GCE. Physics B (Advancing Physics) Mark Scheme for June 2011

Transcription

1 GCE Physics B (Advancing Physics) Advanced Subsidiary GCE Unit G49: Understanding Processes/Experimentation and Data Handing Mark Scheme for June 0 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 0 Any enquiries about publications should be addressed to: OCR Publications PO Box 5050 Annesley NOTTINGHAM NG5 0DL Telephone: Facsimile: publications@ocr.org.uk

3 G49 Mark Scheme June 0 (a) kg m s - (); N m and W s () (a) 0-6 () 0 () Three equal-length arrows (by eye) joined tip-to-tail () Forming a (closed equilateral) triangle () 4 increasing amplitude Deduct one mark for each extra tick. increasing frequency increasing intensity increasing wavelength increasing width 5 (a) d = 0 - m/400 = m () n = d sin sin = n /d No marks for first order sin = m/ If you see 8.7, it must be right = () m = 0.65 Allow () m for using the value of d from (a) = 9 ()m ()e 6 (a) F = 850 kg (7 m s - /5 s)= 50 N 500 N()m ()e P = Fv = 00 N 7 m s - = W = W() 7 displacement = {(5-) + 7 } = 9 =.9/4 paces () 8 (a) bearing = 60 arctan(7/(5-)) = 60 - arctan(0.58) = = 0 st mark is for calculation of the angle and the nd is for correctly reporting it. loop = ½ and cm = 50 cm / 5 () Appropriate test proposed: can be assumed if an appropriate test is carried out correctly () proposed test carried out correctly on all data sets() conclusion (yes, to precision of data given) () Section A total: Allow any clear indication of direction, e.g. N 0. W, including diagram with correct angle labelled. For scale drawing, allow 5 paces at 8 Allow 0. W of N or 59.7 N of W or either angle labelled on the diagram. Allow alternative valid approach, e.g. 5 half-wavelengths = 50 cm so = 50 cm/(5 0.5) = 0 cm Should calculate, for all data pairs, either f /T (4.4, 4.5, 4.7) or f/ T (.79,.80,.8) or their inverses (0.0694, 0.069, ) and (0.64, 0.6, 0.6). Allow conclusion No only if candidate indicates that calculated constant shows a distinct trend. Max mark for answers involving graphs.

4 G49 Mark Scheme June 0 9 (a) v = 0 initially () flat is not enough without reference to 0 W > T (and then W =T) and then T > W () Do not penalise for statements or idea of T increasing. Because W is decreasing/it is ejecting gas () tangent drawn at t = 6.0 s with t () st mark is independent of the others Uses v/ t () e.g. gradient allow rounding (this is a show that question) Answer in range 9 to m s - () F res =ma=6.9 kg 0 m s - =69 N or W=6.9 kg 9.8 N kg - = 68N 69 N () so T = F res + W must be about double W () (c) Starts curving up sooner() Curves diverge continually () Total: 0 0 (a) Energy needed to liberate electrons (); Higher frequency/lower wavelength means higher energy photons (); light provides energy in packets (); violet photons are energetic enough to liberate electrons, while red are not (); 4 greater intensity = more photons (); one photon liberates one electron (); more photons more electrons produced (); in wave model, red light will emit if you wait long enough but this does not happen (so wave model is wrong) () E = hf = J s Hz = J (); comparison of calculated value with given threshold () (c) No electrons produced below.7 ( 0-9 J )(); Above this, (extra) energy supplied goes to electron () (d) Any reasonable application/use involving detection of light or measurement of its intensity (); limitation e.g. limited range of wavelengths detectable (not red end of spectrum), need for clean potassium surface () Total: 0 Use own acceleration or 0 m s - Allow algebraic approach ma = T- mg T = ma + mg And a g so T = mg Allow curve starting at zero. Judge by eye One mark for each point. QWC is organise information clearly. The 4 th mark would not be awarded for a confused answer which does not link quantum behaviour with red and violet light. ORA: calculate f min = J/ J s= Hz(); Reject reference to direct proportion. E.g. solar panel, measuring light level, automatic switch.

5 G49 Mark Scheme June 0 (a) (70 /60 ) 65 days()m; = 70.97()e ( 7 days) period = /40 = 556 minutes ()m ()e 7.0 implies evaluation. Allow rounding of intermediate calculation days 555 minutes. Accept 600 minutes for marks half d = opposite side of right-angled triangle with Working may be on a labelled drawing, possibly on Fig... vertex 5 () st mark for recognising the triangle, second for the algebra. 0.5 d/r = sin(5 ) d = R sin (5 ) () d =.4 0 m sin (5 ) =.6 0 m c =.6 0 m/( 60s) = m s - ()m ()e (iii) suggestion (); explanation () Suggestion: estimate for R too low () this makes d too low which lowers the value for c () Suggestion time too large () because it s hard to measure/only an estimate() Total: 0 (a) horiz: u cos vert: u sin () both needed. Using s = ut + ½at (); s= 0 (); u = vert component of u = u sin (); a = -g () 0 = (u sin )t - ½gt u sin = ½gt () t = u sin g = 8.0 m s - sin(50 )/9.8 m s - =.5 s ()s ()e (c) Throw at smaller angle (); collisions with sides of buckets () Any three points Allow alternative valid approaches, with choice of equation (); a = -g (); other conditions with respect to. u, v, s, t (); Use of invalid equation = zero marks Allow other methods: choice of valid equation and rearrangement as necessary(); substitution (); evaluation ().5 s or. s gets marks automatically Allow any feasible strategy for (); second mark needs a possible physical explanation. Allow e.g lower u () so less energy to dissipate () Total: 9 Section B total: 9

6 G49 Mark Scheme June 0 (a) distance travelled better defined / using similar visual stimulus to start and stop timing / student A s method requires doing more than one thing at a time higher chance of error/ larger distance travelled, so time longer and therefore less uncertain. suggestion(); correction () (c).4/.44.9/.94 Any plausible reason. Allow reading of text to imply B makes repeated measurements of a single pass up the tank. e.g. starting stop watch when wave generated, not at end (); allow to reach end before starting timing (); or measuring depth with ruler with 0 not at end (); correction by subtraction, etc. () Both correct for the mark. Allow or 4 s.f. only. (d) (iii) (iv) Each correct point () best fit line () v gd v gd (so v against d has gradient g) Gradient from graph calculated ()m () e % () Vertically above minor division gridline and not above half-way between minor divisions. Allow e.c.f. from. Judge best fit line by eye. Rearranged equation is enough for the mark. Accept values from 9. to 0. m s - Allow.% or any number of sf percentage/fractional uncertainty in t is significantly greater than in L or d () (iii) v = 0.6 m/( ) s =.8 m s - () g = v /d = (.8 m s - ) /0.0 m = 6. m s - () % uncertainty = (0.5 m s m s - ) 00/0.5 m s - = 40% () Total: 5 Independent marking point. Allow ecf from v to calculate g. e.g. only considering a single journey (omission of the ) gives g =.58 m s -, leading to an uncertainty of 85% Must use 0.0 m in calculation of g. or s.f. only (correct % uncertainty = 40% to or s.f.) 4

7 G49 Mark Scheme June 0 4 (a) Many uncontrolled variables owtte () Can quote e.g. may have different size/widths test for tyre of type A () Accept either way round A or A All values (significantly) > other two tests (c) (iii) Allow any reasoned suggestion; one mark for possible cause, one for explanation giving right direction variation is in rd s.f./uncertainty is about 0.0 N (); s.f. would lose significant information/4 s.f. not justified as you should round to the size of the uncertainty () e.g. pressed harder onto rollers() so friction increased () e.g. fault in inflation pressure meter () causing it to read too low () / systematic error in time taken to stop the wheel () giving time values too short () st mark for appreciation that the variation in a test is in the last figure quoted; nd mark for justifying this. (significantly)> test or test (); does not fit data trend down the column () (d) Type B at 80Ncm - (high pressure) () because the (rolling) friction is lower () Can credit the idea of it being an outlier with reference to the other values horizontally () and vertically () Total: 5

8 G49 Mark Scheme June 0 5 (a) Assumption that the Sun s rays are parallel (); Knew angle was 0 at Syene (); deduced 7 latitude difference between Syene & Alexandria owtte (); Any four points. Or Sun directly overhead knew time to travel at known speed from S to A (); 4 deduced distance from speed or time of travel (); use of 700 stadia per degree/realised distance was 7/60 of circumference of Earth (); calculation /7 = stadia () Any reasonable disadvantage related to lack of repeatability/consistency () QWC is select and use a form and style of writing appropriate to purpose and to complex subject matter ; 4 th mark would not be awarded if the story is not clearly conveyed. Allow bulleted lists. E.g. differences in terrain or weather conditions or day length will affect speed of caravan. (c) 60 m () 80 m () max = m (60 /6 ) = m ( m) () min = m (60 /8 ) = m ( m)() Comparison with m. () Penalise one mark for > sf. Penalise one mark for max and min values in wrong place Third mark is independent of first two marks. (iii) (angle) in 7 = 4%/ (stadion) 5% is in 0 () angle is a far greater source of uncertainty () (d) True distance is less than the one he used (); so the final circumference is too big () (ecf); Estimate uncertainty from the diagram 5-8% () Uncertainty in much less than uncertainty in angle, so will have less effect on the calculated value () Total: 4 Section C total: 40 st mark for comparing uncertainties in angle and stadion; nd for conclusion Accept either approach 6

9 OCR (Oxford Cambridge and RSA Examinations) Hills Road Cambridge CB EU OCR Customer Contact Centre 4 9 Qualifications (General) Telephone: Facsimile: general.qualifications@ocr.org.uk For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee Registered in England Registered Office; Hills Road, Cambridge, CB EU Registered Company Number: OCR is an exempt Charity OCR (Oxford Cambridge and RSA Examinations) Head office Telephone: Facsimile: OCR 0