GCE Further Mathematics B (MEI) Y421/01: Mechanics major Advanced GCE Mark Scheme for November 2020

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Oxford Cambridge and RSA Examinations GCE Further Mathematics B (MEI) Y421/01: Mechanics major Advanced GCE Mark Scheme for November 2020Oxford Cambridge and RSA Examinations OCR (Oxford Cambrid ... ge 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 2020Y421/01 Mark Scheme November 2020 Text Instructions Annotations and abbreviations Annotation in scoris Meaning and BOD Benefit of doubt FT Follow through ISW Ignore subsequent working M0, M1 Method mark awarded 0, 1 A0, A1 Accuracy mark awarded 0, 1 B0, B1 Independent mark awarded 0, 1 E Explanation mark 1 SC Special case ^ Omission sign MR Misread BP Blank page Highlighting Other abbreviations in mark scheme Meaning E1 Mark for explaining a result or establishing a given result dep* Mark dependent on a previous mark, indicated by *. The * may be omitted if only previous M mark. cao Correct answer only oe Or equivalent rot Rounded or truncated soi Seen or implied www Without wrong working AG Answer given awrt Anything which rounds to BC By Calculator DR This indicates that the instruction In this question you must show detailed reasoning appears in the question.Y421/01 Mark Scheme November 2020 3 Subject-specific Marking Instructions for AS Level Mathematics B (MEI) a Annotations must be used during your marking. For a response awarded zero (or full) marks a single appropriate annotation (cross, tick, M0 or ^) is sufficient, but not required. For responses that are not awarded either 0 or full marks, you must make it clear how you have arrived at the mark you have awarded and all responses must have enough annotation for a reviewer to decide if the mark awarded is correct without having to mark it independently. It is vital that you annotate standardisation scripts fully to show how the marks have been awarded. Award NR (No Response) - if there is nothing written at all in the answer space and no attempt elsewhere in the script - OR if there is a comment which does not in any way relate to the question (e.g. ‘can’t do’, ‘don’t know’) - OR if there is a mark (e.g. a dash, a question mark, a picture) which isn’t an attempt at the question. Note: Award 0 marks only for an attempt that earns no credit (including copying out the question). If a candidate uses the answer space for one question to answer another, for example using the space for 8(b) to answer 8(a), then give benefit of doubt unless it is ambiguous for which part it is intended. b An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not always be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed. When a candidate adopts a method which does not correspond to the mark scheme, escalate the question to your Team Leader who will decide on a course of action with the Principal Examiner. If you are in any doubt whatsoever you should contact your Team Leader.Y421/01 Mark Scheme November 2020 c The following types of marks are available. M A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified. A method mark may usually be implied by a correct answer unless the question includes the DR statement, the command words “Determine” or “Show that”, or some other indication that the method must be given explicitly. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. B Mark for a correct result or statement independent of Method marks. E A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument. d When a part of a question has two or more ‘method’ steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation ‘dep*’ is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given. e The abbreviation FT implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only – differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, what is acceptable will be detailed in the mark scheme. If this is not the case, please escalate the question to your Team Leader who will decide on a course of action with the Principal Examiner. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be ‘follow through’. In such cases you must ensure that you refer back to the answer of the previous part question even if this is not shown within the image zone. You may find it easier to mark follow through questions candidate-by-candidate rather than question-by-question.Y421/01 Mark Scheme November 2020 3 f Unless units are specifically requested, there is no penalty for wrong or missing units as long as the answer is numerically correct and expressed either in SI or in the units of the question. (e.g. lengths will be assumed to be in metres unless in a particular question all the lengths are in km, when this would be assumed to be the unspecified unit.) We are usually quite flexible about the accuracy to which the final answer is expressed; over-specification is usually only penalised where the scheme explicitly says so. • When a value is given in the paper only accept an answer correct to at least as many significant figures as the given value. • When a value is not given in the paper accept any answer that agrees with the correct value to 2 s.f. unless a different level of accuracy has been asked for in the question, or the mark scheme specifies an acceptable range. NB for Specification A the rubric specifies 3 s.f. as standard, so this statement reads “3 s.f” Follow through should be used so that only one mark in any question is lost for each distinct accuracy error. Candidates using a value of 9.80, 9.81 or 10 for g should usually be penalised for any final accuracy marks which do not agree to the value found with 9.8 which is given in the rubric. g Rules for replaced work and multiple attempts: • If one attempt is clearly indicated as the one to mark, or only one is left uncrossed out, then mark that attempt and ignore the others. • If more than one attempt is left not crossed out, then mark the last attempt unless it only repeats part of the first attempt or is substantially less complete. • if a candidate crosses out all of their attempts, the assessor should attempt to mark the crossed out answer(s) as above and award marks appropriately. h For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate’s data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A or B mark in the question. Marks designated as cao may be awarded as long as there are no other errors. If a candidate corrects the misread in a later part, do not continue to follow through. E marks are lost unless, by chance, the given results are established by equivalent working. Note that a miscopy of the candidate’s own working is not a misread but an accuracy error. i If a calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers provided that there is nothing in the wording of the question specifying that analytical methods are required such as the bold “In this question you must show detailed reasoning”, or the command words “Show” and “Determine. Where an answer is wrong but there is some evidence of method, allow appropriate method marks. Wrong answers with no supporting method score zero. If in doubt, consult your Team Leader. j If in any case the scheme operates with considerable unfairness consult your Team Leader.Y421/01 Mark Scheme November 2020 Question Answer Marks AOs Guidance 1 At lowest point, PE = B1 1.1 Correct use of PE x is the extension in the string EPE = B1 1.1 Correct use of M1 3.3 Use of conservation of energy (correct number of terms) (because ) as x > 0 A1 2.3 BC – as a minimum must state x > 0 if no explicit rejection of –0.906… seen x = 1.298005445… or x = –0.906005… (Provided the string has not been stretched beyond its elastic limit the) length of string is 4.30 (m) A1 2.2b Awrt 4.30 [5] 2 , and B2 1.2 1.2 B1 for any one correct M1 2.1 Using their and to obtain an equation in M, L and T M1 1.1a Setting up all three equations A1 1.1 [5] 3 (a) and M1 1.1 Setting up equations for a and b Allow sign errors only A1 1.1 A1ft 1.1 oe e.g. 3.16 (ft their F) 3.16227766… Direction is below the horizontal A1 1.1 oe to the downward vertical In radians: 1.25 or 0.322 [4] (± + )0.5 3 g x ( ) ( ) 2 75 2 3 x 2 x2a λ ( ) 2 75 0.5 3 0 x6 − + = g x 2 75 29.4 88.2 0 x x − − = x =1.29... x ≠ −0.90... 1 [ ] T f = − [ ] MLT C = −2 [σ ] = ML−1 T MLT L ML − − − 1 2 1 = ( )α γ β ( ) [ f C ],[ ] [σ ] − = − + − = + = 2 1, 0, 0 α α β γ α γ 1 1 , 1, 2 2 α β γ = = − = − F i j = + a b a + − = 2 3 0 2 1 0 + + = b F i j = − 3 F = 10 71.6 18.4Y421/01 Mark Scheme November 2020 3 Question Answer Marks AOs Guidance 3 (b) e.g. taking moments about O: M1 1.1a Taking moments about any point with correct number of terms 5 A1 1.1 anti-clockwise A1 2.5 oe (could be seen on a diagram) [3] 4 (a) M1* 1.1 Attempt at differentiation – at least one component correct Setting t = 0 M1dep* 1.1 Substituting t = 0 into their v A1 1.1 [3] 4 (b) M1* 3.4 Setting up scalar product with their initial v and v and equating to zero M1dep* 1.1 Correct use of scalar product to form a linear equation in T Allow in terms of t A1 1.1 [3] 4 (c) B1 1.1 B1ft 1.1 Follow through their value of T Displacement vector M1 1.1 Difference in their r values Distance is 125 A1 1.1 [4] 3 2 2 2 2 1 1 3 ( ) + − − ( ) ( ) ( ) v i j = + − 20 10 10 ( t) v i j = + 20 10 20 20 0 10 10 10T         ⋅ = −    ( 400 10 10 10 0T+ − =) 5T= t = = + 0, 5 95 r i j t = = + 5, 105 20 r i j (105 20 5 95 100 75 i j i j i j + − + = − ) ( )Y421/01 Mark Scheme November 2020 5 (a) Driving force = B1 1.2 M1 3.3 Applying N,II with correct number of terms – allow a for dv/dt A1 2.2a AG – sufficient working must be shown as answer given [3] 5 (b) B1 1.1 Must explicitly verify initial conditions M1* 1.1 Attempt at differentiation – must see terms and A1 1.1 Correct derivative (allow unsimplified) or from M1dep* 1.1 Simplify to single fraction before taking reciprocal A1 1.1 AG – sufficient working must be shown as answer given [5] 60000 v 60000 d 1500 900 vd v t − = d 3 d 200 5 3 40 d 5 d v v v v v v t t − = ⇒ = − 40 3 0, 24ln (0) 24ln1 0 40 5 v t = = − = =       40 3 d 24ln ... 40 5 d t t v v v   = − ⇒ =     − 1 40 40 v       − (40 − v)−2 d 1 24 ( 1 40 40 )( )( ) 2 3 d 5 40 40 t v v v −     =   − − − −             − d 24 3 ( 1) d 40 5 t v v − = − − − t = − − − 24ln 40 24ln 40 ( v v ) 53 ( ) d 24 3 3 d ... d 40 5 5 40 d t v v v v v t = − = ⇒ = − − d 5 40 ( ) 3 d 40 d 3 5 d v v v v v t v t − = ⇒ = −Y421/01 Mark Scheme November 2020 3 5 (b) ALT 3 d 5 40 d d 5 d 3 40 v v v v t v t = − ⇒ = ∫ ∫ −v B1 Separate variables correctly 5 40 1 d 3 40 t v = − − ∫ −v M1* Rewrite in an integratable form and attempt to integrate (must contain a log term of 40 – v ) 53 t v v c = − + − + ( 40ln 40 ( ) ( )) A1 + this mark c not required for t v c = = ⇒ = 0, 0 40ln 40 M1dep* Use given initial conditions to find c ( ) ( ) 5 40ln 40 40ln 40 3 40 3 24ln 40 5 t v v t v v = − − − ⇒ = − − A1 AG [5] 6 (a) B1 3.3 Correct equation for the time from release to first bounce Or correct speed at the first impact for B1M1 M1* 3.4 Use of with u = 0 to find the speed at the first impact oe suvat equation ( ) A1ft 1.1 Correct expression for the speed after first impact in terms of their M1* 3.4 Use of with s = 0 to find the time between first and second impact A1 1.1 M1dep* 2.1 Setting up an infinite series (at least three terms on the lhs) equal to 6 M1 3.1a Correct use of the sum to infinity formula for a GP of the form 2 u eu e u 0 1 2 + + + ... Dependent on all previous M marks A1 2.2a [8] 78.4 9.8 4 = ⇒ = 12 ( )t t 1 1 2 ( ) t1 v gt 1 1 = v u at = + v1 v1 = 39.2 v get 2 1 = v2 t1 v e 2 = 39.2 0 ( 1 2 2 ) 1 2 2 = − get t gt 2 s ut at = + 0.5 t2 t e 2 = 8 2 4 8 8 ... 6 + + + = e e 4 8 1 ... 6 4 8 6 ( ) 1 1 e e e e   + + + = ⇒ + =     − 8 2 1 0.2 e e e = − ⇒ = ( )Y421/01 Mark Scheme November 2020 6 (b) 23.52 = M1 3.3 Use of impulse = change in momentum Allow with no value of e substituted A1 1.1 [2] 7 (a) Hooke’s law: B1 1.2 Where e is the extension of the string when P is in equilibrium M1 3.3 Resolving parallel to the plane with P in equilibrium Allow sin/cos confusion A1 2.2a M1* 3.3 NII parallel to the plane (with correct number of terms) M1dep* 3.4 Use of Hooke’s law in NII with correct extension A1 2.2a AG – sufficient working must be shown as answer givn [6] m get gt ( 1 1 − − ( )) ( ) 23.52 0.5 9.8 0.2 (4) 9.8(4) m = = + 3 6 mge T a = T mg = sin30 2 2 mge mg e a a = ⇒ = 2 2 d sin30 d x mg T m t − = ( ) 2 2 3 d sin30 6 d mg x mg a x m a t − + = 2 2 2 2 d 3 3 d 0 0 d d 2 6 6 2 x mg mga mgx x gx m t t a a a − + + = ⇒ + =Y421/01 Mark Scheme November 2020 3 7 (b) B1 1.2a Or as the motion starts at the extreme point of the motion A and B are arbitrary constants M1 3.4 Use initial conditions to find A and B (or just A) A1 1.1 Slack when M1 3.1b Substituting into their equation for x A1 2.2a [5] 7 (c) M1 3.4 Use of with their values or differentiation of their x A1 1.1 [2] cos sin 2 2 g g x A t B t a a = + cos g2 x A t a = 0, 2 2 0, 0 0 t x a A a t x B = = ⇒ = = = ⇒ =  2 cos g2 x a t a = 1 cos 2 2 g x a t a   = − ⇒ = −       x a = − 2 2 2 2 3 3 g a t t a g π π = ⇒ = v a a 2 = − − 2ga ((2 )2 2 ( ) ) v A x 2 2 2 2 = − ω ( ) 3 2 ga v =Y421/01 Mark Scheme November 2020 8 (a) B1 1.2 oe e.g. a correct equation for a line that could be rotated about the y-axis M1 2.1 Use of with their y Limits not required M1 1.1 Use of and attempt at integration Limits not required M1 1.1 Use of correct limits and attempt to solve for A1 1.1 AG SC B0M1M0M1A0 for those who use y = x [5] 8 (b) M1* 2.1 Table of values idea – correct number of terms (dimensionally consistent) (1 1 3 3 π π π π r r r x r r r r 3 2 + = + (2 )) G 3 2 ( 34r ) 2 2 ( ( )) A1 A1 1.1 1.1 Correct LHS Correct RHS mass of the composite is the centre of body from the vertex A1 1.1 oe (e.g. 33 28 r from the base) M1dep* 3.1b Use of correct angle and their to find horizontal distance distance of centre of mass from vertex Or 51 28 tan r r θ r − = Slant height of cone is B1 1.1 Or comparison with 45 or tan 1 θ = (dependent on first M mark) does not topple A1 3.2a [7] r y x h = 2 0 d h r Vx x x x h π   =   ∫   Vx xy x = π∫ 2d 2 4 2 2 0 1 3 4 h r x r hx π h π   =     1 2 3 V r h = π 2 4 2 2 1 0 ... 3 4 r h r hx x π h π   = − ⇒ =       x 3h4 x = xG G 51 28 x r = 51 cos 45 2 51 56 28 x x r r   = ⇒ =           xG r 2 51 2 2 56 r r < ⇒Y421/01 Mark Scheme November 2020 3 8 (c) Moment of frictional force about any point of contact with the horizontal floor is zero and so has no effect on the stability of the toy B1 2.4 Or equivalent [1] 9 (a) M1 3.1b Taking moments about A for the rod – correct number of terms B1 1.1 … B1 1.1 oe e.g. 2 sin 2 aT θ A1 1.1 oe [4] 9 (b) M1* 3.3 Resolving vertically and horizontally at the ring – correct number of terms. Allow this mark if only one direction stated correctly is the normal contact force between the ring and the rod A1 3.3 is the frictional force between the ring and the rod M1dep* 3.4 Use of with their and Allow equals least value of tan gives greatest distance of the ring from A A1 3.1b B1 3.1a oe e.g. stating the angle or sin 0.8 θ = A1 2.2a [6] (8 cos ... W a )( θ ) = = (2 cos sin 2 sin cos a T a T θ θ θ θ )( ) + ( )( ) T W = 2 cosecθ RC R W T C = + sinθ F T C = cosθ FC F R W W C C ≤ ⇒ ≤ µ θ µ 2 cot 3 ( ) F R ≤ µ FC RC 3 2 cot tan 2 3 µ θ θ µ ≤ ⇒ ≥ 2 tan 3 θ µ = 2.4 4 cos cos 0.6 a a = ⇒ = θ θ 2 4 1 3 3 2 µ µ = ⇒ =Y421/01 Mark Scheme November 2020 10 (a) At highest point, PE and KE = 0 At angle PE KE B1 1.1 Correct mechanical energy at either A or - candidates may calculate initial speed of P as for this mark PE zero at A M1* 3.3 Use of conservation of energy – correct number of terms A1 1.1 M1* 3.3 NII radially – correct number of terms – condone a for acceleration Allow r for radius M1dep* 3.4 Substitute their expression for to get expression for R in terms of m, g and M1 1.1 Equate R with and solve to find Dependent on all previous M marks A1 1.1 Allow in degrees [7] 10 (b) M1* 3.4 Use of with r = a and their v A1 1.1 oe M1dep* 3.1a Use of correct double-angle identity A1 2.2a [4] = mg a (2 ) θ, = − mga(1 cos , θ ) 1 2 2 = mv θ 2 ga 2 1 cos ( ) 1 2 2 mga mga = − + θ mv v ga 2 = + 2 1 cos ( θ ) 2 cos mv R mg a θ− = ( R mg = + 2 3cosθ ) v2 θ (2 3cos ) 7 ... 2 mg + = ⇒ = θ θ mg 7 2 mg θ π 3 θ = θ = 60 v r = ω a ga 2 2 ω θ = + 2 1 cos ( ) 2 2 2 4 cos 2 a ga θ ω   =     cos 1 cos 2 2 1 ( ) 2 θ θ = + 2 2 4 cos 2 cos 2 2 g g a a θ θ ω ω     = ⇒ =         k = 2Y421/01 Mark Scheme November 2020 3 10 (c) M1* 2.1 Differentiate with respect to t M1dep* 3.4 Substitute their value of into their expression for angular acceleration A1 2.2a [3] 11 (a) M1* 3.3 Use of conservation of linear momentum – correct number of terms m is the mass of A and B, is the component of the velocity of A parallel to the line of centres after impact and is the equivalent component for B M1* 3.3 Use of Newton’s experimental law – correct number of terms and consistent with conservation of linear momentum A1 1.1 M1dep* 3.4 Use of tan ratio for with their A1 2.2a AG – sufficient working must be shown as answer given [5] 11 (b) The component of the velocity of A perpendicular to the line of centres does not change B1 3.5b [1] d 1 2 sin d 2 2 g t a ω θ ω     = −           ω d sin 2 cos d 6 6 g g t a a ω π π       = −               θ 3 2 g a θ = − mu mv mv cosα = + 1 2 v1 v2 1 2 1 cos 3 v v u − = − α 1 1 cos 3 v u = α 1 sin tan u v α β = β v1 sin tan tan 3tan 1 cos u u3 α β β α α = ⇒ =Y421/01 Mark Scheme November 2020 11 (c) M1 3.1b Use of a correct compound-angle formula for and substitute given result from (a) A1 1.1 [2] 11 (d) M1* 3.1b Attempt to differentiate using quotient rule A1 1.1 Correct derivative equated to zero M1dep* 1.1 Find value of or M1 1.1 Substitute their value for or into their expression for - dependent on both previous M marks A1 1.1 [5] ( ) ( )( ) 3tan tan tan tan 1 3tan tan α α γ β α α α − = − = + tan(β α ± ) 2 2tan tan 1 3tan α γ α = + ( )( ) ( )( ) ( ) 2 2 2 2 2 d sec d 1 3tan 2sec 2tan 6tan sec 0 1 3tan γ γ α α α α α α α     =   + − = + 2 2 1 1 3tan 6tan 0 tan 3 + − = ⇒ = α α α tanα tan2α π 6 α = ( 33 ) 2 2 tan ... 3 1 3 3 γ γ = ⇒ =   +     α tanα tanγ 3 tan 3 γ = π 6 γ =Y421/01 Mark Scheme November 2020 3 12 (a) B1 3.1b is the length of the string B1 3.1b oe x is the radius of the horizontal circle M1* 3.3 Resolving vertically for P – correct number of terms with components of R and T Condone use of the same angle for this mark A1 1.1 T is the tension in the string, R is the normal contact force and is the angle between the horizontal and the normal contact force M1* 3.3 NII horizontally – correct number of terms – allow any form for radius Condone use of the same angle A1 1.1 oe or B1 1.1 Either correct expression for or sin in terms of or stating a correct relationship between θ and α e.g. 1 2 θ α π = − 2 M1dep* 3.4 Eliminate from both equations Dependent on both previous M marks A1 2.2a AG [9] 2 2 2 r r l rl l r = + − ⇒ = 2 cos 2 cos α α l sin x x r 2 cos sin l α = ⇒ = α α R T mg sin cos θ α = + θ R T m r cos sin 2 cos sin θ α α α ω + = ( ) 2 2 cos sin cos r 2cos sin r α α θ = = α α 2 2 cos 2 sin r r 1 2cos r α θ α = − = − cosθ θ α ( ) ( ) 2 2 1 2cos cos 2cos sin sin 2 cos sin R T mg R T mr α α α α α ω α α − = + + = θ ( ) ( ) 2 2 2 2 2 2 2 2 2 2 cos 2 cos 1 2cos 2 cos 2 cos cos 2 cos 2 cos 2 cos 2 cos T mr R R mr R mg R R mr R mg R mg mr ω α α α ω α α α α ω α α ω α = − ⇒ − = − + − = − + = +Y421/01 Mark Scheme November 2020 12 (b) M1* 3.4 Use given expression for R to find expression for T in terms of and M1dep* 3.4 Setting T > 0 A1 2.2a and [3] T m r g r = − − 2 cosα ω ω α ( 2 2 2 2 cos ) r m g , , ,α ω 2 2 2 T r g r > ⇒ − − > 0 ω ω α 2 cos 0 g r < − ω α 2 2 (1 2cos ) k1 = 1 k2 = −2OCR (Oxford Cambridge and RSA Examinations) The Triangle Building Shaftesbury Road Cambridge CB2 8EA [Show More]

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