Part A
Worked solutions
■■■■■■■
Unit
Area of study 1:
Motion in one and two
dimensions
3
■■■■■■■■■
Forces in action
1. u = 0, x = 10 m, t = 4.0 s, a = ?
x = ut + 12 at 2
1
2
Part A — WORKED SOLUTIONS
4. The stationary car is pushed forward by the other vehicle.
As a result, the seat pushes the body of an occupant
forward. This happens almost instantaneously. However,
without a headrest, there is nothing to push the occupant’s
head forward quickly. The head remains at rest until
pulled forward by the spine (Newton’s First Law of
Motion). The head applies an equal and opposite force to
the spine, (Newton’s Third Law of Motion) potentially
causing serious injuries.
5. To say that the passenger is thrown forward implies that a
force accelerates the passenger. The car slows down
rapidly in most collisions as a result of a large external
force. The passenger continues to move at the original
speed of the car while the car slows down.
6. (a)
(b)
Chapter 1
⇒ 10 =
Jacaranda Physics 2, 3rd Edition TSK
× a × 16
⇒ 10 = 8a
⇒ a = 1.25 m s –2
Over first 5.0 m
u = 0, x = 5.0 m, a = 1.25 m s−2
v=?
v2 = u2 + 2ax
= 2 × 1.25 × 5.0
⇒ v = 3.5 m s−1
2. 10 m s−2 down
3.
(c)
7. (a) F (down)
(b) C (perpendicular to surface)
(c) X (Constant velocity must be the result of a zero net
force.)
8.
(a)
Δv = v − u
⇒ Δv = 50 km h−1
30
tan θ =
40
⇒ θ = 37°
Δv
a=
Δt
50 km h S 37°E
2.0 s
= 25 km h−1 s−1 S 37°E
25 km h –1
a=
S 37°E
1s
(25 ÷ 3.6) m s –1
=
S 37°E
1s
= 6.9 m s−2 S 37°E
=
(b)
(a) W = mg
= 4.0 × 10
= 40 N
(b) TH = 12 cos 30°
= 10 N
(c) Net vertical force = 0
⇒ N + T sin 30° = W
⇒ N + 12 sin 30° = 40
⇒ N = 40 − 12 sin 30°
= 34 N
9. The matching reaction to the gravitational pull of Earth
on you is the gravitational pull of you on the Earth.
–1
Jacaranda Physics 2 TSK
2
© John Wiley & Sons Australia, Ltd 2009
10. The net force must be in the direction in which the
elevator is speeding up.
(a)
(i) Normal reaction force is equal to your weight
and the net force is zero.
(ii) Normal reaction force is equal to your weight.
The speed is not changing so the net force must
be zero.
(iii) Normal reaction force is greater than your
weight because the speed of the elevator is
increasing in an upwards direction.
(iv) Normal reaction force is less than your weight
because the net force must be down to make the
elevator decrease its upward speed.
(b) You feel the sensation of weight only if there is an
upward push on you by an object like the ground, a
floor, a chair or a bed. The apparent weight that you
feel is the size of this upward push. If the elevator is
accelerating upwards, the normal reaction force is
greater than your weight and you feel heavier. If the
elevator is accelerating downwards, the normal
reaction force must be less than your weight and you
feel lighter.
Δv
11. (a) a =
t
–2.0 – 5.0
=
0.20
= −35 m s−2
= 35 m s−2 opposite to the initial direction of the
dodgem car.
(b) Fnet = ma
= 200 × 35
= 7.0 × 103 N
12. (a) Braking distance = area under graph over last 20 s
= 12 × 20 s × 20 m s –1
(d) a =
Δv
Δt
⇒ Forward force = 24 000 N
= 2.4 × 104 N
–20 m s –1
20 s
= – 1.0 m s –2
=
13. (a)
Fnet = ma
= −4.0 × 104 × 1.0
= −40 000 N
Additional frictional force = 40 000 − 8000
= 32 000 N
= 3.2 × 104 N
(b) N = W cos 20°
= mg cos 20°
= 1500 × 10 × cos 20°
= 1.4 × 104 N
(c) Zero (Car is at rest.)
(d)
Fnet = 0
⇒ mg sin 20° − R = 0
⇒ R = mg sin 20°
= 1500 × 10 × sin 20°
= 5.1 × 103 N
14. (a) No direction. The net force is zero.
(b)
= 200 m
(b) Total distance travelled by train (and cyclist)
= area under graph
=
1
2
× 50 s × 20 m s –1 + 50 s × 20 m s –1 + 200 m
= 1700 m
distance travelled
time taken
1700 m
=
120 s
= 14 m s –1
constant speed =
(c) a =
15.
Δv
Δt
⇒ mg sin 30° − (air resistance + friction) = 0
air resistance + friction = mg sin 30°
= 60 × 10 × sin 30°
= 300 N
20 m s –1
50 s
= 0.4 m s –2
=
Fnet = ma
= 4.0 × 104 × 0.40
= 16 000 N
Total frictional forces = 8000 N
⇒ Forward force − 8000 N = 16 000 N
Jacaranda Physics 2 TSK
3
© John Wiley & Sons Australia, Ltd 2009
16.
(a) Fnet = ma
= 70 × 2.0
= 140 N north
⇒ tension − frictional forces = 140
⇒ tension − 240 = 140
⇒ tension = 380 N north
(b) Fnet = ma
= 350 × 2.0
= 700 N north
⇒ thrust − tension − frictional forces = 700
⇒ thrust − 380 − 600 = 700
⇒ thrust = 1680 N north (1.7 × 103 N north)
18. (a) Fnet = m
−2.7
0.10
= 5.4 N opposite to the original direction of
motion
Impulse on billiard ball = mΔv
= 0.200 × −2.7
= 0.54 N s
opposite to the original direction of motion
The table doesn’t move because the net force acting
on it is zero. It is zero because friction applied to the
table by the floor is enough to match the force
applied by the ball.
The net force on the bullet is large because the forces
acting on it, other than the one exerted by the rifle,
are small. The net force on the rifle is zero if the
shooter exerts a force on it to balance its weight and
the force applied by the bullet. Therefore, the rifle
does not accelerate.
pf = pi
⇒ 4.0 kg × vrifle + 0.020 kg × 300 m s−1 = 0
⇒ 4.0vrifle = −6.0
⇒ vrifle = −1.5 m s−1
−1
⇒ recoil speed is 1.5 m s
Δp = impulse
= area under graph
= 10 N × 6 s
= 60 kg m s−1
Δp = impulse
= area under graph
= 10 N × 6 s + 5 N × 6 s
= 90 kg m s−1
⇒ pf − pi = 90 kg m s−1
⇒ pf = 90 kg m s−1
= 0.200 ×
(b)
(c)
19. (a)
(b)
The net force on the magpie is zero. Resolving vertically:
2T sin θ = mg
mg
⇒T =
2 sin θ
0.04
tan θ =
2.0
⇒ θ = 1.146°
4.0 × 10
⇒T =
2 × sin 1.146°
= 1000 N
This answer is based on the assumption that the wire has
zero mass and is perfectly flexible.
20. (a)
(b)
17.
(c)
Let the mass of the globe be m and assume that the mass
of the wire is negligible. Resolving vertically:
T cos θ = mg
⇒ T cos θ = 10m
Resolving horizontally:
Fnet = ma
⇒ T sin θ = ma
⇒ T sin θ = 1.5m
Divide equation (2) by equation (1).
sin θ 1.5
⇒
=
cos θ 10
⇒ tan θ = 0.15
21. In each of cases (a), (b) and (c), the change in momentum
is fixed. An increase in the time interval during which the
momentum changes results in a smaller force applied to
the occupants or cyclist as Δp = FΔt.
22. (a) Assigning original direction of motion of car as
positive:
(i) ⇒ Impulse = mΔv
= 70 (0 − 14)
= −980 N s
(ii) Impulse = mΔv
= 70 (0 − 14)
= −980 N s
⇒ θ = 8.5°
Jacaranda Physics 2 TSK
Δv
Δt
4
© John Wiley & Sons Australia, Ltd 2009
x = 0.50 m, u = 14 m s−1,
v = 0, a = ?
v2 = u2 + 2ax
⇒ 0 = 196 + 2a × 0.50
⇒ a = −196 m s−2
(iv)
x = 0.025 m, u = 14 m s−1,
v = 0, a = ?
v2 = u2 + 2ax
⇒ 0 = 196 + 2a × 0.025
–196
⇒a =
0.05
= −3920 m s−2
(b) g = 10 m s−2
acceleration of driver = −196 m s−2
196
=
g
10
= 19.6 g
acceleration of passenger’s head = −3920 m s−2
3920
=
g
10
= 392 g
(c) The seatbelts allow the change in momentum of the
occupant to take place over a longer time interval,
thus reducing the force applied to the occupant.
Without a seatbelt, the occupant continues to move
forward, colliding with the interior of the car. This
‘secondary’ collision causes the change in
momentum of the occupant to take place in a much
smaller time interval than would be the case with a
seatbelt. The force applied to the occupant is
therefore much greater.
(b) The vertically downward momentum of the coal
decreases to zero because there is an upward net force
acting on it when it strikes the cart. The total
momentum of the Earth−coal system has not changed.
(c) pi = 750 × 2.0
= 1500 kg m s−1 south
When the coal falls from the cart it has a horizontal
velocity of 2.0 m s−1 south.
⇒ 500 v + 250 × 2.0 = 1500 kg m s−1 south
⇒ 500 v = (1500 − 500) kg m s−1 south
1000
⇒v=
kg m s –1 south
500
= 2.0 m s−1 south
5. (a) p = mv
= 70 × 2.0 east
= 140 kg m s−1 east
(b) Three seconds before impact, Dean is 6 m from
Melita because he is gliding at 2.0 m s−1 east.
Taking Dean’s position as the origin:
m x + m2 x2
xcm = 1 1
m1 + m2
(iii)
70 × 0 + 50 × 6
120
= 2.5 m
The centre of mass is 2.5 m east of Dean.
(c) Before the collision, the centre of mass travels 3.5 m
(from Dean to Melita) in 3.0 s.
Δx
v=
Δt
3.5 m east
=
3.0 s
=
(d)
Chapter 2
Collisions
(e)
1. Impulse is equal to the change in momentum or words to
that effect.
2. To increase the time interval during which the momentum
of the vehicle’s occupants changes during a collision.
3. No. The system of the two cars is not isolated. There are
unbalanced frictional forces acting on the cars during and
immediately after the collison.
4. (a) The system of the coal and railway cart can be
considered to be isolated if the horizontal motion
only is considered.
pi = 500 × 3.0 + 0
(initial horizontal momentum of the coal is zero)
= 1500 kg m s−1 south
pf = 750 v
where v = final velocity of cart
pf = p i
⇒ 750 v = 1500 south
⇒ v = 2.0 m s−1 south
Jacaranda Physics 2 TSK
6. (a)
(b)
(c)
5
= 1.2 m s –1 east
The momentum of the centre of mass remains
constant.
⇒ The common velocity of Melita and Dean after the
collision is 1.2 m s−1 east.
Impulse on Melita = her change in momentum
= pf − pi
= 70 kg × 1.2 m s−1 east − 0
= 60 N s east
Assigning east as positive
pi = 1500 × −20 + 2000 × 20
= 10 000 kg m s−1
pf = pi
⇒ 3500v = 10 000
⇒ v = 2.9 m s−1 east (2.86 m s−1 east)
Impulse on truck = its change in momentum
= pf − pi
= 2000 × 2.86 − 2000 × 20
= −3.4 × 104
= 3.4 × 104 N s west
Δvcar = v − u = 2.9 − (−20)
= 22.9 m s−1
Δvtruck = v − u = 2.9 − 20
= −17.1 m s−1
© John Wiley & Sons Australia, Ltd 2009
⇒ The car experiences the greatest (in magnitude)
change of velocity.
(d)
Δpcar = −Δptruck
since the total change in momentum is zero. This can
be verified.
Δpcar = pf − pi
= 2000 × 2.86 − 2000 × 20
= −3.4 × 104
= 3.4 × 104 N s west
(e) Each vehicle experiences the same force (in
magnitude) (Newton’s Third Law of Motion).
7. Student responses will vary. In most situations, an
occupant in a larger car is safer. However, the design of
the particular car, the nature of the collision and several
other factors influence the likely effect of an accident on
occupants. The following points should be made.
• The forces applied to each car by the other are equal in
magnitude and opposite in direction.
• The change in velocity of each car is dependent on the
mass of the car. Assuming that the sum of the forces
other than that applied by the other car is zero, the
change in velocity is inversely proportional to the mass
of the car.
• Assuming that you are properly restrained and that the
collision is head-on, your change in velocity (and
therefore the deceleration you are subjected to) is less if
you are in a heavier car.
• The body continues to move in the original direction
and at the original speed of your car until an
unbalanced force acts on you. If you are not restrained,
the unbalanced force will be provided by the
windscreen or part of the interior of the car, which has
already slowed down. A smaller car will have slowed
down more, so the impulse applied to you (mΔv) will
be greater.
8. (a) Ek = 12 mv 2
=
1
2
W = ΔEk
=
(b)
2
Fav x = 9722
9722
0.70
= 1.4 × 104 N
(c) An estimate of the depression of the dashboard by the
driver needs to be made. If the dashboard is
depressed by 5 mm, then
9722
Fav =
0.005
⇒ Fav =
= 2 × 106 N (approx.)
10. (a) Fspring + weight = 0
⇒ Fspring = −mg
= 1.0 × 10
= 10 N up
F
(b) k =
x
= gradient of F vs x graph
=
15 N
0.40 m
= 38 N m –1 (37.5 N m –1 )
(c) A — the spring with the greatest spring constant
(d) W = mg
= 0.500 × 10
= 5.0 N
x = 0.50 m (from graph)
Work done = area under F vs x graph
= 12 × 0.50 × 5.0
= 1.3 J (1.25 J)
(e) Greatest strain energy occurs in the spring with the
F vs x graph of greatest area.
At maximum extension
Area A = 12 × 0.20 × 25 = 2.5 J
= 1.8 × 105 J
(b) 1.8 × 105 J
(c) x = 0.40 m, u = 20 m s−1, v = 0 m s−1
v 2 = u 2 + 2as
Area B =
2
v −u
2x
0 − 202
=
2 × 0.4
–400
=
0.8
= 500 m s –2
a=
⎛ 60 ⎞
× 70 × ⎜
⎟
⎝ 3.6 ⎠
= 9.7 × 103 J
× 900 × 202
2
1
2
1
2
× 0.40 × 15 = 3.0 J
Area C = 1.25 J (from (d))
Spring B has greatest strain energy.
11. (a) 2.3 × 104 N
(b) 4.0 × 103 N
12. (a)
F = ma
= 900 × 500
= 4.5 × 105 N opposite to the initial direction of
motion of the car
9. (a) Assume that the sum of the forces other than that of
the seatbelt applied to the driver is zero.
Jacaranda Physics 2 TSK
6
© John Wiley & Sons Australia, Ltd 2009
(b) ΔEgp = Area under graph of force vs height
= m × area under graph in 12(a)
= 150 × 12
= 1.8 × 103 J
(c) Work = ΔEgp
= 1.8 × 103 J
13. Let Egp = 0 at ground
Air resistance can be assumed to be negligible.
initial Egp (car)
(a)
initial Egp (cricket ball)
=
mgh (car)
mgh (cricket ball)
=
M car
M cricket ball
=
1600 kg
0.160 kg
⇒ v 2 = 250
⇒ v = 16 m s −1
(c) WN = 0
since there is no displacement in the direction of the
force.
(d) Work done by net force = change in kinetic energy
⇒ Fnet x = ΔEk
⇒ (mg sin 30° − friction) × x = 12 mv 2
⇒ (300 − friction) × 25 =
ΔEgp (car)
Ek (car)
(b)
=
Ek (cricket ball) ΔEgp (cricket ball)
(b)
mg Δh (car)
mg Δh (cricket ball)
1600 kg
=
0.160 kg
= 10 000
=
1 mv 2 (car)
2
2
1 mv
2
⇒
(cricket ball)
2
1600 vcar
2
0.160 vcricket
ball
=
2.475 × 2
2.5
= 1.4 m s−1
(c) F = kx, where x = compression
When x = 10 cm, F = 22 N
F
k=
x
22 N
=
0.10 m
Ek (car)
Ek (cricket ball)
= 10000
= 220 N m –1
2
vcar
=
1
2
× 450 × (2.0) 2
900 × 2
3.0 × 107
= 7.7 × 10−3 m
(b) Work done = strain energy
= 900 J
(c) 2.0 m s−1. If the rubber bumper obeys Hooke’s Law
as it expands to its original shape, all of the energy
will be returned to the kinetic energy of the dodgem
car.
= 7500 J
ΔEk = Wgravity
= Wgravity (since initial Ek = 0)
⇒ 12 × 60v 2 = 7500
⇒ 30v 2 = 7500
Jacaranda Physics 2 TSK
mv 2
⇒x =
= 60 × 10 × 25 sin 30°
⇒
1
2
k = gradient of F vs compression graph
300 kN
=
0.01 m
= 3.0 × 107 N m−1
⇒ 12 × 3.0 × 107 × x 2 = 900
(a) W = Fx
where x = displacement in direction of force
⇒ Wgravity = mg Δh
1
mv 2
2
16. (a) ΔEk =
= 900 J
⇒ Maximum strain energy = 900 J
⇒ 12 kx 2 = 900 J
14.
(b)
= 2.5 J (2.475 J)
ΔEk = 2.4755
Ek = 2.475 J (since initial Ek = 0)
1
⇒ 2 mv2 = 2.475 J
⇒v =
=1
2
vcricket
ball
vcar
⇒
=1
vcricket ball
⇒
× 60 × (7.2)2
⇒ 300 − friction = 62
⇒ friction = 238 N
15. (a) Energy stored
= energy under force vs compression graph
= area under force vs length graph as length changes
from 20 cm to 5 cm
= 12 × 0.15 m × 33 N
= 10000
(c)
1
2
7
© John Wiley & Sons Australia, Ltd 2009
20. (a) Assign east as positive.
pi = mu + m × (−20)
pf = 2m × 5
⇒ mu − 20m = 10m
⇒ u − 20 = 10
⇒ u = 30 m s−1
= 30 m s−1 east
(b) Eki = 12 m(30) 2 + 12 m(20) 2
17. (a) The springs are inside each other as shown in the
diagram below.
Ekf =
1
2
× 2 m × (5) 2
Ekf
=
Eki
1
2
m × 25
m(900 + 400)
⎛ 1 ⎞
= 0.038 ⎜ ⎟
⎝ 26 ⎠
pi = mv − mv
=0
⇒ pf = 0
⇒v =0
(b) As the collision is perfectly elastic, the kinetic energy
of each of the two cars has not changed. As the force
on each car in the collision is equal (by Newton’s
Third Law) and their masses are equal, they have
both accelerated equally in the collision. Their speeds
must then be equal and unchanged: that is, 60 km h−1.
22. (a) Ek = ΔEgp
21. (a)
(b) The longest spring is the only one compressed for the
first 30 cm of compression.
20 N
F
k = gradient = =
x 0.30 m
–1
= 67 N m
(c) When the springs are compressed between 30 cm and
40 cm, the longest spring and the middle spring are
compressed. Let the spring constants be kl and km
respectively.
F = kl x + km x
⇒ F = (kl + km) x
gradient = kl + km
10 N
=
0.10 m
kl + km = 100 N m−1
When springs are compressed between 40 cm and
50 cm, all three springs are compressed. Let the
spring constant of the shortest spring be ks.
F = kl + km x + ks x
⇒ F = (kl + km + ks) x
gradient = kl + km + ks
20 N
=
0.10 m
⇒ kl + km + ks = 200 N m−1
but kl + km = 100 N m−1
⇒ ks + 100 = 200
⇒ ks = 100 N m−1
18. (a) No. The total kinetic energy of the ball and the
ground is less after the collision than it was before
the collision.
(b) Sound, along with some heating of the ball, provide
evidence that some of the ball’s initial kinetic energy
is transformed.
(c) Yes, assuming that the ball−ground system is an
isolated system
19. (a) Noise, heating and any damage to the car indicate
that some of the initial kinetic energy is transformed.
(b) Yes. On an icy road, friction is small enough for the
system of the two cars to be considered isolated.
Jacaranda Physics 2 TSK
= 60 × 10 × 30
= 1.8 × 104 J
(b) At the instant that her head touches the water her
kinetic energy is zero. Therefore, all of the
gravitational potential energy lost has been
transferred as strain energy in the cord.
Strain energy = mgΔh
= 60 × 10 × (50 − 1.70)
= 28 980 J
⇒ strain energy = 2.9 × 104 J
23. (a) Assign north as positive.
pi = 0.200 × 2.0 + 0
= 0.400 kg m s−1
pf = 0.200vw + 0.200 × 1.7
pf = pi
⇒ 0.200vw + 0.34 = 0.400
⇒ vw = 0.30 m s−1
= 0.30 m s−1 north
(b) Eki =
1
2
× 0.200 × (2.0)2 + 0
= 0.40 J
Ekf =
1
2
× 0.200 × (1.7) 2 +
1
2
× 0.200 × (0.30) 2
= 0.298 J
0.298 100%
×
0.40
1
= 75%
Percentage returned =
(c) (i) pi = 0.200 × 2.0
= 0.40 kg m s−1
8
© John Wiley & Sons Australia, Ltd 2009
pf = 0.200 × −0.5 + 0.200 × 2.5
= 0.40 kg m s−1
⇒ momentum is conserved.
(ii) The player’s claim suggests that the total kinetic
energy of balls after the collision is greater than
the total kinetic energy of the balls before the
collision. This is not possible unless energy is
transferred to the system other than by the cue.
Eki = 12 × 0.200 × (2.0)2
Chapter 3
Projectile and
circular motion
1.
= 0.40 J
Ekf = 12 × 0.200 × (0.5) 2 + 12 × 0.200 × (2.5)2
= 0.65 J
24.
2. When a basketball falls from rest, there is initially no air
resistance. As it accelerates downwards due to the Earth’s
gravitational pull, the air resistance increases. The
magnitude of the net force, and subsequently its
acceleration, decreases. The air resistance continues to
increase as the acceleration continues downwards.
Because the air resistance is small compared to the
basketball’s weight, the basketball will not reach its
terminal velocity (unless dropped from an aeroplane or
helicopter in flight!).
3. (a)
pi = pf
⇒ m1u1 − m2u2 = − m1v1 + m2 v2
⇒ m1u1 + m1v1 = m2 v2 + m2u2
⇒ m1 (u1 + v1 ) = m2 (v2 + u2 )
⇒
⇒
⇒
⇒
m1 v2 + u2
=
m2 u1 + v1
Eki = Ekf
1
+ 2 m2u22 = 12 m1v12 + 12 m2 v22
m1u12 − m1v12 = m2 v22 − m2u22
m1 (u12 − v12 ) = m2 (v22 − u22 )
m
v2 − u 2
⇒ 1 = 22 22
m2
u1 − v1
1
2
(1)
m1u12
The velocity of the projectile at any point is found by
the gradient of the x vs t graph.
(b)
(2)
Combining (1) and (2)
v2 + u2 v22 − u22
=
u1 + v1 u12 − v12
⇒
(v − u2 )(v2 + u2 )
v2 + u2
= 2
(u1 − v1 )(u1 + v1 )
u1 + v1
The acceleration of the projectile at any point is
found by the gradient of the v vs t graph.
⇒ u1 − v1 = v2 − u2
⇒ u2 − u1 = v2 − v1
(c)
This graph shows the velocity is constant at
−9.8 m s−1. The negative sign indicates the
acceleration is towards the Earth.
Jacaranda Physics 2 TSK
9
© John Wiley & Sons Australia, Ltd 2009
4. (a) Vertical component:
vv = 20 sin 50° = 15 m s−1
Horizontal component:
vh = 20 cos 50° = 13 m s−1
(b) Vertical component:
vv = 11 cos 23° = 10 m s−1
Horizontal component:
vh = 11 sin 23° = 4.3 m s−1
(c) Vertical component:
vv = 5 m s−1
Horizontal component:
vh = 5 sin 0° = 0 m s−1
(d) Vertical component:
vv = 10 sin 0° = 0 km h−1
Horizontal component:
vh = 10 km h−1
(e) Vertical component:
vv = 33 cos 60° or 33 sin 30° = 17 m s−1
Horizontal component:
vh = 33 sin 60° or 33 cos 30° = 29 m s−1
x = ut + 12 at 2
150 = 0t + 12 (10)t 2
150 = 5t 2
30 = t 2
⇒ t = 30 = 5.5 s
(take the positive square root)
(b) v = ?
v 2 = u 2 + 2ax
v 2 = 02 + 2(10) (150)
v 2 = 3000
v = 3000 = 55 m s −1
9. u = 18 m s−1
a = −10 m s−2 (consider up as positive)
(a) Method 1: (consider whole motion)
v = −18 m s−1 (due to symmetry)
t=?
v = u + at
−18 = 18 − 10t
10t = 36
t = 3.6 s
Method 2: (consider half motion to top of flight)
v=0
t=?
v = u + at
0 = 18 − 10t
10t = 18
t = 1.8 s
∴ for whole motion
t = 1.8 × 2
t = 3.6 s
(b) Consider 1st half of motion:
v = 0 m s −1
5. As no forces are acting in the horizontal direction, there
can be no horizontal acceleration. Therefore, the
horizontal component of velocity must remain constant.
6. The time of a projectile’s flight is the time it takes to hit
the ground. Therefore, the projectile cannot take longer to
complete one part of its motion than the other. Time is the
only useful variable that is a scalar and is the same in
both the vertical and horizontal directions.
7. (a)
(i) Air resistance will cause the horizontal
component of the motion to decrease. As the
horizontal component decreases, so will the
horizontal component of air resistance.
(ii) Vertical air resistance will increase as the parcel
speeds up due to gravity. Its acceleration will
decrease while the vertical component of air
resistance continues to increase until it reaches
the magnitude of the parcel’s weight. The
vertical component of velocity will then remain
constant until the parcel reaches the ground.
(b) Air resistance increases as the speed of the parcel
increases.
(i) The horizontal component.
During the first 2 seconds of its fall, the
horizontal speed is greater than the vertical
speed. The vertical speed is less than 20 m s−1
for at least the first 2 seconds.
(ii) The vertical component.
The horizontal component of speed is always
20 m s−1 or less. After about 2 seconds, the
vertical component is greater.
x=?
2
v = u 2 + 2ax
02 = 182 + 2(−10) x
20 x = 182
x=
324
= 16 m
20
(c)
10. m = 500 kg
x = 10 m
a = 10 m s−2
u = 0 m s−1
8. x = 150 m (consider down as positive)
u = 0 m s−1
a = 10 m s−2 (due to gravity)
Jacaranda Physics 2 TSK
t =?
(a)
10
© John Wiley & Sons Australia, Ltd 2009
(consider down as positive)
(a)
v=?
v2 = u2 + 2ax
v2 = 02 + 2(10)(10)
v2 = 200
v = 14 m s−1
(b)
t=?
11. (a) The car stopped moving because the brakes caused a
force to be exerted on the car which opposed its
motion. However, no such force was exerted on the
tissue box, so it continued to move in a straight line
with constant speed as stated in Newton’s first law.
−1
(b) u = 100 km h
(as car is travelling at the same speed)
x = ut + 12 at 2
1000
m s −1
3600
= 28 m s −1
= 100 ×
10 = 0t + 12 (10)t 2
10 = 5t 2
(c)
10
=2
5
t = 1.4 s
(c) horizontally:
u = 0.5 m s−1
a = 0 m s−2 (no forces in this direction)
t = 1.4 s (linking factor between horizontal and
vertical components)
x=?
t2 =
Vertical
Horizontal
u = 28 m s−1
u=0
−2
a = 10 m s
x = 2.5 m s−1
x=?
Must find time, as it is the linking factor. Using
horizontal component:
x
t=
u
2.5
t=
28
x = ut + 12 at 2
t = 8.9 × 10−2 s
(or 9 × 10−2 s if used exact value of u)
Using this with vertical information:
x = ut + 12 at 2
x = 0.5 (1.4) + 12 (0) (1.4)2
x = 0.7 m
(d)
x = 0 (8.9 × 10−2 ) + 12 (10) (8.9 × 10−2 ) 2
x = 0.040 m (or 0.041 m if using t = 9 × 10−2 s).
(d) During sudden accelerations, objects could become
projectiles, moving through the interior of the car,
which could injure the occupants of the car.
12.
Horizontal
u = 7.0 sin 45°
= 4.9 m s−1
a = −10 m s−2
v = −4.9 m s−1
(due to symmetry)
t=?
u = 7.0 cos 45°
= 4.9 m s−1
t = 0.98 s
(from vertical)
x=?
x = ut
x = 4.9 × 0.98
x = 4.8 m
v = u + at
(or 4.9 if maintaining more than
2 sig. fig. through working)
−4.9 = 4.9 − 10t
Set up means you will be at a
10t = 9.8
distance of
t = 0.98 s
0.5 × 11 = 5.5 m ⇒ your friend
will not make the jump and you
will be squashed!
13. Let v = initial velocity and θ be the angle of projection of
the javelin.
Then range = v cos θ × t.
The time is given by the vertical components,
vv = −v sin θ, uv = v sin θ, a = − g.
v 2 = vv2 + vh2
v 2 = 142 + 0.52
v = 142 + 0.52
v = 14 m s −1
14
tan θ =
0.5
⇒ θ = 88°
(e) (i) While attached to magnet, a = 0 m s−2.
⇒ Fnet = ma
Fnet = 500 × 0
Fnet = 0 N
(ii) While falling, a = 10 m s−2 downwards.
⇒ Fnet = ma
Fnet = 500 × 10
Fnet = 5000 N downwards
Jacaranda Physics 2 TSK
Vertical
11
© John Wiley & Sons Australia, Ltd 2009
v = u + at
−v sin θ = v sin θ − gt
t=
= 6.9 m s−1
x = ut
x = 6.9 × 0.69
x = 4.8 m
(c) Horizontal component:
xtop to goal = 7.0 m − 4.8 m
= 2.2 m
u = 6.9 m s−1
t=?
x
t=
u
2.2
t=
6.9
t = 0.32 s
(d) Vertical component:
u = 0 m s−1
t = 0.32 s
a = 10 m s−2
x=?
x = ut + 12 at 2
2 v sin θ
g
Now range = v cos θ ×
2 v sin θ
g
2 v2
sin θ cos θ
g
but 2 sin θ cos θ = sin 2θ so
=
v2
sin 2θ
g
v and g are constant and sin 2θ is a maximum when
2θ = 90°. So the maximum range is given when
θ = 45°.
14. (a) x = 2 m
t = 0.42 s
v=?
x
2.0
= 4.8 m s –1 (4.76)
v= =
t 0.42
(b)
range =
x = 0 (0.32) + 12 (10) (0.32)2
x = 0.51 m
⇒ Final height = 2.4 − 0.51
= 1.9 m
Therefore the ball goes into the net.
16. (a) Vertical component:
u = 50 sin 35° km h−1
1000
= 29 ×
= 8.0 m s −1
3600
a = −10 m s−2
v = 0 m s−1
t=?
v = u + at
0 = 8.0 − 10t
t = 0.80 s
(b) Vertically:
x=?
4.76
5
⇒ θ = 18°
(c) Vertical: (consider up as positive)
u = 5 sin 18° = 1.545 m s−1
a = −10 m s−2
v = 0 m s−1
x=?
v2 = u2 + 2ax
0 = (1.545)2 + 2(−10)x
x = 0.12 m
15. (a) Vertical component:
u = 9.8 sin 45°
= 6.9 m s−1
a = −10 m s−1
v = 0 m s−1 (at top of flight)
t=?
v = u + at
0 = 6.9 − 10t
10t = 6.9
t = 0.69 s
x =?
(b) Vertically:
v2 = u2 + 2ax
02 = 6.92 + 2( −10) x
20x = 6.92
x = 2.4 m
Horizontally:
x=?
u = 9.8 cos 45°
cos θ =
Jacaranda Physics 2 TSK
v 2 = u 2 + 2ax
02 = 8.02 + 2(–10) x
20 x = 8.02
x = 3.2 m
Horizontally: x = ?
u = 50 cos 35° ×
1000
3600
= 11 m s −1
t = 0.80 s
x = ut
x = 11 × 0.80
x = 8.8 m
(equals 9.1 m if use more than 2 sig. fig. in
calculations)
12
© John Wiley & Sons Australia, Ltd 2009
v = u + at
= 8.2 m s−1
Final velocity:
(c) Vertical component:
x = 3.2 + 0.8 = 4.0 m
a = 10 m s –2
u = 0 m s –1
t=?
x = ut + 12 at 2
4.0 = 0t + 12 (10)t 2
v = (12) 2 + (8.2)2
4.0 = 5t 2
= 15 m s −1
t = 0.89 s
(0.90 s if use more than 2 sig. fig. for x)
(d) Horizontal component:
u = 11 m s−1
ttotal = 0.80 + 0.89
= 1.7 s
x =?
x = ut
x = 11 × 1.7
x = 19 m
17. Making use of the hint provided.
Before highest point:
Initial speed = 13.9 m s−1 (50 km h−1)
Vertically, u = 13.9 sin 30°
= 6.95 m s−1
v=0
8.2
12
⇒ θ = 34°
Final velocity = 15 m s−1 at 34° to the horizontal.
tan θ =
18.
Horizontal
u = u sin 28°
a = −10 m s
t=
x = 2.5 m
u = u cos 28°
t=?
–2
2.5
2u cos 28°
t=
–1
x
2.5
=
u 2u cos 28°
v=0ms
Apply the equation v = u + at to the vertical component of
the first half of the gymnast’s motion.
⎛
⎞
2.5
0 = u sin 28° + (−10) ⎜
⎟
⎝ 2u cos 28° ⎠
a = −10
v = u + at
⇒ 0 = 6.95 −10t
⇒ t = 0.695 s
Vertically, v2 = u2 + 2ax
where x = maximum height above ramp edge
⇒ 0 = 48.3 −20x
⇒ x = 2.4 m
After highest point:
Vertically, x = 2.4 m + 1.7 m
= 4.1 m
u=0
a = 10
x = ut + 12 at2
⎛
⎞
25
0 = u sin 28° − ⎜
⎟
⎝ 2u cos 28° ⎠
⎛
⎞
25
⎜
⎟ = u sin 28°
⎝ 2u cos 28° ⎠
25 = 2u 2 sin 28° cos 28°
u=
12.5
sin 28° cos 28°
u = 5.5 m s –1
19.
2
⇒ 4.1 = 5t
⇒ t = 0.82
(a) Range = ut
where u = horizontal velocity (constant) and
t = 0.695 + 0.82
= 1.5 s
⇒ Range = 13.9 cos 30° × 1.5
= 18 m
(b) Final horizontal velocity = 13.9 cos 30°
= 12 m s−1
Vertically, (use downward flight only):
u=0
a = 10
t = 0.82
Jacaranda Physics 2 TSK
Vertical
Vertical
Horizontal
u = 7 sin θ m s –1
x = 3.0 m
a = −10 m s –2
u = 7 cos θ m s –1
t =?
t=
3.0
14 cos θ
t=
x
3.0
=
u 7 cos θ
v = 0 m s –1
where θ = angle to horizontal
v = u + at
⎛ 3.0 ⎞
0 = 7 sin θ + (−10) ⎜
⎟
⎝ 14 cos θ ⎠
30
0 = 7 sin θ −
14 cos θ
13
© John Wiley & Sons Australia, Ltd 2009
m = 35 kg
a = 0.05 m s−2
Fnet = Fc = ma
= 35 × 0.05
= 1.75 N
towards the centre of the circle
(1.7 N if using more than 2 sig. fig. for a)
(c) m = 1500 kg
a = 0.05 m s−2
Fnet = Fc = ma
= 1500 × 0.05
= 75 N
towards the centre of the circle
(d) To move along the same path, the child and the train
require the same acceleration. As the mass of the
child and the train are different, different forces are
needed to produce identical accelerations.
23. (a) r = 65 cm = 0.65 m
m = 0.12 kg
(b)
30
= 7 sin θ
14 cos θ
30
= sin θ cos θ
14 × 7
0.31 = sin θ cos θ
⇒ 2 sin θ cos θ = 0.62
⇒ sin 2θ = 0.62
θ = 19°
20. (a) r = 120 m
v = 6.0 km h−1
= 1.7 m s−1
a=?
v 2 (1.7) 2
a=
=
r
120
(0.024)
a = 0.02 m s−2
towards the centre of the circle
(b) a = 0.024 m s−2
m = 65 kg
F=?
Fc = Fnet = ma
= 65 × 0.024
= 1.6 N
towards the centre of the circle
21.
T = 5.2 s
a=?
4π 2 r
a= 2
T
4π 2 (0.65)
=
(5.2)2
= 0.95 m s –2
towards the centre of the circle
(b) a = 0.95 m s−2
m = 0.120 kg
F=?
Fnet = Fc = ma
= 0.12 × 0.95
= 0.11 N
towards the centre of the circle
(c)
T = 35 s
rN = 2.5 m
rL = 3.2 m
aN =
=
4π 2 r
T2
4π 2 (2.5)
(35) 2
= 0.08 m s –2
aL =
=
4π 2 r
T2
4π 2 (3.2)
(35) 2
= 0.10 m s –2
∴ Lucy experiences the greatest centripetal acceleration.
22. r = 350 m
v = 15 km h −1 ×
1000
3600
= 4.2 m s –1
v 2 (4.2) 2
=
= 0.050 m s –2
350
r
towards the centre of the circle
(a) a =
Jacaranda Physics 2 TSK
14
© John Wiley & Sons Australia, Ltd 2009
24. Newton’s first law states that an object will continue to
move in a straight line with constant speed unless an
unbalanced force acts on it. Therefore, the mass will
continue to move forwards without a propelling force,
once in motion. The centripetal force acts to change the
direction of the mass, not its speed.
27.
r = 2.0 m
T = 12 N
(a) Fc = T cos 10°
= 12 N (11.8 N)
(b) m = 0.200 kg
mv 2
= Fc = 11.8
r
0.200 v 2
⇒
= 11.8
2.0
25. To go around a bend, a motorcyclist needs a horizontal
force acting on the bike towards the centre of the curve of
the bend. This is provided by the road acting on the tyres.
The force of the road on the tyres needs to act through the
centre of mass of the cyclist, otherwise the force will act
to tip the bike over. As the horizontal component of this
force is acting towards the centre of the curve, the
motorcyclist must lean into the curve to avoid falling off.
⇒v =
11.8 × 2.0
0.200
= 11 m s −1 (10.9 m s −1 )
26.
(c)
m = 0.0500 kg T = 0.800 s
r = 1.50 cos 6.03° m
=
T2
⎛ v2 ⎞
⎜⎜ or ⎟⎟
r ⎠
⎝
r = 4.5 m
mv 2
r
(90) (4.2)2
=
4.5
= 3.5 × 102 N
toward the centre of the circle
(b) 3.5 × 102 N toward the centre of the circle (as
frictional forces are causing centripetal motion)
(c)
v = 4.2 m s −1
F = 350 × 90%
= 315 N
r =?
F=
2
4π × 1.50 cos 6.03°
(0.800) 2
= 92.0 m s −2
(c) Fc =
=
towards the centre of the circle
4π 2 r
T2
4 × 0.0500 × π 2 × 1.50 cos 6.03°
(0.800) 2
= 4.60 N towards the centre of the circle
(d) = Fnet
= ma
mv 2
r
90 (4.2) 2
315 =
r
90 (4.2)2
r=
315
r = 5.0 m
The radius will increase to 5.0 m.
= 0.0500 × 92.0
= 4.60 N towards the centre of the circle
F=
Fc
cos 6.03°
4.60
=
cos 6.03°
= 4.63 N
(e) T =
Jacaranda Physics 2 TSK
T2
0.200 × 4π 2 × 2.0
⇒11.8 =
T2
0.200 × 4π 2 × 2.0
11.8
= 1.2 s
28. (a) m = 90 kg
1000
v = 15 km h −1 ×
3600
−1
= 4.2 m s
2πr
T
2π × 1.50 cos 6.03°
=
0.800
= 11.7 m s −1
4π 2 r
m4 π 2 r
⇒T =
(a) v =
(b) a =
Fc =
15
© John Wiley & Sons Australia, Ltd 2009
31. (a)
29.
(b)
mv 2
r
= 360 + R sin 20°
Fnet =
Fnet
mv 2
= 360 + R sin 20°
r
but R cos 20° = mg
10m
⇒R =
cos 20°
mv 2
10m sin 20°
⇒
= 360 +
r
cos 20°
= 360 + 10 m tan 20°
⇒
(i) Fc = Fnet = W − N
As N = 0:
Fc = W = mg
Fc = 800 × 10
Fc = 8 × 103 N downwards
mv 2
Fc =
(ii)
r
Fc r
2
⇒v =
m
v=
8 × 103 × 4.0
800
v = 6.3 m s −1
32. (a) Using KEB = PEA:
1 2
v = g Δh
2
m (9.0) 2
= 360 + 3.64 m
10
⇒ 8.1m = 360 + 3.64 m
⇒ 4.46 m = 360
⇒ m = 81 kg
The mass of the bicycle is 20 kg.
∴ The mass of the cyclist is 61 kg.
30. v = 30 m s−1, r = 12 m
v2
a=
r
302
=
12
= 75 m s −2
The horizontal component of the normal reaction force
must therefore provide a centripetal acceleration of
75 m s−2.
N sin θ = 75 m
v 2 = 2 g Δh
⇒
v 2 = 2 × 10 × 10
v = 4.5 m s −1
mv 2
r
65 × 4.52
Fc =
4.0
(b) Fc =
Fc = 3.3 × 102 N upwards
(c) Fnet = Fc = T − W
W = mg = 65 × 10 = 650 N
T = Fc + W = 3.3 × 102 + 65 × 102
T = 9.8 × 102 N
N cos θ = mg
75
So by dividing, tan θ =
9
⇒ θ = 82°
The problem with this road is that you would have to
drive at 30 m s−1 to avoid slipping off!
But
Jacaranda Physics 2 TSK
16
© John Wiley & Sons Australia, Ltd 2009
33. (a) m = 10 kg
g = 9.70 N kg−1
W=?
W = mg
= 10 × 9.70
= 97 N downwards
(c)
rVenus = 6.05 × 106 m
(d)
35.
m =?
G = 6.67 × 10−11 N m 2 kg −2
kg
mperson = 70 kg
M =
G = 6.67 × 10−11 N m 2 kg −2
g=?
W =?
36.
GM
r2
6.67 × 10−11 × 5.98 × 1024
=
(6.38 × 106 )2
M sun = 1.98 × 1030 kg
rEarth’s orbit = 1.50 × 1011 m
G = 6.67 × 10−11 N m 2 kg −2
F =?
−1
F =
2
= 6.9 × 10 N
=
M Mars = 6.42 × 1023 kg
rMars = 3.40 × 106 m
GMm
r2
6.67 × 10−11 × 1.98 × 1030 × 5.98 × 1024
(1.50 × 1011 ) 2
= 3.5 × 1022 N
GM
37. r = rEarth + h
= 6.38 × 106 + 3.55 × 105 m
= 6.74 × 106 m
T = 92 min
= 5520 s
r2
g = 3.70 N kg −1
W = mg
W = 2.6 × 102 N
Jacaranda Physics 2 TSK
6.67 × 10−11
M Earth = 5.98 × 1024 kg
W = mg = 70 × 9.80
(ii)
4.3(5.0 × 105 ) 2
M = 1.6 × 1022 kg
g=
g=
GM
r2
gr 2
⇒M =
G
rEarth = 6.38 × 10 m
(i)
r = 5.0 × 105 m
g = 4.3 N kg −1
6
(b)
W = mg
W = 5 × 101 N
g=
= 9.80 N kg
GM
r2
g = 0.7 N kg −1
(ii)
r = 6.41 × 10 m
(ii)
g=
(i)
6
(i)
M Pluto = 1 × 1022 kg
rPluto = 1 × 106 m
6.67 × 10−11 × 5.98 × 1024
9.70
M Earth = 5.98 × 10
W = mg
W = 6.2 × 102 N
g = 9.70 N kg−1
r=?
MEarth = 5.98 × 1024 kg
G = 6.67 × 10−11 N m2 kg−2
GM
g= 2
r
GM
r=
g
24
GM
r2
g = 8.87 N kg −1
(ii)
(c)
34. (a)
g=
(i)
(b) Consider up to be positive.
Fnet = W + Fup
⇒ 0 = W + Fup
⇒ Fup = −W
= −(−97 N)
= 97 N
r=
M Venus = 4.87 × 1024 kg
17
© John Wiley & Sons Australia, Ltd 2009
(a) a = ?
a=
=
39. One possible answer.
4π 2 r
T2
4π 2 (6.74 × 106 )
(5520)2
= 8.73 m s −2
(b) g = ?
GM
g= 2
r
6.67 × 10−11 × 5.98 × 1024
=
(6.74 × 106 ) 2
a = 10 m s −2
r = 200 m
= 8.78 N kg −1
(c) (i) The centripetal acceleration of the space station
is caused by acceleration due to gravity (i.e. g).
Therefore, the two answers should be the same.
(ii) Discrepancies between the numbers are due to
the rounding off of data.
(d) Mss = 1200 tonnes
= 1.2 × 106 kg
g = 8.78 N kg−1
W=?
W = mg
= 1.2 × 106 × 8.78
= 1.1 × 107 N
(e)
r = 10 m
M astro = 270 kg
T =?
a=
T=
F =?
GMm
F= 2
r
6.67 × 10−11 × 1.2 × 106 × 270
=
(10) 2
TVenus = 1.94 × 107 s
TSaturn = 9.30 × 108 s
r3
= 2.2 × 10−4 N
T = 24 h
T
= 8.64 × 104 s
2
=
⇒r =
r = 6.38 × 106 m
a=?
=
4π 2 r
a
4π 2 (200)
10
= 28 s
On Earth, the ground pushes up on us. In the space
station, the outer wall pushes the person in (centripetal
force). This means that the person would feel as though
the wall is the ground, and the direction opposite to the
centripetal force is down.
40. Gravity holds solar systems together. As gravity is the
only force acting on a planet, it is the net force. It is
directed toward the sun. Whenever the net force on an
object is toward the centre of a circle, the object will
experience centripetal motion.
41.
M sun = 1.98 × 1030 kg
G = 6.67 × 10−11 N m 2 kg −2
a=
T2
=
M ss = 1.2 × 106 kg
38.
4π 2 r
GM
4π 2
3
GMT 2
4π 2
2
2
⎛T
⎞ 3 ⎛ 9.30 × 108 ⎞ 3
r
∴ Saturn = ⎜ Saturn ⎟ = ⎜⎜
7 ⎟
⎟
rVenus ⎝ TVenus ⎠
⎝ 1.94 × 10 ⎠
= 13
4π 2 r
T2
4π 2 (6.38 × 106 )
42.
(8.64 × 104 )2
= 0.034 m s −2
In Victoria, the radius of the circular path would be
smaller. If r decreases, acceleration also decreases.
Therefore, in Victoria, we experience less centripetal
acceleration due to the Earth’s motion than people on the
Equator.
Jacaranda Physics 2 TSK
18
© John Wiley & Sons Australia, Ltd 2009
M Moon = 7.35 × 1022 kg
24
M Earth = 5.98 × 10
45. A geostationary satellite cannot remain above Melbourne
as the net force on the satellite (the gravitational force) is
towards the centre of the Earth, not towards the centre of
the cirle mapped out by Melbourne as it rotates.
kg
rfrom Earth = x
rfrom Moon = 3.3 × 108 − x
F
:
m
Area under a g vs x graph has the same units as
F
W
x=
(i.e. energy per kg).
m
m
46. As W = Fx, and g =
FMoon = FEarth
GM Moon m
rm2
7.35 × 1022
(3.3 × 108 − x)2
=
=
GM Earth m
rE2
5.98 × 1024
47. (a) Work/kg = area under graph
x2
from 400 km to 600 km
≈ 12 (8.7 + 8.2) 2 × 105
(distance in metres)
= 1.7 × 106 J kg−1
⇒ Work = 1.7 × 106 × 800
= 1.4 × 109 J (approx.)
(b)
r = 6.0 × 105 + 6.38 × 106
(7.35 × 1922 ) x2 = (5.98 × 1024 ) (3.38 × 108 − x)2
(7.35 × 1022 ) x2 = (5.98 × 1024 ) ((3.38 × 108 )2
− 2(3.38 × 108 ) x + x2 )
x2 = 81.36 (1.14 × 1017 −
(6.76 × 108 ) x + x 2 )
= 6.98 × 106 m
0 = 80.36 x2 − (5.5 × 1010 ) x + 9.28 × 1018
x=
M Earth = 5.98 × 1024 kg
2
−b ± b − 4ac
2a
T =?
r3
5.5 × 1010 ± (5.5 × 1010 )2 − 4(80.36) (9.28 × 1018 )
=
2(80.36)
T
x = 3.02 × 108 m is correct because the spacecraft must be
between the Earth and the Moon.
T=
43. According to Newton’s first law, an object will move in a
straight line, with constant speed, unless an unbalanced
force is acting on it. As gravity acts at right angles to
the satellite’s velocity, it does not change the speed of the
satellite; rather it changes its direction. This causes the
satellite to move around the Earth. Because the force of
gravity and the speed of the satellite remain constant, so
mv 2
must its radius as r =
; therefore it cannot move
F
closer to the Earth.
48. (a)
T
r=
3
GMm
, G = 6.67 × 10−11 N m 2 kg −2 ,
r2
= 5.98 × 1024 kg, M satellite = 2400 kg
F=
F=
6.67 × 10−11 × 5.98 × 1024 × 2400
(8.38 × 106 ) 2
= 1.4 × 104 N
Answers will vary if F = mg is used, taking the value
of g from the graph.
(b) Loss of GPE = area under graph from 8.38 × 106 m to
7.18 × 106 m.
This is approximately a trapezium.
A = 12 (a + b) h
GMT 2
4π 2
6.67 × 10−11 × 5.98 × 1024 × (8.68 × 104 )2
= 12 (5.25 + 6.75) × 1.2 × 106
4π 2
= 1.7 × 1010 J.
r = 4.2 × 107 m
Jacaranda Physics 2 TSK
6.67 × 10−11 × 5.98 × 1024
= 8.38 × 106
4π 2
3
(6.98 × 106 )3 4π 2
r = 6.38 × 106 + 2 × 106
GM
r=
r 3 4π 2
GM
r3
M Earth
= 8.64 × 104 s
r=?
=
4π 2
= constant for any satellite of the Earth.
T2
(d) (a) would be halved and (b) would remain the same.
(c)
T = 24 h
2
GM
T = 5.8 × 103 s
M Earth = 5.98 × 1024 kg
r3
=
T=
= 3.83 × 108 , 3.02 × 108
44.
2
19
© John Wiley & Sons Australia, Ltd 2009
(c) It has gained 1.7 × 1010 J of kinetic energy.
EK = 12 mv 2
Initially
50. (a) r = rEarth + h
= 6.38 × 106 + 3.60 × 105
= 6.74 × 106 m
GM
g= 2
r
6.67 × 10−11 × 5.98 × 1024
=
(6.74 × 106 )2
= 12 × 2400 × 69002
= 5.7 × 1010
After falling EK = 5.7 × 1010 + 1.7 × 1010
= 7.4 × 1010 J
1
2
1
mv 2
2
2
= 8.78 N kg −1
= 7.4 × 1010
(b) W = mg
= 70 × 8.78
= 615 N
(c) Zero, because both the space shuttle and the astronaut
are in orbit around the Earth and have the same
acceleration (centripetal acceleration). There is no
normal reaction acting on the astronaut.
× 2400 × v = 7.4 × 1010
v = 7.9 × 103 m s −1
49. Weightlessness can only be experienced by an object if it
has no mass or is in a zero gravitational field (i.e. g = 0)
since W = mg.
Apparent weightlessness is experienced by an object if it
is in free fall, that is, there is no normal reaction force
acting on it. Any object in orbit around the Earth is in free
fall and experiences apparent weightlessness. But g ≠ 0.
Jacaranda Physics 2 TSK
20
© John Wiley & Sons Australia, Ltd 2009
Unit
Area of study 2:
Electronics and photonics
3
Jacaranda Physics 2, 3rd Edition TSK
Part A — WORKED SOLUTIONS
■■■■■■■■■
V =
Electronics
1. (i) (a)
1
=
Reff
=
=
1
30
+
=
1
20
(d)
Reff = 12 Ω
Reff = R1 + R2
= 12 + 12
= 24 Ω
(b) Reff = 6 + 12
= 18 Ω
1
1
1
=
+
(c)
Reff
30 + 10 40
=
=
(iv) (a)
2
40
1
20
(b)
Reff = 20 Ω
(d) Reff = 5 + 12 + 3
= 20 Ω
V
(ii) (a) I =
R
= 12
24
(d)
6Ω
0.50 A
3Ω
V
I=
R
= 93
= 0.25 A
V
(d) I =
R
60
= 20
(b)
Jacaranda Physics 2 TSK
=
2. (a)
30
× 5 V = 5.0 V
30 + 10
15
4
= 3.7 V
3Ω
10 Ω
V = IR
V = IR
=3×3
=3×5
= 9.0 V
= 15 V
20 Ω
30 Ω
V = 60 − 15 − 9
V = 36 V
= 36 V
12 Ω
20 Ω
0.50 A
V
I =
R
6
= 20
= 0.50 A
V
(c) I =
R
5
= 20
(iii) (a)
V =
5
4
10 Ω
I = 0.13 A
(c)
= 0.50 A
V
(b) I =
R
= 189
= 3A
12 Ω
V = IR
= 0.5 × 12
= 6.0 V
6Ω
V = IR
= 0.5 × 6
= 3.0 V
10
×5
30 + 10
= 1.3 V
5
60
1
12
40 Ω
30 Ω
10 Ω
(c)
Chapter 4
Vout
30 Ω
V
I=
R
6
= 30
= 0.30 A
20 Ω
V
I=
R
6
= 20
= 0.20 A
30 Ω
V
I=
R
6
= 30
= 0.30 A
30 Ω
I = 0.13 A
= 0.20 A
40 Ω
V
I=
R
5
= 40
10 Ω
V
I=
R
15
= 10
20 Ω
V
I=
R
= 36
20
= 3.0 A
= 1.5 A
R2
=
Vin
R1 + R2
= 1.8 A
= 0.13 A
30 Ω
V
I=
R
= 36
30
= 1.2 A
2.2
× 6.0
2.2 + 2.2
= 3.0 V
1
1
1
=
+
Reff 4.4 4.4
=
30 Ω
V = 12 − 6
= 6.0 V
20 Ω
V = 12 − 6
= 6.0 V
20 Ω
V=9−3
= 6.0 V
30 Ω
V=9−3
= 6.0 V
(b)
2
4.4
1
=
2.2
= 2.2 kΩ
=
Reff
so Vout = 3.0 V
21
© John Wiley & Sons Australia, Ltd 2009
Vout 2.0
=
=
Vin
6.0
3.
so
⇒
R2
=
R1 + R2
1
3
1
3
R2
=1
3.0 + R2 3
⇒ 3R2 = 3.0 + R2
2R2 = 3.0
3.0
R2 =
= 1.5 kΩ
2
(b)
4. (a)
(c)
R2 Vin
R1 + R2
Vin = 10 V
R1 = 60 Ω
R2 = 40 Ω
40 Ω × 10V
Vout =
(60 Ω + 40 Ω)
= 4.0 V
R2 Vin
(b) Vout =
R1 + R2 + R3
Vout =
(d)
7. (a)
(b)
Vin = 9 V
R1 = 20 Ω
R2 = 30 Ω
R3 = 40 Ω
30 Ω × 9V
⇒ Vout =
90 Ω
= 3.0 V
4 R × 20 V
(c) Vout =
R + 4R
8. (a)
Vout =
R2 Vin
R1 + R2
2×9V
4+2
= 3.0 V
(b) If the temperature is 80°C, R2 = 1 kΩ.
R1 = ?
Vin = 9 V
Vout = 6.0 V
1× 9 V
⇒ 6.0 V =
R1 + 1
⇒ 6R1 + 6 = 9
⇒ 6R1 = 3
⇒ R1 = 0.5 kΩ
= 500 Ω
9. (a) 70°C from graph
(b) 8.0 kΩ from graph
=
⎛ 4R ⎞
=⎜
⎟ 20 V
⎝ 5R ⎠
= 0.8 × 20 V
= 16 V
5. As the resistance decreases, so, too, does the voltage drop
across the variable resistor.
R V
6. (a)
Vout = 2 in
R1 + R2
Vout = 4.0 V
Vin = 6.0 V
R1 = 3.0 kΩ
Jacaranda Physics 2 TSK
R2 = R
R kΩ × 6.0V
⇒ 4.0 V =
(3.0 + R) kΩ
⇒ 12 + 4R = 6R
⇒ 2R = 12
⇒R=6
R = 6.0 kΩ
R V
Vout = 2 in
R1 + R2
12 R
⇒ 6.0 =
R + 660
⇒ 6R + 6 × 660 = 12R
⇒ 6R + 3960 = 12R
⇒ 6R = 3960
⇒ R = 660 Ω
R V
Vout = 2 in
R1 + R2
10 R
⇒ 2.5 =
10 + R
⇒ 25 + 2.5R = 10R
⇒ 7.5R = 25
⇒ R = 3.3 kΩ
R V
Vout = 2 in
R1 + R2
5×9
⇒6 =
R+5
⇒ 6R + 30 = 45
⇒ 6R = 15
⇒ R = 2.5 kΩ
(i) R = 5.0 kΩ (by reading graph)
(ii) R = 1.0 kΩ (by reading graph)
(i) 25°C (by reading graph)
(ii) 50°C (by reading graph)
If the temperature is 40°C, R2 = 2 kΩ
R1 = 4 kΩ
22
© John Wiley & Sons Australia, Ltd 2009
T = 90°C
Rthermistor = 8.0 kΩ (from (b))
(c)
(c) VR = IR
I = 4.0 × 10−3 A
Vout
Rthermistor
=
Vin
Rvariable + Rthermistor
(d)
R=
3
8.0
=
9 Rvar + 8.0
4.0 × 10−3
= 1337.5 Ω
3Rvar = 48
Rvar = 16 kΩ
(e) The output voltage drops. This is because the
proportion of the resistance in the circuit taken by the
resistor decreases, so the voltage drop decreases.
13.
10. (a) 5000 Ω (from graph)
(b) 200°C (from graph)
14.
(c) T = 200°C, Rthermistor = 400 Ω (from (b))
15.
Rthermistor
=
× Vin
Rthermistor + Rvariable
⇒
Vout
( Rthermistor + Rvariable ) = Rthermistor
Vin
⇒
⎛
Vout
V ⎞
Rvariable = Rthermistor ⎜1 − out ⎟
Vin
Vin ⎠
⎝
16.
= 1.3 × 103 Ω or 1.3 kΩ
(d) −6.0 V as the diode is now reverse biased
(e) I = 0
(f) VR = 0
The distortion is due to clipping. The flat batteries reduce
the maximum value for Vout. In Jasper’s case the
maximum value has fallen below the value produced by
the amplification of a strong radio signal.
Output voltage divided by input voltage.
Gradient of linear section of the transfer characteristic.
The output voltage of an inverting amplifier increases as
the input voltage decreases, as long as the input voltage is
in the linear range.
Characteristic (a)
(i) Gain = gradient of linear section of graph
5 −1
=
1 − (−1)
=
⎛V
⎞
Rvariable = Rthermistor ⎜ in − 1⎟
⎝ Vout
⎠
⎛ 12
⎞
= 400 ⎜ − 1⎟
⎝8
⎠
= 200 Ω
4
2
=2
(ii) Maximum output voltage = 5.0 V (from graph)
Minimum output voltage = 1.0 V (from graph)
(iii) Non-inverting (positive gradient in linear section)
Characteristic (b)
0.5 − 5.5
(i) Gain =
[3.0 − (−3.0)] × 10−3
11. VR = 12 − Vdiode
= 12 − 0.7
= 11.3 V
V
I =
R
11.3
=
1.0 × 103
=
−5.0
6.0 × 10−3
= 830 (take positive value for gain)
(ii) Maximum output voltage = 5.5 V
Minimum output voltage = 0.5 V
(iii) Inverting (negative gradient)
Characteristic (c)
1−5
−4
(i)
=
(4 − 1) × 10−3 3 × 10−3
= 11.3 × 10−3 A
= 11.3 mA
= 1.3 × 103
(take positive value for gain)
(ii) Maximum output voltage = 5.0 V
Minimum output voltage = 1.0 V
(iii) Inverting (negative gradient)
Δ Vout
17. (a) Gain =
Δ Vin
or 1.13 × 10−2 A.
12. (a) Read value from graph.
Answer is approximately 0.65 V.
Accept 0.6 V to 0.7 V.
(b) VR = E − VD
= 6.0 V − 0.65 V
= 5.35 V
Jacaranda Physics 2 TSK
5.35 V
=
3Rvar + 24 = 72
Vout
V
I
8
0.4
= 20
=
23
© John Wiley & Sons Australia, Ltd 2009
(b) Non-inverting. An increase in Vin leads to an increase
in Vout.
(c)
R2
Vin
R1 + R2
(b) Vout =
(i)
=
3000
×6
3000 + 2000
=
3
5
×6
= 3.6 V
(c) As illumination increases, the resistance of the LDR
decreases, resulting in an increase in Vout.
5. (a)
(ii)
Relative light
LDR Current
LDR Voltage
LDR Resistance
intensity IL
(μA)
(V)
(kΩ)
Log RL
Log IL
0.6
350
38.7
111
2.04
−0.22
0.4
220
39.0
177
2.25
−0.40
0.2
90
39.1
434
2.64
−0.70
0.1
40
39.0
975
2.99
−1.00
0.05
20
39.0
1950
3.29
−1.30
(b)
(iii)
(c) log RL = 1.17 log IL + 1.80
RL = IL1.17 × 101.80
RL = 63 IL1.17
(iv)
V
is not constant.
R
6. (a) LEDs are semiconductor diodes designed to emit
light when they conduct a current. They contain a
semiconductor chip with p-type material connected to
the anode by a fine wire and n-type material
connected to the cathode through contact with a
metal base.
(b) A diode is forward biased when the p-type material
— the anode — is connected to a higher voltage than
the cathode. In this situation, current will flow if the
voltage difference is greater than a minimum value,
which is determined by the nature of the
semiconducting material and, hence, the colour. A
diode is reverse biased, and blocks the flow of
current if the anode is connected to a lower voltage
than the cathode.
(c)
(d) An LDR is not ohmic because
Chapter 5
Introductory
photonics
1. A transducer is a device that transforms energy from one
form to another.
2. The resistance (in an LDR) is a function of the intensity
of light shining on the LDR — the more light, the less
resistance.
3. Some cameras contain light meters that use voltage
dividers containing an LDR. The intensity of the light
affects the resistance of the LDR, which changes the
voltage to the light meter.
4. (a) 0.4 lux (from graph)
Jacaranda Physics 2 TSK
7. Indicator lights on electronic equipment — they need to
last for the life of the equipment, be compact and energy
efficient.
24
© John Wiley & Sons Australia, Ltd 2009
Some clock and cash register displays for the same
reasons.
Stop lights in cars.
(b) VR = 12 − 1.7
= 10.3 V
V
R= R
I
10.3
=
0.02
= 515 Ω
8. LEDs emit light according to the choice of impurities in
the semiconductor. All diodes emit some light but the
choice of impurities and the structure of the diode
maximise the light produced.
= 5.2 × 102 Ω
9. Limiting resistors are needed because a forward-biased
diode has a very low resistance. This can result in a
current large enough to destroy the diode.
(c) VR = 24 − 1.7
= 22.3 V
V
R= R
I
22.3
=
0.02
= 1115 Ω
10. (a)
(i) 1.6 V (from graph)
(ii) VR = 6.0 − 1.6
= 4.4 V
V
(iii) R =
I
4.4
=
0.02
= 220 Ω
(b) (i) 6.0 V (reverse-biased diode with 6.0 V supply)
(ii) 0 A
(iii) 0 V (V = IR)
= 1.1 k Ω
(d) VR = 50 − 1.7
= 48.3 V
V
I
48.3
=
0.02
= 2415
R=
11. (a) 1.7 V (or 1.8 V) (from graph)
(b) V = IR
= 0.06 × 500
= 30 V
(c) emf = 30 + 1.7 (or 30 + 1.8)
= 31.7 V (or 31.8 V)
= 2.4 × 103 Ω
15. (a)
12. (a) The diode is reverse biased and therefore nonconducting.
(b) 2.5 V — the voltage must drop across the diode as
the voltage across the resistor is given by V = IR
(= 0).
Resistor has 15 − 2 = 13 V across it.
V
13
I = = 100
= 0.13 A
R
(b) The LED might burn out because the current is too
high.
13. (a) V = 9.0 − 1.8
= 7.2 V
V
(b) R =
I
7.2
=
0.04
= 180 Ω
16. A photodetector converts light signals into electrical
signals.
17. (a) A photodiode is one that is sensitive to light. It can be
used to measure light intensity. If reverse biased, a
photodiode begins to conduct when light shines on it.
Alternatively, it can be used as a source of electric
current when light shines on it.
(b) Phototransistors operate like photodiodes but they
amplify the collector–base current, making them
more sensitive than photodiodes.
(c) Photodiodes have a faster response time.
(d) Phototransistors are more sensitive.
(e) Photodiodes respond more quickly than
phototransistors, increasing the bandwidth of photonic
systems that use them instead of phototransistors.
14. (a) VR = 9.0 − 1.7
= 7.3 V
V
R= R
I
7.3
=
0.02
= 365 Ω
= 3.7 × 102 Ω
Jacaranda Physics 2 TSK
25
© John Wiley & Sons Australia, Ltd 2009
(d) Metal wires have a higher density of charge carriers
(free electrons) near the surface when carrying an
alternating current. This makes the resistance of the
wire greater for alternating current than direct
current, causing loss of signal power.
(e) The skin effect is the result of electromagnetic
effects.
20. (a) Optical intensity modulation occurs when the
intensity of a light source is affected by the varying
electrical signal from a transducer.
(b) The signal from a microphone is a current of changing
magnitude. This signal can be used to modulate the
light emitted from a suitably biased LED.
21. (a) Modulation: changing the intensity of the carrier
wave to replicate the amplitude variation of the sound
signal.
Demodulation: Separating the sound signal from the
modulated carrier wave.
(b) LED.
(c) LDR, photodiode or phototransistor.
18. (a) Phototransistors have a base current that is
proportional to the intensity of the light incident on it,
so the collector current is also proportional to the
intensity of the light. In a normal transistor, the base
current is produced by a voltage divider and a signal
from an external device. Phototransistors usually do
not have a base terminal as transistors do.
(b) The base current in a phototransistor is produced by
light shining on the collector–base junction giving
sufficient energy for some electrons in the depletion
region to escape their atoms and form a small current.
19. (a) Bandwidth is the amount of data that can be
transmitted in a fixed amount of time.
(b) 15 000 Hz — the system bandwidth is equal to the
maximum frequency it can carry
(c) Optical fibres can carry up to 1 terabps (= 1 × 1016
binary digits per second).
Telephone wires can carry up to 100 Mbps (= 1 × 108
binary digits per second).
1016
Optical fibre can carry
= 108 times as much
8
10
data per second as metal wire.
Jacaranda Physics 2 TSK
26
© John Wiley & Sons Australia, Ltd 2009
Unit
3
Detailed study 3.1:
Einstein’s special relativity
Part A — WORKED SOLUTIONS
■■■■■■■■■
14. Question 13 gives an example of where a head-on collision
can make a crash in a car travelling at 50 km h−1 as
dangerous as one travelling at 100 km h−1.
15. (a) The Earth’s direction changes by only about 1° each
day so, from our point of view, it is travelling in a
straight line with constant speed. The acceleration
due to the revolution of the Earth around the Sun is
about 6 × 10−2 m s−2 which is tiny. According to the
principle of relativity, if we are not accelerating, we
cannot feel the movement of the Earth no matter what
our speed is.
(b) Daily rotation about its axis and precession of the
equinoxes (the motion of the Earth’s axis), which has
a period of about 22 000 years
(c) The accelerations are so tiny that they are difficult to
detect. Therefore, for most purposes, the Earth can be
considered to be an inertial reference frame.
16. (a) The motion of the ball when viewed from the deck of
the ship would depend on the speed of the ship.
However, if the ship is an inertial reference frame,
the speed of the reference frame should not affect
objects within that frame.
(b) At the base of the mast. If the ship is stationary the
ball would hit the deck at the bottom of the mast.
According to the principle of relativity, the ball must
fall the same regardless of the speed of the ship.
Δv
17. (a) a =
t
100
=
10
= 10 km h −1s −1
Δv
(b) a =
t
( − 100 − −200)
=
10
= 10 km h −1 s −1
(c) It is invariant because it is independent of the
reference frame in which it is measured.
18. (a) 2.9979 × 108 m s−1
(b) v = 1.00 × 105 + 2.9979 × 108
v = 2.9989 × 108 m s−1
19. Yes, because it would increase the time difference
between the two journeys, making it easier to detect.
20. Parallel to the Earth’s motion:
2L
t=
⎛
v2 ⎞
c ⎜⎜1 − 2 ⎟⎟
c ⎠
⎝
Chapter 6
Relativity before
Einstein
1. An oscillating electric field induces an oscillating
magnetic field, which in turn induces an electric field.
These fields move at c. No medium is required for the
electromagnetic wave.
2. Maxwell’s theory of electromagnetic waves produced a
model consistent with the properties of light. Young’s
double-slit experiment; the speed of light is slower in
water than in air; light is polarised.
3. When Maxwell’s equations are solved for the velocity of
an electromagnetic wave, the answer is a constant, equal
to 3 × 108 m s−1.
Apparent wind is the effect of air on an observer moving
through air. It is the result of the motion of the air relative
to the Earth (wind) plus the motion of the observer
relative to the Earth.
4. (a) 6.24 p.m. and 30 seconds
(b) d = vt
= 9.9 × 1010 m
= 9.9 × 107 km
5. A. Those at rest in the aether would measure the speed of
light to be c. Those moving away from the light source
would measure c − v, those approaching the light source
c + v.
6. A frame of reference is the environment from which
motion is observed and measured. For the purpose of
measurement, it consists of a system of clocks and rulers,
all moving with a constant velocity.
7. The value of the velocity depends on the reference frame
in which it is measured.
8. Both clocks read the same time at any instant. When a
long distance separates them, they do not appear to tell
the same time but they do when the time required for light
from both clocks to reach the observer is taken into
account.
9. Inertial reference frames are those that are not subject to
acceleration. Non-inertial frames are accelerating.
10. The principle of relativity states that all inertial reference
frames are equal and that all laws of physics will be the
same regardless of the reference frame. If the speed of
light was relative to the aether, the laws of
electromagnetism in the reference frame of the aether
would be different from other reference frames.
11. There are no indications of movement with constant
velocity. While accelerating, however, a force is required
and you feel this as a push from your seat.
12. Each passenger is measuring the velocity relative to
different reference frames: one in motion with the car; the
other moving at 50 km h−1 relative to the car.
13. v = 50 − −50 = 100 km h−1
Jacaranda Physics 2 TSK
Jacaranda Physics 2, 3rd Edition TSK
= 2×
⎛ 1 − (3.0 × 104 )2 ⎞
100
× 108 ⎜⎜
8 2 ⎟
⎟
3.0
⎝ (3.0 × 10 ) ⎠
= 6.666 666 773 × 10−7 s
27
© John Wiley & Sons Australia, Ltd 2009
Perpendicular to the Earth’s motion:
t=
4. The principle of relativity did not appear to hold for the
theory of light and electromagnetism. Light and
electromagnetic waves were thought to be disturbances in
a stationary aether, suggesting that the speed of the Earth
could be measured relative to the aether. This is a
violation of the principle of relativity.
5. It would seem to be impossible. An object travelling at a
speed of 0.99c toward a stationary observer emits light in
all directions. The light travels at speed c. According to
Newtonian physics, the observer would measure the
speed of the light from the approaching object to be
1.99c. However, the observer actually measures the speed
of light to be c.
6. c
7. Einstein’s second postulate states that the speed of light is
the same for all observers. The Michelson–Morley
experiment worked on the basis that the speed of light
was relative to the aether, which the Earth was moving
through. If this were the case then the light would take the
two paths in different times, however, as the speed of
light is the same for all observers in all directions, it does
not matter which path the light takes, it will take the same
time, if the paths have equal length. This would mean that
the interferometer would detect no changes in the
interference pattern as it was rotated.
8. Events that are simultaneous in one reference frame are
not simultaneous in all reference frames.
9. Time dilation is the slowing of time for objects in motion
relative to the observer. This is significant for only very
great speeds. However, it can be detected with very
sensitive instruments when comparing times on board
aeroplanes with the time on the ground.
10. Depth. Only the length in the direction of the motion is
contracted.
11. The clock in motion relative to you.
12. False. By the principle of relativity, the situation must be
symmetrical so any two inertial reference frames are
indistinguishable. Both observers see the other frame as
moving and, therefore, both see the other’s time running
slow.
13. Time stops for a photon; it does not age and, therefore, it
does not decay or change in any way.
14. Three examples of time dilation effects are seen: when
fast muons fall to earth, when atomic clocks travel around
the world on commercial airliners, and in GPS satellite
navigation systems, which would be incorrect unless time
dilation was accounted for.
15. The length of the returning twin is not contracted. As
shown in the ‘parking spot’ paradox on page 160, the
length contraction is due to a disagreement regarding
what is simultaneous. Measuring lengths involves
measuring where the two ends of an object are at one
point in time. When the twin returns to the same reference
frame as his brother, they again agree on simultaneity and
therefore on length.
16. Refer to diagram in text, p. 154.
17. The principle of relativity states that the laws of physics
are the same for observers in all inertial reference frames.
If all clocks in a reference frame at rest relative to the
observer ticked at the same rate, they must all tick at the
2
v
2L
1− 2
c
c
= 2×
⎛ 1 − (3.0 × 104 ) 2 ⎞
100
× 108 ⎜⎜
⎟
8 2 ⎟
3.0
⎝ (3.0 × 10 ) ⎠
= 6.666 666 7 × 10−7 s
Assuming that c and v are determined precisely, the
difference is 7.3 × 10−8 s.
Despite the large size of this apparatus, the time
difference is still tiny.
2L
.
21. The time of the parallel journey is t =
⎛ v2 ⎞
c ⎜⎜1 − 2 ⎟⎟
⎝ c ⎠
2L
v2
1− 2 .
c
c
Dividing the parallel time by the perpendicular time gives
1
. Therefore, if the parallel journey was
⎛ v2 ⎞
⎜⎜1 − 2 ⎟⎟
⎝ c ⎠
The time of the perpendicular journey is t =
⎛ v2 ⎞
⎜⎜1 − 2 ⎟⎟ , as long as the perpendicular journey and light
⎝ c ⎠
travelled at the same speed in both directions, the time for
the two light journeys would be the same.
22. The speed of sound is relative to whatever medium it is
passing through. Without the medium there is no sound.
The sound must then be relative to its medium and no
laws of physics will change if the speed of sound is
measured by different observers. Light, however, is
dependent on the speed of electric and magnetic fields,
which travel at the speed of light. If the speed of light was
relative then the laws determining the size of the electric
and magnetic fields would be different for different
observers.
23. Absolute means that the speed of light was not relative
but had the same value in all cases.
Chapter 7
A new relativity
1. (a) The laws of physics are the same in all inertial
reference frames and light speed is constant in
vacuum, for all observers.
(b) In previous physics, the laws of electromagnetism
were not the same in all inertial reference frames —
light speed depended on the motion of the source and
the receiver, so was not the same for all observers.
2. Einstein rejected the luminiferous aether altogether.
3. (a) Newton’s Laws are an excellent approximation at
speeds lower than light speed.
(b) Newton’s Laws are accurate enough for most
purposes. They are easier to use and learn.
Jacaranda Physics 2 TSK
28
© John Wiley & Sons Australia, Ltd 2009
same rate if the reference frame is set in motion. The
length of the tick is not absolute but the idea of time
passing at a certain rate within a reference frame must be
maintained or the laws of physics will be broken.
18. See pages 152−54 and 157−58.
19. t0 is the proper time; that is, the time as measured in the
reference frame of the event. t is the time as measured in
a different inertial reference frame.
1
20. γ =
⎛
v2 ⎞
−
1
⎟
⎜⎜
c 2 ⎟⎠
⎝
24. (a) 5 minutes
(b)
23. (a)
(b)
⎛ v2 ⎞
⎜⎜1 − 2 ⎟⎟
⎝ c ⎠
= 0.714 L0
= Lγ
= 4.2 × γ
= 3.0 light-years
d
v
3.0c
=
0.7c
= 4.3 years
(b) t =
d
v
4.2c
=
0.7c
= 6 years
(c) t =
L0
γ
80
=
1.40
= 57 m
= 0.01 × 1.048 28
= 0.010 482 8 minutes
1
f′=
t
= 95.4 beats per minute. (It beats more slowly.)
Unchanged
It would be contracted to 60% of its proper length.
c
1
γ=
⎛ v2 ⎞
⎜⎜1 − 2 ⎟⎟
⎝ c ⎠
= 7.0888
t
t0 =
γ
20
=
7.0888
= 2.82 s
L
L= 0
γ
5
=
7.0888
= 0.7053 m
Jacaranda Physics 2 TSK
⎛ v2 ⎞
⎜⎜1 − 2 ⎟⎟
⎝ c ⎠
1
25. (a) γ =
2⎞
⎛
⎛ 30
5⎞
⎜ 1 − ⎜ × 10 ⎟ ⎟
⎜
⎝ 3
⎠ ⎟⎠
⎝
= 1.000 000 5
1
21. (a) γ =
⎛ v2 ⎞
⎜⎜1 − 2 ⎟⎟
⎝ c ⎠
1
=
⎛ ⎛ 0.3c ⎞2 ⎞
⎜1 − ⎜
⎟
⎜ ⎝ c ⎟⎠ ⎟
⎝
⎠
= 1.048 28
1
(b) T =
f
= 1 100 min
t = t0 γ
22. (a)
(b)
(c)
t0
⎛ ⎛ ⎛ t ⎞2 ⎞ ⎞
∴ v = ⎜ c2 ⎜1 − ⎜ 0 ⎟ ⎟ ⎟
⎜ ⎜ ⎝ t ⎠ ⎟⎟
⎠⎠
⎝ ⎝
= 0.78c
1
=
t =
26. (a) L =
(b) d = ct , L + vt = ct
L
∴t=
c−v
cL
d =
c−v
= 190 m
(c)
L + vt = ct
t=
L
(c − v)
= 6.3 × 10−7 s
(d)
29
L0
c
80
=
× 108
3
= 2.67 × 10−7 s
t=
© John Wiley & Sons Australia, Ltd 2009
32. C. They would require more force to accelerate due to the
increase in their inertial mass.
33. Energy and mass are equivalent. We cannot talk about an
increase in energy without an increase in mass, and vice
versa.
34. (a) m = m0γ
= 70 × 1.67
= 117 kg
(b) E = m0γc2 − m0c2
= 4.2 × 1018 J
(It takes a lot of energy to reach this speed.)
35. E = mc2
= 6.0 × 1024 × (3 × 108)2
= 5.4 × 1041 J
36. m = m0(γ − 1)
= 5.98 × 1024 × 0.000 000 5
= 2.99 × 1018 kg
37. Ek = (γ − 1)m0 c2
L0
γ
100
=
2.29
= 43.6 m
28. No matter how fast an object is moving it can send out a
light beam that travels at c. or If v is greater than c, γ
involves the square root of a negative number. At the
speed of light, m becomes infinite, time stops and the
length becomes zero. Therefore, matter cannot travel at
light speed because it would take up zero volume and
have infinite mass.
29. Ek = (γ − 1)m0c2
⎛
⎞
1
− 1⎟ × 1000 × (3 × 108 )2
(a) = ⎜
⎜ 1 − 0.12
⎟
⎝
⎠
27. L =
= 4.5 × 1017 J
⎛
⎞
1
− 1⎟ × 1000 × (3 × 108 )2
(b) = ⎜
⎜ 1 − 0.52
⎟
⎝
⎠
⎛
⎞
1
=⎜
− 1⎟ × 10 000 × (3 × 108 ) 2
⎜ 1 − 0.62
⎟
⎝
⎠
= 1.4 × 1019 J
= 2.3 × 1020 J
⎛
⎞
1
(c) = ⎜
− 1⎟ × 1000 × (3 × 108 )2
⎜ 1 − 0.82
⎟
⎝
⎠
Ek = (γ − 1)m0c2 so γ =
38.
1
= 6.0 × 1019 J
2
⎛
⎛v⎞ ⎞
⎜1 − ⎜ ⎟ ⎟
⎜
⎝ c ⎠ ⎟⎠
⎝
⎛
⎞
1
− 1⎟ × 1000 × (3 × 108 ) 2
(d) = ⎜
⎜ 1 − 0.92
⎟
⎝
⎠
=
Ek
+1
m0c2
Ek
+1
m0c2
2
⎛
1
⎛v⎞ ⎞
⎜1 − ⎜ ⎟ ⎟ =
2
⎜
⎟
⎝c⎠ ⎠ ⎛ E
⎞
⎝
k
+
1
⎜⎜
⎟⎟
2
⎝ m0c
⎠
20
= 1.2 × 10 J
30. Sketch a graph of energy versus speed using your answers
to the previous question.
⎛
⎜
⎜
v = ⎜1 −
⎜
⎜⎜
⎝
⎞
⎟
⎟
1
×c
2⎟
⎛ Ek
⎞ ⎟
+ 1⎟⎟ ⎟
⎜⎜
2
⎟
⎝ m0c
⎠ ⎠
= 1.0 × 104 m s −1
39. E = mc2
= 0.25 × (3 × 108)2 = 2.25 × 1016J.
40. All of the chemical products of combustion must add up
to less than the mass of the initial coal because energy has
been released.
41. It would take a lot of energy to separate the Moon from
the Earth. This would increase the mass of the Moon.
31. As seen in the previous question, even accelerating 1000 kg
to 0.9c involves enormous amounts of energy. This is not
currently feasible.
Jacaranda Physics 2 TSK
30
© John Wiley & Sons Australia, Ltd 2009
Unit
Detailed study 3.2:
Materials and their use in
structures
3
■■■■■■■■■
Part A — WORKED SOLUTIONS
(b) B
(c) Toughness is given by the area under the σ vs ε
graph. B is therefore the tougher material.
(d) B exhibits more plastic behaviour and is therefore
more ductile.
(e) A
Chapter 8
Investigating
materials
1. (a) Reading from the graph,
Δl = 0.60 mm
(b) Young’s modulus is determined from the slope of
the σ vs ε graph. However, in this case use the
relationship σ = Y × ε:
σ F×L
Y = =
ε A × Δl
=
Jacaranda Physics 2, 3rd Edition TSK
=
= 3.2 MPa
50 × 10−6 × 0.6 × 10−3
(b) ε =
= 500 × 106 Pa
= 500 MPa
π(0.100)
2
8. From:
F
σ=
A
=Y ×ε
= 1.3 × 105 Pa
Y × Δl
L
F×L
⇒ Δl =
A×Y
=
The column causes the larger stress.
3. (a) Strain = ε
Δl
=
L
100
=
10 000
= 0.01
=
15 × 103 × 5
π × (0.001)2 × 200 × 109
= 0.119 m
≅ 120 mm
F
(b) Stress = σ =
A
50 × 103
=
π × (0.004) 2
F
,
A
where A = 12.7 × 12.7 × 10−6 m2 ≅ 161 × 10−6 m2,
Δl
and strain = ε =
= , where L = 5.08 × 10−2 m
L
9. (a) Using stress = σ =
= 9.9 × 108 Pa
σ
ε
80 MPa
=
0.002
= 4 × 104 MPa
4. Y =
= 40 GPa
5. (a) Young’s modulus is given by the slope of the σ vs ε
graph. A has the larger Young’s modulus as it is the
steepest.
Jacaranda Physics 2 TSK
Δl 0.05 × 10−3
=
= 1 × 10−5
5
L
7. σ = Y × ε
= 110 × 109 × 3 × 10−4
= 3.3 × 107 Pa
= 33 MPa
2. Estimate the area of your feet. Taking a mass of 60 kg
supported on two feet, each approximately 25 cm × 6 cm,
the stress would be
60 × 10
= 2 × 104 Pa.
2 × 0.25 × 0.06
4000
π × 0.1002
= 3.2 × 106 Pa
150 × 0.100
The stress in the column is
F
A
100 × 103
6. (a) Stress = σ =
31
Stress (MPa)
Strain
0
0
449
1.97 × 10−3
673
2.95 × 10−3
898
3.94 × 10−3
999
4.92 × 10−3
1170
5.91 × 10−3
© John Wiley & Sons Australia, Ltd 2009
12. (a)
(b)
Y =
(b) It is ductile because of its noticeable plastic behaviour.
(c) From the slope of the linear portion of the graph
24 MPa
= 200 MPa.
before yielding, Y =
12 × 10−2
(d) When stretched to twice its original length, ΔL = L;
L
that is, ε = = 1 = 100%.
L
The stored energy is found from the area under
the σ vs ε graph.
Stored energy ≅ 20 × 106 × 1 = 2 × 107 J m−3
13. The stiffness of fishing line (a) is constant but the
stiffness of (b) and (c) changes. When the stress–strain
graph is steeper, the fishing line is stiffer. In each case the
fisherman will feel the same increasing force, assuming
that the stress in each case is applied at a constant rate.
14. Energy = 12 σ × ε
680 MPa
3 × 10−3
= 227 × 103 MPa
= 227 GPa
(c) A strain of 0.5% = 0.005 is beyond the elastic limit.
Therefore use the graph rather than the relationship
σ = Y × ε. From the graph σ ≅ 1030 MPa.
10. (a)
54 × 106 × 2 × 10−3
5
−3
4
= 1.1 × 10 J m
F
15. (a) Using stress = σ = ,
A
=
1
2
×
2
⎛ 12.5 × 10−3 ⎞ 2
2
−6
where A = π ⎜⎜
⎟⎟ m ≅ 122 × 10 m ,
2
⎝
⎠
Δl
, where L = 50.00 × 10−3 m
and strain = ε =
L
Stress (MPa)
Area under curve A = area under curve B.
(b) As A and B are equally tough, the areas under the
graphs must be equal. This shows that A is stiffer by
a factor of 2.
11. σ = Y × ε
= 110 × 103 MPa × 5 × 10−4
= 55 MPa
Jacaranda Physics 2 TSK
32
Strain
0.0
0.0000
36.7
0.0004
109
0.0014
182
0.0026
254
0.0036
272
0.0150
287
0.0400
291
0.0600
291
0.0800
275
0.1030
© John Wiley & Sons Australia, Ltd 2009
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
(x)
(b) (i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
(x)
(b) and (c)
(i) σY =
31.2 × 103
= 254 MPa
122 × 10−6
(ii) The ultimate tensile strength is given by the
largest stress = 291 MPa.
(iii) Breaking strength = stress at fracture = 275 MPa
16. (a) The minimum force to cause plastic deformation will
be just larger than the force at yield.
That is,
F = σY × A
= 55 × 106 × π × (1.5 × 10−3)2
≅ 3.9 × 102 N
(b)
Δl = ε × L
σ ×L
= Y
Y
55 × 106 × 40 × 10−3
=
76 × 109
(d)
Compression
Tension
Tension
Tension
Compression
Tension
Compression
Nylon — tough, elastic, stiff
Steel — strong, hard, stiff
Skin — elastic (decreases with age), soft, tough
Plastic — tough, stiff
Rubber — elastic, tough
Plastic — tough
Nylon — elastic
Rubber — tough, elastic
Plastic — plastic, tough
Putty or silicon — depends on material and
purpose
20. (a)
ΣF = 0
Στ ≠ 0
21.
= 2.9 × 10−5 m
22.
= 2.9 × 10−2 mm
(c) F = σ × A = 125 × 106 × π × 0.00152 = 884 N
σ 90 × 106
17. ε = ×
= 1.2 × 10−3
Y 75 × 109
Δl = ε × L = 1.2 × 10−3 × 1 = 1.2 × 10−3 m = 1.2 mm
F
6 × 103
=
= 850 MPa
18. (a) σ =
A π × (1.5 × 10−3 )2
23.
Δl 0.4
=
= 8 × 10−4
L 500
σ 850 × 106
= 1.1 × 1012 Pa
(c) Y = =
ε
8 × 10−4
(d) From the area under the graph of force vs extension,
strain energy = 12 × 6.0 × 103 × 0.4 × 10−3
(b) ε =
24.
ΣF = 0
(b) No. Rotational equlibrium is not satisfied.
Because of the lever arm, Vince will get maximum effect
when the pedals are horizontal and he pushes vertically
down. The force he applies will have no effect when the
pedals are vertical because the lever arm is zero and
consequently the torque is zero.
To pick the rod up, a force equal to 500 + 300 + 800 N
must be applied. If the rod is not going to rotate it must be
picked up at the CM where rotational equilibrium would
be satisfied i.e. Στ = 0. Assuming the CM is x cm from
the bucket on the left and clockwise torques are positive:
Στ = 0:300 × 1.5 − 800 × x = 0
1.5
x = 300 ×
= 0.56 m
800
As Sam moves beyond the left of the fulcrum, the seesaw
will rotate anticlockwise. This will happen when the
torque from Sam about the fulcrum is larger than the
torque of the bag.
(a) The reaction from the left abutment decreases and the
reaction from the right abutment increases.
(b)
= 1.2 J
(e) Total energy =
1
2
× 6.0 × 103 × 0.4 × 10−3
= 1.2 J
19. (a)
(i) Tension
(ii) Shear
(iii) Tension
Jacaranda Physics 2 TSK
ΣF = 0: RL + RR = 12 tonne
(1)
Στ = 0: taking torques about the left hand support
12 × 16 − 20RR = 0
(2)
33
© John Wiley & Sons Australia, Ltd 2009
28. This question once again relies on applying the equations
of equilibrium.
ΣF = 0: 2000 + 800 + 600 + 200 − RL − RR = 0
that is, RL + RR = 3600 N (1)
Στ = 0: taking torques about the left hand support
800 × 2.0 + 2000 × 3.0 + 600 × 4.0 + 200 × 5.0 − RR × 6.0
= 0 (2)
16
= 9.6 tonne
20
Substituting in (1), RL = 2.4 tonne.
25. (a) The wall resists the loads from the balcony and the
person with an upwards force and an anticlockwise
torque.
(b) As the person moves towards the wall, the total
reaction remains constant. However, the torque on
the wall decreases.
RR = 12 ×
29. Pirate Bill will tip into the water when the overturning
torque caused by his weight is greater than the stabilising
torque of the plank’s own weight. This will occur when
the pirate is x m past the edge of the boat. The CM of the
plank is 1 m inside the edge of the boat.
Στ = 0: −500 × +800 × 1 = 0
800
x=
= 1.6 m
500
26.
30. If the tension in the cable is T, the force can be found by
considering vertical equilibrium.
2T sin 8° = 4 kN
T = l4.4 kN
Equilibrium must be satisfied at each joint if the tent is to
stay up. The forces can be simply resolved into their
vertical and horizontal components. Let the force in the
guy = T and the force in the pole = P:
(a) ΣFhorizontal : 500 − T cos 60° = 0
31. No, the wire did not return to its original length. The
force–extension graph shows that the wire was loaded
beyond its elastic limit. When it was unloaded, some of
the energy had been used to permanently deform the wire.
500
cos 60°
T = 1000 N tension
(b) ΣFvertical : T sin 60° + P = 0
T =
1000 sin 60° + P = 0
P = −866 N
i.e. compression
The guys create a downward force on the poles.
(c) The poles would sink into the ground until the ground
can resist the downward force from the poles.
27. Taking torques about the left support:
1.8 W − 20(5 − 1.8) = 0
72
1.8
W = 35.6 tonnes ≅ 36 tonnes
W =
Jacaranda Physics 2 TSK
34
© John Wiley & Sons Australia, Ltd 2009
Unit
Detailed study 3.3:
Further electronics
3
Jacaranda Physics 2, 3rd Edition TSK
Part A — WORKED SOLUTIONS
■■■■■■■■■
Chapter 9
(iv) VRMS =
Electronics at work
=
(b) Vp − p = max value − min value
(b)
= 5 V − (−5 V)
= 10.0 V
= 6.0 cm × 2 ms cm −1
= 12 ms
2
5V
(ii) f =
2
= 3.5 V
(iii) Vpeak = 1.5 cm on screen
= 1.5 cm × 5 mV cm
= 7.5 mV
1
=
(iv) VRMS =
4.0 × 10−3 s
= 250 Hz
=
V
= RMS
R
3.5 V
=
100 Ω
3. (a)
(i) T = time to complete one cycle
= 20 ms
(ii) f =
(i) T = time to complete 1 cycle
= 6.0 cm on screen
=
−1
= 6.0 s
=
1
(ii) f =
T
1
20 ms
1
2 × 10−2 s
(iii) Vpeak = 1.5 cm × 1 V cm −1
1
6.0 s
= 0.17 Hz
= 1.5 V
(iv) VRMS =
(iii) Vpeak = 1.5 cm on screen
= 1.5 cm × 1 V cm
1
T
= 50 Hz
=
−1
=
= 1.5 V
Jacaranda Physics 2 TSK
2
7.5 mV
= 4 cm × 5 ms cm −1
= 35 mA
= 6.0 cm × 1 s cm
Vpeak
2
= 5.3 mV
= 3.5 × 10−2 A
2. (a)
1
1.2 × 10−2 s
= 83 Hz
1
T
(f) I RMS
1
T
=
(d) T = time to complete 1 cycle
= 4.0 ms
(e) f =
(i) T = time to complete 1 cycle
= 6.0 cm on screen
Vpeak
=
2
1.5 V
2
= 1.1 V
1. (a) Vpeak = 5.0 V
(c) VRMS =
Vpeak
Vpeak
2
1.5 V
2
= 1.1 V
35
© John Wiley & Sons Australia, Ltd 2009
(b) R = 33 kΩ
= 3.3 × 104 Ω
C = 100 μF
= 10−4 F
τ = 3.3 × 104 Ω × 10−4 F
= 3.3 s
(c) R = 68 kΩ
= 6.8 × 104 Ω
C = 10 μF
= 10−5 F
τ = 6.8 × 104 Ω × 10−5 F
= 6.8 × 10−1 s
= 0.68 s
(d) R = 470 kΩ
= 4.7 × 105 Ω
4. (a) Connect the probes to the terminals marked ‘Com’
and ‘V’. Set the dial on the multimeter to an
appropriate voltage scale and connect the probes in
parallel across the part of the circuit being measured.
(b) Never measure mains voltage with a multimeter
because it could destroy the multimeter and kill or
harm the person holding it.
5. (a) 0.1 μF = 0.1 × 10−6 F
C = 0.47 μF
= 4.7 × 10−7 F
τ = 4.7 × 105 Ω × 4.7 × 10−7 F
= 0.22 s
8. (a) 1.8 V ± 0.2 V (by reading graph)
(b) The time constant for an RC circuit is the time it
takes the capacitor to reach 0.63 of its final voltage
when charging. Assume that the final voltage is
approximately 10 V. Time constant is the time at
which the voltage is 6.3 V.
τ = 5.0 ms (by reading graph)
(c)
τ = RC
τ
⇒C =
R
5.0 × 10−3 s
=
100 Ω
= 1.0 × 10−7 F
(b) 220 μF = 220 × 10−6 F
= 2.20 × 102 × 10−6 F
= 2.2 × 10−4 F
(c) 100 pF = 1.0 × 102 × 10−12 F
= 1.0 × 10−10 F
(d) 140 nF = 1.4 × 102 × 10−9 F
= 1.4 × 10−7 F
6. (a) Time constant for an RC circuit is the time it takes
the capacitor to reach 0.63 of its final voltage when
charging, or 0.37 of its initial voltage when discharging.
(b) After one time constant, the voltage will fall to 0.37
of its initial voltage.
0.37 × 10 V = 3.7 V
7. (a) τ = RC
R = 390 Ω
= 5.0 × 10−5 F
(or 50 μF)
τ
(d) R =
C
5.0 × 10−3 s
=
1.0 × 10−7 F
= 5.0 × 104 Ω
(or 50 kΩ)
9. Rectification is the conversion of an AC voltage signal
into a DC voltage signal.
10. Only the positive or the negative part of the signal will be
passed — not both.
11. (a) A diode is an electronic device that allows current to
flow through it in one direction.
(b) About 0.7 V when forward biased
(c) One
= 3.9 × 102 Ω
C = 100 μF
= 100 × 10−6 F
= 10−4 F
τ = 3.9 × 102 Ω × 10−4 F
= 3.9 × 10−2 s
(τ = 39 ms)
Jacaranda Physics 2 TSK
36
© John Wiley & Sons Australia, Ltd 2009
17. (a) τ = RC
12. (a) It smooths the output voltage of the half-wave
rectifier.
(b) When the diode ‘blocks’ the voltage, the capacitor
discharges through the load resistor.
(c) It becomes smoother.
13. (a) Bridge rectifier
= 1 × 106 Ω × 1 × 10−7 F
= 1 × 10−1 s
= 0.1 s
1.0 V is approximately 0.63 of 1.5 V; that is, it takes
approximately one time constant for the voltage
across the capacitor to reach 1.0 V.
(b) The capacitor is considered to be fully charged after
five time constants = 0.5 s.
(c) When fully charged, the voltage across the capacitor
equals the emf of the cell = 1.5 V.
(d) Q = CV
= 0.1 × 10−6 F × 1.5 V
= 1.5 × 10−7 C
Centre-tap rectifier
(e)
(b) Bridge rectifier,
Centre-tap rectifier,
⇒ VR = 1.5 V − 1.0 V = 0.5 V
18. (a) The peak output voltage equals the peak input
voltage minus the voltage drop across the diode. The
maximum voltage drop across a silicon diode is
approximately 0.7 V.
Vpeak output = Vpeak input − Vdiode
Vout = 12.0 − 1.4
= 10.6 V
Vout = 12.0 − 0.7
= 11.3 V
= 4.0 V − 0.7 V
= 3.3 V
14. Vr(p − p) = 6.3 − 5.8
= 0.5 V
V T
15. Vr(p − p) = max
RC
1
12.5 ×
100
=
RC
0.125
=
RC
(a) Vr(p − p) =
(b)
(c) (i) If C = 10 μF
τ = RC
0.125
3
10 × 10 × 100 × 10
−6
= 0.13 V
(c) Vr(p − p)
= 1 × 103 Ω × 1 × 10−5 F
0.125
(b) Vr(p − p) =
3
= 1 × 10−2 s
The capacitor therefore has a little more than one
time constant to discharge before the next pulse
arrives.
The voltage will fall to slightly less than
0.37 × 3.3 V = 1.2 V.
−6
3.3 × 10 × 100 × 10
= 0.38 V
0.125
=
3
10 × 10 × 5.0 × 10−6
= 2.5 V
(d) Vr(p − p) =
E = VR + VC
0.125
25 × 10 × 20 × 10−6
3
= 0.25 V
(e) Vr(p − p) =
0.125
100 × 103 × 25 × 10−6
= 0.050 V
16. 4.2 V AC
Jacaranda Physics 2 TSK
37
© John Wiley & Sons Australia, Ltd 2009
(ii) If C = 100 μF
τ = RC
(e) τ = RC
= 500 Ω × 2.0 × 10−5 F
= 1 × 103 Ω × 1 × 10−4 F
= 0.010 s
= 10 ms
(f) The capacitor discharges for slightly more than one
time constant between peaks, so Vout will fall to 0.37
of 12.0 V or 4.4 V.
−1
= 1 × 10 s
The capacitor has approximately 0.1 of a time
constant to discharge, so the voltage will fall
very little.
(g) Use a larger value resistor.
Use a larger value capacitor.
20. (a) τ = RC
19. (a) Vpeak = 2 VRMS
= 2.0 × 104 Ω × 1.0 × 10−5 F
= 2 × 9.0 V
= 0.20 s
= 13 V
(= 200 ms)
(b) Yes. A capacitor is considered to be fully charged (or
discharged) after five time constants. The input
voltage is at a high or low value for one second each,
which is five time constants.
(c)
(b) f = 50 Hz (input and output have same frequency)
1
⇒T =
f
1
=
50 Hz
= 0.02 s
= 20 ms
(c) Maximum output voltage
= 12.7 V − 0.7 V
= 12.0 V
(d)
(d) VD max = 0.7 V
When reverse biased, VD is the same as Vin.
Vin = VC + VR
⇒ VR = Vin − VC
21. • The AC supply to your house
• A battery going flat
• Variations in the load resistance
22. (a) Voltage regulation is the maintenance of a steady
voltage supply.
(b) Most electronic devices require a steady voltage
supply.
Jacaranda Physics 2 TSK
38
© John Wiley & Sons Australia, Ltd 2009
(c) Between the power supply and the load
(d) The output voltage is a little less than the input
voltage.
23. Zener diodes are designed to ‘breakdown’ in a reliable
and non-destructive way so that they can be used in
reverse to maintain a fixed voltage across their terminals.
24. The breakdown voltage of a zener diode is the reverse
bias voltage at which the junction breaks down and can
conduct large currents.
25. The zener diode is in series with a limiting resistor (to
limit the current through the diode) and a voltage supply.
It is placed in parallel with the load to provide a regulated
(steady) voltage across the load.
26.
(a) 20 V
(b) 0.20 A
(c) 20 V
(d) 0.40 A
(e) 0.20 A
27.
(a) 40 V to 70 V
(b) Imax = 1.75 × 10−2 A (17.5 mA),
Imin = 0.010 A (10 mA)
(c) I = 3.0 × 10−3 A (3.0 mA)
(d) Imax = 1.45 × 10−2 A (14.5 mA),
Imin = 0.007 A (7 mA)
28.
(a) 12 V
(b) 8 V
(c) 108 mA
(d) 120 Ω
(e) 74 Ω
Jacaranda Physics 2 TSK
29. (a) Read value from graph.
Answer is approximately 0.65 V.
Accept 0.6 V to 0.7 V.
(b) VR = E − VD
= 6.0 V − 0.65 V
= 5.35 V
(c) VR = IR
I = 4.0 × 10−3 A
R=
=
V
I
5.35 V
4.0 × 10−3
= 1337.5 Ω
(e)
(f)
(g)
30. (a)
(b)
39
= 1.3 × 103 Ω or 1.3 kΩ
− 6.0 V as the diode is now reverse biased
I=0
VR = 0
Transistors and diodes
Heat sinks conduct, convect or radiate the thermal
energy away from the device.
© John Wiley & Sons Australia, Ltd 2009
Unit
4
Area of study 1:
Electric power
Jacaranda Physics 2, 3rd Edition TSK
Part A — WORKED SOLUTIONS
■■■■■■■■■
Chapter 10
Magnetism – fields
and forces
8.
1. If the magnet attracts the piece of metal, the metal must
be a magnetic material, which may or may not be a
magnet. If the magnet repels the piece of metal from
some angles, the metal must be a magnet with similar
strength to the known magnet. Otherwise, if the piece of
metal is a magnet, either much stronger or weaker than
the known magnet, the two magnets will attract but the
strength of attraction will depend on the part of the metal
the pole of the magnet is touching.
2. As the material cools, the magnetic domains are aligned
by the Earth’s magnetic field.
3. There is a magnetic field at both ends which induces a
field of the same orientation in the iron nail. Opposite
poles attract and so the nail moves to the magnet.
4. The magnetic pole in the southern hemisphere is a north
magnetic pole.
5. Each iron rod is magnetised in the same orientation.
Adjacent north ends repel each other.
6. (a)
9.
10.
11.
(a) The direction of the magnetic field beneath the wire
is north ⇒ No change.
(b) The direction of the magnetic field beneath the wire
is south ⇒ Opposite direction.
(a) Out of page
(b) Out of page
(c) Out of page
(d) Up the page
(e) To the right
(a) Up the page
(b) to the left
(c) Down to the right
(d) Into the page
(e) Out of page
(a) W: Out of page; X, Y, Z: Into the page
(b) W, X: To the right; Y, Z: To the left
Use the left-hand rule or the right-hand-slap rule.
(a) Into page
(b) Into page
(c) South
(d) Out of page
(e) North
(f) East
(g) South-east
(h) South
12. (a)
(b)
(c)
(b)
(d) Yes, by Newton’s third law, the force by A on B is
equal and opposite to the force by B on A.
13. F = nIlB
= 1 × 4.5 × 0.05 × 0.30
= 0.07 N
14. F = nIlB
= 1 × 2.4 × 0.30 × 0.25
= 0.18 N
F = nIlB
15.
7. Use the right-hand-grip rule.
Jacaranda Physics 2 TSK
40
but
l = 2πr
⇒
F = 500 × 15 × 10−3 × (2π × 1.5 × 10−2 ) × 2.0
= 1.4 N
© John Wiley & Sons Australia, Ltd 2009
16. F = nIlB
(b) Yes, just reverse the direction of the current.
(c) Yes, just increase the voltage of the DC battery,
which increases the size of the current.
27. (a) The commutator would counter the change in voltage
and consistent movement would not be possible.
(b) The motor would turn if it were turning at the same
rate as the AC frequency and if the voltage changed
direction when the coil was perpendicular to the field.
This would be very unlikely.
28. (a) The two ways are series and parallel.
(b) A series-wound motor produces a very large turning
effect when it is started. This makes this sort of motor
usual for trains and trams. Parallel-wound or shuntwound motors maintain a steady speed with different
loads and are used to drive such machines as lathes.
= 1 × 1.8 × 8.0 × 10−2 × 40 × 10−3
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
= 5.8 × 10−3 N
Use several coils of wire wrapped around a matchbox
floating in a dish of water. Supply a DC current.
Fire an electron into a space. The direction it deflects can
be used to determine the direction of the magnetic field,
except where the electron is undeflected. The direction of
the deflection is the direction of the force, the initial
direction of the electron gives the opposite direction of
the conventional current, and the hand rule (left-hand rule
or right-hand-slap rule) gives the direction of the magnetic
field. If the electron is undeflected, the magnetic field is
either parallel or anti-parallel to the electron direction.
By travelling parallel or anti-parallel to the magnetic
field.
An east–west power line would experience an up–down
force because the current reverses direction at 50 Hz. For
most lines, this would be too high a frequency to produce
noticeable movement in the line. But there would be a
particular length for which 50 Hz would be the resonant
frequency.
No, not if the electron is at rest relative to the magnet.
(a) As the electrons approach the wire they experience a
downwards magnetic field. They initially experience
a force to the right.
(b) The force on the electron is away from the wire.
(a) The electron experiences a force in a westerly
direction and moves in a circle.
(b) The electron moves in a circle and moves up, at the
same time producing a spiral path.
A DC current flows through a coil in a magnetic field.
The field exerts forces of the same size, but of opposite
direction on the opposite sides of the coil. These forces
initially make the coil rotate, but the direction of the
current needs to be reversed if the coil is to continue to
rotate. This reversal is done by a commutator twice each
cycle when the coil is at right angles to the field.
(a) The magnet provides a magnetic force that interacts
with the electric current to produce a force.
(b) The brushes prevent tangling of wires by allowing
current to flow from the stationary wires coming
from the battery to the rotating coils of wires.
(c) The commutator reverses the current, twice every
cycle when the coil is at right angles to the field.
(d) Each turn experiences the magnetic force, so the
more turns the larger the force and the faster the
rotation.
(a) If the coil is in a position at right angles to that in the
figure on page 252, then either there will be no
electrical connection between the brushes and the coil
and so no current, or possibly the commutator
touches both brushes allowing the current to bypass
the coil, that is, to short out and possibly burn out the
brushes and the commutator. This can be overcome
by using extra sets of coils at angles to the first, each
with it own set of sections on the commutator. The
figure on page 253 has three sets of coils.
Jacaranda Physics 2 TSK
Chapter 11
Generating
electricity
1. Magnetic flux is the amount of magnetic field passing
through an area.
2. He used these to increase the amount of magnetic flux
passing through the iron core.
3. (a) Φ = Bperpendicular × A
= 3.0 T × 0.050 m 2
= 0.15 Wb
(b) Φ = 0.4 T × 4.5 cm 2
= 0.4 T × 4.5 × 10−4 m 2
= 1.8 × 10−4 Wb
(c) Φ = 0.025 T × 12 × 10−4 m 2 × 50
= 1.5 × 10−3 Wb
4.
5. The loss in gravitational potential energy as the metal rod
fell was not fully converted into kinetic energy. Some of
it was converted into electrical potential energy of the
separated charges.
6. The falling magnet induces circular currents in the metal
tube whose magnetic fields produce repulsion from below
and attraction from above. This is an effective air
resistance as the size of the induced currents depends on
the speed of the falling magnet — a condition for
terminal velocity.
41
© John Wiley & Sons Australia, Ltd 2009
(c) ΔΦ = − 35 − 60
= − 95 Wb
7. (a) As the rod falls through the magnetic field, the falling
electrons are pushed to the left end of the rod, making
the right end positive. This establishes a difference in
voltage between the ends. The rod will continue to
fall, gaining speed. The electrons will have a greater
vertical speed and will experience a greater sideways
force. This will make the difference in voltage
greater. As long as the rod accelerates, the voltage
difference will increase.
(b) This process differs from the charging of a capacitor
in that the force needed to separate the charges is
from within the rod, rather than from an external
battery, and the voltage difference will continue to
increase as the rod speeds up until the rod reaches a
terminal velocity. The rate at which the voltage
increases will also depend on the acceleration and is
likely to be constant. In a capacitor, the initial
increase is rapid, followed by a levelling off.
ΔΦ
Δt
−95
= −1 ×
2.5
= 38 V
13. (a) Circumference of loop = 10 cm
C
r=
= 1.59 cm
2π
ΔΦ = − 0.60 T × (π × 0.01592 ) m 2
ε = −N
ε = −N
= −1 ×
9. (a) Anticlockwise current seen from above
(b) No
(c) Yes, clockwise current
10. (a) Anticlockwise current as seen from the left
(b) Clockwise current produced
(b) ΔΦ = −1.0 T × (0.26 m)2
= 0.0676 Wb
11. (a) Yes; as the coil starts to leave the magnetic field an
anticlockwise current will be induced in order to
produce a magnetic field out of the page.
(b) Same as (a)
(c) No, there is no change in flux.
(d) Same as (c)
ΔΦ
Δt
0.0676
= −1 ×
0.5
= − 0.1352
ε=−N
A = πr 2
−2 2
= π × (5.0 × 10 ) m
ε
R
0.1352
=
2.5
= 0.054 A
I =
2
= 7.895 × 10−3 m 2
ΔΦ = Φ final − Φ initial
= 0 − BA
(c) ΔA = π(0.04)2 − π(0.08)2
−3
= 0 − 0.40 × 7.85 × 10
ΔΦ
ε=−N
Δt
−1 × − 0.40 × 7.85 × 10−3
=
0.2
= 0.016 V
(b) ΔΦ = 35 − 60
= − 25 Wb
= 1.51 × 10−2 m 2
ΔΦ = 2.0 × 1.51 × 10−2 Wb
ΔΦ
ε=−N
Δt
2.0 × 1.51 × 10−2
= −1 ×
0.8
= − 0.037 75 V
ε
R
0.037 75
=
0.2
= 0.19 A
ΔΦ
Δt
−25
= −1 ×
1.5
= 17 V
I =
ε=−N
Jacaranda Physics 2 TSK
−0.60 × π × 0.01592
0.3
= 1.59 × 10−3 V
ε
I =
R
1.59 × 10−3
=
0.4
= 0.004 A
8. Clockwise current to produce a magnetic field into the
page
12. (a)
ΔΦ
Δt
42
© John Wiley & Sons Australia, Ltd 2009
14. (a) f = 50 Hz
T = 0.02 s
(c) ΔΦ = BA
= 0.1 × 10−3 T × 3.0 × 107
Time for a quarter of a turn = 0.005
Φinitial = 0
= 3000 Wb
ΔΦ
ε=−N
Δt
3000
= −1 ×
1
= − 3000 V
ε (magnitude) = 3000 V
Φ final = 2.5 T × 0.042
= 0.10 Wb
ΔΦ
ε=−N
Δt
= − 100 ×
0.10
0.005
18. (a)
= 2000 V
(b)
ε = −N
Φ initial = 0
Φ final = 0
⇒ ΔΦ = 0
⇒
ε=0
ΔΦ
Δt
(b)
15. The rotating coil of the motor is in the magnetic field and
so produces an induced EMF. If the motor is connected to
a battery, this induced EMF will oppose the battery and is
called a back EMF. If the motor is disconnected from the
battery and forced to turn, an alternating EMF will be
produced which will be rectified by the commutator.
(c) The loss in gravitational potential energy as the
magnet fell was not fully converted into kinetic
energy — some went into electrical energy. The
magnet fell slower than it would have without the
coil.
(d) The change in magnetic flux is zero, so the two areas
under the graph cancel each other out.
(e) If the magnet is accelerating, the current below the
coil, where it is travelling faster, will produce a larger
current, but for a shorter time.
ΔΦ
19.
ε = −N
Δt
ΔΦ
⇒
IR =
(magnitude only)
Δt
Q
ΔΦ
⇒
R=
where Q = charge
Δt
Δt
ΔΦ
Q=
⇒
R
16. (a) Simplify the induced EMF equation to:
BA
ε Δt
ε=
, rearranging B =
,
Δt
A
volt × sec
so tesla =
metre2
(b) Voltage = Current × Resistance (V = IR), and
Charge
Q
QR
Current =
( I = ), so ε =
.
Time
Δt
Δt
Substituting into the expression from
coulomb × ohm
QRΔt
QR
B=
, so tesla =
=
Δt × A
A
metre 2
(c) Magnetic flux = Magnetic field × Area (Φ = BA)
QR
Φ = BA =
× A = QR, so weber =
A
coulomb × ohm.
=
3.0 T × 1.6 × 10−3 m 2
0.2 Ω
= 0.024 C
20. In both cases, there is an induced EMF; however, plastic
is an insulator, so the electrons are not free to move as in
the case of a conductor. In the wire there is a current, but
there is no current in the plastic.
17. (a) v = 6000 m s−1 t = 1.0 s
distance = vt
= 6000 m
21. If the approaching north end of the magnet to a coil
produced a magnetic field with the south end towards the
magnet, the magnet would be attracted and speed up, and
so induce an even stronger field whose south end would
make the magnet travel even faster. Thus, kinetic energy
would be created from nothing.
(b) Area = L × W
= 6000 m × 5000 m
= 3.0 × 107 m
Jacaranda Physics 2 TSK
(i) Anticlockwise from above
(ii) Zero current
(iii) Clockwise current from above
43
© John Wiley & Sons Australia, Ltd 2009
22. (a) The faster the motor is turning, the quicker the coil is
cutting the magnetic field in the motor, so the larger
the back EMF.
(b) At slow speeds there is a small back EMF, so the
current will be higher than when the motor is
spinning faster.
(c) The low resistance of the coils will produce a very
large current which will overheat the motor.
23. Distance covered by axle in one second
= vt
3. (a) N prim =
× Nsec
240
× 50
12
= 1000 turns
P
V
20 × 5
=
12
= 8.3 A
(b) I =
(c) Pprim = Vprim I prim = 20 × 5 W
20 × 5
240
= 0.42 A
4. A transformer does not work with constant DC voltage,
such as that supplied by a battery or a regulated power
supply, because there is no change in the magnetic flux.
I prim =
= 50 m 2
ΔΦ
ε=−N
Δt
40 × 10−6 T × 50 m 2
= −1 ×
1
−3
= 2.0 × 10 V
5. To enable the direction of the magnetic field to change as
quickly as the current changes direction.
24. 8.9 V
25. (a) 28 m s
(b) 35 Hz
(c) 50 mV
(d) 100 mV
(e) 35 mV
26. 6.4 V
6. (a) Step down transformer
(b) 48 turns
7. (a) 125 V
(b) 37.5 W
(c) 0.625 A
8. (a) Voltage across generator = 250 VRMS
P
I =
V
50 000
=
250
= 200 A
Chapter 12
Transmission
of power
Vsec
N
= sec
Vprim N prim
(b) Ploss = I 2 R
2000
× 240
100
= 4800 V
N
= sec × Vprim
N prim
= (200)2 × 0.3
⇒ Vsec =
2. (a) Vsec
Vsec
=
= 120 km h −1 × 1 s
120
=
3.6
= 33.33 m
Area covered = L × W
= 33.33 × 1.5
1.
Vprim
= 12 kW
(c) V = IR
= 200 × 0.3
= 60 V
(d) V = 250 − 60
= 190 V
(e)
6
× 240
300
= 4.8 V
N prim
× Vsec
(b) Vprim =
Nsec
=
300
×9
6
= 450 V
=
Jacaranda Physics 2 TSK
44
© John Wiley & Sons Australia, Ltd 2009
P
V
50 000
=
250 × 20
= 10 A
V2
P
(240)2
=
20 000
= 2.88 Ω
Effective resistance of workshop and house (in
parallel):
1
1
1
=
+
Reff
2.88 6.0
I =
(b)
⇒ R=
Ploss = I 2 R
= (10) 2 × 0.3
= 30 W
(c) V = IR
⇒ Reff = 1.95 Ω
(c) Total resistance = 1.95 + 0.20
= 2.15 Ω
= (10)2 × 0.3
= 3.0 V
(d) V = 250 − 3
1.95
× 240
2.15
= 218 V
V =
= 247 V
V 240
+
= 6.0 Ω
40
I
(b) R = 6.0 + 0.2 = 6.2 Ω
(c) V = 6.2 × 240 = 232 V
(d) (i) Calculate resistance of workshop:
P = VI
9. (a) R =
=
Jacaranda Physics 2 TSK
(e)
(f)
10. (a)
(b)
(c)
(d)
11. (a)
(b)
V2
R
45
Yes
Increase
667 A
178 kW
267 V
330 kV, 220 mW
25%
0.25%
© John Wiley & Sons Australia, Ltd 2009
Unit
4
■■■■■■■■■
Area of study 2:
Interactions of light and
matter
Part A — WORKED SOLUTIONS
(b) Two waves are exactly out of phase when one wave
is exactly half a cycle ahead of another; for example,
a wave crest from one source coincides with a wave
trough from another source. Destructive interference
happens when the addition of waves results in zero
disturbance, commonly as a result of two waves
arriving at a point in space exactly out of phase, with
crests meeting troughs. A node is a point, or a line,
where destructive interference occurs, that is, where
there is no wave disturbance.
Chapter 13
Light – waves and
particles
1. (a)
(b)
5. (a) A: antinode, B: node, C: antinode
(b) A: bobbing up and down
B: stationary
C: bobbing up and down
(c) A: 0, A is on central antinode at an equal distance
from both sources
3λ
, B is on second nodal line
B:
2
C: λ, C is on the first antinodal line
(d) A: bright, B: dark, C: bright
The wavelets at the edge of the slits form a wavefront
which represents the bending of diffracting waves
around the edge of the slit.
2. (a)
6. (a) The first minima occur where the path difference for
light rays travelling from different places on the
λ
opening reaches . This will be a region where the
2
intensity of the light will be diminished due to
destructive interference.
(b)
(b) The same number of wavefronts pass X and Y each
minute, since wavefronts cannot disappear nor appear
between those points. The wave frequency is the
same in both materials.
(c) Wave speed is greater in material B. Since the wave
frequency is the same in materials A and B, the
greater speed will be seen as a greater wavelength.
This occurs in material B.
3. Superposition means the adding together of the effects of
waves coinciding at a particular location. For example,
the bright and dark bands in the pattern produced when
light passes through two slits to a screen are caused by the
addition of the effects of waves travelling from each of
the two slits to each point on the screen.
4. (a) Two waves are in phase when corresponding points
on the waves are synchronised; for example, wave
crests leaving two slits at the same time, or wave
troughs arriving at a certain place at the same time.
Waves experience constructive interference when
they are in phase at a point in space, and continue to
be in phase as the waves pass through that point. The
crests arrive together and the troughs arrive together.
This point is said to be on an antinode, which is a
point, or a line, where maximum wave disturbance
occurs.
Jacaranda Physics 2 TSK
Jacaranda Physics 2, 3rd Edition TSK
(c)
7. (a) Diffraction depends on wavelength or colour of light.
White light is a mixture of the different colours in the
spectrum, so when it diffracts when passing through a
small slit, the colours that make up the white light
form slightly different diffraction patterns.
46
© John Wiley & Sons Australia, Ltd 2009
(b) The positions of the minima are given by
mλ
. The red end of the spectrum has the
sin θ =
d
longest wavelengths, therefore θ for any particular
minimum occurs at a greater angle than for blue light
and other parts of the visible spectrum. This means
that the bright fringes in an interference pattern will
be reddened on their outer edges.
8. See table at bottom of page.
9. (a) Ek = 12 mv 2
⇒v=
=
10. Ephoton =
=
=
= 6.21 eV − 5.1 eV
= 1.11 eV
To stop an electron with 1.11 eV of kinetic energy
requires a stopping voltage of 1.11 V.
11. (a) Minimum energy required to raise the electron
energy to zero is 4.5 eV.
(b) In each case Ek of electron is given by Ephoton + initial
electron energy
(i) Ek = 5.9 eV − 4.7 eV = 1.2 eV
Electron is ejected with kinetic energy of
1.2 eV.
(ii) Ek = 5.9 eV − 5.3 eV = 0.6 eV
Electron is ejected with kinetic energy of
0.6 eV.
(iii) Ek = 5.9 eV − 5.9 eV = 0.0 eV
Electron is ejected with zero kinetic energy.
(iv) Energy of photon not sufficient to raise
electron energy to zero, so electron is not
ejected.
12. (a) The maximum current occurs when the accelerating
voltage causes all ejected electrons to be collected at
the anode. The voltage required for this is greater
than zero because some electrons leave at an angle
and their parabolic path may miss the anode at lower
voltages. When the voltage opposes the motion
towards the cathode, electrons travelling towards the
anode slow down. When the magnitude of the voltage
is large enough, the electrons reverse direction and so
do not contribute to the current. At a high enough
retarding voltage, all electrons turn around before
reaching the anode so the current is zero.
(b) Increasing the intensity without changing the
frequency would increase the number of photons per
second reaching the cathode but not change their
energy. As a result, the photoelectrons will have the
same energy spread, and therefore the same stopping
voltage, but there will be more electrons ejected per
second, resulting in a higher photocurrent.
9.1 × 10−31 kg
= 5.3 × 105 m s −1
(b)
(c)
(d)
Infra-red from CO2 laser
Red helium–neon laser
Yellow sodium lamp
UV from eximer laser
X-rays from aluminium
Jacaranda Physics 2 TSK
10.6 μm
633 nm
589 nm
0.193 μm
0.988 nm
9.932 × 10−19 J
1.60 × 10−19 J eV −1
= 6.21 eV
Maximum Ek, photoelectron = Ephoton − W
2 × 0.8 × 1.60 × 10−19 J eV −1
Wavelength
200 × 10−9 m
= 9.932 × 10−19 J
2 Ek
m
Source
hc
λ
6.6262 × 10−34 J s × 2.9979 × 108 m s −1
Frequency
13
2.83 × 10 Hz
4.74 × 1014 Hz
5.09 × 1014 Hz
1.55 × 1015 Hz
3.03 × 1017 Hz
47
Energy
−20
1.87 × 10 J, 0.117 eV
3.14 × 10−19 J, 1.96 eV
3.37 × 10−19 J, 2.11 eV
1.03 × 10−18 J, 6.42 eV
2.01 × 10−16 J, 1.25 keV
Momentum
6.25 × 10−29 kg m s−1
1.05 × 10−27 kg m s−1
1.125 × 10−27 kg m s−1
3.43 × 10−27 kg m s−1
6.69 × 10−25 kg m s−1
© John Wiley & Sons Australia, Ltd 2009
the effect of a single photon. The pattern of arrangement
of these spots demonstrates the wave characteristics of
the light.
15. Neon atoms have well defined energy levels, and the
wavelengths of light emitted by neon atoms correspond to
atoms jumping from higher to lower energy levels. The
sharpness of the wavelengths in the spectrum indicates
the sharpness of the possible energies of the atoms.
16. The magnitude of the energy change of the electrons is
equal to the photon energy, hf.
Ephoton = 6.6262 × 10−34 J s × 4.59 × 1014 Hz
(c) Increasing the frequency of the light would increase
the energy of each photon, resulting in higher energy
photoelectrons and a greater stopping voltage. If the
intensity is unchanged, the energy per second
reaching the cathode is not changed, but since each
photon has a greater energy, this means that there are
fewer photons per second reaching the cathode and
the photocurrent will be reduced.
= 3.04 × 10−19 J
=
3.04 × 10−19 J
1.60 × 10−19 J eV −1
= 1.90 eV
ΔEelectron = −3.04 × 10−19 J
= − 1.90 eV
17.
(d) There will be the same number of photons per second
and they have the same energy but it requires a
different energy to eject an electron. This will change
the maximum Ek of the electrons and therefore the
stopping voltage but since the same number of
photons reach the cathode each second, the maximum
photocurrent will be unchanged. Two answers are
shown. One is for a material with a greater work
function (lower maximum Ek and therefore lower
stopping voltage) and the other is for a material with
a smaller work function (higher maximum Ek and
therefore higher stopping voltage).
Each wavefront forms a set of centres for circular
wavelets. The wavelets closest to the boundary will reach
the material where they will travel at higher speed first.
These wavelets will form the new wavefront in this
material as shown in the diagram, at a different direction
to the incoming wavefront.
18. (a) λ =
=
c
f
3.00 × 108 m s −1
4.59 × 1014 Hz
= 6.53 × 10−7 m
(b) 4.35 × 10−7 m
19.
Since violet light experiences the greater change in
direction, its change in speed must also be greater. The
speed in air
. Since
refractive index is equal to the ratio
speed in flint
this ratio is greater for violet light, violet light travels
more slowly than red light in flint.
13. The frequency of the light determines the photon energy,
and so determines energy delivered to each electron and
the maximum kinetic energy of the electrons. The
maximum kinetic energy also depends on the surface
because this determines the energy supplied to allow the
electron to escape. The wave model predicts that
electrons would be ejected below this threshold
frequency. It would just take a longer time for the energy
to accumulate for electrons to escape.
14. The individual points showing where a chemical reaction
of the film emulsion has occurred, would be the sign of
Jacaranda Physics 2 TSK
20. Constructive: nλ, n = 0, 1, 2, . . ., 0, 1.06 μm, 2.12 μm,
3.18 μm . . .
Destructive: ( n + 12 )λ, n = 0, 1, 2, . . ., 0.53 μm,
1.59 μm, 2.65 μm . . .
48
© John Wiley & Sons Australia, Ltd 2009
21. (a) (i) The bright band corresponds to constructive
interference where crests from the two slits arrive
together, and troughs in the light waves arrive
together.
(ii) The dark band is where destructive interference
occurs. At all times, the sum of the waves from
the two slits is zero, including crests from one
slit coinciding with troughs from the other.
(b) (i) 0
λ
(ii) = 317 nm
2
(iii) 2λ = 1266 nm
(c) The wavelength is smaller so the bright fringes in the
interference pattern would be closer together. As path
difference for the bright fringes equal nλ, a smaller
wavelength means that each bright fringe would be
moved closer to the central bright band.
(d) The increase in distance between the slits would
result in the bright bands being closer together.
(e) Moving the screen away would result in the
interference pattern spreading out, increasing the
distance between the bright bands.
22. Circular wavefronts can be used when the slit width is
extremely narrow in comparison with the wavelength of
the light. Then diffraction produces effectively circular
wavefronts.
23. Each stripe on the soap film corresponds to a certain path
difference for light reflected from the front and back of
the film, resulting in destructive interference for a
particular set of wavelengths. The stripes are
approximately horizontal because the part of the film that
has a certain thickness is a horizontal slice through the
film. As the soap film drains, the part of the film with this
thickness moves downwards, so each stripe does too.
24. The centre of the pattern consists of red and blue light,
hence the pattern would appear magenta coloured. Red
light diffracts more than blue light and so there would be
a point of the centre where there would be red light but no
blue light as it would be at a minimum in intensity. Thus
the pattern would be magenta in the middle with a red
fringe on either side.
25. For green light at, say, 515 nm:
hc
Ephoton =
λ
6.6262 × 10−34 J s × 2.9979 × 108 m s −1
=
515 × 10−9 m
⇒
Eblue photon
rate of emission of red photons
=
rate of emission of blue photons Ered photon
=
λ red photon
λ blue photon
= 1.33
27. Electron energy = 4.0 × 10−19 J
=
4.0 × 10−19 J
1.60 × 10−19 J eV −1
= 2.5 eV
(a) A retarding voltage of 1.0 V would reduce the kinetic
energy by 1.0 eV to 1.5 eV, or 2.4 × 10−19 J.
(b) A retarding voltage of 2.5 V is required to completely
transform the 2.5 eV of kinetic energy into electric
potential energy, that is, to stop the electron.
(c) A stopping voltage of 4.3 V means that the highest
kinetic energy of electrons is 4.3 eV, or 6.9 × 10−19 J.
(d)
28. (a) Maximum Ek = Ephoton − W
hc
Ephoton =
λ
6.6262 × 10−34 J s × 2.9979 × 108 m s −1
=
254 × 10−9 nm
= 7.82 × 10−19 J
= 4.88 eV
Maximum Ek = 4.89 eV − 2.30 eV
= 2.59 eV, or 4.14 × 10−19 J
(b) To transform this kinetic energy into electric
potential energy would require 2.59 V.
(c) and (d)
= 3.86 × 10−19 J
So the detection limit of 2 × 10−17 J is equivalent to:
2 × 10−17 J
= 52 photons.
3.86 × 10−19 J
P = energy per second
26.
= rate of photon emission × energy of one photon
Pblue = Pred
⇒ rate of emission of blue photons × Eblue photon = rate of
emission of red photons × Ered photon
Jacaranda Physics 2 TSK
49
© John Wiley & Sons Australia, Ltd 2009
4. Atoms and molecules in the outer part of the Sun absorb
at characteristic wavelengths, reducing the intensity of
those colours in the light which reaches us.
5. Emission lines are produced when electrons return from
an excited state to a lower energy state. The energy is
released in the form of photons of particular frequencies.
Absorption lines are produced when light from a
continuous spectrum passes through a gas. This light
excites some of the electrons in the atoms making up the
gas, so photons with the energies allowed by the atoms
will be removed from the continuous spectrum. As the
energy required to raise an electron to a more excited
state is equal to the energy released when the electrons
drop back to the lower state, the emission lines and
absorption lines for a particular element will be the same.
6. Possible answers include refracting the light through a
prism. Spectral yellow will remain yellow whereas a
mixture of green and red light will separate into two
beams.
h
7. (a) λ =
p
h
=
mv
6.63 × 10−34
=
1.67 × 10−27 × 3.0 × 107
29.
(a) Threshold frequency is where the graph crosses the
frequency axis, that is, where the maximum Ek of the
electrons, and the stopping voltage, is zero:
f0 = 4.6 × 1014 Hz
c
(b) λ =
f
=
(c)
(d)
30. (a)
(b)
2.9979 × 108 m s −1
4.6 × 1014 Hz
= 6.5 × 10−7 m
W = hf0
= 6.6262 × 10−34 J s × 4.6 × 1014 Hz
= 3.0 × 10−19 J = 1.9 eV
or, use the intercept with the stopping voltage graph.
Planck’s constant is the gradient of the graph:
6.6 × 10−34 J s, or 4.1 × 10−15 eV s
Red light is diffracted more as it passes through a
narrow slit because it has a longer wavelength. As a
result its diffraction pattern will be broader.
The edge of the pattern will be reached by white
light, but minus the colour which experiences
destructive interference there. The violet light has the
narrower interference pattern, so its first node occurs
closest to the centre of the pattern. The remaining
light will therefore have a yellow tinge.
= 1.3 × 10−14 m
h
(b) λ =
p
h
=
2mE
=
2 × 9.11 × 10−31 × 1.6 × 10−19 × 54
= 1.7 × 10−10 m
h
(c) λ =
p
h
=
mv
Chapter 14
Matter – particles
and waves
=
6.63 × 10−34
0.2 × 50
= 6.6 × 10−35 m
8. (a) Ek = − ΔEp
1. The cathode rays are deflected by both magnetic and
electric fields, and have mass and charge, unlike
electromagnetic radiation.
2. There is an electric field between the plates. When a
negative particle enters the electric field, it is attracted to
the positive plate and repelled from the negative plate.
3. A fire glows with a continuous range of wavelengths, and
in a red fire, the red wavelengths have the greatest
intensity. Neon in a discharge tube glows red because the
electrons in the neon atoms are excited to particular
energies. When the electrons return to the ground state
they produce light of a few fixed wavelengths, mainly in
the red part of the spectrum.
Jacaranda Physics 2 TSK
6.63 × 10−34
= − Vqelectron
= − (5 kV × − 1 e)
= 5 keV
= 5 keV × 1.60 × 10 −16 J keV −1
= 8.0 × 10−16 J
(b) Ephoton = 8.0 × 10−16 J
=
50
hc
λ
© John Wiley & Sons Australia, Ltd 2009
⇒ λ =
=
hc
Ephoton
13. (a) λ =
6.63× 10−34 J s × 3 × 108 m s −1
=
= 9.11 × 10−31 × 2.5 × 106
=
= 2.3 × 10−24 kg m s −1
h
λ=
p
6.63 × 10−34
2.3 × 10−24
λ
≈ 1 . As the
d
wavelength is a little greater than the atomic spacing,
diffraction will be significant.
11. (a) 3000 eV, 4.8 × 10−16 J
(b) 2.96 × 10−23 Ns, 2.2 × 10−11 m
(c) λ/w = 0.045, thus the scientist would not expect to
observe any diffraction effects since the wavelength
of the electrons is too small.
(d) The scientist should make the accelerating voltage
smaller. This would reduce the energy of the
electrons and hence reduce the momentum and hence
increase the wavelength.
(e) 2.2 × 10−11 m, 2.96 × 10−23 Ns. To obtain the same
diffraction pattern the wavelength and hence the
momentum of the photons much be the same as the
electrons used in the previous experiment.
(f) 8.88 × 10−15 J, 5.55 × 104 eV
c = fλ
12.
c
⇒λ =
f
(b) Significant diffraction occurs when
=
14.
2 × 9.11 × 10−31 × 1000 × 1.60 × 10−19
6.63 × 10−34
2 × 1.67 × 10−27 × 1000 × 1.60 × 10−19
= 9.1 × 10−13 m
h
λ =
p
h
=
2mE
h
=
2mqV
⇒V =
=
h2
2mqλ2
(6.63 × 10−34 )2
2 × 9.1 × 10−31 × 1.60 × 10−19 × (2.0 × 10−10 )2
= 38 V
15. For a 10 eV electron:
h
λ=
p
h
=
2mE
3.0 × 108
4.5 × 1014
= 6.7 × 10−7 m
h
p =
λ
h
⇒v =
λm
6.63 × 10−34
=
6.7 × 10−7 × 9.1 × 10−31
=
6.63 × 10−34
2 × 9.1 × 10−31 × 10 × 1.60 × 10−19
= 3.9 × 10−10 m
For a 10 eV photon:
hc
E=
λ
hc
⇒ λ=
E
= 1.1 × 103 m s −1
Jacaranda Physics 2 TSK
6.63 × 10−34
= 3.9 × 10−11 m for the electron
Using m = 1.67 × 10−27 kg for the proton,
h
λ=
p
h
=
2mE
= 2.9 × 10−10 m
=
h
2mE
Since E and h are constant, the larger the value of m,
the smaller the wavelength. So the proton will have
the shorter wavelength.
h
(b) λ =
p
h
=
2mE
8.0 × 10−16 J
= 2.5 × 10−10 m
9. Light exhibits both wave and particle behaviour,
depending on the experiment you are performing at the
time. For further discussion of this issue argue with your
friends and teacher!
10. (a) p = mv
=
h
p
51
© John Wiley & Sons Australia, Ltd 2009
=
6.63 × 10−34 × 3.0 × 108
19. (a) and (b)
10 × 1.60 × 10−19
= 1.2 × 10−7 m
The electron has the smaller wavelength.
16. The magnitude of the energy change of the electrons is
equal to the photon energy, hf.
Ephoton = 6.6262 × 10−34 J s × 4.59 × 1014 Hz
= 3.04 × 10−19 J
=
3.04 × 10−19 J
1.60 × 10−19 J eV −1
= 1.90 eV
ΔEelectron = −3.04 × 10−19 J
= −1.90 eV
17. (a) Light of this wavelength corresponds to a photon
energy of
hc 6.63 × 10−34 × 3.0 × 108
E=
=
= 4.7 × 10−19 J
λ
420 × 10−9
Electrons in helium absorb these photons indicating
that there is an energy level within the helium atom
4.7 × 10−19 J above the ground state of the atom.
(b) The increase in energy of the electron is equal to the
energy of the absorbed photon:
ΔE = 4.7 × 10−19 J = 2.94 eV
(c) E1 = 12.8 − 12.1 = 0.7 eV
E2 = 12.8 − 10.2 = 2.6 eV
E3 = 12.8 eV
E4 = 12.1 − 10.2 = 1.9 eV
E5 = 12.1 eV
E6 = 10.2 eV
18. See table at bottom of page.
For red light:
Convert E (eV) to E (J) by multiplying by 1.60 × 10−19.
E (J)
Calculate p =
.
c
E (J)
Calculate f =
.
h
c
Calculate λ = .
f
For electron:
Convert E (eV) to E (J) by multiplying by 1.60 × 10−19.
Calculate p = 2mE .
h
Calculate λ = .
p
Follow the above in reverse for the blue light and
electron.
(d) The least energy is 0.7 eV. The photon with this
energy has a wavelength of
hc
λ=
E
6.63 × 10−34 × 3 × 108
=
1.60 × 10−19 × 0.7
= 1.78 × 10−6 m (1.8 × 10−6 m)
The greatest energy photon has energy 12.8 eV:
hc
λ=
E
6.63 × 10−34 × 3 × 108
=
1.60 × 10−19 × 12.8
= 9.71 × 10−8 m (9.7 × 10−8 m)
λ (nm)
f (Hz)
E (J)
E (eV)
p (Ns)
Red light
633
4.73 × 1014
3.14 × 10−19
1.96
1.05 × 10−27
Electron
0.877
—
3.14 × 10−19
1.96
7.56 × 10−25
Blue light
405
7.41 × 1014
4.91 × 10−19
3.07
1.64 × 10−27
Electron
405
—
1.47 × 10−24
9.19 × 10−6
1.64 × 10−27
Jacaranda Physics 2 TSK
52
© John Wiley & Sons Australia, Ltd 2009
20. (a) and (b)
Third excited state:
Ephoton = − ΔEatom
= − 0.77 eV − ( −5.12 eV)
= 4.35 eV
= 6.96 × 10−19 J
λ=
=
= − 3.01 eV − (−5.12 eV)
= 2.11 eV
= 3.38 × 10−19 J
=
hc
Ephoton
6.63 × 10−34 J s × 3 × 108 m s −1
3.38 × 10−19 J
= 5.88 × 10−7 m
Second excited state:
Ephoton = − ΔEatom
= − 1.37 eV − (−5.12 eV)
= 3.75 eV
= 6.0 × 10
hc
λ=
Ephoton
=
−19
Ephoton
6.63 × 10−34 J s × 3 × 108 m s −1
6.96 × 10−19 J
= 2.86 × 10−7 m
The energy change from the first excited state is
responsible for the yellow glow. A comparison with the
answers to question 10 helps here. Yellow light is
between red and blue light in the spectrum, that is
between 405 nm and 633 nm. The first excited state is the
only one that produces a wavelength in this range.
21. At room temperature, virtually all hydrogen atoms are in
their ground state. As a result, the absorption spectrum
shows only lines corresponding to absorption by atoms in
the ground state. The emission spectrum is formed when
electrons which are in excited states due, for example, to
the energy supplied by a flame or an electron in an
electric discharge emit light as they drop to lower energy
levels. Since there is no restriction on which levels they
can fall to, series of spectra are seen which correspond to
any final state of the atom, not just the ground state. The
0.0122 nm UV radiation corresponds to the transition
between the ground and first excited state, so it will
appear in both emission and absorption spectra. The
656 nm light corresponds to a transition between the first
and second excited states of hydrogen. It will appear in
the emission spectrum but not the absorption spectrum.
22. There is no wave equation for matter of the type v = fλ
since the speed of matter, unlike light, does not depend on
the medium through which it moves. For photons, the
energy is ‘locked-up’ in the frequency variable f: E = hf.
For matter, the energy is ‘locked-up’ in the speed, which
2
unlike light is variable: E = 12 mv . It serves no purpose
(c) Ephoton = −ΔEatom = Eatom initial − Eatom final
First excited state:
Ephoton = − ΔEatom
λ=
hc
to develop a frequency for matter. However, the
de Broglie wavelength is useful as it determines the
extent to which both light and matter behave like a wave
and determines the extent to which both matter and light
behaves like a particle.
J
6.3 × 10−34 J s × 3 × 108 m s −1
6.0 × 10−19 J
= 3.32 × 10−7 m
Jacaranda Physics 2 TSK
53
© John Wiley & Sons Australia, Ltd 2009
Unit
Detailed study 3.1:
Synchrotron and its
applications
4
■■■■■■■■■
Jacaranda Physics 2, 3rd Edition TSK
Part A — WORKED SOLUTIONS
V
d
480
=
0.12
= 4.0 × 103 V m−1 from the bottom plate to the top
plate
(b) F = Eq
= 4.0 × 103 × 1.6 × 10−19
= 6.4 × 10−16 N
(c) W = qV
= 480 eV
= 7.7 × 10−17 J
(d) Increase in kinetic energy = work done
= 7.7 × 10−17 J
Chapter 15
5. (a) E =
The Australian
synchrotron
1. Similarities: Both accelerate electrons to high speeds in a
straight line using a potential difference.
Differences: The cathode ray tube uses a single pair of
charged plates, while a linear accelerator uses a series of
charged plates or tubes whose polarities are reversed
as the electron passes from one pair of plates, or one tube,
to the next. This enables the linac to give the electron
energy that would be impractical with a cathode ray tube
due to the high voltages that would be required.
2. The filament is a wire that is heated with an electric
current. This heating gives some of the electrons in the
filament sufficient energy to escape the wire with the
assistance of the potential difference between the charged
plates.
3. A voltage can be used to accelerate a charged particle
because where there is a voltage there is an electric field.
An electric field exerts a force on charged particles and
forces result in acceleration of the particles. A particle
that is neutral, however, is not influenced by the electric
field.
4. (a)
1
2
6. (a)
(b)
(c)
(b)
(d)
= 7.7 × 10−17 J
× v = 7.7 × 10−17 J
⇒ v = 1.3 × 107 m s−1
V
E=
d
480
=
0.24
= 2.0 × 103 V m−1 from the bottom plate to the top
plate
F = Eq
= 2.0 × 103 × 1.6 × 10−19
= 3.2 × 10−16 N
W = qV
= 480 eV
= 7.7 × 10−17 J
Increase in kinetic energy = work done
= 7.7 × 10−17 J
F = Eq
1
2
Vq
=
d
200 × 1.60 × 10−19
=
0.10
= 3.2 × 10−16 N
× 9.1 × 10
1
mv 2
2
−31
2
× 9.1 × 10
1
mv 2
2
−31
2
= 7.7 × 10−17 J
× v = 7.7 × 10−17 J
v = 1.3 × 107 m s−1
⇒
7. (a) For an electron:
(i) V = 1000 V
E = Vq
= 1000 × 1.6 × 10−19
(c)
= 1.6 × 10−16 J
1
2
⇒
2 × 3.2 × 10−16 = 6.4 × 10−16 N
Jacaranda Physics 2 TSK
× 9.1 × 10
54
1
2
−31
mv 2 = 1.6 × 10−16 J
× v 2 = 1.6 × 10−16 J
v = 1.9 × 107 m s−1
© John Wiley & Sons Australia, Ltd 2009
(ii) V = 10 000 V
(ii) V = 10 000 V
E = Vq
= 10000 × 1.6 × 10
E = Vq
−19
= 10000 × 3.2 × 10−19
= 1.6 × 10−15 J
1
2
× 9.1 × 10
1
mv 2
2
−31
2
= 3.2 × 10−15 J
= 1.6 × 10−15 J
× v = 1.6 × 10−15 J
1
2
v = 5.9 × 107 m s−1
⇒
(iii) V = 100 000 V
v = 9.8 × 105 m s−1
E = Vq
−19
= 1 000 000 × 3.2 × 10−19
= 1.6 × 10−14 J
1
2
× 9.1 × 10
1
2
−31
= 3.2 × 10−14 J
mv 2 = 1.6 × 10−14 J
× v 2 = 1.6 × 10−14 J
1
2
v = 1.9 × 108 m s−1
⇒
(b) For a proton:
(i) V = 1000 V
V =
= 1000 × 1.6 × 10−19
= 1.6 × 10−16 J
1
2
× 1.67 × 10
1
2
× 1.67 × 10
⇒
(iii) V = 100 000 V
J
× v 2 = 1.6 × 10−15 J
v = 1.4 × 106 m s−1
E = Vq
= 1.6 × 10−14 J
× 1.67 × 10
= 1.6 × 10−14 J
× v = 1.6 × 10−14 J
⇒
v = 4.4 × 106 m s−1
(c) For an alpha particle:
(i) V = 1000 V
E = Vq
= 3.2 × 10−16 J
1
2
× 6.64 × 10
⇒
Jacaranda Physics 2 TSK
× 9.1 × 10−31 × (3.0 × 107 ) 2
1.6 × 10−19
1
2
× 1.67 × 10−27 × (3.0 × 107 ) 2
1.6 × 10−19
1
2
× 6.64 × 10−27 × (3.0 × 107 )2
3.2 × 10−19
= 5.1 × 1016 m s −2
10. (a) F = Bqv
= 0.25 × 1.60 × 10−19 × 5.0 × 106
= 2.0 × 10−13 N
F
(b) a =
(the electron is travelling at speeds where
m
relativistic effects are not important)
= 1000 × 3.2 × 10−19
1
mv 2
2
−27
2
1
2
= 9.3 × 106 V
9. (a) F = Bqv
= 2.4 × 1.60 × 10−19 × 1.2 × 105
= 4.6 × 10−14 N
(b) Assuming the electron to be in a synchrotron with the
magnetic field coming from above, the electron will
be deflected towards the south (using the right-handslap rule or equivalent), resulting in a circular arc for
the period that the electron is in the magnetic field.
F
(the electron is travelling at speeds where
(c) a =
m
relativistic effects are not important)
4.6 × 10−14 N
=
9.1 × 10−31
= 100 000 × 1.6 × 10−19
1
2
v = 3.1 × 106 m s−1
q
(c) V =
mv 2 = 1.6 × 10−15 J
1
mv 2
2
−27
2
× v = 3.2 × 10−14 J
= 4.7 × 106 V
= 10000 × 1.6 × 10−19
1
2
−27
= 3.2 × 10−14 J
mv 2
(b) V =
E = Vq
= 1.6 × 10
1
mv 2
2
−2
2
= 2.6 × 103 V
v = 4.4 × 105 m s−1
−15
1
2
(a) V =
= 1.6 × 10−16 J
× v = 1.6 × 10−16 J
⇒
(ii) V = 10 000 V
× 6.64 × 10
⇒
2
1
8. 2 mv = qV
E = Vq
1
mv 2
2
−27
2
= 3.2 × 10−15 J
× v = 3.2 × 10−15 J
⇒
(iii) V = 100 000 V
E = Vq
= 100 000 × 1.6 × 10
× 6.64 × 10
1
mv 2
2
−2
2
= 3.2 × 10−16 J
× v = 3.2 × 10−16 J
v = 3.1 × 105 m s−1
55
© John Wiley & Sons Australia, Ltd 2009
=
2.0 × 10−13 N
9.1 × 10
16. B =
−31
= 2.2 × 10−17 m s −2
F
(c) a =
(the proton is travelling at speeds where
m
relativistic effects are not important)
2.0 × 10−14 N
=
1.67 × 10−27
=
17.
18.
= 1.2 × 1013 m s −2
11. (a) Down the page
(b) Out of the page
(c) Down the page perpendicular to the velocity of the
proton
19.
12.
13. Out of the page
20.
mv 2
14. F =
r
mv 2
Bqv =
r
mv
r=
Bq
(a) r =
22.
9.1 × 10−31 × 3.0 × 107
4.0 × 1.6 × 10−19
= 4.3 × 10−5 m
(b) r =
1.67 × 10−27 × 3.0 × 107
4.0 × 1.6 × 10−19
= 7.8 × 10−2 m
(c) r =
6.64 × 10−27 × 3.0 × 107
4.0 × 3.2 × 10−19
= 0.16 m
15.
mv 2
r
mv 2
Bqv =
r
mv
B=
qr
F=
=
23.
9.1 × 10−31 × 3.0 × 107
1.6 × 10−19 × 0.10
= 1.7 × 10−3 T perpendicular to the velocity of the
electron
Jacaranda Physics 2 TSK
56
p
qr
1.0 × 10−18
1.6 × 10−19 × 1000
= 6.3 × 10−3 T
p = Bqr
= 2.0 × 1.6 × 10−19 × 34.4
= 1.1 × 10−17 kg m s−1
Synchrotrons are designed to produce intense beams of
electromagnetic radiation by accelerating charged particles
(synchrotron radiation). In a synchrotron, electrons are
accelerated to near light speeds and their circular motion
ensures they are accelerating and emitting synchrotron
radiation.
Linac: accelerates electrons to 99.995% of the speed of
light.
Circular booster: increases the energy of the electrons
greatly so that they travel at 99.999994% of the speed of
light.
Storage ring: uses magnets in straight-line sections joined
by curved sections to deflect electrons into a circular path
and emit synchrotron radiation at a tangent to the path.
Beamlines: where the synchrotron radiation is directed
onto an experiment.
An electron gains speed in the electron gun, the linac, the
circular booster and the straight sections of the storage
ring. It loses speed in the curved magnets of the storage
ring.
As these photons have energy less than 100 keV, they
will undergo Thompson scattering. This results in Bragg
diffraction.
nλ = 2d sin θ
nλ
⇒ sin θ =
2d
0.125 × 10−9 × n
=
2 × 0.35 × 10−9
= 0.179n
When n = 1, θ = 10.3°;
n = 2, θ = 21.0°;
n = 3, θ = 32.5°;
n = 4, θ = 45.7°;
n = 5, θ = 63.5°.
Higher values of n mean that 0.179n will be greater than
1. There is no angle whose sine is greater than 1, so
photons are only detected at these five angles.
(a)
nλ = 2d sin θ
nλ
⇒
= sin θ
2d
⇒ sin 15.4° = 0.266
nλ
= 0.266 when n = 1.
So
2d
nλ
= 0.531 ⇒ θ = 32°.
For n = 2,
2d
For n = 3, θ = 53°.
For n = 4 and above, there is no solution.
© John Wiley & Sons Australia, Ltd 2009
(b)
nλ = 2d sin θ, n = 1, θ = 22.5° (half the detector
angle), λ = 0.154 nm
1 × 0.154 × 10−9 = 2 d sin 22.5°
d = 2.01 × 10−10 m
nλ = 2d sin θ, n = 2, θ = 50° (half the detector
angle), λ = 0.154 nm
2 × 0.154 × 10−9 = 2 d sin 50°
d = 2.01 × 10−10 m
(iii) (2.0 ± 0.1) × 10−10 m. Add up the five values,
divide by 5. Use the range of values to estimate
the uncertainty.
(c) For RbCl:
(i) 20°, 24°, 46°, 53°, 87°
The angles 20° and 46° correspond to the
shorter wavelength photon as the angles are
smaller than the corresponding order of peaks
for the other series of peaks in accordance with
Bragg’s Law.
(ii) nλ = 2d sin θ, n = 1, θ = 10° (half the detector
angle), λ = 0.138 nm
1 × 0.138 × 10−9 = 2 d sin 10°
d = 3.97 × 10−10 m
nλ = 2d sin θ, n = 2, θ = 23° (half the detector
angle), λ = 0.138 nm
2 × 0.138 × 10−9 = 2 d sin 23°
d = 3.53 × 10−10 m
nλ = 2d sin θ, n = 1, θ = 12° (half the detector
angle), λ = 0.154 nm
1 × 0.154 × 10−9 = 2 d sin 12°
d = 3.70 × 10−10 m
nλ = 2d sin θ, n = 2, θ = 26.5° (half the detector
angle), λ = 0.154 nm
2 × 0.154 × 10−9 = 2 d sin 26.5°
d = 3.45 × 10−10 m
nλ = 2d sin θ, n = 3, θ = 43.5° (half the detector
angle), λ = 0.154 nm
3 × 0.154 × 10−9 = 2 d sin 43.5°
d = 3.36 × 10−10 m
(iii) (3.6 ± 0.4) × 10−10 m Add up the five values,
divide by 5. Use the range of values to estimate
the uncertainty.
27. (a) To calculate the angles for the three principal planes
of each crystal, use Bragg’s Law,
nλ = 2d sin θ.
nλ
⇒ sin θ =
2d
⎛ nλ ⎞
⇒ θ = sin −1 = ⎜
⎟
⎝ 2d ⎠
Calculate θ for each crystal spacing for n = 1, 2, 3 . . .
for valid values of sin θ (that is, where sin θ ≤ 1).
(b) To construct the diffraction pattern, the intensity of
peaks relative to the first peak can be plotted against
the detector angle for each crystal. This can be done
by setting the first peak for D1 at 100. The
subsequent heights are found by multiplying the
number in brackets by 100.
λ
= 0.266
2d
⇒ λ = 1.2 × 10−10 m
24. As nλ = 2d sin θ and d and λ are unknown, we have one
equation with two unknowns. There are many possible
wavelengths and crystal spacings that would produce
these diffraction angles.
25. nλ = 2d sin θ
For n = 1,
λ = 2 × 543.09 nm × sin 10.00°
= 188.6 nm
26. (a) For NaCl:
(i) 28°, 34°, 60°, 67°, 112°
The angles 28° and 60° correspond to the
shorter wavelength photon as the angles are
smaller than the corresponding order of peaks
for the other series of peaks in accordance with
Bragg’s Law.
(ii) nλ = 2d sin θ, n = 1, θ = 14° (half the detector
angle), λ = 0.138 nm
1 × 0.138 × 10−9 = 2 d sin 14°
d = 2.85 × 10−10 m
nλ = 2d sin θ, n = 2, θ = 30° (half the detector
angle), λ = 0.138 nm
2 × 0.138 × 10−9 = 2 d sin 30°
d = 2.76 × 10−10 m
nλ = 2d sin θ, n = 1, θ = 17° (half the detector
angle), λ = 0.154 nm
1 × 0.154 × 10−9 = 2 d sin 17°
d = 2.63 × 10−10 m
nλ = 2d sin θ, n = 2, θ = 33.5° (half the detector
angle), λ = 0.154 nm
2 × 0.154 × 10−9 = 2 d sin 33.5°
d = 2.79 × 10−10 m
nλ = 2d sin θ, n = 3, θ = 56° (half the detector
angle), λ = 0.154 nm
3 × 0.154 × 10−9 = 2 d sin 56°
d = 2.79 × 10−10 m
(iii) (2.8 ± 0.2) × 10−10 m. Add up the five values,
divide by 5. Use the range of values to estimate
the uncertainty.
(b) For LiCl:
(i) 41°, 45°, 87°, 100°
The angles 41° and 87° correspond to the
shorter wavelength photon as the angles are
smaller than the corresponding order of peaks
for the other series of peaks in accordance with
Bragg’s Law.
(ii) nλ = 2d sin θ, n = 1, θ = 20.5° (half the detector
angle), λ = 0.138 nm
1 × 0.138 × 10−9 = 2 d sin 20.5°
d = 1.97 × 10−10 m
nλ = 2d sin θ, n = 2, θ = 43.5° (half the detector
angle), λ = 0.138 nm
2 × 0.138 × 10−9 = 2 d sin 43.5°
d = 2.00 × 10−10 m
Jacaranda Physics 2 TSK
57
© John Wiley & Sons Australia, Ltd 2009
For example, for Algodonite, the number in
brackets for D2 is 0.4, so set the first peak for D2 to
40. For D3, the number in brackets is 0.2, so set the
first peak for D3 to 20.
The detector angle is twice the Bragg angle.
The graph can be drawn freehand with the height
values on the Y-axis and the detector angles on the
X-axis, with the graph coming back down to the
X-axis after each point. Alternatively, the graph can
be drawn in a spreadsheet program such as Microsoft
Excel. For Excel you need to enter dummy data
points either side of each proper point to bring the
graph down to the X-axis, for example for data point
(38.6, 100), include two extra points (38,0) and
(39,0). When finished, sort your data, do an ‘X−Y
scatter’ graph and select the graph type that has
straight lines between the points.
(b)
Aluminium: Detector Angles
D1
Height
D2
Height
D3
Height
32.4
100
65.0
90
37.6
70
67.7
50
80.3
35
113.4
25
150.4
17
(a)
Algodonite: Bragg Angles (degrees)
D1
D2
D3
19.3
18.2
17.0
41.4
38.6
35.8
82.6
69.2
61.2
(a)
Aurostibite: Bragg Angles (degrees)
(b)
Algodonite: Detector Angles
D1
Height
D2
Height
D3
Height
38.6
100
36.3
40
34.0
20
82.8
50
77.1
20
71.5
10
165.2
25
138.4
10
122.5
5
D1
D2
D3
19.1
12.7
11.4
40.9
26.2
23.3
78.9
41.4
36.3
62.0
52.2
80.8
(b)
Aurostibite: Detector Angles
D1
Height
D2
Height
D3
Height
38.2
100
25.5
75
22.8
70
81.7
50
52.4
37
46.5
35
157.8
25
82.9
18.0
72.6
17
123.9
9
104.3
8
161.7
4
(a)
Aluminium: Bragg Angles (degrees)
D1
D2
D3
16.2
32.5
18.8
33.9
40.1
56.7
75.2
Jacaranda Physics 2 TSK
58
© John Wiley & Sons Australia, Ltd 2009
(a)
Bilibinskite: Bragg Angles (degrees)
D1
D2
D3
16.1
32.3
18.7
33.7
39.9
56.3
74.2
(b)
(a)
Bilibinskite : Detector Angles
D1
Height
D2
Height
D3
Height
32.2
100
64.5
75
37.4
70
D1
D2
D3
67.4
50
79.8
35
18.6
31.4
37.7
112.7
25
148.4
17
39.7
Diamond: Bragg Angles (degrees)
73.2
(b)
Diamond: Detector Angles
D1
Height
D2
Height
D3
Height
37.2
100
62.9
25
75.4
16
79.3
50
146.5
25
(a)
Copper: Bragg Angles (degrees)
D1
D2
D3
18.4
21.3
31.0
39.0
46.7
70.9
28. X-rays travel in straight lines and are not charged. They
diffract as though they had a wavelength less than that of
visible light.
29. (a) 1.2 × 10−10 = 0.12 × 10−9 = 0.12 nm
c
(b) f =
λ
3.0 × 108
=
1.2 × 10−10
(b)
Copper: Detector Angles
D1
Height
D2
Height
D3
Height
36.7
100
42.7
46
61.9
20
78.1
50
93.3
37
141.7
25
Jacaranda Physics 2 TSK
= 2.5 × 1018 Hz
59
© John Wiley & Sons Australia, Ltd 2009
(c) E = hf
= 6.63 × 10−34 × 2.5 × 1018
= 1.7 × 10−15 J
= 10 keV
h
(d) p =
λ
6.63 × 10−34
=
1.2 × 10−10
(ii) λ =
(iii) λ =
31.
= 2.0 × 1018 Hz
3.00 × 108
32.
23 × 10−9
= 1.3 × 1016 Hz
(iii) f =
3.0 × 108
57 × 10−9
33.
= 5.3 × 1015 Hz
(b) λ =
c
f
(i) λ =
34.
3.0 × 108
3.2 × 1018
= 9.4 × 10−11 m
Jacaranda Physics 2 TSK
7.5 × 1019
= 4.0 × 10−12 m
= 5.5 × 10−24 kg m s −1
c
30. (a) f =
λ
3.0 × 108
(i) f =
0.15 × 10−9
(ii) f =
3.0 × 108
60
3.0 × 108
5.7 × 1017
= 5.3 × 10−10 m
Compton scattering occurs when a beam of photons
collides with electrons. When a beam of photons is shone
through a material, some of the photons are scattered by
electrons in the material. These photons have longer
wavelength and lower frequency than the photons in the
beam, indicating that they have lost energy. The
scattering can be treated like a collision between two
particles, where momentum and energy are conserved.
Compton scattering requires high energy (>100 keV)
photons.
Eincoming photon = Escattered photon + Ek scattered electron
Ek scattered electron = hfincoming photon − hfscattered photon
= 6.63 × 10−34 (2.2 − 1.4) × 1018
= 5.3 × 10−16 J
= 3.3 keV
Thompson scattering occurs with photons with energy
less than 100 keV. The photons are scattered without any
change in energy, unlike the loss of energy that occurs in
Compton scattering.
Reflection of X-rays is really an interference effect.
Wavelets of X-rays are scattered by each atom in the
crystal plane. Wavelets from adjacent atoms overlap and
reinforce at an angle equal to the incident angle.
© John Wiley & Sons Australia, Ltd 2009
Unit
4
Detailed study 3.2:
Photonics
Jacaranda Physics 2, 3rd Edition TSK
Part A — WORKED SOLUTIONS
■■■■■■■■■
Chapter 16
Light and lightemitting devices
6.
1. A photon
2. (a) Electricity heating the filament — the electrons gain
sufficient energy from the hot surroundings that they
emit visible light when returning to the ground state.
(b) The high temperatures produced in the burning of the
candle provide electrons in the gas surrounding the
candle with sufficient energy that they emit visible
light when returning to ground state.
(c) Fusion reactions in the core produce very high
temperatures, causing the Sun to behave as an
incandescent light source.
(d) The atoms are vibrating sufficiently due to the
thermal energy that collisions between them result in
the emission of photons across the visible spectrum.
(e) A current of electrons is passed through a low density
gas. Atoms that are hit by the electrons are excited.
When the electrons in these atoms return to lower
energy states they emit photons. Fluorescent tubes
produce light in the UV wavelengths. A phosphor
coating on the tube absorbs the UV light and emits
photons in the visible wavelengths, producing a
whitish light.
(f) LEDs emit light when they are forward biased
because as conducting electrons fall into holes in the
p-type material, they release energy in the form of
photons.
3. Lasers produce coherent light. The input power to the
laser excites a high percentage of the atoms of an optical
resonator to a high energy state. This is known as a
population inversion. The atoms are then stimulated to
release their energy (in the form of photons) by
interacting with photons of the same energy. This is
called stimulated emission. The process is also called
amplification because it takes one photon and produces
many identical photons.
4. Remote sensing is the analysis of photons that have come
from a distant place in order to gain an understanding of
that place.
(a) By analysing the infra-red radiation coming from the
surface of a planet. The wavelength with the peak
intensity indicates the average surface temperature.
(b) Water in the atmosphere of a planet reduces the
intensity of some wavelengths of electromagnetic
radiation coming from the surface of the planet.
(c) The absorption lines in the spectra from stars reveal
the elements present in the stars’ atmospheres.
5. (a) Mercury
(b) A current of electrons is passed through a low density
gas. Atoms that are hit by the electrons are excited.
Jacaranda Physics 2 TSK
7.
8.
9.
10.
11.
12.
13.
14.
61
When the electrons in these atoms return to lower
energy states they emit photons. A phosphor coating
on the tube absorbs the UV light and emits photons in
the visible wavelengths, producing a whitish light.
Not all diodes emit visible light because the energy
released goes into increasing the temperature of the
crystal lattice.
(a) Because the energy levels of atoms split when they
come close to other atoms, as is the case with the
atoms of a solid. The many splittings that occur in the
energy levels due to an atom being in a solid result in
an energy band of many energy levels close together,
rather than a specific energy level.
(b) The valence band in a solid is the highest natural
energy band of a material and contains the valence
electrons.
(c) The conduction band is above the valence band. The
energy levels of different atoms join to form a
common band in which it is easy for an electron to
pass between atoms — conduction.
(d) The difference in energy between the conduction and
valence bands (the gap energy) is greater for
insulators than for semiconductors.
(a) The band gap is the difference in energy between the
conduction and valence bands.
(b) The band gap determines the wavelength of photon
hc
.
emitted by an LED by E =
λ
The wavelength and frequency of a released photon is
related to the gap energy of the LED by the formula
Eg = hf = hc/λ. The materials used in constructing the
LED are manipulated to produce a desired wavelength of
photon.
1
Eg ∝
λ
Blue, green, red
2.14 eV
6.9 × 10−7 m
hc
E=
λ
hc
⇒λ=
E
4.14 × 10−15 × 3.00 × 108
=
2.09
−7
= 5.94 × 10 m
= 594 nm
hc
E=
λ
4.14 × 10−15 × 3.00 × 108
=
930 × 10−9
= 1.34 eV
© John Wiley & Sons Australia, Ltd 2009
E=
15.
=
hc
λ
4.14 × 10−15 × 3.00 × 108
(c)
1500 × 10−9
= 0.828 eV
16. (a) Coherent photons have the same wavelength and are
in phase.
(b) Light bulbs produce photons that are out of phase
because their emission is spontaneous and therefore
at random.
17. (a) Light amplification by stimulated emission of
radiation
(b) One photon results in the emission of multiple
identical photons.
(c) The atoms in a gas are excited by the power supply of
the laser and are stimulated to emit a photon by a
photon of the correct energy. The photons that are
travelling along the length of the laser resonator are
reflected back and forward between two mirrors,
resulting in the stimulated emission of many coherent
photons. One of the mirrors allows some of the
photons to pass through, forming the coherent laser
beam. The laser light is parallel because the photons
travelling in other directions escape the resonator
without being amplified.
18. • The beam does not spread out as much.
• There is a smaller range of wavelengths produced.
• The light has a greater intensity.
5. (a)
An example of 2 modes
in a optic fibre
(b) A low-order mode makes few internal reflections; a
high-order mode makes many internal reflections.
6. (a) A single-mode step-index optical fibre is a very thin
core with thick cladding that only has one possible
mode for light travelling along it.
(b) A single-mode step-index optical fibre is a very thin
core with thick cladding; a multimode step-index
optical fibre has a thicker core and thinner cladding.
(c) A multimode step-index optical fibre has a core of
constant refractive index; a graded-index optical fibre
has a refractive index that decreases from the centre
to the edge of the core.
7. (a) Dispersion is the spreading out of a light signal or
pulse as it travels down an optical fibre.
(b) Modal dispersion occurs when there is more than one
possible path (mode) for the light to take. Some of
the modes are longer than others, resulting in the
spreading of the signal over the length of the fibre.
(c) As dispersion results in the pulse spreading out,
fewer pulses per second can be sent, which means a
lower bandwidth.
(d) Use a smaller diameter fibre; use a graded-index
fibre; use a single-mode fibre.
8. (a) A graded-index optical fibre has a refractive index
that decreases from the centre to the edge of the core.
(b)
Chapter 17
Fibre-optic
waveguides and
systems
1. The core and the cladding that surrounds it
2. 500 MHz
3. A core surrounded by cladding of lower refractive index.
This is covered by a protective layer called a jacket.
n1 sin θc = n2 sin 90°
4. (a)
⇒
(b)
Jacaranda Physics 2 TSK
⎛n ⎞
θc = sin −1 ⎜ 2 ⎟
⎝ n1 ⎠
⎛ 1.53 ⎞
= sin −1 ⎜
⎟
⎝ 1.58 ⎠
= 75.5°
⎛n ⎞
θc = sin −1 ⎜ 2 ⎟
⎝ n1 ⎠
⎛ 1.46 ⎞
= sin −1 ⎜
⎟
⎝ 1.48 ⎠
= 80.6°
(c) Modes that take the light on a longer path and hence
closer to the edge of the fibre spend more time in
lower refractive index fibre where the light travels
faster. Those that take the light closer to the centre of
the core and therefore on a shorter path travel more
slowly, reducing the spreading due to modal effects.
9. (a) If there is only one mode, then all light signals will
pass through the fibre in the same time.
(b) Material dispersion
(c) Material dispersion spreads the pulse so that the
number of pulses per second that can be sent is
reduced — lower bandwidth.
(d) The modal dispersion is a much larger effect.
⎛n ⎞
θc = sin −1 ⎜ 2 ⎟
⎝ n1 ⎠
⎛ 1.46 ⎞
= sin −1 ⎜
⎟
⎝ 1.47 ⎠
= 83.3°
62
© John Wiley & Sons Australia, Ltd 2009
Attenuation
(2)
length of fibre
Pin and Pout can be calculated by rearranging formula 1 to
Pout
= 10− attenuation/10
Pin
10. (a) NA = n12 − n22
and Fibre attenuation =
= 1.482 − 1.462
= 0.242
α = sin −1 NA
= 14°
(b) NA = n12 − n22
Fibre
attenuation
per km
Attenuation
(dB km−1)
(dB)
P in
(mW)
Pout
(mW)
Length
(km)
= 0.171
7.5
1.5
20
α = sin −1 NA
4.0
0.15
0.5
14
28
5.0
0.005
2.0
30
15
0.51
0.41
0.98
0.92
2.3
2.0
0.046
0.55
2
= 1.47 − 1.46
2
= 9.9°
(c) NA = n12 − n22
= 1.582 − 1.532
= 0.394
0.35
0.94
12
17. n1 sin θ c = n2 sin 90°
α = sin −1 NA
⎛n ⎞
θ c = sin −1 ⎜ 2 ⎟
⎝ n1 ⎠
⎛ 1.33 ⎞
= sin −1 ⎜
⎟
⎝ 1.50 ⎠
= 62.5°
= 23°
11. (a) The loss of optical power per kilometre along the
fibre
(b) Scattering, due to imperfections in the fibre resulting
in some of the light travelling in different directions;
absorption, when imperfections in the fibre absorb
some of the light, turning it into thermal energy.
(c) The unit of attenuation is the dB km−1.
12. (a) Rayleigh scattering is the partial reflection of light
when it passes from one refractive index to another.
(b) Variations in composition and structure of the fibre
(c) It increases with frequency.
13. (a) Microphone: changes sound waves into an electrical
signal
(b) LED: changes electrical signal into a light signal
(c) Modulator: imposes the electrical signal onto the
light carrier
(d) Optical fibre: carries the light signal from one place
to another
(e) Photodiode: changes the light signal into an electrical
signal
(f) Multiplexer: sends a number of signals along the
fibre at one time
14. Pulse coded modulation involves converting the analogue
electrical signal into a pulse modulated electrical signal.
To do this, the voltage of the signal is measured
(sampled) thousands of times per second. These sample
voltages are then converted into binary form as a series of
pulses.
15. Optical fibre sensors can be embedded in structures and
skins to carry information from across the structure to
central monitoring equipment.
16. The missing values in the table are calculated using the
formulae
P
Attenuation = − 10 log10 out
(1)
Pin
Jacaranda Physics 2 TSK
7.0
18. (a) NA = n12 − n22
= 1.502 − 1.462
(b)
19. (a)
(b)
(c)
= 0.344
α = sin−1 NA
= 20°
0.96 μm (read directly from the graph)
1.7 dB km−1 (read directly from the graph)
There would be less optical loss due to Rayleigh
scattering.
20. (a) NA = n12 − n22
= 1.52 − 1.42
(b)
(c)
21. (a)
(b)
22. (a)
(b)
(c)
63
= 0.539
α = sin−1 NA
= 33°
Use a fibre with a smaller diameter. As it is a stepindex fibre, the option of a graded-index fibre is ruled
out. It is a short-distance fibre, so a multimode fibre
would probably be used.
2 dB km−1 (read directly from the graph)
Approximately 1300 nm or >1500 nm, where the
fibre attenuation is least
The optical fibre represented in figure 17.22 is better
because it has a smaller minimum attenuation than
the one described in question 21.
1.0–1.7 µm (read directly from the graph)
The increase in attenuation at a wavelength of about
1.4 μm (1400 nm) in both materials is due to the
presence of hydroxyl (OH−) ions. The difference in
© John Wiley & Sons Australia, Ltd 2009
25. (a) Infra-red
(b) 1.4458 to 1.4470
(c)
c
v=
n
attenuation between the materials at this wavelength
occurs because they contain different amounts of
hydroxyl ions.
P
23. (a) Attenuation = − 10 log out
Pin
=
0.020
= − 10 log
0.50
= 14 dB
(b) Attenuation per kilometre =
= 2.0718 × 108 m s −1
v=
14
2.0
= 7.0 dB km −1
24. (a) Attenuation = − 10 log
= − 10 log
=
Pout
Pin
2.9979 × 108
1.4458
(d) The range of wavelengths produced by the LED will
be transmitted through the optical fibre with a range
of speeds, spreading the pulses and hence causing
dispersion.
(e) This is material dispersion.
(f) Material dispersion is reduced by minimising the
range of frequencies in the optical signal.
0.048
2.5
17
= 14 km
1.2
Jacaranda Physics 2 TSK
c
n
= 2.0735 = 108 m s −1
= 17 dB
(b)
2.9979 × 108
1.4470
64
© John Wiley & Sons Australia, Ltd 2009
Unit
Detailed study 3.3:
Sound
4
Jacaranda Physics 2, 3rd Edition TSK
Part A — WORKED SOLUTIONS
■■■■■■■■■
10. v = 340 m s−1
f = 256 Hz
v = fλ
(a)
Chapter 18
Background sound
distance travelled
time taken
996 m
=
3.0 s
⇒λ =
1. speed =
=
v
f
340 m s −1
256 Hz
= 1.33 m
v
(b) λ =
f
= 332 m s −1
2. A wave travels a distance of one wavelength in a time of
one period.
3. No, generally it is only the medium that determines the
speed of the wave.
4. The band may appear to be ‘out of step’ with the music
because the observer sees the band before the
corresponding sound is heard. This is because light
travels at 3.0 × 108 m s−1 and sound travels at
approximately 340 m s−1.
5. High frequency sounds have the same speed as low
frequency sounds.
6. f = 256 Hz
1
T =
f
1
=
256 Hz
1.50 × 103 m s −1
256 Hz
= 5.86 m
=
11.
−1
f (Hz)
λ (m)
335
500
0.67
300
12
1500
5000
60
24
340
1000
0.34
260
440
0.59
v (m s )
= 3.91 × 10−3 s
200 wavelengths take 200 periods to pass a point.
⇒ t = 200T
= 0.78 s
7. Assume that the flash is seen almost instantly.
d = vt
25
0.30
2.5
12.
= 335 m s −1 × 5.0 s
= 1675 m
= 1.7 × 103 m
λ
8.
v =
T
⇒ λ = vT
13. (a)
= 340 m s −1 × 3.0 × 10−3 s
= 1.02 m
λ
9. v =
T
1.32 m
=
4.0 × 10−3 s
(b)
= 330 m s −1
Jacaranda Physics 2 TSK
65
© John Wiley & Sons Australia, Ltd 2009
(c)
(b) f =
1
T
1
=
4 × 10−3 s
= 250 Hz
v
(c) λ =
f
(d)
330 m s −1
250 Hz
= 1.32 m
16. (a) T = time of one complete cycle
=
14. (a) wavelength
= length of one complete cycle on the graph
= 0.80 m
(b) T = ?
= 8 cm × 2 m s cm −1
= 16 m s (1.6 × 10−2 s)
1
(b) f =
T
1
=
1.6 × 10−2 s
= 62.5 Hz
(c) v = f λ
= 62.5 Hz × 5.28 m
v = 340 m s −1
λ = 0.80 m
v
f =
λ
340 m s −1
=
0.80 m
= 425 Hz
T =
=
(d)
1
f
1
425 Hz
= 330 m s −1
(i) If the frequency is doubled, the period is halved.
If the loudness stays the same, assume that the
amplitude remains constant.
= 2.35 × 10−3 s
(c)
(ii) If the loudness is increased, the amplitude
increases.
Explanation
When a particle is at the centre of a compression (or
rarefaction), it is at its ‘mean’ or ‘rest’ position. As
the wave moves forwards, the particle will move
forward from its rest position. It will return to its rest
position after half a period.
(d)
17. power =
energy 2.0 J
=
time
100s
= 2.0 × 10−2 W
18. intensity =
=
15. (a) Period is given by the ‘length’ of one complete cycle
on the graph.
T = 4 ms
Jacaranda Physics 2 TSK
power
area
4.0 × 10−8 W
0.080 m 2
= 5.0 × 10−7 W m −2
66
© John Wiley & Sons Australia, Ltd 2009
19. power = intensity × area
= 4.5 × 10−5 W m−2 × 2.0 m2
= 9.0 × 10−5 W
20.
I0 = 3.0 × 10−3 W m−2
r = 10 m
Itotal = 5 × I0
= 5 × 3.0 × 10−3 W m−2
= 1.5 × 10−2 W m−2
21. (a) 1.6 × 10−5 W m−2
(b) 4.0 × 10−6 W m−2
(c) 2.5 × 10−7 W m−2
(d) 1.0 × 10−18 W m−2
I
22. (a) L = 10 log10
I0
= 10 log10
L = 10 log10
⇒ ΔL = 10 log10
= 10 log10 0.5
= − 3.0 dB
25. A = 5.0 × 10−5 m 2
E=?
t = 3 min = 180 s
I = 1.0 × 10−2 W m −2
P
A
⇒ P = IA
E = Pt
I =
= IAt
= 1.0 × 10−2 W m −2 × 5.0 × 10−5 m 2 × 180 s
I
1.0 × 10−17 W m −2
26.
5.0 × 10−10 W m −2
1.0 × 10
−17
Wm
−2
27.
= 27 d B
(b) L = 10 log10
3.2 × 10−7 W m −2
1.0 × 10−12 W m −2
= 55 d B
(c) L = 10 log10
4.9 × 10−3 W m −2
28.
1.0 × 10−12 W m −2
= 97 d B
(d) L = 10 log10
0.5 I1
I1
1.8 × 10−9 W m −2
1.0 × 10−12 W m −2
29.
= 9.0 × 10−5 J
The threshold of hearing is the lowest sound intensity that
can be detected by an ear at a particular frequency.
Refer to the figure on page 438.
(a) (i) 36 dB
(ii) 2 dB
(iii) −3 dB
(iv) 10 dB
(b) approx. 3000 Hz
(c) approx. −6 dB
Refer to the figure on page 438.
(a) approx. 300 Hz–10 000 Hz
(b) approx. 6000 Hz
(c) approx. 250 Hz–400 Hz
(a)
= 33 d B
23. (a) I
⎛L
⎞
− 12 ⎟
⎜
⎠
= 10⎝ 10
(b)
⎛ 7.0
⎞
− 12 ⎟
⎜
⎠
= 10⎝ 10
= 5.0 × 10−12 W m −2
(b) I
(c)
⎛ 25
⎞
− 12 ⎟
⎜
⎠
= 10⎝ 10
= 3.2 × 10−10 W m −2
(c) I
30. λ = 2 × nodal separation
= 2 × 0.75 m
= 1.50 m
31.
⎛ 54
⎞
− 12 ⎟
⎜
⎠
= 10⎝ 10
= 2.5 × 10−7 W m −2
(d) I
⎛ 115
⎞
− 12 ⎟
⎜
⎠
= 10⎝ 10
= 0.32 W m
24.
ΔL = 10 log10
(a)
−2
(b)
I2
I1
(c)
If I 2 = 0.5 I1
Jacaranda Physics 2 TSK
(d)
67
© John Wiley & Sons Australia, Ltd 2009
32. (a) f = 330 Hz
The regions of maximum sound intensity are caused
by constructive interference of the waves produced
by the two speakers.
(b) λ = 2 × distance between adjacent nodes or antinodes.
⇒
λ = 2 × (4.0 m − 3.5 m)
= 1.0 m
(c)
v = fλ
= 330 Hz × 1.0 m
36.
= 330 m s −1
String Nodes Anti-nodes
33. (a) f = 4.0 Hz
λ = 1.2 m
amplitude = 10 cm
v=fλ
= 4.0 Hz × 1.2 m
= 4.8 m s−1
(b) nodal separation = λ ÷ 2
= 1.2 m ÷ 2
= 0.60 m
(c) The maximum displacement of the standing wave
= 2 × the amplitude of the individual waves
= 2 × 10 cm = 20 cm
(d) The wavelength of a standing wave is equal to the
wavelength of the individual waves.
λ = 1.2 m
(e) The string is straight 2 times per period, that is, it will
be straight 2f times per second.
2 × 4 = 8 times
The string is straight 8 times per second.
34. (a) v = 250 m s−1
length of string = 1.0 m
For the longest standing wave in a string fixed at both
ends,
λ = 2L
= 2 × 1.0 m
= 2.0 m
(b) v = f λ
v
f =
λ
250 m s −1
=
2.0 m
λ
Tone Harmonic
B
2
1
2L
f0
First
D
5
4
f3 = 4 f0
Fourth
C
3
2
L
f1 = 2 f0
Second
A
4
3
2L
f2 = 3 f 0
3
Third
37. (a) For strings,
f1 = 2 f 0
f1
2
500 Hz
=
2
= 250 Hz
(b) The second harmonic = 2f0
⇒
f0 = 516 Hz ÷ 2
= 258 Hz
38. 810 Hz = 3f0 (third harmonic)
⇒ f0 = 810 Hz ÷ 3
= 270 Hz
f 4 = 1400 Hz
39. (a)
⇒ f0 =
f4 = 5 f0
1400 Hz
5
= 280 Hz
(b) 2nd harmonic frequency = 2f0
= 2 × 280 Hz
= 560 Hz
(c) f2 = 3f0
= 3 × 280 Hz
= 840 Hz
40. (a) (i) V = 340 m s−1
λ = 2L
= 2 × 40 cm
= 80 cm
= 0.80 m
v
(ii) f 0 =
λ
340 m s −1
h
=
0.80 m
⇒ f0 =
= 125 Hz
(c) For strings,
f1 = 2f0
= 2 × 125 Hz
= 250 Hz
(d) The second resonant frequency above the
fundamental corresponds to the third harmonic,
since f2 = 3f0:
f2 = 3 × 125 = 375 Hz
35. (a) f0 = 240 Hz
f1 = 2f0
= 2 × 240 Hz
= 480 Hz
Jacaranda Physics 2 TSK
(b) f3 = 4f0
= 4 × 240 Hz
= 960 Hz
(c) The third harmonic = 3f0
= 3 × 240 Hz
= 720 Hz
(d) The 22nd harmonic = 22f0
= 22 × 240 Hz
= 5280 Hz
= 425 Hz
68
© John Wiley & Sons Australia, Ltd 2009
(b) (i) λ = 2L
= 2 × 60 cm
= 1.20 m
v
(ii) f 0 =
λ
340 m s −1
=
1.20 m
= 283 Hz
(c) (i) λ = 2L
= 2 × 1.21 m
= 2.42 m
v
(ii) f 0 =
λ
340 m s −1
=
2.42 m
= 140 Hz
(d) (i) λ = 2L
= 2 × 1.00 m
= 2.00 m
v
(ii) f 0 =
λ
340 m s −1
=
2.00 m
= 170 Hz
(e) (i) λ = 4L
= 4 × 0.50 m
= 2.0 m
v
(ii) f 0 =
λ
340 m s −1
=
2.0 m
= 170 Hz
(f) (i) λ = 4L
= 4 × 0.25 m
= 1.0 m
v
(ii) f 0 =
λ
340 m s −1
=
1.0 m
= 340 Hz
(g) (i) λ = 4L
= 4 × 0.125 m
= 0.50 m
v
(ii) f 0 =
λ
340 m s −1
=
0.50 m
= 680 Hz
(h) (i) λ = 4L
= 4 × 0.17 m
= 0.68 m
Jacaranda Physics 2 TSK
v
λ
340 m s −1
=
0.68 m
= 500 Hz
(ii) f 0 =
41.
Pipe
Nodes
B
D
3
2
Antinodes
2
1
A
4
3
C
5
4
Pipe
Nodes
Antinodes
λ
L
2L
2L
3
L
2
Resonant
frequency
f1 = 2 f 0
Harmonic
f0
Second
First
f2 = 3 f 0
Third
f3 = 4 f 0
Fourth
Resonant
frequency
Harmonic
42.
λ
4L
Fifth
f2 = 5 f 0
5
C
1
1
4L
f0
First
4L
A
2
2
Third
f1 = 3 f 0
3
4L
B
4
4
Seventh
f3 = 7 f 0
7
43. (a) The pipe is closed at one end as there is a pressure
antinode at the closed end of a pipe when it is
sustaining a standing wave.
(b) 0.70 m. There is a pressure node at the open end of a
pipe. Beyond the open end there will be little
variation in pressure.
(c) λ = 0.40 m (one complete cycle on the graph)
v
(d) f =
λ
340 m s −1
=
0.40 m
= 850 Hz
(e) For this pipe, the longest wavelength is given by:
λ = 4L
= 4 × 0.7 m
D
3
3
= 2.8 m
v
f0 =
λ
340 m s −1
=
2.8 m
= 121 Hz
The frequency is seven times the fundamental, so it is
the seventh harmonic. Only odd numbered harmonics
occur with a pipe closed at one end, so this is the
third resonant frequency above the fundamental.
(f) See previous answer.
f0 = 121 Hz
69
© John Wiley & Sons Australia, Ltd 2009
(g)
This is the first resonant frequency above the
fundamental, that is, the third harmonic.
Variation from
Normal air pressure
(b) The wavelength is given by four times the distance
between neighbouring nodes and antinodes:
4L
3
4 × 0.50 m
=
3
= 0.67 m
λ=
44. (a) Draw pressure variation diagram.
v
λ
340 m s −1
=
0.67 m
(c) f =
Three antinodes suggests the third harmonic.
(b) λ = 2 × distance between adjacent nodes.
L
distance between adjacent nodes =
3
2L
⇒ λ=
3
2 × 0.85 m
=
3
= 0.57 m (0.567 m)
v
(c) f =
λ
340 m s −1
=
0.567 m
= 600 Hz
(d)
= 5.1 × 102 Hz
(d) f 0 =
5.1 × 102 Hz
3
= 1.7 × 102 Hz
(e) The third resonant frequency above the fundamental
is the seventh harmonic as only odd-numbered
harmonics are possible.
f3 = 7 f0
= 7 × 1.7 × 102 Hz
= 1.2 × 103 Hz
(f)
(e)
46.
An example of 2 modes
in a optic fibre
(f) T =
=
1
f
1
600 Hz
= 1.7 × 10−3 s
(g) This is the third harmonic, therefore f = 600 Hz.
45. (a)
λ = 2 × distance between adjacent antinodes
= 2 × (0.506 m − 0.170 m)
= 2 × 0.336 m
= 0.672 m
f = 500 Hz
v = fλ
= 500 Hz × 0.672 m
= 336 m s −1
Jacaranda Physics 2 TSK
70
© John Wiley & Sons Australia, Ltd 2009
Chapter 19
Sounding
good
1. A transducer is a device that transforms energy from one
form to another.
2. The resistance of carbon dust inside the microphone
changes with the pressure exerted on it. The pressure
changes when the diaphragm responds to sound waves,
producing a signal.
10. (a) A baffle is a cabinet or partition that reduces the
interference of the sound waves produced from the
front and back of the speaker.
(b) It improves the fidelity of the sound production.
(c) The sound waves from the back of the speaker cone
have no access to those in front of the speaker cone.
They have an infinitely long journey to take before
interference can occur.
3. They need a power supply.
4. A crystal microphone uses the piezoelectric property of
crystals. They can produce a small electric current that is
proportional to the pressure applied to them.
5. The diaphragm is connected to a coil that is wrapped
around a permanent magnet. As the diaphragm responds
to pressure changes in the air, it moves the coil back and
forth over the magnet, generating an electric current by
electromagnetic induction.
11. The 10 000 Hz sound will be diffracted less than the
200 Hz sound because it has a smaller wavelength.
For the 10 000 Hz sound:
v
λ=
f
340
=
10000
= 0.034 m
The first minimum from the 10 000 Hz sound will occur
at
1.22λ
sin θ =
w
1.22 × 0.034
⇒ θ = sin −1
0.05
= 56° from the normal to the speaker.
For the 200 Hz sound:
v
λ=
f
340
=
200
= 1.7 m
The first minimum from the 200 Hz sound would occur at
1.22λ
sin θ =
w
1.22 × 1.7
⇒
θ = sin −1
0.05
which has no solution.
Therefore there is no minimum in front of the speaker.
6. (a) This is the range of human hearing, so frequencies
outside of this are irrelevant for most uses.
(b) The decibel (dB) is a unit for sound level intensity,
which is what the microphone is designed to respond
to.
(c) Approximately 10 000 Hz (read directly from the
graph)
(d) 30 Hz and 60 Hz correspond to natural frequencies of
the diaphragm of the microphone.
(e) When f = 35 Hz, the frequency response is −12 dB.
When f = 60 Hz, the frequency response is −4 dB.
The difference is 8 dB.
7. (a) A
(b) B
8. A loudspeaker converts electrical energy into sound
energy and is, therefore, a transducer.
9. Dynamic loudspeakers involve a cone attached to a
moveable voice coil wrapped around the central pole of a
circular magnet. The signal from the amplifier passes
through the coil, which results in a force between it and
the magnet proportional to the current. The changing
current results in a changing force, causing the speaker to
move in time and magnitude with the electrical signal.
Jacaranda Physics 2 TSK
71
© John Wiley & Sons Australia, Ltd 2009
v
f
340
=
10000
= 3.4 cm
12. (a) λ =
(b) The first minimum from the 10 000 Hz sound will
occur at
1.22λ
sin θ =
w
1.22 × 0.034
⇒
θ = sin −1
0.05
= 56° from the normal to the speaker.
13. (a) The lower curve belongs to a tweeter. Tweeters
respond best to high frequencies as shown in the
lower curve.
(b) The top curve belongs to a woofer. Woofers respond
best to low frequency sounds as shown in the top
curve.
14. (a) Diffraction is the spreading out of waves as they pass
through an opening or around an obstacle.
(b) High frequency sounds are diffracted less as they
leave a loudspeaker, so will be more audible from
directly in front of the speaker. Low frequency
sounds will diffract much more when leaving the
loudspeaker.
v
f
340
=
200
= 1.7 m
15. (a) λ =
18.
(b) There won’t be much difference as λ is greater than
the gap width and significant diffraction occurs.
(c) The drill will be much louder at A than at B as it
produces sound with a much shorter wavelength and
little diffraction occurs.
19.
v
f
340
=
1250
= 0.272 m
20.
16. λ =
21.
1.22λ
w
1.22λ
w=
sin θ
1.22 × 0.272
=
sin 65°
= 0.366 m
= 0.37 m
sin θ =
⇒
17. (a) For 1500 Hz:
Jacaranda Physics 2 TSK
v
f
343
=
1500
= 0.229 m
For 8500 Hz:
v
λ=
f
343
=
8500
= 0.040 m
(b) For 1500 Hz:
1.22 λ
sin θ =
w
⎛ 1.22 × 0.229 ⎞
⇒ θ = sin −1 ⎜
⎟
0.30
⎝
⎠
= 68.6°
For 8500 Hz:
1.22 λ
sin θ =
w
⎛ 1.22 × 0.040 ⎞
⇒ θ = sin −1 ⎜
⎟
0.30
⎝
⎠
= 9.36°
1.22 λ
(c) w =
sin θ
1.22 × 0.040
=
sin 68.6°
= 0.052 m
(a) Less diffraction
(b) More diffraction
(c) More diffraction
(d) Less diffraction
The fan shape distributes the high frequency sounds more
effectively, because they spread less than the low
frequencies due to diffraction.
Anne is correct because the diffraction of sound from the
speaker increases as the diameter of the speaker
decreases.
(a) Ferric oxide, chromium dioxide
(b) A recording head is an electromagnet formed from a
ring of ferromagnetic material wrapped in a coil of
wire that the signal passes through. A small gap in
the ring produces the north and south poles of the
magnet, to which the magnetic tape is exposed.
(c) The signal must be amplified before going to the
recording head to provide sufficient current to
magnetise the tape.
(d) As the voltage of the signal changes, so does the
strength of the magnet.
(e) The erase head has a wider gap and receives a
constant signal of high frequency alternating current
to jumble the magnetic fields stored on the tape.
λ=
72
© John Wiley & Sons Australia, Ltd 2009
22. Guitar pickups contain a permanent magnet wrapped in a
coil of wire. When the ferromagnetic strings vibrate, they
affect the magnetic field passing through the coil,
inducing an electric current in the coil.
(f) Without the erase head, a recording would be a
combination of any previous recordings and the new
recording.
(g) As the tape moves, the changing magnetic field
induces a current in the playback head.
(h) The playback head produces an electrical signal that
is too weak to drive a loudspeaker, so it is amplified.
Jacaranda Physics 2 TSK
73
© John Wiley & Sons Australia, Ltd 2009