GCE Further Mathematics B (MEI) Y421/01: Mechanics major Advanced GCE Mark Scheme for Autumn 2021

Document Content and Description Below

Oxford Cambridge and RSA Examinations GCE Further Mathematics B (MEI) Y421/01: Mechanics major 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 2021Y421/01 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.Y421/01 Mark Scheme October 2021 Question Answer Marks AOs Guidance 1 J = 0.25(4.2 − (−5)) M1 3.3 Use of Impulse = change in momentum J = 0.02F M1 3.3 Use of Impulse = Ft F = 2.3 = 115 (N) 0.02 A1 1.1 cao [3] 2 10mx = 1(3m) + 2(5m) + 5(2m) M1 1.1 Use of x∑ mi = ∑ ximi x = 2.3 A1 1.1 cao 10my = 2(3m) + (−2)(5m) + 3(2m) M1 1.1 Use of y∑ mi = ∑ yimi y = 0.2 A1 1.1 cao [4] 3 (a) T = 4g B1 1.1 Resolve vertically (possibly implied by subsequent working) λ (0.02) = 4g 0.3 M1 3.3 Use of Hooke’s law with their 4g λ = 588(N) A1 1.1 cao oe e.g. 60g [3] 3 (b) e.g. spring stretched beyond its elastic limit e.g. Hooke’s law no longer applies B1 2.2b oe (any correct equivalent statement for why the extension of the spring may not be 0.1 m) [1]Y421/01 Mark Scheme October 2021 Question Answer Marks AOs Guidance 4 DR A = ∫ 1(4 − x 2) − 3 x dx = 4x − 1 x3 − 2x 23  1 0   3  0 M1* 2.1 Correct integral expression for the area and attempt to integrate (at least two terms correct) Ignore limits for first two M marks A = 4 − 1 − 2 = 5 3 3 A1 1.1 SC M1 A0 if correct integral and value seen but with no intermediate working 1 3 3  2 1 4 6 5 1 Ax = ∫0 4x − x − 3x 2 dx = 2x − x − x 2    4 5  0 M1* 1.1 Correct integral expression for Ax and attempt to integrate (at least two terms correct) Ax = 2 − 1 − 6 = 11 4 5 20 A1 1.1 SC M1 A0 if correct integral and value seen but with no intermediate working x = Ax = 11 20 A 5 3 M1dep* 1.1 Correct use of x = Ax A Dependent on both previous M marks = 33 A1 2.2a oe This mark can be awarded even if the two previous A marks were not awarded 100 [6]Y421/01 Mark Scheme October 2021 Question Answer Marks AOs Guidance 5 Let wA and wB be the horizontal components of the velocity of A and B after collision For reference: wA = 1 wB = 2.5 B1 1.2 M1 3.3 Use of conservation of linear momentum (parallel to the line of centres) – correct number of terms 2(6) + 4(0) = 2wA + 4(2.5) A1 1.1 Allow with wB instead of 2.5 M1 3.3 Use of Newton’s experimental law (parallel to the line of centres) – correct number of terms wA − 2.5 = −e(6 − 0) A1 1.1 Use of NEL must be consistent with CLM – allow with wB instead of 2.5 and possibly their wA e = 0.25 A1 1.1 [6]Y421/01 Mark Scheme October 2021 Question Answer Marks AOs Guidance 6 (a) [F] = MLT−2 B1 1.2 [1] 6 (b) [G] = M−1L3T−2 B1 May use F = Gm1m2 to obtain the d 2 dimensions of G [1] 6 (c) G =(6.67 ×10−11 )× 0.454 × 1 (0.305)3 M1 3.1a SC B1 for G =(6.67 ×10−11 )× 1 ×(0.305)3 0.454 = 4.17×10−12 G =1.07×10−9 (lb–1 ft3 s–2) A1 1.1 awrt 1.07×10−9 [2] 6 (d)  kGM  (M −1L3T−2 )M  r  = L M1 2.1 Attempt to calculate the dimension of either kGM or its square root with r [k ] = 1 and two other terms correct  kGM   = LT−1 r    A1 1.1 Or kGM  = L2T−2  r  [v] = LT−1 so the formula is dimensionally consistent A1 2.2a Or allow showing consistency for v2 = kGM r [3]Y421/01 Mark Scheme October 2021 Question Answer Marks AOs Guidance 6 (e) k (6.67 ×10−11 )(5.97 ×1024 ) 11186 = 6 371000 M1 3.4 Allow to 3 sf or better (allow 5015 to 5017 inclusive) k ≈ 2 A1 1.1 k = 2.0019677... 2(6.67 ×10−11 )(6.39 ×1023 ) v = 3 389 500 M1 1.1 v = 5015 (m s-1) A1 2.2a If using k = 2.0019677... expect to see 5017.346122… [4] 7 (a) Driving force of engine is kmg v B1 1.1 kmg − mg = mv dv v dx M1 3.3 Use of N2L, correct number of terms, allow D (oe) for kmg and a (oe) for the v acceleration AG – sufficient working must be shown kg − gv = v2 dv ⇒ v2 dv = (k − v) g dx dx A1 2.2a [3] as answer givenY421/01 Mark Scheme October 2021 Question Answer Marks AOs Guidance 7 (b) gx = k 2 ln   k  − kv − 1 v2 k −v  2   x = 0,v = 0 ⇒ g(0) = k 2 ln  k  − k (0) − 1 (0)2 so k − 0  2   initial conditions are consistent with given equation   g dx = k 2  1  k( k − v)−2   − k − v dv  k     k − v    dx −kv + v2 − k 2 + kv + k 2 g = dv (k − v) v2 = g (k − v) dx ⇒ v2 dv = (k − v) g dv dx B1 M1* A1 M1dep* A1 [5] 1.1 2.1 1.1 1.1 2.2a Attempt to differentiate using chain rule cao oe e.g.  −k − dv   2  k − v   dx   dv dv g = k  k  (k− v)2  − k dx − v dx   Correct method to obtain an expression for dx as a single fraction or as a single dv fraction with dv dx  k 2 − k 2 + kv − kv + v2  dv e.g. g =   k − v   dx AG – sufficient working required as answer given Or equivalent (e.g. solving using separation of variables)Y421/01 Mark Scheme October 2021 Question Answer Marks AOs Guidance 7 (c) Work done by engine is kmgt kgmt = 1 mV 2 + mgx 2 kgt = 1 V 2 + k 2 ln  k  − kV − 1 V 2 2  k −V  2   kgt = k 2 ln k  − kV ⇒ t = k ln k  − V  k − V  g  k − V  g B1 M1* M1dep* A1 [4] 1.1 3.3 3.4 2.2a Use work-energy principle – correct number of terms Use given result from (b) in work-energy equation to eliminate x AG – sufficient working required as answer given SC if correctly found by solving kmg − mg = m dv this can score 3/4 max. v dt 8 (a) B1 [1] 1.2 All remaining forces adding on correctly (with arrows to indicate directions) to the figure in the Printed Answer Booklet 8 (b) FD + R C =W R D = FC F = 1 R and F = 1 R D 3 D C 3 C 1 F + R = W ⇒ 1 R + R = W 3 C C 9 C C R = 9 W C 10 M1* A1 B1 M1dep* A1 [5] 3.3 1.1 3.4 3.4 1.1 Resolve horizontally and vertically (correct number of terms in both equations) Where RC is the normal contact force at C, etc. Correct use of F = µ R at C and D Combine results to get an equation in RC onlyY421/01 Mark Scheme October 2021 Question Answer Marks AOs Guidance 8 (c) M1* 3.1b Taking moments about D (or any other equivalent point) – correct number of terms (r + hsinθ)W + (r + 2h cosθ) FC = (r + 2hsinθ)RC A1 1.1 oe (r + hsinθ )W +(r + 2h cosθ ) 3 W    10   = (r + 2h sinθ ) 9 W    10   M1dep* 3.4 Substitute expressions for FC and RC r = h(2sinθ −1.5cosθ ) A1 1.1 2hsinθ −1.5hcosθ > 0 M1 2.3 Setting their expression for r > 0 4sinθ − 3cosθ > 0 ⇒ tanθ > 3 4 A1 2.2a AG [6]Y421/01 Mark Scheme October 2021 Question Answer Marks AOs Guidance 9 (a) x = −g sinα , y = −g cosα B1 2.1 M1* 3.4 Attempt to integrate (twice) and use of initial conditions x = 5cosθ − gtsinα , y = 5sinθ − gt cosα A1 1.1 x = 5t cosθ − 0.5gt2 sinα y = 5t sinθ − 0.5gt2 cosα A1 1.1 Or M1 for use of s = ut + 1 at 2 parallel 2 to line of greatest slope and then A1 for correct expression for x Similarly M1 A1 for correct expression for y (following SUVAT perpendicular to slope) y = 0 ⇒ t = ... M1dep* 3.3 Sets y = 0 and solve for t t = 10sinθ g cosα A1 1.1  10sinθ   10sinθ 2 x = 5 g cosα cosθ − 0.5g  g cosα  sinα     M1 3.4 Substitute expression for t into equation for x Dependent on both previous M marks x = 50sinθ (cosθ cosα − sinθ sinα ) g cos2 α ⇒ OR = 50sinθ cos(θ + α ) g cos2 α A1 2.2a AG [8]Y421/01 Mark Scheme October 2021 Question Answer Marks AOs Guidance 9 (b) sinθ cos(θ + α ) = 1 (sin (2θ + α ) − sinα ) 2 M1 1.1 Use of given identity to re-write numerator from (a) as a difference of two sines R max occurs when sin(2θ + α ) =1 OR = 25 (sin(2θ + α ) − sinα ) g cos2 α A1 1.1 R = 25 (1 − sinα ) A1 3.1a Use of correct trig. identity and setting sin(2θ + α ) equal to 1 – oe e.g. R = 25 max g (1 + sinα ) max g (1 − sin 2α) [3] 9 (c) 25 = 1.8 or 25(1− sinα ) = 1.8 g (1 + sinα ) g(1 − sin2 α ) M1* 3.4 Setting their expression equal to 1.8 Expression must only contain sinα terms 25 = 1.8 ⇒ sinα = ... g (1 + sinα ) M1dep* 1.1 Attempting to solve for sinα or α - for reference sinα = 184 or α = 24.660053... 441 (or 0.430399… in radians) If solving a 3TQ in sine then must solve using a correct method θ = 45 − 0.5α M1 3.1a Follow through their α θ = 32.7 A1 1.1 32.6699733… or 0.5701986… (in radians) [4]Y421/01 Mark Scheme October 2021 Question Answer Marks AOs Guidance 10 (a) [At B,] KE = 1 mu2 , PE = 0 2 B1 1.1 Note that the reference level for zero GPE might be taken at C [ At θ , ] KE = 1 mv2 , PE = mga (1− cosθ ) 2 B1 1.1 M1* 3.3 Use of conservation of energy – correct number of terms 1 mu2 = 1 mv2 + mga (1 − cosθ ) 2 2 A1 1.1 cao mv2 R − mg cosθ = a M1* 3.3 N2L radially with correct number of terms and weight resolved R − mg cosθ = m(u2 − 2ga (1 − cosθ )) a M1dep* 3.4 Substitute an expression for v 2  u2  R = m 3g cosθ − 2g + a  A1 1.1 [7]Y421/01 Mark Scheme October 2021 Question Answer Marks AOs Guidance 10 (b) Before collision at C, 1 mu2 = 1 mv2 + mga 2 2 After collision at C, speed of P is e u2 − 2ga 1 B ( 2 ) mv 2 = mga + 1 m e u − 2ga 2 2 2 vB2 = 2ga + e2 (u2 − 2ga) 1 mv 2 − 1 mv 2 = Fb 2 B 2 A m(2ga + e2 (u2 − 2ga))− 2bF ≥ 0 Fb ≤ mga + 1 me2u2 − me2ga 2 ⇒ Fb ≤ 1 m e2u2 + 2(1 − e2 )ga, so k = 2 2   M1 A1 M1 M1 M1 A1 [6] 3.4 1.1 3.1b 3.1b 2.5 2.2a Substituting θ = π into their 2 conservation of energy equation from (a) Conservation of energy to find an expression for the speed of P at B Work-energy principle for motion between B and A Set vA ≥ 0 and substitute for vB2 k need not be stated explicitly Where vB is the speed of P at B 11 (a) 4V = 4vA + 3vB vA − vB = −eV v = A V (4 − 3e) and v = 4V (1+ e) 7 B 7 M1* A1 M1* A1 M1dep* A1 [6] 3.3 1.1 3.3 1.1 1.1 1.1 Conservation of linear momentum with correct number of terms cao Newton’s experimental law with correct number of terms Must be consistent with CLM Solve the simultaneous equations to find both speeds Where vA is the speed of A after 1st impact and similarly for vBY421/01 Mark Scheme October 2021 Question Answer Marks AOs Guidance 11 (b) Let θ be the angle subtended by A in time t Where r is the radius of the circular groove For A, t = rθ V (4 − 3e) 7 M1 3.1b Use of s = ut with their vA and s = rθ For B, t = 2π r + rθ 4V (1+ e) 7 M1 1.1 Use of s = ut with their vB and s = 2π r + rθ 2π + θ = θ 4V (1 + e) V (4 − 3e) M1 3.4 Equate expressions for t to form an equation in terms of θ ,V and e θ =2π (4 − 3e) 7e A1 2.2a AG [4] Alternative method Where r is the radius of the circular groove ALT: vB− v A = 4V (1+ e) − V (4 − 3e) = eV 7 7 M1* Difference in speeds calculated Time for B to catch up to A is 2π r eV M1dep* Using their eV 2π r  V (4 − 3e)  2π r dA =   = (4 − 3e) eV  7  7e M1 Where d A is the distance travelled by A θ =2π r (4 − 3e) = 2π (4 − 3e) 7er 7e A1 AGY421/01 Mark Scheme October 2021 Question Answer Marks AOs Guidance 11 (c) (i) 3w + 4w = 12V (1+ e) + 4V (4 − 3e) B A 7 7 M1* 3.3 CLM correct number of terms using their expressions from (a) Where wA is the speed of A after the second collision w − w = −e  4 V (1+ e) − 1V (4 − 3e) B A   7 7  M1* 3.3 NEL correct number of terms 3wB + 4wA = 4V and wB − wA = −e2V A1 1.1 oe M1dep* 1.1 Solve simultaneously for wB w = 4 B V (1− e2) 7 A1 1.1 cao For reference: w = 1 A V (4 + 3e2 ) 7 [5] 11 (c) (ii) If the collision is perfectly elastic (e = 1) B is brought to rest by the second collision and A is moving with speed V (which is the situation before the first collision) B1 3.5a oe correct statement [1] 12 (a) PE = −mg (l + e) (while P is at rest) B1 1.1 Where e is the extension in the string Taking the horizontal through O as the reference level for zero GPE 12mge2 EPE = 2l B1 1.1 6mge2 − mg(l + e )= 0 l M1* 3.3 Conservation of energy with correct number of terms 6e2 − el − l2 = 0 (3e + l)(2e − l ) = 0 M1dep* 1.1a Solving three-term quadratic in e e = l ⇒ length of string is 1 l + l = 3 l 2 2 2 A1 2.2a AG [5]Y421/01 Mark Scheme October 2021 Question Answer Marks AOs Guidance 12 (b) mg − T = mx M1 3.3 N2L vertically with correct number of terms mg − 12mgx = mx l M1 3.4 Use of Hooke’s law and substitute for T in N2L x + 12g x = g so x + ω2 x = g where ω2 = 12g l l A1 2.2a AG [3] 12 (c) x = y + g ⇒ y + ω 2 y = 0 ω2 M1 1.1 Use given substitution to form differential equation in y Dependent on all previous M marks y = Acosωt + B sinωt A1ft 1.2 Correctly solves their differential equation in y x = Acosωt + B sin ωt + g 2ω A1 1.1 oe e.g. x = Acosωt + B sin ωt + l 12 t = 0, x = 0 ⇒ A = − g ω 2 M1 3.4 Use correct initial conditions in their expression for x 1 mv 2 = mgl 2 P M1* 3.1b Use conservation of energy to find speed vP of P at time t = 0 vP = 2gl A1 1.1 t = 0, x = 2gl ⇒ B = 2gl ω M1dep* 3.4 Use initial speed in an expression for x x = − g cosωt + 2gl sin ωt + g ω2 ω ω2 A1 1.1 oe e.g. x = l (1− cosωt + 24 sin ωt) 12 12 l (1− cosωt + 24 sin ωt ) = 0 M1 3.1b Sets x = 0 and replaces ω2 = 12lg cosωt − 24 sinωt =1 so k = 24 A1 2.2a k need not be stated explicitly [10]OCR (Oxford Cambridge and RSA Examinations) The Triangle Building Shaftesbury Road Cambridge CB2 8EA [Show More]

Last updated: 2 years ago

Preview 1 out of 19 pages