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EDEXCEL GCSE COMBINED SCIENCE PHYSICS 3F 2025 PREDICTED MARK SCHEME resource for june 2025 exam revision

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EDEXCEL GCSE COMBINED SCIENCE PHYSICS 3F 2025 PREDICTED MARK SCHEME resource for june 2025 exam revision

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PAPER 2025

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, Edexcel Physics 1PF 2025 Predicted Question Paper




Mark Scheme
(Results)



2025 Predicted Paper


Pearson Edexcel GCSE
Paper 3: Physics 1 (1SC0/1PF) –
Foundation Tier

, Edexcel Physics 1PF 2025 Predicted Question Paper

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.

, Edexcel Physics 1PF 2025 Predicted Question Paper


Find the comprehensive elaborations for the question set provided
beneath the mark scheme for Question 01.

Question 1



Question Answers Extra information Mark



01.1 Correct Answer: B. Light 1




Question Answers Extra information Mark


01.2 Calculation: Wavelength Calculation: 2
Given: Wavelength (λ) is inversely
Frequency (ff) = 340 Hz proportional to frequency (f)
Speed (vv) = 340 m/s for a given speed (v). Longer
Formula: λ=vf\lambda = wavelengths correspond to
\frac{v}{f} lower frequencies.
λ=340340=1 m\lambda =
\frac{340}{340} = 1 \, \text{m}
Answer: λ=1 m\lambda = 1 \,
\text{m}




Question Answers Extra information Mark

Wave Diagram: Amplitude measures the
01.3 Correct labels: wave's energy; larger 2
amplitudes indicate more
Amplitude: Distance from
energy.
baseline to crest (4 cm).
Wavelength is the
Wavelength: Distance
distance between
between consecutive
identical points in
crests or troughs (8 cm).
consecutive cycles, such
as crest-to-crest.


Question Answers Extra information Mark




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