Practical Skills - Mark schemes and Examiner’s Report

Q1.(a) Measurement of at least 30 fringe widths

(check that candidate has not miscounted) e.g. 30 fringe widths = 40 mm or 41 mm.

Correct answer of 1.3 or 1.4 mm quoted to 2sf with unit (one mark)ORCorrect answer 1.33 – 1.37 mm to 3sf with correct unit (two marks).

If candidate quotes value in range 1.33 – 1.37 mm to 3sf they achieve both the 2nd and 3rd marks (A quote to 3sf is justified in terms of uncertainty if a large number of fringe widths have been measured).For 2nd & 3rd marks allow ecf from incorrect measurement in 1st mark.If the printing process in your centre alters the scale of this diagram, measure the values on your printed question papers and mark the scripts accordingly. Send details to the moderator.If a candidate is visually impaired and using a modified paper that alters the scale of this diagram, measure the values on the printed question paper and mark the script accordingly.

3

(b) 1 mark for intermediate step where candidate doesn’t get correct final answer. i.e. calculating % uncertainty of total measurement (i.e. % uncertainty in w)

OR for both marks:

Uncertainty in w = ± 0.03 mm

(Full 2 marks for correct answer with unit – No unit no mark unless a correct intermediate step has been completed which will have been credited for 1 mark as explained above)

Uncertainty in measurement of multiple fringes is ± 1 mm (precision of ruler used).E.g. for length 41 mm ± 1 mm % uncertainty = 1/41 x 100 = 2.4 % Uncertainty in w (single fringe)= 2.4 x 1.4/100 = ±0.03 mm Simply quoting 0.03 - NO marks No penalty for omission of ±

2

(c) (i) (Using w = 1.40 mm)

Wavelength = 5.60 × 10–7 m Allow ecf for value of w from (b)Consistent unit required for the mark. No sf penalty.

1

(ii) (Intermediate step) % uncertainty = 5.8% (From %uncty s = 3.3%, w = 2.4%, D = 0.1% % uncertainty in wavelength = 3.3 +2.4+0.1 = 5.8%).Allow ecf from (b)

1

(iii) (Using wavelength = 5.60 × 10–7 m)

Uncertainty in wavelength = ± 3.2 × 10–8 (m) or ± 32 (nm) or ± 3.2 × 10–5 (mm)Allow ecf from (c)(i) & (ii)No sf penaltyIf the value is consistent with the wavelength quoted in (c)(i), allow the numerical answer without the unit, otherwise a unit is required.

1[8]

Q2.

(a) w from

3 sf answer 2

(b) 1

2allow ecf in 2 for wrong w but not for POT error

2

(c) 1 allow ecf for wrong d in 1

2

number of fringes seen = 13 (integer only)3 allow 12 or 14 fringes

3

(d) close jaws using ratchet

confirm that instrument reads zero 2

(e) mean = 0.57(0) mm; uncertainty = 0.5 × range

2

[11]

Q3.(a) 180 degrees

accept ° for degrees

OR

π radians condone c or ‘rad’ for radianreject ‘half a cycle’treat ‘π radians in phase’ as talk out

1

(b) (idea that) sets of combining waves do not have the same amplitude

condone ‘waves do not have same intensity’ or ‘same energy’ or ‘some energy is absorbed on reflection’ or ‘same power’ or ‘same strength’ or idea that non point source or non point receiver would lead to imperfect cancellationcondone the idea that the waves may not be monochromaticignore ‘some waves travel further’ or ‘waves do not perfectly cancel out’reject ‘waves may not be 180° out of phase’

1

(c) valid use of a set square or protractor against TR (to ensure perpendicular) 1

measure x at two different points [at each end of M] and adjust until [make sure] both distances are the same 2

OR

use of set square to align M with the perpendicular line earns 2

if method used does not allow continuous variation in x then award maximum 1 mark

OR

align graph paper with TR 1

align M with grid lines on graph paper 2 both marks can be earned for suitable sketch showing a viable procedure involving one or more recognisable set squares or protractors; the sketch may also show a recognisable ruler, eg

allow use of scale on set square to measure the perpendicular distances don’t penalise incorrect reference to the set square, eg as ‘triangular ruler’, as long as the sketch shows a recognisable set square

2

(d) Gmax line ruled through bottom of n = 3 error bar and through top of n = 11 error bar 1

Gmin line ruled through top of n = 5 error bar and through bottom of n = 13 error bar 2

Gmax and Gmin calculated from valid y step divided by valid x step; both n steps ≥ 6 3

allow 1 mm tolerance when judging intersection of gradient lines with error barsignore any unit given with Gmax or Gmin; penalise power of ten error in 01.5

12 = 1 MAX if (either) line is thicker than half a grid square or of variable width or not continuous;expect Gmax = 3.2(1) × 10–2 and Gmin = 2.5 (2.49) × 10–2

3

(e)

AND

result in range 2.8(0) to 2.9(0) × 10–2 (m) 1 2

OR

award one mark for

2.7(0) to 3.0(0) × 10–2 (m) 12 penalise 1 mark for a power of ten errorreject 1 sf 3 × 10–2 (m)

if a best fit line is drawn between the Gmax and Gmin lines and the gradient of this is calculated award 1 mark for λ in range 2.8(0) to 3.0(0) × 10–2(m)

2

(f) uncertainty in λ = Gmax – λ

OR

λ – Gmin

OR

1

percentage uncertainty = (uncertainty/λ)×100 2

result in range 11(.0) % to 14(.0) % 3 1 can be earned by showing a valid uncertainty then dividing by λecf their λ, Gmax and Gmin for 1 and 2 allow λ found from best fit line

12

allow × 100 where ∆λ is any plausible uncertainty for 2 numerical answer without valid working can only earn 3

3

(g) (states) calculate the (vertical) intercept 1

OR

outlines a valid calculation method to calculate y 1

determine the intercept for both lines and calculate average value 2

OR

determine the (vertical) intercept of the line of best fit (between Gmax and Gmin) 2

draw the line of best fit (between Gmax and Gmin); perform calculation to find intercept earns 12

2

(h) result reduced not affected increasedGmax Gmin λ y

general marker question

allow any distinguishing mark as long as only one per rowfor and X in same row ignore Xfor and in same row give no markignore any crossed-out response

4

alternative approach: single best fit line drawn on Figure 4

(d) G calculated from y step divided by x step;

n step ≥ 6 3 MAX 1

(e) λ in range 2.8(0) to 2.9(0) × 10–2 MAX 1

(f) percentage uncertainty in λ = × 100

AND

result in range 11(.0) % to 14(.0) % MAX 1

(g) calculate intercept

OR

outlines a valid calculation method to find y MAX 1

(h) as main schemeno ecf possible

4

alternative approach: non-crossing lines for Gmax and Gmin on Figure 4: includes lines that meet but do not cross

(d) Gmax and Gmin calculated from y step divided by x step; both n steps ≥ 6 3

MAX 1

(e) to (h) as main scheme1

[18]

Q4.

(a)

(must see this step either separately or in substitution for

condone i and r for θ etc.)

2

(b) idea implied that (XZ) × (WY) = n × (WX) × (YZ) is of form y = mx (+ c);

plot (XZ) × (WY) against (WX) × (YZ)

calculate gradient to find n (false plot loses both marks) 2

[must mention XZ, WX, YZ and WY for full credit: bland ‘plot sin θ1 against sin θ2 and calculate gradient to find n’= 1 MAX]

[alternative method is to plot XZ against WX to find G1 and plot YZ against

WY to find G2 1 ; evaluate to find n 2]2

(c) upper limit of (XZ) range [largest value] is suitable 1

largest XZ value ≈ length of block (114)

[largest WX value ≈ diagonal distance (131) across block / used

(approximately) largest value of XZ [WX] available]2

lower limit of (XZ or YZ) range [smallest value] is not suitable3

smallest YZ [XZ] values have large percentage uncertainty / are unreliable] 4 (reject idea these values are too close to zero)

smallest WX value ≈ width of block (65)5

[statement that range is suitable plus quantitative comment comparing length of block (114) with 98 (the range of XZ data) or covers more than 85% of available range] 12

equivalent statement regarding WX: compares available range (131 to 65 = 66) with 63 (the range of WX data) 12 = 2 MAX

statement that range is suitable plus simple qualitative comment relating range to the block, e.g. ‘a large fraction / part of the available XZ [WX] range is covered’ 12 = 1 MAX (bland ‘range is large / wide’ is not enough)]

MAX 3[7]

Q5.(a) n changes by 4 units, 2 units, 1 unit for each change in 100 nm

OR

this is not half-life behaviour because graph has false origin for n

OR

the magnitude of n does not halve every interval1

(b) Sensible long (> 8 cm) tangent drawn, correct read-off for points from triangle at least half length of line and readings taken

Substitution correct

(–) (1.5 ± 0.2 ) x 104 and m-1 Condone power of ten error in first two marks

3

(c) Column heading correct

All calculations correct

appropriate (3) sfs

1 / λ2 / 10-12 m-

2

11.1

8.16

6.25

4.94

4.00

3.31

2.78

Accept if calculated in nm-2 instead of m-2

11.1 × 10-6 nm-2 etcUnits as follows: 1 / λ2 / m-2. Alternative acceptable labelling includes 1 / λ2 (m-2), 1 / λ2 in m-2. The 10-12 can be in the body of the table or at top of column.

3

(d) Graph axes labelled correctly and sensible axes

Plots correct to within half a square

Best-fit line by eye Suitably large graph scale (do not award if scale on axis could have been doubled) Scale must be sensible divisions which can be easily read. eg scales in multiples of 3, 6, 7, 9 etc are unsatisfactory.2nd mark is independent mark i.e. if candidates have used an unsuitable scale they can still achieve marks for accurately

plotting the points.The line of best fit should have an approximately equal distribution of points on either side of the line.Check bottom 3 plots.

3

(e) Intercept correct to within half a square [1.6014]

1

(f) The value of refractive index at infinite / very long wavelength1

(g) states that log n = log c + d log λ

plot log n versus log λ

d is the gradient of the graph 3

[15]

Q6.(a) 1.43, 1.29

1

(b) Both plotted points to nearest mm

Best line of fit to points The line should be a straight line with approximately an equal number of points on either side of the line.

2

(c) (i) Large triangle drawn (at least 8 cm × 8 cm)

Correct values read from graph

Gradient value in range (0.618 to 0.652) × 10–6 to 2 or 3 sf Allow the 2nd mark for incorrect numerical values read ignoring incorrect power of 10.Incorrect power of 10 is penalised in gradient value.

3

(ii) Same figure quoted for gradient but with correct unit1

(d) (i) Straight line (through origin)

(directly) proportional 2

(ii) Evidence of substituting data from the table / graph into w = mD/s + c (from y = mx + c)

Computation of correct value for c (i.e. value of w when D/s = 0) with correct unit.

Should be approximately 0.1 × 10–3 m, depending on the exact lobf drawn.

2

(iii) w1

(e) Any reference to either width of slits OR single slit diffraction 1

[13]

Q7.implementingaccuracy: T from nT where n or Σn ≥ 3 *

L in range 95 to 105 cm * * (1)

tabulation:readings:

y / m T / s (1)5 sets, y range ≥ 20 cm (1) (1) (1)

significant figures: all T to 0.01 s * s to 1.0 mm * * (1)

all values tabulated to 3 s.f. (1)

quality: at least 4 points to ± 2 mm of straight line, suitably scaled graph (1)

analysis axes:

marked / no unit (vertical), (horizontal) (1)

scale:points:line:

suitable (e.g. 8 × 8) (1)5 plotted correctly (check at least one) (1)straight, positive gradient, best-fit (1)

G: result:

Δ suitable size (1) to 3 s.f, in s (1)in range 1.90 to 2.10 s m–1/2 [1.80 to 2.20 s m–1/2 (1)] (1) (1)

(16)

evaluating

(i) check that (horizontal) distance between vertical strings is the same at two places (1) check that the string is aligned with the longer pendulum [the string passes through the hole in the stripboard without touching the sides] (1)

(ii) wide range, even distribution (1)in order to define the shape of the graph (1)

(iii) unlikely to significantly improve the evidence (don’t allow additional readings improve the best-fit line):when y is small T is very large: (1)it is difficult to judge exactly when the pendulums are moving in phase (1)

(6)

[22]

Q8.(a) resistance / Ω 0.98 1.20 1.50 1.76 2.03 3.00 (1) (1)

[deduct one mark for each incorrect value]2

(b) (i) sensible scales chosen (1)points plotted correctly [deduct one mark for each mistake] (1) (1)line of best fit (1)

(ii) 0.90 Ω (1)

(iii) 0.22 Ω (1)0.38 Ω (1)

(iv) 1.12 W (1)6.0 W (1)

max 8

(c) resistance increases with increasing temperature (1)increase in heat dissipation for 1.0 A to 2.0 A is greater than for 0 to 1.0 A (1)and so a greater corresponding rise in temperature (1)

max 2[12]

Q9.(a) (i) Voltmeter across terminals with nothing else connected to battery / no

additional load. 1

(ii) This will give zero / virtually no current 1

(b) (i)

Answer must clearly show power: εI and VI, with I cancelling out to give formula stated in the question

1

(ii) Voltmeter connected across cell terminals

Switch open, voltmeter records εSwitch closed, voltmeter records VBoth statements required for mark

Candidates who put the voltmeter in the wrong place can still achieve the second mark providing they give a detailed description which makes it clear that:To measure emf, the voltmeter should be placed across the cell with the external resistor disconnectedAndTo measure V, the voltmeter should be connected across the external resistor when a current is being supplied by the cell

2

(c) Vary external resistor and measure new value of V, for at least 7 different values of external resistor

Precautions - switch off between readings / take repeat readings (to check that emf or internal resistance not changed significantly)

2

(d) Efficiency increases as external resistance increases

ExplanationEfficiency = Power in R / total power generatedI2R / I2(R + r) = R / (R + r)So as R increases the ratio becomes larger or ratio of power in load to power in internal resistance increases

Explanation in terms of V and ε is acceptable2

[9]

Q10.(a) peak (to peak) voltage = 6.3(0) (V)

accept any answer that rounds to 6.3 Vdo not allow power of ten errors, eg 0.0063 V

1

(b) period = 8 divisions

(= 8 × 0.5 × 10–3 (s))

= 4 ms 1

= 250 (Hz) 2 award both marks if 250 Hz seenaccept 4.0(0) ms for 1 but reject 4.05, 3.95 etcecf2 for wrong period

2

(c) any valid approach leading to RC in range

2.1 ×10–4 to 3.4 ×10–4 or 3 ×10–4 (s)

OR

their T in 02.2 × 0.069 ± 10 % 12

1 mark can be awarded for use of any validapproach in which RC is seen withsubstitutions or with rearranged equations, eg

OR

1.75 × 10–4 = RC × ln 2

OR

valid approaches;reads off t when C starts to discharge and t at a lower value of V:

rearranges to calculate RCfor ecf 2 ∆t used must correspond to interpretation of time base used in determining the frequency in (b); there is no ecf for misinterpretation of the voltage scaleORreads off t when C starts to charge and t at a higher value of V:

rearranges to calculate RC etcOR

determines half-life t0.5 and finds RC from for ecf 2 t0.5 used must correspond to etcORuses idea that during discharge V falls to 0.37V0 in one time constant: determines suitable V and reads off RC directlyfor ecf 2 time interval used must correspond to etcOR

uses idea that during charging V rises to 0.63V0 in one time constant: determines suitable V and reads off RC directlyreject idea that V falls to zero in 5RC

2

(d) qualitative comment

idea that the waveform will stretch horizontally 1

quantitative comment

by a factor of

OR

half a cycle now covers 10 (horizontal) divisions on the screen 2 (and also earns 1 )

(so the) resolution of the time axis has increased 3 (and also earns 1 )

measuring larger distance / across more divisions from the screen reduces (percentage) uncertainty in reading the time (constant / interval / half life) 4

for 1 look for reference to time axis or direction waveform is re-scaledaccept ‘graph is longer/stretched’ or ‘will not see whole cycle’ or ‘fewer cycles shown’ or ‘period takes more space’ or ‘distance being measured is larger’ or ‘time per division is less’ or ‘larger in x direction’ or ‘time is stretched’reject ‘waveform becomes larger’ or ‘may not see whole cycle’ or ‘measuring larger time’for 2 there needs to be valid quantitative detailaward 12 for ‘half a cycle now fills the screen’ or ‘half a cycle is displayed’ as these clearly recognise the stretching is along the time axis and ‘half’ is quantitativeaccept ‘new distance (on screen corresponding to half life or time constant) is 2.5 × answer shown in working for (c)’the candidate who realises that half a wave now covers the complete width of the screen cannot claim this is a disadvantage; they would still be able to bring either half cycle into view by using the X-shift and find RC for 3 uses technical language correctlyignore (but do not penalise) ‘times are more precise’ or ‘more accurate’reject ‘smaller resolution’ or ‘lower resolution’for 4 there needs to be a qualifying explanation for the comment about uncertaintyreject ‘advantage because the (time) scale is easier to read’

3 MAX

(e) valid sketch on Figure 7 showing discharge time to 0 V reduced and charging time to peak voltage reduced (see below) 1

connecting resistor in parallel with R halves [reduces by 50%] circuit [total] resistance [time constant] 2

do not insist on seeing second discharge although if shown this must look correct

2

(f) amendment to Figure 8 showing waveform across R with approximately the correct shape, amplitude ± V and the correct phase

correct waveform shown while signal generator output is low (0 V): only the complete negative half cycle needs to be shown but if second negative half cycle is included it must be correct 1

correct waveform shown while voltage across signal generator output is high; condone no signal or signal = 0 V before going to –V for the first time 2

don’t insist on seeing vertical lines

2

(g) reduce the (sensitivity of) (Y-voltage)) gain 1

(change) to 2 V division–1 2 (and earns 1 )

adjust the Y (vertical) shift 3 ‘make (Y-) gain smaller’ or ‘increase the volts per division’ or ‘reduce the Y-resolution’ are acceptable substitutes for ‘reduce the (Y-)gain’increase the (Y-) gain to 2 V division–1 2 not 1

reduce the (Y-) gain to 0.5 V division–1 1 not 2 ignore any comment about time base or ‘X-gain’if all positive waveform is given for (f) allow sensible comment about triggering/stability control, egwaveform may not be stable 1 ; adjust triggering 2

2 MAX[14]

Q11.(a) circuit diagram to show:

wide end of conducting strip to – of battery, narrow end to + (1)voltmeter between wide end and probe (1)

2

(b) resistance gradient increases as x increases (1)because strip becomes narrower (as x increases) (1)current constant throughout strip (1)voltage gradient = current × resistance gradient, sovoltage gradient increases as x increases (1)

4

(c) (i) (2l – x) ln (2l – x)(0.700) (–0.357)0.60(0) –0.5110.53(0) –0.6350.47(0) –0.7550.44(0) –0.8210.42(0) –0.868

1st column correct to 2 s.f. (1)2nd column correct to 3.s.f. (1) (1)(only 4 values correct, (1))

(ii) suitable scales (1)axes labelled and units included (1)5 points correctly plotted (1)acceptable straight line (1)straight line confirms equation because equation is of formy = mx + c with negative gradient (1)

(iii) gradient = (–) = (–) 15.4 (V) (1)

1.44 V1 = 15.4 gives V1 = 11 V (1) (10.7 ± 0.2 V) [alternative: V = V1 when x = l and ln (2l – x) (= ln 0.4) = 0.92 (1)at ln (2l – x) = 0.92, graph gives V1 = 11 V (1) (10.7 ± 0.2 V)]

10[16]

Q12.(a) 0.5 mm [0.05 cm, 0.0005 m]

only acceptable answers1

(b) 8.65 mm [0.865 cm, 0.00865 m] 1

the micrometer reads zero when the jaws are closed 2only 3sf answers are acceptable for 1accept no zero error for 2

2

(c) L = (403 − 289 = ) 114 mm 1

(d) absolute uncertainty = 1 mm 1

percentage uncertainty = × 100 = 0.88%2accept 2 mm for ab. uncertainty 1

allow ecf for wrong L and / or wrong ΔLaccept 1.75%

2

(e) should move wire directly over / closer to scale on the ruler to avoid parallax error

both statement and explanation required for this mark1

(f) five values of R/L correct, recorded to 3 sf [last row to 3sf or 4sf]; accept values in Ω cm−1

mean based on first four rows only; result 9.94 Ω m−1 [9.94 × 10−2 Ω cm−1]

L/cm R/Ω (R/L)Ωm−1

81.6 8.10 9.93

72.2 7.19 9.96

63.7 6.31 9.91

58.7 5.85 9.97

44.1 4.70 10.66 (10.7)2

(g) cross-sectional area = 1

resistivity from × A, correct substitution of result from 01.6 2

1.10 × 10−63

Ω m 4

resistivity from × earns 12

allow 2 if value is not based on mean or on a mean from all five rows of table in 01.6condone 1.12 × 10−6 for 3 if fifth row in 01.6 was not rejectedwithhold 3 for POT error

4[13]

Q13.(a) (i) T = 40(ms) (1)

Hz (1)(allow C.E. for value of T)

(ii) peak voltage (= 3 × 15) = 45 (V) (1)

rms voltage =32 V (1)(31.8 V)

4

(b) (i) Irms = = 59mA (1)

(58.9mA)(use of 32 V gives 59(.2) mA)(allow C.E. for value of Vrms from (a))

(ii) Vrms = 59 × 10–3 × 90 = 5.3(1) V (1)

(allow C.E. for value of Irms from (i)) [or V2 =V1 ]2

(c) Vpeak = 5.31× =7.5(1) (V) (1)best choice: 5 V per division (1)

(allow C.E. for incorrect Vrms and for suitable reason)

reason: others would give too large or too small a trace (1)3

[9]

Q14.(a) Capacitor must not lose charge through the meter

1

(b) Position on scale can be marked / easier to read quickly etc 1

(c) Initial current = = 60.0 μA

100 μA or 200 μA (250 probably gives too low a reading)

Give max 1 mark if 65 μA (from 2.6) used and 100 μA meter chosen2

(d) 0.05 V 1

(e) Total charge = 6.0 x 680 x 10-6 (C) (= 4.08 mC)

Time = 4.08 x 10-3 / 60.0 x 10-6 = 68 s

Hence 6 readings 3

(f) Recognition that total charge = 65 t μC and final pd = 0.098 t

so C = 65μ / 0.098

660 μF Allow 663 μF

2

(g) (yes) because it could lie within 646 – 714 to be in tolerance

OR

it is 97.5 % of quoted value which is within 5% 1

(h) Suitable circuit drawn

Charge C then discharge through R and record V or I at 5 or 10 s intervals

Plot ln V or ln I versus time

gradient is 1 / RC

OR

Suitable circuit drawn

Charge C then discharge through R and record V or I at 5 or 10 s intervals

Use V or I versus time data to deduce half-time to discharge

1 / RC = ln 2 / t½ quoted

OR

Suitable circuit drawn

Charge C then discharge through R and record V or I at 5 or 10 s intervals

Plot V or I against t and find time T for V or I to fall to 0.37 of initial value

T = CR

Either A or V requiredFor 2nd mark, credit use of datalogger for recording V or I.

4[15]

Q15.planning

(a) sensible key factor e.g. p.d. across paper, that, when varied, leads to the determination of resistance: candidate then goes on to estimate the thickness of the paint layer on strip [only allow direct measurement of resistance if the investigation is of how either width or length of a rectangular strip affects the resistance of the paper] (1)

(b) correct measuring instrument given [allow circuit diagram] (1)

(c) dimensions of paper constant when resistance measured [to see how a certain dimension influences the resistance, width (if length varied)/length (if width varied)] (1)

(d) check that current through paper does not exceed 200 mA (1)

(e) sensible qualitative prediction given: thickness can only be estimated due to uncertainty in resistivity (1)

(f) thickness of layer (assuming uniform coating) in range 10–7 to 10–11 m (1)

[or (e) sensible qualitative prediction given: R ∝ l or R ∝ w–1]

(g) reasonable physics reasoning given in support: similarity with behaviour of a metallic conductor (1)

(h) use of VlI to find R [use of repeated readings to reduce uncertainty in measurement of dimension] (1)

(i) calculating possible range of thickness using limiting values of resistivity / assessing the uncertainty in result [plotting graph of results to check relationship] (1)

(j) any other sensible measure, e.g. maintain steady temperature (1)[max 8]

Q16.(a) (i) diagram to show:

(long) wire fixed at one end (1) mass / weight at other end (1) measuring scale (1) mark on wire, or means to measure extension (1)

max 3

[alternative for two vertical wires:

two wires fixed to rigid support (1) mass / weight at end of one wire (1) other wire kept taut (1) spirit level and micrometer or sliding vernier scale (1)]

(ii) measurements: length of the wire between clamp and mark (1) diameter of the wire (1) extension of the wire (1) for a known mass (1)

max 3

(iii) length measured by metre rule (1) diameter measured by micrometer (1) at several positions and mean taken (1) (known) mass added and extension measured by noting movement of fixed mark against vernier scale (or any suitable alternative) (1) repeat readings for increasing (or decreasing) load (1)

max 5

(iv) graph of mass added / force against extension (1)

gradient gives (1)

correct use of data in where A is cross-sectional area (1)

[if no graph drawn, then mean of readings and correct use of data to give 2max) (1)

max 2(13)

The Quality of Written Communication marks are awarded for the quality of answers to this question.

(b) (i) for steel (use of gives) e = (1)

e = (1)

= 5.0 × 10–3 m (1)

(ii) extension for brass would be 10 × 10-3(m) (or twice that of steel) (1) end A is lower by 5 mm √ (allow C.E. from (i))

max 3[16]

Q17.planning(to determine the illumination at points along axis of the lamp)measure distance from LDR to lamp using metre rule / tape

1find resistance of LDR using ohmmeter [or VI method explained,or potential divider method explained: must be from diagram]

1

read illumination from (calibration) graph1

(light meter method, scores 2 / 3; single circuit method orinterchanged meters, scores 1 / 3)

diagram:sensible diagram to include lamp, LDR (light meter) and suitablemeans of resistance measurement: symbols must be correct

1distance, d, shown or means to measure it

1check for correct prediction by repeating at different distances

1plot a suitable graph to test prediction of either student,with method of testing relationship explained (1)(repeated calculations of L / d2 and check for consistency is acceptable)

possible approaches for student A to test inverse-square variation:

possible approaches for student B to test exponential variation:

control:eliminate stray illumination by using blackout [or by collimating beam](subtracting background illumination not acceptable)[or maintain intensity of spotlight by use of appropriate circuit]

1

difficulties:any two of the following: (look for difficulty + how to overcome = 2)

reduce uncertainty in the graph (1)by taking extra readings where the illumination changes

most rapidly with distance (1)

reduce uncertainty in R (V and I) (1)by taking extra readings and averaging (1)

overcome uncertainty in d (1)by measuring from specified point on lamp (1)[ensure that the LDR only moves along axis of the lamp][allow any other good relevant physics]

4

(12 possible marks for) max 8[8]

Q18.(a) Background radiation

Background count = 20 count/minute unit required Ignore any –ve sign for background countMust be written to 2 sf

2

(b) Correct line of best fit The line must be a straight line (as instructed), with approximatelyan equal number of points on either side of the line.

1

(c) Triangle drawn with smallest side at least 8 cm (or 8 grid squares)correct values read from graph gradient = – 0.00698 (± 0.00030) min–1 must have –ve sign and must be to 2 or 3 sf

Gradient must lie within limits stated. No ecf from incorrectly readvalues unless it falls within stated limits. No unit penalty.

3

(d) Recognises gradient = (–) λ orUses gradient for value of λ = 7.0 (± 0.30) × 10–3 minute–1 T½ = 99 minutes to 2 or 3 sf

For 1st mark accept evidence that value of gradient has been substituted into correct formula for half life. No penalty for missing –ve sign. Allow ect from incorrect gradient value. Unit penalty if half life has been calculated in different unit (to minutes stated in question)

2

(e) Random 1

(f) (i) Uncertainty = ( ± √429) = ± 21No sf penalty

Details of calculation not required. Marks can be awarded for correct numerical answers. Also no penalty for quoting uncertainty or % uncertainty without ‘±’.

1

(ii) % uncertainty = ± 4.9% No sf penalty. (Note that % uncertainty in total count is same as % uncertainty in corresponding count rate.)

Accept also 4.8% (number achieved keeping all sig figs in calculator)No penalty for omitting % or ‘±’.No sf penalty

1

(iii) % uncertainty for 84 counts is ± 10.9% Taking data over larger time period / larger total count will have smaller percentage uncertainty.

Accept ±11%No penalty for omission of ± sign. No sf penalty for estimated % uncertainties.

2[13]

Q19.(a) (mark should be at the equilibrium position) since this is where the mass moves with

greatest speed [transit time is least] 1

(b) (i) mean time for 20T (from sum of times ÷5) = 22.7 (s)1 (minimum 3sf)

uncertainty (from half of the range) = 0.3 (s) 2 (accept trailing zeros here)

percentage uncertainty

(allow full credit for conversion from 20T to T, e.g. 1.135 = 1 0.015 =2 ecf for incorrect 1 and / or 2 earns 3

3

(ii) natural frequency

(ecf for wrong mean 20T; accept ≥ 4 sf)1

(c) (i) linear scale with at least 3 evenly-spaced convenient values (i.e. not difficult multiples) marked; the intervals between 1 Hz marks must be 40 ± 2 mm (100 ± 5 mm corresponds to 2.5 Hz)

(ecf for wrong natural frequency: 100 ± 5 mm

corresponds to Hz)1

(ii) 4 mm [allow ± 0.2 mm] 1

(d) (i) student decreased intervals [smaller gaps] between [increase frequency / density of] readings (around peak / where A is maximum)

[student took more / many / multiple readings (around peak) ](reject bland ‘repeated readings’ idea; ignore ideas about using data loggers with high sample rates)

2

(ii) new curve starting within ± 1 mm of A = 4 mm, f = 0 Hz with peak to right of that in Figure 3(expect maximum amplitude shown to be less than for 2 spring system but don’t penalise if this is not the case; likewise, the degree of damping need not be the same (can be sharper or less pronounced)

Peak at value given in (b)(ii); expect 1.25 Hz so peak should be directly over 50 ± 5 mm but take account of wrongly-marked scale

2[11]

Q20.(a) D could not be measured with enough precision; [can only resolve to 1 sf / 2 dp

(and 3 sf / 4 dp needed) / needs to measure to 0.0001 mm] example given to correctly illustrate this point, eg 0.0855 mm would be read as 0.09 mm

same D would be produced for different α example given to correctly illustrate this point, eg when α = 12° / 14° / 16°

there would be a large percentage uncertainty [percentage error] in D example given to correctly illustrate this point, eg when α = 8° percentage uncertainty is 47% (tolerate answers using ∆D = 0.01 mm or 0.02 mm)

α / °D / mm

spreadsheet to 0.01 mm

% uncertainty(∆D = 0.01 mm)

2 0.0855 0.09 11.7%

4 0.0428 0.04 23.4%

6 0.0285 0.03 35.1%

8 0.0214 0.02 46.8%

10 0.0171 0.02 58.5%

12 0.0143 0.01 70.2%

14 0.0122 0.01 81.9%

16 0.0107 0.01 93.6%

max 4

(b) argument is not sensible because (larger value of D leads to) very small values of α (hence) α cannot be measured accurately [uncertainty would be very large]

2

(c) × 100 (working must show 0.0859 in denominator, or 0 / 2)

= 0.466% or 0.47% only (ie 0.5% is worth 1 max)2

[8]

Q21.

(a) (i) 5.1 and 7.1 Exact answers only

1

(ii) Both plotted points to nearest mm Best line of fit to points

The line should be a straight line with approximately an equal number of points on either side of the line

2

(iii) Large triangle drawn at least 8 cm × 8 cm

Correct values read from graph Gradient value in range 0.190 to 0.210 to 2 or 3 sf

3

(iv) (R = ) = 5.0 Ω Must have unit

Allow ecf from gradient valueNo sf penalty

1

(b) (i) 5.04 (Ω) or 5.0 (Ω) ઙ

(Allow also 5.06 Ω or 5.1 Ω, obtained by intermediate rounding up of 3.502)

From R = 1

(ii) (Uncertainty in V = 0.29% ) Uncertainty in V2 = 0.57%, 0.58% or 0.6%

From uncertainty in V = 0.01 / 3.50 × 100%Uncertainty in P = 2.1% From uncertainty in P = 0.05 / 2.43 × 100% = 2.1%Uncertainty in R =2.6%, 2.7% or 3% Answer to 1 or 2 sf only

2.1 % + uncty in V2 (0.6%) = 2.7% Allow ecf from incorrect uncertainty for V2 or P

3

(iii) (Absolute) uncertainty in R is ( ± ) 0.14 or just 0.1 Ω (using 2.6%) (or 0.15 or 0.2 Ω using 3%)

Must have unit (Ω)Must be to 1 or 2 sf and must be consistent with sf used from (ii)No penalty for omitting ± sign

1

(iv) Works out possible range of values of R based on uncertainty in (iii), e.g. R is in range 5.0 to 5.2 Ω using uncertainty of ± 0.1 Ω

No credit for statement to effect that the values are or are not consistent, without any reference to uncertaintyAllow ecf from (iii)

Value from (a)(iv) is within the calculated range (or not depending on figures, allowing ecf)

Allow ecf from (a)(iv)2

[14]

Q22.

(a) λ [the gradient] = (−) 0.015

N½ from (−)

46.2(1) slides (accept 46 but do not penalise '47 slides needed to halve V')

[λ = 0.015 or use of ratio

determination of V0 = 424(.1) mV; ln(V0/2) = 5.36 [5.357]

= 46(.0) slides (accept 46.2, '47 slides needed to halve V' etc ]3

(b) (i) (student must measure or calculate) thickness of slide, t; half-valuethickness = N½ × t [= result from (a) × t]

1

(ii) procedure: measure the thickness of multiple slides (either singly or in a stack) and calculate average thickness [divide by number of slides] (reject bland 'repeat and average')

[measure the thickness at different points on the slide, and average by number of readings or measure the thickness of different slides and average]

1

(iii) procedure: close jaws and check reading ( = zero) ['check for zero error']

(reject idea of measuring 'known' dimension and checking reading or that 1 micrometer is 'zeroed'/'set to zero'/'zero calibrated' before use’)

1

(c) t from = 1.19 mm (3 sf only) 1

(d) n = = 1.47, no unit (3 sf preferred but tolerate 4 sf, do not

penalise here and in part a for sf) 1

(e) (i)/(ii) Δ (R2 − R0) = Δ(R2 − R1) = 0.08 mm 1

(iii) P2 − 0 = % uncertainty in (R2 − R0) = 100 × = 0.56(0)% [0.6%] and

P2 − 1 = % uncertainty in (R2 − R1) = 100 × = 0.82(4)% [0.8%]

working must be shown; allow ecf from i/ii but only if working is correct

Pn = % uncertainty in n = (P2 − 0) + (P2 − 1 ) = 1.38(4)% (accept 1.4 %)

for ecf from i/ii working in iii must be valid; for AE in iii allow ecf in final calculation

[max and min values calculated, eg nmin = , nmax =

;

difference = ½ range () convert to % = 1.38 (± 0.02)% ()]2

[11]

Q23.(a) 5.31, 6.38

Exact answers only1

(b) Both points correctly plotted to the nearest mmWell drawn straight line of best fit.

The orange LED point (4.80, 1.54) is anomalous. The line should follow the trend of the points (ignoring the anomalous point) with an even scatter of points on either side of the line.

2

(c) (i) Triangle drawn with smallest side at least 8 cm in length.Correct readings taken from the line for the triangle. Gradient in range 0.44 to 0.46 (0.435 to 0.464) × 10-14 quoted to 2 or 3 significant figures

The size of the triangle can be implied by readings taken from the line.The third mark is independent of the other two: error carried forward for incorrect readings (or for a poor line of best fit) which give a gradient out of range is not allowed. Unit not required for the mark.

3

(ii) Possible marking points:The anomalous point makes the value less reliable.(However) the (other) points are close to the line of best fit. Suggesting that the value is reliable.

If the candidate has not ignored the anomalous point when drawing the line of best fit accept:The points are not close to the line of best fit so the value is not reliablefor one mark only

2

(d) (i) Recognition that the gradient = h/e h = 0.45 × 10-14 × 1.60 × 10-19 = (6.95 to 7.44) × 10-34 Js

Allow ecf from (c)(i) for second mark (including wrong exponent)Final answer must be in range, have correct exponent, correct unit and be quoted to 2 or 3 sf

3

(ii) ((7.2 × 10-34 - 6.63 × 10-34)/ 6.63 × 10-34) × 100%calculated correctly

Allow ecf from (d)(i): expected answer 8.6%

Allow giving 7.9%

No sf penalty1

(iii) ± 1.1% or + 1%± is required here as it is explicit in the question

1

(iv) %uncertainty in f = 8.6 – 1.1 = 7.5%∴ δf = ± 0.075 × 3.19 × 1014 = ± 2.39 × 1013 Hz

Allow ecf from (d)(ii)Final answer: 2 or 3 sf with unit but ± symbol not required

316

Q24.(a) the travelling microscope won’t interfere with / change the path / interrupt / affect the

stream [flow] of water / affect the reading (being taken)[vernier callipers will interfere with etc]

(reject ‘cannot grip / clamp the flow’)1

(b) (i) straight best-fit line drawn passing within ± 2 mm of 1st and 5th points, 3rd and 4th points to be either side of line;

attempt to measure the gradient (i.e. using from the line or from two of the plotted points if these lie on the line; do not penalise for small steps, false read-off(s) (including failure to take account of false origin) or for calculation error

n = –4 (integer value only, e.g. reject –4.0) 2

(ii) k = 10intercept [antilog of (log s) intercept]

[take values of log s and log d and evaluate 10(log s – (–4)log d) ]

(‘log k = intercept’ is insufficient)1

(iii) units of k = cm5 [accept m5 or mm5; allow ecf for wrong or non-integer value for n, eg ecf for cm(1–n)

1[5]

Q25.(a) extension of wire Q = 2.7 (mm)

ignore any precision given eg ± 0.1 mmif > 2 sf condone if this rounds to 2.7

1

(b) mass = 5.8 (kg) allow ce for incorrect 0.1.1 (only look at 01.1 if answer here is incorrect)allow ± 0.1 kg

1

(c) 0.51 (mm) ignore any precision given eg ± 0.005 mm

1

(d) method 1:

for 1 expect to see some substitution of numerical data

correct use of diameter for 2; ignore power of ten error; expect CSA = 2.0(4) × 10−7; allow ce from 01.3 (eg for d = 0.37 mm CSA = 1.0(8) × 10−7 m2)

penalise use of g = 10 N kg−1

value of ∆l must correspond to Figure 2 value of m; answers to 01.1 and 01.2 are acceptableexpect l = 1.82 m but condone 182 etc; accept mixed units for l and ∆l

MAX 3

method 2:

evidence of from Figure 2 to calculate gradient 1expect gradient between 0.45 to 0.48 mm kg−1

2 3

missing g loses 3substitution of l = 1.82 m 4

condone 182 etc 4

cross-sectional area from 5correct use of diameter for 2; ignore power of ten error; expect CSA = 2.0(4) × 10−7; allow ce from 01.3 (eg for d = 0.37 mm CSA = 1.0(8) × 10−7 m2)

MAX 3

result in range 1.84 × 1011 to 1.91 × 1011 5condone 1.9 × 1011

5 mark requires correct working and no power of ten errors: allow ce for error(s) in 01.1, 01.2 and for false/incorrect CSA (eg for d = 0.37 mm allow result in range 3.49 × 1011 to 3.63 × 1011, 3.5 × 1011 or 3.6 × 1011)

1

(e) (smaller diameter) produces larger extensions 1reduces (percentage) uncertainty (in extension and in result for Young Modulus) 2

(smaller diameter) increases (percentage) uncertainty in diameter or cross sectional area is smaller or increases (percentage) uncertainty in cross sectional area 3increases (percentage) uncertainty (in result for Young Modulus) 4

(smaller diameter) increases likelihood of wire reaching limit of proportionality or of wire snapping or reduces range of readings 5increases (percentage) uncertainty (in result for Young Modulus) 6

outcome and correct consequence for 2 marks, ie 1 followed by 2, 3 followed by 4 etcdna ‘error’ for ‘uncertainty’no mark for consequence if outcome not sensible, eg ‘it gets longer and reduces uncertainty’ earns no mark for ‘diameter smaller so uncertainty greater’ award 1 (need to see further mention of uncertainty to earn 2)

MAX 4[11]

Q26.(a)

h/mm ln(h/mm)

381 5.943

336 5.817 or 5.818

Exact answers only.

1

(b) Both points plotted within 1 mm An accurate best fit straight line drawn with an even scatter of points on either side of the line

2

(c) Triangle drawn with smallest side at least 8 cm in length (or 8 grid squares) and correct values read from the line of best fit Correct answer for gradient in the range -0.0140 to -0.0138

Note: correct answer marks:One mark for the minus sign plus one mark for a value in the range 0.0136 to 0.0140 expressed to 2 or 3 sf

3

(d) H = 665 or a correctly calculated value from the intercept on the graph and the unit quoted for H quoted as mm λ = candidate’s answer to part (c) without the minus sign Unit for λ = s-1

No sf penalty3

(e) There could be a systematic error in the measurement of h Consideration of the effect of this on the natural log values that are being plotted An explanation involving recognition that a (small) change in gradient is likely and this would result in a change in λ

Alternative (simplistic) answer for 2 marks maxThere could be a systematic error in the measurement of h This would be unlikely to affect the gradient of the line since h is numerically large and so the value found for λ would not be affected.

3

(f) There could be random errors associated with the use of a stopclock for measuring time By taking repeat readings these should have been minimised so the value found for λ would not be affected.

2[14]

Q27.

(a) (accept bald answer for 1 mark)1

(b) 0.01 mm (condone 0.005 mm) 1

(c) uncertainty in 3s [in s] = 0.02 [2 × answer for (b)] or 0/2

percentage uncertainty in (use of R2 (use of R1 is required; accept 1 sf 0.5%)

[for precision = 0.005 mm, % uncertainty in

(use of R2 and R1 is required; accept 1 sf 0.2% but reject 0.3%)2

(d) evidence of suitable working, e.g. d from 2s - (R3 - R2) or from

5s - (R3 - R1) or from

d = 0.84 mm

[allow ecf for incorrect s: the candidate in (a) who evaluates the distance between the edges of adjacent holes will get s = 0.59 mm; they get the correct result for d

using 2

[6]

Q28.(a) (i) maximum distance from 1st to 5th image = 139 mm (allow 138 to 140mm)

Any other correctly measured distance(s), to provide additional data, but do not allow the distance from 1st to 2nd image (as this is too small).

Alternatively allow a repeat measurement of the maximum distance for this mark.

E.g. 1st to 4th image distance = 79 mm, 1st to 3rd image distance = 34 mm (allow ± 1mm on these values)If a candidate measures distances which do not go back to the first image the initial velocity, u, will not be zero. Use of the ‘suvat’ equations is then considerably more complicated. A mark can still be awarded, provided the candidate illustrates how g can be calculated or correctly calculates g from this value in (a)(ii).If the printing process in your centre alters the scale of this diagram, measure the values on your printed papers and mark the scripts accordingly. (Send details to the moderator).If the candidate is visually impaired and is using a modified paper that alters the scale of this diagram, measure the values on the printed paper and mark the script accordingly.

2

(ii) Evidence of correct number of time intervals to match the distance used.(e.g Using distance from 1st to 5th marble with 4 time intervals, time = 0.174 s)

Value of g = 9.18 or 9.2 m s–2 (based on 139 mm and 0.174 s)Allow ecf from value of time in (a),and ecf from incorrect measurements in (b), but to allow ecf candidate must show workings of the calculation and not just state a value for g

(Allow also 918 or 920 cm s–2 and 9180 or 9200 mm s –2) Answer must have correct unit

Allow this mark based on a correct calculation from just one measurement of distance in (a)(i)No sf penalty

2

(b) So image is sharp / less blurred / image is well definedORAny answer referring to ‘motion blurring’

Allow reference to ‘image’ or ‘marble’1

[5]

Q29.(a) Both tm values correct: 0.404, 0.429

ANDBoth tm

2 values correct: 0.163, 0.184 Exact values required for the mark.

1

(b) Both plotted points to nearest mm Best line of fit to points

The line should be a straight line with approximately an equal number of points on either side of the line.

2

(c) Large triangle drawn (at least 8 cm × 8 cm) Correct values read from graph Gradient value in range 0.190 to 0.222

Allow 2 or 3 sf for gradient3

(d) g = 9.71 (ms–2) or correct value from gradient value in (c) .

(The answer must be in the range 9.0 to 10.5 (ms–2)).Allow 2 or 3 sf.Unit not required

1

(e)

OR correct computation using value from (d) If the candidate’s value is exactly 9.81, then a statement that there is no (or zero) percentage difference is acceptable.No sf penalty.NB. Allow an answer from a calculation with either the candidate’s value or the accepted value as the denominator in the equation.

1

(f) 0.001 s (half the spread)(Must have unit).

1

(g) g = 2s/tm2 = 2 × 0.300/0.2452 = 10.0 (or 10.00) ms−2

Unit required and 3 or 4sf for the last mark.3

(h) % uncertainty in s = 0.33 and% uncertainty in tm = 0.41

Allow ecf from part (f).

% uncertainty in g= 0.33 + (2 × 0.41) = 1.15

Allow ecf at each stage of calculation.

Uncertainty in g= 10.0 × 1.15/100 = 0.12 m s−2 or 0.1 m s−2

Allow ecf from part (g).

(allow 1 or 2 sf only)(Must have unit for 3rd mark).

3

(i) (a) Use spherical objects of different mass and determine mass with balance Annotate the script with the appropriate letter at the point where the mark has been achieved.

(b) Would need same diameter spherical objects for fair comparison (same air resistance etc)

(c) Time spherical object falling through same height and compare timesAlternative for (c):i.e. repeat whole of experiment, plot extracted values of g against mass. Horizontal line expected, concluding acceleration same for different masses.

3[18]

Q30.(a) (i) W = 2mg cosφ ∴ m = W/(2g cosφ)

The question says show that, so the candidates must write down both steps.

1

(ii) Well drawn straight line of best fit.The line should follow the trend of the points with an even scatter of points on either side of the line.

1

(b) (i) Triangle drawn with smallest side at least 8 cm in length. Correct readings taken from the line for the triangle Gradient in the range 0.45 to 0.49 (0.445 to 0.494) quoted to 2 or 3 significant figures

The size of the triangle can be identified from readings taken from the line.The third mark is independent of the other two: error carried forward for incorrect readings (or for a poor line of best fit) which give a gradient out of range is not allowed.

3

(ii) Candidate’s answer for gradient in (b)(i) correctly multiplied by g (expected answer 4.6)N

No s.f. penalty.The second mark is for the unit and can be awarded if the numerical answer is incorrect.

2

(c) δx% = 0.2 and δy% = 0.5δ(x/y)% = δx% + δy% = 0.2 + 0.5 = 0.7 Use of δ(x/y.)2% = 2 × δ(x/y)(%)

Final answer is (±) 1.4 (%) which automatically gains all three marksOtherwiseAccept only 1 s.f. for 1st and/or 2nd marks.The third mark is for the method, not the final answer

3

(d) (i) Systematic errors in measurements are errors which show a pattern or a bias or a trend

Some acceptable alternatives• A systematic error is one which deviates by a fixed

amount from the true value of a measurement• An error which has the same value in all readings• A difference between the true value of a quantity and

the indicated value caused by a fault in the measuring device

• Accept a good example of systematic error.1

(ii) y would be largerbecause angle θ would be smaller or because friction would be opposing the increasing weight of m

2[13]

Q31.(a) smooth curve of decreasing positive gradient through all 5 points

shaky or fuzzy line does not gain mark1

(b) sensible tangent drawn at b = 25; correct read−offs for points (± 1 mm) from triangle with step sizes at least 8 × 8 1

substitution correct 2

G = 0.11(2) 3change in d divided by change in b for 3; don’t penalise if change in d is given in m acceptable range if d is in mm 0.109 to 0.116 for d in m adjust accordingly; accept only 0.11 for 2 sf; accept ≥ 3 sf for 3

3

(c) d in range 14.25 to 14.30 mm 1

substitution correct 2

R in range 1.34 to 1.38 m 3

accept result for R in mm; no ecf for incorrect or out of range G

3[7]

Q32.(a) Straight line of best fit passing through all error bars

Look for reasonable distribution of points on either side

1

(b) h0 = 165 ± 2 mm1

(c) Clear attempt to determine gradient 1

Correct readoffs (within ½ square) for points on line more than 6 cm apart and correct substitution into gradient equation

1

h0k gradient =( -) 0.862 mm K-1 and negative sign quoted

Condone negative sign Accept range -0.95 to -0.85

1

(d) k =

= 5.2 x 10-3 Allow ecf from candidate values

1

K-1 Accept range 0.0055 to 0.0049

1

(e) for h = 8000 mm, d-1 = 1

d = 1.8 x 10-3 mm 1

(f) Little confidence in this answer because One of It is too far to take extrapolation OR This is a very small diameter

1[10]

Q33.(a) p0 = 198.4 (cm)

only acceptable answer1

(b) all x values correct and recorded to nearest mm

n p/cm x/cm

0 198.4 174.6

2 157.0 133.2

4 125.4 101.6

6 101.3 77.5

9 75.4 51.6

13 53.8 30.0

allow ecf for x = p – 23.8 if p0 ≠ 198.4penalise 2 sf x = 30 for n = 13

1

(c) six values of ln(x/cm) recorded consistently i.e. all to (minimum) 2 dp; confirm that value of ln(x) for n = 6 corresponds to tabulated value of x 1

vertical axis labelled ln(x/cm) i.e. bracket required;

suitable vertical scale (points should cover at least half the grid with a frequency of not less than 5 cm) 2

points plotted for n = 0, 2, 4, 6, 9 and 13;

check n = 6 point is plotted within half a grid square of tabulated position;

suitable continuous ruled line of negative gradient from n = 0 to (at least) n = 13 3

expected data:

n p/cm x/cm ln(x/cm)

0 198.4 174.6 5.162

6 101.3 77.5 4.350

for n = 0, x = 0, ignore missing or incorrect ln(x) and ignore

missing/wrongly-plotted pointfor 2 vertical axis should be labelled ln(x/cm) (note that bracket is essential); expect vertical scale to start at 3 with major divisions of 0.2for 3 a suitable line must pass through all points if these have been correctly calculated;for any errant plotted points the line must be the best line in the opinion of the marker;line must not be thicker than half a grid square and width must not vary;points must not be thicker than half a grid square (reject any dots or blobs)

3

(d) graph is linear and has negative gradient allow ‘straight line’ for ‘linear’; statement must be confirmed by Figure 4allow ‘negative slope’ or ‘slopes downwards’ for ‘negative gradient’no ecf for non-linear graph

1

(e) gradient triangle for Figure 4;

correct read offs (± 1 mm) for all points or for both steps in triangle 1

expected gradient result is –0.135

for gradient between –0.139 to –0.133 (allow this intermediate answer shown as a fraction)

award two marks for minimum 3 sf x when n = 20 in range 11.2 to 12.2 (cm) 23

OR

one mark for x when n = 20 in range 10.8 to 12.7 (cm) 23

OR (if gradient out of range)

marker uses candidate’s gradient (which must be negative) and (marker must read off) intercept on Figure 4 to calculate x when n = 20

minimum 3 sf result in range ± 4% 23

OR

1 mark minimum 3 sf result in range ± 8% 23

(theoretical result for x when n = 20 is 11.7(3) cm)for 1 allow 1 mark for sufficient evidence of working and a valid calculation of the gradient of a linear graph even if graph has a positive gradientfor 2 and 3

give no credit if graph drawn has a positive gradientallow 1 mark for using a positive value for the (negative) gradient in the calculation for x when n = 20 (this leads to 2592 cm); result must be in range ± 4% 23 allow ‘similar triangles’ method;

eg 1 ln x20 = 2.46 2 ; x20 = 11.7(0) cm 3 allow ecf x when n = 20 based on Figure 4 if scales used enable value to be read directly using an extrapolated line; do not allow such working to extend beyond the grid into the margin 1 ; value in range 11.2 to 12.1 cm 23 = 1 MAX

3

(f) valid procedure 1

described 1

explained 2 ;

valid procedure 2

described 3

explained 4 explanation mark (2 ) is only awarded when it is relevant to a correct procedure (1); one procedure/explanation allowed per responseno credit for conflicting statements or wrong physicsany two from:repeat experiment and average calculated (p) 1to reduce (impact of) random [human] error 2and/orrepeat readings to detect anomalies 1so these can be discarded (before averaging) 2and/orview air track at right angles [at eye level] 1to reduce [eliminate] (impact of) parallax error 2and/orrepeat experiment with track direction reversed and average calculated (p) 1to account for the effect of non-level bench 2and/oruse video (camera) technology [or a motion sensor linked to a data logger or laser ranger] to view [record] the position of the glider as it reaches the top of the track 1to reduce (impact of) random [human] error [to identify and eliminate anomalous results] 2reject any suggestion that involves changing the glider, its initial position on the track or the air track itself including the position of the scale

4[13]

Q34.(a) Clear identification of distance from centre of sphere to right hand end of mark, or to

near r.h.end of mark 1

(b) 0.393 (s) Accept 0.39 (s)

1

(c) For 10 oscillations percentage uncertainty = = 0.00637 ≡ 0.64%

same for the ¼ period 2

(d) Identifies anomaly [0.701] and calculates mean distance = 0.759 (m)Allow 1 max if anomaly included in calculation giving 0.750 (m)

2

(e) Largest to smallest variation = 0.026 (m)

Absolute uncertainty = 0.013 (m)1

(f) Use of g = leading to

9.83 (m s-2) Allow 9.98 (m s-2) if 0.39 usedEcf if anomaly included in mean in (d)

percentage uncertainty in distance = 1.7%

Total percentage uncertainty =

1.7 + 2 x 0.64 = 3.0%

Absolute uncertainty = 0.30 (m s-2)

[g = 10.0 ± 0.3 m s-2]Expressed sf must be consistent with uncertainty calculations

3

(g) suggests one change

Sensible comment about change or its impact on uncertainty

eg

Use pointed mass not sphere

Because this will give better defined mark OR because the distance determination has most impact on uncertainty

OR

Time more swings / oscillations

As this reduces the percentage uncertainty in timing

OR

longer / heavier bar would take a greater time to swing to the vertical increasing t and s and reducing the percentage uncertainty in each

If data logger proposed, it must be clear what sensors are involved and how the data are used.

2

(h)

plot graph of s against t 2 or √s against t

calculate the gradient

the gradient is g / 2 or √(g / 2) Accept: plot s against t 2 / 2 or plot 2s against t 2:calculate the gradientin both cases gradient gives gAllow 1 max for answer that evaluates g for each data point and averages

3[15]

Q35.(a) θ1 = 61.0 ± 0.5 °C

reject 2 sf θ1

1

(b) sensible tangent drawn at t = 190 s; correct read-offs for points (± 1 mm) from triangle with step sizes at least 8 × 81

G1 = −9.57 × 10−2 3for 3 insist on correct sign and POT; accept result in range 1.05 × 10−1 to −9.0 × 10−2

3

(c) substitution correct leading to θR = 17.3 ± 2.0 °C allow ECF

1

(d) θ0 − θR correctly evaluated to ± 1 °C for θ0 at suitable reference time 1

evaluates 2

evaluates θ from + θ0 ; time constant deduced from graph with evidence of working (read offs to both axes are required) 3

time for object to reach room temperature in range 1900 to 2000 s 4

example for 1: θ0 = 89 °C at t = 0 gives θ0 −θR = 89 − 21 = 68 °C

allow ecf for failure to take account of θR in 1

example for 2: ; allow ecf for failure to take account of θR in 1

example for 3 : θ = 25 + 21 = 46; time constant = 390 sexample for 4 : time to reach room temperature = 5 × 390 = 1950 s; no ecf for errors in 1 or in 3

4

(e) the starting temperature was lower 1

the starting temperature was 86.5 °C compared to 89.0 °C 2

the room temperature was higher 3

the draught was less 4

the water had only cooled to 38.0 °C after 600 s 5

the sample rate of the data logger was lower 6

samples were recorded every 20 s (rate for original experiment was much higher) 7

other approaches are possibleallow ± 0.3 °C for any temperature quoted for 2 or for 5

MAX 5[14]

Examiner reports

Q1.As anticipated this question provided good discrimination.

(a) Most students attempted to count and measure a large number of fringe widths. A common error was to measure from the first to the fiftieth fringe, but count this as 50 fringe widths rather than 49.

(b) Weaker responses were unsure how to calculate the uncertainty in a single fringe width from the uncertainty in the total measurement.

(c) (i),(ii) & (iii) Many candidates failed to score on any part of this question, despite part (i) being a straightforward substitution into the formula quoted. The uncertainty calculations in (ii) & (iii) were only successfully completed by the most able candidates.

Q3.This question addressed the ideas behind assessed practical activity 2. Some of the graphical techniques described in the Practical Handbook were also tested. The numerical work in questions (d) to (f) (and in the multiple-choice part (h)) was usually where students scored the bulk of their marks.

Questions (a) and (b) exposed poor understanding of superposition. Although there were plenty of correct answers to (a) and very few students forgot to supply any unit (we expected π radians or 180°), there were a variety of unsuccessful alternatives and some students clearly confused path difference with phase difference. Most students seemed to think that the resultant amplitude depended only on the phase difference between the sets of superposing waves, and very few considered the possibility that the waves travelling via M would arrive at R with less amplitude than those travelling directly from T. A small but significant number of students blamed microwave background radiation for the non-zero minimum reading on the ammeter.

Examiners found that many students ‘improvised’ their answer to question (c). Developing strategies for encouraging students to think about the practicalities of arranging apparatus, and challenging their reasoning, would pay dividends in future. A simple approach such as measuring, at different points, the distance between M and the line between T and R, and checking these were equal could earn 1 mark; the sensible use of a set-square earned the second mark. Frequent suggestions that non-standard science equipment, such as tri-squares could be used, perhaps indicated that some students had only had limited opportunities to perform practical work. A common misconception was that the ends of rulers can be assumed to be square; several students stated that aligning the end of the ruler, or the graduations along the edge, with the line between T and R, ensured that perpendicular distances would be measured. One really positive aspect of the work seen was the good standard of sketches students used to illustrate their answers; some of these, by themselves, were sufficient to earn both marks.

Answers to question (d) were much stronger, with many students demonstrating an understanding of how the error bars could be used to establish the maximum and minimum gradients. Most knew they should use large steps for the gradient calculations, and recognised in question (e) that their mean gradient gave λ.

In question (f), many students appeared to have taken a guess, often based on the size of the error bars, about the uncertainty they should use. The better students used a variety of valid ways, based around ‘best gradient − worst gradient’, of judging the uncertainty. However, full credit proved elusive because the final answer was often compromised by

the truncation, in (d), of the values for maximum and minimum gradients.

Many students seemed to have rushed question (g), thinking they were being asked to explain that y was the vertical intercept. However, examiners wanted to know how Figure 4, which did not allow a direct measurement, should be used. Suggesting a calculation method could earn a mark, but this was denied when any suggested algebra was wrong; otherwise, some consideration of an average of the maximum and minimum intercepts also gained credit.

Question (h) provided some respite for students who found the descriptive writing challenging.

Q4.

In part (a) most showed that to earn a mark but some

missed a key step in the manipulation that followed in arriving at

In part (b) the majority saw we expected a simple application of the y = mx + c idea and gained both marks.

Part (c) proved challenging for the candidates, most of whom made only limited progress by making an observation without any supporting explanation, e.g. by comparing the largest XZ value with the length of the block or by computing the diagonal distance across the face of the block and comparing this with the largest WX value. We wanted the candidates to discuss whether the limiting values of the data in the table showed if a suitable range had been used. Clearly the upper limit was ideal but the lower limit raised the question of whether such values could be measured without attracting a significant percentage uncertainty.

Better answers seen compared the range of the XZ data with the length of the block but we saw very few responses that earned full credit.

Several candidates used each row to calculate six values of refractive index and decided on the suitability of the data looking at the spread of results. Others judged the quality of the data based on the intervals between the values in each column. These approaches earned no credit.

Few could score more than 5 out of 7. A/B candidates scored between 3 and 5 while E/U candidates usually scored 1 or 2.

Q6.(a) Correctly answered by almost all candidates.

(b) As usual in this question a small proportion of candidates failed to accurately plot both points and an even greater proportion were unable to draw an acceptable line of best fit.

(c) (i) A large proportion of candidates ignored the powers of 10 on the graph axes.

(ii) A good discriminator, with only the most able candidates giving the correct unit.

(d) (i) A straightforward question with all but the weakest responses scoring both marks.

(ii) A good discriminator, with only the most able candidates correctly using 'y = mx + c' and substituting data from the table or graph to find intercept c.

(iii) Only the most able candidates correctly identified the systematic error to be in w.

(e) Not well understood by many candidates, with only a small proportion referring to single slit diffraction.

Q8.In part (a) the great majority of candidates calculated correctly the required values of resistance and quoted their answers to three significant figures. Those candidates who obtained incorrect values of resistance were awarded marks if they plotted their points correctly on the graph.

In part (b)(i) almost all candidates chose acceptable scales and marked axes clearly. The labelling of the axes on the question paper helped candidates. With few exceptions, candidates plotted their points clearly and correctly. About one-third of candidates attempted to draw a straight line of best fit through their points. About one-half of such candidates persuaded the line to pass through the origin, giving little thought to the fact that they were suggesting that the resistance of the filament is zero when the current is zero. Other candidates drew very poor lines of best fit through the points and lost marks unnecessarily in parts (b)(i) and (b)(ii). In part (b)(iii) the majority of candidates were able to calculate the required changes in resistance and quoted answers to two significant figures. The calculations in part (b)(iv) gave weaker candidates some difficulty since some used the wrong formula whilst others did not know the correct unit for power.

Very few candidates scored the full two marks in part (c). The question clearly asks candidates why the change in resistance of the filament is less for a current change of 0 to 1.0 A than for a current change of 1.0 to 2.0 A. Most candidates gained one mark for stating that the resistance of the filament increases with increasing temperature. However, very few candidates referred to the fact that the increase in heat dissipation for 1.0 to 2.0 A is greater than for 0 to 1.0 A, leading to a greater corresponding rise in temperature.

Q9.(a) (i) Students had to make it clear that the voltmeter ‘alone’ should be connected

across the cell.

(ii) A correct explanation was given by a large proportion of students.

(b) (i) Answered well by the more able students.

(ii) A proportion of students seemed to understand how to use the voltmeter but failed to show the correct position on the circuit diagram.

(c) This question discriminated well. Many students failed to give sufficient detail as required by the mark scheme for the first marking point. The second marking point proved to be more accessible, with a greater proportion of students able to suggest an appropriate precaution.

(d) As anticipated this proved to be very demanding, with only the more able students successfully stating and explaining why efficiency would increase as external resistance increases.

Q10.This question addressed the ideas behind assessed practical activity 9. Students who had

seen the second set of specimen questions should have been well prepared for the parts relating to the use of the oscilloscope. It was clear from the descriptive parts of this question, however, that many students had not had direct experience of using an oscilloscope, as shown by the lack of much, if any, use of correct terminology to describe the controls.

Even if some students thought the waveform shown in Figure 1 showed an ac signal, the instruction to determine the peak to peak voltage should have made question (a) straightforward. However, a disappointing number of students found a variety of ways to get this wrong and were similarly unsuccessful with question (b).

Those students who obtained 6.3 V and 250 Hz were well-placed to score in question (c)

and a popular and successful approach was to use ; a transparent, viable attempt along these lines could earn a mark, even if the final answer fell outside the required range. Others read off the time interval for V to fall to 37% when the capacitor was discharging, but poor communication often made it harder to give any credit, if the final result obtained by this method was unsuccessful. Those using the idea that the capacitor would completely discharge in five time constants were almost invariably unsuccessful due to difficulty in accurately pinpointing the time for 5RC from Figure 5.

In question (d), most students knew what would happen to the waveform displayed in the oscilloscope when the time-base was adjusted, but often struggled to give a convincing explanation. Stating that the ‘waveform would not fit’ left examiners looking for further evidence to decide whether the student was referring to vertical or horizontal direction. Some students referred to the ‘wavelength increasing’, or equivocated by suggested that the waveform ‘might not fit’. While better answers suggesting that either the charging or discharging parts (but not both) would now be displayed, very few students took a quantitative approach, e.g. that the discharge curve would now be 2.5 times wider. Here and elsewhere in the paper, some students explained that uncertainty would be reduced, but without supporting reasoning this gained no credit.

Question (e) discriminated well, with better students stating that the resistance, and hence the time constant, is halved, and spotted the opportunity to give the quantitative detail suggested by the information in the question. The sketches on Figure 7 showed that some students thought that adding the extra resistor would change the amplitude or the period of the waveform; relatively few students supplied the sketch examiners were looking for.

In the sketch for question (f), examiners wanted to see a complete half cycle while the signal generator output was V, and a complete half cycle when the output was zero. Students were generally more successful with the latter, but often failed to consider how Figures 8a and 8b could be used in combination to deduce the other part of the waveform.

In question (g), “reduce Y-gain”, or “increase volt per division” could score but an equivocal “change the Y-gain” could not. Once again, extra credit was available for quantitative detail such as “set the gain to 2 V per division”. Some students clearly failed to appreciate that Figure 8c showed a graph and was not the way the waveform would appear on the oscilloscope; these students stated that the waveform was already fully visible, so no changes were necessary.

Q11.The circuit diagrams produced in part (a) were generally disappointing. Few candidates showed the voltmeter correctly connected to the tapered conducting paper and many had the positive pole of the battery connected to the broad end of the strip and the negative pole to the narrow end. The voltmeter was often shown in parallel with the full length of the conducting strip or else connected between the probe and the narrow end of the conducting strip.

Equally disappointing were the explanations of the non-linearity of voltage with distance along the strip in part (b). Most candidates realized that the increase of potential was non-linear because the strip was tapered, but few scored any further marks. Candidates who knew the correct relationship between resistance, length and cross-sectional area often failed to mention that the cross-sectional area was the product of the thickness and the width of the strip, although others did realize that the increase of resistance was due to the decrease of area as well as the increase of length. Only a few candidates stated that the current was constant or that the pd was proportional to the resistance.

In the graphical section of part (c) most candidates completed the table satisfactorily and proceeded to plot an adequate graph. Some candidates failed to score full marks as a result of careless errors such as omitting the unit of potential from the graph axis or failing to choose suitable scales. Candidates were usually able to relate the given equation to y = mx + c and thereby demonstrate that their graph confirmed to the given equation. Both marks were usually gained in part (iii) although there were some who plotted the potential on the x-axis thereby often losing the final mark because they failed to realize that the gradient was now equal to 1/(1.44 Vl).

Q13.The calculation to determine frequency in part (a) surprisingly brought difficulties to the surface. Many candidates simply did not know how to obtain frequency from the period, but many others had trouble in converting the waveform to an actual value of T. Calculating the rms voltage in part (ii) was usually carried out correctly, although many candidates just used the voltage sensitivity of 15 V per division as the peak voltage.

Although the majority of candidates had part (b) correct, a significant number became completely confused and converted back to peak values, under the impression that they were rms values.

Part (c) introduced a new type of problem for candidates, where they had to select a particular voltage sensitivity to suit the output voltage across the resistor. The general failing, even for good candidates, was failing to convert the answer of part (b) back to peak voltage, thereby losing a mark. Given that the answer to part (b) was treated as a consequential error in part (c), most candidates then chose the correct voltage sensitivity, but failed to support their choice by stating that the other sensitivities would either give a trace that was off the screen or else a trace that was too small for any meaningful measurements to be made.

Q16.Almost all candidates gained reasonable marks on part (a)(i) even though some of the descriptions were lacking in detail. Most of the diagrams were reasonably drawn with the aid of a ruler. Candidates who drew freehand usually produced inferior diagrams which failed to gain all the available marks. A variety of methods were shown, usually two wires hanging vertically, linked together by means of some vernier arrangement and a spirit level. Also shown was a horizontal wire on a bench. Although this is not such an accurate method, it was accepted but many candidates showed the mark on the wire as being about half way along. This of course is only acceptable if the length of the wire is measured to that point, but this was usually overlooked in the description. The least satisfactory method was suspending a single wire with a ruler alongside although this did gain some marks. There were an alarming number of diagrams which showed a completely unrelated length of wire, a ruler, an isolated hook with a mass attached and a micrometer. Needless to say, such efforts gained no marks.

In part (ii) candidates could have saved themselves considerable time and effort by reading the question carefully and just listing the measurement they would make. Many candidates listed the area of cross section as a measurement. This was not acceptable since area is a derived quantity and it is the diameter which is measured. Many

candidates also listed the ‘width’ of the wire, which again was not accepted.

The descriptions in part (iii) were, on the whole, quite reasonable, although most effort seemed to go into describing how the length of the wire and its diameter were measured and not giving sufficient attention to the experiment, i.e. measuring the extension for each mass added and increasing the total mass to a certain value. There were very few references to repeating the readings while unloading. This particular section of the question was also used to award the quality of written communication marks and most candidates scored well on this.

The descriptions in part (iv) of how to use the measurements to give the Young modulus was reasonably done with about 50% of the candidates drawing a graph of force vs extension or stress vs strain and using the gradient accordingly. Candidates who only used one set of values to give one value of the Young modulus were not awarded all the available marks.

The calculation in part (b)(i) was performed satisfactorily, with the majority of candidates calculating the correct extension for the steel wire. Marks were lost in part (ii) when the answer was given without any reasoning.

Q17.Candidates were required to design an experiment to test which of two theories concerning the way that the intensity of a spotlamp varies with distance from the lamp.

Most candidates illustrated their answer with a diagram with suitable labelling of the distance between the lamp and the LDR. Other diagram rarely earned full credit. Many candidates gave schematic views that failed to show whether a suitable working arrangement had been given. Even when correct circuits were drawn the symbol used for the LDR was generally wrong. Some accounts included spurious detail in the LDR circuit such as a series resistor. Other diagrams gave a voltmeter and / or ammeter wrongly connected or used a single circuit to supply both the LDR and spotlamp. Candidates who showed an ohmmeter often incorrectly included a power supply in the circuit. Very few diagrams were drawn to support the statement, usually given later, that the power output of the spotlamp was maintained at a constant level.

Thereafter, progress was generally better with most describing a sensible procedure to find the resistance of the LDR at different distances from the spotlamp. Some saw the calibration graph as evidence that the resistance was directly proportional to the illumination but a significant proportion correctly explained that they could incorporate the resistance values with the graph to find the illumination at certain distances from the spotlamp.

Most candidates recognised that to resolve the argument about whether the variation of illumination followed an inverse-square or exponential pattern, a graph should be drawn. However the use to which the graph would be put was not usually specified correctly and bland statements about ‘straight line’ or ‘linear’ graphs were not accepted.

Many candidates stated suitable control measures for the experiment could include using black-out or enclosing the apparatus in a light-proof tube: provided suitable reasoning was given this was accepted. Keeping the power output of the spotlamp constant was another acceptable answer but once again, justification of why and how this was to be done was expected. Some candidates thought that a parallel could be drawn with radioactivity measurement and stated that a background light level could be measured beforehand to be subtracted from the measured illumination.

Discussion of potential difficulties and how these were overcome tended to be blanket statements about ‘repeating and averaging’ measurements and if reasoning was attached this was often ‘to improve accuracy’. Many candidates stated that their objective was to obtain ‘reliable data’: the use of bland unconvincing jargon does not inspire confidence

and is unlikely to gain credit. No candidates thought to discuss how the distribution of data might be varied to improve the definition of the graph, but many wrote about the need to repeat readings to improve accuracy. Candidates should understand that repeating and averaging only reduces random error, thus improving precision but accuracy is improved only when systematic error is reduced.

It should not, however, be assumed that all candidates are incapable of writing carefully and sensibly about their ideas. There were many good accounts and about 25% earned at least 6 out of the possible 8 marks. This compares with only 10% at PHA3 / P.

Q18.For part (a), most candidates identified background radiation and were able to calculate the value of 20 counts/minute. Unfortunately, a large proportion of candidates lost a mark for omitting the unit.

In part (b) a proportion of candidates were unable to draw a satisfactory line of best fit.

In answer to part (c), most candidates drew an appropriate triangle (or used data points sufficiently far apart), and were able to correctly read off and process the data values. The gradient value must have a negative sign, be within the specified limits and quoted to two or three significant figures.

Part (d) was answered well by a large proportion of candidates who correctly calculate the decay constant from the gradient. The half life must have been quoted in minutes.

Part (e) provided an easy question, correctly answered by almost all candidates.

For parts (f)(i) & (ii) most candidates were able to correctly calculate the uncertainty and percentage uncertainty from the data provided.

Although many students were able to calculate the appropriate percentage uncertainty in part (g)(iii), a much small proportion appreciated the significance in terms of larger time/larger total count having a lower percentage uncertainty.

Q19.Encouraging numbers knew to say in (a) that the fiducial mark is placed at the equilibrium position because this is where ‘the transit time is least’ or ‘the speed of the mass is greatest’ but it was common to find statements such as ‘this is where the mass must pass through in every oscillation’ which attracted no credit.

Part (b)(i) and (b)(ii) were routine calculations in which candidates were usually successful although errors for the frequency arose due to rounding down earlier in the calculation. Full credit could be earned if the correct result for percentage uncertainty was reached using T rather than 20T but we withheld a mark in (b)(ii) if no unit was seen.

Part (c)(i) was very good and we took account of an incorrect (b)(ii) answer that led to an unexpected scale. We expected the scale to be linear and at least three convenient intervals marked. For the expected of 0.88 Hz result in (b)(ii) the 1 Hz intervals on the axis should have been 40 mm apart.

Many stated in (c)(ii) that the initial amplitude of X was 4 mm but a common error was to read off the amplitude at resonance (95 mm).

Part (d) was the most discriminating part of this question. In (d)(i) many grasped the idea behind the question but simply saying ‘take more readings’ was insufficient and only gained one mark. Some candidates avoided any ambiguity by saying ‘the intervals between the frequency readings around the peak were reduced’ to earn full credit. A common but unsuccessful idea was to assume a data logger was in use and to say that

the sample rate was increased around the peak.

In (d)(ii) saw that removing one spring doubled the stiffness and increased the resonant frequency by a factor of . The new curve now had a resonant peak at about 1.25 Hz which for an axis with the expected calibration was 50 mm along the axis. Some spoiled their answer by starting the curve at 0 Hz with amplitude less than 4 mm.

Unlike last year the full mark range was utilised and the question discriminated very well. Those at the A/B boundary usually earned between 7 and 9 out of 11 and E/U candidates often scoring between 4 and 6.

Q20.The answers to part (a) were, in many cases, improvised, but were none the worse for that.Candidates saw that a key issue was to do with the precision of the proposed microscope and took note of the instruction that they should illustrate their answer using suitable calculations. It was also obvious that here, more candidates (in contrast to question 1 (d) of Section A Part 1) took note of the mark subtotal and geared their answer to making two relevant points, each suitably illustrated.

Those who thought carefully about the problem posed in part (b) realised that measures to reduce uncertainty in one variable can have a detrimental effect on another and saw that the proposed change in D would make α far too small to be measured reliably.

It was interesting to compare the answers to part (c) with question 2 from the 2009 paper which also required candidates to calculate percentage uncertainty. As last year, there are still many that cannot do this type of calculation in a wholly transparent way and a lot of scripts were seen where an apparently correct answer was arrived at by an incorrect method; where the value 0.0855 was given in the denominator of the working, no credit was given.

Q21.(a) (i) Correctly answered by almost all students.

(ii) As usual in this question a small proportion of students failed to accurately plot both points and an even greater proportion were unable to draw an acceptable line of best fit.

(iii) The most common error by weaker students was misreading data from the graph. Most students were able to calculate a gradient value within the allowed range.

(iv) A large proportion of students correctly calculated the value of R. A small proportion of students lost credit by failing to quote the unit.

As anticipated this proved to be a demanding question, with only the very best students achieving full marks on all four parts.

(b) (1) Most students successfully calculated the value of R from the data in the table.

(2) Only the more able students successfully calculated the percentage uncertainty in R. The most common mistakes were failure to double the percentage error in V, and correctly add this to the percentage uncertainty in P. Credit was also lost by the significant figure penalty on the final answer.

(3) A relatively easy calculation but made more demanding by the requirements

for correct units and significant figures.

(4) This was the most demanding part of the question, and required students to use the uncertainty value calculated to work out the possible range of values of R to decide whether the two values were compatible.

Q22.Part (a) discriminated in favour of the more mathematically inclined and many completely successful answers were seen. There were two approaches to determining the number of slides equivalent to the half thickness, each requiring the constant λ to be found by measuring the gradient to be calculated so that. Some, borrowing from the idea of half life,

calculated the half thickness from while others calculated ln ( and arrived at

their result by . A number of candidates tried the latter approach but made the mistake of assuming that they should determine N for the interval over which ln(V0) halved and deduced that 202 slides were required.

In (b) those who wrote carefully generally obtained credit but careless writing was still seen, eg in (ii) it was common to find 'take multiple readings and calculate an average', the implication being that the same reading is taken repeatedly, which is not sensible; the readings taken had to be at different points on the slide or over a stack of several slides. In (iii) where a procedure to check for zero error on the micrometer was expected, several wrote 'check the reading with nothing in it' and another popular but unsuccessful idea was that the micrometer 'should be zeroed before use'; candidates who can not see why this is wrong should consider the action of the tare button on an electronic balance. The idea that the micrometer should be used to measure an object of known thickness was accepted. Candidates at the A/B boundary usually earned at least four out of six marks while E/U candidates generally earned two.

In (c) the idea that the slide thickness = was a popular distracter but a correct result for the refractive index was produced by nearly all.

In (e)(i) and (e)(ii) some missed the unit or gave the uncertainty as 0.02 m while in (e)(iii) some added the uncertainties before dividing by 1.47 but an encouraging number produced completely correct solutions.

Candidates at the A/B boundary typically earned four out of five marks and E/U candidates typically earned three.

Q23.(a) An easy mark achieved by most candidates.

(b) Point plotting was done accurately by most candidates. Some candidates failed to recognise the anomalous point for the orange LED, and tried to include this in their line of best fit.

(c) (i) As with ISA P, many candidates still failed to achieve the full 3 marks for determination of the gradient. The most common error is misreading of one of the data points, and this then usually gives a gradient value out of tolerance, hence losing another mark. As with ISA P ‘triangles’ generally met the 8cm criteria.

(c) (ii) Most candidates referred to closeness of the line of best fit to five of the points.

Only a small proportion of candidates made appropriate reference to the anomalous point.

(d) (i)/(ii) Discriminated well with only the more able candidates scoring all 4 marks.

(iii) A straightforward uncertainty calculation answered well by most candidates.

(iv) Another discriminating question with only the higher grade candidates able to score all 3 marks.

Q24.Although all saw the intention of the question in part (a) it was difficult to see enough merit in the response, seen many times, that ‘callipers cannot grip water’. To earn the mark we wanted a clear indication that using the callipers would distort the flow, changing the diameter in the process.

In (b)(i) virtually all drew an acceptable line and attempted a valid gradient calculation to find the integer n. Many predictably forgot that n was an integer or omitted the negative sign. Some calculated the log s and log d steps but then anti-logged these before calculating the ratio.Many correct answers were seen to (b)(ii). Those who wrote k = 10log k needed to have clearly explained that log k was the intercept otherwise the mark was withheld. Those who said k = 100.52 by mistakenly failing to notice the false origin in Figure 2 were not penalised. The many who claimed k = eintercept gained no credit.In (b)(iii) we expected cm5 but we did not penalise for m5 etc or for any logical consequence of a non-integer result for n in (b)(i).

This was another discriminating question with those at the A/B boundary typically earning 3 or 4 out of 5 and E/U candidates scoring between 1 and 3.

Q25.This question was about the determination of the Young modulus of the metal of a wire, one of the six required AS practicals. It gave students the opportunity to demonstrate familiarity with a range of practical equipment and techniques.The work seen suggested that the majority of students were familiar with the experiment and could process the raw data. Problems arose when students were asked to comment on the effect of changing the dimensions of the wire, for example.

(a) Vernier scales are a common feature of the equipment used to measure extension in a Young Modulus determination. Many students were unable to give the correct answer, however, with 2.8 mm a fairly common error.

(b) Students were allowed to carry an error forward from (a) and many students were able to read the scale correctly. Where the answer to (a) fell outside the graph some students attempted to extrapolate or carry out a y = mx + c calculation; these students may have benefited from checking their answer to (a) first.

(c) Most students were able to correctly compensate for the zero error in the calliper reading, although adding it to obtain 0.37 mm was fairly common. Some students also attempted to use the 20.14 mm value seen at the top of figure 3.

(d) This calculation was reasonably well carried out with the majority of students getting at least 3 marks out of 4. Although students should be encouraged to use the gradient of figure 2 to help calculate a value of E, the use of individual points was acceptable on this occasion. It was common to see the values obtained in the previous three sections used to work out the cross sectional area, and then the strain and stress separately, before combining them to calculate E. The two

common errors in this approach were the use of the diameter for the radius, and missing out "g" in the calculation of the force. Several students also made powers of ten errors in the cross-sectional area calculation, or mixed their units in the calculation of strain. This commonly led to the fourth mark being lost.Students should be encouraged to set out their answers logically so that it is easier to check for errors. Some students inevitably lost marks if they made an error attempting to perform the calculation in one go. Over reliance on the data sheet also led to problems with some students using F = kΔl, and using the Boltzmann constant for k.Any errors from (a), (b) or (c) were carried forward so that no student was denied the opportunity to earn subsequent marks.

(e) When guiding students with their responses to questions such as this it is important to emphasise the need for clarity in expression, as examiners cannot credit ambiguous or vague answers. To gain credit, answers required an outcome (e.g. the extension increases) and consequence (the percentage uncertainty in E decreases). It was common for students to miss out the word 'percentage' but on this occasion this was not penalised. There were several different correct approaches but each was accompanied by its own common errors. For example, students who simply stated that the wire would get longer, rather than the extension would increase, failed to get the mark. Students who based an answer on the increased percentage uncertainty in the measurement of the diameter often failed to go on to state how this would affect the percentage uncertainty in the value of E. It was also relatively common to see incorrect physics, for example answers claiming that the value of E would be affected.

Q26.(a) Correctly answered by almost all candidates.

(b) It was rare to see misplotted points on this graph. The line of best fit was more difficult due to a deliberate scattering of points around the line, and a proportion of candidates were unable to draw an acceptable line.

(c) A relatively straightforward gradient calculation, although weaker responses often missed the negative sign.

(d) The most able candidates were able to score the full three marks but a proportion of weaker responses often quoted an incorrect unit or were unsure of the value of λ.

(e) This was a difficult question, requiring an understanding of how a systematic error in h would affect the log values. Only the most able students accessed the full three marks.

(f) Accessible across the ability range, although weaker responses were often unsure as to how this error might affect the value of λ.

Q27.Many were successful in part (a) although a few found the distance between the edges of adjacent holes (despite this they were still able to go on and successfully attempt part (d)).

Success with part (b) was almost universal, candidates clearly taking the cue from the precision implied in the readings for R1 R2 and R3. Instances were seen where candidates gave 0.005 mm as the precision and these answers gained full credit.

Part (c) was a different matter: most chose to use the ratio of their previous answers to

find the percentage uncertainty, e.g. and only very few saw that the method

required was . We gave a mark if the answer given showed 2 × precision given in (b) in the numerator but otherwise very few were able to score.

Many correct answers were seen to part (d) although the solutions were often too convoluted to follow.

A/B candidates typically scored 3 or 4 out of 6 but E/U graders often struggled to score more than 1.

Q28.This question discriminated well, with only the most able candidates scoring four or five marks.

(a) (i) Some candidates attempted to measure the largest distance, between the initial and 4th position. A tolerance of +/-1 mm was allowed on their measurement.

Fewer candidates attempted a second measurement, although the question clearly referred to measurements

.

(ii) This proved to be much more difficult, and candidates had to compute the correct time interval for the distance measured, and then use the correct formula. Only the most able candidates were able to compute a correct value for g.

(b) Fewer than half of the candidates appreciated how the sharpness of the image would depend on the duration of the flash.

Q29.(a) Correctly answered by almost all candidates.

(b) As usual in this question a small proportion of candidates failed to accurately plot both points and an even greater proportion were unable to draw an acceptable line of best fit.

(c) Less marks were lost on this question than in previous years, most likely because these were mostly A2 candidates re-sitting the paper. Most candidates were able to calculate a gradient value within the allowed range.

(d) This was straightforward for nearly all candidates.

(e) A familiar question, well answered by most candidates.

Parts (f), (g) and (h) discriminated well, with only the most able candidates scoring five marks or more.

(f) The easiest part of the question, requiring use of ‘uncertainty = 0.5 × range’

(g) Although this was considered to be straightforward, many candidates failed to score the full three marks. Common errors were incorrect substitution into the formula and missing the unit in the final answer.

(h) This was the most discriminating part of the question, with candidates often scoring no marks. The process of adding percentage uncertainties to calculate the

percentage uncertainty in the calculated value of g, and then converting back to an absolute uncertainty with suitable significant figures and unit was beyond most candidates.

(i) This proved to be difficult, and only the most able candidates scored more than one mark. An easy mark was often lost by lack of reference to measuring the mass of the spheres with a balance. Many candidates failed to realise that for a fair comparison the spherical objects would need to have the same diameter so that air resistance was the same.

Q30.(a) (i) Answered well by most candidates. The explanation had to demonstrate that

candidates understood the resolving process.

(ii) Correctly answered by most candidates, with only a small proportion of inappropriate lines of best fit.

(b) (i) This question features on all ISA’s but many candidates still fail to achieve the full 3 marks. The most common error was again misreading of one of the data points, and this then usually gives a gradient value out of tolerance, hence losing another mark. Triangles were generally large enough to meet the 8 cm criteria, with fewer lost marks on this point than in previous years.

(ii) Answered reasonably well by most candidates – by multiplying their gradient by ‘g’.

A significant proportion of candidates lost the mark awarded for the unit. (Some candidates using ‘n’ instead of ‘N’ !).

(c) This question discriminated well, with only the most able candidates scoring all three marks. Significant figure errors and failing to double the combined uncertainty in x and y were the most common mistakes.

(d) (i) A surprising number of candidates failed to give an adequate description of a systematic error.

(ii) A large proportion of candidates scored the first mark, explaining that y would be larger. Only the most able candidates were able to give an adequate explanation, usually relating to friction opposing the increasing weight of m.

Q33.Those students who did not fall back on a generic response in question (f) gained a significant advantage. The numerical and graphical parts of question 3 were generally done well.

Questions (a) and (b) were quite accessible for those students who took in all the information from Figures 1, 3 and 2. The data students added to Table 2 should give the p and x data to a consistent and appropriate number of decimal places. Thus, for n = 13 examiners wanted to see x = 30.0; a mark was withheld from the many students who missed this point.

In question (c), most students earned a mark for working out the ln (x/cm) values correctly, and recording these in a consistent fashion to at least 2 decimal places. The second marking point was much more discriminating, and was forfeited by many students who included the origin, thus compromising the vertical scale. Others did not show the bracket around the axis label correctly, or spaced their values too widely along the axis. The point plotting was generally fine, but some students lost the mark by using ‘blobs’

rather than clearly-defined points, or by drawing a thick or non-ruled line. Those producing a curve could not score here.

In question (d), examiners wanted students to point out that their graph was a straight line, and that it had a negative gradient. It was clear that large numbers of students seemed to think that any straight line graph represents direct or inverse proportion. Examiners insisted on seeing ‘linear’ or ‘straight line’, and any mention of proportion (of any sort) was rejected.

Once back to numerical work in question (e), students were on surer ground and most measured a gradient before successfully going on to find x when n = 20. Students who feel that this type of question is testing maths and not physics should acquaint themselves with section 6 of the specification.

Question (f) highlighted once again how ill-equipped some students were when it came to writing with authority about practical procedures to reduce uncertainty. Some immediately seized on the idea that (percentage) uncertainty could be reduced by either increasing the raw measurements, or by reducing the intervals between the markings on a scale. Neither of these things made any sense in the context of the question, so suggestions such as “increase the length of the air track”, and “use a vernier scale”, gained no credit. Some students failed to gain credit here because they suggested the use of lasers or computers to measure distances, or simply gave vague statements about some procedure which ‘will reduce uncertainty’. Students who stuck with what, in the circumstances, would be an obvious and effective strategy, such as repeating and averaging to reduce the impact of random error, were on the safest ground. Whenever data-logging is suggested, examiners expect to see an appropriate sensor identified. In the context of this experiment a motion sensor would be fine, but a light gate would not. The use of video cameras, used increasingly in schools and colleges, to pinpoint p as the glider comes to rest, was accepted either as a means to reduce the impact of random error, or to avoid parallax error.