Part A - John Wileycatalogimages.johnwiley.com.au/Attachment/07314/0731408195/JP1_3E... · Jacaranda Physics 1, 3rd edition, ... Part A — WORKED SOLUTIONS Chapter 1 Radiation from

Part AWorked solutions

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Jacaranda Physics 1, 3rd edition, Teacher Support Kit 2 © John Wiley & Sons Australia, Ltd 2009

Unit

1 ■■■■■■■■■

Area of study 1: Nuclear physics and radioactivity

Jacaranda Physics 1, 3rd edition, Teacher Support Kit Part A — WORKED SOLUTIONS

Chapter 1

Radiation from the nucleus 1. (a) 30 protons, 36 neutrons

(b) 90, 140 (c) 20, 25 (d) 14, 17

2. (a) 42 He

(b) 137 N

(c) 23491Pa

3. (a) Gold: 79 protons, 118 neutrons (b) Bismuth: 83 protons, 127 neutrons (c) Lead: 82 protons, 128 neutrons

4. The number of nucleons (protons + neutrons) does not determine the element a particular atom may be. It is the number of protons that determines an element.

5. (a) A radioisotope is an unstable isotope of an atom that can become more stable by emitting radiation from its nucleus.

6. α and β particles and γ-rays are emitted from the nucleus. 7. (a) γ, β−, α (b) α, β−, γ 8. (a) β−: a neutron has transformed into a proton and a β−

is released. (b) β−: a neutron has transformed into a proton and a β−

is released. (c) α: the released particle has two protons and two

neutrons. 9. α decay

(a)

( )

226 4 22288 2 86Ra α Rn energy

radon→ + +

(b)

( )

214 4 21084 2 82Po α Pb energy

lead→ + +

(c)

( )

241 4 23795 2 93Am α Np energy

neptunium→ + +

10. β− decay (a)

( )

060 6027 281Co e N energy

nickel−→ + +

(b)

( )

090 9038 391Sr e Y energy

yttrium−→ + +

(c)

( )

032 3215 161P e S energy

sulfur−→ + +

11. 24 2412 12Mg* γ Mg→ +

12. (a) 4Z 4A 2 A 2X α D energyz −

−→ + +

(b) 0 ZZA 1 A+1X e E energy−→ + +

(c) Z ZA AX* X λ→ +

(d) 27 2 28 113 1 14 0AI H Si n energy+ → + +

(e) 22 4 25 111 2 12 1Na He Mg H energy+ → + +

13. 1 eV = 1.6 × 10−19 J 6

6 19

13

1.71 MeV = 1.71 × 10 eV

= 1.7 × 10 eV × 1.6 × 10 J/eV

= 2.74 × 10 J

−

−

14. 090 9039 401Y e Zr.−→ + See the following figure.

15. β− decay occurs when a neutron in the nucleus is trans-

formed into a proton. In order for the neutral neutron to gain a positive charge and become a proton it must also release a particle with a negative charge equal in size to that of the proton. The particle released is an electron.

16. Safety precautions needed for isotopes producing α radiation: • Store in sealed cardboard boxes. • Handle with gloves. Safety precautions needed for isotopes producing β radiation: • Store in aluminium containers with walls several

millimetres thick. • Shield against, with aluminium of at least several

millimetres thick. Safety precautions needed for isotopes producing γ radiation: • Store in thick (several centimetres) lead containers. • Shield against, using lead which is several centimetres

thick. 17. It would be more dangerous to stand next to a source of

high activity (decays per second) and short half-life since exposure would be significant in a short time frame.

18. 238 20692 82U Pb + ?→

The release of β particles does not result in a change in mass number. As mass number changes from 238 to 206 ⇒ α decay is involved. 238 − 206 = 32, therefore a change of 32 nucleons. Each α particle released reduces mass number by 4, therefore 8 α particles released.


Each α particle released reduces atomic number by 2, therefore 8 α particles reduces atomic number by 16, ⇒ new atomic number of 76. This is 6 too few for 82Pb, therefore 6 extra protons result from β decay.

Therefore, to change from 23892 U to 206

82 Pb , the atom

releases 8 α particles and 6 β particles. 19. (a) 7 α decays and 4 β decays, (b) 6 α decays and 4 β

decays 20. No, unless the alpha source is right up against the badge.

Although there may be some exposure from Radon gas which is more likely in a mine.

21. Astatine 218 22. No, the total changes in atomic number and mass number

are identical. 23. Approximately 2 hours: the time at which half of the

protons remain. This can be read directly from the graph. 24.

25. 1

2= 5730 yearst

1 1 12 2 2

1 1 12 4 81 g g g g

2 3t t t

→ → →

⇒ 3 half-lives to get from 1 g to 18 g

∴ 3 × 5730 years = 17 190 years. 26. 1

2120 s 2 mint = =

Method 1: 2 pm → 1 μg = 1 × 20 μg 1:58 → 2 μg = 1 × 21 μg 1:56 → 4 μg = 1 × 22 μg 1:54 → 8 μg = 1 × 23 μg 1:52 → 16 μg = 1 × 24 μg x half-lives before 2 pm, amount is 1 × 2x μg ⇒ at 1:30 pm, 15 half-lives before 2 pm, amount is 1 × 215 μg = 33 mg Method 2:

A0 = ? 12

F 0 2

tt

A A

−

=

AF = 1 μg = 1 × 10−6 g 3026

01 10 2A−−× =

t = 30 min 2−15 × 1 × 10−6 = A0

12

2 mint = A0 = 33 mg

27. (a) 27.0, 13.5, 6.75, 3.375, 1.69 kBq (b) The graph should have a similar shape to the one in

question 24.

(c) (i) 23 kBq (ii) 25 kBq (d) The activity will decrease over time and eventually

the alarm will be on permanently. (e) The alarm could be redesigned to reset to the new

current level every time the battery is changed. (f) ‘Disposal: Landfill’ Being an α emitter, the smoke

detector is quite safe. Alternatively the Americium could be recycled and reprocessed.

(g) Am 241 → Np 237 → Pa 233 → U 233 → Th 229 → Ra 225 → Ac 225 → Fr 221 → At 217 → Bi

213 → Tl 209 → Pb 209 → Bi 209 → Tl 205. Note: this is the fourth now extinct decay chain mentioned in the text.

28. 3 half lives in 6.0 hours give a half life of 2.0 hours. 29. 4 half lives means the activity is reduced by a factor of

(½)4 = 1/16 to 16 GBq. 3 half lives in 6.0 hours give a half life of 2.0 hours.

30. Free radicals and ions are very chemically reactive and may result in new chemical reactions taking place. For example, the production of H+ and OH− ions may react with molecules either causing damage to DNA (genetic mutations may be passed on) or to mechanisms for controlling cell division (forming cancer), or interfering with the production of molecules necessary for the life of a cell (this may cause the cell to die).

31. Energy absorbed = absorbed dose × mass = 3 × 10−3 × 30 = 9 × 10−2 J

32. Absorbed dose energy absorbedmass

=

2

3

9 1060

1.5 10 Gy1.5 mGy

−

−

×=

= ×=

33. If γ radiation ⇒ quality factor = 1 dose equivalent = absorbed dose × quality factor

= 3 × 10−3 × 1 = 3 × 10−3 Sv 34. Quality factor = 20

dose equivalent = absorbed dose × quality factor = 1.5 × 10−3 × 20 = 30 × 10−3 Sv = 30 mSv 35. Alpha (α) particles cause a lot of localised damage. They

give much of their energy quickly to nearby atoms (i.e. they are stopped over short distances).

36. Using table 4.3, p. 89: >40 Sv will cause death within 48 h. As γ radiation ⇒ quality factor = 1 ⇒ dose eq. = abs. dose energy absorbed = absorbed dose × mass

= 40 × 80 = 3200 J 37. Dose equivalent takes into account the different types of

damage caused by different forms of ionising radiation. 38. In its early stages, the foetus consists only of a few cells

which divide to form the rest of the baby. If these cells are damaged, the foetus may not be able to form properly.


39. Suggested lifestyle changes might include not flying in planes, avoiding certain medical procedures (e.g. X-rays), living closer to sea level, not smoking and staying away from smokers.

40. Assuming a body mass of 60 kg, gives 60 × 2.0 × 10−3 Joules. 1.0 MeV = 106 × 1.6 × 10−19 Joules. So the number of β particles per year is (60 × 2.0 × 10−3)/ (106 × 1.6 × 10−19) = 7.5 × 1011 of β particles per year, which is 24 000 of β particles per second.

41. Cancer cells are more rapidly growing than normal tissue. Cells are more vulnerable to radiation damage when they are dividing. This means that more cancer cells are killed than normal cells.

42. Yes, α particles will be absorbed by the air in the detector and will not enter the atmosphere in the room.

43. γ radiation is more penetrating, so the sterilisation will be more uniform across a sample.


Unit

1 ■■■■■■■■■

Area of study 2: Electricity


Chapter 2

Current electricity 1. After leaving the negative terminal, the electron is able to

move freely through the conductor. When it reaches the globe, the type of material through which it is travelling (most likely tungsten) changes. The path becomes narrower and the electron makes repeated collisions with atoms in the filament, converting electrical potential energy into internal energy, increasing the temperature of the filament. After leaving the filament, the electron travels freely back to the cell’s positive terminal.

In the cell, the electron is absorbed into the chemicals in the cell while other electrons are forced out of the chemicals and into the circuit.

2. Conventional current is in the direction that positive charges would flow if they were free to do so. Electron current is in the direction that electrons would flow, i.e. from the negative terminal of a cell.

3. Direct current: the charge carriers move in only one direction.

Alternating current: the charge carriers periodically change direction (backwards and forwards).

4. Dry cells, photovoltaic cells, capacitors, power packs, thermocouples.

5. Q = It = 2.5 × 15 = 37.5 C

6. 4545 mA10000.045 A

=

=

7. 2.3 × 10−4 A = 2.3 × 10−4 × 103

= 2.3 × 10−1

= 0.23 mA 8. 450 μA = 450 × 10−6

= 4.5 × 10−4 A 9. No. Electrons cannot be destroyed. They transform energy

as they pass through the filament of the globe. Current electricity is the movement of charged particles along a path. Current is not used up in a light globe. The current flowing into a globe is equal to the current flowing out of it.

10. (a) 160 mA (b) 16 mA (c) 1.6 A 11. Q = It I = 3.5 A t = 20 × 60 = 1200 s Q = 3.5 × 1200 = 4.2 × 103 C

12.

15500.30 A

=

=

=

QIt

13. Halve the current ⇒ halve the velocity. Therefore, 8.0 × 10−5 m s−1.

14. energy

1.05 J0.70 C 1.5 V

EQ

=

=

=

15. Potential difference Energy Charge 3.3 V 32 J 9.6 C 6.0 V 4 J 670 mC 9.0 V 31.5 J 3.5 C 12 V 1.02 J 85 mC 4.5 V 12 J 2.7 C 240 V 7.5 kJ 31.25 C

16. (a) 2.3 V

(b) 11.5 V

(c) 23 V 17. Energy = QV Energy = 5.7 × 10−6 × 6.0 Energy = 3.42 × 10−5 J 18. Energy = QV

EnergyQV

⇒ =

43.6 10

6.0

−×=

= 6.0 × 10−5 C = 60 μC

19. (a) PIV

=

60240

=

= 0.25 A

(b) PIV

=

4012

=

= 3.3 A

(c) PIV

=

6.36.0

=

= 1.05 A

(d) PIV

=

1200240

=

= 5.0 A


20. (a) Each coulomb of charge passing through the heater will transform 240 J of energy.

(b) Energy = QV Energy = 25 × 240 Energy = 6.0 × 103 J 21. Energy = VIt Energy = 12 × 2.0 × 30 × 60 Energy = 43 200 J Energy = 4.3 × 104 J 22. P = VI P = 12 × 2 P = 24 W 23. Energy = Pt

EnergytP

⇒ =

45.4 10600

90 s

−×=

=

24. P = VI P = 240 × 10 P = 2400 P = 2.4 kW 25. V = 240 V I = 3.3 A Energy = 3.2 × 104 W t = ? Energy = VIt

Energy⇒ =t

VI

43.2 10240 3.340.4 s

×=×

=

26. Energy

9.0J6.0C1.5V

=

=

=

EQ

27. Energy = VIt = 6.0 V × 3.0 A × 60 s = 1080 J or 1.08 × 103 J 28. Potential difference Current Resistance

32 V 8.0 A 4.0 Ω 48.4 V 22 mA 2.2 kΩ 12 V 2.0 A 6.0 Ω 240 V 3.0 × 10−3 A 8.0 × 104 Ω 9.0 V 6.0 A 1.5 Ω 1.5 V 45 mA 33.3 Ω

29. (a) orange = 3 white = 9 black = 0 gold = ±5% R = 39 Ω ± 5% = 39 ± 2 Ω

(b) green = 5 blue = 6 orange = 3 silver = ±10% R = 56 000 Ω ± 10% = 56 000 ± 5600 Ω (c) violet = 7 green = 5 yellow = 4 gold = ±5% R = 750 000 Ω ± 5% = 750 ± 37.5 kΩ

30. (a) 2

3 2

6 2

43.142 (1.63 10 m)

42.1 10 m

−

−

=

× ×=

= ×

πdA

(b) ρ

ρ

=

⇒ =

LRA

RAL

(c) 6 2

7

5.00 2.1 10 m41.5 m

2.5 10 m

ρ−

−

Ω × ×=

= × Ω

(d) R ∝ L double L ⇒ double R Therefore, 10.0 Ω.

(e) 2

2

14

1

∝ =

⇒ ∝

π dR AA

Rd

If d is halved ⇒ R is multiplied by 4. Therefore, 20.0 Ω. 31. (a) Non-ohmic. The graph of I vs V is non-linear. (b) 0

(c)

0.50(infinity)

VRI

=

=

= ∞

(d) 20 mAI = 0.65VV⇒ =

0.650.02032.5

VRI

=

=

= Ω

32. P = VI or P = I2R = (0.30 A)2 × 5.0 Ω = 0.45 W


33. 2

2

=

⇒ =

VPR

VRP

(a) 2(240)

60960

=

= Ω

R

(b) 26.0

6.35.7

=

= Ω

R

(c) 212

403.6

=

= Ω

R

34. 2

24048

1200 W

=

=

=

VPR

Chapter 3

Circuit analysis 1. (a) Junction: a connection between two or more con-

ducting paths. (b) Open circuit: a circuit in which the conducting path

is broken so that no current flows. (c) Voltage drop: the amount of electrical potential

energy converted in a load for every coulomb of charge passing through it.

(d) Earth: a connection of part of the circuit to the earth, making that part of the circuit be at zero volt.

(e) Series: circuit elements connected one after the other. (f) Parallel: circuit elements connected side by side. (g) Short circuit: the bypassing of a circuit element by

the connection of a conductor in parallel with the element.

2. (a) Into = 3.0 A Away = 1.5 A + Ia ⇒ Ia = 1.5 A away from junction. (b) Into = 2.5 A + Ib Away = 7.3 A ⇒ Ib = 4.8 A into the junction. (c) Into = 1.3 A + 4.2 A + Ic Away = 2.9 A + 3.7 A ⇒ Ic = 6.6 A + 5.5 A = 1.1 A into the junction. 3. Va = 9.0 V Vb = 9.0 V Vc = Vb − Vbc = 9.0 − 5.0 V = 4.0 V Vd = 0.0 V (earthed)

4. (a) Reff = R1 + R2 = 2.7 + 9.8 = 12.5 Ω (b) Reff = R1 + R2 + R3 = 12 + 20 + 30 = 62 Ω (c) Reff = R1 + R2 + R3 = 1.2 + 3.2 + 11 = 15.4 kΩ 5. Ia = 1.0 A (series circuit) V1 = IR1 = 1.0 × 4.0 = 4.0 V V2 = 2.0 V (E = V1 + V2)

22

2.01.02.0

VRI

=

=

= Ω

6. (a) In a series circuit, Ia = Ib Ia = 2.0 A ⇒ Ib = 2.0 A (b) Vbc = IR2 Vbc = 2.0 × 30 Vbc = 60 V

(c) ab1

202.010

=

=

= Ω

VRI

(d) Reff = R1 + R2 Reff = 10 Ω + 30 Ω Reff = 40 Ω (e) E = Vab + Vbc E = 20 V + 60 V E = 80 V 7. (a) Reff = R1 + R2 Reff = 20 + 30 Reff = 50 Ω

(b) eff

10050

2.0 A

=

=

=

EIR

(c) 20 Ω ⇒ V = IR = 2.0 × 20 = 40 V 30 Ω ⇒ V = IR = 2.0 × 30 = 60 V


(d)

8. (a) Reff = R1 + R2 + R3 Reff = 3.0 + 5.0 + 4.0 Reff = 12.0 Ω

(b) eff

1212.01.0 A

=

=

=

EIR

(c) 3.0 Ω ⇒ V = IR = 1.0 × 3.0 = 3.0 V 5.0 Ω ⇒ V = IR = 1.0 × 5.0 = 5.0 V 4.0 Ω ⇒ V = IR = 1.0 × 4.0 = 4.0 V (d) VT = V1 + V2 + V3 = 3.0 V + 5.0 V + 4.0 V = 12.0 V 9. (a) Reff = 1000 + 1500 = 2500 Ω

eff

2525000.010 A

=

=

=

EIR

(b) First person V1 = IR1 = 0.010 × 1000 = 10 V Second person V2 = IR2 = 0.010 × 1500 = 15 V

10. (a) eff 1 2

1 1 1

1 130 202 3

60 60

R R R= +

= +

= +

560

=

eff60 125

⇒ = = ΩR

(b) eff 1 2 3

1 1 1 1

1 1 15 10 306 3 1

30 30 301030

= + +

= + +

= + +

=

R R R R

eff30 3.010

⇒ = = ΩR

(c) eff 1 2 3

1 1 1 1

1 1 115 60 604 1 1

60 60 606

60

= + +

= + +

= + +

=

R R R R

eff60 106

⇒ = = ΩR

(d) eff 1 2 3

1 1 1 1

1 1 120 50 8020 8 5400 400 40033400

= + +

= + +

= + +

=

R R R R

eff40033

12.1

R⇒ =

= Ω

11. Ia = 1.5 A V1 = 6.0 V V2 = 6.0 V

11

1

6.0 V6.01.0 A

VIR

=

=Ω

=

2 a 1

22

2

1.5 A 1.0 A0.5 A

6.00.5

I I I

VRI

= −= −=

=

=

= 12 Ω 12. (a) Vab = I1R1 = 0.20 A × 60 Ω = 12 V


(b) Vcd = 12 V (c) E = 12 V (d) I2 = IT − I1 I2 = 0.50 A − 0.20 A I2 = 0.30 A

(e) cd2

2

12V0.30 A40

=

=

= Ω

VRI

13. (a) eff 1 2

1 1 1

1 110 102

10

R R R= +

= +

=

⇒ Reff = 5.0 Ω

(b) effeff

155.03.0 A

=

=

=

EIR

(c) 11

15101.5 A

=

=

=

EIR

Similarly I2 = 1.5 A

14. (a) eff

1 1 1 160 30 201 2 360 60 606

60

= + +

= + +

=

R

eff606

10

R⇒ =

= Ω

(b) eff

90109.0 A

=

=

=

EIR

(c) 60 Ω ⇒ = VIR

90 V60

=Ω

= 1.5 A

30 Ω ⇒ = VIR

90 V303.0 A

=Ω

=

20 Ω ⇒ = VIR

90 V204.5 A

=Ω

=

15. (a) 1 11

6.0= Ω ⇒ = VR IR

36 V6.06.0 A

=Ω

=

2 22

18= Ω ⇒ = VR IR

36 V182.0 A

=Ω

=

3 33

9.0= Ω⇒ = VR IR

36 V9.04.0 A

=Ω

=

(b) T 1 2 3

6.0 A 2.0 A 4.0 A12 A

I I I I= + += + +=

(c) eff

36 V12 A3.0

ERI

=

=

= Ω

16. (a) G1, G2, G3 (b) G1, G3 (c) G1 17.

eff 1 2

1 1 1

1 10

= +

= +

= ∞

R R R

R

eff1 0⇒ = =∞

R


18. (a) 6 mA (b) 140 V

(c)

3140 V

16 10 A8.75 k

VRI

−

=

=×

= Ω

19. (a) V = IR = 6.0 × 10−3 A × 5.0 × 103 Ω = 30 V (b) 6.0 mA (c) 100 V (from graph) (d) VT = V1 + V2

= 100 V + 30 V = 130 V 20. (a) V = IR = 20 × 10−3 A × 5.0 × 103 Ω = 100 V (b) 100 V (c) 6 mA (d) IT = 20 mA + 6 mA = 26 mA 21. (a) The emf of a power supply is a measure of the

amount of electrical potential energy provided to the circuit for every coulomb of charge passing through the supply.

A voltage drop is a measure of the amount of energy converted in a load for every coulomb of charge passing through it.

(b) The emf of a battery is greater than its terminal voltage when there is a current flowing through the battery and it has an internal resistance.

22. (a) Internal resistance in a cell is caused by: • a build-up of the products of the chemical reaction

that produces electricity • the conductivity of the electrolyte in the cell • the temperature of the cell.

(b) As a cell gets older, its internal resistance rises. 23. E = Vext − Ir

I = 0 ⇒ E = Vext

⇒ E = 9.0 V Vext = 8.0 V when I = 1.0 A E = Vext − Ir

⇒ 9.0 V = 8.0 V − 1.0 × r ⇒ r = 1.0 Ω

24.

(a) eff

eff

4.5

1.54.50.33 A

= Ω

=

=

=

REI

R

(b) Vext = IR = 0.33 × 4.0 = 1.33 V 25. (a) Eeff = 3.0 V reff = r1 + r2 = 0.5 + 0.5 = 1.0 Ω (b)

eff 1 2

1.5 V1 1 1

1 10.5 0.5

E

r r r

=

= +

= +

eff 0.25r⇒ = Ω 26. (a) eff 1 2 3

eff

25 15 1050

10500.20 A

R R R R

EIR

= + += + += Ω

=

=

=

(b) 250.2 255.0 V

150.2 153.0 V

100.2 102.0 V

V IR

V IR

V IR

Ω ⇒ == ×=

Ω ⇒ == ×=

Ω ⇒ == ×=

(c) 255.0 0.21.0 W

153.0 0.20.6 W

102.0 0.20.4 W

P VI

P VI

P VI

Ω ⇒ == ×=

Ω ⇒ == ×=

Ω ⇒ == ×=

(d) PT = P1 + P2 + P3 = 1.0 + 0.6 + 0.4 = 2.0 W


27. (a) eff

1 1 1 125 15 106 10 15

150 150 15031

150

= + +

= + +

=

R

eff15031

R⇒ =

= 4.8 Ω

eff

10 311 150

2.1 A

=

= ×

=

EIR

(b) V across each resistor = 10 V

(c) 25 Ω ⇒ =2VP

R

2(10 V)25

4.0 W

=Ω

=

15 Ω ⇒ =2VP

R

2(10 V)15

6.7 W

=Ω

=

10 Ω ⇒ =2VP

R

2(10 V)10

10 W

=Ω

=

(d) PT = P1 + P2 + P3 = 4.0 + 6.7 + 10 = 21.7 W 28. The voltage provided by the supply is constant. The

power provided by the supply is equal to the current through the supply multiplied by the voltage provided. So the larger the current, the larger the power provided. Henrietta is correct since a smaller resistance gives a larger current.

29.

The circuit consumes/dissipates more power when the

light is bright since the current flowing through the globe is greater and for the circuit P = EI.

30.

A current will flow through the globe only when both

switches are up or down.

Chapter 4

Using electrical energy 1. • It is an alternating current. • The voltage varies between +340 V and −340 V. • The voltage oscillates 50 times per second. • The voltage provides heating effects equivalent to a DC

voltage of 240 V. 2. An overload is when too many appliances are connected

to a circuit. The overall effect is to draw too much current. This can cause wires to melt or materials surrounding the wires to catch fire.

3. Active ⎯ brown; neutral ⎯ blue; earth ⎯ green-and- yellow striped

4. Active ⎯ red; neutral ⎯ black; earth ⎯ green. Colourblind electricians can’t distinguish between red and green.

5.

6. The neutral wire is used to provide an insulated path for

the electricity to return to the generator. The earth wire connects the metal case of the appliance to earth so that if a live active wire accidentally contacts the outer case, a low resistance path for the electricity to return to earth is provided.

7. The earth wire is used in household lighting circuits when the light fittings have a metal case. Plastic and glass fittings do not need to be earthed.

8. Energy in kW-h = power rating (in kW) × time (h) time = 365 × 12 = 4380 h power rating = 0.300 kW Energy = 0.300 kW × 4380 h = 1314 kW-h Cost = energy × rate = 1314 × 11.87 = 15 597.18 cents = $155.97 = $156 (to three significant figures)


9. V = 240 V R = 24 Ω

(a) 2

2

Power

24024

2400 W2.4 kW

=

=

==

VR

(b)

24024

10 A

=

=

=

VIR

(c) Energy = P (kW) × t (h) = 2.4 × 5 = 12 kW-h Cost = energy × rate = 12 kW-h × $0.12 = $1.44 10. (a) Product Energy kW-h Computer 127 Computer monitor 28.9 Cordless phone equipment 32.4 Modem 29.8 DVD player 21.0 Television 54.3 VCR 51.7 (b) 498 kg CO2 11. An electric shock is a violent disturbance of the nervous

system caused by an electrical discharge or current through the body. Electrocution is death resulting from an electric shock.

12. Breaks or cuts in the skin; water 13. Nerve impulses are electrical in nature. The size of the

current will influence the size and type of muscle contraction.

14. Fibrillation is the disorganised rapid contraction of separate parts of the heart so that it pumps no blood.

15. Your muscles would contract and you could grip onto the victim and not be able to let go.

16. The longer the time of exposure, the more severe the shock.

17. Double insulation is a way to protect the user of hand-held appliances. There are two separate layers of insulation between the functional parts of the appliance and the user.

18. V = IR

LRA

ρ=

ρ⇒ = I LV

A

I = 500 A A = 2.0 × 10−4 m2 L = 2.0 × 10−8 m ρ = 2.65 × 10−8 Ω m

8 2

4

3

500 2.65 10 2.0 102.0 10

1.3 10 V

V− −

−

−

× × × ×⇒ =

×

= ×

19. Voltage doesn’t flow, current does. The paper should have reported that there was a voltage drop across his body of 50 000 V.

It is the size of the current that determines the amount of injury. It is not the voltage that kills.

20. Lead−acid batteries can be recharged. They can give a high current with little drop in output voltage.

21. The loom contains wires which form a part of every circuit in an automobile. The loom is connected to the positive terminal of the battery. The other side of the load is connected to the car body which is connected to the negative terminal of the battery.

22. (a) Vext = E − Ir = 12 − 50 × 0.04 = 10 V (b) P = EI = 12 × 50 = 600 W (c) P = I2r = 502 × 0.04 = 100 W

(d) capacitycurrent

80501.6 h (1 hour 36 minutes)

t =

=

=

23. (a) capacityTimecurrent

1204000.30 h18 min

=

=

==

(b) Q = It t = 18 × 10 = 1080 s = 400 × 1080 = 432 000 = 4.32 × 105 C (c) The chemical reaction that produces the emf

produces products that increase the internal resistance of the battery. Also, the chemicals involved in the reaction are used up.

24. (a) =P VI

⇒ = PIV

360 W12 V

30 A

=

=

(b)

6

Energy 4.0 3600 s 14 400 s360 14 400

5.2 10 J

P t t= × = × == ×

= ×


(c) Current capacity current time30 A 4.0 h120 A-h

= ×= ×=

25. 12 V0.2

Er

== Ω

(a) ext

12 25 0.212 5.07.0 V

V E Ir= −= − ×= −=

(b) 2

2

14440

3.6

VPRVRP

=

=

=

= Ω

(c) 7.0 V3.6?

7.03.61.9 A

VRI

VIR

== Ω=

=

=

=

(d) 7.0 1.913.3 W

P VI== ×=

(e) The lights dim because internal resistance and high current reduce the external voltage in the battery.


Unit

2 ■■■■■■■■■

Area of study 1: Motion


Chapter 5

Analysing movement 1. (b) displacement, (d) velocity and (e) acceleration. In

each case, both magnitude and direction are required for a complete description.

2. (a) distance = 5 znotters + 12 znotters = 17 znotters (b)

magnitude of displacement = 2 25 12+ = 13 znotters

tan θ = 512

⇒ θ = 23° Displacement is 13 znotters at an angle of 23° north

of west (270° + 23° = 293° True).

(c) average velocity = displacementtime

= 13 znotters6.5 znitters

at 23° north of west

= 2 znotter znitter−1 at 23° north of west

average speed = displacementtime

= 17 znotters6.5 znitters

= 2.6 znotter znitter−1

(d) vat

Δ=Δ

Units are therefore 1

2znotter znitter znotter znitterznitter

−−=

3. 1

1

100 km100 km h1 h

100 000 m3600 s

27.8 m s

−

−

=

=

=

(or simply divide by 3.6)

4. 1

13600

1

1

1.5 m1.5 m s1 s

0.0015 kmh

3600 0.0015 km h

5.4 km h

−

−

−

=

=

= ×

=

(or simply multiply by 3.6)

5. (a) 1

1

55 miles × 1.6 km/mile55 miles h1 h

88 km h

−

−

=

=

(b) 1

1

88 000 m88 km h 3600 s

24 m s

−

−

=

=

6. (a) Event

(m)

Average speed (m s−1) Calculations

100 10.3 100 m/9.74 s 200 10.4 200 m/19.32 s 400 9.26 400 m/43.18 s 800 7.91 800 m/101.11 s

(1 min 41.11 s = 101.11 s) 1500 7.28 1500 m/206.00 s

(3 min 26.00 s = 206 s) 3000 6.81 3000 m/440.67 s

(7 min 20.67 s = 440.67 s) 5000 6.60 5000 m/759.36 s

(12 min 37.35 s = 757.35 s) 10 000 6.34 10 000 m/1582.75 s

(26 min 17.53 s = 1577.53 s)

(b) The fact that the average speed during the 100 m event is similar to that during the 200 m event is due to the fact that the acceleration from rest to the maximum speed takes place over a significant fraction of the time taken for the 100 m event. Even though the maximum speed of the athlete is greater during the 100 m event, the average speed is not.

(c) average speed = distance travelledtime interval

distance travelledtime intervalaverage speed

⇒ =

142 200 m

7.28 m s5797 s96 min 37 s1 h 36 min 37 s

−=

===

(d) Only Asafa Powell’s. His event is the only one that involves only straight line motion.


7. (a) average speed

1

distance travelledtime interval

3000 m204.537 s

14.67 m s−

=

=

=

(b) time interval

1

distance travelledaverage speed

151 000 m14.67 m s10 293 s2.86 h

−

=

=

==

(c) time interval

1

distance travelledaverage speed151 km

80 km h1.9 h

−

=

=

=

(d) i(i) average speed

1

distance travelledtime speed

302 km4.0 h

76 km h−

=

=

=

(ii) av

10 km h

xvt

−

Δ=Δ

=

since Δx (displacement) = 0

8. Time for tortoise

1

distance travelledaverage speed1000 m

0.075 m s13 333 s

−

=

=

=

Time for hare at maximum speed

1

distance travelledaverage speed

1000 m20 m s50 s

−

=

=

=

In a tied race, the hare must have napped for 13 333 s − 50 s = 13 283 s = 3.7 h (approx. 3 h 41 min)

9. average speed

1 2 3

1


(not 1)3

120 120 120 m20 30 60 s

360 m110 s

3.3 m s

v v v

−

=

+ +

+ +=+ +

=

=

10. (a) Predictions will vary but likely response is 90 km h−1 — incorrectly obtained by

1 180 km h 100 km h2

− −+

(b) average speed

300 30080 100

1


300 300

89 km h−

=

+=+

=

11. Only a finite time Δt can be measured with a stopwatch.

Thus, only the average velocity vt

ΔΔ

can be determined.

Instantaneous velocity is the velocity at an instant of time. There is no time interval to measure with a stopwatch.

12. (a) B, C (b) B, D (c) A, E (d) A, E (e) D

13. A: Constant negative acceleration with an initial positive velocity.

B: Constant positive acceleration from rest. C: A constant positive velocity, followed by an interval

of constant negative acceleration until a negative velocity equal in magnitude to the initial velocity is reached. The velocity then remains constant.

14. (a)

(b)

(c)


15.

16. The instantaneous velocity is the same as the average

velocity at the mid-point of the time interval during which the motion takes place.

17. (i) change

in speed (ii) change in velocity

(a) −40 km h−1 40 km h−1 south (or −40 km h−1 north)

(b) −20 m s−1 −20 m s−1 in original direction (or +20 m s−1 opposite to original direction)

(c) +5 m s−1 −55 m s−1 in original direction (or +55 m s−1 opposite to original direction)

18. Yes, there is an acceleration. Even though the speed has not changed, the velocity has changed.

The magnitude of Δv = 141 m s−1. Its direction is south-west. The acceleration is therefore not zero.

19. For a car that accelerates from rest to 60 km h−1

(17 m s−1) in say 5 seconds: 1

2

17 m s5 s

3 m s

vat

−

−

Δ= =Δ

=

20. vav for a 100 m sprint in, say 9.8 s = 10.2 m s−1

estimate vmax = 12 m s−1 is reached after 2 seconds. 1

2

12 m s2 s

6 m s

vat

−

−

Δ= =Δ

=

21. (a) a = 6.0 m s−2

u = 17 m s−1

v = 28 m s−1 t = ? v = u + at

⇒ 28 = 17 + 6.0t

⇒ t = 28 176.0−

= 1.8 s (b) a = 2.0 m s−2

u = 0 v = 10 m s−1 t = ? v = u + at

⇒ 10 = 0 + 2.0t

⇒ t = 102.0

= 5.0 s 22. a = 10 m s−2

x = 36 m u = 0 (a)

212

?t

x ut at

=

= +

⇒ 36 = 0 + 5t2

365

t⇒ =

= 2.7 s (b) v = ?

v2 = u2 + 2ax = 0 + 2 × 10 × 36 = 720

720v⇒ = = 27 m s−1

23. x = 12 m t = 2 s v = 0 (a) u = ?

2

12 22

u vx t

u

+=

⇒ = ×

112 m su −⇒ =


(b) 21

212

2

?

12 0 4

2 12

6.0 m s

a

x vt at

aa

a

=

= −

⇒ = − × ×

⇒ = −

= −

24. u = 100 km h−1 = 27.7 m s−1

v = 0 (a) x = 48 m a = ? v2 = u2 + 2ax ⇒ 0 = (27.7)2 + 2 × a × 48

2

2

(27.7)96

8.0 m s

a

−

⇒ = −

= −

(b) x = 48 m t = ?

227.748

2

u vx t

t

+=

⇒ =

96

27.73.5 s

t⇒ =

=

(c) The reaction time of the driver needs to be known so that the distance travelled between the instant that the branch is seen and the instant that the brakes are applied can be determined. An estimate of 0.2 s would be reasonable for the reaction time. At a constant speed of 100 km h−1 (27.7 m s−1) the car would travel a distance of 27.7 m s−1 × 0.2 s = 5.5 m. The total distance required to stop is therefore 5.5 m (reacting distance) + 48 m (braking distance) = 53.5 m. The car would not stop in time.

25. In order to make the leap, the dancer must rise and fall in 0.5 s.

Ignoring air resistance, the fall takes the same amount of time as the rise, −0.25 s.

The time taken for the dancer to rise to a height of 80 cm can be calculated:

v = 0, x = 0.80 m, a = −10 m s−2

t = ? 21

220.80 5

0.805

0.4s

x vt at

t

t

= −

⇒ =

⇒ =

=

The time taken for the dancer to fall from a height of 80 cm can be calculated.

u = 0, x = 0.80 m, a = 10 m s−2

t = ? 21

2x ut at= −

20.80 5

0.805

0.4s

t

t

=

⇒ =

=

That is, without knowing the ‘take off’ speed of the dancer it can be shown that the leap would take 0.8 s. The leap is not possible.

26. It is important to remember that at the instant that the Rolls Royce rolls off the truck it is moving in the same direction as the truck.

After one minute it has moved a distance of 1000 m (a constant speed of 60 km h−1 is equal to 1 km min−1).

During the driver’s reaction time the distance moved by the truck is 8.3 m (the distance moved in 0.5 s at a constant speed of 60 km h−1 = 16.67 m s−1 × 0.5 s).

The braking distance of the truck is 25 m. The total distance moved by the truck is 1000 m +

8.3 m + 25 m = 1033 m. The distance moved by the Rolls Royce is 240 m (in

the same direction as that of the truck). The Rolls Royce is therefore 793 m behind the stopped truck.

27.

(a) The balls collide when the tennis ball is at the top of

its path at a time tc. For the tennis ball: v = 0 a = −10 m s−2 (taking up as positive) x = h t = tc u = ? x = vt + 1

2 at2

⇒ h = 5t2 (1) (since a = −10 m s−2) For the golf ball: u = 0 a = 10 m s−2 (taking down as positive) x = 100 − h t = tc

x = ut + 12 at2

⇒ 100 − h = 5t2 (2) Add (1) and (2). ⇒ 100 = 10t2

⇒ t = 10 = 3.15 s For the tennis ball: v = u + at ⇒ 0 = u − 10 × 3.16 ⇒ u = 32 m s−1


(b) Substitute in (1) ⇒ h = 5(3.16)2 = 50 m The balls meet 50 m from the ground.

28. (a) B (b) A, D, E (the intervals in which the gradient is

positive) (c) 40 s (the first instant at which the skateboard rider

moves in a southerly direction) (d) 20 m north (change in position after 80 s = 20 m

north of starting point) (e) 260 m (80 m in northerly direction, followed by

120 m in a southerly direction, followed by 60 m in a northerly direction)

(f) D (the gradient is increasing) (g) E (the gradient is decreasing)

(h) average speed

1


260 m80 s

3.3 m s−

=

=

=

(i)

1

1

gradient120 m20 s6.0 m s

6.0 m s south

v

−

−

=−=

= −

=

(j) v = gradient at time t = 65 s

To find the gradient a tangent must be drawn on the curve at t = 65 s. Two convenient points need to be drawn on the tangent (e.g. A and B).

( )–1

risegradientrun

18 ( 48) approx.70 60

3 m s

=

− − −=−

=

The direction is north. 29. (a) B, D, F (gradient = 0 during these sections. That is,

there is no change in velocity.) (b) Displacement = total area under graph.

This is most easily calculated by dividing the graphs into triangles and rectangles as shown in the following figure.

Total area

1 1 12 2 2

1 1 12 2 2

10 1 10 1 10 1 10 1 10

1 5 1 5 1 10 1 10 1

5 10 5 5 10 2.5 2.5 10 520 m

= × × + × + × × − × × − ×

− × × + × × + × − × ×

= + + − − − + + += +

(c) av

1

20 m80 s

0.25 m s

xvt

−

Δ=Δ+=

=

(d) 30 s (the instant that the velocity becomes negative) (e) It didn’t. (The negative displacement that occurs

between 30 s and 55 s is not as great as the positive displacement between 0 s and 30 s.)

(f) C, G (when the gradient is negative) (g) The first half of interval C, the first half of interval E

and interval G. (During these periods, the magnitude of the velocity is decreasing.)

(h) A negative acceleration doesn’t always decrease the speed and a positive acceleration doesn’t always increase the speed. A negative acceleration increases the speed if the velocity is negative and decreases the speed if the velocity is positive. Similarly, a positive acceleration decreases the speed if the velocity is negative and increases the speed if the velocity is positive.

(i) 1

2

gradient

2.0 m s10 s

0.20 m s

a−

−

=

+=

=


(j) av

1

2

1.0 m s20 s

0.050 m s

vat

−

−

Δ=Δ

=

=

(k) The motion of the toy robot can be described in nine different intervals. First 10 seconds: The toy robot started from rest and increased its speed at a constant rate until reaching a speed of 1.0 m s−1 after 10 seconds. 10 s to 20 s: It maintained a constant speed of 1.0 m s−1. 20 s to 30 s: It slowed down at a constant rate. It was at rest for an instant, 30 seconds after starting. 30 s to 40 s: It increased its speed at the same constant rate as the first interval, but in the opposite direction to reach a maximum speed of 1.0 m s−1. 40 s to 50 s: It maintained a constant speed of 1.0 m s−1. 50 s to 55 s: It decelerated to rest at a constant rate. 55 s to 60 s: It increased its speed at a constant rate in the original direction. The acceleration was twice that of the first interval. 60 s to 70 s: It maintained a constant speed of 1.0 m s−1. 70 s to 80 s: It decelerated to rest at a constant rate.

30. (a) 3.0 s (can be read directly from the graph) (b)

1

2

gradient

10 m s=4 s

2.5 m s

a−

−

=

=

(c) Let T = time at which stuntman catches the bus. At time T the displacement of the stuntman is equal to the displacement of the bus.

12

area under stuntman graph area under bus graph4 10 10( 4) 8

⇒ =⇒ × × + − =T T

20 10 40 82 20

T TT

⇒ + − =⇒ =

⇒ T = 10 s (d) Distance = magnitude of displacement

= area under graph = 8T (or 10T − 20) = 80 m 31. (a) A constant speed is reached when the acceleration

becomes zero. The acceleration of the jet ski becomes zero first, after 8.0 s.

(b) (i) Δv = area under acceleration vs time graph = 1

2 8 4× ×

= 16 m s−1

v = 5.0 m s−1

v = 21 m s−1 for jet ski (ii) Dividing the graph into a trapezium and triangle

as shown in the following figure.

Δv = area under acceleration vs time graph

= 12

6 2 6 (10 6) 22+ × + × − ×

= 24 + 4 = 28 m s−1

v = 5.0 m s−1

⇒ v = 33 m s−1 for car (c)

Speed vs time graph for jet ski and car

Chapter 6

Forces in action 1. ‘My mass is 75 kg’, ‘My weight is 750 N’. 2. Vector quantities have magnitude, unit and direction.

Scalar quantities have magnitude and unit. 3. (b) weight, and (c) gravitational field strength. 4. Weight is a vector quantity. On the surface of the Earth

its magnitude would be constant (to 2 sig. fig.) but the direction changes from place to place. There is not enough information given to compare its direction on Earth with its direction on Mars.

5. (a) W = mg = 1400 kg × 10 N kg−1

= 1.4 × 104 N


(b) On Mars: W = mg = 1400 kg × 3.6 N kg−1

= 5.0 × 103 N (c) m = 1400 kg anywhere. It is a measure of the amount

of matter in an object or substance and does not depend on the gravitational field strength.

6. (a) Apple: m 0.1 kg ⇒ W 0.1 kg × 10 N kg−1

1.0 N (b) This book: m 1 kg ⇒ W 1 kg × 10 N kg−1

10 N (c) If physics teacher has mass of about 80 kg: W 80 kg × 10 N kg−1

800 N 7. Assume mass of student is m kg. (a) 10m N (b) 3.6m N (c) m kg If, for example, m = 65 kg: (a) At Earth’s surface W = 65 kg × 10 N kg−1 = 650 N (b) On the surface of Mars W = 65 kg × 3.6 N kg−1

= 234 N (c) m = 65 kg on Mars and anywhere else. It does not

depend on the gravitational field strength. 8.

The forces applied by the Sun and Earth on the person, floor and Earth have not been included. Their direction is not constant.

9. (a)

The net force is 3 N to the east.

(b)

The answer can be obtained by scale vector diagram

or using trigonometry.

i.e. cos 45

100 N100 N cos 45

x

x

° =

⇒ = °

70.7 N= Magnitude of Fnet = 2x Fnet = 1.4 × 102 N east 10. (a)

Answer is best obtained by scale vector diagram. Missing force = 346 N east

(b) The sum of the two forces acting at 30° to the horizontal can be obtained by scale vector diagram or using trigonometry.

cos30200 N

x° =

⇒ x = 200 N cos 30°


= 173.2 N ⇒ sum = 346.4 N east Adding this to the 200 N force to the west gives a

total force of 146.4 N east. In order to obtain a net force of 200 N east, an

additional force of 53.6 N east is required. 11. (a)

(b) The car is in uniform motion. Therefore the net force

must be zero. 12. It describes what happens to an object when acted on by a

net force equal to zero. The word inertia is used to describe the tendency of an object to resist change.

13. The vehicle experiences a non-zero net force that slows it down. No such force acts on you. The net force on you is zero. Therefore, you continue in your state of constant velocity. You are not really thrown forward. You continue your motion while the vehicle slows down.

14. There is an unbalanced force on the bike and its velocity changes. Your inertia keeps you moving forward as there is no unbalanced force to change your motion (apart from gravity).

15. (a)

(b) The direction of the net force on the ball is down the

hill (parallel to the hill) since the ball is speeding up as it rolls down the hill.

(c) Weight force; if it were not, the net force could not be down the hill.

(d) The ball slows to a stop because of the effect of the friction acting on the ball. On a horizontal surface, the normal reaction is equal in magnitude to the weight.

16. The force of the Earth pushes on the tyres in the opposite

direction to the force applied by you pushing the car.

17.

18. (a)

cos 25

200 N200 N cos 25

x

x

F

F

° =

⇒ = °

21.8 10 N= × (b)

cos 60

200 N200 N cos 60

x

x

F

F

° =

⇒ = °

100 N= (c) 0 N 19. Idealisations can be made to allow the use of a simple

mathematical model to solve a physical problem. For example, in order to use simple equations to analyse the motion of a falling ball, the idealisation can be made that the air resistance is insignificant and the ball does not spin.

20. (a) Fnet = ma = 2.2 × 106 kg × 3.0 m s−2

= 6.6 × 106 N (b) Fnet = thrust − weight ⇒ 6.6 × 106 N = thrust − 2.2 × 107 N ⇒ Thrust = 2.9 × 107 N


21. (a) Both experience the same acceleration. The

acceleration is given by netFam

= . The air resistance

on these objects is insignificant when compared to their weight and can be ignored. Therefore, both the net force (weight) and mass of the gold bar are 10 times as great as they are for the bowling ball.

(b) The air resistance in the doormat is significant when compared with its weight. Therefore, the net force on the doormat is less than that of the bowling ball (which has the same mass as the doormat) and the

acceleration of the doormat netFm

is smaller than that

of the bowling ball (and the gold bar). 22. (a) 70 kg × 10 N kg−1 = 700 N

(b) i(i) The upwards force is greater when the jumper is decelerating downwards or, in other words, accelerating upwards.

(ii) The weight is greater than the upwards pull when the jumper is accelerating downwards.

(c) The tension in the bungee cord must be equal in magnitude to the jumper’s weight in order for the speed to be constant, that is, 700 N. This occurs only for an instant during the fall. At this instant the cord is extending and the tension is increasing.

23.

(a) The net force is zero because the motion is uniform. (b) W = mg = 80 kg × 10 N kg−1

= 800 N Wx = W sin 30° = 800 sin 30° = 400 N (c) Fnet = 0 Therefore, the sum of forces parallel to the slope is

zero. ⇒ D − 400 − 10 = 0 ⇒ D = 410 N (d) The sum of the forces perpendicular to the slope is

zero. ⇒ N = Wy = 800 cos 30° = 6.9 × 102 N 24. (a) Since the speed is increasing down the slope, the net

force must be in the same direction, that is, down the slope.

(b)

(c) Wx = mg sin 30° = 60 × 10 × sin 30° = 300 N

(d) Fnet = 300 N down slope + 8 N up slope = 292 N down slope Magnitude of net force = 292 N 25. (a) Fnet = 10 000 N − 2500 N = 7500 N

(b) net

2 2

750012006.3 m s (6.25 m s )

Fam

− −

=

=

=

(c) u = 0, a = 6.25 m s−2, t = 5.0 s v = u + at = 0 + 6.25 × 5.0 = 31 m s−1

(d) 212

2120 6.25 (5.0)

78 m

x ut at= +

= + × ×

=

26. u = 25 m s−1, v = 0, x = 360 m v2 = u2 + 2ax ⇒ 0 = (25)2 + 720a

2

6257200.868 m s

a

−

⇒ = −

= −

Frictional force = net force = ma = 8.0 × 106 × −0.868 = 6.9 × 106 N 27. x = 50 m, u = 12 m s−1, v = 0 m s−1

v2 = u2 + 2ax ⇒ 0 = (12)2 + 100a


2

1441001.44 m s

a

−

⇒ = −

= −

Frictional force = net force = ma = 70 × 1.44 = 1.0 × 102 N (101 N)

28.

The forces acting on the teacher The force N is provided by the bathroom scales. The

reading on the scales will be equal to N. When the teacher is stationary N = mg = 700 N (a) Fnet = 0 since velocity is constant ⇒ N − mg = 0 ⇒ N = mg = 700 N (b) Assign down as negative. a = −2.0 m s−2

N − mg = ma mg = 700 ∴ m = 70 kg (since g = 10 N kg−1) N − 700 = 70 × −2.0 ⇒ N = −140 + 700 = 560 N (c) a = 2.0 m s−2

N − mg = ma ⇒ N − 700 = 70 × 2.0 ⇒ N = 140 + 700 = 840 N 29. Assigning up as positive, the net force on the lift is given

by Fnet = T − W where T is the tension in the cable and W is the weight of the lift and its passengers. W = mg = (480 + 24 × 70) × 10 = 2160 × 10 = 21 600 N ⇒ Fnet = T − 21 600 ⇒ ma = T − 21 600

max

21600

25 000 N

a Tm

T

⇒ = −

=

max

2

25000 216002160

1.6 m s

a

−

−⇒ =

=

30.

The weight mg can be resolved into components that are

parallel to and perpendicular to the slope. Considering the forces parallel to the slope Fnet = mg sin 30° − R ⇒ ma = mg sin 30° − R ⇒ 120 = 600 sin 30° − R ⇒ R = 180 N 31.

The weight mg can be resolved into components that are

parallel to and perpendicular to the slope. Considering the forces parallel to the slope Fnet = mg sin 40° − 180 ⇒ 400 a = 4000 sin 40° − 180

2 2

4000 sin 40 180400

6.0 m s (5.97 m s )− −

° −⇒ =

=

a

32. (a) 2

net2

6 0.75 m s8

56 kg

56 kg 0.75 m s42 N

vat

mF ma

−

−

Δ= = = ΔΔ

==

= ×=

(b)

The weight, mg, can be resolved into compoments that

are parallel to and perpendicular to the slope.


Considering the forces parallel to the slope: Fnet = mg sin θ − 140 ⇒ 42 = 560 sin θ − 140

42 140sin

56019

+⇒ =

⇒ = °

θ

θ

33. (a) Traction (friction) on your blades (b) The component of gravity down the slope and the

force of the ground on your poles if you use them (c) The tension in the rope attached to the handle that

you are desperately holding (d) The traction (friction) as you push back on the

ground (e) The force exerted by the water on your hands, arms,

legs and feet as you push back with your hands and kick

(f) The force exerted by the water on the oar as the oar pushes back on the water

34. The driving force, the forward force applied to the tyres by the road, is a reaction to the force applied backwards to the road by the tyres. The size of the driving force is therefore controlled by the driver’s use of the accelerator. The driving force acts on all four wheels of a four-wheel-drive vehicle, whereas it acts only on the rear wheels of a rear-wheel-drive car. The front wheels of a rear-wheel-drive car are pushed forward as a result of the driving force on the rear wheels. So the force applied to the front wheels cannot be controlled directly by the driver.

35. Student-designed spreadsheet 36.

The forces acting on the yo-yo The acceleration of the yo-yo in the horizontal direction is

the same as the acceleration of the rollerblader. ⇒ ma (horizontal) = T sin 5° (1) If the vertical acceleration is zero mg = T cos 5° (2) Dividing (1) by (2)

sin 5cos5

tan 510

ma Tmg Ta

°=°

⇒ = °

⇒ a = 10 tan 5° = 0.87 m s−2

37. (a) 0 m s−1

(b) 10 m s−2. This is the same as the acceleration throughout the rest of its flight.

(c) 5.0 N down. The net force on the ball throughout its flight (assuming air resistance is negligible) = mg = 0.50 kg × 10 m s−2 down.

38. The friction on the rear wheels is the driving force, which pushes the bicycle forward. The driving force is a reaction to the backward push of the rear wheel on the road, which can be controlled by the cyclist. The friction on the front wheel is a retarding force, which opposes the forward rolling motion of the front wheel, which is pushed forward by the moving bicycle. When the driving force is greater than the retarding force the bicycle accelerates. When the driving force is equal to the retarding force the bicycle maintains a constant speed. If the driving force is less than the retarding force the bicycle decelerates.

39. (from top to bottom) The wall pushes on your palm in the opposite direction. The bicycle pedal pushes up on your foot. You push down on the ground when you are standing. Your body pulls up on the Earth. The broken-down car pushes on you in the opposite

direction. The nail pushes up on the hammer. 40. (a)

(b)

(c)

41. The dinghy Force Action–reaction pair Resistance forces Fdinghy on air and water, Fair and water

on dinghy Weight FEarth on dinghy, Fdinghy on Earth Tension Frope on dinghy, Fdinghy on rope Normal reaction force Fdinghy on water, Fwater on dinghy

The boat Force Action–reaction pair Resistance forces Fboat on air and water, Fair and water on

boat Weight FEarth on boat, Fboat on Earth Tension Fboat on rope, Frope on boat Normal reaction force Fboat on water, Fwater on boat

42. All swimmers move forwards in the water because the water pushes them forwards. This is the unbalanced force that provides the acceleration during each stroke


(Newton’s First Law). The size of the forward force is equal to, and opposite in direction to, the force that the swimmer applies to the water (Newton’s Third Law). A freestyle stroke pushes water back with a greater force than a breaststroke stroke. Therefore the forward force is greater for a freestyle swimmer.

43. The student is correct but needs to make it clear that the two friction forces are not an action–reaction pair. The best way to do this is to identify the two action–reaction pairs involved. If the car is a front-wheel drive, the friction on each of the front tyres is a reaction to the backward push of the front tyres on the road. The rear tyres are being pulled forward. So the friction on the rear tyres is a reaction to the forward push of the rear tyres on the road.

44.

(a) Fnet = 14 N, m = 7.0 kg

net

net

2

14 N, 7.0 kg

14 N7.0 kg

2.0 m s

F mFam

−

= =

=

=

=

(b) Consider the 3.0 kg trolley: Fnet = T

⇒ 3.0 kg × 2.0 m s−2 = T ⇒ T = 6.0 N

(c) Fnet = 14 N − T = 14 N − 6.0 N = 8.0 N (d) net

net

2

14 N4.0 kg

14 N4.0 kg

3.5 m s

Fm

Fam

−

==

=

=

=

45.

Assign direction to the right as positive. (a) Applying Newton’s Second Law to the system of the

two crates: Fnet = ma ⇒ P − friction = 70a

⇒ 420 − (70 × 2.0) = 70a

420 14070−

⇒ =a

= 4.0 m s−2 to the right (b) Applying Newton’s Second Law to the 40 kg crate: Fnet = ma = 40 × 4.0 = 160 N to the right (c) Applying Newton’s Second Law to the 30 kg crate: Fnet = ma = 30 × 4.0 = 120 N Fnet = P + R21 + friction on 30 kg crate ⇒ 120 = 420 + R21 − (30 × 2) ⇒ 120 = 360 + R21 ⇒ R21 = −240 N = 240 N to the left (d) R12 = −R21 = 240 N to the right (e) The net force is still P − friction. The mass is still 70 kg.

net

420 (70 2.0)70

am

⇒ =

− ×=

F

= 4.0 m s−2 to the right It is no easier.

46. (a) 1 14040 km h m s3.6

− −=

2

11.11 03.2

3.47 m s

vt

−

Δ=Δ

−=

=

a

Fnet = ma = 1000 kg (estimate) × 3.47 m s−2

= 3 × 103 N (b) At terminal velocity air resistance = weight = mg = 80 kg × 10 N kg−1

= 8 × 102 N (c) p = mv = 80 kg (estimate) × 10 m s−1 (estimate) = 8 × 102 kg m s−1

(d) p = mv = 800 kg × 60 km h−1

= 800 kg × 16.7 m s−1

= 1 × 104 kg m s−1

(e) Impulse = mΔ v = 70 kg × 8 m s−1

= 6 × 102 kg m s−1 or 6 × 102 N s (f) Impulse = mΔ v = 0.4 kg (estimate) × 5 m s−1

= 2 N s


(g) Δp = mΔ v = 0.2 kg (estimate) × 50 m s−1 (estimate) = 10 kg m s−1

47. Taking down as positive: (a) Δp = mΔ v = 0.060(−6.0 − 8.0) = −0.84 kg m s−1

= 0.84 kg m s−1 up (b) 0.84 kg m s−1 or 0.84 N s down The impulse applied by the tennis ball to the ground

is equal and opposite to the impulse applied by the ground to the tennis ball.

(c) No. The ground does not move a measurable amount. The impulse applied to the ground is 0.84 kg m s−1 but the mass of the Earth is so large that the change in velocity is negligible.

(d) FnetΔt = Δp

net

30.84 Ns up2.0 10 s

pFt

−

Δ⇒ =

Δ

=×

= 4.2 × 102 N up (e) Fnet = N − mg ⇒ 4.2 × 102 = N − 0.06 × 10 = N − 0.6 ⇒ N = 4.2 × 102 + 0.6 = 4.2 × 102 N up 48. (a) m = 75 kg, u = −3.2 m s−1, v = 0 Assigning up as positive: Impulse = mΔv = 75 (0 − −3.2) = 240 N s = 240 N s upwards.

(b) FnetΔt = mΔv ⇒ Fnet × 0.10 = 240 ⇒ Fnet = 2400 N upwards. Fnet = N − mg where N = force ground applies to feet (normal

reaction force) ⇒ 2400 = N − 750 N = 3150 N = 3.2 × 103 N upwards (c) u = 0, v = −3.2 m s−1, a = −10 m s−2

v2 = u2 + 2ax ⇒ (−3.2)2 = 2 × −10 × x

2(3.2)20

0.51 m

x⇒ = −

= −

Height (assuming feet are motionless relative to the basketballer’s centre of mass) is approximately 0.5 m.

49. m = 1400 kg, u = 60 km h−1 = 16.7 m s−1, v = 0, t = 0.080 s (a) Impulse = mΔv = 1400 × (0 − 16.7)

= 2.3 × 104 N s opposite to initial direction of motion of the car (2.34 × 104 N s)

(b) FnetΔt = mΔv

net

42.34 100.080

mt

Δ⇒ =

Δ×=

vF

= 2.9 × 105 N opposite to initial direction of motion of the car (2.93 × 105 N). (c) The deceleration of the driver is the same as the

deceleration of the car because the driver is wearing a seat belt.

2 2

16.70.080

2.1 10 m s

t

−

Δ=Δ

=

= − ×

va

⇒ Deceleration = 2.1 × 102 m s−2

50. The airbags allow the change in momentum (impulse) of the driver’s head to take place over a longer time interval than would be the case if it collided directly with the steering wheel. The average net force on (and the magnitude of the acceleration of) the driver’s head is therefore less.

51. The change in momentum (impulse) on the legs takes place over a longer interval, reducing the force exerted by the ground on the knee joint and muscles, tendons and ligaments in the leg.

52. (a) The impulse is the area under the graph. The area is approximately equal to the area of a triangle with a base of 0.10 s and a height of 3200 N.

Area = 12 × 0.10 × 3200

⇒ Impulse = 160 N s (approximately) (b) Impulse = mΔ v ⇒ 160 = 60Δ v ⇒ Δ v = 2.7 m s−1

but v = 0 ⇒ u = 2.7 m s−1 (approximately) (c) The impulse on the unbelted occupant is greater than

that on the belted occupant (the area under the force versus time graph is clearly greater).

The change in velocity Δ v is the same for both occupants.

Because impulse = mΔ v, the mass of the unbelted occupant must be greater.


(An estimate of the area under the blue curve shows that the mass of the second occupant is approximately 95 kg.)

(d) The graph describing the force on the occupant with the seat belt shows that the force is applied immediately and is applied for a relatively large amount of time compared with the force applied to the occupant without the seat belt. The occupant without the seat belt experiences no immediate force as she or he continues to move forward at the same speed as the car was moving before impact. The force applied to this occupant increases rapidly to a magnitude greater than the force applied to the occupant with the seat belt. The multiple peaks in force on the second occupant can be explained by multiple impacts with the dashboard or other parts of the car.

53. (a) The impulse is the area under the graph. This can be found most easily by counting the small squares.

The area of each small square = 0.005 s × 100 N = 0.5 N s

The number of small squares under the curve is between 190 and 200.

Impulse = no. of squares × 0.5 N s Impulse = 100 N s upwards (approx.) (b) Impulse applied by net force = mΔ v ⇒ Impulse applied by floor + impulse due to weight = mΔ v ⇒ 100 N s − mg × 0.10 = mΔ v ⇒ 100 − 600 × 0.10 = 60Δ v

1

100 6060

0.67 m s−

−⇒ Δ =

=

v

(taking up as positive) u = 0 ⇒ v = 0.67 m s−1

(c) FΔt = Impulse applied by floor where F = average force applied by floor

100 N s0.10 s

1000 N upwards

F⇒ =

=

(d) The normal reaction force is present as the basketballer is initially pushing down on the floor with a force equal to the basketballer’s weight.

54. Bouncing off during collision results in a greater change in momentum of the cars in a similar or smaller time interval. The rate of change in momentum of the cars, and the resulting net force on the passengers, would therefore be greater (FΔt = mΔv). In low-speed collisions with small vehicles (like dodgem cars) this is not a problem. However, in real cars at typical road speeds more injuries would occur.

55. (a) The Earth and the basketball can be considered to be an isolated system. The total change in momentum of the system is zero. The momentum of the basketball changes. Therefore, the momentum of the Earth must change by the same amount but in the opposite direction. The velocity of the Earth must change as a

result. However, the magnitude of the change is very small.

If the mass of the basketball is 0.5 kg and its change in velocity on striking the ground is 5 m s−1:

Δ pb = 0.5 × 5 = 2.5 kg m s−1

The magnitude of the change in momentum of the Earth is given by:

Δ pE = mEΔ vE ⇒ mEΔ vE = 2.5 kg m s−1

mE = 6 × 1024 kg

E 24

25 1

2.56 104 10 m s

v

− −

⇒ Δ =×

= ×

(b) The concrete wall is firmly attached to the Earth, so the car is effectively colliding with the Earth. The Earth (including the wall) and the car can be considered to be an isolated system. The total change in momentum of the system is zero. The momentum of the car changes. Therefore the momentum of the Earth and concrete wall must change by the same amount but in the opposite direction. The velocity of the Earth and concrete wall must change as a result. However, the magnitude of the change is very small. The distance moved by the concrete wall is also immeasurably small because the force applied to it by the car is very small compared to the forces applied by the ground.

If the mass of the car is 1000 kg and its change in velocity on striking the wall is 10 m s−1:

Δ pC = 1000 × 10 = 10 000 kg m s−1

The magnitude of the change in momentum of the Earth and the wall is given by:

Δ pew = mewΔ vew ⇒ mewΔ vew = 10 000 kg m s−1

mew = 6 × 1024 kg

ew 24

21 1

10 0006 10

2 10 m s

v

− −

=×

= ×

Chapter 7

Mechanical interactions 1. By releasing a high-pressure propellant, the astronaut

gains momentum in one direction while the propellant gains the same amount of momentum in the opposite direction. The total change in momentum of the astronaut and the contents of his or her backpack is zero.

2. (a) Provided that friction is negligible, there is no horizontal net force on the system of the mass and the trolley.

⇒ total momentum before = total momentum after


⇒ 2.0 × 0.60 = 4.0 × v ⇒ v = 0.30 m s−1

(b) There is no horizontal net force on the system of the sand and the trolley.

⇒ 4.0 × 0.60 = 2.0 × vsand + 2.0 vtrolley

vsand = 0.60 m s−1

as it falls from the moving trolley ⇒ 4.0 × 0.60 = 2.0 × 0.60 + 2.0 vtrolley

⇒ 2.0 vtrolley = 1.2 ⇒ vtrolley = 0.60 m s−1

3. (a) Total momentum before collision is zero. Total momentum after collision = 0

L C

L L C C

++ 0

p pm v m v

=⇒ =

L

L

1.5 50 1.2 050 1.2

1.5

m

m

⇒ × − × =×

⇒ =

= 40 kg (b) Impulse on Catherine = mCΔ vC

= 50 × 1.2 = 60 N s (c) Impulse on Lauren = −Impulse on Catherine = −60 N s Magnitude of impulse = 60 N s (d) Zero, because there is no external net force acting on

the system of the two girls. (e) They would have greater speeds but still in the same

ratio as before. The total momentum would remain as zero as there are no external horizontal forces acting on the girls.

4. (a) Total momentum before collision is given by

N N L L–1

+ 60 2 + 70 0

120 kg m s

m v m v = × ×

=

Total momentum after collision is given by

N N L L L

L

+ 60 0 + 7070

m v m v vv

= × ×=

Total momentum is conserved

L

L

70 12012070

v

v

⇒ =

⇒ =

= 1.7 m s−1

(b) N NImpulse on Nick60 2

120 N s

m v= Δ= × −= −

Magnitude of impulse 120 N s= (c) Change in momentum = impulse Change in momentum = 120 kg m s−1

(d) Δ pL = −Δ pN since total change in momentum is zero ⇒ magnitude of Luke’s change in momentum = 120 kg m s−1

(e) They would have different speeds but the total momentum would still be conserved as there are no external horizontal forces acting on the boys.

(f) 1i

f N L

120 kg m s( + ) 120

pp m m v

−===

130 120

120130

v

v

⇒ =

⇒ =

= 0.92 m s−1 5. (a) Total momentum before collision is given by: pC + pP

where pC = momentum of car and driver pP = momentum of police car and occupants = 1250 vC + 1500 × 0 Total momentum after collision: (mC + mP) v = 2750 × 7 Total momentum is conserved. ⇒ 1250 vC = 19 250 ⇒ vC = 15 m s−1 (15.4 m s−1) (b) Impulse on police car = mPΔvP = 1500 × 7 = 10 500 N s (1.1 × 104 N s) in the initial direction of motion of the car. (c) Impulse on driver = mdΔvd

= 50 × (v − u)d

= 50 × (7 − 15.4) = −420 N s = 420 N s opposite to the initial direction of motion of the car. (d) FnetΔt = impulse on police car

net10500 N s

0.10 s105000 N

F⇒ =

=

Average net force = 1.1 × 105 N in initial direction of motion of the car.

6. Student-designed spreadsheet 7.

4.0 10 1.560 J

W Fxmgx

=== × ×=

8. None, since there is no displacement in the direction of the force.

9. (a) Zero. The displacement after one complete revolution is zero. The force applied to keep the toy dog is directed towards the centre of the circle. It is therefore perpendicular to the direction of motion at all times. In other words, there is no displacement in the direction of the applied force. Therefore, no work is done on the dog by the girl.

(b) Although the displacement is not zero after half of a full revolution, there is still no work done on the dog by the girl because the applied force is perpendicular to the direction of motion at all times.


10. WorkUnit N m

Fx==

2

2

but 1 N 1 kg m s

Unit kg m s m

−

−

=

⇒ = ×

2 2kg m s−=

212

1 2

2 –2

Kinetic energy

Unit kg (m s )

kg m s

mv−

=

=

=

11. (a) 1 1

21k 2

212

5

750 kg (estimate)

60 km h 16.7 m s

750 (16.7)

1 10 J (approx.)

m

v

E mv

− −

=

= =

=

= × ×

= ×

(b) 1 1

21k 2

212

1

0.1 kg (estimate)

100 km h 28 m s (estimate)

0.1 (28)

4 10 J (approx.)

m

v

E mv

− −

=

= =

=

= × ×

= ×

12. 2 21 1

2 22

12

2000.058 03.6

90 J

kW E

mv mv

= Δ

= −

⎛ ⎞= × × −⎜ ⎟⎝ ⎠

=

13. (a) W = Fx cos θ W = 8.0 × 2.5 × cos 20° W = 19 J (18.79 J) (b) Work done by net force = ΔEk Work done by net force = Ek (since initial Ek = 0) ⇒ Fnet x = Ek ⇒ Ek = (8 cos 20° − 7.2) × 2.5 = 0.79 J (0.794 J) (c) Zero. There is no component of displacement in the

direction of the normal reaction force. 14. (a)

gp

3

70 kg (estimate)2.5 m (estimate)

70 10 2.5

2 10 J (approx.)

mh

E mg h

=Δ =

Δ = Δ

= × ×

= ×

(b)

gp

30 kg (estimate)2 m (estimate)

30 10 2600 J (approx.)

mh

E mg h

=Δ =

Δ = Δ

= × ×=

(c)

gp

70 kg (estimate)0.75 m (estimate)

70 10 0.75500 J (approx.)

mh

E mg h

=Δ =

Δ = Δ

= × ×=

15. m = 20 kg Δh = 1.0 m (a) gp

20 10 1.0200 J

F mg hΔ = Δ

= × ×=

(b)

20 10 1.0200 J

W Fxmgx

=== × ×=

(c)

To push the crate up the ramp with a constant speed,

the applied force must be mg sin θ. W = Fx

but 1.0sin

1.0sin

=

⇒ =

θx

xθ

1.0sinsin

20 10 1.0200 J

⇒ = ×

= × ×=

W mg θθ

(d) It is better to use the ramp. Although the amount of work needed to move the crate is the same, the force that needs to be applied to the crate is less if the ramp is used.

16. So that as little of their kinetic energy as possible is transferred to gravitational potential energy. Subsequently, a greater proportion of their kinetic energy is available to cover the horizontal distance as fast as possible.

17. (a) ΔEk = ΔEgp (magnitude only) ⇒Ek = mgΔh (since initial Ek = 0) = 0.160 × 10 × 2.0 = 3.2 J (b) 32% of 3.2 J = 0.32 × 3.2 J 32% of 3.2 J = 1.0 J (1.024 J) (c) Assuming 100% recovery of stored energy ΔEgp = mgΔh ⇒ 1.024 = 0.160 × 10 × Δh


1.024

0.160 100.64 m

h⇒ Δ =×

=

18. (a) Spring X: k200 0.2040 N

F x= Δ= ×=

Spring Y: k100 0.2020 N

F x= Δ= ×=

(b) Spring X: strain potential energy ( )

( )

2

2

1 k21 200 0.2024.0 J

= Δ

= × ×

=

x

Spring Y: strain potential energy ( )

( )

2

2

1 k21 100 0.2022.0 J

= Δ

= × ×

=

x

19. (a) ΔEk = ΔEgp (magnitude only) = mgΔh = 1.2 × 10 × 20 = 240 J (b) Ek = 240 J (since initial Ek is zero)

212

2

240

240 21.2

mv

v

⇒ =

×⇒ =

⇒ v = 20 m s−1

20. (a) m = 1500 kg, u = 50 km h−1 = 13.9 m s−1

v = 0, x = 0.60 m Work done by average net force = ΔEk ⇒ Fnet x = Ek (initial)

212

net

5

1500 (13.9)0.60

2.4 10 N

F× ×

⇒ =

= ×

(b) Fnet (av) = maav

5

av

2 2

2.4 101500

1.6 10 m s

a

−

×⇒ =

= ×

(c) Work done by average net force = ΔEk ⇒ Fnet x = Ek (initial)

212

net

6

1500 (13.9)0.10

1.449 10 N

F× ×

⇒ =

= ×

netav

61.449 10 N1500

Fam

=

×=

2 29.7 10 m s−= ×

(d) The kinetic energy of the car is transformed into potential energy of the materials in the crumple zone which undergo a permanent change in shape. This leaves a smaller amount of kinetic energy to be transferred to the passengers.

(e) One could argue that a large car is safer. For a given force applied by an obstacle or another vehicle, the deceleration of a large car is less than that of a small car. Therefore, the deceleration of the occupants inside is less.

For example, consider a car of mass 1500 kg coming to rest from 20 m s−1 when a concrete wall applies a force of 48 000 N to the car.

net

2

48000 N1500 kg

32 m s

Fam

−

=

= −

= −

The deceleration of an occupant with a correctly fitted seatbelt would be 32 m s−2. Consider a car of mass 1200 kg coming to rest from the same speed when the same force is applied by the wall.

net

2

48000 N1200 kg

40 m s

Fam

−

=

= −

= −

The deceleration of an occupant with a correctly fitted seatbelt would be 40 m s−2. Without seatbelts, an occupant would strike the interior of a larger car with a smaller relative speed.

Of course, these arguments are not very strong because there are so many other variables related to car design and the nature of the rigid barrier that affect the deceleration of a car.

21. (a) W = Fx W = 270 × 5 W = 1.4 × 103 J (1350 J) (b) ΔEgp = mgΔh ΔEgp = 30 × 10 × 5 sin 30° ΔEgp = 750 J (c)

Forces acting on the trolley W = Fx = mg × 5.0 sin 30°


ΔEgp = work done against the force of gravity = 750 J (d) Fnet = 270 − 300 sin 30° − 120 Fnet = 0 Work done by net force = Fnet x = 0 (e) Work done by net force = ΔEk ⇒ ΔEk = 0 ⇒ v = 0.50 m s−1

22. (a) ΔEk = work done by net force ⇒ Ek = Fx (since net force is constant and initial kinetic energy is zero) = 240 N × 8.0 m = 1.9 × 103 J (b) Fnet = mg sin 30° + friction + air resistance ⇒ 240 = 50 × 10 × sin 30° + friction + air resistance ⇒ friction + air resistance = 240 − 250 = −10 N The sum of the friction force and air resistance is

10 N up the slope. (c) ΔEk = work done by net force = area under net force vs distance graph

1 12 2

k

240 8 + (240 + 120) 8 + 120 4

1920 + 1440 + 2403600 J (since initial kinetic energy is zero)

= × × × × ×

=⇒ =E

(d) ΔEgp = mgΔh = 50 × 10 × 20 sin 30° = 5000 J (e) Some of the gravitational potential energy is

transformed into thermal energy and sound, due to the frictional force and air resistance.

23. (a) 21k 2

212

4

450 (12)

3.2 10 J (32400 J)

E mv=

= × ×

= ×

(b) k gp

5

At B, 32400 + (magnitude)

32400 +32400 + 450 10 20

1.224 10 J

E E

mg h

= Δ

= Δ= × ×

= ×

2 512

5

1.224 10

2 1.224 10450

⇒ = ×

× ×⇒ =

mv

v

= 23 m s−1

k gp

4

At C, 32400 +

32400 +32400 + 450 10 12

8.64 10 J

E E

mg h

= Δ

= Δ= × ×

= ×

2 412

4

8.64 10

2 8.64 10450

⇒ = ×

× ×⇒ =

mv

v

= 20 m s−1 (19.6 m s−1)

(c) At D, Ek = 1.224 × 105 J Maximum height is achieved when Ek = 0 ⇒ ΔEk = 1.224 × 105

⇒ mgΔh = 1.224 × 105

51.224 10

450 10h ×

⇒ Δ =×

⇒ h = 27 m (since initial height is zero) 24. (a) W = area under F versus x graph W = 8.9 × 105 J (b) W = average force opposing motion × x W = 360 × 1000 W = 3.6 × 105 J (c) ΔEk = work done by net force = 8.9 × 105 J + 3.6 × 105 J = 5.3 × 105 J ⇒ Ek = 5.3 × 105 J since initial kinetic energy is zero 2 51

2

5

5.3 10

2 5.3 101200

mv

v

⇒ = ×

× ×⇒ =

= 30 m s−1

25. (a) W = area under load vs displacement graph This area can be estimated by dividing the area into a

number of triangles, trapezia and rectangles. The figure overleaf shows one way in which this can be done.

The unit of area is J since kN × mm = N m = J 1

2 15 35 + 15 (51 35)

15 + 20 20 + 2319 + 20 + 22 372 2

W = × × × +

⎛ ⎞ ⎛ ⎞+ ×⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2080 J2.1 kJ

==


(b) 3

3

2.1 10 J

2.1 101400 100.15 m (approx.)

mg h

h

Δ = ×

×Δ =×

=

26. When Δx = 0.10 m

Strain potential energy ( )

( )

2

2

1 k21 50 0.1020.25J

x= Δ

= × ×

=

At natural length

2

0.25 J1 m 0.252

=

=

kE

v

2

1

1 0.50 0.252

2 0.250.50

1.0 m s−

⇒ × =

×⇒ =

=

v

v

27. (a) The force applied by the spring is proportional to its compression.

(b)

4 1

kk gradient of graph

15000.0801.9 10 N m

F x

−

= Δ=

=

= ×

(c) Work done as the spring expands from maximum compression is equal to the area under the graph.

12 0.080 1500

60 J

W⇒ = × ×

=

(d) Ek gained = work done by spring Ek gained = 60 J (e) Egp gained = Ek lost = 60 J 60

6030 10

mg h

h

⇒ Δ =

⇒ Δ =×

= 0.20 m 28. As the child falls through the air from maximum height,

gravitational potential energy is transformed into kinetic energy.

After the child touches the trampoline after falling through the air, kinetic energy and gravitational potential energy are transformed into strain potential energy until the trampoline is at maximum extension.

After maximum extension, strain potential energy is transformed into gravitational potential energy and kinetic energy until contact is lost with the trampoline.

After contact is lost, kinetic energy is transformed into gravitational potential energy until maximum height is achieved.

29.

The motion of the centre of mass of a springboard diver The energy transformations include the following. A–B Chemical energy is transformed into strain potential

energy of muscles, tendons and ligaments. Strain potential energy of muscles is transformed into kinetic energy which is then transformed into gravitational potential energy as the diver rises for the first time. Gravitational potential energy is transformed into kinetic energy as the diver descends to the end of the springboard.

B–C Kinetic energy and some gravitational potential energy is transformed into strain potential energy of the springboard until the springboard reaches its maximum deflection. The strain potential energy is transformed into kinetic energy and gravitational potential energy until the diver loses contact with the springboard. Kinetic energy is transformed into gravitational potential energy until the diver reaches maximum height.

C–D Gravitational potential energy of the diver is transformed into kinetic energy until the diver strikes the water.

D–E Kinetic energy of the diver is transferred to the water (as kinetic energy) and eventually transformed into thermal energy of the water particles. Some of the diver’s kinetic energy is transformed into sound energy.

30. The figure below shows the motion of the centre of mass of the skateboard.

The motion of a skateboard rider completing an ‘ollie’ (a) The energy transformations can be displayed with a

graph of energy vs time or energy vs position. A sample graph is shown on the next page.


Graph showing energy transformations during an

‘ollie’ As the rider moves down and up the slope,

gravitational potential energy is transformed to kinetic energy and back again to gravitational kinetic energy. However, the total mechanical energy is not quite conserved and the rider needs to provide some additional energy ‘input’ to reach the top of the slope and point C. Further energy input is needed from the rider in order to gain the gravitational potential energy required at point D. Gravitational potential energy is then transformed into kinetic energy as the rider returns to point F and transformed into gravitational potential energy at point G. At points A, D and G the rider’s kinetic energy is zero.

(b) Between points C and D, the skateboard and the rider are in free fall. They are both travelling at the same speed at point C. Even though they have different amounts of kinetic energy due to their different masses, and therefore gain different amounts of gravitational potential energy, they reach the same height. That is,

21

22 2

mv mg h

v g h

= Δ

⇒ = Δ

2

2vhg

⇒ Δ =

The horizontal components of the speed of both the rider and the skateboard are also the same, as long as air resistance is negligible. The rider therefore needs to make little effort to remain in contact with the skateboard. There is some skill involved in ensuring that the frictional forces made possible by the contact are used to turn the skateboard so that it lands on the ramp before the feet or any other part of the rider’s body.

31.

k

212

212

3

4

0

2000.0583.6

4.0 10

2.2 10 W

WPtEt

mvt

−

=ΔΔ=Δ

−=

Δ

⎛ ⎞× × ⎜ ⎟⎝ ⎠=

×

= ×

32.

4.0 10 1.51.2

50 W

WPt

mg ht

=Δ

Δ=Δ

× ×=

=

33. (a)

3

180 10 1.8

3.2 10 J (3240 J)

W Fxmg h

== Δ= × ×

= ×

(b)

32403.0

1.1 kW (1080 W)

WPt

=Δ

=

=

(c) None, since the barbell undergoes no displacement in the direction in which Stefan is applying the force.

34. Force applied to car due to the engine = 570 N + 150 N = 720 N P = Fv at constant speed = 720 × 20 P = 14 400 J s−1

= 14 kW 35. v = 2.0 m s−1

stride length = 1.0 m Man takes 2 strides per second. ΔEgp = 60 × 10 × 0.030 m in each stride = 18 J per stride ΔEgp in 1 second = 36 J

gp

36 J1s

36 W

EP

tΔ

=Δ

=

=

36. (a) A constant speed is achieved when the force applied to the bicycle to overcome friction and air resistance is 6.5 N + 5.7 N = 12.2 N.

P = Fv ⇒ 56 = 12.2v ⇒ v = 4.6 m s−1 (4.59 m s−1) (b) On a slope, the force applied to the bicycle to achieve

a constant speed is greater. (See figure on next page.)


The applied force must balance the frictional force

and air resistance (a total of 12.2 N) and the component of the weight of the bicycle and its rider down the slope.

⇒ Fapplied = 12.2 + mg sin 30° = 12.2 + 750 sin 30° = 387.2 N P = Fv = 387.2 × 4.59 = 1777 W Additional power required = 1777 W − 56 W = 1.7 × 103 W


Unit

2 ■■■■■■■■■

Area of study 2: Wave-like properties of light


Chapter 8

Reflecting light 1.

2. Light rays from the light source strike the object and are

reflected in all directions. Some of the rays enter my eye. 3. (a) Light rays from the Sun reflected off the flower,

travelled through the transparent glass window and entered my eye.

(b) The points on the TV screen emitted coloured light into the room in all directions. The rays entered my eye.

4. At night, people from across the globe can see the Moon, so the light from the Sun is reflected in all directions by the rough surface of the Moon. Some part of the Moon can be seen at all positions of its revolution about the Earth, except during an eclipse.

5.

6. a = 90° − 50° = 40° b = a = 40° c = 90° − b = 90° − 40° = 50° 7. In each diagram, trace the emerging rays back to find

where they appear to be coming from. The image is where the dotted lines meet. The image is the same distance behind the mirror as the object is in front.

8. The image is laterally inverted: TOYOTA. 9. A camera captures an image in a similar way to the eye.

Light rays coming from the object enter the camera lens. They appear to come from the virtual image. Therefore, if it can be seen, it can be photographed.

10. In the first case, each of the emerging rays is parallel to its corresponding incident ray. In the second case, the ray which is parallel to the angled mirror emerges along the same path from which it came.

11.

12. Your image is as far behind the mirror as you are in front

of it. If you move one metre in one second, then so does your image. Image speed is 1.0 m/s. In that one second, the distance between you and your image decreases by two metres, so the speed of approach is 2.0 m/s.

13. (a) 0.80 m (b) 0.80 m (c) 0.80 m, 0.75 m (d) Unchanged

The answers are the same regardless of how far away you

are from the mirror. 14. (a) Focus the camera lens at 4.0 metres away because

your image is 4.0 metres from the camera. (b)

The camera, the friend and the friend’s image form a

right-angled triangle, so Pythagoras’s theorem can be used to determine the distance from the camera to the image. The distance needs to be set at 4.1 m.


15. Use two plane mirrors with the text upside down on the patient’s chest. The patient looks in one mirror at the image of the text in the other mirror.

16. The size of the image is the same regardless of the distance of the object from the mirror.

17. Place the mirror on the floor under your feet. 18. Some light is reflected off the surface of the drop and into

your eye. 19. Seven images. One pair of mirrors produces three images

of the object as shown in the following figure. The third mirror produces an image of the object and an

image of each of the three images produced by the first pair of mirrors.

20. The table can be completed using the ray diagrams shown below it.

Location of object

Location of image Magnification Nature Orientation

beyond C between C and F

< 1 real inverted

at C at C = 1 real inverted between C and F

beyond C

> 1 real inverted

at F no image nil nil nil inside F behind

mirror > 1 virtual upright

21. Using ray diagrams like those shown in the answer to the

previous question, drawn to scale, the location, size, nature and orientation can be found.

22. (a)

(b) Searchlight, torch or headlamp 23. If the top half of a concave mirror is removed, then the

location, nature, orientation and size of the image are unchanged. The image is still the full image of the object, but the brightness is less because it is formed with half the light.


Chapter 9

Refracting light 1. sin 40sin , 29

1.33sin 50sin , 351.33

°= = °

°= = °

θ θ

θ θ

Angle of refraction increases by 6° when the angle of incidence increases by 10°.

2. sin 55sin 55 sin 33 ,sin 33

1.5

n n

n

°° = ° =°

=

3. sin 65sin , 361.55

°= = °θ θ

The block is rectangular, so the opposite sides are

parallel. Therefore, the angle of incidence at the bottom face is the same as the angle of refraction at the top face. Because Snell’s Law applies regardless of the direction of the light ray, the angle of refraction as the light emerges will be the same as the angle with which it entered the block. This means the ray is parallel to the incoming ray, but shifted sideways.

4. (a) The different layers are parallel to each other, so the angle of refraction at the top surface of a liquid will equal the angle of incidence at the bottom surface. This means that:

nair × sin θair = nacetone × sin θacetone = nglycerol × sin θglycerol = ncarbon tet × sin θcarbon tet nair = 1.0, θair = 25°, θacetone = 18.1°, θglycerol = 16.7°, θcarbon tet = 16.8° ⇒ 1.0 × sin 25° = 1.357 × sin θacetone = 1.4746 × sin θglycerol = 1.4601 × sin θcarbon tet ⇒ 1.4601 × sin θcarbon tet = 1.53 × sin θglass All expressions equal each other, so the angles of

refraction will equal 1.0 × sin 25° / refractive index of the medium. So, for acetone the angle is 18°, for glycerol: 16.7°, carbon tetrachloride: 16.8°, and glass: 16.0°. Because the layers are parallel, the light ray will emerge from the glass into the air at the angle it left the air: 25°.

(b) When the light reflects off the mirror, the angle of reflection will equal the angle of incidence, so the light ray will leave the bottom medium at the same angle

that it entered it. The light path up through the layers will be the reverse of the path coming down through the layers. All the angles will be the same.

5. Find the critical angle. c

c

c

1.33 sin sin 90 11sin 0.75

1.3348.59 49

× = ° =

⇒ = =

⇒ = = °

θ

θ

θ

The angle of incidence of 55° is greater than the critical angle, so the light ray will be totally internally reflected and will reflect off the water surface with an angle of reflection of 55°.

6. (a) The wavelet produced at the surface projects into the air as well as into the glass.

(b) The surface cannot both attract and repel the particles at the same time.

7. (a) 1.0 × sin 35° = 1.55 × sin θ

sin 35sin 0.37001.55

°= =θ

θ = 22°. The angle of incidence at the bottom surface is 22° and the angle of refraction is 35°.

(b)

tan 22 cm

5.05.0 tan 22 2.0 cm

d

d

° =

⇒ = × ° =

(c) Angle of incidence at adjacent face = 90 − 22 = 68°. The critical angle is given by

c1sin

1.55=θ

⇒ θc = 40°, so the light ray undergoes total internal reflection.

(d)

Because of symmetry the ray will meet the bottom

surface at an angle of incidence of 22°, so the angle of refraction will be 35°.


(e) If the initial angle was less than 35°, then the ray would meet the adjacent face at a more glancing angle and so be totally reflected. The largest value for the initial angle is 90°. This gives an angle of refraction of 40° and so an angle of incidence of 50° at the adjacent face, which is greater than 40° and is totally reflected. So all rays, regardless of angles, will be totally reflected.

(f) Total internal reflection at the adjacent face does depend on the refractive index. Any values less than 1.414 will allow the light to pass out the adjacent face for some angles. The value 1.414 is the square root of 2, which is sin 45°, which is the case when the angle of refraction at the top face equals the angle of incidence at the adjacent face.

8. (i) (f) (ii) (d) (iii) (b) (iv) (c) (v) (g) (vi) (a) 9. (a) Your refractive index must be the same as the air, the

medium the body is in. Otherwise, images of objects behind you would appear to be distorted.

(b) The light entering the eye is absorbed by the eye, so a person looking in the direction of a so-called invisible person would see a black screen the shape of the retina obscuring the view in front. So to be truly invisible the retina would not be able to absorb light. That is, a truly invisible person would not be able to see.

10. 1.00 × sin 65° = 1.55 × sin θ Angle of refraction = 36° ⇒ Angle of deviation = 65 − 36 = 19° 11.

The angle of refraction is given by sin 30° = 1.4 × sin θ ⇒ θ = 20.9°. Length of the path ray in the glass

5.0cos 20.9

=°

Sideways deflection = length of light path × sin (30.0 − 20.9) = 0.85 cm 12.

The angle of refraction is obtained from sin 40° = 1.5 × sin θ ⇒ θ = 25.4°.

The angle where the normals meet equals 120°, so the angle of incidence at the exit face equals

180 − 120 − 25.4 = 34.6°. The angle of refraction will therefore be 58.4°. The angle of deviation will equal: (angle of incidence − angle of refraction at the first face)

+ (angle of refraction − angle of incidence at the second face), which equals

(40° − 25.4°) + (58.4° − 34.6°) = 34.4°. 13. sin 30° = 1.5 × sin θ ⇒ θ = 19.5° When the ray meets the opposite side of the glass sphere,

the normal at this point passes through the centre of the sphere (i.e. it is a radius). This means that the triangle formed by the points on the glass sphere and the centre of the sphere is an isosceles triangle and that the angle of incidence at the opposite side of the sphere is 19.5°. This situation is the reverse of the entry into the sphere, so the angle of refraction is 30°.

14. c

c

c

2.5 sin sin 90 11sin 0.4

2.50.4 24

× = ° =

= =

= °

θ

θ

θ

15. (a) The critical angle is 45°, so the refractive index

1 1.4.sin 45

= =°

For total internal reflection, n × sin θc ≥ 1. If n < 1.4, then n × sin θc < 1, so the ray will refract

out of the glass. ⇒ n = 1.4 is the minimum value for total internal

reflection to occur for an angle of incidence of 45°. (b) The positions of the rays are inverted. This device is

used twice in binoculars to correct the normally inverted image produced by a telescope.

16. 1.500 × sin 82.0° = n × sin 90° ⇒ n = 1.500 × sin 82.0° = 1.49 17. Above, the diver would see a circle in which was

compressed a 180° view of the world above the water. Towards the edge of the circle, the image would look distorted. Outside the circle, the diver would see the reflected, inverted image of the world beneath the surface of the water due to the total internal reflection of the light rays from beneath the water as they meet the water−air boundary.


18. Critical angle for light travelling from glass into water is given by:

1.55 × sin θc = 1.33 × sin 90°

c1.33sin 0.8581.55

θ = =

⇒ θc = 59.1°. The angle of incidence at the glass−water boundary is 45°, which is less than the critical angle, so the ray is refracted into the water. The angle of refraction, θw, is given by: 1.55 × sin 45° = 1.33 × sin θw ⇒ θw = 55°

19.

At critical angle 1.33 × sin θc = 1.0 × sin 90° ⇒ θc = 48.8°

tan θc = 40r , so r = 40 × tan 48.8°

= 46 cm 20. (a) Path length of reflected ray

6

2cos 82

0.5 1 102cos 82

r

−

= ×°

× ×= ×°

= 7.18 × 10−6 m Path length of straight ray = 2 × r tan 82° = 2 × 0.5 × 1 × 10−6 × tan 82° = 7.11 × 10−6 m Path length difference is 7.18 × 10−6 − 7.11 × 10−6 = 7.0 × 10−8 m

(b) Speed of light in the glass cn

=

c

1sin

1sin 82

θn

n

=

⇒ =°

⇒ Speed of light in the glass

88 1

816

8

3.0 10 3.0 10 m ssin 82

distanceTime differencespeed

7.0 10 2.3 10 s3.0 10

−

−−

×= = ×°

=

×= = ××

This time is small, but in an optical fibre of 1.00 km,

there would be 3

61.00 107.11 10−

××

reflections, equals

1.4 × 108 reflections, corresponding to a total time difference of 3.28 × 10−8 s. This would limit the upper frequency of light pulses sent down the optical fibre. Too high a frequency and the pulses would overlap. The problem can be overcome by using a narrower optical fibre or using an optical fibre whose refractive index gets smaller the greater the distance from the centre.

21. (a) A B C D

1 Angle of Refractive Refractive Angle of

2 incidence index index refraction

3 89 1.0003 1.00028 89.06794

4 89.06794 1.00028 1.00026 89.14124

5 89.14124 1.00026 1.00024 89.22141

6 89.22141 1.00024 1.00022 89.31085

7 89.31085 1.00022 1.0002 89.41379

8 89.41379 1.0002 1.00018 89.53917

9 89.53917 1.00018 1.00016 89.71526

10 89.71526 1.00016 1.00014 #NUM!

(b) In the seventh layer, the light will be totally internally

reflected. With the refractive indices the same, if the initial angle of incidence was 88° the ray would not be reflected. If the initial angle was 89.5°, the reflection would occur in the first layer. If the rate of change in the refractive index was halved to 0.000 01, then the reflection would occur in the 15th layer.

22. Light bends away from the normal, so the refractive index gets smaller with height. This means the temperature rises with height.


23. (a)

The image is real and inverted and located 60 cm

from the lens on the other side from the object. It is about 12 cm high.

(b)

The image is virtual and upright and located 60 cm

from the lens on the same side as the object. It is 18 mm high.

(c)

The image is real and inverted and located 10.5 cm

from the lens on the other side from the object. It is about 2.5 mm high.

24. ‘Accommodation mechanism’ is the strategy that an animal’s eye uses to keep an image sharp and in focus on the retina whether the object is near or distant. There are several different mechanisms. The thickening of the human lens to achieve a sharp image of nearby objects on the retina is an example. Another is the movement of a fixed focal length lens closer to the retina as the object moves further away.

25. (a) You would see an inverted image of the teacher. (b) The trees are further away, so their image is closer to

the focus of the lens; so move the screen closer to the lens.

26. For objective lens u = 5.2 mm, f = 5.0 mm

1 1 1

1 1 15.0 5.2130 mm

u v f

vv

+ =

⇒ = −

⇒ =

This is 20 mm in front of the eyepiece lens.

i

o

1305.2

H vMH u

= =

=

For eyepiece lens u = 20 mm, f = 40 mm

1 1 120 4040 mm

2

vvvMu

= −

⇒ =

= =

130Total magnification 25.2

50

= ×

=

The same result can be obtained using a ray-tracing diagram.

27. (a) u = 4.0 cm, f = 5.0 cm, v = ?

1 15 4

1 20

20Magnification 54

v

vu

= = −−

= = =

The image is located 20 cm behind the lens and is 5 times as big.

(b) u = 3.0 cm, f = 5.0 cm, v = ?

1 15 7.51 1 25 3

7.5Magnification 2.53

v = = − = −−

= =

The image is located 7.5 cm behind the lens and is 2.5 times as big.

28. (a) v = 400 cm, f = 5.00 cm, u = ?

1 400 5.061 1 795 400

u = = =−

The slide is located 5.06 cm from the lens.

(b) Magnification 79vu

= =

The size of each side of the square image is 79 × 35 = 2765 = 2.76 m.

(c) The image is inverted and will be an ‘L’, but upside down and back to front.

(d) A clear image needs to be formed closer to the lens, so the object, that is, the slide, needs to be moved further away from the lens.

29. The slide projector needs to be moved back and the lens moved closer to the slide.


Chapter 10

Seeing colours 1. Refraction: bending of light as it passes from one material

into another. Reflection: light leaves a surface at the same angle it

approached the surface. Dispersion: the spreading of light into colours as it is

refracted. Spectrum: the range of colours into which light can be

broken. Refractive index: a measure of how much a material

slows light down and therefore changes its direction. Chromatic aberration: circumstances, such as the use of

telescopes and cameras, where dispersion distorts the image.

2. (a)

The emerging coloured rays are parallel to each

other. (b)

In a triangle, the rays emerge at different angles and

spread further apart the further they travel, so that on a distant wall the colours are seen separately. With the rectangle, the colours stay the same distance apart regardless of how far they travel.

3. Red light, as it has a lower refractive index

8

r1

8

b8 1

3.00 10Speed in red light1.514

1.98151 108 m s

3.00 10Speed in violet light1.528

1.963 35 10 m s

cn

cn

−

−

×= =

= ×

×= =

= ×

Difference = 1.82 × 106 m s−1

4. Violet light is bent more as it has a greater refractive index.

5.

6.

7. Because the violet light is bent more than the red at both

the front and back surfaces of the lens, the violet light will come to a focus closer to the lens and so have a shorter focal length.

8. (a) In the morning the Sun is in the eastern sky. To see a rainbow, the Sun needs to be behind you so that light that is totally internally reflected from the raindrop enters your eye. You would need to look to the west. At midday it would be very difficult to see a rainbow as the Sun needs to be behind you, so you need to look down, preferably from a plane or a balloon.

(b) In the northern hemisphere the directions are the same as in the southern hemisphere.

9. The Sun would be behind you on the other side of the plane. The rainbow will still be circular, but because you are in the air, you could see a full circle because there may be water droplets below.

10. A periodic wave is a pulse that is repeated at regular intervals.

11. (a) 5 drops in 10 s = > 1 drop in 2 s: T = 2 s

(b) 112

1 s 0.5 HzfT

−= = =

12. 1514

15

1 1 2.08 104.8 10

2.1 10 s

Tf

−

−

= = = ××

= ×

13. f = 6.5 × 1014

(a) 8

87

14

3 10

3 10 4.6 10 m 460 nm6.5 10

v

vf

λ −

= ×

×= = = × =×

(b) 8

87

14

2.0 10

2 10 3.1 10 m 310 nm6.5 10

v

λ −

= ×

×= = × =×

14. For red light: sin θ = nr sin 30° = 1.4742 × 0.5 ⇒ angle of refraction for the red light is 19.8° For blue light: sin θ − nb sin 30° = 1.4810 × 0.5 ⇒ angle of refraction for the blue light is 19.7° The difference is 0.1°.


15. For red light:

cd

c

1 1sin2.40

24.6

θn

θ

= =

⇒ = °

The critical angle for red light is 24.6°. For blue light:

cd

c

1 1sin2.24

24.2

θn

θ

= =

⇒ = °

The critical angle for blue light is 24.2°. 16.

(a) For red light:

sin 45 sin 45sin1.514

27.8

tan 27.85.0

2.64 cm

tan 455.05.00 cm

5.00 2.64 2.36 cm

θn

θa

aa d

a dd

° °= =

⇒ = °

° =

⇒ =+° =

⇒ + =⇒ = − =

For deep blue light:

sin 45 sin 45sin1.528

27.6

tan 27.65.0

2.61 cm

tan 455.05.00 cm

5.00 2.61 2.39 cm

θn

θa

aa d

a dd

° °= =

⇒ = °

° =

⇒ =+° =

⇒ + =⇒ = − =

Red light is shifted sideways by 2.36 cm, and deep blue light by 2.39 cm.

(b) Deep blue light is shifted more, by 0.03 cm. (c) Each colour is shifted more as the angle of incidence

increases for all angles. By calculating the shifts for different colours for various angles of incidence, you can show that the difference between the shifts for a given pair of colours increases for angles of incidence up to about 57°, then decreases.

17. (a) When the Sun is about 40° above the horizon, a small part of the rainbow circle will be above the ground.

(b) When the Sun is on the horizon, half the circle of the rainbow will be above the ground.

18. The rainbow is produced by reflection only. As the light ray enters the raindrop, it is refracted towards the normal. Because a raindrop is approximately spherical, this normal passes through the centre of the raindrop; that is, it is a radius of the sphere. When the refracted ray meets the other side of the drop, the normal at this point will also be a radius. This means that the triangle formed by the centre of the drop and the two points on the surface that the light ray passes through form an isosceles triangle with the two angles at the surface the same. Since light can travel along a light path in either direction, the ray will be refracted out of the drop with an angle of refraction equal to the initial angle of incidence. Only some of the light entering the drops is reflected, but sufficient to see a rainbow, particularly against a dark background.

19. (a)

(b) The order of the colours should be reversed. The

rainbow should be higher in the sky and probably the colours should be further apart due to the greater path for divergence of the colours.

20. If light behaved like a longitudinal wave, it could not be restricted to travelling only in a plane that was not parallel to its direction of propagation.


21. (a) Filters B and C (b) Filters A+B only slightly (c) Red−orange (d) Black — no colour passes through all three filters.

22. (a) 8

r

8 1

8

v

8 1

3.00 101.514

1.98 10 m s

3.00 101.532

1.96 10 m s

v

v

−

−

×=

= ×

×=

= ×

(b) sin 45For red light: sin 27.81.514

θθ °= ⇒ = °

sin 45For violet light: 27.51.532

θθ °= ⇒ = °

For red light, the length of the path, R, through the glass is given by:

10 10cos 27.8 11.31 cmcos 27.8

RR

° = ⇒ = =°

For violet light, the length of the path, V, through the glass is given by:

10 10cos 27.5 11.27 cmcos 27.5

RV

° = ⇒ = =°

(c) Time to travel across the block is the distance the light travels divided by its speed. The red light emerges first in 0.0571 microseconds, while the violet light emerges in 0.0576 microseconds.

(d)

The rays will be parallel to each other and to the initial incident ray.

(e) The red ray ends up ahead even though it travels a greater distance in the glass.

23. Newton carried out this second stage of the experiment to see if the prism would further break up the red light.


Unit

2 ■■■■■■■■■

Detailed study: Astronomy


Chapter 11

Our view of the sky 1. (a) Altitude is the angle that the object makes with a

right angle from the horizon. (b) Azimuth is the angle clockwise from north around

the horizon. (c) Declination is the angle measured north or south

from the celestial equator. (d) Right ascension is the number of hours east of the

vernal equinox. 2. The azimuth is 90 degrees. 3. She sees the meteor rising from near the horizon. The

horizon has an altitude of 0 degrees and the zenith 90 degrees. A change from 10 to 80 degrees describes a motion from low in the sky to high in the sky.

4. At 7 am the altitude of the Sun is 2 degrees and at 10 am is 37 degrees so the altitude has changed by 35 degrees. The azimuth changes from 94 degrees to 66 degrees (a ruler from the centre of the circle can help here) which is a 28 degree change in azimuth.

5. (a) It must reach the same altitude in summer because the Earth’s alignment with the stars does not change significantly within a year. Over long periods of time the altitude will change due to the precession of the equinoxes.

(b) In winter, when the ecliptic is low in the sky during the day but high in the sky at night

6. (a) 10 am: 71 degrees azimuth, 38 degrees altitude; noon: 40 degrees azimuth, 59 degrees altitude; 6 pm: 265 degrees azimuth, 25 degrees altitude

(b) A (21 h 47 min, −16°) B (21 h 32 min, −5.6°) C (21 h 4 min, −11.3°)

7. They are all potentially visible at some time in the year, but Capella, Deneb and Vega would all be so low in the northern sky that they may not be visible in practice.

8. Procyon will reach its highest point 54 minutes after Sirius but is nearly 22 degrees further north.

9. (a) Formalhaut, with a declination of −29 degrees 45 minutes, is about 8 degrees from the zenith at its maximum altitude.

(b) Betelgeuse reaches about 5 degrees from the zenith at its maximum altitude.

10. (a) Formalhaut has both the RA closest to 0 and declination closest to 38 degrees, so it would be highest in the sky.

(b) Alpha Centuri, Beta Centuri, Alpha Crucis, Beta Crucis, Spica, Arcturus are all high in the Melbourne sky at this time.

11. (a) DEC = 0. Therefore the star is above the equator and will take 6 hours to reach the horizon. It will reach the horizon at 8 pm + 6 hours, which is 2 am. Stars to the north of this star would have set earlier, and stars

to the south later. Stars with declination less than −52 degrees never set in Melbourne.

(b) Yes. To be visible from Melbourne it would need to have a declination of less than 90° either side of −38°. This means that all stars south of DEC 52° are potentially visible from Melbourne. In practice, stars near the northern horizon are not readily visible, but with a declination of −16.7° Sirius is only 21.3° from the zenith when at its highest point in the sky.

(c) 20 minutes. Canopus arrives first because it has the smaller RA.

(d) Canopus as it is closer to the South Celestial Pole. 12. A year is the period during which the Sun moves a full

circle against the background of stars in the sky, returning to the same place against the stars.

13. The Earth’s axis is tilted. This causes the length of the day to be longer in summer than in winter and the Sun to rise higher in the sky in summer than winter.

14. (a) One year (b) One day 15. The daytime would be very long, as would the night.

Stars in the sky on the night side would take a long time to noticeably change position, unlike on Earth where rotation shows obvious changes in star position in under an hour. On Venus such changes would take weeks.

16. Let’s consider that the period of rotation is equal to the period of revolution. In the situation that the rotation is not retrograde (such as with our Moon whose rotation and revolution periods are equal) the same side of Venus would always face the Sun so it would be in perpetual daylight and the other side in perpetual night. The retrograde motion, however, results in 2 days per year. The difference between the sidereal day (period of rotation) of Earth and the Solar day, the time from noon to noon is only a few minutes on Earth due to the large difference between the diurnal and annual periods. With Venus, however, the difference between the solar and sidereal days is months.

Venus shown at eight different places in its orbit of the Sun. The line is a stake in the surface of the planet to show rotation. Although the planet only rotates once in the revolution, there are two periods in the revolution when the stake side is facing the Sun.


17. The seasons would be more extreme. The Sun would rise higher in the sky in summer and not as high in winter. The days would be longer in summer and shorter in winter. Winter would probably be colder and summer hotter as a result.

18. It takes about twelve months for Orion to return to the its high point at midnight. It will take about a quarter of this (three months) before the constellation is setting at midnight.

19. Depending on the time in the Moon’s cycle, it can be seen either in the day or in the night, usually part of both, sometimes neither. Day and night on Earth are due to the Earth’s rotation. Meanwhile, the Moon is revolving around the Earth about once a month.

If we start observing when the Moon is high in the sky at

midnight, one week later the Moon will just be rising at midnight. A week later again and it will be rising near dawn and will be in the sky all day, but invisible as a ‘new’ Moon.

20. Geocentric means that the Earth is at the centre. 21. Ptolemy used epicycles to explain retrograde motion. 22. The geocentric model is consistent with casual

observations of the rising and setting of the Sun and stars. Even with careful observation, the expected parallex of

the stars was not observed. Until Galileo, the laws of physics suggested that the

Moon would not continue to orbit the Earth if it was in motion about the Sun.

Religious beliefs claimed that the Earth did not move. 23. The geocentric model attributed the changing seasons to

changes in the Sun’s orbit of the Earth. This meant that the orbits of the planets also had to change at the same time as the Sun.

24. As the Moon and the Sun revolve around the Earth, half of the Moon is lit up at any time. The phase depends on how much of this lit side is visible from the Earth. The Sun revolves around the Earth once a day but the Moon revolves slightly slower so that the phase from Earth changes.

25. Epicycles were a device that changed the speed of the planets relative to the Earth at different parts of their orbits, for example, for part of Mars’ motion along its epicycle, it was moving in one direction while for the other part it was moving in the other direction. (See following figure.)

26. Heliocentric means that the Sun is at the centre. 27. The Sun is no longer considered to be the centre of the

universe but is itself in motion around the Milky Way galaxy.

28. In a heliocentric model, the motion of the other planets against the background of stars changes with the relative position of the planets, for example:

Earth and Mars moving in same direction. Earth moves

faster than Mars so those on Earth see Mars in retrograde motion.

From this position Mars is seen to be moving ‘forward’ against the background stars.

29. Copernicus made the orbits elliptical, fully eliminating the need for epicycles.

30. The phases of the Moon and due to the relative positions of the Earth, Moon and Sun. The full cycle of phases


occurs as the Moon orbits the Earth (see diagram in text, p. 240).

31. Such a large change in people’s perspective takes a long time. People have had time to accept the change. There have also been many new discoveries that make sense only if the Earth revolves around the Sun, such as Newton’s Law of Gravitation, our new understanding of our position in a vast universe, the Coriolis effect causing weather systems, space travel and so on.

Chapter 12

Telescopes 1. (a) The planets appeared as discs through the telescope,

while the stars were still just points of light. (b) The magnification made the images large enough to

see as discs. The light-gathering power would only have made the planets appear brighter, and the resolution enables smaller features to be observed.

2. Stars are separated by tremendous distances. Even at the speed that stars are travelling it takes a long time for them to significantly change their position in the sky. This can be detected either by very precise mapping of the stars over a long period of time or by shortening the time required to observe the movement by magnifying the distance it moves through using a telescope.

3. (a) Diameter = 15 cm, f = 90 cm

f-numberdiameter90 cm15 cm6

f=

=

=

(b) Brighter because it has a smaller f-number 4. Most of the smaller or more distant bodies are very dim.

The large diameter telescopes collect more light from these objects, making them more visible.

5. They emitted brief, equally spaced pulses of radiation. 6. Galileo observed that Venus had phases consistent with it

revolving around the Sun, Jupiter had moons revolving around it that were not left behind as it moved, the Moon had mountains and valleys like Earth, the Sun had spots, and the Milky Way was a myriad stars. All these data conflicted with the Aristotelian view that the Moon and everything beyond it were different in nature to the Earth. The data were also consistent with a heliocentric solar system, rather than the geocentric one of Aristotle.

7. Kepler published his first two theories of planetary motion and Galileo built his first telescopes and observed the sky with them.

8. Planets were seen as lights in the sky that moved against the background of stars. Galileo showed them to be disks with features on them that were similar in scale and shape to the Earth. Following Galileo’s observations, it became understood that the Earth was a planet.

9. There was some question over whether the Moon would move with the Earth if the Earth was in motion. Either Jupiter or the Earth or both must be in motion, so the observation that Jupiter had moons that moved with it

confirmed that the motion of the moon around the Earth did not mean the Earth must be stationary.

10. Planets, pulsars, black holes, galaxies, nebulae, asteroids, quasars, white dwarfs

11. Enhanced light collection, resolution and magnification 12. A telescope that detects X-ray wavelengths. It would need

to be outside the Earth’s atmosphere. 13. Telescopes aid vision by magnifying and collecting light,

and improving resolution. (a) Magnification makes small objects large enough to

see, collecting light enables dim objects to be seen, and improved resolution enables small details to be distinguished.

(b) Features on planets and moons are only visible because of magnification. For example, the rings of Saturn would not have been visible to the naked eye as they are too small when viewed from Earth.

(c) Planets and bodies in the outer solar system, such as Pluto, that are dim because of their distance from the Sun would not be visible without considerable light collecting power. Asteroids, many comets, distant stars and galaxies all rely on light-collecting power for their discovery.

(d) The moons of Pluto, and any other objects that are located close together, will only be visible as individual objects if the resolution is sufficient to separate them.

14. Pulsars, quasars, et cetera. These objects emit very strongly in the radio frequencies but not as strongly at visible wavelengths.

15. Photographic film, because the longer it is exposed, the fainter the objects it can detect. The eye, once adapted to darkness, does not improve its sensitivity to faint objects with time.

16. Electronic sensors enable computer enhancement, such as colouring particular features, produce an instant image, do not require film removal and development, and are more sensitive to light.

17. The first telescopes suffered from chromatic aberration, where images were surrounded by coloured fringes, and spherical aberration, where images are blurred because spherical lenses do not have a single focal point. Chromatic aberration is solved using achromatic lenses with more than one type of glass, or by using mirrors instead of lenses. Spherical aberration is solved by using parabolic mirrors and lenses.

18. Can view the sky at wavelengths not visible to the human eye, can be used as radar to measure distances and map the surfaces of planets, can see through gas clouds that absorb light

19. One with a larger diameter objective, assuming that the optics of both telescopes are equal

20. (a) Diameter of star disc = 1.5 minutes of arc. The Dawes limit is equal to the radius of the image of the star

⇒ Dawes limit = 0.75 minutes of arc. (b) Yes. The separation of the two stars is greater than

the Dawes limit, so they will be resolved as two stars


in this telescope. Although the images are resolved, they will overlap.

21. (a) Equatorial mounts have the ability of tracking objects by adjusting only one axis.

(b) The rotation of the Earth about its axis (c) Move the telescope about its polar axis, anti-

clockwise looking south 22. Objects whose position is not affected by the rotation of

the Earth are easily located looking left and right and up and down. An equatorial mount is difficult to manoeuvre to different targets. Observing the heavens that are rotating about a polar axis is much simpler with the equatorial mount.

23. Galileo allowed light to strike only the central region of the objective lens, so that the aberration was small.

24. (a) The light-gathering power of a telescope is proportional to the area of its objective lens or mirror.

2

2HST

22

KECK

A 4.52 m4

A 78.54 m4

πd

πd

= =

= =

The Keck telescope therefore has 78.54 ,4.52

approximately 17 times the light-gathering power. (b) The relative brightness of the images can be

compared by calculating the f-number of each of the telescopes. The lower the f-number, the brighter the image.

58 mf-number (HST) 242.4 m17.5 mf-number (KECK) 1.810 m

fdfd

= = =

= = =

So, Jupiter will be brighter when viewed through Keck.

(c) Hubble is in orbit so it is not affected by the atmosphere, which distorts the light and absorbs some of the radiation, restricting the performance of telescopes on Earth.


Unit

2 ■■■■■■■■■

Detailed study: Astrophysics


Chapter 13

The nature of stars 1. The Sun 2. The spectrum of the Sun resembles the spectra of stars. It

has similar composition to the stars. The distances to many stars can be accurately measured and their luminosities, temperatures, masses and diameters are similar to that of the Sun. The theory of how the Sun works predicts stages in its evolution that other stars are currently displaying.

3. (a) SA = 4πr2

= 4π × (1.5 × 1011)2

= 2.83 × 1023 m2

(b) A = πr2

= π × (6.38 × 106)2

= 1.28 × 1014 m2

Fraction 14

23

–10

1.28 102.83 10

4.52 10

×=×

= ×

(c) Power = 3.86 × 1026 × 4.52 × 10−10

= 1.74 × 1017 W Energy = 1.74 × 1017 × 24 × 60 × 60 = 1.5 × 1022 J (d) 1.5 × 1022 J spread over 5.12 × 1014 m2

∴ (1.5 × 1022) ÷ (5.12 × 1014) J m−2

= 2.9 × 107 J

4. 34

SS 334

E E38 3

3 3

6

(6.955 10 )(6378.1 10 )

1.30 10

rVV r

ππ

=

×=×

= ×

5. The radius of Rigel is 560 times the radius of the Sun. 6. The total mass in all of the major planets plus Pluto is

2.67 × 1027 kg. The mass of the Sun is 1.99 × 1030 kg. The Sun is therefore 1.99 × 1030/2.67 × 1027 = 745 times the mass of all of the planets put together.

7. Atoms found in the Sun’s atmosphere absorb light from the continuous spectrum produced by the dense gas of the Sun beneath. Absorbed light is at the specific frequencies permitted by the electron energy levels of the atoms in the Sun’s atmosphere.

8. Ancient observers thought that the stars were all the same distance from us, fixed on a sphere that revolved around the Earth. The stars were not distant Suns. We now know that the Sun is a star, that the stars are distributed throughout a vast universe and that the Earth is not the centre of this universe.

9. Universe, super cluster of galaxies, cluster of galaxies, galaxy, solar system, Sun, Earth

10. Student-generated timeline 11. By the lines in its spectrum 12. m = 5 is brighter than m = 5.1. 13. Luminosity does not change, it is a property of the star.

Absolute magnitude does not change. Apparent magnitude is reduced.

14. rm = 3.84 × 108

Tm = 27.3d

2 3

24G

rMT

π=

2 8 3

11 2

24

4 (3.84 10 )6.67 10 (27.3 24 60 60)

6.0 10 kg

π−

×=× × × × ×

= ×

15. (a) Sun

30

8 343

3

Sun

water

1.99 10(6.96 10 )

1409 kg m140910001.41

mV

ρ

π

ρρ

−

=

×=× ×

=

=

=

(b) WD

30

5 343

9 3

1.99 10(4.9 10 )

4.0 10 kg m

mV

ρ

π−

=

×=× ×

= ×

(c) The Sun will have less mass that it does now partly due to the mass defect from the billions of years of fusion reactions, but particularly because it will eject large volumes of material into space that will form a planetary nebula.

(d) The same as the density of a neutron, 1017 kg m−3

(e)

3017

343

1.4 1.99 1010

mV

r

ρ

π

=

× ×⇒ =

r = 18 806 m = 1.88 × 104 m 16. Blue stars are very hot, much hotter than yellow stars, so

Beta Centauri is a much hotter star than Alpha Centauri. Beta Centauri must be much further from us than Alpha Centauri, because hot blue stars are massive and have high luminosity.

17. Population II stars formed from clouds of hydrogen and helium and contain very little of any other element. The


heavy elements that make up the Earth are absent, so Earth-like planets would not form around Population II stars.

18. Many of the spectral lines of type O and type A stars are the same, but type A stars have additional spectral lines.

19. Fusion reactions involving fusion of light nuclei, usually hydrogen.

20. Hydrogen and helium 21. (a) P = 3.86 × 1026 J s−1

From 1 reaction E = 26.76 MeV = 26.76 × 1.6 × 10−13 J = 4.2816 × 10−12 J

Number of reactions per second = 3.86 × 1026 ÷ (4.2816 × 10−12) = 9.02 × 1037

(b) 10% of 1057 = 1056

Number of seconds = 1056 ÷ (9.02 × 1037) = 1.11 × 1018 s = 1 × 1018 s = 4 × 1010 years 22. (a) 3 4 7

2 2 4He He + + 1.59 MeVγ+ → Be

(b) 2

13

8 2

–30

c1.59 1.6 10

(3 10 )

2.83 10 kg

Em

−

=

× ×=×

= ×

(c) Some helium is present in the gas cloud that the star formed from; most of the helium would be formed from earlier fusion of hydrogen in the star.

(d) 42 He

23. (a) Nothing else could produce so much energy for so long, the composition of the star is as predicted for fusion, neutrinos are detected in the numbers expected.

(b) Gravity could only sustain the Sun’s energy for about 15 million years. This is not nearly long enough to sustain processes of geological and biological evolution as witnessed on Earth.

24. Hydrogen and helium are very light and easily gain enough energy to escape the Earth, except for the hydrogen that is bound in compounds like water.

25. Student research 26. White dwarf, black dwarf, neutron star, pulsar, black

hole. 27. Hydrogen and helium, because the nebula comes from the

outer layers of stars that have not undergone extensive nuclear fission.

28. Massive stars have shorter lives than small stars because the greater gravitational force increases the rate of fusion in the core. She is not correct.

29. (b) The position of a star on the main sequence is determined by its mass, so assuming its mass is stable, it will not move along the main sequence.

(c) The Sun is a fairly small star that will become a red giant late in its life. It will not become a supergiant because it is a medium-mass star.

(d) Those above the main sequence are fusing elements heavier than hydrogen.

(e) Very massive stars move to the right of the H–R diagram when hydrogen burning ceases in their cores so that they cool.

(f) The bottom of the H–R diagram has stars that are still hot but dim, white dwarfs.

(g) They have the same temperature, but not brightness, so they are not at the same place on the H–R diagram.

(h) It is fusing hydrogen. (i) Blue

30. Look for stars that have motions that can only be explained by a companion with the gravitational pull of a black hole. Look for the high-energy X-rays that are emitted by matter as it falls into a black hole.

31. Student internet research 32. Description should include the following ‘stages’: nebula,

main sequence star, red giant, white dwarf, black dwarf. 33. A galaxy is a collection of many millions of stars that are

bound together by gravity. 34. Astronomers observed objects they called nebulae in the

sky which may have been nearby clouds of gas or distant galaxies of stars. When they were able to measure the distance to these nebulae they found that some were clouds of gas in our own galaxy, but there were also many at tremendous distances, well beyond any of the stars and other features found in our galaxy. Until this discovery astronomers considered that our galaxy might contain all of the matter in the universe.

35. (a) θ = 0.155 d = ?

d = 1θ

= 10.155

= 6.45 pc (b) 1 pc = 3.086 × 1016 m 6.45 pc = 1.99 × 1017 m (c) 1 LY = 9.46 × 1015

distance = 17

151.99 109.46 10

××

= 21 light-years 36. θ = 0.05 arcsec

1

10.0520 pc

dθ

=

=

=

37. The standard method of parallax uses the Earth’s movement around the Sun to provide a baseline. This can be extended by using the motion of the Sun through the galaxy over a number of years. The distance traveled by the Sun in this time is significantly greater than the distance it travels in its orbit around the Sun.

38. The moving cluster method can be used to measure the distance to stars in clusters where the stars are not too


close together, relatively near to Earth and are either moving toward or away from Earth.

39. Cepheid variable stars vary in brightness with a period that is related to its luminosity. This enables the luminosity of a Cepheid variable to be determined and then its distance measured by observing its brightness from Earth.

40. Both stars have the same luminosity but the first is closer, making it appear brighter. As brightness obeys an inverse square law, a ratio of 8 in brightness corresponds to a ratio of √8 in distance, so the second star and its galaxy are √8 = 2.8 times as far away as the first.

Chapter 14

Cosmology 1. That most galaxies were receding and the more distant,

the faster their recession. 2. The expansion of the universe stretches light waves so

that their wavelengths become longer. The longer the wavelength, the more the light is shifted towards the red end of the spectrum.

3. Andromeda is moving towards the Milky Way due to gravity. The light is blue shifted by the Doppler effect, which causes light in front of a moving object to have a shorter wavelength.

4. Hubble’s Law states that the recession speed of galaxies is proportional to their distance from us.

5. The constant has not really changed, only our ability to measure it. Astronomers have greatly improved the accuracy of their measurements of distant galaxies. They have also greatly increased the distance to the furthest galaxies measured, enabling a much more accurate determination of the gradient of the velocity–distance graph.

6.

7. The wavelength of light from a star moving away from us

is stretched because the star is closer to us when it emits the start of the wavelength than when it gets to the end. As all parts of the wavelength of the light leave the star at the speed of light the wavelength is longer than it would be if the star’s distance from Earth was not changing. Distant galaxies emit light that travels the long journey through space to reach us, taking millions, if not billions of years. During this time the universe expands,

stretching the light wave as it travels. The further the galaxy, the longer the light travels and therefore the greater it is stretched.

8. The light from a star or galaxy has a nearly continuous spectrum, however, some gases present in the atmospheres of stars absorb some of the light. The absorption lines that result have specific wavelengths according to the element that absorbed the light. In redshift these lines are noticed to have a longer wavelength that for the gas at rest and for blueshift the lines have a shorter wavelength.

9. The greater the speed that the star is moving away from us, the greater the redshift.

10. Infra-red. The protostars are in clouds of gas that absorb visible light. Some of the infra-red radiation can pass through the clouds of dust.

11. Regions close to the Sun were too hot for the more volatile materials to condense, so when the sun began to shine the Sun’s radiation blew these materials away from the inner solar system. The rocky materials condensed to form the inner planets and asteroids. Further from the Sun, the gases and ice remained, so the planets contain much greater quantities of these gases and ice.

12. Spiral galaxies formed from rotating clouds of gas. Rotating objects tend to flatten at the poles.

13. Solar system, quasar, galaxy, cluster, supercluster, universe

14. (a) Population I, because these stars are younger and therefore contain the elements necessary for life on Earth.

(b) In the spiral arms, because this is where the young stars are. Only old Population II stars are in the halo.

(c) Elliptical galaxies are unlikely to be the home to advanced life forms because they are dominated by Population II stars.

15. Collisions between galaxies and gravitational interaction with neighbouring galaxies may determine the type of galaxy formed.

16. The central core and the halo contain old stars but new stars are always forming in the spiral arms of galaxies.

17. Any motion of gas and dust outside the plane of the galaxy is damped by interaction with other matter as it passes through the plane of the galaxy. This interaction is not sufficient to absorb the momentum of stars. As a result stars that formed before the galaxy collapsed into a disc still orbit the galactic centre on whatever plane of revolution they were on when they formed. Stars that formed in the disc of the galaxy are relatively young and remain in their orbit in the plane of the galaxy.

18. The Earth is made from elements such as iron, aluminium, oxygen and carbon. All of these elements were absent from the hydrogen and helium that was produced in the big bang. It was only stars that formed these elements so the solar system is the product of material that was once made by earlier stars that were already in the Milky Way before the Solar System formed.

19. The stars within galaxies are not static but in motion around the galactic centre. This orbital motion is stable and prevents the star ‘sinking’ to the centre of the galaxy,


in the same way that the motion of the Moon keeps it in orbit about the Earth.

20. Shapley measured the distances to globular clusters and found that they formed a sphere with its centre in the direction of Sagittarius. From that he deduced that the centre of the galaxy was in the direction of Sagittarius.

21. Spiral 22. 1 × 1011 23. Hubble had to show that the nebula was beyond the edge

of the Milky Way. To do this he needed to know where the edge was.

24. A quasar is an object found at a great distance that emits intense radiation across the spectrum. They are believed to be the result of enormous black holes drawing in vast amounts of matter. As the matter is ripped apart on its way into the black hole, it emits radiation which powers the quasar. They are found at the centres of galaxies in the early universe.

25. Student construction 26. Depends on galaxies chosen. 27. This is only true as long as most of the light emitted by

the galaxy is of shorter wavelength than red. However, a galaxy can be so redshifted that most of the light it emits in longer than red and is infra-red.

28. Very distant objects have their light so red-shifted that it is in the infra-red or radio parts of the spectrum.

29. Galaxy shape Number Percentage Spiral (inc. all disc galaxies)

3 = 3/43 = 7%

Elliptical (inc. all galaxies that are regular but not disc)

28 = 28/43 = 65%

Irregular (anything else) 12 = 12/43 = 28% Note: The membership of some galaxies in the local Group is disputable and new galaxies are discovered from time to time so the numbers in the table may change with time and the reference used.

30. Based on the local group most galaxies are elliptical and irregular, with relatively few being spirals.

31. No. Gravity holds the Earth in orbit around the Sun. Expansion only pulls galaxies apart that are sufficiently distant that the gravity between them is small.

32. Before the big bang theory was developed, scientists believed that the universe had existed forever and was essentially unchanging.

33. Normal explosions start with a mass that rapidly expands into the space around it. The big bang started with nothing, no mass and no space to expand into.

34. As the recession is due to the expansion of space, the further objects are apart the faster they will recede from each other. Two hundred light-years will expand twice as much as one hundred light-years in the same time, so an object two hundred light-years away will move away at double the speed.

35. Galaxies receding at a rate dependent on their distance from us, increase in density of galaxies in the past, presence of cosmic microwave background radiation, proportion of hydrogen and helium in the universe

36. According to the big bang theory, matter would have first appeared as neutrons, which decayed into protons and electrons. This provides the hydrogen. In the very early universe the pressure and temperature were sufficient for protons to fuse with neutrons to form deuterium, tritium and helium. These reactions occur from interactions between individual protons and neutrons. The pressure and temperature of the universe had diminished sufficiently before heavier elements could be created.

37. Heavier elements were formed in stars. The heaviest elements are only formed in supernova explosions, which also distribute the elements throughout the galaxy.

38. The cosmic microwave background had been predicted by theorists as a necessary outcome of the big bang. There was no other explanation for its existence. Theories that successfully predict new discoveries are highly regarded by scientists because these theories do not just explain the observed but also what is yet to be observed, making them far more powerful as theories and much more likely to be correct.

39. It was a violation of the steady state cosmological principle. The big bang theory requires evolution of the universe over time.

40. Telescopes detect electromagnetic radiation. Parts of the universe are too far away for light from them to reach the Earth within the age of the universe, so we cannot detect them. They are outside the visible universe.

41. The rapid expansion of the universe meant that material had to clump together early on, or else it would not have had time to clump to form galaxies, clusters and superclusters. Before the tiny variations were found, the cosmic microwave background seemed completely uniform and another explanation would have been needed. This was another example of the theory predicting a phenomenon that was later discovered.

42. Hubble estimated the age of the universe to be 1 ,H

where

H is the velocity of galaxy recession divided by the distance to the galaxy. If the distances are twice what Hubble thought then H will be halved. Halving H doubles the estimated age of the universe.

43. Galaxies rotate in a way that suggests the presence of more mass than is observed. Also, current theories of the universe suggest that it has just enough mass to avoid collapse; if this is so there must be much more mass than is visible.

44. Inflation theory states that there was a period of enormous expansion in the first moments of the universe. It is popular with cosmologists because it helps to explain features observed in the universe today, such as the uniformity of the cosmic microwave background radiation.

45. Student research 46. Student investigation


47. (a) Hubble’s constant 1

6 1

–1 –1

gradient

1000 km s2 10 Mpc

500 km s Mpc

−

−

=

=×

=

(b) The biggest reason for the difference from today’s figure is that Hubble used an incorrect value of the distance to other galaxies.

48. (a) The steady state theory claimed that the universe has existed forever, and has always looked much the same as it does now. The steady state universe constantly creates new matter so there should be ancient galaxies and young galaxies. The big bang theory says that the universe had a beginning, when virtually all of the matter in the universe was created. It has expanded with time so its density has decreased. Early in the universe, all of the galaxies were young, now they are all old.

(b) Looking back in time we see different features, such as quasars and a greater density of galaxies. The steady state theory has no explanation for the cosmic microwave background radiation, and all galaxies and other objects in the universe seem to be less than 15 billion years old.

(c) The big bang theory suggests that the universe was smaller, more active and more dense in the past. It even suggests a time before which the universe does not exist. The universe in this view is not the same for all time as required by the cosmological principle of the steady state theorists.

49. Debate. 50.

51. The big bang theory readily explains the major

observations of the universe; steady state does not explain anything that the big bang cannot.

52. Scientific accounts are based on measurements and tested hypotheses. They are adapted as new data come to light. Ancient accounts are preserved in spite of scientific evidence against them. They have more to do with morals and values, whereas science attempts to deal objectively with observations.

53. Often simulations of galaxy formation start with individual particles and the forces, such as gravity, that act between them. In this way astronomers can test hypotheses they have about the formation of galaxies and other structures. As the simulation runs, astronomers can observe whether or not the particles move to form structures similar to those observed in the universe.

54. It is impossible to get a cloud of gas sufficient to form a galaxy and wait long enough to see if the galaxy forms as expected. Simulations allow astrophysicists to perform experiments with simulated clouds of gas and can run the simulation in a time that allows them to see what would have occurred over billions of years.


Unit

2 ■■■■■■■■■

Detailed study: Energy from the nucleus


Chapter 15

Nuclear energy 1. The nucleons are held together in the nucleus by a force

of attraction called strong nuclear force, which is strong enough to overcome the electrostatic force of repulsion.

2. (a) Fusion is the joining together of two nuclei to form a larger, more stable nucleus. Fission is the splitting of a single nucleus to form two smaller, more stable nuclei.

(b) Fusion 3. In both cases, more stable nuclei than the original are

being created. Excess energy is then released. 4. From the graph (see figure 8.4, p. 200), binding energy

per nucleon for 235U is = 7.8 MeV. ⇒ binding energy of 235U = 235 × 7.7 MeV

= 1810 Binding energy after decay

= binding energy of 141Ba + binding energy of 92Kr = 141 × 8.4 + 92 × 8.7 = 1184.4 + 800.4 = 1984.8

⇒ Energy released = 1985 − 1810 = 175 = 200 MeV

(1 significant figure is appropriate considering data are required from graph.) Therefore, it is close to the accepted value.

5. Energy is released when fusing two small nuclei together because a larger, more stable nucleus results with a greater binding energy per nucleon.

6. The neutron knocks an electron from an atom. 7. A flat mass has a much lower volume/surface area than a

sphere. That is, for the same volume and mass as a sphere, more of the flat sheet is exposed to the air. This allows many more free neutrons to escape rather than sustain a chain reaction.

8. Kinetic energy of a neutron and a fused nucleus and some gamma radiation.

9. Because they have no charge and can easily enter a nucleus.

10. A chain reaction occurs when more than one of the neutrons produced in a fission reaction triggers another fission reaction.

11. Similarities: Each energy source is used to heat water, produce steam and turn turbines. The kinetic energy of the moving turbine in each plant is converted into electrical energy (via a generator). Electrical energy is distributed by other plants to the community via transformers and transmission lines.

Differences: Nuclear power is the initial energy source. A nuclear power plant does not produce CO2. Nuclear waste can be stored until harm to the environment is reduced.

12. Control rods absorb neutrons that may split other uranium atoms. By absorbing these neutrons, the chain reaction can be slowed.

13. The fuel for conventional reactors is uranium-235. The plutonium-239 used in fast breeder reactors is derived from uranium-238, which is more plentiful than uranium-235. Also, because plutonium-239 absorbs fast-moving neutrons, there is no need for a moderator to slow them down.

14. The fuel in ‘fast breeder’ reactors (plutonium-239) readily absorbs fast-moving neutrons, whereas uranium-235 reactors using uranium-235 as fuel require a moderator to slow the neutrons down. Slow-moving neutrons are more likely to be captured by uranium-235.

15. The list might include, for example: Benefits: Good supply of fuel (in some locations), reduced production of greenhouse gases Risks: Possible accidents causing large-scale environ-mental and human devastation, radioactivity of waste products

16. Answers should be based on information contained on pp. 348–52 of the text.

17. The mass of the two isotopes U235 and U238 are very close, differing by about 1%.

18. Boron is a neutron absorber and is also used in control rods.

19. The speed of particles in a substance is a measure of its temperature, the higher the temperature, the faster the particles are moving. The neutrons are slowed down by the moderator to a speed comparable to that of the uranium atoms in the fuel rods because of the temperature they are at.

20. Control rods regulate the nuclear reaction. They absorb neutrons so when they are pulled out the reaction rate increases; when they are dropped in, the reaction rate slows or is stopped.

21. Suggestions of location should be areas which are: • geologically stable • away from high density populations. Safety precautions should include: • secure location • inability for waste to leak – e.g. use of Synrock – dry area • space between waste and perimeter fence • discussion of shielding and half-life.

22. Depleted Uranium is the leftover Uranium from the enrichment process. It has a higher concentration of U238, being almost pure. It is a very dense metal and is


used as a medical radiation shield. Its high density and high melting point means that it is also used in the tips of armour piercing missiles. There are concerns about the radiation exposure to soldiers and the non-combatants from exploded shell fragments. However the material is weakly radioactive, but it has a very long half-life.

23. What does the phrase ‘reprocessing of spent fuel rods’ mean? A spent fuel rod contains a mixture of fission fragments, uranium and transuranic elements such as Plutonium. Reprocessing is the chemical separation of these for either disposal as waste or re-use as a nuclear fuel.

24. Fissionable material is contained as two or more subcritical masses within a bomb. This allows safe transportation. These subcritical masses are brought together through the detonation of a small conventional bomb to reach critical mass (needed for a chain reaction to be sustained).

25. Atmospheric testing: The effects may remain for months/years as the radioactive fallout (with long half-lives) returns to the earth. Underground testing: As the ground is a good shield, the effects are not as significant. There is little fall-out above ground level.

26. Personal reflections of students

27. A neutron bomb releases a significant amount of its energy as neutrons, which damage tissue, but not infrastructure, however, while neutron bombs have about one-tenth of the explosive power of a nuclear bomb, this is still considerably greater than conventional weapons. The neutrons are produced by the fusion reaction within a thermonuclear bomb, but are allowed to escape the reaction vessel in a neutron bomb.

28. Kinetic energy of fission fragments and neutrons. 29. Fusion has minimal waste problems and a largely

inexhaustible source of fuel, for example, sea water. The disadvantage is that after about 60 years of research a sustainable fusion reaction has still not been achieved.

30. Amount of energy in joules = 103 × 106 × 60 × 60 × 24 joules. Amount of energy in eV = (103 × 106 × 60 × 60 × 24)/(1.60 × 10−19) Number of fusion reactions = ((103 × 106 × 60 × 60 × 24)/(1.60 × 10−19))/(4.0 × 106) Mass of Deuterium = (2.00 × 1.66 × 10−27) × ((103 × 106 × 60 × 60 × 24)/ (1.60 × 10−19))/(4.0 × 106) Mass of Deuterium = 0.45 kg


Unit

2 ■■■■■■■■■

Detailed study: Investigations: flight


Chapter 16

Flight investigations 1. The centre of pressure is the point on an aircraft’s wing at

which the lift forces are considered to be acting. The centre of gravity is the point on any object at which the weight force can be considered to be acting.

2. (a) ↑ 100 N, net force, accelerates up (b) → 100 N, decelerates (c) ↑ 200 N + ← 10 N, accelerates up and forwards (d) ↓ 200 N + ← 10 N, accelerates down and forwards

3. (a) Constant speed ∴ thrust = drag drag = 25 kN (b) F = 25 000 N

v = 200 m s−1

P = Fv = 25 000 × 200 = 5 000 000 = 5.0 × 106 W 4. In fluids, the particles experience more freedom than in

solids, they are bound together less tightly, and the spaces between them are larger. This is especially so in gases.

5. v1 = 9.3 cm s−1 v2 = 13 cm s−1

A1 = 61 cm2 A2 = ? v1A1 = v2A2

A2 1 1

2

2

9.3 6113

44 cm

v Av

=

×=

=

6. v1 = 2.1 m s−1 v2 = ? d1 = 0.15 m d2 = 0.45 m

2 21 2

1 2

21 1

2 22

2

2

1

4 4

2.1 0.15 0.45

0.23 m s

d dv v

v dvd

π π

−

=

=

×=

=

7. (a) 1 1 2 2

1 12

2

2 1

1 12

1

but 2

2

v A v Av AAv

v vv AA

v

=

⇒ =

=

⇒ =

1

2A

=

The area is halved.

(b) 12

22 1

2

21

2

1

2

2

2

2

AA

rr

rr

r

ππ

=

=

=

=

The radius would be reduced by a factor of 12

or

0.71. 8. 1 1 2 2

2 21 2

1 2

2 21 1 1 2

22 12

12

4 410

10

10

v A v A

d dv v

v d v d

dd

dd

π π

=

=

=

=

=

Decrease in diameter by a factor of 110

or 0.32

increases air speed by a factor of 10. 9. As the speed of a fluid increases, the pressure decreases.

This is Bernoulli’s principle in simple terms. 10. In the term 1

2 ρv2, ρ = density, measured in kg m−3, and

v = fluid speed, measured in m s−1. Therefore the units of 12 ρv2 are kg m−3 × (m s−1)2 = kg m−3 m2 s−2 = kg m−1 s−2 .

N is equivalent to kg m s−2, so N m−2 is equivalent to kg m s−2 × m−2 or kg m−1 s−2 as above.

11. In the term ρgh, ρ = density, measured in kg m−3, g = acceleration due to gravity, measured in m s−2, and h = height, measured in m. Therefore the units of ρgh are kg m−3 × m s−2 × m = kg m−1 s−2. N is equivalent to kg m s−2, so N m−2 is equivalent to kg m s−2 × m−2 or kg m−1 s−2 as above.

12. An aerofoil is curved on the top and flat on the bottom. It is shaped so that the air travelling over the top surface is sped up, reducing the air pressure below normal pressure. This results in an upwards force, known as the lift force, that acts from the region of normal air pressure below the aerofoil towards the region of lower pressure above the aerofoil.


13.

14. As an aerofoil travels through the air, it deflects the airflow

downwards. This is downwash. The reaction to this action is an upward push on the wing by the air. This provides about 15% of the lift needed by a cruising aircraft.

15. Wing-tip vortices are caused by the higher-pressure air under the wing flowing around the wingtip to the region of lower-pressure air on the top of the wing.

16. (a)

(b) Roughly 108 km h−1

(c) A high lift-to-drag ratio means an aircraft experiences maximum lift and minimum drag. This corresponds to an optimum cruising speed for the aircraft and determines its maximum range and endurance.

(d) Stalling occurs at low airspeed due to an increase in drag. This corresponds to the left-hand region of the graph.

(e) In a stall, the airflow over the wing breaks down and becomes turbulent. Lift is lost and the aircraft could drop in an uncontrolled manner.

17. Glide ratio = 9 : 1 loss of altitude = 1200 m

(a) glide distanceGlide ratioloss of altitude

glide distance glide ratio loss of altitude9 120010 800 m (10.8 km)

=

= ×= ×=

(b) Lift-to-drag ratio = glide ratio = 9 : 1

18. (a) glide distanceGlide ratioloss of altitude12 000

80015 :1

=

=

=

(b) Lift-to-drag ratio glide ratio15 :1

==

19. glide distanceGlide ratioloss of altitudeglide distanceLoss of altitude

glide ratio3600 m

4090 m

Final altitude 0Initial altitude 90 m

=

=

=

===

20. The function of the tailplane in an aircraft is to provide an extra force that can be called into action when required to correct any out-of-balance of the aircraft.

21. (a) Taking torques about the centre of gravity,

net

clockwise anticlockwise

w w T T

T

T

3

0

15000 1.1 9.315000 1.1

9.31.8 10 N downwards

I d L dL

L

τ =τ = τ

× = ×× = ×

×=

= ×

(b) The force that depends on the mass of the plane, the

weight, mg, acts through the centre of gravity. When torques are taken about the centre of gravity, the torque due to the weight is zero.

22. The location of an aircraft’s centre of gravity can be altered by repositioning the load it carries, for example, fuel, passengers and cargo. In this case, one or all of these would need to be shifted rearwards.

23. τnet = 0 W1 = 10 000 g W2 = 15 000 g d1 = 12 m d2 = ? 1 1 2 2

2

2

10000 g 12 15000 g10000 12

150008.0

W d W dd

d

=× =

×=

=

The second container should be located 8.0 m in front of the plane’s centre of gravity.


24. The additional torque due to the weight of the added cargo Wc has to be compensated for by a decrease in the torque due to the weight of released fuel Wf.

added cargo released fuel

c c f f

f

f

τ τ

200 3.6 4.0200 3.6

4.0180 kg.

W d W dg M g

M

=

=× = ×

×=

=

25. The longitudinal axis of an aircraft runs from the front end to the rear end through the centre of gravity. The vertical axis runs vertically downwards through the aircraft’s body, through the centre of gravity. The lateral axis runs parallel with a line from wing tip to wing tip and also through the centre of gravity.

26.

Type of motion

Axis about which motion

occurs

Aircraft control surface

responsible for motion

yaw vertical rudder

pitch lateral elevators

roll longitudinal ailerons


Unit

2 ■■■■■■■■■

Detailed study: Investigations: sustainable energy sources


Chapter 17

Investigating energy alternatives 1. Renewable sources of energy replace themselves

continuously. Non-renewable sources of energy rely on fuels that don’t replace themselves.

2. Energy sources are described as renewable if they replace themselves continually. Energy sources described as sustainable are those that have a long term future because they are not limited by supply and environmental factors.

3. Without a Greenhouse Effect to prevent too much energy leaving the Earth’s atmosphere, the temperature at and near the Earth’s surface would be too low for the survival of living things.

4. A high-grade form of energy can be efficiently converted into other useful forms of energy. Low-grade energy cannot be converted into another useful source of energy.

5. Internal energy is classified as a low-grade form of energy because it cannot be converted efficiently into other useful forms of energy.

6. In any energy conversion, some energy is always degraded into lower grade forms.

7. Estimate the area of the body facing the Sun. For example,

2

1.5 m 0.5 m

0.75 m

1200 W

A

Qt

= ×

=

=Δ

where Q is the quantity of energy received by an area of 1 m2.

Δt = 30 × 60 s = 1800 s ⇒ Q = 1200 W × 1800 s ⇒ Energy received by body = 1200 W × 1800 s × 0.75 = 1.6 × 107 J = 16 MJ A major assumption is that the sky is clear. The value

seems high but one must remember that: • some of the incident energy is reflected by the body

and not absorbed • while radiant energy is being absorbed, the body is also

cooling by conduction, convection and/or evaporation. 8. Energy used 110 0.5

55 kJenergy usedPowertime interval

= ×=

=

55 kJ55 s

1.0 kW

=

=

9. Answers will vary from school to school. Assumptions might include:

• average power consumption for lighting and heating = 50 kW

• hours of operation per day = 7 • days of operation per year = 200 For example, Energy = 50 000 × 7 × 60 × 60 × 200 = 2.5 × 1011 J Cost of lighting: Assume 1 kW power consumption. Energy used in one year = 1 kW × (7 × 200) h = 1400 kW-h Cost = 1400 × $0.17 = $238 10. Energy input in one hour = 10 × 3.3 × 107 J = 3.3 × 108 J Mechanical energy output in one hour = 25% of 3.3 × 108 J = 8.25 × 107 J

7

4

energy outputPower outputtime

8.25 10 J3600 s

2.3 10 W23 kW

=

×=

= ×=

11. 4500 MWEnergy collected 0.15

30 000 MW

=

=

10

8 2

Each square metre collects 80 Warea needed 30 000 MW 80 W

3.0 10 80

3.8 10 m

∴ = ÷

= × ÷

= ×

12. 2

2

2

Insolation 1.2 kW m

Area 2000 mEfficiency 18%

Incident energy 1.2 20002400 kW

Energy converted 0.18 2400432 kW

4.3 10 kW

−=

=== ×== ×=

= ×

13. The increase of salt concentration with depth.


14. (a) 2 1

2

2

2

Insolation 6.5 kW-h m day6.5 kW-h

1 m 24 h

(6.5 24) kW m

0.27 kW m

− −

−

−

=

=×

= ÷

=

(b) 0.27 1000%1.37 119.7% (20%)

= ×

=

15. (a)

(b) The shapes are mostly as expected. Insolation greater

during summer and loss during winter. Less insolations the further from the equator. Unexpected variation in southern hemisphere due to factors other than latitude, for example, location with respect to land mass and ocean.

16. The flow of water through a turbine is used to generate electricity in a hydroelectric power station. The speed of the water is determined mainly by the height the water has fallen through.

(a) Flow = 2 ML min−1

h = 180 m

6

9 1

9

7

4

Energy conversion

2 10 9.8 180

3.53 10 J min

3.53 10 J60 s

5.9 10 W

5.9 10 kW or 59 MW

mgh

−

=

= × × ×

= ×

×=

= ×

= ×

(b) Some of the available energy is transferred as heat to the tunnel, turbine and surrounding air as a result of friction and magnetic effects.

17. Student research. Answers should include the size of the power plant (which will affect the cost) and the huge amount of time it would take to build. Because of its likely weight, it would have to be built in space.

18. Incident radiation= 1.37 kW m−2

Area of panel = 8 km × 8 km = 8000 m × 8000 m = 6.4 × 107 m2

Incident energy = 1.37 × 6.4 × 107

= 8.768 × 107 kW Receiving energy rate = 8.768 × 107 × 0.18 × 0.65 = 1.0 × 107 kW = 10 000 MW

19. (a) 2

2

3 2

65 m Area4

3318 m

3.3 10 m

dd π= =

=

= ×

(b) mV

=ρ

= 1.22 kg m−3

V = A × thickness = 3318 × 1 = 3318 m3

m = ρ × V = 1.22 × 3318 = 4048 kg = 4.0 × 103 kg (c) P = 1

2 ρv3πr2

= 12 ρv3 × area

= 12 × 1.22 × 83 × 3318

= 1.0 × 106 W = 1.0 MW (d) P ∝ v3

If v is reduced by 12 , v3 is reduced by ( 1

2 )3 = 18

Power is reduced by 18

(e) Power output = 1.04 × 106 W × 0.35 Power output = 3.6 × 105 W Power output = 360 kW 20. Student answers will vary. 21. Student research 22. Student research 23. Student research


Unit

2 ■■■■■■■■■

Detailed study: Medical physics


Chapter 18

Waves and radioactivity in medical diagnosis and treatment 1. Ultrasound is sound above 20 000 Hz frequency, whereas

sound for normal hearing in humans is of frequency below 20 000 Hz; 20 000 Hz is at the limit of human hearing.

2. High frequency is not used to scan internal organs because it doesn’t reach them. High-frequency ultrasound is absorbed by living tissue more than low-frequency ultrasound. High-frequency ultrasound is used only on regions close to the skin.

3. Z = ρv, where Z = 1.59 × 106 kg m−2 s−1 v = 1570 m s−1

⇒ ρ = 1.59 × 106 ÷ 1570 = 1.01 × 103 kg m−3

The density of blood is 1.01 × 103 kg m−3 (or 1.01 g cm−3).

4. (a) Z = ρv, where ρ = 1060 kg m−3

v = 1540 m s−1

⇒ Z = 1060 × 1540 kg m−2 s−1

= 1.63 × 106 kg m−2 s−1

The acoustic impedance of soft tissue is 1.63 × 10 kg m−2 s−1.

(b) Z = ρv, where ρ = 1600 kg m−3

v = 4080 m s−1

⇒ Z = 1600 × 4080 kg m−2 s−1

= 6.53 × 106 kg m−2 s−1

The acoustic impedance of bone is 6.53 × 106 kg m−2 s−1.

(c) Z1 = Zliver = ρv, where ρ = 1050 kg m−3

v = 1570 m s−1

⇒ Z1 = 1050 × 1570 kg m−2 s−1

= 1.6485 × 106 kg m−2 s−1 Z2 = Zmuscle = ρv, where ρ = 1075 kg m−3

v = 1590 m s−1

⇒ Z2 = 1075 × 1590 kg m−2 s−1

= 1.709 25 × 106 kg m−2 s−1

Ir ÷ Io = (Z2 − Z1)2 ÷ (Z2 + Z1)2, where Z1 = 1.6485 × 106 kg m−2 s−1

Z2 = 1.709 25 × 106 kg m−2 s−1

= 3.27 × 10−4

Note that if Z1 and Z2 are calculated to only three significant figures, the answer obtained is 3.19 × 10−4.

5. (a) There is the most reflection at the skin–air interface (as Ir/Io has the largest value).

(b) There is the least reflection at the brain–fat interface (as Ir/Io has the smallest value).

(c) The greatest amount of absorption occurs where there is least reflection. This is at the brain–fat interface.

(d) For the fat–bone interface, Ir/Io = 0.029, where Io = 60 mW cm−2.

∴ Ir = 1.74 mW cm−2 The intensity of the reflected signal is 1.74 mW cm−2. (e) For the fat–muscle interface, Ir/Io = 0.011, where Io = 80 mW cm−2. ∴ Ir = 0.88 mW cm−2

Amount of signal travelling into the muscle = 80 − 0.88 mW cm−2

= 79.12 mW cm−2

(f) Some of the energy is converted to heat as the ultrasound travels through the muscle and the rest of the energy travels through the muscle to the next interface, where part is reflected and part is transmitted.

6. If the ultrasound travels through fat to liver, the same percentage of ultrasound will be reflected as when the ultrasound travels through liver to fat. Because of the squared term in the equation to find Ir/Io, it does not matter whether fat or liver is assigned the value Z1 — the ratio will remain the same.

7. A full bladder pushes the lower intestine, which contains a lot of gas, out of the way and lifts the uterus to a good position for taking an image. If the lower intestine were in the way, ultrasound would be reflected from the gas and it would not be possible to obtain an image of the foetus.

8. Z1 = 1000 kg m−3 × 1500 m s−1 (for aqueous humour) = 1.5 × 106 kg m−2 s−1 Z2 = 1140 kg m−3 × 1620 m s−1 (for lens) = 1.85 × 106 kg m−2 s−1

Ir ÷ Io = (Z2 − Z1)2 ÷ (Z2 + Z1)2 = (0.35 × 106 kg m−2 s−1)2 ÷ (3.35 × 106 kg m−2 s−1)2

= 0.109 ⇒ Ir = 0.0109 Io = 0.0109 × 15 mW cm−2

= 0.16 mW cm−2

The intensity of the reflected beam is 0.16 mW cm−2. 9. (a) Most of the ultrasound would be reflected from the

interface between air and skin as the difference in acoustic impedance between air and skin is very large.

(b) The optimum acoustic impedance of the gel would be the same as that of the skin. Then all the incident ultrasound would be transmitted into the skin.

The acoustic impedance of the gel = 1010 × 1540 = 1.56 × 106 kg m−2 s−1


(c) Z = ρv, where Z = 1.56 × 106 kg m−2 s−1

ρ = 1200 kg m−3

∴ v = 1300 m s−1

The speed of ultrasound in the gel is 1300 m s−1. 10. When subjected to an alternating potential difference, the

piezoelectric crystal vibrates. This vibration is rapid and produces signals in the ultrasound frequency range. In this way, ultrasound waves are produced. When ultrasound strikes a piezoelectric crystal, the pressure changes, due to compressions and rarefactions, cause the crystal to vibrate. The vibration generates an alternating potential difference across the crystal. In this way, the ultrasound waves are received.

11. An ultrasound A-scan is a range measuring system in which an ultrasonic pulse is sent into the body in one line and the time for the pulse to be reflected from an interface in the body is measured. The intensities of the reflected signals are plotted as a function of time and from this information the position of features inside the body can be found. For a B-scan, a pulse is sent into the body along one line, and the reflected pulses are recorded as dots, the brightest corresponding to the signal of greatest intensity. By sending pulses into the body at different angles, a series of B-scans can be used to build up a 2-D picture of a cross-section through the body.

12. (a) Ultrasound signal is sent through the heel bone. The speed and attenuation of the transmitted ultrasound are measured. These data are then compared with information about healthy heel bone.

(b) This method has limited effectiveness in the diagnosis of osteoporosis as it is not done at the site of possible fractures — i.e. the hip or spine. It can be used as an indicator of the need for further tests using X-rays if an abnormal result is obtained.

13. (a) Speed of sound = 1540 m s−1

Distance travelled by pulse = 700 mm = 7 × 10−1 m

Time for pulse to travel this distance

4

0.7 s15404.5 10 s−

=

= ×

∴ minimum time between pulses = 4.5 × 10−4 s (b) A faster pulse rate would mean that the time between

pulses was shorter than 4.5 × 10−4 s. A new pulse would be sent into the body before the reflected pulse was received, leading to interference between the reflected pulse and the incoming pulse and lack of clarity of the position of the image.

14. A continuous signal is directed at the foetal heart and the reflected signal is detected by a separate receiver. The output is electronically filtered so that only the difference in frequency due to the Doppler shift is amplified. This difference is in the audible range. The tone will vary with the speed of movement of the heart and hence with the heartbeats, which can therefore be monitored.

15. (a) Electrons are emitted from a heated filament in a highly evacuated tube. The heated filament is the cathode. A very high potential difference (maybe over 100 000 volts) is applied between the cathode and the tungsten anode. The very fast electrons strike

the anode and are absorbed, and some of their energy is converted into X-rays, which are sent in a direction determined by the angle of the tungsten target.

(b) The X-rays are passed through a filter so that the

low-frequency rays are observed, leaving the beam with a higher proportion of the more penetrating, high-frequency rays.

16. (a) Attenuation, or reduction in intensity, of X-rays depends on the atomic density of the material encountered. Atomic density refers to the number of protons in the nuclei of the atoms encountered by the X-rays. Bone has a high atomic density and so will absorb the X-rays readily and show up clearly in an X-ray image. Soft tissue has a moderate atomic density and hence will not absorb X-rays as well as bone. Soft tissue such as muscle and skin will show up very faintly on an X-ray image. Air, on the other hand, has a low atomic density, so air in the body will not absorb X-rays and will appear black on an X-ray image.

(b) The bone would be white, the muscle would be light grey and translucent in appearance and the air would be black. Bone would absorb X-rays, because bone has a high atomic density as outlined in part (a), and so the X-rays would not reach the photographic film, which would show up as white behind the bone. Some X-rays would penetrate the muscle, with a medium atomic density, and reach the film, resulting in the light grey colour behind muscle. X-rays would not be stopped by air and these X-rays would reach the photographic film, resulting in a black image.

17. (a) Hard X-rays have a higher frequency and are more penetrating than soft X-rays.

(b) Hard X-rays are preferred because they penetrate the body and are absorbed by bone. Soft X-rays are absorbed by the skin before they reach bone and expose the patient to X-rays which have no value in obtaining an image.

18. Conventional X-ray images are produced by a stationary X-ray tube. CAT scans are produced by an X-ray tube that is rotated around while the patient is slowly moved through a gantry.


19. CAT scan X-rays Ultrasound Soft tissue accurately imaged, resulting in fine detail being shown.

Soft tissue does not absorb conventional X-rays to any noticeable extent, so soft tissue will not show up, unless a contrast medium is used.

Ultrasound will not clearly distinguish one type of soft tissue from another.

Images of slices, close together, through the body can be taken.

Conventional X-rays cannot be used to image a slice of the body.

Ultrasound cannot accurately image slices close together.

The images of slices can be built up into a 3-dimensional image.

X-rays cannot do this as slices cannot be imaged.

Ultrasound cannot accurately image slices close together so a 3-dimensional image built up from ultrasound is not very clear.

The brain can be imaged through the bony skull.

Conventional X-rays would show up the skull rather than what was inside the skull.

Ultrasound would be reflected from the skull and not image what was underneath, unless the ultrasound was taken through a gap in the skull.

CAT scans provide a clear image of the lungs.

X-rays provide less detailed images than CAT scans.

Ultrasound cannot image the lungs because the air–tissue boundary reflects sound waves.

Complicated bone structure is imaged and a 3-dimensional image may be obtained.

Bone absorbs most of the rays giving a 2-dimensional image of the part of the bone facing the X-rays, and so the complex structure cannot be seen.

Ultrasound is reflected from bone, and so no useful image is formed.

20. (a) In a coherent bundle the fibres are in the same position relative to one another at each end of the fibre.

(b) A coherent bundle of optical fibres is used to transmit the reflected light back from the inside of the body to the eye or camera outside the body. It is necessary that an accurate image is produced and hence each fibre must remain in the same position relative to the other fibres. In this way light from the far right of the object will emerge from the body and be at the far right of the image; this matching of light beams will occur for each optic fibre. The result will be an image which is exactly the same as the object inside the body.

(c) It is dark inside the body. Hence, the inside of the body will not be visible unless light is passed from a powerful light source along an optical fibre, reflected off the inside of the body and transmitted back to the outside. The internal organs are not necessarily good reflectors of light, so the source must be bright to reflect as much light as possible to see the organs.

(d) If the fibres in the bundle are narrow and have a large core to cladding ratio, light reflected from many points on the object will be transmitted. Hence, the resulting image will be made of many points and will therefore be clearer.

21. The endoscope can be inserted into the oesophagus through the mouth. The tumour in the oesophagus can be viewed because light is reflected off the oesophagus and the tumour. A small tool on the end of the endoscope can be used to remove a sample of the tumour. This sample is then removed from the oesophagus by this tool and sent for examination. ‘Taking a biopsy’ means removing a sample of the tumour.

22. The high intensity of the light makes lasers suitable for heating, cutting or vaporising tissue. The narrowness of a laser beam allows surgeons to target small areas, minimising damage to surrounding tissue. The narrow range of wavelengths for each particular laser allows the selection of a laser for a particular purpose of location in the body.

23. Advantages of keyhole surgery over other methods include:

• less invasive than open surgery • less risk of infection • less swelling and scarring • quicker recovery time. 24.

The graph should show that the activity for carbon-11 halves every 20 minutes while the activity for bromine-75 halves every 100 minutes.

25. (a) Sixteen days is two half-lives. After one half-life, 12

of the original amount remains. After two half-lives, 14 of the original amount remains.

14 of original amount = 6.0 mg

⇒ the original amount = 24 mg 24 mg of iodine-131 was used in the treatment.


(b) After another 16 days, two more half-lives will have past.

The amount left = 14 of 6.0 mg = 1.5 mg.

(c) Iodine-131 emits beta and gamma radiation, exposing the patient to more radiation than iodine-123, which decays by gamma radiation only. Iodine-131 has a longer half-life than iodine-123, exposing the patient to radiation for longer than necessary.

26. (a) In one half-life the activity will drop to 2.0 MBq, and in another half-life the activity will drop to 1 MBq. Hence, the isotope will take two half-lives, or 4.0 minutes to reach an activity of 1.0 MBq.

(b) Halving the activity of each half-life, we see it will take four half-lives for the activity to reach 0.25 MBq. The time involved will be 4 × 2.0 minutes or 8.0 minutes.

27. This radioisotope is not suitable for use in medical diagnosis as the half-life is too long. After waiting a suitable time for the radiopharmaceutical to accumulate in the target organ, the rate of emission will be low, because the half-life is 100 days. Hence, a clear image will not be able to be made. A larger dose of radioisotope could be given to increase the rate of emission, but this will mean more radioisotope will either remain in the patient’s body or be excreted into the sewerage system.

28. The amount of emissions produced may have decreased below a useful level for producing an image before the radiopharmaceutical has accumulated in the target organ. Once the radiopharmaceutical has accumulated in the target organ, a short half-life will mean that the emission rate will be too low by the time the examination has finished.

29. (a) Example

Isotope Where it is used Justification Iodine-123 To examine the

thyroid gland It is taken up by the thyroid gland. The amount of γ radiation given out is a measure of whether or not the thyroid gland is functioning normally, because the uptake of a normal thyroid gland is known.

Technetium-99m To examine bone Polyphosphate ions labelled with technetium-99m accumulate in bone and the gamma rays produced show the flow of blood. High blood flow and hence high γ-ray activity are called ‘hot spots’ and are often associated with disease.

(b) Radioisotopes used in medical imaging are taken into the body and the γ-rays which are produced are

detected outside the body. If α particles are produced, they will penetrate some of the internal tissue before being absorbed and causing ionisation, which is damaging. They will not be detected outside the body as they will be absorbed before they emerge. Hence, they would be useless for imaging; they are also harmful.

30. (a) A study to measure the volume of blood in the body uses a radioactive tracer which mixes with the blood.

(b) A study to detect blockage in the lungs uses a tracer which is trapped in the fine capillaries in the lung. If the tracer cannot become trapped, it may be because the lung is blocked.

31. Technetium-99m has a relatively short half-life (6 hours), it emits γ-rays only when it decays, and it readily attaches to different compounds to form radioactive tracers. These different compounds, when labelled, are metabolised by a number of different organs and hence technetium-99m can be used to image many organs.

32. (a) The top study shows both lungs because they are functioning normally. The radioisotope will reach the capillaries around the alveoli in the perfusion study, and the radioisotope will reach the spaces in the alveoli in the ventilation study. The bottom studies show only the right lung in the perfusion study. Both lungs are visible in the ventilation study.

(b) In the bottom study, the radioisotope does not reach the capillaries because they are blocked in the left lung. (This is a front view of the lungs.) The perfusion study shows the blocked capillaries. The radioisotope can reach the spaces in the alveoli in the ventilation study, and so both the lungs show up on the image. The ventilation study would show blocked alveoli, not blocked capillaries.

33. (a) The X-ray shows the bone as whiter on the outside. The inside of the bone is a similar shade to the surrounding tissue. The bone scan shows the bone brighter at the ends of the long bones in the legs — possibly due to these areas being where the bone grows.

The X-ray shows a distinct break in one bone. The bone scan shows white patches on bones in the spine, ribs, shoulder and upper legs of the skeleton, which has tumours.

(b) The X-ray produces an image because X-rays are absorbed by bone and tissue through which they pass. The bone is denser than the surrounding tissue and so bone absorbs more X-rays. A shadow of the bone forms on the X-ray photo and this shadow shows a break in the bone.

In the bone scan, a radiopharmaceutical is injected into the bloodstream and accumulates in bone. Where there is increased blood flow more γ radiation from the decaying radio-isotope is detected. The areas of high blood flow show up as white spots on the scan and often indicate tumours.

34. Xenon should not be used in preference to krypton for investigations of lung function. In the investigation the radioactive gas would be inhaled and accumulate in the lungs. The γ-rays produced as the radioisotope decays


would be detected with a gamma camera. As the radioisotope collects in the lungs very quickly, krypton-81m with a short half-life of 13 seconds would be adequate. A small dose would produce significant activity as the half-life is very small. If xenon-133 were used, a larger dose would be needed to produce the same activity and any residual radioisotope, which was not excreted, would remain in the body for many days because the half-life is 5.3 days. Xenon-133 would also produce β particles as it decayed and these would penetrate the tissue with which they were in contact and cause damage to the tissue.

35. The radioisotopes used in PET scans decay by the emission of positrons — positively charged β particles. Those not used in PET scans emit other types of radiation.

36. (a) A positron is a positively charged β particle. (b) Positrons may be obtained when a proton dis-

integrates into a neutron and a positron. (c) Positron–electron interaction results in the

annihilation of the pair and the production of two γ-rays of energy 0.51 MeV travelling in opposite directions. In medical diagnosis, the pairs of γ-rays produced inside the body are detected and their location determined from knowledge about the attenuation of the gamma rays as they pass through tissue. By locating the source of γ-rays, the source of the radiopharmaceutical can be determined. Often a different amount of radiopharmaceutical from what is expected indicates disease.

37. Sample answer: The radioisotope fluorine-18 replaces a hydrogen atom on some molecules of β-D-glucose and the radio-pharmaceutical so formed is called FDG. This radiopharmaceutical is injected into the bloodstream. The molecules are of a suitable size to reach the brain. Fluorine-18 decays, with a half-life of 109.8 minutes, emitting positrons. After travelling a short distance from the place of emission, a positron encounters an electron and the pair of particles annihilate one another producing two γ photons of energy 0.51 MeV. These travel in opposite directions from the site of annihilation and emerge from the head, to be detected by the gamma cameras surrounding the patient’s head. The intensities of pairs of γ-rays is measured, and by comparison with known attenuation for γ-rays travelling through tissue, the site of the annihilation can be determined. About half a million γ-ray pairs are needed to make a useful image, and so a computer has a valuable role in analysing the collected data. The concentration of glucose in the brain for healthy brain function is known. Tumours require more oxygen and hence more glucose. The site of a tumour could show up as a site where more γ-rays than expected were detected, because more positron emitting radioisotopes circulated there.

It is important to give the patient a dose of the radiopharmaceutical that is large enough to last for the duration of the test but not so large that it will remain in the patient’s body longer than necessary and expose the patient to unnecessary radiation. The radiographer should avoid contact with the γ-radiation produced or with the positrons emitted from the fluorine-18 by using a shield to absorb any harmful radiation and by moving away from the region where radiation would be produced.

38. Protons align themselves so that they are parallel to a strong external magnetic field.

39. Hydrogen is the most abundant of the elements in the human body with nuclei that align themselves with an external magnetic field.

40. The intensity of the pulse released by the protons is analysed to provide information about the proton density in the tissue. The duration of the pulse released is analysed because it provides an indication of how easily the protons release their energy and, therefore, the type of hydrogen compounds in the tissue.

41. MRI distinguishes well between tissues and cancerous tumours because they have different proton densities.

42. A printed image of the organs should be made. The activity in the diseased organ may be greater than in the healthy organ, as the bone scans show in the figure at the bottom of page 426. On the other hand, the activity may be less in the diseased organ due to blockage of the radioisotope from the diseased area, as shown in the figure at the top of page 426. Make sure you compare the images you find.

43. The answer is best presented in a table. Use ideas including spatial clarity, time for examination, cost and mobility of machine, best method for imaging organs, and patient comfort and safety.


Appendix

2 ■■■■■■■■■

Revision questions


page 449 A2.1 (a) Gigawatt, megavolt, kilojoule, decimetre,

centimetre, milliampere, microgram, nanometre (b) (i) 1500 mA = 1500 × 10−3 A = 1.5 A (ii) 750 g = 750 × 10−3 kg = 0.750 kg (iii) 250 GW = 250 × 109 W = 250 000 000 000 W (2.5 × 1011 W) (iv) 0.52 km = 0.52 × 103 m = 520 m (v) 600 nm = 600 × 10−9 m = 0.000 000 6 m (6.00 × 10−7 m) (vi) 150 μs = 150 × 10−6 s = 0.000 15 s (1.5 × 10−4 s) (vii) 5 cm = 5 × 10−2 m = 0.05 m (viii) 50 MV = 50 × 106 V = 50 000 000 V (5.0 × 107 V) (ix) 12 dm = 12 × 10−1 m = 1.2 m

(c)

1

2

m sunit of accelerations

m s

vat

−

−

Δ=Δ

⇒ =

=

(d) (i)

1

Nunit ofkg

N kg

F mgFgm

g

−

=

⇒ =

⇒ =

=

(ii) N kg−1 = kg m s−2 × kg−1

= m s−2

page 451 A2.2 (a) 637 000 m = 6.37 × 105 m (b) 300 000 000 m s−1 = 3.0 × 108 m s−1

(c) 0.000 000 000 3 m = 3 × 10−10 m page 452 A2.3 (a) (i) 4 (ii) 2 (iii) 4 (iv) 6 (v) 2 (vi) 3 (b) (i) A = lw = 30.5 m × 15.24 m = 464.82 m2

Answer should be expressed as 465 m2 as least accurate quantity (30.5 m) has three significant figures.

(ii) P = 2(l + w) = 2(96.3 m + 72.42 m) = 2 × 168.72 m

The value of l + w should be rounded off to one decimal place as the minimum number of decimal places in the data is one.

⇒ P = 2 × 168.7 m = 337.4 m

(c)

1

distance travelledAverage speedtime taken

2 64 m46.52 s

8.644 107 m s

π

−

=

× ×=

=

(using the π button on calculator) Answer should be expressed to two significant

figures only as the least accurate quantity (64 m) has two significant figures.

⇒ average speed = 8.6 m s−1

page 454 A2.4 (a)

(b)


(c)

(a) Area = Area A + Area B = 1

2 × 7.5 s × 5 m s−1 + 12 × 2.5 s × −5 m s−1

= 12.5 m (b) Area = 1

2 × 0.010 s × 20 N

= 0.10 N s (c) Area of each small square = 0.005 m × 0.5 N = 0.0025 N m Total number of squares is approximately 270. Approximate area = 270 × 0.0025 N m = 0.68 N m page 455 A2.5 (a) (i) P ∝ V ⇒ if voltage is tripled, power is tripled ⇒ P = 60 W (ii) P ∝ I ⇒ if current is doubled, power is doubled ⇒ P = 40 W (iii) P ∝ VI if both V and I are tripled, P increases by a

factor of nine ⇒ P = 180 W (iv) P ∝ VI if V is doubled and I is halved, P remains

the same ⇒ P = 20 W (b) (i) Ek ∝ mv2

Ek = kmv2 where k = constant of proportionality

12k⇒ =

(ii) If v is tripled, v2 increases by a factor of 9 ⇒ Ek increases by a factor of 9.

(iii) 21k 2

k

k

kAA

B kB

kA

kB

2

21.4

E mv

Evm

v E

Evv E

EE

=

⇒ =

⇒ ∝

⇒ =

=

==

(c) 1ρV

∝

⇒ ρV = constant ⇒ if V is divided by 3, ρ triples ⇒ ρ = 3 × 1.4 × 10−3 g cm−3

= 4.2 × 10−3 g cm−3

page 457 A2.6

(a) (A + B)2 = A2 + B2 (Pythagoras’ theorem) = (24)2 + (10)2

= 676 ⇒ A + B = 26 units

10tan24

θ =

⇒ θ = 23° Sum is 26 units in a direction 23° east of north. (b) (C + D)2 = C2 − D2 (Pythagoras’ theorem) = (20)2 − (12)2

= 256 ⇒ C + D = 16 units Sum is 16 units south.


page 458 A2.7

Δv = v + (−u) Because the magnitudes of u and v are equal and

the angle between them is 60°, the triangle produced by adding the vectors is equilateral. Therefore the magnitude of Δv is 20 m s−1.

page 458 A2.8

QH = 50 cos 65° (since cos 65° = H

50Q )

= 21 units

QV = 50 sin 65° (since sin 65° = V

50Q

)

= 45 units ⇒ horizontal component is 21 units; vertical component is 45 units.

page 459 A2.9

In each case, the vectors are added ‘tail’ to ‘head’.

(c) Fnet2 = 22 + 22

= 8 ⇒ Fnet = 2.8 N. (h) The triangle produced is equilateral. Therefore, Fnet = 400 N.


page 460 A2.10

page 463 A2.11 This is an example of exponential growth. The number of grains on the nth square is given by 2n − 1. Therefore, the number of grains on the 64th square would be 263. That is 9.2 × 1018 grains. The total number of grains on the board after the nth square is filled is given by 2n − 1. When the chessboard is entirely filled the total number of grains would be 264 − 1, that is, 1.8 × 1019 grains.

A reasonable estimate of the number of grains of wheat to make up a gram is 15. The mass of the grains on the

chessboard is, therefore, 19

181.8 10 1.2 10 ,15

g× = × or

1.2 × 1015 kg. The total world production of wheat in 1998 was about 6 × 1011 kg, so the king would have found it somewhat difficult to honour the wish.

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