GCE Physics B (Advancing Physics) Mark Scheme for June Unit G492: Understanding Processes/Experimentation and Data Handing

Transcription

1 GCE Physics B (Advancing Physics) Unit G49: Understanding Processes/Experimentation and Data Handing Advanced Subsidiary GCE Mark Scheme for June 04 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 candidates of all ages and abilities. OCR qualifications include AS/A Levels, Diplomas, GCSEs, Cambridge Nationals, Cambridge Technicals, 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 04

3 . These are the annotations, (including abbreviations), including those used in scoris, which are used when marking Annotation Meaning Blank Page this annotation must be used on all blank pages within an answer booklet (structured or unstructured) and on each page of an additional object where there is no candidate response. Benefit of doubt given Contradiction Incorrect response Error carried forward Follow through Not answered question Benefit of doubt not given Power of 0 error Omission mark Rounding error Error in number of significant figures Penalised only in 4(iii) Correct response Arithmetic error Wrong physics or equation

4 . Abbreviations, annotations and conventions used in the detailed Mark Scheme (to include abbreviations and subject-specific conventions). Annotation Meaning / alternative and acceptable answers for the same marking point () Separates marking points reject Answers which are not worthy of credit not IGNORE ALLOW Answers which are not worthy of credit Statements which are irrelevant Answers that can be accepted ( ) Words which are not essential to gain credit ecf AW ORA Underlined words must be present in answer to score a mark Error carried forward Alternative wording Or reverse argument. The following questions should be annotated with ticks, crosses and carets to show where marks have been awarded in the body of the text: 8bii 0bii c d aii aiv 4biii 4c 4. Do not penalise excessive significant figures anywhere except Q4(iii). Unit errors are covered in Q9(i).

5 J s - kg m s - B C (c) A (d) C nd box and no others. 5 (c) B and H E: arrow at o clock (); G: arrow between 0 and o clock () or clockwise combination. allow one mark if both E and H are mirror images of correct arrows Needs to show the addition of three tip-to-tail arrows forming equilateral triangle (by eye). Reject a right-angled triangle; reject arrows in star formation. 6 7 speed: evidence of scale drawing or trig. / Pythagoras (); answer of 6 m s - (); direction: angle of 5 or 55 correctly indicated (); E of S () or equivalent, e.g. bearing 45 ; 5 m s - (); (towards the) west or equivalent, e.g. bearing 70 () m = d sin(8. )(); d = m / sin(8. ) = m ( 0-6 m) ()m; ()e largest θ = 90 / sin θ = so λ =d sin θ / = m/ = 60 nm Section A total 4 a bald 6 m s - gets both marks (± for scale drawing) labelled angle on diagram in range 7 gets both marks 4 th mark not awarded if no angle/incorrect angle given ecf from speed (If wind not E-W, any southerlyish wind with westerly component of 5 m s - is acceptable) allow ecf from part Watch for m sin(8. ) = m which gets 0. d = 0-6 m gives 667 nm 4

6 From diagram, radius = 4mm/70mm x 5 = 7 m (); accept radius range 6 9 m 8 (i) Area = π 7 = 46 m () Ecf own radius for nd mark but watch out for incorrect method In s, V =.5 m s m = 50 m (); m= Vρ = 50 m 00 kg m - = kg () using area of triangular base which gets 0. Accept ecf own area in 46 m gives 55 m (); and kg () (i) E k = ½ kg (.5 m s - ) = J ( 4 MJ) (); friction / electrical resistance in turbine/generator & wiring (); causes turbines/generators/some energy dissipated as heat/sound (); turbine blades don t trap all water/water passing through turbine circle not slowed down to 0 (); so water retains some kinetic energy (); 4 Needs evidence of calculation E = ½ (.5) = J or a calculated answer (with no working) of. MJ (from 0 6 kg) or 4.0 MJ (from unrounded answer to aii) QWC: 4 marks not awarded unless two separate factors are clearly and separately discussed. This means that 4 marks does require a clear structure, e.g. columns, paragraphs. 9 (i) (iii) (i) (iii) front turbines will disturb flow to one at back (); so less energy extracted at back () Total 9 f = E/h = 0-5 J/ J s = Hz (); λ = c/f = m s - / Hz = m = 66 pm() output energy in s = 0.5/00 0 J = 60 J(); N = 60 J/ 0-5 J = () most of the kw is dissipated inside the instrument and would damage it/start a fire () v = (E/m) = ( J/ kg) = ( J kg - ) = m s - ()m; ()e λ = h/mv = J s/( kg m s - ) = m = ( m) () E k v v 4 v 4 = (to half of answer in (i)) (); λ /v λ by the same ratio as the velocities () Total 0 Or: recall of E = hc/λ (); substitution & evaluation () Watch for use of 50% which loses st mark but can get nd mark for FT (gives ) answer needs to recognise that this power would damage the tube Watch for incorrect use of λ from ai gives.09 x 0 7 Using 0 7 m s - gives 7 pm Accept recalculation (i) () & () E k v λ without numbers gets () Reject a bald answer doubles without explanation.. 5

7 accept cm for first marking point 66 cm = ½ λ and so λ =. m (); 0 (i) f = v/ λ = 40 m s - no ecf on λ /. m = 58/60 Hz () (i) (i) N and A labelled and alternating(); Antinode at each end of drawing/list (); loops e.g. ANANANA or NANANAN () same f needed (); λ v (both increase) so flute needs to be longer () v must become 40 m s - (98/8) = 49 m s - (); percentage increase in v = 00% (9/40) =.6% (); use of f α v (λ is unchanged) () Gradient is steepest at the origin (); therefore greater acceleration and so greater force () board has stopped accelerating the diver upwards/starts to decelerate due to gravity/has maximum (upwards) velocity this is the highest point because v = 0 / changes direction at this point Total 0 Candidates may intend equal spacing which is hard to draw so do not penalise inequality here. e.g. a correct unlabelled diagram would get mp & mp no mark for longer flute without explanation or ratio of vs = ( ratio of Ts) =.06 (); which is a.6% increase (); watch for stage-by-stage calculation with rounding between stages resulting in e.g..% which is OK Must refer to gradient at the start/at the origin recognising that this is the end of the upwards push earns this mark (c) (i) distance risen (from board) to highest point of dive allow distance between B and C / distance travelled in this time (c) (d) area above CD =½ base height ()m; ½ 9. m s - (.64s s) = 4. m ()e force most at start and drops all the time (); larger v larger resistive force () Total 9 allow equivalent using equations of uniformly accelerated motion, with ()m ()e Mp describes the change in a or F Mp is a mechanism in terms of resistance 6

8 (i). to. squares = 4. to 4.6 V cycle in 4 divisions T = 8 ms () ; f=/t = 5 Hz () If 8 seen assume 8 ms meant and award mp 0. 0 ns = x 0-9 s () (c) Advantages: Great range of voltages can be used for many different input p.d.s (); Excellent time resolution can analyse fast-changing signals (); Values of amplitude and period can be easily obtained wave form easy to see and () Can plot rapid changes owtte because electrons have small mass () Award marks only for correct explanations of advantages & disadvantages Must have one advantage and one disadvantage. Do not award marks for unexplained quotes from the article, e.g. screen display has limited precision NOT low inertia (In article) (d) Disadvantages: Only repeated signals can be displayed because it needs the signal to be redrawn on the screen continually (); Screen resolution limited because lacks minor divisions () large computer memory (); needed because large amounts of data captured by cro sweep(); sensors with rapid response times (); needed because cro signal changes rapidly/has small time resolution () memory/processors with rapid response times (); needed because sensor output changes rapidly/has small time resolution () Total 0 4 Any two pairs. st mark is for describing a necessary development, which could be quoted from the article. nd mark must be an interpretation of the article, not a straight quote; it. must be an explanation in terms of what a cro trace could capture and which the digital storage oscilloscope must copy. 7

9 (i) 4000 Wh / 5 Wh kg - = 686 kg (> 500 kg) ()m; ()e Car: mass increase will reduce the acceleration/increase the force needed for the (same) acceleration (); also increases energy required to accelerate / climb hills (); Forklift: makes it more stable / less prone to tip () Idea of low centre of mass () (iii) 0% of 50kW = 0kW = J in s ()m; ()e (iv) decreased range or increased frequency of re-charging (); battery will overheat (); this will increase self-discharge () Do NOT credit any reference to fuel Accept increasing braking distance / increasing friction Accept correct discussion of moments. out 4 marking points QWC is clear organisation and use of appropriate terms from marking points (i) a = 7 m s - / 5.4 s =.75 m s - (); F = ma = 70 kg.75 m s - = 050 N ( kn) () E k = ½mv = kg (7 m s - ) = J (); P = J / 5.4 s = W ( ½ of 49 kw) () or v mean = 7 m s - / =.5 m s - (); P = Fv = 050 N.5 m s - = W () but must use mean velocity otherwise zero marks Using kn gives W with this approach. (iii) t = 0 km / 80 km hour - =.5 hours (); 6 kwh /.5 hours = 0.67kW = kw () Total 5 8

10 accept other reasonable suggestions 4 (i) (variation in) human reaction time is the limiting factor Repeats are varying at the first decimal place uncertainty in end and start time is now divided by 0/reduces Allow can measure longer time interval with greater accuracy % uncertainty in T values (iii) same percentage uncertainty/divide 0. by 0 () (c) (iv) (i) (iii) T min =.09 s & =. s so.88 < T <. (); T mean =. and T =.. = 0.0 () T values of 0.6 s & 0.6 s (); correct plotting () 4π T = L (); g relate to line equation y = mx (+ c) () best-fit line drawn (); gradient correctly determined using triangle with L at least 0. m (); g = 4π /gradient calculated () advantages: (much) longer T reduces uncertainty T (); % uncertainty in L decreases () more precise determination of e.g. central point of swing (); greater range of lengths to increase reliability/validity (); disadvantages: practical difficulty of working with m pendulum in lab.(); uncertainty in length L will increase as L longer than metre rule (); long pendulum more susceptible to draughts/air resistance/wobble () Total 5 both are method marks accept T mean to upper or lower limit T accept doubling percentage error in T; percentage error in T = 0.9 (); percentage error in T doubles to.8 T = (.8.)/00 = 0.0 = 0.0 to s.f. () Ignore incorrect / absent values for T First mark for algebraic manipulation. Allow nd mp for showing gradient = T /L (= 4π /g) if data for points not on line used, this mark is not awarded Penalise > sf any allow all advantages or all disadvantages Wobble includes departure from D motion, i.e. acting as conical pendulum 9

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