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Heinemann Physics Content and Contexts Units 2A and 2B
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Contents
Unit 2A
Aerospace physics
Medical physics: atoms in action
3
5
Chapter 1 Measurement and data
1.1 Measurement and units
1.2 Data
1.3 Graphical analysis of data
Chapter review
7
7
8
10
Chapter 2 Describing motion
2.1 Describing motion in a straight line
2.2 Speed, velocity and acceleration
2.3 Graphing motion: position, velocity and acceleration
2.4 Equations of motion
Chapter review
14
15
17
20
23
Chapter 3 Forces and their effects
3.1 Force as a vector
3.2 Vector techniques
3.3 Newton’s first law of motion
3.4 Newton’s second law of motion
3.5 Newton’s third law of motion
Chapter review
27
28
34
35
37
40
Chapter 4 Energy and momentum
4.1 The relationship between momentum and force
4.2 Conservation of momentum
4.3 Work
4.4 Mechanical energy
4.5 Energy transformation and power
4.6 Elastic and inelastic collisions
Chapter review
45
46
47
48
50
51
52
Chapter 5 Nuclear energy
5.1 Atoms, isotopes and radioisotopes
5.2 Alpha, beta and gamma radiation
5.3 Properties of alpha, beta and gamma radiation
5.4 Half-life and activity of radioisotopes
5.5 Splitting the atom: nuclear fission
5.6 Nuclear fission weapons
5.7 Nuclear reactors
Chapter review
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Heinemann Physics Content and Contexts Units 2A and 2B
ISBN 978 1 7408 1791 2
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Heinemann Physics Content and Contexts Units 2A and 2B
Unit 2B
Controlling our environment
Electricity at home
64
66
Chapter 6 Heating and cooling
6.1 Heat: a historical perspective
6.2 Kinetic theory
6.3 Heat and temperature
6.4 Specific heat capacity
6.5 Latent heat
6.6 Evaporation: heat energy in context
6.7 Conduction and convection
6.8 Radiation
Chapter review
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68
68
70
70
71
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Chapter 7 Electricity
7.1 Electrical charge
7.2 Electric forces and fields
7.3 Electric current, EMF and electric potential
7.4 Resistance, ohmic and non-ohmic conductors
7.5 Electrical energy and power
7.6 Simple electric circuits
7.7 Circuit elements in parallel
7.8 Cells, batteries and other sources of EMF
Chapter review
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Heinemann Physics Content and Contexts Units 2A and 2B
ISBN 978 1 7408 1791 2
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Heinemann Physics Content and Contexts Units 2A and 2B
Unit 2A
Aerospace physics
E1
E2
E3
E4
E5
E6
E7
E8
E9
E10
E11
E12
E13
E14
E15
E16
E17
The vector diagram above shows how a jet engine provides an aircraft with forward thrust. Ff is
the forward push of air on the jet, Fa is the push of the jet on the air, and Fd is air resistance.
The sudden change in relative air speed causes a large and sudden change in the lift generated by
air speed past the wing. This can be enough to cause a stall.
The aerofoil shape is only part of the lift generated by the wing. The transfer of momentum from
air incident on the surface of the wing also contributes large amounts of lift. The amount of lift
from this source is determined by the angle of attack. A plane flying upside down will maintain
an angle of attack sufficient to generate the lift required.
Lift coefficient, L = C1 × ½ ρv × 2A; therefore a doubling of air speed will produce four times the
lift than at the original speed.
Reducing drag for a glider can be achieved by making it small and light, having a small frontal
area, a very smooth outer skin, and long and narrow wings.
Broad wings provide lift at low speeds for take-off and, with flaps extended, extra drag for
landing. At high speeds, swept-back wings reduce drag and lift, which is proportional to speed.
The surface was pushed off by the increased pressure under the surface of the wing or by a high
angle of attack.
The sheets of paper come together due to the creation of a low pressure region between the sheets
as a result of the faster air flow.
Taking off into the wind increases relative air speed past the wings and increases lift.
The wings are not generating any lift as a result of relative air speed. For the same reason,
helicopters use considerably more fuel when hovering.
The air over the top of the coin creates an area of low pressure. If there is a little air under the
coin, it will lift through this area of low pressure.
According to Bernoulli’s equation: P1 + ½ ρv12 = P2 + ½ ρv22
So P1 – P2 = ½ ρ(v12 − v22) = 1.4 × 104 N m–2
So Fnet = 1.4 × 104 × 100 = 1.4 x 106 N
Considering the force exerted due to the air intake in one second:
ΔP = m2v2 – m1v1 = (mP × 200) – {(mP + 150) × 200} = 3 × 104 kg m s-1
This is the impulse, or FΔt so in one second : F = 30 kN
The change in momentum resulting from the air and burnt fuel being expelled from the engines in
one second is:
m(air+burnt fuel)v(expelled) − m(air+burnt fuel)v(coasting) = (152 × 450) − (152 × 200) = 38 000 kg m s-1 = FΔt
So the force exerted on plane is 38 kN.
Fnet = 38 kN-30 kN = 8kN
P = Fv = 8000 × 200 = 1.6 MW
The change in relative position of the large mass of fuel affected the aircraft’s angle of attack,
changing the amount of lift generated by the wing. Moving the fuel allowed the pilot to restore
the angle of attack without the need for large rear elevators.
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Heinemann Physics Content and Contexts Units 2A and 2B
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Heinemann Physics Content and Contexts Units 2A and 2B
E18
E19 Other aspects of the aircraft must be designed around the position of the centre of gravity to
maintain stability. Other considerations include frontal area and increased drag for the wider
wing, compared with decreased glide ratio for delta wings.
E20 Because the fuel is distributed to balance the weight about the centre of gravity, we can calculate
the mass of fuel in each tank:
τ1 = τ2 so 2m1g = 5m2g and m1 + m2 = 3000 kg
m1 
5
m2
2
5
m2  m2  3000 so m2= 857.1 kg, m1 = 2142.9 kg.
2
2142.9
 428.6 kg of fuel needs to be pumped to
If the fuel from m2 is emptied then to balance,
5
the rear tank.
E21
E22 In order for the torques to offset each other:
τ1= τ2 so (d2 × 25 000) = (20 × 15 000)
So d2 = 12m, therefore the second container must be placed 12 metres forward of the aircraft’s
centre of gravity.
E23 To compensate for the shift in the load, a lift is applied from the rear of the aircraft, i.e. 40 m
behind the centre of gravity, so:
2.5 × 15 000 = 40 L
Therefore the lift required is 937 N.
E24 Altering the angle of attack increases lift from the main wing.
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Heinemann Physics Content and Contexts Units 2A and 2B
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Page 4 of 85
Heinemann Physics Content and Contexts Units 2A and 2B
Medical physics: atoms in action
E1
a
b
E2
E3
99
42
E4
E5
E6
E7
E8
E9
E10
E11
E12
E13
E14
E15
E16
E17
Technetium-99m is the most commonly used radioisotope.
99
It is the daughter nucleus of the atom molybdenum, 42
Mo . Hospitals produce technetium
from samples of molybdenum using a technetium generator.
0
Mo99
43Tc  1 
A short half-life (within hours), it must not emit alpha or beta
Ionising radiation can produce ionised water molecules and highly reactive hydroxide ions in the
cells. These can damage parts of the cell including DNA.
Gamma radiation will leave the body and be detected by the camera; alpha and beta cannot leave
the body but will reach nearby cells and destroy them.
Cells in reproductive organs, the eye and the bowel because they reproduce relatively quickly.
To give healthy cells a chance to recover between treatments.
To locate the exact position of a tumour and monitor health of nearby organs.
A radioactive isotope attached to a carrier substance that will aggregate in a particular part of the
body.
The effect that ionising radiation has on DNA is that it can cause the mutation of the cell as it
reproduces; the mutated cell can become cancerous. Enough ionising radiation can kill such a cell
or at least damage it so that it can’t reproduce.
a
Using the quality factor of 20 for alpha radiation from Table mp.2 in the student book:
dose equivalent = absorbed dose × 20 = 10 mSv
b
Using the quality factor of 1 for beta radiation:
dose equivalent = absorbed dose × 1 = 0.50 mSv
c
Again using the quality factor of 1 for gamma radiation:
dose equivalent = absorbed dose × 1 = 0.50 mSv
a
dose equivalent = absorbed dose × quality factor so the dose equivalent is 200 × 10−3 × 1 =
200 mSv
b
radiation energy absorbed = absorbed dose × mass of tissue = 200 × 10−3 × 80 = 16 J.
a
B; multiplying each absorbed dose by the corresponding quality factors from Table mp.2 in
the student book gives dose equivalents of 200 mGy, 400 mGy and 50 mGy for each
absorbed dose listed, so the most damaging dose is that of 20 mGy of alpha radiation.
b
A; these have already been converted into dose equivalents, so the most damaging is the
largest equivalent dose, which is the 200mSv of gamma radiation.
D; technetium-99m is the most appropriate for use as a radioactive tracer for the reasons stated on
p.29 of the student book.
a
Based on a daily radiation exposure of 1000 mSv, an astronaut would reach the annual
Earthly background dose in:
(2 × 10−3 × 24)/1000 × 10−3 = 0.48 hours or about 3 minutes
b
To exceed the maximum dose for workers, the astronauts will need to be in space for:
50 × 10-3 × 24 = 1.2 hours
c
Based on the cosmonaut receiving 1000 mSv of radiation per day, over 439 days he will be
exposed to:
439 × 1000 = 4.39 × 103 mSv
To minimise exposure to radiation, you could live in a weatherboard house at low altitude near
the equator, not have any X-rays taken and avoid travel by plane.
a
To deliver a dose of 0.40 Gy, the source must be left in the uterus for:
36/0.4 = 90 hours
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Heinemann Physics Content and Contexts Units 2A and 2B
ISBN 978 1 7408 1791 2
Page 5 of 85
Heinemann Physics Content and Contexts Units 2A and 2B
b
Beta particles have enough penetrating ability to irradiate the tumour, and the long half-life
means that the activity of the sample would not decrease noticeably during the course of the
treatment.
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Heinemann Physics Content and Contexts Units 2A and 2B
ISBN 978 1 7408 1791 2
Page 6 of 85
Heinemann Physics Content and Contexts Units 2A and 2B
Chapter 1 Measurement and data
1.1 Measurement and units
1
2
a
Area: m2 – derived.
b Electric charge: Q = I × t – derived.
c
Temperature – fundamental.
d Electric current – fundamental.
e
Force: F = m × a – derived.
f
Mass – fundamental.
The distance from Darwin to Alice Springs is:
1.4  10 6
 1.4  10 3 km
3
10
3
The distance from Darwin to Alice Springs is:
1.4  10 6
 1.4  108 cm
2
10
4
The distance from Darwin to Alice Springs is:
1.4  10 6
 1.4 Mm
10 6
5
The mass of the Earth is: 5.98 × 1024 × 103 = 5.98 × 1027 g
6
The mass of the Earth is: 5.98  10 24 
7
8
9
10
The mass of the Earth is: 5.98 × 1024 × 103 × 106 = 5.98 × 1033 μg
Volume of block = l × w × h = 100 × 50 × 130 mm3 = 6.5 × 105 mm3
Volume of block = l × w × h = 0.1 × 0.05 × 0.13 m3 = 6.5 x 10−4 m3
Surface area of the block:
2lw+2lw+2wh = (2 × 0.1 × 0.05) + (2 × 0.1 × 0.13) + (2 × 0.05 × 0.13)
= 0.049 m2 or 4.9 × 102 m2
-1
90/3.6 m s = 25 m s-1
178 cm s−1 = 1.78 m s−1
1.78 × 3.6 km h−1 = 6.4 km h−1
11
12
10 3
 5.98  1018 Gg
10 9
1.2 Data
1
2
Assuming a 1 cm scale, the absolute uncertainty is 0.5 cm.
The relative uncertainty for height is:
0.5
 100  2.4%
21
The relative uncertainty for width is:
0.5
 100  1.8%
28
3
4
5
Area of screen = h × w = 21 × 28= 588 cm2
The percentage uncertainty in the area of screen is the sum of the percentage uncertainties for
height and width:
2.4% + 1.8% = 4.2 %
Absolute uncertainty in area of screen = 4.2% of 588 = 24.5 cm2
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Heinemann Physics Content and Contexts Units 2A and 2B
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Page 7 of 85
Heinemann Physics Content and Contexts Units 2A and 2B
6
The percentage uncertainty in the diameter of cylinder is:
0.05
 100  0.20%
24.8
7
The percentage uncertainty in the length of cylinder is:
0.5
 100  0.37%
135
8
To calculate the percentage uncertainty in volume of cylinder we consider the following formula:
V  r 2 h
9
So we add the percentage uncertainties of each:
percentage uncertainty of height + percentage uncertainty of radius + percentage uncertainty of
radius = 0.37 +0.20+0.20 = 0.77%
V  r 2 h , so the volume of the cylinder =   (24.8 / 2) 2  135  65 212 mm 3
Absolute uncertainty = 0.77/100 × 65212 = 502 mm3
1.3 Graphical analysis of data
1
2
3
4
5
If the believed relationship is correct, then k will be the gradient of the line of best fit shown on
the graph and P0 will be the vertical intercept.
The gradient of line of best fit is 0.47, the vertical intercept is 127. So the linear equation is:
P = 0.47T + 127
k= 0.47 kPa°C-1 and P0=127 kPa
To find the experimental value for absolute zero, find the value of T when P is zero. Substituting
into the above equation:
0 = 0.47T +127
So T  
1.27
 270  C
0.47
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Heinemann Physics Content and Contexts Units 2A and 2B
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Heinemann Physics Content and Contexts Units 2A and 2B
6
7
To test this relationship, students should plot a graph showing R on the vertical axis and √l on the
horizontal axis.
8
Resistance, R (kPa)
Light intensity (lux)
9.4
12.8
20.5
27.6
29.9
43.8
57.8
0.5
1.2
4.5
9.0
10.7
25.9
44.6
(
)
0.7
1.1
2.1
3.0
3.3
5.1
6.8
9
10
11
The gradient of the graph is 8.0 kΩ lux-0.5.
The vertical intercept is 3.7 kΩ.
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Heinemann Physics Content and Contexts Units 2A and 2B
ISBN 978 1 7408 1791 2
Page 9 of 85
Heinemann Physics Content and Contexts Units 2A and 2B
12
The equation expressing the relationship between resistance and light intensity from this data is:
13
From the equation above, the values for the constants are: a = 8 kΩ lux-0.5 and b = 3.7 kΩ.
R  8.0 l  3.7
Chapter 1 Review
1
Area of floor of the classroom = l × w = 11.4 × 9.5 = 108.3 m2
2
a
0.05
 100  0.526%
9.5
0.05
 100  0.439%
Percentage uncertainty in length =
11.4
Percentage uncertainty in width 
Percentage uncertainty in area calculation = 0.526 + 0.439 = 0.965%, or approximately
0.97%.
b
Absolute uncertainty in area =
0.97
 108.3  1.05 m 2
100
3
Volume of the classroom= l × w × h = 11.4 × 9.5 × 3.2 = 346.6 m3
4
a
Percentage uncertainty in height
0.05
 100  1.563%
3.2
Percentage uncertainty in volume calculation = 0.526 +0.439 +1.563 = 2.5%
b
5
6
Absolute uncertainty in volume calculation 
2.5
 346.6  8.76 m3, or approximately 8.7
100
m3
c
The values possible may range from plus 8.7 m3 to minus 8.7 m3 about the calculated value
of 346.6 m3. That is, between (346.6 + 8.7) = 356.3 m3 to (346.6 − 8.7) = 338.9 m3.
3
1 m is 1000 L, so the volume of the classroom in litres is:
346.6 m3 = 346.6 × 1000 L = 3.47 x 105 L
1
1 foot = 0.3048 m, so 1 m 
feet and 1 m3 =
0.3048
3
 1 

 = 35.3 cubic feet
 0.3048 
8
9
10
So 346.6 m3 = 346.6 × 35.3 = 12235 cubic feet, or approximately 1.22 × 104 ft3
Betty is accurate because she hits the target evenly, but she is not precise because she didn't hit
the bull's eye.
Andrew is precise, with all shots hitting within a small region, but is not accurate.
David is both precise and accurate, with each shot hitting the bull's eye.
Celia lacks accuracy and precision.
11
Speed = distance/time =
12
To conclude whether the motorist should be fined for speeding, we need to examine the absolute
uncertainty in the speed calculation:
7
150
= 23.6 m s−1 or 117.4 km h-1
4. 6
5
 100  3.33%
150
0.2
 100  4.35%
Percentage uncertainty in timing measurement =
4.6
Percentage uncertainty in distance measurement =
Percentage uncertainty in speed calculation = 3.33 + 4.35 = 7.68%
Absolute uncertainty in speed measurement=
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7.68
 117.4  9 km h -1
100
Heinemann Physics Content and Contexts Units 2A and 2B
ISBN 978 1 7408 1791 2
Page 10 of 85
Heinemann Physics Content and Contexts Units 2A and 2B
13
14
15
Because of this level of uncertainty, the motorist could have been travelling at (117.4−9) km h−1,
which is 108.4 km h-1, so the fine is not justified in this case.
Percentage uncertainty in measured speed of the car is 7.68%.
Absolute uncertainty is 9 km h-1, leading to the range of possible speeds from 108–126 km h-1.
The motorist was probably speeding based on this figure, but they cannot be fined as there is a
chance they were not.
16
17
To obtain a linear plot, you need to graph sin(i) on the vertical axis and sin(r) on the horizontal
axis, because sin(i) = n sin(r)
18
i (degrees)
12
25
33
52
70
r(degrees)
10
19
26
37
48
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sin(i)
0.20
0.42
0.54
0.79
0.94
sin(r)
0.17
0.33
0.44
0.60
0.74
Heinemann Physics Content and Contexts Units 2A and 2B
ISBN 978 1 7408 1791 2
Page 11 of 85
Heinemann Physics Content and Contexts Units 2A and 2B
19
20
21
22
23
24
sin(i) = 1.29 sin(r) – 0.01
The experimental value of n = 1.29 ≈ 1.3.
The established relationship between v and s is v2 = u2 + 2as.
a
As a result, we need to plot v2 on the vertical axis
b
and s (displacement) on the horizontal axis to produce a straight line graph.
a
In a graph of v2 against s, the gradient is 2a.
b
The vertical intercept is u2.
We need to create a new data table to plot v2 against displacement, u:
Speed, v (ms-1)
20.0
22.5
25.2
25.8
29.5
v2
400.0
506.3
635.0
665.6
870.3
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s (m)
40
60
80
100
120
Heinemann Physics Content and Contexts Units 2A and 2B
ISBN 978 1 7408 1791 2
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Heinemann Physics Content and Contexts Units 2A and 2B
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26
27
The equation for the line of best fit: v2 = 5.50s + 175
a
Slope of the line of best fit = 5.50 m s−2
b
Vertical intercept of the line of best fit = 175 m s−2
28
The gradient of the line of best fit is 2a. a 
29
30
The initial speed of the car is 13.23 m s−1.
5.50
 2.75 m s−2
2

 1.7097

Therefore:     1.7097  5.37 N C -1


c
 12.331
Therefore:   c  12.331  3.70  10 9 N C -1 V m s -1
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Heinemann Physics Content and Contexts Units 2A and 2B
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Heinemann Physics Content and Contexts Units 2A and 2B
Chapter 2 Describing motion
2.1 Describing motion in a straight line
4
In moving from A to B, the ant travels a displacement of: 40 − 0 = +40 cm and a distance
of 40 cm.
b
In moving from C to B, the ant travels a displacement of: 40 − 50 = −10 cm and a distance
of 10 cm.
c
In moving from C to D, the ant travels a displacement of: 70 − 50 = +20 cm and a distance
of 20 cm.
d
In moving from C to E and then to D, the ant travels a displacement of (70 − 50= +20 cm
and a distance of: 50 + 30 = 80 cm
a
The distance travelled is: 50 +30 = 80 km
b
The displacement = final position – initial position = 20 km north.
a
The displacement from ground floor to basement is 10 m down.
b
The displacement from basement to the top floor is 60 m up.
c
Total distance travelled = 10 m +10 m + 50 m= 70 m
d
The displacement is the change in position, which is 50 m up.
Displacement is the only vector listed here.
5
a
1
2
3
a
The resultant vector is 25 m east.
b
6
The resultant vector is 20 m west.
a
D represents s1 + s2 = 10 m south.
b
D as s1 + s2 = s2 + s1
c
C depicts 3s2 = 30 m north
d
A depicts −s1, which is 20 m north.
7
a
s1 + s2 = 10 m south
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Heinemann Physics Content and Contexts Units 2A and 2B
ISBN 978 1 7408 1791 2
Page 14 of 85
Heinemann Physics Content and Contexts Units 2A and 2B
b
s2 + 2s1 = zero
a
b
c
Distance travelled = 10 + 4 + 15 + 5 + 5 = 39 steps
The treasure is buried one step west of the clothes line.
The displacement is 1 step west of the clothes line.
8
2.2 Speed, velocity and acceleration
1
a
b
c
d
If you walk 100 m in 50 seconds then your speed would be 2 m s-1.
A snail may crawl at about 1 mm s-1.
A cricket ball may travel at about 10 m s-1.
A bowling ball may travel 30 m in about 6 seconds, giving an estimated speed of:
30
 5 m s -1
6
2
a
Average speed =
distance 2.5  60

 15 km h -1
time
10
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Heinemann Physics Content and Contexts Units 2A and 2B
ISBN 978 1 7408 1791 2
Page 15 of 85
Heinemann Physics Content and Contexts Units 2A and 2B
3
1000
 4.2 m s -1
3600
b
Average speed = 15 
c
No, she would probably be travelling faster or slower than this average speed. It depends on
the traffic and the terrain.
a
b
c
d
distance 400

 22.2 m s -1
time
18
change in velocity 120
Average acceleration 

 6.7 km h -1s -1
time taken
18
120
Final velocity reached  120 km h -1 
 33.3 m s -1
3.6
33.3
 1.85 m s -2
So average acceleration =
18
distance
Average speed =
time
Average speed 
So distance travelled = average speed × time
The driver would have travelled
120
 0.60  20 m during their reaction time.
3 .6
4
a
10 m s-1 east minus 10 m s-1 east = 10 m s-1 plus 10 m s-1 west = 5 m s-1 west
bn
b
12 m s-1 west minus 8 m s-1 west = 12 m s-1 west plus 8 m s-1 east = 4 m s-1 west
a
b
change in speed = final speed − initial speed = 25 − 15 = 10 m s-1
change in velocity = final – initial velocity = 15 m s-1 west – 25 m s-1 east = 40 m s-1 west
5
c
6
a
b
change in velocity
40

 800 m s -2 west
time taken
0.05
change in velocity 60
Average acceleration 

 12 km h -1 s -1 south
time taken
5
change in velocity 60  3.6
Average acceleration 

 3.3 m s -2 south
time taken
5
Acceleration during contact 
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Heinemann Physics Content and Contexts Units 2A and 2B
ISBN 978 1 7408 1791 2
Page 16 of 85
Heinemann Physics Content and Contexts Units 2A and 2B
7
8
9
10
11
12
a
Kieren Perkins travelled a distance of:
30 × 50 = 1500 m
b
Average speed 
c
His displacement is zero because he ends the race where he starts it.
d
Average velocity 
distance travelled
1500

 1.7 m s -1
time taken
(14  60)  42
change in displaceme nt
and his displacement is zero, so the
time taken
average velocity is also zero.
To travel 100 km with an average speed of 50 km h-1, the trip would take 2 hours. Christopher has
already travelled for 2 hours to reach 50 km at the halfway point of his journey. It is now
impossible for him to average a speed of 50 km h-1.
B; In a steady slope as shown in (a), the ball will steadily increase its speed, resulting in a
constant acceleration.
C; In a curved ramp as shown in (b), the speed of the ball will increase but its rate of increase will
lessen as the gradient of the slope reduces throughout the journey.
To check your estimate:
The number of metres in a light year = distance light travels in one year
= speed of light (in m s-1) × number of seconds in one year
= 3.0 × 108 × 24 × 60 × 60 × 365.25
= 9.5 × 1015 m
Assuming you are sitting 3 m away from the TV set, then the signal will take:
time taken 
distance travelled
3

 110 8 seconds
8
average speed
3 10
2.3 Graphing motion: position, velocity and acceleration
1
2
3
4
5
a
b
The dancer started 4 m in the positive direction, or +4 m.
The dancer is at rest when the gradient of the position/time graph is zero, i.e. during
sections A and C.
c
The dancer moves in a positive direction when the gradient of the graph is positive, i.e.
through section B.
d
The dancer moves with a negative velocity when the gradient of the graph is negative, i.e.
through section D.
The car initially moves in a positive direction and travels 8 m in 2 s. It then stops for 2 s. The car
then reverses direction for 5 s, passing back through its starting point after 8 s. It travels a further
2 m in a negative direction before stopping after 9 s.
a
After 2 s the car is located at +8 m.
b
After 4 s it is still located at +8 m.
c
After 6 s the car is found at +4 m.
d
After 10 s it is located at −2 m.
The car returned to its starting point when it returned to the zero position on the graph, i.e. after 8
seconds.
a
The velocity of the car during the first 2 seconds is given by the gradient of the graph:
gradient 
b
rise 8
  4 m s -1
run 2
After 3 seconds the gradient is zero, therefore the velocity of the car is zero.
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Heinemann Physics Content and Contexts Units 2A and 2B
c
The velocity of the car during the period from 4–8 seconds is given by the gradient of the
graph:
gradient 
6
7
d
e
a
b
a
b
The velocity is still constant at −2 m s−1.
The velocity is still constant at −2 m s−1.
During the 10 s of motion, the car has travelled a distance of: 8 + 10 = 18 m
Its displacement is −2m.
The cyclist travels with a constant velocity in a positive direction for the first 30 s,
travelling 150 m during this time. Then the cyclist speeds up for 10 s, travelling a further
150 m. The cyclist maintains this increased speed for the final 10 s, covering another 200 m
in this time.
The velocity of the cyclist during the first 30 s is given by the gradient of the graph:
gradient 
c
rise 150

 5 m s -1
run
30
During the final 10 s, the cyclist’s velocity is:
gradient 
d
rise  2  8  10


 2 m s -1
run
94
5
rise 200

 20 m s -1
run
10
To calculate the instantaneous velocity at 35 s, draw a tangent to the curve at 35s and find
the gradient of this line:
rise 400

 13 m s -1
run
30
change in displaceme nt 150
average velocity 

 15 m s -1
time taken
10
gradient 
e
8
a
b
c
d
9
a
b
10
c
d
e
f
a
b
B; this graph could depict a car steadily coming to a halt at traffic lights.
A; the horizontal graph represents a constant velocity.
C; the constant gradient of this graph indicates a body moving with constant acceleration.
D; this graph depicts a greater initial acceleration from rest (due to steeper gradient),
followed by a lessening of acceleration as the driver changes gears and continues forward.
The dog is initially running north at 1 m s−1.
It is now increasing in speed from 1 m s−1 to 3 m s−1 (accelerating at +1m s−2) while
running north.
The dog is running north but slowing to a stop or decelerating at −1m s−2.
The dog is stationary.
Accelerating from rest to 1 m s-1 while running south.
Running south at 1 m s−1.
We can calculate the displacement of the dog by calculating the area underneath the graph
after 2 s:
area = 1 × w = 2 m north
Adding each area section underneath the graph up to 7s gives us the displacement:
1
 1

6    4  2    11  6  4  0.5  10.5 m
2
 2

c
To calculate the displacement after 10 s, we need to calculate the area beneath the 8-10 s
section of the graph and subtract this from 10.5 m:
1

10.5   11  (11)  9.0 m north
2

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Heinemann Physics Content and Contexts Units 2A and 2B
11
12
a
b
The train reaches its cruising speed (constant velocity) when the gradient of the graph is
zero, i.e. after 80 seconds.
To calculate the instantaneous acceleration, we need to calculate the gradient of a tangent at
this point:
50
 1.25 m s -2
40
40
gradient of the tangent 
 0.5 m s -2
80
gradient of the tangent 
c
d
13
a
To find the displacement, we need to find the area beneath the graph. We can estimate that
there are 50 squares in which each represents 10 × 10= 100 m. Therefore the displacement
is approximately equal to 50 × 100 = 5000 m.
For the first period of constant acceleration, during the first 4 seconds:
average acceleration 
b
c
change in velocity 8
  2 m s -2
time taken
4
The bus starts gaining ground on the bicycle after 4 seconds.
We can work out when the bus overtakes the bicycle by trial and error in calculating the
displacement (area underneath graph) for the bus and the bicycle for different times. When
these are equal, the bus is about to overtake the bicycle.
For t = 10 s, the displacement of the bus is 8 × 10 = 80 m and the displacement of the
1
2


1
2


bicycle is:   4  8   (6  8)    4  4   (2  4)  80 m
d
e
The bicycle has travelled 80 m when it is passed by the bus.
To calculate the average velocity of the bus in the first 8 s, we need to work out its
displacement. Calculating the area underneath the graph:
1
2


1

2

change in displaceme nt 56

 7 m s -1
This means that the average velocity =
time taken
8
displacement =   4  8   4  8    4  4   56 m
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Heinemann Physics Content and Contexts Units 2A and 2B
14
a
b
The change in velocity is given by the area underneath the graph:
change in velocity = (2 × 4) + (1 × 4) = +12 m s-1
2.4 Equations of motion
1
a
b
u = 0, t = 8.0s,v = 16 m s-1, a = ?
Using the relationship: v = u + at
16 = 8a
a = +2.0 ms-2
Calculate the displacement of the car first:
1
s  ut  at 2
2
1 2 1
s   at    2  8  8  64 m
2
 2
So v av 
2
c
a
change in displaceme nt 64

 8.0 ms -1
time taken
8
As above, the car travels 64 m.
u = 0, s = 400 m, t = 16 s, a = ?
1
s  ut  at 2
2
1
400  a  16 2
2
800
a
 3.1 m s -1
256
b
s = 400 m, t = 16 s, u = 0, v = ?
(u  v)t
2
16v
400 
2
s
c
3
a
So v = 50 m s−1
Converting the final speed into km h−1:
50 m s−1 = 50 × 3.6 = 180 km h−1
u = 75 km h−1, v = 0
u = 75 km h−1 = 75/3.6 = 21 m s−1
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Heinemann Physics Content and Contexts Units 2A and 2B
b
u = 21 m s−1, t = 0.25 s, s = ?
Because there is zero acceleration and the car is travelling at a constant speed, we can say
that:
s  ut  21 0.25  5.2 m
c
u = 21 m s−1, a = −6.0 m s−2, s = ? ridden
v2 = u2 + 2as
0 = 212 + (2 × −6)
s
d
4
a
b
c
d
5
a
b
441
 37 m
12
Initially, while the driver is reacting to apply the brakes, the car travels 5.2 m. It travels
another 37 m in slowing to a stop, so in total, the car travels 37 + 5.2 = 42.2 m after the
driver first notices the danger.
u = 0, s = 4.0 m (halfway down the ramp), a = 2.0 ms−2, v = ?
v2 = u2 + 2as
v2= 2 × 2 × 4
So v = +4.0 ms−1
u = 0, s = 8.0 m, a = 2.0 m s-1, v = ?
v2 = u2 + 2as
v2 = 2 × 2 × 8 = 32
So v ≈ +5.7 m s-1
u = 0, s = 4.0 m (halfway down the ramp), a = 2.0 m s−1, v (halfway down) = 4.0 m s−1, t = ?
v = u + at
4 = 0 + 2t
2t = 4
So t = 2 s for the ball to roll the first 4 m down the ramp.
u = 4.0 m s-1 (halfway down the ramp), v = 5.7 m s−1, s = 4.0 m, a = 2.0 ms−2
1
s  ut  at 2
2
1

4  4t    2t 2 
2

0 = t2 + 4t − 4
So t = 0.83 s
u = 12 m s−1(cyclist), a = 1.5 m s−2(bus)
Considering the bus; it has an initial velocity of zero and we are concerned with when its
final velocity reaches 12 m s−1.
v = u + at
12 = 0 + 1.5t
So t = 8.0 s
Set the displacement of the cyclist as equal to that of the bus and solve for t:
12t 
c
6
a
b
1
 1.5t 2
2
0 = 0.75 t2 − 12t
0 = 3t2 − 48t
t = 16 s
So, it takes 16 s for the bus to catch up to the cyclist.
The cyclist has travelled:
12 × 16 = 192 m in this time.
C; this graph depicts the constantly increasing velocity of the falling ball.
D; this graph shows the constant acceleration experienced by the ball.
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Heinemann Physics Content and Contexts Units 2A and 2B
c
s = 5.0 m, u = 0, a =-−9.8 m s-2, t = ?
s  ut 
d
7
a
5 = 4.9t2
t ≈ 1.0 s
v2= u2 +2as = 2 × 9.8 × 5.0 = 98
v ≈ 9.9 m s-1
u = 0, t = 1.0 s, a = −9.8 m s−2, v = ?
v = u + at = 9.8 m s-1
s  ut 
c
s = ?, t = 0.5s, a = −9.8 ms−2
d
a
b
c
d
e
a
1 2 1
at   9.8  0.5  0.5  1.2 m
2
2
If the book falls a total of 4.9 m, with it falling 1.2 m in the first 0.5 s, then it must fall:
4.9 – 1.2 = 3.7 m in the second 0.5 s.
If the cork takes 4.0 s to complete its entire journey, then it takes 2.0 s to reach its
maximum height.
a = –9.8 ms–2, t = 2.0 s, v = 0, u = ?, s = ?
v = u + at
0 = u – (9.8 × 2)
u = 19.6 m s-1
Now we can substitute this value for u to calculate s, the vertical displacement:
s  ut 
9
1 2 1
at   9.8  1  1  4.9 m
2
2
b
s  ut 
8
1 2
at
2
1 2
at  (19.6  2)  (4.9  2 2 )  19.6 m
2
As shown above, the cork is initially travelling at 19.6 m s–1 up.
By symmetry, if the cork’s initial velocity is 19.6 m s–1 up, then its final velocity is 19.6 m
s–1 down.
Because acceleration is constant at –9.8 m s–2, this is the acceleration experienced at each
point:
i
9.8 ms–2 down
ii
9.8 ms–2 down
iii
9.8 ms–2 down
s = 15.0 m, a = -9.8 m s–2, v =0 (at the top of the flight)
Firstly, we will calculate the initial velocity of the football:
v2= u2 +2as
0 = u2 – (9.8 × 2 × 15)
u = 17.1 m s–1
Now we can substitute this value to calculate time taken:
v = u + at
0 = 17.1 −9.8t
t = 1.7 s
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Heinemann Physics Content and Contexts Units 2A and 2B
b
10
The ruckman should time his tap after the ball has reached the top of its journey and then
fallen to be 4 m above the ground.
At s = 4 m, we can calculate the velocity of the football:
v2= u2 +2as = 17.12 + (2 × −9.8 × 4)
v −14.6 m s−1
Now we can substitute to find the time a which the ball is located 4 m above the ground:
v = u + at
−14.6 = 17.1 – 9.8t
t = 3.2 s.
The ruckman should time his tap to 3.2 s after the bounce.
The balloon is travelling downwards at a constant velocity. The time taken for the balloon to
reach the ground is given by:
time 
t
change in displaceme nt
velocity
80
 10 s
8
The coin falls with an initial velocity of 8 m s-1 downwards and a constant acceleration of 9.8 m s2
. Using v2= u2 +2as:
v2 = 82 + (2 × 9.8 × 80)
v ≈ 40.4 m s-1 downwards
Because v = u + at:
40.4 = 8 + 9.8t
t = 3.3 s
This means that the coin is ahead of the balloon by:
10 – 3.3= 6.7 s.
Chapter 2 Review
1
2
3
4
5
6
7
D; the acceleration of the ball is constant due to gravity.
v = 0, a = −9.8 m s−2, t = 1.5 s, u = ?
v = u + at
0 = u + (−9.8 × 1.5)
So u = 0 + (9.8 × 1.5) = 14.7 m s-1 upwards.
v2= u2 +2as, so:
0 = 14.72 + (2 × −9.8s)
Therefore the vertical displacement is 11.0 m.
C; the area under a velocity–time graph is displacement.
a
The motorcyclist travels in a northerly direction between 10 and 25 seconds.
b
The motorcyclist travels in a southerly direction between 30 and 45 seconds.
c
The motorcyclist is stationary from 1–10 seconds, 25–30 seconds and 45–60 seconds.
The motorcyclist passes back through the intersection at the t-intercept, that is at 42.5 seconds.
The gradient to the graph at each point gives us the instantaneous velocity.
a
b
rise 300

 20 m s -1 north
run
15
rise  600

 40 m s -1 south
At 35 seconds: gradient 
run
15
At 15 seconds: gradient 
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Heinemann Physics Content and Contexts Units 2A and 2B
8
a
Because average velocity 
change in displaceme nt
:
time taken
- 300
 5 m s -1 , or 5 m s -1 south
60
distance travelled 900
Average speed 

 15 m s -1
time taken
60
average velocity 
b
9
10
s = 10 m, u = 10 m s−1, v = 0, a = ?
v2 = u2 +2as
0 = 102 + 20a
Therefore a = −5 m s−2 and the correct response is C.
v = u + at
0 = 10 + −5t
So the time taken to stop is 2.0 seconds.
distance travelled 15  5  10

 15 km h -1
time taken
2
change in displaceme nt 20
v av 

 10 km h -1 north
time taken
2
10
vav 
 2.8 m s -1 north
3.6
Average speed of the athlete 
11
12
a
b
13
a
u = 0, t = 1 s, s = 2.0 m, a = ?
1
s  ut  at 2
2
1
2 a
2
b
c
Therefore a = 4.0 m s−2.
The speed at the end of the first second is:
v = u + at
So v = 0 + 4 × 1 = 4.0 m s-1.
Distance travelled after two seconds of motion:
1
s  ut  at 2
2
1
s  0   4  2 2  8.0 m
2
Because the jet ski travelled 2.0 m during the first second of motion, it has travelled 6.0 m
during the second second of motion.
14
a
b
distance travelled
time taken
2  1.8  1.6  1.4  1.2

(5  1)  50
 80 cm s -1  0.80 m s -1
distance travelled
Average speed in section B 
time taken
1111

 50 cm s -1  0.50 m s -1
4 / 50
Average speed in section A 
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Heinemann Physics Content and Contexts Units 2A and 2B
c
Average speed over the journey 
(12  9)
 66.7cm s -1  0.67 m s -1
50
15
a
distance travelled
time taken
distance travelled
1.6

 80 cm s -1
time taken
(1  50)
distance travelled
1.4
The instantaneous speed at the dot after X =

 70 cm s -1
time taken
(1  50)
The instantaneous speed at the dot before X =
Therefore we can estimate the instantaneous speed at X to be 75 cm s−1 or 0.70 m s−1.
16
change in velocity
time taken
2
Initial velocity in section A =
 100 cm s -1
(1  50)
1.2
Final velocity in section A =
 60 cm s -1
(1  50)
change in velocity  40

Therefore acceleration in section A =
time taken
0.1
 400 cm s -1 or 0.4 m s -1
b
Acceleration through section A =
a
Considering the shot-put:
s = ut +
1 2
1
at and 60 = ×9.8t2
2
2
So t = 3.5 s
Considering the 100 g mass:
v2= u2 +2as
v2= 102 + (2 × 9.8 × 70) and v = 38.3 m s−1
Substituting in this final velocity, we can calculate the time taken to reach the ground as:
v = u + at
38.3 = 10 + 9.8t
t = 2.89s
To determine when the mass will overtake the shot-put:
b
17
For the shot-put: s = ut +
1 2 1
at =
× 9.8t2 = 4.9t2
2
2
From the starting level of the mass: s = 4.9t2 + 10 (as it is dropped 10 m above the shot put).
s = ut +
18
1 2
1
at = 10t +
× 9.8 × t2 = 10t +4.9t2
2
2
The mass will overtake the shot put when:
10t + 4.9t2 = 4.9t2 + 10, so the mass overtakes after 1.0 s.
Displacement is the area below the graph ≈ 58 squares × 2 = 116 m.
Average velocity =
19
20
116
 10.5 m s -1 north
11
Because the area below the graph remains positive, the cyclist always travels north so the answer
is A.
a
At 1 s, acceleration is zero because the gradient of graph is zero.
b
At 5s, acceleration = gradient of graph =
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rise 12

≈ 2.0 m s-2 north
run
6
Heinemann Physics Content and Contexts Units 2A and 2B
ISBN 978 1 7408 1791 2
Page 25 of 85
Heinemann Physics Content and Contexts Units 2A and 2B
c
21
At 10 s, gradient of graph =
rise 14

≈ 7.0 m s-2
run
2
The skater will cover a distance of:
5 × 5 × 60 = 1500 m.
Assume a radius of the wheel of 4 cm:
Then C = 2  r = 2  × 4 = 25 cm = 0.25 m
Based on this estimate, the roller-blade wheel will make
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1500
= 6000 rotations.
0.25
Heinemann Physics Content and Contexts Units 2A and 2B
ISBN 978 1 7408 1791 2
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Heinemann Physics Content and Contexts Units 2A and 2B
Chapter 3 Forces and their effects
3.1 Force as a vector
1
2
3
a
A scalar is a quantity that is completely defined by its magnitude, whereas a magnitude and
a direction are essential for a vector quantity to be fully defined.
b
Scalar: mass, distance travelled, average speed, time
Vector: displacement, velocity, acceleration, force
Your answer to this may vary depending on the sequence your class has followed.
c
Average speed is a scalar quantity—defined as the rate of change of distance (a scalar).
Average velocity is a vector quantity—defined as the rate of change of displacement (a
vector). For the same motion, these quantities can be different, e.g. running around an oval
and returning to the same point. The average speed will be a certain value, and the average
velocity will be zero.
a
Estimate of the force required to use a stapler: 10 N.
b
Estimate of the force required to kick a beach ball: 100 N.
c
Estimate of the force required to lift a school bag: 50–100 N (depending upon mass of the
bag and its contents).
d
Estimate of the force needed to open this textbook: 1 N
Based on the fact that it takes a force of about 10 N to hold a kilogram of sugar, you may be able
to exert a much larger force, of around 1000 N, to pull on an anchored rope. If all members of a
ten person tug-of-war team are of similar strength, then they should be able to exert about 10 000
N.
These estimates may vary widely.
4
5
6
140°T and S40°E, 200°T and S20°W and 280°T and N80°W are the same. So the correct
response is: B, C and D.
a
N60°E is equivalent to 60°T (as the direction is given in a clockwise direction from north).
b
N40°W is 40° less than a full circle revolution, that is : 320°T.
c
S60°W; because the true measurement begins from north, the true bearing is (180 + 60)=
240°T.
d
SE; this is equivalent to halfway between the south and east directions, that is : (90 + 45) =
135°T
e
NNE; this is one quarter of the way between north and east, or: 90 ÷ 4 = 22.5°T
a
1
1
2 F  2  30 N east  75 N east
2
2
b
−6F = 6 × 30 N west = 180 N west, as the negative sign reverses the direction of the vector.
c

1
1
F   30 N east  15 N west, as again the negative sign reverses the direction of the
2
2
vector
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7
a
b
c
According to the scale, 1 cm represents 20 N. Therefore the vector shown as a force on the
ball is 3.5 × 20 = 70 N right.
The lift force shown on the suitcase is 2.5 × 20 = 50 N up.
The weight force on the dog is 4.0 × 20 = 80 N down.
3.2 Vector techniques
1
a
Looking at the scale diagram:
F1 + F2 = 35 N north 45° east
b
The resultant vector here is 2F1 = 50 N north.
c
Looking at the scale diagram: 2F1 + F2 = 56 N north 27° east.
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d
2
Again, looking at the scale diagram: 2F1 + 2F2 = 71 N north 45° east
It can be seen through the diagram below that reversing the order of addition of vectors makes no
difference to the resultant vector:
3
a
b
c
d
e
4
B; because the three vectors, F1, F2 and F3, are added tip to tail to return to their starting
point, equalling zero.
C; because the addition of vectors F1 and F2 produce vector F3.
A; because F1 plus the reverse of F2 (−F2) equals the vector F3; also D because F1 plus F3
equals –F2; and F because F1 plus –F2 equals –F3.
E; because F1 plus –F3 plus –F2 equals zero; and G for the same reason.
F2 + F3 = F1 is identical to the vector statement F1 – F2 – F3 = 0, so the answers here are E
and G from part (d) above.
a
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Heinemann Physics Content and Contexts Units 2A and 2B
Studying the vector diagram, it can be seen that Hugh applies a force of 50 N north on the
chair, whilst Elisa and Rachel combine to apply a force of 50 N south on the chair. The
resultant force on the chair is zero.
b
The forces from part (a) were in balance and the resultant force was zero. Rachel now
applies an additional force of 10 N south 60° west. This is now the resultant force.
c
If Rachel lets go, the chair will move in the direction of the sum of the forces provided by
Hugh and Elisa. Using a scale diagram, the direction that the chair will move is N 60° east.
5
a
Looking at the scale diagram, the magnitude of the resultant force is 5 N. To calculate the
direction we can use trigonometry:
tan  
4
3
So θ = 53.1. The resultant force is 5 N (90 + 53)T, or 5 N 143T.
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b
From the diagram, the magnitude of the resultant force is 100 N.
tan  
4
3
So θ = 53.1. The resultant force is 100 N (270 + 53)T or 100 N 323°T
c
From the diagram, it can be seen that the resultant force is 7.1 N north 15° east.
d
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From the diagram, the magnitude of the resultant force is 2.5 N and using trigonometry,
tan  
2
1 .5
So θ = 53.1. The resultant force is 2.5 N (20 + 53°)T or 2.5 N 73°T
6
Each person pulls with a force that has a horizontal component that will pull the car. There are
also vertical component forces, but these are equal in both directions and cancel each other out.
Therefore the magnitude of the pulling force is given by 400 × cos40 × 2 = 613 N in a direction
that bisects the two ropes.
7
a
From the right-angled triangle shown, considering the vertical component of force, Fv:
Fv
100
Fv  cos 60  100  50 N south
cos 60 
Considering the horizontal component, Fh:
sin 60 
Fh
100
Fh = 87 N east
b
As this vector only has a vertical component, the force is 60 N north.
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c
From the right-angled triangle shown, considering the vertical component of force, Fv:
Fv
300
So Fv  cos 20  300  282 N south
cos 20 
Considering the horizontal component, Fh:
sin 20 
Fh
300
So Fh = 103 N east.
d
From the right angled triangle shown, considering the vertical component of force, Fv:
Fv
sin
3  10 5
So Fv  sin 30  3  10 5  1.5  10 5 N up
sin 30 
Considering the horizontal component, Fh:
Fh
cos
3  10 5
So Fv  cos 30  3  10 5  2.6  10 5 N horizontal
cos 30 
8
Considering the vertical component of the pulling force on the tree, Fv:
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FV
300
So Fv  sin 30  300  150 N upwards.
sin 30 
The horizontal component, Fh: may be found as:
cos 30 
Fh
300
So Fh = 260 N horizontal.
9
By studying the diagram, it can be seen that the weight force can be considered as having a
component, Fy, perpendicular to the incline, and another component, Fx, running along the incline
and pulling the block down the incline. By trigonometry:
sin 30° =
Fx
, so Fx = 100 N acting down the incline.
200
10
By studying the diagram it can be seen that the addition of the two forces produces a right-angled
triangle. We can use Pythagoras’ Theorem to calculate the magnitude of the resultant force:
a2 + b2 = c2
682 + 512 = c2
c = 85 N
3.3 Newton’s first law of motion
1
Aristotle felt that the natural state for any object was at rest in its natural place. This meant that
any moving body would come to rest of its own accord. Galileo introduced the idea that friction
was a force that could be added to other forces that act on a moving body, but it was Newton who
explained that the moving object should continue to travel with constant velocity unless a net
force is acting. Aristotle could argue that because a coin flicked across a concrete path will soon
slow to a stop, it has come to rest of its own accord because its natural state is at rest. Newton
could argue that if instead the coin was a puck flung across a field of ice, it would travel much
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2
3
4
5
6
further before coming to rest, because the force of friction slowing the motion is greatly reduced
on this surface.
No force acts on the person. In accordance with Newton’s first law of motion, the bus slows, and
the standing passenger will continue to move with constant velocity unless acted on by a force—
usually the passenger will lose his or her footing and fall forward.
Hard leather soles provide little or no grip on the ice—there is no friction. As a consequence, no
force can be applied to begin walking, and the net force on the person will be zero. To be
propelled with ice skates, the blade of the skate has to be dug into the surface of the ice, allowing
the skater a fixed point from which to ‘push off’.
20 N in a forward direction, so that the net force will be zero.
To exactly balance the other forces, lift = 50 kN up, and drag = 12 kN west.
a
The boy must balance the force of friction in order for the cart to travel with constant
velocity; so he must apply a force of 25 N.
b
The horizontal component of force required is still 25 N.
c
The diagram shows that the horizontal component of the pulling force along the rope is 25
N. By trigonometry:
cos 30 
So
7
8
25
F
25
 29 N at 30 to the horizontal
cos 30
When the car or aircraft slows suddenly, a passenger will continue to travel with the same
velocity as before, until being acted on by an unbalanced force. The purpose of the seatbelt is to
supply that force, but across the body to reduce the effects of the force.
A—gravitation; B—electric force; C—friction between the tyres and the road; D—tension in the
wire.
3.4 Newton’s second law of motion
1
According to Newton’s second law:
 F  ma
It follows that the net force acting on the ball is:
57  10 3 
2
30  0
 244 N
7  10 3
Firstly we can calculate the acceleration of the car in coming to rest from 60 km h−1 in 150 m.
Using v2 = u2 + 2as
2
 60 
0= 
 + 300a
 3 .6 
Therefore a = −0.93 m s−2
Substituting this value for acceleration into Newton’s second law:
 F  ma
it follows that the average force required of the brakes to bring the car to rest is:
1200 × −0.93 = 1.1 × 103 N opposing the motion.
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3
The 1.5 kg shot-put is the larger of the two masses, and so for the same applied force, its
acceleration will be lower. This means a lower speed on leaving the athlete’s arm, so it cannot be
thrown as far as the lighter 1.0 kg ball.
4
a
b
Because the two forces act at right angles, we can use Pythagoras’ Theorem to calculate the
size of the net force acting:
a2 + b2 = c2
c2 = 1002 + 1252
c ≈ 160 N
We can use trigonometry to calculate the angle:
tan  
c
125
100
θ = 141°
We can say that the ball will travel 141°T.
Using Newton’s second law is:
 F  ma
it follows that the acceleration of the ball is:
a
F 160

 213 m s -2 141T (but this is for a very short time).
m 0.75
5
We know that by Newton’s second law is:
6
Therefore if a car has to overcome a drag force opposing motion of 800 N, the motor must
provide a force 800 N larger than the net force to overcome this, that is:
Fm – 800 = ma
Fm = (900 × 2) + 800 = 2600 N in the direction of the acceleration.
On the Earth’s surface, the gravitational field strength is almost the same at every point, so the
weight of a stationary object is proportional to its mass (W = mg, where g is constant). Scales
make use of this fact and are calibrated in kilograms. Scales would be inappropriate for use on the
Moon as they measure weight, not mass. A set of bathroom scales would only register a 72 kg
man’s weight as:
 F  ma
72
 12 kg
6
7
8
This is because the gravitational field strength on the Moon is only one sixth that on Earth.
The mass of the hammer remains the same everywhere, so this is still m = 1500 g. Its weight has
altered because W = mg = 1.5 × 3.6 = 5.4 N towards the Martian surface.
By trigonometry the horizontal force acting on the skier, Fh, is:
1015 × cos10° = 1000 N
Considering the drag force opposing motion of 400 N, the net force is:
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1000 − 400 = 600 N in the direction of motion.
Using Newton’s second law:
 F  ma
600 = 60a
So a = 10 m s−2 horizontally.
9
As the parachutist leaves the aircraft, his or her weight will be the net force acting, accelerating at
9.8 m s−2, but as the speed increases, the drag force (air resistance) opposing the motion also
increases until it equals the weight. At this point in time, the net force will be zero and the
parachutist will travel with a constant speed.
3.5 Newton’s third law of motion
1
a
a
b
Hitting a ping-pong ball with a bat: the action force is the force from the bat onto the ball.
The reaction force is the force that the ball exerts on the bat in the opposite direction.
A pine cone falling from the top of a tree to hit the ground: the action force is the weight of
pine cone, i.e. gravitation pulling it to the Earth. The reaction force is that of the Earth
being pulled toward the pine cone with the same force.
Letting go of an untied balloon filled with air: the action force is the air under pressure in
the balloon pushing backwards on the air escaping from the balloon. The reaction force is
the air escaping from the balloon pushing forward on the air inside the balloon.
140 N acts on the boat in the opposite direction to the leaping fisherman.
Considering the 40 kg boat, by Newton’s second law:
c
140 = 40a
So a= 3.5 m s−2 in the opposite direction to the fisherman.
Considering the 70 kg fisherman,: by Newton’s second law:
b
c
2
3
a
b
 F  ma
 F  ma
140 = 70a
a = 2.0 m s−2
Using v = u + at, then v = 0 + (2 × 0.5) = 1.0 m s−1.
The boat is moving with an acceleration of 3.5 m s−2, so:
v = u + at = 0 + (3.5 × 0.5) = 1.75 m s−1 in the opposite direction to the fisherman.
If no other forces act, there will be an action/reaction force pair in which the action force is
the force on the tool kit, and the reaction force will act on the astronaut in the opposite
direction. If the kit is thrown directly away from the ship, hopefully the reaction force will
propel him back to the craft.
By Newton’s second law:
net force = ma = 2.5 × 8 = 20 N
This force acts on each mass, but in opposite directions.
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c
Considering the astronaut: F = 20 N, m = 100 kg, a = ?
By Newton’s second law:
a
d
So to calculate his velocity:
v = u + at
v = 0 + (0.2 × 1.0) = 0.2 m s−1 back towards the mother ship.
Distance travelled = v × t, so the time taken to return to the ship is:
t
4
6
d 100

 500 s or 8 min 20 s
v 0.2
The bowl and the jack collide head on. The bowl exerts a force on the jack, which exerts an equal
and opposite force back on the bowl. By Newton’s second law, the force on the jack is:
F = ma = 1 × 25= 25 N north
This means that the force on the bowl is 25 N south:
a
5
F
20

 0.2 m s -2
m 100
F
 12.5 m s -2 south
m
The force of gravitation on the speaker will be 49 N downwards and the normal force exerted by
the bookshelf on the speaker will be 49 N upwards. The reaction force to the weight of the
speaker will be a force on the Earth of 49 N acting towards the speaker.
a
Forces acting on the competitor: weight vertically down, normal force up and perpendicular
to the incline, friction up the incline
b
The forces that act on the competitor are: weight W = mg = 65 × 9.8 = 637 N downwards,
and the normal reaction force, N. Studying the diagram it can be seen that:
cos 50 
c
d
N
637
So N = 637 × cos50° = 409 N up and perpendicular to the track.
Referring to the diagram, the component of the weight force down the slope is:
ΣF = 637 × sin50° = 488 N along the incline.
By Newton’s second law:
 F  ma
 F  488  7.5 m s
a
m
-2
65
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7
8
9
10
In order to travel with a constant velocity down the hill, the cyclist needs to apply sufficient
braking force to balance out the component of her weight acting down the incline, such that the
net force acting is zero. The weight of the cyclist and her bike = mg = 110 × 9.8 = 1078 N
downwards.
The component of this weight acting down the slope = 1078 sin15 = 280 N. Thus, the cyclist
needs to apply a braking force of 280 N to travel with constant velocity.
As the lift accelerates, the normal force that you experience from the floor increases—this is your
apparent weight. The normal force has to balance your weight and provide extra force to
accelerate you upwards:
N = ΣF – W
The same situation occurs when the lift comes to rest while ‘going down’ as the net force must
still be upwards.
a
When the lift is at rest, Paul’s apparent weight is equivalent to his actual weight:
weight = mg = 80 × 9.8 = 784 N down.
b
When the lift is accelerating upwards at 2.2 m s−2, the net force is acting upwards. This
means that the force of the lift, FL, minus his weight is the net force acting:
FL – W = ma
FL – 784 = 80 × 2.2
So the force of the lift is 960 N up, and Paul’s apparent weight is 960 N down.
c
If the lift is travelling at a constant velocity, the force of the lift balances the weight force,
and so Paul’s apparent weight is 784 N down.
d
If the lift is slowing upwards, the net force is acting downwards. This means that:
W – FL = ma
So FL = (80 × 9.8) – (80 × 2.2) = 608 N.
Therefore Paul’s apparent weight is 608 N down.
e
If the lift is accelerating downwards, the net force is acting downwards. This means that
Paul’s apparent weight is again 608 N down.
a
b
The net force on the rope is the weight of the student minus the weight of the sandbag:
Net force = (50 × 9.8) – (30 × 9.8) = 196 N down on the side of the student.
We can say that, considering the student:
490 – T (tension in the rope) = ΣF = ma
T = 490 – ma
So T = 490 – 50a
Now considering the sandbag:
T – 294= ΣF = ma
T = (30 × 9.8) + 30a
T = 294 + 30a
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c
Now, eliminating T:
490 – 50a = 294 + 30a
80a = 196
So a = 2.45 m s−2 down to the side of the student.
Substituting into the first equation to find T:
T = 490 – (50 × 2.45) = 368 N
Chapter 3 Review
1
2
3
A person trips because their foot stops momentarily, and the rest of the body continues with
constant velocity. The constant velocity of a satellite in deep space far from the influence of any
gravitational fields. A racing car will leave the race track at a bend and continue in a straight line
if it hits an oil patch, because of a lack of friction.
If James needs to push the trolley with a force of 30 N to maintain its speed, he must be
overcoming a drag force of 30 N. In order for the trolley to accelerate at 0.5 m s−2; we can say
that:
FJ (James’ force) − 30 = ma
Therefore FJ= (35 × 0.5) + 30 = 47.5 N
To calculate the acceleration of the bowling ball, we use Newton’s second law:
ΣF = ma
a
4
5
6
 F  25  1.67 m s
m
-2
15
Now we can calculate the velocity of the ball:
v = u + at = 0 + 1.67 × 4 ≈ 6.7 m s−1
Kicking the tyre of a car hurts because you apply a force to the tyre (action) and the tyre will
apply a reaction force to your foot (reaction). Hot gases are forced out of a jet engine (action) and
the gases push the engine forward (reaction). The weight of any object can be considered an
action force—the Earth pulls on the body; the reaction force acts on the Earth—the object pulls
on the Earth.
a
u = 0, a = 2.0 m s−2, t = 2.5 s, v = ?
Using v = u + at:
v = 0 + 2 × 2.5 = 5.0 m s−1
b
i
When the lift is stationary, Jane’s apparent weight is her actual weight:
W = mg = 539 N down.
ii
When accelerating upwards:
FL − 539 = ma
So FL = (55 × 2) + 539 = 649 N down.
iii
When the lift is travelling with constant speed, the net force is zero and so Jane’s
apparent weight is again equal to her weight = 539 N down
iv
When the lift is accelerating downwards, the net force is acting downwards so we can
say that:
539 – FL = ma
So FL= 539 – (2 × 55) = 429 N down.
a
The weight of the astronaut on Earth = mg = 85 × 9.8 = 833 N down.
b
The weight of the astronaut on the Moon = mgM = 85 × 1.6 = 136 N down.
c
The weight of the astronaut on Mars = mgMA = 85 × 3.6 = 306 N down.
d
The weight of the astronaut during free fall, considering g = 9.8 m s−2 = mg = 85 × 9.8 =
833 N down.
e
If we were to place the astronaut over a set of scales while in free fall, they would read zero
as there is no normal force.
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f
7
8
The apparent weight, F, of the astronaut during lift-off:
F – 833 = ma
So F = (85 × 5) + 833 ≈ 1260 N down.
A block is struck with a sharp blow so that it overcomes the grip of the blocks with which it is in
contact, and so it is ejected from the pile. The other blocks experience no horizontal net force, and
so stay in the vertical stack.
a
Looking at the diagram, we can resolve the pushing force into horizontal and vertical
components:
cos 20 
Fh
120
So Fh = 120 cos20 = 113 N horizontally.
sin 20 
Fv
120
So Fv = 120 sin20 = 41 N down.
If the horizontal pushing force of 113 N keeps the trolley travelling at a constant velocity,
then the net force must be zero and therefore the drag force is 113 N to the south.
c
Because the vertical forces are balanced, the normal reaction force is:
N = W + Fv = 196 + 41 = 237 N upwards.
d
When the trolley is pulled, the vertical component of the applied force is upwards rather
than downwards, and so a smaller (upward) normal force is needed—helping the wheels to
rotate more freely.
Considering the first second:
b
9
a
F 0.5

 5.0 m s -2
M 0.1
So after 1 second, the velocity of the glider is:
v = u + at = 0 +(5 × 1) = 5.0 m s−1
Considering the second second:
a
10
11
F 0.2

 2.0 m s -2
m 0.1
The velocity is now:
v = u + at = 5 + (2 × 1) = 7.0 m s−1
B; the puck moves with constant velocity—there is no friction, so no other horizontal force exists
otherwise the forces would not be balanced.
a
We can calculate the acceleration of the trolleys using Newton’s second law:
 F  ma
120 = (30 + 50)a
a
120
 1.5 m s -2 in the direction of the force.
80
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b
12
a
b
Considering the contact force experienced by the second trolley:
F = ma = (50 × 1.5) = 75 N
1:1 because the force experienced by the girl and the skateboard are equal.
By Newton’s second law because the forces are equal, the ratio of acceleration experienced
by the girl and the skateboard varies inversely with the mass of each:
a
1
m
Because the girl’s mass is four times that of the skateboard, the ratio is 1:4.
We could calculate the velocity of the girl and her skateboard using the relationship:
v = u + at
this means that as the initial velocity is zero, v  a and so once again this ratio is 1:4.
T (tension) − W = ma
100 − (12 × 9.8) = 12a
c
13
So a 
14
17.6
 1.47 m s -2
12
Therefore to maintain a tension of less than 100 N in the rope, the bucket must accelerate at
greater than 1.47 m s−2 downwards. If the acceleration falls below this, the rope will snap.
In calculating the maximum possible mass that may be caught:
T – W = ma
50 − 9.8m = 2.2m
So 12m = 50
Therefore the heaviest fish that could be caught without breaking the line is 4.2 kg.
15
If the slope has a grade of 1 in 5, then:
tan  
1
5
So θ = 11.3
Considering the diagram, the component of weight down the road is:
F = mg sin11.3 = 9.8m sin11.3 = 1.96 m
Now, by Newton’s second law we know that:
a
F 1.92m

 1.96 m s -2
m
m
Therefore, using v2 = u2 − 2as:
v2= 2 × 1.96 × 100
So v = 19.8 m s−1 = (19.8 × 3.6) km h−1 = 71.3 km h−1
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16
Studying the diagram, we can see that the component of the weight of the car down the slope is:
F = mg sinθ
By Newton’s second law, if the acceleration is
mg sin   m 
17
1
g ; cancelling out m and g:
2
sinθ = 0.5 and so the angle of the track is 30.
The component of Eddie’s weight force minus friction is the resultant force, so:
mg sin35° − 250 = ma
(70 × 9.8 × sin35°) − 250 = 70a
393.5 − 250 = 143.5 = 70a
So a 
18
1
g , then:
2
143.5
 2.05 m s -2 down the incline.
70
a
b
19
Because the 10 kg mass is the heaviest, it will pull the 5 kg mass upwards. We will take the
upwards direction as positive. Considering the 5 kg mass:
T − W = ma
T − (5 × 9.8) = 5a
T = 5a + 49
Considering the 10 kg mass:
T − W = ma
T = 98 − 10a
Combining these equations:
98 − 10a = 5a +49
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So 15a = 49 and a = 3.27 m s−2.
Substituting to find the tension in the rope:
T = 5a + 49 = (5 × 3.27) + 49 = 65.3 N upwards.
20
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Chapter 4 Energy and momentum
4.1 The relationship between momentum and force
1
2
b
Momentum is: p = mv – the product of mass and velocity:
p = mv = 20 × 5 = 100 kg m s−1
p = mv = 20 × 0.05 = 1 kg m s−1
c
p = mv = 20 
a
b
Initial momentum, p = mv1 = 8 × 3 = 24 kg m s−1
Final momentum, p = mv2 = 8 × 8 = 64 kg m s−1
c
Change in momentum per second =
a
5
= 28 kg m s−1
3.6
40
= 8 kg m s−1
5
d
e
3
4
Change in momentum = impulse experienced by object = 40 N s
Impulse = FΔt
40 = 5F
F=8N
Momentum of medicine ball 1= mv = 4.5 × 3.5 = 15.75 kg m s−1
Momentum of medicine ball 2 = mv = 2.5 × 6.8 = 17 kg m s−1
So the 2.5 kg medicine ball has greater momentum.
a
Momentum, p = mv = 4.5 × 9.1 = 41 kg m s−1
Momentum, p = mv = 0.250 ×
c
Firstly we need to calculate the final velocity of the object:
v = u + at = 0 + 9.8 × 15 = 147 m s−1
Momentum, p = mv = 3.4 × 147= 500 kg m s−1
Change in momentum = impulse:
p = FΔt = 45 × 3.5 ≈ 158 kg m s−1
Momentum upon leaving the racquet = mv = 0.065 × 61 = 4.0 kg m s−1
Change in momentum = impulse = FΔt
Therefore 4.0 = 0.032F
The average force applied by the racquet on the ball is 125 N.
To calculate the impulse the ball experiences, we need to calculate its change in
momentum. Its initial momentum is zero and its final momentum is:
p = mv = 0.200 × 45 = 9.0 kg m s−1
Thus the impulse is 9.0 N s.
I = FΔt
So the net average force acting on the ball during contact time:
d
5
a
b
6
a
b
F
c
7
a
b
8
3 .5
= 0.24 kg m s−1
3 .6
b
a
9 .0
 180 N in the direction of the ball’s travel.
0.05
By Newton’s third law, the average net force acting on the bat during contact time is 180 N
opposite the direction of the ball’s travel.
Reading from the graph, the maximum force acting on the athlete’s foot during contact time
is 1200 N.
Impulse is the area underneath the graph.
We can estimate this as = 33 squares × 10 ×10-3 × 200 = 66 N s.
Change in momentum = mΔv = 0.025 × 50 = 1.25 kg m s−1 in the opposite direction of the
flight of the arrow.
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b
c
9
10
Impulse is 1.25 N s opposite the direction of flight of the arrow.
v = 0, u = 50 m s−1, s = 0.02 m, a = ?
Using v2 = u2 +2as:
0 = 502 + 2 × 0.02a
0.04a = −2500 and a = −6.25 × 104 m s−2
By Newton’s second law:
F = ma = 0.025 × 6.25 × 104 = 1.56 × 103 N in the opposite direction of the flight of the
arrow.
a
The design of a crash helmet is based on the idea of impulse. The stopping time of the
impact is increased by the collapsing shell of the helmet. Because impulse = FΔt, it follows
that increasing the length of time of the impact will reduce the impact force.
b
In the case of a rigid shell, the time over which the impact occurs would be reduced and as
a result the impact force would be increased, so this design would not be successful.
We need to include mass and initial velocity, and then calculate Δp = mΔv. Estimate the time of
impact in order to calculate an estimate of the impact force using: F 
P
(units: N).
t
4.2 Conservation of momentum
1
2
3
4
m1 = 0.1 kg, u1 = 2 m s−1, m2 = 0.1 kg, u2 = 0
By the law of conservation of momentum:
m1u1 + m2u2 = m1v1 + m2v2
0.1 × 2 + 0 = 0 + 0.1v2
Hence the black ball moves off with a velocity of 2.0 m s−1 in the direction of the white ball’s
original motion.
m1 = 50 kg, u1 = 5 m s−1, m2 =4 kg, u2=1.0 m s−1
By the law of conservation of momentum:
m1u1 + m2u2= m1v1 + m2v2
50 × 5 + 4 × 1 = (m1+m2)v (as the girl and skateboard move off with a common velocity)
Therefore:
254 = 54v
v = 4.7 m s−1 in the same direction.
By the law of conservation of momentum:
m1u1 + m2u2 = m1v1 + m2v2
(70 × 2.5) + 0 = 0 + 400v
400v = 175
The boat moves off at 0.44 m s−1 in the opposite direction.
The railway car and stationary train become coupled together after the collision:
m1u1 + m2u2 = (m1 + m2)v
(2000 × 2) + 0 = (2000 + m2) × 0.3
4000 = 600 + 0.3m2
The mass of the train, m2 =
5
6
3400
 11.3 tonnes.
0. 3
m1u1 + m2u2 = (m1 + m2)v
(4 × 4.5) + 0 = (4 + 2)v
6v = 18
Therefore the combined velocity is 3.0 m s−1 – which is response B.
a
Only if Superman is fixed to the ground, so that the momentum of the truck is transferred
directly to the ground.
b
If we estimate the truck to have a mass of 2000 kg and an initial velocity of 80 km h-1, and
Superman to have a mass of 100 kg, then the combined velocity of Superman and the truck
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after the collision would be:
m1u1 + m2u2 = (m1 + m2)v
2000 
7
Therefore v ≈ 21.2 m s−1 or about 76 km h−1.
mC = 1100 kg, uC = ?, mF= 2200 kg, uF = 50 km h−1, v= 0
m1u1 + m2u2 = 0 (as both cars come to a stop after the collision):
1100 × u1 = 2200 x 50
u1 
8
9
10
80
 2100v
3.6
2200  50
 100 km h -1 in the opposite direction to the four wheel drive.
1100
So if the speed limit was 70 km h−1 then yes, the first car was speeding.
If the first object is the apple and the second is the arrow, then: m1 = 0.1 kg, u1 = 0, m2 = 0.08 kg,
u2 = 35 m s−1, v1 = ?, v2 = 25 m s−1
By the law of conservation of momentum:
m1u1 + m2u2 = m1v1 + m2v2
0 + 0.08 × 35 = 0.1v1 + 0.08 × 25
2.8 = 0.1v1 + 2
v1 = 8.0 m s-1
So the arrow will fly off Master Tell’s head at 8.0 m s−1
The momentum of the 5 kg of expelling exhaust gases is equal to the momentum given to the
remainder of the space shuttle (1000 – 5 = 9995 kg), so:
5 × 6000 = 9995v
The velocity of the space shuttle is 3.0 m s-1 in the opposite direction to the exhaust gases.
a
The rocket loses 50 kg of fuel over two seconds, so we will use the average mass of the
rocket as 225 kg. Then the momentum of the exhaust gases is equal to the momentum of
the rocket:
50 × 180 = 225v
So the velocity of the rocket after the initial acceleration is:
v
b
9000
 40 m s -1
225
Impulse = change in momentum = FΔt
upward force 
c
50  180
 4.5  10 3 N up
2
Net force = upward force − weight force
F = (4.5 × 103) – (225 × 10) = 4500 – 2250 = 2250 N
F = ma so the acceleration of the rocket =
2250
 10 m s -2 upwards.
225
4.3 Work
1
2
3
The force required to lift a 4.5 kg mass = mg = 4.5 × 9.8 = 44.1 N. The work done in lifting the
mass 6.0 m is:
W = Fs = 44.1 × 6 = 265 J
The work done in climbing the hill = Fs; where the force to be overcome is weight:
work done = 60 × 9.8 × 250 ≈ 1.5 × 105 J
No, the bushwalker also does work travelling against frictional forces that were not calculated in
the answer above.
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4
5
6
7
8
9
Statement (i) is incorrect because it mentions distance rather than displacement. Statement (iii) is
incorrect because work done is affected by the angle at which a force acts. Statement (ii) is
correct, which is response D.
a
The work done in lifting one box into the tray is:
W = Fs = 10 × 9.8 × 1.5 = 147 J
b
The work done is equal to the change in energy the removalist has used, which is 147 J.
c
The work done in lifting all of the boxes is:
W = 147 + (10 × 9.8 × 1.8) + (10 × 9.8 × 2.1) + (10 × 9.8 × 2.4) + (10 × 9.8 × 2.7)
W = 1029 J ≈ 1.0 × 103 J
The correct response is B, as the weight of the lift is (500 × 9.8 = 4900 N) and the motor won’t
convert all supplied energy to useful work.
No work is being done, since there is no change in vertical position.
a
W = Fs = 100 × 9.8 × 2.4 ≈ 2.4 x 103 J
b
Nil, as there is no change in position.
a
b
Calculating the component of tension in the rope that acts in the horizontal direction of
motion by the crate:
cos35° =
10
F
60
F = 60 cos35° = 49.1 N
Therefore work done = Fs = 49.1 × 4 = 197 J
c
Work done by the rope = 197 J
Work done by friction = −10 × 4 = −40 J
Total work done = 157 J
d
Energy lost as heat and sound = work done by friction = 40 J
The work done on the rubber band is given by the area beneath the graph. We can estimate this
as:
W
1
 (2  12)  0.2  1.4 J
2
4.4 Mechanical energy
1
a
b
1 2 1
mv   1 2.5 2  3.1 J
2
2
1 2 1
Kinetic energy, E K  mv   0.005  400 2  400 J
2
2
Kinetic energy, E K 
2
2
c
1
1
 75 
5
Kinetic energy, E K  mv 2   1200  
  2.6  10 J
2
2
 2.6 
a
b
Gravitational potential energy, EP = mgh = 1 × 9.8 × 5 = 49 J
Gravitational potential energy, EP = mgh = 0.105 × 9.8 × 400 = 412 J
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3
4
c
Gravitational potential energy, EP = mgh = 1200 × 9.8 × 10 ≈ 1.2 × 105 J
a
EP (initial position) = mgh = 0.1 × 9.8 × 2.5 ≈ 2.5 J
b
EP (final position) = mgh = 0.1 × 9.8 × 1.8 ≈ 1.8 J
c
EP (initial − final position) = mgΔh = 0.1 × 9.8 × (2.5 − 1.8) = 0.69 J
Cricket ball 1 has the following energy:
EK 
1 2 1
mv   0.15  25 2  46.9 J
2
2
Cricket ball 2 has the following energy:
EP = mgh = 0.15 × 9.8 × 14 = 20.6 J
Cricket ball 3 has the following energy:
EK  EP 
5
1 2
1

mv  mgh    0.15  10 2   (0.15  9.8  10)  22.2 J
2
2

Therefore the correct response is A.
Work done is equal to the change in kinetic energy; so:
1 2
mv
2
1
100
50 F   900 
2
3.6
Fs 
6
50F = 3.47 × 105 and F ≈ 6.9 x 103 N
a
Just before the impact, the potential energy of the steel ball has been converted to kinetic
energy:
1 2
mv  mgh
2
b
v2 = 2gh = 2 × 9.8 × 1.25 = 24.5
So v ≈ 4.95 m s-1
The kinetic energy just before impact is:
EK 
1 2 1
mv   0.08  4.952  0.98 J
2
2
c
7
The change in gravitational potential energy is equivalent to the kinetic energy just before
impact, which is 0.98 J.
Calculating the potential energy of the high-jumper at the highest point:
EP= mgh = 65 × 9.8 × 2.13 = 1357 J
If the 0.3 m mat is compressed by 0.18 m, then the high-jumper’s lowest point during the jump is:
0.30 − 0.18 = 0.12 m, so:
change in potential energy = mgΔh = 65 × 9.8 × (2.13 − 0.12) = 1280 J
Therefore work done = Fs = 1280
F
8
1280
 7.1 10 3 N
0.18
In this case, work done = 1357 J as the high-jumper lands on the ground, and:
Fs = 1357
1357
 2.7  10 4 N
0.05
1
1
Kinetic energy  mv 2   0.08  802  256 J
2
2
F
9
10
Work done = Fs
256 = 0.24F
F = 1.07 x 103 N
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4.5 Energy transformation and power
1
2
3
4
5
Car slows to rest: kinetic energy → heat energy
High-jumper leaps: kinetic energy → gravitational potential energy
Swimmer dives off board: elastic energy → kinetic energy → gravitational potential energy
→ kinetic energy
d
Athlete’s foot on track: kinetic energy → heat/sound energy
Swimmer diving into pool of water: elastic potential energy → kinetic energy → gravitational
potential energy → kinetic energy (+ heat/sound energy) → heat energy (+ sound energy, kinetic
energy of water rebounding)
a
EP = mgΔh = 46 × 9.8 × 12 = 5410 J ≈ 5.4 × 103 J
a
b
c
Power 
a
b
W = ΔEP = mgΔh = 15 × 0.5 × 9.8 × 1 = 73.5 J
This is the total gravitational potential energy of the shot-puts: 73.5 J.
c
Power 
d
It will be greater—the coach has to lift his own body, also there are heat energy losses to
the environment.
Potential energy of one shot-put half way to the ground:
EP = mgh = 0.5 × 9.8 × 0.5 = 2.45 J ≈ 2.6 J
Kinetic energy of the shot-put half way to the ground is half of the original potential
energy, which is 2.6 J.
It is transferred to the ground and transformed into other energy forms.
At the end of the vine, Tarzan’s total energy will be potential energy. This is equivalent to
his initial kinetic energy when he is running at his fastest speed:
a
b
6
work 5410

 451 W  4.5  10 2 W
t
12
b
c
a
work 73.5

 0.61 W
t
120
1 2
mv  mgh
2
1 2
v  gh
2
v 2 2  9.2  9.2
h

 4.3 m
2g
9.8
b
7
We have not considered the fact that the length of rope may be insufficient to allow height
change, air resistance and other factors leading to energy transfer/loss to the surrounding
environment.
In this case, the initial kinetic energy must provide sufficient energy to allow for the potential
energy of the high jump and a final kinetic energy:
1
1
2
2
mvi  mgh  mv f
2
2
Cancelling the mass from each side of the equation:
1 2
1 2
vi  gh  v f
2
2
1 2
1

So vi  (9.8  1.80)    0.50  0.50   17.77
2
2

vi2= 35.53, so v ≈ 6.0 m s−1
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8
a
Ignoring air resistance, all of the gravitational potential energy will be converted to kinetic
energy as the apple falls:
mgh = mgh 
b
Cancelling the mass from each side:
v2 = 2gh = 2 × 9.8 × 5 = √98 = 9.9 m s−1
If the apple falls to the ground with a speed of 3.0 m s−1, then kinetic energy at impact:
EK 
9
1 2 1
mv   0.1 32  0.45 J
2
2
The initial gravitational potential energy is:
EP = mgh = 0.1 × 9.8 × 5 = 4.90 J
Therefore energy lost = 4.90 − 0.45 = 4.45 J
Work done = energy lost = Fs; hence 4.45 = 5F
So F = 0.89 N
Work done = Fs = 500 × 0.50 = 250 J
If 485 J was actually required to do this work, then the efficiency of the engine is:
efficiency 
10
1 2
mv
2
250
 100  52%
485
Work done = Fs = 450 × 10 = 4500 J
work
 5000
t
4500
t 
 0.90 s
5000
Power 
So it will take the engine 0.90 s to mover the crate.
4.6 Elastic and inelastic collisions
1
C; in an elastic collision, both kinetic energy and momentum are conserved.
2
a
b
3
c
d
a
b
c
4
a
1 2 1
mv   0.2  9 2  8.1 J
2
2
1
1
1
1
2
2
m1v1  m2 v2   0.2  32   0.1 12 2  8.1 J
EK (after the collision) =
2
2
2
2
EK (before the collision) =
This collision is elastic since there is no loss of kinetic energy during the collision.
No. Truly elastic collisions only occur at the atomic level.
The total momentum before the collision is zero because the cars have the same mass and
have velocities equal in magnitude but in opposite directions.
Because momentum is conserved, the total momentum after the collision is still zero.
m1u1 + m2u2 = 0 = (m1 + m2)v
0= 3000v hence v after the collision is zero.
Total kinetic energy before collision =
2
1
1
1 
 72  
2
2
5
mu1  mu2  1500  
   2  6.0  10 J
2
2
2 
3
.
6

 
5
b
c
a
b
c
Because the cars are at rest after the collision, the total kinetic energy after is zero.
Because the kinetic energy has been lost as sound, heat, etc. the collision is inelastic.
All of the original kinetic energy is lost = 6.0 × 105 J
It is converted into other forms of energy , such as heat and sound, during the collision.
Because all of the energy has been lost, efficiency is 0%.
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6
Initial kinetic energy =
1
1
1
2
2
mu1  mu 2   4  3 2  2  36 J
2
2
2
20 J of this energy is lost during the collision, leaving 16 J. Therefore after the collision the balls
have the following total kinetic energy:
E K  16 J 
7
1 2
mv  2  mv 2
2
v2 = 4
So the final velocity is 2.0 m s−1 left, 2.0 m s−1 right.
a
Kinetic energy before impact = initial gravitational potential energy of the plasticine = mgh
= 0.200 × 9.8 × 2.0 = 3.92 J
b
Velocity of the plasticine before impact:
v12 = u2 + 2as = 2 × 9.8 × 2 = 39.2
v1 ≈ 6.26 m s−1
The initial momentum of the plasticine = mv1 = 0.2 × 6.26 = 1.25 kg m s−1
Velocity after impact: 1.25 = Mv2 = (0.2 + 0.4)v2 = 0.6v2
v2 = 2.09 m s−1
Therefore kinetic energy after the impact =
8
Initial energy of the ball =
1
1
2
mv2  mv 2   0.6  2.09 2  1.31 J
2
2
1
1
2
mv2   0.1 10 2  5.0 J
2
2
We need to find the displacement such that the area beneath the graph is 5 J. By trial and error,
this can be found to exist when the displacement is approximately 7 mm:
i.e. for a displacement of 7 mm area, the energy = 0.5 × 7 × 10−3 × 1400 = 4.9 J
9
a
b
c
1
 0.04  400  8.0 J
2
1
The area under the graph   0.03  300  4.5 J
2
The area under the graph 
Energy is converted into heat during this process.
Chapter 4 Review
1
By the law of conservation of momentum:
m1u1 + m2u2 = v(m1 + m2)
55 × 5 + 70 × 0 = (55 + 70)v
So v 
2
By the law of conservation of momentum:
m1u1 + m2u2 = m1v1 + m2v2
(300 × 2) + (0.5 × 2) = (−0.5 × 98) +300v
601 = −49 + 300v
v
3
275
 2.2 m s−1 in direction of the moving player.
125
650
 2.2 m s−1 north
300
By the law of conservation of momentum:
m1u1 + m2u2 = m1v1 + m2v2
0.15 × 3 = 0.15v + 0.1 × 1.2
0.45 = 0.15v + 0.12
0.15v = 0.33
So v = 2.2 m s−1 in the same direction.
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4
5
6
7
8
9
This is true, but due to the much larger mass of the Earth, there is negligible movement but there
is a change. We could estimate: Δp(human) = Δp(Earth), so for a 60 kg person, jumping at 0.5 m
s−1:
Δp (human) = 30 kg m s−1 = Δp(Earth) = mEv = 5.97 × 1024v
So v ≈ 5.0 × 10−24 m s−1
Velocity of the cars after the impact when they are coupled together:
m1u1 + m2u2 = (m1 + m2)v
1350 × 15 = (1350 + 1520)v
So velocity = 7.1 m s−1
I = Δp = mΔv = 1520 × 7.1 ≈ 1.1 × 104 N s
Considering the acceleration of the cars locked together:
v2 = u2 + 2as
0 = 7.12 + (2 × 5.2a)
a = −4.85 m s−2
So the net force acting = ma = (1350 + 1520) × 4.85 = 1.4 × 104 N in the opposite direction to the
first car’s motion.
Just before hitting the ground, the stone has kinetic energy equivalent to its initial potential
energy = mgh = 3 × 9.8 × 5 = 147 J. So the correct response is D.
Initially the kinetic energy of the mass is 100 J. Therefore:
100 
10
So v = 10 m s−1 and the correct response is B.
This kinetic energy is converted to potential energy at the top of the flight of the mass, therefore:
100 = mgh = 2 × 9.8 × h
h
11
12
100
 5.1 m
19.6
This is response A.
t = 1.3 s, a = 9.8 m s−2, u = 0, v = ?
Using v = u + at:
v = 9.8 × 1.3 ≈ 13 m s−1
Using s = ut + =
s
13
1 2 1
mv   2v 2  v 2
2
2
1 2
at :
2
1
 9.8  1.32  8.3 m
2
14
800 L falls every second, so in 10 s: 800 × 10 = 8000 L of water falls, this is:
weight = mg = 8000 × 9.8 = 7.8 × 104 N
W = Fs = 7.8 × 104 × 8.3 = 6.5 × 105 J
15
P
16
Gravitational potential energy + kinetic energy → kinetic energy (+ heat/sound energy) → kinetic
energy of ground + heat energy + sound energy + gravitational potential energy of rebounding
water
work done 6.5 105

 6.5 10 4 W
t
10
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Heinemann Physics Content and Contexts Units 2A and 2B
17
At point A, the roller coaster has potential and kinetic energy:
mghA +
1
1
mvA2 = 30 × 9.8m +
× m × 42 = 294m +8m = 302m J
2
2
At point B, all of this energy is kinetic, so:
1
mvB2 = 302m
2
vB2 = 604 and vB = 24.6 m s−1
At point C:
302m = mghC +
1
1
1
mvC2 = 9.8 × 25m +
mvC2 = 245m +
mvC2 = m(245 + 0.5vC2)
2
2
2
Cancelling m from both sides:
302 = 245 +0.5vC2
Therefore vC2 = 114 and vC ≈ 10.7 m s−1.
At point D:
302m = mghD +
1
1
1
mvD2 = 9.8 × 12m +
mvD2 = 117.6m +
mvD2 = m(117.6 + 0.5vD2)
2
2
2
Cancelling m from both sides:
302 = 117.6 + 0.5 vD2, therefore vD2 = 2 × 184.4 and vD ≈ 19.2 m s−1
18
19
If the roller coaster reaches C with no kinetic energy, then:
total energy at C = mgh = 9.8 × 25 × m = 245m
The total energy at A = mghA +
1
1
mvA2 = 30 × 9.8m +
× m × 42 = 294m + 8m = 302m
2
2
Therefore (302 – 245)m = 57m J has been lost between A and C.
245m
 100  81%
302m
20
Percentage efficiency =
21
m1u1 + m2u2 = (m1 + m2)v so:
(300 × 2) − (100 × 2) = 400v
600 − 200 = 400v = 400
v is 1.0 m s−1 east
22
Energy in =
23
The collision is inelastic because kinetic energy is lost.
1
1
× 300 × 22 +
× 100 × 22 = 800 J
2
2
1
Energy out = × 400 × 12 = 200 J
2
200
 100  25%
Therefore percentage efficiency =
800
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Heinemann Physics Content and Contexts Units 2A and 2B
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Heinemann Physics Content and Contexts Units 2A and 2B
24
In one second, ΔEP = mgh = 30 × 9.8 × 1.2 = 352.8 J
P
work done 352.8

 352.8 W
t
1
However, the machine is only 75% efficient, so the machine requires an input of:
352.8
 470 W
0.75
25
Initial potential energy is converted to kinetic energy:
mgh =
1
mv2
2
0.40 × 9.8 × 0.3 =
26
27
28
1
× 0.40 × v2
2
So v2 = 5.88 and v≈ 2.42 m s−1
Work = EP(initial) – EP(final) = 0.4 × 9.8 × 0.3 − 0.4 × 9.8 × 0.14 = 0.63 J
Magnitude of thrust equals magnitude of drag and the net force is zero.
a
FΔt = mΔv
F × 5 × 10−3 = 0.057 ×
29
90
3 .6
b
So F = 285 N
By Newton’s third law, the average force is 285 N.
a
Impulse, I = Δp = mΔv = 0.16 ×
b
Calculating the average force on the fielder’s head, F:
FΔt = mΔv
So F × 5 × 10−3 = 8.9
200
= 8.89 N s
3 .6
8 .9
=1.78 × 103 N
3
6  10
F 1780

Average acceleration, a =
= 494 m s−2
m
3.6
F=
c
30
a
We need to calculate the velocity of the diver as they hit the water:
v2 = u2 +2as
v2 = 0 + 2 × 9.8 × 36 = 705.6
So v ≈ 26.56 ms−1
Vertical momentum = mv = 50 × 26.6 = 1.33 × 103 kg m s−1
b
ΔEK =
c
d
Possible bruising, sprains or broken bones.
Possibly as the answer to (b) is equivalent to 600 kg mass. Multiply this by ten, say, and the
forces are dangerously large.
1
mv2 = work = Fs
2
1
1
mv2 =
× 50 × 26.562 = F × 3
2
2
17635
 5880 N
F=
3
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Heinemann Physics Content and Contexts Units 2A and 2B
ISBN 978 1 7408 1791 2
Page 55 of 85
Heinemann Physics Content and Contexts Units 2A and 2B
Chapter 5 Nuclear energy
5.1 Atoms, isotopes and radioisotopes
1
2
3
4
5
6
45
Ca has 20 protons, 25 neutrons and 45 nucleons.
a
197
Au has 79 protons, 118 neutrons and 197 nucleons.
b
235
U has 92 protons, 143 neutrons and 235 nucleons.
c
230
Th has 90 protons, 140 neutrons and 230 nucleons.
d
a
zinc
b carbon
c
silicon
d gallium
a
Cobalt-60 has 27 protons and 33 neutrons.
b
Plutonium-239 has 94 protons and 145 neutrons.
c
Carbon-14 has 6 protons and 8 neutrons.
A radioisotope is an unstable isotope. At some time, it will spontaneously eject radiation in the
form of alpha particles, beta particles or gamma rays from the nucleus. Three isotopes that are not
radioisotopes could be any three stable isotopes, e.g. carbon-12, lead-206 and bismuth-209.
Polonium-210 and uranium-238. These have atomic numbers of 84 and 92 respectively; and
every isotope beyond bismuth (Z = 83) in the Periodic table is radioactive.
a
Volume of a sphere =
4 3
r
3
4
 (1.07 × 10−15)3 = 5.13 × 10−45 m
3
mass
1.67  10 27
density =
; so the density of a proton =
 3.25 1017 kg m-3
 45
volume
5.13  10
So the volume of a proton (assuming it is a sphere) =
b
c
d
e
7
1 cm3 = 1 x 10−6 m3; so the mass of 1 cm3 = 3.25 × 1017 × 1 × 10−6 = 3.25 × 1011 kg (325
million tonnes)
It would take 325 million cars to balance 1 cm3 of nuclear material.
The density of normal matter is far lower than the density of an atomic nucleus.
Volume of nucleus of gold atom =
4 3 4
r =  × (6.2 × 10−15)3 = 1.0 × 10−42 m3
3
3
4 3 4
r =  × (1.3 × 10−10)3 = 9.20 × 10−30 m3
3
3
Volume of nucleus
1.0  10 42

 1.1  10 13
30
Volume of atom
9.20  10
Volume of gold atom =
8
Using ratios of the radius of the atom compared to the radius of the nucleus:
x
1.3  10 10

0.01 6.2  10 15
Therefore the radius of the sphere would need to be:
x
9
a
b
0.011.3 10 10
 210 m to be consistent with this scale.
6.2 10 15
There are no differences.
Krypton-89 has five more neutrons in its nucleus.
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Heinemann Physics Content and Contexts Units 2A and 2B
10
28
Al , produced in the process:
The radioisotope produced is 13
27
13
28
Al  01n13
Al
5.2 Alpha, beta and gamma radiation
1
2
3
4
a
the nucleus
b the nucleus
c
the nucleus
For example, an α
two protons and two neutrons, and
it is a relatively heavy, positively charged particle; a β particle is an electron, negatively charged
and lighter; γ
d no charge.
a
beta particle
b proton
c
alpha particle
d neutron
a
The emission of an α particle reduces the atomic number by 2 and the mass number by 4, so
X is lead: 214
82 Pb .
b
The emission of an α particle reduces the atomic number by 2 and the atomic number by 4,
so X is thorium: 231
90Th .
c
The emission of a β particle increases the atomic number by 1, so X is actinium:
228
89
Ac .
198
80
5
6
7
8
9
d
The emission of a β particle increases the atomic number by 1, so X is mercury: Hg .
a
X is an α particle.
b
X is a β particle.
c
X is a β particle.
d
X is an α particle.
e
X is a γ particle.
The mass number of the radioisotope is: 10 + 1 – 4 = 7; and the atomic number is: 5 – 2 = 3; so X
is lithium-7: 37 Li .
a
14
7
N  24 178 O 11p ; hence X must be a proton.
b
27
13
27
Al  01n12
Mg 11H ; hence X is a neutron.
c
14
7
N  01n146C 11p ; hence X is a neutron.
d
a
b
c
a
23
11
26
Na 24 12
Mg 11H ; hence X is an alpha particle.
On the decay side, there are seven protons, seven neutrons and one electron.
A neutron has decayed from the parent nucleus to become a proton.
1
0
n11p  10 e  v
b
c
d
Reading just to the right of the Z scale shows stable nuclides of calcium-20 at: 40Ca, 42Ca,
43
Ca, 44Ca, 46Ca, 48Ca.
One stable nuclide.
It is a beta emitter.
48
48
0
19 K  20 Ca  1 e ; the daughter nucleus is stable.
e
For
f
This radioisotope has: 217 − 87 = 130 neutrons, so from Fig. 5.9 in the student book this is
an alpha emitter.
217
4
213
213
4
209
87 Fr  2   85 At and 85 At  2   83 Bi ; bismuth-209 is a stable nuclide.
g
48
19
K:
n 29

 1.53 ; for
p 19
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48
20
Ca :
n 28

 1.40.
p 20
Heinemann Physics Content and Contexts Units 2A and 2B
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Page 57 of 85
Heinemann Physics Content and Contexts Units 2A and 2B
10
a
Absorption process:
b
Decay process:
198
79
197
79
Au  01n198
79 Au
Au 193
80 Hg  
5.3 Properties of alpha, beta and gamma radiation
1
2
3
4
5
6
7
Increasing penetrating ability: α, β, γ.
Increasing ionising ability: γ, β, α.
B; emissions with less ionising ability have more penetrating ability.
Gamma radiation is most suitable since its penetrating ability will enable it to reach the tumour.
A beta emitter would be best suited because its penetrating ability would enable it to irradiate a
small volume of tissue around the source. Alpha radiation would not penetrate the tumour at all,
and gamma radiation would pass out of the body, irradiating some healthy cells along the way.
Beta and gamma radiation. Alpha particles are soon absorbed by air, but beta and gamma
radiation can travel through air for metres.
a
Energy = 8.8 × 106 × 1.6 × 10−19 = 1.4 x 10−12 J
b
Energy = 0.42 × 106 × 1.6 × 10−19 = 6.7 x 10−14 J
c
Energy = 500 × 103 × 1.6 × 10−19 = 8.0 x 10−14 J
a
Passing through 1 cm of air, the alpha particles will lose: 100,000 × 34 = 3.4 × 106 eV.
a
b
8
9
10
Distance travelled before energy lost 
5.6
 1.6 cm
3.4
Gamma rays have far greater penetrating ability because 0.81 MeV > 700 keV, so the correct
response is D.
Alpha radiation will be most damaging due to its high ionising ability; the correct response is A.
If a form of radiation interacts with and ionises matter easily, it will lose its energy very quickly
and so penetrate only a small distance.
5.4 Half-life and activity of radioisotopes
1
2
After 1 hour, 50 mg exists, and after another hour, 25 mg exists, therefore the response is C.
a
After 15 mins: 20 g → 10 g remains
b
After 30 mins: 20 g → 10 g → 5 g remains
c
After 45 mins: 20 g → 10 g → 5 g → 2.5 g remains
d
After 1.5 hours (90 mins): 20 g → 10 g → 5 g → 2.5 g → 1.25 g → 0.625 g → 0.31 g
remains
3
a
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Heinemann Physics Content and Contexts Units 2A and 2B
4
5
6
7
8
9
10
b
After 15 mins, activity is ≈ 235 Bq.
c
The half-life is 20 min.
d
After 60 mins, activity is approximately 50 Bq.
6000 Bq → 3000 Bq → 1500 Bq → 750 Bq → 375 Bq
If the decay occurs over 1 hour, then the half-life must be 15 minutes.
1 chance in 2
Enough must be produced to allow for four half-life decays in the transport time. Therefore if 12
μg is required, the hospital needs to order: 12 × 2 × 2 × 2 × 2 = 192 μg.
a
From 100% → 50% → 25% → 12.5% → 6.25% → 3.125% → 1.56% → 0.78% → 0.39%
→ 0.19% → 0.097%
So 10 half-lives have elapsed.
b
This needs to be stored for 10 x 24,000; so about ~240 000 years to be considered safe to
handle.
a
The sample with greater activity will be that with the smaller half life: uranium-235.
b
It has a much shorter half-life than uranium-238 and so has decayed much more rapidly
since the formation of the Earth.
A time period of 5 days is 120 hours, which equates to 8 half-lives of sodium-24. Therefore after
5 days:
10 million Bq → 5 million Bq → 2.5 million Bq → 1.25 million Bq → 6.25 × 105 Bq → 3.12 ×
105 Bq → 1.56 × 105 → 7.8 × 104 Bq →3.9× 104 Bq
a
Over time, the radioisotopes transmute by a series of decays to form lead-206, which is
stable. The percentage of lead in the sample will increase over time.
b
Po-214 has such a short half-life (160 μs) that when Bi-214 nuclei decay to Po-214, they
almost instantaneously transmute to Pb-210.
t
11
 1  t0.5
Using the equation: A  A0  
2
24
 1  15
A  30     9.9 g remaining
2
t
12
 1  t0.5
Using the equation: A  A0  
2
t
 1 8
52  137    52= 137 x
2
t
1
log 0.379  log  
8 2
8 log 0.379
So t 
 11.2 hours
log 0.5
t
14
13
 1  t0.5
 1  5.3
A  A0   100   16% will remain after 14 years.
2
2
14
The decay begins immediately, but half of the material will still be present after 28 years.
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Heinemann Physics Content and Contexts Units 2A and 2B
5.5 Splitting the atom: nuclear fission
2
3
4
a
n = (235 + 1) − (148 + 85) = 3
b
n = (235 + 1) − (142 + 90) = 4
c
n = (235 + 1) − (127 + 104) = 5
x = (130 + 106 + 4 × 1 − 1) = 239; y = (94 − 54) = 40
B
D
5
a
6
b
About 20 times more energy is released in the fission reaction.
E = mc2
1.33 × 106 × 1.6 × 10−19 = m(3.0 × 108)2
1
m
7
8
d
a
b
c
2.110 13
 2.36 10 30 kg
9 1016
3.1 10 -28
Percentage mass decrease=
× 100 = 0.079%
3.92  10 25
For a 5 kg sample:
mass loss = 0.00079 × 5 = 3.95 x 10−3 kg
Therefore energy released is:
E = mc2 = 3.95 × 10−3 × (3 × 108)2 = 3.56 × 1014 J
Looking at the left-hand side of the equation:
mass before the fission = 1.67495 × 10−27 + 3.90305 × 10−25 = 3.91980 x 10−25 kg
mass after the fission = 2.45698 × 10−25 + 1.41045 × 10−25 + (3 × 1.67495 × 10−27) =
3.91768 × 10−25 kg
So the decrease in mass is (3.91980 × 10−25 kg 3.91768 × 10−25 kg) = 2.12 × 10−28 kg
E = mc2
2.12 × 10−28 × (3.0 × 108)2 = 1.91 × 10−11 J
0.60 W = 0.6 Js−1
Number of fission reactions occurring each second =
10
3.1 10 11
 196 MeV
1.6  10 19
Studying the equation, X must be an alpha particle.
a
Looking at the left-hand side of the equation:
mass before the fission = 1.67495 × 10−27 + 3.90305 × 10−25 = 3.91980 × 10−25 kg
mass after the fission = 2.389992 × 10−25 +1.47653 × 10−25 + (3 × 1.67495 × 10−27) =
3.91670 × 10−25 kg
So the decrease in mass is (3.91670 × 10−25 kg − 3.91980 × 10−25 kg) = 3.1 × 10−28 kg
b
E = mc2
3.10 × 10−28 × (3.0 × 108)2 = 2.8 × 10−11 J
c
9
E = mc2 = 3.48 × 10 − 28 × (3.0 × 108)2 = 3.1 × 10−11 J =
a
b
0. 6
= 3.14 × 1010
1.91 10 11
In fissile nuclei, the nuclear forces of attraction are just stronger than the electrostatic forces
of repulsion.
The additional energy causes the nucleus to break apart.
5.6 Nuclear fission weapons
1
2
3
4
B
It doesn’t have a high enough concentration of the fissile isotope, uranium-235.
D
a
Energy released = 200 × 106 × 1.6 × 10−19 = 3.2 ×10−11 J for each fission.
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Heinemann Physics Content and Contexts Units 2A and 2B
5
6
7
8
b
Total energy = 3.2 × 10−11 × 1024 = 3.2 × 1013 J
c
TNT equivalent of energy =
3.2 1013
= 8000 tonnes
4 109
First, a neutron causes fission to occur in a uranium-235 nucleus, thus releasing two or three more
neutrons. These then go on and induce fission in more uranium-235 nuclei, each resulting in the
release of two or three neutrons and so on. The chain reaction grows very rapidly and energy is
released in each fission reaction.
As a result of its shape, a very high proportion of neutrons are able to escape from the material,
and so the chain reaction dies out.
a
It is carried as two or more subcritical masses.
b
The subcritical masses are forced together to form a combined supercritical mass.
E = mc2, therefore:
4 × 1015 = m x (3 × 108)2
m=
4  1015
= 0.044 kg
9  1016
5.7 Nuclear reactors
1
2
3
4
5
6
7
8
9
10
11
B
D
a
The fission process in the reactor core produces heat. This heat energy is conducted into the
coolant which is flowing through the core. The energy is used to produce steam, which
drives a turbine to generate electricity.
b
The difference is that the heat energy that makes the steam is produced by burning coal
instead of a nuclear-fission reaction.
c
They both use steam to turn a turbine to generate electricity.
The nucleus is too heavy. When a neutron collides with a lead nucleus, the neutron will keep
almost all of its energy, and so not slow down sufficiently to be captured by a fissile nucleus.
a
The chain reaction will be self-sustaining, i.e. critical, and a steady release of energy will
result.
b
The chain reaction will die out because it is subcritical. This will lead to a decrease in the
amount of energy produced.
c
The chain reaction will grow, causing an increasing amount of energy to be produced. This
may be dangerous and could result in an explosion.
a
Fast neutrons are most unlikely to be captured by the nuclei.
b
Slow neutrons are likely to be absorbed by the nuclei and cause fission.
a
It results in the uranium-238 changing into plutonium-239.
b
Plutonium is highly radioactive and has a half-life of about 24 000 years.
a
plutonium-239
b
They rely on fast, high energy neutrons to induce fission in plutonium nuclei.
c
They produce more of their own fuel, plutonium-239, when neutrons are absorbed by
uranium-238 nuclei.
d
Fast breeder reactors do not have moderators.
Since only one neutron is required to sustain the chain reaction, the remaining neutrons are able
to breed more plutonium.
Over a period of months, the fissile nuclei in the fuel rods become depleted, the number of
fissions decreases, and so fewer neutrons are flying around in the core. In order to maintain the
chain reaction, the control rods must be gradually withdrawn.
Tritium as it has greater mass and is more stable.
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Heinemann Physics Content and Contexts Units 2A and 2B
12
This is the equation for the fusion of deuterium to produce tritium. A proton is also produced in
the process.
Chapter 5 Review
1
2
Looking at the left-hand side of the equation:
mass before the fission = 1.67495 × 10−27 + 3.90305 × 10−25 = 3.91980 × 10−25 kg
mass after the fission: 2.33979 × 10−25 + 1.52451 × 10−25 + (3 × 1.67495 × 10−27) = 3.91455 ×
10−25 kg
So the decrease in mass is (3.91980 × 10−25 kg − 3.91455 × 10−25 kg) = 5.25 × 10−28 kg
E = mc2
5.25 × 10−28 × (3.0 × 108)2 = 4.73 × 10−11 J
5.25 10 28
 0.13%
3.9198 10 25
3
Percentage of initial mass lost =
4
5
6
0.13 % of 20 kg = 0.0013 × 20 = 26 g
Amount of energy released = mc2 = 0.026 × (3× 108)2 = 2.3 × 1015 J
The absorption process is:
U  01n 239
92 U
238
92
7
The decay process is:
0
U  239
93 U  1    and finally
239
92
8
9
13
14
15
16
17
0
Np 239
94 Pu  1   
From the graph, the half-life is 60 s.
After 5 half-lives, the amount remaining is:
t
10
11
12
239
93
 1  t0.5
A  A0   = 150 ×
2
26
26
0
11 Na12 Mg  1   
5
1
   4.7 g
2
B
The smaller piece has more surface area per unit volume, loses a higher proportion of fission
neutrons, resulting in the chain reaction dying out. In the larger mass, a smaller proportion of
neutrons is lost and so it is capable of spontaneously exploding.
4
2
He  49Be 126C  01n
Uranium-235 has a shorter half-life and so is decaying at a slightly faster rate than Uranium-238.
Therefore the proportion of uranium-235 will be decreasing.
Leukaemia, tumours, radiation sickness and probable death within months.
Yes, genetic problems could arise in future generations.
To extract heat from the reactor core. This energy is used to produce steam, which is used to drive
turbines and generate electricity.
1250
 100  42%
2  1500
18
Efficiency =
19
20
B; the sample with the shorter half-life is more active.
Initially, bismuth-211 had four times the activity of bismuth-215. That is, if the activity of
bismuth-211 is 4x and that of bismuth-215 is x, then after four half-lives, the activity of bismuth211 is
21
x
x
; whilst after just one half-life, the activity of bismuth-215 is . It then follows that
4
2
after 8 minutes, the bismuth-215 sample will have twice the activity of bismuth-211.
3 (from the diagram).
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Heinemann Physics Content and Contexts Units 2A and 2B
22
The energy of the fission is used as kinetic energy. Therefore:
1  10 6  1.6  10 -19 
v2 
23
24
25
26
27
28
29
30
31
32
1 2
mv
2
2  1.6  10 13
1.675  10 27
So v ≈ 1.4 × 107 m s−1
The fission neutrons are moving too fast to induce further fission in uranium-235 nuclei, and so a
moderator is used to slow them down.
The moderating nuclei must be relatively light so that the incident neutrons lose some of their
energy upon colliding.
Their fuel, plutonium-239, is highly fissile when struck by fast moving neutrons.
In the core of nuclear reactors, when uranium-238 nuclei absorb neutrons and decay to form
plutonium-239.
Electrons that are emitted from the nucleus of an unstable atom.
7 protons, 9 neutrons and 16 nucleons
16
7
N 168O  10 e  
No, its half-life is too short.
α, β, β, α, α, α
The nuclear equation for the fusion of deuterium atoms to tritium is:
2
1
H 13H 11p
The mass of the reactants is:
4 × 1.67 × 10−27 = 6.68 × 10−27 kg
Therefore in 100 g there is an energy release of:
33
0.1
 4  10 6  6.0  10 25 MeV
 27
6.68  10
1
1
2
0
1 H 1 H 1 H  1 e
H 11H 23He
2
1
3
2
34
He  23He24 He 11p11p
The mass of three alpha-particles = 6 × 1.67 × 10−27 = 1.0 × 10−26 kg
energy released by 100 g =
0.1
× 12.98 × 106 × 1.6 × 10−19 = 2.08 × 1013 J
 26
1 10
Because 1 day = 24 × 60 × 60 = 86 400 seconds:
work done 2.08 1013
power 

 240 MW
time
86400
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Heinemann Physics Content and Contexts Units 2A and 2B
Unit 2B
Controlling our environment
T
x
E1
Temperature gradient =
E2
35  Tin
0.15
So the temperature inside the house is 22ºC. Note that we have assumed here that it is warmer
outside the house than inside.
The thermal resistances of the two cubes are:
86.7 
RPb 
x
0.02
x
0.02

 0.1166 respectively.

 0.1416 and R Ag 
2
kA 429  0.02 2
kA 353  0.02
So the total thermal resistance of the two cubes is: 0.1416 + 0.1166 = 0.2582
T 100  0
I

 390 W
R
0
.
2582
T
The thermal current is:
E3
If we let T be the temperature at the boundary between the two metals, and realise that the thermal
current through each of the metals must be the same (i.e. 390 W), then for the lead cube we have:
T 0
390 
0.1416
So T = 54.8ºC, assuming that the lead cube was exposed to the 0ºC temperature wall.
Using Wien’s displacement law, we have:
2.898  10 3
T
2.898  10 3 2.898  10 3
T

 4830 K
 max
600  10 9
 max 
E4
Using Stefan–Boltzmann’s law:
I  eσAT 4
e
E5
I
12

 0.78
4
8
σAT
5.67  10  4    0.03 2  (120  273.15) 4
The heat of vaporisation of water is 2.25 ×106 J kg–1, so the energy required to boil 0.8 L
(0.8 kg) of water is:
2.25 × 106 × 0.8 = 1.8 × 106 J.
Since the water boils away in ten minutes, this allows us to determine the rate of energy input to
the water as:
1.8 10 6
 3000 W
(10  60)
This is the thermal current that flows through the copper (i.e. 3000 W). We can then determine
the temperature on the element side of the copper (whose thermal conductivity is 401 W m–1 K–1):
T kA(Te  Tw )

x
x
Ix
3000  0.003

 1.27
Te – 100 =
kA 401  (  0.75 2 )
I = kA
E6
Te = 101.3°C
Vbreath = 4.0 – 1.2 = 2.8 L = 2.8  10–3 m3
Vsnorkel = Vbreath = 2.8  10–3 m3
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Heinemann Physics Content and Contexts Units 2A and 2B
lsnorkel =
Vsnorkel

r 2
2.8  10 3
 2.5  10 2 

  
2


2
 5.7m
The maximum length for the deepest possible breathing is 5.7 m. (Repeat using Vbreath = 0.9 
10–3 m3 for an average shallow breath and the maximum length is 1.8 m.)
E7
E8
E9
P1V1 180  12 1 7.8
= 3.25 bar


P2
1
2.4
PV
180  12
a
V2 = 1 1 =
= 2.2  103 L
1
P2
PV
180  12
b
V2 = 1 1 =
= 1.1  103 L
2
P2
PV
2.10  3.20
V1 = 2 2 =
= 0.039 L
172
P1
P2 =
Calculate the volume of air remaining in the tank before it expands:
Vtank = V0 – V1 = 8.0 – 0.039 = 7.961 L
Calculate the pressure of the air remaining in the tank after it expands:
P2 =
P1V1 172  7.961
=
= 171 bars
8.0
P2
E10 Heat flows from a body of higher temperature to a body of lower temperature. When we sit in the
Sun, sunlight is warmer than our bodies and heat moves from the sunlight to our cooler bodies.
When we sit in the shade, our bodies are (generally) warmer than the shaded air and heat flows
from our bodies into the cooler air.
E11 Vwater = Al = 1.6  (1  10–3) = 1.6  10–3 m3
mwater = waterVwater = 1000  1.6  10–3 = 1.6 kg
Therefore heat lost:
Q = mLv = 1.6  2.25  106 = 3.6  106 J
E12 A wetsuit needs to be a snug fit so that the water that has been heated by the body does not move
around. The movement of the warmed water away from the skin, by either the direct effects of
motion or by convection, results in the removal of heat from the surface of the body.
E13 a
b
I = kA
T
15
= 0.58  A 
x
0.8  10 3
I
= 1.1  104 J s–1 m–2
A
T
15
I = kA
= 0.024  A 
x
2.0  10 3
I
= 1.8  102 J s–1 m–2
A
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Heinemann Physics Content and Contexts Units 2A and 2B
Electricity at home
E1
E2
E3
E4
E5
E6
E7
E8
E9
C
The two wires carrying electricity from the power lines to our houses are the active and neutral
wires. Power points also have an earth wire.
To make the circuit safe, by ensuring no current flows into the circuit.
Ensure neutral is 0 V.
The red wire should be connected to the brown as these are both active. The black should be
connected to the blue as these are both neutral wires. The earth wires are correct. Peter has
interchanged the active and neutral wires. The toaster will work, but it is not safe as it remains
live when off.
This is dangerous because in the situation of an active wire making contact with the outside of the
appliance, the casing of the appliance could become live.
Double insulation adds another barrier layer between the active wire and a person, making the
appliance safer to use
Hopefully the electrical contact between the person and the wire will not be good through a
finger, and if the skin is dry it has a level of resistance. In comparison, touching a live wire with a
pair of pliers provides a good contact and very low resistance, resulting in a severe shock.
Using Ohm’s law: V = IR:
current flowing 
V
240

 2.4 mA
R 100  10 3
E10 In using an appliance outdoors, be wary of any pools of water and ensure you are insulated by
wearing footwear.
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Heinemann Physics Content and Contexts Units 2A and 2B
Chapter 6 Heating and cooling
6.1 Heat: a historical perspective
1
2
3
4
5
6
7
8
9
10
a
b
Heat is being drawn away from the hand and is flowing into the cold material.
The particles within the hand have more energy. The energy is being transferred to the cold
surface, evening up the temperature of each. The energy transfer will cease when the
temperature of both is the same.
Heat was present as caloric energy. Thompson could not confirm the existence of caloric energy.
Heat was still produced without any breaking of material to release caloric energy.
Heat was being produced as long as work was being done. Thompson suggested that a transfer of
energy was required.
The other person’s hand will feel cold. It is at a lower temperature and heat energy is transferred
from the warmer hand.
The energy is mainly converted to heat. Some will be converted to sound energy.
Mechanical work is being done on each face, increasing the internal energy of the ice and
converting it into water.
a
Chemical potential energy → heat and light energy → heat energy transferred to pot →
heat energy transferred to water
b
Chemical potential energy → electrical energy → heat energy transferred to water
Energy has been transformed at a faster rate than the body is able to release to its surroundings.
Power is the rate at which work is done, so if a light bulb is rated 60 W, then 60 J of energy is
transformed to heat and light each second.
Power =
work done (E ) 350 10 3

 583 W
t
10  60
If half of the energy from the fire reaches the billy then the fire must develop double the power to
boil the water = 1166 W ≈1.2 Kw.
6.2 Kinetic theory
1
2
3
4
5
6
7
8
9
10
A; as there are electrical forces of attraction and repulsion between particles in a substance.
B; this is the only response that is correct.
B; heating a substance will increase its internal energy.
Diagram should show movement about fixed points. It could be the same as Fig.6.7 (a) in the
student book, but should show longer arrows in each direction to display greater movement when
temperature has increased.
This energy increases the potential energy of the particles.
No. There is a distribution of speeds. The average of the distribution of kinetic energies indicates
the temperature of the substance.
True. Heat energy will be transferred to the surrounding air and as a result the temperature of the
drink will decrease.
False. Refrigerators transfer energy away from the interior, reducing the average kinetic energy
and consequently the temperature.
True. Particles from the hot water will collide with those in the cold water, transferring kinetic
energy as they do so. The overall average energy of each particle increases.
True. Energy is always transferred from the higher temperature to the lower (heating). Less
energy (cooling) cannot be transferred.
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Heinemann Physics Content and Contexts Units 2A and 2B
6.3 Heat and temperature
1
2
3
a
i
27°C = 27 + 273= 300 K
ii
−27°C = −27+273 = 246 K
iii
500°C = 500 + 273 = 773 K
b
i
0 K = 0 − 273 = −273°C
ii
500 K = 500 − 273 = 227°C
iii
1000 K = 1000 − 273 = 727°C
Because the size of one degree Kelvin and one degree Celsius are equivalent, the interval is 100
K.
By Guy-Lussac’s law, we know that:
P1 P2

T1 T2
P2
103  10 3

25  273 4  273
4
5
6
7
P2 = 95.7 kPa
a
Before the size of one graduation can be established, there must be a fixed amount to be
divided.
b
Duplicate the conditions under which the fixed point was established, e.g. for the Celsius
scale it would be the freezing and boiling points of water.
A; as per Fig.6.17 in the student book.
A clinical thermometer needs only to be accurate for a narrow band of temperatures around 37°C.
It has a thin glass bulb and only reads a small section of the total Celsius scale.
By Charles’ law we know that:
V1 V2

T1 T2
V2
18

273 25  273
8
9
V2 = 19.65 cm3
So the change in volume = 19.65 – 18 = 1.65 cm3
Negative has no meaning on the Kelvin scale. It is an absolute scale beginning at zero.
Using the combined gas law, we know that:
P1V1 P2V2

T1
T2
P 15
110 103  90
 2
100  273
250  273
10
P2 = 925.4 kPa
So the pressure increase = 925.4 – 90 = 815.4 kPa ≈ 815 kPa.
Water becomes denser when it approaches freezing. This results in water just above freezing
point sinking and leaving liquid water below a frozen surface. (In this particular case, the pressure
exerted by the ice above will also have contributed to keeping the water from freezing—effects of
pressure are detailed later in this chapter.)
6.4 Specific heat capacity
1
2
Heat energy transferred (where specific heat capacity, c for water is 4200 J kg−1K−1):
ΔQ = mcΔT = 0.1 × 4200 × (20 − 15) = 0.1 × 4200 × 5 = 2.1 kJ
Heat energy required: ΔQ = mcΔT = 0.15 × 4200 × (50 − 10) = 25.2 kJ
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Heinemann Physics Content and Contexts Units 2A and 2B
3
4
a
Water requires more energy to be heated to a certain temperature compared to aluminium
because it has a much greater specific heat capacity.
b
Al: water = 900 : 4200 or 1 : 4.67 or 10 : 47
c
Because temperature rise is inversely proportional to c, specific heat capacity for equivalent
energy input and mass, the temperature rise will be in the ratio Al : water of 4.67 : 1 or
47°C : 10°C.
The heat lost by the copper block is equal to the heat absorbed by the water, in order for the two
to reach thermal equilibrium:
ΔQC = ΔQW
mC cC ΔTC = mW cW ΔTW
2  390  (100  T )  5  4200  (T  20)
780(100  T )  21 000(T  20)
78000  780T  21000T  420000)
498000  21780T
5
6
7
8
T = 22.9°C ≈ 23°C
B is the best answer. D could apply over a small temperature range.
Energy supplied from kettle = power × time = 2000 × 5 × 60 = 600 000 J
If all of this energy is used to heat the water, then we can calculate the rise in temperature using:
ΔQ = mcΔT
So 600000 = 3 × 4200ΔT = 12600 ΔT
ΔT = 47.6°C
So the final temperature of the water will be 20 + 47.6 ≈ 68°C, so the water won’t boil. A
comparison of the energy available with the energy required to boil the water would also confirm
this response.
The wasted energy is: 200 − 180 = 20 W. If this is converted to thermal energy, the amount of
thermal energy produced in 10 minutes = 20 × 10 × 60 = 12 kJ
ΔQ = mcΔT so using the 12 kJ energy calculated above, 12 000 = m × 440 × 80 (where the
specific heat capacity of iron is 440 J kg−1K−1) so the mass of iron able to be heated by this
amount is:
12000
= 0.34 kg
35200
9
10
11
12
Mercury’s low specific heat capacity is an advantage as it will not take much energy to increase
its temperature and for it to expand. A higher specific heat capacity would lead to a slow response
time, assuming that the rate of heating is the same.
The ocean near Brisbane has a higher specific heat capacity than the deserts around Alice
Springs. This moderates the extremes of hot and cold evident between day and night in Central
Australia.
Estimating the energy required to heat water to make a cup of coffee: if we use 300 mL or 0.3 kg
of water and the water is initially at 15°C then:
ΔQ = mcΔT
So ΔQ =0.3 × 4200 × 85 = 100 kJ energy would be required
Estimating the energy required to heat water to have a bath: if we use 68 L or 68 kg of water to
have a shower and the water is initially at 15°C and we heat it to 40°C then:
ΔQ = mcΔT
So ΔQ = 68 × 4200 × 25 = 7140 kJ (Answers to these questions will vary according to
estimations used in calculations.)
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Heinemann Physics Content and Contexts Units 2A and 2B
6.5 Latent heat
1
2
3
4
In this initial section of the graph, the thermal energy is being used to changing the state of the
mercury by melting it.
−39°C from the graph
357°C from the graph
The energy required to melt mercury can be found using Q = mLf.
From the graph, the energy required to melt the 0.01 kg sample is 126 J:
Lf 
5
The energy required to vaporise mercury can be found using Q = mLf.
From the graph, the energy required to vaporise the 0.01 kg sample is:
3520 − 670 = 2850 J
Lv =
6
7
8
9
10
126
 1.26  10 4 J kg -1
0.01
2850
0.01
Lv = 2.85 × 105 J kg−1
It is the potential energy required to overcome the intermolecular forces binding the sulfur in
solid form and get to a point where the particles are free to move randomly within a fixed
volume.
Energy is needed to overcome the weak intermolecular forces in a liquid, allowing particles to
move freely with no fixed volume.
As the temperature has not changed, the kinetic energy and therefore the speed remains the same.
There is a difference in the total energy of each molecule. The potential energy has increased.
Q = mLv
Q = 0.1 × 22.5 × 105 = 225 kJ
Q = energy required to heat water to 100°C + energy required to evaporate the water
Q = mcΔT + mLv = 0.05 × 4200 × 80 + 0.05 × 22.5 × 105 = 1.29 × 105 J
6.6 Evaporation: heat energy in context
1
2
3
4
5
6
7
8
B; as seen in Fig 6.28 in the student book.
a
Water molecules with higher kinetic energy evaporate, reducing the average kinetic energy
of the water remaining and hence reducing the temperature.
b
The rate of evaporation will increase with increased temperature, and decreased pressure
and humidity. Speed of the vehicle allows good air circulation and the good surface area of
the bag to the air will also assist.
a
Perspiration allows evaporative cooling. Evaporating water molecules with high kinetic
energy remove large amounts of energy, reducing the average kinetic energy of particles
within the skin and cooling the blood flowing through.
b
High humidity slows evaporation, reducing the rate at which energy will be transferred
away from the body.
Evaporation of the moisture transfers energy away from the body.
Molecules with high kinetic energy evaporate and reduce the average kinetic energy of the
remaining liquid, lowering the temperature. A breeze will increase the rate of evaporation.
Refrigerant fluid in the pipes evaporates, removing energy from inside the refrigerator.
To aid natural convection inside the refrigerator. The cooler air is denser and will fall to the
bottom of the refrigerator, cooling the whole interior.
Metal fins increase the surface area exposed to the outside air, increasing the transfer of heat
away from the refrigerator.
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Heinemann Physics Content and Contexts Units 2A and 2B
9
10
No, because heat from inside the refrigerator is being returned to the room via the pipes and fins
on the rear of the unit.
80% of the energy used by the runner is converted to heat:
0.8 × 11000 × 103 = 8.80 × 106 J
To calculate the mass of water this is used to evaporate:
8.80 × 106 = mLv = 22.5 × 105 m
m
8.80 10 6
 3.9 kg  3.9 litres
22.5 105
6.7 Conduction and convection
1
2
a
b
c
a
fibreglass, paper, polystyrene
copper (silver too expensive)
fibreglass, wood
The rate at which thermal energy is lost is:
Q kAT

t
L
Therefore for the copper pipe:
Q 380  A  80

 10.1 MW per m 2  10.1 MW m -2
3
t
3  10
b
c
3
a
Surrounding air will act as an insulating layer if it is still, so doesn’t carry heat away by
convection.
Put pipes inside the walls rather than outside the house and put insulating material around
the pipes to further reduce heat loss.
Area of window pane = l × w = 2 × 1.4 = 2.8 m2
heat energy lost in one hour =
kAT
1 2.8  15
 60  60 
 3600  21.6 MJ
L
0.007
b
4
5
6
7
8
9
10
Rate of heat loss from the window will be reduced by: double glazing, use of pelmets and
thick curtains, which trap a layer of air around or within the window.
The ‘burn’ occurs due to the rapid rate of heat transfer via conduction, due to the large
temperature difference.
Plastic and rubber have a low thermal conductivity and are insulators. Metals, being conductors,
have a high thermal conductivity and therefore transfer heat energy away from your hand faster,
cooling your hand when you touch them.
A fan to assist convection (forced convection) gives more even heat energy distribution than
conventional natural convection.
Summer: want heat upwards and cool air down. Winter: want cool up and hot air down.
No, because there are no free molecules to establish convection in solids.
a
The ‘effective’ temperature drops by approximately 1°C for every 1 km h−1 of wind speed.
So in a 30 km h−1 wind, the effective temperature drops by 30°C and the temperature will
feel like: 15−30 = −15°C.
b
Conventional thermometers shielded against convective losses.
c
To allow for an effective temperature measurement, cover thermometer bulb in wet cloth.
Cooling of cloth by convective transfer will also cool bulb and be recorded.
The air columns are hot air rising; this is why they are called ‘thermals’.
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Heinemann Physics Content and Contexts Units 2A and 2B
6.8 Radiation
1
2
3
4
5
6
7
8
9
10
By the Stefan-Boltzmann equation, we can calculate the rate of transfer of radiant heat from the
person’s body as:
Q  eA(T1O) 4  (T1S ) 4  0.70  (5.67 10) 8 1.5  (34  273) 4  (15  273) 4  119 W Q
Light, reflective colours have low radiant heat energy losses.
The person will radiate heat energy faster, assisting in transferring heat energy away from the
body and reducing overheating due to living in a hot climate.
a
In contrast to the situation above, having fair skin and light hair means people will have a
lower emissivity. Therefore in a cooler climate this would be a useful characteristic to
reduce radiant heat energy losses.
b
Increase radiant gains when Sun is visible—it is important to maximise these gains for
wellbeing. As exposure time is limited this becomes the most important evolutionary
consideration of heat energy loss.
Light, shiny surfaces radiate slowest, which corresponds to the gloss white beaker. This will
retain its heat the longest. The matt black beaker will lose its heat the fastest.
The newer style wetsuits will be more efficient, since lighter colours both absorb and emit radiant
heat energy slower.
The black car will heat up and cool down fastest.
In summer, foil will reduce radiant energy gains. In winter it will not reduce conductive losses.
Radiant energy does not require heating of the environment; hence it is potentially more efficient
to heat the surfaces of a room. However, convective losses need to be considered if it is to be
more efficient.
To appear on infra-red photographs, the leaves must have high emissivity as there is high radiant
heat transfer.
Chapter 6 Review
1
2
3
4
5
6
7
A; chemical energy of the match is converted to thermal energy.
D; the liquid expands more quickly.
Yes. Temperature is a measure of the average kinetic energy of an object.
Two fixed points are required, then a scale may be graduated. The scale is only absolute if
repeatable and based on one point being an absolute zero.
C; the average of the two temperatures.
ΔQ =mcΔT = 100 × 4200 × (35 − 20) = 6.3 MJ
The power is the rate at which work is done. Therefore:
energy to heat bath water = power × time taken
6.3 × 106 = 5.6 × 103t
t
8
9
6.3 10 6
 1125 seconds = 18.75 min ≈19 min.
5.6 103
For first situation:
Q = mcΔT
4000 = cP × 1 × 2 = 2cP
cP = 2000 Jkg−1K−1
Now substituting this value for the specific heat capacity for paraffin to calculate the energy in
the second situation:
Q = 5 × 2000 × 1 = 10000 = 10 kJ of energy required; which is response D.
Greater specific heat capacity of water than air means more heat is needed from the body to heat
water directly in contact with the body.
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10
11
12
13
14
15
The mystery metal is heated, and then this heat is transferred to the copper calorimeter and its
water. Therefore:
mcΔT(metal) = mcΔT(copper) + mcΔT(water)
0.1 × c × (75 − 25) = 0.07 × 390 × (25 − 20) + 0.2 × 4200 × (25− 20)
5c =136.5 + 4200 = 4336.5
c = 867 ≈ 870 J kg−1K−1
Therefore, according to Table 6.2 in the student book, the metal is probably aluminium, with a
specific heat capacity of 900 J kg−1K−1.
C; glaciers can still remain solid throughout the summer months.
Energy supplies latent heat of fusion in order to change the state of the ice, but no additional
energy to heat the water.
Q = mcΔT + mLf = 0.1 × 2100 × 4 + 0.1 × 3.34 × 105 = 34.2 kJ
Doesn’t matter—temperature won’t increase until all the water has boiled.
Heat energy from the steam is used to melt the ice and heat the water to a temperature, T. This
means:
mS Lv + mS c (100 − T) = mI Lf + mI c T
225000  420(100  T )  167000  2100T
225000  42000  420T  167000  2100T
16
17
18
19
20
100000 = 2520T
T = 39.7°C
Energy absorbed from boiling water:
QW = mcΔT = 0.1 × 10−3 × 4200 × 63 = 26.46 J
Energy absorbed by steam:
QS =mLv + mcΔT = 0.1 × 10−3 × 22.5 × 105 + 0.1 × 10−3 × 4200 x 63 ≈ 251 J
Therefore, steam provides 225 J more energy (or 9.5 times more).
Q = mL
5.0 × 105 = 22.5 × 105 × m
m = 0.22 kg or 220 mL
Energy must be removed to cool the water to zero degrees and then freeze it, and then further cool
it to −10°C.
mcΔT (cool water) + mLf + mcΔT (cool ice)
0.50 × 4200 × 20 + 0.50 × 3.34 × 105 + 0.50 × 2100 × 10 = 2.2 × 105 J
B; an increase in pressure will raise the boiling point of water.
Energy emitted by the kettle in one minute = 60 × 2.5 × 103 = 1.5 × 105 J
If this energy is used to vaporise the water, then:
1.5 × 105 = 2.25 x 105m
m
21
22
1.5 10 5
 0.067 kg or 67 g
2.25  105
Therefore, 33 g of water will remain and the kettle won’t boil dry.
The top section where the freezing unit is located is cooled directly as refrigerant fluid in the
pipes evaporates, removing heat energy. The cooler air is denser and falls to the bottom of the
refrigerator, cooling the middle section of the interior.
Using Boyle’s law:
V2 P2

V1 P1
P
V2  1  V1
P2
105
V2 
 10.5  6.9 cm 3
160
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Heinemann Physics Content and Contexts Units 2A and 2B
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
Tank needs to allow hot water to be drawn off at the top and cold water to enter at the bottom.
Convective currents move water to the top as it heats. Heating tank should lie below storage tank
to assist this circulation.
Snow has low thermal conductivity and acts as an effective insulator.
Conductive heat losses increase in water. Air has a lower thermal conductivity.
a
Both are the same temperature.
b
Can has a higher conductivity, but its size may mean that it has considerably more energy
to transfer than the paper label.
D; the smallest gradient suggests that it both cools and therefore heats the slowest.
C; as paraffin has a lower heat capacity than water, its temperature increase will be greater for the
equivalent amount of energy absorption
a
Cold air is denser so largely remains inside the chest.
b
Conduction and convection currents will carry energy into the top section of the freezer,
warming the air and the food. Therefore it is important for health and safety reasons that the
food is stored only to a certain level in the freezer to ensure it does not begin to defrost.
C; these are the correct responses, because metal is a better conductor of heat than wood, and
conduction and convection are impossible without a fluid to flow, such as in a vacuum.
C; a shiny black surface is a good emitter of radiation, but a shiny white surface is not.
A; from the graph, each response is true.
Turbine: kinetic → electrical
Hot water: electrical → heat
Battery: chemical potential → electrical
Light: electrical → light
High-grade energy: kinetic, electrical, chemical potential (and light if the wavelength is towards
the visible or higher).
Combined efficiency is 0.75 × 0.45 = 0.3375 or 33(.75)%.
If 32 MJ energy heats the water, then the mass that could be heated is:
106 = m × 4200 × 50 = 210 000
m ≈ 152 kg or 152 L
Combined efficiency of the turbine, storage battery and light is:
0.45 × 0.9 × 0.5 = 0.2025 or 20(.25)%
Power required by the turbine to keep the 30 W globe operating at its normal level:
P=
39
30
= 148 W
0.2025
By Guy-Lussac’s law:
P1 P2

T1 T2
P2 T2  10  273


 0.814
P1 T1
50  273
40
So the percentage reduction is 18.6%
By the combined gas law:
P2 T2 V1
 
P1 T1 V2
P2
400  273 450


130 100  273 60
P2 = 1759 kPa
Hence the increase in pressure is 1759 − 130 = 1629 kPa
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Chapter 7 Electricity
7.1 Electrical charge
1
2
3
4
5
6
7
8
9
10
Benjamin Franklin’s experiment was so dangerous because he was providing lightning with an
easy path to himself.
Almost none of our modern technology would exist!
Many different motors could be listed, for example a food processor, sewing machine, electric
drill, electric can opener etc. The average household may have 20–30 devices.
Both electrostatic and magnetic experiments show repulsion and attraction. Magnetism is
relatively permanent, electrostatic effects are temporary. Magnetic poles cannot be isolated,
whereas opposite charges can.
Early experimenters probably concluded there were two types of electric charge because charges
either repel or attract, there are no other alternatives.
Strip A and B are oppositely charged as they attract each other. Strips B and C are the same as
they repel. Therefore strip A and C must be oppositely charged and will attract.
An excess of the fluid could create one type of charge, and a deficit could create the other.
Charge leaks from sharp points.
Lightning seeks the easiest route to ground and will take a path through the high point (tree). You
should seek shelter inside a building or stay as low as possible and try to insulate yourself from
the ground.
Positive and negative can cancel each other, as do the effects of opposite charges; colours do not.
7.2 Electric forces and fields
1
a
By Coulomb’s law, the force, F, between two charges is:
F=
2
Therefore if one of the charges is doubled to 2q, F is doubled to 2F.
b
If both of the charges are doubled, then F increases by a factor of (2 × 2 = 4) 4F.
c
If one of the charges is negative, the force becomes negative as well, to –F.
d
If R is halved then F is increased by a factor of (2 × 2 = 4) 4F.
By Coulomb’s law:
F=
3
4
kq q
R2
kq1 q 2 9 109 11
=
 9.0 105 N or 900 kN
2
2
100
R
One coulomb of charge is an enormous amount of charge, and in practise a 1 C charge would not
exist. If it did then it would not be possible to put two charges near each other due to the size of
the repulsive force acting.
a
By Coulomb’s law:
kq1 q 2 9  10 9  5  10 6  5  10 6
F=
=
= 0.35 N
0.8 2
R2
b
5
a
The like charges will repel and therefore move to the furthest part of the dome, thereby
increasing the effective distance between them.
The force on a charge in an electric field:
F1 = q1 E = 2 × 10−6 × 5000 = 0.01 N down
and F2 = q2 E = 5 × 10−6 × 5000 = 0.025 N up.
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Heinemann Physics Content and Contexts Units 2A and 2B
b
Charge on the sphere:
q
F 110 3

 0.2 C
E
5000
(The charge is negative because it experiences an upwards force in the electric field.)
7.3 Electric current, EMF and electrical potential
1
2
3
a
Using the value of 5 mA of current as typical for use in a calculator:
Q = It = 5 × 10−3 × 10 × 60 = 3 C
b
Using 200 A as the current in the starter motor of a car:
Q = It = 200 × 5 = 1000 C
c
Using value of 400 mA as current in a light bulb:
Q = It = 400 × 10−3 × 60 × 60 = 1440 C
No, these are values of charge. The energy required is dependent upon the energy carried by the
charges in the current.
To find the current flowing up the belt to the dome:
I
4
a
Q 50  1012  1.6  10 19

 3A
t
3
This would represent a current:
I
Q 1.7  10 26  1.6 10 19

 2.7 10 7 A
t
1
b
5
6
7
No, this is not the actual electric current flowing through the hose because there are the
same numbers of protons moving, so the net current is zero.
Spark plugs operate at a much higher voltage than a car battery, in the order of 15 000 V.
D; the other terminal must be +12 V.
Energy lost, E = qV
The potential of the battery is:
V
8
9
By the definition of voltage, 1 volt = 1 joule per coulomb, each charge from a 9 V battery
possesses 9 J of energy.
E = qV
q
10
E 100

 20 V
q
5
E 2000

 167 C
V
12
B and C
7.4 Resistance, ohmic and non-ohmic conductors
1
a
b
2
a
b
c
The voltmeter must measure the potential difference across the light bulb, so it could be
placed in position M2 or M3.
The ammeter measures the current flowing through a point of the circuit. This could be
placed in position M1 or M4.
Because there is not a linear relationship between I and V, this conductor is non-ohmic.
Reading off the graph, a current of 0.5 A is drawn when a 10 V potential difference is
applied.
15 V would be required to double the current to 1.0 A.
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Heinemann Physics Content and Contexts Units 2A and 2B
By Ohm’s law, for 10 V:
d
R
V 10

 20 
I 0.5
For 20 V:
R
3
V 20

 13.3 
I 1.5
By Ohm’s law:
a
R
V 2. 5

 0.71 
I 3.5
b
4
Because the current has doubled when the voltage is doubled, a linear relationship exists
between I and V, and so the conductor is ohmic.
At 5 V:
R
V
5

 25 
I 200  10 3
and at 10 V:
R
5
V
10

 20 
I 500  10 3
Therefore both girls are right as these are the resistances for the two voltages.
Because the resistor is ohmic, it has a constant resistance. We can calculate this as:
R
V
5

 111.1 
I 45  10 3
Therefore for 8 V:
I
6
a
V
8

 72 mA
R 111.1
We know that the resistor is ohmic. By Ohm’s law: V = IR. For 6 V:
6  IR
So R 
6
I
For 10 V we can say that:
10 = IR = (I+2) ×
10  6 
6
I
12
I
12
4
I
I = 3.0 A
Therefore the resistance =
7
6
=2.0 Ω
I
b
Current drawn at 10 V = I + 2 = 5 A
a
The resistance of a conductor =
b
the resistance, to 1.6 Ω.
If the cross-sectional area is doubled, then the resistance is halved to 0.4 Ω.
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L
. So if the length of the conductor doubles, then so does
A
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Heinemann Physics Content and Contexts Units 2A and 2B
d 2
d 
The area of a circle of diameter d =    
4
2
2
8
d 2
 2d 
So if the diameter of the wire is doubled, A =  
 
1
 2 
2
So the area has increased by a factor of 4. Therefore the resistance will be
9
a
1
as great = 0.11 Ω.
4
R A PA

RC
PB
RA
2.8 10 8

0.068 1.7 10 8
b
10
RA = 0.11 Ω
V = IR so across the 0.11 Ω resistor, V = 10 × 0.11 = 1.1 V.
Because resistance, R =
V
, the I–V graph with the lower slope will equate to greater resistance.
I
If the graph were a V–I graph, this would not be the case.
7.5 Electrical energy and power
1
2
3
4
a
By the definition of voltage as energy per coulomb, 4.5 J is given to each coulomb of
charge.
b
This is electrical potential energy: 4.5 J.
c
No, much of the energy will be lost to other forms, such as heat and sound.
Each coulomb of charge receives 1.5 J of energy, so each electron receives:
1.5 × 1.6 × 10−19 J = 2.4 × 10−19 J
a
Power, P = IV = 3 × 0.2 = 0.6 W
b
Power, P = IV = 200 × 12 = 2400 W
c
Power, P = IV = 240 × 3 = 720 W
a
Rearranging the power equation:
P 60

 250 mA
V 240
P 1200
I 
5A
V
240
P 90
I 
 7.5 A
V 12
I
b
c
5
a
Rearranging the power equation:
P 100

 25 V
I
4
P 200  10 3
V 
 8.7 V
b
I
23  10 3
P 7500
V 
 417 V
c
I
18
P 500 10 6
I 
 21 kA
V 2.4 10 3
V
6
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Heinemann Physics Content and Contexts Units 2A and 2B
7
Energy used in 1 week = 5 × 7 × 24 × 60 × 60 = 3.0 MJ or 0.005 × 7 × 24 = 0.84 kW h
If the cost per kW h is 12 cents, then the cost for the electricity used to run the clock for a year is:
0.12 × 0.84 × 52 = $5.24
8
a
P 18

 1.5 A
V 12
P 18
I 
 0.075 A
V 240
I
b
9
10
AC mains voltage oscillates, continually changing direction 50 times per second. DC EMF
always pushes charges in one direction. A portable battery operated radio, phone or torch use DC
EMF, while appliances plugged into the household mains, such as TVs, computers or vacuum
cleaners all use AC voltage.
Less current is needed, which means there is less power lost in the transmission lines in the
transportation process.
7.6 Simple electric circuits
1
2
3
4
5
6
7
Kirchoff’s laws state that: in any electrical circuit the sum of all currents flowing into any point is
equal to the sum of the currents flowing out of it and the sum of all EMF values around a circuit
is equal to the sum of all potential drops around the circuit
We would assume that 2 V has been lost across the internal resistance of the battery.
By Kirchoff’s first law we can say that:
2.5 +1.0 − 4.2 + x = 0
−0.7 + x = 0
x = 0.7 A
a
The current is constant in a series circuit, so this remains 0.25 A.
b
By Kirchoff’s second law:
4.5 − 2.1 = voltage through second bulb = 2.4 V
a
They will work correctly wired in series.
b
Yes, if one light fails, then they will both go out.
In series, the total resistance in a circuit is the sum of resistances in the circuit. This means each
resistor has a resistance of 17 Ω.
By Ohm’s law:
R1 
V1 10

 25 
I 1 0.4
and for the second resistor:
R2 
V2 10

 20 
I 2 0.5
Therefore the effective resistance is:
25 + 20 = 45 Ω
The current flowing is given by Ohm’s law:
I
8
a
V 10

 0.22 A
R 45
Because the resistors are wired in series, their effective resistance is the sum of their
resistances. Therefore current flowing from the battery is:
I
V
5

 0.01 A
R 100  400
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Heinemann Physics Content and Contexts Units 2A and 2B
b
9
a
By Ohm’s law, V = IR so the voltage across the first resistor is:
V1 = IR = 0.01 × 400 = 4 V
and the voltage across the second resistor is:
V2 = 0.01 × 100 = 1 V
To sketch the I–V graph of the resistors combined in series, we need to add the voltages
used by each resistor for constant current values. In doing this, we can produce a table of
values that can be used to plot the linear relationship:
Current, I (A)
0
1
2
4
b
10
R1 =
Voltage through R1
0
5
10
20
Voltage through R2
0
2.5
5
10
Total voltage, V (V)
0
7.5
15
30
10
5
= 5 Ω and R2= = 2.5 Ω
2
2
So the effective resistance is 5 + 2.5 = 7.5 Ω.
For a constant current, such as in the case of a series circuit, voltage is proportional to resistance.
Therefore if the smaller resistor drawing 20 V has a resistance of 5 Ω, then the larger resistor has
resistance = 3 × 5 = 15 Ω.
7.7 Circuit elements in parallel
1
a
2
b
a
b
In a parallel circuit, current is split through the circuit paths. Hence current through the
other bulb = 0.55 – 0.25 A = 0.30 A.
In a simple parallel circuit, the voltage drop across both elements is the same: 3 V.
Effective resistance in series = R1 + R2 = 10 + 10 = 20 Ω
In a parallel circuit, effective resistance is given by:
1
1
1
1 1
2


  
Re R1 R2 10 10 10
Re = 5.0 Ω
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Heinemann Physics Content and Contexts Units 2A and 2B
3
a
The combined characteristics of the I–V graph with two resistors connected in parallel is
found by adding the values of current for constant voltages. In doing this we can generate a
table of data as follows and plot the graph:
Current through R1
(A)
0
1
2
b
Current through R2
(A)
0
2
4
Total current, I (A)
0
3
6
Voltage, V (V)
0
5
10
R1 = 5 Ω and R2 = 2.5 Ω, so:
1
1
1 1 1


 
Re R1 R2 5 2.5
Re = 1.7 Ω
4
1
1
1
2



34 R1 R2 R
5
R = 34 × 2 = 68 Ω
When connected in parallel, the potential difference across each resistor will be 10 V. So the
current flowing through each will be 0.4 A and 0.5 A. Therefore the total current in the circuit is
0.4 + 0.5 A = 0.9 A. By Ohm’s law, the effective resistance is:
R
6
a
b
c
V 10

 11.1 
I 0.9
1
1
1
1
1
1
2
3



 


Re R1 R2 20 10 20 20 20
20
Re 
 6.7 
3
By Ohm’s law: V=IR = 3 x 6.7 = 20 V across the pair of resistors
The 3 A of current is split through the two paths in the circuit so that it is inversely
proportional to resistance, so the current flowing through the 20 Ω resistor is 1 A, and 2 A
flows through the 10 Ω resistor.
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Heinemann Physics Content and Contexts Units 2A and 2B
7
8
Power = IV, so in the 20 Ω resistor, power = 1 × 20 = 20 W and in the 10 Ω resistor, power = 2 ×
20 = 40 W.
a
The effective resistance in series is:
Re = R1 + R2 + R3 = 900 + 1500 + 2000 = 4400 Ω = 4.4 kΩ
b
The effective resistance in parallel is:
1
1
1
1
1
1
1






Re R1 R2 R3 900 1500 2000
Re = 440 Ω = 0.44 kΩ
9
a
Power =
5
b
10
V2
R
10 2
 20 
R
If the voltage doubles, then power will increase by a factor of: 2 + 2 = 4. Therefore power
would increase to 20 W.
V2
Power =
R
2
P2  V2   240 
  
  1.19
P1  V1   220 
2
Therefore, the hairdryer uses about 20% more power, so Betty is correct.
7.8 Cells, batteries and other sources of EMF
1
2
3
4
5
6
For the dry cell to work, the terminals must have different electronegativities.
The leads need to be connected from + to + and – to –. If the connection is the wrong way round,
the large current could burn the leads or cause the battery to explode.
A short circuit is a path of zero resistance. If batteries are connected head to tail to create a short
circuit, they will go flat quickly and could heat up to dangerous levels.
Total EMF = 9 + (4 × 1.5) = 15 V
a
EMF = 4 × 1.5 = 6 V
b
The torch operates with half of the current through the cells and so the batteries last twice
as long.
When the large current flows through the battery as the starter motor commences operation, there
is a voltage drop across the internal resistance of the battery, which lowers the useful voltage of
the battery and dims the headlights.
7
V  EMF  IRi  12  (100  0.02)  12  2  10 V
8
V  EMF  IRi  1.3  0.2r
9
internal resistance, r = 1
V = EMF Ω
To calculate the maximum current possible, set the voltage to zero:
10
0  6  0.5I
I = 12 A
In lowering the resistance, a greater current will flow, but because power = IV, the voltage will
decrease with an increase in current.
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Heinemann Physics Content and Contexts Units 2A and 2B
Chapter 7 Review
1
2
3
7
8
B; a voltmeter is always connected in parallel across a component.
A; an ammeter should always be placed in series with a component.
Alf is wrong because EMF is a potential energy, which exists even without current. Bert is wrong
because EMF is not the same as the output voltage of the cell, which will vary depending on
current.
They would go flat.
Because all of the batteries are connected in series, total voltage = 1.5 + 1.5 + 9 = 12 V
Taking into account the internal resistances of the batteries:
output voltage = 12 × I(0.3 + 0.3 + 5) × 0.1 = 12 − 0.56 = 11.44 V ≈ 11.4 V
The 9 V battery will go flat before the others.
Charges escape at the same rate.
9
Number of electrons being moved per second =
4
5
6
10
11
12
13
14
15
16
17
18
19
20
2  10 6
 1.2  1013
19
1.9  10
These are being carried off the dome.
Power = IV = 400000 × 2 × 10−6 = 0.8 W
Both the Van de Graaff and the jack-in-the-box have potential energy due to ‘compression’.
Although the charges have considerable energy, there is not enough charge to sustain a current.
400 kV = 400000 JC−1 = 400000 × 1.6 × 10−19 J per electron = 6.4 × 10−14 J
Charges separate on the paper due to the charged rod. Opposite charges on the near side
experience a stronger force than similar charges on the far side, so the attractive forces dominate.
Problems involved with the use of solar cells: cost of the cells, cloud cover limiting electricity
produced, seasonal changes to angle of sun and no electricity generation at night. Possibilities
from use of solar cells: a cleaner, more sustainable source of electricity to reduce greenhouse gas
production.
The cell operating in bright sunshine has an output current of 1.2 A for an output voltage of 0.5
V:
Power = IV = 1.2 × 0.5 = 0.6 W
If resistance is lowered to increase the output current, then voltage is decreased so there is no gain
in power.
If voltage is 0.6 V, then current becomes 0.4 A:
power = IV = 0.4 × 0.6 = 0.24 W
By Coulomb’s law, the force between the two balls is proportional to the charge on each. If ball C
is uncharged, half of the charge of A will be transferred to C, hence the magnitude of the
attractive force now between A and B is halved to 0.02 N.
q
, so if it is brought close to ball B, charge will transfer such that C has
2
q
q
q
a charge of  . This means that the charge on A is  and the charge on B is  , therefore
4
2
2
Ball C has a charge of 
the attractive force between A and B is one quarter the original force:
0.02
 0.005 N
4
21
a
b
Josh has placed the voltmeter and ammeter in incorrect positions. The lamp doesn’t light
because the voltmeter has very high resistance and uses most of the energy in the circuit.
Because the ammeter should be placed in series in the circuit to measure current at a point,
it has a very low resistance. Therefore if Josh places this in parallel in the circuit, he will
probably burn out the ammeter because the current will flow through this path of low
resistance in preference to through the lamp.
Copyright © Pearson Australia 2009
(a division of Pearson Australia Group Pty Ltd)
Heinemann Physics Content and Contexts Units 2A and 2B
ISBN 978 1 7408 1791 2
Page 83 of 85
Heinemann Physics Content and Contexts Units 2A and 2B
c
22
23
24
25
The voltmeter should be placed in position M2 to measure potential difference, and the
ammeter should be placed at M1 to measure current.
a
Current through X will be equivalent to the current through A2, which is 1.3 A.
b
Current through Y is also equivalent to this: 1.3 A.
c
Current through Z = total current through other branches of the parallel circuit:
I = 4.5 − 1.3 = 3.2 A
a
Because V2 measures the voltage across X, this is 4.9 V.
b
The total voltage available for each branch of the parallel circuit is 6.0 V, so if 4.9 V are
used across X, then 6 − 4.9 = 1.1 V is the potential difference across Y.
c
Because Z is the only circuit component in this branch of the parallel circuit, it uses the full
6.0 V.
Reading from the graph, I = 2 A
By Ohm’s law:
R
V
I
At 50 V:
R
50
 50 
1
At 100 V:
R
100
 50 
2
At 150 V:
R
150
 55 
2.7
At 200 V:
R
26
27
28
200
 67 
3
Power = IV = 2.7 × 150 = 405 W
The resistance would be greater at 200 V than 100 V because the element is hotter.
Power (input) = IV = 6 × 0.25 = 1.5 W
Power (output) =
Efficiency =
29
4
 0 .8 W
5
0.8
 100  53%
1.5
Power (input) = IV = 8 × 0.3 = 2.4 W
Efficiency = 60 %
Power (output) = 0.60 × 2.4 = 1.44 W
Therefore:
4
 1.44
t
30
31
t = 2.8 s
Energy, E = VQ
10 = 6Q
Q = 1.7 C
Maximum current likely to be used can be calculated using:
Power = IV
10000 = 240I
I = 42 A
Copyright © Pearson Australia 2009
(a division of Pearson Australia Group Pty Ltd)
Heinemann Physics Content and Contexts Units 2A and 2B
ISBN 978 1 7408 1791 2
Page 84 of 85
Heinemann Physics Content and Contexts Units 2A and 2B
32
33
34
35
36
37
38
39
40
Voltage drop for the minimum load:
P = IV
120 = 240I
I = 0.5 A
Therefore V = IR = 0.5 × 0.5 = 0.25 V
For maximum load:
V (drop due to resistance in cables) = IR = 42 × 0.5 = 21 V.
So voltage at house switchboard = 240 − 21 = 219 V
These bulbs are equivalent in power to four 60 W, 60 V globes.
P = IV
Total power of new circuit = 60 + 60 + 60 + 60 = 240 W = 240I
So the current is 1 A, the same as in the old circuit.
Mary is correct, the power bill will be unchanged.
A key disadvantage is that if one globe blows, they will all go out if they are connected in series.
In parallel:
1
1
1 1 1 11  7 18


  

Re R1 R2 7 11
77
77
Re = 4.3 Ω
The greatest resistance is achieved by wiring the three resistors in series:
R = 4.7 × 103 × 3 =14.1 kΩ
The least resistance is achieved by connecting the resistors in parallel:
1
1
1
1
3




 1.57 k
Re R1 R2 R3 4.7  10 3
V 40

 4.0 A
a
Current in the 10 Ω resistor =
R 10
b
Because the total current is 9 V, the current through the second resistor is:
I = 9 − 4 = 5.0 A
We can calculate the resistance of the second resistor using Ohm’s law:
V = IR
40 = 5R
R = 8Ω
Copyright © Pearson Australia 2009
(a division of Pearson Australia Group Pty Ltd)
Heinemann Physics Content and Contexts Units 2A and 2B
ISBN 978 1 7408 1791 2
Page 85 of 85