GCE Further Mathematics A Y533/01: Mechanics Advanced Subsidiary GCE Mark Scheme for Autumn 2021

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Oxford Cambridge and RSA Examinations GCE Further Mathematics A Y533/01: Mechanics Advanced Subsidiary 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 2021Y533/01 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.Y533/01 Mark Scheme October 2021 Question Answer Marks AO Guidance 1 (a) ω = 2π / 0.84 soi M1 1.1 Correct formula for angular velocity used awrt 7.48 rad s–1 A1 1.1 �50 21 �� [2] 1 (b) v = 2.8× “7.48…” or 2π×2.8 / 0.84 M1 1.1 Correct formula for speed used FT their value for ω if used awrt 20.9 m s–1 A1 1.1 �20 3 �� [2] 1 (c) a = “20.9…”2 / 2.8 or 2.8× “7.48…”2 or “20.9…” × “7.48…” M1 1.1 Any correct formula for acceleration used FT their value for v if used awrt 157 (or 156) ms –2 A1 1.1 156 if rounded values used. �1000 63 � 2� [2] 1 (d) ...towards O B1 1.2 Any indication that the acceleration is towards the centre of the circle [1]Y533/01 Mark Scheme October 2021 Question Answer Marks AO Guidance 2 (a) D = 15000 / 20 = 750 B1 3.4 “P = Fv” used in the solution D – R = 800×0.4 M1 3.3 Use of NII with a driving force (might be incorrectly derived from power), R and correct ma term. R = 750 – 320 = 430 A1 1.1 AG [3] 2 (b) Need 15000 / vmax = “430” M1 3.4 Driving force = resistive force and “P = Fv” vmax = 34.9 so max speed is 34.9 ms–1 (3 sf) A1 1.1 [2] 2 (c) D – R – 800g × sinα = 800 × 0.15 (= 15000 / v – 60v – 1568 = 120) M1 3.1b NII with a driving force, R, a component of weight (condone incorrect component) and correct ma term. 60v2 + 1688v – 15000 = 0 M1 3.1a Reduction to 3 term quadratic equation (must be equation) 7.10 or –35.2 A1 1.1 BC (condone 7.09 from incorrect rounding for this mark) Both roots must be seen for this mark Since v > 0, speed is 7.10 ms–1 (3 sf) A1FT 2.3 FT their quadratic, if one positive and one negative root (ie if ac < 0) for selecting their positive root with valid reason given. SC1 if A0A0 for 7.10 ms–1 with no justification [4] Question Answer Marks AO Guidance 3 (a) Cons of Momentum: 0.5 × 3.15 = 0.5vA + 0.8 × 2vA M1 1.1 Or 0.5 × 3.15 = 0.5 × ½vB + 0.8 × vB vA = 0.75 A1 1.1 vB = 1.5 So vB = 2vA = 1.5 A1 1.1 vA = ½vB = 0.75 [3]Y533/01 Mark Scheme October 2021 3 (b) � = (±) "1.5" − "0.75" 3.15 − 0 M1 1.1 Speed of separation over speed of approach. Using their values from 3(a) provided c.o.m. used (and in subsequent questions) 5 21 or awrt 0.238 A1 1.1 [2] 3 (c) Because e is the ratio of two speeds (in ms–1) (the units cancel and so) it is a dimensionless quantity. B1 2.4 oe [1] 3 (d) Initial KE = ½ × 0.5 × 3.152 M1 1.1 3969 1600 = 2.48... Correct KE calc Or change/gain of KE of B = 0.8 × “1.5”2 Final KE = ½ × 0.5 × “0.75”2 + ½ × 0.8 × “1.5”2 M1 1.1 333 320 = 1.04... KE calculation with correct m and their u and 2u Change/loss of KE of A = ±½ × 0.5 × “0.75”2 ∓ ½ × 0.5 ×3.152 KE Loss = 2.48... – 1.04... = 1.44 J A1 1.1 FT their speeds if positive. 36 25 = 1.44 2.34 – 0.9 = 1.44J NB Must be positive value for the amount lost [3] 3 (e) Not perfectly elastic since KE is lost oe B1 2.4 eg e ≠ 1 oe (but just e = 0.238... is insufficient) [1] 3 (f) Change in B’s momentum = 0.8 × “1.5” M1 1.1 Using impulse = change in momentum (condone sign error) Or by finding the change in A’s momentum: 0.5 × 0.75 – 0.5 × 3.15 (±)1.2 Ns or kgms–1 A1 1.1 Impulse on B (Hence impulse B exerts on A is (±)1.2 Ns) = (±)1.2 Ns in the opposite direction to A’s original direction of motion A1 1.1 This statement oe needed for full marks in the opposite direction to A’s original motion [3]Y533/01 Mark Scheme October 2021 Question Answer Marks AO Guidance 4 (a) (i) Gain in KE = ½ × 4.2 × 4.52 (J) M1 1.1 Correct formula for KE used. Can be implied by awrt 42.5 Work done by force = 35 × 2.4 (J) M1 1.1 Correct formula for WD by force used. Can be implied by awrt 84.0 Do not allow the assumption that the resistance is constant, e.g. by use of suvat, also in part (ii) Energy lost = 84.0 – 42.5 = awrt 41.5 J A1 1.1 SC2 if using suvat to find correct average resistance and hence total energy lost. [3] 4 (a) (ii) R = 41.5 / 2.4 M1 3.1b Their energy loss divided by 2.4 So average resistive force is awrt 17.3 N A1 1.1 SC1 only for 17.3N, if using suvat/N2L [2] 4 (b) (i) Other resistive forces (eg air resistance) can be ignored. B1 3.3 “No friction” is not a valid answer here [1] 4 (b) (ii) Need ½ × 4.2 × 4.52 = 4.2gh M1 2.2b Equating KE with PE (4.2 may be missing on both sides). If “resistive force” term included then M0 unless recovered. h = 1.033... A1 1.1 Distance = 1.033 / sin20° = awrt 3.02 m A1 1.1 Alternative method: a = –gsin20° M1 Correctly deducing the acceleration up the slope. 02 = 4.52 + 2×–gsin20°×s M1 Using a suvat equation, or equations, which lead(s) to s from a and u given with v = 0 and consistent signs Distance = awrt 3.02 m A1 [3]Y533/01 Mark Scheme October 2021 Question Answer Marks AO Guidance 5 (a) [r] = L, [m] = M and [U] = LT–1 B1 2.1 Correct dimensions for other parameters (U, r and m) soi (no need for them to be used for this mark to be awarded). [�] = ���2�� M1 1.1 Comparing dimensions, realising that 2 is dimensionless and rearranging Could be done by dimensional analysis e.g. [�] = ������ and equate indices using� = �2���oe ∴ [G] = (LT–1)2LM–1 = L3M –1T –2 A1 2.2a AG [3] 5 (b) [�] = (MLT-2�)/T = ML2�−3 B1 3.3 Using P = WD/t oe Need LT−1 = ���2��−3������−2��� B1 3.3 Realising condition for equation to be dimensionally correct and substituting in dimensions. ft errors in [P] and/or [W] here and in subsequent method marks provided M, L and T appear at least twice on the RHS M: α + β = 0, L: 1 = 2α + β M1 3.4 Comparing to obtain equations in α and β α = 1, β = –1 A1 1.1 T: –1 = –3α – 2β + γ M1 3.4 Comparing to obtain equation in γ γ = 0 A1 1.1 [6] 5 (c) Because γ = 0, the modelled minimum launch speed V does not depend on the time t for which the engines operate... B1ft 3.5a ie the modified model predicts that V does not vary when t varies Or appropriate comment from their result, e.g. if γ = -1, then V is inversely proportional to t [1]Y533/01 Mark Scheme October 2021 Question Answer Marks AO Guidance 6 (a) I = mu => u = I/m B1 3.1b Use of Impulse = change of momentum Init PE = ��� − ������ � 3 M1 1.1 (= ½mgr). Attempt to use ‘mgh’ to find initial PE. Could use eg edge as zero PE level (so init PE = –½mgr) but must be clear and signs consistent 1 2 ��2 + 1 2 ��� = 1 2 ��2 + ��� M1 1.1 Conservation of energy; KE & PE considered on both sides �2 = �2 − �� ⟹ � = ���22 − �� A1 1.1 oe e.g. � = ��2−��2�� Alternative method u = I/m B1 ��� = ���cos � 3 (= 1 2 ���) M1 1 2 ��2 = 1 2 ��2 − 1 2 ��� M1 Subtract gain in PE �2 = �2 − �� ⟹ � = ���22 − �� A1 [4] (b) 1 2 ��2 = ��ℎ ⟹ ℎ = 1 2� ���22 − ��� = 2��2 �2 − � 2 B1 1.1 oe e.g. ℎ = �22−��22��� [1] (c) Consider the case where h → 0 M1 3.1b e.g. �2 �2 = �� maximum possible value of m is � √�� A1 3.5b [2] (d) Work done against R = � �� 2 + � 3� � M1 3.4 1 2 ��2 + 1 2 ��� = 1 2 ��2 + ��� + � �� 2 + � 3� � or �2 2� + 1 2 ��� = 1 2 ��2 + ���+ � �� 2 + � 3� � Or 1 2 ��2 = 1 2 ��2 − 1 2 ��� − � �� 2 + � 3�� M1 3.4 Revising the energy equation (condone incorrect initial energy from (a)) to include an energy loss term (work done against R). Could already be in terms of I rather than u. Could be expressed as an inequality at this stage: eg 1 2 ��2 + 1 2 ��� > ���+ � �� 2 + � 3 � � Need v > 0 so � > ��2��+ 5��� 3 � A1 1.1 AG [3]Y533/01 Mark Scheme October 2021 (e) [I]=MLT-1 And [RHS]= (M2LT-2L+ MLMLT-2)1/2 M1 1.1 Attempt dimensional analysis on both sides. Hence [RHS]=MLT-1=[I] so the inequality is dimensionally consistent A1 2.2a [2]OCR (Oxford Cambridge and RSA Examinations) The Triangle Building Shaftesbury Road Cambridge CB2 8EA OCR Customer Contact Centre [Show More]

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