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4 a) If h = 1 _ 2 g t2 ⇒ t2 = 2 h ___ g = __ 2 g × h A graph of t2 against h therefore should be a straight line through the origin of gradient equal to 2 __ g . b) h/cm 40 60 80 100 120 t/s 0.30 0.38 0.42 0.47 0.51 t 2 /s 2 0.090 0.144 0.176 0.221 0.260 Your graph should be as in Figure A.1. t 2 /s 2 0.30 0.25 0.20 0.15 0.10 0.05 0 0 Figure A.1 � 0.20 0.40 0.60 h/m 0.80 1.00 1.20 Full Answers to Review Questions: 2 A guide to practical work Table A.4 � c) The graph is a straight line but does not pass through the origin. This suggests a small systematic error. There has possibly been a systematic error in measuring the distance, h (perhaps caused by measuring to the wrong place each time) or, more likely, there has been a systematic error in t due to time delays in releasing the sphere or opening the trap door. d) Gradient = 0.250 s2 − 0.015 s2 _______________ 2 –1 = 0.206 s m 1.14 m − 0.00 m 2 __ g = 0.206 ⇒ g = 2 _____ = 9.7 m s–2 0.206 e) Percentage difference = (9.8 – 9.7) m s–2 _____________ 9.8 m s –2 × 100% = 1% f) i) The values of h have been recorded only to the nearest centimetre when they would have, presumably, been set to a precision of 1 mm. The values should therefore have been recorded as 40.0 cm, 60.0 cm, etc. to reflect this. ii) Repeat timings should definitely have been recorded, with at least three values for each height. More readings, with larger values of h if possible, would also improve the results. Edexcel Physics for AS © Hodder Education 2009 4
3 Rectilinear motion 1 a) a = b) s = v − u _____ t v + u _____ 2 (40 − 0) m s−1 = ____________ 80 s = 0.50 m s−2 – the answer is B. (40 + 0) × t = _______ 2 m s−1 × 80 s = 1600 m – the answer is B. 2 The gradient of a displacement–time graph represents the velocity – the answer is D. 3 The area under a velocity–time graph represents the displacement – the answer is B. total displacement 4 Average velocity = ________________ time Instantaneous velocity = δx ___ at an instant δt 5 a) Acceleration is the change of velocity in unit time: a 5 Dv ___ Dt b) i) Average acceleration ≈ 3 m s−2 ii) Gear changes are likely to affect the acceleration. 6 s = 1 _ 2 (u + v)t v = u + at s = ut + 1 _ 2 at 2 v 2 = u 2 + 2as 7 a) v = 12 m s −1 b) s = 160 m 8 b) For a height of about 2 m, the time is about 0.6 s. The % uncertainty in h is (±0.001 m)/(2.000 m) × 100% = 0.05%; for t, the value is (0.01 s)/(0.60 s) × 100% = 1.7%. Therefore, time has the greater effect on the uncertainty in g. 9 a) s = u t + 1 _ 2 at2 = 0 + 1 _ 2 × 9.8 m s –2 × (2.2 s) 2 = 24 m b) v = u + at = 0 + 9.8 m s −2 × 2.2 s = 22 m s −1 10 a) 7.3 m b) 1.2 s c) 7.6 m s −1 (downward) 11 a) Vertical component = 10 m s −1 Horizontal component = 17 m s −1 b) t = 2.0 s c) x = 36 m 12 a) A = stationary; B = uniform velocity; C = uniform velocity in the opposite direction to B; D = increasing velocity (acceleration) b) A = constant velocity; B = uniform acceleration; C = uniform deceleration; D = increasing acceleration. 13 a) 20.2 m s −2 b) 1800 m c) 10 m s −1 Full Answers to Review Questions: 3 Rectilinear motion Edexcel Physics for AS © Hodder Education 2009 5
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