GCE Further Mathematics A Y543/01: Mechanics Advanced GCE Mark Scheme for Autumn 2021

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Oxford Cambridge and RSA Examinations GCE Further Mathematics A Y543/01: Mechanics Advanced GCE Mark Scheme for Autumn 2021Oxford Cambridge and RSA Examinations 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 2021Y543/01 Mark Scheme October 2021 Annotations and abbreviations Annotation in RM assessor 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 SC Special case ^ Omission sign MR Misread BP Blank Page Seen Highlighting Other abbreviations in mark scheme Meaning dep* Mark dependent on a previous mark, indicated by *. The * may be omitted if only one 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 question included the instruction: In this question you must show detailed reasoning.Y543/01 Mark Scheme October 2021 Question Answer Marks AO Guidance 1 Initial Elastic PE = 24 0.92 2 0.6 × = × B1 1.1 Use of 2 x2l λ with attempt at finding extension (ie not just x = 1.5) 16.2 J Final Elastic PE = 24 0.42 2 0.6 × = × B1 1.1 Use of 2 x2l λ with attempt at finding extension (ie not just x = 1) 3.2 J Increase in PE = 0.4g×2.5 M1 1.1 Attempt at use of “mgh” to find the increase of gravitational PE from initial position to ceiling 9.8 J “16.2” = “3.2” + ½×0.4v2 + “9.8” M1 1.1 Attempt at conservation of energy with consideration of KE and their PE 8.624 J v2 = 16 => speed is 4 m s–1 A1 1.1 Not ±. Units required. [5]Y543/01 Mark Scheme October 2021 Question Answer Marks AO Guidance 2 (a) I = mv – mu = 2(–3i + j – (5i + 16j)) M1 1.1 Correct use of formula (award if mu – mv) or using the cosine rule on vectors u,v,I to reach |I| = 34 = 2(–8i – 15j) A1 1.1 Allow 16i + 30j I = − + − 2 ( 8) ( 15) 2 2 M1 1.1 or ( 16) ( 30) − + − 2 2 oe = = 2 289 34 A1 1.1 16 1 cos 34 1 θ = =− × × I.i I i M1 1.1 Attempting to use the dot product of I and i to find the required angle or use of ordinary trigonometry eg 30 tan 16 θ =− − 1 8 cos 118.1 17 θ = = ° − − or 2.06 rad A1 1.1 [6] 2 (b) Init KE = × × + 12 2 5 16 ( ) 2 2 M1 1.1 281 J Final KE = × × − + 12 2 ( 3) 1 ( ) 2 2 M1 1.1 10 J Loss = 281 – 10 = 271 J A1 1.1 [3]Y543/01 Mark Scheme October 2021 Question Answer Marks AO Guidance 3 (a) [F] = MLT–2 B1 1.1 [ ][ ][ ] [ ] 2 2 d ML T 2 MLT d L v m v v mv x x −   −   = = =   B1 2.1 Correctly finding the dimensions of both sides is sufficient for B1B1; an explicit conclusion is not necessary. [2] 3 (b) Only quantities with the same dimensions can be added (or subtracted) [so [a2] = [x2] which means that [a] = [x]] B1 2.4 [1] 3 (c) [ ] ( ) 1 1 k M L LT − 2 2 1 2 = − M1 2.2a Use of formula for v to derive dimensional equation for [k] 12 1 [ ] M T k = − A1 1.1 Alternative solution 12 12 2 2 2 2 vm v km a x k a x − = − ⇒ = − so the units of k are 12 kg s−1 M1 Use of formula for v to derive units of k. 12 1 [ ] M T k = − A1 [2] 3 (d) 1 1 d 1 2 2 2 2 ( 2 ) ( ) d 2 v km x a x x − − = − − M1 1.1 Use of chain rule to differentiate v wrt x 1 1 d 2 2 2 2 ( ) d v km x a x x − − = − − 1 1 1 1 2 2 2 2 2 2 2 2 dd 1 ( ) ( 2 ) ( ) 2 v F mv x m km a x km x a x − − − ∴ = = × − − − M1 1.1 Use of formula for F with m, v and their d d vx substituted in. ∴ = − F k x 2 A1 1.1 [3]Y543/01 Mark Scheme October 2021 Question Answer Marks AO Guidance 4 (a) KE of P 1 2 2 = mv B1 1.2 SSU – change C to R if a better reflection of candidate solutions  C mg sinθ = M1 3.3 Balancing forces in the vertical. C must be resolved In this solution, C is the normal contact force between P and the cone and θ is the semi-vertical angle of the cone ↔ = C ma cosθ M1 3.3 NII in the horizontal using a resolved component of C 2 cos sin a v g rg θ θ = = M1 3.4 Eliminating C (and m) between the two equations and using a correct form for a May see v gh 2 = here and used later PE of P (exceeds that of Q by) 2 cos 2 tan sin r r v mgh mg mg mg mv g θ θ θ = = = = soi M1 3.4 Using the relationship to find the (excess) PE of P in terms of m and v (and possibly g) only h is the vertical height of P above Q So total ME of P exceeds that of Q by 2 2 2 1 3 2 2 = + = mv mv mv J A1 2.2a AG. Or total ME of Q = 0 but some justification of excess for PE at least must be seen in the solution Use R instead of C? [6] 4 (b) One of: - We have assumed that the radius of the circle which P moves in is the same as the radius of the cone at that level - Q is at V [neither of which is quite true if P and Q do not have a negligible radius] B1 3.5b Also accept e.g. - CofM of P lies on the edge of the cone - CofM of Q lies at V V is the vertex of the cone [1] 4 (c) Resistance to the motion of P should be included in the model. B1 3.5c eg air resistance. Allow friction. [1]Y543/01 Mark Scheme October 2021 Question Answer Marks AO Guidance 5 (a) 2 2 2 1 ( 1) d d 3 ( 1) d d 3( 1) F t k v v k F ma t t t t ∝ + ∴ = = = ⇒ = + + B1 3.1b AG [1] 5 (b) 2 1 d 3 (1 ) 3(1 ) k k v x u t t − ∴ = = + + + ⌠⌡ M1 3.1b Separating variables correctly and integrating to 1 C + t ; award if “+ u” missing May use + c instead of u t = 0, v = 0 ⇒ k = 3u M1 3.1b Substituting initial values to determine a relationship between k and u. NB The units of k are N s2 or kg m but these are not required. t = 1, v = 2 ⇒ 2 3(1 1) k u − = + + M1 3.1b Substituting t = 1 to determine a second relationship between k and u oe. 4 4, 12 4 1 u k v t ⇒ = = ⇒ = − + oe A1 1.1 eg 4 1 t v t = + [4] 5 (c) d 4 4 4 4ln(1 ) d 1 x x t t c t t = − ⇒ = − + + + M1 1.1 For integrating their ‘v’ to reach an expression involving � ln(1 + �) oe Can be awarded even if no “+ c” t x c x t t = = ⇒ = = − + + 0, 1 1 so 4 4ln(1 ) 1 A1 1.1 [2] 5 (d) 95% of vT = × = 0.95 4 3.8 B1 2.2a 4 3.8 3.8 4 1 v t = ⇒ = − + M1 3.1b Setting their v to their 3.8 in the appropriate equation 4 0.2 1 20 19 1 t t t ⇒ = ⇒ + = ⇒ = + A1 1.1Y543/01 Mark Scheme October 2021 Question Answer Marks AO Guidance so x = × − + + 4 19 4ln(1 19) 1 M1 1.1 Substituting their t into the equation for x x = − 77 4ln 20 so distance moved is 76 4ln 20 − m or awrt 64 m A1 1.1 [5]Y543/01 Mark Scheme October 2021 Question Answer Marks AO Guidance 6 (a) 20 = 4u ⇒ u = 5 B1 1.1 Initial energy = 12 × × 4 52 B1 1.1 = 50 Assuming zero PE level at initial level of P Energy at θ θ = × × + × − 12 4 4 0.8(1 cos ) v g 2 M1 1.1 Attempt to derive total ME at general or specific angle 2 2 2 15.68 50 17.16 v v + = ⇒ = A1 1.1 Equating energies to derive a value for v2 v = 4.142... Radial: 2 17.16 r 0.8 0.8 v a = = M1 3.1b Correct form for centripetal acceleration and use of v2 21.45 ar = Tangential: sin ma mg t 3 π = − M1 3.1b NII for tangential direction with weight resolved (– not necessary) 3 8.4870... t 2 g a = − = − a = − + =       2 20 3 429 g 2    )2 23.067... so the magnitude of the acceleration is 23.1 m s–2 (3 sf) A1 1.1 [7] 6 (b) Radial: 2 4 4 cos 0.8 v T g − = θ M1 2.1 NII for radial direction. T could be set to 0. Correct form of ar . v g 2 2 = − × − 5 2 0.8(1 cos ) θ M1 2.1 v2 in terms of cosθ from conservation of energy 2 v = + 9.32 15.68cosθ 7.84cos 9.32 15.68cos 9.32 cos 23.52 113.3 or 1.98 rads θ θ θ θ − = + ∴ = − ∴ = ° A1 3.2a [3]Y543/01 Mark Scheme October 2021 Question Answer Marks AO Guidance 7 (a) u u Ax Bx = = − 3, 2 B1 3.3 Resolving horizontal components of uA and uB. Accept �� = 5 cos � and �� = −4 cos � 3 but must have opposite signs or directions indicated on diagram. Signs may be reversed throughout 3 2 0 m m m v m A B × + × − = + × A Ax B M1 3.4 Conservation of momentum May be seen in (b) 2 3 B Ax m A v m = − A1 1.1 0 3 2 vAx e − = − − or 5 v e Ax = − M1 3.4 Restitution 0 3 ( ) 2 2 3 3 2 5 5 B A B A m m m e m − − = = − − − 2 3 3 0 0 5 5 2 B B A A m m e m m ≥ ⇒ − ≥ ⇒ ≥ A1 2.1 AG 2 3 1 1 4 5 5 B B A A m m e m m ≤ ⇒ − ≤ ⇒ ≤ A1 2.1 AG [6] 7 (b) Total initial KE = 1 1 2 5 6 4 73 2 2 2 2 × × + × × = B1 1.1 v u v u Ay Ay By By = = = , 2 3 M1 3.4 Perpendicular components found and unchanged 3 vAx = − M1 3.4 Using their formula for vAx from (a). NB If method mark for conservation of momentum not seen in (a) then award M1 in (a) if either 3 2 m m m v A B × + × − = A Ax or 2 3 6 2 2 × + × − = vAx seen here If method mark for restitution not seen in (a) then award M1 in (a) if seen here. ‘ KE Loss = 73 2 (3 4 ) 6 2 3 12 J − × × + + × × = ( 1 1 2 2 2 2 ( )2) A1 1.1 [4]Y543/01 Mark Scheme October 2021 Question Answer Marks AO Guidance 8 (a) 12a M x m Ma mx 12 x M m M m × + × + = = + + B1 1.1 AG. www [1] 8 (b) 3 3 a M y m Ma my y M m M m × + × + = = + + B1 1.1 [1] 8 (c) If P is at O, x 12Ma M m = + and y 3Ma M m = + B1ft 3.3 FT their expression for �� Alternative: B1 for correct expressions for �̅, �� M1: forming 2 inequalities with 2a and 6a (must be right way around) M1: simplifying or manipulating both inequalities so that they can be combined or compared A1: fully correct and conclusion www 12 y a M M m m M < ⇒ < + ⇒ > 2 3 2 2 M1 3.4 x a M M m m M < ⇒ < + ⇒ > 6 12 6 6 M1 3.4 Conclusion: 1 m M > 2 A1 2.4 AG. [4] 8 (d) 12 12 Ma m ak x M m + × = + used B1 3.3 12 12 Ma m ak 6a M m + × = + M1 3.4 Their x equated to 6a Ignore working with y 2 m M k m − = oe A1 1.1 k = − 12( ) 1 Mm Ignore working with affects final answer y unless this [3] 8 (e) 3 1 m M k = ⇒ = 2 6 OC B1 3.3 kOC = = 18 3 0.16 3 3 6 2 32 Ma M ak y M M + × = + M1 3.4 Substituting y = 6ak and 32 m M = into their yY543/01 Mark Scheme October 2021 Question Answer Marks AO Guidance 6 18 2 2 2 5 9 OA a ak y a = ⇒ = ⇒ = + a k A1 3.4 4 0.2 kOA = = 18  (k changes from 1 to 0 and k k OA OC > so) lamina topples over edge OA. A1 2.2a www [4]OCR (Oxford Cambridge and RSA Examinations) The Triangle Building Shaftesbury Road Cambridge CB2 8EA OCR Customer Contact Centre [Show More]

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