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November 2023

Pearson Edexcel International GCSE

In Physics (4PH1) Paper 2P

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November 2023

Question Paper Log Number P73429A

Publications Code 4PH1_2P_MS_

All the material in this publication is copyright

© Pearson Education Ltd 2023

Question number Answer Notes Marks 1 (a) all five rows correct = 3 marks row cannot score if both boxes ticked allow any reasonable indication other than a tick

3 - 4 rows correct = 2 marks 1 - 2 rows correct = 1 mark (b) high temperature; ignore references to repulsion or attraction

(so that) nuclei will have high(er) speed or high(er) {kinetic energy/KE}; condone particles, atoms, molecules for nuclei high pressure; plus one from:

• (so that nuclei will be) closer together / eq;
• (so that nuclei are) more likely to collide/ eq; e.g. closely packed e.g. collide more frequently Total for Question 1 = 7 marks

Question number Answer Notes Marks 2 (a) (i) substitution OR rearrangement; 2 evaluation; e.g. 19 = force × 0.55 OR force = moment/distance (force =) 35 (N) allow 34.5, 34.54, etc.

• 1 POT error (ii) A (0.25 m); 1 B is incorrect because this is not a perpendicular distance C is incorrect because this is the distance between weight and force F D is incorrect because this is the distance to force F (b) (i) force X has the shorter distance to the CoG; Note: no credit for repeating question I.e. CoG is closer to support A condone reference to force A accept ‘weight’ for CoG

moments of two forces must be equal/eq; allow RA i.e. (total) clockwise moment = (total) anti- clockwise moment so force X must be larger; DOP on either (ii) any three from: 3 force X decreases; force Y increases; change by the same amount; allow idea of change with distance is linear total force remains the same; Total for Question 2 = 9 marks

(b) (i) substitution into given formula; 3 rearrangement; evaluation; - 1 POT error e.g. 54 000 = 0.56 × c × 22 c = 54 000 / 0.56 × 22 (c =) 4400 (J/kg °C) allow 4383… allow any reasonable rounding ignore rounding error after 3sf (ii) any two from: ignore references to KE or sound

MP1. heating of beaker/heating of thermometer; allow idea of ‘heat goes to beaker/thermometer’ MP2. heating of surroundings/heating of insulation; allow idea of ‘heat going to surroundings or insulaton’ MP3. insulation not perfect; e.g. insulation conducts or radiates MP4. hole(s) in lid for thermometer/heater leave gaps; allow idea of heat ‘escaping’ through lid or via evaporation Total for Question 3 = 11 marks

Question number Answer Notes Marks 4 (a) substitution into given formula; ignore units 3 rearrangement; evaluation; 2 marks max. if time not converted correctly to seconds e.g. 40 (J), 2400 (J) e.g. 16 = energy / 2.5 (× 3600) energy = 16 × 2.5 (× 3600) (energy =) 140 000 (J) allow 144 000 (J) 2.5 × 60 × 60 or 9000 seen for 1 mark if no other mark awarded (b) any six from: 6 MP1. it steps up or steps down the voltage; allow “increases”, “decreases” or “changes” MP2. current in (primary) coil produces magnetic field; MP3. current is changing/alternating; MP4. causing a (changing) magnetic field in the core ; e.g. making the core magnetised MP5. the core strengthens the magnetic field; MP6. idea that iron is a soft magnetic material I.e. can gain and lose its magnetism easily allow “concentrates” for “strengthens” MP7. field lines interact with (secondary) coil; idea of core linking the two coils with magnetic field MP8. which induces a voltage in the secondary coil; condone “induces a current” MP9. transformer won’t work with (steady) d.c.; ignore idea of “works with a.c.” repetition of stem (c) less turns (on the primary coil) / eq; ignore any reference to secondary ignore references to increasing current, power or voltage of the primary coil

condone ‘coils’ for ‘turns’ Total for Question 4 = 10 marks

Question number Answer Notes Marks 6 (a) idea that (total) momentum before = (total) momentum after (event); ignore ‘momentum is conserved’

(b) (i) zero/0/nought/nothing; 1 (ii) recall of momentum = mass × velocity; allow standard symbols e.g. p = m × v

substitution; evaluation; - 1 POT error e.g. p = mv p = 2.6 × 10−^8 × 26 000 p = 6.8 × 10−^4 (kg m/s) allow 6.76 × 10 −^4 (iii) 6.8 × 10−^4 (kg m/s); allow ecf from (ii) 2 right; (iv) substitution into ‘F=ma’; ignore units 4 rearrangement; evaluation; −1 for POT error answer given to 2s.f.; independent mark e.g. 2.6 (×10−^3 )= 1.2 × acceleration acceleration 2.6 (×10−^3 ) / 1. (acceleration =) 2.16… × 10−^3 (m/s^2 ) (acceleration =) 2.2 × 10−^3 (m/s^2 ) (c) any two from: ignore idea of simple yes/no

MP1. idea of tiny amount of fuel ‘consumed’ per second; MP2. any attempt of calculation of time to run out of xenon seen ;

MP3. correct calculation of 7.575... × 106 s; accept 88 or 87.68..

days, 2104 hours, 1.26..

× 105 minutes , 12.5..

weeks, 0.24... years MP4. idea that ‘burn’ is for a long time; MP5. idea that low acceleration for long time gives high speed change; MP6. mass of spacecraft will be larger so acceleration is even smaller; Total for Question 6 = 13 marks

Question number Answer Notes Marks 7 (a) microphone; 1 (b) (i) determination of number of squares for one period; allow 5.2 − 5.5 squares 2 use of timebase to determine appropriate period in seconds; ‘5 x 0.002’ or ‘0.01’ scores 1 ‘8 x 0.002’ or ‘0.016’ scores 1 accept 0.0104 − 0.0110 (s) for 2 marks e.g. period = 8/1.5 = 5.3 squares period = (5.3 × 0.002 =) 0.011(s) (ii) use of f = 1 / T to evaluate frequency; allow ecf from (i) 1 e.g. frequency = (1 / 0.011 =) 91 (Hz) allow 90.9-96.2 (Hz) (c) (i) energy (transferred) = charge × voltage; allow rearrangements and standard symbols e.g E = Q × V

allow W for energy ignore C for charge (ii) amplitude of signal in volts determined; 3 evaluation of effective voltage; ecf amplitude in volts evaluation of energy transferred; reject 6.3 × 10 −5^ / √

• 1 POT error e.g. amplitude = 2 squares × 5 = 10 V effective voltage = 10 / √2 = 7.1 V energy transferred = (7.1 × 6.3 × 10 −^5 =) 4.5 × 10 −^4 (J) allow 4.4 × 10 −^4 − 4.5 × 10 −^4 (J) 8.9... × 10 −4^ scores 2 Total for Question 7 = 8 marks

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