Solutions576

Solutions

Chapter 1 Nuclear physics and radioactivity 1.1 Atoms, isotopes and radioisotopes1 a1 a 20 protons, 25 neutrons, 45 nucleonsb 79 protons, 118 neutrons, 197 nucleons c 92 protons, 143 neutrons, 235 nucleons d 90 protons, 140 neutrons, 230 nucleons 2 a2 a 27 protons, 33 neutrons b 94 protons, 145 neutrons c 6 protons, 8 neutrons3 A radioisotope is an unstable isotope. At some time, it will spontaneously eject radiation in the form of alpha particles, beta particles or gamma rays from the nucleus. Three isotopes that are not radioisotopes could be any three stable isotopes, e.g. carbon-12, lead-206 and bismuth-209.4 Yes, a natural isotope can be radioactive. For example, every isotope of uranium is radioactive.5 Polonium-210 and uranium-238. These have atomic numbers of 84 and 92 respectively; and every isotope beyond bismuth (Z = 83) in the periodic table is radioactive.6 a6 a 88 protons, 138 neutrons, 226 nucleonsb 88 protons, 140 neutrons, 228 nucleonsc Ra-228 is slightly denser.7 1.1 × 10−13 8 210 m 9 a9 a There are no differences.b 89Kr has 5 more neutrons in its nucleus.1010 28Al

1.2 Radioactivity and how it is detected1 a1 a the nucleus b the nucleus c the nucleus2 2 α is a helium nucleus, β is an electron, γ is electromagnetic radiation. 3 a3 a beta particle b proton c alpha particle d neutron 4 a4 a Z = 82, A = 214, lead b Z = 90, A = 231, thorium c Z = 89, A = 228, actinium d Z = 80, A = 198, mercury5 a5 a α b β− c β− d α e γ 6 lithium-77 a7 a proton b neutron c neutron d alpha particle8 a8 a 7 protons, 7 neutrons, 1 electron, 1 antineutrino b A neutron has decayed. c 10n→1

1p + –10e + ν

d Kinetic energy carried by the beta particle, antineutrino and nitrogen-14 nucleus.9 a9 a 40Ca, 42Ca, 43Ca, 44Ca, 46Ca, 48Ca b one c betad 48

19K → 4820Ca + –1

0β, 4820Ca is stable

e K: 1.53, Ca: 1.40 f alpha emitter g 217

87Fr→21385At + 42α→209

83Bi + 42α; bismuth-209 is stable10 a10 a 197

79Au + 10n → 19879Au b 198

79Au → 19880Hg + –1

0β

1.3 Properties of alpha, beta and gamma radiation1 a1 a α, β, γ b γ, β, α 2 B3 Gamma radiation is most suitable since its penetrating ability will enable it to reach the tumour.4 A beta emitter would be best suited because its penetrating ability would enable it to irradiate a small volume of tissue around the source. Alpha radiation would not penetrate the tumour at all, and gamma radiation would pass out of the body irradiating some healthy cells along the way.5 a 5 a 5.3 × 107 eV b 4.0 × 108 eV c 2.9 × 108 eV6 a6 a 1.4 × 10−12 J b 6.7 × 10−14 J c 8.0 × 10−14 J7 a7 a 3.4 × 106 eV b 1.65 cm8 D 9 A 1010 D

1.4 Half-life and activity of radioisotopes1 C 2 a2 a 10 g b 5 g c 2.5 g d 0.31 g3 a

010 20 30 40 50

Time (min)

400

350

300

250200

150

100

50

Cou

nt r

ate

(Bq

)

b ~235 Bq c 20 min d 50 Bq 4 15 min 5 0.5 6 192 μg 7 a7 a 10 half-lives b ~240 000 years8 a8 a uranium-235 b It has a much shorter half-life than uranium-238 and so has decayed much more rapidly since the formation of the Earth.9 a From the activity graph, the time that the activity takes to halve from 800 Bq to 400 Bq is 10 minutes. This is the half-life. b 50 Bq10 a10 a Over time, the radioisotopes transmute by a series of decays to form lead-206, which is stable. The percentage of lead in the sample will increase over time.b 214Po has such a short half-life (160 μs) that when 214Bi nuclei decay to 214Po, they almost instantaneously transmute to 210Pb.

577Solutions

1.5 Radiation dose and its effect on humans1 a1 a A, B, C, E b C, D, E2 a2 a 10 μSv b 0.50 μSv c 0.50 μSv3 a3 a 2.0 × 10−4 Sv b 1.6 × 10−2 J 4 a4 a B b A5 a 5 a 2 days b b 50 days c c 807 mSv6 Minimise your exposure to medical procedures involving ionising radiation, keep air travel to a minimum, and live in a weatherboard house at low altitude near the equator.7 They must have received massive doses of at least 8 Sv.8 a 90 hoursb Beta particles have enough penetrating ability to irradiate the tumour, and the long half-life means that the activity of the sample would not decrease noticeably during the course of treatment.9 D 1010 Effective dose = ∑(dose equivalent × W) = (35 × 0.20) + (35 × 0.05) + (50 × 0.12) = 15 mSv

Chapter 1 review 1 a1 a 17 protons, 18 neutrons, 35 nucleons b 88 protons, 138 neutrons, 226 nucleons 2 a2 a X-rays, infrared radiation, microwaves, gamma rays b alpha particle c alpha particle d beta particlese α, β, γ, X-rays f gamma radiation 3 a3 a x = 82, y = 208 b x = 78, y = 176 4 a4 a β− b α c β+

5 a5 a 4.8 × 10–10 J b 1.3 MeV6 a6 a 3.75 g b 218

84Po → 21482Pb + 42α c lead-214, bismuth-214,

polonium-214 d Lead-214; it has the longest half-life and so is the slowest to decay. This means that the atoms of this isotope remain in the sample for the longest time.7 a7 a 14

6C→147N + –1

0e + ν b 4 per minutec ~11 500 years (2 half-lives)8 a8 a 2.2 × 10−14 J b 6 hours c 2.0 × 106 Bq d 0.5 × 106 Bq9 a9 a 24

11Na → 2412Mg + –1

0e + γ, magnesium-24 is produced b Gamma rays can penetrate a great thickness of air, but beta particles can penetrate only a few metres. c β d Kinetic energy carried by the gamma ray, beta particle and magnesium-24 nucleus.10 10 Her blood cells have been affected; this is a somatic effect.1111 7 alpha, 4 beta12 a12 a 3.0 × 1010 nuclei b 1.5 × 1010 nuclei c 7.5 × 109 nuclei d 9.4 × 108 nuclei1313 D 1414 lithium-615 a 15 a The half-lives of the samples are equal. b The activity of A is twice that of B.16 a16 a A proton has transformed into a neutron, a positron and a neutrino.

b Kinetic energy carried by the neutrino, positron and carbon-12 nucleus.1717 3.1 × 1019 Bq18 a18 a ~3 μSv b about 4 times background19 19 Effective dose for patient A = 1080 μSv. This is greater than the dose received by patient B, so patient A is at a slightly greater risk of developing cancer from this exposure.

Area of study review (Nuclear physics and radioactivity)1 Cs-137: 55 protons, 82 neutrons, 137 nucleonsI-131: 53 protons, 78 neutrons, 131 nucleons2 2 Beam ii is gamma since it is undeflected by magnetic fields. Beam i is alpha and beam iii is beta. Beta is much lighter and is deflected more strongly.3 a 3 a 2.4 × 10–10 J b gamma ray4 a nucleus b nucleus c nucleus5 5

6 6 Radioactive isotopes of the same element will have identical chemical properties to the non-radioactive isotope.7 Each atom of gold-198 has one more neutron in its nucleus meaning that these atoms have a slightly higher density.8 73Li + 11H → 2(4

2He)9 a9 a 8.8 MeV b 0.42 MeV c 0.50 MeV10 a 10 a

1111 ~320 Bq 1212 10 min 1313 100 Bq 1414 B1515 Gamma rays, having no electric charge, and moving at the speed of light, are the most highly penetrating form of radiation. Gamma rays interact with matter infrequently when they collide directly with a nucleus or electron. The low density of an atom makes this a relatively unlikely occurrence. Hence, gamma rays pass through matter very easily.

Time (G years)

No. of K nuclei

No. of Ar nuclei

Ratio K : Ar

0 1000 0 –

1.3 500 500 1 : 1

2.6 250 750 1 : 3

3.9 125 875 1 : 7

1000

800

600

400

200

0

Act

ivity

(Bq

)

Time (min)0 5 10 15 20 25 30

Solutions578

1616 Alpha particles travel through air at a relatively slow speed and with a double positive charge. They interact with just about every atom that they encounter. The alpha particle ionises many thousands of these atoms. Each interaction causes the alpha particle to slow down a little, and eventually it will be able to pick up some loose electrons to become a helium atom. This takes place within a centimetre or two in air. As a consequence, the air becomes quite ionised, and the alpha particles are said to have a high ionising ability. But alpha particles don’t get very far in the air, so they have a poor penetrating ability.1717 1.0 × 1020 Bq 1818 1 p, 2 n 1919 helium 2020 B 2121 A neutron has transformed into a proton and an electron. 2222 185

79Au → 18177Ir + 42α 2323 79 p, 106 n

2424 No, the alpha particles would not penetrate into the living cells of your hand.2525 Gamma rays have the ability to penetrate the skull and pass through the target region. The half-life of 5.3 years means that the source would retain its activity for a number of years and would not need replacing very often.2626 300 μSv 2727 60 s 2828 2.3 g 2929 26

11Na → 2612Mg + –1

0β + ν30 30 radiation sickness, hair loss3131 testes 3232 C 3333 1.5 × 103 μSv3434 These bombarding electrons may not be product of radioactive decay. 3535 The electrons would be strongly repelled by the electron clouds of the atom.3636 B 3737 After 8 minutes, 211Bi will have half the activity of 215Bi.3838 a a 380 J b 5.0 Sv c c nausea, hair loss, leukaemia, possibly death 39 39 1.2 × 10−5 Sv 4040 This represents about 15 times background radiation.

Chapter 2 Concepts in electricity2.1 …lectric charge1 He was providing lightning with an easy path to him. 2 Almost none of our modern technology would exist! 3 About 20–304 Both show repulsion and attraction. Magnetism is relatively permanent, electrostatic effects are temporary. Magnetic poles cannot be isolated, opposite charges can. 5 Charges either repel or attract, there are no other alternatives. 6 They will attract.7 An excess of the fluid creates one type and a deficit the other.8 Charge leaks from sharp points.

9 Lightning will strike tall objects and may jump to nearby objects. Find a spot away from tall objects.1010 Positive and negative can cancel each other, as do the effects of opposite charges; colours do not.

2.2 …lectric forces and fields1 a1 a 2F b 4F c –F d 4F 2 900 kN 3 It is not!4 a4 a 0.35 N b Charges are further apart.5 a5 a 0.01 N down, 0.025 N up b – 0.2 μC

2.3 …lectric current, …MF and electrical potential1 a1 a 3 A b 0.5 A c 0.008 A 2 a2 a 5 C b 300 C c 18 000 C3 a3 a 9 A right b 2 A left 4 a4 a 3 C b 1000 C c 1440 C 5 No 6 3 μA down belt 7 a7 a 2.7 × 107 A b No, I = 0 8 Spark plugs have a p.d. of 15 000 V9 a i9 a i 4 V ii ii 4 V iii iii 2 V b ib i 10 A ii ii 1 A iii iii 1 A 1010 D1111 20 V 1212 9 J 1313 167 C 1414 the gravitational potential energy of the water

2.4 Resistance, ohmic and non-ohmic conductors1 a1 a M2 or M3 b M1 or M42 a2 a 3 A b 50 V c assumed constant resistanced 2.5 Ω at all voltages graphed3 a3 a 0.71 Ω b ohmic 4 Both right at different voltages.5 72 mA 6 a6 a 2 Ω b 5 A 7 a7 a 1.6 Ω b 0.2 Ω8 a8 a non-ohmic b 0.5 A c 15 V d id i 20 Ω iiii 13.3 Ω9 Cathryn, lower slope means less current at given voltage.1010 a 0.11 Ω b 1.1 V

2.5 …lectrical energy and power1 a1 a 4.5 J b 4.5 J c No, some lost as heat. 2 2.4 × 10−19 J3 a3 a 0.6 W b 2400 W c 720 W 4 a4 a 250 mA b 5 A c 7.5 A5 a5 a 25 V b 8.7 V c 417 V6 21 kA 7 0.84 kW h, 3.0 MJ, $6.508 a8 a 1.5 A b 0.075 A 9 AC oscillates, DC steady1010 Less current is needed.

Chapter 2 review1 Negative: electrons; orbit nucleus. Positive: protons; in nucleus. It can lose/gain electrons. 2 Ensure no p.d. and hence no spark.3 i3 i A +, B – iiii same iiiiii same iviv neutral 4 No.5 No, all the likes would collect together well away from the others. Atoms depend on the mutual repulsion of the electrons and their attraction to the nucleus.6 90 N 7 9000 N8 4 × 105 N C−1 9 0.001 s, comes as pulses.

579Solutions

1010 Conducts in electrostatic experiments, not at mains and below.1111 240 J 12 a12 a 3 V b 0 V c 3 V13 a13 a in the filament b same14 a14 a 18 C b 54 J c the battery15 a15 a 48 Ω b R increases with temperature c 1200 W16 a16 a 0.125 Ω b 1.25 V c 12.5 W 17 a17 a no b 40 Ω, 27 Ω c 1.6 W, 9.6 W18 a18 a 2 MW b 5 A 19 a19 a 0.079 Ω b no (twice)2020 440 kW h, 1600 MJ

Chapter 3 Electric circuits3.1 Simple electric circuits1 Itot = 0 at a point. Etot = ΔVtot in a circuit.2 2 V drop in circuit 3 +0.7 A 4 a4 a 0.25 A b 2.4 V5 a5 a yes b yes 6 17 Ω 7 0.22 A, 45 Ω8 a8 a 0.01 A b 4 V and 1 V9 a9 a through (15V, 2A) b b 7.5 Ω 1010 15 Ω

3.2 Circuit elements in parallel1 a1 a 0.30 A b 3 V 2 a2 a 20 Ω b 5 Ω3 a3 a through (10 V, 6 A) b 1.7 Ω 4 68 Ω 5 0.9 A, 11 Ω6 a6 a 6.7 Ω b 20 V c c 1 A (in 20 Ω) and 2 A7 20 W (in 20 Ω), 40 W 8 a8 a 4.4 kΩ b 0.44 kΩ9 a9 a A1 = 7 A, A2 = 5 A, A3 = 2 A b 50 W and 20 W respectively c 1.4 Ω d 70 W (same as previous total)1010 Betty is correct.

3.3 Cells, batteries and other sources of …MF1 Need different electronegativities. 2 + to + and – to –3 No resistance across terminals. 4 15 V5 a5 a 6 V b batteries last longer6 The heavy current causes a p.d. within the battery.7 10 V 8 1 Ω 9 12 A 1010 Increased I leads to lower V.11 a11 a 0.4 W b 0.33 V c No, most power is available at 0.42 V.12 a12 a 1.6 V b 5 Ω c 30 mW, 120 mW, 110 mWd 2 mW, 72 mW, 242 mW

3.4 Household electricity1 C 2 Neutral and earth are common.3 Make the circuit safe. 4 Ensure neutral is 0 V. 5 Yes. Live when off. 6 Case could become live. 7 safer 8 Less contact, more resistance. 9 2.4 mA 1010 Ensure better ground contact.

Chapter 3 review1 Car body is the other conductor. 2 12 V 3 no

4 a4 a 2 A b 60 Ω c 1920 W 5 5 Ω 6 a6 a 100 ma b 15 Ωc no definite current 7 a7 a 220 mA b no, much more8 a8 a 400 mA b 800 mA c 1.6 A 9 a9 a 15.4 kΩ b 63 kΩ1010 0.1 mho, 0.2 mho 1111 4.5 A12 a12 a Voltmeter has very high resistance.b burn out ammeter c M1 = A, M2 = V13 a13 a 1.3 A b 1.3 A c 3.2 A 14 a14 a 4.9 V b 1.1 V c 6.0 V1515 A1 = 3.2 A, A2 = 0, V2 = 0 1616 A1 = 3.2 A, A2 = 0, V1 = 6.0 V, V2 = 6.0 V17 a17 a 10 Ω b 25 Ω 18 a18 a 4 A b 8 Ω 1919 0.25 Ω20 a20 a 320 W b 232 V

Area of study review (Electricity)1 1 B 2 2 A 3 a3 a 1.25 × 1019 b 1.60 × 10−17 J c The electrical energy is converted into heat energy in the wire. d The EMF of a battery is defined as the number of joules of electrical energy the battery adds to each unit of charge passing through it. Therefore 100 V = 100 J C−1. This means that the battery adds 100 J to each coulomb of charge passing through it. e 200 W f 200 W g 200 W h Answers are the same because power is the energy given to each unit of charge (volt) times the number of charges that flow in the wire each second (amperes).4 4 Charges have Ep not Ek.5 a 5 a Charges escape at the same rate. b 1.2 × 1013 off c 0.80 W d d 6.4 × 10−14 J6 a 6 a 50 Ω b 1.44 × 10−18 J c 1.62 W d 1.1 × 1019

7 B 8 a8 a 100 Ω b 1.0 A c ic i 80 V iiii 10 V iiiiii 8.0 V d 98 V e 100 W 9 C 10 a10 a 60 Ω b 2.0 A c I1 = 1.20 A, I2 = 0.60 A, I3 = 0.20 A d 240 W e 240 W 11 a11 a 6.7 kΩ b 2.5 kΩ 12 a 12 a 12 V b 11.4 V c The 9 V battery will go flat before the others.13 a 13 a 2.0 W b 0.67 A 1414 2 A 1515 50 Ω, 50 Ω, 55 Ω, 67 Ω 1616 409 W 1717 Element is hotter.18 a18 a 200 Ω b 2.0 A c c V1 = 200 V, V2 = 160 V, V3 = 36 V, V4 = 396 V d 800 W e 800 W f 396 V g 400 V19 a19 a I = 0.27 A. We would not expect twice the current at 4 V. If the curve is extended to 4 V we could expect around 0.31 A, but it might well be that the globe would have burnt out before 4 V is reached, as the graph implies that it is probably a 3 V globe. b At 1 V, R = 5 Ω; at 3 V, R = 10 Ω. c This is not an ohmic conductor because the I–V graph is non-linear. The graph of an ohmic conductor must be a straight line because the current is directly proportional to the voltage applied. d 0.9 W

Solutions580

20 a20 a The device is non-ohmic. The purpose of this device is to limit the current through a particular section of the circuit to a constant value regardless of the voltage across that part of the circuit. b The resistance of the device increases with voltage. c V1 = 150 V, V2 = 100 V d 0.30 W e 0.20 W f 0.50 W 2121 53% 2222 2.8 s 2323 42 A 2424 0.25 V 2525 219 V 2626 60 V, 60 W 2727 1 A, same 2828 Mary 2929 4.3 Ω 3030 a a 12 V b b 8 V c c 16 Ω3131 A 3232 D 3333 C 3434 A 3535 a 50 Ω b 15 Ω 36 a36 a 20 mA b 100 mA 3737 A 3838 D 3939 A 40 a40 a 8.0 V b 7.3 V

Chapter 4 Aspects of motion4.1 Describing motion in a straight line1 a1 a +40 cm, 40 cm assuming movement to the right is positive b –10 cm, 10 cm c +20 cm, 20 cm d +20 cm, 80 cm2 a2 a 80 km b 20 km north3 a3 a 10 m down b 60 m up c 70 m d 50 m up4 displacement 5 a5 a D b D c C d A 6 a6 a 39 steps b 1 step west of the clothes linec 1 step west of the clothes line7 a7 a ~2 m s−1 b ~1 mm s−1 c ~10 m s−1 d ~5 m s−1

8 a8 a 15 km h−1 b 4.2 m s−1 c No, she would probably be travelling faster or slower than this average speed at different times. It depends on the traffic and the terrain.9 a9 a 22.2 m s−1 b 6.7 km h−1 s−1 c 1.85 m s−2 d 20 m10 a 10 a –10 m s−1 b 40 m s−1 west c 800 m s−2

11 a11 a 12 km h−1 s−1 b 3.3 m s−2 south12 a12 a 1500 m b 1.7 m s−1 c 0 d 0

4.2 Graphing motion: position, velocity and acceleration1 a1 a +4 m b A, C c B d D e 0.4 m s–1

2 The car initially moves in a positive direction and travels 8 m in 2 s. It then stops for 2 s. The car then reverses direction for 5 s, passing back through its starting point after 8 s. It travels a further 2 m in a negative direction before stopping after 9 s.3 a3 a +8 m b +8 m c +4 m d –2 m 4 8 s5 a5 a +4 m s−1 b 0 c –2 m s−1 d –2 m s−1 e –2 m s−1

6 a6 a 18 m b –2 m7 a7 a The cyclist travels with a constant velocity in a positive direction for the first 30 s, travelling 150 m during this time. Then the cyclist speeds up for 10 s, travelling a further 150 m. Finally the cyclist maintains this increased speed for the final 10 s, covering another 200 m in this time. b +5 m s−1 c +20 m s−1

d approx. +13 m s−1 e +15 m s−1 8 a8 a B b A c C d D

9 a9 a running north at 1 m s−1 b increasing speed from 1 m s−1 to 3 m s−1 while running north c running north but slowing to a stop d stationarye accelerating from rest to 1 m s−1 while running south f running south at 1 m s−1

10 a10 a 2 m north b 10.5 m north c 9 m north11 a 11 a 2 m s–1 north b b 1.7 m s–1 north1212

13 a13 a 80 s b ~1.3 m s−2 c ~0.5 m s−2 d ~4.9 km14 a14 a +2 m s−2 b 4 s c 10 s d 80 m e +7 m s−1

15 a15 a

b 12 m s−1

4.3 …quations of motion1 a1 a +2.0 m s−2 b +8.0 m s−1 c 64 m 2 a2 a +3.1 m s−2 b 50 m s−1 c 180 km h−1

3 a3 a 40 m s−2 b 1120 m c 580 km h−1 d 80 m s−1

e 124 m s−1

4 a4 a –99.4 m s−2 b 0.284 s c 19.9 m s−1 5 Cars have greater accelerations when they are travelling slowly (i.e. when they are in a low gear). When they are travelling fast, they may have a high speed, but this speed does not increase rapidly when the throttle is pushed.6 D 7 a7 a +0.40 m s−2 b 28 m c 5.3 m s−1 8 a8 a 21 m s−1 b 5.3 m c 36 m d 42 m9 a9 a 4.0 m s−1 b 5.7 m s−1 c 2.0 s d 0.85 s10 a10 a 8.0 s b 16 s c c 192 m

4.4 Vertical motion under gravity 1 a1 a The brick would fall much faster than the paper clip. b They would fall together. c They fall together.

Pos

ition

(m)

10

2

34

5

67

8

9

10

11

Time (s)1 2 3 4 5 6 7 8 9 10 11 12

Acc

eler

atio

n (m

s–2

)

0

1

2

Time (s)

2 4 6 8 10 12 14 16 18 20

581Solutions

2 The bubbles rise because they are an air element and are returning to their natural place in the atmosphere.3 B 4 A, D 5 a i5 a i 9.8 m s−1 iiii 20 m s−1 iiiiii 29 m s−1

b i i 14 m s−1 iiii 20 m s−1 iiiiii 24 m s−1 c 12 m s−1

6

7 a7 a 3.9 m s−1 b 0.78 m c 0.20 m d 0.58 m8 a8 a 2.0 s b 20 m c 20 m s−1 d 20 m s−1

e i 9.8 m s−2 down iiii 9.8 m s−2 down iiiiii 9.8 m s−2 down9 a9 a 1.7 s b 3.2 s 1010 6.7 s

Chapter 4 review 1 D 2 15 m s−1 up 3 11 m 4 4 Aristotle’s theories did not explain why light objects fell equally as fast as heavy ones or why a solid (earth element) such as wood floated on water.5 a5 a from 10 s to 25 s b from 30 s to 45 s c from 0 s to 10 s, 25 s to 30 s, and 45 s to 60 s6 after 42.5 s 7 a7 a 20 m s−1 north b 40 m s−1 south8 a8 a 5 m s−1 b 15 m s−1 9 C 1010 2.0 s 1111 15 km h−1

12 a12 a 10 km h−1 north b 2.8 m s−1 north13 a13 a 4.0 m s−2 b 4.0 m s−1 c 6.0 m14 a 14 a 0.80 m s−1 b 0.50 m s−1 c 0.67 m s−1

15 a15 a 0.75 m s−1 b 8.0 m s−2 16 a16 a 3.5 s b 2.9 s1717 1.0 s 1818 ~10 m s−1 north 1919 A20 a20 a 0 b ~2.0 m s−2 north c 7.0 m s−2 south

Chapter 5 Newton’s laws5.1 Force as a vector1 1 B, D 2 a2 a 150 N up b 40 N west c 0 d 14 N NW3 a3 a 10 N b 10 N c 50–100 N d 400–800 N4 B, C, D5 a5 a 60°T b 320°T c 240°T d 135°T e 22.5°T6 a6 a 60 N right b 50 N up c 70 N down7 a7 a 1 N west b 100 N 143°T8 610 N in a direction that bisects the two ropes9 a9 a 50 N south, 87 N east b 60 N north, 0 N east

c 282 N south, 103 N eastd 1.5 × 105 N up, 2.6 × 105 N horizontal1010 150 N horizontal , 260 N upwards

5.2 Newton’s first law of motion1 Aristotle felt that the natural state for any object was at rest in its natural place. This meant that any moving body would come to rest of its own accord. Galileo introduced the idea that friction was a force that could be added to other forces that act on a moving body, but it was Newton who explained that the moving object should continue to travel with constant velocity unless a net force is acting. 2 a A b D c D3 3 No forwards force acts on the person. In accordance with Newton’s first law of motion, the bus slows and the standing passenger will continue to move with constant velocity unless acted on by a force—usually the passenger will lose their footing and fall forwards.4 20 N in a forwards direction, so that the net force will be zero5 lift = 50 kN up, drag = 12 kN west 6 a6 a 25 N b 25 N horizontallyc 29 N at 30° to the horizontal7 Their inertia would cause them to move forwards along the aisle as the plane came to a stop. 8 a8 a gravitation b electric force c friction between the tyres and the road d tension in the wire9 a9 a If the cloth is pulled quickly, the force on the glasses acts for a short time only. This force does not overcome the tendency of the glasses to stay where they are i.e. their inertia. b Using full glasses makes the trick easier because the force will have less effect on the glasses due to their greater mass. The inertia of the full glasses is greater than that of empty glasses.1010 The fully laden semi-trailer will have more difficulty pulling up. Its large mass means that more force is required to bring it to a stop.

5.3 Newton’s second law of motion1 a U b B c B d U 2 a 4.3 × 103 m s−2 b b 260 N3 The 1.5 kg shot-put is the larger of the two masses, and so for the same applied force, its acceleration will be lower. This means a lower speed on leaving the athlete’s arm, and so it cannot be thrown as far as the lighter 1.0 kg ball.4 a4 a 160 N b 141°T c 210 m s−2, 141°T (but this is for a very short time) 5 2.6 × 103 N in the direction of travel

a b c

d e

d v

t t

t t

t

x

v a

Solutions582

6 a6 a 50 kg b 490 N c 20 N south d 0.31 m s−2 7 m = 1.5 kg, Fg = 5.4 N8 6.7 × 103 N opposing the motion9

As the parachutist leaves the aircraft, his or her weight will be the net force acting, accelerating at 9.8 m s−2. As the speed increases, the drag force (air resistance) opposing the motion also increases until it equals the weight. At this point in time, the net force will be zero and the parachutist will travel with a constant speed.10 a10 a 1.6 m s−2 b 0.80 m s−1 c 0.20 m s−2

5.4 Newton’s third law of motion1 a1 a Force that the racquet exerts on the ball; force that the ball exerts on the racquet. b Force of attraction (gravity) that the Earth exerts on the pine cone; force of attraction (gravity) that the pine cone exerts on the Earth. c Downwards force that the pine cone exerts on the ground; upwards force that the ground exerts on the pine cone. d Force of attraction (gravity) that the Sun exerts on the Earth; force of attraction (gravity) that the Earth exerts on the Sun.2 a2 a 140 N in opposite direction to leaping fisherman b 3.5 m s−2 in opposite direction to fisherman c fisherman: 1.0 m s−1, boat: 1.8 m s−1

3 a3 a If no other forces act, there will be an action—reaction force pair in which the action force is the force on the tool kit, and the reaction force will act on the astronaut in the opposite direction. If the tool kit is thrown directly away from the ship, hopefully the reaction force will propel him back to the craft. b 20 N act on each body, but in opposite directions. c 0.10 m s−1 d 100 s or 1 min 40 s4 12.5 m s−2 south 5 a B b C6 a6 a Weight acts vertically down, normal force is perpendicular to the incline, drag and friction act up the incline. b Fg = 640 N downwards, FN = 410 N up and perpendicular to the track c ∑F = 490 N along the incline d 7.5 m s−2 along the track 7 280 N8 The normal force will be smallest at A and become progressively larger as he travels through points B and C.

9 Tania is correct. The forces are equal in size but opposite in direction, but they are both acting on the lunchbox and so cannot be an action—reaction pair in the sense of Newton’s third law.10 a10 a 196 N down on the side of the studentb 2.45 m s−2 down to the side of the student c 367.5 N

Chapter 5 review1 C 2 48 N 3 5.0 m s−1

4 a 590 N b Drag force is larger than gravitational force. c Drag force is equal to gravitational force.5 a5 a 5.0 m s−1 b ib i 540 N down iiii 540 N down6 a6 a 85 kg b 85 kg c 306 N down7 A block is struck with a sharp blow so that it overcomes the grip of the blocks with which it is in contact, and it is quickly ejected from the pile. The other blocks experience a frictional force that acts for a very short time and they stay in the vertical stack. 8 a8 a 110 N horizontally, 41 N down b 110 N to the south c FN = 240 N upwards d When the trolley is pulled, the vertical component of the applied force is upwards rather than downwards, and so a smaller (upwards) normal force is needed—helping the wheels to rotate more freely. 9 7.0 m s−1 1010 B11 a11 a 1.5 m s−2 in the direction of the force b 75 N12 a12 a 1 b 0.25 c 0.251313 To maintain a tension of less than 100 N in the rope, the bucket must accelerate at greater than 1.5 m s−2 downwards. If the acceleration falls below this, the rope will snap.14 a 14 a No b Both forces are acting on the water tank. Action–reaction forces act on different objects. 1515 71 km h−1 1616 30° 1717 3.4 m s−2 down the incline1818 a

b

Net force on parachutist700 N

0 Time

force dueto racquet

direction of travel

drag forcedue toair resistance

F

F

583Solutions

1919 3.3 m s−2 down, on the side of the 10 kg mass; 65 N2020 a Julia is correct. The marble and the billiard ball exert equal and opposite forces on each other as described by Newton’s third law. b Julia is correct. The floor and the basketball exert equal and opposite forces on each other as described by Newton’s third law.

Chapter 6 Momentum, energy, work and power6.1 The relationship between momentum and force1 a1 a 100 kg m s−1 b 1.0 kg m s−1 c 28 kg m s−1

2 a2 a 24 kg m s−1 b 64 kg m s−1 c 8.0 kg m s−1 d 40 N s e 8.0 N 3 ball of mass 2.5 kg4 a4 a 41 kg m s−1 b 0.24 kg m s−1 c 500 kg m s−1 d 160 N s5 a5 a 4.0 kg m s−1 b 124 N6 a6 a 9.0 N s b 180 N in the direction of the ball’s travel c 180 N opposite the direction of the ball’s travel 7 a7 a 1200 N b 63 N s8 a8 a 1.25 kg m s−1 opposite direction of flight b 1.25 N s opposite direction of flight c 1.6 × 103 N opposite direction of flight9 a9 a Based on idea of impulse. Stopping time increased by collapsing shell reduces impact force. b Stopping time would be reduced and force increased, therefore not as successful.1010 Need to include mass and initial velocity then Δp = mΔv. Impulse = Δp = FΔt (units: N s). Estimate stopping time then F = Δp/Δt (units: N).

6.2 Conservation of momentum1 2.0 m s−1 in direction of white ball’s original motion2 4.7 m s−1 in the same direction3 0.44 m s−1 in the opposite direction4 11(.33) tonnes 5 B6 a6 a Only if Superman is fixed to the ground so that the momentum of the truck is transferred directly to the ground.b Final velocity = ptruck/total mass of Superman + truck7 vcar = 100 km h−1 so yes! 8 8.0 m s−1

9 3 m s−1 opposite the direction of the exhaust gases10 a10 a Rocket is losing 50 kg over this 2 s period so use average mass of rocket is 225 kg and v = 40 m s−1 up. b 4.5 × 103 N c 10 m s−2

6.3 Work1 2.6(5) × 102 J 2 1.5 × 105 J3 Also travelled horizontally against frictional forces.4 D since i mentions distance rather than displacement5 a5 a 1.5 × 102 J b >1.5 × 102 J c 1.0(29) × 103 J

6 B, since the motor won’t convert all supplied energy to useful work7 No work, since there is no change in position (don’t need to read the scale). 8 a8 a 2.4 × 103 J b nil9 a9 a

b b 2.0 × 102 J c c 1.6 × 102 J d d 40 J 10 10 ~1.3 J

6.4 Mechanical energy1 a1 a 3(.125) J b 4 × 102 J c 2.6 × 105 J2 a2 a 49 J b 412 J c 1.2 × 105 J3 a3 a 2.5 J b 1.8 J c –0.69 J4 A 5 6.9 × 103 N 6 a6 a –0.98 J b 0.98 J c 4.9(5) m s−1

7 A: k = 4000 N m−1 (stiffest), B: k = 1300 N m−1, C: k = 570 N m−1

8 8 A: 1.3 J, B: 3.7 J, C: 8.7 J9 a9 a 3.13 J b 12.5 J c 28.1 J 1010 9.13 cm

6.5 Energy transformation and power1 a1 a kinetic energy to heat b Ek to Us to Ek to Ug c Us to Ek to Ug d kinetic to heat/sound2 elastic potential → kinetic → gravitational potential → kinetic (+ heat/sound) → heat (+ sound, kinetic of water rebounding)3 a3 a 5.4 × 103 J b 4.5 × 102 W 4 a4 a 73(.5) J b 73 J c 0.61 Wd coach lifting own body, losses to surrounding environment5 a5 a 2.4(5) J b 2.4(5) J c It is transferred to the ground and transformed into other energy forms such as heat and sound.6 a6 a 4.3(2) m b Length of rope insufficient to allow height change, air resistance and other factors leading to transfer/loss to the surrounding environment.7 6.0 m s−1 8 a8 a 9.9 m s−1 b 8.9 × 10−1 N9 a9 a 5.0 J b 8.2 m s−1 c 6.1 cm10 a10 a 75 J b 2.2 m s−1 c 0.26 m

Chapter 6 review1 5.0 J 2 5.0 J 3 10 m s−1

4 No. Using the area under the graph the work done is 1.25 J or one-quarter of the total work required.5 2.5(6) × 102 J 6 1.1 × 103 N7 a7 a 2.4 × 104 J b gravitational potential energy → kinetic energy → elastic potential energy (and heat)

FN

FT

Ff

Fg

35

Solutions584

c 6.9 × 103 J d 1.7 × 104 J e The gravitational potential energy and Ek values will diminish with each bounce. The total mechanical energy is not conserved as this is not an isolated system. This is because some of the work done in stretching the rope is converted to heat.8 D (147 J) 9 B (10 m s−1) (B) 1010 A (5.1 m) 1111 a = 0, so ΣF = 0 1212 2970 N = 3.0 × 103 N 1313 1715 N = 1.7 × 103 N1414 Mandeep; Newton’s first and second laws: no net force so no acceleration, but constant velocity maintained.1515 3998 = 4.00 × 103 J1616 9995 = 1.0 × 104 watts1717 B: 25 m s−1, C: 11 m s−1, D: 19 m s−1

1818 Total mechanical energy is constant.

1919 57 × mass J 2020 81%2121 2.2 m s−1 in direction of moving player 2222 2.2 m s−1 north 2323 2.2 m s−1 in same direction2424 negligible movement but there is a change (v = 2.7 × 1021 m s−1)2525 7.1 m s−1 2626 1.1 × 104 N s 2727 14 × 103 N

Area of study review (Motion)1 10.2 s 2 510 m 3 9.8 m s−2 down4 The weight will travel three times as far during the 2nd second as during the first. 5 10 m 6 5.0 s 7 60 m 8 4.99 s 9 4.42 s 1010 platinum sphere 24.4 m s−1 down; lead sphere 31.7 m s−1 down 1111 Yes, air resistance would affect the spheres over the distance. The sphere with the larger diameter would be affected more. 12 a12 a 0.10 m s−1 b b 0.30 m s−1 c c 0.50 m s−1 13 a13 a 0.10 m s−1 b 0.30 m s−1 c 0.50 m s−1 1414 The marble is moving with constant acceleration. 1515 0.375 m s−2 west 1616 375 N west 1717 575 N west 1818 575 N east 1919 70.6 km h−1 2020 600 N 2121 800 N2222 4.0 m s−2 up the incline 2323 1.6 × 103 N2424 a 3.3 m s−2 clockwise; i.e. 3.3 m s−2 up for the 10 kg mass and 3.3 m s−2 down for the 20 kg mass b 1.3 × 102 N

25 a25 a 300 N b 4.0 kJ c 1.0 kJ d 3.0 kJ e 3.0 kJ f 17.3 m s−1 g The frictional force is retarding the motion of the force and producing heat.26 a26 a 1.6 MJ b 1.6 × 104 N c 16.4 kN d 1.64 MJ e 328 kW f 40 kJ2727 140 J 2828 80 J 2929 6.4 m s−1 3030 140 J3131 a a 40 J b 250 J c 5.0 cm d The elastic potential energy stored in the spring is transferred back to the trolley as kinetic energy when the spring starts to regain its original shape. e The spring is elastic. This means it can retain its original shape after the compression force has been removed. f pinball machines, watches, computer keyboards etc.32 a32 a The section from 2.0 cm to 5.0 cm, because the cube loses kinetic energy in this section. b 7.1 m s−1 c 3.0 J d The kinetic energy has been converted into heat and sound. e 100 N3333 100 m 34 a34 a 9.1 m s−1 b 8.0 m s−1 35 a35 a A, B, C b D c E, F, G d A, E e C, G36 a36 a 160 m b 2.0 m s−1 37 a37 a 1.2 m s−1 north b 6.0 m s−1 north c 3.0 m s−1 south3838 2.4 m s−1 3939 0.63 J 4040 Energy is converted into sound and heat when the mass collides with the pencil.4141 0.69 m s−2 4242 5.7 × 102 N 4343 2.5 × 102 m4444 Kinetic energy is converted into heat.4545 On Earth W = 1.96 × 105 N; on Moon W = 3.2 × 104 N4646 a The chair will obtain an initial velocity from the push, and then slow down to a stop due to sliding friction. b Castors will involve rolling friction, which is less than sliding friction, and hence the chair will travel further before coming to a stop. c In both cases the force of the initial push causes the chair to accelerate while being pushed, which is an application of Newton’s first and second laws. While the person pushes on the chair (action) the chair will push back on the person (reaction) with an equal and opposite force being an example of Newton’s third law.4747 The steady force applied by the engine is equal and opposite to the combined resistance forces such as air resistance, friction between the wheels and track etc. The net resultant force on the carriages is zero, and according to Newton’s first and second laws, constant velocity is the result.48 a48 a 85 N m−1 b 43 J c Estimate area under force–extension graph.4949 10 kg m s–1 5050 10 kg m s–1 5151 0.059 m s–1

Ene

rgy

x m

ass

(J)

Vertical displacement (m)0 5 10 15 20 25 30

300

250

200

150

100

50

0

Total = 302 m

D (12, 184)

D (12, 117.6)

C (25, 245)

A (30, 294)

C (25, 57)

A (30, 8)

B (0, 302)

Ep

Ek

B (0, 0)

585Solutions

Chapter 7 The nature of waves7.1 Introducing waves1 A particle carrying its kinetic energy with it, or a wave transferring energy but not matter.2 A continuous wave has a vibrating item at its source; a pulse is caused by a single input of energy.3 a3 a continuous b pulse c continuous d pulse e pulse4 a4 a transverse b In a longitudinal wave the particle vibrates parallel to the direction of travel of the wave. In a transverse wave the vibration is perpendicular to the direction of travel.5 a5 a transverse b pulsec c

6 a6 a longitudinal b continuousc c

7 D8 a8 a 3 b ib i right iiii left iiiiii right9 Answers will vary.1010 For a transverse wave, where a water wave travels parallel to the surface the water, particles must be able to move up and down. Only the particles near the surface can do this. Below the surface, the wave wouldn’t possess enough energy for particles to lift the weight of the water above.

7.2 Representing wave features1 a1 a f = 0.80 Hz, T = 1.2(5) s b f = 2.8 Hz, T = 0.36 s2 0.28 m s−1

3 It diminishes. Amplitude indicates energy. Wave energy is spread over a larger area as the wave moves further from the source, hence the amplitude is smaller.4 F = 8 Hz, T = 0.125 s5 5 A 6 D 7 B8 See See Heinemann Physics 11Heinemann Physics 11 Pearson Reader. Pearson Reader.9 a9 a 0.80 m, 20 cm b Q: up, S: up c c

10 a10 a A = 0.5 × 10−8 cm, T = 2.0 ms, f = 500 Hzb ib i 0 iiii 0.5 × 10−8 cm iiiiii –0.5 × 10−8 cm

c c

d The y-axis is actually showing displacement parallel to the direction of travel of the wave, but when showing this on a graph the y-axis is always perpendicular to the x-axis. Therefore this reminds us of a transverse wave.

7.3 Waves and wave interactions 1 a1 a halved b no change 2 0.60 m s−1 3 5.0 Hz4 0.040 m 5 D 6 D7 They have identical frequency (or wavelength), and they are in phase.8 8

9

1010

0.40

20

10

–10

–20

0.80 1.00 1.20 1.40 1.60P

Q

R

S

Distance along spring (m)Par

ticle

dis

pla

cem

ent

(cm

)

Time (ms)

Dis

pla

cem

ent

of Q

(x 1

0–8 c

m) 0.5

– 0.5

1 2 3 4

wavelength = 4 cmamplitude = 2 cm

Am

plit

ude

(cm

)

Distance (cm)

1

1

2

2 3 4 5 6 7 8

1

2

Distance (cm)

Dis

pla

cem

ent

(cm

)

wavelength = 4 cmamplitude = 1.7 cm

1 2 3 4 5 6 7 8

Solutions586

Chapter 7 review1

2

3

4 1D: wave in string or spring, 2D: water surface waves, 3D: sound waves5 string = transverse, air = longitudinal (sound wave)6 2 cm 7 4 cm 8 0.15 m s−1 9 f = 3.7(5) Hz, T = 0.27 s 1010 C, D 1111 a 4.0 s b 1.95 × 10−3 s12 a12 a 2.9 Hz b 250 Hz c 100 Hz13 13 D1414 A 1515 f = 86 Hz, T = 1.2 × 10−2 s1616 A1717

1818

1919

2020

Chapter 8 Models for light8.1 Modelling simple light properties1 shadows, eclipses, can’t see around corners2 a2 a 3.6 m b 0.66 m c 18 m3 B (after D)4 a4 a A reflective material is deposited onto one side of a piece of flat glass. This is usually covered by a protective layer of paint. b Some reflection occurs from the front of the glass of a mirror as well as from the silvered layer. The images produced are slightly offset and seen as separate.5 a 5 a upright mirror i = 50°, r = 50°, horizontal mirror i = 40°, r = 40° b b upright mirror i = 30°, r = 30°, horizontal mirror i = 60°, r = 60°6 Looking out of a window you can see your own image and the scene through the window. The two sets of images overlap.7 a7 a diffuse: v, vi regular: i, ii, iii, iv

10

10

20 30 40 50 60 70 80 90

Distance (cm)

Dis

pla

cem

ent

(cm

) t = 0 s

P Q

10

10

20 30 40 50 60 70 80 90

Distance (cm)

Dis

pla

cem

ent

(cm

)

t = 5.5 sP

0.25

10

0.50 0.75 1.0 1.25 1.50 1.75 2.0

Time (s)

Dis

pla

cem

ent

of

Q (c

m)

1 2 3 4 5 6 7 8

1.5

wave A

Dis

pla

cem

ent

(cm

)

Distance (cm)

1 2 3 4 5 6 7 8

1.5

wave B

Dis

pla

cem

ent

(cm

)

Distance (cm)

1 2 3 4 5 6 7 8

2.6wave A and wave B

Dis

pla

cem

ent

(cm

)Distance (cm)

1 2 3 4

1.5

wave A

Dis

pla

cem

ent

(cm

)

Time (s)

587Solutions

b The paper will produce diffuse reflection, and so light from the same point will be scattered in different directions.8 B, D9 The ocean can be considered to be made up of many layers of water from the surface down to the ocean floor. Each layer will reflect and absorb a little light, allowing the rest to be transmitted. Eventually, all the light will have been absorbed before it can reach the floor of the ocean.1010

8.2 Refraction of light1 a i1 a i D iiii B iiiiii A iviv E v Cb It slows down since it is refracted towards the normal.2 a2 a A, C b B, D c C d A3 As the wavefront enters the faster medium the wavelength will increase and, as a result, the direction of travel will bend away from the normal.4 a4 a The light rays travelling from your feet are refracted at the air–water boundary and appear to have come from a higher position, therefore you are shortened.b The path of light from a section of the sky is curved because of layers of air at different temperatures. Light enters the eye, travelling slightly upwards. Therefore the ‘sky’ is seen on the road ahead. This is interpreted as a puddle of water reflecting the sky.5 C 6 a6 a 1.5 b 2.26 × 108 m s−1

7 a7 a

7 b7 b 1.5 c 1.5 d 2.0 × 108 m s−1 8 a8 a 1.81 b 35°9 a9 a 58.0° b 39.6° c 58.0° d 18.4°1010 a = 25.4°, b = 25.4°, c = = 28.9°

8.3 Critical angle, TIR and …MR1 a1 a no b yes c yes d no 2 1A, 2C, 3C, 4D3 a3 a 24.4° b 38.7° c 48.8° d 62.5° 4 1.46 5 36.9° 6 a6 a 10−2 m b 300 nm7 a7 a Answers will vary. b b travels at speed of light travels at speed of light (3 (3 × 10 108 m s m s-1-1) in a vacuum, does not require a medium ) in a vacuum, does not require a medium in which to travelin which to travel8 C 9 a9 a magenta b white c white d white1010 Blending of colours of the spectrum to form white.

8.4 Dispersion and polarisation of light waves1 a1 a red b blue c blue2 If light was corpuscular, the orientation of the polarising material shouldn’t affect the amount of light that can pass through.3

4 If they are polarising, holding the lenses parallel to one another and rotating one lens 90° relative to the other should block any light from passing through.5 n(diamond) > n(glass) therefore more dispersion as light enters and leaves diamond. ic(glass) ≈ 42°, ic(diamond) ≈ 24°. Therefore light is more likely to strike the diamond–air boundary at an angle greater than the critical angle and totally internally reflect, spreading the spectrum even further.6 Only light that has its electric field aligned with the filter will be able to get through, therefore polarised light emerges from the filter.7 a7 a violet b violet c 25.4° d 24.5° e 0.9° 8 red 43.4°9 The pathway is determined by the wave property, frequency.1010 The first filter reduces the intensity by half, the second filter will reduce the intensity further, but not completely block the light.

Chapter 8 review1 a1 a speeds up b away2 Take measurements of corresponding angles of incidence and refraction. Graph sin i versus sin r to determine the refractive index.3 a3 a radio waves, microwaves, infrared, visible light, ultraviolet light, X-rays, gamma rays b All EMR travels at 3 × 108 m s−1 and all involve electric and magnetic fields varying at right angles to one another.4 Speed of light is calculated as 2.25 × 108 m s−1, so medium is not air as this is too slow. 5 a5 a 1 b 2 6 a6 a 32.0° b 53.7° c 21.7° d 1.97 × 108 m s−1

eye

red

violetwhite light

0.100

0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

sin i

sin

r

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

Solutions588

7 a7 a radio waves, microwaves, visible light, X-raysb X-rays, visible light, microwaves, radio wavesc radio waves, microwaves, visible light, X-rays8 Blue light is refracted most by water and therefore scattered most, making the water appear blue. 9 C1010 Light reflected from horizontal surfaces is quite polarised, so sunglasses can filter out a large proportion of reflected light. 11 a11 a There is refraction by layers of the atmosphere.b See Figure 8.21.12 12 reflection, refraction, total internal reflection1313 Black: no colours of light are reflected; grey: all colours partially reflected.1414 A 15 a15 a 19.5° b 19.1° c 0.4° d 1.96 × 108 m s−1

1616 2.1 × 108 m s−1 17 a17 a particle or ray b electromagnetic wave c ray d electromagnetic wave1818 1.17 19 a i19 a i 24.4° iiii 41.8° iiiiii 48.8°b Closer n values result in a larger critical angle.2020 It is refracted twice, first towards the normal on entering the glass and then away from the normal on exiting the glass. It emerges parallel to its original direction of travel.

Chapter 9 Mirrors, lenses and optical systems9.1 Geometrical optics and plane mirrors1 Any two of: linear propagation, reflection, paths in lenses and mirrors, and intensity reducing with the square of the distance from source.2 When the dimensions of the physical systems are much greater than the wavelength of light.3 If light is particle-like, sharp-edged bright areas of illumination would occur opposite the pinhole.4 a upright b same size c virtual5 nature (real or virtual), orientation, position, size6 a6 a They all produce images which are virtual, upright and same size. b A virtual image is produced behind the mirror (see Figure 9.3). 7 a7 a 5.5 m toward the mirror b 3 m s−1

8 a8 a 0.8 m—exactly half her height b 0.75 m above the ground 9 b9 b An inverted image of the ‘S’ is seen.10 b10 b The image would become inverted as there is now only one mirror.

9.2 Applications of curved mirrors: concave mirrors1 a1 a Both mirrors are curved, usually spherical. Concave mirrors are like the inside of a spoon (e.g. make-up mirrors). Convex mirrors are like the back of a spoon (e.g. security mirrors). b The mirrors are a part of a sphere. It is the radius of this sphere. c f = 25 cm d See Figure 9.10.

2 a2 a See Figure 9.8. b torch, headlights c Spherical aberration is the blurring of an image near the edge of a mirror because most spherical concave mirrors are not capable of bringing parallel light rays to an exact focus.3 a3 a virtual b 72 cm c 6 mm4 real, inverted, diminished, v = 30 cm5 f = 2 cm 6 a6 a 40 cm b 10 cm7 4 mm, virtual, upright, enlarged8 80 cm; less than 40 cm from the pole of the mirror9 u = 26.7 cm or 53 cm. A real or virtual, upright image can occur with M = 3.1010 f = 6.7 cm

9.3 Convex mirrors1 35 cm 2 See Figure 9.15.3 If a wide field of view is required. Both mirrors produce upright and virtual images. The convex mirror will produce a diminished image.4 a4 a The image of the ant is virtual, upright and slightly diminished; v ~–1 mm.b Reflected rays diverge in a convex mirror. 5 B6 shaving and make-up: concave, f ≈ 50 cm;shop security: convex, f ≈ 70 cm; rear-view mirror: convex, M ≈ 1

40

7 –3.3 cm; –2.5 cm; –1.4 cm8 upright, virtual, diminished, v = –2.7 m, M = 0.099 0.4 m 1010 2 m

9.4 Refraction and lenses1 a1 a

b See Figure 9.25.2 Rays must actually intersect to form a real image. Ray paths through convex lens only allows ray from above principal axis to intersect below principal axis. Therefore image is inverted.3 a3 a B b A4 Yes, if different radius of curvature. Higher n requires longer radius of curvature.5 a5 a Focal length increases since light no longer bent as severely, since change of speed is less. b Since light speeds up on entering air from water, rays are refracted away from the normal and a beam of light would be diverged.

principal axis

optical centre

optical axis

FF

F = principal foci

589Solutions

6 20 cm 7 B 8 D9 a real, inverted, enlarged b real, inverted, diminished c virtual, upright, enlarged10 a10 a 0.6 m b 10 cm

9.5 Concave lenses1 See Figure 9.25.2 Rays are diverged, hence no real intersection of rays.3 C 4 b4 b virtual c 10 cm d 1 cm5 b5 b virtual c –7.5 cm c 5 cm6 a6 a –30 cm b 3 c 15 mm 8 a8 a 22.5 cm b 202.5 cm9 a9 a Image is no longer real, it is virtual.b Image is now on the same side of the lens as the object, and at a distance of 30 cm is twice as close as previously.c Image is half the original size. 1010 3.87 cm

9.6 Optical systems1 spherical and chromatic aberration 2 equal to3 a3 a 4 cm b 4 mm 4 a4 a 5.2 cm b away from film or sensor5 i a5 i a low b low c high ii aii a high b high c lowiii aiii a low b low c high6 It has no spherical or chromatic aberration.7 a7 a 50 cm, 10 cm b objective 50 cm, eyepiece 10 cm c 60 cm d inverted8 Concave mirrors can produce real images. Photographic arrangement replaces eyepiece lens.9 a9 a 20 cm from objective lens, 5 cm from eyepiece lensb 30 cm from eyepiece lens

Chapter 9 review1 D 2 B 3 a3 a 6 m b See Figure 9.3.4 Parallel rays not converged to a perfect focus, resulting in an image that is blurred near the edges.5 All images are virtual and upright but diminished in size. The mirror will give a very wide field of view.6 a6 a v = 3 m b Hi = 0.5 m c real, inverted, diminished7 a7 a virtual, upright, magnified b v = –120 cm c c M = 4 d Hi = 8 mm8 a8 a v = –8.6 cm b virtual, upright and diminishedc M = 0.439 rear-view mirror, shop-security mirrors1010 Object can be inside F to produce virtual image (at 1

2 f), or between C and F to produce real image.1111 radius of curvature, refractive index12 b12 b virtual c –6.7 cm d 5 cm 1313 hypermetropic 1414 C15 a15 a v = 120 cm, real, inverted, 6 cm highb v = 72 cm, real, inverted, 2 cm highc v = –13.3 cm, virtual, upright, 3.3 cm diameter1616 A, C, D17 a17 a Microscope objective has shorter focal length.b Microscope forms a virtual, enlarged image. Refracting telescope forms a real, inverted image.

1818 at infinity 19 a19 a upright b 0.01 c 0.1 m d –0.10 m20 a20 a Either u = 10 cm, v = 40 cm, or u = 40 cm, v = 10 cm b 8 cm or 0.5 cm

Area of study review (Wave-like properties of light)1 A wave is a periodic disturbance that travels with a velocity, v, and continuously transfers energy through a medium without any movement of the medium as a whole. Wave motion involves the transfer of energy without transfer of the medium.2 C 3 B 4 B 5 a 0.02 s b 1.0 cm c 200 cm d 100 m s−1

6 a6 a 0.19 m b 1.33 s c no d temperature of the water 7 A 8 When light passes from a one transparent medium into another transparent medium its speed is altered. It is this change of speed that leads to the bending of light rays as they pass through media of different optical densities. This phenomenon is referred to as refraction of light.9 2.42 10 a10 a 29° b 42°11 a11 a 1.49 b 2.01 × 108 m s−1 c 42.15° d Total internal reflection will occur. e Medium 1 is more optically dense than medium 2 because it has a higher absolute refractive index.1212 1.5613 a13 a 1.94 × 108 m s−1 b 3.09 × 10−11 s c 5.08 × 1014 Hz 1414 D 1515 D 1616 C 1717 D 1818 Sound waves are longitudinal, all the others are transverse waves.1919 a Medium determines wave speed, so speed is unaltered, v. b λ/22020 In a given medium the velocity of light, v, is set. From f = vλ we know that the frequency, f, varies inversely with the wavelength, λ. For example, a higher frequency note will have a shorter wavelength.21 21 2500 m s−1 2222 790 nm 2323 B 2424 C 2525 C 2626 This phenomenon, referred to as dispersion, occurs because each frequency of light (colour) is refracted a slightly different amount as it passes through the glass.2727 Concave mirrors form inverted real images of objects placed beyond the principal focus. If the object is between the principal focus and the mirror, the image is virtual, upright and magnified. Convex mirrors produce only upright virtual images of objects placed in front of them. The images are diminished.

Solutions590

2828

Since the image is to be projected onto a screen it must be real; u = 3 m, f = 2.5 m and, hence, the mirror is concave.2929 a v = 20 cm, Hi = 1.0 cm, the image is real, inverted and the same size as the object. b v = 30 cm, Hi = 2.0 cm, the image is real, inverted and magnified. c The image cannot be located. d v = –10 cm, Hi = 2.0 cm, the image is virtual, upright and magnified.3030 E3131

Since the image is to be projected onto a screen it must be real; u = 2.25 m and v = 11.25 m, f = 1.88 m.3232 v = –0.20 m, Hi = 1.5 cm; hence the image will be upright, virtual, 1.5 cm high, and appear to be 20 cm behind the mirror.3333 The focal length of a convex lens is determined by its index of refraction and its radius of curvature.3434 C 35 a35 a i 8.0 cm iiii real, inverted and same size b virtual, upright, magnified36 a36 a

b The image is virtual, upright and diminished, apparently 0.75 m behind the mirror.3737 A and C 3838 D 3939 D 4040 a a u = 361 cm or 3.61 m b M = –0.024 c real, inverted, diminished d If u is very large, then using the lens equation v = 8.40 cm, so the lens will need to be moved a distance of 0.20 cm towards the film.4141 Chromatic aberration occurs because higher frequency colours are refracted slightly more by a medium than colours with lower frequency.4242 Polarised waves consist of transverse waves vibrating in one particular plane.

4343 A 4444 Lenses refract different frequencies of light through slightly different angles. White light would therefore be spread out into its component colours.4545 Polarising filters only allow light with an electric field varying in a particular orientation to pass through. Other orientations are blocked. We say that the orientation of the transverse wave in only one direction is called polarisation.4646 A luminous body emits a stream of particles that travel in straight lines at high speed. These particles are small and do not interact with each other. These particles travel at different speeds in different media.4747 B 4848 A 4949 A 5050 D 5151 C

Chapter 10 Astronomy10.1 Motion in the heavens 1 C 2 D 3 A 4 SE, rising5 27° above horizon in Brisbane, 38° in Melbourne6 a C b D 7 a Pollux b b Fomalhaut 8 a 6 h 23 min, –53° b 16 h 27 min, –26° 9 Stars rotate 361° per day. Orion will be 7° W of N.1010 Aquarius, 7 h before Orion

10.2 The Sun, the Moon and the planets 1 The Sun has reached its highest or lowest noon point.2 Dates are the same but reversed.3 No, the appropriate pole tilts to the Sun. 4 77½° summer, 30½° winter5 Area is inversely proportional to the sine of the angle (75½° and 28½°), and the ratio of areas is 2.03.6 Ecliptic is the actual path of the Sun; Zodiac is the band of stars through which the ecliptic passes.7 If Moon–Earth distance is 30 cm, Earth’s diameter is 1.0 cm, Moon’s diameter is 1.7 mm.8 Full moon rises at sunset, opposite the Sun. New moon rises with the Sun. This is because the full moon phase occurs when the Moon is opposite the Sun (with Earth in between), while the new Moon phase occurs when the Moon is almost in line with the Sun (between Sun and Earth).9 a, b, c midnight1010 Earth would not move in the Moon sky but would rotate and go through phases (29½ Earth days).

10.3 Understanding our world 1 2290 km 2 Venus is inside the Earth’s orbit. 3 C 4 a It won’t; it is in line with the Sun. b opposition; no, only the superior planets

object

screen

(image)

9.0 m

mirror (concave)

v

u

3.0 m 1.0 m

0.75 m

F

object

screen

(image)

12 m

mirror (concave)

v

u

591Solutions

5 235 m.6 6 Brahe found no evidence of parallax motion of the stars. Brahe had the planets circling the Sun.7 a 7 a Mercury, 102.1%; Earth, 100.1%; Mars, 100.5% b 2 mm, 0.1 mm and 0.5 mm8 a 8 a 2.8 years b 4 AU 9 These implied that the Sun and the Moon were not perfect.1010 Relationship between the phases and the apparent size.

10.4 The telescope: from Galileo to Hubble 1 a 1 a B b C 2 2 12.5° 3 Yes. Light gathering power 2.25 times that of the other.4 No hole because light unfocused; image a little duller.5 Telescope has to counter Earth’s rotation. One rotation in 23 h 56 min.6 a The eye is unable to resolve planets. b Pupil is larger, and diffraction pattern smaller.7 θ = 0.003 arcsec, so diameter over 40 m—far larger than the biggest telescopes.8 250 m 9 It can use UV and IR, which are absorbed by the atmosphere, and gather light for longer.1010 In light-gathering power it is like a 16 m telescope. Resolution is like that of a 200 m telescope.

10.5 New ways of seeing 1 1 Very faint, difficult to tell from a star or comet. Sky maps drawn by hand and needed long, careful observation. Outer planets move very slowly through the stars.2 Neptune was predicted from its gravitational effect on Uranus.3 Long exposures enabled very faint objects to be seen, recorded and compared over time.4 electromagnetic radiation outside visible light; UV wavelengths <400 nm, IR wavelengths >700 nm5 They did, but the atmosphere strongly absorbs these wavelengths.6 Maybe, but any heat source would create shimmer in a telescope.7 Radio waves are recorded electronically, day and night.8 Advantages: not looking through the atmosphere, continuous operation. Disadvantages: cost, difficult repairs and adjustments.9 X-rays are absorbed by the atmosphere. Stars emit the X-rays, people don’t.1010 Basically the same, both use a curved mirror to focus. Radio waves are >>1000 times longer. Optical image is

formed in a two-dimensional plane; radio telescope has to scan the object.

Chapter 10 review 1 Many answers!2 Stars appear to rotate on the inside of a huge sphere.3 Diurnal is nightly rotation due to Earth rotation. Annual is due to Earth’s motion around orbit (1° per night).4 52°; higher in Brisbane, lower in Hobart5 due east and west everywhere6 No night. In eclipse all stars rotated around the SCP by 180°.7 Opposite halves of the sky visible.8 8 min; actual difference about 7 min9 a ia i A iiii N iiiiii A iviv P v P vivi P b Orion 12 h, star v 3 h, star vi 21 h10 a 10 a Arcturus b Betelgeuse c Alpha Centauri d Aldebaran 11 a 11 a RA 6 h 43 min, dec. –17° b RA 1 h 36 min, dec. –58° c RA 18 h 35 min, dec. +39° d RA 5 h 12 min, dec. –8°1212 It varies with position in time zone and time of year (±1 h).1313 A 1414 Altitude of Sun is 52° at equinoxes. The 23½° tilt of the Earth is added or subtracted.1515 Concentric, but tilted at 23½°. Tilt rotates around the sphere once every 26 000 years.1616 A 1717 A full moon is on the opposite side of the ecliptic to the Sun and so tilted the other way.1818 a Io b i b i Lunar eclipse is much more likely. ii ii Io appears about 5 times larger than the Sun and so a total solar eclipse is more likely and will last longer.1919 23°, about 1½ hours 2020 5.5° 21 21 30.3 km s–1 2222 The one with the 20× magnification gives the larger image, but the one with the 26 mm objective would have a brighter image.23 a23 a 44 times b 975× 2424 a 8 × 30 b 7 × 50 by 2.8 times2525 They record data using either photographic film or electronic charge-coupled devices (sensors).2626 Both are electromagnetic waves, but radio waves are over 1000 times longer.

Chapter 11 AstrophysicsNote that while many of these questions require detailed answers, there is room here for only very brief answers. More complete answers are provided in the Heinemann Physics 11 Pearson Reader.

Solutions592

11.1 The stars— how far, how bright?1 a 1 a apparent movement of close objects relative to distant ones b yes c No, stars were very far away.2 a 2 a 3600 arcseconds b 1800 arcseconds c 4 km3 Calculated speed is 94 km h–1, so car is not travelling perpendicular to line of sight.4 a It is the distance to a star with parallax angle of 1 arcsecond: d = 1/p. Used because it is simply determined from parallax measurement. b 2.3 pc, 7.4 l.y., 4.7 × 105 AU, 7.0 × 1016 m5 Hipparchus called brightest stars ‘first magnitude’ (+1) and dimmest ‘sixth magnitude’ (+6). Brighter ones were then ‘below’ +1.6 a 6 a 2 W m–² b all c c Mostly IR with a little visible light.7 a 2640 W m–² b 50.7 W m–² 8 Deneb by 84 times9 9 D

11.2 Our favourite star 1 3600 km2 Physics of laws of heat transfer was well known.3 Known processes could only explain a few thousand years of Sun’s life.4 4 Nuclear forces are 104 times greater so 108 times more nuclear energy is produced.5 a 2.3 × 10–19 J, 1 × 10–12 J b about 4 million timesc 20 tonnes6 1.6 × 10–9 kg, 7 g 7 There is a very small theoretical difference.8 Hydrostatic equilibrium—see full answers.9 Conduction then radiation—see full answers.10 a 10 a C b A

11.3 We know stars by their light 1 Radio, IR, UV and X-rays are used. The atmosphere blocks much of it.2 continuous (temperature), emission (composition), absorption (gases in space)3 Temperature gives rate of radiation per m², which is compared to luminosity to give surface area.4 a a D b A5 Determines type of star and hence gives luminosity, which is compared to apparent magnitude.6 Binary star mass found from Newton’s laws. Gives mass–luminosity relationship for all main sequence stars.7 C8 The same mass–luminosity and H-R diagram relations hold for all stars and these depend on the fusion processes.9 Comparing ‘clusters’ and H-R position enables estimates of lifetime.

1010 It is the explosion of a very large star. Requires huge gravitational forces.

11.4 Whole new worlds 1 Hubble found Cepheids in them, and so found absolute magnitude and hence distance.2 approx 15 Milky Ways; 5º, about 10 full moons; very far away, so too faint3 Centre deduced from globular clusters. Dust clouds obscured the centre.4 Distance to dimmer Cepheid is twice distance to other.5 a 5 a 2 × 1012 pc3 b 20 pc3 per star c 2.7 pc d About 1 pc, because stars in outer reaches are much more sparse.6 On approach the frequency is higher, then gets lower—as if engines are slowing.7 30 000 km s–1 away from us 8 2 × 10–18 s–1 9 6300 km s–1, 2% of c1010 Looking far back in time to early universe. Quasars are not seen at closer distances.

11.5 The expanding universe 1 1 Gravity should collapse a non-infinite universe, but it appeared unchanging; Olbers’s paradox2 2 Sky should be bright because there should be stars in all directions. Expanding universe implied light from stars >14 billion light-years away would not reach us.3 3 Steady state (universe infinite, eternal) and big bang (universe expanding). In steady state theory, matter is continuously created.4 4 4 × 1031 molecules; would not detect formation of a few new ones 5 5 0.5 cm s–1 and 1 cm s–1 6 6 No meaningful answer possible!7 7 Radio waves detected were the left-over radiation from the big bang.8 8 Wavelength has increased due to expansion of space.9 9 Irregularities necessary for the formation of galaxies.1010 Implies age is 2 billion years—less than the age of the Earth.

Chapter 11 review 1 It is the apparent movement of stars as the Earth moves around the Sun. Galileo said stars must be very far away.2 2 85 km s–1, about 3 times Earth’s orbital speed3 3 18 km 4 a4 a 32 b 1005 a5 a 0.87 AU b 250 W m–2 c 1.75L6 6 Radiation emitted increases very strongly with T and increases in frequency (into visible).7 7 colour → temperature → radiation m–²; L → total radiation and thus → surface area → radius8 a8 a ~0.8 b ~4500 K, reddish9 9 1093 = 1.3 million

593Solutions

1010 a 5000 years bb 10 billion years 1111 C1212 Convection currents bring the heat from the core. Surface changes over minutes.1313 Spectra show lines similar to spectra of Sun and known elements.1414 temperature, elements and their state, pressure, magnetic field1515 Sun is close to centre of H-R diagram, but most stars are actually cooler.1616 1000 times (approx.)1717 the middle of the main sequence (approx.)1818 See full answer.1919 nebulae remnants and pulsars2020 Period is related to luminosity, so distance can be found. Also very bright.2121 a about 1.9 pc average; central stars are closer, outer ones are further apart b b Sun–stars approx 1 pc; a little closer than average2222 34 m s–1 (or 122 km h–1) toward us2323 3.5 km s–1 (4.4 km s–1 SMC); a negligible speed astronomically.2424 Blueshifts imply motion toward us—only relatively close galaxies. Expansion is on the very large scale.2525 Position on the H-R diagram and nuclear processes; the cosmic microwave background radiation. All estimates around 14 billion years.

Chapter 12 Energy from the nucleus12.1 Splitting the atom—nuclear fission1 1 B 2 2 D 3 a 3 a 3 b b 4 c c 5 4 4 x = 239, y = 40 5 a 5 a In fissile nuclei, the forces that act between the nucleons are not as large. b b The energy of the additional neutron makes the nucleus wobble and the electrostatic forces cause the nucleus to break apart.6 a 6 a 8.3 × 10–16 J b 4.9 × 108 eV c 8.4 × 10–11 J7 a 7 a 3.1 × 10−11 J, 196 MeV b b About 20 times more energy is released in the fission reaction.8 8 2.36 × 10−30 kg 9 9 alpha particle10 a 10 a 1.91 × 10−11 J b b 1.19 × 108 eV c c 3.1 × 1010

12.2 Aspects of fission1 1 B 2 2 It doesn’t have a high enough concentration of the fissile isotope, uranium-235. 3 3 B4 4 238U itself is not very fissile. However, when 238U absorbs a neutron it decays to form 239Pu, which is fissile.5 5 When a slow neutron strikes the nucleus, the nucleus is made to wobble slightly, but not enough to disturb the balance between the repulsive electrostatic and attractive nuclear forces. However, when a fast neutron strikes the nucleus, it has enough energy to make the nucleus wobble and elongate. This upsets the delicate balance

between the strong and electrostatic forces and the electrostatic forces cause the nucleus to separate.6 6 239Pu, a smaller amount of this isotope is capable of exploding.7 7 First, a neutron causes fission to occur in a uranium-235 nucleus, thus releasing 2 or 3 more neutrons. These then go on and induce fission in more uranium-235 nuclei, each resulting in the release of 2 or 3 neutrons and so on. The chain reaction grows very rapidly and energy is released in each fission reaction.8 8 As a result of its shape, a very high proportion of neutrons can escape from the material, and so the chain reaction dies out.9 a fertile b fissile1010 It must be a low percentage of U-235.

12.3 Nuclear fission reactors1 1 B 2 2 D 3 a 3 a The fission process in the reactor core produces heat. This heat energy is conducted into the coolant which is flowing through the core. The energy is used to produce steam which drives a turbine to generate electricity. b b The difference is that the heat energy that makes the steam is produced by burning coal instead of a nuclear fission reaction. c c They both use steam to turn a turbine to generate electricity. 4 4 The nucleus is too heavy. When a neutron collides with a lead nucleus, the neutron will keep almost all of its energy, and so not slow down sufficiently to be captured by a fissile nucleus. 5 a 5 a The chain reaction will be self-sustaining, i.e. critical, and a steady release of energy will result. b b The chain reaction will die out because it is subcritical. This will lead to a decrease in the amount of energy produced. c c The chain reaction will grow, causing an increasing amount of energy to be produced. This may be dangerous and could result in an explosion.6 a 6 a Fast neutrons are most unlikely to be captured by the nuclei. b b Slow neutrons are likely to be absorbed by the nuclei and cause fission.7 a 7 a It results in the uranium-238 transmuting into plutonium-239. b b Plutonium is highly radioactive and has a half-life of about 24 000 years.8 a 8 a plutonium-239 b b It relies on fast, high-energy neutrons to induce fission in plutonium nuclei. c c It produces, or breeds, more of its own fuel, plutonium-239, when neutrons are absorbed by uranium-238 nuclei. d d Fast breeder reactors do not have moderators.9 9 Since only one neutron is required to sustain the chain reaction, the remaining neutrons are able to breed more plutonium.

Solutions594

10 10 Over a period of months, the fissile nuclei in the fuel rods become depleted, the number of fissions decreases, and so fewer neutrons are flying around in the core. In order to maintain the chain reaction, the control rods must be gradually withdrawn.11 a 11 a about 100 000 years b b The fission fragments, which have relatively short half-lives, have decayed.12 a 12 a The uranium stays in the spent fuel rod, which is initially stored in water, then kept underground in storage tanks. b b The fuel rod is initially stored in water, then the uranium is extracted from the fuel rod. It then undergoes enrichment and fabrication before being used as fuel again. c c More energy is extracted from the fuel and a much smaller quantity of high-level waste is produced.

12.4 Nuclear fusion1 1 D 2 2 C 3 water in lakes and oceans 4 4 achieving temperatures high enough for fusion to take place and containing the extremely hot fusion material within the reactor5 5 The tokamak style of reactor has a torus-shaped core where the plasma of the fusion fuel is located. This fusion material is held away from the reactor wall by magnetic fields. The heat energy produced is carried from the reactor core by lithium and water. Inertial confinement is a technique in which a pellet of fusion fuel is struck by high-intensity laser beams, creating temperatures high enough to sustain nuclear fusion. 6 a 6 a No—the hydrogen nuclei would not be travelling fast enough. b b The nuclei would only experience the electrostatic force of repulsion as they approached. c c When the nuclei are widely separated, only the electrostatic force will be acting. Then when the nuclei move close together, the strong nuclear force of attraction will overwhelm the electrostatic force and bring the nucleons together. 7 7 Inside a massive star, the gravitational pressure and so the temperatures will be greater than inside a star the size of the Sun. As a result, fusion will take place at a greater rate, causing these stars to shine more brightly, but also causing them to run out of nuclear fuel more quickly. 8 a 8 a The hydrogen and helium-3 nuclei have more mass than the products of the reaction. b b The total mass of the products is less than that of the original nuclei. The difference in mass is the energy released. c c 3.4 × 10−12 J d d 3.7 × 10−29 kg e e It has decayed into a neutron, a positron and a neutrino. 9 9 The fuel that is used for fusion is readily available and easily extracted from water. The waste products of fusion are not highly radioactive.

10 10 They use magnetic fields to confine the plasma in which the fusion process is taking place to the middle of the doughnut-shaped reactor core—away from the wall of the reactor.

Chapter 12 review1 1 C 2 a 2 a uranium-238 b b uranium-235 c c The concentration of the fissile uranium-235 is too low.3 3 2 4 4 2.7 × 103 tonnes 5 5 B 6 6 They are mostly absorbed by uranium-238 nuclei, absorbed by the control rods, or escape from the reactor core.7 7 The uranium deposits would have spontaneously exploded millions of years ago.8 a 8 a sliced b b two hemispherical pieces c c a 2 kg lump d d The material is carried as small subcritical pieces, then combined at the time of detonation.9 a 9 a The kinetic energy of the fission fragments. b b The heat energy is removed from the reactor core by a coolant and is used to create steam, which is then used to turn a turbine and generate electricity.10 a 10 a 10n + 238

92U → 23992U, 238

92U → 23893Np + –1

0e, 239

93Np → 23994Pu + –1

0eb b It has a relatively short half-life (24 400 years) and so has completely decayed since the formation of the Earth.11 a 11 a There are about the same number of protons and neutrons. b b There are more neutrons than protons. c c The forces are relatively large, so the nuclide is very stable. d d The forces are not as large as those on proton X. This nuclide is less stable.12 a 12 a The oceans are so vast that any radioactivity leaks would be greatly diluted. The radioisotopes also have relatively short half-lives. b b The radioactivity could leak and cause damage to fish and other marine life.13 a 13 a The control rods absorb neutrons and control the rate of the fission process. b b The coolant, after being heated by the fuel rods, is used to produce steam. This is in turn used to drive a turbine around and generate electricity. c c It can reach higher temperatures when under pressure.14 a 14 a The nucleus is not rigid. Rather, it can move and wobble like a drop of water. b b The neutron causes the nucleus to wobble around and neck in the middle. At this point, the repulsive forces overwhelm the attractive forces and the nucleus splits.15 a 15 a 9.991 kg b b 99.91%16 a 16 a slow b b Iron is not fissile. c c the decrease in the binding energy during fission d d They also cause fission and so produce a chain reaction.17 a 17 a The reactants have more mass than the fusion products. b b The difference in potential energy

595Solutions

represented by the missing mass has appeared as the kinetic energy gained by the fusion products. c c As the nuclides approach, an electrostatic force of repulsion is acting on them, but their speed enables them to overcome this force and fuse together. At this point, electrostatic forces of repulsion and larger nuclear forces of attraction are acting on the protons. The neutrons experience only strong nuclear forces of attraction.18 a 18 a 4.49 × 10–11 J b b 280 MeV c c Fission fragments are usually radioactive. 19 19 D20 a 20 a Electrostatic forces of repulsion act on the protons. They do not have enough energy to overcome this force and fuse together and so the strong nuclear force does not come into effect. These protons have not jumped the energy barrier. b b Electrostatic forces of repulsion act on the protons initially, but they have enough energy to push past these forces and get close enough for the strong nuclear forces of attraction to take effect. This force enables the nucleons to fuse together. These protons have overcome the energy barrier.

Chapter 13 Investigation: flight 13.1 The four forces of flight1

Fa and Fd are the air resistance or drag of the jet engine itself and the plane, respectively. Ff is the forward thrust of the jet. The plane will move forward as long as Ff is greater than Fa + Fd .2 The sudden change in relative air speed causes a large and sudden change in the lift generated by air speed past the wing. This can be enough to cause a stall.3 The aerofoil shape is only part of the lift generated by the wing. The transfer of momentum from air incident on the surface of the wing also contributes large amounts of lift. The amount of lift from this source is determined by the angle of attack. A plane flying upside down will maintain an angle of attack sufficient to generate the lift required.4 Lift will be four times that at the original speed.5 small and light, small frontal area, very smooth outer skin, long and narrow wings6 Broad wings provide lift at low speeds for takeoff and, with flaps extended, extra drag for landing. At high speeds swept back wings reduce drag and lift, which is proportional to speed.

7 The surface was pushed off by the increased pressure under the surface of the wing or from a high angle of attack.8 The sheets of paper come together due to the creation of a low pressure region between the sheets as a result of the faster air flow.9 Taking off into the wind increases relative air speed past the wings and increases lift.10 10 The wings are not generating any lift as a result of relative air speed. For the same reason, helicopters use considerably more fuel when hovering.11 11 The air over the top of the coin creates an area of low pressure. A little air under the coin and it will lift through this area of low pressure.12 12 Fnet = 1.4 kN 1313 30 kN 1414 38 kN 1515 8 kN 1616 1.6 × 106 W

13.2 Modelling forces in flight1 The change in relative position of the large mass of fuel affected the aircraft’s angle of attack, changing the amount of lift generated by the wing. Moving the fuel allowed the pilot to restore the angle of attack without the need for large rear elevators.2

3 Other aspects of the aircraft must be designed around the position of the centre of gravity to maintain stability. Other considerations include frontal area and increased drag for the wider wing compared with decreased glide ratio for delta wings.4 612 kg

Fd

Ff

Fa

a

b

centre of gravity centre of gravity

centre of side

centreof side

Solutions596

5 5

6 12 metres forward of the aircraft’s centre of gravity7 937 N8 Altering the angle of attack increases lift from the main wing.

Chapter 14 Investigations: sustainable energy sources14.1 …nergy transformations1 a 1 a electrical to sound b kinetic to gravitational potential c gravitational potential to kinetic (and sound) d light to electrical2 a 2 a high grade, high grade b b high grade, high grade c c high grade, high grade d d low grade (depending on wavelength), high grade3 3 A isA is chemical potential B isB is heat energy C isC is kinetic D isD is electrical4 4 250%5 5 Additional heat energy is gathered from the surrounding environment. (Note: While the heat pump may still be powered by fossil fuels, the efficiency of heat pumps makes them a valuable contributor to energy conservation.)6 6 39%7 7 Most of the remaining energy is transformed into heat energy and lost to the surrounding environment through the cooling system.8 8 360 J9 9 It is mainly transformed into heat.10 10 It is harder to harness into effective work.11 11 The energy is transferred to the ground and transformed into heat energy. New energy-efficient cars recycle this energy to drive generators that are charging batteries, or to power devices such as superchargers.12 12 2.52 kJ13 13 It has been mainly transformed into heat; the material the kettle is made of will get hotter.

Chapter 15 Medical physics15.1 Ultrasound and how it is made1 The period, T, is the time interval for one vibration or cycle to be completed. The frequency of a wave, f, is the number of vibrations or cycles that are completed per second (Hz). The wavelength is the distance between successive points having the same displacement and moving in the same direction; that is, points that are in phase. The amplitude, A, is the value of the maximum displacement of a particle from its mean position. 2 a i2 a i vocal cords and air in the larynx vibrating iiii small speaker disc vibrating iiiiii the sides of the cicada’s thorax (body) vibrating b All sounds must have a vibrating item at their source.3 Diagnostic ultrasound involves images taken of the inside of the body, and therapeutic ultrasound involves treatments for specific illnesses such as kidney stones.4 a4 a 6.16 × 10−4 m b 6.24 × 10−4 m c 5.88 × 10−4 m d 6.28 × 10−4 m 5 2.9 × 10–7 s 6 a 5.00 MHz b 5.00 MHz7 1.70 × 10−2 m8 The piezoelectric crystal layer (the disc) is alternately compressed and stretched slightly by the arriving sound wave. The compression results in a net negative charge occurring close to one electrode, leaving the other plate with a slight positive charge. The stretching of the crystal layer results in a potential difference across the plates of the opposite polarity. An alternating electrical signal is produced.9 Rather than a whole item vibrating while maintaining a constant thickness, the piezoelectric crystal expands and contracts to produce sound waves. 1010 1.7 mm

15.2 Ultrasound interactions1 6.400 × 106 kg m−2 s−1 2 2 0.005 J3 Very high frequencies are more readily absorbed by the body and therefore cannot penetrate deeply enough and reflect back to the transducer. 4 Penetration is ~15 cm so it is suitable.5 There is too much reflection from bone (ribs) and therefore insufficient energy reaches the tissue behind.6 a6 a 1.01 × 103 kg m−3 b 9.39 × 102 kg m−3 c 1.07 × 103 kg m−3

7 The gas–soft tissue boundary has a large difference in acoustic impedance values; therefore, strong reflection occurs. 8 From Table 15.4, only 1% would penetrate.9 a9 a 0.99 or 99% b 0.00744 or 0.744% c 3.17 × 10−4 or 0.0317%

aileron down—liftincreased

aileron up—liftreduced

weight

upwards force

sideways force

downwards force

597Solutions

15.3 Scanning techniques1 The gallstone is so dense that ultrasound waves reflect and there is little penetration beyond it.2 The pulses may be reflected from different depths. Therefore, more absorption will occur for the ones that have travelled through more tissue.3 3 Remove skin defects, improve blood supply, reshape corneas.4 Smaller surgical site therefore faster recovery.5 The Doppler effect is the change in the wavelength of a periodic wave that is reflected from a moving object. The presence of plaque in the artery would be indicated by irregular blood-flow speeds in the region.6 The A-scan, since only the timing of the emitted and reflected pulse is utilised.7 No. This is a very high frequency and would not penetrate the 10 cm or so that is required to image the heart.8 Ultrasound cannot image items that lie behind or inside bone or air sacs within the body, due to the large reflections at these items. Layers of soft tissue that are very thin may not be detected as separate layers. 9 Many tiny transducers are placed in the one probe; they are triggered one after another to produce a sequence of images.1010 Equipment that produces two-dimensional images is far more common and cheaper for hospitals to purchase. The images produced are more than adequate for diagnosis. Three-dimensional images don’t necessarily greatly improve diagnostic capabilities and the machines are very expensive.

15.4 Diagnostic X-rays1 a1 a Target: electron strike results in electron deceleration and hence X-ray production. b Filament: source of released electrons. c Electric field: accelerates electrons to high speeds towards the target. d Lead cover: absorbs stray X-rays to protect workers. e Aluminium filter: filters/removes higher energy X-rays from the emergent beam. 2 a2 a 3.0 × 1018 Hz b 3.0 × 1017 Hz 3 3.3 × 10−15 J4 to accelerate electrons to a high enough speed so that their sudden deceleration is large enough to produce an X-ray5 2.4 × 1019 Hz 6 a6 a 100 000 eV b 1.6 × 10−14 J7 a7 a yes b no c yes d no8 Large atomic number (i.e. heavy atom) and a very high melting point.9 a9 a The average and the maximum X-ray energy would be reduced. b The line spectra will have different frequency (energy) values.

1010 No, each electron is given 10 000 eV. X-ray photon energies lie between 30 and 50 keV.

15.5 Radiotherapy, radioisotopes in medicine and P…T1 a1 a technetium-99m b as the mother molybdenum 99

42Mo2 99

42Mo → 9943Tc + –1

0β3 It must have a short half-life (within hours), it must not emit alpha or beta radiation, it must emit gamma radiation and be available in the highest possible activity.4 Ionising radiation can produce ionised water molecules in the cells and highly reactive hydroxide ions that can damage parts of the cell including DNA. 5 Gamma radiation will leave the body and be detected by camera; alpha and beta cannot leave the body but will reach nearby cells and destroy them.6 Cells in reproductive organs, the eye and the bowel need protecting, because they reproduce relatively quickly.7 to give healthy cells a chance to recover between treatments8 to locate the exact position of a tumour and monitor health of nearby organs9 a radioactive isotope attached to a carrier substance that will aggregate in a particular part of the body1010 The effect that ionising radiation has on DNA is that it can cause the mutation of the cell’s DNA as it reproduces; the mutant cell can be cancerous. Enough ionising radiation can kill such a cell or at least damage it so that it can’t reproduce.

Chapter 15 review 1 If the gel is not used 99% of the sound wave would be reflected at the air–skin boundary.2 They should be sources of alpha and beta radiation of sufficient energy to penetrate the diseased section of the body. It is convenient if the isotope is also a weak source of gamma radiation so that its presence in the body can be monitored by gamma cameras.3 a3 a 1.47 × 10−3 m b 1.57 × 10−3 m c 3.70 × 10−3 m4 4 Any device that converts energy from one form into another. An ultrasound transducer can convert electrical energy into a sound wave and vice versa.5 These cells are reproducing at a faster rate than normal cells.6 When certain crystals are placed under stress (stretched or compressed) they acquire equal and opposite charges on opposite faces. Hence an electrical potential difference develops between the surfaces.7 C 8 A 9 C10 a10 a 2.6 × 10−15 J b 1.7 × 104 eV

Solutions598

11 a11 a the level of efficiency with which sound waves are able to pass through a given mediumb 1.63 × 106 kg m−2 s−1

1212 A short half-life means that a patient’s exposures are limited and they are not emitting radiation when they leave hospital.1313 the collision of the electron with the heavy positive target1414 gamma (food irradiation), X-rays (medical imaging), ultraviolet (tanning booths), visible (sight), infrared (radiator-style heaters), microwaves (communication), radio waves (remote control garage doors)1515 3.0 × 10−13 to 3.0 × 10−8 m1616 electrons moving between shells in the atoms of the target material1717 soft (low energy), since they do not need to be deeply penetrating18 a18 a HVT is the thickness of a material that is required to reduce the intensity of an X-ray beam to half of its original value. b Heterogeneous beams include X-rays of a range of energy values. On their way through the body the lower energy X-rays are absorbed into the tissue, the beam is hardened and therefore it becomes more penetrating—hence the HVT increases with depth.1919 Benign tumours are self-contained and non-spreading. Malignant tumours are invasive and often cells can be carried through the body.2020 absorption and scattering2121 0.07 J 2222 true2323 The Australian Nuclear Science and Technology Organisation, ANSTO2424 D 2525 A 2626 D 2727 B 2828 D 2929 C, D 3030 B