data-logger or correctly positioned motion sensor connected to data-logger

1

measurement of an appropriate distance to determine v (at Q) e.g. rule to measure an appropriate length or length of a card to interrupt light beam or distance from motion sensor to Q

1

Method of analysis

plot a graph of v2 against sin θ 1

relationship valid if a straight line produced (not passing through the origin) 1

+= − ×gradient

2B mg

ms

or

=g +×-intercept

2B my

Bs

1


May/June 2017




D1 use cushion/foam/sandbox for falling body (B)

D2 (sin) θ determined using trigonometry relationship using marked lengths

D3 appropriate equation to determine v (at Q) e.g. =

2svt

D4 repeat experiment for each θ and average v or t

D5 use of balance to measure mass of wooden block m and falling body B and rule to measure s

D6 =

+2-intercept .BsgyB m

D7 clean surfaces of blocks/inclined plane/ensure surface of the plane is smooth

D8 keep B and m constant or keep mass of block and mass of falling body constant

D9 method to ensure that wooden block starts at the same position P, e.g. put a mark on the block or align front or back of block

D10 method to prevent plane slipping so that angle being measured remains the same, e.g. a mass as a stop


May/June 2017



2(a) gradient = − 2

v

y-intercept = 2Lv

1

2(b) 0.80 ± 0.01

0.77 ± 0.01

0.73 ± 0.01

0.70 ± 0.01

0.66 ± 0.01

0.62 ± 0.01

First mark for all values of t correct. Second mark for uncertainties correct.

2


1

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

1


May/June 2017



2(c)(ii) Line of best fit drawn. If points are plotted correctly then upper end of line should pass between (4.8, 0.76) and (5.6, 0.76) and lower end of line should pass between (17.6, 0.64) and (18.8, 0.64). Line should not be from first to last plot.

1


1

2(c)(iii) Gradient determined with a triangle that is at least half the length of the drawn line. Gradient must be negative.

1


1

2(c)(iv) y-intercept read-off y-axis to less than half small square or determined by substitution into y = mx + c. 1


1

2(d)(i) v determined from gradient and units for v and L correct with correct power of ten.

= − = −2 2

gradient 2(c)(iii)v

1

L determined from y-intercept and v and L given to 2 or 3 significant figures. Correct substitution of numbers must be seen.

= × = × = − = −-intercept (c)(iv)-intercept (c)(iv)

2 2 gradient (c)(iii)v v yL y

1


May/June 2017



2(d)(ii) % uncertainty in v = % uncertainty in gradient 1

% uncertainty in L = % uncertainty in y-intercept + % uncertainty in gradient or % uncertainty in L = % uncertainty in y-intercept + % uncertainty in v Correct substitution of numbers must be seen. Maximum/minimum methods:


mingradientyL y v


yL y v

1

9702 s17 ms_all

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