GCE Further Mathematics B (MEI) Y431/01: Mechanics minor Advanced GCE Mark Scheme for Autumn 2021

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Oxford Cambridge and RSA Examinations GCE Further Mathematics B (MEI) Y431/01: Mechanics minor 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 2021Y431/01 Mark Scheme October 2021 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.Y431/01 Paper Mark Scheme October 2021 Question Answer Marks AOs Guidance 1 (a) MLT−2 B1 1.2 [1] (b) [ ] RHS M LT ( ) 1 2 L − = M1 3.4 Using given formula with [m] = M and [r] = L and [v] = LT–1 = ��2�−2�−1 = ���−2 = [���] A1 2.2a must see (��−1)2 expanded [2] (c) 1.1 0.9 2 ÷ M1 1.1 Using correct formula with 1.1 and 0.9 =1.34444 so 34.4% A1 2.2b 34.44444… [2] (d) 1016 1609 60 × ÷ 2 M1 1.1 Condone denominator which is not squared e.g 60 �� 603 for the M mark = 454 N A1 1.1 454.0955… [2]Y431/01 Mark Scheme October 2021 Question Answer Marks AOs Guidance 2 Let the components of the force at A be �� (parallel to floor) and �� (parallel to wall). �� = 20 B1 1.1 �� = �702 − 202 = √4500 = 30√5 M1 3.1b Using Pythagoras to find the normal contact force at A ⟹ � = �� = 30√5or 67 A1 1.1 Accept exact or to at least 2 sf 67.082039… � = �� �� = 20 30√5 = 2 3√5 = 2√5 15 or 0.30 B1 3.4 Using F R = µ - accept any equivalent exact form or 0.30 (2 sf or better) 0.29814… 2√5 15 e.g. Taking moments about A: �� cos � = ��2� sin � M1* 3.3 Taking moments about A (or B etc.) – correct number of terms. Allow cos/sin errors but must reflect ratio of distances. 34 ⇒ = tan 5 θ M1dep* 1.1 Substituting �� = 20 and their value for W and then obtain a value for tan ⟹ � = 59° A1 1.1 2 sf or better 59.19301… [7]Y431/01 Paper Mark Scheme October 2021 Question Answer Marks AOs Guidance 3 P gv = 65 B1 3.1b Use of P Fv = (either one) P g v = + 40 3 ( ) B1 1.1 65 40 3 gv g v = + ( ) M1 3.4 Equating their two expressions for P – with at least one correct equation � = 4.8 ��−1 A1 1.1 � = 3060 � A1 1.1 or 3.06 kW (but must state kW in this case) 3057.6 [5]Y431/01 Mark Scheme October 2021 Question Answer Marks AOs Guidance 4 (a) 1 2 � × 7.22 = ��ℎ M1 3.4 Must use energy method as in question. directed h = 2.64489 so maximum height is 4.24 m A1 1.1 4.2448979… [2] (b) Let the amount of work done per metre against air resistance be W 1 2 � × 7.22 − 2.5� = 2.5�� M1 3.1b Work-energy principle. All three terms present. Condone sign errors. W m = 0.568 A1 1.1 AG – sufficient working must be shown [2] (c) Let v be the speed of the ball just before impact with ground 1 2 4.1 4.1 mg W mv − = 2 M1 B1 3.3 1.1 Work-energy principle: all three terms present (or all four if starting from when ball leaves the hand). 4.1W with W m = 0.568 12 2 mv m v = ⇒ = 37.8512 8.70ms-1 A1 1.1 8.7007126… [3] (d) Let V be the speed of the ball just after impact with ground. 1 2 ��2 − 2.8� = 2.8�� M1 3.3 Work-energy principle - all three terms present V = 7.6197 A1 1.1 Coefficient of restitution = 7.6197 8.7007 = 0.876 A1ft 3.4 FT their answer to (c) 0.87576318… [3] (e) − + = mv mV 12 M1 3.3 Using impulse = change in momentum. Correct number of terms but allow sign errors. FT their values for v and V 12 0.735 m = = v V + A1 1.1 [2]Y431/01 Paper Mark Scheme October 2021 Question Answer Marks AOs Guidance 5 (a) Let the coordinates of centre of mass be ( ) x y z , , z =1.5 B1 1.1 210 �� ��̅� = 120 �2 5� + 90 �8.5 9 � M1 1.1 Any correct equation ratio of masses of the constituent , using correct parts. e.g. could also have 2 7 28 42 3.5 8.5         +     x = 5 A1 1.1 y = 6.5 A1 1.1 [4] (b) 1 θmin = arctan 6.5 or θmax = arctan 6.5 5 M1 3.1b tan 1 y θ = or tan x y θ = - condone reciprocal fractions for this mark θmin = 8.75 A1 1.1 8.746162… � ��� = 37.6 A1 1.1 37.568592… [3] (c) Let the thresholds for breaking equilibrium be sliding and toppling be Ps and Pt sin30 s n30 i R P + s s ° = ⇒ = − mg R mg P ° M1* 3.3 Resolve vertically – correct number of terms (allow sin/cos errors) Ps cos30° − ° = = Fmax µ (mg Ps sin30 ) M1* 3.4 Resolve horizontally and use of F R = µ Ps = ° + µmg µ scos30 in 30° A1 1.1 oe 14 sin30 cos30 10 5 Pt t ° + ° = P mg M1* 3.1b Moment – correct number of terms. Condone sin/cos errors (and sign errors) 14sin30 cos30 5 t 10 mg P = ° + ° A1 1.1 oeY431/01 Mark Scheme October 2021 Question Answer Marks AOs Guidance P P s t > , so 5 cos30 sin30 14sin30 co10 s30 µmg mg µ > ° + ° ° + ° µ µ mg (14sin30 cos10 5 ° + ° > ° + ° 30 cos30 ) ( mg sin30 ) µ (9sin30 cos30 cos3 ° + ° > ° 10 ) 05 M1dep* 2.1 Dependent on all previous M marks � > 5 9 tan30° + 10 � > 5 3√3 + 10 �� ���� = 5 3√3 + 10 = 50 − 15√3 73 A1 2.2a Accept any equivalent exact form, or 0.329 (or better) Need not be explicitly stated if inequality is present. 0.3290306… [7] (d) Either If the angle.�, were smaller then tan� would be smaller … M1 2.1 … so µmin would be larger. A1 2.2a Or If the angle were smaller then P would have a larger horizontal component and a smaller anticlockwise turning effect … M1 2.1 … so µmin would be larger. A1 2.2a [2]Y431/01 Paper Mark Scheme October 2021 Question Answer Marks AOs Guidance 6 (a) Let the block move x metres before coming to rest. 12 2 mv Fx − = 0 M1 1.1 Work-energy principle – correct number of terms Since block is sliding, F = = F m max µ g M1 3.4 Use of F R = µ 2 12 2 0 v2 v mgx g m µ x µ − = ⇒ = A1 1.1 N.B. answer given. Alternative method: Since block is sliding, F = = F m max µ g M1 3.4 µ = − = −µ mg a g m M1 1.1 So 02 = �2 + 2 (−��) � ⟹ � = �2 2�� A1 1.1 [3] (b) mu mv mv = + S 8 B M1* 3.3 Conservation of linear momentum – correct number of terms (allow sign errors) v v u B S − = 0.8 M1* 3.3 Newton’s experimental law – must be consistent with CoLM (so signs of �� in the two equations must be different) � = � � + 8 (0.8� + ��) = 9�� + 6.4� M1dep* 3.4 Attempt at eliminating either variable – dependent on both previous M marks ⇒ S = −0.6uv and v u B = 0.2 A1 1.1 Ignore incorrect signs. S has speed 0.6u towards the wall B has speed 0.2u away from the wall A1 2.4 Both correct. Accept other appropriate descriptions of direction (e.g. ‘opposite to original direction of travel’, etc.) [5]Y431/01 Mark Scheme October 2021 Question Answer Marks AOs Guidance (c) Each time S returns for impact it has 3 5 the speed it had previously; therefore after impact, the block will have also have 3 5 the speed it had just after the previous impact … M1 3.5a Argument using their value of vS from (b) – must relate this value to B … so by part (a), the block will move only ( ) 53 2 = 25 9 of the distance moved after the previous impact. A1 2.2a Must reference result in part (a) (or convincingly explain where the squaring comes from) May consider that the ratio of successive distances travelled by B is v2 [2] (d) After first impact, speed of block is 2× = 0.211. 24 2. B1ft 3.4 Follow through their value of vB from (b) So 17 ( ) 2 1 2.24 1.792 2 9.8 x = = ⋅ ⋅ M1 1.1 Using given result in (a) to find distance travelled after first collision 9 1 25 1.792 2.8 1 n xn ∞ = = = − ∑ (m) A1 2.2a AG derive required result Use of infinite sum of a GP to [3]OCR (Oxford Cambridge and RSA Examinations) The Triangle Building Shaftesbury Road Cambridge CB2 8EA OCR Customer Contact Centre [Show More]

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