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

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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 [Show More]

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