English Essay

1076 WordsMar 1, 20155 Pages
General Certificate of Secondary Education 2011 Additional Mathematics Paper 2 Mechanics and Statistics [G0302] THURSDAY 19 MAY, MORNING MARK SCHEME 6483.01 1 y AVAILABLE MARKS PN x (i) Resolve vertically: ⇒ 10 cos a = 6 Resolve vertically: Resolve vertically: ⇒ a = 53.1º ⇒ cos α = 6 = 0.6 10 MW1 W1 MW1 W1 M1 M1 M1 M1 4 4 (ii) Resolve horizontally Resolve vertically: ⇒ P = 10 sin 53.1º Resolve vertically: ⇒ P = 10  0.8 Resolve vertically: ⇒ P = 8 2 (i) suitable example for population for subset of a population (ii) situation involving cost/time . . . (iii) situation involving destruction . . . 3 (i) P B W1 (26g, 34g) W1 (15g, R) (ii) ⇒ R = 26g + 15g + 34g ⇒ R = 75gN = 750N (iii) Let x = distance AP Take moments about A ⇒ 750x = 150  3 + 340  6 ⇒ x = 3.32 ⇒ 26x + 15x +AP = 3.32m MW1 M1 W1 W1 6 6483.01 2 [Turn over 4 (i) f.d. = 18/3 f.d. = 6 (ii) f.d. = 2/4 f.d. = 1/2 : 7 = f/2 8 = f/2 5 = f/5 f = 14 f = 16 f = 25 MW1 for 6 MW1 for 14, 16 MW1 for 25 M1W1 calc W1 plot AVAILABLE MARKS 10 8 Frequency density 6 4 2 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 Number of months 6 5 (i) Using s = ut + 1 2 at 2 1 2 at 2 ⇒ – 16i + 24j = MW1 ⇒ – 16i + 24j = 8a ⇒ – q6i + 24a = (–16i + 24j)/8 ⇒ – q6i + 24a = (–2i + 3j) m/s2 (ii) Using v = u + at ⇒ – q6i + 24v = 0 + (–2i + 3j)  4 ⇒ – q6i + 24v = (–8i + 12j) m/s (iii) Using F = ma ⇒ – q6i + 24F = 3(–2i + 3j) ⇒ – q6i + 24F = (–6i + 9j)N ⇒ – q6i + 24F = P + Q ⇒ – q6i + 24–6i + 9j = (10i – 5j) + Q ⇒– Q = (–6i + 9j) – (10i – 5j) ⇒– Q = (–16i + 14j)N W1 MW1 W1 MW1 MW1 W1 7 6483.01 3 6 (a) Total­number­of­students­=­389 ­ ­ ­ ­ ­ ­ 195 − 109 × 10 = 27 median = 20 + 122 (or­194.5­instead­of­195) ­ MW1­(20) MW1­(195) MW1­(other­values)

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