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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) (b) 0.000 036, move decimal point 5 places Volume of sphere = 4πr 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) Density, ρ = m ⇒ 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 r = 25.4 = 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 cos θ = 3 x x = 3 = 5.2 cm cos 55 1.5 (b) 50°

1 4 cm

0.5 x tan 50° = x 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° sin θ = x 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

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 tan θ = 8 ⇒ θ = 63.4° θ 4 4 km 50

18 By Pythagoras 50 km 25 R = 1002 + 502 = 112 km 40 V tan θ = 50 ⇒ 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. The answer is C. so gradient × d = 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. This is the same as vector x + (– vector y). 3 T = 2π m 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.

M13_IBPH_SB_IBGLB_9021_SL01.indd 4 22/07/2014 16:38 (b) Uncertainty = (max – min) = (0.26 – 0.19) 2 2 1.8 = 0.04 cm2 (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. The biggest uncertainty is in the measurement (iii) of time; this could be improved by repeating the measurement several times and taking )

2 3.5 the average. (cm 2 3.0 D

2.5

2.0 2.3 1.5

1.0 10 0.5

0 0123456789101112 n

From graph, gradient of best ﬁ t line = 2.3 = 0.23 cm2 10 gradient of steepest line = (3.1 – 0.2) 11 = 0.26 cm2 gradient of least steep line = (2.4 – 0.3) 11 = 0.19 cm2 5

M13_IBPH_SB_IBGLB_9021_SL01.indd 5 22/07/2014 16:38 Worked solutions

Chapter 2 4 N Exercises

100 km 100 000 1 (a) → = 27.8 m s–1 1 hour 60 × 60 θ 20 km 20 000 (b) → = 5.6 m s–1 –1 1 hour 60 × 60 4 m s 2 (a) speed = distance = 400 = 4.2 m s–1 time 96 W E (b) total displacement = 0 m –1 1 m s average velocity = 0 m s–1 (c) After 48 s runner will be half way around, S travelling south. The speed is constant, relative velocity = 42 + 1 = 4.1 ms–1 so the magnitude of the velocity will be tan = 1 4.2 m s–1. θ 4 = 14° instantaneous velocity = 4.2 m s–1 (minus θ

sign indicates travelling south) –2 5 5 m s 1 (d) After 24 s the runner will be 4 way round v? 6.35 m 100 m s = 100 m u = 0 m s–1 v = ? a = 5 m s–2 6.35 m t = – displacement = 63.52 + 63.52 Use v2 = u2 + 2 as = 90 m v2 = 2 as = 2 × 5 × 100 = 1000 –1 3 bird v = 1000 = 31.6 m s

–2 10 6 5 m s

20 m s v? 20 –1 200 m 10 m s θ –1 s = 200 m 20 m s u = 20 m s–1 v =? In this case we must subtract the velocity of a = 5 m s–2 the car t = – velocity = 202 + 102 = 22.4 m s–1 2 2 10 Use v = u + 2as tan θ = 20 v2 = 202 + 2 × 5 × 200 = 2400 θ = tan–1 0.5 v = 2400 = 49 m s–1 θ = 26.6°

M14_IBPH_SB_IBGLB_9021_SL02.indd 1 17/07/2014 16:25 –1 –2 –1 7 u? 10 m s 20 m s v = –20 m s a = –10 m s–2 t = 5 s t = ? s = – Use a = v – u t u = ? t = v – u v = 20 m s–1 a a = 10 m s–2 = –20 – 20 = 4 s –10 t = 5 s 11 a Use a = v – u u = v – at = 20 – 10 × 5 ⇒ –1 t 25 m s u = –30 m s–1 t = 10 s 8 t = 2 s v 25 –2 10 m s 20

–1 30 m s 15

s = ? 10 u = 30 m s–1 5 v = – 0 a = –10 m s–2 (acceleration is negative) 0 1 2 3 4 5 6 7 8 9 10 t = 2 s t 1 1 1 Use s = ut + at2 = 30 × 2 – 10 × 22 distance travelled = area = 2 × 10 × 25 = 125 m 2 2 25 = 60 – 20 = 40 m acceleration = gradient = = 2.5 m s–2 10 12 velocity/m s–1 9 areas cancel 10

–2 10 m s 6.5 cm 9 3 6 time/s

–10 v positive acceleration of 10 m s–2 for 3 s 3 s = 0.65 m 10 followed by negative acceleration of m s–2 u = 0 m s–1 3 for 6 s v = ? displacement = area under graph = 10 × 3 a = –10 m s–2 2 = 15 m t = – Use v2 = u2 + 2as 13 constant acceleration constant acceleration but lower than 1st part v2 = 0 – 2 × 10 × 0.65 = 13 constant velocity v = 13 = 3.6 m s–1 a , v

10 , s

–2 –1 10 m s 20 m s s

v s = 0 m u = 20 m s–1 a 2 t

M14_IBPH_SB_IBGLB_9021_SL02.indd 2 17/07/2014 16:25 14 16 displacement

negative acceleration no vertical motion time a , v ,

s time still moving downwards a but positive acceleration

velocity t The gradient of the displacement–time graph starts from zero and gets more negative until it v reaches a constant value.

–1 s 30 m s

15 displacement 60° A From vertical components C s = 0 u = 30 × sin 60° = 26 m s–1 v = –26 m s–1 B D time a = –10 m s–2 velocity t = ? Using a = v – u , t = v – u t a = –26 – 26 = 5.2 s A B C D time –10 From horizontal components s = vt = 30 × cos 60° × 5.2 = 78 m

A The gradient starts from zero and becomes –1 18 20 m s more negative as the ball falls. 5 m B The ball bounces and the velocity suddenly θ changes to a positive value with slightly Vertical motion less magnitude. s = 5 m As the ball rises it slows down until it stops at C. u = 20 × sin θ The velocity then becomes negative as it falls. v = 0 Note the gradient of all the diagonal parts is the a = –10 m s–2 same; this is because g is constant. t = – The velocity just after the ball leaves the ground (a) v 2 = u2 + 2as is the same as the velocity just before it hits 0 = u2 – 2 × 10 × 5 the ground. ⇒ u = 100 = 10 m s–1 1 ∴ 20 × sin θ = 10, sin θ = 2 , θ = 30°

M14_IBPH_SB_IBGLB_9021_SL02.indd 3 17/07/2014 16:25 v – u v – u (b) a = ⇒ time to reach wall t = Vertical components cancel, so resultant t a = 10 N to the right t = 0 – 10 = 1 s –10 (b) In this time, horizontal displacement = v t h 3 N 3 = 20 × cos 30° × 1 = 17.3 m θ 5 N 5 19 200 m Resultant = 52 + 32 = 5.8 N

–1 3 –1 y Angle θ = tan = 31° 200 m s 5

22 (a) 30° Using horizontal components 30° 40 N 40 N d d 200 40 N v = ⇒ t = = = 1 s t v 200 60 N 60 N Using vertical components F s = y F u = 0

v = – Horizontal forces cancel Convenient to take down as positive in this F = 40 N example. –1 (b) a = 10 m s 30° 30°

t = 1 s 40 N 40N 40N 1 2 1 2 s = ut + 2at ⇒ y = 0 × 1 + 2 × 1060 × N 1 60 N y = 5 m F F

–1 20 20 m s

Horizontal forces cancel Vertical components of upward force θ = 2 × 40 × cos 30° = 69 N

maximum distance is when θ = 45° 23 (a) 40 N F Using vertical components 50 N 20° 40 N s = 0 θ 60 N

–1 20 N u = 20 × sin 45° = 14.14 m s 20° –1 v = –20 × sin 45° = –14.14 m s 50 N a = –10 m s–2 t = ? Vertical components cancel: v – u v – u a = ⇒ t = 50 × sin 20° – 50 × sin 20° = 0 t a Horizontal components: = –14.14 – 14.14 = 2.8 s –10 20 – 2 × 50 × cos 20° = –74 N Using horizontal components (b) F Range = v × t = 14.14 × 2.8 = 39.6 m 50 N 40 N

20° 40 N θ 60 N 21 (a) 10 N 20 N 20°

50 N Horizontal components: 60 – 40 = 20 N 10 N Resultant = 402 + 202 = 45 N 40 Angle θ = tan–1 = 63.4° 20

10 N 4

M14_IBPH_SB_IBGLB_9021_SL02.indd 4 17/07/2014 16:25 20 N 24 (a) 8.66 N (b) F3 1 N 30°

30° 60 N 10 N Horizontally Parallel to slope: F = 20 × cos 30° Resultant = 10 × sin 30° – 1 2 F = 17.3 N = 5 – 1 = 4 N down slope. 2 Vertically Perpendicular to slope: F + 20 × sin 30° = 60 Resultant = 8.66 – 10 × cos 30° = 0 3 F = 60 – 10 = 50 N Resultant force = 4 N down the slope. 3 26 (b) 4 N 6 N 30° 30°

60° T 30°

4 N 10 N Vertical R = 4 × cos 30° + 6 × cos 60° – 4 = 3.46 + 3 – 4 = 2.46 N (up) (a) Since forces are balanced Horizontal R = 6 × sin 60 – 4 × sin 30° horizontal resultant = 0 = 5.20 – 2 = 3.2 N (right) ⇒ F = T sin 30° (b) Vertical resultant = 0 2.46 ⇒ 10 = T cos 30° 10 θ (c) T = = 11.5 N 3.2 cos 30° (d) F = 11.5 × sin 30° = 5.8 N Resultant = 3.22 + 2.462 = 4 N 2.46 Angle θ = tan–1 = 37.6° 27 N 3.2 F 25 (a) 6 N 90°

30° 45° 30° 45° 50 N 6 N

(a) Parallel to ramp F = 50 × sin 30° F1 (b) Perpendicular to ramp Taking components along the line of F 1 N = 50 × cos 30° 2 × 6 × cos 45° = F 1 (c) F = 25 N F = 8.49 N 1 N = 43.3 N 5

M14_IBPH_SB_IBGLB_9021_SL02.indd 5 17/07/2014 16:25 28 Resultant F: 1000 – F = 2.5 ⇒ F = 997.5 N 80° 80° Friction = 997.5 N T T knot 33 a T = ma = 10 × a 10 kg 600 N T (a) Vertical components 2T × cos 80° = 600 N T (b) Horizontal components a 50 – T = ma = 5a 5 kg T = 50 – 5a T × sin 80° = T × sin 80° 600 (c) T = = 1728 N 50 N 2 × cos 80° (a) Equating the equations for T: 29 0.2 kg –1 before 10 m s 10a = 50 – 5a 15a = 50 –1 –2 after 5 m s a = 3.3 m s momentum before = 0.2 × 10 = 2 N s (b) Using the equation at the top of the momentum after = 0.2 × –5 = –1 N s diagram impulse = change in momentum T = 10a = 10 × 3.3 = 33 N = final – initial 34 = –1 – 2 = –3 Ns

30 0.067 kg –1 10 m s

12 kN

–1 50 m s momentum before = 0.067 × 10 = 0.67 N s 1000 kg momentum after = 0.067 × –50 = –3.35 N s 10 kN impulse = change in momentum = –3.35 – 0.67 Resultant force = 2 kN F 2000 ⇒ a = = = 2 m s–2 = –4.02 N s m 1000

0.1 N 35 31 N

–2 0.5 m s

W = mg = 65 kg 0.006 × 10 = 0.06 N Upward force = 0.1 – 0.06 650 N = 0.04 N F F = ma = 65 × 0.5 = 32.5 N F = ma ⇒ a = m Resultant force = 32.5 N = 0.04 = 6.7 m s–2 ⇒ N – 650 = 32.5 0.006 N = 682.5 N 32 Acceleration = v – u = 0.1 – 0 = 0.05 m s–2 t 2

50 kg 1000 N F Resultant F = ma = 50 × 0.05 = 2.5 N 6

M14_IBPH_SB_IBGLB_9021_SL02.indd 6 17/07/2014 16:25 –1 –1 36 (b) 5 m s 5 m s U m m

–1 v 1 m s

3 m m 4000 cm Before collision momentum = –5 m + 5 m = 0 After collision momentum = –mv + m W = 2.5 N Conservation of momentum ⇒ 0 = –mv + m (a) U = weight of fluid displaced m = mv –1 volume displaced = 4000 × 10–6 m3 v = 1 m s (Could be + or – depending on mass displaced = ρU = 1000 × 0.004 which ball you take.) = 4 kg (c) Upthrust = 40 N 100 kg Resultant F = 40 – 2.5 = 37.5 N up. (b) a = F = 37.5 = 150 m s–2 m 0.25 2 kg Friction will act against the motion of the ball

37 Must throw the hammer away from the ship.

(a) Newton 1: Since the velocity of the gas changes there must be an unbalanced vm vn force on the gas. Newton 3: If the rocket exerts a force on the gas the gas must exert a force on the rocket. This force is unbalanced so the rocket accelerates. Must travel 2 m in 2 min

(b) Same as 37(a) but replace gas with water 2 –1 ⇒ vm = = 0.017 m s and rocket with boat. 120 momentum before = momentum after Skateboard replaces rocket, person (c) 0 = –100 × 0.017 + 2 × v replaces gas. h v = 1.7 = 0.85 m s–1 Sounds possible. (d) The ball accelerates up due to upthrust. h 2 If the water pushes ball up then ball pushes 39 F water down so the reading on balance 5 N will increase.

–1 38 (a) 10 m s m m –1 v 1 m s m t 0.35 s Before collision momentum = m × 10 + m × 0 (a) Impulse = area After collision momentum = mv + m × 1 1 = 2 × 5 × 0.35 Conservation of momentum ⇒ 10 m = mv + m = 0.875 N s ⇒ 9 m = mv v = 9 m s–1 7

M14_IBPH_SB_IBGLB_9021_SL02.indd 7 17/07/2014 16:25 (b) Impulse = change of momentum = mΔv (c) F mΔv = 0.875 Ns 8 N so if m = 0.02 kg 0.02 × Δv = 0.875 0.875 4 N Δv = = 44 m s–1 0.02 40 Impulse = area = 0.5 × 0.35 × 5 = 0.875 N s Change in velocity = Δmv = 0.875 = 43.75 m s–1 2 cm 4 cm x m 0.02 1 = final velocity since initial was zero (d) Work done = area under graph = 2 × 4 × 2

1 2 = 4 J From conservation of energy 2mv = mgh 2 2 (e) Red area = 2 × 1 (4 + 8) = 12 J so h = v = 43.75 = 7.7 m 2 2g 2 × 9.8 45 F 41 150 N

Ben 150° 30°

10 m x Use 150° since this is the angle between 5 8 10 direction of F and displacement. 1 1 Area = 2 × 300 × 5 + 300 × 3 + 2 × 300 × 2 (a) Work done = 150 cos 150° × 10 m = 750 + 900 + 300 = –1300 J = 1950 J

(b) Dog is doing the work. –1 46 5 m s m 42 Since displacement = 0, no work is done.

150 N 180° 50 m

2 m m Work done = 150 cos 180° × 2 v2 = –300 J

44 At top of cliff the stone has 1 2 PE + KE = 2 mv1 + mgh 6 cm 1 2 At the bottom the stone has only KE = 2 mv2 Conservation of energy 8 cm 1 2 ⇒ 2 m × 5 + m × 10 × 50 1 2 = 2 mv2 2 cm 2 1 2 v2 = 2 × (2 × 5 + 10 × 50) = 1025 –1 v2 = 32 m s

(a) Spring is stretched 2 cm. (b) F = kx F = 2 × 2 = 4 N 8

M14_IBPH_SB_IBGLB_9021_SL02.indd 8 17/07/2014 16:25 47 49 –1 0.25 kg 2 m s

0.2 kg h

5 m (a) Original KE = final PE 1 2 2 mv = mgh 2 2 h = v = 2 = 0.2 m 2g 2 × 10 2 Δx (b) If v = 4 m s–1, h = 4 = 0.8 m 2 × 10 50 (a) Work done = mgh = 2 × 9.8 × 100 (a) When ball hits spring = 1.96 kJ KE = original PE (b) Efficiency = (useful work/work in) × 100% = mgh 45 = 1.96 × 103 × 100 = 0.25 × 10 × 5 E = 12.5 J E = 4.36 kJ (b) Ball loses all its energy so work done = 51 (a) Useful work = gain in KE loss of energy. Convert velocity to ms–1 = 100 × 1000 Work done = 12.5 J (60 × 60) = 27.7 ms–1 1 2 (c) Energy given to spring = kx 1 2 1 2 2 KE = 2 mv = 2 × 100 × 27.7 1 2 12.5 = 2 × 250 000 × x = 3.86 × 105 J x2 = 0.0001 (b) 60 = 3.86 × 105 × 100 x = 0.01 m = 1 cm E (Note: We have ignored the loss of PE by E = 6.43 × 105 J = 0.643 MJ the ball as it squashes the spring; this is (c) 36 MJ per litre so 0.643 = 1.8 × 10–2 l very small.) 36 52 0.01 kg 0.01 kg 48 15 N

0.1 kg m m 0.05 m (a) Work done to compress spring = elastic PE (a) Work done = F × d 1 2 of spring = 2 kx = 15 × 0.05 1 2 = 2 × 0.1 × 0.05 = 0.75 J = 1.25 × 10–4 J (b) Work done will increase the PE of the ball (b) KE gained by balls = 1.25 × 10–4 J Increase in PE = 0.75 J = mgh Since balls are the same must get 1 –4 0.75 = 0.1 × 10 × h 2 × 1.25 × 10 J each h = 0.75 m = 6.25 × 10–5 J 1 2 –5 (c) KE = 2 mv = 6.25 × 10 J 2 × 6.25 × 10–5 v2 = 0.01 v = 0.1 m s–1 9

M14_IBPH_SB_IBGLB_9021_SL02.indd 9 17/07/2014 16:25 53 10 m s–1 15 m s–1 57 1000 N 20 m s–1

1000 N

2 kg 10 kg If velocity constant, forces are balanced so forward force = 1000 N In 1 s the car moves 20 m so work done v = 1000 × 20 = 20 000 J 12 kg Power = work done per second = 20 kW

58 (a) Momentum before = 2 × 10 – 10 × 15 = 20 – 150 = –130 N s Momentum after = 12 × v Power in = 100 W 2 m Conservation of momentum: –130 = 12v v = –10.83 m s–1

1 2 1 2 10 kg (b) KE before = 2 × 2 × 10 + 2 × 10 × 15 = 100 + 1125 = 122 J 1 2 Useful work = mgh = 10 × 10 × 2 = 200 J KE after = 2 × 12 × 10.83 = 703.7 J Energy loss = 521.3 J power = work = 200 = 50 W time 4 54 10 m s–1 15 m s–1 efficiency = power out × 100% = 50 × 100% power in 100 = 50% 15 m s–1 10 m s–1 59 efficiency = energy out = E = 70 energy in 60 100 E = 60 × 70 = 42 kJ If collision is elastic the velocities swap. 100 55 60 300 N 300 N

2000 N 2 m (a) Constant velocity ⇒ forces balanced ⇒ forward force = 300 N power = force × velocity = 300 × 80 000 3600 = 6.67 kW Work done lifting weight = 2000 × 2 = 4000 J (b) efficiency = power out = 6.67 = 60 power = work done = 4000 = 800 W power in power in 100 time 5 power in = 11.1 kW

56 50 kg

50 m

PE loss = mgh = 50 × 10 × 50 = 25 000 J power = energy = 25 0000 = 1000 W time 25 10

M14_IBPH_SB_IBGLB_9021_SL02.indd 10 17/07/2014 16:26 (b) (i) from the gradient a = 0.8 = 1.6 m s–2 Practice questions 0.5 (ii) distance = area 1 = 2 × 0.5 × 0.8 + 0.8 × 11 1 18 m s–1 18 m s–1 18 m s–1 1 S + 2 × 0.5 × 0.8 = 9.2 m s x u u 0.5 s 0.5 s P P P P

11.0 s 4.5 m s–2 time, 6.0 s time, t ) –1 0.80 s (a) (i) Car S travels at constant velocity so 0.70

sS = 18 × t 0.60 velocity (m (ii) Police car has constant acceleration so 0.50 1 2 s = ut + 2 at 0.40 0.8 m s–1 1 2 sP = 0 × 6.0 + 2 × 4.5 × 6.0 0.30 = 81 m 0.20 v – u (iii) Using a = ⇒ v = at + u t 0.10 = 4.5 × 6.0 + 0 = 27 m s–1 0.00 Could use v2 = u2 + 2as 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.011.012.0 ⇒ time (s) v = 2 × 4.5 × 81 = 27 m s–1 but not good practice to use s since if you (iii) Minimum work = gain in PE = mgh calculated it wrong in part (ii) then this = 250 × 10 × 9.2 would be wrong. = 23 000 J However, since errors are not carried (iv) power = work done = 23 000 = 1916 W time 12 forward you would not lose marks if you ≈ 1.9 kW did this. (v) efficiency = power out × 100% (iv) Police car travels at constant velocity power in = 1.9 × 100% = 38% from 6.0 s until the cars meet at time t. 5 So time at constant velocity = (t – 6.0) (c) On the original graph, velocity changed ⇒ x = 27 (t – 6.0) instantly ΔV (b) The police car catches up (draws level) with ⇒ a = = ∞ 0 car S when they have travelled the same This cannot happen; the changes happen distance. over time. Distance travelled by S in time t = 18t

Distance travelled by P = 81 + 27 (t – 6.0) ) –1 0.80 So when they meet 18t = 81 + 27 (t – 6.0) s 0.70 18t = 81 + 27t – 162 0.60

18t – 27t = 81 – 162 velocity (m 0.50 t = 9.0 s 0.40 2 (a) Mass can be defined in two ways: 0.30 1) In terms of the force experienced by a 0.20 mass in a gravitational field 0.10 F = mg so m = F gravitational mass G 0.00 2) In terms of the acceleration 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.011.012.0 experienced when a constant force is time (s) exerted on the mass. F = ma so m = F inertial mass a 11

M14_IBPH_SB_IBGLB_9021_SL02.indd 11 17/07/2014 16:26 (d) (i) Since velocity is constant the force On the way down PE is converted first to must be balanced. KE then to heat.

T 3 velocity = 0 m s–1 at top

h T = W –1 8 m s

10 m s–1

(ii) Since elevator is going up and slowing down acceleration is down so W > T.

v2 – u2 (a) (i) Using v2 = u2 + 2as ⇒ s = W > T 2a u = 8 m s–1 v = 0 m s–1 a = –10 m s–2 W d = h 02 – 82 W should be the same in each diagram, it h = = 3.2 m –2 × 10 is T that changes. Good idea to write which v – u v – u (ii) Using a = ⇒ t = force is bigger in case diagram isn’t clear. t a t = –8 = 0.8 s (e) The reading on the scales is the upward –10 force on the person; this is bigger than W (b) (alternative method) when accelerating up but equal to W when Time to reach sea = 3.0 s 1 2 velocity is constant. using s = ut + 2 at u = 8.0 m s–1 t = 3.0 s a = –10 m s–2 1 2 s = 8 × 3 – 2 × 10 × 3 W

reading on scales = 24 – 45 = –21 m i.e. 21 m below start. So the cliff = 21 m high

4 (a) Newton’s third law: If body A exerts a force on body B then body B must exert an equal 0.00 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.011.012.0 and opposite force on body A. time (s) (b) Law of conservation of momentum: for a (f) 0 → 0.5 s system of isolated bodies (i.e. no external Electrical energy changes to PE + KE forces acting) the total momentum is 0.5 → 11.5 s constant. Electrical energy changes to PE 11.5 → 12.0 s KE + electrical energy changes to PE 12

M14_IBPH_SB_IBGLB_9021_SL02.indd 12 17/07/2014 16:26 (c) (i) Applying conservation of momentum FBA A B FAB momentum before = momentum after 800 × 5.0 = (800 + 1200) v

4000 = 2000 v Should be the same length acting through v = 2.0 m s–1 the centre of the spheres. (ii) Initial KE = 1 × 800 × 52 = 10 000 J (change in mv) mv – 0 2 (d) (i) F = B = B Final KE = 1 × 2000 × 22 = 4000 J AB time t 2 mv F = B Loss of KE = 10 000 – 4000 = 6000 J AB t v (c) When the trucks collide heat and sound are produced. Before A B 6 (a) (i) PE = mgh vA vB mg = weight of man = 700 N height from A to B = 30 × sin 40° After A B = 19.3 m

(ii) FBA = rate of change of momentum (change in mv) mv – mv = A = A time t 30 m 30 × sin40° m(v – v) F = A BA t 40° (e) According to Newton’s third law A FAB = –FBA mv –m(v – v) gain in PE = 700 × 19.3 = 13 500 J = B = A t t (ii) If 48 people go up per minute the total mv = –mv + mv B A increase in PE = 48 × 13 500 mv + mv = mv B A = 6.48 × 105 J (f) If KE conserved then initial KE = fi nal KE (iii) Assume that people stand still on the 1 2 1 2 1 2 2 mv = 2 mvA + 2 mvB escalator and that they all weigh 700 N. v2 = v 2 + v 2 A B (b) (i) power = work done = gain in PE From conservation of momentum we know time time that v = v + v 6.2 × 105 A B = = 1 × 104 W 60 If vA is at rest then vA = 0 and B travels at v P then v = v out B effi ciency = ⇒ P = 15 kW P in So v2 = 02 + v2 and v = 0 + v in which means that this is a possible (ii) The escalator is a continuous band; it outcome goes up on the outside and down on In fact it is the only solution. the inside.

5 (a) Linear momentum = mass × velocity (c) Since the effi ciency will be less than 100% due to friction etc. the power in will be (b) 5 m s–1 1200 kg 800 kg greater than useful work done. B A Unless they are small children running up

immediately before collision the escalator. v B A

immediately after collision 13

M14_IBPH_SB_IBGLB_9021_SL02.indd 13 17/07/2014 16:26 7 (a) The total momentum of a system of isolated bodies is always constant. Challenge yourself

(b) (i) 58 g 2 g 56 g 1 First draw a diagram

140 m s–1 v

Conserving momentum –10 m s–2 56 × 0 + 2 × 140 = 58 × v 20 m s–2 280 = v = 4.8 m s–1 58 (ii) 58 g 4.8 m s–1 58 g y

F 30° 2.8 m x As block slows work done against friction = average F × distance moved Taking components of the motion in direction of force Horizontal Work done against friction will equal the x = 20 × sin 30° × t 1 2 1 2 Vertical KE lost = 2 mv = 2 × 0.058 × 4.8 1 2 = 0.7 J y = 20 × cos 30° × t – 2 × 10 × t y We also know that = –tan θ (negative because So average F × 2.8 = 0.7 x average force = 0.24 N the value of y is negative) Dividing (c) (i) Assuming vertical component of 2 y = 20 × cos 30° × t – 0.5 × 10 × t = –tan 30° velocity is uniform we can use x 20 × sin 30° × t 20 × sin 30° × t 1 2 t = 4.62 s s = ut + 2 at so x = 46.2 m u = 0 so 2s = 2 × 0.85 = 0.41 s a 10 y = –26.7 m Horizontal velocity is constant Distance down slope = 46.22 + –26.72 = 53 m = 4.3 m s–1 2 Horizontal distance = vt = 4.3 × 0.41 4 m s–1 = 1.8 m 6 m s–1 45° (ii) θ table clay block 0.2 kg 0.5 kg v path Taking components of the momentum Horizontal: 0.2 × 6 = 0.2 × 4 × cos 45° + 0.5 × v cos θ 0.85 m So v cos θ = 1.27 Vertical: 0.2 × 4 × sin 45° = 0.5 × v sin θ ground So v sin θ = 1.13 v sin 1.13 θ = = tan θ v cos θ 1.27 θ = –41.7° v = 1.27 = 1.7 m s–1 cos(–41.7°) 14

M14_IBPH_SB_IBGLB_9021_SL02.indd 14 17/07/2014 16:26 Worked solutions

Chapter 3 6 Exercises 4 kg A 1 (a) 1 mole of copper has mass = 63.54 g = 0.06354 kg mass M Density, ρ = so V = B volume ρ V = 0.063 54 = 7.123 × 10–6 m3 8920 (b) 1 mole contains 6.022 × 1023 atoms (from 5 m s–1 A 5 m defi nition) 3 m (c) If the volume of 6.022 × 1023 atoms is 4 kg 7.123 × 10–6 m3 then the volume of 1 atom B –6 = 7.123 × 10 m3 = 1.183 × 10–29 m3 6.022 × 1023 Loss of PE = gain in KE + WD against friction mass 1 2 2 Density, ρ = 4 × 9.8 × 3 = 2 × 4 × 5 + W volume W = 117.7 – 50 = 67.7 J Volume = 10 cm3 work = force × distance in direction of force –6 3 = 10 × 10 m distance travelled = 5 m Density = 2700 kg m–3 67.7 = F × 5 F = 13.5 N mass = V × ρ = 10 × 10–6 × 2700 mass = 2.7 × 10–2 kg = 27 g 7

3 (a) Block has PE = mgh = 10 × 9.8 × 40 = 3.92 × 103 J 30 cm (b) This PE will all be converted to heat so heat to fl oor + block = 3.92 × 103 J 10 cm 12 cm 4 Mass of car in example 1 = 1000 kg so if v = 60 m s–1 1 2 1 2 KE = 2 mv = 2 × 1000 × 60 = 1.8 × 106 J

5 If the speed is constant then rate of change of melting ice boiling water unknown temperature PE = gain in energy of surroundings A change in height of 20 cm is equivalent to a mgΔh = mgv change in temperature of 100°C Δt = 75 × 9.8 × 50 100 = 5°C cm–1 20 4 = 3.7 × 10 J The unknown temperature is 2 cm above zero; this is equivalent to 2 cm × 5°C cm–1 = 10°C L – L Alternatively using T 0 × 100 L100 – L0 T = 12 – 10 × 100 = 10°C 30 – 10

M15_IBPH_SB_IBGLB_9021_SL03.indd 1 16/07/2014 10:07 3 (b) Energy added = 3 × 105 J = mc 8 (a) Average KE of air molecules = 2 kT Δθ 3 × 105 (This assumes air is an ideal gas which it so Δθ = = 667°C 0.5 × 900 isn’t, but it gives an approximate answer) so if initial temperature = 20°C temperature in K = 273 + 20 = 293 K 3 –23 final temperature = 687°C Average KE = 2 × 1.38 × 10 × 293 –21 1 2 1 2 = 6 × 10 J 14 (a) Initial KE = 2 mv = 2 × 1500 × 20 (b) molar mass of air = 29 g mol−1 = 3 × 105 J 29 23 Final KE = 0 J so KE lost = 3 × 105 J mass of 1 molecule = 6 × 10 = 4.8 × 10–23 g = 4.8 × 10–26 kg (b) 75% of 3 × 105 J (c) KE = 1 mv2 = 75 × 3 × 105 = 2.25 × 105 J 2 100 2KE 2 × 6 × 10–21 Q v = = = 500 m s–1 (c) Q = mcΔθ ⇒ Δθ = m 4.8 × 10–26 mc 2.25 × 105 9 From definition Q = C = = 51°C Δθ 10 × 440 Heat lost = 210 × 103 × 2 15 (a) 8 litre/min = 8 kg/min since 1 litre has a Q = 420 kJ mass of 1 kg. 10 (a) A 1 kW heater will deliver 103 J per second So in 10 minutes 80 kg of water is used. so if it’s on for 1 hour: (b) Using Q = mcΔθ 3 6 heat delivered = 60 × 60 × 10 = 3.6 × 10 J Q = 80 × 4200 × (50 – 10) Q (b) From definition C = = 1.34 × 107 J Δθ the room is heated from 10°C to 20°C 16 From definition Q = ml so Δθ = 10°C (fusion since water is turning into ice) 6 So for the room, C = 3.6 × 10 Heat released, Q = 1 × 106 × 3.35 × 105 J 10 = 3.35 × 1011 J = 3.6 × 105 J/°C (c) Some heat will be lost to the outside. 17 To change 400 g of water at 100°C into steam requires 0.4 × 2.27 × 106 = 9.08 × 105 J 11 From table, C = 380 J/kg°C copper If power of heater = 800 W So Q = 0.25 × 380 × (160 – 20) energy E 9.08 × 105 4 then since P = , t = = = 1.33 × 10 J time P 800 M 3 12 (a) Density, ρ = , M = ρ × V t = 1.135 × 10 s = 19 min. V 3 –6 3 18 1 litre = 1000 cm = 1000 × 10 m 2 2 cm 1000 m = 10–3 m3 M = 1000 × 10–3 = 1 kg (a) Volume = 1000 × 2 × 10–2 = 20 m3 (b) Q = mc∆θ = 1 × 4200 × (100 – 20) M Density, ρ = ⇒ M = Vρ = 20 × 920 = 3.36 × 105 J V 4 (c) 1 kW ⇒ 1000 J per second = 1.84 × 10 kg 5 So time taken = 3.36 × 10 = 336 s (b) Q = ml = 1.84 × 104 × 3.35 × 105 1000 = 6.16 × 109 J energy or power = 9 time (c) P = Q = 6.16 × 10 = 3.42 × 105 W energy t 5 × 60 × 60 13 (a) power = so energy = power × time 5 time (d) Power per m2 = 3.42 × 10 1000 energy = 500 × 10 × 60 (time in seconds) = 3.42 × 102 Wm–2 = 3 × 105 J

M15_IBPH_SB_IBGLB_9021_SL03.indd 2 16/07/2014 10:07 19 Practice questions 3 3

V1 = 500 cm V2 = 500 cm

P2 = 250 kPa P2 = ? θ (°C) 1 T1 = 300 K I2 = 350 K

17 P V P V 1 1 = 2 2 T1 T2 250 × 500 = P2 × 500 300 350 150 s P = 250 × 350 = 292 k Pa 2 300 0 X 20

V1 = 2 m

n = 5 mol

T = 293 K –15

t (s) (a) PV = nRT 15 165 200 P = nRT = 5 × 8.31 × 293 V 2 (a) Ice melts when temperature is constant P = 6 kPa 0°C. All melted at 165 s. (b) If half of gas leaks, n = 2.5 mol (b) Heat goes to increase PE not KE so P = 3 kPa temperature remains constant. (c) (i) For last part of graph, water is heated 21 from 0 to 15°C in 30 s Q = mcΔθ ⇒ heat supplied = 0.25 × 4200 × 15 = 1.79 × 104 J V2 = ? 4 3 T = 250 K Q 1.79 × 10 V = 250 m 2 Power = = 1 P = 100 kPa t 30 2 T1 = 300 K

P1 = 150 kPa = 525 W ≈ 530 W (ii) Time to heat ice from –15 to 0°C = 15 s P1 V1 P2 V2 150 × 250 100 × V2 = ⇒ = Q = power × t = 530 × 15 = 7950 J T1 T2 300 250 Q = mc V = 150 × 250 × 250 Δθ 2 300 × 100 C = 7950 = 2.1 × 103 J kg–1 3 0.25 × 15 V2 = 312.5 cm (iii) Takes 150 s to melt 0.25 kg of ice 22 Heat given Q = 530 × 150 = 79 500 J = mL L = 79 500 = 3.2 × 105 J kg–1 V1 = V 0.25 T = T 3 1 V =V 250 = ½ m V 2 (a) When a liquid evaporates the molecules P = 100 kPa 1 2 1 T =T 300 = 2 KT 1 2 with most energy escape from the surface, P =P 150 = ? KPcm 1 2 resulting in a reduction in the average KE P V P V P × ½V and hence temperature. If heat is added 1 1 = 2 2 100 × V = 2 ⇒ temperature will remain constant. T1 T2 T 2T P = 2 × 100 = 400 kPa 2 ½ 3

M15_IBPH_SB_IBGLB_9021_SL03.indd 3 16/07/2014 10:07 (b) Blowing across the surface reduces (ii) Total energy generated = 2.2 × 106 J humidity of surrounding air; increased 50% lost in evaporation = 1.1 × 106 J temperature of liquid; increased surface This energy goes to latent heat of area of liquid vaporization Q = mL

6 (c) Heat lost when water turns into ice m = Q = 1.1 × 10 = 487 g L 2.26 × 106 water ice (iii) Wind Skin temperature Humidity 350 g –5°C 25°C Air temperature Area of skin Clothing –1 86 J s 4 (a) (i) Constant speed so resistive force = = mcΔθ + mL + mcΔθ component of mg acting down the water water–ice ice slope 0.35 × 4200 × 25 + 0.35 × 3.3 × 105 + 0.35 × 2.1 × 103 = 156 000 J 15° Power = Q 15° t mg t = Q = 156 000 = 1800 s P 86 mg sin 15° = 960 × 9.8 × 0.259 = 2.4 kN 3 (a) In this context thermal energy is the internal 1 2 1 2 energy of the molecules of the runner. (ii) KE = 2mv = 2 × 960 × 9 = 39 kJ This can be KE and PE. Increased thermal (b) Work done = average force × distance energy will increase the average KE of the Work done against braking force molecules which increases the temperature, = loss of KE = 39 kJ = average force × 15 m in other words the runner becomes hot. average force = 39 000 = 2.6 kN (b) (i) Energy generated = power × time 15 Energy given to brakes = 39 kJ = 1200 × 3600 = 2.2 × 106 J (c) This causes the brakes to get hot so KE (ii) Q = mcΔθ lost = thermal energy gained = mcΔT Q 2.2 × 106 Δθ = = = 7.5 K Two brakes so total mass = 10.4 kg mc 70 × 4200 39000 = 10.4 × 900 × ΔT ΔT = 4.2 K This assumes no heat lost and all KE 70 kg converted to heat not sound.

5 (a) (i) The molecules of an ideal gas are

–1 considered to be small perfectly elastic 1200 J s × 1800 spheres moving in random motion with (c) Convection no forces between them. Small and Conduction elastic is mentioned in the question so Radiation This is no longer on the syllabus. 1. Motion is random (d) (i) The molecules with greatest KE leave 2. No forces between molecules the surface resulting in a decrease in except when colliding average KE and hence temperature. 4

M15_IBPH_SB_IBGLB_9021_SL03.indd 4 16/07/2014 10:07 (ii) The molecules of an ideal gas have no forces between them so changing their Challenge yourself position does not require work to be done; gas molecules therefore have no 1 PE; this implies that the internal energy 100 kPa n 100 kPa of a gas is related to the average KE 300 k 300 k V V of the molecules. If energy is added to the gas, temperature increases so we P P see that temperature is related to the 400 k 300 k average KE. V V n n (b) (i) Using PV = nRT 1 2 T = 290 K When first filled and joined we can eattr the two P = 4.8 × 105 Pa flasks as one container. Applying the ideal gas V = 9.2 × 10–4 m3 equation, PV = nRT, we get 100 × 2V = nR × 300 PV 4.8 × 105 × 9.2 × 10–4 After one flask is heated we have to treat them n = = RT 8.3 × 290 separately but since they are connected the = 0.18 mol pressure is the same.

(ii) If temperature constant P1V1 = P2V2 PV = n1R × 400 5 –4 –4 4.8 × 10 × 9.2 × 10 = P2 × 2.3 × 10 PV = n2R × 300 P = 9.2 × 4.8 × 105 = 19 × 105 Pa The total number of moles n is the same before 2 2.3 ( ) P P and after so (iii) If volume is constant 1 = 2 T T n = n + n P = 19 × 105 Pa 1 2 1 2 1 substituting gives 200V = PV + PV T1 = 290 K 300R 400R 300R 2 1 1 P2 = ? = + P 3 (400 300) T2 = 420 K P = 19 × 105 × 420 = 2.8 × 106 Pa P = 114.3 kPa 2 290 (c) P

(b)(iii) (c)

(b)(ii) A

0 0 V

M15_IBPH_SB_IBGLB_9021_SL03.indd 5 16/07/2014 10:08 Worked solutions

Chapter 4 Gain in PE = mgh = 0.2 × 9.8 × 2 × 0.5 Exercises = 1.96 J 1 2 1 2 KE at bottom = 2mv = 2 × 0.2 × 10 1 (a) distance = 2πr = 2π × 5 = 31.4 m = 10 J (b) displacement = 0 m KE at top = 10 – 1.96 = 8.04 J

–1 (c) speed = 2 m s v = 2KE = 2 × 8.04 = 9 m s–1 m 0.2 t = d = 31.4 = 15.7 s speed 2 (b) (d) f = 1 = 1 = 6.4 × 10–2 Hz T 15.7 mg T (e) ω = 2πf = 0.4 rad s–1 mv2 v2 At the top T + mg = (f) a = = ω2r = 0.8 m s–2 r r 2 2 2 1000 × 30 000 2 mv 0.2 × 9 2 Centripetal force = mv = ( 3600 ) T = – mg = – 0.2 × 9.8 r 50 r ( 0.5 ) = 1389 N = 30 N

3 6

1 m 0.2 kg 50 N M m 2 Maximum centripetal force = 50 N = mv r 50 × 1 v2 = ⇒ v = 15.8 m s–1 0.2 r 4 From Newton’s universal law F = GMm r2 mg But the weight of an object = mg mg = GMm r2 acceleration due to gravity, g = GM r2

–11 22 for the Moon g = 6.67 × 10 × 7.32 × 10 (1.74 × 106)2 Minimum speed when 1mv2 = mgh, 2 g = 1.61 m s–2 1 2 2 so 2mv = mg(2r), v = 4gr 7 v = 4gr = 4 × 9.8 × 5 = 14 m s–1

17 5 1.89 × 10 kg 200 g

50 cm 71 492 km Using the equation for gravitational field strength of a spherical object g = GM r2

–11 17 –1 6.67 × 10 × 1.89 × 10 10 m s g = (7.1492 × 107)2 (a) At the top some of the KE has turned to PE = 24.7 N kg–1

M16_IBPH_SB_IBGLB_9021_SL04.indd 1 17/07/2014 15:27 8 14 A and E are at the same height so no potential difference. 1000 km

24 5.97 × 10 kg 15 No work is done since there is no change in potential.

16 8 3.8 × 10 m 6367 km 8 8 3.7 × 10 m 0.1 × 10 m 24 22 7367 km ME = 6.0 × 10 kg Mm = 7.4 × 10 kg rE A rm Using the formula for the gravitational field due Earth Moon GM to a spherical mass g = where r = 7367 km GM GM r2 (a) Potential at A = E + m as shown rE rm –11 24 6 × 1024 7.4 × 1022 g = 6.67 × 10 × 5.97 × 10 = 6.67 × 10–11 + (7.367 × 106)2 (3.7 × 108 0.1 × 108 ) g = 7.34 N kg–1 = 6.67 × 10–11 × (1.6 × 1016 + 7.4 × 1015) = 1.6 MJ kg–1 9 B 1000 kg 100 kg (b) PE of 2000 kg rocket = v × m = 1.6 × 106 × 2000 = 3.1 × 109 J

1 m (c) V (d) Field strength 6 m zero when x gradient = 0 Since both masses are on the same side of B then the two fields are in the same direction. Note: V is not zero here The field vectors will therefore simply add. Using the equation for the field due to a sphere (d) Field strength zero when gradient = 0 g = GM Note: V is not zero here. r2 –11 Field due to 1000 kg = 6.67 × 10 × 1000 12 17

3 = 6.67 × 10–8 N kg–1 1738 × 10 m –11 Field due to 100 kg = 6.67 × 10 × 100 62 = 0.018 × 10–8 N kg–1 22 Total field = 6.67 + 0.018 = 6.69 × 10–8 N kg–1 mass = 7.4 × 10 kg V = 2GM 10 A escape R 100 kg 100 kg –11 22 = 2 × 6.67 × 10 × 7.4 × 10 1738 × 103 If masses are equal the fields will be equal and = 2.38 × 103 m s–1 opposite as shown. 18 Hydrogen is a small atom so its mean velocity Resultant field = 0N kg–1 would be much higher than air molecules; some –1 11 VC = ghC = 10 × 14 = 140 J kg hydrogen atoms would be travelling faster than –1 VD = ghD = 10 × 11 = 110 J kg the escape velocity. Potential difference between C and D 19 To be black hole V = 3 × 108 m s–1 = 140 – 110 = 30 J kg–1 escape = 2GM 12 Work done from D to C = ΔV × m = 30 × 3 R R = 2GM = 90 J (3 × 108)2 13 PE = mgh = 3 × 10 × 8 = 240 J 2 × 6.67 × 10–11 × 2 × 1030 = 8 2 = 3 km 2 (3 × 10 )

M16_IBPH_SB_IBGLB_9021_SL04.indd 2 17/07/2014 15:27 20 V = 2GM where R = distance from rocket 24 (a) KE = GMm = 5.9 × 1010 J escape R 2r to centre of Earth = (6400 + 100 000) km (b) PE = –GMm = –1.2 × 1011 J r –11 24 10 V = 2 × 6.67 × 10 × 6 × 10 = 2.74 km s–1 (c) Total = KE + PE = –6.1 × 10 J escape 1.064 × 108 21 radius/ period radius3/ period2 1010km days (1010km)3 days2 Practice questions 5.79 88 194.104539 7744 1 (a) Velocity is a vector so as the direction of 10.8 224.7 1259.712 50490.09 the car changes, velocity must change. 15 365.3 3375 133444.09 Acceleration is the rate of change of velocity so if velocity changes the car must accelerate. 22.8 687 11852.352 471969 (b) (ii) Weight and the normal force both act 77.8 4330 470910.952 18748900 downwards. Not centripetal force; this 143 10700 2924207 114490000 is the resultant. 288 30600 23887872 936360000 450 59800 91125000 3576040000

0.45 m ) 2 4 000 000 000

(days 3 500 000 000 2 0.8 m 3 000 000 000 period 2 500 000 000 F 2 000 000 000

1 500 000 000

1 000 000 000 0.45 m (iii) If no energy loss then loss of PE = gain 500 000 000 in KE. 0 0.8 m 1 2 0 20 000 000 60 000 000 100 000 000 mgΔh = 2mv –500 000 000 40 000 000 80 000 000 1 2 radius3 (1010km3) ⇒ 10 × (0.8 – 0.35) = 2v ⇒ v = 9 = 3 m s–1 F This is plotted from the data in Chapter 12. W The scale is very big so the smaller numbers are (iv) We know that if moving in a circle 2 N F = mv not visible. r 0.05 32 F = × 22 For a TV satellite T = 1 day = 24 × 60 × 60 ⇒ 0.35/2 = 86 400 s Now this force is caused by normal T 2 4 2 T 2GM From Kepler’s law = π ⇒ r 3 = r 3 GM 4π2 force and weight 2 –11 24 86 400 × 6.67 × 10 × 6 × 10 22 = 2 = 7.6 × 10 4π ⇒ 2.6 = N + W W r = 4.2 × 107 m ~ 7R Use logs to find 3 E √ where W = 0.5 N N 23 Orbit radius = 6400 + 400 km = 6.8 × 106 m so 2.6 = N + 0.5 3 2 6 3 2 N = 2.6 – 0.5 = 2.1 N T 2 = r 4π = (6.8 × 10 ) × 4π = 3.1 × 107 GM 6.67 × 10–11 × 6 × 1024 2 (a) Coefficient of friction is defined by the T = 5.57 × 103 s = 1.5 hours equation F = μR so μ is the ratio of friction force . normal reaction force 3

M16_IBPH_SB_IBGLB_9021_SL04.indd 3 17/07/2014 15:27 (b) (i) Since the person is not moving relative To reach the moon it must have enough to the wall, the friction would be static KE so that it reaches the position of friction. zero field, a distancer = 0.8 from the

(i) friction planet. From here to the moon it will be attracted by the moon’s field Loss of KE = gain in PE 1 2 If final KE = 0 then loss = 2mv Gain in PE = change from planet to 0.8 (normal) reaction from planet from the graph ΔV = 4.4 × 107 so ΔPE = 4.4 × 107 × 1500 = 6.6 × 1010 J Original KE = 6.6 × 1010 J

weight 4

6 42 (c) (i) The minimum speed is such that friction R (10 m) 0 5 10 15 20 25 30 35 40 45 50 = weight 0 mg = μR –1 So R = mg = 80 × 10 = 2000 N μ 0.4 –2 (ii) The body is moving in a circle so the

) –3 unbalanced force = centripetal force –1 kg 2 J R = mv 7 –4 r (10

V –5 v = Rr = 2000 × 6 = 12 m s–1 m 80 –6 3 (a) Gravitational potential is the amount of work done per unit mass in taking a small test –7 mass from infinity (a place of zero potential) –8 to the point in question. (a) (i) V at surface = –6.3 × 107 J kg–1 (b) (i) If field strength = 0 then field strength (ii) Height 3.6 × 107 m = 36 × 106 m of planet is equal and opposite to field So R = (36 + 6) × 106 m strength of moon. R = 42 × 106 m planet moon 0.8 0.2 So from graph V = –1.0 × 107 J kg–1 r 0 0.2 0.4 0.6 0.8 1.0 (b) As satellite leaves the Earth, KE → PE so if 0 final KE = 0 then original KE = gain in PE –1 From the graph ΔV = (6.3 – 1) × 107 J kg–1 –2 7 –1 V (×107 J kg–1) ΔV = 4.4 = 5.3 × 10 J kg –3 so ΔPE = 1 × 104 × 5.3 × 107 = 5.3 × 1011 J –4 So minimum KE = 5.3 × 1011 J –5 Note: ﬁeld strength = 0 when gradient = 0 (c) The rocket doesn’t stop when it reaches the orbit; it must have enough velocity to stay in 2 GM GM M r 2 p = m p = p = 0.8 = 16 2 2 ⇒ 2 2 orbit rp rm Mm rm 0.2 2 mv = GMm (ii) As satellite travels from planet its r r2 KE → PE During the early stages, whilst the rocket is in the atmosphere, energy is lost due to air 4 resistance.

M16_IBPH_SB_IBGLB_9021_SL04.indd 4 17/07/2014 15:27 5 (a) The force is directed towards the centre so (b) Mass is travelling in a circle so centripetal is perpendicular to the direction of motion. force = horizontal component of tension mv2 Work done is the force x distance moved = Tsinθ in the direction of the motion which is zero r Since no vertical acceleration weight = since there is no motion towards the centre. vertical component of tension Alternatively one could argue that since the mg = Tcos speed and distance to centre are constant θ v2 Dividing gives tanθ = there is no change in either KE or PE therefore gr no exchange of energy so no work done. v = gr × tanθ = 9.8 × 0.33 × tan30° (b) (i) The centripetal force is provided by = 1.4 ms–1 gravitational attraction between the masses so 2 mv = GMm r r2 Challenge yourself v = Gm r (ii) Total energy = KE + PE 1 On the ﬂ at track 50 m KE = 1mv2 = 1m GM GMm 2 2 r 2r R PE = –GMm r μR Total energy = GMm – –GMm = –GMm mg 2r r 2r (c) The total energy is –GMm so if r is increased 2r R = mg so R = 10 000 N total energy becomes less negative i.e. μR = 0.8 × 10 000 = 8000 N mv2 bigger. If the energy has increased then Circular motion so = μR r work has been done. To do this the engines v2 = μRr = 8000 × 50 = 400 must be fi red in the direction of motion so m 1000 –1 –1 rocket moves in the direction of the force. v = 20 m s = 72 km h On a banked track

6 (a) (i) 50 m

30° R

vertical μR mg tension 45° resultant Notice the friction is acting down the slope weight r = 0.33m since the force on the car is acting upwards. Vertical components: R cos 45° = mg + μR sin 45° (ii) The ball is not in equilibrium since mg R = 30° it is not at rest or travelling at (cos 45° – μ sin 45°) 10 000 = vertical = 7.1 × 104 N constant velocity. The (cos 45° – 0.8 sin 45°) forces are therefore not Horizontal components: balanced as can be shown mv2 tension = R sin 45° + μR cos 45° by adding together the r v2 = R(sin 45° + 0.8 cos 45°) × r forces on the diagram. resultant m weight–1 r = –10.33m v = 67ms = 241 km h 5

M16_IBPH_SB_IBGLB_9021_SL04.indd 5 17/07/2014 15:27 Worked solutions

Chapter 5 4 Exercises acceleration is constant here so not 3 cm simple harmonic motion A A A 1 (a) acceleration is constant here so not simple harmonic motion O 5 s time

B B B 5 If time period = 10 s (b) U then f = 1 = 0.1 Hz U 10 So ω = 2πf = 0.1 × 2π = 0.2π Using the equation for simple harmonic motion x = x cos ω t W 0 We want to know what the time is when x = 1 m U is proportionalW to weight of fluid 1 = 2 cos(0.2πt) displaced; this is proportional to the 0.5 = cos(0.2πt) distance the rod is pushed under the water So cos–1 (0.5) = 0.2πt = 1.047 so this is simple harmonic motion. t = 1.047 = 1.67 s (c) The tennis ball does not have an 0.2π acceleration and displacement so this is 6 not simple harmonic motion. v (d) A bouncing ball has constant acceleration –1 (g) except when bouncing so this is not +0.5 m s f = 1Hz simple harmonic motion. –1 0.5 m s t 0.5 s 1 s 1.5 s 2 (a) Frequency, f = number of swings per second –1 –0.5 m s 20 swings in 12 s (assuming complete swings) 20 = 1.67 Hz From the equation for simple harmonic motion 12 v = –v sin ω t (b) Angular frequency, ω = 2πf = 10.49 rads/s 0 v0 = the maximum velocity 3 We can see from the graph that 0.5 s after travelling upwards at 0.5 m s–1, the mass will have a velocity of 0.5 m s–1 downwards. x0

3.0 cm

time (s)

If released from the top then x = x0cosω t After 1.55 s x = 3.0 × cos(2π × 0.2 × 1.55) = –1.1 cm 1

M17_IBPH_SB_IBGLB_9021_SL05.indd 1 17/07/2014 15:26 7 1 2 2 (b) maximum KE = 2 mω x0 1 2 2 = 2 × 0.1 × (3π) × 0.04 = 7.1 × 10–3 J (c) maximum PE = maximum KE = 7.1 × 10–3 J

1 2 2 2 t = 5 s (d) KE = 2 mω (x0 – x ) 1 2 2 2 = 2 × 0.1 × (3π) (0.04 – 0.02 ) = 5.3 × 10–3 J (e) At any given instant, PE + KE = 7.1 × 10–3 J so PE = 7.1 × 10–3 – 5.3 × 10–3 J 2 m = 1.8 × 10–3 J (a) Maximum velocity = ω x 0 11 f = 1 = 1 = 5.6 × 10–4 Hz T (30 × 60) = 2πf x0 1 –4 3 –1 = 2π × × 2 v = f λ = 5.6 × 10 × 500 × 10 = 280 m s 5 = 2.5 m s–1 12 (a) v

2 v/2 (b) Maximum acceleration = ω x0 0.4 m 2 2 = π × 2 (5 ) = 3.16 m s–2

8 0.2 m

v = f so is proportional to v f = 2 Hz λ λ

λ1 v1 λ1 0.4 = = 2 so λ2 = = = 0.2 m λ2 v2 2 2 (b) Inverted since after knot medium is more 5 cm dense (c) Because some of the energy is in the reflected wave Using the equation 2 2 v = f v = ω x0 – x 13 λ 1 = 2π × 2 0.052 – 0.012 f = = 2 Hz 0.5 = 4π 0.0024 λ = 0.6 m = 0.62 m s–1 v = 2 × 0.6 = 1.2 m s–1

9 As it passes through the equilibrium position its 14 (a) M = 1.2 × 10–3 kg m–1 T = 40 N velocity is maximum so use V = T = 40 = 182.6 m s–1 M 1.2 × 10–3 v = ω x max 0 (b) L = 63.5 cm = 2π × 1 × x = 1 m s–1 2 0 λ = 2L = 127 cm = 1.27 m 2 1 v 182.6 x0 = = = 0.32 m V = f λ ⇒ f = = = 143.8 Hz 2π π λ 1.27 10 m = 0.1 kg 15 v = f T = 1 T μ 2l μ x0 = 0.04 m f = 1.5 Hz λ = 2l 10–3 μ = 1.2 × = 0.6 × 10–3 kg m–1 (a) ω = 2πf = 3π rad s–1 2 T = f 2 × 4l 2 × μ = 5002 × 4 × 0.32 × 0.6 × 10–3 = 54 N 2

M17_IBPH_SB_IBGLB_9021_SL05.indd 2 17/07/2014 15:26 sin i v 0.3 16 I = 1 m 1 1 1 = = sin i2 v2 0.5 0.5 sin i2 = sin 20° × 0.3 f1 = 650 Hz i2 = 35° C 19 6 cm A f2

6.2 cm

I2 = 80 cm B v = T μ l (a) λ = 2 cm T and μ are constant so f is proportional to v path difference = 6.2 – 6 = 0.2 cm f l 650 80 1 = 1 = path difference 0.2 f l f 100 phase angle = × 2π = 2 2 2 λ 2.0 f = 812.5 Hz 1 2 = × 2π = π 10 5 17 path difference 8 – 7 1 (b) × 2π = = × 2π = π λ 2 2 path difference 11.5 – 10 wave slows so (c) × 2π = 2 –1 refracts towards λ 0.5 m s 30° 1.5 3 normal = × 2π = π 2 2 These answers are incorrect in some editions.

20 i2

–1 0.4 m s λ = 4 × 50 = 200 cm 50 cm

(a) f = v = 0.5 = 1.7 Hz λ 0.3 v 0.4 (b) λ = = = 0.24 m f 1.7 sin i v v = f λ (c) 1 = 1 sin i v f = v = 340 = 170 Hz 2 2 λ 2 sin 30° 0.5 = 21 sin i1 0.4 0.4 sin i1 = sin 30° × = 0.4 0.5 I1 i1 = 24° 18 shallow 20°

–1 0.3 m s

wave speeds up so –1 0.5 m s refracts away from (a) First harmonic when l = 1λ normal 1 4 v 340 deep λ = = = 1.328 m f 256 3 l1 = 33.2 cm

M17_IBPH_SB_IBGLB_9021_SL05.indd 3 17/07/2014 15:26 3 (b) Second harmonic when l2 = 4λ = 99.6 cm 25 5 Third harmonic when l3 = 4λ = 166.0 cm The third harmonic won’t be heard.

22 40° air

–1 30 m s water r

eeek (1000 Hz) Using Snell’s law sin i = refractive index sin r sin 40° = 1.33 (from table) sin r sin 40° –1 sin r = = 0.48 40 m s 1.33 r = 29° (a) Would hear lower note on the way down Remember to change your calculator to and higher note on the way up. degrees. (b) Maximum on the way up cf 26 f = 0 = 340 × 1000 = 1097 Hz 1 c – v 340 – 30 (c) Minimum on the way down cf f = 0 = 330 × 1000 = 895 Hz 2 c + v 330 + 40 40° (d) Would hear higher note on the way down air and lower note on the way up. diamond r 23 v 500 Hz

Using Snell’s law sin i = refractive index sin r sin 40° = 2.42 sin r Frequency increases when plane approaches sin 40° c × f sin r = = 0.266 f = 0 2.42 1 c – v r = 15° If frequency received = 20 000 Hz = 340 × 500 340 – v 340 × 500 27 340 – v = = 8.5; v = 340 – 8.5 8 20 000 air 3 × 10 m/s v = 331.5 m s–1 glass c 24 g

f0 = 300 Hz –1 20 m s c Refractive index when light travels from medium 1 → medium 2 = velocity in medium 1 velocity in medium 2 8 If moving towards source 1.50 = 3 × 10 c + v 340 + 20 c f = × f = × 300 g 1 c 0 340 8 c = 3 × 10 = 2 × 108 m s–1 f = 317.6 Hz g 4 1 1.5

M17_IBPH_SB_IBGLB_9021_SL05.indd 4 17/07/2014 15:27 28 30 θ2 50 µm 30° θ1 water n1 70° D

i 2 sin 70° sin 70° (a) = 1.5; sin θ1 = ; θ1 = 38.8° sin θ1 1.5 air n2 (b) θ2 + θ1 = 90°; θ2 = 90 – 38.8 = 51.2° 1 1 (c) sin c = ⇒ c = sin–1 = 41.8° n (n) sin i n (d) Ray will be totally internally reflected 1 = 2 50 sin i n (e) tan 38.8 = ; D = 62.2 μm 2 1 D sin 30° 1 = D sin i2 1.33 sin i = sin 30° × 1.33 2 38.8°

I2 = 42° 50 µm 29 air plastic

20° 31 30° θ 0.05 mm y θ

n1 = 1 n2

5 m λ = 550 nm λ 550 × 10–9 sin i n θ = = = 0.011 rad 1 = 2 b 0.05 × 10–3 sin i n 2 1 y 2θ = = 0.022 ⇒ y = 0.022 × 5 = 0.11 m n = 1 × sin 30° = 1.46 5 2 sin 20° Now surrounded by water: 32

θ 5 cm water plastic θ

I2 4 m 0.05 30° 2θ = = 0.0125 4 θ = 0.006 25 n = 1.33 n λ λ 550 × 10–9 1 2 θ = ⇒ b = b θ 0.006 25 b = 8.8 × 10–5 m = 88 μm

33 4 × 1012 sin i n 1 = 2 θ θ x sin i2 n1 1.33 4 cm sin i = sin 30° × = 27° λ = 570 mm 2 1.46 5 i2 = 27°

M17_IBPH_SB_IBGLB_9021_SL05.indd 5 17/07/2014 15:27 The angle θ between the two diffraction 39 Δλ = 589.6 – 589 = 0.6 nm patterns can be found using R = λ = mN; m = 1 (the order) –9 Δλ 1.22λ 1.22 × 570 × 10 –5 θ = = = 1.74 × 10 rad 589 b 4 × 10–2 N = = 982 lines x 0.6 The angle subtended by the stars θ = 4 × 1012 1 –5 12 7 40 Constructive when 2t = (m + 2)λ x = 1.74 × 10 × 4 × 10 = 7 × 10 m 1 minimum when m = 0 ⇒ 2t = 2λ 1 1 –9 34 t = 4λ = 4 × 600 × 10 m = 150 nm λ = 600 mm 41 Destructive interference t = λ 1 cm θ θ –9 4 t = 380 × 10 = 95 nm d 4 200 km 42 (a) Water has lower refractive index so no Angle subtended at camera phase change. 1 × 10–2 θ = Oil n = 1.5 200 × 103 = 5 × 10–8 rad Water n = 1.3 1.22λ Diffracting angle θ = λ 580 b (b) λ = air = = 387 nm b = λ oil n 1.5 θ (c) 2t = (m + 1)λ for constructive interference 1.22 × 600 × 10–9 2 = = 14.6 m 1 1 5 × 10–8 minimum when m = 0; 2t = 2λ ⇒ t = 4λ 1 t = 4 × 387 = 97 nm 35 1 m 43 Since the change at both boundaries is from

θ θ 0.01 cm less dense → more dense ⇒ no phase change on refl ection 5 mm λ = 600 nm ⇒ t = λ for destructive interference. 4 1.22λ 1.22 × 600 × 10–9 580 414 Diffracting angle θ = = –3 λ = = 414 nm; t = = 104 nm b 5 × 10 coating 1.4 4 = 1.5 × 10–4 rad 44 Angle subtended at eye by pixels,

0.01 λ1 = 690 nm θ = = 1 × 10–4 rad 1 –4 The angle subtended is less than 1.5 × 10 so 8 –1 λ = 650 nm 3 × 10 m s 0

you will not be able to resolve them. f0 = c/λ 14 = 4.615 × 10 Hz 36 d = 0.01 × 10–3 m c 14 –9 F1 = = 4.348 × 10 Hz λ = 600 × 10 m λ1 D = 1.5 m Δf = (4.615 – 4.348) × 1014 = 2.67 × 1013 Hz D 600 × 10–9 × 1.5 b = λ = = 9 cm v c × 2.67 × 1013 d 0.01 × 10–3 = F ⇒ v = = 0.06 c c 1 4.348 × 1014 37 d = 0.01 × 10–3 m 45 0 λ = 400 × 10–9 m D = 1.5 m λD 400 × 10–9 × 1.5 b = = = 6 cm 6 –1 d 0.01 × 10–3 2 × 10 m s 38 (a) 300 lines/mm 658 nm separation of lines = 1 = 3.3 × 10–3 mm 300 = 3.3 μm (b) d sin θ = nλ, n = 1 700 × 10–9 sin θ = λ = = 0.212 d 3.3 × 10–6 v Δλ = λ0 θ = 12.24° c 2 × 106 6 Δλ = × 658 = 4.38 nm 3 × 108

M17_IBPH_SB_IBGLB_9021_SL05.indd 6 17/07/2014 15:27 46 (b) (i) The line should be parallel to D. 60° (ii) We can see that the wavelength gets shorter ⇒ velocity is less in medium R. I0 A C E

F medium I medium R At fi rst polarizer intensity transmitted I = 50% of I = 0 F 0 2 At second polarizer intensity transmitted 2 = I0 cos θ but light incident on the second polarizer B D I 1 = 0 so light transmitted = × cos2 60° 2 2 1 1 I = × = 0 2 4 8 Could also tell from the way the wave bends. V Ratio = I = 3.0 = 2.0 VR 1.5 Practice questions (c) (i) The sign of velocity changes ⇒ direction changes ⇒ body is oscillating. 1 v (m s–1) wavelength 6 M

t (m s) amplitude 3 displacement (mm)

(ii) Time period = 3 m s F = 1 = 1 = 330 Hz T 0.003 distance / m (iii) Maximum displacement is when Note: the graph in some editions of the book velocity is zero – think of a pendulum; it is not the same as the original, which was stops at the top. much easier. (iv) To fi nd area either count squares or Sound is a longitudinal wave. You may be (a) make a triangle that is a bit higher than confused by the graph which looks transverse the top. but remember this is a graph not the wave. Squares are rather small so I prefer to (b) (i) Wavelength = 0.5 mm 1 calculate 2 × 7 × 0.0015 = 5.25 mm (ii) Amplitude = 0.5 mm (iii) Speed = 330 ms–1 7 m s 2 (a) A ray shows the direction of a wave and a wavefront is a line joining points that are in phase. A ray is perpendicular to 1.5 m s a wavefront. (v) Area under v–t graph is displacement. 7

M17_IBPH_SB_IBGLB_9021_SL05.indd 7 17/07/2014 15:27 In this case it is displacement between 1200 –1 (c) (i) From gradient vp = = 9.6 km s two times when velocity = zero 125 (ii) v = 1200 = 5.8 km s–1 i.e. 2 × amplitude. s 206 (d) (i) P S

P wave is fastest so gets to the detector first.

(ii) L3 is closest because the signal arrives first. (iii) 1. First pulse arrives first. 3 (a) (i) The speed of a wave is the distance 2. Separation of pulses is shorter. travelled by the wave profile per 3. Amplitude of pulses is bigger. unit time. (iv) Measure on the graph where the d horizontal distance between the lines is 68 s, 42 s and 27 s (approximately)

So if wave progressed distance d in time t, v = d t displacement (v) (ii) Velocity = time where displacement is the distance moved 1 2 in a certain direction. However light 3 spreads out in all directions. earthquake (b) (i) Displacement is how far a point on a ~ 1060 km, 650 km, 420 km wave is moved from its original position. L L L For example on a water wave how far 1 2 3 Closest to L furthest from L up or down a point is relative to the 3 1 original flat surface of the water. (e) (i) antinode (ii)

node

In a longitudinal wave, e.g. a wave in antinode a slinky spring, the displacement is in the same direction as the wave but in a transverse wave, e.g. water, the (ii) If the standing wave is as shown then displacement is perpendicular to the wavelength = 2 × height of building direction. λ = 2 × 280 = 560 m v 3400 displacement v= fλ so f = = displacement λ 560 = 6.1 Hz (about 6 Hz) direction direction longitudinal transverse

M17_IBPH_SB_IBGLB_9021_SL05.indd 8 17/07/2014 15:27 So if the earthquake has a frequency of Since angle θ is 0.25 3 × 0.25 6 Hz, the building will be forced to vibrate θ = λ = ⇒ λ = = 5 × 10–4 mm at its own natural frequency so will have a 3 1500 1500 = 500 m large amplitude vibration (resonance). μ Note: You probably don’t need to do part e 4 (a) Superposition is what happens when in the core; this is now only part of the EM two waves coincide: the displacements option. of waves add vectorially to produce a 5 (a) (i) (ii) resultant wave. direction of wave (b) longitudinal

wave X direction of wave transverse T (b) In the wave progresses 1 cycle is 1λ 4 4 4 wave Y λ

A M If X and Y are added they give the resultant. (c) (i) Two sources are said to be coherent if they have the same frequency, similar amplitude and a constant phase N difference. Wave moves left so trough is about to

(ii) To get interference the light from S1 and reach M. S must overlap; this only happens if 2 (c) (i) λ = 5.0 cm; v = 10 m s–1 the light is diffracted which means the v 10 v = fλ ⇒ f = = = 2.0 Hz slits must be narrow. λ 5 1 1 1 (d) (i) For maximum intensity the path (ii) In 4T wave moves 4λ = 4 × 5.0 difference = nλ where n is a whole = 1.25 cm number. (d) When two waves coincide the resultant (ii) displacement at any point is equal to the I vector sum of the individual displacements. maximum at P and 0; question states no

other maximum between A1 A2 P and 0

A + A 1 2 O P distance If waves that have the same frequency (e) and a constant phase overlap then, due to superposition, they will add or cancel out. This is called interference. 0.25 mm θ (e) (i) If the path difference = nλ then there

3 mm θ will be constructive interference. n is an integer. S X (ii) Since angles are small θ = 2 λ d 1.5 m 9

M17_IBPH_SB_IBGLB_9021_SL05.indd 9 17/07/2014 15:28 y (iii) Again small angles so ϕ = n 7 (a) (i) (ii) D A A (f) (i) θ = 2.7 × 10–3 rad d = 1.4 mm S X θ = 2 where S X = 8λ d 2 d 1.4 × 10–3 × 2.7 × 10–3 so λ = θ = 8 8 N = 473 nm (ii) From geometry: ϕ = θ = 2.7 × 10–3 rad D = 1.5 m y ϕ = n ⇒ y = ϕ × D = 4 mm A N D n (b) (i) Frequency of 1st pipe = 512 Hz spacing = 4 = 0.5 mm 8 Wavelength of wave = v = 325 f 512 6 (a) = 63.5 cm 63.5 pipe is 1λ so length = = 31.7 cm 2 2 v V (ii) The length of a closed pipe is shorter than an open one of the same frequency so if the organ pipes are closed they take up less space. Vt 8 (a) (i) When light passes through the aperture vt of the lens it will be diffracted (spread (b) In time t distance moved by source = vt out). distance moved by sound = Vt So all waves produced in time t are (ii) squashed into a distance Vt – vt ahead of the source, so number of complete cycles

produced = f0t Vt – vt V so λ = but f = A circular opening leads to a circular f t λ 0 V × f diffraction pattern, as shown. so apparent frequency f = V = 0 1 λ V – v (b) 17 mm A bit confusing using V and v. V (c) Using the formula Δλ = λ λ = 550 nm c V 0.004 = × 600 α 4 μm 3 × 108 V

(i) angle is small so 4 × 10–6 α = = 2.4 × 10–4 rad 17 × 10–5 1.22 (ii) If α = λ then resolved. V = 0.004 × 3 × 108 = 2 × 103 m s–1 d 600 –9 So d = 1.22 × 550 × 10 = 2.8 mm = 2 km s–1 2.4 × 10–4

M17_IBPH_SB_IBGLB_9021_SL05.indd 10 17/07/2014 15:28 2 x 2 Equation of a wave = A sin ω t – π Challenge yourself ( λ ) This wave travels from left to right, since points to the right of the origin lag behind the origin. 1 F 1 Equation of a wave travelling from right to left F2 2 x = A sin ω t + π 2 cm ( λ ) If these waves superpose the resultant X displacement is given by 2 x 2 x A sin ω t – π + A sin ω t + π ( λ ) ( λ ) a + b a – b 10 cm (h) But sin a + sin b = 2 sin cos ( 2 ) ( 2 ) displacement = 2A sin ω t – 2πx + ω t + 2πx (( 2 2λ ) ( 2 2λ )) cos ω t – 2πx – ω t + 2πx (( 2 2λ ) ( 2 2λ )) 2 x y = 2A sin (ωt) cos π , since cos(–a) = cos(a) ( λ ) This displacement is zero when cos 2πx = 0 W ( λ ) 2πx π 3π W which is when = , etc. λ 2 2 When floating the forces are balanced so when x = λ, 3λ etc. 4 4 F = W 1 These points are separated by λ. F1 = buoyant force = weight of fluid displaced = 2

hAρw g

W = mg = hAρw g When pushed down there is extra buoyant force

= xAρw g

Resultant upward force = –xAρw g (if x taken to be in the positive direction this is –) xA g acceleration = – ρw m

but m = hAρw (from the condition of equilibrium) –xA g a = ρw hAρw a = – g x (h) This implies simple harmonic motion so g a = –ω 2x, so ω 2 = h 2 ω = π = g T h T = 2π h = 0.6 s g

M17_IBPH_SB_IBGLB_9021_SL05.indd 11 17/07/2014 15:28 Worked solutions

Chapter 6 5 μ Exercises 0.2 C + m 1 E = 40 NC–1 = 0.01 kg 0.5 N/C q = 5 × 10–6 C F –6 –4 (a) From deﬁ nition of E, E = ⇒ F = Eq F = Eq = 40 × 5 × 10 = 2 × 10 N q μ μ 2 F = 3 × 10–5 N so F = 0.5 × 0.2 C = 0.1 N q = –1.5 × 10–6 C (b) From Newton’s second law F = ma F 3 × 10–5 –6 –1 F 0.1 × 10 –5 –2 E = = –6 = 20 N C a = = = 1 × 10 ms in the q –1.5 × 10 m 0.01 (direction is south, opposite to force since direction of the ﬁ eld. charge is negative) 6 0.2 m 3 a = 100 ms–2 + q = –1.6 × 10–19 C + +

m = 9.1 × 10–31 kg + + 50 μC + F = ma = 9.1 × 10–29 N –29 + + E = F = 9.1 × 10 = 5.7 × 10–10 NC–1 + q 1.6 × 10–19

9 –6 4 v = kQ = 9 × 10 × 50 × 10 = 2.25 × 106 v 10 cm r 0.2 9 –6 7 0.4 m from the sphere V = 9 × 10 × 50 × 10 0.4 = 1.13 × 106 V potential difference = (2.25 – 1.13) × 106 μ 2 C = 1.13 × 106 V

(a) Using the equation for electrical ﬁ eld 8 strength of a sphere B E = kQ r2 9 –6 E = 9 × 10 × 2 × 10 = 1.8 × 106 NC–1 (10 × 10–2)2 F A D (b) 10 cm from the sphere r = 20 cm Q1 Q2 9 –6 so E = 9 × 10 × 2 × 10 20 V (20 × 10–2)2 10 V –40 V E = 4.5 × 105 NC–1 0 V –30 V –20 V C (c) Using the equation E = F ⇒ F = Eq q –10 V So F = 0.1 × 10–6 × 4.5 × 105 1 m = 4.5 × 10–2 N E ε (d) Relative permittivity = = 4.5 so ε = 4.5ε ε 0 0 1 (a) Q is negative since the potential near it is F is proportional to ε 2 negative. If sphere is surrounded by concrete F (b) A positive charge would move to a position F = air = 0.045 = 0.01 N 4.5 4.5 of lower potential, i.e. towards Q2 1

M18_IBPH_SB_IBGLB_9021_SL06.indd 1 22/07/2014 16:39 9 Field strength is greatest where the potential 16 VB = 5 V

gradient is greatest (equipotential closest) so VD = 0 V position F potential difference = 5 V → 10 (a) A – C | 0 – –20| = 20 V 17 PE = V0q = 5 × 3 = 15 J → (b) C – E |–20 – –10| = 10 V 18 Change in potential from C → B = 5 – 3 = 2V (c) B – E → |–10 – –10| = 0 V Work done = ΔV × q = 2 × 2 = 4 J 11 (a) C → A; ΔV = 20 V; work = ΔVq = 20 × 2 19 VAB = 5 – 1 = 4 V = 40 J Work done = ΔV × q = 4 × –2 = –8 J (b) E → C; ΔV = –10 V; work = ΔVq = –10 × 2 20 (a) It would accelerate downwards. = –20 J (b) gain in KE = loss in PE = ΔV × q (c) B → E; ΔV = 0 V; work = 0 J = 4 × 3 12 = 12 J

B 0.5 m 21 VAB = 5 – 1 = 4 V gain in KE = 4 eV

22 VCD = 3 – 0 = 3 V WD moving electron = 3 eV 23 (a) ρ = M so V = M F A D V ρ Q1 Q2 0.0635 20 V = 7.1 × 10–6 m 3 10 V –40 V 8960 (b) 1 mole contains 6 × 1023 molecules 0 V –30 V 6 × 1023 –20 V C atoms per unit volume = 7.1 × 10–6 –10 V = 8.5 × 1028 m –3 1 m one electron per atom so electrons per unit E volume, n = 8.5 × 1028 m –3 each small square = 0.1 m (c) I = nAve ⇒ v = I ΔV nAe Field strength = A = πr2 Δx v = 1 The potential difference across D = 10 V (using (8.5 × 1028 × π × (0.5 × 10–3)2 × 1.6 × 10–19) the two nearest lines) = 9.4 × 10–5 ms–1 Distance between lines = 0.2 m ρ 24 R = L 10 A E = = 50 Vm–1 ρ 0.2 A = L = 1.1 × 10–6 × 2 = 4 × 10–7 m 2 Only an estimate since ﬁ eld not uniform R 5 π 2 A –4 A = r r = π = 3.7 × 10 m 13 At point A potential = 0 V = V1 + V2 So diameter = 3.7 × 10–4 m kQ kQ –9 Q 0 = 1 + 2 = 9 × 109 1 × 10 + 2 r r ( 0.5 1.5) ρL 2000 1 2 25 R = = 1.7 × 10–8 × = 10.8 Ω A π × (0.1 × 10–2)2 –1.5 –9 Q2 = × 1 × 10 = –3nC 0.5 26 9 V 14 (a) E → A; ΔV = 10 V; work done = –10 eV 3 mA (b) C → F; ΔV = 50 V; work done = –50 eV R (c) A → C; ΔV = –20 V; work done = 20 eV Using Ohm’s law V = IR 15 V = 1 V A V 9 R = = –3 VC = 3 V I 3 × 10 potential difference = 2 V = 3 × 103 = 3 kΩ 2

M18_IBPH_SB_IBGLB_9021_SL06.indd 2 22/07/2014 16:39 27 V 31 23 Ω

1 μA Ω 300 k I Ω Using Ohm’s law 1 V = IR –6 3 V = 1 × 10 × 300 × 10 12 V –3 = 300 × 10 Total resistance in the circuit = 24 Ω = 0.3 V Using Ohm’s law V = IR V 12 28 I = = = 0.5 A 12 V R 24 Using Ohm’s law again the pd across the 23 Ω I Ω 600 resistor = IR = 0.5 × 23 = 11.5 V Using Ohm’s law 32 20 Ω V = IR V 12 I = = 5 A R 600 I = 0.02 = 20 mA (a) Energy per second is power. Using the equation P = I2R 29 Using Ohm’s law P = 52 × 20 = 25 × 20 V = IR P = 500 W R = V Therefore 500 J is converted in 1 second. I (b) In 1 minute, 500 × 60 = 3 × 104 J will be V (V) I (mA) V/I kΩ released. 1.0 0.01 100 10.0 0.10 100 33 25.0 1.00 25 0.25 A Ω 30 11

0.5 Ω 0.5 A Using P = I2R The power dissipated in the internal resistance R = 0.252 × 0.5 = 0.031 W 6 V Using Ohm’s law V = IR 34 4 W pd across 11 Ω resistor = 0.5 × 11 = 5.5 V

This means pd across R = 0.5 V 0.5 A

0.5 V

R 0.5 A 9 V Power delivered by battery = IV Using Ohm’s law again R = V = 1 Ω I = 9 × 0.5 = 4.5 W If 4 W are dissipated in the external resistance 0.5 W must be dissipated in the internal resistance.

M18_IBPH_SB_IBGLB_9021_SL06.indd 3 22/07/2014 16:39 35 30 m/s 40 8 Ω 8 Ω

1000 kg

1 2 14 Ω 2 Ω (a) KE = 2 mv 1 2 = 2 × 1000 × 30 The top two and the bottom two are in series so = 450 kJ they simply add. (b) Ignoring friction etc. the power of the car This circuit can then be simpliﬁ ed: 16 Ω = energy gained = 450 000 = 37.5 kW energy taken 12 (c) Using P = IV 37 500 = I × 300 16 Ω I = 125 A Two equal resistors in parallel have a combined 1 36 No energy is lost, no heat produced, motor is resistance of 2 of one of them so: Ω 100% efﬁ cient, no friction Rtotal = 8

37 100 W 41 4 Ω 8 Ω 16 Ω

220 V These are in series so the resistances simply add (a) Using P = IV Ω RT = 4 + 8 + 16 = 28 I = P = 100 = 0.45 A V 220 42 16 Ω (b) If 20% of 100 W is converted to light, Ω 20 × 100 = 20 W are converted. 8 100 That’s 20 J per second. 4 Ω

38 (a) Using P = IV These 3 are in parallel so I = I + I + I P 1000 RT R1 R2 R3 I = = = 4.5 A 1 1 1 V 220 = + + 16 8 4 (b) If the power is 1 kW then the heater = 1 + 2 + 4 = 7 releases 1000 J per second. In 5 hours, 16 16 16 16 5 × 60 × 60 × 1000 = 1.8 × 107 J are 16 Ω RT = released. 7 43 6 V 39 16 Ω

A 8 Ω 10 Ω 2 Ω These resistors are in parallel so: I = I + I RT R1 R2 = 1 + 1 = 1 + 2 16 8 16 16 V 1 = 3 R 16 T Total resistance = 12 Ω 16 R = Ω so using Ohm’s law I = V = 6 = 0.5 A T 3 R 12 0.5 A ﬂ ows through the 10 Ω resistor so V = IR = 0.5 × 10 = 5 V 4

M18_IBPH_SB_IBGLB_9021_SL06.indd 4 22/07/2014 16:39 44 6 V 47 6 V

A 1 kΩ 1 kΩ 4 Ω

2 kΩ 2 Ω 2 Ω V

Without meter pd = 3 V V Resistance of 1 kΩ plus meter I 1 1 3 2 Ω = + = ⇒ R = = 0.67 kΩ The two 2 resistors are in series so add up to R 1 2 2 3 give 4 Ω. This combination is in parallel with the Total resistance = 1.67 kΩ 4 Ω resistor so the total resistance = 2 Ω. Current in whole circuit Using Ohm’s law for the whole circuit I = V = V = 6 × 103 = 3.6 mA V 6 R (R + R ) 1.67 I = = = 3 A total 1 2 R 2 pd across meter = IR = 3.6 × 10–3 × 0.67 × 103 The pd across the two 2 Ω resistors = 6 V. This = 2.4 V will be dropped equally across them so pd Difference = 3 – 2.4 = 0.6 V across each = 3 V % difference = 0.6 × 100% = 20% 45 6 V ( 3 ) 48 6 V V Ω 4 A 0.5 Ω 2 Ω 1 Ω 2 Ω 2 Ω A Without meter R = 3 Ω The pd across the 4 Ω resistor is the same as I = V = 6 = 2 A the battery, 6 V. R 3 The pd across the two 2 Ω resistors is also 6 V. With meter R = 3.5 Ω They are in series so total resistance = 4 Ω I = 6 = 1.7 A 3.5 Using Ohm’s law I = V = 6 = 1.5 A R 4 Difference = 0.3 A 46 6 V % difference = 0.3 × 100% = 15% ( 2 )

49 2 Ω V A 12 V I 4 Ω 1 I AB 4 Ω I2 1 Ω 6 V The voltmeter reads the pd across the battery Ω = 6 V 1 The resistors are in parallel so total resistance Kirchhoff’s ﬁ rst law = 2 Ω V 6 I + I = I (1) Using Ohm’s law I = = = 3 A 1 2 R 2 Kirchhoff’s second law to outer loop

12 = 2I1 + I (2) 5

M18_IBPH_SB_IBGLB_9021_SL06.indd 5 22/07/2014 16:39 Inner loop Fill out the rest to make sure consistent:

6 = I2 + I (3) 6 V 12 V 3 V 6 V 2 Ω 1 Ω Need to ﬁ nd I so substitute for I2 in (3) 3 A

6 = (I – I1) + I = 2I – I1 A V multiply by 2; 12 = 4I – 2I1 add this to (2) (12 = I + 2I ); 1.5 V 1.5 V 1 1 Ω 1 Ω 24 = 5I I = 4.8 A 3 V61.5 A V 1.5 A 2 Ω 2 Ω so VAB = 1 × 4.8 = 4.8 V Check: 51 I = (12 – I) = 3.6 A 2 kΩ 1 2

I2 = 4.8 – 3.6 = 1.2 A Fill in all the unknowns to see if consistent: 1 kΩ 7.2 V 12 V 2 Ω 1 V 3.6 A AB Total of parallel combination 1 = 1 + 1 = 2.5, R 2 0.5 Ω 1.2 A so R = 0.4 k 1.2 V 6 V Ω Ω 1 Rtotal = 0.9 k Current = V = 1 = 1.11 mA 4.8 A 4.8 V R 900 1 Ω pd across 0.4 kΩ = 1.11 × 10–3 × 0.4 × 103 = 0.44 V 50 2 Ω 1 Ω 12 V 6 V Or R2 0.4 Vo = × Vi = × 1 = 0.44 V A R1 + R2 0.9 V ρ 52 Resistance of nichrome = L Ω Ω A 1 1 –6 1 –3 2 Ω = 1.5 × 10 × π × (0.05 × 10 ) = 191 Length divided by 3 gives: 2 Ω 2 Ω 4 1 MΩ Assume meters are all ideal. Ω Ω Ω First ﬁ nd total resistance. 100 143 48 Total for parallel combination at the bottom left 1 V I = 1 + 1 = 1, so R = 1Ω R 2 2 Add the series resistors R Ignore the 1 MΩ as it will draw hardly any Ω current. Rtotal = 2 + 1 + 2 + 1 = 6 Total emf = 18 V Total resistance = 291 Ω V 1 Current = 18 = 3 A current I = = = 3.44 mA 6 R 291 Ammeter reading = 3 A –3 V143 = IR = 3.44 × 10 × 143 = 0.49 V pd across parallel combination = 1 × 3 = 3 V Current through top branch = 3 A 2 pd across 1 Ω resistor = 3 × 1 = 1.5 V 2 Voltmeter reading = 1.5 V 6

M18_IBPH_SB_IBGLB_9021_SL06.indd 6 22/07/2014 16:39 –3 –19 53 ﬁeld into page 56 F = Bqv = 5 × 10 × 1.6 × 10 × 500 (vertically down) = 4 × 10–19 N

N 0.5 m 2 A 57 B

500 V

20 μT

(a) Using the formula F = BIL –6 F = 20 × 10 × 2 × 0.5 r = 10 cm F = 2 × 10–5 N

thuMb (b) (a) KE = Vq = 500 × 1.6 × 10–19 = 8 × 10–17 J (b) KE = 1 mv2 2

–17 v = 2KE = 2 × 8 × 10 = 1.3 × 107 ms–1 m 9.1 × 10–31 seCond (north–south) First (vertically downwards) (c) Moving in a circle so 2 Bqv = mv r Using Fleming’s left hand rule, force is to B = mv the east qr

–31 7 wire with current 9.1 × 10 × 1.3 × 10 54 = –19 out of page (up) 1.6 × 10 × 0.1 = 7.4 × 10–4 T

58 100 ms–1 B = 10 μT v 30° B I = 0.5 A L = 1 m q + 5 mT

(a) Using formula F = BIL F = 10 × 10–6 × 0.5 × 1 m F = Bqv sin θ F = 5 × 10–6 N = 5 × 10–3 × 1.6 × 10–19 × 100 × sin 30° (b) First (directed north) = 4 × 10–20 N

59 B = 50 μT thuMb seCond (upwards) 20 cm

–1 20 m s

(a) emf = BLv Using Fleming’s left hand rule (probably best = 50 × 10–6 × 0.2 × 20 using your own ), force is west = 2 × 10–4 V

–4 55 Use Fleming’s left hand rule: (b) Using Ohm’s law 2 × 10 V V 2 × 10–4 (a) (b) (c) I = = I R 2 B = 1 × 10–4 A I Ω 2 7

M18_IBPH_SB_IBGLB_9021_SL06.indd 7 22/07/2014 16:39 (c) Power dissipated = I 2R = (1 × 10–4)2 × 2 (b) Component of ﬁ eld perpendicular to plane = 2 × 10–8 W (J/s) of coil = B cos 30° (d) Work done = energy dissipated = 2 × 10–8 J Flux enclosed = BAN cos 30° –5 2 –1 = 1.3 × 10 Tm (e) Velocity of wire = 20 m s so moves 20 m in Δ (c) emf = rate of change of ﬂ ux = B 1 s. Δt –5 (f) Work done = F × d = (1.5 – 1.3) × 10 = 0.67 μV 3 work 2 × 10–8 I F = = = 1 × 10–9 N 62 I = 0 ⇒ I = I × 2 distance 20 rms 2 0 rms (Since velocity is constant the forces are I0 = 110 × 2 = 156 V balanced.) 63 Vrms = 220 V; P = Vrms Irms = 4000 W 60 –4 2 Area = 2 × 10 m 50 turns I = 4000 = 18 A rms 220

64 –4 2 A = 5 × 10 m

B = 50 mT B = 100 μT

(a) Flux enclosed by each coil = A × B 500 turns = 2 × 10–4 × 100 × 10–6 = 2 × 10–8 T m2 Since there are 50 turns, total fl ux (a) (i) 50 rev s–1 = 50 × 2π rad s–1 = 50 × 2 × 10–8 = 1 × 10–6 T m2 = 100π rad s–1 μ ε ω (b) If fl ux density changed to 50 T then fl ux (ii) max = BAN = 3.9 V enclosed = 0.5 × 10–6 μT m2 3.9 (iii) ε = = 2.8 V Δ rms 2 Rate of change of fl ux = B = 1.0 – 0.5 Δt 2 (b) 1 the angular velocity so 1 the ε = 1.4 V = 0.25 μT m–2 s –1 2 2 rms V 2 V 2 2002 (c) Induced emf = rate of change of fl ux 65 P = rms ⇒ R = rms = = 48.4 Ω R P 1000 = 0.25 μV 66 Vp = 220 V

61 0.02 m 50 turns Vs = 4.5 V Np Vp (a) If Np = 500, = Ns Vs 500 220 0.03 m 500 μT = Ns 4.5 N = 500 × 4.5 = 10.3 turns (10) s 220 (b) P = IV = 0.45 × 4.5 = 2 W

30° B (c) If 100% efﬁ cient power in = power out

30° Ip Vp = Is Vs

Ip × 220 = 0.45 × 4.5

Ip = 9.2 mA (a) Flux enclosed = BAN (d) If charger not charging then no power = 500 × 10–6 × 0.02 × 0.03 × 50 out, which implies no power in, so no = 1.5 × 10–5 T m2 current ﬂ ows. This is for a 100% ideal transformer.

M18_IBPH_SB_IBGLB_9021_SL06.indd 8 22/07/2014 16:39 67 72 (a) In series 1 = 1 + 1 = 1 + 1 = 3 Power C C C 4 8 8 station 1 2 C = 8 = 2.67 μF 3 50 kV (b) In parallel C = C + C = 12 μF 500 MW 1 2

73 2 μF Transformer

4 μF

100 kV 6 μF 8 Ω

Transformer 6 V

220 V Capacitance of the capacitors in parallel is Town 2 + 6 = 8 μF 1 = 1 + 1 = 1 + 1 = 3 (a) 500 MW at 100 kV; P = VI C C C 4 8 8 P 500 × 106 1 2 I = = = 5 × 103 A 8 V 100 × 103 C = μF 3 2 3 2 (b) Power loss = I R = (5 × 10 ) × R = 200 MW Total charge = CV = 8 × 6 = 16 μC 3 (c) 200 × 100% = 40% 500 Charge on the 4 μF capacitor is equal to total (d) Power delivered = 500 – 200 = 300 MW charge so V = Q = 16 = 4 V C 4 (e) Available to town = 300 MW 74 μ μ 6 4 F 12 F (f) P = VI; I = 300 × 10 = 1.36 MA 220

68 0.1 m 2 μF

0.005 m

12 V ε A –12 π 2 C = 0 = 8.85 × 10 × × 0.05 = 1.39 × 10–11 F d 0.005 Total of the capacitors in series is 1 = 1 + 1 0.1 m 69 C C1 C2 = 1 + 1 = 4 4 12 12 0.001 m C = 3 μF Charge on the capacitors in series is ε ε A –12 π 2 C = r 0 = 4 × 8.85 × 10 × × 0.05 μ d 0.001 CV = 3 × 12 = 36 C μ = 2.78 × 10–10 F Charge on the 4 F capacitor is the same as the total charge on the capacitors in series = 36 μC 70 pd across the 4 μF capacitor = Q = 36 = 9 V C 4 0.01 m 1 2 1 –6 2 –4 75 E = 2 CV = 2 × 5 × 10 × 9 = 2.03 × 10 J 0.0001 m ε A 0.12 0.2 m 76 (a) C = 0 = 8.85 × 10–12 × π × d 0.002 ε ε A 5 × 8.85 × 10–12 × 2 × 0.01 C = r 0 = –10 d 0.0001 = 1.39 × 10 F = 8.85 × 10–9 F (b) Q = CV = 1.39 × 10–10 × 6 = 8.34 × 10–10 C (c) E = 1 QV = 1 × 8.34 × 10–10 × 6 71 Q = CV = 2 × 10–6 × 6 = 1.2 × 10–5 C 2 2 = 2.5 × 10–9 J 9

M18_IBPH_SB_IBGLB_9021_SL06.indd 9 22/07/2014 16:39 (d) If isolated, charge will remain the same but new capacitance is 1 previous 2 Practice questions C = 6.95 × 10–11 F Q = 8.34 × 10–10 C 1 E = 8.1 × 103 J 2 E = Q = 5 × 10–9 J Q = 5.8 × 103 C 2C r Extra energy stored is gained from work done pulling plates apart. ε 0.2 V 1.4 V –19 77 50 electrons so Q = 50 × 1.6 × 10 C R = 6 Ω = 8 × 10–18 C –18 Q 8 × 10 –11 1.2 V V = = –9 = 8 × 10 V C 100 × 10 3 (i) emf = energy per coulomb = 8.1 × 10 78 (a) τ = CR = 5 × 10–3 × 10 × 103 = 50 s 5.8 × 103 = 1.4 V (b) Q = CV = 5 × 10–3 × 10 = 50 mC –t (ii) If ε = 1.4 V and pd across R = 1.2 V RC (c) When discharging V = V0e –20 then pd across r = 0.2 V V = 10 × e 50 = 6.7 V so 1.4 = 0.2 + 1.2, as shown (d) When starting to discharge pd across If you go up 1.4 V you must come down R = 10 V 1.4 V. V 10 I = = = 1 mA Applying Ohm’s law to R R 10 × 103 –t V 1.2 (e) When discharging I = I e RC I = = = 0.2 A 0 R 6 ln I = –t Applying Ohm’s law to r ( I ) RC V 0.2 V 0 r = = = 1.0 Ω I 0.2 A I0 If I = 3 2 (iii) Charge fl owing = 5.8 × 10 C ln 1 = –t Potential difference across R = energy ( 2 ) RC converted to heat per unit charge t = RC × ln 2 = 35 s So energy converted in R = pd × Q 3 3 79 (a) time constant = RC = 1 × 10 × 10–6 = 1.2 × 5.8 × 10 = 6.9 × 10 J = 1 × 10–5 s, so the capacitor will be fully (iv) Current is made up of electron fl ow, as charged after 1 s electrons fl ow through the metal they (b) Initially pd across C = pd across R interact (collide) with the metal atoms, giving them energy. This is rather like Vc = 5 V 5 = IR the way a rubber ball gives energy to the steps as it falls down the stairs. I = 5 0.5 × 106 Increased vibration of the atomic lattice I = 1 × 10–5 A results in an increase in temperature. (c) RC = 0.5 × 106 × 10 × 10–6 = 5 s –t –2 2 (a) V RC 5 V = Voe = 5 × e = 3.35 V

0 I 0 10

M18_IBPH_SB_IBGLB_9021_SL06.indd 10 22/07/2014 16:39 V (b) (i) Resistance = ; this can be found by If we look at circuit 2 we see that I dividing V by I. the pd across the internal resistance (ii) Non-ohmic since the graph is not is 0.4 V. linear. An ohmic resistor should have a Now I = 0.2 A constant value for R. So R = V = 0.4 = 2.0 Ω I 0.2 (c) I = 120 mA V = 6.0 V (c) (i) Apply Ohm’s law: I = 0.18 A, V = 0.6 V V 6.9 (i) R = = = 50 Ω V 0.4 I 120 × 10–3 R = = = 3.3 Ω I 0.2 (ii) 24 V (ii) Ohm’s law again: I = 0.2 A, V = 2.6 V R = V = 2.6 = 13 Ω 120 mA I 0.2 (d) Resistance is different because temperature of bulb is greater in c (ii) 18 V 6 V (e) V If a resistor R is connected in series then the total pd across R and bulb = 24 V. pd across bulb = 6 V so pd across R = 18 V pd across R = 18 V R = V I = 18 = 150 Ω 120 × 10–3 0 I 3 (a) (i) This means that if the bulb is 0 connected to 3 V then 0.6 W of power (f) The circuit is the same as this: is dissipated. I (ii) P = IV ⇒ I = P = 0.6 = 0.2 A V 3 Ω (b) (i) Minimum value is when R is maximum; 12 this can only be zero if maximum value of R is inﬁ nite. 3 V Maximum value of current is when R Ω Ω is zero. Ideally the pd across the bulb 12 4 would then be 3 V but it isn’t because the circuit and battery have resistance. Ω 0.18 A First calculate resistance of the 12 and (ii) A 4 Ω resistors in parallel: 3.0 V 0.6 V V 1

R 12 4

maximum value 0.20 A A

3.0 V 2.6 V 1 1 1 1 + 3 4 2 V = + = = RT 12 4 12 12 R = 12 = 3 Ω 0.4 V T 4 R minimum value 11

M18_IBPH_SB_IBGLB_9021_SL06.indd 11 22/07/2014 16:39 Now add the other 12 Ω V(V) V when I = 0 R = 12 + 3 = 15 Ω 1.5 I

12 Ω

3 Ω

Total R = 15 Ω; apply Ohm’s law to the V 3 whole combination → I = = = 0.2 A 0.42 R 15 Applying Ohm’s law to the 12 Ω resistor → V = IR = 0.2 × 12 = 2.4 V so the pd across the bulb must be

3 – 2.4 = 0.6 V 0.90 I(A)

4 (a) (i) Power from cell = emf × current = EI (d) (i) When I is zero there will be no pd E r across r, so V = E ⇒ E = 1.5 V (ii) If the resistance R is very small then I V V = 0 so current can be found from the intercept on the I axis ≈ 1.3 A R (iii) When R = 0 pd across r = 1.5 V energy converted emf is V 1.5 charge so r = = = 1.2 Ω I 1.3 so emf × current = energy × charge charge time (e) If R = r then R must equal 1.2 Ω energy = = power 1.5 V time (ii) Power dissipated in cell = power 1.2 Ω dissipated in r = I 2r (iii) Power dissipated in external circuit = I 2R = IV Ω 1.2 (b) From the law of conservation of energy: So total R = 2.4 Ω power from cell = power dissipated in circuit Ohm’s law → I = V = 1.5 A 2 R 2.4 EI = I r + VI ⇒ E = V + Ir 2 Power = I 2R = 1.5 × 1.2 = 0.48 W (c) r ( 2.4 ) 5 (a) Gravitational ﬁ eld strength is the force experienced per unit mass by a small test V mass placed in the ﬁ eld. A R 2 (b) Should be GM = g0R From Newton’s law the force on a mass m variable resistor on the surface of the Earth F = GMm R2 Field strength g = F = GMm ⇒ g = GM 0 m R2m 0 R2 ⇒ GM = g R2 12 0

M18_IBPH_SB_IBGLB_9021_SL06.indd 12 22/07/2014 16:39 (c) Use Fleming’s right hand rule Error carried forward: In IB questions you (d) F = B cos θ × V × e generally don’t get penalized for carrying an error forward so if you got part (e) wrong B θ cos and used your answer in part (g), then you θ Component of B perpendicular should get the marks for (g). to wire 6 (a) (i) The emf induced in a conductor placed in a magnetic ﬁ eld is directly proportional F direction of current to the rate of change of the ﬂ ux it encloses.

velocity out of page (ii) The loop encloses B fi eld as shown. If the current changes then the enclosed (e) E = energy converted from mechanical fl ux fi eld will change so, according to work to electrical PE per unit charge. Faraday, emf will be induced. Work done on an electron in pushing it along the wire = force × distance current-carrying wire = B cos θ × ev × L Work done per unit charge = WD e = B cos θ × vL This is the correct answer since question says deduce from (d). However a more obvious solution uses Faraday’s law. E = rate of fl ux cut = B cos θ × area swept out per second (b) When gradient of B versus t is maximum ε = B cos θ × vL since wire moves a distance then is maximum. B ∝ I v in 1 second. I so flux varies as current

(f) For an orbiting body the gravitational force Bϕ ε = d = centripetal force dt 2 so GMm = mv R2 R (iii) When coil is further from the wire, the B (where R is the orbit radius) ﬁ eld will be less so ﬂ ux enclosed will be GM smaller. so v = ϕ R As a result dN is less so ε is less. From the question we know that for the dt 2 (c) Advantage – does not need to be in contact Earth’s surface GM = g0R0 with the wire. (where R0 is the Earth’s radius) = 10 × (6.4 × 106)2 Disadvantage – distance from the wire So GM = 4.1 × 1014 N m2 k g –1 should be known. Height = 3 × 105 m V 7 0 so R = 3 × 105 + 6.4 × 106 = 6.7 × 106 m 4.1 × 1014 so V = = 6.1 × 107 I 6.7 × 106 0 = 7.8 × 103 m s–1 (g) From answer to (e), E = B cos θ × vL R so E = 6.3 × 10–6 cos 20° × 7.8 × 103 × L Power = Vrms Irms = 1000 V v0 I0 L = 1000 where V = and I = 6.3 × 10–6 cos 20° × 7.8 × 103 rms 2 rms 2 = 2.2 × 104 m 13

M18_IBPH_SB_IBGLB_9021_SL06.indd 13 22/07/2014 16:39 v I 2 Power = 0 0 rearranging, we fi nd r3 = kQ × 2L 2 mg The answer is A. 9 –10 2 = (9 × 10 × (5.5 × 10 ) × 2 × 0.5) (10 × 10–6 × 10) 8 For an ideal transformer no power is lost. = 2.7 × 10–5 power in = power out and r = 3 cm VpIp = VsIs The answer is C. 2 A vacuum cleaner has an electric motor, which This is always true only if the transformer is consists of a coil rotating in a magnetic fi eld. ideal; however, none of the other answers When a coil rotates in a magnetic fi eld an emf makes any sense. will be induced in it that opposes the change producing it. This means that the induced emf will oppose the current fl owing through the coil. If there were no resistance this emf would equal Challenge yourself the applied emf and no current would fl ow. When the motor starts there is no induced emf 1 The situation looks like this: (back emf) opposing the current so the current is much larger than when running. This can cause the circuit breaker in the house to switch off. To prevent this a variable resistor could be θ placed in series with the motor; this is reduced as the motor starts to rotate.

3 12 V 4 μF L

T θ 2 μF 2 μF

F When connected to the battery Q = CV = 24 μC 1 2 –6 2 r Energy stored = 2 CV = 0.5 × 2 × 10 × 12 μ W = 144 J When connected to the other capacitor the total Taking components: charge is conserved and the pd across each is θ vertical T cos = mg the same. θ horizontal T sin = F (the electric force) Q = Q + Q F 1 2 Dividing gives tan θ = V = V = V mg 1 2 So Q = C V + C V = C V + C V but the angle is small so we can approximate: 1 1 2 2 1 2 24 = 4V + 2V tan θ ≈ (r/2) = r L 2L V = 4 V kQ Q kQ2 We also know that F = 1 2 = , Energy stored = 1 C V 2 + 1 C V 2 r2 r2 2 1 2 2 –6 2 –6 2 with each sphere taking half of the total charge = 0.5 × 4 × 10 × 4 + 0.5 × 2 × 10 × 4 μ (5.5 × 10–10 C) = 48 J μ 2 2 Change = 96 J so r = (kQ /r ) 2L mg

M18_IBPH_SB_IBGLB_9021_SL06.indd 14 22/07/2014 16:39 Worked solutions

Chapter 7 5 Exercises –0∙54 eV –0∙85 eV

1 (a) KE = V e = 0.6 × 1.6 × 10–19 max s –1∙51 eV = 9.6 × 10–20 J (b) λ = 422 nm ➞ f = c = 7.1 × 1014 Hz λ –3∙39 eV ϕ ⇒ ϕ (c) KEmax = hf – = hf – KEmax = 6.63 × 10–34 × 7.1 × 1014 – 9.6 × 10–20 = 3.7 × 10–19 J ϕ –19 (d) ϕ = hf ⇒ f = = 3.7 × 10 0 0 h 6.67 × 10–34 = 5.6 × 1014 Hz –13∙6 eV

c 15 10 possible transitions as shown. 2 (a) f = λ = 2.1 × 10 Hz λ = 144 nm, ϕ = 4.3 eV 6 –0∙54 eV E = hf = 6.63 × 10–34 × 2.1 × 1015 –0∙85 eV = 1.38 × 10–18 J 1eV = 1.6 × 10–19 J –1∙51 eV 1.38 × 10–18 E(eV) = –34 = 8.6 eV 6.63 × 10 –3∙39 eV

(b) KEmax = hf – hf0 = 8.6 – 4.3 = 4.3 eV

Vs = 4.3 V (d) hf = 4.3 × 1.6 × 10–19 0 –13∙6 eV –19 f = 4.3 × 1.6 × 10 0 6.63 × 10–34 The maximum energy is released from the 15 biggest transition. f0 = 1.0 × 10 Hz 3 No electrons emitted since i.e. from –0.54 ➞ –13.6 eV 7.1 × 1014 < 1.0 × 1015 Energy released = 13.6 – 0.54 V = 13.06 eV 4 KE = hf – ϕ, KE = 1.4 eV, ϕ = 5.0 eV max max This is equivalent to 13.06 × 1.6 × 10-19 J ϕ = 2.09 × 10–18 J hf = KEmax + = 1.4 + 5.0 = 6.4 eV Using the equation E = hf –19 E 2.09 × 10–18 hf = 6.4 × 1.6 × 10–19 J ⇒ f = 6.4 × 1.6 × 10 f = = 0 0 h h 6.63 × 10–34 15 = 1.5 × 1015 Hz f = 3.15 × 10 Hz

M19_IBPH_SB_IBGLB_9021_SL07.indd 1 22/07/2014 16:40 –19 –18 7 –0∙54 eV This is 13.6 × 1.6 × 10 J = 2.18 × 10 J –0∙85 eV Using E = hf

–18 f = E = 2.18 × 10 = 3.28 × 1015 Hz –1∙51 eV h 6.63 × 10–34 ε n2h2 9 r = 0 πme2 –3∙39 eV Use n =1 to ﬁ nd radius of atom when in lowest energy state

–12 –34 2 r = 8.85 × 10 × 1 × (6.63 × 10 ) π × 9.1 × 10–31 × (1.6 × 10–19)2 –11 –13∙6 eV = 5.3 × 10 m The minimum energy is given by the smallest 10 E = –13.6/n2 eV transition. ΔE = –13.6 1 – 1 (n2 n2 ) –0.54 ➞ –0.85 2 1 = –13.6 1 – 1 = –10.2 eV E = 0.85 – 0.54 = 0.31 eV (22 1 ) Δ –19 ΔE = hf, so f = E = (10.2 × 1.6 × 10 ) This is equivalent to 0.31 × 1.6 × 10– 19 J h 6.63 × 10–34 = 2.5 × 1015 Hz = 4.96 × 10–20J 11 (a) KE = 100 eV, V = 100 V Using E = hf (b) 1eV = 1.6 × 10–19 J E 4.96 × 10–20 f = = = 7.44 × 1013 Hz 100 eV = 100 × 1.6 × 10–19 J h 6.63 × 10–34 KE = 1.6 × 10–17 J 8 –0∙54 eV (c) KE = 1 mv2 ➞ v = 2 × KE –0∙85 eV 2 m –17 = 2 × 1.6 × 10 = 5.9 × 106 m s–1 9.1 × 10–31 –1∙51 eV Momentum p = mv = 9.1 × 10–31 × 5.9 × 106 = 5.4 × 10–24 kg m s–1 –3∙39 eV –34 λ = h = 6.63 × 10 = 1.2 × 10–10 m p 5.4 × 10–24 12 Momentum of car = 1000 × 15 = 1.5 × 104 N s

λ = h = 4.4 × 10–38 m p –13∙6 eV Can’t pass a car through such a small opening. To remove an electron from the lowest energy –15 level it must be given 13.6 eV 13 Size of nucleus is about 10 m Δx ≈ 10–15 m O eV ΔxΔp > h/4π

Δp > h ÷ 10–15 4π Δp ≈ 5 × 10–20 N s –13∙6 eV 2 –20 2 KE = 1 mv2 = p = (5 × 10 ) 2 2m 2 × 9.1 × 10–31 = 1.4 × 10–9 J = 8.6 GeV (far too much) 2

M19_IBPH_SB_IBGLB_9021_SL07.indd 2 22/07/2014 16:40 14 KE = 1 mv2 = 1 × 1000 × 202 = 2 × 105 J (a) KE of alpha particle = 7.7 MeV 2 2 = 7.7 × 106 × 1.6 × 10–19 J mass equivalent from E = mc2 = 1.23 × 10–12 J 2 × 105 m = = 2.2 × 10–12 kg 1 2KE (3 × 108)2 KE = mv2 ⇒ v = 2 m 15 (a) E = Vq = 500 eV –12 = 2 × 1.23 × 10 (b) 1 eV = 1.6 × 10–19 J 6.7 × 10–27 v = 1.9 × 107 m s–1 500 eV = 500 × 1.6 × 10–19 = 8 × 10–17 J –17 (b) When alpha particle is closest to nucleus, (c) mass equivalent = 8 × 10 (3 × 108)2 KE = electrical PE –34 = 8.9 × 10 kg ⇒ 1.23 × 10–12 = kQq 8 2 –34 (3 × 10 ) r (d) mass equivalent = 8.9 × 10 × 9 –18 –19 1.6 × 10–19 9 × 10 × 2.1 × 10 × 3.2 × 10 r = –12 = 500 eVc–2 1.23 × 10 r = 4.9 × 10–15 m 16 (a) 35 protons + neutrons 21 17 protons Z Symbol A Mass (u) number of neutrons = 35 – 17 = 18 92 U 233 233.0396

(b) 58 protons + neutrons (a) Z is the number of protons = 92 p 28 protons A – Z is the number of neutrons = 233 – 92 number of neutrons = 58 – 28 = 30 = 141 n (c) 204 protons + neutrons (b) Total mass (u) 82 protons = 92 × 1.007 82 + 141 × 1.008 66 number of neutrons = 204 – 82 = 122 = 234.9405 u

54 (c) Mass defect = mass (parts) – mass 17 26Fe has 26 protons. Each proton has charge + 1.6 × 10–19 C (nucleus) ⇒ Charge of nucleus = 26 × 1.6 × 10–19 C = 234.9405 – 233.0396 = 1.9009 u = 4.16 × 10–18 C (d) Total binding energy = 1.9009 × 931.5 Total number of protons + neutrons = 54 = 1770.6 MeV A proton and a neutron have a mass (e) There are 233 nucleons so ≈ –27 1.67 × 10 kg each 1770.6 BE/nucleon = = 7.6 MeV/nucleon Approximate mass of nucleus 233 = 54 × 1.67 × 10–17 = 9.0 × 10–26 kg 22 10 18 92 protons + 143 neutrons ⇒ atomic mass = 235 8

235 So nuclear symbol is 92U 6 238 19 92U has 92 protons and 238 – 92 = 146 neutrons

binding energy per nucleon (MeV) 4 A different isotope would have 92 protons but a different number of neutrons, e.g. 145. 2 20 Charge of alpha particle = 2e = 2 × 1.6 × 10–19 C = 3.2 × 10–19 C 0 0 50 100 150 200 250 A 3

M19_IBPH_SB_IBGLB_9021_SL07.indd 3 22/07/2014 16:40 212 ➞ 208 4 23 92 Po 92 Pb + 2 He t = 12 min

–0.14 × 12 E released = [mass (Po) – (mass (Pb) + mass At = 40 × e = 7.45 Bq

(He)] × 931.5 MeV –λt 31 At = A0e ; A0 = 20 Bq = 8.95 MeV A ln t = –λt; A = 15.7 Bq ( A ) t 24 The proposed decay equation is 0 A 1 1 λ = ln 0 × = ln ( 2.0 ) × 213 ➞ 213 β+ ν ( A ) t 15.7 10 84 Po 85 At + + ¯ t Energy released = 2.4 × 10–2 year–1 = [mass (Po) – mass (At)] × 931.5 MeV In 2 t 1 = = 28.6 years = –73 keV 2 λ 32 (a) 5.27 years = 5.27 × 365 × 24 × 60 × 60 This is negative (meaning energy would need = 1.66 × 108 s to be supplied to make it happen) so won’t In 2 happen naturally. (b) λ = = 4.17 × 10–9 s–1 t 1 2 139 ➞ 139 β ν (c) 60Co has mass no 60 25 56 Ba 57 La + + ¯ ⇒ 60 g contains 6.02 × 1023 atoms Energy released = [mass (Ba) – mass (La)] × 931.5 MeV ⇒ 1 g contains 1 × 6.02 × 1023 = 1.0 × 1022 atoms = 2.32 MeV 60 (d) Activity = dN = – λN Δ γ dt 26 E = energy = 4.17 × 10–9 × 1.0 × 1022 = 4.17 × 1013 s –1 5.485 – 5.443 = 0.042 MeV (e) If activity = 50 Bq then sample contains 50 ΔE 0.042 × 106 × 1.6 × 10–19 = 1.2 × 10–12 g E = hf; f = = 4.17 × 103 h 6.63 × 10–34 33 (a) 2 H + 2 H ➞ 3 He + 1 n = 1.0 × 1019 Hz 1 1 2 0 Change in mass 27 Half-life is 4 s, so 16 s is 4 half-lives, so sample = [2.014 101 + 2.014 101] will halve 4 times. – [3.016 029 + 1.008 664] Original sample contained 200 g so after 4 half- = 0.003 509 u lives will contain Energy released = 0.003 509 × 931.5 1 1 1 1 200 × 2 × 2 × 2 × 2 = 12.5 g = 3.268 MeV (b) 2 H + 2 H ➞ 3 H + 1 p (1 H is a proton) 28 Rate of decay halves each half-life. 1 1 1 1 1 Half-life is 15 s so 42 s is 3 half-lives. Change in mass If activity was 100 decay/s then after 42 s it = [2 × 2.014 101] – [3.016 049 + 1.007 825] will be = 0.004 328 u 100 × 1 × 1 × 1 = 12.5 decay/s 2 2 2 Energy released = 0.004328 × 931.5 1 1 1 1 1 14 29 16 is 2 × 2 × 2 × 2 so if the amount of C has = 4.032 MeV 1 14 reduced to the wood is 4 half-lives of C old 2 3 ➞ 4 1 16 (c) 1 H + 2 He 2 He + 1 p = 4 × 6000 = 24 000 years. Change in mass

30 A = 40 Bq; t 1 = 5 mins 0 2 = [2.014 101 + 3.016 029] λ = ln 2 = 0.14 min–1 – [4.002 603 + 1.007 825] 5 –λt = 0.019 702 u At = A0e Energy released = 0.019 702 × 931.5 4 = 18.35 MeV

M19_IBPH_SB_IBGLB_9021_SL07.indd 4 22/07/2014 16:40 236 ➞ 100 126 1 ➞ 34 92 U 42 Mo + 50 Sn + x0 n 39 p + p p + p + p¯¯¯ To balance nucleon number No ➝ Baryon 1 + 1 ➞ 1 + 1 + –1 236 = 100 + 126 + x × 1 Lepton 0 + 0 ➞ 0 + 0 + 0 x = 10 No ➝ Charge 1 + 1 ➞ 1 + 1 – 1 To calculate energy released, ﬁ rst ﬁ nd change in Not ok mass: 40 p + p ➞ p + p + π° 236.045 563 – [99.907 476 + 125.907 653 + 10 × 1.008 664] Baryon 1 + 1 ➞ 1 + 1 + 0 = 0.1438 u Lepton 0 + 0 ➞ 0 + 0 + 0 Energy released = 0.1438 × 931.5 = 133.9 MeV Charge 1 + 1 ➞ 1 + 1 + 0 35 233 U ➞ 138 Ba + 86 Kr + 9n 92 56 36 0 Seems ok Change in mass 41 p + p¯¯¯ ➞ π° + π° = 233.039 628 – [137.905 233 + 85.910 615 + 9 × 1.008 664] = 0.1458 u Baryon 1 + –1 ➞ 0 + 0

Energy released = 0.1458 × 931.5 MeV Lepton 0 + 0 ➞ 0 + 0 = 135.8 MeV Charge 1 + –1 ➞ 0 + 0 – + 36 – e e + e e Seems ok

42 e– + e+ ➞ γ + γ electron absorbs positron emits photon and is photon and is Baryon 0 + 0 ➞ 0 + 0 deﬂected deﬂected Lepton 1 –1 ➞ 0 + 0 37 + e+ e A e+ e– ➞ B Charge –1 –1 0 + 0 γ Seems ok γ 43 e– + e+ ➞ n + γ e– e+ + e+ e No ➝ Baryon 0 + 0 ➞ 1 + 0

electron and positron No ➝ Lepton 1 –1 ➞ 0 + 0 positron emits photon annihilate to form photon that is absorbed by which forms an electron Charge –1 + 1 ➞ 0 + 0 another positron positron pair. e– and e+ can be swapped in all cases to give Not ok an equally valid answer 44 p + p¯¯¯ ➞ n + 38 p + e– ➞ n + No ➝ Baryon 1 + –1 ➞ 1 + 0 Baryon 1 + 0 ➞ 1 + 0 No ➝ Lepton 0 + 0 ➞ 0 + –1 Lepton 0 + 1 ➞ 0 + 1 Charge +1 –1 ➞ 0 + 0 Charge +1 – 1 ➞ 0 + 0 Not ok Seems ok

M19_IBPH_SB_IBGLB_9021_SL07.indd 5 22/07/2014 16:40 – 45 (a) π : charge –1, strangeness = 0 e 50 e– Mesons are quark–antiquark combinations

u¯¯ d X – 2 – 1 ➞ total change = –1 3 3 (b) Ω–: charge –1, strangeness –3 ➞ s s s v v (c) Ξ–: charge –1, strangeness –2 ➞ s s d e e charge –1 –1 –1 Weak interaction but no exchange of charge so 3 3 3 0 (d) Ξ°: charge 0, strangeness –2 X is a Z s s u charge –1 –1 –1 3 3 3 Practice questions 46 neutron proton d d d u 1 (a) (i) When white light passes through a gas u u photons can excite atomic electrons Down changes to up into higher energy levels. When this happens the photon is absorbed 47 (a) (b) resulting in a dark line in the spectrum. d u After absorption the electron will go d red u blue red green back to its original level, re-emitting the photon in a random direction. blue antired red antigreen gas absorbs red

spectrum with 48 c absorption line s white light in red

+ X e

Y (ii) To produce a spectrum the light can If one arrow points into vertex the other must be passed through a diffraction grating; point out so Y arrow points out, this is forwards this will produce interference maxima in time so Y is a neutrino. for each colour at a different angle.

Charge exchange is positive so X is a W+.

49 c¯ s¯

– X e

Y Y arrow must point in so is antineutrino.

Charge exchange is negative so X is a W–.

M19_IBPH_SB_IBGLB_9021_SL07.indd 6 22/07/2014 16:40 c (d) The temperature of the core of the Sun is (b) (i) E = hf where f = λ so high that all atoms would be ionized; –34 8 so E = hc = 6.63 × 10 × 3 × 10 λ 588 × 10–9 this means that chemical reactions such = 3.38 × 10–19 J as burning could not take place. Spectral analysis of the Sun shows that it is mostly (ii) According to the previous calculation H and He so the reaction taking place is absorption of a 588 nm photon will fusion not ﬁ ssion. The amount of energy give an atomic electron 3.38 × 10–19 J produced is also of the right order of of energy. This would correspond to a magnitude. change from the –5.8 to the –2.42 level. (e) (i) Balancing the nucleon numbers (atomic 5.8 – 2.42 = 3.38 mass number) 4 + 4 + 4 = 12 energy/10–19 J Balancing proton number (atomic 0 number) 2 + 2 + 2 = 6 (ii) The energy released is due to loss in –1.59 mass –2.42 –3.00 Mass defect = 3 × mass of He – mass of C 3 × 6.648 325 × 10–27 – 1.993 200 0 × 10–26 –5.80 = 1.2975 × 10–29 kg This is equivalent to mc2 energy –7.64 = 1.2975 × 10–29 × (3 × 108)2

= 1.17 × 10–12 J (To answer this simply try subtracting (f) (i) The other particle emitted in beta decay 3.38 from each level.) is an antineutrino. (c) (i) The Bohr model assumes that electrons (ii) During beta decay the energy released orbit the nucleus, like planets orbiting (change in binding energy) is shared the Sun. To explain the line spectrum of between the daughter, beta particle hydrogen, the electrons can only exist and neutrino. The daughter has a much in certain stable orbits deﬁ ned by their bigger mass than the other two so angular momentum. Absorption of a doesn’t receive much energy, resulting photon of light causes the electrons to in most energy being shared between change to a larger orbit. the beta particle and neutrino. (ii) The Schrödinger model considers (iii) Using the decay equation N = N e–λt the atomic electrons to behave as o where the decay constant λ = In 2 waves trapped in the potential well of half-life the nucleus. Like standing waves in = In 2 = 0.845 s–1 0.82 a string, the ‘electron wave’ can only So fraction remaining after 10 s = N have certain discrete wavelengths and N0 –0.845 ×10 therefore discrete energies. = e = 0.000 213, which is 0.02% (iv) During beta decay a neutron decays into a proton + electron The quark content of a neutron is ddu and a proton is uud so a down quark has changed into an up quark 7

M19_IBPH_SB_IBGLB_9021_SL07.indd 7 22/07/2014 16:40 2 (a) Rate of decay is a nuclear process so is not (d) The alpha decay could leave the nucleus in affected by temperature or pressure of the an excited state, leading to the emission of sample. However the rate of decay does a γ photon. depend on how many nuclei are present. (e) (i) Fusion is when two small nuclei join to form a larger nucleus with higher Property Increase Decrease Stays the binding energy. This results in the same release of energy. Temperature ✓ Pressure ✓ Amount ✓

226 ➞ 4 α 222 (b) (i) 89Ra 2 + 86Rn From text given it is α decay so we

know A – 4, Z – 2 binding energy per nucleon (ii) (loss of mass)c2 = energy released ⇒ mass (Ra) > mass (α + Rn) E = [mass(Ra) – mass (α + Rn)]c2 Δm = 226.0254 – (222.0176 + 4.0026) = 0.0052 u A 1 u ⇒ 931.5 MeV (ii) Nuclear force is very short range, so to So E = 0.0052 × 931.5 = 4.84 MeV fuse, nuclei must get very close. But c. (i) V V M m Rn M m α nuclei are positive, so repel each other.

after + + before The momentum of the particles before To get them close they must move the decay = 0 since the bodies are very fast. This can be achieved if the isolated momentum is conserved so temperature is high. momentum after decay = 0 To increase the number of collisions, This means that the momentum of the the density of nuclei should be high. nucleus is equal and opposite to the This is achieved by increasing pressure. momentum of the alpha particle. In other words they move in opposite 3 (a) (i) Fission is when a large nucleus splits directions. into two smaller ones of roughly equal size. (ii) 222 × vRn = –4 × vα ⇒ vα 222 = – = –55.5 vRn 4 1 2 Radioactive decay is when the nucleus (iii) KEα = 2 mvα emits a small particle (α, β, γ) KE = 1 mv 2 but m = 222 × m and Rn 2 Rn Rn 4 α v v = α R 55.5 2 1 222 vα substituting KE = ( ) m × ( ) 235 1 90 142 1 Rn α (ii) U + n ➞ Sr + Xe + 4 n 2 4 55.5 92 0 38 54 0 = 0.018 × 1 mv 2 2 α Calculate how many neutrons form the

so KERn < KEα change in A: 235 + 1 = 90 + 142 + 4 8

M19_IBPH_SB_IBGLB_9021_SL07.indd 8 22/07/2014 16:40 (iii) During beta decay a neutron ➞ proton So mass required so the number of nucleons is unchanged = 1.1 × 1016 × 3.9 × 10–25 but the number of protons increases = 4.1 × 10–9 kg by 1. 4 (a) (i) A nucleon is a proton or neutron; these ➞ A is unchanged, Z Z + 1 are the particles that make up the nucleus. (b) 102 MeV 65 MeV 90 142 (ii) Nuclear binding energy is the energy required to pull a nucleus apart or the Sr Xe energy released when it is formed form (i) Sr has KE = 102 MeV. This is its constituent nucleons. 102 × 106 × 1.6 × 10–19 J (b) and (c) = 1.63 × 10–11 J mass of Sr = 90 × mass of nucleon 9 F

–17 (MeV)

= 90 × 1.7 × 10 kg (approximately) E 8 –25 U = 1.53 × 10 kg 7 KE = 1 mv2 so v = 2KE 2 m 6 = 1.46 × 107 m s–1 5 momentum = mass × velocity = 2.2 × 10–18 Ns 4 (ii) Momentum of two parts is not the 3 same because the four neutrons will 2 also have momentum. H 1 (iii) Sr 0 4 neutrons 0 25 50 75 100 125 150 175 200 225 250 A Xe (d) BE = (mass of parts – mass of nucleus)c2 Since we don’t know which way the neutrons go it is diffi cult to say which of If mass is in u then can convert to MeV by the arrows should be biggest or their multiplying by 931.5 MeV 3 exact direction. 2He has 2 protons and 1 neutron The way shown is with the neutrons on BE = [(Mass neutron + 2 × Mass proton) – the side of the Xe. Mass He] × 931.5 MeV (c) (i) Energy released = 198 MeV = [(1.00867 + 2 × 1.00728) – 3.01603] = 198 × 106 × 1.6 × 10–19 × 931.5 MeV = 3.17 × 10–11 J = 0.0072 × 931.5 = 6.7 MeV 6.7 –12 BE/nucleon = = 2.2 MeV 25% of this = 7.9 × 10 J 3 2 2 ➞ 3 1 (ii) Q = mcΔT = 0.25 × 4200 × 80 (e) 1H + 1H 2He + 0n = 8.4 × 104 J (i) This is a fusion reaction. (iii) The number of fi ssions to heat the (ii) When two small nuclei fuse the binding 4 water = 8.4 × 10 = 1.1 × 1016 fi ssions energy increases. This means energy 7.9 × 10–12 must be released. Each nucleus has mass 3.9 × 10–25 kg

M19_IBPH_SB_IBGLB_9021_SL07.indd 9 22/07/2014 16:40 hc 5 (b) (i) Photon energy = hf = λ –34 8 = 6.63 × 10 × 3 × 10 = 3.02 × 10–19 J 658 × 10–9 19 = 3.02 × 10 = 1.89 eV 1.6 × 10–19 (ii) A transition from n = 2 ➞ n = 3 is equal

binding energy per nucleon to 1.89 eV so light of this wavelength He will excite electrons from n = 2 ➞ n = 3 and therefore be absorbed. (iii) The Bohr model has the electrons

H orbiting the nucleus in circular orbits. The Schrödinger model has the position of the electrons deﬁ ned by a wave A function resulting in electron probability (a) The de Broglie hypothesis states that all distributions that are not circular. particles have a wave associated with them. The Schrödinger model predicts that This wave gives the probability of ﬁ nding different energy changes have different the particle: its wavelength is related to the probabilities; the Bohr model does not. momentum of the particle by the formula 7 (a) According to the wave model the energy λ = h where h = Planck constant. p in the wave is related to the amplitude, not (b) (i) Gain is KE = eV = 850 eV the frequency. This means that the KE of or in joules photoelectrons should be dependent on the 850 × 1.6 × 10–19 = 1.4 × 10–16 J intensity. (ii) KE = 1 mv 2 so v = 2KE × momentum, However KE is dependent on frequency, 2 m not intensity. p = mv = m 2KE = 2m × KE m This can be explained if we consider light to = (2 × 9.1 × 10–31 × 1.4 × 10–16) be made up of photons. Each photon has = 1.6 × 10–23 Ns energy = hf. This is in the data book. A photoelectron is emitted when the atom absorbs a photon so the KE of a You need to know what is in the data photoelectron is related to f. book in case you need to use a value. –34 Intensity is related to the number of (iii) λ = h = 6.6 × 10 = 4.1 × 10–11 m p 1.6 × 10–23 photons, so increased intensity increases 6 (a) An electron can move from ground state to the number of photoelectrons, not their KE. a higher energy level if (b) (i) From x intercept, threshold frequency n = 3 ______–1.51 eV = 3.8 × 1014 Hz

n = 2 ______–3.40 eV (ii) From gradient, Planck constant –19 n = 1 ______–13.68 eV = (4 – 1) × 1.6 × 10 (13.5 – 6) × 1014 1. It absorbs a photon. = 6.4 × 10–34 J s 2. It gains energy as the gas is heated. (iii) Work function = y intercept = 1.5 eV

M19_IBPH_SB_IBGLB_9021_SL07.indd 10 22/07/2014 16:40 2 (c) So Bqv = mv KE 4 r (14, 4) ⇒ r = mv ⇒ r ∝ m Bq 3 (b) m1 = 20 u r1 = 15 cm 2 m r If r ∝ m then 1 = 1 m r (6, 1) 2 1 r2 = 16.5 cm 20 u 15 0 = 2 4 6 8 10 12 14 m2 16.5 14 –1 3∙8 × 10 Hz m2 = 22 u 1∙5 eV –2 (c) Neon has proton number = 10 This would be given; you are not expected –3 to remember these numbers. –4 so 20 u nucleus has 10 p, 10 n ϕ KE = hf – 22 u nucleus has 10 p, 12 n y = mx + c (They are isotopes.) gradient = h y intercept = – ϕ 10 (a) (i) μ– ➞ e– + γ An electron is a lepton and has an 8 (a) 40 K ➞ 40Ar + β+ + ν 19 18 electron lepton number 1; the μ is a The proton number has gone down by 1 muon with muon lepton number 2. but the nucleon number is These don’t balance. constant ➝ p+ ➝ no + β+ + ν Alternatively according to the standard (b) Rock contains 1.2 × 10–6 g of K and model leptons do not interact across 7.0 × 10–6 g of Ar generations. Originally all of this was K so the original (ii) p + n ➞ p + π0 amount of K = (7.0 + 1.2) × 10–6 p and n are baryons so have baryon = 8.2 × 10–6 g number 1. (c) (i) λ = ln 2 = ln 2 = 5.3 × 10–10 year–1 1 9 π0 t 2 1.3 × 10 is a meson so has baryon number 0 –6 (ii) N0 = 8.2 × 10 g ➞ baryon number not conserved N = 1.2 × 10–6 g (iii) p ➞ π+ + π– N = N e–λt 0 In this example the charge doesn’t N t = –1 ln N = 1 ln 0 λ λ ( ) balance. N0 N t = 1 × ln 8.2 Charge on left = +1 5.3 × 10–10 1.2 Charge on right = +1 – 1 = 0 = 3.6 × 109 years Also baryon number doesn’t balance mv2 9 (a) Circular path ➞ F = ≠ r +1 0 + 0 This is due to magnetic force = Bqv (b) The gluon is the exchange particle of the strong reaction between quarks. The meson is the exchange particle between nucleons so this could also be 2 mv the answer. r

M19_IBPH_SB_IBGLB_9021_SL07.indd 11 22/07/2014 16:40 11 K– + p ➞ K0 + K+ + X A is a π+ meson (as stated in the question) K– su¯ B must be an antiparticle as it points back K+ us¯ in time and the other particle on the vertex K0 ds¯ is a muon so this must be an antimuon (a) The K is a hadron since it is made of two neutrino. quarks. (b) (i) The W boson has charge (+ or –). (b) proton – duu (ii) The W boson has rest mass but the This you have to remember I am afraid. photon has zero rest mass. (c) balancing quarks su¯ + duu ➞ ds¯ + us¯ + sss 14 (a) (i) An elementary particle cannot be split strange 1 ➞ –1–1+1+1+1 into anything smaller. up –1+1+1 ➞ +1 (ii) An antiparticle has the same rest mass down –1+1+1 ➞ +1 as a particle but opposite charge. A lepton is a group of fundamental 12 (a) (i) The force between quarks is the strong particles that do not take part in strong force. interactions, e.g. an electron (–) and its (ii) The exchange particle of the strong antiparticle the positron (+). force is the gluon. (b) Coming into the interaction there is an (b) baryon baryon electron and a positron: going out there are ↓ ↓ two gamma photons. ν¯ + p ➞ n + e+ γ ↑ ↑ e– antilepton antilepton

baryon number 0 + 1 = 1 + 0 conserved

lepton number –1 + 0 = 0 + –1 conserved e+ γ

charge 0 + 1 = 0 + 1 conserved To complete the diagram each vertex must baryon antilepton have two straight lines and a wavy line so ↓ ↓ there must be a straight line joining the ν ➞ + + p n + e vertices. ↑ ↑ γ lepton baryon e– lepton number +1 + 0 ≠ 0 + –1 not conserved

baryon number 0 + 1 = 1 + 0 conserved

charge 0 + 1 = 0 + 1 conserved e+ γ

Second interaction will not happen because lepton number is not conserved. This particle must be an electron (because of the way the arrow is drawn; if drawn 13 (a) B the other way this would be a positron) + w to make the lepton numbers balance. In A this interaction this electron is the virtual u μ+ particle.

M19_IBPH_SB_IBGLB_9021_SL07.indd 12 22/07/2014 16:40 (c) (i) π+ has an up and an antidown quark. 2 22Na ➞ 22Ne + e+ (ii) In this interaction the baryon number is Energy released = (mass (Na) – 11 × mass (e)) – not conserved. [(mass (Ne) – 10 × mass (e)) + mass (e)] = m(Na) –2m – m(Ne) (d) Two particles with spin 1 cannot occupy the e 2 = 21.994 434 – 2 × 0.000 548 6 – 21.991 383 same quantum state. = 0.001 95 (e) Quarks have spin 1 so must obey the Pauli 2 = 0.001 95 × 931.5 MeV = 1.82 MeV exclusion principle. The introduction of colour as a property of quarks allows them to exist in baryons and mesons without violating the principle.

Challenge yourself

1 Proton number = 29 so Binding energy = (29 × mass (H) + 34 × mass (n)) – mass (Cu) = 29 × 1.0078 + 34 × 1.008 66 – 62.929 599 (you have to look that up) = 0.591 04 u = 0.591 04 × 931.5 MeV = 551 MeV = 551 × 106 × 1.6 × 10–19 J = 8.8 × 10–11 J per atom 1 mole of copper has mass 63 g so 3 g has 3 × 6 × 1023 atoms. 63 So 3 g would require 8.8 × 10–11 × 2.86 × 1022 = 2.5 × 1012 J

M19_IBPH_SB_IBGLB_9021_SL07.indd 13 22/07/2014 16:40 Worked solutions

Chapter 8 (b) Each half-life activity falls by 1 Exercises 2 1 × 1 × 1 × 1 × 1 × 1 × 1 × 1 × 1 × 1 2 2 2 2 2 2 2 2 2 2 = 1 1 1024 light ⇒ 10 half-lives

electrical 10 × 11 months = 110 months = 9.2 years.

239 ➞ 96 136 Thicker arrow ⇒ 5 (a) Pu Zr + Xe + xn heat more heat than balancing nucleon number. light (depends on ⇒ type of bulb) 239 = 96 + 136 + x x = 7 (b) 239Pu ➞ 96Zr + 136Xe + 7n 2 This one probably should be much smaller (c) Energy released is found from the change in mass: light 239.052 158 – [95.908 275 + 135.907 213 electrical + 7 × 1.008 664] = 0.176 u

mechanical heat energy released = Δm × 931.5 eV = 164 MeV heat This is what happens (d) 1 mole Pu = 239 g ⇒ this contains to most of the energy 6.022 × 1023 atoms (e) 1 kg is 1000 moles 3 (a) Power = energy ⇒ energy = power × time 239 time so contains 1000 × 6.022 × 1023 atoms 1 day = 24 × 60 × 60 = 8.64 × 104 s 239 = 2.5 × 1024 atoms energy produced = 1000 × 106 × 8.64 × 104 (f) 164 MeV released per nucleus so = 8.64 × 1013 J 2.5 × 1024 × 164 = 4.1 × 1026 MeV (b) % efﬁ ciency = energy out × 100 = 40% energy in (g) 1 eV = 1.6 × 10–19 J so energy in = energy out × 100 energy released = 4.1 × 1032 × 1.6 × 10–19 40 13 13 = 6.6 × 10 J = 8.64 × 10 × 100 = 2.16 × 1014 J 40 6 If 3% of the fuel is 235U then of 1 kg, 3 is 235U (c) Energy density = energy so 100 mass = 0.03 kg 14 energy released mass = energy = 2.16 × 10 energy density = 6 mass energy density 32.5 × 10 = 9 × 1013 J kg–1 = 6.65 × 106 kg ⇒ energy released = 9 × 1013 × 0.03 (d) 1 tonne = 1000 kg; each truck contains = 2.7 × 1012 J 105 kg 7 (a) energy = power × time 6.6 × 106 number of trucks = = 66.5 3 10 105 = 10 000 × 10 × 60 × 60 = 3.6 × 10 J (67 trucks needed) ↑ ↑ 142 ➞ 142 β– ν 10 000 kW 1 hour 4 (a) 56Ba 57La + + ¯

M20_IBPH_SB_IBGLB_9021_SL08.indd 1 22/07/2014 16:41 (b) 1 kg releases 2.7 × 1012 J (d) Each cell ➞ 0.03 A so 10 batteries in ➞ i.e. 2.7 × 1012 J kg–1 parallel 0.3 A Amount of fuel needed to release 3.6 × 1010 J (e) Each cell produces 0.015 W, so to produce 100 –2 100 W requires = 6667 cells = 1.3 × 10 kg = 13 g 0.015 8 1000 Wm–2

etc. 4 m2

10 electrical

(a) Intensity = power light area power = intensity × area = 1000 × 4 = 4 kW heat absorbed ↓ (b) % efﬁ ciency = power out × 100 = 50 power in 11 (a) Using the formula 1ρπr2v3 = power in the ↑ 2 from Sun wind power = 1 × 1.2 × 3.14 × 542 × 103 power out = 4 × 50 = 2 kW (2000 J s–1.) 2 100 6 (c) In 60 s, energy = 2000 × 60 = 1.2 × 105 J = 5.5 × 10 W = 5.5 MW (b) % efﬁ ciency = power out × 100 = 20% This is given to 1 kg of water so using power in θ Q = mcΔ power out = 20 × 5.5 = 1.1 MW 100 1.2 × 104 = 1 × 4200 × Δθ = 28.6°C (c) Since power is proportional to v3, increasing 9 1000 Wm–2 speed by a factor 1.5 increases power by factor 1.53. So power = 1.1 × 1.53 = 3.7 MW Note: power in the wind = 5.5 × 1.53 = 18.6 MW

12 (a) λ = 0.002 89 = 0.002 89 = 5.8 × 10–6 m T 500

P 4 –8 4 2 (b) = σT = 5.67 × 10 × 500 1 cm A (a) Incident power = area × intensity = 3.54 × 103 W m–2 σ 4 π 2 σ 4 = 1 × 10–4 × 1000 = 0.1 W (c) P = A T = 4 r T = 4π × 0.022 × 3.54 × 103 = 18 W 15% of this is absorbed ⇒ 15 × 0.1 100 (d) At 1m, I = P = 18 × 12 = 1.4 W m–2 = 1.5 × 10–2 W 4πr2 4π (b) P = IV ⇒ 1.5 × 10–2 = I × 0.5 13 (a) Direct from Sun = 68 W m–2 I = 0.03 A radiated from Earth = 358 W m–2 (c) 10 × 0.5 V batteries in series ➞ 5 V convection = 105 W m–2 –2 etc. total = 531 W m (b) 31 W m–2

M20_IBPH_SB_IBGLB_9021_SL08.indd 2 22/07/2014 16:41 (c) Reﬂ ected = 110 W m–2 (b) (i) From the graph: incident = 350 W m–2 350 albedo = 110 = 0.31 350 340 (720, 340) (d) Radiated back to the Earth = 332 W m–2 (K) radiated away 199 W m–2 W

ature 330 total = 531 W m–2

action 100 (e) 41 W m–2 passes straight through 320 200 W –2 300 W 399 W m radiated from Earth 400 W output temper power extr 41 × 100% = 10.3% (399) 310

300 Practice questions 200 300 400 500 600 700 800 input power (W)

1 (a) Fossil fuels are continually being made on Input power = 720 W the sea fl oor as organisms die; however, the so if intensity = 800 W m–2 rate at which they are made is much slower will need an area = 720 = 0.90 m2, 800 than the rate they are used up. You don’t really need to understand (b) (i) Nuclear waste: the process of nuclear what is happening; you just read the fi ssion produces radioactive waste, graph. which is diffi cult to dispose of. (ii) The point (500, 320) is between the (ii) Nuclear weapons: although nuclear 100 W and 200 W lines. fuel cannot be used directly to make 350 bombs, the process of enrichment and raw materials are the same. A country with a nuclear power programme could 340 (K) theoretically be producing weapons. W

ature 330 2 (a) A solar panel is a panel containing water pipes that absorb the Sun’s radiation to action 100

320 200 W heat the water. This hot water can be used 300 W 400 W for showers, washing dishes, etc. output temper power extr A solar cell is a semiconductor device that 310 absorbs light, converting it to electrical potential energy. 300 200 300 400 500 600 700 800 input power (W)

Estimate power extraction ≈ 150 W efﬁ ciency = power out × 100% power in = 150 × 100% = 30% 500

M20_IBPH_SB_IBGLB_9021_SL08.indd 3 22/07/2014 16:41 3 (a) Coal is burnt to produce heat; this heats Also assumes all of the energy of the wind water, which turns to steam; the steam is converted to KE of turbine, this is not the turns a turbine, which turns a generator. case. The generator produces electricity. (d) The main disadvantage is that turbines take Chemical energy ➞ Heat ➞ Kinetic energy ➞ Electrical up a lot of space and must be built in windy places. coal water turbine generator It is not easy to build big towers in windy (b) (i) Although coal is being made from places. dead plants it is classifi ed as non- renewable since it is used faster than it 5 (a) (i) This reaction is a fi ssion reaction. is produced. (ii) The energy liberated is given to the (ii) The heavy elements used in nuclear KE of the products. Increasing the KE fuel are not produced on the Earth, so of the atoms results in an increased nuclear fuel is non-renewable. temperature. (c) (i) To maintain a chain reaction, neutrons (b) This is best shown in a diagram. must be absorbed by further nuclei. This won’t happen if the neutrons are U moving too quickly, so they are slowed U down by the moderator. U (ii) The chain reaction can be slowed down by preventing some of the neutrons Neutrons from the fi rst fi ssion are absorbed from being absorbed by the fuel. The by U nuclei, initiating further fi ssions. control rods absorb neutrons slowing down the rea ction. (c) (i) If the neutrons are travelling too quickly, they will pass through the U nucleus. (d) The main advantage is that the nuclear To allow them to be absorbed they are power station does not produce slowed down by the moderator atoms. greenhouse gases. Another advantage is that there is a lot more nuclear fuel moderator remaining in the Earth than coal.

12 4 (a) Annual energy = 120 T J = 120 × 10 J fast neutron neutron loses energy as it 12 so total power = 120 × 10 collides with moderator atoms seconds in 1 year 12 (ii) The control rods are used to slow = 120 × 10 = 3.8 MW 60 × 60 × 24 × 365 the reaction down. They do this by If 20 turbines, each turbine has power absorbing neutrons, preventing them 235 3.8 = 0.19 MW from being absorbed by U, leading to 20 further ﬁ ssions. (b) Power of turbine = 1 ρπr2 v3 2 (d) Fission of U ➞ KE to products. This causes 6 ⇒ r = 2P = 2 × 0.19 × 10 = 12 m the temperature to increase. The hot fuel is ρπv2 1.2 × π × 93 used to turn water into steam, which drives (c) This is an estimate since the wind speed is a turbine. The turbine turns a generator that not always the same; the average will vary produces electricity. from year to year. This calculation also doesn’t take into 6 (a) Only half of the Earth is exposed to the Sun, account energy loss due to friction etc. which absorbs energy as if it were a disc of area πR2. 4

M20_IBPH_SB_IBGLB_9021_SL08.indd 4 22/07/2014 16:41 π 2 So energy power absorbed = 1400 R (b) (i) P = σT4 = 5.67 × 10–8 × 58004 A If we now calculate the power received per = 6.4 × 107 unit area we get π 2 π 8 2 π 2 A = 4 r = 4 × (7x10 ) 1400 R total area of the Earth = 6.16 × 1018 m 2 π 2 7 18 = 1400 R = 350 W m–2 P = 6.4 × 10 × 6.16 × 10 π 2 4 R = 3.9 × 1026 W (b) (i) Emissivity = power radiated by body (ii) At a distance of 1.5 × 1011 m, the power divided by power radiated by a black has spread over a larger area, so body at the same temperature. From 26 power per unit area = 3.9 × 10 the diagram, we see that power 4π × (1.5 × 1011)2 –2 radiated by atmosphere = 0.7 σT4. = 1400 W m A black body radiates σT4, so emissivity (iii) If average power absorbed per unit = 0.7. area = 240 W m–2 then (ii) Power radiated per unit area = 0.7 σT4 = power in = 240 × area 0.7 × 5.67 × 10–8 × 2424 = 136 W m–2 (iv) power out = area × σT4 (iii) Power in = 245 + 136 = 381 W m–2 power in = power out area × σT4 = 240 × area If in equilibrium, power in = power out σT4 = 240 Wm–2 σT 4 = 381 W m–2 E T = 255 K T = 381 = 286 K E 4 5.67 × 10–8 (c) The radiated radiation is in the infrared (c) (i) Carbon dioxide molecules have a region of the spectrum so is absorbed

natural frequency in the infrared region by the CO2 in the atmosphere. After of the electromagnetic spectrum, this absorption, the molecules re-radiate in all means that infrared radiation will cause directions. A proportion of this returns to the the molecule to oscillate and therefore Earth; this increases the temperature. An be absorbed. The temperature of increase in the Earth’s temperature results the Earth is such that the peak in in more power radiated until equilibrium is the electromagnetic spectrum is in maintained.

the infrared region so a lot of the (d) Burning fossil fuels releases CO2 into

power radiated from the Earth will be the atmosphere, this increase in CO2 absorbed. concentration leads to more absorption (ii) The Sun has a temperature of about of infrared radiation, enhancing the 6000 K so radiates in the visible region, greenhouse effect, resulting in more radiation being re-radiated back to Earth. which does not resonate with the CO2 molecules.

(iii) Burning fossil fuels produces CO2.

Plants absorb CO2. 7 (a) The power emitted per unit surface area of a black body is proportional to the fourth power of its absolute temperature P = σT4 A

M20_IBPH_SB_IBGLB_9021_SL08.indd 5 22/07/2014 16:41 2 Distance to Sun, R = 1.5 × 1011 m Challenge yourself radius of Moon, r = 1.7 × 106 m power from Sun, P = 3.8 × 1026 W 1 3 litres of diesel contains 3 × 36 = 108 MJ of power per unit area at Moon = P 4πR2 energy = 1.34 × 103 W m–2 50% efﬁ cient so 54 MJ is converted to useful albedo of Moon = 0.123, so power absorbed work per unit area = (1 – 0.123) × 1.34 × 103 Work done against air resistance: = 1.18 × 103 W m–2 work done = force × distance power absorbed by the Moon = 3 6 = F × 100 × 10 54 × 10 1.18 × 103 × πr2 = 1.07 × 1016 W 3 = F × 100 × 10 power radiated = εσAT4 F = 540 N emissivity, ε = 0.9 –1 This is the force at 50 km h , force when when equilibrium reached, stopped = 0N power in = power out 540 so average force = = 270 N 16 –8 π 2 4 2 1.07 × 10 = 0.9 × 5.67 × 10 × 4 r × T This gives an average acceleration T4 = 5.77 × 109 T = 276 K = F m = 0.27 m s–2 initial velocity = 50 kmh–1 = 14 m s–1 assuming acceleration is constant v2 = u2 + 2as so 0 = 142 – 2 × 0.27 × s s = 196 = 363 m 0.54

M20_IBPH_SB_IBGLB_9021_SL08.indd 6 22/07/2014 16:41 Worked solutions

Chapter 9 4 γ = 1 = 1.4 Exercises (1 – 0.72) Δt = 2 s –1 Δ ′ γ Δ 1 s’ 8 m s t = t = 1.4 × 2 = 2.8 s 5 γ = 1 = 7.1 (1 – 0.992) A C 0.5 m s–1 Δt = 30 s s Δt′ = γ Δt = 7.1 × 30 = 213 s

6 a) Rocket observer uses same clock so measures proper time = 2 years B (b) γ 1 = 1.67 (1 – 0.82) Δ ′ γ Δ (a) Velocity measured by B = 8 + 0.5 t = t = 1.7 × 2 = 3.3 years –1 = 8.5 m s 7 γ = 1 = 1.4 (b) On train C walks 20 × 0.5 = 10 m (1 – 0.72) L (c) Measured by B, C walks 20 × 8.5 = 170 m L = 0 = 2 = 1.43 m γ 1.4 2 x = 100 m 8 γ = 1 = 7.1 t = 4 × 10–8 s (1 – 0.992) v = 2 × 108 m s–1 (a) d = vt = 0.99 × 3 × 108 × 2 × 10–8 = 5.94 m 1 (a) γ = = 1.34 (b) Δt′ = γ Δt = 7.1 × 2 × 10–8 = 14.2 × 10–8 v2 (1– 2 ) c = 1.42 × 10–7 s ′ γ (b) x = (x – vt) (c) d = vt′ = 0.99 × 3 × 108 × 1.42 × 10–7 = 1.34 × (100 – 2 × 108 × 4 × 10–8) = 42.2 m = 123.4 m 2 × 10–8 ms–1 t′ = γ (t – vx ) c2 8 = 1.34 × (4 × 10–8 – 2 × 10 × 100 ) (3 × 108)2 = –2.44 × 10–7 s

γ 1 3 = = 1.67 42.2 m (1 – 0.82) Event 1 t′ = γ (t – vx ) c2 (d) Proper time is measured in nucleus frame = 1.67 × (4 × 10–6 – 0.8 × 0 ) = 6.68 × 10–6 s (3 × 108)2 since the same clock can be used to Event 2 measure at the start and ﬁ nish. t′ = 1.67 × (4 × 10–6 – 0.8 × 100 ) = 6.22 × 10–6 s (e) Proper length is measured on Earth since (3 × 108)2 the start and ﬁ nish do not move relative to Earth.

M21_IBPH_SB_IBGLB_9021_SL09.indd 1 22/07/2014 16:43 c 9 0.8 12 s’ s 0.8 c 0.7 c B

u′ = u – v ; u = –0.7 c; v = 0.8 c 1 – uv c2 5 Iight hours X u′ = –0.7 c – 0.8 c = – 1.5 c = – 0.96 c 1 – –0.7 c × 0.8 c 1.56 c2 γ = 1 = 1.67 13 x = 100 m (1 – 0.82) t = 4 × 10–8 s ′ (a) t = d = 5 = 6.25 hours x = 123.4 m v 0.8 t′ = –2.44 × 10–7 s L 5 (b) L = 0 = = 3 light hours 2 2 8 –8 2 2 γ 1.67 (ct) – x = (3 × 10 × 4 × 10 ) – 100 = –9.86 × 103 m 2 (c) t = d = 3 = 3.75 hours v 0.8 (ct′)2 – x′2 = (3 × 108 × –2.44 × 10–7)2 – 123.42 10 s’ = –9.87 × 103 m 2 s 0.9 c 0.9 c 14

t (years) t'

7 8

7 6 u′ = u – v ; u = –0.9 c; 1 – uv c2 v = 0.9 c 6 5 u′ = –0.9 c – 0.9 c = –1.8 c = –0.99 c 1 – –0.9 × 0.9 c2 1 + 0.81 5 c2 4 x' 11 s’ s’ 4 3 7 c c 0.5 0.6 3 6 5 2 2 4 s 1 3 1 2 1 0 012 3 4 5 6 7 8 x (light years) u′ = u – v ; u = 0.5 c; v = –0.6 c 1 – uv x = 4 light years c2 t = 6.5 years u′ = 0.5 c – –0.6 c = 1.1 c = 0.85 c 1 – 0.5 × –0.6 c2 1.3 x′ = 1 light years c2 ′ If meteor and ship are the other way round the t = 5 years 1 answer is –0.85 c. γ = = 1.15 (1 – 0.52) x′ = γ ( x – vt) = 1.15 × (4 – 0.5 × 6.5) = 0.86 light years 2

M21_IBPH_SB_IBGLB_9021_SL09.indd 2 22/07/2014 16:43 t′ = γ (t – vx ) = 1.15 × (6.5 – 0.5 × 4) 16 c2 t (years) = 5.2 years t'

2 2 2 2 2 7 S = (ct) − (x) = 4 − 6.5 = –26.3 ly 8 S′ = (ct′)2 − (x′)2 = 12 − 52 = –24 ly2 7 6 15 6 t (years) t' 5

7 5 8 4 x' 4 7 6 3 7 3 6 6 5 5 2 2 4 5 4 x' 1 3 1 4 2 3 7 1 6 0 3 012 3 4 5 6 7 8 5 x 2 (light years) 2 4 Δt′ = 4 years 1 3 γ 1 1 2 = = 1.15 (1 – 0.52) 1 0 Δt′ = γΔt = 1.15 × 3.5 = 4 years 012 3 4 5 6 7 8 x (light years) 17 x′ = 5 light years t (years) t' t′ = 1 year 7 x = 6.3 light years 8 t = 4 years γ = 1 = 1.15 7 6 (1 – 0.52) 6 x′ = γ ( x – vt) = 1.15 × (x – 0.5 × 4) = 5 5 x = ( 5 ) + 2 = 6.3 light years 5 1.15 4 x' vx t′ = γ (t – ) = 1.15 × (t – 0.5 × 6.3) = 1 year 4 c2 3 7 t = ( 1 ) + 3.15 = 4.0 years 6 1.15 3 5 S = (ct)2 − (x)2 = 42 − 6.32 = –23.7 ly2 2 2 4 S′ = (ct′)2 − (x′)2 = 12 − 52 = –24 ly2 1 3 1 2 1 0 011.72 3 4 4.3 5 6 7 8 x (light years) length = 4.3 – 1.7 = 2.6 light years

L0 = 3 light years L L = 0 = 3 = 2.6 light years γ 1.15 3

M21_IBPH_SB_IBGLB_9021_SL09.indd 3 22/07/2014 16:43 18 From the view of the rocket, the Earth moves with velocity 0.5c as shown. t (years) t' T light years = 0.5 T light years + 2 light years 7 8 T = 4 years This is the time for the signal to reach Earth but light 7 6 the signal was sent 4 years after launch so it arrives 8 years after launch. 6 5 The Earth observer can measure both launch

5 and arrival of signal with the same clock, so this T 8 4 x' is the proper time. T = = = 7 years after 0 γ 1.15 4 launch. 3 7 Alternative method: 6 3 When rocket has travelled 4 years measured by 5 2 the rocket, the time for an Earth observer 2 4 = 4 × 1.15 = 4.6 years 1 3 1 2 Distance travelled by the rocket in this time 1 = 2.3 light years 0 Time for signal to arrive = time to point of 012 3 4 5 6 7 8 x (light years) sending + time for signal to travel back Note that the rocket is taken to be S and the = 4.6 + 2.3 = 7 years Earth S′ This is the proper time, since it can be Signal arrives on Earth (t′) 7 years after the measured with the same clock on Earth. Time for rocket observer = γT = 7 × 1.15 rocket launched. This is 8 years as measured 0 on the rocket. = 8 years Maybe you are now convinced about space time diagrams.

t (years) t'

7 8

2 light years 0.5 c 7 6

6 5 arrive 5 leave 4 x' 4 3 7 3 6 5 2 2 4 1 3 1 2 1 0 012 3 4 5 6 7 8 910 0.5 T x (light years) light years 2 light years Rocket departs in June 2003 and arrives back T light years in January 2001. If travelling at 2c the rocket arrival and departure 4 would be simultaneous.

M21_IBPH_SB_IBGLB_9021_SL09.indd 4 22/07/2014 16:43 1 2 2 4 2 2 –2 Can’t check since γ = , which we 23 (a) E = m0 c + p c ; m0 = 938 MeV c –15 E 2 = 9382 + 1502; p = 150 MeV c–1 cannot do, even using complex numbers, since E = 950 MeV it would give a complex answer and complex (b) KE = total energy – rest energy = 950 – 938 time does not have a meaning. = 12 MeV 20 If v = 0.8 c (c) Accelerating pd = 12 MV γ 1 = = 1.67 (d) E = γ m c2 ⇒ γ = E = 950 = 1.013 (1 – 0.82) 0 2 m0c 938 1 m = 100 MeVc–2 ⇒ 1.013 = 0 1 – v2 Energy of particle = rest energy + KE c2 v2 = (1 – 1 ) c2 γ 2 2 E = m0c = 1.67 × 100 = 167 MeV 1.013 2 2 4 2 2 but E = m0 c + p c v = 0.16 c ⇒ 2 2 2 2 4 2 2 p c = E – m0 c = 167 – 100 = 17 889 pc = 17 889 = 134 MeV 24 KE of each electron = 1 MeV p = 134 MeV c–1 Total energy = KE + rest energy or = 2 × 1 + 2 × 0.5 = 3 MeV γ –2 This is 1.5 MeV each if the energy is shared M0V = 1.67 × 100 MeV c × 0.8 c = 134 MeV c–1 equally between them.

–1 –2 ⇒ 25 1 MeVc 21 M0 = 200 MeV c rest energy = 200 MeV

KE = 1GeV – 2 MeV e 45° (a) E = rest energy + KE = 1000 + 200 = 1200 MeV e+ 45° E 2 = m 2c4 + p2c2 0 –1 1 MeVc p2c2 = E 2 – m 2c4 = 12002 – 2002 0 Momentum of photon = E = 2 MeV c–1 ⇒ pc = 1183 MeV; p = 1183 MeVc –1 c (b) KE = (γ – 1) m c2 ⇒ 1000 = (γ – 1) 200 Horizontal component of momentum after 0 = 2 × 1 × cos 45° = 1.41 MeV c–1 γ 1 ⇒ 2 1 2 = 6 = 2 v = (1 – ) c 1 – v 62 So momentum of nucleus c2 = 2 – 1.41 = 0.59 MeV c–1 v = 0.986 c

–2 26 22 M0 = 200 MeV c γ = 1 = 1.67 (1 – 0.82) 0.0042 nm v = 0.8 c

γ 2 0.003 nm 60° (a) KE = ( – 1) m0c = (1.67 – 1) 150 = 100.5 MeV (b) Total energy = rest energy + KE = 200 + 100.5 = 300.5 MeV –34 8 E = hc = 6.63 × 10 × 3 × 10 = 0.414 MeV in –19 –9 2 2 4 2 2 λ 1.6 × 10 × 0.003 × 10 (c) E = m0 c + p c E = hc = 0.296 MeV 2 2 2 2 4 2 2 out λ p c = E – m0 c = 400.5 – 200 pc = 260; p = 260 MeV c–1 Energy transferred to the electron = 0.414 – 0.296 = 0.118 MeV

M21_IBPH_SB_IBGLB_9021_SL09.indd 5 22/07/2014 16:43 27 Photon energy = 14.4 keV = 14.4 × 103 × 1.6 × 10–19 J = 2.3 × 10–15 J Practice questions –15 E = hf ⇒ f = E = 2.3 × 10 = 3.48 × 1018 Hz h 6.63 × 10–34 Δ Δ Δ 1 (a) Proper length and time are lengths and f = g h ⇒ Δf = fg h f c2 c2 times measured by an observer in a frame 18 = 3.48 × 10 × 9.8 × 22.6 of reference that is at rest relative to the 8 2 (3 × 10 ) events being measured. Δ 3 f = 8.56 × 10 Hz (b) (i) If Miguel sees the matches light 28 (a) The rocket is an inertial frame of reference simultaneously then the light from so no different from a stationary frame; the each strike must arrive to him at the wavelength will be the same at each end. same time. But to Carmen, Miguel is (b) The rocket is accelerated; this is the same moving towards B so the light from as if it were in a gravitational ﬁ eld, as B has travelled a shorter distance, so shown: if the lights reach Miguel at the same time, the match A must have been struck ﬁ rst.

A was here when B was here Carmen struck match

a Note: This is the other way round if the events are simultaneous on the station but not on the train.

AB AB (ii) Miguel is at rest relative to A and B so

L0 = 20 m Carmen is moving relative to A and B so L L = 0 ⇒ γ = 2 γ As photon goes from A to B it will lose 2 γ = 1 ⇒ 1 = 1 – u energy and its wavelength will increase, so 1 – u2 γ 2 c2 c2 its frequency decreases. ⇒ u = c 1 – 1 γ 2 29 R = 2GM u = 0.87 c s C2 2 × 6.67 × 10–11 × 2 × 1031 R = = 29 600 m (iii) The measurements are different s (3 × 108)2 = 29.6 km. because they are in different frames of reference; there is no right and wrong. Δt 60 30 (a) Δt = = = 60.0009 s 1 – Rs 1 – 29 600 r 109 2 (a) 60 v (b) Δt = = 71.5 s 1 – 29 600 105 60 (c) Δt = = 84.3 s 1 – 29 600 v = c 6 × 104 Tends towards c but doesn’t pass

same at low speeds

0 0 V 6

M21_IBPH_SB_IBGLB_9021_SL09.indd 6 22/07/2014 16:43 (b) According to the principles of special (iv) We know that observer E sees the light relativity it is not possible for the electron travel from F to R in time T. Since the to exceed the speed of light. This is speed of light is the same for all inertial because the mass tends to go as velocity observers this must be a distance cT. approaches the speed of light so to reach c 2 vT 2 so cT0 = 2 D + ( 2 ) would require ∞ force. cT cT 2 vT 2 but D = 0 so cT = 2 ( 0 ) + ( ) At low speeds there is, however, no 2 2 2 difference between classical and relativistic c2T 2 2 2 squaring ⇒ c2T 2 = 4( 0 + v T ) predictions. 4 4 c2T2 = c2T 2 + v2T 2 6 0 (c) pd = 1.5 × 10 V 2 2 (c2 – v2)T 2 = c2T 2 ⇒ T 2 = ( c – v ) T 2 ⇒ 0 0 2 gain in KE of electrons = 1.5 MeV 2 c T = (1 – v ) T 2 velocity = 0.97 c 0 c2 T (i) Relativistic mass = γ m ⇒ T = 0 0 v2 1 – 2 where γ = 1 = 1 = 4.1 c 1 – v2 1 – 0.972c2 4 (a) According to special relativity, energy and c2 c2 2 Rest mass of e– = 0.5 MeV c–2 mass are equivalent (E = mc ) so the mass of a body at rest can be converted into ⇒ Relativistic mass = 4.1 × 0.5 energy; this is the rest mass energy. = 2.1 MeV c–2 When accelerated, a body gains KE so it Easiest to work in MeV c–2 now has KE + rest mass energy; this is the 2 (ii) Total E = mc = 2.1 MeV total energy of the body. 3 (a) An inertial frame of reference is a co- (b) Total energy of β particle = 2.51 MeV ordinate system covered in clocks within β particle is an electron so has rest mass which Newton’s laws of motion are obeyed. = 0.511 MeVc–2 In other words not accelerating. 2 γ 2 Total energy = mc = m0c 2D distance (b) T0 = ( ) ⇒ γ ⇒ γ c velocity 2.51 = × 0.511 = 4.91 γ D (c) (i) If = 4.91 then 37 cm C v

(c) (i) can ﬁ nd speed of β particles using γ 1 ⇒ 1 = v = c 1 – 2 1 – v2 ( γ ) c2 FR 1 v = c (1 – 4.912 ) = 0.979 c (ii) Distance F–R = vT (ii) 0.979 c = 0.979 × 3 × 108 (iii) From Pythagoras: = 2.94 × 108 m s–1 2 vT 2 path length = 2 × (D + 2 ) distance = 0.37 m t = d = 0.37 = 1.258 ns D v 2.94 × 108 (d) (i) From the β particle’s frame of reference, ½vT the detector and source are moving.

M21_IBPH_SB_IBGLB_9021_SL09.indd 7 22/07/2014 16:43 0.8 c + 0.8 c u v u′ = = 1.6 c 1 – –0.8 c × 0.8 c 1 + 0.64 c2 u′ = 0.98 c

6 (a) A reference frame is a coordinate system (ii) The speed of the detector is the same covered in clocks. It is used by an observer as the speed of the β particle as to measure the time and position of an measured in the laboratory reference event. frame. (b) Classically the light coming from the lamp v = 2.94 × 108 m s–1 will have velocity = c – v (iii) In the frame of reference of the β like throwing something off the back of particle the distance from source to a truck. detector is contracted v L so L = 0 γ L = 37 cm, the distance measured at 0 c rest relative to the detector L = 37 = 7.5 cm 4.91 5 (a) Postulate 1. The laws of physics are the same for all inertial observers. (c) According to Maxwell’s theory the speed of Postulate 2. The speed of light in a vacuum light doesn’t depend on the velocity of the is the same as measured by all inertial source so velocity = c ′ u – v observers. (d) u = cv ; substitute u = c to get 1 – c2 (b) (i) ′ c – v c – v c(c – v) u = cv = v = = c 1 – 2 1 – c – v 0.80 c 0.80 c c c (e) (i) Proper time is the time as measured by AB an observer at rest relative to the event being timed. (ii) T = γT T = 1.5 μs Each spacecraft moves away from the 0 0 observer at 0.8 c; this is not greater γ = 2 T = 3.0 μs T 2 than c. γ = 0 ⇒ 1 = 1 – v 1 – v2 γ 2 c2 To ﬁ nd out how fast they move c2 ⇒ ( 1 ) from each other, we would have to v = c 1 – γ 2 determine the velocity of A relative to B. v = 0.87 c (ii) velocity transform is u′ = u – v 1 – uv 7 (a) (i) The time to travel 52 light years at a c2 52 Let us measure the velocity of A from speed of 0.8 c = = 65 years 0.8 the reference frame of B (ii) The Earth observer measures the proper distance between Earth and the O AB planet distance measured by Amanda c c 0.80 0.80 L 52 = 0 = = 31.2 light years γ ( 5 ) 3 u – velocity of A measured by O = –0.8c v – velocity of B’s reference frame relative to O = 0.8 c 8

M21_IBPH_SB_IBGLB_9021_SL09.indd 8 22/07/2014 16:43 (iii) Time to reach plenet according to 9 (a) Particle A has only rest energy; KE = 0 spacecraft is Particle B has rest energy + KE 31.2 light years = 39 years, (b) (i) Using the relativistic velocity 0.80 c transformation so Amanda is 20 + 39 = 59 years old. ′ ux – v u x = u v 1 – x (b) If we take Amanda’s frame of reference, c2

Earth is moving away so the signal has to ux – velocity of body measured in S ′ ′ travel an extra distance to get to Earth. ux – velocity of body measured in S 0.80c v – relative velocity of two frames of reference planet Earth u′ = 0.96 c + 0.96 c = 0.999 c x 1 + 0.96 c × 0.96 c c2 0.80c γ 2 (ii) Total energy = m0c where 1 1 Earth planet where γ = = = 3.57 1 – v2 1 – (0.96 c)2 c2 c2

31.2 light years 0.8T light years The rest energy of a proton = 938 MeV so total energy of proton in question In the time T for the signal to get to = 3.57 × 938 = 3.35 GeV Earth the signal travelled 31.2 + 0.8 T light years (c) (i) Total energy before collision = 2 × 3.35 = 6.7 GeV The signal travelled at the speed of light so in T years light will have travelled T After collision energy = energy of proton light years + energy of neutron + energy of pion T = 31.2 + 0.8 T = 156 years If pion has 502 GeV (0.502 MeV) then the proton and neutron will have 8 (a) The Schwarzschild radius is the minimum 6.7 – 0.502 = 6.2 GeV distance from the centre of a black hole that (ii) using E 2 =m 2c4 + p2c2 light can escape. 0 –11 31 where E = total pion energy (b) R = 2GM = 2 × 6.7 × 10 × 2 × 10 s c2 (3 × 108)2 pc = pion momentum 4 = 3 × 10 m 2 m0c = pion rest energy 2 2 2 2 (c) (i) Clocks tick slowly near to large masses p c = 502 –140 = 232 400 so the oscillator near the large mass p = 482 MeV c–1 will have a lower frequency than an (d) The proton must have momentum in the identical oscillator on the space station. –x direction to balance the proton and (ii) The time dilation equation is pion plus momentum in the –y direction to Δt Δt = 0 balance the y momentum of the neutron. 1 – Rs r n0 R Δt s = 1 – ( 0 )2 r Δt Δ Δ where t0 = 1 hour and t = 10 hours

Rs = 0.99 + r π

r = 1.01 Rs, which means they would be

0.01 Rs from the event horizon. proton 9

M21_IBPH_SB_IBGLB_9021_SL09.indd 9 22/07/2014 16:43 Worked solutions

Chapter 10

(b) F Exercises

1 0.1 m 0.2 m 0.2 m 600 N

0.8 m 1.0 m

F × 1.8 = 600 × 0.8

F = 267 N 3 N W

1 m (c) 50 N Clockwise: W × 20 Anticlockwise: 3 × 20 F W × 20 = 3 × 20 30 cm W = 3 N Mass = 300 g 5 mm paint tin lid 2 0.1 m Again F is force acting on the lid; force on 0.1 m 0.5 m L the screwdriver is opposite. 50 × 30 = F × 0.5 F = 3000 N

1 N 2 N 3.5 N 4 FA FB 1 m 1 m 5 m Clockwise: 3.5 × L Anticlockwise: 1 × 0.5 + 2 × 0.1 = 0.7 N m 3.5 × L = 0.7 A B L = 0.2 m 800 N 100 N Take torques about B 3 (a) F Clockwise: F × 5 2 m A Anticlockwise: 800 × 4 + 100 × 2.5 0.1 m

500 N FA = 690 N

Vertical forces balance so FA + FB = 900 N F = 210 N F marked is the force on the rock. B Force on crowbar acts in the opposite direction so turns the bar anticlockwise. 500 × 2 = F × 0.1 F = 10 000 N

M22_IBPH_SB_IBGLB_9021_SL10.indd 1 17/07/2014 15:24 5 6 N 7 3 m 0.1 m 0.4 m 0.4 m 0.1 m L 0.5 m F 45° R

A 2 N 8 N B 1.5 m

Drawn at breaking point 500 N 100 N Take torques about A Clockwise: 2 × 0.4 + 8 × L Anticlockwise: 6 × 0.8 = 4.8 N m 4.8 = 0.8 + 8 × L L = 0.5 m (a) Take torques about the wall Moved 0.1 m T × sin 45° × 2.5 = 500 × 3 + 100 × 1.5 T = 1650 6 2.5 × sin 45° 3 m T = 933 N (b) Horizontal forces are balanced R = horizontal component of T 0.5 m F 45° R = 933 cos 45° R = 660 N 1.5 m (c) Vertical forces are balanced 600 N 500 + 100 = 933 × sin 45° + F F = 600 – 660 F = –60 N (downwards)

8 (a) Take torques about the wall RW T × sin 45° × 2.5 = 600 × 1.5 T = 900 2.5 × sin 45° T = 509 N θ (b) Horizontal forces are balanced R = horizontal component of T 5 m

= 509 × cos 45° = 360 N 4 m

F 3 m It’s a 3-4-5 triangle so height = 4 m

Vertical forces are balanced, Rg = 200 N 2

M22_IBPH_SB_IBGLB_9021_SL10.indd 2 17/07/2014 15:24 Take torques about the top 13 Clockwise: Rg × 3 = 200 × 3 = 600 N m Anticlockwise: F × 4 + 200 × 1.5 600 = 4F + 300 F = 75 N 2 m

9 0.5 m 200 N

F = μN = 75 N

(a) ω = 0.25 × 2π = 0.5π = 1.57 rad s–1 Ladder is about to slip so –1 μ × N = 75 (b) red child v = ω r = 2 × 0.5π = π = 3.14 ms μ × 200 = 75 blue child v = ω r = 0.5 × 0.5π = 0.25π μ = 0.375 = 0.79 ms–1 2 –1 (c) F = mω r 10 ω i = 6 rad s α = 2 rad s–2 red child F = 20 × (0.5π)2 × 2 = 98.7 N t = 5 s blue child F = 20 × (0.5π)2 × 0.5 = 24.7 N (a) a = (v – u) t 14 v = u + at

–1 ω f = 6 + 2 × 5 = 16 rad s 20 N (b) s = (v + u)t 2 0.5 m θ = (16 + 6) × 5 = 55 rad 2 Number of revolutions = 55 = 8.75 2π 11 θ = 2π ω = 5 × 2π rad s–1 i –1 2.5 kg ω f = 0 rad s = I (a) v2 = u2 + 2as Γ α 2 2 2 (v2 + u2) I = mr = 2.5 × 0.5 = 0.625 kg m a = 20 2s α = Γ = = 32 rad s–2 2 l 0.625 α = (0 – (10π) ) (2 × 2π) 15 α = −25π = –78.54 rads–2 (b) s = (u + v)t 2 t = 2s = 2 × 2π = 0.4 s 0.5 m (u + v) 10π 12 5 m 2 ms–2 F

(a) α = a = 2 = 0.4 rad s–2 r 5 2.5 kg (b) a = r = 2.5 × 0.4 = 1 m s–2 α –1 ω i = 2π rad s –1 ω f = 0 rad s t = 1s 3

M22_IBPH_SB_IBGLB_9021_SL10.indd 3 17/07/2014 15:24 (ω – ω ) α = f i = –2π rad s–2 (b) Length of string = 1 m t I = mr2 = 2.5 × 0.52 = 0.625 kg m2 Circumference of cylinder = 2π r = 0.126 m Γ = Iα = 0.625 × 2π = 3.93 N m Number of revolutions = 1 = 7.96 0.126 Γ = Fr (c) θ = 7.96 × 2π Γ 3.93 F = = = 7.85 N –1 r 0.5 ω i = 0 rad s (This is just a spinning wheel; much more force α = 500 rad s–2

is required if the bike is being ridden.) 2 2 ω f = ω i + 2α θ = 0 + 2 × 7.96 × 2π × 500 16 20 N = 15920 π ω = 224 rad s–1 pivot f 2 kg 2 kg 19 40 cm 40 cm 40 cm 40 cm 2 kg 20 N 0.02 m I = Σmr2 = 2 × 0.82 + 2 × 0.42 = 1.6 kg m2 Total Γ = 20 × 0.4 + 20 × 0.8 = 24 N m Γ = Iα 15 N 15 N α = Γ = 24 = 15 rad s–2 10 N 10 N l 1.6 Take clockwise as positive since that is direction 17 20 N of initial rotation pivot 2 kg 2 kg (a) Γ = –15 × 0.02 + 10 × 0.02 = –0.1 N m (minus sign means it’s acting anticlockwise) 80 cm 40 cm 40 cm 120 cm 1 20 N (b) I = mr 2 = 0.5 × 2 × 0.022 = 4 × 10–4 kg m2 2 2 2 2 2 I = Σmr = 2 × 0.8 + 2 × 0.4 = 1.6 kg m = Γ = –0.1 = –250 rad s–2 α –4 Total Γ = –20 × 0.8 + 20 × 2 = 24 N m l 4 × 10 –1 (clockwise) (c) ω i = 100 × 2π rad s = I –1 Γ α ω f = 0 rad s Γ 24 –2 α = = = 15 rad s –2 l 1.6 α = –250 rad s (ω – ω ) A couple has the same effect no matter where α = f i t it acts. (ω – ω ) t = f i = (–200π) = 2.51 s –250 18 2 kg α 20 5 m

0.02 m 200 N

200 N

(a) Sum of torques = 200 × 5 – 200 × 2.5 10 N = 500 N m (anticlockwise) (a) I = 1 mr2 = 0.5 × 2 × 0.022 = 4 × 10–4 kg m2 2 (b) I = 1 = 1 = 167 kg m2 3 mL2 3 × 20 × 52 Γ = Fr = 10 × 0.02 = 0.2 N m α = Γ = 500 = 3 rad s–2 Γ = Iα l 167 α = Γ = 0.2 = 500 rad s–2 l 4 × 10–4

M22_IBPH_SB_IBGLB_9021_SL10.indd 4 17/07/2014 15:24 6gh 21 4 m v = 5 a b 6 is less than 10 so the hollow ball will have 3 kg c 5 7 0.02 m lower velocity when it reaches the bottom (it will take more time) (a) KE = 1 Iω2 2 2 24 I = 1 × ML2 = 3 × 4 = 4 kg m2 centre 12 12 ω = 1 × 2π = 2π rad s–1 5 cm KE = 0.5 × 4 × (2π)2 = 79 J 10° 2 (b) I = 1 ML2 = 3 × 4 = 16 kg m2 end 3 3 (a) KE at bottom = PE at top = mgh KE = 0.5 × 16 × (2π)2 = 316 J = 0.5 × 10 × 0.05 = 0.25 J (c) I = 1 Mr2 = 0.5 × 3 × 0.022 (b) since ball is solid 2 10gh = 6 × 10–4 kg m2 v = 7 –4 2 KE = 0.5 × 6 × 10 × (2π) = 0.85 m s–1

= 0.012 J (c) sin 10° = 0.05 L 22 0.5 kg L = 0.29 m (d) u = 0 m s–1 v = 0.85 m s–1 0.45 m s = 0.29 m s = (u + v)t 2 t = 2s = 0.68 s (u + v) 25 ω = 2 × 2π = 4π rad s–1

I = mr2 = 0.5 × 0.452 = 0.101 0.4 kg KE = 1 Iω 2 = 0.5 × 0.101 × (4π)2 = 8 J 2 20π rads–1 (This is just the spinning wheel.) 0.05 m 1 23 I = mr2 = 5 × 10–4 kg m2 2 –4 –2 v L = Iω = 5 × 10 × 20π = π × 10 ω = 3.14 × 10–2 kg m2 s–1 h 26 0.1 m

mgh = 1 mv2 + 1 Iω 2 0.1 m 2 2 10π rads–1 For a hollow sphere I = 2 mr2 3 0.75 kg v If no slipping ω = I = 2 mr2 = 3 × 10–3 kgm2 r 5 2 –3 –2 1 2 1 2 2 v 1 2 2 2 L = Iω = 3 × 10 × 10π = 3π × 10 mv + mr × = mv + mv ( 2 ) 2 –1 2 2 3 r 2 6 = 0.942 kg m s mgh = 5 mv2 6 5

M22_IBPH_SB_IBGLB_9021_SL10.indd 5 17/07/2014 15:24 27 (a) 0.15 m 28 0.15 m π rads–1 2π rads–1 1 kg

1 m

1 2 2 (a) I = mr = 0.5 × 1 × 0.15 2 kg 1 2 = 1.125 × 10–2 kg m2 0.1 m 0.15 m

1 kg

60 kg

I = I + I 1 body arms (b) I = 1.125 × 10–2 + mr2 = 1 × 60 × 0.152 + 2 × 2 × 12 = 4.675 kg m2 2 2 = 1.125 × 10–2 + 0.1 × 0.12 (b) 0.15 m = 1.225 × 10–2 kgm2

0.25 cm 0.5 kg

60 kg (a) L = Iω 1 2 2 I2 = × 60 × 0.15 + 2 × 2 × 0.25 2 2 2 2 I1 = mr = 0.5 × 0.5 = 0.125 kg m v 2 = 0.925 kg m2 ω = = = 4 rad s–1 1 r 0.5 2 –1 L1 = 0.5 kg m s (c) I1ω1 = I2ω2

(b) I1ω1 = I2ω2 4.675 × 2π = 0.925 × ω2 2 2 2 –1 I2 = mr = 0.5 × 0.2 = 0.02 kg m ω2 = 5.1 × 2π rad s L –1 –1 = 1 = 25 rad s 5.1 revolutions s ω2 I2 1 (d) KE before = I ω 2 = 0.5 × 4.675 × (2π)2 speed = ω r = 5 m s–1 2 1 1 1 = 92.3 J 30 1 mole Ar = 40 g 1 KE after = I ω 2 = 0.5 × 0.925 × (10.2π)2 100 2 2 2 number of moles n = moles = 2.5 moles = 475 J 40 U = 3 nRT = 1.5 × 2.5 × 8.31 × 300 = 9.4 kJ (e) Work done pulling her arms in 2 31 Average KE = 3 kT = 1.5 × 1.38 × 10–23 × 400 2 = 8.28 × 10–21 J

M22_IBPH_SB_IBGLB_9021_SL10.indd 6 17/07/2014 15:24 32 P1 = 70 kPa 34 –6 3 V1 = 100 × 10 m –6 3 a) 200 V2 = 50 × 10 m (kP (a) PV = nRT P PV 70 × 103 × 100 × 10–6 T = 1 = = 700 K 150 1 nR 0.01 PV 3 –6 (b) T = 2 = 70 × 10 × 50 × 10 = 350 K 2 nR 0.01 100 (c) ΔU = 3 nR ΔT = 1.5 × 0.01 × 350 = 5.25 J 2 (d) Work done = PΔV = 70 × 103 × 50 × 10–6 = 3.5 J 50 (e) Q = ΔU + W

Work done by gas = –3.5 J 0 Change in internal energy = –5.25 J 0 50 100 150 200 250 V (cm3) Heat loss = 8.75 J 5 (a) Adiabatic so PV 3 = constant 33 5 5 3 3 P1V1 = P2V2 a) 200 5

(kP 3 –6 3 P 200 × 10 × (52 × 10 ) 5 = P × (150 × 10–6) 3 150 2

P2 = 34 kPa 100 Close, given the difﬁ culty in reproducing and reading the graph. (b) Expansion, so the gas does work 50 Work is area under graph = 83 squares As before, each square 0 = 10 × 103 × 10 × 10–6 = 0.1 J 0 50 100 150 200 250 V (cm3) So the gas does work = 8.3 J P V 3 –6 P = 70 × 103 Pa (c) T = 1 1 = 200 × 10 × 52 × 10 = 1040 K 1 1 nR 0.01 V = 150 × 10–6 m3 1 P V 34 × 103 × 150 × 10–6 P = 200 × 103 Pa (d) T = 2 2 = = 510 K 2 2 nR 0.01 3 3 V2 = 50 × 10 m 3 3 –6 (e) ΔU = nRΔT = 1.5 × 0.01 × (510 – 1040) (a) T = PV = 70 × 10 × 150 × 10 = 1050 K 2 nR 0.01 = –7.95 J (b) Work done = area under curve = 70 + 46 (f) Q = ΔU + W = –7.95 + 8.3 = 0.35 J = 116 squares (This should be zero but is not, owing to 3 –6 Each square = 10 × 10 × 10 × 10 = 0.1 J the difﬁ culty in estimating the area under Work done = 11.6 J the graph.) (c) Temperature is constant so no change in internal energy heat loss = work done on gas = 11.6 J

M22_IBPH_SB_IBGLB_9021_SL10.indd 7 17/07/2014 15:24 35 (a) P (c) Gas does work when it expands D C Work done = area under red and black 200 kPa curves = A + B + C + D + E Work done = 320 J 100 kPa A B (d) Work done on gas when it is compressed Work done = area under green and blue

3 3 V curves = A + B + C 250 cm 500 cm P V P V Work done = 135 J (b) (i) A A = B B ⇒ 100 × 250 = 100 × 500 T T 300 T A B B (e) Net work done = E + D = 185 J

TB = 600 K P V P V 37 (ii) B B = C C ⇒ 100 × 500 = 200 × 500 TB TC 600 TC a) T = 1200 K 200 C (kP P P V P V (iii) C C = D D ⇒ 200 × 500 = 200 × 250 TC TD 1200 TD 150 TH TD = 600 K (c) Work done by gas from A ➞ B = area under graph 100 = 100 × 103 × 250 × 10–6 = 25 J

(d) Work done on gas from C ➞ D = area 50 under graph TC = 200 × 103 × 250 × 10–6 = –50 J (e) Net work done by gas = –25 J (25 J of work 0 0 50 100 150 200 250 done on gas) V (cm3)

3 –6 36 P (a) At top point PV = 150 × 10 × 35 × 10 nR 0.01 = 525 K

3 –6 (b) At bottom point PV = 20 × 10 × 180 × 10 E nR 0.01 = 360 K A D 1 – T B C (c) η = C = 1 – 360 = 0.31 (≈0.3) T 525 V H (d) Net work = area enclosed = 14 squares Each area marked with a letter represents = 14 × 10 × 103 × 10 × 10–6 = 1.4 J an energy: A – 50 J, B – 45 J, C – 40 J, D – 35 J, E – 150 J (e) Heat added = work done = area under isothermal expansion = 48 squares = 4.8 J (a) Isothermal expansion (f) = W = 1.4 = 0.29 ( 0.3) Work done = area under red curve η ≈ QH 4.8 = A + B + E 38 Entropy = ΔQ Work done = 245 J T (b) Adiabatic compression 400 K 250 K

Work done = area under green curve = A 500 J Work done = 50 J

M22_IBPH_SB_IBGLB_9021_SL10.indd 8 17/07/2014 15:24 (a) (i) Entropy lost by hot body = –500 42 400 0.4 m = –1.25 J/K 0.6 m (ii) Entropy gained by cold body = +500 250 = +2 J/K

(b) Change in entropy = 2 + (–1.25) = +0.75 J/kg

39 Lifting a load increases PE of load. Motor (a) Volume of wood under water transfers electrical energy to PE. PE of load is = 0.6 × 1 × 1 = 0.6 m3 more ordered than electrical energy in battery. Mass of water displaced = Vρw ⇒ Heat must be lost otherwise entropy would = 0.6 × 103 kg be reduced. Buoyant force = 6000 N 40 Floating, so buoyant force = weight of wood F 10 N = 6000 N

3 1 cm3 150 cm Mass of wood = 600 kg Density of wood = 600 kg m–3 (a) P = F = 10 = 105 Pa A 1 × 10–4 (b) If the cube is pushed under water, mass of water displaced = Vρ = 1 × 103 kg (b) P = F w A Upthrust = 10 000 N

The pressure in the ﬂ uid is the same 10 000 N everywhere: 10 = F 1 150 F = 1500 N

41 T

F 6000 N F + 6000 = 10 000 F = 4000 N

43 W L1

L m = 60 kg 2 3 –3 ρg = 19.3 × 10 kg m V = m = 3.1 × 10–3 m3 ρg

Mass of water displaced = Vρw = 3.1 × 10–3 × 1 × 103 = 3.1 kg If area of cube = A, then weight of water Buoyant force, FB = 31 N displaced = L2 × A × ρw × g To lift ball T = W – FB = 600 – 31 = 569 N Weight of ice = (L1 + L2) × A × ρi × g

M22_IBPH_SB_IBGLB_9021_SL10.indd 9 17/07/2014 15:25 2 The ice is ﬂ oating, so (a) Volume ﬂ ow rate = A1v1 = π × (0.01) × 3 –4 3 –1 L2 × A × ρw × g = (L1 + L2) × A × ρi × g = 9.4 × 10 m s L ρ 0.92 2 = i = = 0.89 (b) Using the continuity equation A1v1 = A2v2 (L + L ) ρ 1.03 1 2 w π × (0.01)2 × 3 = π × (0.03)2 × v This means that 89% is under water. 2 2 0.01 –1 v2 = × 3 = 0.33 m s 44 (0.03) 1 2 1 2 (c) P1 + ρ v1 + ρ gh1 = P2 + ρ v2 + ρ gh2 P1 V1 2 2 No change in height so 5 m P + 1 ρ v 2 = P + 1 ρ v 2 1 2 1 2 2 2 P V 2 2 500 × 103 + 0.5 × 103 × 32 3 2 = P2 + 0.5 × 10 × 0.33 Pressure at depth 5 m = P + ρgh atmos P = (500 + 4.5 – 0.054) × 103 = 504.4 kPa 101 × 103 + 103 × 10 × 5 = 151 × 103 2 = 151 kPa 49 Assume the temperature is constant

P1V1 = P2V2

r1 = 0.015 m 101 × 100 = 151 × V2

3 P1 = 100 kPa V2 = 67 cm –1 v1 = 0.5 ms

45 A1v1 = A2v2

20 × 3 × 1 = 20 × 1 × v2 –1 v2 = 3 m s 20 m 46 Volume ﬂ ow is the same = 1.5 2 = π(0.25) × v2 –1 v2 = 7.6 m s 47 40 cm V2 3 cm 3 mm

r2 = 0.005 m

P2 v = velocity of piston = 0.4 = 0.1 m s–1 1 4 (a) Volume ﬂ ow rate = A v = × (0.015)2 × 0.5 A v = A v 1 1 π 1 1 2 2 –4 3 –1 2 2 = 3.53 × 10 m s π × (0.03) × 0.1 = π × (0.003) × v2 –1 (b) Using continuity equation A v = A v v2 = 10 m s 1 1 2 2 π × (0.015)2 × 0.5 = π × (0.005)2 × v 48 2 2 v = 0.015 × 0.5 = 4.5 m s–1 2 (0.005) (c) P + 1 ρ v 2 + ρ gh = P + 1 ρ v 2 + ρ gh 1 2 1 1 2 2 2 2 3 3 2 3 –1 V1 = 3 ms V2 100 × 10 + 0.5 × 10 × 0.5 + 10 × 10 × 20 3 2 = P2 + 0.5 × 10 × 4.5 3 P2 = (100 + 0.125 + 200 – 10.0125) × 10

P1 = 500 kPa P2 P = 290 kPa r1 = 0.01 m r2 = 0.03 m 2

M22_IBPH_SB_IBGLB_9021_SL10.indd 10 17/07/2014 15:25 2 gr2 50 density of air = 1.3 kgm–3 v = × (ρ – ρ ) t 9 η s f static pressure 2 = 2 × 10 × 0.005 (8000 – 900) 9 0.2 = 1.97 m s–1 9 × v × η 9 × 0.5 × 0.2 54 ρ = t + ρ = + 900 s 2 × gr2 f 2 × 10 × 0.032 stagnation pressure = 50 + 900 –3 3 cm ρs = 950 kg m 55 R = vrρ e η

Turbulent if Re > 1000 v = 1000 × 0.002 = 0.2 m s–1 0.01 × 1000 density of water = 1000 kgm–3 Volume ﬂ ow rate = Av = π × (0.01)2 × 0.2 P – P = ρ gΔh = 1 ρ v2 = 6.3 × 10–5 m3 s–1 2 1 w 2 a 56 2ρ gΔh 2 × 1000 × 10 × 0.03 –1 v = w = = 21.5 m s 1.3 ρa 51 P – P = 1 ρ v2 25 2 1 2 a energy (J) 600 km h–1 = 600 × 1000 = 166.7 m s–1 20 3600 ΔP = 0.5 × 1.3 × 166.72 = 18 kPa 15

52 10

5 4 cm

2 cm 0 0 1 2 3 4 4 cm2 time (s) 1 cm2 If f = 0.5 Hz, the time period = 1 = 2 s 0.5 energy stored Q = 2π A 2 energy lost per cycle gΔh = 1 v 2 1 – 1 2 1 A [( 2 ) ] 25 Q = 2π = 9.8 (25 – 9) v = 2gΔh 1 A 2 1 – 1 57 ( A ) 2 A 1 = 4 A2 2 × 10 × 0.02 v = = 0.163 m s–1 1 42 – 1 B –4 C Volume ﬂ ow rate = A1v1 = 4 × 10 × 0.163 = 6.5 × 10–5 m3 s–1 A D E –3 53 Density of oil = 900 kg m F Density of steel = 8000 kg m–3 (a) D is same length as A so resonance – this Viscosity of oil = 0.2 N s m–2 implies a π phase difference. 2

M22_IBPH_SB_IBGLB_9021_SL10.indd 11 17/07/2014 15:25 B is much shorter so driver has lower (b) L = Iω = 280 × π = 880 kgm2s–1 frequency – it will be in phase. (c) No external torques act so angular F is much longer so driver has higher momentum is conserved. frequency – it will have a π phase difference. Initial angular momentum = ﬁ nal angular C and E will be somewhere in between. momentum

(b) D has highest amplitude as it resonates with I1ω1 = I2ω2 the driver. 2 2 2 I2 = 0.5 × 60 × 2 + 40 × 1 = 160 kg m

I1ω1 880 –1 ω2 = = = 5.50 rad s I2 160 (d) Rotational KE = 1 Iω2 Practice questions 2 Initial KE = 0.5 × 880 × π2 = 1380 J

2 1 (a) The conditions for equilibrium are that the Final KE = 0.5 × 160 × 5.5 = 2420 J sum of the forces acting on the body are Change in KE = 2420 – 1380 = 1040 J zero and the sum of the torques about any (e) The increase in KE is due to the work done point is zero. by the child pulling himself towards the (b) The forces must add to zero so the centre.

resultant force must close the triangle of ) 3 Pa

forces as shown. 5 5 (10 P 4

3 m 1.0 m 2 0.8 70° R 1 P R S 10° 0 F m 0 1 2 3 4 5 6 base of spine 0.5 v (m3) Work done compressing gas = area under line F = 1 × 105 × 3 = 3 × 105 J R The answer is C. S (c) Torque about base = S × sin 70° × 0.8 4 (a) Work done during cycle = area inside cycle 5 6 (d) If in equilibrium torques about base are = 2 × 10 × 8 = 1.6 × 10 J balanced so 8 m3

) A B

Pa 3

clockwise torque = anticlockwise torque 5

S × sin70° × 0.8 = F × sin10° × 0.5 (×10 P 2 2 × 105 F = 0.8 × sin70° = 8.66 ≈ 9 S 0.5 × sin10° 1 2 (a) The radius should be marked as 2 m. D C Moment of inertia = I + I disc child 0 1 2 2 2 4 6 8 10 = md r + mc r 0 2 V(m3) = 0.5 × 60 × 22 + 40 × 22 I = 280 kgm2 12

M22_IBPH_SB_IBGLB_9021_SL10.indd 12 17/07/2014 15:25 (b) 1.8 × 106 J thermal energy ejected (iii) According to ﬁ rst law of So the energy in must equal thermodynamics work done + energy lost Q = ΔU + W ⇒ ΔU = Q – W = (1.6 + 1.8) × 106 = 3.4 × 106 J Gain in internal energy = heat added – work done by gas Efﬁ ciency = work out × 100% energy in ΔU = 8 × 103 – 6 × 103 = 2 × 103 J = 1.6 × 100% = 47% 3.4 6 (a) (i) A ➞ B the volume is getting less ⇒ gas

adiabatic )

Pa A is on a higher isotherm than B 5 5 C so A → B temperature reduced. (10 P adiabatic 4

isothermal 2

V 1 B A

(d) (i) Adiabatic expansion 0 Adiabatic compression 0 0.1 0.2 0.3 0.4 0.5 0.6 V (m3) Isothermal expansion (ii) Temperature of gas goes down and Isothermal compression work is done on gas. 5 (a) Isothermal – the temperature remains First law: Q = ΔU + W constant but heat enters or leaves. If ΔU and W are both negative then Q is Adiabatic – no heat exchanged with negative ⇒ heat lost. surroundings and temperature not constant. (b) Work done from A ➞ B = area under A–B = 5 5 (b) (i) PV = constant 1 × 10 × 0.4 = 0.4 × 10 J T (c) Total work done = area of ‘triangle’ ABC P = 1·2 × 105 Pa = 1 × 4 × 105 × 0.4 = 0.8 × 105 J 2 (d) Useful work done = 8 × 104 J Thermal energy supplied = 120 kJ 3 3 0.050 m 0.10 m 4 3 = 12 × 10 J Q = 8 × 10 J Efﬁ ciency = useful work × 100 = 8 × 100 P is constant so when V increases T energy in 12 must increase ⇒ not isothermal = 67% Heat added ⇒ not adiabatic c. and d. are overestimates since the area is ∴ not adiabatic or isothermal; it is a bit less than the area of the triangle. isobaric 7 (a) Isothermal is not a steep as adiabatic (ii) Work done on gas = PΔV so AC – adiabatic = 1.2 × 105 × 0.05 = 6 × 103 J AB – isothermal

M22_IBPH_SB_IBGLB_9021_SL10.indd 13 17/07/2014 15:25 8 (a) Volume per second = Av = × 0.022 × 0.5 (b) ) π

Pa –4 3 –1 5 7.0 = 6.3 × 10 m s ; C (b) Using continuity equation 6.0 A v = A v

pressure (10 1 1 2 2 5.0 2 2 B π × 0.02 × 0.5 = π × 0.015 × v2; –1 4.0 v2 = 0.9 m s ; (c) Using Bernoulli equation 3.0 P + 1 ρv 2 + ρgh = P + 1 ρv 2 + ρgh ; 1 2 1 1 2 2 2 2 A 2.0 300 × 103 + 0.5 × 103 × 0.52 + 103 × 10 × 0 1.0 2.0 3.0 4.0 5.0 6.0 –3 3 3 2 3 volume (10 m ) = P2 + 0.5 × 10 × 0.9 + 10 × 10 × 5 3 Net work done is area inside the cycle ⇒ P2 = (300 + 0.125 – 0.405 – 50) × 10 (c) Estimate by counting small squares ∼ 150 = 250 kPa;

5 –3 Each square is 0.1 × 10 × 0.1 × 10 = 1 J 9 (a) Using Bernoulli equation so work done = 150 J P + 1 ρ v 2 + ρ gh = P + 1 ρ v 2 + ρ gh 1 2 1 1 2 2 2 2 (d) Adiabatic ⇒ no exchange of heat ⇒ Q = 0 Assume difference in height is negligible First law states Q = ΔU + W P + 1 ρ v 2 = P + 1 ρ v 2 Heat added = increase in internal energy + 1 2 1 2 2 2 work done by gas P – P = 1 ρ (v 2 – v 2) 1 2 2 2 1 In this case Q = 0 = 0.5 × 1.3 × (3402 – 2902) and W is negative since work is done on ΔP = 2.05 × 104 Pa gas (b) Upward force = ΔP × A = 2.05 × 104 × 90 0 = U – W ⇒ Δ = 1.8 × 106 N ⇒ ∆U = W ∴ Internal energy increases ⇒ Temperature increases

M22_IBPH_SB_IBGLB_9021_SL10.indd 14 17/07/2014 15:25 Worked solutions

Chapter 11 4 5 cm Exercises

1 15 cm

(a) u = 5 cm v = ? f = 15 cm 1 + 1 = 1 25 cm 5 v 15 1 = 1 – 1 = 1 – 3 = – 2 Distance = focal length = 25 cm v 15 5 15 15 1 + 1 = 1 v = –7.5 cm u v f (b) Virtual (negative) 1 + 1 = 1 ∞ v f (c) M = v = 7.5 = 1.5 v = f u 5 5 2 30 cm 5 m

5 cm 10 cm

(a) u = 30 cm (a) u = 5 m f = 10 cm (b) u = 500 cm 1 + 1 = 1 v = ? u v f 1 1 1 v = 5 cm + = 1 1 1 30 v 10 + = 500 v 5 1 1 1 3 – 1 2 = – = = 1 1 1 100 – 1 99 v 10 30 30 30 = – = = v 5 500 500 500 v = 15 cm v = 5.05 cm (b) Real (c) M = v = 5.05 = 0.01 (c) M = v = 15 = 0.5 u 500 u 30 (d) If bush is 1 m then image is 0.01 m. 3 u = ? v = 20 cm 6 20 cm f = 5 cm 1 + 1 = 1 u 20 5 5 cm 1 = 1 – 1 = 4 – 1 = 3 u 5 20 20 20 u = 6.67 cm (a) u = 20 cm v = ? f = 5 cm 1 + 1 = 1 20 v 5 1

M23_IBPH_SB_IBGLB_9021_SL11.indd 1 22/07/2014 16:44 1 = 1 – 1 = 4 – 1 = 3 9 400 000 v 5 20 20 20 v = 6.67 cm 3500 (b) M = v = 6.6 = 0.33 u 20 α = 3500 = 8.75 × 10–3 rad 7 (a) f = –0.3 m 400 000 u = 5 m 10 25 cm 1 = 1 + 1 gives 1 = 1 – 1 f u v v f u 1 mm 1 1 1 –50 – 3 –53 = – = = α = 1 = 4 × 10–3 rad v –0.3 5 15 15 250 v = –28.3 cm 11 u

5 m

50 cm 25 cm 5 cm u = ? v = –25 cm (b) Linear magniﬁ cation = v = 0.283 = 0.057 u 5 f = 5 cm So if object is 50 cm as in diagram, image 1 – 1 = 1 will be 50 × 0.057 = 2.85 cm which is about u 25 5 1 = 1 + 1 = 5 + 1 = 6 the same as the diagram. u 5 25 25 25 u = 4.16 cm 8 20 cm 20 cm 10 cm 12 M = 1 + 25 = 1 + 25 = 6 image in ﬁrst lens f 5

40 cm 13 u = 24 cm f = 1r = 10 cm 2 1st lens: 1 = 1 – 1 = 24 – 10 Object is 2f from lens so image distance will be v 10 24 240 2f. Image is real. v = 17.1 cm (real) 2nd lens: M = v = 17.1 = 0.71 Image in 1st lens is object for 2nd but since rays u 24 h will not cross at this image/object it is taken to 0.71 = i h be a virtual object. o h = 0.71 × 2 = 1.42 cm u = –30 cm i f = 20 cm 1 = 1 + 1 gives 1 = 1 – 1 f u v v f u 1 = 1 – 1 = 3 + 2 v 20 –30 60 v = 12 cm Image is real. 14 u = 5 cm f = 1r = 10 cm 2 1 = 1 – 1 = 5 – 10 v 10 5 50 v = –10 cm (virtual) 2

M23_IBPH_SB_IBGLB_9021_SL11.indd 2 22/07/2014 16:44 M = v = 10 = 2 (c) 1 cm 5 cm u 5 3 cm 4.17 cm h 2 = i ho

hi = 2 × 2 = 4 cm 1.5 cm Can be seen from diagram that distance between lenses = 3 + 4.17 cm = 7.17 cm

u v u 18 o o e original object f f o e image in objective 15 v = 30 cm object for eyepiece u = 20 cm 1 cm 16 cm 4 cm final image 1 = 1 + 1 = 1 + 1 = 3 + 2 f u v 20 30 60 25 cm f = 12 cm For the eyepiece: 16 u = 5 cm f = 4 cm v = –25 cm (we know the image is virtual) f = 1 r = –10 cm e 2 1 = 1 – 1 = 1 – 1 = 25 – –4 1 = 1 – 1 = 1 – 1 = –1 – 2 u f v 4 –25 100 v f u –10 5 10 ue = 3.45 cm v = –3.33 cm For the objective: v 3.33 M = = v = 21 – 3.45 = 17.55 cm u 5 o h f = 1 cm 3.33 = i 5 h 1 1 1 1 1 17.55 – 1 o = – = – = h = 3.33 × 2 = 1.33 cm u f v 1 17.55 17.55 i 5 uo = 1.06 cm Linear magnifi cation of objective lens = 17.55 1.06 = 16.6 Linear magnifi cation of eyepiece lens = 25 3.45 = 7.2 Overall angular magnifi cation = 16.6 × 7.2 = 120 1.22 λ f 1.22 × 600 × 10–9 × 2 × 10–2 19 (a) d = 0 = D 1 × 10–2 –6 17 (a) For objective, u = 1.5 cm, f = 1 cm = 1.46 × 10 m λ 1.46 × 10–6 1 = 1 – 1 = 1 – 1 = 3 – 2 = 1 ⇒ v = 3 cm (b) λ = air so d = = 8.1 × 10–7 m v f u 1 1.5 3 3 oil n 1.8

(b) For eyepiece, v = –25 cm, f = 5 cm 20 100 cm 10 cm 1 = 1 + 1 = 5 + 1 = 6 u = 4.17 cm u 5 25 25 25

f (a) Angular magniﬁ cation = o = 100 = 10 fe 10

(b) Distance between lenses = fo + fe = 110 cm 3

M23_IBPH_SB_IBGLB_9021_SL11.indd 3 22/07/2014 16:44 (c) Eye ring is the image of objective in eyepiece 24 Pin = 1 mW Pin 110 cm (a) Attenuation = 10 log 10 (P ) out 10 cm Pout = 0.1 mW; A = 10 log10 (10) = 10 dB

(b) Pout = 0.2 mW; A = 10 log10 (5) = 7 dB

(b) Pin = 1 mW u = 110 cm Attenuation = 10 dB = 10 log 1 10 (P ) f = 10 cm ⇒ out Pout = 0.1 mW 1 1 1 1 1 11 – 1 = – = – = 26 I = 0.1 kW m–2 v f u 10 110 110 0 v = 11 cm from the eyepiece I = 0.8 kW m–2

21 f x = 4 mm 50 cm e

–μx (a) I = I0 e ⇒ log I = –μx e (I ) 0 I μ = 1 log 0 x e ( I ) = 5.6 × 10–2 f Angular magniﬁ cation = o f 0.693 e (b) x = = 12.4 mm 1/2 μ 10 = 50 f = 5 cm f e –2 e 27 I0 = 0.5 kW m 22 400 000 km x = 3 mm

μ 0.693 –1 3 500 km (a) = = 0.693 mm x1/2 –μx –0.693 × 3 (b) I = I0 e = 0.5 × e 3.5 –3 θ = = 8.75 × 10 rad –2 400 = 0.0625 kW m 23 θ 28 (a) 40% reduction ⇒ I = 40 = e–μx 2 I 100 1.7 0 n ⇒ –μ × 6 θ 2 x = 6 mm 0.4 = e a c μ loge (0.4) = – x 1.8 μ 1 n = loge (0.4) 1 x –1 n = 0.153 mm a = 1 n (b) x = 0.693 = 4.5 mm sin c = 2 1/2 0.153 n1 c = sin–1 1.7 = sin–1(0.94) = 71° (1.8) θ 2 = 90° – c = 19° Apply Snell’s law to light entering ﬁ bre: θ sin a n2 θ = sin 2 na θ sin a = sin 19° × 1.8 θ a = 36° 4

M23_IBPH_SB_IBGLB_9021_SL11.indd 4 22/07/2014 16:44 μ 0.693 –1 (b) Time for reﬂ ection from far side of organ 29 b = = 0.385 cm 1.8 = 120 μs μ = 0.693 = 0.198 cm–1 Distance travelled = 120 × 10–6 × 1500 m 3.5 = 18 cm ⇒ Depth = 9 cm muscle Thickness of organ = 9 – 4.5 = 4.5 cm

bone Practice questions

1 (a) (i) The point at which rays that are parallel 1 cm 10 cm 1 cm to the axis converge. 12 cm muscle: (ii)

I0 –μ x –0.198 × 12 converging lens = e m m = e = 0.093 I1 ⇒ 9.3% 2 cm muscle and 10 cm bone: principal axis I μ μ 0 = e–( mxm + bxb) = e–(0.198 × 2 + 0.385 × 10) = 0.014 FO F I2 ⇒ 1.4%

30 Z = ρc; for muscle Z = 1540 x 1060 = 1.63 x 106 kg m–2s–1

for bone Z = 3780 x 1900 = 7.18 x 106 kg m–2s–1 (iii) The image is virtual since rays don’t for fat Z = 1480 x 900 really cross. = 1.33 x 106 kg m–2s–1 (b) f = 6.25 cm 31 Greatest percentage refl ection from greatest u = 5 cm difference in impedance, so bone and fat. 5 cm I Z – Z 2 2 32 r = 2 1 = 7.18 – 1.63 I ( Z + Z ) (7.18 + 1.63) f 0.8 cm 0 2 1 = 0.397 ⇒ 39.7% 6.25 cm 33 (a) (i) Using 1 + 1 = 1 ⇒ 1 + 1 = 1 u v f 5 v 6.25 ⇒ 1 = 1 – 1 = 5 – 6.25 = –1.25 v 6.25 5 5 × 6.25 31.25 ⇒ organ Signal strength v = –25 cm v 25 bone 60 120 200 μ (ii) Linear magnifi cation = = = 5 Time ( s) u 5 So image is 5 × object = 5 × 0.8 Depth of organ can be found from 1st peak = 4.0 cm Time for refl ection = 60 μs Distance travelled = 60 × 10–6 × 1500 = 0.09 m = 9 cm ⇒ Depth = 4.5 cm 5

M23_IBPH_SB_IBGLB_9021_SL11.indd 5 22/07/2014 16:44 2 (a) First draw the red ray that goes through objective lens eyepiece lens the centre of the lens, appearing to come 60 mm from the virtual image at I2. Then draw the dashed line parallel to the axis from the F F F1 F1 image at I1 to the lens L2. Finally, draw the 20 mm blue ray, which comes from the image at 24 mm u I2 and passes through the point in the lens determined by the dashed line. This ray will 240 cross the axis at the focal point. Using 1 = 1 + 1 f u v I 2 1 1 1 L L ⇒ = – 1 2 u f v where f = 60 mm O v = –240 mm (negative since F virtual) 1 = 1 – 1 = (4+1) u 60 –240 240 I 1 u = 240 = 48 mm 5 (b) Magnifi cation in objective = v = 120 u 24 Magnifi cation in eyepiece = v = 240 u 48 Total magnifi cation = 120 × 240 = 25 (b) (i) Object is 1 cm tall and I2 is 2 cm tall so 24 48 linear magnifi cation = 2 4 (a) Ultrasound frequency 1 MHz → 20 MHz (ii) Final image is 6 cm tall so linear You will just have to remember this. magnifi cation of eyepiece = 6 = 3 2 (b) (i) Gel is applied to prevent the ultrasound (c) Total magnifi cation = 2 × 3 = 6 from being refl ected when it passes 3 (a) (i) Using the lens formula 1 = 1 + 1 from air to body. f u v It does this by reducing the impedance ⇒ 1 = 1 – 1 v f u difference. where f = 20 mm Also makes the sensor slide more easily. u = 24 mm (ii) 1 = 1 – 1 = (24–20) = 4 v 20 24 480 480 A D B C d v = 480 = 120 mm 4 (ii) We know the image is real because the O ultrasound value for v is positive; this is because transmitter and receiver the object is further than the focal length from the convex lens. l layer of fat layer of fat (iii) First put the image and object for the and skin and skin eyepiece onto the diagram (can’t do this exactly but it helps to see their A positions). B D C pulse strength (relative units)

0 25 50 75 100 125 150 175 200 225 250 275 300 6 t (μs)

M23_IBPH_SB_IBGLB_9021_SL11.indd 6 22/07/2014 16:44 Each time there is a change in tissue, (ii) Fr om graph x1/2 = 4 mm there is a reﬂ ection. The beam gets I 0 20 weaker as it passes through more tissue. When the beam passes out of 155 the body the reﬂ ection is greater, since I 0

ary units) 10 the change in impedance from tissue to 2 air is large.

/ (arbitr 5 D should be passing from body to air 4 not from organ to body. 0 0 c 4 6 8 9 10 12 3 –1 x x (iii) v = 1.5 × 10 m s 1/2 (mm) The depth of the organ can be found (iii) 20% of 20 = 0.2 × 20 = 4 using peaks A and B. from the graph the thickness required Time for reﬂ ection = 50 – 35 = 15 μs = 9 mm ⇒ time to get to organ = 7.5 μs μ In 2 (iv) Using I = I e– x where μ = Depth of organ = 1.5 × 103 × 7.5 × 10–6 0 8 = 0.087 mm–1 = 1.1 cm If 80% reduction then 20% gets If B is one side of the organ and C the through other, then time to pass from B to C to I 20 –0.087x μ So = = e B is 50 s I0 100 50 Time from B to C = μs = 25 μs ⇒ ⇒ 2 In (0.2) = –0.087x x = 18.5 mm Thickness of organ = 1.5 × 103 × 25 × 10–6 = 3.8 cm 6 (a) The half thickness is the thickness of material that will reduce the intensity of an (c) B-scan gives a 3D image. 1 X-ray beam by 2. (d) Advantage: non-ionizing (b) The half thickness for bone is 1 × the half Disadvantages: small depth penetration, 150 thickness for soft tissue limit to size of objects that can be imaged, t 1 2 s t 1 = blurring of images 2 b 150 5 (a) (i) An X-ray picture shows up bones very μ = ln2 t 1 clearly since X-rays are absorbed more 2 ln 2 by bone than soft tissue; this would μ ( t 1 b ) t 1 b = 2 = 2 s = 150 therefore be the best choice to view a μ t 1 s ln2 2 b broken bone ( t 1 s ) 2 (ii) It is too dangerous to use X rays to μ = μ × 150 = 0.035 × 150 = 5.3 cm–1 view a fetus so non-ionizing radiation b s (c) (i) I = I e–μsx such as ultrasound is used. B A I B = e–0.035 × 5 = 0.84 (b) (i) The half thickness is the thickness of IA

the material required to reduce the –μbx (ii) IC = IBe intensity of an X-ray beam to half of its I C = e–5.3 × 5 = 3.1 × 10–12 original intensity. IB (d) Most of the X-rays pass through the soft tissue but almost all are absorbed by the bone; this means that a shadow of the bone will be cast on a photo plate placed under the leg.

M23_IBPH_SB_IBGLB_9021_SL11.indd 7 22/07/2014 16:44 7 (a) (i) Ultrasound is sound that is higher (d) (i) The time between transmission and frequency than we can hear, over refl ection = 50 μs. This is the time to 20 kHz. reach the stomach and back, so the μ (ii) Ultrasound is produced by applying an time to the stomach = 25 s –6 alternating voltage to a piezo-electric Distance = vt = 1600 × 25 × 10 crystal. This causes molecules to align = 4.0 cm with the fi eld, resulting in vibration. (ii) An A scan simply shows the distance (b) Z = ρc = 2800 × 1.5 × 103 to the organ whereas a B scan gives an = 4.2 × 106 kg m–2 s–1 image. (c) (i) The different parts of the brain are all made of the same material, so there is no difference in impedance, so no refl ections.

I Z – Z 2 6 2 (ii) R = 1 2 = 1.6 × 10 – 430 = 1 I ( Z + Z ) (1.6 × 106 + 430) O 1 2 (iii) The previous answer shows that from air to tissue almost all of the ultrasound will be reﬂ ected. By placing gel between the transmitter and tissue the impedance difference is reduced, resulting in less reﬂ ection.

M23_IBPH_SB_IBGLB_9021_SL11.indd 8 22/07/2014 16:44 Worked solutions

Chapter 12 parallax angle = 0.05 = 0.025 arcsec Exercises 2 d = 1 = 40 pc p 1 1 light year = 9.46 × 1015 m 6 13 16 4 × 10 km = 4 × 10 m 21 10 m 16 4 × 10 11 = 4.2 light years 1.5 × 10 m 9.46 × 1015 p 2 Distance from Sun to Earth = 1.5 × 1011 m 1011 11 p = 1.5 × = 1.5 × 10–10 rad 8 d 1.5 × 10 1021 c = 3 × 10 ; t = = 8 = 500 s c 3 × 10 1 arcsec = 4.8 × 10–6 rad = 8 min 20 s p = 3.13 × 10–5 arcsec 3 Distance to nearest star = 4 × 1013 km (d = 3.2 × 104 pc) 13 t = d = 4 × 10 = 1.3 × 109 hours Photograph scale = 50 mm arcsec so v 30 000 = 1.5 × 105 years Angle between 6 months = 2 p = 6.26 × 10–5 arcsec 4 1pc = 3.26 light years = 3.26 × 9.46 × 1015 m Distance on photograph = 50 × 6.26 × 10–5 Distance to nearest star = 3.18 × 10–3 mm. This is too small to measure. 3 = 4 × 1013 × 10 × 9.46 × 1015 = 1.3 pc 3.26 7 3 4 × 10 km Meissa 11 2 × 1.5 × 10 m Betelgeuse θ

11 Bellatrix 3 × 10 –6 θ = = 7.5 × 10 rad Alnilam 4 × 1016 –6 Mintaka 1 arcsec = 4.8 × 10 rad Alnktak θ = 1.56 arcsec

Saiph Rigel p 2 AU θ p Estimate from size of spot on photograph: Betelgeuse 1 (0.4) d = 1 = 1 = 1.28 pc Meissa 4 (3.5) p 0.78 Bellatrix 2 (1.64) 5 Alnilam 3 (1.7) Alnitak 3 (2) Mintaka 3 (2.23) Saiph 2 (2.09) Rigel 0 (0) 1 mm = 0.05 arcsec actual values (from Wikipedia) in brackets 2 mm = 0.1 arcsec 8 (a) L = 3.839 × 1026 W d = 1.5 × 1011 m photo 1 and 2 photo 1 and 3 L 3 –2 1 telescope rotated 0.1 arcsec 6 months later b = = 1.36 × 10 Wm 4πd2 1

M24_IBPH_SB_IBGLB_9021_SL12.indd 1 16/07/2014 11:55 (b) d = 10 pc = 10 × 3.26 light years 13 spectral type 15 17 = 32.6 × 9.46 × 10 m = 3.1 × 10 m 6 O B A F G K M 10 –10 L 3.839 × 1026 100 R 1000 R (I) b = = 5 2 17 2 10 10 R f. Betelgeuse M2I –8 4πd 4π(3.1 × 10 ) red giants (II) –6 = 3.2 × 10–10 Wm–2 104 e. Cassiopeiae M1III d. Mira M7III –4 26 3 1 R (III) 10 absolute magnitude ( 9 (a) L = 25 L = 25 × 3.839 × 10 W –2 27 = 9.6 × 10 W 102 main sequence ) 0 d = 8.61 light years = 8.61 × 9.46 × 1015 m L (IV) 101 0.1 R a. Beta Pictoris A5V g. Eridani K1V 2 = 8.1 × 1016 m c. Eta Arietis F5V 4 L 9.6 × 1027 100 Sun (V) luminosity ( b = 2 = 16 2

6 M 4πd 4π × (8.1 × 10 ) –1 10 0.01 R ) = 1.2 × 10–7 W m–2 (VI) 8 –2 white dwarfs 10 b. 61 Cygni A K5V 10 (b) Brightness at 10 pc (VII) 10–3 0.001 R 12 (10 × 3.26 × 9.46 × 1015 m = 3.1 × 1017 m) red dwarfs 27 –4 14 9.6 × 10 –9 –2 10 = = 7.9 × 10 W m 40 000 20 000 10 000 5000 2300 4π × (3.1 × 1017)2 temperature (K) 10 L = 5.0 × 1031 W –9 –2 b = 1.4 × 10 W m 14 (a) λ = 400 × 10–9 m L max b = 2.9 × 10–3 4πd2 λ = max T L 5.0 × 1031 19 d = = = 5.3 × 10 m 2.9 × 10–3 4πb 4 × 1.4 × 10–9 T = = 7250 K π 400 × 10–9 15 1 light year = 9.46 × 10 m (b) Diffi cult to fi nd 7250 on scale since it is not 19 5.3 × 1019 m = 5.3 × 10 linear but gives ~ 1 L 9.46 × 1015 = 5.6 × 103 light years 106L

11 supergiants 11 r = 3.1 × 10 m; T = 2800 K 4 luminosity 10 L A = 4πr2 = 1.2 × 1024 m2 4 –8 24 4 L = σAT = 5.6 × 10 × 1.2 × 10 × 2800 102L = 4.2 × 1030 W giants 1L –9 main sequence 12 (a) λmax = 400 × 10 m –3 2.9 × 10 3 10–2L T = –9 = 7.25 × 10 K 400 × 10 white dwarfs (b) Power/m2 = T4 = 5.67 × 10–8 × 72504 10–4L σ O B A F G K M 8 –2 = 1.6 × 10 Wm 30 000 K 7500 K 5000 K 10 000 K 6000 K 3500 K spectral class From HR diagram L ~ L = 3.84 × 1026 W (c) b = 0.5 × 10–12 Wm–2 b = L 4πd 2

26 d = 3.84 × 10 4π × 0.5 × 10–12 d = 7.8 × 1018 m or 826 light years

M24_IBPH_SB_IBGLB_9021_SL12.indd 2 16/07/2014 11:56 15 The spectral type of Beta Pictoris is A5V so This is greater than the Oppenheimer–Volkoff

from the HR diagram limit (3 M ) so Phi Orionis could become a (a) L = 10L black hole (b) r = 2R 21 Δλ = v λ c (c) T = 8000 K λ = 434.0 nm (d) d = L = 10L = 6.86 × 1017 m 4πb 4π × 6.5 × 10–10 Δλ = 479.8 – 434.0 = 45.8 nm Δλ 8 45.8 7 –1 = 22.2 pc v = c × = 3 × 10 × = 3.17 × 10 m s λ 434 3.5 16 L = M = 10 22 λ = 434.0; Δλ = 481.0 – 434.0 = 47.0 nm L (M ) Δλ 8 47 7 –1 1 v = c × = 3 × 10 × = 3.25 × 10 m s M = 103.5 × M λ 434 M = 1.9 M It is further away since it is moving faster. 23 H = recessional velocity 17 From fi gure, 20 days corresponds to a 0 separation luminosity of 6 × 103 L . separation = recessional velocity = 150 = 2.1 Mpc 3 26 H 72 d = L = 6 × 10 × 3.84 × 10 = 1.5 × 1019 m 0 –10 4πb 4π × 8 × 10 24 recessional velocity = H × separation = 72 × 20 d = 490 pc 0 = 1440 km s–1 2 18 Mass of Beta Pictoris = 1.9 M 3H0 –1 25 ρc = using H0 in s Δt = (1.9)–2.5 Δt = 0.2 Δt 8πG (2.33 × 10–18)2 10 = 3 × –11 19 d = L = 10 L = 3.6 × 1025 m 8π × 6.7 × 10 4πb 4π × 2.3 × 10–16 = 9.7 × 10–27 kg m–3 = 1200 Mpc –27 Mass of one hydrogen atom = 1.7 × 10 kg 3 20 so this is equivalent to about 6 atoms per m . spectral type 2.9 × 10–3 26 λ = = 1.06 mm O B A F G K M max 2.73 106 –10 2.9 × 10–3 100 R 1000 R (I) 27 λ = = 9.7 × 10–7 m 105 10 R –8 max 3000 red giants (II) –3 –7 –6 λ(obs)–λ(em) 1.06 × 10 – 9.7 × 10 104 Phi Orionis BOV z = = 9.7 × 10–7 –4 λ(em) 3 1 R (III) 10 absolute magnitude ( 3 –2 = 1.09 × 10 main sequence 102 R 1.06 × 10–3 ) 0 (obs) λ(obs) L (IV) 28 = = = 1093 R λ 9.7 × 10–7 101 0.1 R 2 (em) (em) 1 0 4 –4 10 Sun (V) R(em) = = 9.2 × 10

luminosity ( 1093

6 M –1 10 0.01 R ) (VI) 8 –2 white dwarfs 10 10 (VII) Practice questions 10–3 0.001 R 12 red dwarfs 10–4 14 40 000 20 000 10 000 5000 2300 1 (a) (i) An alternative to temperature is spectral temperature (K) class. Luminosity of Phi Orionis is ≈ 2 × 104 L (ii) An alternative to luminosity is absolute 3.5 L = M = 2 × 104 magnitude. L (M ) 1 M = (2 × 104)3.5 M = 17 M

M24_IBPH_SB_IBGLB_9021_SL12.indd 3 16/07/2014 11:56 (b) A = Main sequence 2 (a) The parallax angle, p is the angle B = Super red giant subtended by a star to the Earth when the C = White dwarf Earth has moved a distance of 1AU (1 Earth D = Main sequence orbit radius). For practical reasons, the

spectral type angle is usually measured when the Earth O B A F G K M is either side of the Sun (times separated 106 –10 (I) by 6 months); this gives an angle 2p. This 105 supergiants B –8 A red giants (II) angle is measured by measuring the angle –6 104 between the star and a very distant star, as –4 shown in the diagram. 3 (III)

10 absolute magnitude ( –2 102 main sequence ) 0 L Sun (IV) 101 2 4 100 (V) luminosity (

6 M –1 C 10 ) (VI) 8 10–2 white dwarfs (VII) D 10 10–3 12 red dwarfs distant star distant star 10–4 14 40 000 20 000 10 000 5000 2300 temperature (K) p (c) B is larger than A because even though B is p d colder it gives out more power (luminosity) L = σAT4 so if L is large and T is small, A must be big.

6 (d) From HR diagram, LB = 10 L –8 –2 bB = 7.0 × 10 W m b = 1.4 × 103 W m–2 A B d = 1.0 AU 2 –8 2 Sun L = 4πbd ⇒ LB = 4π × 7 × 10 × d 3 2 L = 4π × 1.4 × 10 × 1.0 1 AU 1 AU L –8 2 So B = 4π × 7 × 10 × d = 106 (b) If parallax angle = 0.549 arc seconds then L 4π × 1.4 × 103 × 1.02 distance d = 1/0.549 = 1.82 pc 6 3 d = 10 × 1.4 × 10 = 1.4 × 108 AU = 1.82 × 3.26 = 5.94 light years 7 × 10–8 (c) (i) Apparent brightness is the radiant 1 pc = 2.1 × 105 AU 1.4 × 108 power received per unit area at the d = 2.1 × 105 Earth. = 700 pc (ii) b = L 4πd2 At 700 pc, the parallax angle will be too (e) b (L /4πd 2) b = b b small to measure. 2 bs (Ls/4πds ) –9 θ ∼ 7 × 10 rad L b d 2 b d 2 b = b b = b × b 2 ( ) Ls bsds bs ds 1 AU θ = 2.6 × 10–14 × (5.94 × 6.3 × 104)2 8 1.4 × 10 AU = 3.6 × 10–3

M24_IBPH_SB_IBGLB_9021_SL12.indd 4 16/07/2014 11:56 (d) Barnard’s star is (ii) The period of a Cepheid is related to its (i) not hot enough to be a white dwarf luminosity, so if the period is measured its luminosity can be calculated. If (ii) too small to be a red giant the apparent brightness is measured, (see position on HR diagram; Barnard’s we can then fi nd its distance from star is in fact a red dwarf) the Earth. spectral type In this way the distance to distant O B A F G K M 106 –10 galaxies can be measured. 100 R 1000 R (I)

105 10 R –8 ) –2 1.3 red giants (II) A 4 –6

10 m W 1.2

–4 –10 3 1 R (III) 10 absolute magnitude ( (10 1.1 –2 b 2 main sequence

) 10

L 0 (IV) 1.0 101 0.1 R 2 0.9 4 100 Sun (V)

luminosity ( B

6 M 0.8 –1 10 0.01 R ) 0 1 2 3 4 5 6 7 8 9 10 (VI) 8 time (days) –2 white dwarfs 10 10 (VII) (d) (i) L = 7.2 × 1029 W 10–3 12 0.001 R and from the graph red dwarfs 10–4 14 b = 1.25 × 10–10 W m–2 40 000 20 000 10 000 5000 2300 L temperature (K) b = ⇒ d = L 4πd 2 4πb 3 (a) (i) Luminosity is the total power radiated 29 d = 7.2 × 10 = 2.14 × 1019 m from a star. 4π × 1.25 × 10–10 (ii) Apparent brightness is the power (ii) A standard candle is an object of received from a star per m2 by an known luminosity; it can be used to observer on the Earth. calculate distance by measuring its brightness. Since the luminosities of the L = power radiated b = power received Cepheid variables are known, they can m2 be used as standard candles. r b = L 4πr2 4 (a) We know that the Universe is expanding and that all particles of matter are attracted (b) A Cepheid variable has a change in to each other by gravity. This means that luminosity due to its change of size. When the rate of expansion is getting less. The it expands, its surface area increases, so it rate at which the expansion is slowing down radiates more energy, leading to increased depends on the density of the Universe. If luminosity and hence brightness. very dense, it will stop expanding and start (c) (i) The brightness is greatest when the to collapse. If not very dense, it will keep on star is biggest. So it is biggest at expanding forever. The critical density is the 2 days. density beyond which the Universe will stop expanding. 3H 2 3 × (2.7 × 10–18)2 (b) (i) ρ = 0 = o 8πG 8π × 6.7 × 10–11 = 1.3 × 10–26 kg m–3

M24_IBPH_SB_IBGLB_9021_SL12.indd 5 16/07/2014 11:56 (ii) A nucleon has mass = 1.7 × 10–27 kg 6 (a) The peak in the CMBR occurs at about –26 1.07 mm so using Wien’s law we can so number in 1 m3 = 1.3 × 10 1.7 × 10–27 calculate the temperature = 7.7 ~ 8 per m3 –3 T = 2.9 × 10 = 2.7 K 1.07 × 10–3 5 (a) (b) This radiation is the same in all directions R (on a large scale) and has the same spectrum as the radiation from a black body. This is the same type of radiation that would have filled the Universe soon after the Big Bang. At this time the wavelength would have been much shorter but it has expanded as the space it is in has expanded. (c) The fact that light from all distant galaxies luminosity is red-shifted implies that they are moving S away from us. If space is expanding it must

W have been smaller in the past. This supports the idea that the Universe began with a Big Bang.

temperature

(b) The amount of power radiated per m2 depends on the temperature of the star. L = σT4 (A ) So the total power emitted depends on the size and the temperature. As a star grows, it has a bigger surface area so can give out more power even though it is cooler.

M24_IBPH_SB_IBGLB_9021_SL12.indd 6 16/07/2014 11:56