GCE. Mathematics (MEI) Mark Scheme for June Advanced Subsidiary GCE Unit 4761: Mechanics 1. Oxford Cambridge and RSA Examinations

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1 GCE Mathematics (MEI) Advanced Subsidiary GCE Unit 476: Mechanics 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 00 Annesley NOTTINGHAM NG 0DL Telephone: Facsimile: publications@ocr.org.uk

3 476 Mark Scheme June 0 comment You should expect to follow through from one part to another unless the scheme says otherwise but not follow through within a part unless the scheme specifies this. Each script must be viewed as a whole at some stage so that (i) a candidate s writing of letters, digits, symbols on diagrams etc can be better interpreted; (ii) repeated mistakes can be recognised (e.g. calculator in wrong angle mode throughout penalty in the script and FT except given answers). You are advised to set height in scoris, particularly for question 7(ii). Questions and 8(v) also spread onto two pages. Q mark notes v ( 9.8).4 M Use of v u asor complete sequence of correct suvat. Accept sign errors in substitution. A All correct v = 8.6 so 8.6 m s -. A cao [Award all marks if 8.6 seen WWW] Do not condone 8.6.

4 476 Mark Scheme June 0 Q mark comment either for u first: 8 u. M Using s u v t u =.7 so.7m s - A. =.7 + a M Use of any appropriate suvat with their values and correct signs a = 0. so 0. m s - F Sign must be consistent with their u, FT from their value of u Directions of u and a are defined F Establish directions of both u and a in terms of A and B. May be shown by a diagram, eg showing A and B and a line between them together with an arrow to show the positive direction. Without a diagram, the wording must be absolutely clear: eg do not accept left/right, forwards/backwards without a diagram or more explanation. Dependent on both M marks. Or for a first: 8. a M Using s vt at a = 0. so 0. m s - A. = u + 0. M Use of any appropriate suvat with their values and correct signs u =.7 so.7m s - F Sign must be consistent with their a, FT from their value of a Directions of u and a are defined F Establish directions of both u and a in terms of A and B. May be shown by a diagram, eg showing A and B and a line between them together with an arrow to show the positive direction. Without a diagram, the wording must be absolutely clear: eg do not accept left/right, forwards/backwards without a diagram or more explanation. Dependent on both M marks. Or using simultaneous equations Set up one relevant equation with a and u. M Using one of v = u + at, s = ut + ½ at² and v² = u² + as Set up second relevant equation with a and u. M Using another of v = u + at, s = ut + ½ at² and v² = u² + as Solving to find u =.7 so.7m s - A Solving to find a = 0. so 0. m s - F FT from their value of u or a, whichever found first Directions of u and a are defined F Establish directions of both u and a in terms of A and B. May be shown by a diagram, eg showing A and B and a line between them together with an arrow to show the positive direction. Without a diagram, the wording must be absolutely clear: eg do not accept left/right, forwards/backwards without a diagram or more explanation. Dependent on both M marks.

5 476 Mark Scheme June 0 Q mark Notes (i) 6 = M May be implied so y = = 9 and z = 4 = A Both correct [Award for both correct answers seen WW] (ii) a 4 0. a so accn is 0.4 m s - M B A Use of Newton s nd Law in vector form for all cpts of attempted resultant Treat use of wrong vectors as MR. Correct LHS The acceleration may be written as a magnitude in a given direction. Magnitude is 0. ( 0.4) ( ) M FT their values. Condone missing brackets. Condone no signs. =.0944 so.0 m s -, ( s. f.) F Accept.. Accept surd form. Must come from a vector with non-zero components for a 7

6 476 Mark Scheme June 0 Q 4 mark Comment (i) B Any one force in correct direction correctly labelled with arrow or all forces with correct directions and arrows. A force may be replaced by its components if labelled correctly eg mgcos0, mgsin0. B All correct (Accept words for labels and weight as W, mg, 47 (N)) No extra or duplicate forces. Do not allow force and its components unless components are clearly distinguished, eg by broken lines. (ii) Either Up the plane Pcos0 9.8 sin0 = 0 M Attempt to resolve at least one force up plane. Accept mass not weight. No extra forces. If other directions used, all forces must be present but see below for resolving vertically and horizontally. P =.06 so. ( s. f.) A Cao A Accept only error as consistent s c. Or Vertically and horizontally Rcos 0 g, Rsin 0 P Eliminate R M Attempt to resolve all forces both horizontally and vertically and attempt to combine into a single equation. No extra forces. Accept s c. Accept mass not weight. g A Accept only error as consistent s c. P sin 0 cos 0 P. (.s.f.) A Cao Or Triangle of forces Triangle drawn and labelled M All sides must be labelled and in correct orientation; three forces only; condone no arrows P tan 0 g A Oe P. (.s.f.) A Cao 4

7 476 Mark Scheme June 0 Q mark notes Usual notation either consider height: Attempt to substitute for u and a in s ut at M Accept: g as g, ±9.8, ±9.8, ±0; u = 0; s c. y 0sin t 4.9t A Derivation need not be shown Need y = 0 for time of flight T B 0sin giving T ( =.69 ) 4.9 A cao. Any form. May not be explicit. Or Consider time to top Attempt to substitute for u and a in v u at M Accept: g as g, ±9.8, ±9.8, ±0; u = 0; s c. v 0sin 9.8t A Derivation need not be shown Need v = 0 and to double for time of flight T B 0sin giving T ( =.69 ) 4.9 A cao. Any form. May not be explicit. then x 0cos T M Accept s c if consistent with above 0sin so x = 0cos ( = ) 4.9 F FT for their time Condone consistent s c error (which could lead to correct answer here). Required time for sound is x/4 M FT from their x Total time is =.769 so.76 s ( s. f.) A cao following fully correct working throughout question. 8

8 476 Mark Scheme June 0 Q6 mark notes Column vectors may be used throughout; lose mark once if j components put at top or if fraction line included.. Notation used must be clear. Either using suvat: (i) Use of v = u + ta M substitution required. Must be vectors. v = 4i tj A Use of r = (r 0 +) tu + ½ t²a M substitution required. r 0 not required. Must be vectors. + j B May be seen on either side of a meaningful equation for r r = 4ti + ( t²)j A Accept r = j + 4ti ½ t² j oe written in a correct notation. Isw, providing not reduced to scalar: (see c in marking instructions) Or using integration: v a dt M Attempt at integration. Condone no +c. Must be vectors. v = 4i tj A cao r v dt M Integrate their v but must contain components. Must be vectors. + j B May be seen on either side of a meaningful equation for r r = 4ti + ( t²)j A Accept r = j + 4ti ½ t² j oe written in a correct notation. Isw, providing not reduced to scalar: (see e in marking instructions) (ii) v(.) = 4i j B FT their v M Award for arctan attempted oe. FT their values. Allow argument to be ± (their i cpt)/(their j cpt) or Angle is (90 )arctan 4 ± (their j cpt)/(their i cpt). Allow this mark if bearing of position vector attempted. = so 4 ( s. f.) A cao 8 6

9 476 Mark Scheme June 0 Q7 mark notes (i) 0 0 M Use of a suitable triangle to attempt at v/ t for suitable interval. Accept wrong sign. 0 m s - A cao. Allow both marks if correct answer seen. (ii) Signed area under graph M Using the relevant area or other complete method (A) 0 0 A (B) either using areas Signed area t is ( ) (4..4) ( 4) = 0. B Allow Signed area t 6is 8 4 B Total displacement is.8 m B cao but FT from their 0 in part (A) or using suvat From t = 0 to t =.4: 9. B From t = 4. to t = 6:.0 B From t =.4 to t = 4.: 8.4 B Total :.8 Both required and both must be correct. (iii) a = 4t 4 M Differentiate. Do not award for division by t. A a(0.) = so m s - A (iv) Model A gives 4 m s - B May be implied by other working For model B we need v when a = 0 M Using (iii) or an argument based on symmetry or sketch graph that a = 0 when t =. 7 v 4. A so model B is 0. m s - less F Accept values without more or less 4 7

10 476 Mark Scheme June 0 (v) Do not penalise poor notation 6 Displacement is t 4t 0 dt M Limits not required. 0 6 t 7t 0t 0 A Limits not required. Accept terms correct. M Substitute limits = so m. A cao. Accept bottom limit not substituted

11 476 Mark Scheme June 0 Q 8 mark notes (i) N B Condone no units. Do not accept - N. (ii) 0 cos M Attempt to resolve 0 N. Accept s c. No extra forces. = 4.8 so 4. N ( s. f.) A cao but accept 4.. (iii) Resolving vertically M All relevant forces with resolution of 0 N. No extras. Accept s c. R + 0 sin = 0 A All correct. R = so 7. N ( s. f.) A (iv) Newton s nd Law in direction DC M Newton s nd Law with m = 8. Accept F = mga. Attempt at resolving 0 N. Allow 0 N omitted and s c. No extra forces. 0 cos 0 8a A Allow only sign error and s c. a = so.4 m s - ( s. f.) A cao Q8 continued (v) Resolution of weight down the slope B mgsin where m = 8 or 0 or 8, wherever first seen either Newton s nd Law down slope overall sin 0 = 8a M F = ma. Must have 0 N and m = 8. Allow weight not resolved and use of mass. Accept s c and sign errors (including inconsistency between the N and the N). a = 0.69 A cao Newton s nd Law down slope. Force in rod can be taken as tension or thrust. Taking it as tension M F = ma. Must consider the motion of either C or D and include: component of weight, resistance and T. No extra forces. Condone sign errors and s c. Do not condone inconsistent value of mass. T gives For D: sin T 0a F FT only applies to a, and only if direction is consistent. +T if T taken as a thrust ( For C: sin T 8a ) -T if T taken as a thrust T =.888 = -.89 N ( s. f.) A If T taken as thrust, then T = The force is a thrust A Dependent on T correct 9

12 476 Mark Scheme June 0 or Newton s nd Law down slope. Force in rod can be taken as tension or thrust. Taking it as tension T gives M F = ma. Must consider the motion of C and include: component of weight, resistance and T. No extra forces. Condone sign errors and s c. Do not condone inconsistent value of mass. M F = ma. Must consider the motion of D and include: component of weight, resistance and T. No extra forces. Condone sign errors and s c. Do not condone inconsistent value of mass. For C: sin T 8a A Award for either the equation for C or the equation for D correct. -T if T taken as a thrust For D: sin T 0a +T if T taken as a thrust a = 0.69 T =.888 = -.89 N (s.f.) A First of a and T found is correct. If T taken as thrust, then T = F The second of a and T found is FT The force is a thrust A Dependent on T correct then After s: v = + a M Allow sign of a not followed. FT their value of a. Allow change to correct sign of a at this stage. v = so.49 m s - ( s. f.) F FT from magnitude of their a but must be consistent with its direction

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