Cambridge A Level Mathematics 9709 — 2012 Oct/Nov Paper 5 · Variant 3

1 A 30° O 0.6 m B F N A circular object is formed from a uniform semicircular lamina of weight 12 N and a uniform semicircular arc of weight 8 N. The lamina and the arc both have centre O and radius 0.6 m and are joined at the ends of their common diameter AB. The object is freely pivoted to a fixed point at A with AB inclined at 30◦to the vertical. The object is in equilibrium acted on by a horizontal force of magnitude F N applied at the lowest point of the object, and acting in the plane of the object (see diagram). (i) Show that the centre of mass of the object is at O. [3] (ii) Calculate F. [3]

Mark scheme: v = 6.32 ms − 1 A1 [2] (v = 40 ) (ii) 60e/2 = 60(2–e)/2 ± 0.6g M1 Attempt to find equilibrium position Upper ext = 1.1, Lower ext = 0.9 A1 Distance from A = 3.1 m A1 0.6g × 1.1 + 60(2 2 – 0.9 2 )/4 M1 Energy balance, descent from A. cv 2 A1 upper and lower ext = 60 × 1.1 /4 + KE KE = 36.3 J A1 [6] OR KE –0.6(6.32) 2 /2 = 60 × 2 2 /4 M1 Energy balance, descent from A. cv 2 2 A1ft upper and lower ext, answer (i) –60 × 1.1 /4 – 60 × 0.9 /4 – 0.6g × 0.9 KE = 36.3 J A1 GCE A LEVEL – October/November 2012 9709 53 3 (i) t = 2/(25cos70) (= 0.234) B1 y = (25sin70) × 0.234 – g × 0.234 2 /2 M1 y = 5.22 A1 [3] OR y = xtan70 – gx 2 /2(25cos70) 2 B1 y = 2tan70 – g2 2 /2(25cos70) 2 M1 y = 5.22 A1 s = ut + gt 2 /2 Award if seen in (i) (ii) 1.2 = (25sin70)t –gt 2 /2 B1 5t 2 – 23.5t + 1.2 = 0 M1 Solves 3 term quadratic for larger root t = 4.65 s A1 [3] (iii) R = 15 2 sin2α /10 = 20 M1 Or solves (15sinα )t–5t 2 = 0 and 20=(15cosα )t for α α = 31.4 o A1 [2] 4 (i) (0.9/2)/r = tan45 M1 r = 0.45 m A1 [2] (ii) M1 Take moments about A (π 0.9 2 × 0.9+π 0.45 2 h)OG =π 0.9 2 × 0.9(h + 0.45) +π 0.45 2 h × A1 cv(0.45) h/2

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