practice questions answers
TRANSCRIPT
NAS Chemistry Teachers’ Guide © 2005 Nelson Thornes Ltd.
Section 6 –Solutions to Practice Questions
SN A
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 1Mechanics and Radioactivity
Solutions to Practice QuestionsChapter 11 Length = (74.4 + 0.4) mm = 74.8 mm
Width = (32.7 + 0.4) mm = 33.1 mm
Height = (28.9 + 0.4) mm = 29.3 mm
2 Close micrometer and check for any zero error
Use it to measure combined thickness of all 56 pages (not the covers)
Page thickness = measurement/56 [≈ 5 mm/56 ≈ 0.09 mm]
3 E.g. the metre rule may have shrunk or its end may be 3 mm short
4 Percentage uncertainty = 0.05 mm × 100/(1.23 mm) = 4.1 %
5 Uncertainty = ± 4.7 kΩ × 2/100 = ± 94 Ω ≈ ± 100 Ω = ± 0.1 kΩPossible values range from 4.6 kΩ to 4.8 kΩ
Chapter 21 See experiment description on page 4
2 Mass = density × volume
(a) Volume of room = 4 m × 3 m × 2 m = 24 m3
Mass of air = 1.3 kg m–3 × 24 m3 = 31.2 kg = 31 kg
(b) Volume of Earth = 4πr3/3 = 4π × (6.35 × 106 m)3/3 = 8.04 × 1020 m3
Mass of Earth = 5500 kg m–3 × 8.04 × 1020 m3 = 4.42 × 1024 kg
(c) Volume of rod = πr2 × l = π × (0.2 cm)2 × 24 cm = 3.02 cm3
Mass of rod = 8.0 g cm–3 × 3.02 cm3 = 24 g
3 Volume = mass/density = 170 g/(2.7 g cm–3) = 63 cm3
Length3 = 63 cm3
Length of each side = 3.98 cm = 4.0 cm
4 Volume of cube = 5 cm × 5 cm × 5 cm = 125 cm3
Mass = density × volume = 2.5 g cm–3 × 125 cm3 = 313 g
5 Estimates: diameter = 10 cm; thickness = 1 mm; mass = 20 g
Volume of DVD = πr2 × t π × (5 cm)2 × 0.1 cm = 8 cm3
Density = mass/volume = 20 g/(8 cm3) = 2.5 g cm–3 = 2500 kg m–3
Similar to that of glass
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 1Mechanics and Radioactivity
Solutions to Practice Questions6 Volume of mercury = mass/density = 10.1 g/(13.6 g cm–3) = 0.743 cm3
Volume of cylinder = πr2 × l = 0.743 cm3
r = √(0.743 cm3/(π × 10.5 cm)) = 0.150 cm
Internal diameter = 2r = 2 × 0.150 cm = 0.300 cm
Chapter 31 See Table 3.1 on page 7
2 Metre – the distance an electromagnetic wave travels in a vacuum in a time of 1/(299 792 458) s
Using this definition has the advantages that the metre can be reproduced anywhere in the worldand it does not vary with temperature like the original standard bar
A disadvantage is the difficulty imagining the distance travelled by such a fast wave in such a shorttime compared with observing the actual length of the original standard bar
3 A caesium atomic clock makes 9 192 631 770 oscillations every second
So in 1 day
Number of oscillations = 9 192 631 770 s–1 × 24 hour × 3600 s hour–1 = 794 243 384 928 000
4 Examples: 75 kg 32 mm 5.4 m s–1
75 kg = 75 × kg
5 13 Mm/(13 µm) = 13 × 106 m/(13 × 10–6 m) = 1 × 1012
Chapter 41 All quantities, other than base quantities, are called derived quantities
All derived quantities can be produced by suitable combinations of base quantities
2 Speed m s–1
Area m2
Volume m3
3 Density = mass/volume
Units are kg/m3 = kg m–3
4 Homogeneous means the same type
Can only equate or add together quantities which are of the same type
5 e.g. Density = 3 × mass/volume
Speed = distance/(4 × time)
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 1Mechanics and Radioactivity
Solutions to Practice QuestionsChapter 51 Total distance = 3.5 km + 5.5 km = 9.0 km
Final displacement from home = 0 m
2 A scalar is a physical quantity where the magnitude is not associated with any particular direction …a scalar has only size while a vector has both size and direction
Scalars: distance, energy, volume, speed, mass
Vectors: acceleration, weight, displacement, velocity, force
3 Average speed = total distance travelled/total time taken
(a) Average speed = 100 m/(10 s) = 10 m s–1
(b) Average speed = 42 500 m/(2.25 hour × 3600 s hour–1) = 5.25 m s–1
4 Attach a measured length of card centrally to the trolley
Position the light gate so that the card blocks its beam as the trolley passes
Use an electronic timer to record the time interval for which the beam is blocked
Average speed = length of card/recorded time
5 Average speed = total distance travelled/total time taken
Both journeys are the same length L, so total distance is 2L
For each journey, time = distance/speed
Time to work = (L/3) seconds time to home = (L/9) seconds
Total time = (L/3) + (L/9) = (3L/9) + (L/9) = (4L/9) seconds
Average speed = 2L/(4L/9) = 2 × 9/4 = 4.5 m s–1
[OR chose a journey of, say, 90 m ]
[Time to work = 90 m/(3 m s–1) = 30 s ]
[Time to home = 90 m/(9 m s–1) = 10 s ]
[Total time = 30 s + 10 s = 40 s ]
[Average speed = 180 m/(40 s) = 4.5 m s–1 ]
Chapter 61 Average acceleration = change in velocity/time taken
a = (25.1 m s–1 – 3.4 m s–1)/(6.2 s) = 21.7 m s–1/(6.2 s) = 3.5 m s–2
2 Time taken = change in velocity/acceleration
t = (330 m s–1 – 75 m s–1)/(5 m s–2) = 255 m s–1/(5 m s–2) = 51 s
3 Attach a double interrupter card of measured ‘prong’ length x centrally to the trolley
Position the light gate so that the prongs block its beam as the trolley passes
Use an electronic timer to record:
the time intervals for which the beam is blocked t1 t2
the time interval between the interruptions t3
Average velocity = length of prong/recorded time
v1 = x/t1
v2 = x/t2
Acceleration a = (v1 – v2)/t3
4 ‘rate of ’ means ‘divided by time’
so ‘rate of doing work’ means ‘work done divided by time’ (which is ‘power’)
5 Average acceleration = change in velocity/time taken
a = (30 m s–1 × 0 m s–1)/(8 s) = 4 m s–2
Chapter 71 The gradient of a displacement-time graph is the instantaneous velocity
The gradient of a velocity-time graph is the instantaneous acceleration
The area of a velocity-time graph is the change in displacement
The area of an acceleration-time graph is the change in velocity
2 Since body moves 20 m in 15 s and graph is a straight line
(a) (10 s/15 s) × 20 m = 13.3 m
(b) (8 m/20 m) × 15 s = 6.0 s
(c) Average speed = total distance/total time = 20 m/(15 s) = 1.3 m s–1
3 (a) Object is first accelerating, then constant velocity and then decelerating
(b) Stage 1
Acceleration = change in velocity/time = 12 m s–1/(5 s) = 2.4 m s–2
Distance = area under graph up to 5 s = 12
× 12 m s–1 × 5 s = 30 m
Stage 2
Acceleration = 0 m s–2 (since constant velocity)
Distance = area under graph from 5 s to 10 s = 12 m s–1 × 5 s = 60 m
Stage 3
Acceleration = –12 m s–1/(12 s) = –1 m s–2
Distance = area under graph from 10 s to 22 s = 12
× 12 m s–1 × 12 s = 72 m
(c) Total distance = 30 m + 60 m + 72 m = 162 m
Average speed = total distance/total time = 162 m/(22 s) = 7.4 m s–1
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 1Mechanics and Radioactivity
Solutions to Practice Questions
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 1Mechanics and Radioactivity
Solutions to Practice Questions4
Acc
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NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 1Mechanics and Radioactivity
Solutions to Practice Questions
Chapter 81 Record motion of ball in front of a vertical metre rule using a video camera
Replay the video a frame at a time and record displacement from scale at 0.04 s intervals
2
TimeVelo
city
Velo
city
/m s
–1
Time/s
Height of second bounce = either of the shaded areas
Acceleration of gravity = gradient of negative sloping lines
3
The two graphs are the same for the times for which the two balls are in the air
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 1Mechanics and Radioactivity
Solutions to Practice Questions
Time
Dis
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Time
Velo
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Time
Acc
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4 Centre of the track (since both positive and negative displacements)
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Time
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Time
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Chapter 91 Using v = u + at
t = (v – u)/a = (18 m s–1 – 0 m s–1)/(4.5 m s–2) = 4.0 s
2 Using v = u + at
a = (v – u)/t = (0 m s–1 – 18 m s–1)/(4.5 s) = – 4.0 m s–2
Using x = 12
(u + v)t
x = 12
(18 m s–1 + 0 m s–1) × 4.5 s = 41 m
3 Using x = ut + 12
at2
x = (3.6 m s–1 × 4.5 s) + [12
× 1.4 m s–2 × (4.5 s)2] = 16.2 m + 14.2 m = 30.4 m
4 Using x = ut + 1–2 at2
u = (x – 12
at2)/t = 60 m – [12
× 35 m s–2 × (1.6 s)2]/(1.6 s) = (60 m – 44.8 m)/(1.6 s) = 15.2 m/(1.6 s) = 9.5 m s–1
Using v = u + at
v = 9.5 m s–1 + (35 m s–2 × 1.6 s) = 9.5 m s–1 + 56.0 m s–1 = 65.5 m s–1 = 66 m s–1
5
Unit 1Mechanics and Radioactivity
Solutions to Practice Questions
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 1Mechanics and Radioactivity
Solutions to Practice Questions5 (a) Using v2 = u2 + 2ax
a = [v2 – u2]/2x = [(9.0 × 105 m s–1)2 – (1.0 × 105 m s–1)2]/(2 × 0.20 m) = 2.0 × 1012 m s–2
(b) Using x = 12
(u + v)t
t = 2x/(u + v) = 2 × 0.20 m/(10.0 × 105 m s–1) = 4.0 × 10–7 s
Chapter 101 See pages 22 and 23
2 Using x = ut + 1–2 at2
from rest, x = 1–2 at2 = 12
× 9.81 m s–2 × (1.9 s)2 = 17.7 m
3 Using x = 12
at2 (from rest)
t = √(2x/a) = √[2 × 30 m/(9.81 m s–2)] = √(6.12 s2) = 2.5 s
Using v2 = 2ax (from rest)
v2 = 2 × 9.81 m s–2 × 30 m = 588.6 m2 s–2
v = √(588.6 m2 s–2) = 24.3 m s–1
4 Using v2 = u2 + 2ax and taking upwards as positive
u2 = v2 – 2ax = 02 – (2 × –9.81 m s–2 × 200 m) = 3924 m2 s–2
u = √(3924 m2 s–2) = 62.6 m s–1
Using x = 12
at2 (from rest)
Time to rise = time to fall = t = √(2x/a) = √[2 × 200 m/(9.81 m s–2)] = √(40.8 s2) = 6.4 s
Total distance travelled = 200 m + 200 m = 400 m
Final displacement = 0 m
5 (a) Using v = u + at
Speed = at = 150 m s–2 × 6 s = 900 m s–1
Distance = 12
at2 = 12
× 150 m s–2 × (6 s)2 = 2700 m
(b) Rocket then decelerates at 9.81 m s–2
Time to slow down = 900 m s–1/(9.81 m s–2) = 91.7 s
Total time to reach top = 6 s + 91.7 s = 97.7 s
Further distance covered = 450 m s–1 × 91.7 s = 41 284 m
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 1Mechanics and Radioactivity
Solutions to Practice Questions(c)
Chapter 111 A body thrown horizontally from a cliff top takes the same time to reach the bottom as a body
dropped vertically. Provided air resistance is small, the horizontal velocity of a projectile is constantwhile its vertical velocity increases at 9.8 m s–2.
2 Using x = 12
at2 (from rest)
Time to fall t = √(2x/a) = √[2 × 0.85 m/(9.81 m s–2)] = √(0.173 s2) = 0.42 s
Time to fall t = √(2x/a) = √[2 × 2 m/(9.81 m s–2)] = √(0.408 s2) = 0.64 s
3 as Figure 11.3 on page 25 with h = 0.85 m and x = 6.4 m
Using x = 12
at2 (from rest) for the vertical motion
Time to fall t = √(2x/a) = √[2 × 0.85 m/(9.81 m s–2)] = √(0.173 s2) = 0.42 s
Since horizontal speed is constant
Speed = x/t = 6.4 m/(0.42 s) = 15.4 m s–1 = 15 m s–1
4 Using x = 12
at2 (from rest) for the vertical motion
Time to fall t = √(2x/a) = √[2 × 2 m/(9.81 m s–2)] = √(0.408 s2) = 0.64 s
Since horizontal speed is constant
x = speed × t = 400 m s–1 × 0.64 s = 255 m
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NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 1Mechanics and Radioactivity
Solutions to Practice Questions5 Using x =
12
at2 (from rest) for vertical motion of dart (falls 0.4 m vertically from rest)
t = √(2x/a) = √[2 × 0.4 m/(9.81 m s–2)] = 0.29 s
Since horizontal velocity is constant
Velocity = x/t = 3 m/(0.29 s) = 10.5 m s–1
Chapter 121 A force can cause a body to accelerate; either to speed up, to slow down or to change direction
2 The single force that could replace all other forces acting on a body and have the same effect
Maximum resultant force when forces act in same direction = 8 N + 12 N = 20 N
Minimum resultant force when forces act in opposite directions = 12 N – 8 N = 4 N
3 Both bodies have zero acceleration
So resultant force on each body must be zero
4 A body will remain at rest or continue to move with a constant velocity as long as the forces on itare balanced, i.e. the resultant force is zero
5 The inertia of a body is its reluctance to change velocity
Apply the same force to both stationary cans
The empty can will move the most as it has least inertia
Chapter 131 See page 28
2 Set up apparatus as described on page 28
Measure the mass of the trolley
Use a forcemeter to apply a constant force to the trolley and measure the resulting acceleration
Repeat for a range of known masses added to the trolley
Plot a graph of acceleration against 1/mass
A straight line through the origin shows that acceleration is directly proportional to 1/mass, soacceleration is inversely proportional to mass
3 Using F = ma
a = F/m = 24 000 000 N/(2 000 000 kg) = 12 m s–2
4 Using v = u + at
a = (v – u)/t = (40 m s–1 – 0 m s–1)/(10 s) = 4 m s–2
Using F = ma
F = 1200 kg × 4 m s–2 = 4800 N
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 1Mechanics and Radioactivity
Solutions to Practice Questions5 (a) Using F = ma
a = F/m = 150 N/(30 kg) = 5 m s–2
(b) Resultant force F = 150 N – 30 N = 120 N
a = 120 N/(30 kg) = 4 m s–2
Chapter 141 None
2 Both skaters have equal and opposite forces acting on them so move away from each other withaccelerations that depend on their masses; the heavier skater having the smaller acceleration
3 Whenever one body exerts a force on a second body, the second body always exerts a force on thefirst body; hence forces occur only in pairs
4 While body A exerts a force on body B, body B exerts an equal and opposite force on body A
5 A Newton III pair of forces cannot cancel each other as they act on different bodies
Chapter 151 Your weight arises from the gravitational attraction of the Earth pulling on you
The Newton III force that pairs with your weight is the gravitational attraction of you pulling on the Earth
2 Gravitational forces are always attractive while electromagnetic forces can be either attractive orrepulsive
3 Gravitational force of attraction between two masses
Electrostatic force of repulsion between two electrons
Magnetic force of attraction between two opposite poles
4 Gravitational, electromagnetic, strong nuclear and weak nuclear
5 Contact forces arise from electrostatic forces acting over very short distances
Chapter 161 Diagram showing a body with no forces acting on it
2 Similar to Figures 16.2 and 16.3 on page 34
Planet pulls the body down with a gravitational force
Body pulls the planet up with an equal gravitational force
3 While a body A exerts a force on a body B, body B exerts a force on body A. The forces are equal,opposite and of the same type; they have the same line of action and act for the same time
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 1Mechanics and Radioactivity
Solutions to Practice Questions
•••• Earth pushes chair up
•• I push chair down
• Earth pulls me down
•• Chair pushes me up
••• Chair pullsEarth up
•••• Chair pushes Earth down
• I pull Earth up
••• Earth pulls chair down
4 Joe pushes Fred right with a contact force of 40 N
5 See Figures 16.4, 16.5 and 16.6 on page 35
Chapter 171
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 1Mechanics and Radioactivity
Solutions to Practice Questions2 Tension in cables
pulls crane down ••
Tension in cablespulls container up ••
Crane pushes Earth down •••
Container pullsEarth up
Crane pullsEarth up
• ••••
Earth pullscontainer ••••down
Earth pushescrane up •••
Earth pullscrane down •
3 Similarities: equal magnitude, same type, same line of action, act for the same time (any 2)
differences: opposite directions, act on different bodies
4 (a) Acting on same body
(b) Acting in same direction
(c) Different lines of action
(d) Different types of force
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 1Mechanics and Radioactivity
Solutions to Practice Questions5
Both forces act on the same body rather than on different bodies
Forces are of different types (gravitational down and contact up) rather than the same type
Forces produce equilibrium of book, a Newton III pair cannot oppose each other
Chapter 181 Total distance = 800 m + 1200 m + 300 m = 2300 m
As you end up being 500 m north and 1200 m east of your starting point:
Distance = √[(500 m)2 + (1200 m)2] = √(1 690 000 m2) = 1300 m
Angle east of north = tan–1 [1200 m/(500 m)] = tan–1 2.4 = 67°
2 8 m s–1 and 2 m s–1 are two perpendicular velocities
Resultant velocity = √[(8 m s–1)2 + (2 m s–1)2] = √(68 m2 s–2) = 8.2 m s–1
Angle to bank = tan–1 [8 m s–1/(2 m s–1)] = tan–1 4.0 = 76°
3 Resultant force = √[(14 N)2 + (9 N)2] = √(277 N2) = 17 N
Angle to 14 N force = tan–1 [9 N/(14 N)] = tan–1 0.64 = 33°This object is not in equilibrium as it has a resultant force of 17 N acting on it
800 m
1200 m
300 m
finaldisplacement
Book
Table pushes book upwith a contact force
Earth (and table) pullsbook down with
gravitational force
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 1Mechanics and Radioactivity
Solutions to Practice Questions4 Split the SW force of 86 N into two components directed towards the South and the West:
86 N × cos 45° = 60.81 N towards the South
86 N × sin 45° = 60.81 N towards the West
Unbalanced force towards South = 60.81 N – 35 N = 25.81 N
Resultant force = √[(25.81 N)2 + (60.8 N)2] = √(4 363 N2) = 66 N
Angle below East = tan–1 [25.81 N/(60.8 N)] = tan–1 0.42 = 23°
5 When acting in same direction, resultant force = 25 N +14 N = 39 N
When acting in opposite directions, resultant force = 25 N – 14 N = 11 N
When these forces are perpendicular
Resultant force = √[(25 N)2 + (14 N)2] = √(821 N2) = 29 N
Angle to 25 N force = tan–1 [14 N/(25 N)] = tan–1 0.56 = 29°
Chapter 191
2 Vertical component = 220 N × cos 40° = 170 N
Horizontal component = 220 N × cos 50° = 140 N
3 (a) Force component = 85 N × cos 30° = 74 N
(b) Force component = 85 N × cos 55° = 49 N
(c) Force component = 85 N × cos 90° = 0 N
Vertical component
Horizontal component
4 Since pendulum is in equilibrium
Weight W = vertical component of tension = 35 N × cos 20° = 33 N
Force F = horizontal component of tension = 35 N × cos 70° = 12 N
5 Parallel component of tension = 600 N × cos 25° = 544 N
Perpendicular component of tension = 600 N × cos 65° = 254 N
Since moving with a constant velocity parallel to the tow path, resultant force is zero, so:
a force of 544 N must act backwards – produced by the water as the barge pushes it out of its path
a force of 254 N must act away from the bank – produced by an angled rudder
6 Vertical component of 60 N force = 60 N cos 30° = 52.0 N
Vertical component of 100 N force = 100 N cos 45° = 70.7 N
Horizontal component of 60 N force = 60 N cos 60° = 30.0 N
Horizontal component of 100 N force = 100 N cos 45° = 70.7 N
Total vertical force = 70.7 N – 52.0 N = 18.7 N down
Total horizontal force = 30 N + 70.7 N – 50 N = 50.7 N to the right
Resultant force = √[(50.7 N)2 + (18.7 N)2] = √(2920 N2) = 54 N
Angle = tan–1 [18.7 N/(50.7 N)] = tan–1 0.37 = 20° below the right-hand horizontal
For equilibrium, a fourth force of 54 N must act at 20° above the left-hand horizontal
Chapter 201 Upthrust is an upward force that acts on all immersed objects
Upthrust arises from the greater pressure acting on the bottom than on the top of an immersedobject
When in a river, part of the boulder’s weight is already opposed by the upthrust
Force required = weight – upthrust
2 Use apparatus as in Figure 20.2 on page 42
Time how long it takes for ball bearing to pass through each equal length section
Times will decrease but then become constant
Constant times show that the ball bearing has reached its terminal speed
3 Using v2 = u2 + 2ax
From rest v2 = 2ax = 2 × 9.81 m s–2 × 1000 m = 19 620 m2 s–2
v = √(19 620 m2 s–2) = 140 m s–1
Other forces acting are upthrust and drag
As the speed of a raindrop increases so do the drag forces acting on it
Resultant force on the raindrop = weight – (upthrust + drag)
Resultant force on the raindrop decreases as its speed increases, becoming zero before drop’s speedreaches anywhere near 10 m s–1
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 1Mechanics and Radioactivity
Solutions to Practice Questions
4 Air has to travel faster over the curved upper wing surface than the flat lower surface
Air pressure is least where the air travels fastest
The pressure difference produces the upward force known as aerodynamic lift
The upside-down wing on a racing car produces a downward force that improves the grip betweenthe tyres and the track
5 See Figure 20.7 on page 43
Drag = thrust
Lift = weight
Horizontally:
forward components of thrust and lift = backward component of drag
Vertically:
upwards components of lift and drag = weight + downward component of thrust
air pushesaircraft (lift)
Earth pulls aircraft(weight)
air pushes aircraft(thrust)
air pushesaircraft (drag)
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 1Mechanics and Radioactivity
Solutions to Practice Questions
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 1Mechanics and Radioactivity
Solutions to Practice QuestionsChapter 211
2 (a) The ground/starting blocks push the athlete forwards
The force arises as the athlete pushes backwards on the ground/starting blocks
(b) Constant velocity so no acceleration and horizontal forces must balance
Ground pushes athlete
Earth pulls athlete
Air pushes athlete (drag) Ground pushes athlete
Earth pulls book down
Table pushes book up
Book pushes table down
Earth pullstable down
Earth pushes table up
man pushesbox to right
friction with Earth’ssurface pushes boxto left
Earth pulls boxdown (gravitational)
Earth pushes box up(contact)
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 1Mechanics and Radioactivity
Solutions to Practice QuestionsEarth
pulls carAir
pulls car
Groundpushes car
Groundpushes car
Groundpushes car
Trailerpulls car
Car pulls trailer Earth pulls trailer
Friction fromEarth’s surfacepushes trailer
Normal forcesfrom Earth’s surface
pushes trailer
3
4
(a) Forces on box must balance in both horizontal and vertical directions
(b) Force from man pushing box is greater than frictional force from floor on box
5
Lamp is stationary so resultant force on it must be zero
Chapter 221
Moment about centre of wheel = (18 N × 0.18 m) + (18 N × 0.18 m) = 6.5 N m
2 The moment of a force is the product of force and its perpendicular distance from the point aboutwhich the force is acting
A torque is the resultant moment of two or more turning forces
3 (a) Moment = F × perpendicular distance = 20 N × 0.40 m = 8.0 N m
(b) Perpendicular distance from P = 0.60 m × sin 65° = 0.54 m
Moment = F × perpendicular distance = 35 N × 0.54 m = 19 N m
18 N
18 N
Earth pulls lamp
Tension pulls lamp Tension pulls lamp
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 1Mechanics and Radioactivity
Solutions to Practice Questions
4 If a body is in equilibrium, the sum of the moments about any point must be zero
Sum of the moments about:
left-hand support = (35 kN × 24 m) – (10.5 kN × 80 m) = 840 kN m – 840 kN m = 0
right-hand support = (35 kN × 56 m) – (24.5 kN × 80 m) = 1960 kN m – 1960 kN m = 0
centre = (35 kN × 16 m) + (10.5 kN × 40 m) – (24.5 kN × 40 m) = 560 kN m + 420 kN m – 980 kN m = 0
5 For beam to balance, sum of moments about its pivot is zero:
16 N × 25 cm = (a) × 10 cm
(a) = 400 N cm/(10 cm) = 40 N
3 N × (b) = 5 N × 18 cm
(b) = 90 N cm/(3 N) = 30 cm
35 N × 8 cm = 14 N × (c)
(c) = 280 N cm/(14 N) = 20 cm
(d) × 18 cm = 45 N × 16 cm
(d) = 720 N cm/(18 cm) = 40 N
Chapter 231 The point at which all the weight of the body appears to act
See experiment at top of page 48
2 Condition 1: the sum of the forces in any direction is zero
Condition 2: the sum of the moments about any point is zero
Usually condition 1 is applied to two perpendicular directions to produce two equations andcondition 2 used to produce the third equation
3 Foot of ladder is 3 m from base of wall √[(5 m)2 – (4 m)2]
Sum of vertical forces is zero:
R2 – 150 N = 0 … equation (1)
Sum of horizontal forces is zero:
R1 – F = 0 … equation (2)
Sum of moments about point of contact of ladder with floor is zero:
R1 × 4 m = 150 N × 1.5 m
R1 × 4 m = 225 N m … equation (3)
From equation (1), R2 = 150 N
From equation (3), R1 = 225 N m/(4 m) = 56 N
From equation (2), F = R1 = 56 N
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 1Mechanics and Radioactivity
Solutions to Practice Questions
4m
R1
1.5m
150N
F
R2
4 Rule 1: when the three forces are drawn as head-to-tail vectors, they form a closed triangle
Rule 2: all three forces must pass through the same point
Note that the intersection of lines of action of F and W determine the common point through whichthe third force must act
5 W (N) 2.0 3.0 4.0 5.0 6.0 7.0 8.0
x (cm) 52 35 26 21 17 15 13
1/x (m–1) 1.92 2.86 3.85 4.76 5.88 6.67 7.69
Weight of stand = gradient of graph/(0.08 m) = 1.04 N m/(0.08 m) = 13 N
Chapter 241 See second experiment on page 50
Precautions:
balance rule accurately before adding masses
balance screw on its head, using the slot in its head to position it accurately on the rule’s scale
similarly use any slot in 10 g mass to assist in its accurate positioning
use a range of large distances from the pivot
0 1 6(1/x)/m–1
W/N
2 3 4 5 7 8
8
7
6
5
4
3
2
1
0
gradient=Wx=1.04N m
force fromtop hinge
W
F
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 1Mechanics and Radioactivity
Solutions to Practice Questions
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 1Mechanics and Radioactivity
Solutions to Practice Questions2
For vertical equilibrium, vertical component of tension = weight of sphere = 25 N
Vertical component of tension = T × cos 30°T = 25 N/(cos 30°) = 29 N
For horizontal equilibrium, horizontal component of tension = horizontal force F
Horizontal component of tension = T × cos 60°F = 29 N × cos 60° = 14 N
3 Line of action of painter’s weight is 2.4 m from foot of ladder 3 m × 4 m/(5 m)
Sum of vertical forces is zero:
R2 – 150 N – 750 N = 0
R2 – 900 N = 0 … equation (1)
Sum of horizontal forces is zero:
R1 – F = 0 … equation (2)
Sum of moments about point of contact of ladder with floor is zero:
R1 × 4 m = (150 N × 1.5 m) + (750 N × 2.4 m)
R1 × 4 m = 225 N m + 1800 N m
R1 × 4 m = 2025 N m … equation (3)
From equation (1), R2 = 900 N
From equation (3), R1 = 2025 N m/(4 m) = 510 N
From equation (2), F = R1 = 510 N
30˚
tensionin string (T)
weight(25N)
horizontalforce (F)
4m
R1
1.5m
150N
F
R2
2.4m750N
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 1Mechanics and Radioactivity
Solutions to Practice Questions4 Taking moments about P
T2 × 2.2 m = 150 N × 1.2 m = 180 N m
T2 = 180 N m/(2.2 m) = 82 N
For vertical equilibrium
T1 + T2 = 150 N
T1 = 150 N – 82 N = 68 N
5 The figure shows the forces acting on the rod
Clockwise moment about hinge = 90 N × 1.5 m = 135 N m
For equal anticlockwise moment, Tvertical = 135 N m/(1.5 m) = 90 N
Tvertical × 1.5 m = 135 N m
Tvertical = 135 N m/(1.5 m) = 90 N
Angle wire makes with vertical = tan–1 1.5 m/(0.9 m) = tan–1 1.67 = 59°
T × cos 59° = 90 N
T = 90 N/(cos 59°) = 175 N
Horizontal component of tension = T × cos 31° = 175 N × cos 31° = 150 N
For horizontal equilibrium (only two forces have horizontal components):
Horizontal force of hinge = horizontal component of tension = 150 N
For vertical equilibrium:
weight acting down = 90 N
vertical component of tension acting up = 90 N
So vertical force of hinge = 90 N – 90 N = 0 N
0.9 m
90 N
hinge
Fv
Fh
T
1.5 m
P
0.3 mT1
2.2 m 0.5 m
1.2 m 1.0 m
150 N
T2
6 Taking moments about A
T2 × 1.2 m = (100 N × 0.5 m) + (250 N × 0.6 m) + (350 N × 1.0 m)
T2 × 1.2 m = 50 N m + 150 N m + 350 N m = 550 N m
T2 = 550 N m/(1.2 m) = 460 N
For vertical equilibrium
T1 + T2 = 100 N + 250 N + 350 N = 700 N
T1 = 700 N – 460 N = 240 N
Chapter 251 Momentum = mass × velocity
Momentum is a vector quantity
2 Momentum = mass × velocity
(a) Momentum of rugby player = 120 kg × 10 m s–1 = 1200 kg m s–1
(b) Momentum of electron = 9 × 10–31 kg × 2 × 107 m s–1 = 1.8 × 10–23 kg m s–1
(c) Momentum of toy train = 1.6 kg × 0.25 m s–1 = 0.4 kg m s–1
3 Change in momentum = final momentum – initial momentum = mv – mu
(a) Change in momentum of car = (800 kg × 30 m s–1) – (800 kg × 5 m s–1) = 24 000 kg m s–1 – 4000 kg m s–1 = 20 000 kg m s–1
(b) Change in momentum of trolley = (0.8 kg × 0.2 m s–1) – (0.8 kg × 0.8 m s–1) = 0.16 kg m s–1 – 0.64 kg m s–1 = – 0.48 kg m s–1
(c) Change in momentum of ball = (0.05 kg × –5 m s–1) – (0.05 kg × 7 m s–1) = – 0.25 kg m s–1 – 0.35 kg m s–1 = – 0.6 kg m s–1
4 The rate of change in momentum of a body is directly proportional to the resultant force acting onit and takes place in the same direction as the resultant force
Force ∝ rate of change in momentum
F ∝ (mv – mu)/t ∝ m(v – u)/t ∝ ma
F = kma where k is the constant of proportionality
In SI units, the newton is defined so that k = 1, so F = ma
0.5 m
T1
A
100 N
250 N
0.4 m0.1 m
B
350 N
T2
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 1Mechanics and Radioactivity
Solutions to Practice Questions
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 1Mechanics and Radioactivity
Solutions to Practice Questions5 Force = change in momentum/time
(a) Force acting on car = 20 000 kg m s–1/(8 s) = 2500 N
(b) Force acting on trolley = – 0.48 kg m s–1/(3 s) = – 0.16 N
(c) Force acting on ball = – 0.6 kg m s–1/(0.6 s) = –1 N
Chapter 261 Provided no external forces act, the total momentum in any direction remains constant so that the
total momentum after the collision equals the total momentum before the collision
See experiment on page 54
2 Momentum of skateboard = mu = 4 kg × 2 m s–1 = 8 kg m s–1
Combined mass after bag lands on it = 4 kg + 1 kg = 5 kg
Assuming momentum is unchanged
5 kg × v = 8 kg m s–1
v = 8 kg m s–1/(5 kg) = 1.6 m s–1
3 Momentum of bullet = mu = 0.02 kg × 300 m s–1 = 6 kg m s–1
Since block is stationary, total initial momentum (TIM) = 6 kg m s–1
Total final momentum (TFM) = TIM = 6 kg m s–1 (since momentum conserved)
Final speed = TFM/(total mass) = 6 kg m s–1/(4 kg) = 1.5 m s–1
4 TIM = (65 kg × 7 m s–1) + (45 kg × – 6 m s–1)
TIM = 455 kg m s–1 – 270 kg m s–1 = 185 kg m s–1
TFM = TIM = 185 kg m s–1
Combined speed = TFM/(total mass) = 185 kg m s–1/(110 kg) = 1.7 m s–1
In the original direction of the 65 kg skater
5 Prior to a gun being fired, its total momentum is zero (TIM = 0)
As momentum is conserved, total momentum after firing must also be zero (TFM = 0)
Pellet has momentum in the forward (positive) direction
so gun must have momentum in the backward (negative) direction, and so recoils
6 (a) Momentum of stone = mu = 0.1 kg × 6 m s–1 = 0.6 kg m s–1
(b) TFM = TIM = 0.6 kg m s–1
Speed of squirrel = TFM/(total mass) = 0.6 kg m s–1/(0.6 kg) = 1 m s–1
(c) Total momentum must remain at 0.6 kg m s–1
Stone’s new momentum = 0.1 kg × –2 m s–1 = – 0.2 kg m s–1 (as backwards)
Squirrel’s momentum = (0.6 + 0.2) kg m s–1 = 0.8 kg m s–1 (to keep TFM = 0.6 kg m s–1)
Squirrel’s final speed = 0.8 kg m s–1/(0.5 kg) = 1.6 m s–1
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 1Mechanics and Radioactivity
Solutions to Practice QuestionsChapter 271 Impulse = force × time
Unit of impulse from this equation is N s
In base units:
N s = kg m s–2 × s = kg m s–1
The same unit as momentum
2 (a) Impulse = Ft = 60 N × 0.008 s = 0.48 N s
(b) Ft = mv – mu but since u = 0, Ft = mv
v = Ft/m = 0.48 N s/(0.15 kg) = 3.2 m s–1
3 Impulse = area under force-time graph
Impulse = 12
× 6 N × (0.3 s + 0.9 s) = 3.6 N s
Impulse = Ft = mv – mu but since u = 0, Ft = mv
v = Ft/m = 3.6 N s/(0.8 kg) = 4.5 m s–1
4 (a) Change in momentum = mv – mu = m(v – u) = 900 kg (0 m s–1 – 30 m s–1) = –27 000 kg m s–1
(b) Impulse of wall on car = mv – mu = –27 000 N s
(c) Force of wall on car F = impulse/t = –27 000 N s/(0.5 s) = –54 000 N
5 (a) Change in momentum = m(v – u) = 0.3 kg (–2 m s–1 – 8 m s–1) = 0.3 kg × –10 m s–1
= –3 kg m s–1
(b) Impulse of hammer on nail = mv – mu = –3 N s
(c) Force of hammer on nail F = impulse/t = –3 N s/(0.012 s) = –250 N
(d) Connect one of the ‘start’ leads of a digital timer to the nail and the other to the metal hammer head
6 When two bodies collide, they exert equal and opposite forces on each other, F and –F
These forces act for the same length of time t
Therefore the impulses are also equal and opposite, Ft and –Ft
and the change in momentum of one body is equal and opposite to the change in momentum ofthe other
So the overall change in momentum is zero and total momentum is conserved
7 TIM = 3 kg × 4 m s–1 = 12 kg m s–1
TFM = TIM = 12 kg m s–1
Momentum of 2 kg sphere after collision = 2 kg × 4.5 m s–1 = 9 kg m s–1
Momentum of 3 kg sphere = 12 kg m s–1 – 9 kg m s–1 = 3 kg m s–1
Speed of 3 kg sphere = momentum/mass = 3 kg m s–1/(3 kg) = 1 m s–1
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 1Mechanics and Radioactivity
Solutions to Practice QuestionsChapter 281 Work = force × distance moved in the direction of the force
(a) Work = 60 N × 3 m = 180 J
(b) Work = 4000 N × 0.25 m = 1000 J
2 (a)Work done = area under graph up to 60 mm = 12
× 1.5 N × 0.06 m = 0.045 J
(b) Work done = area under graph between 40 mm and 140 mm=
12
× (1.0 N + 3.5 N) × (0.14 m – 0.04 m) = 0.225 J
3 Work done = area under graph
Area up to 3 cm ≈ 12
× 0.03 m × 13.6 N ≈ 0.20 J
Area between 3 cm and 9.5 cm ≈ 0.065 m × 14.2 N ≈ 0.90 J
Total work done ≈ 1.1 J
4 Power is the rate of doing work
(a) Average speed = distance/time = 36 m/(60 s) = 0.60 m s–1
(b) Average upward velocity = upward displacement/time = 21 m/(60 s) = 0.35 m s–1
(c) Total work done against his weight = weight × height = 800 N × 21 m = 16 800 J
(d) Average power = work/time = 16 800 J/(60 s) = 280 W
5 144 km h–1 = 144 000 m h–1/(60 × 60 s h–1) = 40 m s–1
Power = force × velocity
Resistive force = power/velocity = 35 000 W/(40 m s–1) = 875 N
Chapter 291 Energy is a scalar quantity
Base units: J = N m = kg m s–2 m = kg m2 s–2
2 Energy can be stored as either potential energy or kinetic energy
3 types of potential energy: gravitational, electromagnetic and nuclear
elastic potential energy
3 Increase in gravitational potential energy ∆W = mg∆h
(a) ∆W = 0.5 kg × 9.81 N kg–1 × 25 m = 123 J
(b) ∆W = 60 kg × 9.81 N kg–1 × 0.30 m = 177 J
(c) ∆W = 1.2 × 10–6 kg × 9.81 N kg–1 × 0.15 m = 1.8 × 10–6 J
4 Kinetic energy = 12
mv2
Kinetic energy of car = 12
× 900 kg × (20 m s–1)2 = 180 000 J
If all kinetic energy is converted into gravitational potential energy, mgh
Height risen h = kinetic energy/(mg)
h = 180 000 J/(900 kg × 9.81 N kg–1) = 20.4 m
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 1Mechanics and Radioactivity
Solutions to Practice Questions5 (a) Kinetic energy of trolley =
12
× 0.60 kg × (0.85 m s–1)2 = 0.22 J
h = 0.22 J/(0.60 kg × 9.81 N kg–1) = 0.037 m
(b) Kinetic energy of ball = 12
× 0.11 kg × (40 m s–1)2 = 88 J
h = 88 J/(0.11 kg × 9.81 N kg–1) = 82 m
not surprisingly, the worked example gives the same height!
Chapter 301 In this situation, the decrease in the gravitational potential energy corresponds to an increase in the
internal energy of the object and its surroundings due to the frictional forces acting on it
2 The internal energy of a body is the total of the random kinetic and potential energies of all themolecules of that body
Internal energy may be increased by:
mechanical working by hammering
electrical working by the passing of an electric current
heating in a hot fire
3 The energy content of a closed or isolated system remains constant
4 (a) Potential energy lost by falling mass = mg∆h = 1 kg × 9.81 N kg–1 × 0.25 m = 2.45 J
(b) Total kinetic energy just before hitting ground = potential energy lost by falling mass
12
(m1 + m2)v2 = 2.45 J
v2 = 2 × 2.45 J/(1 kg + 4 kg) = 0.98 m2 s–2
v = √(0.98 m2 s–2) = 0.99 m s–1
5 Efficiency = useful output/input
Power of light emitted = (2/100) × 60 W = 1.2 W
The other 58.8 W increases the internal energy of the surroundings
Chapter 311 See experiment on page 54
2 Momentum and energy are conserved in both elastic and inelastic collisions
Elastic collisions also conserve kinetic energy; inelastic collisions do not
The collisions between the molecules of a gas are, on average, elastic
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 1Mechanics and Radioactivity
Solutions to Practice Questions3 TIM = mu = 4 × 104 kg × 3 m s–1 = 1.2 × 105 kg m s–1
TFM = TIM = 1.2 × 105 kg m s–1
Speed after collision = TFM/(total mass) = 1.2 × 105 kg m s–1/(6 × 104 kg) = 2 m s–1
Kinetic energy after collision = 12
× 6 × 104 kg × (2 m s–1)2 = 1.2 × 105 J
Kinetic energy before collision = 12
× 4 × 104 kg × (3 m s–1)2 = 1.8 × 105 J
This shows that the collision was inelastic, 6 × 104 J is spread around, mainly raising the internalenergy of the buffers and surroundings so that energy is still conserved
4 Gravitational potential energy → kinetic energy → gravitational potential energy → kinetic energy→ gravitational potential energy
5 (a) Momentum of bullet = mu = 0.02 kg × 300 m s–1 = 6 kg m s–1
Since block is stationary before collision, TIM = 6 kg m s–1
Since momentum is conserved, TFM = TIM = 6 kg m s–1
Combined speed = TFM/(total mass) = 6 kg m s–1/(4 kg) = 1.5 m s–1
(b) Kinetic energy of bullet = 12
mu2 = 12
× 0.02 kg × (300 m s–1)2 = 900 J
Kinetic energy of block and bullet = 12
Mv2 = 12
× 4 kg × (1.5 m s–1)2 = 4.5 J
Collision is inelastic as kinetic energy is not conserved
99.5% of bullet’s kinetic energy is converted to other forms
6 Motorway crash barriers are designed to absorb the kinetic energy of any vehicle that hits them toprevent the vehicle from bouncing back into the carriageway
Chapter 321 neutron neutral
proton positive
electron negative
2 An atom consists of a very small central nucleus where most of its mass is concentrated and aroundwhich low-mass electrons orbit
Beryllium-8 has 4 protons and 4 neutrons in its nucleus with 4 orbiting electrons
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 1Mechanics and Radioactivity
Solutions to Practice Questions3 Density = mass/volume
An atom is mostly empty space with no mass
Nearly all the mass is concentrated into a very small central volume (very large nuclear density)
The overall density is an average for the whole material, taking into account the empty space
4 (a) number of protons = 82
(b) number of nucleons = 207
(c)number of neutrons = 207 – 82 = 125
5 Isotopes are nuclides with the same number of protons but a different number of neutrons
Lightest (1st) isotope of tin is Sn-108 i.e. (107 + 1)
2nd isotope of tin is Sn-109 i.e. (107 + 2)
3rd isotope of tin is Sn-110 i.e. (107 + 3)
So 25th isotope of tin is Sn-132 i.e. (107 + 25)
Possible symbols: 10950Sn, 110
50Sn ...... 13150Sn, 132
50Sn
Chapter 331 See Figure 33.2 on page 70
The vast majority of alpha particles are deflected very little as they travel through the gold foil
while a tiny minority (about 1 in 8000) are deflected through angles greater than 90°
2 See Figure 33.3 on page 70
The positive nuclei have comparatively large distances between them so most alpha particles aredeflected very little
The closer the path of an alpha particle comes to a nucleus, the more the alpha particle is deflected
Deflections through angles greater than 90° result from almost head-on collisions
3 Diameter of an atom ≈ 3 × 10–10 m
Diameter of a nucleus ≈ 1 × 10–14 m
∴ The diameter of this atom is about 30 000 times greater than the diameter of this nucleus
Area of circle ∝ diameter2
∴ The head-on area of this atom is about 30 0002 times greater than the head-on area of this nucleus
Percentage taken up by nucleus = 100 × 1/(30 000)2 ≈ 0.000 000 1% (which is why so many alphaparticles miss!)
4 Quarks are the particles from which all sub-atomic particles are made
There are six types of quark called up, down, strange, charm, top, bottom
A proton consists of two up quarks and one down quark
A neutron consists of one up quark and two down quarks
A baryon is a particle made up of three quarks
A meson is a particle made up of a quark and an antiquark
5
Chapter 341 When ionised, an atom releases an electron and becomes a positively charged ion
Air can be ionised either by a flame or by the radiation from a radioactive source
2 See second experiment on page 72
3 21584Po → 211
82Pb + 42α22789Ac → 223
87Fr + 42α
4
Smoke particles in air reduce the number of alpha particles reaching ‘the detector’
fewer ionisations then occur within ‘the detector’
so a lower current flows when smoke is present
A reduction in the current is used to trigger the alarm
5 22890Th → 224
88Ra + 42αTo get to lead-212 from radium-224, 12 nucleons must be removed
as each alpha particle removes 4 nucleons (and beta removes none), 3 more alpha decays are required
the proton number of all isotopes of lead is 82
Chapter 351 A GM tube is more sensitive: it can detect any single ionising event that occurs inside the tube while
an ionisation chamber needs a large number of ionising events to produce a measurable current
A GM tube allows each ionising event to be directly registered on a counter
A GM tube detects radiation once it has entered the tube through its walls (it is poor at detectingalpha radiation that is mostly stopped by the walls) while the radioactive source can be placeddirectly inside an ionisation chamber
2 Place source close to GM tube window
Observe count rate as paper placed between source and GM tube – no change, no alpha
Repeat with aluminium instead of paper – reduction in count rate, beta being emitted
Repeat with lead instead of aluminium – no emissions from source detected, no gamma
Alphasource
Alpha particledetector
Air gap
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 1Mechanics and Radioactivity
Solutions to Practice Questions
deep inelastic scattering alpha particle scattering
Incident particles electrons alpha particles
Targets protons/neutrons atoms in foil
Process deflections off smaller parts within nucleons deflections off smaller parts within atoms
Results quarks discovered nucleus discovered
3 In beta-minus decay, a neutron in the nucleus splits into a proton and an electron10n → 1
1p + –10e
The proton stays in the nucleus but the electron is ejected at high speed as a beta-minus particle
4 21684Po → 212
82Pb + 42α212
82Pb → 21282Pb + γ
21282Pb → 212
83Bi + –10β
5
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 1Mechanics and Radioactivity
Solutions to Practice Questions
Pu24194
0
–1Am241
95
42
Np23793
42
Pa23391
0
–1U233
92
42
Th22990
42
Ra22588
0
–1Ac225
89
42
Fr22187
42
At21785
Bi21383
0
–1Po213
84
42
Pb20982
0
–1Bi209
83
42
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 1Mechanics and Radioactivity
Solutions to Practice QuestionsChapter 361 Background radiation is the radiation that is around us all the time
natural radioactivity is associated with isotopes that occur naturally
artificial radioactivity is associated with isotopes produced by man by neutron bombardment
cosmic rays are another important contributor to background radiation
2 Radioactive decay is a random process as it is impossible to predict when an individual atom will decay
A radioactive source contains an extremely large number of atoms
The unpredictable individual decays of such a large number together produce a statistical patternfrom which predictions can be made
3 Activity is the number of decays of a radioactive source per second
Decay constant is the probability of decay per nucleus per second
Activity = λN
Activity = 8.0 × 10–6 s–1 × 3.0 × 1011 = 2.4 × 106 s–1
As nuclei decay, there are fewer and fewer nuclei left to decay so the decay rate decreases
4 Half-life is the average time taken for half the nuclei of that isotope to decay
See experiment on page 78
5 t = ln 2/λt = ln 2/λ = 0.69/(7.84 × 10–10 s–1) = 8.80 × 108 s = 8.80 × 108 s/(60× 60 × 24 × 365.25 s y–1)
= 27.9 y
84 years = 3 × 28 years = 3t
Mass of strontium isotope will have fallen to (12
)3 = 1/8
Mass = 4.5 mg × 1/8 = 0.56 mg
12
12
12
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 2Electricity and Thermal Physics
Solutions to Practice QuestionsChapter 11 An electric current is a flow of electric charge
2 The cell does electrical work by applying a force to the charge carriers in the direction in which
they move
Both electrical and mechanical work transfer energy
3 A cell is a single source of e.m.f. whereas a battery is a group of cells connected together
4 Circuit A – see Figure 4.6 on page 9
Circuit B
Circuit C
Circuit D
The circuits will run out of fuel in order A, C, B
D need not run out of fuel at all, as the engines can be turned off and still produce the same effect!
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 2Electricity and Thermal Physics
Solutions to Practice Questions5 Direct current: the electrons travel in one direction only
Alternating current: the electrons move first one way and then the other; they keep moving
backwards and forwards
Chapter 21 +4 nC on A and – 4 nC on B
2 The duster takes electrons from the surface of the rod
The duster now has a negative charge and the rod an equal positive one
3 Current is a base quantity
Charge is a derived quantity
Charge is measured in coulombs and current in amperes
1 ampere = 1 coulomb per second
4 (a) Q = It = 3 A × 4 s = 12 C
(b) Q = 7 A × (8 × 60) s = 3360 C
(c) Q = 0.25 A × (2 × 60 × 60) s = 1800 C
5 (a) I = Q/t = 700 C/(35 s) = 20 A
(b) I = Q/t = 3600 C/(3 × 60 s) = 20 A
6 Q = It = 4.5 A × (20 × 60) s = 5400 C
Number of electrons = 5400 C/(1.6 × 10–19 C) = 3.4 × 1022
7 Charge = current × time
C = A s
Chapter 31 Draw a large copy of the circuit diagram on a piece of paper
Place components on top of their circuit symbols
Connect components using wires of the correct length to keep the circuit looking like its
circuit diagram
2 Break the circuit at the required point
Insert the ammeter; an additional lead will be needed
Make sure the red terminal of the ammeter is connected nearest to the positive terminal of the
power supply
3 Torch bulb 0.3 A, LED 20 mA, small motor 1 A, buzzer 0.1A, mains lamp 0.25 A, electric kettle 10 A
4 A series circuit is one where the components are all connected in-line, one after the other
The current passes through one component, then through the next, then the next, etc.
Current is not used up by any component (conservation of charge): what goes in comes out
so the current throughout a series circuit is the same
5 The current entering a house equals the current leaving, so company A makes no charge
The charge entering a house equals the charge leaving, so company B makes no charge
Electrical devices in a house remove energy from the supply, so company C charges its customers
Chapter 41 A parallel circuit is one where components are connected across each other
Current flowing into a parallel circuit splits at the junction so that a part of it goes through
each route
Current flowing out of each route of a parallel circuit join together at the junction
2 The sum of the currents entering a point is equal to the sum of the currents leaving that point
Conservation of charge
3
4 In series:
(a)
12 V
2.4 A 1.6 A
0.8 A 0.8 A 0.8 A
1.6 A2.4 A
0.8 A 0.8 A 0.8 A
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 2Electricity and Thermal Physics
Solutions to Practice Questions
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 2Electricity and Thermal Physics
Solutions to Practice Questions(b)
5 Note that currents in series are the same and that currents add up to zero at junctions
Circuit A … I1 = I2 = 1 A + 1 A = 2 A
Circuit B … I3 = I4 = 1.5 A
Circuit C … I5 = I6 = 1.2 A [current through parallel resistor = 0.4 A]
Circuit D … I7 = 20 mA, I8 = 20 mA – 0.1 mA = 19.9 mA, I9 = 0.1 mA
Chapter 51 Adding a resistor to a series circuit increases the total resistance
So that the current flowing through the whole circuit decreases
A circuit can be given additional resistance by:
making part of the wiring thinner
using longer wires
using a material through which electrons find it hard to move
using a material in which there are only a few electrons that can move
2
mA
3 V
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 2Electricity and Thermal Physics
Solutions to Practice Questions3
4
5 An LDR will not pass enough current to run a motor
See Figure 5.7 on page 11
As the level of illumination increases, the resistance of the LDR falls
Current flowing through LDR and relay coil increases sufficiently to close the relay switch
which allows current to flow directly through the motor from the supply
Chapter 61 3 × 1.5 V = 4.5 V
Three cells in parallel will supply energy for a longer time compared to a single cell
[Will also reduce the overall effect of any internal resistance as current splits between the three cells
– see Chapter 17]
2 The average resultant force on the electrons is zero
The power supply pushes the electrons along in their direction of travel
The circuit resistances apply equal forces in the opposite direction
live
neutral
12.4 A
12 4 A
electricfire
12 A
12 A
0.4 A
0.4 A
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 2Electricity and Thermal Physics
Solutions to Practice Questions3
4 V1 = 1.5 V
V2 = 1.5 V
V3 = 3.0 V
V4 = 12 V, V5 = 12 V
V6 = 9.0 V, V7 = 9.0 V
5 The two types of voltage difference are in opposite directions
Chapter 71 Two components that give energy to a circuit: cells (batteries), generators
Two components that take energy away from a circuit: lamps, motors
Voltage differences across components that give energy to a circuit are called e.m.f.s
Voltage differences across components that take energy away from a circuit are called
potential differences
2 (a) W = VQ = 9 V × 15 C = 135 J
(b) W = VQ = VIt = 9 V × 0.5 A × (2 × 60) s = 540 J
3 (a) Energy per second (power) = 36 000 J/(10 × 60 s) = 60 J s–1
(b) 3 A = 3 C s–1
(c) Energy per coulomb (voltage) = 60 J s–1/(3 C s–1) = 20 J C–1
4 P = VI
V = P/I = 0.75 W/(0.3 A) = 2.5 V
5 Q = It = 2.5 A × (10 × 60) s = 1500 C
e.m.f. = W/Q = 300 000 J/(1500 C) = 200 V
V
+
V
+
V
+
V
+
Chapter 81
2
3 V1 = 12 V – 3 V = 9 V
V2 = V3 = 6 V
V4 = Vlamp = 4 V
V5 = 9 V – 4 V = 5 V
4 See Figure 8.6 on page 17
Source of power in the water circuit is the Sun
Source of power in the electric circuit is the battery
Assuming no water either evaporates from the streams or is absorbed into the ground then
the amount of water flowing down mountain equals amount of water evaporating from sea
5 When at A, VX > potential of right-hand terminal of ammeter, so current through ammeter flows
from left to right
When at E, VX = potential of right-hand terminal of ammeter, so no current flows through ammeter
When at B, VX < potential of right-hand terminal of ammeter, so current through ammeter flows
from right to left
12 V 12 V1 kΩ 12 V1 kΩ 12 V1 kΩ
12 V
4V
4V
4V
1 kΩ
1 kΩ
1 kΩ
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 2Electricity and Thermal Physics
Solutions to Practice Questions
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 2Electricity and Thermal Physics
Solutions to Practice QuestionsChapter 91 The voltage at a point in a circuit is the voltage difference between zero and that point
The voltage across a component is the difference in the voltages at its two ends
2 (a) Vlamp = 1.8 V – 0.4 V = 1.4 V
(b) Vresistor = 5.7 V – 2.1 V = 3.6 V
(c) Vwire = 2.1 V – 1.8 V = 0.3 V
3
4 V1 = 1.5 V
V2 = 3.0 V, V3 = 1.5 V
V4 = 3.0 V, V5 = 1.5 V
V6 = 2.0 V, V7 = 4.0 V, V8 = 6.0 V
V9 = 3.0 V, V10 = 6.0 V
V11 = –6.0 V, V12 = –3.0 V, V13 = 3.0 V, V14 = 6.0 V
5 Around any closed loop, the sum of the e.m.f.s is equal to the sum of the p.d.s
6 For energy conservation:
total energy gained by a coulomb going round a complete circuit equals the total energy lost
and as voltage is a measure of the energy transferred per unit charge
total e.m.f. (energy providers) = total p.d. (energy takers)
Chapter 101 Either volt = J C–1 = N m C–1 = kg m s–2 m C–1 = kg m s–2 m (A s)–1 = kg m s–2 m A–1 s–1
= kg m2 s–3 A–1
or volt = W A–1 = J s–1 A–1 = N m s–1 A–1 = kg m s–2 m s–1 A–1 = kg m2 s–3 A–1
ohm = V A–1 = kg m2 s–3 A–1 A–1 = kg m2 s–3 A–2
2 See experiment on page 20
Volta
ge/V
6
5
4
3
2
1
0
Position
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 2Electricity and Thermal Physics
Solutions to Practice Questions3 Resistors are connected in parallel (combined resistance < individual resistance)
1/total = 1/R1 + 1/R2
1/R2 = 1/total – 1/R1 = 1/(120 Ω) – 1/(180 Ω) = 0.002778 Ω–1
R2 = 360 Ω
4 The seven arrangements are
(i) 18 Ω, (ii) 36 Ω, (iii) 54 Ω, (iv) 9 Ω, (v) 6 Ω, (vi) 12 Ω, (vii) 27 Ω
5 Power = I2R
I2 = P/R = 0.810 W/(25 Ω) = 0.0324 A2
I = √(0.0324 A2) = 0.18 A
V = IR = 0.18 A × 25 Ω = 4.5 V
Number of cells = 4.5 V/(1.5 V) = 3
Chapter 111 (i) Behaviour of circuit will be unaffected by presence of voltmeter as resistance of parallel
(voltmeter) combination will be as close as possible to resistance of component
(ii) Behaviour of circuit will be unaffected by presence of ammeter as it does not change the circuit
resistance
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 2Electricity and Thermal Physics
Solutions to Practice Questions2 In Figure 11.1 on page 22:
Voltmeter records the correct voltage for the resistor
Ammeter records the current through both the resistor and the voltmeter
Calculated resistance value = correct V/larger I = lower R
In Figure 11.2 on page 22:
voltmeter records the voltage across both the resistor and the voltmeter
ammeter records the correct current for the resistor
Calculated resistance value = larger V/correct I = higher R
3 First connect the probes of the digital ohmmeter together and note the reading
Connect the probes to the ends of the component and note the new reading
Resistance of the component is the difference in these two readings
4
5 The resistance of a light-dependent resistor decreases as the light intensity falling on it increases
Chapter 121 Number of tennis balls along each 1 m side = 1000 mm/(66 mm) = 15
Total number in 1 m3 box = 15 × 15 × 15 = 3375
2 Volume of wire V = Al = 3 × 10–6 m2 × 30 × 10–2 m = 9 × 10–7 m3
Number of free electrons N = nV = 6.4 × 1028 m–3 × 9 × 10–7 m3 = 5.76 × 1022
Amount of free charge Q = Ne = 5.76 × 1022 × 1.6 × 10–19 C = 9216 C = 9200 C
3 A = 3 C s–1 (I = Q/t)
t = Q/I = 9216 C/(3 C s–1) = 3072 s = 3100 s
Average speed = distance/time = 30 × 10–2 m/(3072 s) = 9.8 × 10–5 m s–1 = 0.1 mm s–1
3 When the switch is closed, electrons throughout the circuit start moving almost straight away
The electromagnetic wave that starts the electrons moving travels around the circuit very quickly (at
the speed of light … 3 × 108 m s–1)
Electrons move in, and give energy to, the lamp almost instantaneously
Delay time t = length of cable/(3 × 108 m s–1)
tungsten filament lamp
NTC thermistor
Temperature
Res
ista
nce
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 2Electricity and Thermal Physics
Solutions to Practice Questions4 I = nAvq
v = I/nAq = 25 A/(7 × 1028 m–3 × 2.5 × 10–6 m2 × 1.6 × 10–19 C) = 8.9 × 10–4 m s–1 = 0.89 mm s–1
5 See experiment on page 25
Chapter 131 Resistivity = resistance × area/length
Units of resistivity = Ω m2/m = Ω m
In base units:
resistance = Ω = kg m2 s–3 A–2 (see solution 10.1)
resistivity = Ω m = kg m2 s–3 A–2 m = kg m3 s–3 A–2
2 Resistivity of tungsten is greater than that of copper
Cross-sectional area of the tungsten filament is smaller than that of the copper connecting wires
As R =ρl/A, filament resistance is greater than that of the same length of copper connecting wire
Since they are in series, the current through both the filament and the connecting wires is the same
Power P = I2R ∝ R, so filament dissipates more power and gets hotter
3 R =ρl/A
A = wt = 2.0 × 10–3 m × 8.5 × 10–3 m = 1.7 × 10–5 m2
R = 1.7 × 10–8 Ω m × 4.0 × 10–2 m/(1.7 × 10–5 m2) = 4 × 10–5 ΩV = IR = 25 × 10–3 A × 4 × 10–5 Ω = 1 × 10–6 V = 1 µV
4 In both components:
lattice vibrations increase with temperature
producing increased carrier obstruction and reduced drift speed v
In the tungsten filament lamp:
current I ∝ v (nAq are constant), so producing a smaller current and a larger resistance
In the NTC thermistor:
charge carrier density n increases massively with temperature
n increases much more than v decreases
current I ∝ nv (Aq are constant), so producing a larger current and a smaller resistance
5 I = V/R = 12 V/(300 Ω) = 0.04 A = 40 mA
Power = V2/R = (12 V)2/(300 Ω) = 0.48 W = 480 mW
Thermistor gains 180 mW (i.e. 480 mW generated – 300 mW dissipated)
(a) Temperature increases
(b) Resistance decreases
Both the current flowing and the power generated within the thermistor will increase
If process continues:
temperature rises more producing further reductions in resistance and increases in current and
power so that ‘thermal runaway’ occurs
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 2Electricity and Thermal Physics
Solutions to Practice Questions6 Resistivity increases with temperature: copper and iron
Resistivity decreases with temperature: carbon, glass, paraffin wax and silicon
Chapter 141 A potential divider circuit consists of a chain of resistors, all connected in series
It divides the voltage from a source in proportion to their resistances
2 Total resistance = 7 kΩ + 3 kΩ = 10 kΩCurrent I = e.m.f./(total resistance) = 9 V/(10 kΩ) = 9 × 10–4 A
VOUT = IR = 9 × 10–4 A × 3 × 103 Ω = 2.7 V
3 VOUT will decrease
Larger current in 7 kΩ resistor so greater p.d. across top resistor
(or effective resistance of bottom resistor is reduced)
4 (a) 1/R = (1/600 Ω) + (1/400 Ω) = 0.0042 Ω–1
R = 1/(0.0042 Ω–1) = 240 Ω(b) Total R = 360 Ω + 240 Ω = 600 Ω(c) I = e.m.f./total R = 12 V/(600 Ω) = 0.02 A = 20 mA
(d) V360 = IR = 0.02 A × 360 Ω = 7.2 V
(e) Vparallel = Vsupply – V360 = 12 V – 7.2 V = 4.8 V
[or V240 = IR = 0.02 A × 240 Ω = 4.8 V]
(f) I600 = V600/R = 4.8 V/(600 Ω) = 0.008 A = 8 mA
(g) I400 = Isupply – I600 = 20 mA – 8 mA = 12 mA
[or I400 = V400/R = 4.8 V/(400 Ω) = 0.012 A = 12 mA]
5 Parallel combination:
1/R = (1/12 Ω) + (1/36 Ω) = 0.111 Ω–1
R = 1/(0.111 Ω–1) = 9 Ω
Circuit:
total R = 18 Ω + 9 Ω = 27 Ω
I = e.m.f./total R = 9 V/(27 Ω) = 0.33 A
I18 = 0.33 A
V18 = IR = 0.33 A × 18 Ω = 6 V
Vparallel = Vsupply – V18 = 9 V – 6 V = 3 V
I12 = V12/R = 3 V/(12 Ω) = 0.25 A
I36 = Isupply – I12 = 0.33 A – 0.25 A = 0.08 A
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 2Electricity and Thermal Physics
Solutions to Practice QuestionsChapter 151 A rheostat is a variable resistance in series with the component (see Figure 15.1 on page 30)
The rheostat controls the current through the lamp
A potentiometer is a variable potential divider (see Figure 15.2 on page 30)
The potentiometer controls the voltage across the lamp
An advantage of using a potentiometer is that the voltage across the lamp can be reduced to zero
A disadvantage of using a potentiometer is that circuit current still flows even when there is zero
current in the lamp
2 Digital voltmeter shows that there is 1.5 V across bottom resistor of potential divider
Torch bulb is connected in parallel with the bottom resistor
Their combined resistance is less so p.d. across them is less than 1.5 V
3 Using a digital voltmeter (infinite resistance):
Total resistance = 2000 Ω + 3000 Ω + 1000 Ω = 6000 ΩI = e.m.f./total R = 12 V/(6000 Ω) = 0.002 A
VBC = IRBC = 0.002 A × 3000 Ω = 6 V
Using analogue voltmeter (resistance = 6000 Ω):
Combined resistance RBC = 1/[(1/3000 Ω) + (1/6000 Ω)] = 2000 ΩTotal resistance = 2000 Ω + 2000 Ω + 1000 Ω = 5000 ΩI = e.m.f./total R = 12 V/(5000 Ω) = 0.0024 A
VBC = IRBC = 0.0024 A × 2000 Ω = 4.8 V
Using both voltmeters:
combined resistance RBC = 1/[(1/3000 Ω) + (1/6000 Ω) + (1/∞ Ω)] = 2000 Ωso both voltmeters read 4.8 V
4 The terminals must be joined together by a wire of negligible resistance [short-circuited]
R1 = Vsupply/Imax = 4 V/(0.5 A) = 8 ΩNote that the current of 0.1 A flows between the output terminals and not through R2
VR1 = Vsupply – Vout = 4 V – 3 V = 1 V
IR1 = VR1/R1 = 1 V/(8 Ω) = 0.125 A
IR2 = IR1 – Iout = 0.125 A – 0.100 A = 0.025 A
R2 = VR2/IR2 = Vout/IR2 = 3 V/(0.025 A) = 120 Ω
5 (a) At 1000 lux, Rldr = 150 Ω so total circuit resistance = 1050 ΩCurrent flowing = 5 V/(1050 Ω) = 0.0048 A
V900 = IR = 0.0048 A × 900 Ω = 4.29 V = 4.3 V
(b) At 90 lux, Rldr = 1500 Ω so total circuit resistance = 2400 ΩCurrent flowing = 5 V/(2400 Ω) = 0.0021 A
V900 = IR = 0.0021 A × 900 Ω = 1.875 V = 1.9 V
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 2Electricity and Thermal Physics
Solutions to Practice QuestionsChapter 161 The current through a component is directly proportional to the voltage across it
See experiment on page 32
2 The resistance of the tungsten filament of a lamp increases as it gets hotter
its current-voltage graph is curved
If the filament is maintained at a constant temperature (e.g. by keeping it immersed in water)
its current-voltage graph is a straight line through the origin
showing that the material from which the filament is made is ohmic
3 With switch S closed:
Vsupply = VR1 + Vdiode
Graph shows that at 50 mA, Vdiode = 0.77 V
So VR1 = 2.5 V – 0.77 V = 1.73 V
R1 = VR1/I = 1.73 V/(0.05 A) = 34.6 Ω = 35 ΩWith switch open:
Vsupply = VR1 + VR2 + Vdiode
Graph shows that at 10 mA, Vdiode = 0.65 V
VR1 = IR1 = 0.01 A × 34.6 Ω = 0.346 V
So VR2 = 2.5 V – (0.65 V + 0.346 V) = 1.504 V
R2 = VR2/I = 1.504 V/(0.01 A) = 150.4 Ω = 150 ΩDiode power when switch S closed = IdiodeVdiode = 0.05 A × 0.77 V = 0.0385 W = 39 mW
4 Voltage across series resistance VR = Vsupply – VLED = 6.0 V – 1.7 V = 4.3 V
R = VR/I = 4.3 V/(20 × 10–3 A) = 215 Ω = 220 Ω
5
Chapter 171 Voltage across resistance = IRexternal = 0.5 A × 12 Ω = 6 V
‘lost volts’ = E – voltage across resistance = 9 V – 6 V = 3 V
Internal resistance r = ‘lost volts’/I = 3 V/(0.5 A) = 6 Ω
Potential difference
Res
ista
nce
2 See experiment on page 35
3 E.m.f. = terminal voltage + ‘lost volts’
E = V + Ir
V = (–r)I + E
Graph of terminal voltage V against current I has gradient equal to –r and intercept equal to E
E.m.f. = intercept = 1.64 V
Internal resistance = –gradient = 1.62 Ω
4 Working resistance R = V/I = 3.5 V/(0.35 A) = 10 Ω
Total circuit resistance = E/I = 12 V/(0.35 A) = 34 Ω
Required series resistance = 34 Ω – (4 Ω + 10 Ω) = 20 Ω
‘lost volts’ = 4.5 V – 3.5 V = 1.0 V
Internal resistance = ‘lost volts’/I = 1.0 V/(0.35 A) = 2.9 Ω
5 Digital voltmeter reads 12 V when switch S is open
Current I = e.m.f./total resistance = 12 V/(2 Ω + 4 Ω) = 12 V/(6 Ω) = 2 A
Terminal voltage = IRexternal = 2 A × 4 Ω = 8 V
Power dissipated in internal resistance Pr = I2r = (2 A)2 × 2 Ω = 8 W
Power dissipated in external circuit PR = I2R= (2 A)2 × 4 Ω =16 W
6 A car battery has a very small internal resistance (10 mΩ)
so that it can supply a large current (e.g. 200 A) to the starter motor
A school e.h.t. supply has a very large internal resistance (5 MΩ)
to prevent it from supplying dangerously large currents
0.30 0.1
Current/A
1.8
1.6
1.4
1.2
1
0.8
0.6
Term
inal
vol
tage
/V
0.2 0.4 0.5 0.6
equation of line: terminal voltage=(–1.62 current)+1.64V = –r I + E
+
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 2Electricity and Thermal Physics
Solutions to Practice Questions
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 2Electricity and Thermal Physics
Solutions to Practice QuestionsChapter 181 Pressure is the force acting per unit area
Pressure = force/area
Pa = N m–2 = kg m s–2 m–2 = kg m–1 s–2
2 (a) Pressure = force/area = 150 N/[π(0.06 m)2] = 1.3 × 104 N m–2
Force on piston = pA = 1.33 × 104 N m–2 × π(0.96 m)2 = 38 000 N
(b) For volume of fluid to remain the same
(Distance moved × area)piston = (distance moved × area)plunger
Distance moved by piston = 0.64 m × π(0.06 m)2/[π(0.96 m)2] = 0.0025 m = 2.5 mm
3 See parts 2 and 3 of experiment ‘Blowing up balloons’ on page 38
Use a pressure gauge to measure the pressure of the gas and a thermometer to measure
its temperature
4 E.m.f. generated = 35 µV °C–1 × 160 °C = 5600 µV = 5.6 mV
Current I = e.m.f./total resistance = 5.6 × 10–3 V/(16 Ω + 4 Ω) = 2.8 × 10–4 A
Power produced P = VI = 5.6 × 10–3 V × 2.8 × 10–4 A = 1.57 × 10–6 W
5 Mark liquid level at 0 °C (ice/water) and 100 °C (steam) using elastic bands
divide interval between these two marks into 100 equal divisions
For this thermometer:
100 °C ≡ (18.0 – 2.0) cm = 16.0 cm
So scale is 0.16 cm °C–1
(a) 35 °C ≡ 35 °C × 0.16 cm °C–1 = 5.6 cm
Alcohol level is (5.6 + 2.0) cm = 7.6 cm above the bulb
(b) – 8 °C ≡ – 8 °C × 0.16 cm °C–1 = –1.28 cm
Alcohol level is (–1.28 + 2.0) cm = 0.72 cm above the bulb
Chapter 191 Macroscopic means large scale
Volume m3
Pressure Pa (or N m–2)
Temperature K
Amount of gas present
2 Boyle’s law: for a fixed mass of gas at constant temperature, the product of the pressure and
volume is constant
See experiment on page 40
Precaution: after each compression, wait for gas to cool before taking readings
3
4 Pressure × volume = 1.8 × 106 Pa × 0.06 m3 = 1.08 × 105 Pa m3
When tap opened:
new pressure × new volume = 1.08 × 105 Pa m3
new volume = 1.08 × 105 Pa m3/(100 × 103 Pa) = 1.08 m3
air leaving cylinder = (1.08 – 0.06) m3 = 1.02 m3 = 1.0 m3
5 As the bubble rises, there is less water pushing down on it from above
so pressure on the fixed mass of gas in the bubble decreases as the bubble rises
and its volume increases so that pV remains constant
Volume of bubble has increased by a factor of 5 (20 mm3/4 mm3)
so pressure at surface must be 1/5 of pressure at bottom of lake
Pressure at bottom of lake = 5 × pressure at surface = 5 × atmospheric pressure
Pressure at bottom is equivalent to 5 × 10 m of water = 50 m of water
Depth of lake = (50 – 10) m = 40 m
Chapter 201 Pressure law: for a fixed mass of gas at constant volume, the pressure is directly proportional to the
Kelvin temperature
See experiment on page 42
Precautions:
submerge as much of the flask as possible in the water
use a short length of tubing to connect flask to pressure gauge
allow time for the gas in the flask to reach the temperature of the water before taking readings
Pre
ssur
e
Volume
Higher temperature
Lower temperature
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 2Electricity and Thermal Physics
Solutions to Practice Questions
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 2Electricity and Thermal Physics
Solutions to Practice Questions2 As p ∝ T then p/T = constant
For the given results:
34 °C is the incorrect temperature
Correct temperature = 111 kPa/(average of other p/T values) = 111 kPa/(0.3506 kPa K–1) = 317 K
= 44 °C
3 Pressure/temperature = 100 kPa/[(273 + 15) K] = 0.347 kPa K–1
New pressure/new temperature = 0.347 kPa K–1
New temperature = new pressure/(0.347 kPa K–1) = (100 + 20) kPa/(0.347 kPa K–1) = 345.6 K
= 72.6 °C
4 V reduces to 1
4V0 when p increases to 4p0 (constant T)
1
4V0 increases to V0 when T increases to 4T (constant p)
4T = 4 × (273 + 27) K = 1200 K = 927 °C
To return to original conditions: temperature must be reduced back to 27 °C at constant volume
5 The critical temperature of a gas is the temperature above which it cannot be liquefied
An ideal gas is one that would obey the gas laws at all temperatures and pressures and would
never liquefy
p is the pressure of the gas in Pa
V is the volume of the gas in m3
n is the number of moles of gas present in mol
R is the molar gas constant in J K–1 mol–1
T is the Kelvin temperature of the gas in K
Pre
ssur
e
0
Volume
V014 V0
12 V0
34 V0
4p0
3p0
2p0
p0
0
C
A
B
Temperature / °C 1 12 29 34 58 78
Pressure / kPa 96 100 106 111 116 123
Temperature / K 274 285 302 307 331 351
p/T / kPa K–1 0.350 0.351 0.351 0.362 0.350 0.350
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 2Electricity and Thermal Physics
Solutions to Practice QuestionsChapter 211 See experiment on page 44
When viewed through the microscope, the smoke particles appear as very small bright dots which
dance around randomly, moving first one way and then immediately another
2 The smoke particles are being bombarded by the gas particles
Although very light, the gas particles are moving very fast and undergo a significant change
in momentum
At any instant, more gas particles will hit one side of the smoke particle than another
creating a resultant force that momentarily pushes the smoke particle in that direction
then the collision imbalance and the direction of the resultant force changes
so the smoke particle continually gets pushed in different directions
3 (a) Use more ball bearings in the Perspex tube
(b) Increase the speed of the motor to make the molecules move faster
(c) Add mass to the cardboard disc
Place a small polystyrene sphere, representing a smoke particle, in with the ball bearings
4 Collisions of gas particles with the container walls exert forces and create pressures on them
Increasing the temperature of a gas causes its particles to move faster; the gas particles collide with
the walls harder and more often, producing a greater pressure
Increasing the volume of a container decreases the packing density of the gas particles within; fewer
collisions per unit wall area occur per unit time and a lower pressure is produced
5 If collisions were inelastic, the average kinetic energy of the gas molecules would decrease
The molecules would slow down and stop moving, just like the ball bearings in the model
The temperature of the gas would fall and it would change into a liquid and then a solid
Chapter 221 For one collision:
Change of momentum = mv – mu = 0.2 kg × (15 – –15) m s–1 = 0.2 kg × 30 m s–1 = 6 kg m s–1
For 600 collisions:
Total change of momentum = 600 × 6 kg m s–1 = 3600 kg m s–1
Average force = total change of momentum/time taken = 3600 kg m s–1/(12 s) = 300 N
600 collisions in 12 s = an average of 1 collision every 20 ms = 5 collisions in 100 ms
Average force acting on wall = total area under the graph/(100 ms)
Forc
e
0Time/ms
20 40 60 80 100
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 2Electricity and Thermal Physics
Solutions to Practice Questions2 p = pressure of the gas
ρ = density of the gas
<c2> = mean square speed of the gas molecules
Left-hand side:
pressure = N m–2 = kg m s–2 m–2 = kg m–1 s–2
Right-hand side:
ρ <c2> = kg m–3 (m s–1)2 = kg m–3 m2 s–2 = kg m–1 s–2
E.g. (any four)
gas consists of a large number of particles in rapid, random motion
all collisions are elastic or particles assumed to be hard elastic spheres
molecular size is negligible compared with the volume occupied by the gas
intermolecular forces are negligible except during collisions
time of collision is negligible compared with time between collisions
3 Since p = ρ <c2>/3
<c2> = 3p/ρ = 3 × 101 × 103 Pa/(0.09 kg m–3) = 3.37 × 106 m2 s–2
Root mean square speed <c> = √(3.37 × 106 m2 s–2) = 1835 m s–1
See Figure 22.3 on page 47
4 (a) Left-hand side:
1
2mv2 = kg (m s–1)2 = kg m2 s–2
Right-hand side:
3kT/2 = J K–1 K = J = N m = kg m s–2 m = kg m2 s–2
(b) Molar gas constant = Avogadro constant × Boltzmann constant
(c) Since 1
2 mv2 = 3kT/2
v2 = 3kT/m so v2 ∝ T
T2/T1 = (847 + 273) K/[(7 + 273) K] = 1120 K/(280 K) = 4
∴ v2 increases by a factor of 4
R.m.s. speed increases by a factor of 2 (i.e. √4)
R.m.s. speed = 2 × 380 m s–1 = 760 m s–1
5 (a) Mean speed = (350 + 420 + 280 + 610 + 680 + 540 + 590 + 490) m s–1/8 = 3960 m s–1/8
= 495 m s–1 = 500 m s–1
(b) Mean velocity = (350 + 420 – 280 + 610 – 680 – 540 + 590 – 490) m s–1/8 = –20 m s–1/8
= –2.5 m s–1
(c) Mean square speed = (3502 + 4202 + 2802 + 6102 + 6802 + 5402 + 5902 + 4902) m2 s–2/8 =
2 091 600 m2 s–2/8 = 261 450 m2 s–2 = 260 000 m2 s–2
(d) Mean square velocity = [3502 + 4202 + (–280)2 + 6102 + (–680)2 + (–540)2 + 5902 +
(–490)2] m2 s–2/8 = 2 091 600 m2 s–2/8 = 261 450 m2 s–2 = 260 000 m2 s–2
(e) Root mean square speed = √(261 450 m2 s–2) = 511 m s–1 = 510 m s–1
(f) Root mean square velocity = √(261 450 m2 s–2) = 511 m s–1 = 510 m s–1
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 2Electricity and Thermal Physics
Solutions to Practice QuestionsChapter 231 Internal energy: the total of the random kinetic and potential energies of all the molecules of that
body
Examples: working on it [e.g. hammering or passing an electric current through it (electrical
working)] or heating it (e.g. placing body in a hot fire)
2 Solid: molecules constantly vibrate about fixed positions
Liquid: molecular vibrations sufficient to allow molecules to interchange positions
Gas: molecules move throughout their container at high speeds
Internal energy is stored as both molecular kinetic energy (due to all kinds of motion including
vibration and spin) and potential energy (in the repulsive fields between the molecules)
3 A spoonful of hot water has more energy per degree of freedom than a bucketful of cold water as it
is at a higher temperature, but it has less total internal energy as it has a much smaller mass
4 Random shuffling of energy quanta between the two bodies will favour movement from hot to cold
as a hot body has more energy per degree of freedom than a cold body
5 The internal energy of an ideal monatomic gas is only kinetic, as there are no forces between
its atoms
The internal energy of a real monatomic gas is both kinetic and potential, as there are forces
between its atoms
Chapter 241 Copper has conduction (free) electrons in its structure
The conduction electrons in the hotter part of the material gain energy and pass this on to the
colder part as they diffuse (move) into it
Quartz has very few conduction electrons in its structure but its atoms are strongly linked together
Large vibrational energy at the hot end is transferred along the atoms to atoms at the cold end
2 From flame to base of pan: direct contact of hot and cold ‘bodies’, convection within the flame and
radiation from it
Through base of pan: conduction
Into water: adjacent to the base by conduction and throughout the water by convection
3 Energy conducts from the hot water to the cooler metal of the ‘radiator’
Energy mainly conducts from the warm ‘radiator’ to the cooler air in contact with it
(Radiation from the ‘radiator’ will be limited as its temperature is fairly low)
The heated air expands and rises, carrying energy into the room by convection
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 2Electricity and Thermal Physics
Solutions to Practice Questions4 Conduction: helpful in allowing energy to reach the contents of a pan through its base; unhelpful
in allowing energy to escape through the walls of a warm house
Convection: helpful in heating a whole room from a single source of heat e.g. one ‘radiator’;
unhelpful when convection currents in the atmosphere result in a ‘bumpy’ flight
Radiation: helpful in allowing the Sun to heat the Earth; unhelpful in radiating heat from a cooking
pot even when it has a lid to prevent evaporation and convection
5 The two bodies in thermal equilibrium are at the same temperature
It is a dynamic situation as energy flows between the two bodies, although at the same rate in
both directions
In thermal equilibrium, all the temperatures are the same and there is no net flow of
internal energy
In steady state, the temperatures are constant but different and there is a steady flow of
internal energy
Chapter 251 Specific heat capacity: energy needed to raise the temperature of 1 kg of that substance by 1 K
without a change of state
Unit: J kg–1 K–1
In base units: J kg–1 K–1 = N m kg–1 K–1 = kg m s–2 m kg–1 K–1 = m2 s–2 K–1
2 See second experiment on page 52
Precautions: lagging, good thermal contact of block with heater and thermometer, measure
maximum temperature reached after heater turned off
Calculation: energy absorbed by block = mass × specific heat capacity × temperature rise
Energy supplied = potential difference × current × time = VIt
Specific heat capacity c = VIt/(mass × temperature rise)
3 Energy supplied = VIt = V2t/R = (5.0 V)2 × 60 s/(25 Ω) = 60 J
mc∆T = 60 J
∆T = 60 J/(mc) = 60 J/(0.030 kg × 380 J kg–1 K–1) = 5.3 K
4 Kinetic energy dissipated = 1
2 mv2 =
1
2 × 900 kg × (30 m s–1)2 = 405 000 J
Energy absorbed by each disc = 1
4 × 405 000 J = 101 250 J
mc∆T = 101 250 J
∆T = 101 250 J/(mc) = 101 250 J/(2.8 kg × 460 J kg–1 K–1) = 78.6 K = 79 K
Either perform second experiment on page 52, or
Place a known mass M of water at a measured temperature θ1 in a calorimeter of known mass m
Increase temperature of disc to a known temperature (e.g. 100 °C using boiling water)
Quickly dry and transfer hot disc to water in calorimeter
Measure maximum final temperature θ2 of liquid
Energy lost by disc = energy gained by water and calorimeter
Specific heat capacity of disc = [M × cwater × (θ2 – θ1)] + [m × ccalorimeter × (θ2 – θ2)]/
[mass of disc × (100 – θ2)]
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 2Electricity and Thermal Physics
Solutions to Practice Questions5 Atomic masses mA:
Aluminium = 4.5 × 10–26 kg
Copper = 1.1 × 10–25 kg
Iron = 9.3 × 10–26 kg
Lead = 3.5 × 10–25 kg
If c ∝ 1/mA then cmA = constant:
For aluminium, 880 J kg–1 K–1 × 4.5 × 10–26 kg = 4.0 × 10–23 J K–1
For copper, 380 J kg–1 K–1 × 1.1 × 10–25 kg = 4.2 × 10–23 J K–1
For iron, 450 J kg–1 K–1 × 9.3 × 10–26 kg = 4.2 × 10–23 J K–1
For lead, 130 J kg–1 K–1 × 3.5 × 10–25 kg = 4.6 × 10–23 J K–1
Since the products are all approximately the same, c ∝ 1/mA
Number of atoms determines energy required
Larger atomic masses result in fewer atoms in each kilogram of substance
Less energy required to raise temperature of fewer atoms by 1 K
Concrete can be raised through a much higher temperature than water and so store more energy
per kilogram
Concrete is cheap and safe to use (best not to use water with electric storage heaters!)
Chapter 261 See experiment on page 54
Precautions: lag the cylinder; make sure heater is fully immersed
Calculation: energy used to vaporise water = mass of water vaporised × specific latent heat
of vaporisation
Energy supplied = potential difference × current × time = VIt
Specific latent heat of vaporisation L = VIt/(mass of water vaporised)
2 Energy required = mL = 0.016 kg × 2.25 × 106 J kg–1 = 36 000 J
Minimum power = energy/time = 36 000 J/(60 s) = 600 W
Actual power will be greater to compensate for energy transferred to the surroundings by radiation,
convection and conduction
3 Power of kettle = VI = 230 V × 8 A = 1840 W
Energy required to heat water = mc∆T = 0.6 kg × 4200 J kg–1 K–1 × (100 – 15) K = 214 200 J
Energy required to heat kettle = 350 J °C–1 × (100 – 15) °C = 29 750 J
Time taken = total energy required/power = (214 200 + 29 750) J/(1840 W) = 132.6 s = 130 s
Energy supplied by kettle in two minutes = 1840 W × 120 s = 220 800 J
Mass of water vaporised = energy supplied/latent heat = 220 800 J/(2.25 × 106 J kg–1) = 0.098 kg
Mass of water remaining in kettle = (600 – 98) g = 502 g = 500 g
4 (a) Energy given out by water = mc∆T = 0.284 kg × 4200 J kg–1 K–1 × 22 K = 26 200 J
mL = 26 200 J
mass of ice that melts = 26 200 J/L = 26 200 J/(330 × 103 J kg–1) = 0.080 kg
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 2Electricity and Thermal Physics
Solutions to Practice Questions5
The lead has solidified
(a) Energy lost by solid lead = mc∆T = 3 kg × 130 J kg–1 K–1 × 6 K = 2340 J
Rate of energy loss = energy loss/time = 2340 J/(10 s) = 234 J s–1 (or W)
(b) Energy loss in 5 minutes = 234 J s–1 × (5 × 60) s = 70 200 J
mL = 70 200 J
Specific latent heat of fusion of lead = 70 200 J/m = 70 200 J/(3 kg) = 23 400 J kg–1
(c) Energy loss in 16 s = 234 J s–1 × 16 s = 3744 J
mc∆T = 3744 J
Specific heat capacity of molten lead = 3744 J/m∆T = 3744 J/(3 kg × 8 K) = 156 J kg–1 K–1
Chapter 271 Working involves a moving force
Heating involves a temperature difference
Mechanical working:
force squashes material and does mechanical work F∆x
Electrical working:
power supply forces electrons through material and does electrical work VI∆t
2 KE = 1
2 mv2 =
1
2 × 0.3 kg × (15 m s–1)2 = 33.75 J = 34 J
Lead is ‘soft’ compared with the hammer head and yields under the hammer blows
Mechanical work is done on the lead as the force compresses it
The lead’s internal energy rises
(The lead is now able to heat the hammer head as it is at a higher temperature but the two are in
contact only for a short period so little energy is transferred)
Total energy from 50 blows = 33.75 J × 50 = 1690 J = 1700 J
mc∆T = 1690 J
Temperature rise = 1690 J/mc = 1690 J/(0.15 kg × 130 J kg–1 K–1) = 87 K
3 Heating is the transfer of energy through a temperature difference from hot to cold
The hot lamp can heat the cold cell but not vice versa
The cold cell is forcing electrons to move through the hot lamp, an example of electrical working
1500 300
Time/s
610
608
606
604
602
600
598
596
594
592
Tem
pera
ture
/K
50 100 200 250 350
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 2Electricity and Thermal Physics
Solutions to Practice Questions4 Electrical work done = VI∆t = 0.12 V × 3.5 A × 30 s = 12.6 J = 13 J
I = nAqv
v = I/nAq = 3.5 A/(1.7 × 1029 m–3 × 1.0 × 10–6 m2 × 1.6 × 10–19 C) = 1.29 × 10–4 m s–1
= 0.13 mm s–1
∆x = v∆t = 1.29 × 10–4 m s–1 × 30 s = 3.9 × 10–3 m = 3.9 mm
Work done = F∆x = 12.6 J
F = 12.6 J/∆x = 12.6 J/(3.9 × 10–3 m) = 3264 N = 3300 N
5 Electric mains supply forces electrons to move through the fire
The mains supply does electrical work on the electrons
The electrons then transfer their energy by collisions to the lattice of the heating element
The mains supply cannot heat the electric fire since it is at a lower temperature than the fire
Some energy will radiate into the room from the glowing element but convection currents will carry
most of it around the room
Chapter 281 ∆U: increase in the internal energy of the system
∆Q: energy transferred to the system by heating
∆W: energy transferred to the system by working
Conservation of energy
2 ∆Q = 0 either when the system is thermally isolated or when the system is at the same temperature
as its surroundings
3 Thermos flasks do not completely isolate their contents from the surroundings
Some energy will always flow through either the walls or the lid
It is impossible to produce a completely isolated system
4 ∆U = 0
The filament has reached a steady temperature
∆W = P∆t = 24 W × 5 s = 120 J
The power supply is working on the filament
∆Q = ∆U – ∆W = 0 – 120 J = –120 J
The filament is heating the surroundings
5 ∆U = ∆Q + ∆W
When a gas rapidly expands, ∆Q = 0 as there is insufficient time for energy to enter or leave the
system
so ∆U = ∆W
As the gas is doing work on its surroundings, ∆W is negative
so ∆U is also negative
The temperature of a rapidly expanding gas decreases
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 2Electricity and Thermal Physics
Solutions to Practice QuestionsChapter 291 Heat engine: a device that takes energy from a hot source, uses some of this to do mechanical work,
and gives the rest to a cold sink
See Figure 29.1 on page 60
2 Efficiency = useful output/input
Power of light emitted = (2/100) × 60 W = 1.2 W
The other 58.8 W increases the internal energy of the surroundings
3 Make source temperature very high and sink temperature very low
Maximum efficiency = 1 – T2/T1 = 1 – 300 K/(673 K) = 0.55
4 In a heat pump, work is done to force energy to flow from cold to hot
In a heat engine, energy flowing from hot to cold is used to do work
Heat pumps are used in refrigerators, freezers and air conditioners
5 Maximum efficiency = 1 – 3 K/(5 000 000 K) = 0.999 999 4
Eventually, the Sun’s temperature will decrease greatly (and that of the Universe will
increase slightly)
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 4Waves and our Universe
Solutions to Practice QuestionsChapter 11 Constant velocity means a constant speed in a constant direction (a straight line) along a circular
path, the direction is changing so the velocity will change
2 By changing its direction
‘Accelerating’ means that the object’s velocity is changing
Velocity is a vector quantity, involving both size (speed) and direction so a change in velocity
involves either a change in speed or a change in direction (as in this case)
3 Direction of velocity is along a tangent
Velocity is constantly changing direction as the object follows a circular path
Acceleration is defined as a change in velocity; so object is accelerating
Change in velocity, and therefore acceleration, is towards the centre of the circle
4 (a) The gravitational pull of the Earth on the Moon
(b) Friction between you and the surface of the roundabout
(c) Friction between the tyres and the road surface
(d) Normal contact force of seat on you
5 Work done = force × distance moved in the direction of the force
A centripetal force is directed towards the centre of the circular path
All movement occurs perpendicular to this force so distance moved in direction of force is zero
Work done = centripetal force × 0 = 0
Chapter 21 Left-hand side F in N = kg m s–2
Right-hand side m in kg
v in m s–1
v 2 in (m s–1)2 = m2 s–2
r in m
So mv2/r in kg m2 s–2 m–1 = kg m s–2 = same as left-hand side and ∴ homogeneous
2 (a) a = v2/r = (7 m s–1)2/(2 m) = 24.5 m s–2 towards the centre of the circle
(b) F = ma = 0.5 kg × 24.5 m s–2 = 12.25 N
(c) Work done = 0 J
No work is done since no component of the tension acts in the direction of motion
3 Period is the time for one complete rotation, measured in seconds
Frequency is the number of complete rotations each second, measured in hertz
Angular speed is the rate at which the central angle changes each second, measured in radians
per second
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 4Waves and our Universe
Solutions to Practice Questions4 108 km h–1 = 108 000 m h–1/(60 × 60 s h–1) = 30 m s–1
Angular speed w = v/r = 30 m s–1/(1000 m) = 0.03 rad s–1
Acceleration a = v2/r = (30 m s–1)2/(1000 m) = 0.9 m s–2
5 (a) Length of tape = speed × time = 0.048 m s–1 × (45 × 60) s = 129.6 m = 130 m
(b) w = v/r
wA = 0.048 m s–1/(0.0255 m) = 1.88 rad s–1 = 1.9 rad s–1
wB = 0.048 m s–1/(0.0105 m) = 4.57 rad s–1 = 4.6 rad s–1
(c) As the tape plays, wA increases while wB decreases
Chapter 31 At bottom of vertical circular path:
tension in the string provides the centripetal force and supports the weight
so tension is a maximum
At top of vertical path:
the weight helps to provide the centripetal force and the tension is less
2 (a) Tbottom – Ttop = 2mg = 7 N – 1 N = 6 N
Weight = mg = 6 N/2 = 3 N
(b) m = 3 N/g = 3 N/(9.81 N kg–1) = 0.30 kg
(c) Centripetal force = 4 N = mv2/r
v = √[4 N × r/m] = √[4 N × 0.7 m/(0.30 kg)] = 3.0 m s–1
3 Fishing line breaks when body is at bottom of vertical circular path
At this point, tension Tmax = mv2/r + mg
15 N = mv2/r + 6 N
so mv2/r = 9 N
v 2 = 9 N × 0.30 m/(6 N ÷ 9.81 N kg–1) = 4.41 m2 s–2
v = √(4.41 m2 s–2) = 2.1 m s–1
4 To be weightless, a body must be in deep space with no gravitational forces acting on it, whereas a
body in free fall experiences no upward forces and therefore can be considered to be weightless
5 Free-body force diagram:
30°
TensionT
Weight
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 4Waves and our Universe
Solutions to Practice Questions(a) For vertical equilibrium:
Force up = force down
T cos 30° = mg
so T = mg/(cos 30°) = 0.51 kg × 9.81 N kg–1/(cos 30°) = 5.8 N
(b) Only one force component acts horizontally, T sin 30° this provides the centripetal force for a
circular motion
Radius of circle = 0.6 m × sin 30° = 0.3 m (found using radius)
T sin 30° = mv2/r
so v = √[rT sin 30°/m] = √[0.3 m × 5.8 N × sin 30°/0.51 kg] = 1.3 m s–1
(c) Period T = 2πr/v = 2π × 0.3 m/(1.3 m s–1) = 1.45 s
Chapter 41 The swing of a pendulum is regular
The swings of a particular pendulum always take the same time (its period)
Two other clock mechanisms: a mass/spring system and a quartz crystal (atomic oscillations)
2 Equilibrium position: where the resultant force on the oscillating body is zero
Displacement: how far, and in what direction, the oscillating body is from its equilibrium position
Amplitude: maximum displacement
3 E.g. place a card on the glider perpendicular to its direction of travel (like a sail)
Use a motion sensor, connected to a data-logger, at one end of the air track to monitor positions of
the glider on the track
Data-logger sends position data to computer that plots time trace
Take centre of air track as zero displacement:
4 Period T = 1/f = 1/(160 Hz) = 0.063 s
5 Amplitude = 0.15 m
11
2 cycles occur each second, frequency = 1.5 Hz
Body has zero speed when at maximum displacement in either direction
Time
Dis
plac
emen
t
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 4Waves and our Universe
Solutions to Practice Questions
Chapter 51 SHM: motion where the acceleration (or force) is directly proportional to the displacement from a
fixed point and always directed towards that point
2 Graph shows that F ∝ – x
3 (a) Zero velocity at A and C
(b) Zero acceleration at B
(c) Maximum velocity to the right at B
(d) Maximum acceleration to the left at C
(e) Maximum kinetic energy at B
4 See Figure 5.4 on page 11
5 (i) k = F/e = 800 × 10–3 kg × 9.81 N kg–1/(4.0 × 10–2 m) = 196.2 N m–1 = 200 N m–1
(ii) ω = √(k/m) = √[196.2 N m–1/(800 × 10–3 kg)] = √(245 s–2) = 15.66 rad s–1 = 16 rad s–1
(iii) T = 2π/ω = 2π/(15.66 rad s–1) = 0.40 s
Chapter 61 Amplitude = 8 cm
ωt = 4πt so angular speed ω = 4π rad s–1
Period T = 2π/ω = 2π/(4π rad s–1) = 0.5 s
Frequency f = 1/T = 1/(0.5 s) = 2 Hz
Displacement
Force
Dis
plac
emen
t/m
0.15
0
–0.15
0 0.5 1.0 1.5 2.0 2.5Time/s
positions of zero speed
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 4Waves and our Universe
Solutions to Practice Questions2
vmax = ωx0 = 4π rad s–1 × 0.08 m = 1.01 m s–1
amax = ω2x0 = (4π rad s–1)2 × 0.08 m = 12.6 m s–2
3 ω = 2π/T = 2π/(π s) = 2 rad s–1
Displacement (in cm) x = x0 cos ωt = 12 cos 2t
Velocity (in cm s–1) v = –ωx0 sin ωt = –24 sin 2t
Acceleration (in cm s–2) a = –ω2x0 cos ωt = –48 cos 2t
4 Oscillations are in phase when they are exactly in step with each other
The phase angle between their oscillations is zero
Oscillations are in antiphase when they are completely out of step with each other
The phase angle between their oscillations is π radians (180°)Their frequencies must be identical
5 (a) See Figure 6.6 on page 13
(b) See Figure 6.5 on page 13
Diagram showing both runners at same point on track
Chapter 71 Left-hand side T in s
Right-hand side 2π has no units
m in kg
k in N m–1 = kg m s–2 m–1 = kg s–2
So m/k in kg (kg s–2)–1 = kg kg–1 s2 = s2
and √(m/k) in √s2 = s = same as left-hand side and ∴ homogeneous
Time/s
Dis
plac
emen
t
8 cm
–8 cm
0
Time/sVelo
city
0
Time/s
Acc
eler
atio
n
0
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 4Waves and our Universe
Solutions to Practice Questions2 Let the spring constant of each spring be k
T = 2π √(m/k) ∝ 1/ √k
(a) 4 springs in series will have a spring constant of 1
4k
1/√ 1
4 = 1/
1
2 = 2
So new period = 2 × old = 2 × 2.4 s = 4.8 s
(b) 4 springs in parallel will have a spring constant of 4k
1/√4 = 1
2
So new period = 1
2 × old =
1
2 × 2.4 s = 1.2 s
3 T = 2π√(l/g) = 2π × √[35 × 10–2 m/(9.81 m s–2)] = 2π × √(0.0357 s2) = 2π × 0.189 s = 1.2 s
4 Measure 20T for a measured length l
Repeat and calculate an average value for T
Change and measure l and repeat measurements of 20T to find average T for this new length
Continue to get at least 6 corresponding values for l and T
Plot a graph of T 2 against l
Since T 2 = 4π 2l/g
Graph will be a straight line through the origin with gradient = 4π2/g
So g = 4π2/gradient
5 A clock requires a regular repeating motion
The period must remain constant even when the amplitude changes
If amplitude is halved, body must travel half the distance in the same time, so its average speed
is halved
Thus its average acceleration is also halved
So a ∝ (–)x, a condition satisfied by simple harmonic motion
Chapter 81 Energy changes continually from potential to kinetic and back to potential at the ends the trolley is
momentarily at rest and has no kinetic energy
Maximum elastic potential energy is stored in the stretched and compressed springs
In the centre the trolley is moving fastest and has maximum kinetic energy
The stored elastic potential energy is at a minimum
The energy has transferred into internal energy and sound
2 (a) Graph similar to the lower one in Figure 8.2 on page 16 but going on for twice as long
i.e. showing two complete cycles
(b) Graph identical to that in Figure 8.3 on page 17
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 4Waves and our Universe
Solutions to Practice Questions3 Potential energy given to system =
1
2kx2 =
1
2 × 250 N m–1 × (8 × 10–2 m)2 = 0.8 J
Kinetic energy of mass at centre = 0.8 J
1
2mvmax
2 = 0.8 J
vmax2 = 2 × 0.8 J/(600 × 10–3 kg) = 2.67 m2 s–2
vmax = √(2.67 m2 s–2) = 1.6 m s–1
Using an ‘average’ speed of 1
2 × 1.6 m s–1 = 0.8 m s–1
T = total distance travelled in one oscillation/ ‘average’ speed = 4 × 8 × 10–2 m/(0.8 m s–1) = 0.39 s
T = 2π√(m/k) = 2π√[600 × 10–3 kg/(250 N m–1)] = 0.31 s
Comment: e.g. since speed varies sinusoidally, average speed is greater than 1
2vmax
so time taken is less
4 As the mass rises:
its gravitational potential energy increases continually
the elastic potential energy in the spring decreases continually
its kinetic energy increases from zero to a maximum half-way up and then decreases to zero at
the top
5 For spring, k = F/e = 0.40 kg × 9.81 N kg–1/(0.06 m) = 65.4 N m–1
Elastic potential energy at bottom = 1
2kx2 =
1
2 × 65.4 N m–1 × (0.12 m)2 = 0.471 J
Elastic potential energy at mid-point = 1
2kx2 =
1
2 × 65.4 N m–1 × (0.06 m)2 = 0.118 J
Gravitational potential energy at mid-point (taking zero at bottom) = mg∆h = 0.40 kg × 9.81 N kg–1
× 0.06 m = 0.235 J
Kinetic energy at mid-point = 0.471 J – (0.118 J + 0.235 J) = 0.118 J
Chapter 91 The frequency at which a free-standing system oscillates after it has been displaced and then released
2 Left-hand side f in s–1
Right-hand side 1/2π has no units
g in m s–2
L in m
so g /L in m s–2 m–1 = s–2
and √(g /L) in √s–2 = s–1 = same as left-hand side and ∴ homogeneous
f 2 = g/4π2L
L = g/4π2f 2 = 9.81 m s–2/[4 × π2 × (1 s–1)2] = 0.248 m
3 Oscillator vibrates at forcing frequency
When this is below or above its natural frequency, the amplitude is low
When this is equal to its natural frequency, the amplitude can be much higher
This ‘high amplitude response’ is known as resonance
Resonance is the large-amplitude oscillations that arise as a result of an oscillatory system being
driven at a frequency equal to its natural frequency
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 4Waves and our Universe
Solutions to Practice Questions4 The rear-view mirror oscillates on the end of its metal support
It has a natural frequency that depends on its mass and stiffness and length of the support
Rotation of engine forces mirror to oscillate at different frequencies
Resonance occurs when engine frequency equals mirror’s natural frequency
Large amplitude oscillations result and blurring of the image occurs
To overcome problem, need to change the natural frequency of the mirror either by changing its
mass or its support system
5 T = 2π√(m/k)
k = 4π2m/ T 2 = 4π2 × (25 + 35) kg/(2 s)2 = 590 N m–1
Uneven distribution of load in rotating inner drum produces forced oscillations of the whole system
Maximum amplitude of oscillation achieved when system driven at its natural frequency
fnatural = 1/T = 1/(2 s) = 0.5 Hz or 30 min–1
Without the block of concrete:
resonant frequency higher as f ∝ 1/√m [T ∝ √m]
resonant amplitude larger as energy transferred into smaller mass
Chapter 101 Use a similar set-up to that in Figure 10.1 on page 20 but without the dipper and with the motor-
driven beam lowered into the water’s surface to produce the continuous plane waves
Change frequency by varying the speed at which the motor on the beam vibrates
‘Freeze’ wave motion using a stroboscope and measure as many wavelengths as possible
Divide length by number to get wavelength
Find the average time taken by a wave to cover a measured distance
Divide distance by average time to get wave speed
Am
plitu
de o
f osc
illat
ions
/cm
00 10 20 30 40 50 60
0.5
1.0
1.5
2.0
2.5
3.0
Inner drum frequency/min–1
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 4Waves and our Universe
Solutions to Practice Questions2 A progressive wave (or travelling wave) is a disturbance that transfers energy away from its source
Amplitude decreases with distance from source since energy is spread out over a larger area
Energy absorbed by material along the path taken
3 Energy flux: the energy that a wave carries perpendicularly through unit area each second
Inverse square law: when a quantity decreases in proportion to the square of the increasing distance
E.g. distance × 2, quantity × 1
4
distance × 3, quantity × 1
9
Energy leaving uniformly from a point source spreads out evenly over the surface of a sphere
Applying the principle of energy conservation,
Energy flux × surface area of sphere = total power emitted
Energy flux = total power emitted/(4πr2) ∝ 1/r2 and therefore an inverse square law
4 If energy flux obeys an inverse square law, a graph of energy flux against 1/distance2 should be a
straight line passing through the origin
Distance/m 0.50 1.00 1.50 2.00 2.50 3.00
Energy flux/W m–2 2.56 0.64 0.28 0.16 0.10 0.07
1/(distance)2 /m–2 4.00 1.00 0.44 0.25 0.16 0.11
Either
since φ = P/4πr2
gradient of graph = P/4πgradient = 2.56 W m–2/(4 m–2) = 0.64 W
P = 4π × gradient = 4π × 0.64 W = 8.04 W
Or
using P = φ × 4πr2
when r = 0.50 m P = 2.56 W m–2 × 4π(0.50 m)2 = 8.04 W
when r = 1.00 m P = 0.64 W m–2 × 4π(1.00 m)2 = 8.04 W
when r = 1.50 m P = 0.28 W m–2 × 4π(1.50 m)2 = 7.92 W
when r = 2.00 m P = 0.16 W m–2 × 4π(2.00 m)2 = 8.04 W
when r = 2.50 m P = 0.10 W m–2 × 4π(2.50 m)2 = 7.85 W
when r = 3.00 m P = 0.07 W m–2 × 4π(3.00 m)2 = 7.92 W
Ene
rgy
flux/
W m
–2
0
1/(Distance)2/m–2
1.0 3.0 4.02.0
3.0
2.5
2.0
1.5
1.0
0.5
0
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 4Waves and our Universe
Solutions to Practice QuestionsLamp transfers only a small amount of its input energy into light
Most of the input energy transferred into internal energy
Actual power of lamp includes both transfers
5 φ = P/4πr2
P = φ × 4πr2 = 1.4 × 103 W m–2 × 4π(1.49 × 1011 m)2 = 3.91 × 1026 W
Chapter 111 Displacements of longitudinal waves are parallel to direction of wave travel while those of
transverse waves are perpendicular to it
2 Observe movement of candle flame in front of a loudspeaker emitting a low frequency as shown in
Figure 10.2 on page 20
3 In an unpolarised wave, the vibrations occur in a large number of planes perpendicular to the
direction of energy propagation
In a plane polarised wave, the vibrations are confined to a single planes perpendicular to the
direction of energy propagation
Visible light and microwaves are transverse waves so can be polarised
Sound is a longitudinal wave so cannot be polarised
4 The microwave beam is polarised
In one orientation, the metal rods absorb the microwave energy (when they are parallel to the
microwave’s oscillating electric field) and only a low signal is received
When turned through 90°, little energy is absorbed by the metal rods and a much larger signal
is received
So as the grille is rotated, the received amplitude falls and rises twice per rotation
5 View reflected light through a rotating single Polaroid filter; if it is polarised, transmitted intensity
will vary twice per rotation
Chapter 121 Speed of light is greater than the speed of sound
Assuming light reaches her instantaneously
Distance = vt = 330 m s–1 × 0.8 s = 264 m
Extra distance travelled by echo = 330 m s–1 × 2.4 s = 792 m
But this is to the hill and back to the woodcutter
So the hill is 1
2 × 792 m = 396 m behind the woodcutter
2 Trace B
Second trace as pulse takes longer to travel the greater distance
Smaller trace as pulse dissipates more energy over the greater distance
A to B = 3 divisions = 3 × 20 ns = 60 ns
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 4Waves and our Universe
Solutions to Practice QuestionsExtra distance travelled by light = vt = 3 × 108 m s–1 × 60 × 10–9 s = 18 m
But this is twice the distance moved by the mirror
So the mirror was moved 1
2 × 18 m = 9 m
3 Wavefront: a line joining all points across adjacent rays that have exactly the same phase
Wavelength: the minimum distance between two in-phase points on a wave
4 Speed c = f λ = 20 Hz × 1.3 × 10–2 m = 0.26 m s–1
Note: period = 1/f = 50 ms, so 12.5 ms = 1
2T
A and B are in phase; A and C (or B and C) are out of phase
5 (a) Frequency f = c/λ = 330 m s–1/(8.5 × 10–2 m) = 3900 Hz
(b) Number of waves in 20 ms = 3900 Hz × 20 × 10–3 s = 77(.6) waves
(c) Length of wave train = ct = 330 m s–1 × 6.5 s = 2100 m
Chapter 131 Diffraction: the spreading out of a wave as it passes through an aperture
Amount of diffraction occurring depends on wavelength
Wavelength of light is much less than that of microwaves so light diffracts much less than microwaves
2 Wavelength of red light is greater than that of blue light
Bass (low frequency) sounds have a longer wavelength that treble (high frequency) sounds so bass
sounds will diffract the most
84 85025
075 100
Time/ms
A B C
Dis
plac
emen
t of A
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 4Waves and our Universe
Solutions to Practice Questions3
The central fringe is twice as wide as the others and much brighter
4 See experiment at top of page 27
Law more difficult to test for microwaves since it is impossible to get a narrow beam of microwaves
because of diffraction and, unlike light, you cannot see the path that microwaves follow
For sound waves:
Tape a loudspeaker connected to a signal generator to a long cardboard tube
Tape a microphone connected to an oscilloscope to another long cardboard tube
Use loudspeaker tube to direct sound waves towards a reflector at a measured incident angle
Vary position of microphone tube until maximum signal is detected
Measure angle of reflection and compare with angle of incidence
Repeat for different incident angles
5
Note: direction of travel changes towards the normal
wavelength reduces to 3
4 (
2
2
1
8) of that in the deep water
[angle of refraction with normal = 25.5°]
A
B
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 4Waves and our Universe
Solutions to Practice QuestionsChapter 141 See experiment at top of page 28
Use differently shaped pulses to tell whether they pass through or reflect from each other
2
1st square resultant displacement = 1 + 2 = 3
2nd square resultant displacement = 1 + –1.5 = –0.5
3rd square resultant displacement = –1 + 1 = 0
4th square resultant displacement = –1 + –0.5 = –1.5
3 Principle of superposition: the resultant displacement at any point is equal to the vector sum of the
displacements of the individual waves at that point at that instant
Constructive superposition is the combination of waves that are in-phase, producing a wave of
increased amplitude
Destructive superposition is the combination of waves that are out-of-phase, producing a wave of
reduced amplitude
4 Coherent sources: sources that, in addition to having identical frequencies, always maintain a
constant phase relationship with each other
Having different frequencies results in waves being emitted with a continually changing phase
relationship; the amplitude at any given point on the superposition pattern varies from maximum
to minimum with this changing phase: the pattern varies with time
5 Both sources are emitting waves with similar amplitudes
Chapter 151 0 rad corresponds to 0λ
π/2 rad corresponds to 1
4λ
π rad corresponds to 1
2λ
3π/2 rad corresponds to 3
4λ
2π rad corresponds to λ
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 4Waves and our Universe
Solutions to Practice Questions2 The loudspeakers are emitting out-of-phase sound waves
They are connected such that the cone of one moves forwards when that of the other moves backwards
Reverse the connections to one of the loudspeakers to get a central maximum
3 Place a double-slit arrangement in front of the single source
Each slit acts as a separate coherent source
4 Using Pythagoras’ theorem
S2P = √[(110 cm)2 + (66 cm + 14 cm)2] = √(18 500 cm2) = 136.0 cm
S1P = √[(110 cm)2 + (66 cm – 14 cm)2] = √(14 804 cm2) = 121.7 cm
Path difference = 136.0 cm – 121.7 cm = 14.3 cm
For 3rd maximum, path difference = 3λWavelength λ = 14.3 cm/3 = 4.8 cm
5 Waves will be coherent since they come from a single coherent source
Distance travelled along top path = 2 × √[(15 cm)2 + (7 cm)2] = 33.1 cm
Distance travelled along bottom path = 2 × √[(15 cm)2 + (10 cm)2] = 36.1 cm
Path difference = 36.1 cm – 33.1 cm = 3.0 cm ≈ 2.9 cm = λSince path difference is 1 wavelength, received signal is a maximum
Chapter 161 Random bursts of unrelated light leave the different parts of the filament at various times so the
filament produces incoherent light
Single slit limits light used to small part of filament
Light used is far more coherent
A laser produces coherent light so a single slit is not required
2 Monochromatic: a single frequency (or wavelength)
See Figure 16.3 on page 33
3 See Figure 16.4 on page 33
Double-slit pattern: all fringes are the same width
Single-slit pattern: central fringe is twice the width of the others
Using white light, the differently coloured fringes occur at different separations and may overlap
See Figure 16.5 on page 33
4 λ = xs/D
s = λD/x = 680 × 10–9 m × 1.8 m/(3.2 × 10–3 m) = 3.8 × 10 –4 m = 0.38 mm
5 λ = xs/D
x = λD/s = 550 × 10–9 m × 2 m/(0.25 × 10–3 m) = 4.4 × 10–3 m
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 4Waves and our Universe
Solutions to Practice QuestionsChapter 171 Stationary wave: a disturbance that does not transfer energy although it does have energy associated
with it
Conditions: two waves with identical frequency and amplitude, moving in opposite directions
2 Node: a point on a stationary wave where the displacement is always zero
Antinode: a point on a stationary wave that oscillates with the maximum amplitude
Nodes: at any given moment, the two displacements are equal and opposite (out-of-phase); resulting
displacement is always zero
Antinodes: at any given moment, the two displacements are equal (in-phase); Resulting displacement
varies from zero to twice the amplitude of the individual waves; position of maximum oscillation
3 Fundamental frequency: the lowest frequency at which a stationary wave occurs for a given system
Left-hand side f in s–1
Right-hand side 1
2 has no units
l in m
T in N = kg m s–2
µ in kg m–1
So T/µ in kg m s–2 kg–1 m = m2 s–2
and √(T/µ) in √m2 s–2 = m s–1
and 1/l × √(T/µ) in m–1 m s–1 = s–1 = same as left-hand side and ∴ homogeneous
Tension increased by a factor of 160/40 = 4
Fundamental frequency increases by a factor of √4 = 2
Fundamental frequency doubles when tension quadruples
4 A progressive wave conveys energy in the direction of travel
A stationary wave does not convey any energy
Along a progressive wave, adjacent points oscillate with a slight phase difference; only points that
are exactly one wavelength apart oscillate in phase
Along a stationary wave, the set of points between adjacent nodes oscillate in phase with each other
and out of phase with all those in adjacent sets
5 λ = c/f = 330 m s–1/(850 Hz) = 0.39 m
Separation of nodes = λ/ 2 = 0.39 m/2 = 0.19 m
T = 1/f = 1/(5 Hz) = 0.2 s
So microphone takes 0.2 s to move 0.19 m between antinodes
Microphone’s speed = distance/time = 0.19 m/(0.2 s) = 0.97 m s–1
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 4Waves and our Universe
Solutions to Practice QuestionsChapter 181 Photoelectric emission: emission of electrons from a surface when illuminated with electromagnetic
radiation of sufficient frequency
See experiment on page 38
2 Threshold frequency: minimum frequency that will cause photoelectric emission from a material
c = fλλ = c/f = 3.0 × 108 m s–1/(0.88 × 1015 Hz) = 3.4 × 10–7 m = 340 nm
3 The larger the wavelength, the lower the photons’ energy
No emission occurs when photon energy is less than the work function
Electrons gain energy from an increase in temperature so less energy then required to remove them
from the surface (smaller work function)
Less energy corresponds to a longer wavelength
4 Photon: a small packet of electromagnetic energy; the smallest amount of light you can get at a
given frequency
(a) Both produce photons with identical energy but the bright source produces more photons each
second than the dim source
(b) Visible source produces lower energy photons than the ultra-violet source, although it produces
them at a greater rate to achieve the same intensity
5 Work function: minimum amount of energy needed to release an electron from the surface of a metal
Ultra-violet photon energy is greater than zinc’s work function so electrons are released
Visible photon energy is smaller than zinc’s work function so no electrons are released
It’s the individual photon energy that matters, not how many there are
Chapter 191 Kinetic energy of freed electron = photon energy – energy required to remove electron
Surface electrons are the easiest to remove so have greatest kinetic energy and move the fastest
2 See experiment on page 40
Measure the voltage VS needed just to stop their emission
Energy = eVS where e = 1.6 × 10–19 C
3 Photon energy = hf = hc/λso hc/λ = φ + maximum kinetic energy
and maximum kinetic energy = hc/λ – φMaximum kinetic energy/10–19 J 3.26 2.56 1.92 1.25 0.58
Incident wavelength/10–7 m 3.00 3.33 3.75 4.29 5.00
(1/incident wavelength)/106 m–1 3.33 3.00 2.67 2.33 2.00
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 4Waves and our Universe
Solutions to Practice Questions
Gradient of graph = hc
hc = (3.26 – 0.58) × 10–19 J/[(3.33 – 2.00) × 106 m–1] = 2.02 × 10–25 J m
h = 2.02 × 10–25 J m/c = 2.02 × 10–25 J m/(3.00 × 108 m s–1) = 6.72 × 10–34 J s
Intercept of graph = – φ– φ = – 3.4 × 10–19 J
φ = 3.4 × 10–19 J
4 Maximum kinetic energy = (hc/λ) – φ = [6.6 × 10–34 J s × 3 × 108 m s–1/(319 × 10–9 m)] –
3.78 × 10–19 J = 6.21 × 10–19 J – 3.78 × 10–19 J = 2.43 × 10–19 J
eVS = 2.43 × 10–19 J
VS = 2.43 × 10–19 J/(1.6 × 10–19 C) = 1.52 V
5 Maximum kinetic energy = (hc/λ) – φSituation 1:
2.4 × 10–19 J = [6.6 × 10–34 J s × 3 × 108 m s–1/(500 × 10–9 m)] – φφ = 3.96 × 10–19 J – 2.4 × 10–19 J = 1.56 × 10–19 J
Situation 2:
9.0 × 10–19 J = (6.6 × 10–34 J s × 3 × 108 m s–1/λ) – 1.56 × 10–19 J
6.6 × 10–34 J s × 3 × 108 m s–1/λ = 9.0 × 10–19 J + 1.56 × 10–19 J = 1.056 × 10–18 J
λ = 6.6 × 10–34 J s × 3 × 108 m s–1/(1.056 × 10–18 J) = 1.88 × 10–7 m = 188 nm
Chapter 201 Electronvolt: the energy transferred to an electron when it moves through a potential difference
of 1 V; 1 eV is equivalent to 1.6 × 10–19 J
E = hc/λ = 6.6 × 10–34 J s × 3 × 108 m s–1/(253 × 10–9 m) = 7.83 × 10–19 J
= 7.83 × 10–19 J/(1.6 × 10–19 J eV–1) = 4.89 eV
2 φ = 1.4 eV = 1.4 eV × 1.6 × 10–19 J eV–1 = 2.24 × 10–19 J
hc/λο = 2.24 × 10–19 J
λο = 6.6 × 10–34 J s × 3 × 108 m s–1/(2.24 × 10–19 J) = 8.84 × 10–7 m = 884 nm
Max
imum
kin
etic
ene
rgy
/10–1
9 J
0
(1/Incident wavelength)/106 m–1
1.0 3.0 3.52.0
4.0
2.0
1.0
3.0
0
–1.0
–2.0
–3.0
–4.00.5 1.5 2.5
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 4Waves and our Universe
Solutions to Practice Questions3 Photon energy = hc/λ = 6.6 × 10–34 J s × 3 × 108 m s–1/(0.4 × 10–6 m) = 4.95 × 10–19 J
φ = 1.94 eV × 1.6 × 10–19 J eV–1 = 3.14 × 10–19 J
Maximum kinetic energy = photon energy – φ = 4.95 × 10–19 J – 3.14 × 10–19 J = 1.81 × 10–19 J
1
2mvmax
2 = 1.81 × 10–19 J
vmax2 = 2 × 1.81 × 10–19 J/(9.1 × 10–31 kg) = 3.99 × 1011 m2 s–2
vmax = √(3.99 × 1011 m2 s–2) = 6.31 × 105 m s–1
4 (a) The intensity
(b) The photon frequency and the material’s work function
5 Photon energy = hc/λ = 6.6 × 10–34 J s × 3 × 108 m s–1/(250 × 10–9 m) = 7.92 × 10–19 J
In electronvolts, photon energy = 7.92 × 10–19 J/(1.6 × 10–19 J eV–1) = 4.95 eV
Maximum kinetic energy = photon energy – φ = 4.95 eV × 0.88 eV = 4.07 eV
So stopping voltage = 4.07 V
Photon energy increases with frequency (E = hf )
Maximum kinetic energy increases with photon energy (maximum kinetic energy = hf – f )
A greater potential difference is needed to stop these more energetic electrons
Chapter 211 Ionisation energy: the energy required to completely free a ground state electron from it’s atom
Ground state: the condition of an atom where all its electrons are in their lowest energy positions
Excited state: the condition of an atom with one or more of its electrons raised above their ground-
state positions
Energy needed = 10.4 eV – 5.5 eV = 4.9 eV = 7.84 × 10–19 J
2 Minimum kinetic energy = 5.14 eV
1
2mvmin
2 = = 5.14 eV × 1.6 × 10–19 J eV–1 = 8.22 × 10–19 J
vmin2 = 2 × 8.22 × 10–19 J/(9.1 × 10–31 kg) = 1.81 × 1012 m2 s–2
vmin = √(1.81 × 1012 m2 s–2) = 1.34 × 106 m s–1
3 Energy released = 3.4 eV – 1.5 eV = 1.9 eV = 3.04 × 10–19 J
λ = hc/E = 6.6 × 10–34 J s × 3 × 108 m s–1/(3.04 × 10–19 J) = 6.51 × 10–7 m = 651 nm
This wavelength is in the visible region
4 Emission spectrum: the range of frequencies emitted by de-exciting atoms
Left-hand side n has no units
λ in m
so nλ in m
Right-hand side sin θ has no units
N in m–1
so (sin θ )/N in 1/m–1 = m = same as left-hand side and ∴ homogeneous
See experiment on page 45
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 4Waves and our Universe
Solutions to Practice Questions5 Bohr model: electrons in discrete orbits
Each orbit has an associated energy state
Schrödinger model: electrons form stationary waves inside the atom
Different modes (number of ‘loops’) of the stationary wave account for the different energy states
Chapter 221 They are polarised and, therefore, must be transverse
2 (a) Microwaves
(b) Infra-red
(c) Ultra-violet
(d) X-rays
Wavelength decreases from left-to-right
(a) 3 cm ≡ microwaves
(b) 600 nm ≡ visible
3 Using c = f λ where c = 3.00 × 108 m s–1
(a) λ = c/f = 3.00 × 108 m s–1/(200 × 103 Hz) = 1500 m
(b) f = c/λ = 3.00 × 108 m s–1/(3.24 m) = 9.26 × 107 Hz = 92.6 MHz
(c) λ = c/f = 3.00 × 108 m s–1/(0.516 × 109 Hz) = 0.581 m
4 E = hf = 6.6 × 10–34 J s × 909 × 103 Hz = 6.0 × 10–28 J
Number of photons leaving each second = power/E = 12 × 103 W/(6.0 × 10–28 J) = 2.0 × 1031 s–1
Photons spread out over an area 4πr2 = 4π × (20 × 103 m)2 = 5.0 × 109 m2
‘Density’ of photons reaching aerial each second = 2.0 × 1031 s–1/(5.0 × 109 m2) = 4.0 × 1021 s–1 m–2
Number incident on aerial each second = 4.0 × 1021 s–1 m–2 × 0.01 m2 = 4.0 × 1019 s–1
5 λmin = c/fmax = 3 × 108 m s–1/(1021 Hz) = 3 × 10–13 m
λmax = c/fmin = 3 × 108 m s–1/(1017 Hz) = 3 × 10–9 m
Kinetic energy = 1
2mv2 =
1
2 × 9.1 × 10–31 kg × (2.3 × 107 m s–1)2 = 2.41 × 10–16 J
hf = 2.41 × 10–16 J
f = 2.41 × 10–16 J/(6.6 × 10–34 J s) = 3.65 × 1017 Hz
Chapter 231 Light displays a wave nature when diffracting and forming superposition patterns, but a particle
nature when behaving as photons (packets of energy) and releasing photoelectrons
Electrons have a set charge and mass and experience forces and accelerations like all other particles,
but high-speed electrons display a wave nature when diffracting through the atomic planes of graphite
2 When accelerated through 1500 V, an electron gains 1500 eV of energy
1500 eV = 1500 eV × 1.6 × 10–19 J eV–1 = 2.4 × 10–16 J
1
2mv2 = 2.4 × 10–16 J
v 2 = 2 × 2.4 × 10–16 J/(9.1 × 10–31 kg) = 5.27 × 1014 m2 s–2
v = √(5.27 × 1014 m2 s–2) = 2.3 × 107 m s–1
Momentum p = mv = 9.1 × 10–31 kg × 2.3 × 107 m s–1 = 2.1 × 10–23 kg m s–1
de Broglie wavelength λ = h/p = 6.6 × 10–34 J s/(2.1 × 10–23 kg m s–1) = 3.2 × 10–11 m
3 nλ = (sin θ)/N (See question 21.4)
Spacing = 1/N = nλ/sin θ = 1 × 3.2 × 10–11 m/(sin 12°) = 1.5 × 10–10 m
Spacing achieved using an atomic grating (as in graphite experiment on page 49)
4 Sketch showing a number of concentric rings around a central spot
Constructive interference where rings and circle appear
Destructive interference in spaces between rings
With a larger accelerating voltage, the electrons will reach a greater speed v
These faster electrons will have a greater momentum p (= mv) and a shorter de Broglie wavelength
λ (= h/p) so therefore diffract less
The diameter of the rings on the screen decreases; the ring pattern closes up
5 Using λ = h/p where p = mass × velocity
(a) p = 60 kg × 2 m s–1 = 120 kg m s–1
λ = h/p = 6.6 × 10–34 J s/(120 kg m s–1) = 5.5 × 10–36 m
(b) p = 7 kg × 3 × 10–3 m s–1 = 2.1 × 10–2 kg m s–1
λ = h/p = 6.6 × 10–34 J s/(2.1 × 10–2 kg m s–1) = 3.1 × 10–32 m
(c) p = 9.1 × 10–31 kg × 6 × 107 m s–1 = 5.5 × 10–23 kg m s–1
λ = h/p = 6.6 × 10–34 J s/(5.5 × 10–23 kg m s–1) = 1.2 × 10–11 m
(d) p = 1 × 10–15 kg × 0.3 × 10–6 m s–1 = 3 × 10–22 kg m s–1
λ = h/p = 6.6 × 10–34 J s/(3 × 10–22 kg m s–1) = 2.2 × 10–12 m
Electron’s wavelength is similar to the lattice spacing while both student wavelengths are very much
shorter than the width of the doorway
Bacterium wavelength much shorter than length so no diffraction expected
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 4Waves and our Universe
Solutions to Practice Questions
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 4Waves and our Universe
Solutions to Practice QuestionsChapter 241 When an electron falls to a lower energy level in an atom, a photon is released
Falls between different levels release photons with different frequencies
This series of discrete frequencies is an emission spectrum
An electron will move up to a higher energy level when its atom absorbs a photon of the
required energy
Promotions between different levels absorb photons with different frequencies (energies)
When white light passes through a gas, certain photons are absorbed and these are missing from
the transmitted light
This series of discrete missing frequencies is an absorption spectrum
For a given element, the frequencies missing from its absorption spectrum correspond to those
present in its emission spectrum
2 Doppler effect: the apparent change of frequency of a wave observed when there is relative motion
between the observer and the wave’s source
Red shift of light from galaxies
The apparent decrease in the light’s frequency shows that these galaxies are moving away from us
As an ambulance approaches, its siren’s frequency appears higher than it actually is as it moves
away, the frequency appears lower
3 ∆ λ/λ = v/c
∆ λ = vλ/c = 1.2 × 107 m s–1 × 410.2 nm/(3.0 × 108 m s–1) = 16.4 nm
Since moving away, wavelength will appear to be longer
Apparent wavelength = 410.2 nm + 16.4 nm = 426.6 nm
4 v = HD
D = v/H = 1.2 × 107 m s–1/(1.7 × 10–18 s–1) = 7.1 × 1024 m
Time = distance/speed
t = 7.1 × 1024 m/(1.2 × 107 m s–1) = 4.2 × 1017 s = 1.3 × 1010 years
Value of H is uncertain
Calculation of t assumes that the speed of the galaxy has not changed
5 A light year is the distance travelled by light in one year
Distance = speed × time
Light year = 3 × 108 m s–1 × (365.25 × 24 × 3600) s = 9.47 × 1015 m
Chapter 251 The universe is currently expanding
Gravitational forces oppose and slow down this expansion
Whether or not the expansion can be halted and reversed depends on the average mass-energy density
If greater than about 10–26 kg m–3, the universe will end up collapsing back in on itself
If less, the universe will continue to expand
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 4Waves and our Universe
Solutions to Practice Questions2 Left-hand side ρo in kg m–3
Right-hand side 3/8 π has no units
H in s–1
H2 in (s–1)2 = s–2
G in N m2 kg–2 = kg m s–2 m2 kg–2 = kg–1 m3 s–2
1/G in (kg–1 m3 s–2)–1 = kg m–3 s2
so H2/G in s–2 kg m–3 s2 = kg m–3 = same as left-hand side and ∴ homogeneous
Value of H is uncertain
3 Density = mass/volume
Assuming Earth is a sphere
V = 4πR3/3 = 4 × π × (6400 × 103 m)3/3 = 1.1 × 1021 m3
r = M/V = 6 × 1024 kg/(1.1 × 1021 m3) = 5500 kg m–3
So density of Earth is about 5.5 × 1029 times greater than the critical density of the universe
4 Visible matter is that which we can observe i.e. that in stars and galaxies
Dark matter is all that which is ‘non-visible’ and undetectable, including that of interstellar dust
Dark matter makes up most of the mass of the universe
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 5Fields, Forces and Synthesis
Solutions to Practice QuestionsChapter 11 Mass is the amount of matter in a body
Mass is a scalar quantity
Weight is the gravitational force of attraction that the Earth exerts on the body
Weight is a vector quantity
2
3 Gravitational field: a region where a gravitational force is exerted on a mass
Gravitational field strength: the force exerted by a gravitational field on each kilogram
4 g = F/m = 32 N/(20 kg) = 1.6 N kg–1
F = mg = 70 kg × 1.6 N kg–1 = 112 N
5 The unit of gravitational field strength = N kg–1
N kg–1 = kg m s–2 kg–1 = m s–2 = the unit of acceleration
Acceleration of Moon = 0.0028 m s–2
(Numerically the same as the Earth’s gravitational field strength at the Moon’s orbit)
Chapter 21 Newton’s law of gravitation: the gravitational force between two bodies is directly proportional to
the product of their masses and inversely proportional to the square of their separation
since F = GmM/r2
G = Fr2/(mM)
so units of G = N m2/(kg × kg) = N m2 kg–2
in base units, N m2 kg–2 = kg m s–2 m2 kg–2 = kg–1 m3 s–2
2 Mass of each sphere = 1
2 × 12 kg = 6 kg
Volume = mass/density = 6 kg/(11 400 kg m–3) = 5.26 × 10–4 m3
4πR3/3 = 5.26 × 10–4 m3
Radius R = 3√(5.26 × 10–4 m3 × 3/4π) = 0.05 m
When touching, centre to centre separation r = 2 × 0.05 m = 0.1 m
F = GmM/r2 = 6.67 × 10–11 N m2 kg–2 × 6 kg × 6 kg/(0.1 m)2 = 2.4 × 10–7 N
5 kg
Spring balancepulls mass up
Earth pullsmass down
3 The 75 g mass is attracted by both masses
Using F = GmM/r2
Force towards 300 g mass = 6.67 × 10–11 N m2 kg–2 × 0.3 kg × 0.075 kg/(0.2 m)2 = 3.75 × 10–11 N
Force towards 500 g mass = 6.67 × 10–11 N m2 kg–2 × 0.5 kg × 0.075 kg/(0.2 m)2 = 6.25 × 10–11 N
Resultant force = (6.25 – 3.75) × 10–11 N = 2.50 × 10–11 N towards the 500 g mass
4
Mass of Schiehallion = volume × density = 1.6 × 109 m3 × 3000 kg m–3 = 4.8 × 1012 kg
Force of attraction = GmM/r2 = 6.67 × 10–11 N m2 kg–2 × 4.8 × 1012 kg × 2 kg/(2 × 103 m)2 =
1.6 × 10–4 N
Tension in string both supports the weight of the bob and opposes the attraction of the mountain:
Vertical component = weight = 2 kg × 9.8 N kg–1 = 19.6 N
Horizontal component = attraction to mountain = 1.6 × 10–4 N
Angle from vertical θ :
tan θ = opposite/adjacent = 1.6 × 10–4 N/(19.6 N) = 8.17 × 10–6
θ = 4.68 × 10–4 °
5 The force on the bob from the mountain is less (since mountain’s centre of mass actually
further away)
The force is also angled downwards
Both of these result in a smaller horizontal force being exerted on the bob so it deflects through a
smaller angle
Chapter 31
The free fall acceleration of a satellite is also its centripetal acceleration
This keeps the satellite on a circular path about the Earth’s centre
Satellite
Earth pulls satellitedown with a
gravitational force
Gravitationalattraction of
Earth(weight)
Gravitationalattraction ofmountain
Tensionθ
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 5Fields, Forces and Synthesis
Solutions to Practice Questions
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 5Fields, Forces and Synthesis
Solutions to Practice QuestionsTherefore the satellite remains a constant distance above the Earth’s surface
The gravitational force always acts towards the centre of the circle
This is at 90° to the satellite’s instantaneous tangential direction of motion
The force therefore does not do any work
No energy is transferred to or from the satellite
Its kinetic energy and speed remain constant
2
Graph is a straight line passing through the origin so T 2 ∝ r3
3 Comparing T 2 = (4π2/GM) × r3 with the equation of a straight line y = mx + c
shows that a graph of T 2 against r3 will be a straight line through the origin (c = 0)
with a gradient (m) of 4π2/GM
For the same central mass M, the value of 4π2/GM is a constant
so gradient is constant for all bodies orbiting the same central mass
Gradient = 6 × 1019 s2/(1.95 × 1038 m3) = 3.1 × 10–19 s2 m–3
4π2/GM = 3.1 × 10–19 s2 m–3
M = 4π2/(6.67 × 10–11 N m2 kg–2 × 3.1 × 10–19 s2 m–3) = 1.9 × 1030 kg
4 Centripetal acceleration rω 2 = g
ω = √(g/r) = √[9.8 N kg–1/(6400 × 103 m)] = 1.24 × 10–3 rad s–1
T = 2π/ω = 2π/(1.24 × 10–3 rad s–1) = 5080 s = 84.6 min
5 Geostationary satellite: orbits above the equator with a period of 24 h and so maintains the same
position above the Earth’s surface
ω = 2π/ T = 2π/(24 × 60 × 60 s) = 7.3 × 10–5 rad s–1
r3ω2 = GM
so r3 = GM/ω2
Since mass of Earth is 6.0 × 1024 kg
r3 = 6.67 × 10–11 N m2 kg–2 × 6.0 × 1024 kg/(7.3 × 10–5 rad s–1)2
r = 3√(7.57 × 1022 m3) = 4.23 × 107 m
Height above surface = (42.3 – 6.4) × 106 m = 3.6 × 107 m
(Per
iod
of o
rbit)
2 /1019
s2
0
(radius of orbit)3/1038 m3
0.5 2
7
5
4
6
3
2
1 1.5
1
0
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 5Fields, Forces and Synthesis
Solutions to Practice QuestionsChapter 41 g = GM/r2
r2 = GM/g = 6.67 × 10–11 N m2 kg–2 × 6.0 × 1024 kg/(0.0028 N kg–1) = 1.4 × 1017 m2
r = √(1.4 × 1017 m2) = 3.8 × 108 m
2 Inverse square law: when a quantity decreases in proportion to the square of the increasing
distance e.g. if distance ×2 then quantity × (1
2)2 (i.e. ×
1
4), if distance ×3 then quantity × (
1
3)2 (i.e. ×
1
9)
3 Radius of satellite orbit = (9.6 + 6.4) × 106 m = 16 × 106 m
Ratio = 16 × 106 m/(6.4 × 106 m) = 2.5 : 1
Using the inverse square law
Gravitational acceleration of satellite = 9.8 m s–2/(2.52) = 1.57 m s–2
Centripetal acceleration = 1.57 m s–2
a = v 2/r
v = √(ar) = √(1.57 m s–2 × 16 × 106 m) = 5009 m s–1
T = 2πr/v = 2π × 16 × 106 m/(5009 m s–1) = 20 070 s = 5.6 h
4 At the surface
g = GM/R2 = 6.67 × 10–11 N m2 kg–2 × 6.0 × 1024 kg/(6400 × 103)2 = 9.77 m s–2
At a height of 10 km
g = GM/R2= 6.67 × 10–11 N m2 kg–2 × 6.0 × 1024 kg/(6410 × 103)2 = 9.74 m s–2
So, as g only falls by 0.3 % over this height, it can be considered to be uniform
5 Work done = force × distance moved
Force (mg) can be considered to remain constant over 5 km
Work done = mgh = 500 kg × 9.8 m s–2 × 5 × 103 m = 2.45 × 107 J
This value is greater since less force is required at 50 km (as g will be reduced)
so less work is done moving the same distance
Chapter 51 Q = CV
V = Q/C = 40 × 10–9 C/(0.1 × 10–6 F) = 0.4 V
2 Coulomb’s law: the electrostatic force between two charges is directly proportional to the product
of their charges and inversely proportional to the square of their separation
F = kqQ/ r2
k = Fr2/q Q
The unit of k is N m2 C–2
In base units, this is kg m s–2 m2 (A s)–2 = kg m3 s–2 A2 s–2 = kg m3 s–4 A2
3 By removing electrons from it
F = kqQ/r2 = 9.0 × 109 N m2 C–2 × 95 × 10–9 C × 106 × 10–9 C/(0.12 m)2 = 6.3 × 10–3 N
Repulsive force since both positive
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 5Fields, Forces and Synthesis
Solutions to Practice Questions4 F = GmM/r2 = 6.67 × 10–11 N m2 kg–2 × 0.4 kg × 0.4 kg/(12 × 10–2 m)2 = 7.4 × 10–10 N
Electrostatic force is 6.3 × 10–3 N/(7.4 × 10–10 N) = 8.5 × 106 times bigger
5
For vertical equilibrium
Tcos θ = mg = 500 × 10–6 kg × 9.81 N kg–1 = 4.9 × 10–3 N
T = 4.9 × 10–3 N/(cos 7°) = 4.94 × 10–3 N
For horizontal equilibrium
Electrostatic repulsion = Tsin θ = 4.94 × 10–3 × sin 7° = 6.02 × 10–4 N
so kqQ/r 2 = 6.02 × 10–4 N
k = 6.02 × 10–4 N × r2/q Q = 6.02 × 10–4 N × (20 × 10–2 m)2/(50 × 10–9 C × 50 × 10–9 C) =
9.6 × 109 N m2 C–2
1/(4πεair) = 9.6 × 109 N m2 C–2
εair = 1/(4π × 9.6 × 109 N m2 C–2) = 8.3 × 10–12 N–1 m–2 C2 = 8.3 × 10–12 F m–1
Chapter 61 Electric field: a region where there are electric forces on charges
Electric field strength: force exerted by an electric field on each coulomb of charge
Its unit is N C–1
2 In base units, N C–1 = kg m s–2 (A s)–1 = kg m s–2 A–1 s–1 = kg m s–3 A–1
The test charge must cause as little distortion as possible to the electric field it is measuring
therefore it has to be very small
3 E = kQ/r 2 = 9.0 × 109 N m2 C–2 × 35 × 10–9 C/(15 × 10–2 m)2 = 14 000 N C–1
(a) At 30 cm
Double the distance means (1
2)2 =
1
4 × the field strength (inverse square law)
E = 1
4 × 14 000 N C–1 = 3500 N C–1
(b) At 45 cm
Treble the distance means (1
3)2 =
1
9 × the field strength
E = 14 000 N C–1/9 = 1600 N C–1
Electrostaticrepulsion
Tensionθ
Weight
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 5Fields, Forces and Synthesis
Solutions to Practice Questions4 See experiment on page 12
(a) See Figure 6.4 on page 13
(b) Same as Figure 6.4 on page 13 but with arrows reversed (i.e. towards charge)
5 Centripetal force provided by the electrostatic attraction between the electron and the proton
This force has no effect on the speed as it always acts at 90° to the direction of travel and so does
no work on the electron
The orbiting electron is continually changing direction and so its velocity (a vector) is continually
changing
Chapter 71 See Figure 7.2 on page 14
A uniform electric field has a constant electric field strength
Occurs in Figure 7.2 where the field lines are parallel and equally spaced
Exists in the central region between the plates
2 Energy transferred = qV = 500 × 10–12 C × 180 V = 9 × 10–8 J
Work done = average force × distance moved = 9 × 10–8 J
Average force = 9 × 10–8 J/distance moved = 9 × 10–8 J/(9 × 10–2 m) = 1 × 10–6 N
3 Equipotential: a line joining points of equal potential energy
See second experiment on page 14
4 See Figure 7.5 on page15
Field lines are always perpendicular to equipotentials
5 E = V/x = 1 × 103 V/(5 × 10–2 m) = 2 × 104 V m–1
F = qE = 1.6 × 10–19 C × 2 × 104 N C–1 = 3.2 × 10–15 N
a = F/m = 3.2 × 10–15 N/(9.1 × 10–31 kg) = 3.5 × 1015 m s–2
Since the electric field is uniform between the plates force on electron remains constant
so its acceleration remains constant
Chapter 81
These equipotentials are circular
The electric field gets weaker with distance from the source
So the same amount of work will move each coulomb through a greater distance
+Q
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 5Fields, Forces and Synthesis
Solutions to Practice Questions2 (a) Work done = 4.5 × 103 J kg–1 × 8 kg = 3.6 × 104 J
(b) The gravitational potential difference between the two points is the same no matter what path
is taken between them
So work done = 3.6 × 104 J
3 Escape speed: the vertical speed with which one body must be projected from the surface of a
second body in order to escape completely from the gravitational field of the second body
1
2v2 = 3.0 × 106 J kg–1
v = √(2 × 3.0 × 106 J kg –1) = 2450 m s–1
4 Similarities:
point masses and point charges both produce radial fields
the force between masses and that between charges both obey inverse square laws
equipotentials can be used for both to join points with the same energy
Differences:
gravitational fields originate from/affect masses while electric fields originate from/affect charges
gravitational fields only have attractive forces while electric fields have both attractive and
repulsive forces
gravitational fields cannot be shielded while electric fields can
5 The gravitational field in a relatively small region near the surface of a spherical body, where the
field lines are almost parallel, can be considered to be uniform
Energy transferred in a uniform electric field = QEx (= QV )
Chapter 91 A capacitor consists of two metal plates separated from each other by an insulating material
A wire from each plate connects the capacitor to the circuit
2 Electrons are removed from one plate of the capacitor and added to the other without any passing
directly through the insulating material between them
3 See Figure 10.1 on page 22
Start the stopclock as the capacitor is connected to the resistor
Record the current every 5 s until it has fallen to 5% of its starting value
The current will decrease with time as the capacitor discharges and the potential difference across
it decreases
4 When a capacitor is charged, a charge Q is displaced from one plate to the other
The charge on the positive plate is +Q while that on the negative plate is –Q
Total charge on capacitor = +Q + (–Q) = zero
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 5Fields, Forces and Synthesis
Solutions to Practice Questions5 (a) Use two or more engines in series
Would force membrane to bend more and so displace more balls
(b) Use a weaker rubber membrane in the reservoir
Would bend more for same engine and so displace more balls
Chapter 101
The supply voltage and the circuit resistance
Initial charging current I = V/R = 12 V/(50 × 103 Ω) = 2.4 × 10–4 A
2
Charge = area under the current-time graph
Area is approximately 18 squares
1 square = 10 µA × 20 s = 200 µC
18 squares = 18 × 200 µC = 3600 µC
3 Capacitance = charge displaced per unit potential difference across its plates
farad = C V–1
but C = A s
and V = J C–1 = N m (A s)–1 = kg m s–2 m A–1 s–1 = kg m2 s–3 A–1
so farad = A s × kg–1 m–2 s3 A = A2 s4 kg–1 m–2
Cur
rent
/µA
0 120
80
0
70
60
50
40
30
20
10
20 40 60 80 100
Time/s
Cur
rent
Time
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 5Fields, Forces and Synthesis
Solutions to Practice Questions4 E.m.f. = initial current × circuit resistance = 80 × 10–6 A × 50 × 103 W = 4.0 V
So potential difference across capacitor when fully charged = 4.0 V
C = Q/V = 3600 µC/(4.0 V) = 900 µF
5 See Figure 10.6 on page 23
Resistance of variable resistor is reduced to keep the current constant as the capacitor charges
Q = CV = 500 µF × 12 V = 6000 µC
Q = It
t = Q/I = 6000 µC/(40 µA) = 150 s
Chapter 111 Current I = VR/R = 6 V/(50 × 103 Ω) = 1.2 × 10–4 A
When shorting wire removed:
capacitor charges and voltage across it increases
voltage across resistor falls as at all times VR + VC = 6 V
current decreases as VR falls and R is constant
2 VR = IR = 30 × 10–6 A × 50 × 103 Ω = 1.5 V
VC = 6 V × 1.5 V = 4.5 V
Q = CVC = 500 × 10–6 F × 4.5 V = 2.25 × 10–3 C
3 Student has calculated the initial current and assumed this remains constant
However, as capacitor charges, current decreases
So capacitor will take much longer than 100 s to charge
4 To charge to the same potential difference requires the same amount of charge
When in series with a larger resistance, the current is less
Since current is the rate of flow of charge, a lower current takes longer to supply the necessary charge
5 Use a circuit to compare VR with a known voltage
Timer operates as long as VR is greater
As capacitor charges, VR falls below the known voltage and timer stops
Time for VR to fall can be adjusted using a variable resistor
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 5Fields, Forces and Synthesis
Solutions to Practice QuestionsChapter 121
Potential difference across the resistor varies with time in the same way as the current
since I = VR /R ∝ VR
2
3
Initial current same for both (same e.m.f. and circuit resistance)
High-value capacitor will displace more charge in reaching the same potential difference
so area under current-time graph will be greater and capacitor will take longer to charge
Cur
rent
Time
high-value C
low-value C
Pot
entia
l diff
eren
ce
Time
Same graph for bothcapacitor and resistor
Pot
entia
l diff
eren
ce
Time
Capacitor
Resistor
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 5Fields, Forces and Synthesis
Solutions to Practice Questions4 The time constant is the time for a discharging capacitor’s charge and potential difference to
decrease to 1/e of its original value
5 Time constant = RC = 100 × 103 Ω × 250 × 10–6 F = 25 s
After 25 s, current = (1/e) × 90 µA = 33 µA
After 50 s, current = (1/e) × 33 µA = 12 µA
Final potential difference = e.m.f. = initial current × circuit resistance = 90 × 10–6 A × 100 × 103 Ω= 9.0 V
Chapter 131 Q = CV = 2200 × 10–6 F × 5.0 V = 0.011 C = 11 mC
Time constant = RC = 15 × 103 Ω × 2200 × 10–6 F = 33 s
Q = Q0 e–t/RC = 11 mC × e–20 s/(33 s) = 6.0 mC
2 Since mass of isotope ∝ number of atoms of that isotope
m = m0 e–λt
t = 14 years = (14 × 365.25 × 24 × 3600) s = 4.4 × 108 s
so m = 4.5 µg × e –7.8 × 10–10 s–1 × 4.4 × 108 s = 3.2 µg
3 1
2 N0
Substituting
1
2 N0 = N0e–λt1–
2
Cancel N0
1
2 = e –λt1–
2
Invert both sides
2 = eλt1–2
Take natural logs of both sides
ln 2 = λt1–2
as required
4
Time / s 0 20 40 60 80 100 120 140 160 180 200
Activity / Bq 820 530 330 210 145 95 59 35 22 14 9
ln(activity / Bq) 6.71 6.27 5.80 5.35 4.98 4.55 4.08 3.56 3.09 2.64 2.20
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 5Fields, Forces and Synthesis
Solutions to Practice Questions
Gradient = (6.71 – 2.20)/(0 s – 200 s) = – 0.0226 s–1
so λ = +0.0226 s–1
t1–2
= ln 2/λ = ln 2/(0.0226 s–1) = 31 s
5 For capacitor discharge, t1–2
= ln 2 × RC = ln 2 × 33 s = 23 s
Chapter 141 k = F/x = 18 N/(0.045 m) = 36 N/(0.09 m) = 400 N m–1
when F = 25 N
x = F/ k = 25 N/(400 N m–1) = 0.0625 m
Energy stored = 1
2Fx =
1
2 × 25 N × 0.0625 m = 0.78 J
2 See experiment on page 30
3 C = Q/V = 10 × 10–6 C/(50 V) = 2 × 10–7 F
W = 1
2QV =
1
2 × 10 × 10–6 C × 50 V = 2.5 × 10 –4 J
4 Energy stored in spring = 1
2Fx =
1
2mgx
Gravitational potential energy lost by mass = mgx
So only half the energy lost by the mass is stored in the spring the remainder is dissipated as
internal energy to the surroundings
5 Comparing V = Q/C with F = kx
1/C is analogous to k
so large C is analogous to small k
A large value capacitor is analogous to a weak spring
ln(a
ctiv
ity/B
q)
0 250
7
050 100 150 200
Time/s
6
5
4
3
2
1
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 5Fields, Forces and Synthesis
Solutions to Practice QuestionsChapter 151 In parallel
Ct = C1 + C2 = 150 µF + 350 µF = 500 µF
in series
1/Ct = 1/C1 + 1/C2 = 1/(150 µF) + 1/(350 µF) = 0.0095 µF–1
Ct = 1/(0.0095 µF–1) = 105 µF
2
Combined capacitance:
(i) 500 µF (ii) 250 µF (iii) 167 µF (iv) 1000 µF (v) 1500 µF (vi) 750 µF (vii) 333 µF
3 12 V
Q = CV = 330 × 10–6 F × 12 V = 3.96 × 10–3 C
W = 1
2CV 2 =
1
2 × 470 × 10–6 F × (12 V)2 = 0.034 J
4 1/Ct = 1/C1 + 1/C2 = 1/(50 µF) + 1/(150 µF) = 0.027 µF–1
Ct = 1/(0.027 µF–1) = 37.5 µF
Q = CV = 37.5 µF × 9 V = 340 µC displaced on each capacitor
For 50 µF:
V = Q/C = 337.5 µC/(50 µF) = 6.75 V
W = 1
2Q2/C =
1
2 × (337.5 × 10–6 C)2/(50 × 10–6 F) = 1.14 × 10 –3 J
For 150 µF:
V = Q/C = 337.5 µC/(150 µF) = 2.25 V
W = 1
2Q 2/C =
1
2 × (337.5 × 10–6 C)2/(150 × 10–6 F) = 3.80 × 10–4 J
(v)
(vi)
(vii)
(i)
(ii)
(iii)
(iv)
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 5Fields, Forces and Synthesis
Solutions to Practice Questions5 Q = CV = 15 × 10–6 F × 50 V = 7.5 × 10–4 C
W = 1
2CV 2 =
1
2 × 15 × 10–6 F × (50 V)2 = 0.019 J = 19 mJ
(a) in parallel
Ct = C1 + C2 = 15 µF + 25 µF = 40 µF
(b) V = Q/C = 7.5 × 10–4 C/(40 × 10–6 F) = 19 V
(c) For 15 µF:
Q = CV = 15 × 10–6 F × 18.75 V = 2.8 × 10 –4 C
For 25 µF:
Q = CV = 25 × 10–6 F × 18.75 V = 4.7 × 10 –4 C
(d) W = 1
2CV2 =
1
2 × 40 × 10–6 F × (18.75 V)2 = 0.007 J = 7 mJ
When the second capacitor is connected, 12 mJ of energy is dissipated in the connecting wires
as the charge redistributes between the two capacitors
Chapter 161 A magnetic field is a region within which there are magnetic forces
Use either iron filings or a small plotting compass – see experiment on page 36
The direction of a magnetic field is that indicated by the north end of the needle of a plotting compass
2 See Figure 16.3 on page 37
3
4 Neutral point: position within overlapping magnetic fields where the resultant field is zero
NS S N
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 5Fields, Forces and Synthesis
Solutions to Practice Questions5
The X’s indicate the neutral points
The X’s indicate the neutral points
N
S
X X
Magnetic North
X
Magnetic North
S
N
X
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 5Fields, Forces and Synthesis
Solutions to Practice QuestionsChapter 171 See Figure 17.2 on page 38
2 A solenoid is a cylindrical current-carrying coil of wire with a large number of turns
See Figure 17.4 on page 39
3 The shape of the magnetic field outside both a solenoid and a bar magnet are similar
The field lines through the centre of a solenoid are like the aligned domains within a bar magnet
A suitable length of steel is placed inside a solenoid and the current switched on
4 Current is flowing anticlockwise around this end of the coil
So polarity is north – see Figure 17.5 on page 39
5 Coils will attract, as they will have opposite polarities
(a) Coils will repel, as they will now have the same polarity
(b) Coils will continue to attract, as they will still have opposite polarities
Chapter 181 Both produce magnetic fields
A bar magnet is long and thin while a magnadur magnet is short and fat
A bar magnet has its poles at its ends while a magnadur magnet has its poles on its large sides
Place two attracting magnadur magnets on opposite sides of an iron yoke – see Figure 18.2 on
page 40
A uniform magnetic field is produced in the space between them
2 See Figure 18.3 on page 40
On the left of the wire, the two magnetic fields are in the same direction and so a strong field results
On the right, the two fields are opposite and so the field is weak
The field lines get distorted around the left of the wire
The effect of the field lines trying to straighten produces a force on the wire towards the right
3
X
N
S
X = position ofneutral point
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 5Fields, Forces and Synthesis
Solutions to Practice Questions4 (a) Force is towards the top of the page
(b) Magnetic field is down the page
(c) Current is up the page
5 See parts 1 and 2 of experiment on page 41
Plot a graph of force against current
Should be a straight line through the origin – see Figure 18.6 on page 41
Force depends also on the length of the wire and the strength of the magnetic field
Chapter 191 Tesla: unit of magnetic flux density, where 1 T produces a force of 1 N on each metre length of wire
carrying a current of 1 A perpendicular to the magnetic field
T = N A–1 m–1 = kg m s–2 A–1 m–1 = kg s–2 A–1
2 Length of coil = N × πd = 300 × π × 4 × 10–2 m = 38 m
F = BIl = 200 × 10–3 T × 5 × 10–3 A × 37.7 m = 0.038 N
The coil is forced in when the current flows in one direction and forced out when the current flows
in the opposite direction
This results in the coil oscillating in and out
3 Current I = P/ V = 9.0 W/(6.0 V) = 1.5 A
Current flows from left to right (from + to – of battery)
F = BIl = 50 × 10–3 T × 1.5 A × 100 × 10–3 m = 0.0075 N
This force acts upwards (FLHR)
If the magnetic field is reversed, the magnetic force acts downwards
With no magnetic force, balance reading = 1.5094 N + 0.0075 N = 1.5169 N
With downward magnetic force, balance reading = 1.5169 N + 0.0075 N = 1.5244 N
4 Turns density n = N/l = 100/(35 × 10–2 m) = 290 m–1
B = µ0nI = 4π × 10–7 N A–2 × 290 m–1 × 4.2 A = 1.5 × 10–3 T
5 B = µ0 I/2π r
I = 2π rB/µ0 = 2π × 15 × 10–2 m × 6 × 10–6 T/(4π × 10–7 N A–2) = 4.5 A
The current will increase the temperature of the wire (electrical work)
Chapter 201 (a) Closing the switch allows a current to flow in the first coil
This sets up a magnetic flux in the iron core
As this magnetic flux builds up through the second coil, a voltage is induced across it
The microvoltmeter gives a reading
(b) The magnetic flux is steady so there is no relative movement
The microvoltmeter gives no reading
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 5Fields, Forces and Synthesis
Solutions to Practice Questions(c) Opening the switch stops the current from flowing in the first coil
The magnetic flux in the iron core collapses
As this magnetic flux decreases through the second coil, a voltage is induced across it
The microvoltmeter gives a reading in the opposite direction
2 Φ = BA
Φvertical = Bvertical × area of floor = 50 × 10–6 T × (9.2 m × 7.5 m) = 3.45 × 10–3 Wb
Φhorizontal = Bhorizontal × area of north-south aspect = 20 × 10–6 T × (7.5 m × 2.4 m) = 3.6 × 10–4 Wb
3 Area of coil = 4.0 cm2 = 4.0 × 10–4 m2
For each turn, Φ = BA = 450 × 10–3 T × 4.0 × 10–4 m2 = 1.8 × 10–4 Wb
Total magnetic flux linkage = NΦ = 200 × 1.8 × 10–4 Wb = 0.036 Wb
If this magnetic flux linkage reduces to zero in 0.25 s
Change in magnetic flux linkage = – 0.036 Wb
Rate of reduction = (–)0.036 Wb/(0.25 s) = (–)0.14 Wb s–1
E.m.f. induced = 0.14 V
4 Initial magnetic flux linkage = NBA = 80 × 150 × 10–3 T × 40 × 10–4 m2 = 0.048 Wb
(a) Final magnetic flux linkage = 0
Change in magnetic flux linkage = – 0.048 Wb
Rate of reduction = (–)0.048 Wb/(50 × 10–3 s) = (–)0.96 Wb s–1
E.m.f. induced = 0.96 V
(b) Final magnetic flux linkage = NBA = 80 × 240 × 10–3 T × 40 × 10–4 m2 = 0.077 Wb
Change in magnetic flux linkage = 0.077 Wb – 0.048 Wb = +0.029 Wb
Rate of increase = 0.029 Wb/(150 × 10–3 s) = 0.19 Wb s–1
E.m.f. induced = (–)0.19 V
(c) Final magnetic flux linkage = NBA = 80 × –150 × 10–3 T × 40 × 10–4 m2 = –0.048 Wb
Change in magnetic flux linkage = – 0.048 Wb – 0.048 Wb = – 0.096 Wb
Rate of decrease = (–)0.096 Wb/(300 × 10–3 s) = (–)0.32 Wb s–1
E.m.f. induced = 0.32 V
E.m.f. (b) will have the opposite polarity to the other two
as the magnetic flux in (b) is increasing while the other two are decreasing
5 Induced e.m.f. = – Blv = – 50 × 10–6 T × 60 × 10–2 m × 12 m s–1 = (–)3.6 × 10–4 V
No current flows in the handlebar-student circuit since:
The motion of the handlebars through the magnetic field forces the electrons to move in a
certain direction
For a current to flow, electrons would have to flow back through the student in the opposite direction
They would have to flow against the force exerted on them by the student’s motion through the
magnetic field
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 5Fields, Forces and Synthesis
Solutions to Practice QuestionsChapter 211 Faraday’s law: the magnitude of the induced e.m.f. in a circuit is directly proportional to the rate of
change of magnetic flux linkage through that circuit
Lenz’s law: any current driven by an induced e.m.f. opposes the change causing it
Lenz’s law is a consequence of the conservation of energy
2 Student B is correct
As the ring moves through the magnetic field, an e.m.f. is induced across it since the ring is
complete, an induced current flows in it
This induced current interacts with the magnetic field to produce a force that is always opposite to
the direction in which the ring is moving
The motion of the ring is always opposed
3 When the electromagnet is on, the spinning disc cuts through its magnetic field
Induced currents flow in the metal disc, setting up forces that oppose the motion of the disc
The disc, and the vehicle, therefore slow down
When parked, the disc is not moving
No currents are induced in it and so there is no opposition to prevent the vehicle from moving
(Electromagnetic braking is only really effective at high speed when the rate of change is greatest)
4 (a) Induced current produces magnetic field that opposes movement of magnet
Solenoid will have north pole on left and south pole on right
(b) For south pole, current is clockwise (see Figure 17.5 on page 39)
Current flows up on nearside and down on farside
Force on bar magnet is to the left i.e. opposing its motion
If switched off:
E.m.f. still induced across solenoid
No current in solenoid so no magnetic field and no opposing force on magnet
5 (a) No current or magnetic field in solenoid as circuit is incomplete
Flux through ring remains at zero and doesn’t change
(b) Circuit completed so current and magnetic field in solenoid increases
Flux through ring increases
(c) Current and magnetic field in solenoid remain constant
Flux through ring remains constant and doesn’t change
Ring leaps up only when there is a change in the flux through it
Change in flux induces an e.m.f. and a current in the ring (a complete circuit)
Current produces a magnetic field which opposes that of the solenoid
Magnetic fields of ring and solenoid repel
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 5Fields, Forces and Synthesis
Solutions to Practice QuestionsChapter 221 Direct current: a current that flows in one direction only
Alternating current: a current whose direction continually reverses
Graphs B and C represent direct currents (both are always positive)
Graphs A and D represent alternating currents (each has both positive and negative values)
For graph A:
period T = 20 ms
frequency f = 1/T = 1/(20 × 10–3 s) = 50 Hz
For graph D:
period T = 10 ms
frequency f = 1/T = 1/(10 × 10–3 s) = 100 Hz
2 Structure:
two coils, called primary and secondary, of insulated wire wound around a soft iron core
Action:
an alternating current in the primary coil sets up an alternating (changing) magnetic field in the core
This changing magnetic field induces an alternating e.m.f. across the secondary
Vs = NsVp/Np = 700 × 2.5 V/140 = 12.5 V
3 See Figure 22.5 (a) and (b) on page 49
Flux through primary passes through core and secondary coil
Since alternating, flux through secondary is continually changing
E.m.f. induced across secondary as flux through it changes
Maximum e.m.f. when flux changes at its greatest rate
Direction of change determines polarity of e.m.f. at that instant
4 Transformers produce an output voltage when flux through secondary changes
Use a.c. input so that flux produced by current in primary changes continually
Using d.c. would only produce an output when current switched on or off
5 Np/Ns = Vp/Vs = 230 V/(9 V) = 25.6
Ps = VsIs = 9 V × 0.5 A = 4.5 W
Minimum Pp = Ps = 4.5 W
Minimum Ip = Pp/Vp = 4.5 W/(230 V) = 0.02 A
Energy will be dissipated within the transformer as internal energy to both the coils and the core
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 5Fields, Forces and Synthesis
Solutions to Practice QuestionsChapter 231 Electrons are produced by thermionic emission
A low-voltage supply powers a heater which heats the cathode
The cathode’s lattice vibrates more vigorously and throws off some of its electrons
A high-voltage supply puts the anode at a much higher positive potential than the cathode
In the space between the electrodes, electrons experience a resultant force towards the anode
which produces an acceleration in the same direction as the resultant force
2 Particle accelerates towards the negative plate
Positive charge experiences a resultant force in same direction as the electric field, from + to –
Maximum kinetic energy gained = qV = 4.8 × 10–19 C × 400 V = 1.92 × 10–16 J
(assuming space between the plates is vacuum) at the negative plate
3 A charge e passing through a potential difference of 1 V gains 1 eV of energy
A charge e passing through a potential difference of 350 V gains 350 eV of energy
A charge 2e passing through a potential difference of 350 V gains 700 eV of energy
So magnitude of charge = 2 × 1.6 × 10–19 C = 3.2 × 10–19 C
700 eV ≡ 700 eV × 1.6 × 10–19 J eV–1 = 1.12 × 10–16 J
1
2mv 2 = 1.12 × 10–16 J
v = √[2 × 1.12 × 10–16 J/(6.6 × 10–27 kg)] = 1.8 × 105 m s–1
4 Since proton is accelerated through a potential difference of 300 V
Energy gained = 300 eV ≡ 300 eV × 1.6 × 10–19 J eV–1 = 4.8 × 10–17 J
1
2mv2 = 4.8 × 10–17 J
v = √[2 × 4.8 × 10–17 J/(1.7 × 10–27 kg)] = 2.4 × 105 m s–1
Alpha particle has twice the charge of a proton
So alpha particle gains twice the energy (600 eV = 9.6 × 10–17 J)
Alpha particle has four times the mass
1
2 × 4m × v 2 = 9.6 × 10–17 J
v = √[2 × 9.6 × 10–17 J/(4 × 1.7 × 10–27 kg)] = 1.7 × 105 m s–1
Proton gains 1
2 × the energy and √2 × the final speed of the alpha particle
5 As a particle accelerates, its kinetic energy 1
2mv2 increases
As a body approaches the speed of light, its mass increases
There is a significant increase in m and very little increase in v
Maximum possible speed = speed of light = 3 × 108 m s–1
A linear accelerator (linac) consists of a straight evacuated tube containing a series of metal
cylinders
See Figure 23.5 on page 53
The voltage across the gap between adjacent cylinders switches repeatedly from positive to negative
Thus a charged particle always has an accelerating electric field across the gap ahead of it
Total energy gained = sum of the energies gained from each gap
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 5Fields, Forces and Synthesis
Solutions to Practice QuestionsChapter 241 Since I = nAqv
F = BIl = BnAqvl
The total number of moving charged particles in the length l = nAl
Force on a single particle = F/nAl = BnAqvl/nAl = Bqv
2 F = Bqv = 120 × 10–3 T × 3.2 × 10–19 C × 1.8 × 107 m s–1 = 6.9 × 10–13 N
Force acts at 90° to both the magnetic field and the velocity of the alpha particle
3 Force acts at 90° to both the magnetic field and the particle’s velocity so no work is done on the
particle and its speed remains constant
Force provides the required centripetal force for the particle’s circular motion
When moving at 70° to field, particle will follow a ‘forward moving’ circular path, a helix
4 (a) Energy gained = 250 eV ≡ 250 eV × 1.6 × 10–19 J eV–1 = 4.0 × 10–17 J
1
2mv2 = 4.0 × 10–17 J
v = √[2 × 4.0 × 10–17 J/(9.1 × 10–31 kg)] = 9.4 × 106 m s–1
(b) Bqv = mv 2/r
r = mv/Bq = 9.1 × 10–31 kg × 9.4 × 106 m s–1/(0.92 × 10–3 T × 1.6 × 10–19 C) = 0.058 m = 5.8 cm
5 A linac is straight while a cyclotron is circular
A linac uses only electric fields while a cyclotron uses both electric and magnetic fields
Advantage:
for high energies, a linac needs to be very long – see page 51
a cyclotron overcomes this problem by making the particles follow circular paths
Disadvantage:
relies on the high-frequency voltage across the ‘dees’ accelerating all particles whatever their speeds
however this only applies if their masses are all the same
only true if particles are being accelerated to significantly below the speed of light
Bqv = mv2/r
r = mv/Bq ∝ v
If the radius at speed v is r
Time spent in a ‘dee’ at speed v = x/t = πr/v
Since r ∝ v, radius at speed 3v is 3r
Time spent in a ‘dee’ at speed 3v = 3πr/3v = πr/v = time spent in a ‘dee’ at speed v
Since time in a ‘dee’ does not depend on speed, an accelerating voltage of constant frequency can
be used
Chapter 251 A cyclotron uses a constant frequency accelerating voltage
A synchrotron uses an accelerating voltage whose frequency is synchronised to take account of the
increasing mass of the particles as they approach the speed of light
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 5Fields, Forces and Synthesis
Solutions to Practice Questions2 In any collision, momentum has to be conserved
Any particles created from a single beam technique must conserve the momentum of the initial beam
Hence the particles produced must have momentum and kinetic energy, so some of the available
energy has to become kinetic rather than being used to create new particles
Two identical beams travelling in opposite directions have equal and opposite momentum
Total momentum before and after any collision is zero
Hence the particles produced have zero momentum and no kinetic energy, so all the available
energy can be used to create new particles
3 Bubble chamber, spark chamber, drift chamber, G-M tube + counter
All particle detectors use the ionisations produced by the charged particles passing through them
Alpha particles produce bright (thick), straight cloud chamber tracks and if from the same decay, all
alpha tracks have the same length
Fast beta particles produce long, thin, straight cloud chamber tracks
Slow beta particles produce short, thicker, tortuous tracks
4 See Figure 25.6 on page 57
All charged particles moving at 90° to a magnetic field follow circular paths
Alpha particles are positive and produce a current in the same direction as the beam
With magnetic field into the page, FLHR shows that initial force on alpha particles is upwards
Alpha particles are deflected very little as they are relatively massive
Beta-minus particles are negative and produce a current in the opposite direction to the beam
With magnetic field into the page, FLHR shows that initial force on beta-minus particles is
downwards
Deflection is much more than for alpha particles as beta-minus particles are much less massive
Gamma radiation is uncharged and so is not deflected
In a perpendicular electric field:
forces act parallel to electric field (rather than at 90° as in a magnetic field)
paths followed by charged particles are therefore parabolic and not circular
force on alpha particle is always towards the negative plate
force on beta-minus particle is always towards the positive plate
no force acts on the gamma radiation since it is uncharged
5 The curved path results from the presence of a uniform magnetic field at 90° to the particle’s path
The track spirals inwards as the particle slows down, due to ionising collisions
and r = mv/Bq ∝ v
+
–
β–
αγ
NAS Physics Teachers’ Guide © 2005 Nelson Thornes Ltd.
Unit 5Fields, Forces and Synthesis
Solutions to Practice QuestionsChapter 261 ∆E = mc∆T = 400 × 10–3 kg × 880 J kg–1 K–1 × (80 – 15) K = 2.3 × 104 J
∆m = ∆E/c2 = 2.3 × 104 J/(3 × 108 m s–1)2 = 2.5 × 10–13 kg
2 Total mass after decay = 3.419 18 × 10–25 kg + 6.644 32 × 10–27 kg = 3.485 623 2 × 10–25 kg
Mass of energy released = 3.485 72 × 10–25 kg – 3.485 623 2 × 10–25 kg = 9.68 × 10–30 kg
Energy released = c2∆m = (3 × 108 m s–1)2 × 9.68 × 10–30 kg = 8.7 × 10–13 J
3 Unified mass unit: one-twelfth of the mass of a carbon-12 atom
1 u = 1.66 × 10–27 kg
Mass of polonium atom = 3.485 72 × 10–25 kg/(1.66 × 10–27 kg u–1) = 210 u
Mass of lead atom = 3.419 18 × 10–25 kg/(1.66 × 10–27 kg u–1) = 206 u
Mass of alpha particle = 6.644 32 × 10–27 kg/(1.66 × 10–27 kg u–1) = 4 u
4 Nuclear fission:
the splitting up of large nuclides into much smaller nuclides – a process that releases energy
Nuclear fusion:
the joining together of light nuclei – a process that also releases energy
5 Mass before = 235.04 u + 1.01 u = 236.05 u
Mass after = 140.91 u + 91.91 u + (3 × 1.01 u) = 235.85 u
Mass of energy released = 236.05 u – 235.85 u = 0.20 u = 0.20 u × 1.66 × 10–27 kg u–1
= 3.32 × 10–28 kg
Energy released in joules = c2∆m = (3 × 108 m s–1)2 × 3.32 × 10–28 kg = 3.0 × 10–11 J
Energy released in electronvolts = 3.0 × 10–11 J/(1.6 × 10–19 J eV–1) = 1.87 × 108 eV
There are a very large number of atoms N in even a tiny sample of uranium-235
so although each individual fission reaction releases only a small amount of energy (3.0 × 10–11 J)
the total energy released is large (N × 3.0 × 10–11 J)
Precautions:
concrete shielding around the nuclear reactor
careful monitoring of power station and its surroundings
suitable disposal of all nuclear waste material