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2. physics

Answers to End-of-chapter questions
Chapter 1
1 a Linear momentum = mass × velocity
[1]
b Momentum is a product of a scalar quantity (mass) and a vector quantity (velocity);
hence it is a vector quantity.
[1]
c v2 = u2 + 2as and u = 0
v = 2 × 3.5 × 40 = 16.7 m s–1 ≈ 17 m s–1
p = mv = 900 × 16.7 ≈ 1.5 × 104
Unit: kg m s–1
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d Initial momentum = final momentum
(4.0 × 2.0) + (4.0 × –3.0) = 8.0v
8.0v = –4.0
4.0
v= −
= –0.50 m s–1
8.0
The objects move to the left with a combined speed of 0.50 m s–1
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2 a i
In an elastic collision, both momentum and kinetic energy are conserved.
ii In an inelastic collision, momentum is conserved but kinetic energy is not.
Some of the kinetic energy is transformed into other forms such as heat.
–1
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b Initial momentum = 0.35 × 2.8 = 0.98 kg m s
Final momentum = –(0.35 × 2.5) = –0.875 kg m s–1 ≈ –0.88 kg m s–1
Change in momentum = –0.875 – 0.98 = –1.855 kg m s–1 ≈ –1.9 kg m s–1
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c Momentum is conserved.
The initial momentum of the ball is equal to the final momentum of the ball and
the momentum of the snooker table.
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3 Momentum = mass × velocity
Principle of conservation of momentum:
The total momentum before an interaction (collision) is equal to the total momentum
after the interaction.
This is true provided there are no external forces acting on the system.
Principle of conservation of energy:
The total energy before a collision is equal to the total energy after the collision.
Kinetic energy is only conserved in elastic collisions.
[1]
4 a Momentum = mass × velocity
[1]
b Total initial momentum = total final momentum
0 = mAvA + mBvB
mAvA = –mBvB
vA
mB
Therefore
= −
vB
mA
COAS Physics 2
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1
Answers to End-of-chapter questions
Chapter 2
1 a i
1 Maximum force = 48 N
2 Contact time = 0.25 s
ii Impulse of the force = area under the graph
Impulse = (6.5 ± 1.0) N s
b i
F
48
=
m
0.50
Acceleration = 96 m s–2
a=
1
2
mv2 =
[1]
1
2
× 0.50 × 132
Kinetic energy = 42.25 J ≈ 42 J
v − u 
c F = ma = m 

 t 
 − 0.80 − 14 
F = 0.50 × 

 0.18 
Force = (–)61 N
∆p
mv − mu
=
∆t
t
(500 × 1.8) − (500 × 0.6)
F=
5.0
Force = 120 N
iv F =
3 a Change in momentum = mv – mu = m(v – u)
Change in momentum = 1100 × (24 – 0) = 2.64 × 104 kg m s–1 ≈ 2.6 × 104 kg m s–1
∆p
2.64 × 10 4
=
∆t
20
Braking force = 1.32 × 103 N ≈ 1.3 kN
b Braking force F =
COAS Physics 2
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distance
3.0
=
= 5.0 s
speed
0.60
s=?
u = 24 m s–1
2
2
v = u + 2as
02 = 242 + (2 × –1.2 × s)
24 2
s=
= 240 m
2.4
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ii The force acts to the right.
c Magnitude of deceleration =
[1]
[1]
The momentum of the steel bar increases.
According to Newton’s second law of motion, there must be a force on the bar
because force = rate of change of momentum.
Or
The bar is accelerating.
The bar must be experiencing a force because force = mass × acceleration.
iii Time =
[1]
[1]
2 a Momentum = mv = 500 × 0.60 = 300
Unit: kg m s–1 or N s
b i
[1]
[1]
[1]
ii Impulse = change in momentum = ∆p
6.5 = (0.50v) – (0.50 × 0) (the initial velocity of the ball is zero)
6.5
v=
= 13 m s–1
0.50
iii Kinetic energy =
[1]
F
1.32 × 10 3
=
= 1.2 m s–1
m
1100
v=0
a = 1.2 m s–1
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Answers to End-of-chapter questions
Chapter 3
1 a i
distance
2 πr
=
time
T
2 × π × 0.015
v=
1 50
v = 4.71 m s–1 ≈ 4.7 m s–1
Speed =
v2
4.712
=
r
0.015
a = 1.48 × 103 m s–2 ≈ 1.5 × 103 m s–2
ii a =
b The tension in the belt is not sufficient to provide the centripetal force.
Hence the belt will slip / will not grip the pulley.
2 a The centripetal force is the net force acting on an object describing a circle.
It is directed towards the centre of the circle.
b i
distance
time
2 × π × 0.15
v=
3.0
v = 0.314 m s–1
mv 2
0.060 × 0.314 2
F=
=
r
0.15
Force = 0.0394 N ≈ 0.039 N
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Speed =
ii The centripetal force on the toy increases with its speed.
The toy falls off because the frictional force between the turntable and the toy
is not sufficient to provide the centripetal force.
3 a Change in potential energy = kinetic energy gained
mgh = 12 mv2
2 × 9.81 × 0.70
v = 3.71 m s ≈ 3.7 m s–1
2 gh =
v=
–1
mv 2
0.050 × 3.712
=
r
1.50
Centripetal force = 0.459 N
T – mg = 0.459
T = 0.459 + (0.050 × 9.81) = 0.95 N
b Centripetal force =
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c The weight is equal to the tension only when the ball is at rest in a vertical position. [1]
The ball is not is not in equilibrium in the vertical position because it has
an upward (centripetal) acceleration.
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Chapter 4
1 a Arrow at S pointing towards the centre of O.
b i
g=
GM
[1]
r2
ii At S: g =
40
52
g = 1.6 N kg–1
COAS Physics 2
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Answers to End-of-chapter questions
iii At C: g =
40
4
2
= 2.5 N kg–1
[1]
1.6 + 2.5
= 2.05 N kg–1
2
Increase in GPE:
∆E = mgh = 3.0 × 103 × 2.05 × 2.0 × 107
≈ 1.2 × 1011 J
iv Average g =
c F = mg =
v2
r
v = gr = 1.6 × 1.0 × 10 8 = 1.26 × 104 m s–1
2 πr
2 π × 1.0 × 10 8
v=
; 1.26 × 104 =
T
T
8
6.28 × 10
Period T =
≈ 5.0 × 104 s
1.26 × 10 4
It is the gravitational force per unit mass.
ii g =
[1]
[1]
mv 2
r
g=
2 a i
[1]
GM
[1]
[1]
[1]
[1]
[1]
[1]
r2
iii g = 4.0 N kg–1 when r = 10 × 106 m (or any suitable pair from the curve)
4.0 × (10 × 10 6 ) 2
gr 2
M=
=
G
6.67 × 10 −11
Mass = 6.0 × 1024 kg
[1]
iv The straight-line graph through the origin.
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–1
v At the surface, g = 9.8 N kg when r = 6400 km
For distance within the Earth (linear graph), r = 64 km
because the field strength is a factor of 100 times smaller.
For distance beyond the surface (curve), r = 64 000 km
1
because the field strength ∝ 2 .
r
b
GM Earth
R12
=
GM Moon
R22
2
M Earth
 R1 
 = 92 = 81
= 
M Moon
R
 2
6.0 × 10 24
Mass of moon = MMoon =
= 7.4 × 1022 kg
81
3 a It is the gravitational force per unit mass.
b i
–1
g = 4.5 N kg
ii g =
COAS Physics 2
GM
r2
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4
Answers to End-of-chapter questions
1
iii g ∝
4 a i
; hence as r increases by a factor of 3, the field strength decreases by
r2
a factor of 32 = 9.
40
Hence g =
= 4.44 N kg–1
9
[1]
Arrows of equal lengths
directed towards the centre of the orbit.
[1]
[1]
ii A centripetal force is required for circular motion.
Hence the gravitational force must be along the line joining the stars.
b F = gravitational force between the stars; G = gravitational constant;
M = mass of each star; R = radius of the orbit.
c i
distance
time
In a time of one period, the distance travelled is equal to the circumference.
2 πR
Hence, v =
T
[1]
[1]
[1]
Speed =
ii F =
F=
iii
[1]
[1]
Mv 2
R
[1]
4 π 2 MR
M × ( 2πR T ) 2
=
R
T2
[1]
GM 2
4R 2
M=
d M=
=
4 π 2 MR
T2
42 × π2 × R × R 2
G ×T 2
[1]
= 16π2
R3
GT 2
16 × π 2 × (0.5 × 1011 ) 3
6.67 × 10 −11 × (100 × 86 400) 2
Mass = 4.0 × 1030 kg
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Chapter 5
1 a i
A motion where the acceleration is directly proportional to the displacement
from the equilibrium position
and the acceleration is directed towards the equilibrium position.
[1]
[1]
ii From the graph, period = 0.25 s
1
1
f=
=
= 4.0 Hz
T
0.25
[1]
iii a = (2πf )2x
a is maximum when x = A (= 5.0 mm)
Maximum acceleration = (2π × 4.0)2 × 5.0 × 10–3
a = 3.16 m s–2 ≈ 3.2 m s–2
[1]
b i
Any three from:
Resonance occurs at the natural frequency of the oscillating system.
At resonance, the natural frequency is equal to the frequency of the driver.
At resonance, the amplitude of the oscillating system is maximum.
At resonance, the oscillating system absorbs maximum energy from the driver.
COAS Physics 2
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Answers to End-of-chapter questions
ii 1 Greater mass leads to smaller amplitude.
This is because resonance will occur at a lower frequency.
2 a i
2 Greater stiffness leads to smaller amplitude.
This is because the resonance will occur at higher frequency.
[1]
[1]
In figure 1, the acceleration a is in the opposite direction to the displacement x.
In figure 2, a ∝ x.
[1]
[1]
ii Gradient of line = (2πf)2
50
(2πf)2 =
= 1000
0.050
1000
f=
= 5.03 Hz ≈ 5 Hz
2π
iii ‘Cosine’ curve with initial amplitude 25.0 mm.
Correct period of 0.2 s shown on graph.
The amplitude decreases (exponentially) with time.
b i
There is an acceleration (or centripetal force) towards the centre of the circle.
This acceleration is constant because the speed of rotation is constant.
ii From Figure 2, acceleration = 50 m s–2 when deflection is 50 mm.
v2
a=
r
v2
50 =
10
v = 500 = 22.4 m s–1 ≈ 22 m s–1
3 a i
Cosine curve.
Correct period of 1.0 s shown on the graph.
Amplitude decreases (exponentially) with respect to time.
ii The amplitude will decrease more rapidly
because of increased damping (or air resistance).
The frequency of oscillation will decrease (or the period is longer)
because the mass is greater.
(a = (2πf )2x; a is smaller because of greater mass; hence f will decrease.)
b Resonance occurs at the natural frequency of the oscillating aeroplane.
At resonance, the amplitude is maximum, as the oscillator absorbs greatest energy.
As frequency is increased from 0 to 1 Hz, the amplitude increases.
The amplitude us maximum at 1.0 Hz.
As the frequency is increased from 1.0 Hz to 2.0 Hz, the amplitude decreases.
4 a i
A motion where the acceleration is directly proportional to the displacement
from the equilibrium position
and the acceleration is directed towards the equilibrium position.
ii It is a straight-line graph through the origin (hence a ∝ x).
The graph has a negative gradient (hence acceleration is always directed
towards the equilibrium position).
b i
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Amplitude = 0.050 m
[1]
ii Gradient of line = (2πf )2
12.5
(2πf )2 =
= 250
0.050
[1]
COAS Physics 2
6
Answers to End-of-chapter questions
250
= 2.52 Hz
2π
1
1
Period =
=
≈ 0.40 s
f
2.52
Cosine curve.
Correct period of 0.40 s shown on graph.
Correct amplitude of 0.05 m shown.
f=
c i
ii Displacement = 0
Any suitable time, e.g. 0.1 s, 0.3 s, 0.5 s, etc.
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Chapter 6
1 E = mc∆θ
E = 600 × 1.1 × 103 × (70 – 30)
Energy = 2.6 × 107 J
2 a i
The internal energy of a system is the sum of the random distribution of kinetic
and potential energies associated with the molecules (or atoms) that make up
the system.
ii The specific heat capacity of a substance is the energy required to raise the
temperature by 1 K (or 1 °C) per unit mass of the substance.
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b Any four from:
The material is heated electrically.
The energy supplied to the material is E = VIt (V = p.d., I = current and t = time)
Measure the mass of the material and the change in temperature.
The specific heat capacity c is determined using the equation VIt = mc∆θ
The material can be insulated (lagged) to minimise heat losses to the surroundings.
[1]
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[1]
[1]
c E = mc∆θ = 2.0 × 920 × (293 – 0)
Energy = 5.39 × 105 J ≈ 540 kJ
[1]
[1]
3 a i
m
m
=
(where m is the mass per unit time, V is the volume per unit time,
V
Av
A is the cross-sectional area and v is the velocity)
m=ρ×A×v
[1]
0.090 = 1000 × 7.5 × 10–5 × v
[1]
v = 1.2 m s–1
ρ=
ii The speed will be doubled because the cross-sectional area is halved.
Speed = 2.4 m s–1
iii Force = rate of change of momentum
∆p
(0.090 × 2.4) − (0.090 × 1.2)
F=
=
∆t
1.0
F = 0.11 N
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[1]
[1]
[1]
iv Direction of force is opposite to the direction of flow of water from shower head. [1]
b i
E = mc∆θ
Dividing both sides of the equation by time t gives:
E
m
=
× c∆θ
t
t
E
Power =
= 0.090 × 4200 × (27 – 15)
t
Power = 4.54 × 103
Unit: watts (W)
ii Energy losses in the pipe from the heater tot he shower head.
COAS Physics 2
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7
Answers to End-of-chapter questions
iii The change in temperature is doubled to:
(27 – 15) × 2 = 24 °C
Maximum temperature = 15 + 24 = 39 °C
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Chapter 7
1 a pV = nRT
280 × 10 3 × 2.1 × 10 −3
pV
n=
=
8.31 × (273 + 15)
RT
n = 0.246 mol ≈ 0.25 mol
b
[1]
[1]
[1]
pV
= constant (V = constant)
T
280 × 10 3 × V
290 × 10 3 × V
=
273 + 15
T
290
T=
× 288 = 298.3 K
280
Temperature = 298.3 – 273 ≈ 25 °C
[1]
[1]
[1]
c Internal energy = kinetic energy ∝ T
298.3
Therefore ratio =
≈ 1.04
288
[1]
[1]
2 a The Celsius and Kelvin scales have the same increment.
Adding (or subtracting) 273 to 500 000 does not make a significant difference.
b i
According to pV = nRT, as pressure (or volume) tends to zero, the temperature
tends to 0 K.
ii Mean kinetic energy of molecules ∝ T
As the temperature tends to 0 K, the kinetic energy tends to zero.
p0
p
p
= constant ⇒
=
T
300
400
400
p=
p0 = 1.33p0
300
pV
ii
= nR or n ∝ p
T
p
300
f=
=
= 0.75
p0
400
c i
[1]
[1]
pV = nRT
1.0 × 10 5 × 1.26 × 10 4
pV
n=
=
8.31 × ( 273 − 43)
RT
5
n = 5.23 × 10 mol ≈ 5.2 × 105 mol
5
–3
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[1]
[1]
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3
ii Mass = 5.23 × 10 × 4.0 × 10 = 2.1 × 10 kg
COAS Physics 2
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pV
= constant
T
1.0 × 10 5 × V
1.0 × 10 3 × 1.0 × 10 6
=
273 + 17
273 − 43
9
1.0 × 10 × 290
V=
= 1.26 × 104 m3 ≈ 1.3 × 104 m3
5
1.0 × 10 × 230
c i
[1]
[1]
3 a When the temperature is 0 K, the pressure or volume of the gas tends to zero.
b
[1]
[1]
[1]
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Answers to End-of-chapter questions
d Internal energy = total kinetic energy ∝ T
The temperature decreases from 17 °C (290 K) to –43 °C (230 K),
290
hence the kinetic energy must decrease by a factor of
= 1.261. Therefore:
230
kinetic energy = 1900 × 1.261 = 1500 MJ
[1]
e i
[1]
Weight acting downwards and greater upthrust/lift acting upwards.
ii Net force = Ma
1.3 × 105 – Mg = Ma
1.3 × 10 5
M=
9.81 + 27
M ≈ 3.5 × 103 kg
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Chapter 8
1 a
b i
Correct field pattern as in figure.
[1]
The field lines touch the plate at right angles.
[1]
The field direction is shown correctly
(away from the plate).
[1]
Correct triangle of forces as shown.
[1]
F
F
=
W
1.0 × 10 −5
F = 1.0 × 10–5 × tan 20° ≈ 3.64 × 10–6 N
tan 20° =
F
3.64 × 10 −6
=
Q
1.2 × 10 −9
E = 3.0 × 103
Unit: V m–1 or N C–1
ii E =
c F=
Qq
r2 =
4 π × 8.85 × 10
r ≈ 6.0 × 10–2 m
COAS Physics 2
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(1.2 × 10 −9 ) 2
−12
[1]
[1]
(Q = q = 1.2 × 10–9 C and F = 3.64 × 10–6 N)
4 πε 0 r 2
[1]
× 3.64 × 10
−6
= 3.56 × 10–3 m2
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9
Answers to End-of-chapter questions
d Correct field pattern as shown.
[1]
The field pattern for the two point charges is symmetrical about the plane midway
between the charges.
2 a Horizontal field lines between the plates.
Field lines equally spaced (to show uniform field).
Correct directions of fields (from +1200 V to 0 V, and from +600 V to 0 V)
V
d
600
E=
= 1.5 × 104 V m–1
0.040
b E=
v2 =
2 × 2.4 × 10 −15 × 0.040
−31
1
2
[1]
[1]
1
2
mv2
= 2.108 × 1014
[1]
[1]
mv2; speed ∝ V
The p.d. is doubled; hence speed =
2 × 1.452 × 107 ≈ 2.05 × 107 m s–1
f Fewer electrons will reach the grid B or plate C.
Hence the current will fall.
3 a Electric field strength is the force experienced per unit positive charge.
COAS Physics 2
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[1]
[1]
9.11 × 10
v = 1.542 × 107 m s–1 ≈ 1.5 × 107 m s–1
e VQ =
[1]
[1]
c F = EQ = 1.5 × 104 × 1.6 × 10–19
Force = 2.4 × 10–15 N
d Work done = kinetic energy; Fd =
[1]
[1]
[1]
[1]
[2]
10
Answers to End-of-chapter questions
b i
Correct field pattern.
[1]
The field pattern is symmetrical (quality mark)
Correct direction of field lines from positive to negative.
[1]
[1]
ii For each charge, E =
Q
4 πε 0 r 2
; r = 4.0 × 10–10 m
1.6 × 10 −19
E=
−12
−10 2
= 9.0 × 109
4 π × 8.85 × 10 × ( 4.0 × 10 )
Total E = 2 × 9.0 × 109 = 1.8 × 1010 (Both fields are in the same direction.)
Unit: V m–1 or N C–1
c i
Equal but opposite forces act on each charge.
ii A: No motions of dipole as forces at each end are equal and opposite.
B: The dipole rotates (clockwise).
The dipole experiences a couple/turning effect.
4 a i
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[1]
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Correct field pattern.
[1]
Correct direction of field lines from positive ion to negative ion.
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ii F =
F=
Qq
4 πε 0 r 2
1.6 × 10 −19 × 1.6 × 10 −19
4 π × 8.85 × 10 −12 × (5.0 × 10 −10 ) 2
Force = 9.2 × 10–10 N
b FH = mHaH and FI = mIaI
The forces are the same.
Simple harmonic motion implies a ∝ –x.
xH
aH
mI
Therefore
=
=
= 127
xI
aI
mH
COAS Physics 2
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Answers to End-of-chapter questions
c i
Correct sine or cosine curve.
[1]
Amplitude is 8.0 × 10–12 m
[1]
1
Correct period of 1.5 × 10–14 s shown. (Period =
≈ 1.5 × 10–14 s.)
f
[1]
ii The molecules are resonating.
This happens when the frequency of the incident infrared radiation is equal to
the natural frequency of oscillation of the molecules.
[1]
[1]
Chapter 9
1 a Arrow(s) between the north and south poles of the magnet.
V
1.5
b Current =
=
= 0.1875 A
R
8.0
F = BIL
F = 1.2 × 10–2 × 0.1875 × 24 = 0.054
Unit: newton (N)
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[1]
[1]
[1]
[1]
c The force will decrease by a factor of four.
[1]
This is because the resistance of the wire increases by a factor of four (R ∝
1
)
A
and reduced the current by a factor of four.
2 a i
Equally spaced (horizontal) electric field lines between the parallel plates.
Correct direction of the electric field (towards the –7.0 kV plate).
ii Work done = energy transformed; eV =
v=
b i
[1]
2eV
=
m
2 × 1.6 × 10 −19 × 7000
9.11 × 10
−31
1
2
mv2
= 4.96 × 107 m s–1 ≈ 5.0 × 107 m s–1
[1]
[1]
[1]
[1]
The arrow (force) is perpendicular tot he path and towards the centre of the arc. [1]
ii Out of the plane of the paper.
Fleming’s left-hand rule gives the direction of the force.
mv 2
iii
= Bev
r
mv
r=
Be
9.11 × 10 −31 × 4.96 × 10 7
r=
= 9.4 × 10–2 m
3.0 × 10 −3 × 1.6 × 10 −19
COAS Physics 2
[1]
[1]
[1]
[1]
[2]
12
Answers to End-of-chapter questions
c Change in direction of magnetic field to change direction of electron beam.
Change magnitude of current in the coil to move the electron beam across.
[1]
[1]
3 a It has a positive charge; it is repelled by the positive charges on the ‘top’ plate.
[1]
b i
Kinetic energy = eV
Kinetic energy = 1.6 × 10–19 × 300 = 4.8 × 10–17 J
1
2
ii
mv2 = 4.7 × 10–17
v=
2 × 4.8 × 10 −17
2.3 × 10
−26
[1]
= 6.46 × 104 m s–1 ≈ 6.5 ≈ 104 m s–1
V
d
600
V
d=
=
= 1.5 × 10–2 m (1.5 cm)
E
4.0 × 10 4
c E=
d i
Semicircle to the right of the hole.
mv 2
ii
= BQv
r
mv
r=
BQ
r=
2.3 × 10 −26 × 6.46 × 10 4
0.17 × 1.6 × 10 −19
Radius = 5.46 × 10–2 m
Distance from hole = 2 × 5.46 × 10–2 m ≈ 0.11 m
force
current × length of conductor in field
The field is at right-angles to the magnetic field.
4 a Magnetic flux density =
b i
[1]
[1]
Arrow towards the centre of the circle.
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
ii The field is out of the plane of the paper.
The direction of motion of the proton gives the direction of the conventional
current when using Fleming’s left-hand rule.
[1]
iii F = Bev
[1]
iv F =
mv 2
r
mv 2
= Bev
r
mv
1.67 × 10 −27 × 1.5 × 10 7
B=
=
er
1.6 × 10 −19 × 60
–3
B ≈ 2.6 × 10
Unit: tesla (T)
mv
∝v
er
Therefore for protons moving at double the speed, the magnetic flux density
should be doubled in order to contain the protons within the tube.
v B=
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
Chapter 10
1 a i
F is towards the open end of the tube.
The direction can be determined using Fleming’s left-hand rule.
COAS Physics 2
[1]
[1]
13
Answers to End-of-chapter questions
ii F = BIw
[1]
–3
iii F = 0.15 × 800 × 25 × 10 N
Force = 3.0 N
b i
[1]
[1]
According to Faraday’s law:
induced e.m.f. = rate of change of magnetic flux linkage.
The rate of change of magnetic flux is directly proportional to the speed v.
[2]
[1]
ii The flux linkage is doubled.
According to Faraday’s law, the induced e.m.f. is also doubled.
[1]
[1]
2 a Faraday’s law: induced e.m.f. = rate of change of magnetic flux linkage
Magnetic flux = magnetic flux density × area × cosθ ; φ = BAcosθ
When field is normal to area, φ = BA
Magnetic flux linkage = number of turns × magnetic flux (= BAN)
[2]
[1]
[1]
[1]
b A sine or cosine curve shown for the voltage against time graph.
Doubling the speed will:
double the maximum induced voltage
because the rate of change of flux is doubled;
double the frequency.
Doubling the number of turns will:
double the maximum induced voltage
because the flux linkage is doubled.
Removing the iron will:
reduce the induced voltage because the magnetic flux density is reduced.
3 a i
Magnetic flux = magnetic flux density × area (normal to field) = BA
ii Flux linkage = number of turns × magnetic flux
Flux linkage = NBx2
∆ ( Nφ )
NBx 2
=
∆t
t
b Faraday’s law: E =
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
x
= NBxv
t
e.m.f. = 1250 × 0.032 × 0.020 × 0.10 = 0.080 V (80 mV)
E = NBx ×
[1]
[1]
Chapter 11
1 a i
Q = VC
Therefore W =
1
2
QV =
1
2
(VC)V =
1
2
CV
2
ii Parabolic shape through the origin.
Accurate plot showing W = 12 × 2.2 × V 2 = 1.1V 2
b i
Time constant = CR
Time constant = 2.2 × 6.8 × 103 = 1.5 × 104 s (5 h 9 min)
ii Energy lost = 12 × 2.2 × 5.02 –
Energy lost = 9.9 J
COAS Physics 2
1
2
× 2.2 × 4.02
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
14
Answers to End-of-chapter questions
iii V = V0e–t/CR
V 
t = –RC ln  
 V0 
[1]
 4.0 
 ≈ 3.3 × 103 s
t = –1.5 × 104 × ln 
 5.0 
9. 9
iv Power =
3.3 × 10
3
[1]
= 3.0 × 10–3 W
[1]
2 a Q0 = VC
Q0 = 5000 × 1.2 × 10–11 = 6.0 × 10–8
Unit: coulomb (C)
b i
[1]
[1]
[1]
Time constant = RC = 1.2 × 1015 × 1.2 × 10–11 = 1.44 × 104 s ≈ 1.4 × 104 s
ii Current =
[1]
5000
V
=
= 4.16 × 10–12 A ≈ 4.2 pA
R 1.2 × 1015
[1]
iii 1.44 × 104 s ≈ 1.4 × 104 s (same as time constant in b I)
–t/CR
iv t = RC; Q = Q0e
= Q0e
Q
= e–1 ≈ 0.37; hence the charge lost is about 63%, which is roughly
Q0
3 a i
[1]
–1
[1]
2
3
.
Parallel field lines between the plates.
Correct field direction (away from the positive charges).
[1]
[1]
[1]
V
; V = Ed = 3.0 × 105 × 1.5 × 103 = 4.5 × 108 V = 450 MV
d
[1]
b The capacitor plates model the base and centre regions of the clouds.
The battery acts as the source of energy; it models the wind.
The resistor is included to account for the slow rate of charging of the clouds.
[1]
[1]
[1]
ii E =
c i
Q
t
I=
[1]
Current =
20
= 0.80 A
25
[1]
V
4.5 × 108
=
I
5 × 0.80
Resistance = 1.13 × 108 Ω ≈ 1.1 × 108 Ω
ii R =
20
Q
=
V
4.5 × 108
Capacitance = 4.44 × 10–8 F ≈ 4.4 × 10–8 F
iii C =
iv Time constant = RC
Time constant = 1.13 × 108 × 4.44 × 10–8 F ≈ 5.0 s
4 a Capacitance =
b i
charge
p.d.
1
2
CV 2 =
1
2
× 0.47 × 10–6 × 112
Energy = 2.84 × 10 J ≈ 2.8 × 10 J
–5
COAS Physics 2
[1]
[1]
[1]
[1]
[1]
[1]
1 Q = VC
Q = 11 × 0.47 × 10–6 = 5.2 × 10–6 C
2 W=
[1]
–5
[1]
[1]
[1]
[1]
15
Answers to End-of-chapter questions
V
11
=
R
2200
Current = 5.0 × 10–3 A
ii 1 I =
c i
[1]
[1]
2 Time constant = CR
Time constant = 2200 × 0.47 × 10–6 = 1.03 × 10–3 s ≈ 1.0 ms
[1]
[1]
The graph should have a constant-ratio property (for a given time interval).
0.70
0.30
Attempt made, e.g.
≈
≈ 0.4
1. 8
0.70
[1]
ii Time constant = 1 ms and Q = Q0e–t/CR
Charge at 2.0 ms = 5.2 × 10–6 × e–2 = 7.04 × 10–7 C
Charge at 1.0 ms = 5.2 × 10–6 × e–1 = 1.91 × 10–6 C
Charge flow = (1.91 – 0.704) × 10–6 C ≈ 1.2 × 10–6 C
Alternatively:
area under graph = charge flow
attempt to find area
correct answer
[1]
[1]
[1]
[1]
[1]
[1]
[1]
Chapter 12
1 Alpha scattering provided evidence for the existence of the nucleus of an atom.
The nucleus contains most of the mass of the atom.
The nucleus must have a very small size (10–14 m), because very few α-particles were
scattered through large angles.
The nucleus must be (positively) charged, because it repelled the positive the
(positively charged) α-particles.
When high-speed electrons are fired at the nuclei, they show a diffraction pattern.
The moving electrons travel as waves in accordance with the de Broglie equation
h
(λ = ).
p
The diffraction is significant when the wavelength of the electrons is similar to
the size of the nucleus.
[1]
[1]
2 a i
[1]
An arrow pointing away from N but in line with N and A.
ii Arrow along a line passing perpendicular to the tangent at the point of closest
approach and directed away from the path of the α-particle.
b i
KE =
v2 =
1
2
mv2 = 8.0 × 10–13 J
2 × 8.0 × 10
6.7 × 10
−13
−27
; v = 1.545 × 107 m s–1 ≈ 1.5 × 107 m s–1
ii p = mv = 6.7 × 10–27 × 1.5 × 107 ≈ 1.0 × 10–19 kg m s–1
c i
p2
so for the same KE, p ∝ m
2m
A proton has a quarter of the mass of an α-particle;
hence for the same kinetic energy, the initial momentum will be half.
Therefore, momentum = 0.50 × 10–19 kg m s–1.
KE =
1
2
mv2 =
ii Any two from: average force is smaller, smaller recoil or less interaction time.
COAS Physics 2
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[1]
[1]
[1]
[1]
[1]
[1]
[1]
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[1]
[2]
16
Answers to End-of-chapter questions
d F=
F=
F=
3 a i
Qq
[1]
4 πε 0 r 2
2e × 79e
[1]
4 πε 0 × (7.3 × 10 −13 ) 2
2 × 79 × (1.6 × 10 −19 ) 2
4 π × 8.85 × 10
−12
× (7.3 × 10
−13 2
)
≈ 6.8 × 10–2 N
[2]
Number of protons = 3
[1]
ii Number of neutrons = 7 – 3 = 4
[1]
iii The number of electrons for a neutral atom is 3. Since the net charge is +e,
this ion must have 2 electrons.
[1]
b i
Isotopes are nuclei of the same element with different numbers of neutrons but
the same number of protons.
ii The molar mass of iron-55 is 55g.
30 × 10 −6
Number of nuclei =
55 × 6.02 × 10 23
= 3.28 × 1017 ≈ 3.3 × 1017
iii A = 55 – 3 = 52; Z = 26 – 1 = 25
Hence the new nucleus can be represented as
4 a i
[1]
[1]
[1]
[1]
52
25 X
[1]
[1]
.
Proton or neutron.
[1]
ii Electron.
[1]
b Proton = (u u d)
Neutron = (u d d)
[1]
[1]
c i
Any 3 from:
charge, momentum, baryon number, strangeness and mass–energy.
ii For proton, B = +1
For neutron, B = +1
Hence baryon number for π+ particle = (+1 + 1) – (+1 + 1) = 0
[3]
[1]
[1]
[1]
Chapter 13
1 a i
These are slow-moving neutrons.
[1]
ii Thermal neutrons have a greater chance of interacting with the
b Binding energy of uranium nucleus = 7.6 × 235 = 1786 MeV
Binding energy of products = 146 × 8.2 + 87 × 8.6 = 1945.4 MeV
Energy released = 1945.4 – 1786 = 159.4 MeV ≈ 159 MeV
2 a Number of protons = 92; number of neutrons = 143
–15
⅓
–15
b Radius = r = 1.41 × 10 × 235 = 8.701 × 10
3.89 × 10 −25
mass
Density =
=
volume
4 π × (8.701 × 10 −15 ) 3
Density ≈ 1.41 × 1017 kg m–3
m
c The mass of the nucleons is greater than the mass of the nucleus.
This difference is equivalent to the binding energy of the nucleus.
COAS Physics 2
235
U nucleus.
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
17
Answers to End-of-chapter questions
d Determine the total mass of the nucleons (143mn + 92mp).
Determine the mass defect ∆m by subtracting the mass of the nucleus from the total
mass of the nucleons.
Calculate the binding energy ∆E of the nucleus using ∆E = ∆mc2.
∆E
Binding energy per nucleon =
235
3 Mass of helium nucleus = 4 × 1.67 × 10–27 = 6.68 × 10–27 kg
1.00
Number of nuclei in 1.00 kg =
= 1.50 × 1026
6.68 × 10 −27
Energy generated = 1.50 × 1026 × (28.4 × 106 × 1.6 × 10–19) ≈ 6.8 × 1014 J
[1]
[1]
[1]
[1]
[1]
[1]
[1]
Chapter 14
192
79 Au →
β–: 192
79 Au →
1 a For β+:
For
0
+1 e
0
−1 e
+
+
192
ν
78 Pt +
192
80 Hg + ν
[1]
[1]
b For β+: initial mass = 191.92147 u
final mass = 191.91824 + 0.00055 = 191.91879 u
The final mass is less than the initial mass, so β+ decay can occur spontaneously.
For β–: initial mass = 191.92147 u
final mass = 191.92141 + 0.00055 = 191.92196 u
The final mass is greater than the initial mass, so β– decay cannot occur naturally.
[1]
[1]
[1]
[1]
[1]
[1]
c For β+: mass loss
[1]
[1]
[1]
[1]
= 191.92147 – 191.91879 = 0.00268 u
= 0.00268 × 1.66 × 10–27 = 4.45 × 10–30 kg
energy emitted = ∆mc2 = 4.45 × 10–30 × (3.0 × 108)2
= 4.00 × 10–13 J
2 a
236
92 U
b i
→
100
40 Zr
+
131
52Te
+ 5 01 n
[1]
No change in the nucleon number. Proton number increases by 1.
ii Nucleon number = 100 and proton number = 44.
3 a A: This is the activity of the source; it is the rate of decay of nuclei.
λ: This is the decay constant; it is the probability of decay of a nucleus
per unit time interval.
N: This is the number of undecayed nuclei in the sample.
b i
Z = 90 and A = 234
6
ii Mass = 2.0 × 10 × 7.0 × 10 = 14 kg
iv λ =
14 000
× 6.02 × 1023 = 3.54 × 1025
238
0.693
9
7
= 4.81 × 10–18 s–1
4.5 × 10 × 3.2 × 10
A = λN = 4.81 × 10–18 × 3.54 × 1025 ≈ 1.7 × 108
Unit: Bq or s–1
c There is a change of state, with water changing into steam (water vapour).
This change of state requires latent heat.
In order to calculate the energy required to heat the steam beyond 393 K,
the specific heat capacity of the steam must be known.
4 a i
29 protons
COAS Physics 2
[1]
[1]
[1]
[1]
[1]
–6
iii Number of nuclei =
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
18
Answers to End-of-chapter questions
ii 34 neutrons
b λ=
c i
0.693
120 × 3.2 × 10 7
[1]
= 1.805 × 10–10 s–1 ≈ 1.8 × 10–10 s–1
Q = VC = 90 × 1.2 × 10–12
Q = 1.08 × 10–10 C ≈ 1.1 × 10–10 C
Q
1.08 × 10 −10
=
e
1.6 × 10 −19
Number = 6.75 × 108 ≈ 7 × 108
ii Number =
[1]
[1]
[1]
[1]
[1]
iii A = λN
A
6.75 × 108
N=
=
λ
1.805 × 10 −10
Number ≈ 3.7 × 1018
[1]
iv 1 year is less then 1% of 120 years, so expect it to be within 1%.
[1]
5 a i
ii
0
−1 e
[1]
[1]
212
83 Bi
→
212
84 Po
+
212
84 Po
→
208
82 Pb
+ 42 He
+ ν
b Place a radiation detector close to the source to determine the count rate.
Place a thin sheet of paper between the detector and source. If the source
emitted α radiation only, the count rate would drop dramatically to almost
the background rate.
If the source emitted β radiation only, the paper would have little effect, but
a sheet of aluminium at least a couple of millimetres thick would reduce the
count rate to close to the background rate.
If the source emits both α and β radiation, the paper will partially reduce the
count rate; the aluminium will further reduce it to close to the background rate.
c i
The molar mass of bismuth is 212 g. 1 mole contains 6.02 × 1023 nuclei.
1.0 × 10 −9
Number of nuclei = N =
× 6.02 × 1023 = 2.84 × 1012
212
A = λN = 0.0115 × 2.84 × 1012
A ≈ 3.27 × 1010 min–1 ≈ 3 × 1010 min–1
ii t½ =
0.693
≈ 60 min
0.0115
iii Smooth curve passing through (0, 33 × 103 min–1), (60 min, 16 × 103 min–1) and
(120 min, 8 × 103 min–1).
[2]
[2]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
Chapter 15
1 a Any seven from:
X-ray source and detectors round patient.
X-ray source is rotated around patient.
X-rays are absorbed by tissue and bone of patient.
X-rays are absorbed more by high-Z matter such as bone (or absorbed less by
low-Z matter such as soft tissues).
The attenuation mechanism is mainly the photoelectric effect.
There is a possibility of using a contrast medium (e.g. barium or iodine).
A computer is used to process the image.
A 3-D image is produced.
COAS Physics 2
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[1]
[1]
[1]
[1]
[1]
[1]
19
Answers to End-of-chapter questions
b The patients are exposed to ionising radiation.
Ionising radiation could cause cancer / damage cells.
It is expensive or time-consuming.
2 Any seven from:
C (cathode) is the negative heater and A (anode) is the positive target (e.g. tungsten).
Electrons are accelerated from C to A.
The high-speed electrons hit the target metal and produce X-ray photons.
(Some of) the kinetic energy of the electrons is transferred into X-ray photons.
About 1% of the incident electron energy is converted into X-rays; the rest is
transferred into heat in the target metal.
The graph of intensity against X-ray photon energy shows characteristic spectral lines
and ‘broad-background’ braking or Bremsstrahlung radiation.
hc
The minimum wavelength λ of the X-rays can be determined using eV =
.
λ
3 a I = I0e–µx
347 = I0e–250 × 0.025
347
I0 =
= 1.78 × 105 W m–2
1.93 × 10 −3
power
b intensity =
cross - sectional area
power = intensity × cross-sectional area = 347 × π × (0.10 × 10–2) 2
power = 1.09 × 10–3 W ≈ 1.1 mW
100
c i power of electron beam = 18 ×
0.15
= 1.2 × 104 W
ii 12 mv2 × 7.5 × 1017 = 1.2 × 104
v=
2 × 1.2 × 10 4
9.11 × 10 −31 × 7.5 × 1017
= 1.87 × 108 m s–1 ≈ 1.9 × 108 m s–1
4 a
Graph as shown in figure.
Six points plotted correctly.
Smooth curve drawn through
the data points.
Additional (seventh) point
plotted correctly.
b i
Thickness = (1.0 ± 0.1) mm (see example dashed lines on figure)
–µx
ii I = I0e
0.5 = e–µ × 1.0
–µ × 1.0 = ln 0.5
ln 0.5
µ=
≈ 0.69
− 1.0
Unit: mm–1 (Answer is 690 if unit is m–1.)
5 Formation of image – any three from:
The X-rays are detected by a film.
High Z (or proton number) means greater attenuation.
COAS Physics 2
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[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
20
Answers to End-of-chapter questions
Reference to the photoelectric effect or energy range around 1 keV – 100 keV
Attenuation (absorption) coefficient ∝ Z3
[1]
[1]
Explanation of contrast material:
X-rays do not show soft tissues well.
Low Z (or proton number) of tissues means less attenuation.
Contrast material has high Z or example given, e.g. barium or iodine.
Taken orally or as an enema, or can be injected.
[1]
[1]
[1]
[1]
Example of type of structure that can be imaged:
Intestines or throat or stomach.
[1]
Chapter 16
1 a Any one from:
Bone growth, blood circulation in lung/brain/liver, function of the heart/brain.
b i
99
99
m
+ −01 e
42 Mo → 43Tc
99
m
→ 99
43Tc
43Tc + γ
+ ν
[1]
[1]
ii Technetium-99m emits γ-ray photons and these will pass through the patient.
Technetium-99m has a short half-life (of 6 h) and hence does not stay inside
the patient for a long time.
iii 1 λ =
0.693
0.693
=
= 2.87 × 10–6 s–1 ≈ 2.9 × 10–6 s–1
t1 2
67 × 3600
2 A = λN
A
600 × 10 6
N=
=
= 2.09 × 1014
λ
2.87 × 10 −6
2.09 × 1014
Mass =
× 99 ≈ 3.4 × 10–8 g
6.02 × 10 23
3 Activity = A0e–λ t = 600 × e −2.87×10
Activity = 440 MBq
2 a i
[1]
−6
×30×3600
When a strong magnetic field is applied to nuclei, they start to precess.
The natural angular frequency of precession of the nuclei is known as
the Larmor frequency.
ii The nuclei resonate when radio frequency waves are directed at the nuclei.
Then these radio waves are switched off, the nuclei take some time to go back
to their lower energy state – this time is referred to as the relaxation time.
The relaxation time depends on the environment of the nuclei.
b It does not use ionising radiation and therefore is not dangerous.
MRI gives better soft-tissue contrast than a CAT scan.
[1]
[1]
[1]
[2]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
c A large superconducting magnet, which produces the strong external magnetic field. [1]
A set of gradient coils, which produce an additional external magnetic field
that varies across the length, width and depth of the patient’s body.
[1]
A radio-frequency (RF) coil that transmits radio-wave pulses into the patient.
[1]
A radio-frequency (RF) coil that detects the signal emitted by the relaxing nuclei. [1]
A computer that controls the gradient coils and RF pulses and which produces and
displays images.
[1]
3 a A is the collimator.
This is made of lead tubes. It ensures that only γ-ray photons travelling along the
axis of the tubes are detected, and thus ensures a sharp image.
B is the scintillation crystal.
COAS Physics 2
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[1]
[1]
21
Answers to End-of-chapter questions
This changes the γ-ray photons into visible light photons.
C is the computer.
This processes the signals from the photomultiplier tubes to form an image.
b i
Number = 39
Number = 19 683 ≈ 2.0 × 104 electrons
ii Current =
Current =
Q
Ne
=
t
t
1.97 × 10 4 × 1.6 × 10 −19
2.0 × 10 −9
Current ≈ 1.6 × 10–6 A
[1]
[1]
[1]
[1]
[1]
[1]
[1]
Chapter 17
1 a Acoustic impedance Z = ρc, where ρ is the density of the material and
c is the speed of ultrasound in the material.
Unit: Z → [kg m–3 × m s–1] → [kg m–2 s–1]
b i
Air–skin boundary:
Ir
(1.6 × 10 6 − 430) 2
( Z 2 − Z1 ) 2
=
=
I0
(1.6 × 10 6 + 430) 2
( Z 2 + Z1 ) 2
Fraction reflected = 0.9989
Fraction transmitted = 1 – 0.9989 ≈ 0.0011 (0.11%)
Gel–skin boundary:
Ir
(1.7 × 10 6 − 1.6 × 10 6 ) 2
( Z 2 − Z1 ) 2
=
=
I0
(1.7 × 10 6 + 1.6 × 10 6 ) 2
( Z 2 + Z1 ) 2
Fraction reflected = 0.00092
Fraction transmitted = 1 – 0.00092 ≈ 0.999 (99.9%)
ii With gel on the skin, there is acoustic matching.
Only a small amount of ultrasound is reflected: most enters the body.
Hence a useful image is produced.
2 a Any five from:
A voltage is applied to the piezoelectric crystal.
This makes the crystal change its shape.
The crystal material is named, e.g. quartz.
An alternating voltage makes the crystal vibrate.
The crystal resonates when the applied voltage has the same frequency as
the natural frequency of the crystal.
The crystal is damped to stop it vibrating when the voltage is switched off.
This is achieved be a backing material/epoxy resin.
b i
Distance 5.4 cm ± 0.1 cm read from graph.
Time interval = 5.4 × 20 µs = 108 µs
Distance = speed × time = 1500 × 108 × 10–6 = 0.162 m
0.162
Size of fetal head =
= 0.081 m = 8.1 cm
2
ii There is greater intensity of reflected ultrasound at the skin.
This is because of the large difference between the acoustic impedances
of air and skin.
The peaks reflected from the fetal head will have very small ‘heights’.
3 a The two factors are: speed of ultrasound and density of the material.
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Answers to End-of-chapter questions
b i
Ir
( Z 2 − Z1 ) 2
=
I0
( Z 2 + Z1 ) 2
Ir
= 4.42 × 10–4 =
I0
 Z 2 − Z1 

 =
 Z 2 + Z1 
 Z 2 − Z1 


 Z 2 + Z1 
2
4.42 × 10 −4 = 0.0210
 Z 2 − Z1 
 worked out from the table are:
The values of 
 Z 2 + Z1 
soft tissue–blood: 0.0124; soft tissue–water: 0.0415; soft tissue–brain: 0.0156
soft tissue–bone: 0.654; soft tissue–muscle: 0.0210
The unknown medium is muscle.
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ii ∆t = 26.5 × 10–6 s
c = 1.54 × 103 m s–1
c∆t
26.5 × 10 −6 × 1.54 × 10 3
Depth =
=
= 0.0204 m
2
2
Depth = 2.04 cm
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iii c = fλ
c
1.54 × 10 3
λ=
=
f
3.5 × 10 6
Wavelength = 4.4 × 10–4 m
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Chapter 18
1 a There are fusion reactions within the core of the star; these release photons.
The radiation pressure from the photons prevents gravitational collapse.
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b A red giant is a star that increases in size when its hydrogen fuel becomes exhausted.[1]
A red giant has enormous size; for a star the mass of the Sun, up to the radius of
the Earth’s orbit.
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It appears red because its surface is cooler than that of our Sun.
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It is brighter than our Sun because it has a greater surface area than our Sun,
and so emits greater power.
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c It first evolves into a super red giant.
As the nuclear fuel runs out, the star rapidly collapses against its neutron core.
This generates a shockwave that explodes the stellar material into space.
This explosion is known as a supernova.
The remnant is either a neutron star or a black hole.
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d There are no fusion reactions within a white dwarf; hence it cannot be a star.
It simply leaks away radiation and energy from past fusion reactions.
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2 a The wavelengths have been redshifted.
Hence, the star must be moving away from us.
b
∆λ
v
=
λ
c
v
1.4
=
= 0.0117
c
119.5
v = 0.0117 × 3.0 × 108
Speed ≈ 3.5 × 106 m s–1
3 a The composition of a star can be determined by examining the spectrum
obtained from it.
The spectrum is compared with the spectra of known elements in the laboratory.
COAS Physics 2
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Answers to End-of-chapter questions
b i
Parallax is the apparent change in the position of a star against the background
of distant stars.
The parallax is obtained from measurements taken when a star is viewed from
different positions (e.g. different positions of the Earth in its orbit).
ii
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The parsec is the distance at which the radius of the Earth’s orbit around the Sun [1]
subtends an angle of 1 arc second.
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4 a Olber’s paradox: In an infinite and static universe
the night sky ought to be bright.
This is because the line of sight always ends up on a star.
Hubble’s law: speed of recession of galaxy ∝ distance of the galaxy from us
All galaxies in the universe are receding from each other.
This implies that the universe must have had zero size / been a singularity
in the finite past.
b i
1 pc = 3.1 × 1016 m
70 000
H0 = 6
10 × 3.1 × 1016
Hubble constant = 2.26 × 10–18 s–1 ≈ 2.3 × 10–18 s–1
1
1
=
H0
2.26 × 10 −18
Age = 4.43 × 1017 s ≈ 4.4 × 1017 s
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ii Age =
iii Size of observable universe = speed of light × age
Size = 3.0 × 108 × 4.43 × 1017
Size ≈ 1.3 × 1026 m
c The universe has expanded at a constant rate.
5 a i
1
parallax (arc seconds)
1
Distance =
0.314 arc seconds
Distance = 3.18 pc
Distance (pc) =
ii 1 pc = 3.1 × 1016 m
Distance = 3.18 × 3.1× 1016 ≈ 9.9 × 1016 m
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Chapter 19
1 a Any six from:
Very high temperatures at the Big Bang.
Quarks and leptons formed.
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Answers to End-of-chapter questions
Temperature of the universe decreases as it expands.
Strong nuclear force takes effect.
Protons, neutrons and pions are formed.
Helium nuclei are formed by fusion.
25% of the mass is helium.
Lots of γ radiation.
The universe becomes transparent to electromagnetic radiation.
b Temperature of 3 K (or 2.7 K).
Agrees with expected cooling of the universe.
2 a The universe is homogeneous and isotropic.
b i
H0 =
7.5 × 10 3
6
16
= 2.419 26 × 10–18 s–1
10 3.1 × 10
1
1
Age =
=
= 4.133 × 1017 s
H0
2.419 × 10 −18
Age =
4.133 × 1017
= 1.3 × 1010 years
365 × 24 × 3600
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ii The universe is flat.
The density of the universe is equal to the critical density.
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iii Curve passing through point P.
The curve returns to the time axis.
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iv The universe has a finite lifetime. It ends up in a Big Crunch.
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3 a
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Open:
universe expands for all time; gradient of the graph is always > 0,
even after infinite time.
Closed: universe collapses back.
Flat:
universe expands for all (finite) time; gradient of the graph is zero
after infinite time.
Some correct reference to critical density (e.g. density > critical density for
an open universe).
(Can score the marks on a fully labelled diagram.)
3H 02
8πG
8π × 6.67 × 10 −11 × 1 × 10 −26
H 02 =
3
H0 = 2.36 × 10–18 s–1 ≈ 2.4 × 10–18 s–1
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b ρ0 =
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