GCE. Mathematics. Mark Scheme for June Advanced GCE Unit 4729: Mechanics 2. Oxford Cambridge and RSA Examinations

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1 GCE Mathematcs Advanced GCE Unt 4729: Mechancs 2 Mark Scheme for June 2011 Oxford Cambrdge and RSA Examnatons

2 OCR (Oxford Cambrdge and RSA) s a leadng UK awardng body, provdng a wde range of qualfcatons to meet the needs of pupls of all ages and abltes. OCR qualfcatons nclude AS/A Levels, Dplomas, GCSEs, OCR Natonals, Functonal Sklls, Key Sklls, Entry Level qualfcatons, NVQs and vocatonal qualfcatons n areas such as IT, busness, languages, teachng/tranng, admnstraton and secretaral sklls. It s also responsble for developng new specfcatons to meet natonal requrements and the needs of students and teachers. OCR s a not-for-proft organsaton; any surplus made s nvested back nto the establshment to help towards the development of qualfcatons and support whch keep pace wth the changng needs of today s socety. Ths mark scheme s publshed as an ad to teachers and students, to ndcate the requrements of the examnaton. It shows the bass on whch marks were awarded by Examners. It does not ndcate the detals of the dscussons whch took place at an Examners meetng before markng commenced. All Examners are nstructed that alternatve correct answers and unexpected approaches n canddates scrpts must be gven marks that farly reflect the relevant knowledge and sklls demonstrated. Mark schemes should be read n conjuncton wth the publshed queston papers and the Report on the Examnaton. OCR wll not enter nto any dscusson or correspondence n connecton wth ths mark scheme. OCR 2011 Any enqures about publcatons should be addressed to: OCR Publcatons PO Box 5050 Annesley NOTTINGHAM NG15 0DL Telephone: Facsmle: E-mal: publcatons@ocr.org.uk

3 1 PE = 70x3g KE change =70x( )/2 PE change + KE change J 20(90 + T) = T = N 70g T = 70.( ) T = N ft Must nclude evaluaton Accept Allow all values to be negatve. Work done = Energy change used ft(cv( )) accept 17.2 Use of v 2 =u 2 + 2as to fnd a AND use of N2 law(4 terms) accept /25 0 = 21000/25 25k 1250gsn2 k = /v= 16.5v v = 35.7 ms -1 ft Use of force = power/speed 3 terms cv(21000/25) ft on cv(k) 3 - (8cos30/3)(8 2 sn60/2) + (4)(8 2 ) = ( sn60/2)(x G ) x G = 2.09 cm tanθ = (2.09/4) θ = 27.6 [5] ft [2] Table of moments dea, may nclude g and/or densty x 27.7 ft cv(x G ) 1

4 4 a b If reversed = e( ) e > 1 mpossble = e( ) e = 0.2 3m 0.2x1.5 = 2m + 0.2x2.9 m = = 0.2v + 0.2x2.9 v = 0.5 e = 0.5/2.9 e = [2] [2] Award f no explct numercal justfcaton May be seen n a Conservaton of momentum Accept wth g ncluded consstently Do not award f g used Impulse = change n momentum Separaton speed not 2.9 Allow 5/29 5 x = (7cos30)t y = (7sn30)t gt 2 /2 y = xtan30 - gx 2 /(2x7 2 cos 2 30) 2x 2 /15 x/ = 0 or 9.8t 2 7t = 0 x = 1.73 m or 3 m 2.6(0) m or 3 3/2 m v 2 = (7sn30) 2 2x9.8x0.6 v = 0.7 ms -1 tanθ = 0.7/(7cos30) θ = 6.59 to horzontal or 83.4 to vertcal Attempt to dfferentate equaton of trajectory tan30 - gx/(7 2 cos 2 30) Substtute x = 3 and equate to tan θ = 6.59 to horzontal or 83.4 to vertcal Attempt to elmnate t y = x/ 3 2x 2 /15 or y = 0.577x 0.133x 2 aef Create a 3 term Q.E. n x or t wth y = 0.6 Solve 3 term Q.E. for x or t Usng v 2 = u 2-2gs wth u a component of 7; can fnd t frst from ther x n (), and then use v = u + at. Use component of 7 Allow 1/ 3 4x/15 or y = x 2

5 6 Rsn30 = 0.3g Rcos30 = 0.3ω 2 x 0.12 ω = 11.9 rads -1 S + Rcos30 = 0.3x2.1 2 /0.2 R = 5.88 S = 1.52 N v P = 11.9x0.12, or h = 0.2/tan30 or 0.12/tan30 or 0.08/tan30 +/-(Q P) = 0.5x0.3(2.1 2 (11.9x0.12) 2 ) + (0.2/tan /tan30) x 0.3g Q-P = +/ J 7 F x 0.8 = 0.6cos60 x 550 F = T x 2 x 0.8/tan30 = 550 x (0.8/sn30 0.6cos60) T = 258 R = 550 Tcos30 Fr = Tsn30 μ = 129/326.6 μ = [5] ft A2ft [5] * * * * dep* [10] R = 5.88 or 0.6g accept v 2 /0.12 for acceleraton cao Resolve and use N2L on sphere Q, 3 terms needed ft cv(r) from () cv(ω) from () Attempt to calculate KE or PE for both partcles KE dfference (ft on cv(ω)) or PE dfference Q P = +/-( ) Attempt at moments Accept 206, cao Moment of T about P T x 2.77 Moment of weght about P 550 x ( ) Accept to 2sf Resolvng vertcally, 3 terms needed Value for T not requred Value for T not requred; accept < or For correct use of F = R, R 550 3

6 T x 0.8/tan x 0.6cos60 = R x 0.8/cos60 R = 550 Tcos30 Solve for T or R T = 258 or R= Fr = Tsn30 μ = 129/326.6 μ = * A2 * * dep* [10] Moments about V, 3 terms needed for two terms correct Resolvng vertcally, 3 terms needed Accept to 2sf Value for T not requred; accept < or For correct use of F = R, R 550 Fr x 1.6cos x (1.6sn sn30) = R x ( sn30) R = 550 Tcos30 Fr = Tsn30 Solvng for at least one of R, Fr, or T Ether R = , or Fr = 129( ), or T=258 μ = 129/326.6 μ = * A2 * * dep* [10] Moments about Q, 3 terms needed for two terms correct Resolvng vertcally, 3 terms needed accept < or Only one needed. Accept to 2sf. For correct use of F = R, R 550 4

7 OCR (Oxford Cambrdge and RSA Examnatons) 1 Hlls Road Cambrdge C 2EU OCR Customer Contact Centre Qualfcatons (General) Telephone: Facsmle: Emal: general.qualfcatons@ocr.org.uk For staff tranng purposes and as part of our qualty assurance programme your call may be recorded or montored Oxford Cambrdge and RSA Examnatons s a Company Lmted by Guarantee Regstered n England Regstered Offce; 1 Hlls Road, Cambrdge, C 2EU Regstered Company Number: OCR s an exempt Charty OCR (Oxford Cambrdge and RSA Examnatons) Head offce Telephone: Facsmle: OCR 2011