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# 剑桥国际A&AS水平物理课程第二版详解

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《剑桥国际AS与A Level物理学课程教科书》第二版是来自剑桥大学出版社的一本权威教材，隶属于英国剑桥大学。该教材旨在通过传播高质量的知识，支持教育、学习和研究在全球范围内达到卓越标准。该书由David Sang、Graham Jones、Gurinder Chadha和Richard Woodside共同编著，专为准备参加剑桥国际AS和A Level物理学考试的学生设计。 这本书的内容涵盖了广泛的物理理论、实验、定律以及应用，确保学生在学术上能够系统地掌握基础和进阶的物理知识。作者们精心编排了章节，包括理论讲解、实例分析和习题解答，旨在培养学生的批判性思维和问题解决能力。此外，书中还可能包含实验指导和现代科技相关的章节，以反映物理学的最新进展。 版权信息表明，未经剑桥大学出版社书面许可，除非符合版权许可协议或根据1988年《版权、设计和专利法》第3章的规定（如在教育文集中复制短篇节选或为设定考试题目目的），否则不得非法复制本书的任何部分。对于英国的教师，如果要复制或电子存储书中的内容，必须遵循上述授权条件。 值得注意的是，书中的答案和其他章节末尾的问题是由作者提供的，旨在作为教学参考，而不是直接提供给学生作为解题答案。此外，由于版权保护，出版社无法对引用的外部网站信息进行持久性和准确性保证，因此书中的价格、时间表等信息应在出版时视为准确，但之后可能会有所变化。 总体而言，《剑桥国际AS与A Level物理学课程教科书》第二版是一本全面且严谨的教材，适合学生在学术生涯中深入理解物理学原理，并准备应对高级水平的考试。对于教师和学生来说，它不仅提供了丰富的学习资源，也体现了剑桥大学出版社在教育领域的重要角色和承诺。

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4

Cambridge International AS Level Physics

Distance and displacement,

scalar and vector

In physics, we are o en concerned with the distance

moved by an object in a particular direction. is is called

its displacement. Figure 1.8 illustrates the di erence

between distance and displacement. It shows the route

followed by walkers as they went from town A to town C.

eir winding route took them through town B, so that

they covered a total distance of 15 km. However, their

displacement was much less than this. eir nishing

position was just 10 km from where they started. To give a

complete statement of their displacement, we need to give

both distance and direction:

displacement = 10 km 30° E of N

Displacement is an example of a vector quantity. A

vector quantity has both magnitude (size) and direction.

Distance, on the other hand, is a scalar quantity. Scalar

quantities have magnitude only.

4

A trolley with a 5.0 cm long card passed through

a single light gate. The time recorded by a digital

timer was 0.40 s. What was the average speed of

the trolley in m s

−1

?

5 Figure 1.7 shows two ticker-tapes. Describe the

motion of the trolleys which produced them.

6 Four methods for determining the speed of a

moving trolley have been described. Each could

be adapted to investigate the motion of a falling

mass. Choose two methods which you think

would be suitable, and write a paragraph for each

to say how you would adapt it for this purpose.

start

a

b

Figure 1.7 Two ticker-tapes; for Question 5.

BOX 1.1: Laboratory measurements of speed (continued)

Start by inspecting the tape. This will give you a

description of the trolley’s movement. Identify the start

of the tape. Then look at the spacing of the dots:

even spacing – constant speed

increasing spacing – increasing speed.

Now you can make some measurements. Measure the

distance of every fi h dot from the start of the tape.

This will give you the trolley’s distance at intervals

of 0.1 s. Put the measurements in a table and draw a

distance–time graph.

Measuring speed using a motion sensor

The motion sensor (Figure 1.6) transmits regular pulses

of ultrasound at the trolley. It detects the reflected

waves and determines the time they took for the trip

to the trolley and back. From this, the computer can

deduce the distance to the trolley from the motion

sensor. It can generate a distance–time graph. You can

determine the speed of the trolley from this graph.

Choosing the best method

Each of these methods for finding the speed of a trolley

has its merits. In choosing a method, you might think

about the following points:

Does the method give an average value of speed

or can it be used to give the speed of the trolley at

di erent points along its journey?

How precisely does the method measure time – to the

nearest millisecond?

How simple and convenient is the method to set up in

the laboratory?

motion

sensor

trolley

computer

Figure 1.6 Using a motion sensor to investigate the motion

of a trolley.

QUESTIONS

5

Chapter 1: Kinematics – describing motion

7

Which of these gives speed, velocity, distance or

displacement? (Look back at the definitions of

these quantities.)

a

The ship sailed south-west for 200 miles.

b

I averaged 7 mph during the marathon.

c

The snail crawled at 2 mm s

−1

along the straight

edge of a bench.

d

The sales representative’s round trip was

420 km.

C

B

A

7 km

8 km

10 km

N

S

W E

Figure 1.8 If you go on a long walk, the distance you travel

will be greater than your displacement. In this example, the

walkers travel a distance of 15 km, but their displacement is

only 10 km, because this is the distance from the start to the

finish of their walk.

Speed and velocity calculations

We can write the equation for velocity in symbols:

v =

s

t

v =

Δs

Δt

e word equation for velocity is:

velocity =

change in displacement

time taken

Note that we are using Δs to mean ‘change in displace-

ment s’. e symbol Δ, Greek letter delta, means ‘change

in’. It does not represent a quantity (in the way that s does);

it is simply a convenient way of representing a change in a

quantity. Another way to write Δs would be s

2

− s

1

, but this

is more time-consuming and less clear.

e equation for velocity, v =

Δs

Δt

, can be rearranged

as follows, depending on which quantity we want to

determine:

change in displacement Δs = v × Δt

change in time Δt =

Δs

v

Note that each of these equations is balanced in

terms of units. For example, consider the equation

for displacement. e units on the right-hand side are

m s

−1

× s, which simplies to m, the correct unit for

displacement.

Note also that we can, of course, use the same

equations to nd speed and distance, that is:

v =

d

t

distance d = v × t

time t =

d

v

Speed and velocity

It is oen important to know both the speed of an

object and the direction in which it is moving. Speed

and direction are combined in another quantity, called

velocity. e velocity of an object can be thought of as

its speed in a particular direction. So, like displacement,

velocity is a vector quantity. Speed is the corresponding

scalar quantity, because it does not have a direction. So,

to give the velocity of something, we have to state the

direction in which it is moving. For example, an aircra

ies with a velocity of 300 m s

−1

due north. Since velocity is

a vector quantity, it is dened in terms of displacement:

velocity =

change in displacement

time taken

Alternatively, we can say that velocity is the rate of change

of an object’s displacement. From now on, you need to be

clear about the distinction between velocity and speed,

and between displacement and distance. Table 1.2 shows

the standard symbols and units for these quantities.

Quantity

Symbol for

quantity

Symbol for unit

distance d m

displacement s, x m

time t s

speed, velocity v m s

−1

Table 1.2 Standard symbols and units. (Take care not to

confuse italic s for displacement with s for seconds. Notice

also that v is used for both speed and velocity.)

QUESTION

6

Cambridge International AS Level Physics

Making the most of units

In Worked example 1 and Worked example 2, units have

been omitted in intermediate steps in the calculations.

However, at times it can be helpful to include units as this

can be a way of checking that you have used the correct

equation; for example, that you have not divided one

quantity by another when you should have multiplied

them. e units of an equation must be balanced, just as the

numerical values on each side of the equation must be equal.

If you take care with units, you should be able to carry

out calculations in non-SI units, such as kilometres per

hour, without having to convert to metres and seconds.

For example, how far does a spacecra travelling at

40 000 km h

−1

travel in one day? Since there are 24 hours in

one day, we have:

distance travelled = 40 000 km h

−1

× 24 h

= 960 000 km

1

A car is travelling at 15 m s

−1

. How far will it travel in

1 hour?

Step 1 It is helpful to start by writing down what you

know and what you want to know:

v

= 15 m s

−1

t = 1 h = 3600 s

d = ?

Step 2 Choose the appropriate version of the

equation and substitute in the values. Remember

to include the units:

d = v × t

= 15 × 3600

= 5.4 × 10

4

m

= 54 km

The car will travel 54 km in 1 hour.

2

The Earth orbits the Sun at a distance of

150 000 000 km. How long does it take light from

the Sun to reach the Earth?

(Speed of light in space = 3.0 × 10

8

m s

−1

.)

Step 1 Start by writing what you know. Take care

with units; it is best to work in m and s. You need to

be able to express numbers in scientific notation

(using powers of 10) and to work with these on your

calculator.

v = 3.0 × 10

8

m s

−1

d = 150 000 000 km

= 150 000 000 000 m

= 1.5 × 10

11

m

Step 2 Substitute the values in the equation for

time:

t =

d

v

=

1.5 × 10

11

3.0 × 10

8

= 500 s

Light takes 500 s (about 8.3 minutes) to travel from

the Sun to the Earth.

Hint: When using a calculator, to calculate the time t,

you press the buttons in the following sequence:

[1.5] [EXP] [11] [÷] [3] [EXP] [8]

or

[1.5] [×10

n

] [11] [÷] [3] [×10

n

] [8]

8

A submarine uses sonar to measure the depth of

water below it. Reflected sound waves are detected

0.40 s aer they are transmitted. How deep is the

water? (Speed of sound in water = 1500 m s

−1

.)

9

The Earth takes one year to orbit the Sun at a

distance of 1.5 × 10

11

m. Calculate its speed. Explain

why this is its average speed and not its velocity.

Displacement–time graphs

We can represent the changing position of a moving object

by drawing a displacement–time graph. e gradient

(slope) of the graph is equal to its velocity (Figure 1.9).

e steeper the slope, the greater the velocity. A graph

like this can also tell us if an object is moving forwards or

backwards. If the gradient is negative, the object’s velocity

is negative – it is moving backwards.

Deducing velocity from a displacement–

time graph

A toy car moves along a straight track. Its displacement at

dierent times is shown in Table 1.3. is data can be used

to draw a displacement–time graph from which we can

deduce the car’s velocity.

Displacement / m

1.0 3.0 5.0 7.0 7.0 7.0

Time / s

0.0 1.0 2.0 3.0 4.0 5.0

Table 1.3 Displacement (s) and time (t) data for a toy car.

WORKED EXAMPLES

QUESTIONS

7

Chapter 1: Kinematics – describing motion

It is useful to look at the data rst, to see the pattern

of the car’s movement. In this case, the displacement

increases steadily at rst, but aer 3.0 s it becomes

constant. In other words, initially the car is moving at a

steady velocity, but then it stops.

Now we can plot the displacement–time graph

(Figure 1.11).

We want to work out the velocity of the car over the

rst 3.0 seconds. We can do this by working out the

gradient of the graph, because:

velocity = gradient of displacement−time graph

s

t

0

0

s

t

low v

high v

0

0

s

t

0

0

s

t

0

0 T

s

t

0

0

The straight line shows that the

object’s velocity is constant.

The slope of this graph is 0.

The displacement s is not changing.

Hence the velocity v = 0.

The object is stationary.

The slope of this graph suddenly

becomes negative. The object is

moving back the way it came.

Its velocity v is negative aer time T.

This displacement–time graph is

curved. The slope is changing.

This means that the object’s

velocity is changing – this

is considered in Chapter 2.

The slope shows which object is

moving faster. The steeper

the slope, the greater the velocity.

s

t

0

0

s

t

low v

high v

0

0

s

t

0

0

s

t

0

0 T

s

t

0

0

The straight line shows that the

object’s velocity is constant.

The slope of this graph is 0.

The displacement s is not changing.

Hence the velocity v = 0.

The object is stationary.

The slope of this graph suddenly

becomes negative. The object is

moving back the way it came.

Its velocity v is negative aer time T.

This displacement–time graph is

curved. The slope is changing.

This means that the object’s

velocity is changing – this

is considered in Chapter 2.

The slope shows which object is

moving faster. The steeper

the slope, the greater the velocity.

s

t

0

0

s

t

low v

high v

0

0

s

t

0

0

s

t

0

0 T

s

t

0

0

The straight line shows that the

object’s velocity is constant.

The slope of this graph is 0.

The displacement s is not changing.

Hence the velocity v = 0.

The object is stationary.

The slope of this graph suddenly

becomes negative. The object is

moving back the way it came.

Its velocity v is negative aer time T.

This displacement–time graph is

curved. The slope is changing.

This means that the object’s

velocity is changing – this

is considered in Chapter 2.

The slope shows which object is

moving faster. The steeper

the slope, the greater the velocity.

s

t

0

0

s

t

low v

high v

0

0

s

t

0

0

s

t

0

0 T

s

t

0

0

The straight line shows that the

object’s velocity is constant.

The slope of this graph is 0.

The displacement s is not changing.

Hence the velocity v = 0.

The object is stationary.

The slope of this graph suddenly

becomes negative. The object is

moving back the way it came.

Its velocity v is negative aer time T.

This displacement–time graph is

curved. The slope is changing.

This means that the object’s

velocity is changing – this

is considered in Chapter 2.

The slope shows which object is

moving faster. The steeper

the slope, the greater the velocity.

s

t

0

0

s

t

low v

high v

0

0

s

t

0

0

s

t

0

0 T

s

t

0

0

The straight line shows that the

object’s velocity is constant.

The slope of this graph is 0.

The displacement s is not changing.

Hence the velocity v = 0.

The object is stationary.

The slope of this graph suddenly

becomes negative. The object is

moving back the way it came.

Its velocity v is negative aer time T.

This displacement–time graph is

curved. The slope is changing.

This means that the object’s

velocity is changing – this

is considered in Chapter 2.

The slope shows which object is

moving faster. The steeper

the slope, the greater the velocity.

Figure 1.9 The slope of a displacement–time (s–t) graph tells

us how fast an object is moving.

s

/

m

t

/

s

8

6

4

2

0

1 2 3 4 50

∆s

gradient = velocity

∆t

Figure 1.11 Displacement–time graph for a toy car; data as

shown in Table 1.3.

We draw a right-angled triangle as shown. To nd the

car’s velocity, we divide the change in displacement by the

change in time. ese are given by the two sides of the

triangle labelled Δs and Δt.

velocity v =

change in displacement

time taken

v =

Δs

Δt

v =

(7.0 − 1.0)

(3.0 − 0)

=

6.0

3.0

= 2.0 ms

−1

If you are used to nding the gradient of a graph, you may

be able to reduce the number of steps in this calculation.

8

Cambridge International AS Level Physics

Combining displacements

e walkers shown in Figure 1.12 are crossing dicult

ground. ey navigate from one prominent point to the

next, travelling in a series of straight lines. From the map,

they can work out the distance that they travel and their

displacement from their starting point:

distance travelled = 25 km

(Lay thread along route on map; measure thread against

map scale.)

displacement = 15 km north-east

(Join starting and nishing points with straight line;

measure line against scale.)

A map is a scale drawing. You can nd your displacement

by measuring the map. But how can you calculate your

displacement? You need to use ideas from geometry and

trigonometry. Worked examples 3 and 4 show how.

Displacement / m

0 85 170 255 340

Time / s

0 1.0 2.0 3.0 4.0

Table 1.4 Displacement (s) and time (t) data for

Question 12.

13

An old car travels due south. The distance it

travels at hourly intervals is shown in Table 1.5.

a

Draw a distance–time graph to represent the

car’s journey.

b

From the graph, deduce the car’s speed in

km h

−1

during the first three hours of the

journey.

c

What is the car’s average speed in km h

−1

during the whole journey?

Time / h Distance / km

0 0

1 23

2 46

3 69

4 84

Table 1.5 Data for Question 13.

START

FINISH

1 2 3 4 5 km

river

ridge

bridge

valley

cairn

Figure 1.12 In rough terrain, walkers head straight for a

prominent landmark.

10

The displacement–time sketch graph in Figure

1.10 represents the journey of a bus. What does

the graph tell you about the journey?

11

Sketch a displacement–time graph to show your

motion for the following event. You are walking at

a constant speed across a field aer jumping o a

gate. Suddenly you see a bull and stop. Your friend

says there’s no danger, so you walk on at a reduced

constant speed. The bull bellows, and you run back

to the gate. Explain how each section of the walk

relates to a section of your graph.

12

Table 1.4 shows the displacement of a racing car at

dierent times as it travels along a straight track

during a speed trial.

a

Determine the car’s velocity.

b Draw a displacement–time graph and use it to

find the car’s velocity.

s

t

0

0

Figure 1.10 For Question 10.

QUESTIONS

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