Orked W Solutions

WORKED SOLUTIONS VISIT… Worked solutions Chapter 1 Volume = πr2 × l Exercises = π × (12.7 × 10–6)2 × 20 × 10–2 = 1.0 × 10–10 m 3 1 (a) 48 000, move decimal point 4 places left (b) Radius of the Earth = 6.378 × 106 m 4 4.8 × 10 (from text) 4 (b) 0.000 036, move decimal point 5 places Volume of sphere = πr 3 3 right 3.6 × 10–5 = 4π × (6.378 × 106)3 = 1.09 × 1021 m 3 (c) 14 500, move decimal point 4 places left 3 m 4 7 ρ = ⇒ m = ρV 1.45 × 10 V (d) 0.000 000 48, move decimal point 7 places Density of air, ρ = 1.2 kg m–3 right 4.8 × 10–7 V = 5 × 10 × 3 = 150 m3 m = 1.2 × 150 = 180 kg 2 (a) 5585 km = 5.585 × 106 m (b) 175 cm = 1.75 m 8 30 cm = 30 × 10–2 m –2 (c) 25.4 μm = 2.54 × 10–5 m 15 cm = 15 × 10 m (d) 100,000 million, million, million km = 105 × 106 × 106 × 106 km = 1022 km 10 cm = 10 × 10–2 m = 1025 m Volume = 30 × 15 × 10 × 10–6 m 3 3 (a) 85 years = 4500 × 10–6 m 3 = 85 × 365 (days in a year) = 4.5 × 10–3 m 2 × 24 (hours in a day) × 60 (min in an hour) m Density, ρ = ⇒ m = ρV × 60 (seconds in a min) V = 1.93 × 104 × 4.5 × 10–3 = 2.68 × 109 s = 86.85 kg (b) 2.5 ms = 2.5 × 10–3 s (c) 4 days = 4 × 24 × 60 × 60 = 3.46 × 105 s 9 From question 6, volume of Earth = 1.09 × 1021 m 3 (d) 2 hours 52 min 59 s From text, mass of Earth = 5.97 × 1024 kg = 2 × 60 × 60 + 52 × 60 + 59 24 Density = M = 5.97 × 10 = 5.48 × 103 kg m–3 = 7200 + 3120 + 59 V 1.09 × 1021 = 10 379 = 1.04 × 104 s 10 The points on the fi rst graph are not far spread 4 (a) 200 g = 0.2 kg from the line so the apples seem to be of almost equal size. On closer inspection the 4th point (b) 0.000 01 g = 1 × 10–5 g = 1 × 10–8 kg seems to be plotted in the wrong position since (c) 2 tonne = 2000 kg 1 it is indicating the mass of 32 apples. 5 Volume = 5 × 10 × 3 = 150 m3 The points on the second graph are more spread out so the apples appear to be slightly 6 (a) Hair is cylindrical more uneven than the fi rst case. On closer inspection the difference between 2 apples and 25.4 mm 3 apples is very small. This is either due to an unusually small apple or a mistake. 20 cm = 20 × 10–2 m 25.4 r = = 12.7 μm = 12.7 × 10–6 m 2 1 M13_IBPH_SB_IBGLB_9021_SL01.indd 1 22/07/2014 16:38 The apples in the last graph are all of about 13 d = 400 ± 1 m the same size but according to the graph one total distance is 4 × 400 = 1600 m apple has no size. This could be because of a total uncertainty is 4 × 1 = 4 m systematic error in counting the apples (unlikely) distance is 1600 ± 4 m or because the balance reading was always 14 T = 11.2 ± 0.1 s about 200 g too low. It is also strange that the time for one swing is 11.2 = 1.12 s measurements start at two apples; why didn’t 10 0.1 the experimenter measure the mass of one uncertainty for one swing is = 0.01 s 10 apple fi rst? Probably some mistake has been time for one swing = 1.12 ± 0.01 s made transferring the data. – Detective work! 15 (a) 11 Taking natural logs of A and B: x A 2.5 log 55° 3 cm 2.0 3 cos θ = x 3 x = = 5.2 cm cos 55 1.5 (b) 50° 1 4 cm 0.5 x x tan 50° = 4 0 x = 4 tan 50 = 4.8 cm –1 –0.5 0 0.5 (c) log B 6 cm Equation of line log A = 2 logB + 0.6 x A = 100.6 B2 A = 4B2 30° x sin θ = 12 L = 0.050 ± 0.001 m 6 % uncertainty = 0.001 = 2% x = 6 sin 30 = 3 cm 0.05 (d) m = 1.132 ± 0.002 kg x % uncertainty = 0.18% 3 cm 20° density = m = 1.132 = 9056 kg m–3 V (0.05)3 sin θ = 3 total % uncertainty = 0.18 + 3 × 2 = 6.18% x 3 6.18 × 9056 = 560 kg m–3 x = = 6 cm 100 sin 30° density = 9056 ± 560 kg m–3 2 M13_IBPH_SB_IBGLB_9021_SL01.indd 2 22/07/2014 16:38 16 (a) 19 Component north = 10 cos 30° N (because it’s next to angle) x 3 cm = 8.66 km 10 km 30° 4 cm W E x = 42 + 32 = 25 20 Component south = 8 cos 20° N x = 5 cm (because it’s next to angle) W E (b) = 7.52 km 8 km 20° x 4 cm S 21 Vertical component = 500 sin 60° 4 cm (not next to angle) m 2 2 = 433 m x = 4 + 4 = 32 500 x = 5.66 cm 60° (c) x 2 cm Practice questions 6 cm 2 2 x = 2 + 6 = 40 Note: You don’t need to know about capacitors to x = 6.32 cm answer this question. (d) 1 (a) Use a convenient scale. 200 x 3 cm (nC) Q 175 2 cm 150 x = 32 + 22 = 13 x = 3.61 cm 125 175 nC 17 By Pythagoras 100 R = 42 + 82 = 80 8 km R R = 8.94 km 75 8 tan θ = ⇒ θ = 63.4° θ 4 4 km 50 18 By Pythagoras 50 km 25 R = 1002 + 502 = 112 km 40 V 50 tan θ = ⇒ θ = 26.6° 0 100 100 km R 0 102030405060 θ V (V) Note: The error bar on 30 nC is too small to plot accurately. 3 M13_IBPH_SB_IBGLB_9021_SL01.indd 3 22/07/2014 16:38 (b) Uncertainty in Q is 10% so error bar for 6 F = 10.0 ± 0.2 N 1 0.2 30 nC is ± 3 nC (about 2 division) and for % uncertainty = × 100% = 2% 180 nC ± 18 nC (about 4 divisions) 10 m = 2.0 ± 0.1 kg (c) Gradient of steepest line shown = 175 40 % uncertainty = 0.1 × 100% = 5% = 4.4 nC V–1 2 (d) Units of capacitance = C V–1 % uncertainty in F = 2 + 5 = 7% m ε A Q ε A Q (e) Q = 0 V ⇒ = 0 where When two numbers are divided their % d V d V is the gradient uncertainties add. gradient × d The answer is C. so = ε A 0 π × 8.1 7 = 6.38 –9 –3 15.9 ⇒ 4.4 × 10 × 0.51 × 10 0.15 To one signifi cant fi gure, this is 6. = 1.5 × 10–11 CV–1 m –1 The answer is D. 2 If distance is between 49.8 cm and 50.2 cm 8 then the mean value = 49.8 + 50.2 = 50 cm 2 V The spread of the measurement y = 50.2 – 49.8 = 0.4 cm 0.4 So the uncertainty is ± = 0.2 cm Vx 2 Measurement is 50 ± 0.2 cm Velocity of x relative to y is found by subtracting The answer is C. the velocity of y from x. m This is the same as vector x + (– vector y). 3 T = 2π k Squaring, T 2 = 4π2 m k ⇒ T 2 ∝ m So T 2 plotted against m will be a straight line. The answer is A. 2 4 Power = I R uncertainty in I = 2% The answer is B. so uncertainty in I 2 = 2 × 2% = 4% uncertainty in R = 10% 9 The mass of an apple is about 100 g so uncertainty in I 2R = 4 + 10 = 14% So its weight = 0.1 × 10 = 1 N The answer is C. This would be a small apple. The answer is C. 5 A zero offset error means that when no current passes through the ammeter the reading is 10 (a) It isn’t possible to draw a straight line that not zero. touches all error bars and passes through If instead of zero the meter reads 0.1 A then all the origin. measurements will be 0.1 A too big. The precision of each reading will not be altered but the readings will not be accurate. Like a football player hitting the post every time – precise but not accurate. The answer is C. 4 M13_IBPH_SB_IBGLB_9021_SL01.indd 4 22/07/2014 16:38 (max – min) (0.26 – 0.19) (b) Uncertainty = = 2 2 1.8 2 = 0.04 cm (cm) D 1.6 So gradient = 0.23 ± 0.04 cm2 1.4 (iv) The unit of the constant is cm2. 1.2 1.0 0.8 Challenge yourself 0.6 1 v = d = 0.05 = 0.83 ms–1 0.4 t 0.06 2 Rearrange the equation: g = 7v 0.2 10h 2 2 0 g = v × 7 = 0.83 × 7 = 8.10 m s–2 0123456789101112 10 × h 10 × 0.06 n Percentage uncertainty in d = 0.5 × 100 = 4% ( 5 ) (c) D = Cnp Percentage uncertainty in t = 0.01 × 100 = 17% (0.06) log10 D = log10C + plog10n Percentage uncertainty in v = 4 + 17 = 21% Plotting log10 D vs log10 n will give a straight line with gradient p (y = mx + c) Percentage uncertainty in h = 0.2 × 100 = 3% ( 6 ) (d) (i) From the error bars, the uncertainty in 2 Percentage uncertainty in g = v × 7 D for n = 7 is ± 0.08 10 × h % uncertainty = 0.08 × 100% = 6.3% = 2 × 21 + 3 = 45% ( 1.26 ) 45 –2 2 Absolute uncertainty = × 8.10 = 3.62 m s uncertainty in D = 2 × 6.3 = 13% (100) (ii) The straight line passes through all the Final value = 8 ± 4 m s–2 error bars shown and the origin.

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