GCE. Physics A. Mark Scheme for June Advanced GCE 2826/01 Unifying Concepts in Physics. Oxford Cambridge and RSA Examinations

Transcription

1 GCE Physics A Advanced GCE 2826/0 Unifying Concepts in Physics Mark Scheme for June 200 Oxford Cambridge and RSA Examinations

2 OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range of qualifications to meet the needs of pupils of all ages and abilities. OCR qualifications include AS/A Levels, Diplomas, GCSEs, OCR Nationals, Functional Skills, Key Skills, Entry Level qualifications, NVQs and vocational qualifications in areas such as IT, business, languages, teaching/training, administration and secretarial skills. It is also responsible for developing new specifications to meet national requirements and the needs of students and teachers. OCR is a not-for-profit organisation; any surplus made is invested back into the establishment to help towards the development of qualifications and support which keep pace with the changing needs of today s society. This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which marks were awarded by Examiners. It does not indicate the details of the discussions which took place at an Examiners meeting before marking commenced. All Examiners are instructed that alternative correct answers and unexpected approaches in candidates scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes should be read in conjunction with the published question papers and the Report on the Examination. OCR will not enter into any discussion or correspondence in connection with this mark scheme. OCR 200 Any enquiries about publications should be addressed to: OCR Publications PO Box 5050 Annesley NOTTINGHAM NG5 0DL Telephone: Facsimile: publications@ocr.org.uk

3 2826/0 Mark Scheme June 200 Candidates are expected to answer all the questions (a) arrows correct () (contact) force from seat and/or chain (tension allowed) () (gravitational) force from Earth / weight () [3] (b) (i) 220 N / 9.8 = 22.4 kg () {} (ii) 2π x 3 / 4.7 = 4.0 m s - () {} (iii) ½mv 2 () = ½ x 22.4 x = 80 J () {2} (iv) F = mv 2 / r () = 22.4 x / 3 = 20 N () horizontally towards the centre of the child s rotation () {3} (v) vertical component 220 N; horizontal component 20 N () θ = tan - (20 / 220) = 28.6 () {2} [9] (c) a triangle of force is for equilibrium () the child is accelerating (so is not in equilibrium) () the tension and the weight alone are the forces acting OR the centrifugal force does not exist OR there is an inward resultant force () [3] [Total: 5] 2 (a) gravitational field () strength of gravitational field is force per unit mass () electric field () electric field strength is force acting per unit positive charge () magnetic field () magnetic field strength is force acting per unit current on a wire of unit length () [6] (at right angles to the field) (b) in a gravitational field force always acts towards mass () in an electric field the force acts (on a positive charge) in a direction towards a negative charge () but away from a positive charge () in a magnetic field the force direction is at right angles to the field () and in a direction given by Fleming s left hand rule () [5] (c) 2 of cannot be a gravitational force because it is far too small () cannot be an electrical force because that would be repulsion () cannot be magnetic because there is no current () [2] [Total: 3] 3. (a)(i) at any junction the algebraic () sum of the currents is zero () {2} (ii) sketch e.g. () {} 7.0 A 4.2 A 2.8 A (iii) charge () {} [4]

4 2826/0 Mark Scheme June 200 (b)(i) energy () {} (ii) 0 Ω and 40 Ω in parallel have total resistance 8 Ω total resistance = 0 Ω () current from supply = 0.30 A () p.d. across 2 Ω resistor = 0.60 V () p.d. across 0 and 40 Ω resistors = 2.4 V () {4} 2 a closed loop consists of the cell, the 2.0 W internal resistor and ONE of the other two resistors () e.g. so 3 V e.m.f. = V 0 + V 2 = 0.6 V V () {2} [7] [Total: ] 4 (a) 38 is the number of protons (in the nucleus) () 90 is the number of protons + neutrons () {2} [2] (b) the proton and the neutron do not have mass exactly equal to u () binding energy alters the total mass () the mass of all the protons and neutrons is larger than the nuclear mass () the difference between them is given by E = Δmc 2 () MAX [3] (c) (i) () {} H + 2 H 2He (ii) mass of left hand side = 2 x = u mass of right hand side = u Δm = u () E = x.66 x 0-27 x (3.0 x 0 8 ) 2 () = 3.82 x 0-2 J = 3.82 x 0-2 /.6 x 0-3 (= 23.9 MeV) () {3} [4] (d) (i) n + 92U 54Xe + 38Sr + 4 0n initial neutron () correct nuclides () numbers all balancing () {3} (ii) left hand side = right hand side = () Δm = u allow e.c.f. from (i) () equivalent to x 0-28 kg E = x 0 - J () = 89 MeV () {4} [7] [Total: 6] 5 sensible choice of answer () relation between physics and use (3) MAX reasons given (3) MAX OVERALL MAXIMUM [5] [Total: 5] 2

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