Practice Questions Answers

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



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)



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



Solutions to Practice Questions



Solutions to Practice Questions4

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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




Time

Dis

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emen

t

Time

Velo

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Acc

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4 Centre of the track (since both positive and negative displacements)


Time

Dis

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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





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



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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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



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




•••• 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



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




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



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




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)






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)



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




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




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

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1

0

gradient=Wx=1.04N m

force fromtop hinge

W

F







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



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


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






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



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



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



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



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



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°


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




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




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



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


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


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!



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






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




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




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


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






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



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)



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



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


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



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



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



Solutions to Practice QuestionsChapter 161 The current through a component is directly proportional to the voltage across it


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


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

+






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


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


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






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



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



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


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



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



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)]



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




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



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



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


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)


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



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



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




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


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




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



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



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



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



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



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



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




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



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



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)



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


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



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



Solutions to Practice QuestionsChapter 181 Photoelectric emission: emission of electrons from a surface when illuminated with electromagnetic

radiation of sufficient frequency


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


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




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



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




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






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



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


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






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



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



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


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



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



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


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



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



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


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



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



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




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


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



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)



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


3

4 Neutral point: position within overlapping magnetic fields where the resultant field is zero

NS S N




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



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


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


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



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



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


(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. (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



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



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



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


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



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



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


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



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

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