Mathematics > A/As Level Mark Scheme > > GCE Further Mathematics B (MEI) Y431/01: Mechanics minor Advanced GCE Mark Scheme for Autumn 2021 (All)
> GCE Further Mathematics B (MEI) Y431/01: Mechanics minor Advanced GCE Mark Scheme for Autumn 2021
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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) [Show More]
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