wph13 01 MSC 20200123 PDF
wph13 01 MSC 20200123 PDF
wph13 01 MSC 20200123 PDF
October 2019
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October 2019
Publications Code WPH13_01_1910_ MS
All the material in this publication is copyright
© Pearson Education Ltd 2019
General Marking Guidance
• All candidates must receive the same treatment. Examiners must mark the first candidate
in exactly the same way as they mark the last.
• Mark schemes should be applied positively. Candidates must be rewarded for what they
have shown they can do rather than penalised for omissions.
• Examiners should mark according to the mark scheme not according to their perception of
where the grade boundaries may lie.
• There is no ceiling on achievement. All marks on the mark scheme should be used
appropriately.
• All the marks on the mark scheme are designed to be awarded. Examiners should always
award full marks if deserved, i.e. if the answer matches the mark scheme. Examiners should
also be prepared to award zero marks if the candidate’s response is not worthy of credit
according to the mark scheme.
• Where some judgement is required, mark schemes will provide the principles by which
marks will be awarded and exemplification may be limited.
• When examiners are in doubt regarding the application of the mark scheme to a candidate’s
response, the team leader must be consulted.
• Crossed out work should be marked UNLESS the candidate has replaced it with an
alternative response.
Mark scheme notes
Underlying principle
The mark scheme will clearly indicate the concept that is being rewarded, backed up by examples. It is
not a set of model answers.
For example:
(iii) Horizontal force of hinge on table top
This has a clear statement of the principle for awarding the mark, supported by some examples
illustrating acceptable boundaries.
3. Significant figures
3.1 Use of an inappropriate number of significant figures in the theory papers will normally
only be penalised in ‘show that’ questions where use of too few significant figures has
resulted in the candidate not demonstrating the validity of the given answer.
3.2 The use of g = 10 m s-2 or 10 N kg-1 instead of 9.81 m s-2 or 9.81 N kg-1 will be penalised by
one mark (but not more than once per clip). Accept 9.8 m s-2 or 9.8 N kg-1
4. Calculations
4.1 Bald (i.e. no working shown) correct answers score full marks unless in a ‘show that’
question.
4.2 If a ‘show that’ question is worth 2 marks then both marks will be available for a reverse
working; if it is worth 3 marks then only 2 will be available.
4.3 use of the formula means that the candidate demonstrates substitution of physically correct
values, although there may be conversion errors e.g. power of 10 error.
4.4 recall of the correct formula will be awarded when the formula is seen or implied by
substitution.
4.5 The mark scheme will show a correctly worked answer for illustration only.
4.6 Example of mark scheme for a calculation:
Use of L × W × H
Example of answer:
6. Graphs
6.1 A mark given for axes requires both axes to be labelled with quantities and units, and drawn
the correct way round.
6.2 Sometimes a separate mark will be given for units or for each axis if the units are complex.
This will be indicated on the mark scheme.
6.3 A mark given for choosing a scale requires that the chosen scale allows all points to be
plotted, spreads plotted points over more than half of each axis and is not an awkward scale
e.g. multiples of 3, 7 etc.
6.4 Points should be plotted to within 1 mm.
• Check the two points furthest from the best line. If both OK award mark.
• If either is 2 mm out do not award mark.
• If both are 1 mm out do not award mark.
• If either is 1 mm out then check another two and award mark if both of these OK,
otherwise no mark.
For a line mark there must be a thin continuous line which is the best-fit line for the candidate’s
results.
Question Answer Mark
Number
1 (a) • Reaction time
Or timer not reset to zero (1) (1)
Teacher – 2 marks
Example of Calculation
Percentage uncertainty = (0.02 / 0.89) × 100 % = 2.247 %
Example of Calculation
s = 330 m s-1 × (0.91 s ÷ 2)
s = 150.2 m
Total for question 1 13
Question Answer Mark
Number
2 (a) • Measure the distance at which the plastic sphere lands with a ruler
Or measure the launch angle with a protractor (1)
• Repeat measurements (for each angle) and calculate the mean d (1)
• Plot a graph of d and θ, and use to find θ for maximum d value
Or continue changingθ until d decreases to find maximum (1)
• Around the maximum d take measurements for smaller changes in angle (1) (4)
0.45
y = 1.3x + 0.02
0.40
0.35
0.30
Potential difference / V
0.25
0.20
0.15
0.10
0.05
0.00
0.00 0.05 0.10 0.15 0.20 0.25 0.30
Length / m
Example of Calculation
V / VT = R / RT
VT = (V × RT) / R
VT = (0.41 V × 0.27 Ω) / 0.070 Ω = 1.58 V
3 (c) • Further readings would make the line of best fit more accurate (1)
• Giving a more accurate value for the p.d. of the rod (at 30cm) (1) (2)
4 (b) • Higher power lamp would have a heating effect on the solution (1)
Or Higher power lamp would increase the temperature of the solution
• Heating would cause expansion of the sucrose solution (1)
Or heating would cause evaporation of the sucrose solution
• Which would change the concentration/density (of sucrose solution) (1) (3)
Accept any named EM radiation with higher frequency than visible light
5 (b) • Use of W = VQ
Or W = eV (1)
• W = (−)2.5 × 10–19 J
Or W = 1.58 eV (1) (2)
Example of Calculation
W = 1.58V × 1.6 × 10–19 C
W = 2.53 × 10–19 J
1/λ 1.2
Mean V
λ / nm / ×106
(Mean) V / V
1.0
/V
m–1
0.8
380 2.63 1.58
2.27 0.6
440 1.10
470 2.13 0.94 0.4
530 1.89 0.66 0.2
570 1.75 0.46 0.0
620 1.61 0.34 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8
1/λ / × 106 m-1
Accept 1/λ in nm–1 or pm–1 at this
stage.
5 (c)(ii) • Calculates gradient using large triangle (1)
• Use of ℎ =
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 × 𝑒𝑒
with their gradient (1)
𝑐𝑐 (1) (3)
• h = (6.2 to 7.0) × 10-34 J s
Example of Calculation
1.4𝑉𝑉−0.2𝑉𝑉
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 = = 1.2 × 10–6 V m–1
(2.50−1.50) ×106
1.2×10–6 × 1.6×10–19
ℎ= 3.0×108
= 6.4 × 10–34 J s