Heinemann Physics Content & Contexts Units 3A & 3B Solutions

Heinemann Physics Content and Contexts Units 3A and 3B

Copyright © Pearson Australia 2010 Heinemann Physics Content and Contexts Units 3A and 3B (a division of Pearson Australia Group Pty Ltd) ISBN 978 1 4425 1140 8

Page 1 of 185

Worked solutions Unit 3A

Contents

Fairground physics ..................................................................................................................... 2

Power distribution and generation ........................................................................................... 8

Chapter 1 Analysing motion.................................................................................................... 11

1.1 Projectile motion .......................................................................................................................... 11

1.2 Circular motion in a horizontal plane ........................................................................................... 17

1.3 Circular motion in a vertical plane ............................................................................................... 20

1.4 Vectors and free-body diagrams ................................................................................................... 24

1.5 Motion in a straight line ............................................................................................................... 31

1.6 Energy and momentum................................................................................................................. 34

Chapter 1 Review ............................................................................................................................... 38

Chapter 2 Applying forces ....................................................................................................... 47

2.1 Gravitational fields ....................................................................................................................... 47

2.2 Satellite motion ............................................................................................................................ 52

2.3 Torque .......................................................................................................................................... 56

2.4 Equilibrium .................................................................................................................................. 58

Chapter 2 Review ............................................................................................................................... 63

Chapter 3 Understanding electromagnetism ......................................................................... 73

3.1 Magnetic fields ............................................................................................................................. 73

3.2 Force on current-carrying conductors ........................................................................................... 73

3.3 Electric motors ............................................................................................................................. 76

3.4 Electric fields in circuits ............................................................................................................... 77

3.5 Electric circuits ............................................................................................................................. 79

Chapter 3 Review ............................................................................................................................... 81

Chapter 4 Generating electricity ............................................................................................ 88

4.1 Magnetic flux and induced currents .............................................................................................. 88

4.2 Induced EMF: Faraday’s law ....................................................................................................... 91

4.3 Electric power generation ............................................................................................................. 94

4.4 Transformers ................................................................................................................................ 98

4.5 Distributing electricity ................................................................................................................ 100

Chapter 4 Review ............................................................................................................................. 105



Page 2 of 185

Fairground physics E1

2

c

2

1 1

c

(6.00 2) m

9.80 m s [(0.0200 9.80)(3.00)]

0.0200 7.67 10 m s

vr

r

v

g v

a

g

a

E2 A. Inside column, as a smaller distance is covered in the same time period, which means that the

centripetal force is less on the small child

E3 Outside column, as a greater distance is covered in the same time period, which means that the

centripetal force will be greater on the older child.

E4 -2

c c c

1

c

1

c

0.0500 9.80 m s

(20.0 50.0) kg (70.0)(4.90 10 )

3.43 10 N

m

m

a F a

F

F

1

attachment

2

attachment

(3.43 10 ) 12

4.12 10 N

F

F

E5 2

- c wall on rider

2

c wall on rider

2

c wall on rider

(60.0)(4.00)60.0 kg

(5.00)

5.00 m 1.92 10 N

mv

r

m

r

F

F

F

1

towards the centre

4.00 m sv

E6 2

- c wall on rider

wall on rider c wall on rider

2.00

10.0 m (2

v

r

g v r

r v

a

a a

1 1

1

1 1

9.80)(10.0) 196

4.47 10 m s

14 m s

1.4 10 m s

v

v



Page 3 of 185

E7

3

down track

2 3

down track

3

down track

1.00 10 kg cosθ

9.80 m s (1.00 10 )(9.80)cos40

θ 50.0 7.51 10 N down the slope

m m

F g

g F

F

E8

down slope down slope

2

2

500.0 N cosθ

(500.0)9.80 m s

(9.80) (cos60 )

θ 60.0 1.02 10 kg

m

m

m

F F g

g

E9

k, gain p, lost

2 212

50.0 m

9.80 m s

2

2(9.80)(

y

y

y

E E

mv m

v

v

h

g g h

g h

1 1

50.0)

3.13 10 m sv

E10

friction p,start p,end

2

friction

friction

(75.0 62.0) m

9.80 m s

900.0 kg (900.0)(9.80)(13.0)

y y

y

E E E m

E m

m E

h g h

g g h

5

friction 1.15 10 JE

E11 1

k,gain p,lost

1 212

22

2

100.0 km h

27.78 m s

9.80 m s 2

(27.78)

2(9

y

y

y

v E E

v mv m

v

g h

g hg

h

1

.80)

3.94 10 my h

E12 Same maximum speed, as the mass of the car and riders cancels out of the equation.



Page 4 of 185

E13 1

p,gain k,lost

2 212

2

2

19.0 m s

9.80 m s

2

(19.0)

2(9.80

v E E

m mv

v

g g h

hg

h

1

)

1.84 10 m h

E14

E15 2

1

2 22

6.00 m s 2

(6.00)9.80 m s

2 2(9.80)

1.84 m

mvv m

v

gr

g rg

r

E16 2

2

1 1

10.0 m 1.5

9.80 m s 1.5 (1.50)(10.0)(9.80)

1.21 10 m s

mvm

v

v

r gr

g rg

E17 2

1

c

3 23

c

4

c

15.0 m s

(1.00 10 )(15.0)1.00 10 kg

(20.0)

20.0 m 1.13 10 N

mvv

m

Fr

F

r F

E18 A 2.00g ride can be created if the accelerating force is vertical as opposed to horizontal, for

100.0% 90.0% 81.0% 72.9% 65.6% 59.0%



Page 5 of 185

example a 2.00g force can be experienced at the bottom of a curve.

E19

small c, total c, small c, large

22large1 small

small c, total

small large

large c, total

2.00 m

0.500 m s

(50.0)(0.5010.0 m

mvmvv

r F F F

Fr r

r F2 2

1

large c, total

1

c, total

0) (50.0)(2.00)

(2.00) (10.0)

2.00 m s (6.25 (20.0)

50.0 kg 2.63 10 N

v

m

F

F

E20

small c, total c, small c, large

22large1 small

small c, total

small large

2

large c, total

2.00 m

1.00 m s

(1.00)10.0 m

(2.00

vvv

r a a a

ar r

r a2

1

large c, total

1 2

c, total

(0.600)

) (10.0)

0.600 m s (0.500) (0.0036)

5.36 10 m s

v

a

a

E21 a

1 1

yellow

yellow yellow yellow purple purple yellow purple yellow purple

pu

13.0 km h 3.61 m s momentum is conserved, so:

420.0 kg

m m m m

v

u

u u

uyellow purple purple1 1

rple yellow purple

y purple

purple yellow purple

7.00 km h 1.94 m s

(420.0)(3.61) (500.0)( 1.94)500.0 kg

(920.0

ellow

m m

m

m

u uv

v

1 1 1

yellow purple yellow purple

)

920.0 kg 5.92 10 m s or 2.13 km hm

v



Page 6 of 185

b

1 1

yellow

yellow yellow yellow purple purple yelloow purple yellow purple


420.0 kg

m m m m

u

u u v

uyellow yellow purple purple1 1


yellow purple


6.00 km h 1.67 m s

(420.0)(1.67) (500.0)( 1500.0 kg

m m

m

m

u uv

v

1 1 1

yellow purple yellow purple

.67)

(920.0)

920.0 kg 1.45 10 m s or 5.22 km hm

v

E22

1 1

orange

orange orange orange silver silver silver orange silver

s


410.0 kg m

orangem m m

m

u

u u v

orange silver orange silver orange orange

ilver silver

silver

1

orange silver silver

350.0 kg

(760.0)(0.00) (40.00 m s

m m

m

v uu

v u

1 1 1

orange silver silver

10.0)(2.78)

(350.0)

760.0 kg 3.26 m s or 1.17 10 km hm

u

E23

1 1

gold

gold gold gold green green green gold green gold


530.0 kg

800.0 kg

green

m m m m

m

u

u u v

gold gold green green

green gold

green gold

1

green green gold

green gold

(530.0)(4.17) (800.0)(0.00)0.00 m s

(1330.0)

13

m m

m

m

u uv

u v

1 1

green gold30.0 kg 1.66 m s or 5.98 km h

v



Page 7 of 185

E24

1 1

turquoise

turquoise final turquois

In the direction:

6.00 km h 1.67 m s momentum in the direction is conserved, so:

425.0 kg

x

x

x

m m

ρ

u

e turquoise

final

2 1

final

(425)(1.67)

7.08 10 kg m s

As

x

x

ρ

ρ

u

2 1

white

the final direction is south-east, then the momentum in the direction

must be equal to that in the direction.

7.08 10 kg m s momentum in the direc

z

x

z

ρ

white final white white

finalwhite

white

tion is conserved, so:

675.0 kg

(7.08

z

z

m m

m

ρ

ρ

u

u2

1

white

10 )

(675.0)

1.05 m su



Page 8 of 185

Power distribution and generation E1

10691 m 002

10001

00210652 m

2

10002

m 10652

4cable

22

8

cable

2

2cable8

Al

.R.L

).(

).)(.(R

.r

r

L

A

LR.

ρρρ

E2 a

V 10061

2

10015

2 V 10015

4rms

3

rms

peakrms

3peak

.V

).(V

VV.V

b i

3 s ss

p p

33

p 3

9.00 10 V turns ratio

(9.00 10 )15.0 10 V turns ratio

(15.0 10 )

turns ratio 0.600 :1 or 1:1.67

N VV

N V

V

ii The current in the secondary coil of a step-down transformer is always greater than

the current in the primary coil, as current and voltage are inversely proportional in a

transformation.

E3

total ac r cb fs sp dl

3

total

3

total

(1.00 10 ) (10.0) (720.0) (50.0) (20.0) (30.0)

1.83 10 W

P P P P P P P

P

P

Substituting into P = VI:

3 totaltotal total

house

3

house total

total

1.83 10 W

(1.83 10 )240.0 V

(240.0)

7.63 A

PP I

V

V I

I

This is too large to run all these appliances at once; the extension cord on the power pack could

heat up too much.



Page 9 of 185

E4 a

gen3

gen TL

gen

3

gen TL

1

TL

20.0 10 W

(20.0 10 )250.0 V

(250.0)

8.00 10 A

PP I

V

V I

I

1 2

TL loss TL TL

1 2

TL loss

3

loss

8.00 10 A

1.20 (8.00 10 ) (1.20)

7.68 10 W

I P I R

R P

P

b

3 lossloss drop

TL

31

TL drop 1

1

drop

7.68 10 W

(7.68 10 )8.00 10 A

(8.00 10 )

9.60 10 V

PP V

I

I V

V

start TL end TL start TL drop TL

1 1

drop TL end TL

2

end TL

250.0 V

9.60 10 V (250.0) (9.60 10 )

1.54 10 V

V V V V

V V

V

c They should be able to use some appliances; however, lights would not be as bright,

motors would not go as fast. Electronic devices may not function at all.

d

3 ss TL

s

33

s TL 3

TL

20.0 10 W

(20.0 10 )6.00 10 V

(6.00 10 )

3.33 A

PP I

V

V I

I

2

TL loss TL TL

2

TL loss

1

loss

3.33 A

1.20 (3.33) (1.20)

1.33 10 W

I P I R

R P

P



Page 10 of 185

e

1 lossloss drop

TL

1

TL drop

drop

1.33 10 W

(1.33 10 )3.33 A

(3.33)

4.00 V

PP V

I

I V

V

3

start TL end TL start TL drop TL

3

drop TL end TL

end TL

6.00 10 V

4.00 V (6.00 10 ) (4.00)

5.99

V V V V

V V

V

36 10 V

s ss

p p

3

p 3

240.0 V turns ratio

(240.0)5.996 10 V turns ratio

(5.996 10 )

turns ratio 0.0400 :1 or

N VV

N V

V

1:25.0



Page 11 of 185

Chapter 1 Analysing motion

1.1 Projectile motion

1 a

212

212

-2

4.90 m ( 4.90) (0) ( 9.80)

2( 4.90)9.80 m s

( 9.80)

1.00 s

y

y

t t

t t

t

t

s u a

s

g

b

1

(20.0)(1.00)

20.0 m s 20.0 m

1.00 s

x x

x x

- v t

t

s

v s

c The only force acting on the ball is gravity, therefore the only acceleration is

g = –9.80 m s–2.

d

1 1

2

2 2 2 2

(0) ( 9.80)(0.800)

0 m s 7.84 m s

9.80 m s

0.800 s (20.0) ( 7.84)

y

y y

x y

t

t v v v

v u a

u v

g

1 18.4 m sv

e

1 1

2

2 2 2 2

(0) ( 9.80)(1.00)

0 m s 9.80 m s

9.80 m s

1.00 s (20.0) ( 9.80)

y

y y

x y

t

t v v v

av u

u v

g

1 22.3 m sv



Page 12 of 185

2 a Horizontal velocity remains constant, so vx = 10.0 m s–1.

b

2 2 2

1 1

2 1

2 (0) 2( 9.80)( 1.00)

0 m s 4.43 m s

9.80 m s 4.43 m s down

1.00 m

y y y

y y

y

y

s

v u as

u v

g v

c

2 2 2 2

1 1

1

(10.0) ( 4.43)

4.43 m s 10.9 m s

10.0 m s

4.430.800 s sin 0.4048

10.9

x y

y

x

t

v v v

v v

v

1

23.9

10.9 m s 23.9 down from horizontal

v

d

212

212

2

1.00 m ( 1.00) (0) ( 9.80)

2( 1.00)9.80 m s

( 9.80)

0.452

y

y

t t

t t

t

t

s u a

s

g

s

e

0.452 s (10.0)(0.452)

10.0 m 4.52 m

x x

x x

t t

s v

v s

f

Fgravity



Page 13 of 185

3 a

1

cos30.0

cos30.0 (28.0)(0.8660)

24.2 m s

x

x

x

v

v

v

v

v

b 124.2 m sx

v

c 124.2 m sx

v

4 a

1

sin30.0

sin30.0 (28.0)(0.5000)

14.0 m s up

y

y

x

v

v

v v

v

b

1 1

2

(14.0) ( 9.80)(1.00)

14.0 m s 4.20 m s up

9.80 m s

1.00 s

y

y y

t

t

v u a

u v

g

c

1 1

2 1

(14.0) ( 9.80)(2.00)

14.0 m s 5.60 m s

9.80 m s 5.60 m s down

2.00 s

y

y y

y

t

t

av u

u v

g v

5 a

1

2

1

(0 14.0)14.0 m s

( 9.80)

9.80 m s 1.43 s

0 m s

y y

y y

y

y

t

t

t

a

v u a

v uu

g

v



Page 14 of 185

b 21

2

1 212

2

14.0 m s (14.0)(1.43) ( 9.80)(1.43)

9.80 m s 10.0 m

1.43 s

y y

y y

y

t t

t

s u a

u s

g s

c The acceleration of the ball is constant, due to gravity = –9.80 m s–2.

6 a At its maximum height, this is the point at which the ball has zero vertical velocity, while

maintaining its horizontal velocity.

b 124.2 m sx

v

c -1s m 431.t

d

7 a

1

1

2

14.0 m s

( 14.0) (14.0)14.0 m s

( 9.80)

9.80 m s 2.86 s

y y

y y

y

y

t

t

t

t

gv u

v uu

g

v

g

b

2 2 2 2

1 1

1

( 14.0) (24.2)

14.0 m s 19.8 m s

24.2 m s

14.00.800 s sin 0.7071

19.8

x y

y

x

t

v v v

v v

v

1

45.0

19.8 m s 45.0 down from horizontal

v

Fgravity



Page 15 of 185

c

1

2.86 s (24.2)(2.86)

24.2 m s 69.3 m

x x

x x

t t

s v

v s

8 C

9 a

k p

k final

p k initial k final

1 21k final2

212 k 2

At maximum height loss of is due to ain in

16.0 J

0.2500 kg

16.0 m s

9.80 m s

E g E

E

m E E E

u mgh mu E

mu Eg h

21final 2

(0.2500)(16.0) (16.0)

(0.2500)(9.80)

6.53 m

mg

h

b 2 2

2

1 2

2 1

2

6.53 m 2

0 m s (0) 2( 9.80)(6.53)

9.80 m s 11.3 m s

y

v u as

s u v as

v u

g u

c

(11.3)cos 0.7071

(16.0)

45.0

x

v

v

d

1 1

2

2 2

(11.3) ( 9.80)(1.00)

11.3 m s 1.51 m s up

9.80 m s

1.00 s (1.51) (11.3)

y

y y

t

t v

g

v u a

u v

1 11.4 m sv



Page 16 of 185

e 21

2

1 212

2

11.3 m s (11.3)(1.00) ( 9.80)(1.00)

9.80 m s 6.41 m

1.00 s

y y

y y

y

t t

t

s u a

u s

g s

2 2 2 2

(11.3)(1.00) 11.3 m

(6 41) (6.41) (11.3) tan 0.5669

(11 3)

13.1 m 29.5 up from ho

x x

y x

v t

.

.

s

s s s

s rizontal

f

1

2

1

( 11.3) (11.3)11.3 m s

( 9.80)

9.80 m s 2.31 s

11.3 m s

y y

y y

y

y

t

t

t

v u g

v uu

g

g

v

g

1

2.31 s (11.3)(2.31)

11.3 m s 26.1 m

x x

x x

t t

s v

v s

10 a

2 1

1

1.50 s (0) ( 9.80)(1.50)

9.80 m s 14.7 m s

0 m s

y y

y y

y

y

t

t t

v u g

u v g

g u

v

1(14.7) 22.9 m s

sin sin40.0

y

uv



Page 17 of 185

b 21

2

1 212

2

14.7 m s (14.7)(1.50) ( 9.80)(1.50)

9.80 m s 11.0 m

1.50 s

y y

y y

y

t t

t

s u a

u s

g s

c 2 2

1 2

2 1

2

14.7 m s (14.7) 2( 9.80)( 10.0)

9.80 m s 20.3 m s

10.0 m

y y y

y y

y

y

v u as

u v

g v

s

( 20.3) (14.7)

( 9.8)

3.57 s

y y

y y

t

t

t

v ua

v u

a

1.2 Circular motion in a horizontal plane

1 a A, D

b She has continued to travel in a straight line, while the car has turned, so the right side of

the cabin is actually accelerating towards her.

2 a 18.00 m sv

b 18.00 m s southv

c

2 21

c

2

c

(8.00)8.00 m s

(9.20)

9.20 m 6.96 m s east

vv

r

r

a

a

3 a 2 3 2

1

c

3

c

3

(1.20 10 )(8.00)8.00 m s

(9.20)

9.20 m 8.35 10 N east

1.20 10 kg

mvv

r

r

m

F

F



Page 18 of 185

b The force that causes the centripetal force is the reaction of the sideways frictional force of

the car’s tyres on the road, that is the sideways force of friction of the road on the car’s

tyres.

4 a 18.00 m s northv

b west

5 The car would probably skid off the road as the centripetal force required would increase to a

value greater than the force of friction could provide.

6 a

2 21

c

2

c

(2.00)2.00 m s

(1.50)

1.50 m 2.67 m s towards the centre

vv

r

r

a

a

b The forces are unbalanced as she is accelerating. According to Newton’s first law an

unbalanced force will cause an object to change its motion, in this case the direction of the

motion is changing, not the magnitude.

c 2 2

1

c

2

c

(50.0)(2.00)2.00 m s

(1.50)

1.50 m 1.33 10 N towards the centre

50.0 kg

mvv

r

r

m

F

F

d The sideways reaction force of the skate on the ice, which is the sideways force of the ice

on the skate.

7 a

1 11 12.00 rev s 5.00 10 s

(2.00)f T

f

b

1

1

1

2 2 (0.800)5.00 10 s

(5.00 10 )

10.1 m s

rT v

T

v

c

2 21

c

2 2

c

(10.1)10.1 m s

(0.800)

0.800 m 1.26 10 m s towards the centre

vv

r

r

a

a



Page 19 of 185

d 2 2

1

c

2

c

(2.50)(10.1)10.1 m s

(0.800)

0.800 m 3.16 10 N towards the centre

2.50 kg

mvv

r

r

m

F

F

e The force causing the centripetal acceleration of the ball is the tension force of the cable on

the ball.

f The ball would continue in a straight line that is tangential to the circular path at the point

at which the wire breaks.

8 a

2.40 m cos

cos (2.40)(cos 60.0 )

1.20 m

rl

l

r l

r

b The forces acting on Ella are gravity and the tension force of the rope on her.

c Ella’s acceleration is towards the centre of rotation about the pole.

d

g2

c

c

2

c

9.80 m s tan

(30.0)(9.80)30.0 kg

tan (tan 60.0 )

1.70 10 N towards the centre

mm

Fg

F

gF

F

e 2

2

c c

2

c

1

1.70 10 N

(1.70 10 )(1.20)1.20 m

(30.0)

30.0 kg 2.61 m s

mv

r

rr v

m

m v

F F

F

9 a

2 3 21

c

c

3

(1.20 10 )(18.0)18.0 m s

(80.0)

80.0 m 4.86 kN towards the centre

1.20 10 kg

mvv

r

r

m

F

F



Page 20 of 185

b 3

g

3

c

(1.20 10 )(9.80)tan 2.42

(4.86 10 )

67.5

(90.0 ) (67.5 )

22.5

F

F

10 The driver will have to turn the car’s tyres down the track to enable the horizontal component of

the sideways frictional force to help turn the car. The combined centripetal force of the banked

track and the horizontal component of the sideways frictional force will enable the car to turn at

this higher speed while maintaining the same radius as before.

1.3 Circular motion in a vertical plane

1 a The acceleration is towards the centre of the circular path of the yo-yo.

b At the bottom of the circular path the tension in the string is greatest.

c At the top of the circular path the tension in the string is lowest.

d At the bottom of the circular path where the tension in the string is greatest.

2 2

g c

2

1

9.80 m s critical speed when

1.50 m

( 1.50)( 9.80)

3.83 m s

g

vr

r

v r

v

a a

g

g

3 a The force of gravity and the reaction force of the road on the car.

Fg

Fc



Page 21 of 185

b

road on car c g

2

road on car

21

road on car

800.0 kg ( )

10.0 m ( )

(800.0)(4.00)14.4 km h ( (800.0)( 9.80))

( 10.0)

4.00 m s

m

mvr m

r

v

v

F F F

F g

F

1 3 3

road on car

2 3

road on car

( 1.28 10 ) (7.84 10 )

9.80 m s 6.56 10 N upwards

F

g F

c Yes it is possible, however it is her apparent weight she was ‘feeling’ not her mass, which

doesn’t change. The force that the seat applies to her is less as she goes over the hump,

therefore she feels like she is lighter on the seat.

d 2

g c

2

1


10.0 m

( 10.0)( 9.80)

9.90 m s

g

vr

r

v r

v

a a

g

g

1 35.6 km hv

4 a 1

k at Y k at X p gain

2 21 12 2

2 2 21 12 2

2.00 m s

50.0 m

9.80 m s 2( ) 2( (2.00) (9.80)(50.0))

500.0 kg

y y

y

u E E E

s mv mu mg s

v u g s

m

g

1 31.4 m sv

b 1

k at Z k at X p gain

2 21 12 2

2 2 21 12 2

2.00 m s

20.0 m

9.80 m s 2( ) 2( (2.00) (9.80)(20.0))

500.0 kg

y y

y

u E E E

s mv mu mg s

v u g s

m

g

1 19.9 m sv



Page 22 of 185

c

track on cart c g

2

track on cart

21

track on cart

500.0 kg ( )

15.0 m ( )

(500.0)(19.9)19.9 m s ( (500.0)( 9.80))

( 15.0)

m

mvr m

r

v

g

F F F

F

F

g 2 4 3

track on cart

3

track on cart

3

track on cart

9.80 m s ( 1.32 10 ) (4.90 10 )

8.30 10 N

8.30 10 N do

F

F

F wnwards

5 2

g c

2

1


15.0 m

( 15.0)( 9.80)

12.1 m s

g

vr

r

v r

v

a a

g

g

6

seat on pilot c g

2

seat on pilot

2-1

seat on pilot

80.0 kg ( )

100.0 m ( )

(80.0)(35.0)35.0 m s ( (80.0)( 9.80))

( 100.0)

m

mvr m

r

v

F F F

F g

F

g -2 2 2

seat on pilot

2

seat on pilot

2

seat on pilot

9.80 m s ( 9.80 10 ) (7.84 10 )

1.96 10 N

1.96 10 N dow

F

F

F nwards

7 2

g c

2

1


100.0 m

( 100.0)( 9.80)

31.3 m s

g

vr

r

v r

v

a a

g

g



Page 23 of 185

8

2

seat on pilot

2

seat on pilot

2

9.80 m s 9 ( )

100.0 m ( )

9 ( ( 9.80)) ( ( 9.80))(100.0)

vr

r

v

g a g

a g

1

8 ( ( 9.80))(100.0)

88.5 m s

v

v

9 a

g

2

4.00 kg (4.00)( 9.80)

9.80 m s 39.2 N

m m

T F g

g T

b

k p

-2 212

-1

4.00 kg

9.80 m s

2.00 m 2(9.80)(2.00)

6.26 m s

y

y

m E E

mv m s

s v

v

g g

c g

2

2

( )

( )

(4.00)(6.26) ( (4.00)( 9.80))

(2.00)

mvm

r

T F F

T g

T

T 1 1

2

(7.84 10 ) (3.92 10 )

1.18 10 N

T

10 The wire is more likely to break when the ball is moving through position X as the tension in the

wire is three times the tension it had when it was stationary at point X.



Page 24 of 185

1.4 Vectors and free-body diagrams

1 a

1 1

1 1

( 15.0) ( 20.0)

15.0 m s west 15.0 m s 35.0

20.0 m s west 20.0 m s

x

x

v u

u v u

v1 35.0 m s west v u

b

1 1

1 1

( 15.0) ( 20.0)

15.0 m s west 15.0 m s 5.0

20.0 m s east 20.0 m s

x

x

v u

u v u

v1 5.0 m s east v u

c

1 2

1 1 2

2 1

( 10.0) ( 12.0)

10.0 N up 10.0 N 2.0

12.0 N down 12.0 N

y y

y y y

y y y

F F

F F F

F F F 2 2.0 N down

2

a

1

1 1

( ) ( 20.0) ( 15.0)

15.0 m s west ( ) 5.0

15.0 m s east 15.0 m s

x

x

v u

u v u

u 1

1 1

( ) 5.0 m s west

20.0 m s west 20.0 m s x

v u

v

b

1

1 1

( ) ( 20.0) ( 15.0)

15.0 m s west ( ) 35.0

15.0 m s east 15.0 m s

x

x

v u

u v u

u 1

1 1

( ) 35.0 m s east

20.0 m s east 20.0 m s x

v u

v



Page 25 of 185

c

2 1

1 2 1

1

( ) ( 12.0) ( 10.0)

10.0 N up 10.0 N ( ) 22.0

10.0 N down 10.0 N

y y

y y y

y

F F

F F F

F F 2 1

2

( ) 22.0 N down

12.0 N down 12.0 N

y y

y

F

F

3 a 2 2 2 2

2 2 2

1

17.5 m s south ( )

70.0 m s west ( ) (17.5) (70.0)

( ) 7.22 10 m s

x x z x z

z x z

x z

a a a a a

a a a

a a 2

(70.0) tanθ

(17.5)

θ tan(4.00)

θ

z

x

a

a

1 2

76.0

( ) 7.22 10 m s south76.0 westx z

a a

b 1 2 2 2

1 2 2

1

15.0 m s west ( )

23.0 m s north ( ) (15.0) (23.0)

( ) 2.75 10 m s

x x z x z

z x z

x z

v v v v v

v v v

v v 1

(23.0) tanθ

(15.0)

θ tan(1.53)

θ

z

x

v

v

1 1

56.9

( ) 2.75 10 m s north 56.9 westx z

v v



Page 26 of 185

c 2 2 2

2 2

1

15.0 N up ( )

20.0 N west ( ) (15.0) (20.0)

( ) 2.50 10

y y x y x

x y x

y x

F F F F F

F F F

F F N

(15.0) tanθ

(20.0)

θ tan(0.750)

θ

y

x

F

F

1

36.9

( ) 2.50 10 N 36.9 up from horizontal to the westy x

F F

4

4 a

22 2 2

2 2 2

2 1 2

19.0 m s west ( ) ( )

19.0 m s east ( ) (20.0) (19.0)

20.0 m s south ( ) 2.76 10 m s

x z x z x

x z x

z z x

v v v v v

v v v

v v v

( ) (19.0) tanθ

(20.0)

θ tan(0.950)

x

z

v

v

1 2

θ 43.5

( ) 2.76 10 m s south 43.5 eastz x

v v

b

21 2 2

1 2 2

1 1

11.0 kg m s north ( ) ( )

11.0 kg m s south ( ) (30.0) (11.0)

30.0 kg m s east ( ) 3.20 10

x z x z x

x z x

z z x

p p p p p

p p p

p p p 1kg m s

(30.0) tanθ

( ) (11.0)

θ tan(2.73)

z

x

p

p

1 1

θ 69.9

( ) 3.20 10 kg m s south 69.9 eastz x

p p



Page 27 of 185



Page 28 of 185

c 2

2 2

2 2

10.0 N down ( ) ( )

10.0 N up ( ) (12.0) (10.0)

12.0 N right ( )

y x y x y

y x y

x x y

F F F F F

F F F

F F F11.56 10 N

( ) (10.0) tanθ

(12.0)

θ tan(0.833)

y

x

F

F

1

θ 39.8

( ) 1.56 10 N 39.8 up from

hor

x y

F F

izontal to the right

5 a

x x

x

2 1

x

z z

west cos35.0

(255)(0.819)

2.09 10 m s west

south sin35.0

v v v

v

v

v v v

z

2 1

z

(255)(0.574)

1.46 10 m s south

v

v

b

1 1

east sin67.5

(0.250)(0.924)

2.31 10 kg m s east

north cos67.5

x x

x

x

z z

p p p

p

p

p p p

2 1

(0.250)(0.383)

9.57 10 kg m s north

z

z

p

p



Page 29 of 185

c

1

left cos27.5

(100.0)(0.8874)

8.87 10 N left

down sin27.5

x x

x

x

y y

F F F

F

F

F F F

1

(100.0)(0.462)

4.62 10 N down

y

y

F

F

6 a

1 2 2 2

plane plane wind plane wind

1 2 2

wind plane wind

100.0 m s south ( )

25.0 m s west ( ) (100.0) (25.0)

v v v v v

v v v

2 1

plane wind

wind

plane

( ) 1.03 10 m s

(25.0) tanθ

(100.0)

θ tan(0.250)

v v

v

v

2 1

plane wind

θ 14.0

( ) 1.03 10 m s south 14.0 west

v v

b The plane should steer south 14.0° east to maintain a southerly path.

c

1 2 2 2

plane plane plane south wind

1 2 2

wind plane south

100.0 m s south 14.0 east ( )

25.0 m s west (100.0) (25.0)

v v v v

v v

1 1

plane south 9.68 10 m s south v

7 a

1

right cos60.0

(50.0)(0.500)

2.50 10 N right

x x

x

x

F F F

F

F

b

Fground on trolley

Fgravity on trolley

Fpush on trolley



Page 30 of 185

c

, push , friction

, push

50.0 N 60.0 below horizon to rt

( 25.0) ( 10.0)

25.0 N right

x x x

x

x

F F F F

F

F 1

, friction

, friction

1.50 10 N right

10.0 N left

10.0 N sin60.0

x

x

x y

F

F

F F F

1

(50.0)(0.866)

4.33 10 N down

y

y

F

F

2 2 2

resultant

1 2 1 2

resultant (1.50 10 ) (4.33 10 )

x y

F F F

F

F1

resultant

1

1

4.58 10 N

(4.33 10 ) tanθ

(1.50 10 )

θ

y

x

F

F

1

1

resultant

tan (2.887)

θ 70.8

4.58 10 N 70.8 down from

F

horizontal to the right

8 a

1

1

1

16.0 m s speed -

16.0 m s (16.0 16.0)

0.00 m s

x x x x

x x

x

u v v u

v v

v



Page 31 of 185

b 1

1 2 2 2

1 2 2

16.0 m s north speed ( )

16.0 m s south ( )

16.0 m s east (16.0) (16.0)

u v v u

u v v u

v v

1 1 2.26 10 m s

(16.0) tanθ

(16.0)

θ tan(1.00)

v

u

v

1 1

θ 45.0

2.26 10 m s south 45.0 east

v

9

10

1.5 Motion in a straight line

1 a A to B: Displacement 40 cm to the right, Distance 40 cm

b C to B: Displacement 10 cm to the left, Distance 10 cm

c C to D: Displacement 20 cm to the right, Distance 20 cm

d C to E and then to D: Displacement 20 cm to the right, Distance 80 cm

2 a Distance 80.0 km

b Displacement 20.0 km north

3 a 10 m down

b 60 m up

c 70 m

d 50 m up

Fclub on ball

Fgravity on ball

Ftee on ball

Fair resistance on ball

Fgravity on ball



Page 32 of 185

4 displacement

5 a D

b D

c C

d A

6 a 39 steps

b 1 step west of the clothes line

c 1 step west

7 a

1

av

1 1

av

(33.3) (0)0 m s

2 2

120.0 33.3 m s 16.7 m s3.6

v uu v

v v

b

1

av

1 1 1

av

(120.0) (0)0 km h

(18.0)

120 km h 6.67 km h s forwards

18.0 s

t

t

v uu a

v a

c

1

av

1 2

(33.3) (0)0 m s

(18.0)

120 33.3 m s 1.85 m s3.6

18.0 s

t

t

v uu a

v v

d

133.3 m s (33.3)(0.600)

0.600 s 20.0 m

v s v t

t s

8 a

1

1 1

25.0 m s (15.0) (25.0)

15.0 m s 10.0 m s

u v v u

v v



Page 33 of 185

b

1

1 1

1 1

25.0 m s east ( ) ( 15.0) ( 25.0)

25.0 m s west 10.0 m s

25.0 m s 10.0 m s east

v

v v u

u v v u

u

u v

1

1

15.0 m s west

15.0 m s

v

v

c

av

1 2

av

1 2 2

av

( 15.0) ( 25.0)

(0.0500)

25.0 m s 2.00 10

15.0 m s 2.00 10 m s east

0.0500 s

t

t

v ua

u a

v a

9 a

1

av

1 1 1

av

(0) (60.0)60.0 km h north

(5.00)

0 km h 12.0 km h s north

5.00 s

t

t

v uu a

v a

b

1

av

1 2

av

(0) (16.7)60.0 16.7 m s 3.6 (5.00)

0 m s 3.33 m s north

5.00 s

t

t

v uu a

v a

10 a

3

50.0 m laps (50.0)(30)

laps 30 1.50 10 m

l s l

s

b

33

av

1

av

(1.50 10 )1.50 10 m

(878)

14:38 1.71 m s

878 s

ss v

t

t v

t

c 0 m

d 0 m s–1



Page 34 of 185

1.6 Energy and momentum

1 a

2

d

2 2

d

5

d

8.00 10 kg

9.80 m s (8.00 10 )(9.80)(90.0)

90.0 m 7.06 10 J

y y

y

m W Fs mgs

g W

s W

b 2 21

total k p 2

2 2 2 21total 2

5

total

1

8.00 10 kg

9.80 m s (8.00 10 )(2.00) (8.00 10 )(9.80)(50.0)

50.0 m 3.94 10 J

2.00 m s

y

y

m E E E mv mgs

g E

s E

v

c

2

2 2

1 4

8.00 10 kg

9.80 m s (8.00 10 )(9.80)(2.00)

2.00 m s 1.596 10 W

m P Fv mgv

g P

v P

2 a 2 21 1

total k1 k2 1 1 2 22 2

1 1 2 1 21 11 total 2 2

1

1 total

2

2.00 10 kg (2.00 10 )( 9.00) (1.00 10 )(0)

9.00 m s 8.10 J

1.00 1

E E E m u m u

m E

u E

m

1

1

2

0 kg

0 m su

b 2 21 1

total k1 k2 1 1 2 22 2

1 1 2 1 21 11 total 2 2

1

1 total

2

2.00 10 kg (2.00 10 )( 3.00) (1.00 10 )( 12.0)

3.00 m s 8.10 J

1.

E E E m v m v

m E

v E

m

1

1

2

00 10 kg

12.0 m sv

c This collision is elastic as no kinetic energy is lost in the collision.

d This is an unrealistic situation as is all macroscopic collisions some kinetic energy is

always lost in the form of heat or sound.



Page 35 of 185

3 a

1

2 1

3

8.00 10 kg

9.80 m s (8.00 10 )(9.80)(5.00)

5.00 m 3.92 10 J

yp

p

yp

m E mg s

g E

s E

b 1 2 2 2

2 1

-1

12

8.00 10 kg 2 (0) 2(5.00)(5.00)

5.00 m s (5.00 10 )

5.00 m 7.07 m s

y

y

k

m v u as

a v

s v

E

2 1 212

3

(8.00 10 )(7.07)

2.00 10 Jk

mv

E

c Some energy potential energy would have been lost to heat or sound energy as the

firefighter slid down the pole, therefore there would be less energy converted into kinetic

energy.

d The work done is equal to the gain in energy of the firefighter, equal to 2.00 × 103 J.

e

p heat p k

k heat

3 3 3

3 3

Assume that all lost energy is transferred into heat.

3.92 10 J (3.92 10 ) (2.00 10 )

2.00 10 J 1.92 10 J

E E E E

E E

4 D

5 a

1

1 1

10.0 m s Δ (8.00) (10.0)

8.00 m s Δ 2.00 m s

u v v u

v v

b

1

1 1

Δ ( 8.00) ( 10.0)

10.0 m s Δ 18.0

8.00 m s Δ 18.0 m s up

v v u

u v

v v

c

3

1

3 1 1

Δ (80.0 10 )( 18.0)

Δ 18.0 m s Δ 1.44

80.0 10 m s Δ 1.44 kg m s up

m

m

p v

v p

p



Page 36 of 185

d Impulse is equal to change in momentum = 1.44 kg m s–1 up.

6 a

av net

1

av net

av net

( 1.44)

(0.0500)

Δ 1.44 kg m s 28.8

0.0500 s 28.8 N up

t

t

pF

p F

F

b

av av net wt

3

av av

3

av

( )

28.8 N ( 28.8) ( 80 10 )( 9.80)

80.0 10 kg 29.6

9.8

m

g

F F F

F F

F

1

av0 m s 29.6 N up F

c According to Newton’s third law, the force of the ball on the floor is 29.6 N down

7 a 3

sc

1 4

sc

1 4 1

sc

3

(1.00 10 )( 10.0)

36.0 km h 1.00 10

10.0 m s 1.00 10 kg m s east

1.00 10 kg

m

m

p v

v p

p

b 3

wag

1 4

wag

1 4 1

wag

3

(2.00 10 )( 5.0)

18.0 km h 1.00 10

5.0 m s 1.00 10 kg m s west

2.00 10 kg

m

m

p v

v p

p

c

4 4

total sc wag

4 1 -1

sc total

4 1

wag

( 1.00 10 ) ( 1.00 10 )

1.00 10 kg m s 0 kg m s

1.00 10 kg m s

p p p

p p

p

8 a

before after

1

before after

1

combined

0 kg m s 0

0 m sv

p p

p p

b The initial momentum has gone into changing the momentum of the other vehicle.



Page 37 of 185

c 3

sc

1 4

sc

1 4 1

sc

( ) (1.00 10 )[(0) ( 10.0)]

10.0 m s 1.00 10

0 m s 1.00 10 kg m s west

1.

m

m

p v u

u p

v p

300 10 kg

d

kg 10 002

east s m kg 10001 s m 0

10001 s m 005

005010002

3

1-4wag

1-

4wag

1-

3wag

.m

.

..

)].())[(.()(m

pv

pu

uvp

9 Mary was correct, as the momentum before is equal to the momentum after, so the momentum of

the railway tanker and water (combined) will be equal to the sum of the momentum of the tanker

and water (separated). The sum of the masses of the water and tanker will be the same after as it

was before therefore the speed of the tanker and water will be the same after as it was before.

10 a

girl before

1 2

girl girl before

2 1

girl before

(48.0)( 4.00)

4.00 m s 1.92 10

48.0 kg 1.92 10 kg m s to the rig

m

m

p u

u p

p ht

b

before after sb sb girl girl girl sb girl sb

1

girl girl sb

girl gir

4.00 m s (2.00 (0) (48.0 ( 4.00) (50.0)

48.0 kg

m m m

) )

m

p p u u v

u v

v 1

l sb

1

sb

sb

3.84 m s to the right

0 m s

2.00 kgm

u

c

before after girl sb girl sb sb sb girl girl

1

girl sb girl sb girl sb sb sb girl girl

girl sb

3.84 m s

50.0 kg

m m m

m m m

m

p p u v v

u u v v

sb

1 1

girl sb

girl sb

(50.0 ( 3.84) (48.0 ( 3.84)

(2.00)

3.84 m s 3.84 m s

48.0 kg 3.84 m

) )

m

v

v v

v 1

sb

s to the right

2.00 kgm



Page 38 of 185

Chapter 1 Review

1 a

1

1 1

3.50 m s (3.00) (3.50)

3.00 m s 0.50 m s

u v v u

v v

b

1 1

1

(3.00) ( 3.50)

3.50 m s 6.50 m s upwards

3.00 m s

v v u

u v

v

2 a The force increases as the bounce continues to the point where the springs are stretched to

their maximum, then decreases as the bounce continues to the point where Hannah leaves

the trampoline. The force is usually named the reaction force.

b D

3 a 1 2 2

2 2 2 21

av

2

av

16.7 m s 2

(0) (16.7)0 m s

2 2(15.0)

15.0 m 9.26 m s

vu u as

v uv a

s

s a

b

1

2

av

1

16.7 m s

9.26 m s (16.7) ( 9.26)(1.50)

1.50 s 2.78 m s

t

t

u v u a

a v

v

c

av

2

av av

120.0 kg (120.0)( 9.26)

9.26 m s 1.11 kN

m m

F a

a F



Page 39 of 185

4 a 21

2

1 212

2

for ball X

0 m s ( 2.00) (0) ( 9.80)

2(-2.00)9.80 m s

( 9.80)

2.00 m 0

y

y

t t

t t

t

t

s u a

u

g

s .639 s

s 6390 m 002

809

-2.00)2 s m 809

8090-2.00) s m 0

Y ballfor

2-

2

211-

2

21

.t.

).(

(t.

t).(t)((

tt

y

y

s

g

u

aus

b

1 1

2

2 2 2

for ball X (0) ( 9.80)(0.639)

0 m s 6.26 m s

9.80 m s

0.639 s ( 6.26) (5.00

y y

y y

y x

t

t

g

v u a

u v

v v v 2

1 1

)

5.00 m s 8.01 m s x

v v

c

1 1

2

2 2 2

for ball Y (0) ( 9.80)(0.639)

0 m s 6.26 m s

9.80 m s

0.639 s ( 6.26) (7.50

y y

y y

y x

t

t

v u a

u v

g

v v v 2

1 1

)

7.50 m s 9.77 m s x

v v

d

1

- for ball X (5.00)(0.639)

5.00 m s 3.19 m for ball X

0.639 s

for ball Y (7.50)(0.639

x x

x x

x x

t

t

t

s v

v s

s v

1

)

7.50 m s 4.79 m for ball Y

(4.79) (3.19) 1.60 m

x x

x

v s

s



Page 40 of 185

5 2 2

-

1 2 2

1

2

2

0 m s 2 (0) 2( 9.80)(4.00)

4.00 m 8.85 m s

9.80 m s

y y y

y y y y

y y

v

v u gs

v u gs

s u

g

(8.85) sin 0.885

(10.0)

62.3

y

u

v

6

1

1

10.0 m s cos

62.3 cos (10.0)(cos 62.3 )

4.65 m s

x

x

x

vv

v

v v

v

7 D

8

1

1

1

( 8.85) (8.85)8.85 m s

( 9.80)

8.85 m s 1.81 s

4.65 m s

y y

y y

y

y

x

t

t

t

av u

v uu

a

v

v

(4.65)(1.81)

8.40 m

x x

x

t

s v

s

9 a

1

cos

cos (8.00)(cos 60.0 )

4.00 m s

x

x

x

v

v

v v

v

b

1

sin

sin (8.00)(sin 60.0 )

6.92 m s

y

y

x

v

v

v v

v



Page 41 of 185

c

1

1

2

(0) (6.92)6.92 m s

( 9.80)

0 m s 0.707 s

9.80 m s

y y

y y

y

y

t

t

t

v ua

v uu

a

v

g

d 21

2

1 212

2

total

6.92 m s (6.92)(0.707) ( 9.80)(0.707)

0.707 s 2.457 m

9.80 m s

y y

y y

y

t t

t

s u a

u s

s

g s

total

(1.50) (2.457)

3.95 m

y

y

s

e 2 2

1 2 2

1

2

2

6.92 m s 2 (6.92) 2( 9.80)( 1.50)

1.50 m 8.80 m s

9.80 m s

y y y

y y y y

y y

v

v

v

s

u as

u u as

g

( 8.80) (6.92)

( 9.80)

1.60 s

y y

y y

t

t

t

v ua

v u

a

f

1

x x4.00 m s (4.00)(1.60)

1.60 s 6.42 m

x

x

t

t

v s v

s

10 14.00 m sx

v

11

1 2 21 1k 2 2

k

4.00 m s (2.00)(4.00)

2.00 kg 16.0 J

x E mv

m E

v

12 29.80 m s a g



Page 42 of 185

13 C

14

1 5 2 5 2

a

3

a

7.50 m s 3.78 10 (3.78 10 )(7.50)

2.13 10 N in the opposite direction of the motion

v

v F

F

15

g3

a 3

a max

g 3

a max

2

(2.00)(9.80) 2.13 10 N

(2.13 10 )

2.00 kg 9.22 10

9.80 m s

m

FF

F

F

F

g

This answer tells us that the force of air resistance is insignificant when compared to the force

due to gravity on the ball.

16 a . 2 2

2 2

1

before

2

9.80 m s (0) 2( 9.80)( 10.0)

10.0 m 14.0 m s

0.200 kg

(0.

y y y

y

y y

m

mu

v u gs

g v

s v

p

1

before

1

before

200)( 14.0)

2.80 kg m s

2.80 kg m s downwards

p

p

b C

after

1 1

after

1

after

(0.200)(10.0)

10.0 m s 2.00 kg m s

0.200 kg 2.00 kg m s upwards

mu

v

m

p

p

p

c D

after before

1

after

1 1

before

2.00 kg m s (2.00) ( 2.80)

2.80 kg m s 4.80 kg m s upwards

I

I

I

p p p

p

p

d

av 3

1 3

av

3

(4.80)

(1.00 10 )

4.80 kg m s upwards 4.80 10 N upwards

1.00 10 s

I

t

I

t

F

F



Page 43 of 185

17 a

1 2 2 2 21 1 1 1k 2 2 2 2

1 2

k

0 m s (50.0)(3.00) (50.0)(0)

3.00 m s 2.25 10 J

u E mv mu

v E

b J 10252doneWork 2k .E

c

2

1

200.0 N cos (200.0)(cos 60.0 ) 100.0 N

50.0 kg

(100.0)10.0 s 2.00 m s

(50.0)

0 m s

(0) (2.00)(10.0)

x

m

tm

u

t

F F F

Fa

v u a

1

2 21 1k ideal 2 2

4

k ideal

lost

20.0 m s

(50.0)(20.0)

1.00 10 J

E mv

E

E

v

4 2

to heat k ideal k

3

lost to heat

(1.00 10 ) (2.25 10 )

9.78 10 J lost

E E

E

18 a

44

total

3

(1.00 10 )1.00 10 J

(10.0)

10.0 s 1.00 10 W

EE P

t

t P

b

33

lost

2

(9.78 10 )9.78 10 J

(10.0)

10.0 s 9.78 10 W

EE P

t

t P

c

22

k

1

(2.25 10 )2.25 10 J

(10.0)

10.0 s 2.25 10 W

EE P

t

t P

19 a 1

total before 0 kg m sp



Page 44 of 185

b

1

boy boy after boy

2 1

boy boy after

4.00 m s (50.0)(4.00)

50.0 kg 2.00 10 kg m s

m

m

v p v

p

c 2 1

sled after 2.00 10 kg m s p

20

2 1

sled after sled after

2

sled

2.00 10 kg m s

( 2.00 10 )200.0 kg

(200.0)

mv

m v

p p

1 1.00 m sv

21 1

sled before after

sled sled sled boy boy boy sled boy sled

sled sled boy boy

boy boy sled

boy sled

1.00 m s

200.0 kg

(250.0 kg

u

m m u m u m v

m u m um v

m

p p

1 1

boy boy sled

00.0)( 1.00) (50.0)( 4.40)

(200.0 50.0)

4.40 m s 1.68 m su v

22 a

sled after before sled sled sled sled

1

sled

1

sled

50.0 kg

4.40 m s (200.0)( 1.68) (200.0)( 1.00)

1.68 m s 1.36 10

m m v m u

u

v

p p p

p

p 2 1 kg m s

b

boy after before boy boy boy boy

1

boy

-1 2 1

boy

50.0 kg

4.40 m s (50.0)( 1.68) (50.0)( 4.40)

1.68 m s 1.36 10 kg m s

m m v m u

u

v

p p p

p

p

23



Page 45 of 185

before after

1 1 1 2 2 1 2 1 2

1 1 1 2 21 1 2

1 2

0.3000 kg

(0.3000)(2.00) (0.1000)( 2.00)2.00 m s

(0.300

m m u m u m v

m u m uu v

m

p p

1

2 1 2

1

2

0 0.1000)

0.1000 kg 1.00 m s east

2.00 m s

m v

u

24 C 2 21 1

k before 1 1 2 22 2

2 21 11 k before 2 2

1

1 k before

0.3000 kg (0.3000)(2.00) (0.1000)( 2.00)

2.00 m s 0.8

E m u m u

m E

u E

2

1 212 k after 1 2 1 22

1 211 2 k after 2

k after

00 J

0.1000 kg

2.00 m s

1.00 m s (0.4000)(1.00)

0.200

m

u E m v

v E

E

J

(0.200) 100 %Eff 25.0%

(0.800)

25 B

26 a A

b D

c C

27

1 distance 2 2 (20.0)10.0 m s

(10.0)

20.0 m 12.6 s

rv T

v v

r T

28 a

2 21

c

2

c

(10.0)10.0 m s

(20.0)

20.0 m 5.00 m s west

vv

r

r

a

a

b

2

c c c

3

c

5.00 m s west (1510)(5.00)

1510 kg 7.55 10 N west

m

m

a F a

F

c east N 10557 3c .F

29 C



Page 46 of 185

30 a i

seat on boy c g

2 2

seat on boy

2 2

seat on boy

( )

(50.0)(5.00)50.0 kg ( ) ( (50.0)( 9.80))

( 10.0)

10.0 m ( 1.25 10 ) (4.90 10 )

mvm m

F F F

F gr

r F

g 2 2

seat on boy9.80 m s 3.65 10 N upwards F

ii

seat on boy c g

2 2

seat on boy

2 2

seat on boy

( )

(50.0)(5.00)50.0 kg ( ) ( (50.0)( 9.80))

(10.0)

10.0 m (1.25 10 ) (4.90 10 )

9

mvm m

F F F

F gr

r F

g 2 2

seat on boy.80 m s 6.15 10 N upwards F

b D



Page 47 of 185

Chapter 2 Applying forces

2.1 Gravitational fields

1 a

11

a oa g 2 2

12

o g

(6.67 10 )(0.1000)(0.2000)0.1000 kg

(0.500)

0.2000 kg 5.34 10 N attraction

0.500 m

Gm mm

r

m

r

F

F

b

11 24 424 E s

E g 2 3 6 2

4 5

s g

3

6

E

(6.67 10 )(5.98 10 )(2.00 10 )5.98 10 kg

(600.0 10 6.37 10 )

2.00 10 kg 1.64 10 N attraction

alt 600.0 10 m

6.37 10 m

Gm mm

r

m

r

F

F

c 11 24 22

24 E sE g 2 8 2

22 20

m g

8

E m

(6.67 10 )(5.98 10 )(7.35 10 )5.98 10 kg

(3.84 10 )


3.84 10 m

Gm mm

r

m

r

F

F

d 11 27 31

p e27

p g 2 11 2

31 47

e g

11

p

(6.67 10 )(1.67 10 )(9.11 10 )1.67 10 kg

(5.30 10 )


5.30 10 me

Gm mm

r

m

r

F

F



Page 48 of 185

2 a

a m a mg1 g1 g22 2

1 2

2 2

g2 a m g1 1 a m g2 2

2 2

1 moon g1 1 g2 2

160 N

40 N

Gm m Gm m

r r

Gm m r Gm m r

r r r r

FF F

F F F

F F

g12 2

2 1

g2

2 2

2 1

(160)

(40)

r r

r r

F

F

2 2

2 1

2 moon

4

2

r r

r r

b 2 2

g1 g1 1 g2 2

2

g1 1

g2 g2 2

2

2

moon1 moon g2 2

moon

2 moon

160 N

40 N

(160)

(4 )

4

r r

r

r

rr r

r

r r

F

F F F

F F

F

2

moong2 2

moon

g2

(160)

(16)

10 N attraction

r

r

F

F

3 11 16 23

16 P MP M-P 2 6 2

15 15

D M-P

23

M

6 D MM P M-D 2

(6.67 10 )(1.08 10 )(6.42 10 )1.08 10 kg

(9.40 10 )


6.42 10 kg

9.40 10 m

Gm mm

r

m

m

Gm mr

r

F

F

F11 15 23

7 2

7 14

M D M-D

15

M-P

14

M-D

(6.67 10 )(1.80 10 )(6.42 10 )

(2.35 10 )

2.35 10 m 1.40 10 N attraction

(5.23 10 ) 37.5

(1.40 10 )

r

F

F

F



Page 49 of 185

4

E s E sg2 g1 g1 g22 2

1 2

2 2

g1 1 g2 2

0.01

Gm m Gm m

r r

r r

F F F F

F F

g12 2

2 1

g2

g12 2

2 1

g1

2

2

0.01

1

r r

r r

r

F

F

F

F

2

1

2 E

00

10

r

r r

5 a 11

a x xa a-x 2 2

b

11

a x xa x b-x 2 2

b x

(6.67 10 )( ) kg

(0.5 )

0.01 kg

(6.67 10 )(0.01 )0.5 m

(0.5 )

0.5 m

Gm m M mm M

r R

m M

Gm m M mr R

r R

r R

F

F

11 2

a-x x

2 11

b-x x

a-x

b-x

(6.67 10 )( ) (0.5 )

(0.5 ) (6.67 10 )(0.01 )

1 100

0.01

M m R

R M m

F

F

F

F



Page 50 of 185

b

a-xa

b-x

11 2

xb 2 11

x

2

a x 2

b x

kg 8100

(6.67 10 )( ) [(1 ) ]0.01 kg 8100

( ) (6.67 10 )(0.01 )

[(1 ) ] m 8100

( ) (0.01)

(1 ) m

m M

M m x Rm M

xR M m

x Rr xR

xR

r x R

F

F

2

2

[(1 ) ] 81.0

( )

[(1 ) ] 9.00

( )

9.00

10.0

x R

xR

x R

xR

xR R xR

xR R

10.0

0.100

distance 0.100

Rx

R

x

R

6

a-xa

b-x

11 2

xb 2 11

x

2

a x 2

b x

kg 1

(6.67 10 )( ) [(1 ) ]0.01 kg 1

( ) (6.67 10 )(0.01 )

[(1 ) ] m 1

( ) (0.01)

(1 ) m

m M

M m x Rm M

xR M m

x Rr xR

xR

r x R

F

F

2

2

[(1 ) ] 0.01

( )

[(1 ) ] 0.01

( )

0.01

1.01

x R

xR

x R

xR

xR R xR

xR R

1.01

0.990

distance 0.990

Rx

R

x

R



Page 51 of 185

7 11 23

11 2 2

M 2 6 2

23 1

M M

6

M

26

S S 2

(6 67 10 )(3.30 10 )6 67 10 N kg m

(2.44 10 )

3.30 10 kg 3.70 N kg

2.44 10 m

(6 67 105.69 10 kg

Gm .G .

r

m

r

Gm .m

r

g

g

g11 26

7 2

7 -1

S S

27

J

11 277

J J 2 7 2

)(5.69 10 )

(6.03 10 )

6.03 10 m 10.4 N kg

1.90 10 kg

(6 67 10 )(1.90 10 )7.15 10 m

(7.15 10 )

r

m

Gm .r

r

g

g

1

J 24.8 N kgg

8 a

a wt M

1

M wt M

80.0 kg (80.0)(3.70)

3.70 N kg 296 N

m m

F g

g F

b

a wt S

1

S wt S

80.0 kg (80.0)(10.4)

10.4 N kg 835 N

m m

F g

g F

c

a wt J

-1

J wt J

80.0 kg (80.0)(24.8)

24.8 N kg 1980 N

m m

F g

g F

9 Saturn’s mass is approximately 100 times that of Earth, while the radius of Saturn is only 10

times that of earth. When the radius of the planet is squared the factor of 10 becomes 102, which

is enough to cancel out the factor of 100 by which the mass of Saturn is greater.



Page 52 of 185

10 24

E E M

22 E s MM 2 2

8 E M

2 2

5.98 10 kg

7.35 10 kg ( )

3.84 10 m ( )

s

m

Gm m Gm mm

x r x

m mr

x r x

F F

2

M

2

E

8 22

24

8

( )

(3.84 10 ) (7.35 10 )

(5.98 10 )

3.84 10 0.110

1.110

mr x

x m

x

x

x x

8

8

8

3.84 10

3.84 10

1.110

3.46 10 m

x

x

x

2.2 Satellite motion

1 D

2 As the satellite does not change its energy while orbiting around the Earth, it doesn’t change its

height above the surface of the Earth so its gravitational potential energy does not change, and its

speed doesn’t change so its kinetic energy doesn’t change.

3

3 3

s net

1 2

net

2.30 10 kg (2.30 10 )(0.220)

0.220 N kg 5.06 10 N

m m

F g

g F

4 The source of this force is the gravitational attraction of the Earth on the satellite.

5 a 11 26

26 NN 2 8 2

T

8 2

T

11 2 2

(6.67 10 )(1.02 10 )1.02 10 kg

(3.55 10 )

3.55 10 m 0.0540 m s

6.67 10 N m kg

Gmm

r

r

G

g

g



Page 53 of 185

b

22

8 8

T

3 1

0.0540 m s

3.55 10 m (0.0540)(3.55 10 )

4.38 10 m s

v

r

r v r

v

g g

g

c

3 1

88

T 3

8

distance4.38 10 m s

distance (2 (3.55 10 ))3.55 10 m

(4.38 10 )

(5.10 10 )

24 60 6

v vT

r Tv

T(

0)

5.90 days

T

6 a 6

6

6

9 1

T

6

T

distance (2 (1.22 10 ))1.38 10 s

(1.38 10 )

1.22 10 m 5.55 km s

1.22 10 km

T vT

r v

r

b

2 3 23 1

c 9

9 2

T c

(5.55 10 )5.55 10 m s

(1.22 10 )

1.22 10 m 2.53 m s

vv

r

r

a

a

7

2 2

c c

11 2 2 Sc 2

2 2 9 29 c

S 11

2.53 10 m s

6.67 10 N m kg

(2.53 10 )(1.22 10 )1.22 10 m

(6.67 10 )

GmG

r

rr m

G

a a g

a

a

26

S 5.65 10 kgm



Page 54 of 185

8 a 7

c g

224 s V s

V 2

2

11 -2 2 V

2.10 10 s

4.87 10 kg

(2 )6.67 10 N kg m

T v

m v Gm mm

r r

GmrG

T r

F

2 2

V

2

2 11 24 7 2

V3 32 2

3 27

(4 )

(6.67 10 )(4.87 10 )(2.10 10 )

4 4

3.63 10

Gmr

T r

Gm Tr

r

9 1.54 10 mr

b

97

7

9 2 1

distance (2 (1.54 10 ))2.10 10 s

(2.10 10 )

1.54 10 m 4.60 10 m s

T vT

r v

c

2 2 22 1

c 9

9 4 2

c

(4.60 10 )4.60 10 m s

(1.54 10 )

1.54 10 m 1.37 10 m s

vv

r

r

a

a

9 a i 8

8 4 1AA A

A

A

88 3 1H

H H

H

H

2 2 (1.37 10 )1.37 10 m 1.65 10 m s

(0.602)(24 60 60)

0.602 day

2 2 (3.77 10 )3.77 10 m 9.97 10 m s

(2.75)(24 60 60)

2.75 days

rr v

T

T

rr v

T

T

4

A

3

H

1.65 10 1.66

9.97 10

v

v



Page 55 of 185

ii

2 4 28 2A

A A 8

A

4 1

A

2 3 28 2H

H H 8

H

3 1

H

(1.65 10 )1.37 10 m 1.99 m s

(1.37 10 )

1.65 10 m s

(9.97 10 )3.77 10 m 0.264 m s

(3.77 10 )

9.97 10 m s

vr

r

v

vr

r

v

a

a

A

H

1.99 7.58

0.264

a

a

b 3 3

8 A TA 2 2

A T

3 24 2 T A

A T 3

A

9 3 4 29

T T 8 3

1.37 10 m

5.20 10 s

(1.20 10 ) (5.20 10 )1.20 10 m

(1.37 10 )

r rr

T T

r TT T

r

r T

12

T

6

T

T

1.82 10

(1.35 10 ) s

(24 60 60)

15.6 days

T

T

T

10

6 AE

B

A B

E E

2 2

E o

2

o

6.37 10 m 1.2

1.2

1.2

r

Gm Gm

r r

r

g

g

g g

2

E

o E

6

o

6

o

1.2

1.2

1.2 (6.37 10 )

6.98 10 m

r

r r

r

r



Page 56 of 185

2.3 Torque

1 a .The axis of rotation is the tap spindle; the lever arm is approximately 0.04 m

b The axis of rotation is the axle of the wheel; the lever arm is approximately 1 m

c The axis of rotation is the end of the tweezers; the lever arms are approximately 0.07 m

d The axis of rotation is the place in which the screwdriver contacts the edge of the tin; the

lever arm is approximately 0.2 m

2 a The line of application of the force is a larger perpendicular distance from the hinges (pivot

point) when the force is applied to the handle than when it is applied to the centre of the

door.

b Using a long crowbar with a small rock as a pivot a large force can be applied to the large

rock if a small effort arm is used with a long effort arm, a ratio of load arm to effort arm of

1:10 will multiply the force you apply by ten times.

3

H

2

2

40.0 kg

9.80 m s (40.0)(9.80)(2.25)

2.25 m 8.82 10 N m

m Fr

r

g

4

cw cw acw

acw

2

400.0 N m

(400.0)1.60 m

(1.60)

2.50 10 N upwards

rr

τ τ τ

τF

F

5 a 3

w

2 3

4

2.50 10 kg

9.80 m s (2.50 10 )(9.80)(2.00)

2.00 m 4.90 10 N m

m Fr

r

g

b Cranes use counter-weights on the other side of the pivot point to the load to provide an

opposing torque to balance the torque due to the load.

6 a

w

2

1.00 kg

9.80 m s (1.00)(9.80)(0.500)

0.500 m 4.90 N m

m Fr

r

g

b

w

2

1.00 kg

9.80 m s (1.00)(9.80)(1.00)

1.00 m 9.80 N m

m Fr

r

g



Page 57 of 185

c

w

2

1.00 kg cos (1.00)(cos 60.0 )

9.80 m s 0.500 m

(1.

m r r

r

Fr

g

00)(9.80)(0.500)

4.90 N m

7 The weights provide a large counter torque should the performer overbalance. Only a small

movement of the pole is enough to balance the torque provided by the performer overbalancing.

8 The bench will not work successfully. The supports should be moved so that the centre of gravity

is between the supports or bolts could be used to attach the left hand support to the bench-top.

9 a The weight of the bag will produce a torque about a pivot point around the base of the

spine, which will tend to rotate the torso to the right. To compensate for this the person

must lean to the left, or by holding the left arm out from the body to move it farther from

the pivot point.

b

w

2

14.0 kg

9.80 m s (14.0)(9.80)(0.3)

assume 0.3 m 40 N m

m Fr

r

g

10 a 3 3

w wt

2 4

wt

3.50 10 kg (3.50 10 )(9.80)

9.80 m s 3.43 10 N

m m

F g

g F

b As the perpendicular distance from the line of action of the load to the pivot point does not

change, then the torque does not change.

c 4

wt

4

5

3.43 10 N

assume 15.0 m (3.43 10 )(15.0)

5.15 10 N m

Fr

r

F

F

r 60.0°

r



Page 58 of 185

2.4 Equilibrium

1 A, B, D

2

p T cable w1 w2

2

T cable

w1 T cable

w2

0

50.0 kg 4 ( ) 0

9.80 m s 4 ( 600) ( 850) [(50.0)( 9.80)] 0

600 N 4 1940 0

850 N

y

m m

F

F F F g

g F

F F

F2

T cable

(1940) 4.85 10 N

4 F

3

girder pillar load girder

2

pillar

load pil

0

5000 kg 2 8 ( ) ( ) 0

9.80 m s 2 8 (20 000)( 9.80) (5000)( 9.80) 0

20 000 kg 2

y

m m m

m

F

F g g

g F

F 6

lar

65

pillar

1.96 10 0

(1.96 10 ) 9.80 10 N

2

F

4

girl T vert girl beam

2

T vert

beam

0

60.0 kg 2 ( ) ( ) 0

9.80 m s 2 (60.0)( 9.80) (30.0)( 9.80) 0

30.0 kg 2

y

m m m

m

F

F g g

g F

F2

T vert

22

T vert

T vert

T

2

T

8.82 10 0

(8.82 10 ) 4.41 10 N

2

sin

(4.41 10 )

sin 5

F

F

F

F 25.06 10 N

5° FT vert

FT



Page 59 of 185

5

w1 cw acw

1 w1 1 w2 2

w1 12 w2

2

w2

200 N

1.2 m

(200)(1.2)1.5 m

(1.5)

160 N

r r r

rr

r

τ τF

F F

FF

F



Page 60 of 185

6

beam beam 1 light 2

1

1

light

2

2

2.0 kg

0.25 m (2.0)(9.80)(0.25) (5.0)(9.80)(0.5)

5.0 kg 2.94 10 N m

0.5 m

9.80 m s

m m r m r

r

m

r

τ

τ

τ

7 a

2 cw wt 1

cw bridge 1

acw T 2

acw T

10 sin 30 m

2

2 10 si

r r

m r

r

F

g

F

F n 30

b

T vert T

T horiz T

sin 30

cos 30

F F

F F

c

2 cw acw

1 wt 1 T 2

bridge bridge 1 T

bridge 12

T

10 sin 30 m

5.0 m 2

700 kg 2 10 sin 30

9.80 m s

2 10 sin

r

r r r

m m r

m rg

g

F F

F

gF

3

T

(700)(9.80)(5.0)

30 2 10 sin 30

3.43 10 N

F

Fwt

FT r2

r1

30°



Page 61 of 185

8

2 cw acw

3 wt train 1 wt beam 2 Y 3

beam train 1 beam 2 Y 3

train 1

10 m

20 m

5000 kg

5000 kg

r

r r r r

m m r m r r

m r

g g

F F F

F

F 3

Y 3 beam 2

train

3

Y 1

2

(30.6 10 )(20) (5000)(9.80)(10)

(5000)(9.80)

30.6 10 N 2.49 m

9.80 m s

r m r

m

r

g

g

g

F

9 a

ladder wt ladder wt ladder 1

1 wt ladder ladder 1

wt ladder

16 kg

2.4 cos 65 (16)(9.80)(2.4 cos 65 )

159 N m

m r

r m r

τ

τ

τ

F

g

b

person wt person wt person 1

1 wt person wt person 1

wt person

50 kg

1.2 cos 65 (50)(9.80)(1.2 cos 65 )

248 N m

m r

r m r

τ

τ

τ

F

g

c

person wt person wt person 1

1 wt person wt person 1

wt person

50 kg

3.6 cos 65 (50)(9.80)(3.6 cos 65 )

745 N m

m r

r m r

τ

τ

τ

F

g

X Y

Fwt beam

Fwt train

r1 r2

r3

65°

r1

65°

r1

65°

r1



Page 62 of 185

10 a

1

2

3 painter 1 platform 2 R T 3

painter

platform R T

1.50 m Take moments about left-hand side:

2.50 m

5.00 m

20 kg

cw acw

r

r

r m r m r r

mm

g g F

gF

1 platform 2

3

painter R T

2

1

2

(70)(9.80)(1.50) (20)(9.80)(2.50)

(5.00)

70 kg 304 N up

9.80 m s

3.50 m Take moments about right-hand side:

2.50 m

r m r

r

m

g

r

r

g

F

3 painter 1 platform 2 L T 3

painter 1 platform 2

L T

3

5.00 m

(70)(9.80)(3.50) (20)(9.80)(2.50)

(5.00)

cw acw

r m r m r r

m r m r

r

g g F

g gF

L T 578 N upF

b

2

3 cw acw

L T painter 1 platform 2 R T 3

R T 3 platfor

R T 1

2.50 m Take moments about left-hand side:

5.00 m

557 N up

325 N up

r

r

m r m r r

r mr

F g g F

FF

m 2

painter

painter 1

platform

2

(325)(5.00) (20)(9.80)(2.50)

(70)(9.80)

70 kg 1.65 m from left-hand side

20 kg

9.80 m s

r

m

m r

m

g

g

g

11 a

T horiz T

T horiz

T vert T

T vert

cos 60 (800) cos 60

400 N

sin 60 (800) sin 60

693 N

F F

F

F F

F



Page 63 of 185

b

1

2 cw acw

T horiz T 2 T horiz 1

T 2T horiz

1

0.85 m Take moments about base of pole:

1.0 m

400 N

(400)(1

r

r

r r

r

r

F F F

FF

T horiz

.0)

(0.85)

471 NF

Chapter 2 Review

1

20 1 2

2

11 21 2611 2 2 1 2

20

21 8

D

2.79 10 N

(6.67 10 )(1.05 10 )(5.69 10 )6.67 10 N m kg

(2.79 10 )

1.05 10 kg 3.78 10 m fro

Gm m

r

Gm mG r

m r

F

F F

26

S

m Saturn's centre of mass

5.69 10 kgm

2 D

3 a D

b B

c C

d A

e A

4

X X

XX

2 2

X X

2 2

X X

2 2

X

2 2

(0.8 )

(0.2 )

(0.64) 16

(0.04)

M m

mM

M m

mM

M m

M M

m m X

M

m

Gm mGm m

r r

mm

r r

m r R

m r R

m

m

F F



Page 64 of 185

5 a

11 2611 2 2 N

2 7 2

26 1 1

N

(6.67 10 )(1.02 10 )6.67 10 N m kg

(2.48 10 )

1.02 10 kg 1.11 10 N kg

GmG

r

m

g

g

b C

6

11 2 2

c g

224 M E M

E 2

224 E

M 2

2 38

E

6.67 10 N m kg

5.98 10 kg

(2 )5.98 10 kg

4 43.84 10 m

G

m v Gm mm

r r

Gmrm

T r

rr T

Gm

F F

2 8 3

11 24

8

(3.84 10 )

(6.67 10 )(5.98 10 )

(2.37 10 ) s

(24 60 60)

27.4 days

T

T

7 a

11 2 - 2

c g

227 J LL

J 2

10 J

J

6.67 10 N m kg

1.90 10 kg

1.10 10 m

G

Gm mm vm

r r

Gmr v

r

Gmv

r

F F

11 27

10

3 1

(6.67 10 )(1.90 10 )

(1.10 10 )

3.39 10 m sv



Page 65 of 185

b

11 2711 2 2 J

2 10 2

27 3 2

J

10

(6.67 10 )(1.90 10 )6.67 10 N m kg

(1.10 10 )

1.90 10 kg 1.05 10 m s towards Jupiter

1.10 10 m

GmG

r

m

r

g

g

c

3 1

1010

3

7

2 3.39 10 m s

2 2 (1.10 10 )1.10 10 m

(3.39 10 )

(2.04 10 ) 236 days

(24 60 60)

rv v

T

rr T

v

T

8 a C

b The satellite will always be positioned above the same location on Earth therefore radio

and TV signals can be exchanged with the satellite from any location on Earth that has a

line of sight view of the satellite

c 11 2 2

c g

224 s E s

E 2

2

2

2 11

E32

6.67 10 N m kg

5.98 10 kg

(2 )24 60 60 s

(6.67 10 )(

4

E

G

m v Gm mm

r r

GmrT

T r

Gm Tr

F F

24 2

32

7

5.98 10 )(24 60 60)

4

4.23 10 mr



Page 66 of 185

9 a

11 2 2

c g

223 s M s

M 2

26 M

2

2

M32

6.67 10 N m kg

3.30 10 kg

(2 )5.07 10 s

(6.67

4

G

m v Gm mm

r r

GmrT

T r

Gm Tr

F F

11 23 6 2

32

8

10 )(3.30 10 )(5.07 10 )

4

2.43 10 mr

b

86

6

8 2 1

2 2 (2.43 10 ) 5.07 10 s

(5.07 10 )

2.43 10 m 3.01 10 m s

rT v

T

r v

c

2 2 22 1

8

8 4 2

(3.01 10 )3.01 10 m s

(2.43 10 )

2.43 10 m 3.73 10 m s

vv a

r

r a

10 a The work done to increase the kinetic energy if the rock is equal to the area under the

curve from

3.00 × 106 m to 2.50 × 106 m

d

6

d

7

d

Area approximate number of squares area of 1 square

(5.5) (10)(0.5 10 )

2.75 10 J

W

W

W

b

1

k k1 d

2 2 71 1k d2 2

7 7

d k

1000 m s

20.0 kg (20.0)(1000) (2.75 10 )

2.75 10 J 3.75 10 J

u E E W

m E mv W

W E

c

7 21k k 2

k

3 1

3.75 10 J

2 2(3.75 10)20.0 kg

(20.0)

1.94 10 m s

E E mv

Em v

m

v

d From the graph, at 2.50 × 106 m, F = 70 N

1

70 N

(70)20.0 kg 3.50 N kg

(20.0)

mg

m gm

F F

F



Page 67 of 185

11 a

c g

2

E

2

2

E

2

2 32

E

2 2 3

1 1 E

2 2 3

2 E 2

2 3 3 3

1 1

2 3 3 3

2 2

1

2

2

4

4

4

( )

(2 ) 8

1 1C

8 8

s s

s s

s s

s s

s

s

Gm mmv

r r

rm

Gm mT

r r

rT

Gm

T r Gm

T Gm r

T r R R

T r R R

T

T

F F

b

c g

2

E

2

2 E

2

1 2E

2

2 1 E

2

1 2

2

2 1

1

2

(2 )

( )

2 B

s s

s s

s s

s s

s

s

Gm mmv

r r

Gmv

r

v rGm

v r Gm

v r R

v r R

v

v

F F

c

D4

2

2

1

2

2

21

22

2

1

E

22

21

E

2

1

2E

c

s

s

s

s

s

s

s

ss

s

)R(

)R(

r

r

Gm

r

r

Gm

r

Gm

a

a

a

a

a

a

a

ga



Page 68 of 185

12

c g

2

S

2

2

S

2

2 3 2 11 3

S 2 11 2

30

S

2

4 4 (1.50 10 )

(6.67 10 )(365.25 24 60 60)

2.01 10 kg

Gm mmv

r r

rm

Gm mT

r r

rm

GT

m

F F

13 B

14 3 3

N CC 2 2

N C

2 3 2 3

C NC N 3 3

C

N N

6.40 days

(6.40) (49000)19600 km

(19600)

49000 km 25.3 days

r rT

T T

T rr T

r

r T

15

c g

2

P

2

2

P

2

2 3 2 3 3

P 2 11 2

22

P

2

4 4 (19600 10 )

(6.67 10 )(6.40 24 60 60)

1.46 10 kg

Gm mmv

r r

rm

Gm mT

r r

rm

GT

m

F F



Page 69 of 185

16

5

ISS E c g

26 E

ISS 2

7 2 EO2D E

27 ISS O2DE

O2D 2

O2D ISS E

(3.80 10 ) m

6.75 10 m

(3.60 10 ) m

4.24 10 m

r r

Gm mmvr

r r

Gmr r v

r

v rGmr

v r Gm

F F

2 7

ISS O2D

2 6

O2D ISS

ISS

O2D

(4.24 10 )

(6.75 10 )

6.28 2.51

v r

v r

v

v

17 6

ISS c g

27 E

O2D 2

2 32

E

2 2 3

ISS ISS E

2 2

O2D E

6.75 10 m

4.24 10 m

4

4

4

r

Gm mmvr

r r

rT

Gm

T r Gm

T Gm

F F

3

O2D

2 3 6 3

ISS ISS

2 3 7 3

O2D O2D

3 2ISS

O2D

(6.75 10 )

(4.24 10 )

4.04 10 6.36 10

r

T r

T r

T

T

18 6

ISS c

7 EO2D 2

2

ISS O2DE

2

O2D ISS E

6.75 10 m

4.24 10 m

r

Gmr

r

rGm

r Gm

a g

a

a

a

2 7 2

ISS O2D

2 6 2

O2D ISS

ISS

O2D

(4.24 10 )

(6.75 10 )

39.4

r

r

a

a

a

a



Page 70 of 185

19

20 a C

b The forces must be on different objects and be equal in magnitude.

21

cw

contents

445 N m

(445)1.60 m

(1.60)

278 N up

r

rr

τ τ

τ

F

F

F

22

3 3

4

contact

5.00 10 N (5.00 10 )(3.00)

3.00 m 1.50 10 N m

r

r

τ

τ

F F

The wrecker could hit the wall higher up to increase the radius.

23 E

2

(3.00)(cos 65.0 )

1.27 m (7.50)(9.80)(1.27)

9.80 m s 93.2 N m

7.50 kg

r r mgr

r

m

τ

τ

τ

F

g

24

2

(2.0)(cos 65.0 )

0.85 m (60)(9.80)(0.85)

9.80 m s 4.99 N m

60 kg

r r mgr

r

m

τ

τ

g τ

F

25 B

3 3

load

2 4

4.50 10 kg (4.50 10 )(9.80)

9.80 m s 4.41 10 N

m m

F g

g F

Fground on ball

Fgravity on ball

r

r



Page 71 of 185

26 B

27

T1vert T1horiz T1vert

T1vert T1horiz

2 2 2

T1 T1vert T1horiz

2 2

T1

T1

(1.5)(9.80) tan 50 (14.7)(1.19)

14.7 N 17.5 N

(14.7) (17.5)

22.9

m

gF F F

F F

F F F

F

F

T2 T1horiz

N

17.4 N F F

28

1

2 cw acw

3 1 1 2 2 3 3 beam 4 RHS 5

beam

(6.00 9.80) N taking moments about LHS

(10.0 9.80) N

(13.0 9.80) N

(2.00 9.80) N

F

F

F r r r r r

F

τ τ

F F F F F

1 1 2 2 3 3 beam 4RHS

5

RHS

(58.8)(2.0) (98.0)(5.0) (124.4)(7.0) (19.6)(5.0)

(10.0)

r r r r

r

F F F FF

F

2

RHS

cw acw

1.60 10 N upwards

taking moments about RHS

τ τ

F

1 1 2 2 3 3 beam 4 LHS 5

1 1 2 2 3 3 beam 4LHS

5

LHS

(58.8)(8

r r r r r

r r r r

r

F F F F F

F F F FF

F

2

LHS

.0) (98.0)(5.0) (124.4)(3.0) (19.6)(5.0)

(10.0)

1.44 10 N upwards

F



Page 72 of 185

29 a

3

load cw acw

1 load 1 c-w 2

33 load 1

load 2 3

c-w

1.50 10 kg

25.0 m

(1.50 10 )(25.0)20.0 10 kg

(20.0 10 )

m

r r r

rm r

τ τ

F F

F

F

2 1.88 mr

b This reduces the torque acting on the crane making it less likely that the crane will topple

over.

30 a

1

250 N (250)(1.0)

1.0 m 250 N m

r

r

τ

τ

F F

b This torque would cause the barrier to bend slightly to the left causing tension at Y and

compression at X, as concrete can withstand more compression than tension it is more

likely to crack at position Y.

r1 r2

Fload Fc-w

pivot



Page 73 of 185

Chapter 3 Understanding electromagnetism

3.1 Magnetic fields

1 B

2 C

3 a North

b North-east

c East

4 a South

b North

c Zero due to the two wires, but the Earth’s magnetic field still exists and will be directed

north.

5 At point R, but only if the combined field from m and n are balanced by the Earth’s field.

6 a South

b South

c South, but only if it greater than the Earth’s field at that point.

7 a B, into the page

b 3B, into the page

c Zero

8 A

9 South

10 South

3.2 Force on current-carrying conductors

1 a

5

5 1

100.0 m

80.0 A W E (80.0)(100.0)(5.34 10 )

5.34 10 T 4.27 10 N upwards

l l

F I B

I F

B F

b

5

5 1

100.0 m

50.0 A E W (50.0)(100.0)(5.34 10 )

5.34 10 T 2.67 10 N downwards

l l

F I B

I F

B F



Page 74 of 185

2

wt

5

B

wt

B

5

(50.0)(9.80)100.0 m

(100.0)(100.0)(5.34 10 )

100.0 A 918

5.34 10 T

50.0 kg

mgl

l

m

F

F I B

FI

F

B

3 B

4 a

3

3 4

0.0500 m

2.00 A into page (2.00)(0.0500)(2.00 10 )

2.00 10 T 2.00 10 N north

l l

F I B

I F

B F

b

3

3 4

0.0500 m

1.00 A out of page (1.00)(0.0500)(2.00 10 )

2.00 10 T 1.00 10 N south

l l

F I B

I F

B F

5 a

3

3

3

1.00 10 m

3.00 A into page (3.00)(1.00 10 )(0.500)

0.500 T 1.50 10 N west

l l

F I B

I F

B F

b

3

3

3

1.00 10 m

3.00 A out of page (3.00)(1.00 10 )(1.00)

1.00 T 3.00 10 N east

l l

F I B

I F

B F

6 a

33

3

3

(8.00 10 )2.00 10 m

(2.00 10 )(0.100)

8.00 10 N south 40.0 A into the page

0.100 T

ll

FI

B

F I

B



Page 75 of 185

b

23

3

2

(2.00 10 )2.00 10 m

(2.00 10 )(0.500)

2.00 10 N north 20.0 A out of the page

0.500 T

ll

FI

B

F I

B

7 a

3 3

2 1

5.00 A

4.00 10 T south (5.00)(4.00 10 )

2.00 10 N m west

Fl I

F

F

B

B

b

3 3

2 1

5.00 A

4.00 10 T north (5.00)(4.00 10 )

2.00 10 N m east

Fl I

F

F

B

B

8 a

3 3

3 1

2.0 A

1.0 10 T north-west (2.0)(1.0 10 )

2.0 10 N m north-east

Fl I

F

F

B

B

b

3 3

3 1

1.0 A

1.0 10 T north-west (1.0)(1.0 10 )

1.0 10 N m south-west

Fl I

F

F

B

B

9 Magnetic flux due to wire N at point M is south.



Page 76 of 185

10 a 3 3

3

3 3 3 3

3

90.0 sin (1.00 10 )(sin 90.0) 1.00 10 T

2.00 10 m

1.00 10 A (1.00 10 )(2.00 10 )(1.00 10 )

1.00 10 T

l l

B B

F I B

I F

B 92.00 10 N F

b

2

0 sin (1.00)(sin 0) 0 T

5.00 10 m 0 N

1.00 A

1.00 T

l

B B

F

I

B

c

1 2

3

3 2

1

30.0 sin (1.00 10 )(sin 30.0) 5.00 10 T

1.00 10 m

5.00 A (5.00)(1.00 10 )(5.00 10 )

1.00 10 T

l l

B B

F I B

I F

B 42.50 10 N F

3.3 Electric motors

1

2

PS

0.100 T (2.00)(0.05)(0.100)

0.0500 m 1.00 10 N into the page

2.00 A

Il

l

I

B F B

F

2

2

QR

0.100 T (2.00)(0.05)(0.100)

0.0500 m 1.00 10 N out of the page

2.00 A

Il

l

I

B F B

F

3 0 N as the field and current are parallel.

4 Anticlockwise

5 D

6 a Down

b Up

7 Anticlockwise

8 a Down

b Up

c 0 N m



Page 77 of 185

9 C

10 The commutator reverses the direction of the current through the coil of the motor at a particular

point. This enables the resultant torque on the coil at that point keep the motor rotating in a

constant direction.

3.4 Electric fields in circuits

1 a F doubles

b F quadruples

c F becomes attractive

d F quadruples

2

1 21 2

9

1 2

9 2 2 5

1.00 C

(9.00 10 )(1.00)(1.00)1.00 C

(100.0)

9.00 10 N m C 9.00 10 N repulsion

100m

kq qq

r

q

k

r

F

F

F

3 a

6 1 21 2

9 6 66

1 2

9 2 2 1

5.00 10 C

(9.00 10 )(5.00 10 )(5.00 10 )5.00 10 C

(0.800)

9.00 10 N m C 3.52 10 N repulsion

0.800 m

kq qq

r

q

k

r

F

F

F

b As the charges on the Van de Graaff machine are mobile, and are of the same sign, they

will repel each other so that they move to opposite sides of the dome, this will increase the

distance between the ‘centre’ of the charges.

4 a 3 1 3 6

1 1

6 8

1 1

6

2

2 2

5.00 10 N C (5.00 10 )( 2.00 10 )

2.00 10 C 1.00 10 N downwards

5.00 10 C

q

q

q

q

E F E

F

F E 3 6

8

2

8

2

(5.00 10 )( 5.00 10 )

2.50 10

2.50 10 N upwards

F

F

b



Page 78 of 185

33 1 1

3

3 1

( 1.00 10 )5.00 10 N C

(5.00 10 )

1.00 10 N 2.00 10 C

q

q

FE

E

F

5 a

3 35 10 A (5.00 10 )(10.0 60)

(10.0 60) 3 C

I q I t

t q

b

3

200 A (200)(5.00)

5.00 s 1 10 C

I q I t

t q

c

3 3

3

400 10 A (400 10 )(60 60)

(60 60) s 1.4 10 C

I q I t

t q

6

12

12 1919

6

50 10

(50 10 )( 1.60 10 )1.60 10 C

(3)

3 s 3 10 A

e ee

e

N qqN I

t t

q I

t I

7 The potential difference of a car battery to your hand is 12 V, which causes an insufficient

electric field to cause a current to flow through the air to your skin. The spark in a spark plug

results from a potential difference of thousands of volts, which will cause current to flow through

air to your hand.

8

d

dd

Assume 100% efficiency

5.00 C

(100.0)100.0 J

(5.00)

20.0 V

q W q V

WW V

q

V

9

d

d


1.00 C (9.00)(1.00)

9.00 V 9.00 J

q W q V

V W



Page 79 of 185

10

3

d d

3

d

2


2.00 10 J

2.00 10 )12.0 V

(12.0)

1.67 10 C

W W q V

W (V q

V

q

3.5 Electric circuits

1 Either there is some internal resistance in the battery or there is some form of resistance in the

circuit, which may be due to corroded connections.

2

Current in current out 0

( 2.50) ( 1.00) ( 4.20) 0

0.70

0.70 A out of the point

I

I

I

3 a 0.25 A

b 2.40 V

4 a Yes

b If one bulb breaks the other will go out too.

5 a

1

2

2

5.00 V

(5.00)400.0

(400.0 100.0)

100.0 1.00 10 A

V V IR

VR I

R

R I

b

2 2

1 1

2

1 2 2

2

1.00 10 A (1.00 10 )(400.0) 4.00 V

400.0 (1.00 10 )(100.0) 1.00 V

100.0

I V IR

R V IR

R

6 a Lamp A gets brighter

b Lamp C turns off

c Current increases

d Potential difference across lamp B increases

e Potential difference across lamp C decreases

f Total power increases

7 a R1, R4 and R5

b R2 and R3



Page 80 of 185

c R1, R4 and R5

8

R1 () R2 () Vout (V)

1000 1000 10

3000 1000 5.0

400 100 4.0

900 100 2.0

2.0 3.0 12

9 a D

b F

c D

d A

10 a

4

3 3

A

3

A

3 3 4

B

4

4 3

C

3

C

3 3 4

Total

1 1 11.33 10

(10 10 ) (30 10 )

7.5 10

R (7.5 10 ) (5 10 ) 1.25 10

1 1 11.13 10

(1.25 10 ) (30 10 )

8.8 10

(8.8 10 ) (40 10 ) 4.9 10

R

R

R

R

R

b

T T

4

4

T

4 3

T 40

(10.0)2.05 10 A

(4.9 10 )

(2.05 10 )(40 10 )

8.2

T

V I R

VI

R

V I R

V V

c

5

30 3

30

4 5

5 T 30

4

5

(10) (8.2) 1.8 V

(1.8)6.02 10 A

(30 10 )

(2.05 10 ) (6.02 10 )

1.4 10 A

V

VI

R

I I I

I



Page 81 of 185

d 4 3

5 5 5

1

5

1

10 5

10

(1.4 10 )(5 10 )

7.2 10 V

(1.8) (7.2 10 )

1.1 V

V I R

V

V V V

V

Chapter 3 Review

1 B, C

2 B

3 A

4 a R4

b R2 and R3

c R4

d E2

5 a All four in series.

b Two in series that are connected to two in parallel.

c All four in parallel.

d One resistor that is connected to three in parallel.

6 a

1

T

T

T

T

T

T

Circuit a:

Δ (12.0)Δ 12.0 V 7.50 10 A

(16.0)

Circuit b:

Δ (12.0) 1.20 A

(10.0)

Circuit c:

Δ (12.0) 12.0 A

(1.00)

VV I

R

VI

R

VI

R

T

T

Circuit d:

Δ (12.0) 2.25 A

(5.33)

VI

R



Page 82 of 185

b

1

T 1 2 3 4

T 1 2 3 4

T1 2 3 4

T 1 2

Circuit a:

7.50 10 A

Circuit b:

1.20 A, 0.60 A

Circuit c:

12.0 3.00 A

4 4

Circuit d:

2.25 A,

I I I I I

I I I I I

II I I I

I I I

1T3 4

2.257.50 10 A

3 3

II I

7

T A B

T

T

T

1 T A

11 4

3

4

4

( )4 4

R R R R R

R R

V VI

R R

V VV I R R

R

V V

8 a

T 3

T

T

T

A

B

(12.0)

(200.0 10 )

60.0

(60.0)

4 4

15.0

15.0

3 45.0

VR

I

R

RR

R

R R

R R

b

2 3 2

B B

B

(200.0 10 ) (45.0)

1.80 W

P I R

P



Page 83 of 185

9

Therm T out

Therm

outT 3

3

T

ThermTherm 3

T

3

Therm

3

(6) (1)

5 V

(1)

(1 10 )

1 10 A

(5)

(1 10 )

5 10

from graph at 5 10 temperature is 20 C

V V V

V

VI

R

I

VR

I

R

10 a 2

Xmax

X max

X

XT

X

TY

Y T Y

Y

AB X Y

AB

(25)(100)

50 V

(50)0.50 A

(100)

(0.50)0.25 A

2 2

(0.25)(100)

25 V

(50) (25 )

75 V

VP

R

V P R

V

VI

R

II

V I R

V

V V V

V



Page 84 of 185

b

P Y Z

P

T X P

T

2 2

ABT

T

T

1 1 1 1 10.02

(100) (100)

50

(100) (50 )

150

(75)

(150)

37.5 W

R R R

R

R R R

R

VP

R

P

11

R1 () R2 () Switch Vout (V)

1000 2000 Open 60

2000 4000 Open 60

4000 2000 Open 50

8000 5000 Closed 0

12 a

A 230

25

555

V 685

31925

V 319

10941

A 931

9412

25

94612

9421010

942

340520

1

5

1

10

11

20

520

P20

P

10P

10

10T10

T

TT

T

PT

P

P

.I

)(

).(

R

VI

.V

).()(VVV

.V

))(.(RIV

.I

).(

)(

R

VI

.R

).()(RR

.R

.)(R



Page 85 of 185

b

P 10

P

(25) (19.3)

5.7 V

V V V

V

13 a Y

b 3

2 1

2 1

4 1

4

3

( ) (39 10 0)slope

( ) (75 0)

slope 5.2 10 A V

1 1

slope (5.2 10 )

1.9 10

y y

x x

VR

I

R

c A 1040graph thefrom then V 60When 3TY

IIV

d

3

P 3 3

X X

2

P

3 2

P P

P

Battery Y P

Battery

1 1 1 11.04 10

(1.9 10 ) (1.9 10 )

9.6 10

(40 10 )(9.6 10 )

38 V

(60) (38)

98 V

RR R

R

V IR

V

V V V

V

e

3

T T T

T

(98)(40 10 )

3.9 W

P V I

P

14 a Into the page

b Out of the page

c Out of the page

d Out of the page

15 5.00 × 10–5 T south

16 Into the page



Page 86 of 185

17

5 6

T E C

4

T

( 5.00 10 ) ( 50 10 )

1.00 10 T north

B B B

B

18

2 2 5 2 6 2

T E C

5

T

( 5.00 10 ) ( 50 10 )

7.07 10 T

B B B

B

19 North-west

20 C

21 a

3 3 3

9

(1.00 10 )(5.00 10 )(1.00 10 )

5.00 10 N into the page

l

F I B

F

b

2

3

(2.00)(1.00 10 )(0.100)


l

F I B

F

c

3

2

(5.00)(10.0 10 )(1.00)


l

F I B

F

22 a Out of the page

b Into the page

23 Zero

24 a Attraction

b Attraction

c Repulsion

25 Zero

26

1c (1.00)(2.00) 2.00 N ml

F

IB



Page 87 of 185

27

1.0 A (1.0)(0.50)(0.20)

0.50 m 0.10 N

0.20 T

l

l

I F I B

F

B

28 Anticlockwise

29 D



Page 88 of 185

Chapter 4 Generating electricity

4.1 Magnetic flux and induced currents

1 3

0.00

6

0.00

3

45.0

6

45.0

3

60.0

6

60.0

cos (2.00 10 )(0.0400 0.0400)(cos0.00 )

3.20 10 Wb

cos (2.00 10 )(0.0400 0.0400)(cos45.0 )

2.26 10 Wb

cos (2.00 10 )(0.0400 0.0400)(cos60.0 )

1.60 10 W

A

A

A

B

B

B

3

90.0

90.0

b

cos (2.00 10 )(0.0400 0.0400)(cos90.0 )

0 Wb

A

B

2 6 6

0 to 45 45.0 0.00

7

0 to 45

6 6

0 to 60 60.0 0.00

6

0 to 60

6

45 to 90 90.0 45.0

6

45 to 90

(2.26 10 ) (3.20 10 )

9.37 10 Wb

(1.60 10 ) (3.20 10 )

1.60 10 Wb

(0) (2.26 10 )

2.26 10 Wb

6

0 to 90 90.0 0.00

6

0 to 90

(0) (3.20 10 )

3.20 10 Wb

3 a

f i f i

6

6

cos cos

(0)(0.0400 0.0400)(cos0.00 ) (3.20 10 )

3.20 10 Wb

A A

B B



Page 89 of 185

b

f i f i

6 6

6

cos cos

( 3.20 10 ) (3.20 10 )

6.40 10 Wb

A A

B B

c

f i f i

3 6

6 6

6

cos cos

(4.00 10 )(0.0400 0.0400)(cos0.00 ) (3.20 10 )

(6.40 10 ) (3.20 10 )

3.20 10 Wb

A A

B B

d

f i f i

3 6

6 6

6

cos cos

(1.00 10 )(0.0400 0.0400)(cos0.00 ) (3.20 10 )

(1.60 10 ) (3.20 10 )

1.60 10 Wb

A A

B B

4 a Zero

b Negative

c Positive

d Negative

5 There must be a changing magnetic flux in the conductor that makes the coil, and the coil must

be part of a complete circuit.

6 As S is closed a current in Y grows, which deflects the galvanometer needle to the right, and then

drops to zero.

While S is closed, no current flows.

As S is opened a larger current in Y grows, which deflects the galvanometer needle to the left,

and then drops to zero.

7 As the current in X steadily decreases the current in Y is constant and deflects the galvanometer

needle to the left.

As the current in X steadily increases the current in Y is constant and deflects the galvanometer

needle to the right.

8 a

2 2 3 2 2

3 5

4.00 10 m (2.00 10 ) (4.00 10 )

2.00 10 T 1.01 10 Wb

r A r

B B

B

b Zero



Page 90 of 185

9 a

5

f i

5

(0) (1.01 10 )

1.01 10 Wb

b

5

3

2 1

( 1.01 10 )

(1.00 10 )

1.01 10 Wb s

t

t

c 4.00 mA flowing from Y to X through the milliammeter.

10 a 5 5

f i

5

5

3

2 1

( 1.01 10 ) (1.01 10 )

2.02 10 Wb

( 2.02 10 )

(1.00 10 )

2.02 10 Wb s

This is double the change of flux

induced current is 8.00 mA from X to Y

t

t

b 2 2 3 2 2

3 6

6

f i

6

6

3

3 1

2.00 10 m (2.00 10 ) (2.00 10 )

2.00 10 T 2.51 10 Wb

(0) (2.51 10 )

2.51 10 Wb

( 2.51 10 )

(1.00 10 )

2.51 10 Wb s

This is one-qu

r A r

t

t

B B

B

arter the change of flux




Page 91 of 185

c 5

f i

5

5

3

3 1

(0) (1.01 10 )

1.01 10 Wb

( 1.01 10 )

(2.00 10 )

5.03 10 Wb s

This is half the change of flux


t

t

4.2 Induced EMF: Faraday’s law

1 a

3 3

6

2.00 10 T (2.00 10 )(0.02 0.03)

1.2 10 Wb

A

B B

b Zero

c

66

i 3

5

f

3

(0) (1.2 10 )1.2 10 Wb EMF

(40 10 )

0 Wb EMF 3.0 10 V

40 10 s

t

t

d

55

5

(3.0 10 )EMF 3.0 10 V

(1.50)

1.50 2.0 10 A

VI

R

R I

2 a 3

i f i f i

4 3 4

f

2 5

4 2

80.0 10 T

0 T (0)(10.0 10 ) (80.0 10 )(10.0 10 )

10.0 cm 8.00 10 Wb

10.0 10 m

A A

A

A

B B B

B



Page 92 of 185

b

55

3

3 3

( 8.00 10 )8.00 10 Wb EMF

(20 10 )

20 10 s EMF 4.00 10 V

t

t

c

55

3

3

( 8.00 10 )8.00 10 Wb EMF (500)

(20 10 )

20 10 s EMF 2.00 V

500

Nt

t

N

3 a

3

f i f i

2 3 4 3 4

i

4 2 4 5

i

2

f

5.00 10 T

50.0 cm (5.00 10 )(250.0 10 ) (5.00 10 )(50.0 10 )

50.0 10 m (1.25 10 ) (2.50 10 )

250.0 cm

A A

A

A

A

B B B

4

4 2

f

1.00 10 Wb

250.0 10 mA

b

44

3

(1.00 10 )1.00 10 Wb EMF (30)

(0.500)

0.500 s EMF 6.00 10 V

30

Nt

t

N

4 a

5 5 2

3

4.42 10 T (4.42 10 ) (4.00)

4.00 m 2.22 10 Wb

A

r

B B

b

33

2

(2.22 10 )2.22 10 Wb EMF (1)

(0.125)

8.00 Hz EMF 1.78 10 V

0.125 s

Nt

f

t

5 D



Page 93 of 185

6

3 2 3 2

1

23 2

1.60 10 m (1.60 10 ) 4.00 10 m

2.5 m s

(4.00 10 )EMF 5.0 10 V 1.60 10 s

(2.5)

EMF

A l A

v

st

v

Nt

3 2

5

5

f

(5.0 10 )(1.60 10 )

1

8.00 10 Wb

(8.00 10 )

(1.60

EMF t

N

A

B

3

2

10 )

5.00 10 T

B

7 C

8 a Out of the page

b Into the page

c Out of the page

9 a Positive

b Positive

c Negative



Page 94 of 185

4.3 Electric power generation

1

3 3 -4

0.00

-4 2 5

0.00

3 -4

15.0

5.00 10 T cos (5.00 10 )(20.0 10 )(cos0.00 )

20.0 10 m 1.00 10 Wb

cos (5.00 10 )(20.0 10 )(cos15.0 )

A

A

A

B B

B

6

15.0

3 -4

30.0

6

30.0

9.66 10 Wb

cos (5.00 10 )(20.0 10 )(cos30.0 )

8.66 10 Wb

A

B

3 -4

45.0

6

45.0

3 -4

60.0

cos (5.00 10 )(20.0 10 )(cos45.0 )

7.07 10 Wb

cos (5.00 10 )(20.0 10 )(cos60.0

A

A

B

B

6

60.0

3 -4

75.0

6

75.0

)

5.00 10 Wb

cos (5.00 10 )(20.0 10 )(cos75.0 )

2.590 10 Wb

A

B

3 -4

90.0

90.0

cos (5.00 10 )(20.0 10 )(cos90.0 )

0 Wb

A

B



Page 95 of 185

2 6 5

3 0 15 15.0 0.00

3

-4 2 4 10 15

3

15 30 30.0

(9.66 10 ) (1.00 10 )5.00 10 T

(1.00 10 )

20.0 10 m 3.41 10 Wb s

1.00 10 s

t t

At

t

t

B

6 6

15.0

3

4 115 30

6 6

30 45 45.0 30.0

3

(8.66 10 ) (9.66 10 )

(1.00 10 )

9.99 10 Wb s

(7.07 10 ) (8.66 10 )

(1.00 10 )

t

t

t t

3 130 45

6 6

45 60 60.0 45.0

3

45

1.59 10 Wb s

(5.00 10 ) (7.07 10 )

(1.00 10 )

t

t t

3 160

6 6

60 75 75.0 60.0

3

3 160 75

2.07 10 Wb s

(2.59 10 ) (5.00 10 )

(1.00 10 )

2.41 10 Wb s

t

t t

t

6

75 90 90.0 75.0

3

3 175 90

(0) (2.59 10 )

(1.00 10 )

2.59 10 Wb s

t t

t

3 The rate of change of flux increases as the angle increases.



Page 96 of 185

4

4 20 15

4 215 30

EMF (100)( 3.41 10 ) 3.41 10 V

100

EMF (100)( 9.99 10 ) 9.99 10 V

NΔt

N

Δt

3 130 45

45 60

EMF (100)( 1.59 10 ) 1.59 10 V

EMF (100)

Δt

Δt

3 1

3 160 75

( 2.07 10 ) 2.07 10 V

EMF (100)( 2.41 10 ) 2.41 10 V

Δt

3 175 90 EMF (100)( 2.59 10 ) 2.59 10 VΔt

5 a At 90° as the rate of change of flux is maximum at this point, the wire is moving

perpendicular to the lines of flux at this point.

b

4 2 4 2

3 2

23 1 1

3

20.0 10 m 20.0 10 4.47 10 m

5.00 10 T 2.236 10 m2

2 2 (2.236 10 )24.0 10 s 5.85 10 m s

(24.0 10 )

100

A l

lr

rT v

T

N

B

1 3 2 EMF 2 2(100)(5.85 10 )(5.00 10 )(4.47 10 )

EMF 2.62 V

Nv l

B

6 B



Page 97 of 185

7 a C

b D

c C

d B

e D

8

3

2

3 3PP RMS

1 11000 0.0200 s

(50.0)

(0.0200)50.0 Hz 5.00 10 s

4 4

10.0 10 m

(8000)8.00 10 V 5.66 10 V

2 2

N Tf

Tf t

r

VV V

3 3

f i

2

i

EMF

EMF (5.66 10 )(5.00 10 )

(1000)

(0) 2.83 10

Nt

t

N

A

B

2

2 2

2.83 10

(10.0 10 )

0.900 T

B

B

9 a

P RMS

P

2 2 (240)

339 V

V V

V

b

P-P P

P-P

2 2 (339)

679 V

V V

V

c

RMSRMS RMS

P RMS

P

(240)240 V 2.40 A

(100)

100

2 2 (2.40)

3.39 A

VV I

R

R

I I

I



Page 98 of 185

d

RMSRMS RMS

(240)240 V 2.40 A

(100)

100

VV I

R

R

10 a 2

RMSRMS

2 2

RMS

240 V

(240)600 W

(600)

96.0

VV P

R

VP R

P

R

b

V 339

24022

P

RMSP

V

)(VV

c

1P P P2

P

P

P

339 V

2 2(600)600 W

(339)

3.54 A

V P V I

PP I

V

I

4.4 Transformers

1 a t

NV

B

1p

b t

NV

B

2s

c 2

1

s

p

N

N

V

V

2 a A, B, D

b A, B, D

3 a A

b B, D

4 Power losses occur when electrical energy is converted into heat energy in the copper windings

and in the iron core. Energy losses in the core are due to eddy currents.



Page 99 of 185

5 a B

b D

c A

6 a

p p

p

s s

p s

p s

p

s s

8.00 V

(8.00)(200)20 turns

(20)

200 turns 80.0 V

V NV

V N

V NN V

N

N V

b

p p p s s

p p

p s

s

s s

Assume 100% efficient

8.00 V

(8.00)(2.00)2.00 A

(80.0)

80.0 V 0.200 A

V V I V I

V II I

V

V I

c

s s s s

s s

Assume 100% efficient

80.0 V (80.0)(0.200)

0.200 A 16.0 W

V P V I

I P

7 a

p p

p

s s

p s

p s

p

s s

240 V

(800)(12.0)800 turns

(240)

12.0 V 40.0 turns

V NV

V N

N VN N

V

V V

b

p p p s s

s ss p

p

s p

240 V

(12.0)(2.00)2.00 A

(240)

12.0 V 0.100 A

V V I V I

V II I

V

V I

c

p p p p

p p

240 V (240)(0.100)

0.100 A 24.0 W

V P V I

I P

8 The security light would not operate. In order for an EMF to be generated in the secondary coil a



Page 100 of 185

changing magnetic flux is required, a constant DC supply will create a constant field, therefore

no EMF is induced in the secondary coil.

9 There is no power consumed in the primary coil during this time. This is because the change in

flux in the transformer core is not causing any current in the secondary coil, so no energy is lost

from the magnetic field. The change in flux in the primary coil will induce a back EMF, which is

equal in magnitude and opposite in direction to the applied EMF if it is a perfect transformer. In

reality there is some energy loss in the eddy currents in the core so some energy is used from the

field and less energy is available to create the back EMF in the primary coil. This slight

imbalance in the applied EMF and the back EMF results in a small current flowing in the primary

coil and therefore some small power consumption occurs. This is why transformers should be

unplugged when they are not being used.

Assume 2 W consumed by primary coil:

(2)(10 60)

1200 J

EP

t

E P t

E

E

4.5 Distributing electricity

1 a By transforming to higher voltages the current decreases, this allows thinner cables to be

used. Also the power lost by the cables is reduced significantly as RIP 2loss

b The corona effect limits the voltage for power transmission. Differences in potential of

1000 V per centimetre will cause current to flow through air, At 500 kV this means that

any ground source must be at least 500 cm away from the active wire. This becomes

problematic for the design of the transmission towers and insulators used.

2 a

6ps6

ps tl 3

ps

3 3

ps tl

(500 10 )500 10 W

(250 10 )

250 10 V 2.00 10 A

PP I

V

V I

b

6ps6

ps tl 3

ps

3 3

ps tl

(500 10 )500 10 W

(500 10 )

500 10 V 1.00 10 A

PP I

V

V I



Page 101 of 185

3 a 2 3 2

tl loss tl tl

3 7

tl loss

7

lossloss 6

ps

10.0 (2.00 10 ) (10.0)

2.00 10 A 4.00 10 W

(4.00 10 ) % 100 100

(500 10 )

R P I R

I P

PP

P

loss % 8.00%P

b 2 3 2

tl loss tl tl

3 7

tl loss

7

lossloss 6

ps

10.0 (1.00 10 ) (10.0)

1.00 10 A 1.00 10 W

(1.00 10 ) % 100 100

(500 10 )

R P I R

I P

PP

P

loss % 2.00%P

4

3 1tl 1 2

1

2 1 12 12 2 2

2 1 1

2 3 2

loss tl tl

1.00 10 A

2 2 0.5

(2 ) 4

(1.00 10 ) (5.0)

I RA r

R Rr r r

P I R

6

loss

6

lossloss 6

ps

loss

5.00 10 W

(5.00 10 ) % 100 100

(500 10 )

% 1.00%

P

PP

P

P

5 a

3spg3

spg tl

spg

spg tl

(5.00 10 )5.00 10 W

(500)

500 V 10.0 A

PP I

V

V I

b

2 2

tl loss tl tl

2

tl loss

4.00 (10.0) (4.00)

10.0 A 4.00 10 W

R P I R

I P



Page 102 of 185

c

2

lossloss 3

spg

loss

(4.00 10 ) % 100 100

(5.00 10 )

% 8.00%

PP

P

P

d

tl tl

tl

house spg tl

2

house

10.0 A (10.0)(4.00)

4.00 40.0 V

(5.00 10 ) (40.0

I V IR

R V

V V V

V

2

house

)

4.60 10 VV

6 a

3spg3

spg tl 3

spg

3

spg tl

(5.00 10 )5.00 10 W

(5.00 10 )

5.00 10 V 1.00 A

PP I

V

V I

b 2 2

tl loss tl tl

tl loss

lossloss 3

spg

4.00 (1.00) (4.00)

1.00 A 4.00 W

(4.00) % 100 100

(5.00 10 )

R P I R

I P

PP

P

loss % 0.0800%P

c

tl tl

tl

house spg tl

3

house

1.00 A (1.00)(4.00)

4.00 4.0 V

(5.00 10 ) (4.0)

I V IR

R V

V V V

V

3

house 4.996 10 VV

7 a

1.00 kW (1.00)(2.00) 2.00 kW h

2.00 h cost rate (2.00) 0 14 $0.28

P E P t

t E ( . )



Page 103 of 185

b

3 3 2

2

80.0 10 kW (80.0 10 )(0.500) 4.00 10 kW h

Δ 0.500 h cost rate (4.00 10 )(0 14) $0.0056

P E P t

t E .

c

3 3250.0 10 kW (250.0 10 )(12.0) 3.00 kW h

Δ 12.0 h cost rate (3.00)(0 14) $0.42

P E P t

t E .

d

3 36.00 10 kW (6.00 10 )(7 24) 1.01 kW h

7 24 h cost rate (1.01)(0 14) $0.14

P E P t

Δt E .

e

3 33.00 10 kW (3.00 10 )(365.25 24) 26.3 kW h

365.25 24 h cost rate (26.3)(0 14) $3.68

P E P t

Δt E .

8 a 6

6 towntown tl

town

6

town tl

6

tl tl

(500.0 10 )500.0 10 W

(250.0)

250.0 V 2.00 10 A

Δ (2.00 10 )(2.00)

PP I

V

V I

V I R

6 Δ 4.00 10 V

This would be impossible.

V

b 6

6 towntown tl 3

town

3 3

town tl

3

tl tl

(500.0 10 )500.0 10 W

(100.0 10 )

100.0 10 V 5.00 10 A

Δ (5.00 10 )(2.00)

PP I

V

V I

V I R

4

4 3

ps town

5

ps

Δ 1.00 10 V

(1.00 10 ) (100.0 10 )

1.10 10 V

V

V ΔV V

V



Page 104 of 185

c

3 2 3 2

tl loss tl tl

7

tl loss

5.00 10 A (5.00 10 ) (1.00)

1.00 2.50 10 W

I P I R

R P

9 a

3spg3

wt tl

tl

tl tl

(150.0 10 )150.0 10 W

(10 000)

10 000 V 15.0 A

PP I

V

V I

b

tl tl

t2 t1

3

t2

Δ (15.0)(2.00)

Δ 30.0 V

Δ (10 000) (30.0)

9.97 10 V

V I R

V

V V V

V

c 2 2

tl loss tl tl

2

tl loss

15.0 A (15.0) (2.00)

2.00 4.50 10 W

This would not be a great problem.

I P I R

R P

10 a

3spg3

wt tl

tl

tl tl

(150.0 10 )150.0 10 W

(1000)

1000 V 150.0 A

PP I

V

V I

b

tl tl

t2 t1

2

t2

Δ (150.0)(2.00)

Δ 300.0 V

Δ (1000) (300.0)

7.00 10 V

V I R

V

V V V

V



Page 105 of 185

c 2 2

tl loss tl tl

4

tl loss

town wt loss

town

150.0 A (150.0) (2.00)

2.00 4.50 10 W

I P I R

R P

P P P

P

3 4

5

town

(150.0 10 ) (4.50 10 )

1.05 10 WP

d No, as this results in a significant loss of power over the length of the transmission line

(30%).

Chapter 4 Review

1 a

4 2

4 4 4 44

1 3

4 3

2

3

40.0 10 m EMF

(16.0 10 )(40.0 10 ) (8.00 10 )(40.0 10 )8.00 10 T EMF

(1.00 10 )

16.0 10 T EMF 3.20 10 V

1.00 10 s

1.00

At

B

B

t

R

3

3(3.20 10 ) 3.20 10 A clockwise

1.00

VI

R

b

4 2

4 4 4 44

1 3

4 3

2

3

40.0 10 m EMF

( 8.00 10 )(40.0 10 ) (8.00 10 )(40.0 10 )8.00 10 T EMF

(2.00 10 )

8.00 10 T EMF 3.20 10 V

2.00 10 s

1.00

At

B

B

t

R

3

3(3.20 10 ) 3.20 10 A counterclockwise

1.00

VI

R



Page 106 of 185

c

4 2

4 4 4 44

1 3

4 3

2

3

40.0 10 m EMF

(4.00 10 )(40.0 10 ) (8.00 10 )(40.0 10 )8.00 10 T EMF

(1.00 10 )

4.00 10 T EMF 1.60 10 V

1.00 10 s

1.00

At

B

B

t

R

3

3(1.60 10 ) 1.60 10 A clockwise

1.00

VI

R

2 a

4 2 4 2

4 2 1

2

40.0 10 m 40.0 10 6.32 10 m

28.00 10 T 2 (6.32 10 ) 1.99 10 m

2

1 1100 Hz 1.00 10 s

(100)

A l A

lB C r

f Tf

11 1

2

4 2 1

(1.99 10 ) 1.99 10 m s

(1.00 10 )

EMF 2 2(8.00 10 )(6.32 10 )(1.99 10 )

EM

Cv

T

Blv

3F 2.01 10 V

b

33

peak

3

peak

(2.01 10 )EMF 2.01 10 V

(1.00)

1.00 2.01 10 A

VI

R

R I

3 a current halved to 1.00 mA, period doubled to 20 ms = A

b current same as 2.01 mA, period halved to 5 ms = C

c current halved to 1.00 mA, period same as 10 ms = B

4 a To the left, as the soft iron core is induced to become a temporary magnet by the

permanent magnet’s field.

b To the left, attraction.

c To the right, repulsion.



Page 107 of 185

5

2

3 2 2

3

2

2

4.00 10 m EMF

(0) (20.0 10 ) (4.00 10 )20.0 10 T EMF (40)

(0.100)

40 turns EMF 4.02 10 V

0.100 s

(4.02 10 )2.00

r Nt

B

N

t

VR I

R

22.01 10 A from Y to X

(2.00)

6 A, C

7 The direction would be from X to Y as according to Lenz’s law the EMF will be induced in a

direction that causes a current that creates a magnetic field that opposes the change that is causing

the current. Current from X to Y will cause a north pole at the top and a south at the bottom of

the coil, which will oppose the north at the top of and the south at the bottom of the permanent

magnet.

8 a 3 3 2

2 3

1

33

10.0 10 T EMF (10.0 10 )(20.0 10 )(2.00)

20.0 10 m EMF 4.00 10 V

2.00 m s

(4.00 10 )1.00 4.00 10 A from X to Y

(1.00)

B Blv

l

v

VR I

R

b

3 3 2 3

2 6

3

10.0 10 T (4.00 10 )(20.0 10 )(10.0 10 )

20.0 10 m 8.00 10 N to the left

4.00 10 A

B IlB

l

I

F

F

9 There is no induced current as there isn’t a complete circuit, as the switch is open.

10

2

2 2

1.00 T (1.00)(5.00 10 )(1.00)

5.00 10 m 5.00 10 N

1.00 A

B IlB

l

I

F

F

11 To the right



Page 108 of 185

12

2

2 2

1.00 T (1.00)(1.00 10 )(1.00)

1.00 10 m 1.00 10 N

1.00 A

B IlB

l

I

F

F

13 To the left

14 a

5 5

3

1

5.00 10 T EMF (5.00 10 )(8.00)(4.00)

8.00 m EMF 1.60 10 V

4.00 m s W

B Blv

l

v

b Zero current is induced in the loop, as both vertical sides of the loop have an EMF induced

in the same direction (downwards). This means that each side produces equal and opposing

EMFs so no current flows in the loop.

15

2

5

5 1

4

40.0 m EMF

(1.00 10 ) 40.01.00 10 T s EMF (1)

(1.00)

1 turn EMF 4.00 10 V

8.00

A Nt

B

N

R

45(4.00 10 )

5.00 10 A(8.00)

VI

R

16 a

3 3 3 3

i

3 6

i

3

100.0 10 m (1.00 10 )(100.0 10 )(50.0 10 )

50.0 10 m 5.00 10 Wb

1.00 10 T

l B A

w

B

b

3 3

f

3

f

3

100.0 10 m (1.00 10 )(0)

50.0 10 m 0 Wb

1.00 10 T

l B A

w

B



Page 109 of 185

c

6

i

6

f 3

3

3

5.00 10 Wb EMF

(0) 5.00 100 Wb EMF (1)

(2.00 10 )

1 EMF 2.50 10 V

2.00 10 s

Nt

N

Δt

d 3

3

3

(2.50 10 )2.50 10 V

(2.00)

2.00 1.25 10 A

VV I

R

R I

e No the current will stop. For an EMF to be induced, the flux must be changing in the loop,

if it is not changing then no EMF is induced and therefore no current will flow.

17 a This is a quarter of the time so the EMF and therefore the current will increase by a factor

of four = 2.00 × 10–4 A.

b

6

meter total

2

coil

6

2

595 Δ (50.0 10 )(595 5.00)

5.00 Δ 3.00 10 V

50.0 10 A

(0) ( )100 EMF

3.00 10

R V IR

R V

I

B rN N N

t t

r

2

2

2

2 2

1

EMF m

(3.00 10 )(2.00)Δ 2.00 s

(100) (3.00 10 )

2.12 10 T

tB

N r

t B

B



Page 110 of 185

18 a

b

peak

peak RMS

RMS

(0.900)0.900 V

2 2

0.636 V

VV V

V

c Period halved to 5 ms, Vpeak doubles to 1.8 V, VRMS becomes 1.3 V.

19 a

RMS p RMS p RMS p RMS s RMS s

RMS p RMS p

RMS p RMS s

RMS s

RMS s RMS s

14.0 V

(14.0)(3.00)3.00 A

(42.0)

42.0 V 1.00 A

V V I V I

V II I

V

V I

b

RMS p p

RMS p

RMS s s

RMS p s

s p

RMS s

RMS s p

14.0 V

(14.0)(30)30

(42.0)

42.0 V 10 turns

V NV

V N

V NN N

V

V N

c

RMS p s p RMS p RMS p

RMS p s

14.0 V (14.0)(3.00)

3.00 A 42.0 W

V P P V I

I P

20 a C

b A

c B

21 C



Page 111 of 185

22 C

23 a

3 2

3

1 12.0 10 s 5.0 10 Hz

(2.0 10 )T f

T

b

peak

peak RMS

RMS

(25)25 V

2 2

17.7 V

VV V

V

c

peak p-p peak

p-p

25 V 2 2(25)

50 V

V V V

V

24 a

peak

peak RMS

RMS RMS

RMS RMS RMS

RMS

(15)15 A

2 2

17.7 V 10.6 A

(17.7)(10.6)

188 W

II I

V I

P V I

P

b

peak peak peak peak

peak peak

15 A (15)(25)

25 V 375 W

I P V I

V P

c

RMSRMS

RMS

RMS

(17.7)10.6 A

(10.6)

17.7 V 1.67

VI R

I

V R

25 C

26 2 2(30)

15 (15)

30 V 60

VR P

R

V P W

Set C is equivalent to 60 W.



Page 112 of 185

27 a

peak p-p peak

p-p

peak

RMS

RMS

12.6 V 2 2(12.6)

25.2 V

(12.6)

2 2

8.91 V

V V V

V

VV

V

b

peak

peak RMS

RMS

(25.2)25.2 V

2 2

17.8 V

VV V

V

c

RMS peak RMS

peak

peak 2 2

peak 1 1

2

16.0 V 2 2 (16.0)

22.6 V

V V V

V

V f

V f

ff

1 peak 2

peak 1

2

(50.0)(22.6)

(12.6)

89.8 Hz

V

V

f

d

peak 2 2peak

peak 1 1

32 peak 1

peak 2 3

1

peak 2

12.6 V

(60.0 10 )(12.6)

(80.0 10 )

9.45 V

V BV

V B

B VV

B

V



Page 113 of 185

28 a

tr

480.0

(480.0)240.0 V 2.00 A

(240.0)

(2.00)(2.00) 4.00 V

(240.0) (4.0

P W P VI

PV I

V

V IR

V

0) 236.0 V

The machine should work satisfactorily.

b By dropping the voltage by a factor of 10 the current is increased by a factor of 10. This

would result in a significant power loss over the cable.

c

trans out machine cable

s s

p

20.0 (20.0)(2.00) 40.0 V

2.00

(24.0) (40.0) 64.0 V

I A V IR

R

V V V

N V

N V

p

(64.0)0.267

(240.0)

d

loss20.0 (40.0)(20.0) 800.0 W

40.0 V

I A P VI

V

29 Appliances with built-in transformers or motors that require AC will not function correctly and

could burn out. At full load there would be a power loss of about 555 W, or about 66 V

difference in potential which would only leave about 173 V potential at the farmhouse.



Page 114 of 185

30 He needs a step-down transformer with a turns ratio of 5 : 1.

low

cable

p gen cable

(500)500 W 0.417 A

(1200)

Δ (0.417)(8.00) 3.33 V

Δ (1200) (3.33) 1197 V

PP I

V

V IR

V V V

s p

1 1 (1197) 239 V

5 5V V

half

cable

p gen cable

(1000)1000 W 0.917 A

(1200)

Δ (0.917)(8.00) 7.33 V

Δ (1200) (7.33) 1193 V

PP I

V

V IR

V V V

s p

1 1 (1193) 239 V

5 5V V

full

cable

p gen cable

(2000)2000 W 1.67 A

(1200)

Δ (1.67)(8.00) 13.3 V

Δ (1200) (13.3) 1187 V

PP I

V

V IR

V V V

s p

1 1 (1187) 237 V

5 5V V

This set-up would suit his purposes well.



Page 115 of 185

Worked solutions Unit 3B

Contents

The sounds of music 116

The search for understanding 120

Chapter 5 Wave properties and light 129

5.1 Wave properties 129

5.2 Wave behaviour 129

5.3 Wave interactions 131

5.4 Electromagnetic radiation 133

5.5 Electromagnetic radiation and matter 136

5.6 X-rays 139

Chapter 5 Review 141

Chapter 6 Matter, relativity and astronomy 152

6.1 Extending our model of matter 152

6.2 Einstein’s special theory of relativity 154

6.3 To the stars 156

6.4 Fundamentals of astronomy 157

6.5 Hubble’s universe 158


Chapter 7 Electric and magnetic fields 164

7.1 Force on charges in magnetic fields 164

7.2 Particle accelerators 164

7.3 Synchrotons 167

7.4 Mass spectrometry 169


Chapter 8 Working in physics 175

8.1 Measurements and units 175

8.2 Data 176

8.3 Graphical analysis of data 178

8.4 Writing scientific reports 180




Page 116 of 185

The sounds of music E1

0

12 2

0 12

12

42.0 dB 10log

10 W m (42.0) 10log10

(4.20) log10

IL L

I

II

I

4.20

12

4.20 12 8 2

1010

10 10 1.58 10 W m

I

I

E2

0

12 2

0 12

9.80

1 12

2

98.0 dB 10log

10 W m (98.0) 10log10

3.00 m 1010

5.00 m 10

IL L

I

II

Ir

r I

9.80 12 3 2

2 2

1 1 2 2

2 3 2

1 12 2 2

2

10 6.31 10 W m

(6.31 10 )(3.00 )

(5.00 )

I r I r

I rI

r

3 2

2 2.27 10 W m

I

E3

12

2

2

2

112

3

2

13

2

1

2

1

2

1 3

ff

T

Lf

3T

Lf

T

Lf

T

LfTT

1

1

2

1



Page 117 of 185

E4 2

2 1

2

2

(30.0 10 )0.400 m 7.50 10 kg m

(0.400)

120.0 N

1 130.0 10 kg

2 2 0.400) (7.50 10 )

mL

L

T

T (120.0)m f

L (

50.0 Hzf

E5 a

air sound

2 1

sound

26.0 C 331 0.6 331 0.6(26.0)

3.47 10 m s

T v T

v

1

2

1

1

2 2

3.47 10 )100.0 Hz

2 2(100.0)

1.73 m

vL f

L

v (f L

f

L

b

1

2

1

1

4 4

3.47 10 )100.0 Hz

4 4(100.0)

0.867 m

vL f

L

v (f L

f

L

E6

2

1

1

3 1

(3.40 10 )0.400 m

2 2(0.400)

425 Hz

3 3(425)

vL f

L

f

f f

3

3 1.28 10 Hzf

E7

1

60.0 Hz

346)assume 346 m s 5.77 m

(60.0)

size of shell 6 m

vf f

v (v

f



Page 118 of 185

E8

air sound

2 1

sound

2.00 C 331 0.6 331 0.6(2.00)

3.32 10 m s

T v T

v

return

to prey

2

to prey to prey

0.0300 s

0.0150 s (3.32 10 )(0.0150)

4.98 m

st v

t

t s v t

s

E9 To ensure that the air column inside their instrument maintains a constant temperature, this will

ensure that the speed of sound in the instrument is also constant and therefore the resonant

frequencies are in tune with the standard frequencies expected by the conductor.

E10 She should loosen the string slightly.

E11 2

1 21 1

1 2

2

1 12 1

1 1

2

1

1

100

0.50.5 100

0.5

0 100

1

100

R

R

R

R

m mm m E

m m

m mm m E

m m

.5mE

.5m

E

2 ( 0.33)

11.1%RE

E12 2

1 21

1 2

2

2

2

1.00 100

(1.00) (1.33)1.33 100

(1.00) (1.33)

0 33 100

2 33

R

R

R

n nn E

n n

n E

.E

.

2 100 ( 0.142)

2.01%

R

R

E

E

E13 Yuki should either loosen or tighten her violin string and listen to see if the number of beats

increases or decreases. What she should be trying to achieve is to decrease the number of beats

per second until no beats are heard. At this point the frequency of the tuning fork and of the

violin string are the same.

E14 B



Page 119 of 185

E15 C

I would look for a low minimum power rating to avoid ‘clipping’ that could damage the tweeters.

It is unlikely that I would turn up the speakers to maximum so I could get speakers that are less

than the maximum power of the amplifier, so I would choose speakers C.

E16

2

80.0 W and

8.00

(80.0)

(8.00)

3.16 A

P P VI V IR

R P I R

PI

R

I



Page 120 of 185

The search for understanding The questions in this context are largely based on opinion and discussion, with an emphasis on critical

thinking and the formation of supported arguments. In most cases, the following answers consist only of

lists of some relevant points which do not necessarily support only one conclusion.

It is important to note that although other information may be relevant if introduced to, or by,

students, the suggested responses are based only on information that is either presented within the

context or available to the general public.

E1 Before Thales theories may have suggested that the gods (or spirits; mythical creatures; magic)

created the structure and nature of the Earth and the heavens.

E2 a The presence of condensation on cold objects, when organic matter is crushed it often

contains liquids. Water itself, and the large number of other fluids. Death—animals lose

fluid, including blood, as they die and break down, plants appear dry as they die. Mud

loses water before becoming dust and disappearing in the wind. The clouds contain water

(rain) and the air contains water (condensation). Ice to water to steam is an example of

water taking different forms.

b Yes as it attempts to explain observations.

E3 a It illuminates objects, it cannot be seen as it passes by our eyes, it can be reflected from

shiny objects and bend as it passes through transparent objects, shadows with distinct

edges, so light is blocked from a straight path.

b i It illuminates objects – all three filaments interact.

It cannot be seen as it passes by our eyes – the filament from the object is not

present

It can be reflected from shiny objects – the filaments reflect off the shiny surface.

It bends as it passes through transparent objects – the filaments are bent by the

substance

When an object passes in front of what is being looked at, the filament could easily

be cut to now be shorter (to see the closer object) and go no further (leave a shadow

behind the object)

ii It illuminates objects – particles from your eye strike the object.

It cannot be seen as it passes by our eyes – the particles pass by our eyes.

It can be reflected from shiny objects – the particles bounce off the shiny surface.

It bends as it passes through transparent objects – the particles change direction as

they enter the substance.

Particles directed at an object will not hit any part of a surface directly behind that

object (shadows)

iii It illuminates objects – waves from a source strike the object then enter our eye.

It cannot be seen as it passes by our eyes – the wave from the object passes by our

eyes.

It can be reflected from shiny objects – the waves reflect off the shiny surface.

It bends as it passes through transparent objects – the waves change direction as they

enter the substance.

Waves do not pass beyond objects in their path (shadows); they do over longer

distances, but other factors could then come into play



Page 121 of 185

E4 a As rocks are broken down, the smaller grains tend to become more and more similar in

size (they resemble distinct particles). Some substances can be made by mixing other

substances, suggesting that the final form of matter is not the same as the material from

which it is built. Heating/burning changes substances, suggesting that different structures

of the same material produce different properties

b The four distinct elements could be made up of either particles or continuous matter. The

four elements could be four types of atom, instead of one as proposed. The four elements

could be four basic ways of structuring the proposed identical atoms

E5 a Both Newton and Hooke describe light as being all of the same type regardless of the

source. Both Newton and Hooke can explain reflection and mirrors, and refraction. Al-

Haytham suggests that light slows down when it enters a more dense medium, Hooke’s

waves would slow down as they enter the greater resistance of a more dense medium, and

Newton’s particles speed up as they are affected by the attractive forces at a boundary with

another medium

b Al-Haytham’s work is very successful as it applies to the practical considerations of a wide

variety of mirrors. Observations can be successfully explained. The explanation is only

theory.

c Australian society is essentially of European origin our history is traditionally European

(and European American) only. Al Haytham was not European and so not part of the

debates of academics that tend to be discussed and become widely known. Communication

between people in different parts of the world was not as quick and easy as it is now; al-

Haytham’s ideas may simply never have been taken to Europe. Europe was in the midst of

the Middle Ages and was less concerned with scientific advances and was lax in the

keeping of written records; Al-Haytham could have been known about without any written

record. Along with the ideas of women not being taken seriously, the male European

scholars who wrote our scientific history may not have considered ideas from the Middle

East to be worthy of inspection.

E6 Filaments are instantaneous, particles and waves travel and so have a speed. Calculation of the

speed of light as it moves into a more dense medium is important in deciding between waves,

which should slow down, and particles, which should speed up.

E7 a Waves – as a wave moves out from a point source, the radius of the curve increases; the

same wave energy is spread over a larger surface and so must be less at each point.

Particles – a spray of particles leaving a point would consist of slightly different directions;

as they move further from their origin, the difference in direction remains unchanged while

the particles spread further apart, and so there are less particles within the same area.

b Determining the distance to a known light source or determining the intensity of a light

source at a known distance. Investigating the relative intensities of the stars, when other

measurements are made to indicate relative distances.

E8 According to Newton, reflection is due to repulsive forces at the boundary and refraction is due

to attractive forces at the boundary. How is it possible for the same particles to be both repelled

and attracted by the forces at the same boundary?

E9 The speed of light in different media and/or as it crosses boundaries. A medium through which

light travels (the ether) or diffraction (bending around edges).



Page 122 of 185

E10 a Both the wave and particle theories explain current observations. The particle theory does

not require the missing diffraction effects or ether in order to be valid (note that because

something has not been observed, it does not mean that it does not exist)

b Newton is well-respected and has presented a comprehensive theory, including an

explanation of colour and the development of the successful reflecting telescope

E11 The diffraction effect may be very small, due to very small wavelengths, and so is unable to be

measured with available resources.

E12 Roemer’s use of a very, very large distance minimises the relative effects of small measurement

errors.

E13 Two theories: continuous or indivisible particles. Atoms have been proposed since 440 BCE.

Boyle, in 1661, explains gases as atoms and molecules moving around. Newton includes light as

being particles (the same general idea although with no mass). There is no evidence to dispute the

continuous theory as there has been no direct observation of atoms

E14 Lavoisier’s accurate weights suggest a simple movement of matter from one place to another,

with the total amount remaining constant.

E15 a Refraction; possible explanation for diffraction and colour; possible explanation for

interference

b Newton’s theory was suggested partly because of an absence of observed diffraction. If the

observed pattern is of interference, then light cannot be made up of particles; particles are

not able to interfere and cancel each other out

E16 Dalton’s explanation is both comprehensive and successful when put to practical use (chemical

reactions). It is not as easy to explain whole number ratio combinations when using continuous

matter, but it is still a possibility. There has been no direct observation of the atom

E17 a Supporting the wave theory:

Diffraction of light, interference of light, Fresnel’s mathematical basis for wave theory

predicts the observations of light, electric and magnetic fields are suggested as a possible

means for the propagation of waves, light slows down when travelling in a more dense

medium, Maxwell’s equations regarding electromagnetic waves match the properties of

light, Hertz produced electromagnetic waves that have the same properties as light.

Discrediting the particle theory:

Interference patterns, Michelson’s interferometer.

b There is a lot of support for waves, but more important is the existence of discrediting

evidence regarding particles. The conclusion that light cannot be made up of particles does

not automatically mean that it has a wave nature.

E18 Fresnel’s model was derived from observations independent of wave theory. A mathematical

basis provides a method for predicting outcomes with greater precision than observation alone.

E19 a Seemingly separate theories and phenomena may be linked; for example, the properties of

light and matter may be related

b Phenomena that we see as separate may be intrinsically linked. There may be a single

theory that is responsible for all that we observe (a grand unified theory)



Page 123 of 185

E20 No medium (e.g. ether) for the transmission of light waves has been found. Lines of electric and

magnetic force could not be observed directly, but would only be detected via their effects on

electric charges and/or magnetic materials—no medium for the propagation of waves can be

found using traditional methods. The moving electric and magnetic fields of electromagnetic

waves could produce their own medium through which to travel; that is, no pre-existing medium

is necessary

E21 The wave theory suggests that light travels more slowly in a more dense medium; this is

supported by Foucault’s measurements. The particle theory suggests that light speeds up as it

crosses into another medium, due to the attractive forces at the boundary; this is not supported by

Foucault’s measurements.

E22 No particles were observed. Particles don’t generally cause things to glow and the glowing of the

gas implies a gain of energy (rays/waves). The glowing appears to be caused by the electric

current, and there is no indication that electricity involves particles. The shadow indicates straight

line paths, which could be either waves or particles

E23 a The properties of electric current are known well enough for practical uses of electricity to

be successful, e.g. light bulbs and street lighting. The nature of electric current is unknown

and the cause of electric current is unknown

b Health risks are unknown if phenomena are not understood; for example, X-rays. Other

effects are unknown if phenomena are not understood; for example, nuclear explosions.

Electricity was used successfully for a long time before it was understood; its properties

were clear, only the underlying process was missing. Even when a phenomenon is

understood, risks and unknowns still exist; for example, it’s clear how mobile phones work

but their long-term effect on the human brain is not known. The development of

applications of phenomena is often a step in the process of understanding the underlying

principles. We have been using our knowledge of electric charge in many ways, including

electricity and electronics as well as electromagnetism, but we don’t know what causes

charges to attract and repel each other. What would our lives be like if we continued to

refuse to work with electric charges?

E24 Michelson’s equipment was more precise and more accurate, therefore allowing fewer and

smaller errors. Michelson removed the effects of human error based on reaction times. Michelson

was the first to achieve what others had consistently failed to do; that is, measure the speed of

light over an observable distance. Michelson’s measurements were taken on Earth, and thus

could not have been affected by light travelling through space. Michelson’s measurements were

clearly of the light itself, rather than some other possible characteristic of space.



Page 124 of 185

E25 The light will only reflect from the octagonal mirror to the distant mirrors and back, and then

from the other side of the octagonal mirror to the telescope, when the octagonal mirror is in the

position shown in figure su.15 at both times. The spinning will therefore create pulses of light

that only reflect to the distant mirrors when the octagonal mirror is in the orientation shown, and

these pulses of light will only then reflect to the telescope when the octagonal mirror is again in

the shown orientation.

Assuming that a pulse of light is sent to the distant mirrors, the orientation shown in Figure su.15

will only occur again (for the light to reflect to the telescope) if, during the time taken for the

light to reach the distant mirrors and return, the octagonal mirror:

does not move (which would prove nothing)

moves one-eighth of a rotation

moves two-eighths of a rotation

moves three-eighths of a rotation

moves four-eighths of a rotation etc.

Therefore, if the octagonal mirror spins any eighth of a rotation during the time taken for the light

to return from the distant mirrors, the pulse of light will return with the octagonal mirror in the

correct position to reflect it to the telescope. Any rotation other than an eighth will not reflect the

light to the telescope.

Importantly, each pulse of light will reach the telescope should any full eighth of a rotation occur

in the time it takes for the light to travel. The rate of rotation used will vary however, making it

difficult to determine the time taken. For example, if it were to actually take 1 s for the pulse of

light to reach the telescope, then all of the following would give similar results:

one-eighth of a rotation every second (i.e. 8 s per rotation)

two-eighths of a rotation every second (i.e. 4 s per rotation)

three-eighths of a rotation every second (i.e. 2.7 s per rotation)

four-eighths of a rotation every second (i.e. 2 s per rotation) etc.

A one-eighth rotation will have been achieved when the rate of rotation has been reduced until

the slowest rate that allows every pulse of light to reach the telescope is identified. This is the rate

at which the octagonal mirror moves around one face only. At no point will there be a face in the

correct position if a slower rate of rotation is used. Reaching this point means that both the rate of

rotation and the eighth of a rotation used are known, giving a single possible value for the time

taken for the light to travel its path.

E26 Maxwell produced mathematical equations based on electromagnetic properties, and found that

the predicted properties of electromagnetic waves matched the properties of light. Hertz produced

electromagnetic waves experimentally, using oscillating currents, and found that the properties of

the electromagnetic waves matched the properties of light. Two different approaches, one

theoretical and one experimental, that independently produce the same results tend to validate the

underlying theory, rather than allowing the results to be explained by factors specific to one set of

conditions.



Page 125 of 185

E27 Matter:

particles/atoms consistently explain the properties of matter

Boyle’s theory of matter, as atoms moving around and colliding, successfully explains

observations regarding the nature of matter

Lavoisier’s theory of the conservation of matter shows matter to be a constant presence

Dalton’s work with ratios of atoms in combination continues to be supported by the field

of chemistry

no direct observation of atoms has occurred—the evidence is circumstantial only.

The particle theory of matter has an abundance of support, but remains theory.

Light:

that light has a speed supports both the particle and wave theories, but not the filament

theory

reflection can be explained by both wave and particle theories

mirrors and lenses can be explained by both wave and particle theories

Kepler’s inverse square law of light intensity and distance can be explained by both wave

and particle theories

Foucault’s determination of the change in the speed of light in refraction supports the wave

theory over the particle theory

diffraction effects support the wave theory over the particle theory

interference effects can be successfully explained by the wave theory only

Maxwell’s equations can explain the properties of light by assuming that it consists of

electromagnetic waves

light has the same properties as the electromagnetic waves produced by Hertz

Michelson and Morley’s failure to detect the ether supports the particle theory, but Faraday

suggests that electric and magnetic fields provide a means of propagating light waves

without any other medium.

It is clear that light must have a wave nature.

E28 a The wave theory of light explains all of the observations so far, only the wave theory

successfully explains interference effects.

b Thomson’s plum pudding model of the atom provides support for the atomic theory of

matter. The identification of electrons as distinct and identical particles provides evidence

of the existence of fundamental particles. The particle theory of atoms is able to explain all

of the observations so far.

c Results are reproducible. Forms of radiation can be used in experiments. There appear to

be no adverse effects. The properties of the two identified forms of radiation have been

identified. The cause of the radiation is not known.

d The properties of electric current are well understood. Practical uses have been

successfully implemented for some time already. The discovery of the electron allows an

explanation of the processes of electric charge and electricity.



Page 126 of 185

e i Spectral lines, both emission and absorption, the photoelectric effect and the origin

of radiation.

ii Spectral lines are clearly related to light, and light is understood. The photoelectric

effect is related to light and electricity, both of which are understood. Radiation is

clearly related to atoms, and atoms are quite well understood. The details of the

properties of these phenomena are known, with only the underlying theory needing

to be identified

E29 Light is understood and matter is understood, however, the observations yet to be explained are

related to phenomena that are understood. Other fields like motion and gravitation have been

mastered, leading to feats of architecture and engineering. Experiments have been producing

more accurate measurements, rather than new information. It is disturbing that with so much that

is so clear, there are no real clues as to the underlying principles of spectral lines, the

photoelectric effect and radiation.

E30 From the 1600s, gaps of 20–50 years between advances were not uncommon. These gaps are

now more like 5–10 years, with annual advances occurring. The increase in the frequency of

advances may be due to:

greater investment in research at universities

improved access to colleagues to share ideas and argue theories

greater literacy rates leading to more academics doing research

greater literacy rates leading to more widespread participation in the discussion of ideas

improved design of equipment, based on previous advances, which can more easily test

theories

a recognition of the possible economic advantage in understanding, such as the widespread

use of electricity, promoting interest in research.

E31 Lenard shows that the current wave theory of light not only cannot explain the photoelectric

effect, but that the observations recorded are in fact in opposition to this theory.

E32 a Planck’s theory relates to the energy possessed by objects, and works mathematically

Einstein’s theory relates to the nature of light, and explains observations logically

b Very different approaches related to very different phenomena provide validation for a

common underlying theory; the results cannot be explained away as a function of a

particular type of experiment or selective observations.

E33 Planck first proposed the quantum, E = hf. Planck proposed a mathematical property, rather than

suggesting that it was an actual physical property of matter. Einstein built his ideas from the idea

of quanta originally proposed by Planck. Einstein determined that quanta were physical

properties of light energy. Einstein proposed a comprehensive theory using quanta to explain the

previously unexplained photoelectric effect.

E34 Relativistic effects are only noticeable when velocities near the speed of light are involved. We

do not experience these speeds in our lives. Note that, although we are unaware of it, satellites

travel at high speeds relative to us. Data transmissions must therefore account for relativistic

effects; that is, we use special relativity everyday as we watch TV, use a GPS etc.

E35 Rather than providing an answer, Einstein’s E = mc2 provides a new dilemma of light having

both wave and particle properties. The debate over wave or particle may be irrelevant; if energy

and mass are interchangeable then waves and particles would be different forms of the same

thing.



Page 127 of 185

E36 Thomson’s plum pudding model of the atom

Cathode rays

Electric charge and electricity

The photoelectric effect

Nuclear radiation

E37 Rutherford’s model of the atom needed the concept of an atom, and the suggestion that atoms

contained electrons. The plum pudding model itself has little to do with Rutherford’s experiment

or the Solar System model of the atom that followed.

E38 Bohr’s model of the atom needed the concept of an atom, and the knowledge that atoms

contained electrons. Determining the different energies of the electrons need not have involved

the concept of a nucleus. The model of electrons orbiting a positive nucleus provides a structure

for how the atom possesses the electron energy levels that had been identified. Bohr was working

with Rutherford and so knew of Rutherford’s Solar System model.

E39 Einstein’s photoelectrons receive their energy from photons, where a photon is either completely

absorbed as part of an absorption spectrum or leaves the atom unaffected; that is, either the exact

quantity of energy needed to jump a level is contained within the photon or no transfer of energy

will occur.

The electrons in the Franck–Hertz experiment receive their energy from other electrons, which

are able to give up only part of their energy and continue on their way; that is, any electron with

energy above the minimum needed is capable of transferring some of its energy to an electron

within the atom and then being re-emitted with the energy that remains.

E40 a Matter and energy are interchangeable (under the right conditions), and thus light and

matter are able to demonstrate both wave and particle properties at different times.

b Light must contain wave properties to undergo interference and particle properties to

create the photoelectric effect. The Compton effect suggests the particle property of

momentum for photons of light. de Broglie provides a workable theory and equation for

determining the wavelength of particles of matter. Electron microscopes are currently in

use, and use the wave properties of an electron in the same way as a normal microscope

uses the wave properties of visible light.

E41 Science traditionally deals with observable and testable facts. Precision and accuracy are

generally presented as essential. Schools (and the media) tend to teach science as ‘the right

answer’, rather than probable answers with necessary uncertainty.

E42 Quantum mechanics dominates at the micro level; that is, subatomic particles.

Newtonian physics dominates at the macro level; that is, observable to the average person. For

example, it doesn’t matter that an electron is only probably at a given position within an atom as

long as the nail contains enough atoms to provide enough friction to hold the wall in place

E43 We live in the macro world where classical Newtonian physics rules. Matter waves, probabilities

and uncertainties all come into play at such small magnitudes that we don’t notice them. The

field of electronics relies on the theories of quantum mechanics, and is widespread in our lives

during this age of computers.



Page 128 of 185

E44 Einstein’s work was based on thought experiments and relativity. E = mc2 was proposed as a

consequence of travelling near the speed of light, which we do not do. The nucleus of the atom

was not discovered until 6 years after Einstein’s theory, with protons and neutrons yet to be

identified. What was happening to the nuclei of atoms involved in nuclear reactions was difficult

to detect even in 1938.

E45 Even in 1927, the theories proposed (and since confirmed) are very abstract and difficult to

comprehend. More recent theories are more complicated still, generally requiring knowledge of

mathematics far greater than that of the average Year 11 or 12 student.



Page 129 of 185

Chapter 5 Wave properties and light

5.1 Wave properties

1 Both involve the transfer of energy from one place to another without the net movement of

particles.

2 Mechanical waves involve the physical transfer of vibration from one particle to another particle

within the medium. The bonds between solid particles are more stiff than the bonds between gas

particles and so the energy will be transferred more rapidly in a solid. The particles in a solid are

also closer than the particles in a gas and so the interactions between particles occur faster.

3

3

(40)40 vibrations

(0.250)

0.250 s 160 Hz

1 1 6.25 10 s

160

NN f

t

t f

Tf

4

1

3 Hz

1.80 m (3)(0.60)

0.60 m 1.8 m s

cf f

l c f

c

5 Decrease the frequency, as this would allow more time between each wave, therefore increasing

the length of each wave.

6 C

7 C, E

8 A

9 B

10 a Sound energy to kinetic energy of the microphone to electrical energy of the signal.

b Maximum pressure variation occurs at: 0.50 s, 1.50 s, 2.50 s, 3.50 s, 4.50 s and 5.50 s.

5.2 Wave behaviour

1 a

3 3

1 1

4 0.25 1.0 10 s (340)(1.0 10 )

340 m s 3.4 10 m

t v t

v

b 13.4 10 m

c 1340 m sv



Page 130 of 185

2 D. The maximum reflected signal strength occurs for 30.0°, so at 40.0° the amplitude would be

less, but the frequency would remain the same.

3 a

3

3

1 1

(340)2.00 10 Hz

(2.00 10 )

340 m s 1.70 10 m

vf

f

v

b 11.70 10 m

c 1340 m sv

4 C

5 2 1

2

1

880 Hz 331 (0.60 20.0) 3.43 10 m s

(3.43 10 )

(880)

3.90 10 m

f v

v

f

6 2 1

2

1

880 Hz 331 (0.60 30.0) 3.49 10 m s

(3.49 10 )

(880)

3.97 10 m

f v

v

f

7

1

2

22 1 2

cool 2

1

2 1

warm

sin50.0

sin

sin (3.49 10 )(sin 50.0 )3.43 10 m s sin

(3.43 10 )

3.49 10 m s sin 0.7794

vii

r v

v iv r

v

v r

51.2r

8

1

2

22 1 1

cool 2

2

2 1

warm

sin90.0

sin

sin (3.43 10 )(sin 90.0 )3.43 10 m s sin

(3.49 10 )

3.49 10 m s sin 0.98289

vir

r v

v rv i

v

v i

79.4i



Page 131 of 185

9 a

2

2 1 1

(3.40 10 )1000 Hz

(1000)

3.40 10 m s 3.40 10 m

vf

f

v

b Since the aperture is approximately the same as the wavelength appreciable diffraction will

occur, resulting in significant sound energy will arrive at positions P and R

10 a

23

3

2 1 2

(3.40 10 )4.0 10 Hz

(4.0 10 )

3.40 10 m s 8.50 10 m

vf

f

v

b Since the aperture is greater than the wavelength less diffraction will occur, resulting in

less sound energy arriving at positions P and R

c Since less diffraction occurs, more sound energy arrives at point Q

5.3 Wave interactions

1 a True

b False

c True

d False

2 When the glass is exposed to sound of the same frequency as its natural frequency of vibration,

resonance will occur. The amplitude of the vibrations then will increase if sufficient energy is

supplied, the amplification from resonance will cause the glass to shatter.

3 The sound box of a guitar is tuned to resonate in the range of frequencies being produced by the

guitar strings. Resonance within the sounding box amplifies the sound.

4 As a result of the superposition of two waves of equal amplitude and frequency travelling in

opposite directions in the same medium.

5 a C

b B

c C

6 a At the centre.

b One quarter of the way along the string.

c One sixth of the way along the string.

7 a

n

1

1 1

-1 2

1

1 2

1(300)5.00 10 m

2(5.00 10 )

300 m s 3.00 10 Hz

nvn f

L

L f

v f



Page 132 of 185

b

n

1

2 1

1 2

2

2 2

2(300)5.00 10 m

2(5.00 10 )

300 m s 6.00 10 Hz

nvn f

L

L f

v f

c

2

1 n 1

2

3

2

3

3.00 10 Hz

3 (3.00 10 )

9.00 10 Hz

f f n f

f

f

8 Resonance in air columns of a particular length is due to an incoming sound wave with the exact

frequency required to match a resonant frequency of the pipe. This results in reflection of waves

arriving at the ends of the column. The reflected waves are either phase shifted by 180° (open

end) or not phase shifted (closed end). These reflected waves are superimposed on the incoming

waves to produce a standing wave pattern. This amplifies the sound and results in resonance.

9 a

2

1

2 1

1

2 2(45.0 10 )1

1

45.0 10 m 9.00 10 m

Ln

n

L

b

2

1

2 1

1

2 2(45.0 10 )2

2

45.0 10 m 4.50 10 m

Ln

n

L

c

n

1

3 1

1 3

3

3 2

3(330)4.50 10 m

2(4.50 10 )

330 m s 1.10 10 Hz

nvn f

L

L f

v f



Page 133 of 185

10 a

n

1

1 1

1 2

1

1 4

1(330)7.50 10 m

4(7.50 10 )

330 m s 1.10 10 Hz

nvn f

L

L f

v f

b

2

1 n 1

2

3

2

3

1.10 10 Hz

3 (1.10 10 )

3.30 10 Hz

f f n f

f

f

c 2

1 n 1

2

5

2

5

n 1

1.10 10 Hz

5 (1.10 10 )

5.50 10 Hz

f f n f

f

f

f n f

2

7

2

7

7 (1.10 10 )

7.70 10 Hz

f

f

5.4 Electromagnetic radiation

1 a Radio waves – transmitting music and information over large distances

Infrared – remote control devices

Visible light – seeing

Ultraviolet – sterilising surfaces

X-ray – imaging broken bones

b Speed in a vacuum – 3.00 × 108 m s–1.

Electric and magnetic field components.

2 a

814

14

8 1 7

(3.00 10 )5.60 10 Hz

(5.60 10 )

3.00 10 m s 5.36 10 m

cf

f

c



Page 134 of 185

b 14 34 14

34 19

19

19

5.60 10 Hz (6.63 10 )(5.60 10 )

6.63 10 J s 3.71 10 J

(3.71 10 ) 2.32 eV

(1.60 10 )

f E hf

h E

E

3 A–E

4 a i

eV 951 V 951 maxk 0 .E.V

ii

19

k max 0

19

0 k max

19

k max

1.60 10 C

1.95 V ( 1.60 10 )( 1.95)

3.12 10 J

e E eV

V E

E

b

19 21k (max) max2

19k (max)19

k (max) max 31

31 5 1

max

1.60 10 C

2 2(3.12 10 )3.12 10 J

(9.11 10 )

9.11 10 kg 8.28 10 m s

e E mv

EE v

m

m v

5 C

6 a False

b True

c False

d True

7 E

8 B, E

9 a

88 1

0 9

9 14

0

(3.00 10 )3.00 10 m s

(652 10 )

652 10 m 4.60 10 Hz

cc f

f

b 15 15 14

0

14

0

4.14 10 eV s (4.14 10 )(4.60 10 )

4.60 10 Hz 1.90 eV

h W hf

f W



Page 135 of 185

c 15 8

15

ph 9

9

orange ph

(4.14 10 )(3.00 10 )4.14 10 eV s

(620 10 )

620 10 m 2.00 eV

hch E

E

d

ph k (max) ph

k (max)

2.00 eV (2.00) (1.90)

1.90 eV 0.0851 eV

E E E W

W E

Note: be sure to use exact numbers to calculate the difference, not the rounded numbers.

e 19 20

k (max) k (max)

31 21k (max) max2

20k (max)

max

0.0851 eV (0.0851)(1.60 10 ) 1.36 10 J

9.11 10 kg

2 2(1.36 10 )

(9.11

E E

m E mv

Ev

m

31

5 1

max

31 5

max max

10 )

1.73 10 m s

(9.11 10 )(1.73 10 )

v

mv

25 1

max 1.58 10 kg m s

10 a 20

k (max) k (max) 0

2019

0 19

2

0

1.36 10 J

(1.36 10 )1.60 10 C

( 1.60 10 )

8.51 10 V

E E eV

e V

V

b The photoelectrons do not have sufficient kinetic energy to reach the anode.

c 34

k (max)

14

0

34 14 19

0 19

19

6.63 10 J s

5.20 10 Hz

(6.63 10 )(5.20 10 ) (3.07 10 )1.90 eV

( 1.60 10 )

3.07 10 J

h E hf W

f eV hf W

hf WW V

e

W

0

19

0.236 V

1.60 10 C

V

e



Page 136 of 185

5.5 Electromagnetic radiation and matter

1 a 14 34 14 19

3 1 3 1

14 34 14 19

2 1 2 1

34 19

3 2 3 1 2 1

6.00 10 Hz (6.63 10 )(6.00 10 ) 3.98 10 J

4.00 10 Hz (6.63 10 )(4.00 10 ) 2.65 10 J

6.63 10 J s (3.98 10 )

f E hf

f E hf

h E E E

19 19

8 1

34 8

19

3 2

6

(2.65 10 ) 1.33 10 J

3.00 10 m s

(6.63 10 )(3.00 10 )

(1.33 10 )

1.50 10 m

c

hc

E

b D

2 a

19

2 2 1

19

1

34 18

8 1

3.39 eV ( ) (1.60 10 )

13.6 eV ( 3.39) ( 13.6) (1.60 10 )

6.63 10 J s 1.63 10 J

3.00 10 m s

E E E E

E E

h E

c

photon

18

34

15

(1.63 10 )

(6.63 10 )

2.46 10 Hz

E hf E

Ef

h

f

b

19

3 3 1

19

1

34 18

8 1

1.51 eV ( ) (1.60 10 )

13.6 eV ( 1.51) ( 13.6) (1.60 10 )

6.63 10 J s 1.93 10 J

3.00 10 m s

E E E E

E E

h E

c

photon

34 8

18

7

(6.63 10 )(3.00 10 )

(1.93 10 )

1.03 10 m

hcE E

hc

E



Page 137 of 185

c

1

1

0 eV ( )

13.6 eV (0) ( 13.6)

13.6 eV

E E E E

E E

E

3 a

4 4 1

3

1

3

0.88 eV ( )

1.51 eV ( 0.88) ( 13.6)

13.6 eV 12.7 eV not enough energy

(

E E E E

E E

E E

E E E

1)

( 1.51) ( 13.6)

12.1 eV sufficient energy for 1 to 3

E

E n

b No, as photons cannot give off a portion of their energy to the electron, it must transfer all

of its energy.

c It would eject the electron causing the atom to become ionised. The ejected electron would

have (14.0 – 13.6) = 0.40 eV of kinetic energy.

4 a

4 4 1

1

0.88 eV ( )

13.6 eV ( 0.88) ( 13.6)

12.7 eV to 4 level

E E E E

E E

E n

b

4 ph 4 1

3 ph 4 2

2 ph 4 3

1 ph 3 1

0.88 eV ( ) 12.72 eV

1.51 eV ( ) 2.51 eV

3.39 eV ( ) 0.63 eV

13.6 eV ( ) 12.09

E E E E

E E E E

E E E E

E E E E

ph 3 2

ph 2 1

eV

( ) 1.88 eV

( ) 10.21 eV

E E E

E E E

5 Any excess energy above the ionisation energy is retained by the ejected electron in the form of

kinetic energy, which may be any value, as kinetic energy is not quantised.



Page 138 of 185

6

19

3 3 1

19

1

34 19

8 1

3.19 eV ( ) (1.60 10 )

0 eV (3.19) (0) (1.60 10 )

6.63 10 J s 5.10 10 J

3.00 10 m s

E E E E

E E

h E

c

photon

34 8

19

7

(6.63 10 )(3.00 10 )

(5.10 10 )

3.90 10 m

hcE E

hc

E

7

5-1 5-2 5-3 5-4

4-1 4-2 4-3

3-1 3-2

2-1

, , ,

, ,

,

10 different photon energies

E E E E

E E E

E E

E

8 They will be able to emit more frequencies of light than they can absorb, if the sodium atoms are

absorbing light in a low-energy state. Emitting light will result from transitions from higher

energy levels to any lower energy level including the ground state; however, absorption can only

occur from the ground state to a higher level. This is assuming that the valence electrons in the

sodium atom are in the ground state to start with and are not in excited states due to heating.

9

19

7 7 1

19

2

34 19

8 1

Assume 0.1 eV ( ) (1.60 10 )

3.4 eV ( 0.1) ( 3.4) (1.60 10 )

6.63 10 J s 5.28 10 J

3.00 10 m s

E E E E

E E

h E

c

photon

34 8

19

7

(6.63 10 )(3.00 10 )

(5.28 10 )

3.77 10 m

hcE E

hc

E

377 nm which is outside the visible range (UV)

10 De Broglie proposed a model of the atom in which the electrons were viewed as matter waves

with resonant wavelengths that determined the circumference of the energy levels. This is similar

to the resonant wavelengths that fit the length of a violin string.



Page 139 of 185

5.6 X-rays

1 a

3 3

k (max)

3 19

k (max)

14

k (max)

70.0 10 V 70.0 10 eV

(70.0 10 ) (1.60 10 )

1.12 10 J

V E

E

E

b

31 21e k (max) max2

14k (max)14

k (max) max 31

8 1

max

9.11 10 kg

2 2(1.12 10 )1.12 10 J

(9.11 10 )

1.57 10 m s

m E mv

EE v

m

v

c

ph (max) k (max)

3 14

ph (max) 70.0 10 eV or 1.12 10 J

E E

E

2 a

ph (max) k (max)

3 14

ph (max) 150 10 eV or 2.40 10 J

E E

E

b 34

ph (max) max

14ph (max)14

ph (max) max 34

19

max

6.63 10 J s

(2.40 10 )2.40 10 J

(6.63 10 )

3.62 10 Hz

h E hf

EE f

h

f

c

819

max min 19

max

8 1 12

min

(3.00 10 )3.62 10 Hz

(3.62 10 )

3.00 10 m s 8.29 10 m

cf

f

c



Page 140 of 185

3 a 34

ph (max) max

15ph (max)15

ph (max) max 34

8 1 18

max

6.63 10 J s

(5.60 10 )5.60 10 J

(6.63 10 )

3.00 10 m s 8.45 10 Hz

h E hf

EE f

h

c f

8

min 18

max

11

min

(3.00 10 )

(8.45 10 )

3.55 10 m

c

f

b This is the wavelength of the most energetic photons that can be produced by the most

energetic bombarding electrons. A smaller wavelength would have to come from a more

energetic bombarding electron, which cannot be produced by this accelerating potential

difference.

4 Hz 10458 18max .f

5 The two peaks in the spectrum are the line spectrum caused by the bombarding electrons ejecting

inner orbital electrons from the molybdenum atoms. When these inner electrons are ejected they

are replaced by outer electrons in transitions that produce high-energy x-ray photons.

6 a 34 8

11

peak 11

8 1 15

peak

34

peak

(6.63 10 )(3.00 10 )The peak is around 3.1 10 m

(3.1 10 )

3.00 10 m s 6.42 10 J

6.63 10 J s 4.01

hcE

c E

h E

410 eV

b The peak in the spectrum is due the bombarding electrons ejecting inner orbital electrons

from the barium atoms. When these inner electrons are ejected they are replaced by outer

electrons in transitions that produce characteristic high energy x-ray photons of a particular

wavelength due to the specific energy transition that occurs in barium, but not in any other

atom.

c This is the wavelength of the most energetic photons that can be produced by the most

energetic bombarding electrons. A smaller wavelength would have to come from a more

energetic bombarding electron.

d 11

34 834

max 11

8 1 15

max

The smallest wavelength is around 2.5 10 m.

(6.63 10 )(3.00 10 )6.63 10 J s

(2.5 10 )

3.00 10 m s 7.96 10 J

hch E

c E

4

max 4.97 10 eVE



Page 141 of 185

7 a

b

8 Tungsten is used because it has a very high melting point temperature. A great deal of heat is

produced in the anode of an X-ray machine. Copper is used because it is a very good conductor

of electricity and has a very high conductivity of heat. It conducts excess electrons away from the

surface and it conducts excess heat away from the target site.

9 B

10 X-rays from an X-ray tube are produced in a single burst, while those generated by a synchrotron

can be generated continuously for hours. X-rays from an X-ray tube can pass through lighter

atoms, while those produced in a synchrotron are more likely to interact with these atoms.

Synchrotron X-rays are 100 million times brighter than X-rays from traditional sources.

Chapter 5 Review

1 4.0 m

2 D

3 A

4 C

5 C, D

6

1

2

(1)(346)Assume 346 m s

4 4(0.85)

0.85 m 1.02 10 Hz

nvv f

L

L f

Photon frequency (Hz)

Inte

nsi

ty

1.20 × 1019

Photon energy (J)

Inte

nsi

ty

7.96 × 10-15



Page 142 of 185

7

1

2

(3)(346)Assume 346 m s

4 4(0.85)

0.85 m 3.05 10 Hz

nvv f

L

L f

8 a i

34 89

9

34 19

8 1

(6.63 10 )(3.00 10 )580 10 m

(580 10 )

6.63 10 J s 3.43 10 J

3.00 10 m s

hcE

h E

c

ii

19

19

(3.43 10 )2.14 eV

(1.60 10 )E

b

349

9

34 27 1

(6.63 10 )580 10 m

(580 10 )

6.63 10 J s 1.14 10 kg m s

h

h

c

19 totalph 19

ph

21

ph

(500)3.43 10 J

(3.43 10 )

500 W 1.46 10 photons

EE N

E

P N

9 a

34 89

9

34 19

8 1

(6.63 10 )(3.00 10 )432 10 m

(432 10 )

6.63 10 J s 4.60 10 J

3.00 10 m s

hcE

h E

c

b

349

9

34 27 1

(6.63 10 )432 10 m

(432 10 )

6.63 10 J s 1.53 10 kg m s

hρ

h ρ

c



Page 143 of 185

19 totalph 19

ph

20

ph

(230)4.60 10 J

(4.60 10 )

230 W 5.00 10 photons

EE N

E

P N



Page 144 of 185

d

20 19ph ph19

ph ele

20 3

ph ele

0.001 (5.00 10 )(4.60 10 )(0.001)4.60 10 J

100 100

5.00 10 2.30 10 W

%eff 0.001%

N EE P

N P

10 A

11 a The wave model predicts that light of any frequency will emit photoelectrons from a

metallic surface if given sufficient time.

b The wave model predicts that the energy delivered to the electrons by a light beam of

constant intensity will be proportional to time. This suggests that a low intensity light

beam should take longer to eject photoelectrons from a metallic surface.

c According to the wave model of light, a higher intensity beam will deliver more energy to

the electrons and will therefore produce photoelectrons with higher kinetic energy than a

lower intensity beam.

12

a

20 2034

14 14

(7.0 10 2.0 10 )slope 6.7 10 J s

(6.06 10 5.31 10 )

y

x

b The slope of the line is very close to Planck’s constant.

c When the kinetic energy of the ejected electrons is zero the frequency of the incoming

photons is 5.0 × 1014 Hz, this is the threshold frequency for rubidium.



Page 145 of 185

d

89

9

8 1 14

(3.00 10 )680 10 m

(680 10 )

3.00 10 m s 4.41 10 Hz

cf

c f

This is below the threshold frequency so it will not eject photoelectrons from rubidium.

13 a

34 34 14

0

1914

0 19

19

6.63 10 J s (6.63 10 )(5.0 10 )

(3.32 10 )5.0 10 Hz

(1.60 10 )

2.07 eV or 3.32 10 J

h W hf

f W

W

b 34 34 14

ph

14 19

ph

19 19

k ph

6.63 10 J s (6.63 10 )(5.60 10 )

5.60 10 Hz 3.71 10 J

(3.71 10 ) (3.32 10 )

h E hf

f E

E E W

20

k 3.98 10 JE

c

20 21k (max) k (max) max2

20k (max)31

max 31

5 1

max

3.98 10 J

2 2(3.98 10 )9.11 10 kg

(9.11 10 )

2.96 10 m s

E E mv

Em v

m

v

31 5

max max

25 1

max

(9.11 10 )(2.96 10 )

2.69 10 kg m s

mv

d 20

k (max) k (max) 0

2019

0 19

1

0

3.98 10 J

(3.98 10 )1.60 10 C

( 1.60 10 )

2.49 10 V

E E eV

e V

V

14 C



Page 146 of 185

15

1919

3 11 3 1 34

15

2 3 1

1919

3 23 3 2 34

( 3.7) ( 10.4) (1.60 10 )(1.60 10 )10.4 eV

(6.63 10 )

5.5 eV 1.62 10 Hz

( 3.7) ( 5.5) (1.60 10 )(1.60 10 )3.7 eV

(6.63 10 )

En f

h

n f

En f

h

14

3 2

1919

2 12 1 34

15

2 1

4.34 10 Hz

( 5.5) ( 10.4) (1.60 10 )(1.60 10 )

(6.63 10 )

1.18 10 Hz

f

Ef

h

f

16 There is no shortest wavelength, as this corresponds to the highest energy photon. Any photon

over the ionisation level will result in ionisation and any remaining energy will be kept by the

electron in the form of kinetic energy.

17 When the excited electron returns to the ground state it emits a photon of the same energy

(10.2 eV) as it absorbed, however, the direction in which the photon is emitted is random. Some

photons may travel in the same direction as the incoming photons, but when resolved in to the

absorption spectrum, these few photons would be far less intense than the surrounding photons.

The effect would be a black line at the frequency of light corresponding to the 10.2 eV photon.

18 a 34 8

8 1

ph 6

34 20

ph

6 20

ph 20

(6.63 10 )(3.00 10 )3.00 10 m s

(3.00 10 )

6.63 10 J s 6.63 10 J

(60.0)3.00 10 m 9.05 10 photons

(6.63 10 )

60.0 W

hcc E

h E

N

P

b Infrared

19 a 14 34 14

ph

34 19

ph

21

ph 19

5.20 10 Hz (6.63 10 )(5.20 10 )

6.63 10 J s 3.45 10 J

(1000.0)1000 W 2.90 10 photons

(3.45 10 )

f E hf

h E

P N

b W1000P

20 B

21 C

22 A



Page 147 of 185

23

19

ph k max

34 834 19

0 9

9 19

1.60 10 C

(6.63 10 )(3.00 10 )6.63 10 J s ( 1.60 10 )( 1.21)

(200 10 )

200 10 m 8.019 10 J o

e W E E

hch W eV

W

8 1

0

r 5.01 eV

3.00 10 m s

1.21 V

c

V

24 Only certain frequencies of light will cause photoelectrons to be emitted from a surface.

There is no time delay between the absorption of photons of different intensities and the

emission of a photoelectron.

The maximum kinetic energy of the ejected photoelectrons is the same for different light

intensities of the same frequency.

25 a

19

4 4 1

19

1

34 18

8 1

1.6 eV ( ) (1.60 10 )

10.4 eV ( 1.6) ( 10.4) (1.60 10 )

6.63 10 J s 1.41 10 J

3.00 10 m s

E E E E

E E

h E

c

1815

34

87

15

(1.41 10 ) 2.12 10 Hz

(6.63 10 )

(3.00 10 ) 1.41 10 m

(2.12 10 )

Ef

h

c

f

b

19

2 2 1

19

1

34 19

8 1

5.5 eV ( ) (1.60 10 )

10.4 eV ( 5.5) ( 10.4) (1.60 10 )

6.63 10 J s 7.84 10 J

3.00 10 m s

E E E E

E E

h E

c

1915

34

87

15

(7.84 10 ) 1.18 10 Hz

(6.63 10 )

(3.00 10 ) 2.54 10 m

(1.18 10 )

Ef

h

c

f



Page 148 of 185



Page 149 of 185

c

19

4 4 3

19

3

34 19

8 1

5.5 eV ( ) (1.60 10 )

10.4 eV ( 1.6) ( 3.7) (1.60 10 )

6.63 10 J s 3.36 10 J

3.00 10 m s

E E E E

E E

h E

c

1914

34

87

14

(3.36 10 ) 5.07 10 Hz

(6.63 10 )

(3.00 10 ) 5.92 10 m

(5.07 10 )

Ef

h

c

f

26 a 17

k65.0 V 65.0 eV or 1.04 10 JV E

b

31 21e k (max) max2

17k (max)17

k (max) max 31

6 1

max

9.11 10 kg

2 2(1.04 10 )1.04 10 J

(9.11 10 )

4.78 10 m s

m E mv

EE v

m

v

c 34

31

e 31 6

max

34 10

6 1

max

(6.63 10 )9.11 10 kg

(9.11 10 )(4.78 10 )

6.63 10 J s 1.52 10 m

4.78 10 m s

hm

mv

h

v

27 a

19

4 4 1

19

1

18

1.60 eV ( ) (1.60 10 )

10.4 eV ( 1.60) ( 10.4) (1.60 10 )

1.41 10 J

E E E E

E E

E

b 34 8

18

ph 18

ph

8 1 7

34

(6.63 10 )(3.00 10 )1.41 10 J

(1.41 10 )

3.00 10 m s 1.41 10 m

6.63 10 J s

hcE

E

c

h

c J 10661or eV 410 18min

..E

28 a 19

beam e

18

e

7.00 eV (7.00) (1.60 10 )

1.12 10 J

E E

E



Page 150 of 185

b

19

2 2 1

19

1

18

3

5.50 eV ( ) (1.60 10 )

10.4 eV ( 5.50) ( 10.4) (1.60 10 )

0.78 10 J can reach this level

3.70

E E E E

E E

E

E

19

3 1

19

1

18

4

eV ( ) (1.60 10 )

10.4 eV ( 3.70) ( 10.4) (1.60 10 )

1.07 10 J can reach this level

1.60 eV

E E E

E E

E

E

19

4 1

19

1

18

( ) (1.60 10 )

10.4 eV ( 1.60) ( 10.4) (1.60 10 )

1.41 10 J this is too high

E E E

E E

E

Therefore the mercury atom could be excited to the n = 3 level.

c

19

2 2 1

19

1

18

3

5.50 eV ( ) (1.60 10 )

10.4 eV ( 5.50) ( 10.4) (1.60 10 )

0.78 10 J or 4.90 eV

3.70 eV

E E E E

E E

E

E

19

3 1

19

1

18

3

( ) (1.60 10 )

10.4 eV ( 3.70) ( 10.4) (1.60 10 )

1.07 10 J or 6.70 eV

3.70 eV

E E E

E E

E

E

19

3 2

19

2

19

( ) (1.60 10 )

5.50 eV ( 3.70) ( 5.50) (1.60 10 )

2.88 10 J or 1.80 eV

E E E

E E

E

d 34 8

18

ph 18

ph

8 1 7

34

(6.63 10 )(3.00 10 )1.07 10 J

(1.07 10 )

3.00 10 m s 1.86 10 m

6.63 10 J s

hcE

E

c

h



Page 151 of 185

29 34 8

e 19 19

ph 1-

1-

8 1 8

34

(6.63 10 )(3.00 10 )30.4 eV

(30.4)(1.60 10 ) (10.4)(1.60 10 )

10.4 eV

3.00 10 m s 3.05 10 m

6.63 10 J s

hcE

E E

E

c

h

30 a 19

beam e

19

e

4.00 eV (4.00) (1.60 10 )

6.40 10 J

E E

E

This is less than the energy required to promote an electron to the n = 2 level so with no

electrons being promoted no photons can be emitted.

b This is sufficient energy to ionise the mercury atoms. The liberated electrons are then free

to conduct current along the tube.

c

e

ph

34 819

ph 19

ph

8 1 7

6.20 eV is sufficient to promote one of mercury's electron to 2.

4.90 eV

(6.63 10 )(3.00 10 )7.84 10 J

(7.84 10 )

3.00 10 m s 2.54 10 m

E n

E

hcE

E

c

h

346.63 10 J s



Page 152 of 185

Chapter 6 Matter, relativity and astronomy

6.1 Extending our model of matter

1 Strong nuclear 1038 : electromagnetic 1036 : weak nuclear 1032 : gravitational 1

2 a 7particle size 5.00 10 m

b 8

8 1

19

19 11

(3.00 10 )3.00 10 m s

(3.00 10 )

3.00 10 Hz 1.00 10 m

cc

f

f

c

8 1

k

34 834

27 8 212

1 15

p

3.00 10 m s

(6.63 10 )(3.00 10 )6.63 10 J s

(1.6726 10 )(0.650 3.00 10 )

0.650 m s 6.265 10 m

1.6726 1

hcc

E

f

v c

m

270 kg

3 a electrostatic repulsion between a proton and proton (not within a nucleus).

photon

p+

p+ p+

p+

Time

Position



Page 153 of 185

b

c

4 a

27 2

p

27 27 27 8 2

p

10

1.6726 10 kg

1.6726 10 kg (1.6726 10 1.6726 10 )(3.00 10 )

3.0107 10 J

m E mc

m E

E

b 27 2

n

27 27 27 8 2

n

10

1.6749 10 kg

1.6749 10 kg (1.6749 10 1.6749 10 )(3.00 10 )

3.0148 10 J

m E mc

m E

E

Time

Position

meson

n

p+ n

p+

Time

Position

pion

p+

p+ n

n



Page 154 of 185

5 a 31 2

k ph pair phe

31 6 19 31 8 2

k

6 14

ph k

9.11 10 kg

9.11 10 kg (1.20 10 )(1 60 10 ) (2 9.11 10 )(3.00 10 )

1.20 10 eV 2.80 10 J

e

m E E E E mc

m E .

E E

14

k 1.40 10 J each particleE

b 27 2

k ph pairp

27 23 34 27 8 2

kp

23 16

k

34

1.6726 10 kg

1.6726 10 kg (4.541 10 )(6.63 10 ) (2 1.6726 10 )(3 00 10 )

4.541 10 Hz 3.00 10 J

6.63 10 J s

m E E E hf mc

m E .

f E

h

16


c

27 2

k ph pairn

34 827 27 8 2

n k 16

16 11

k

34

1.6749 10 kg

(6.63 10 )(3.00 10 )1.6749 10 kg (2 1.6749 10 )(3 00 10 )

(6.059 10 )

6.059 10 m 2.68 10 J

6.63 10 J s

hcm E E E mc

m E .

E

h

11


6 a down, down, up

b anti-up, anti-up, anti-down

c up, down, strange

d charm, anti-down

7 a proton

b rho-minus

c kaon-minus

d sigma-minus

6.2 Einstein’s special theory of relativity

1 Galileo realised that a force changes motion, not caused it.

2 a 6

6 2 1

equ

1

pole

2 2 (6.37 10 )6.37 10 m 4.63 10 m s

(24 60 60)

0 m s

rr v

T

v



Page 155 of 185

b 11

11 4 2 1

equ

4 1

pole

2 2 (1.50 10 )1.50 10 m 2.99 10 4.63 10 m s

(365.25 24 60 60)

2.99 10 m s

rr v

T

v

c The person at the equator has acceleration towards the centre of the Earth, both people also

have an acceleration towards the Sun.

3 11

11 8 1

11

8 8

8

2 2(1.50 10 )1.50 10 m 2.27 10 m s

(22 60)

1.50 10 m

3 00 10 ) (2.27 10 ) %diff 100

3 00 10 )

%

rr v

t

t

( .

( .

diff 24.2%

4 The GPS picks up its signal from a satellite that is stationary in the Earth’s frame of reference.

5 a

snd-you snd-gnd gnd-you

1 1

snd-gnd snd-you

1

gnd-you

( 346) ( 30.0)

346 m s 376 m s

30.0 m s

v v v

v v

v

b


1 - 1

snd-gnd snd-you

1

gnd-you

( 346) ( 40.0)

346 m s 306 m s

40.0 m s

v v v

v v

v

c


1 1

snd-gnd snd-you

1

gnd-you

( 346) (0)

346 m s 346 m s

0 m s

v v v

v v

v

d


1 1

snd-gnd snd-you

1

gnd-you

( 346) ( 30)

346 m s 376 m s

30 m s

v v v

v v

v

6 A, D



Page 156 of 185

7 a If there was an apparent force on you towards one of the walls of the room, or is a small

pendulum maintains an angle to the vertical.

b The motion of the merry-go-round is not constant as it is changing direction constantly, it

is accelerating towards the centre of the circular path.

8 a

ball-Earth ball-train train-Earth

ball-train ball-Earth

ball-Earth

(5.00)5.00 m ( 10.0)

(0.200)

0.200 s ( 25.0)t

v v v

s v

v

1 1

train-Earth ball-Earth

( 10.0)

10.0 m s 15.0 m s

v v

b

ball-Earth ball-Earth

1

ball-Earth ball-Earth

ball-Earth

15.0 m s ( 15.0)(0.200)

0.200 s 3.00 m

t

t

s v

v s

s

c s 2000.t

9 a 150.0 m sc

b 150.0 m sc

10 A

6.3 To the stars

1 3600 arc sec

2 1800 arc sec

3

11800

23

11800

5 radiustan( )

distance

(1 10 )distance 1 10 m

tan( )

c

4 Hipparchus called the brightest stars ‘first magnitude’ (+1) and the dimmest ‘sixth magnitude’

(+6), brighter stars discovered after Hipparchus had to then have negative values relative to

Hipparchus’s brightest.

5 The physics of heat transfer from hot bodies was well known, so accurate predictions could be

made.

6 Nuclear forces are 104 times greater than electromagnetic forces, therefore there are 108 times

more energy produced in nuclear reactions than chemical reactions.

distance

5 c radius

0.5 arc sec



Page 157 of 185

7 Hydrostatic equilibrium: in which the inward pressure from the weight of the matter is balanced

by the outwards pressure of the radiation released by the nuclear reactions in the core.

Assumptions about the mass, temperature, heat content, ability to transfer heat, and the pressure

gradient and more.

8 Fusion occurs in the inner 0.25R of the Sun, energy flows from this region by radiative diffusion

and then convection. EM radiation ‘bounces’ from particle to particle transferring heat energy in

the process out to about 0.7R. Then hot gases rise from this region to the upper regions by

convection. The hot gases on the surface radiate their energy out into space by EM radiation,

including in a direction towards the Earth.

9 C

10 A

6.4 Fundamentals of astronomy

1 C

2 In Brisbane the SCP would be 27° above the horizon in the south, while in Perth the SCP is 32°

above the horizon in the south.

3 a C

b D

4 a Pollux

b Formalhaut

5 Stars appear to move 1° further from their position at the same time in each subsequent night.

Orion will be 7° further west of due north after one week.

6 Aquarius rose first, about 7 hours before Orion.

7 C

8 a It won’t be able to be seen as it will be in line with the Sun and Earth.

b It is in a position known as ‘opposition’ only the superior planets can be seen in this

position.

9 a

(0.387)Mercury: 100 100 102.1%

(0.379)

(1.000)Earth: 100 100 100.1%

(0.999)

(1.524)Mars: 100 100 100.5%

(1.517)

l

w

l

w

l

w



Page 158 of 185

b

2 3

2 4

2 4

2.1 2.1Mercury: Δ (10 10 ) 2.1 10 m

100 100

0.1 0.1Earth: Δ (10 10 ) 1 10 m

100 100

0.5 0.5Mars: Δ (10 10 ) 5 10 m

100 100

l w

l w

l w

10 a 3 3

E AE 2 2

E A

3 2 3 2

A EE A 3 3

E

A A

1.00 AU

(2.00) (1.00)1.00 year

(1.00)

2.00 AU 2.83 Earth years

R RR

T T

R TT T

R

R T

b 3 3

E AE 2 2

E A

3 2 3 2

E A33

E A 2 3

E

A A

1.00 AU

(1.00) (8.00)1.00 year

(1.00)

8.00 year 4.00 AU

R RR

T T

R TT R

T

T R

6.5 Hubble’s universe

1 About 15 times the Milky Way galaxy away.

It would have an angular diameter of about 5°.

It is too far away so it is very faint.

2 The faint Cepheid is about twice the distance away than the brighter Cepheid.

3 a 3 2 3 2 3

3 12 3

25 10 pc (25 10 ) (1 10 )

1 10 pc 1.96 10 pc

r V A h r h

h V

b 12

3

9

1.96 1019.6 pc per star

100 10V



Page 159 of 185

c 3 34

3

3 3

19.6 pc per star

3V 3(19.6)

4 4

1.67 pc

from one star to another is 2 r

distance 2(1.67) 3.3

V V r

r

r

5 pc

d The nearest stars to the Sun are about 1 pc away. The figure calculated above is an average

distance as the inner stars are closer while the outer stars are sparser.

4

8

galaxy

7 1

galaxy

(0.10)(3.00 10 )

3.00 10 m s

v c

v

5 1 1

0 galaxy 0

3 1

galaxy

6 1

galaxy

70 km s Mpc (70)(90)

90 Mpc 6.3 10 km s

6.3 10 m s

H v H d

d v

v

6

galaxy 8

(6.3 10 ) 100 2.1% of speed of light

(3.00 10 )v

6 By looking at distant objects astronomers are looking far back in time to the early universe. We

see distant objects as they were earlier in their life-cycle.

7 Steady State Theory: The Universe is infinite and eternal

Big Bang: The Universe is expanding.

Essentially the Big Bang theory is different to the Steady State theory in that the Steady State

theory requires that matter is being continuously created.

Experiments on the large scales of distance and time that characterise the universe are extremely

difficult to create.

8 There is no way to tell.

9 The wavelength of the radiation has increased due to the expansion of space.

10 This value leads to an age of the Universe of about 2 billion years. This is less than the age of the

Earth.

Chapter 6 Review

1 Strong nuclear force – gluon

Electromagnetic force – photon

Weak nuclear force - W+, W- and Z0

Gravitational force – graviton



Page 160 of 185

2 a m 10021 10 .

b

8 1 21k 2

34

34 827

p 10 27

10

3.00 10 m s

26.63 10 J s

2(6.63 10 )(3.00 10 )1.6726 10 kg

(1.02 10 )(1.6726 10 )

1.02 10 m

hcc E mv

hch v

m

m v

6 1.53 10 mv

3 31 2 21

total k pair 2e

31 31 8 2 31 8 21total 2e

8 1 14

total

9.11 10 kg

9.11 10 kg (2 9.11 10 )(1.50 10 ) (2 9.11 10 )(3.00 10 )

3.00 10 m s (2.05 10 ) 1.

m E E E mv mc

m E

c E (

13

8 1 13

total

64 10 )

1.50 10 m s 1.84 10 Jv E

4 Bosons - mediate the strong nuclear, electromagnetic and weak nuclear forces.

Leptons – interact via the weak nuclear force, and if charged will interact via photons, each

lepton has a corresponding neutrino.

Mesons – have a baryon number of zero, are made up of a quark and an anti-quark pair.

Baryons – have a baryon number of one, or negative one, if it is the antiparticle. Is made up of

three quarks.

5 Proton – up, up, down.

Neutron – up, down, down.

Anti-proton – anti-up, anti-up, anti-down.

Anti-neutron – anti-up, anti-down, anti-down.

6 A, C

7 C

8 A, B, C

9 A, C

10 B

11 B

12 A

13 When stopped at the station (i) and when travelling at constant speed (iii) there will be no

difference, when accelerating (ii) you would feel a force acting on you from the train pushing you

in the direction that the train is accelerating.

14 In Perth the celestial equator is 58° above the northern horizon.

In Albany it would be less than this and in Kununurra it would be greater than this.

15 a Arcturus

b Betelgeuse

c Alpha Centauri

d Aldebaran



Page 161 of 185

16 a RA 6 h 43 min, dec. –17°

b RA 1 h 36 min, dec. –58°

c RA 18 h 35 min, dec. +39°

d RA 5 h 12 min, dec. –8°

17 6

1 1 2

1

1 1 1 2 2

66 1 1

2 2 6

2

152.1 10 km Area Area

29.3 km s

(152 1 10 )(29.3)147.1 10 km

(147.1 10 )

r

v rv r v

rv .r v

r

1

2 30.3 km sv

18 3 3 3 6 34

63 E EE 3 3 3 34

E E E E3

(109 ) (1.30 10 )109 1.30 10

R R R RR R

R R R R

19 30 30 6 37

coal

26

3711coal

26

2 10 kg (2 10 )(30 10 ) 6 10 J

4 10 W

(6 10 ) Δ 1.5 10 s

(4 10 )

Δ

m E

P

Et

P

t

34.8 10 years

20 30 30 14 45

H

26

4518coal

26

2 10 kg (2 10 )(6 10 ) 1.2 10 J

4 10 W

(1.2 10 ) Δ 3 10 s

(4 10 )

Δ 9

m E

P

Et

P

t

5 billion years

21 The spectra from these stars show lines similar to the spectra from our Sun and known elements.

22 The period of the Cepheid variables is related to luminosity, so distance can be found, they are

also very bright.



Page 162 of 185

23 a 3 2 3 2 3

3 11 3

113

9

343

15 10 pc (15 10 ) (1 10 )

1 10 pc 7.07 10 pc

(7.07 10 ) 7.07 pc per star

(100 10 )

r V A h r h

h V

V

V r

3 33V 3(7.07)

4 4

1.19 pc

From one star to another is 2 distance 2(1 19 2.4 pc

r

r

r . )

The stars closer to the centre will be closer together, while those stars further out from the

centre will be spaced further apart.

b The distance from our Sun to its nearest neighbour is about 1 pc, this is less than half of the

average distance between stars in the Milky Way.

24 Temperature

Elements and their state

Pressure

Magnetic field

25

1

sound car sound

1 1

car

346 m s (0.10)(346)

34.6 m s or 124 km h towards us

v v v

v

26 1 1 3

0 LMC 0

3 1

LMC LMC

3

SMC

70 km s Mpc (70)(50 10 )

50 kpc 50 10 Mpc 3.5 km s

63 kpc 63 10 Mpc

H v H d

d v

d

3

SMC 0

1

SMC

(70)(63 10 )

4.4 km s

v H d

v

These are very small velocities on a galactic scale.



Page 163 of 185

27 Blueshifts imply that the galaxies are moving towards us, they must be moving towards us at a

rate that is greater than rate at which the Universe is expanding.

28 The age of the Universe can also be found using the position on the H-R diagram and nuclear

decay processes. All of these methods estimate an age of about 14 billion years.



Page 164 of 185

Chapter 7 Electric and magnetic fields

7.1 Force on charges in magnetic fields

1 C

2 A

3 a South

b C

4 A

5 a North

b A

6 A particle with no charge, like a neutron, as it will not experience any force due to moving in the

magnetic field, therefore its path will not change.

7 C, D

8 A and B as the particle must have the opposite charge, or have the field in the opposite direction,

to experience a force in the opposite direction.

9 a South

b i 2F

ii 2F

iii 4F

10 a 2F north

b Greater radius

7.2 Particle accelerators

1 B

2 a Electrons leave the hot cathode of the evacuated tube and accelerate towards a positively

charged anode. The electrons can be deflected as they pass through an electric field

produced by a pair of oppositely charged parallel plates and a magnetic field generated by

an electromagnet. They can be detected as they hit a fluorescent screen at the rear of the

tube.

b The electrons are accelerated away from the negative cathode and towards a positively

charged anode.



Page 165 of 185

3 The standing-wave linear accelerator consists of a large number of drift tubes, each separated by

a gap. Electrons enter the cylinder and are accelerated towards the first drift tube by an electric

field. An alternating potential difference is applied to each tube and is timed such that the

electrons are accelerated across each gap between the drift tubes. Inside the drift tubes, they

travel at a constant velocity as they are shielded from the effects of the electric field. The

particles pick up more energy each time they leave the drift tubes, until they are accelerated out

of the linac.

4 A circular accelerator, such as a cyclotron, can be used to accelerate particles within a more

compact space than the equivalent operations of a very long linear accelerator.

5 a 3

k d

19 212

19 331

31

10 10 V

1.60 10 C

2 2(1.60 10 )(10 10 )9.11 10 kg

(9.11 10 )

V E W

e mv e V

e Vm v

m

v

7 110 m s

b

31 719

19

31 4

7 1

1 50 T

(9.11 10 )( 10 )1.60 10 C

(1.60 10 )(1 50)

9.11 10 kg 10 m

10 m s

mv. r

e

e r.

m r

v

BB

6 a

b The radius of the electron’s path is dependent on its velocity and the magnitude of the

magnetic field that is acting.

7 a

2

2

4 1

500 V

(500)3.5 10 m

(3.5 10 )

10 V m

VV E

d

d E

E

–



Page 166 of 185

b 4 1

E B

3

4

3

1.43 10 V m

1.50 10 T

(1.43 10 )

(1.50 10 )

E

qE qv

Ev

F F

B B

B

6 1 9.52 10 m sv

8 3

k d

19 212

19 331

31

2 50 10 V

1.60 10 C

2 2(1.60 10 )(2 50 10 )9.11 10 kg

(9.11 10 )

V . E W

e mv e V

e V .m v

m

v

7 110 m s

9 2

c

219

31 631

19 2

6 1

4 60 10 m

1.60 10 C

(9.11 10 )( 10 )9.11 10 kg

(1.60 10 )(4 60 10 )

10 m s

r . F F

mve qv

r

m .

v

B

B

B

B 410 T

10 a 19 19 6 3

3 15

6 1

1.60 10 C (1.60 10 )( 10 )(8.60 10 )

8.60 10 T 9.63 10 N

10 m s

e F qv

F

v

B

B

B

B

b 2

15

c

2 31 6 231

15

c

6 1 3

9.63 10 N

9.11 10 )( 10 )9.11 10 kg

(9.63 10 )

10 m s 4.63 10 m

mvF F F

r

mv (m r

F

v r

B B



Page 167 of 185

7.3 Synchrotons

1 a i The linac consists of an electron gun, a vacuum system, focusing elements and RF

(radio-frequency) cavities. Electrons escape from the electron gun as they boil off

the heated filament of the assembly and then accelerate across the 100 keV potential

difference. The electron beam travels through the ultra-high vacuum within the

linac, to prevent energy loss through the interaction with air particles. As the

electrons travel, focusing elements act on the beam to ensure it doesn’t collide with

the walls of the vacuum tube. RF (radio-frequency) cavities throughout the linac

produce intense electromagnetic radiation of several hundred megahertz

perpendicular to the electron beam. The RF radiation propagates through the linac as

a travelling wave. Electrons are timed in pulses so that they travel through the linac

in bunches, which are accelerated by the RF radiation. As a result, electrons are

accelerated to close to the speed of light throughout their journey through the linac.

ii Each time the charged particles travel around the circular booster ring they receive

an additional energy burst from a radio-frequency (RF) chamber.

iii In the storage ring of the synchrotron, electrons revolve around for hours at a time at

speeds close to the speed of light. A series of magnets make them bend in an arc as

they travel through the ring. It is when the electrons change direction that they emit

synchrotron radiation.

iv The beamline is the path taken by synchrotron light as it exits the storage ring

towards an experimental station (or endstation).

b i The strength of the magnetic field used in the circular booster ring is periodically

increased as the velocity of the electrons increases to account for energy losses due

to the increasing effects of relativity which result from the increased velocity and

relativistic mass by this stage.

ii Electrons move from a heated filament inside the electron gun within the linac.

iii RF cavities accelerate electrons within the linac.

iv Insertion devices are located in the straight section of the storage ring.

2 a The precise configuration of the bending, focussing and steering magnets found in the

storage ring.

b The specification of the lattice sets the parameters for the synchrotron light produced.

3 a Due to the presence of an oscillating electromagnetic field produced by the transformers

extra energy is given to the electrons.

b The particles would gradually lose their energy through collisions with other atoms and the

production of synchrotron light. Their orbital speed would be slowed and they would cease

to produce synchrotron light.

c By replenishing the energy lost via a burst of energy from the RF cavity, the charged

particles continue to move at the same speed in a path of constant radius in the storage

ring. The particles of a cyclotron increase their energy with each revolution and so their

radius of orbit increases each revolution.

4 To minimise energy losses through collisions between electrons and gas molecules.

5 B



Page 168 of 185

6 An undulator consists of some hundred low-power magnetic poles aligned closely together in

rows. The effect of the undulator is to deflect the electrons more gently, thus producing a much

narrower beam of brighter synchrotron radiation that is enhanced at specific wavelengths. This is

in contrast to a bending magnet which produces a continuous broad cone of much less bright

synchrotron light. The output from a wiggler is radiation that is not as bright as from the

undulator, but brighter than that from the bending magnet. The wiggler forms a broad band of

incoherent synchrotron light.

7 a 3 GeV

b Pharmaceutical development, mining and mineral exploration, manufacturing

microstructures etc.

8 a 3

k d

19 212

19 331

31

120 10 V

1.60 10 C

2 2(1.60 10 )(120 10 )9.11 10 kg

(9.11 10 )

V E W

e mv e V

e Vm v

m

v

8 110 m s

b For electrons being accelerated across a potential difference of 2.5 keV, the effects of

relativity will come into play. The effective mass of the electrons will be greater than their

rest mass. As a result, they will not reach the velocity calculated in part a (for such a

potential difference the electrons will reach about 80% of the calculated velocity).

9 a

31 819

19

31 3

1 50 T

(9.11 10 )(0 999 9855 10 )1.60 10 C

(1.60 10 )(1 50)

9.11 10 kg 10 m

998 55% of

mv. r

e

e r.

m r

v

BB

0.9999999855c

b At velocities close to the speed of light the effects of relativity have a large impact on the

radius of the electron beam. Because the mass of the electrons is about 6000 times greater

than their rest mass it follows that the expected path radius will also be some 6000 times

greater than predicted in part a. Because the bending magnets are only found in sections of

the storage ring the actual path is greater still.

10 Similarities include: Same basic design, a linac, booster ring, storage ring and a number of

beamlines and a third generation source.

Differences include: Delta power output is 1.5 GeV compared to the proposed 3 GeV for the

Australian Synchrotron. Delta has less beamlines, and is half the size of the Australian

Synchrotron. Delta is oval shaped while the Australian Synchrotron is symmetrical through all

axes.



Page 169 of 185

7.4 Mass spectrometry

1

27

p

27 3

19

3 1 5

19 1

1.67 10 kg

(1.67 10 )(1.50 10 )0.600 T

(1.60 10 )(0.600)

1.50 10 m s 2.61 10 m

1.60 10 C

mvm r

q

r

v r

q

B

B

2 27 21

p 2

19 35 1

27

273

1.67 10 kg

2 2(1.60 10 )(2.55 10 )0.450 T 6.99 10 m s

(1.67 10 )

(1.67 10 )(6.992.55 10 V

m mv q V

q Vv

m

mvV r

q

B

B

5

19

19 2

10 )

(1.60 10 )(0.600)

1.60 10 C 1.22 10 mq r

3

O

2

2 2 24

2

19 1

20.850 m and

27.20 10 T

21.10 10 V

1.60 10 C

rq q Vr v v

m m

rq q V

m m

r q q VV

m m

rq

B

BB

B

2 2

2 2 2 19 2 2

4

26

2

(0.850) (1.60 10 )(7.20 10 )

2 2(1.10 10 )

2.72 10 kg

qV

m

r qm

V

m

B

B

4 Positive

5 The direction of the magnetic field would need to be reversed and the voltage used to accelerate

the ions would need to be reversed too.



Page 170 of 185

6 27 21

p 2

2 27 6 21 119 2 2

19

6 1 4

1.67 10 kg

(1.67 10 )(2.00 10 )1.60 10 C

(1.60 10 )

2.00 10 m s 2.09 10 V

m mv q V

mvq V

q

v V

7

27

p

27 6

19

6 1 2

19 1

1.67 10 kg

(1.67 10 )(2.00 10 )0.555 m

(1.60 10 )(0.555)

2.00 10 m s 3.76 10 T

1.60 10 C

mvm

qr

r

v

q

B

B

B

8 No. If both particles were accelerated with the same (but opposite) accelerating potential then

they would have very different speeds due to their very different masses. The velocity selector

would only allow significant numbers of only one of them to pass into the spectrograph. If

significant numbers of each particle were allowed into the spectrograph at the same velocity then

the radii of the two pathways would be very different due to the same (but opposite) magnetic

flux density on the different masses.



Page 171 of 185

9 If the separation is 0.330 mm for a half turn, the difference in radius between the two paths would

be 0.165 mm.

2

2

2 2

2 2

2

2 2

NCOCO

CO N

-3

N N

N N

CO

-3

N3

CO N N

28.0106 u

( 0.165 10 )28.0134 u

( 0.165 10 )(28.0134)0.165 10 m

(28.0106)

rrvm

q m m

r mm r

m

rr r r

B

2 2

2

3

N N

3 3

N

28.0106 28.0134 4.6222 10

2.8 10 4.6222 10

r r

r

2

2

3

N -3

N

(4.6222 10 )

(2.8 10 )

1.65 m

r

r

This is the larger radius of the two particles; therefore the radius of the mass spectrometer must

be greater than, or equal to this value.

Chapter 7 Review

1 B

2 The radius of curvature of both electron beams is the same therefore the velocity of both beams

must be equivalent Using the right-hand rule Bx must be out of the page for the electron beam to

bend up the page, and By must be into the page for the electron beam to bend down the page.

3 C

4 A

5 19 3

3 1

E 2

19 14

E

2

(1.60 10 )(15.0 10 )15.0 10 V m

(12.0 10 )

1.60 10 C 2.00 10 N

12.0 10 m

q VV

d

q

d

F

F

6 B

r

-

C

O

0

.

3

3

0



Page 172 of 185

7 3

k d

19 212

19 331

31

28 0 10 V

1.60 10 C

2 2(1.60 10 )(28 0 10 )9.11 10 kg

(9.11 10 )

V . E W

e mv e V

e V .m v

m

v

7 110 m s

8 3

3

2

2 5 1

(28 0 10 )28 0 10 V

(20.0 10 )

20.0 10 m 10 V m

V .V . E

d

d E

9 D

10 6 1

c B

2

31 631

e 19

19

e

4.20 10 m s

1.20 T r

v (9.11 10 )(4.20 10 )9.11 10 kg

(1.60 10 )(1.20)

1.60 10 C

v

mvqv

mm r

q

q

F F

B B

B

5 1.99 10 mr

11 34 8

11

11

34 15

154

19

(6.63 10 )(3.00 10 )λ 6 80 10 m

(6 80 10 )

6.63 10 J s 10 J

101.83 10 eV

10

hc. E

.

h E

E

12 A

13

10

k ph

34 88 1 19

k 10

16

k

34

9.80 10 m

(6.63 10 )(3.00 10 )3.00 10 m s (4.20)(1.60 10 )

(9.80 10 )

4.20 eV 2.02 10 J

6.63 10 J s

hcE E W W

c E

W E

h



Page 173 of 185

14

15 3

E B

3

37 1

7 3

19

3.00 10 V

1.60 10 T

(3.00 10 )v 3.23 10 m s

(3.23 10 )(1.60 10 )

1 60 10 C

V

q Vqv

d

Vd

v

q . d

F F

B B

B

25.81 10 m

16 2

c B

24

11 1 11 2 4

6.00 10 m

1.50 10 T

1.76 10 C kg (1.76 10 ) (6.00 10 )(1.50 10 )

r

mvqv

r

qr qqv r

m m m

F F

B B

BB

6 1 1.58 10 m sv

17

2

1 21

1 2

3

2 11

2

3 1

2 1

5

(155 10 )(5)3

(3)

155 10 m 2.58 10 m

r rvm

q m m

r mm r

m

r r

B

1 1 2(2.58 10 ) 5.17 10 md

This assumes that the particles in the mass spectrograph have a semicircular path

electrons



Page 174 of 185

18

3

He

2 2 2

2

19 1

240.2 10 m and

20.160 T

2500.0 V

1.60 10 C

rq q Vr v v

m m

rq q V

m m

r q q VV

m m

q

B

BB

B

2 2

2 2 3 2 19 2

27

2

(40.2 10 ) (1.60 10 )(0.160)

2 2(500.0)

6.62 10

r qV

m

r qm

V

m

B

B

kg

19

O

2

2 2 23

2

219 1

20.850 m and

27.20 10 T

211.0 10 eV

1.60 10 C

rq Eqr v v

m m

rq Eq

m m

r q EqE

m m

r qq

B

BB

B

B 2

2 2 2 19 2 2

3

26

2

(0.850) (1.60 10 )(7.20 10 )

2 2(11.0 10 )

2.72 10 kg

Em

r qm

E

m

B



Page 175 of 185

Chapter 8 Working in physics

8.1 Measurements and units

1 a Acceleration is derived.

b Magnetic field intensity is derived.

c Force is derived.

d Density is derived.

e Time is fundamental.

f Torque is derived.

2 2

2

2

width: 945 cm 945 10 m 9.45 m

length: 468 cm 468 10 m 4.68 m

height: 255 cm 255 10 m 2.55 m

3 2 3

2 3

2 3

width: 945 cm 945 10 m 9.45 10 mm

length: 468 cm 468 10 m 4.68 10 mm

height: 255 cm 255 10 m 2.55 10 mm

4 2 6

2 6

2 6

width: 945 cm 945 10 m 9.45 10 μm

length: 468 cm 468 10 m 4.68 10 μm

height: 255 cm 255 10 m 2.55 10 μm

5

2 3

width 9.45 m

length 4.68 m (9.45)(4.68)(2.55)

height 2.55 m 1.13 10 m

V w L h

V

V

6 3

3 3 3

7 2

width 9.45 10 mm

length 4.68 10 mm (9.45 10 )(4.68 10 )

4.42 10 mm

A w L

A

A

7

fuel prior fuel prior fuel prior

1

fuel prior

7682 L conversion factor

conversion factor 0.803 kg L (7682)(0.803)

V m V

m

3

fuel prior 6.17 10 kgm



Page 176 of 185

8 3

fuel prior fuel to add fuel required fuel prior

4 4 3

fuel required fuel to add

6.17 10 kg

2.23 10 kg (2.23 10 ) (6.17 10 )

m m m m

m m

4

fuel to add 1.61 10 kgm

9 4

4 fuel to addfuel to add fuel to add

1 4

fuel to add

1.61 101.61 10 kg

conversion factor 0.803

conversion factor 0.803 kg L 2.01 10 L

mm V

V

10

1

1

JJ s

s

E hf

Eh

f

8.2 Data

1

%022y(%)uncertaint

100824

50y(%)uncertaint

.

.

.

Note that the value 0.5 and 100 are exact numbers.

2

0.5uncertainty (%) 100

35

uncertainty (%) 1.4%

Note that the value 0.5 and 100 are exact numbers.

3 2diameter uncertainty (%) 2.02% π

length uncertainty (%) 1.4% uncertainty (%) 2(2.02) (1.4)

V r L

V

uncertainty (%) 5.4%V



Page 177 of 185

4 2

2 3 2 2

3

uncertainty (%) 5.4% π

length 35 10 m π(12.4 10 ) (35 10 )

diameter 24.8 10 m

V V r L

V

4 3

3

4

1.7 10 m

radius 12.4 10 m

5.4 uncertainty (1.7 10 )

100

V

V

V

6 3 uncertainty 9.2 10 m

5

50mass uncertainty (%) 100

1200

mass uncertainty (%) 4.2%

1speed uncertainty (%) 100

18

speed uncertainty (%) 5.6%

3radius uncertainty (%) 100

46

radius uncertainty (%) 6.5%

Note that the values 50, 1, 3 and 100 are exact numbers.

6 2

21

3

1200 kg

(1200)(18)18 m s

(46)

46 8.5 10 N

mvm F

r

v F

r m F

7 2

mass uncertainty (%) 4.2%

speed uncertainty (%) 5.6% uncertainty (%) (4.2) 2(5.6) (6.5)

radius uncertainty (%) 6.5%

mvF

r

F

uncertainty (%) 21.8%F

8

3

3 3

21.8 uncertainty (%) 21.8% uncertainty (8.5 10 )

100

8.5 10 N uncertainty 1.8 10 N

F F

F F

9 Answers depend on your experimental data.



Page 178 of 185

10 Answers depend on your experimental data.

8.3 Graphical analysis of data

1

There seem to be some data points that should be excluded.

2

cxmy

bxaF

Therefore they should plot F against x

3

Applied force (F) (N) Square root of length ( )x12( )m

10.0 0.056

20.0 0.119

30.0 0.158

40.0 0.263

50.0 0.291

60.0 0.350

70.0 0.408

80.0 0.467

90.0 0.526

100.0 0.584

110.0 0.643



Page 179 of 185

4

Note the datum excluded.

5

9410 169

9405069.531

.xF

.xy

Hence, gradient =170 N m–1

6

9410 169

9405069.531

.xF

.xy

Hence, y-axis intercept =0.941 N

7

169 0.941F x

8 a = 170, b = 0.941

9 2

c

2

c

0

m vF

r

mF v

r

y m x c

1.3.10

Hence, we need to plot Fc against v2 to get a linear graph.

Applied force (F) vs square root of length (x)



Page 180 of 185

10

r

m slope

Hence, the slope is the ratio between m and r with the units of kg m–1.

8.4 Writing scientific reports

1 Aim, Apparatus, Method, Results, Analysis, Conclusion

2 a False

b True

c False

d True

e True

f False

3 1 A pendulum was set up as shown in the diagram.

2 The mass of a full set of slotted masses was determined.

3 The set of slotted masses were attached as the pendulum bob.

4 The pendulum string was made such that the masses were clear of the ground as it swung.

5 The length of the pendulum was measured from the retort arm to the base of the set of

slotted masses.

6 The pendulum bob was pulled to one side such that the bob was displaced by 2 cm

horizontally from its original position.

7 The stopwatch was started as the bob was released.

8 The time was recorded for 20 oscillations of the pendulum bob.

9 The period for each oscillation was calculated by dividing the time in step 8 by 20.

10 Steps 8 and 9 were repeated twice more to determine an average period for one oscillation.

11 Steps 6–10 were repeated for displacements of 4, 6, 8 and 12 cm.

Chapter 8 Review

1 3 3 3 3

9 3

569 mm 569( 10 ) m

569 10 m

V V

V

2 3

3

3

3 1

2.80 10 N

(2.80 10 )2.60 A

(2.60)(4.30 10 )

4.30 10 m 2.50 10 T

Il

I

l

F F B

B

B



Page 181 of 185

3 3

3

3

3

(0.10 10 )2.80 10 N : % uncertainty 100 3.57%

(2.80 10 )

(0.05)2.60 A : % uncertainty 100 1.92%

(2.60)

4.30 10 m : % uncert

I I

l l

F F

3

3

(0.20 10 )ainty 100 4.65%

(4.30 10 )

4

: % uncert ( % uncert) ( % uncert) ( % uncert)

: % uncert 3.57) (1.92) (4.65) 10.1%

I l

(

B F

B

5

1

2

10.1: abs uncert (2 50 10 )

100

: abs uncert 2.54 10 T

.

B

B

6

2 2

1 2 3 2

2 2 2 6

2 4 2

ρρ

(m s ) (kg m )

m s kg m

kg m s

Bv B v

B

B

B

7 Lachlan

8 Joshua

9 Tom

10 Charlie

11

1

lane 1

lane 1 lane 1

lane 1

lane 8

(2.00)346 m s

(346)

2.00 m 0.00578 s

25.0 m

(25.0)

(346)

sv t

v

s t

s

st

v

lane 8 0.0723 st



Page 182 of 185

12

1

lane 1

lane 1

(50.00)346 m s

(346)

50.00 m 0.145 s

sv t

v

s t

2 2 2

lane 8

2 2

lane 8

lane 8

(50.00) (25.00)

(50.00) (25.00)

55.90 m

s

s

s

lane 8

lane 8

(55.90)

(346)

0.162 s

st

v

t

13

lane 1 swimmer lane 1 lane 1 timer lane 1 swimmer

lane 1 timer lane 1

lane 8 swimmer

0.00578 s (0.145) (0.00578)

0.145 s 0.1387

0.0723 s

t t t t

t t s

t

lane 8 lane 8 timer lane 8 swimmer

lane 8 timer lane 8

(0.162) (0.0723)

0.162 s 0.0893

t t t

t t s

With a greater difference in time between the swimmer starting and the timer timing, lane 1 has

an advantage.

14

lane 1 timer 1 lane 1 timer 2lane 1 timer 1 lane 1

lane 1 timer 2 lane 1

lane 8 timer 1

(35.05) (35.03)35.05 0.20 s average

2 2

35.03 0.20s average 35.04 0.40

35.

t tt t

t t s

t

lane 8 timer 1 lane 8 timer 2lane 8

lane 8 timer 2 lane 8

(35.05) (35.06)05 0.20 s average

2 2

35.06 0.20 s average 35.055 0.40 s

t tt

t t

Lane 1 probably won the race.

15 2 2

2 2

2

2

v u as

v as u

y mx c

a v2

b s

c 2a

d u2



Page 183 of 185

16

Displacement (s) (m)

Speed (v) (m s–1) Speed squared (v2)

(m2 s–2)

40.0 20.0 400.0

60.0 22.5 506.3

80.0 25.2 635.0

100.0 25.8 665.6 exclude

120.0 29.5 870.3

17

18 y = 5.93x + 158.39

19 a 5.93 m s–2

b 158.39 m2 s–2

20

2

2 5.93

(5.93)2.97 m s

2

a

a

21 2

1 1

158.39

158.39 1.26 10 m s

u

u



Page 184 of 185

22

Load force (FL) (kN) Length (l) (m) Stress (F m–2) Strain

0.0 50.00 0.00 0.00

2.2 50.04 7.29 × 107 8.00 × 10-4

6.7 50.12 2.22 × 108 2.40 × 10-3

11.1 50.25 3.68 × 108 5.00 × 10-3

15.6 50.41 5.17 × 108 8.20 × 10-3

16.7 51.50 5.53 × 108 3.00 × 10-2

17.6 54.00 5.83 × 108 8.00 × 10-2

17.8 56.00 5.90 × 108 1.20 × 10-1

17.8 58.00 5.90 × 108 1.60 × 10-1

16.9 60.2 5.60 × 108 2.04 × 10-1

23

24 Young’s modulus = 6.24 × 1010 N m–2

25

2

2 1 0

lT

k

lT

k

T lk

y m x c

Therefore, graph T2 ion the y-axis and l on the x-axis to graph a straight line.



Page 185 of 185

26 Slope of the line represents kl

27

28

2

14.9046

0.204 m s

k

k

29 Mass has no affect on the period of a pendulum.

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Heinemann Physics Content & Contexts Units 3A & 3B Solutions

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