Mathematics > MARK SCHEMES > OCR GCE Further Mathematics A Y543/01: Mechanics Advanced GCE Mark Scheme for Autumn 2021 (All)
OCR GCE Further Mathematics A Y543/01: Mechanics Advanced GCE Mark Scheme for Autumn 2021
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Mark Scheme October 2021 Question Answer Marks AO Guidance 2 (a) I = mv – mu = 2(–3i + j – (5i + 16j)) M1 1.1 Correct use of formula (award if mu – mv) or using the cosine rule on vectors ... u,v,I to reach |I| = 34 = 2(–8i – 15j) A1 1.1 Allow 16i + 30j I = − + − 2 ( 8) ( 15) 2 2 M1 1.1 or ( 16) ( 30) − + − 2 2 oe = = 2 289 34 A1 1.1 16 1 cos 34 1 θ = =− × × I.i I i M1 1.1 Attempting to use the dot product of I and i to find the required angle or use of ordinary trigonometry eg 30 tan 16 θ =− − 1 8 cos 118.1 17 θ = = ° − − or 2.06 rad A1 1.1 [6] 2 (b) Init KE = × × + 12 2 5 16 ( ) 2 2 M1 1.1 281 J Final KE = × × − + 12 2 ( 3) 1 ( ) 2 2 M1 1.1 10 J Loss = 281 – 10 = 271 J A1 1.1 [3]Y543/01 Mark Scheme October 2021 Question Answer Marks AO Guidance 3 (a) [F] = MLT–2 B1 1.1 [ ][ ][ ] [ ] 2 2 d ML T 2 MLT d L v m v v mv x x − − = = = B1 2.1 Correctly finding the dimensions of both sides is sufficient for B1B1; an explicit conclusion is not necessary. [2] 3 (b) Only quantities with the same dimensions can be added (or subtracted) [so [a2] = [x2] which means that [a] = [x]] B1 2.4 [1] 3 (c) [ ] ( ) 1 1 k M L LT − 2 2 1 2 = − M1 2.2a Use of formula for v to derive dimensional equation for [k] 12 1 [ ] M T k = − A1 1.1 Alternative solution 12 12 2 2 2 2 vm v km a x k a x − = − ⇒ = − so the units of k are 12 kg s−1 M1 Use of formula for v to derive units of k. 12 1 [ ] M T k = − A1 [2] 3 (d) 1 1 d 1 2 2 2 2 ( 2 ) ( ) d 2 v km x a x x − − = − − M1 1.1 Use of chain rule to differentiate v wrt x 1 1 d 2 2 2 2 ( ) d v km x a x x − − = − − 1 1 1 1 2 2 2 2 2 2 2 2 dd 1 ( ) ( 2 ) ( ) 2 v F mv x m km a x km x a x − − − ∴ = = × − − − M1 1.1 Use of formula for F with m, v and their d d vx substituted in. ∴ = − F k x 2 A1 1.1 [3]Y543/01 Mark Scheme October 2021 Question Answer Marks AO Guidance 4 (a) KE of P 1 2 2 = mv B1 1.2 SSU – change C to R if a better reflection of candidate solutions C mg sinθ = M1 3.3 Balancing forces in the vertical. C must be resolved In this solution, C is the normal contact force between P and the cone and θ is the se [Show More]
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