nelson thornes answer to practice problems

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

Acc

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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

plac

emen

t

Time

Velo

city

Time

Acc

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


Time

Dis

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Time

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n

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

6

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4

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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

For vertical equilibrium

T1 + T2 = 100 N + 250 N + 350 N = 700 N

T1 = 700 N – 460 N = 240 N

Chapter 251 Momentum = mass × velocity

Momentum is a vector quantity

2 Momentum = mass × velocity

(a) Momentum of rugby player = 120 kg × 10 m s–1 = 1200 kg m s–1

(b) Momentum of electron = 9 × 10–31 kg × 2 × 107 m s–1 = 1.8 × 10–23 kg m s–1

(c) Momentum of toy train = 1.6 kg × 0.25 m s–1 = 0.4 kg m s–1

3 Change in momentum = final momentum – initial momentum = mv – mu

(a) Change in momentum of car = (800 kg × 30 m s–1) – (800 kg × 5 m s–1) = 24 000 kg m s–1 – 4000 kg m s–1 = 20 000 kg m s–1

(b) Change in momentum of trolley = (0.8 kg × 0.2 m s–1) – (0.8 kg × 0.8 m s–1) = 0.16 kg m s–1 – 0.64 kg m s–1 = – 0.48 kg m s–1

(c) Change in momentum of ball = (0.05 kg × –5 m s–1) – (0.05 kg × 7 m s–1) = – 0.25 kg m s–1 – 0.35 kg m s–1 = – 0.6 kg m s–1

4 The rate of change in momentum of a body is directly proportional to the resultant force acting onit and takes place in the same direction as the resultant force

Force ∝ rate of change in momentum

F ∝ (mv – mu)/t ∝ m(v – u)/t ∝ ma

F = kma where k is the constant of proportionality

In SI units, the newton is defined so that k = 1, so F = ma

0.5 m

T1

A

100 N

250 N

0.4 m0.1 m

B

350 N

T2






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°

2 See Figure 33.3 on page 70

The positive nuclei have comparatively large distances between them so most alpha particles aredeflected very little

The closer the path of an alpha particle comes to a nucleus, the more the alpha particle is deflected

Deflections through angles greater than 90° result from almost head-on collisions

3 Diameter of an atom ≈ 3 × 10–10 m

Diameter of a nucleus ≈ 1 × 10–14 m

∴ The diameter of this atom is about 30 000 times greater than the diameter of this nucleus

Area of circle ∝ diameter2

∴ The head-on area of this atom is about 30 0002 times greater than the head-on area of this nucleus

Percentage taken up by nucleus = 100 × 1/(30 000)2 ≈ 0.000 000 1% (which is why so many alphaparticles miss!)

4 Quarks are the particles from which all sub-atomic particles are made

There are six types of quark called up, down, strange, charm, top, bottom

A proton consists of two up quarks and one down quark

A neutron consists of one up quark and two down quarks

A baryon is a particle made up of three quarks

A meson is a particle made up of a quark and an antiquark

5

Chapter 341 When ionised, an atom releases an electron and becomes a positively charged ion

Air can be ionised either by a flame or by the radiation from a radioactive source

2 See second experiment on page 72

3 21584Po → 211

82Pb + 42α22789Ac → 223

87Fr + 42α

4

Smoke particles in air reduce the number of alpha particles reaching ‘the detector’

fewer ionisations then occur within ‘the detector’

so a lower current flows when smoke is present

A reduction in the current is used to trigger the alarm

5 22890Th → 224

88Ra + 42αTo get to lead-212 from radium-224, 12 nucleons must be removed

as each alpha particle removes 4 nucleons (and beta removes none), 3 more alpha decays are required

the proton number of all isotopes of lead is 82

Chapter 351 A GM tube is more sensitive: it can detect any single ionising event that occurs inside the tube while

an ionisation chamber needs a large number of ionising events to produce a measurable current

A GM tube allows each ionising event to be directly registered on a counter

A GM tube detects radiation once it has entered the tube through its walls (it is poor at detectingalpha radiation that is mostly stopped by the walls) while the radioactive source can be placeddirectly inside an ionisation chamber

2 Place source close to GM tube window

Observe count rate as paper placed between source and GM tube – no change, no alpha

Repeat with aluminium instead of paper – reduction in count rate, beta being emitted

Repeat with lead instead of aluminium – no emissions from source detected, no gamma

Alphasource

Alpha particledetector

Air gap




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

See experiment on page 78

5 t = ln 2/λt = ln 2/λ = 0.69/(7.84 × 10–10 s–1) = 8.80 × 108 s = 8.80 × 108 s/(60× 60 × 24 × 365.25 s y–1)

= 27.9 y

84 years = 3 × 28 years = 3t

Mass of strontium isotope will have fallen to (12

)3 = 1/8

Mass = 4.5 mg × 1/8 = 0.56 mg

12

12

12

nelson thornes answer to practice problems

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