Mathematics > MARK SCHEME > OCR GCE Further Mathematics A Y533/01: Mechanics Advanced Subsidiary GCE Mark Scheme for Autumn 2021 (All)
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 [Show More]
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