Physics (AS & A Level) · May / June 2016 · Paper 22 · Question 1
9702
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1(a)[1]
acceleration = change in velocity / time (taken) or rate of change of velocity [1]
1(b)(i)[2]
v = 0 + at or v = at (a = 36 / 19 =) 1.9 (1.8947) m s–2 [2]
1(b)(ii)[1]
s = ½(u + v)t or s = v2 / 2a or s = ½at2 = ½ × 36 × 19 = 362 / (2 × 1.89) = ½ × 1.89 × 192 = 340 m (342 m / 343 m / 341 m) [1]
1(b)(iii)[4]
1. (∆KE =) ½ × 95 × (36) 2 = 62 000 (61 560) J [2]
2. (∆PE =) 95 × 9.81 × 340 sin 40° or 95 × 9.81 × 218.5 = 200 000 J [2]
1(b)(iv)[2]
work done (by frictional force) = ∆PE – ∆KE or work done = 200 000 – 62 000 (values from 1b(iii) 1. and 2.) (frictional force = 138 000 / 340 =) 410 (406) N [420 N if full figures used] [2]
1(b)(v)[2]
–ma = mg sin 20° – f or ma = –mg sin 20° + f –95 × 3.0 = 95 × 3.36 – f f = 600 (604) N [2]
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