Heinemann Physics Content & Contexts Units 3A & 3B Solutions
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Transcript of 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
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 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
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 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.
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 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%
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 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
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 6 of 185
b
1 1
yellow
yellow yellow yellow purple purple yelloow purple yellow purple
6.00 km h 1.67 m s momentum is conserved, so:
420.0 kg
m m m m
u
u u v
uyellow yellow purple purple1 1
purple yellow purple
yellow purple
purple 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
10.0 km h 2.78 m s momentum is conserved, so:
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
15.0 km h 4.17 m s momentum is conserved, so:
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
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 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
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 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.
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 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
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 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
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 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
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 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
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 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
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 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
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 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
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 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
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 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
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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
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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
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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
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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
9.80 m s critical speed when
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
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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
9.80 m s critical speed when
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
9.80 m s critical speed when
100.0 m
( 100.0)( 9.80)
31.3 m s
g
vr
r
v r
v
a a
g
g
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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.
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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.
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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
Heinemann Physics Content and Contexts Units 3A and 3B
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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.
Heinemann Physics Content and Contexts Units 3A and 3B
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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
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 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
Heinemann Physics Content and Contexts Units 3A and 3B
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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
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 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
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 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
Heinemann Physics Content and Contexts Units 3A and 3B
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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
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 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
Heinemann Physics Content and Contexts Units 3A and 3B
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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
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 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
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 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
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 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 )
7.35 10 kg 1.99 10 N attraction
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 )
9.11 10 kg 3.61 10 N attraction
5.30 10 me
Gm mm
r
m
r
F
F
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 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 )
1.80 10 kg 5.23 10 N attraction
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
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 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
Heinemann Physics Content and Contexts Units 3A and 3B
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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
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 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.
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
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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
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
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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
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
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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
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 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
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
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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
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 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
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 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
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 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
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 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°
Heinemann Physics Content and Contexts Units 3A and 3B
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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
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 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
Heinemann Physics Content and Contexts Units 3A and 3B
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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
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 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
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 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
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 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
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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
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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
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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
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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
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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
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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
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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
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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
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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.
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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
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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
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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
Assume 100% efficiency
1.00 C (9.00)(1.00)
9.00 V 9.00 J
q W q V
V W
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Page 79 of 185
10
3
d d
3
d
2
Assume 100% efficiency
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
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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
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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
Heinemann Physics Content and Contexts Units 3A and 3B
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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
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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
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
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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
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
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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
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
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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)
2.00 10 N into the page
l
F I B
F
c
3
2
(5.00)(10.0 10 )(1.00)
5.00 10 N into the page
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
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
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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
Heinemann Physics Content and Contexts Units 3A and 3B
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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
Heinemann Physics Content and Contexts Units 3A and 3B
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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
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
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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
induced current is 1.00 mA from X to Y
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
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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
induced current is 2.00 mA from X to Y
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
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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
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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
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 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
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 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
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 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.
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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
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 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
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
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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.
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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
Heinemann Physics Content and Contexts Units 3A and 3B
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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
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
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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
Heinemann Physics Content and Contexts Units 3A and 3B
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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 ( . )
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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
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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
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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
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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.
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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
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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
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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
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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
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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.
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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
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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.
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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.
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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 6 Review 159
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 7 Review 171
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
Chapter 8 Review 180
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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
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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
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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
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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
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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
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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).
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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)
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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.
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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.
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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.
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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.
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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.
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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.
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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
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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
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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
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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
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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
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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
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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
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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
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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.
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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.
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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
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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
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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
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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
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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
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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.
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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
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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
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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
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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
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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
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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
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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
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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
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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
k 1.50 10 J each particleE
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
k 1.34 10 J each particleE
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
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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
snd-you snd-gnd gnd-you
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
snd-you snd-gnd gnd-you
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
snd-you snd-gnd gnd-you
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
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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
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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
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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
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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
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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
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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.
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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.
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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.
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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.
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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
–
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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
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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
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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.
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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.
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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.
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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
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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
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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
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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
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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
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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
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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.
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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
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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)
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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
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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
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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
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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
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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.
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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.