I'm privately giving edexcel A2 physics and chemistry exams on june. The problem is unit 6 (6CH06 for chem and 6PH06 for physics) appears nowhere in the 'entry detail form', except in the single check box, that includes all 6 units, but i have already givin AS papers. Edexcel 2016* - Bio B1 - Bio. The Student Room - A free website that's full of students and tutors to discuss work with - with free revision resources. A have pdfs of a different type of OCR chemistry and physics full A level books.

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Embed Edexcel Physics A2 Answers (Student's Book). By MissCutielicious Chapter 1 1.1.1 Momentum ST Answers 1)a) 5.4 x 107 kg m s-1 b) 4750 kg m s-1 2)a) If the batsman only changed the ball’s direction very slightly, this is a smaller momentum change than in the worked example, so less force would be needed. B) If the ball is returned back in the direction of the bowler, this would be a greater momentum change than in the worked example, so more force would be needed. 3) 6600 N 4) a)1025 N s b) The airbag takes a longer time to remove the passenger’s momentum. This means that it applies less force to the passenger, reducing injury. 5) a) Mass = 1.15 kg –1 b) ∆p = 5.50 or 5.74 kg m s /N s c) F = 458 or 478 N depending on ∆p above d) Handle mass/weight/ head weight/force exerted by user (handle) neglected e) ∆t goes up/∆p goes up ⇒ less force, less effective/more force, more effective 1.1.2 Collisions ST Answers 1) 0.031 m s-1 2)a) 0.2 m s-1 b) 100 N 3) In order to move forward, the boy must experience a force from the boat. There is an equal and opposite force from him onto the boat which moves the boat backwards away from him.

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4)a) insert artwork as per below – note that at the bottom right of this pic, all arrowheads meet which is why it looks a bit rubbish ptotal = 1430 kg m s-1 vafter = 4.77 m s-1 b) twaterfall = 37 s tbank = 4 s so they would land safely on the riverbank. 5) a) mass × velocity Words or defined symbols; NOT ft b) iii c) ii, or i and ii.Line 2 OR 1& 2 d) No (net) external forces/no friction/drag.

(he assumes the force exerted by the other trolley is the resultant force) e) Suitable collision described and specific equipment to measure velocities e.g. Light gates. Measure velocities before and after collision. Describes how velocities calculated e.g.

How light gates used. Measure masses / use known masses/equal masses.

Calculate initial and final moment a and compare OR for equal trolleys in inelastic collision. 1.1.3 Energy in collisions ST Answers 1)a) 3.35 m s-1 b) 7.13 J 2)a) 8.73 x 10-13 J b) 5 460 000 eV c) 5.46 MeV 3) Non-contact collisions are always elastic, and no external forces act.

1.1.4 Real collisions ST Answers 1) 0.46% of the speed of light; 1.38 x 106 m s-1 2) It ends up in the other corner pocket, moving at 4.24 m s-1 at 45º to the original line. 1.2.1 Angular displacement ST Answers 1) 720 º b) π/5 rad; 0.63 rad -1 2) 3.46 rad s b) 4.71 rad s-1 c) 8.2 rad s-1 -1 3) 18.8 rad s 4) ω = 7.27 x 10-5 rad s-1; v = 465 m s-1 5)a) 36.6 cm b) The error is proportionately the same 1.2.2 Centripetal force ST Answers 1) 2530 N 2)a) 2.54 N b) W = 736 N (much smaller) c) At the pole, the reaction force would be 736 N; at the equator, it would be 733 N 1.2.3 All the fun of the fair ST Answers 1)a) estimates in range 500 – 1500 kg b) estimates in range 10 – 30 m c) e.g. 1000 kg & 20 m gives v = 18.1 m s-1 Chapter 2 2.1.1 Electric fields ST Answers 1) 4.8 x 10-17 N 2) 8000 V m-1 3) 4.8 x 1011m s-2 4)a) 1 x 105 V m-1 c) +3000 V b) 1.62 x 107 m s-1 0V d) Acceleration is in the opposite direction because the proton’s charge is the opposite sign to an electron; and the acceleration is less as the proton mass is greater than an electron’s. 2.1.2 Uniform and radial fields ST Answers 1) Net field = zero, so the net force = zero.

2) insert aw, as per fig 2.1.6 sketch of two plus charges, except this one should be two minus charges. The lines shape is identical, but all arrows must be reversed. 3)a) 600 V b) No difference 4) Charge concentrates near points or spikes. 5) 5.13 x 1011 V m-1 2.1.3 Coulomb’s Law ST Answers 1) 2.12 x 10-6 N 2)a) 1.14 x 1017 V m-1 b) 0.0364 N 3) The measurements for r must be taken to the centres of the spheres. An error here would lead to a disparity in the results, compared with Coulomb’s Law.

Past paper and marking scheme downloads for all subjects: Useful websites: All subjects: IGCSE Edexcel Textbook Answers: All textbooks: Chemistry: Physics: Mathematics A student book 1: Mathematics A student book 2: Mathematics B book: Nelson Thornes Textbook answers: National Textbooks of all subjects up to grade 11: Past papers and mark schemes: AQA & OCR past papers: Physics Edexcel Physics Textbooks and Solutions/Answers:. Edexcel AS Physics Students’ book. Pearson – Longman.

Edexcel AS Physics Students’ book. Pearson – Longman. ANSWERS:. Edexcel A2 Physics Student’s book. Answers:.

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Edexcel Physics AS Hodder textbooks answers: Animations:.