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PEARSON EDEXGEL INTERNATIONAL AS / A LEVEL
皮尔森爱德思格尔国际A/A级考试

PHYSICS STUDENT BOOK 1 物理学生用书 1

MILES HUDSON 米尔斯-胡森

PDF COMPILED BY SAAD pdf 由 saad 编制

PEARSON EDEXCEL INTERNATIONAL AS/A LEVEL PHYSICS
皮尔森爱德思国际A/A级物理

Student Book 1 学生用书 1
Miles Hudson 迈尔斯-哈德森
Published by Pearson Education Limited, 80 Strand, London, WC2R ORL.
培生教育有限公司出版,80 Strand, London, WC2R ORL。

www.pearsonglobalschools.com
Copies of official specifications for all Pearson Edexcel qualifications may be found on the website: https://qualifications.pearson.com
所有 Pearson Edexcel 资格证书的官方规格副本均可在网站上找到:https://qualifications.pearson.com
Text © Pearson Education Limited 2018
文本 © 培生教育有限公司 2018

Designed by Tech-Set Ltd, Gateshead, UK
由英国盖茨黑德 Tech-Set 有限公司设计

Edited by Kate Blackham and Jane Read
编辑:凯特-布莱克姆和简-雷德

Typeset by Tech-Set Ltd, Gateshead, UK
由英国盖茨黑德 Tech-Set 有限公司排版

Original illustrations © Pearson Education Limited 2018
原创插图 © 培生教育有限公司 2018

Cover design by Pearson Education Limited 2018
封面设计:培生教育有限公司 2018

Picture research by Aptara, Inc
Aptara 公司的图片研究

Cover photo © RUSSELL CROMAN/SCIENCE PHOTO LIBRARY
封面照片 © RUSSELL CROMAN/SCIENCE PHOTO LIBRARY

Inside front cover photo: Dmitry Lobanov
封面内页照片:德米特里-洛巴诺夫

The right of Miles Hudson to be identified as author of this work has been asserted by him in accordance with the Copyright, Designs and Patents Act 1988.
迈尔斯-赫德森已根据 1988 年《版权、外观设计和专利法》申明了其作为该作品作者的权利。
First published 2018 2018 年首次出版
21201918
10987654321
British Library Cataloguing in Publication Data
大英图书馆出版数据编目

A catalogue record for this book is available from the British Library
大英图书馆可提供此书的目录记录

ISBN 9781292244877
All rights reserved. No part of this publication may be reproduced in any form or by any means (including photocopying or storing it in any medium by electronic means and whether or not transiently or incidentally to some other use of this publication) without the written permission of the copyright owner, except in accordance with the provisions of the Copyright, Designs and Patents Act 1988 or under the terms of a licence issued by the Copyright Licensing Agency, Barnard’s Inn, 86 Fetter Lane, London EC4A 1EN (www.cla.co.uk). Applications for the copyright owner’s written permission should be addressed to the publisher.
保留所有权利。未经版权所有者的书面许可,不得以任何形式或通过任何方式复制本出版物的任何部分(包括影印或通过电子手段将其存储在任何介质中,无论是否短暂复制或附带于本出版物的其他用途),除非符合《1988 年版权、设计和专利法》的规定,或根据版权许可机构(Copyright Licensing Agency, Barnard's Inn, 86 Fetter Lane, London EC4A 1EN)颁发的许可条款(www.cla.co.uk)。版权所有者的书面许可申请应向出版商提出。
Printed by Neografia in Slovakia
由斯洛伐克 Neografia 印刷

Endorsement statement 认可声明

In order to ensure that this resource offers high-quality support for the associated Pearson qualification, it has been through a review process by the awarding body. This process confirmed that this resource fully covers the teaching and learning content of the specification at which it is aimed. It also confirms that it demonstrates an appropriate balance between the development of subject skills, knowledge and understanding, in addition to preparation for assessment.
为确保本资料能为相关培生资格证书提供高质量的支持,本资料已通过授证机构的审查程序。审查结果表明,本教材完全涵盖了其所针对的规范的教学内容。它还证实,除了为评估做准备外,它还在学科技能、知识和理解能力的发展之间实现了适当的平衡。
Endorsement does not cover any guidance on assessment activities or processes (e.g. practice questions or advice on how to answer assessment questions) included in the resource, nor does it prescribe any particular approach to the teaching or delivery of a related course.
认可不包括资源中包含的任何有关评估活动或过程的指导(如练习题或有关如何回答评估问题的建议),也不对相关课程的教学或实施规定任何特定的方法。
While the publishers have made every attempt to ensure that advice on the qualification and its assessment is accurate, the official specification and associated assessment guidance materials are the only authoritative source of information and should always be referred to for definitive guidance.
尽管出版商已尽力确保有关资格及其评估的建议准确无误,但官方规范和相关评估指导材料是唯一权威的信息来源,应始终参考以获得明确的指导。
Pearson examiners have not contributed to any sections in this resource relevant to examination papers for which they have responsibility.
培生主考官没有参与本资料中与其负责的试卷相关的任何部分。
Examiners will not use endorsed resources as a source of material for any assessment set by Pearson. Endorsement of a resource does not mean that the resource is required to achieve this Pearson qualification, nor does it mean that it is the only suitable material available to support the qualification, and any resource lists produced by the awarding body shall include this and other appropriate resources.
考官不得将认可的资源作为培生设置的任何评估的材料来源。对某一资源的认可并不意味着该资源是获得该培生资格证书的必备资源,也不意味着该资源是支持该资格证书的唯一合适材料。

Acknowledgements 致谢

Logos 徽标

Logo on page 168 from LIFEPAK® 1000 DEFIBRILLATOR brochure, http://www. physio-control.com/uploadedFiles/Physio85/Contents/Emergency_Medical_Care/ Products/Brochures/LP1000_Brochure 20w 20Rechargable 20Battery_3303851_C. pdf, copyright © 2012 Physio-Control, Inc.
LIFEPAK® 1000 除颤器手册第 168 页上的徽标,http://wwwphysio-control.com/uploadedFiles/Physio85/Contents/Emergency_Medical_Care/ Products/Brochures/LP1000_Brochure 20w 20Rechargable 20Battery_3303851_C. pdf, copyright © 2012 Physio-Control, Inc.

Tables 表格

Table on page 84 from Mountaineering: The Freedom of the Hills, 6th Edition, published by Mountaineers Books, Seattle and Quiller Publishing, Shrewsbury 148, copyright © 1997. Illustration and table reprinted with permission of the publisher; Table on page 84 adapted from ‘The 2013/14 catalogue of DMM International’, www.dmmwales.com. Reproduced with permission from DMM International Ltd; Table on page 144 from ‘Electrical Power Annual’ report, December 2013, The United States Energy Information Association Source: U. S. Energy Information Administration (Dec 2013).
登山运动》第 84 页的表格:第 6 版,由西雅图登山者图书公司和什鲁斯伯里 148 号 Quiller 出版社出版,版权所有 © 1997 年。插图和表格经出版商许可转载;第 84 页上的表格改编自 "DMM 国际公司 2013/14 年目录",www.dmmwales.com。经 DMM 国际有限公司许可转载;第 144 页的表格摘自 "电力年报",2013 年 12 月,美国能源信息协会:美国能源信息署(2013 年 12 月)。

Text 文本

Extract on page 54 from manufacture hockey goalkeeping equipment, 8 July 2014, http://www.obo.co.nz/the-o-lab, OBO. Reproduced by permission; Extract on page 70 from ‘What is a Plimsoll line?’, http://oceanservice.noaa.gov/facts/plimsoll-line. html. Source: NOAA’s National Ocean Service; Extract on page 96 from USGC FAQs, http://www.usgs.gov/faq. Source: U.S. Geological Survey Department of the Interior/USGS; Extract on page 114 adapted from ‘Tuning the Marimba Bar and Resonator’ http://lafavre.us/tuning-marimba, copyright © 2007 Jeffrey La Favre; Extract on page 126 from ‘Forensic Glass Comparison: Background Information Used in Data Interpretation’, publication number 09-04 of the Laboratory Division of the Federal Bureau of Investigation, Vol.11, No 2 (Maureen Bottrell),April 2009, Source: Federal Bureau of Investigation, Quantico, Virginia, USA; Extract on page 168 from LIFEPAK® 1000 DEFIBRILLATOR brochure, pp. 56-57, http://www. physiocontrol.com /uploadedFiles/Physio85/Contents/Emergency_Medical_ Care/Products/Brochures/LP1000_Brochure%20w%20Rechargable%20 Battery_3303851_C.pdf, copyright © 2012 Physio-Control, Inc. ; Extract on page 190 from ‘Rover Team Working to Diagnose Electrical Issue’, 2 November 2013, http://www.jpl.nasa.gov/news/news.php?feature=3958, Source: NASA.
第 54 页摘自制造曲棍球守门员设备,2014 年 7 月 8 日,http://www.obo.co.nz/the-o-lab,OBO。经许可转载;摘自 "什么是 Plimsoll 线?",http://oceanservice.noaa.gov/facts/plimsoll-line. html,第 70 页。资料来源:美国国家海洋和大气管理局(NOAANOAA's National Ocean Service;摘自 USGC FAQs 第 96 页,http://www.usgs.gov/faq。来源:美国地质调查局美国地质调查局内政部/美国地质调查局;第 114 页摘录自 "马林巴琴琴杆和谐振器的调音"http://lafavre.us/tuning-marimba,版权 © 2007 年 Jeffrey La Favre;第 126 页摘录自 "法医玻璃比较:数据解释中使用的背景信息",联邦调查局实验室部门第 09-04 号出版物,第 11 卷,第 2 期(Maureen Bottrell),2009 年 4 月,资料来源:联邦调查局,匡提科:美国弗吉尼亚州匡提科联邦调查局;摘自 LIFEPAK® 1000 去纤颤器手册第 168 页,第 56-57 页,http://wwwphysiocontrol.com /uploadedFiles/Physio85/Contents/Emergency_Medical_ Care/Products/Brochures/LP1000_Brochure%20w%20Rechargable%20 Battery_3303851_C.pdf, copyright © 2012 Physio-Control, Inc.; 摘自《漫游者团队努力诊断电气问题》第 190 页,2013 年 11 月 2 日,http://www.jpl.nasa.gov/news/news.php?feature=3958,资料来源:美国国家航空航天局。
Every effort has been made to contact copyright holders of material reproduced in this book. Any omissions will be rectified in subsequent printings if notice is given to the publishers.
我们已尽力与本书所转载资料的版权持有者取得联系。如果出版商收到通知,任何遗漏都将在后续版本中得到纠正。
For Photo and Figure Acknowedgements please see page 214
照片和图表致谢请参见第 214 页

COURSE STRUCTURE … iv 课程结构 ... iv
ABOUT THIS BOOK … vi
关于本书 ... vi

PRACTICAL SKILLS … viii 实践技能 ... viii
ASSESSMENT OVERVIEW … x x x\mathbf{x}
评估概述 ... x x x\mathbf{x}

WORKING AS A PHYSICIST … 2
担任物理学家...... 2

TOPIC 1 … 8 主题 1 ... 8
TOPIC 2 … 58 主题 2 ... 58
TOPIC 3 … 88 主题 3 ... 88
TOPIC 4 … 148 专题 4 ... 148
MATHS SKILLS … 194 数学技能 ... 194
PREPARING FOR YOUR EXAMS … 200
准备考试...... 200

SAMPLE EXAM ANSWERS … 202
考试答案示例 ... 202

COMMAND WORDS … 205 命令词...... 205
GLOSSARY … 207 术语表 ... 207
INDEX … 211 索引 ... 211

WORKING AS A PHYSICIST 物理学家

1 STANDARD UNITS IN PHYSICS
物理 1 个标准单位

2 ESTIMATION 2 估算

TOPIC 1 MECHANICS 专题 1 力学

1A MOTION 1A 动议
1 VELOCITY AND ACCELERATION
1 速度和加速度

2 MOTION GRAPHS 2 个运动图表
3 ADDING FORCES 15 3 加力 15
4 MOMENTS 17 4 瞬间 17
5 NEWTON’S LAWS 5 牛顿定律
OF MOTION 运动
6 KINEMATICS EQUATIONS 23
6 运动学方程 23

7 RESOLVING VECTORS 26 7 解析向量 26
8 PROJECTILES 28 8 枚射弹 28
THINKING BIGGER: 想得更远
THE BATTLE OF AGRA 30
阿格拉战役 30

EXAM PRACTICE 32 考试练习 32
1B ENERGY 34 1B 能源 34
1 GRAVITATIONAL POTENTIAL AND KINETIC ENERGIES
1 重力势能和动能

36
2 WORK AND POWER 39
2 工作与权力 39

THINKING BIGGER: 想得更远
THE MECHANICS 技术
OF SOCCER 足球
42
EXAM PRACTICE … 44 考试练习 ... 44
1C MOMENTUM 46
1 MOMENTUM 1 力量 48
2 CONSERVATION OF 2 保护
LINEAR MOMENTUM 线动量 50
1C MOMENTUM 46 1 MOMENTUM 48 2 CONSERVATION OF LINEAR MOMENTUM 50| 1C MOMENTUM | 46 | | :--- | :--- | | 1 MOMENTUM | 48 | | 2 CONSERVATION OF | | | LINEAR MOMENTUM | 50 |
THINKING BIGGER: 想得更远
SAVING HOCKEY GOALKEEPERS
拯救曲棍球守门员

EXAM PRACTICE … 56 考试练习 ... 56
TOPIC 2 主题 2
MATERIALS 材料
2A FLUIDS … 58 2a 流体 ... 58
1 FLUIDS, DENSITY AND 1 流体、密度和
UPTHRUST … 60 推力 ... 60
2 FLUID MOVEMENT … 63
2 流体运动 ... 63

3 VISCOSITY … 65 3 粘度 ... 65
4 TERMINAL VELOCITY … 67
4 终点速度 ... 67

THINKING BIGGER: 想得更远
THE PLIMSOLL LINE … 70
普利姆索尔线...... 70

EXAM PRACTICE … 72 考试练习 ... 72
2B SOLID MATERIAL 2b 固体材料
PROPERTIES … 74 属性 ... 74
1 HOOKE’S LAW … 76
1 胡克定律...... 76

2 STRESS, STRAIN AND 2 应力、应变和
THE YOUNG MODULUS … 79
3 STRESS-STRAIN GRAPHS … 81
3 应力应变图 ... 81

THINKING BIGGER 大处着眼
GET ROPED IN … 84
陷入困境...... 84

EXAM PRACTICE … 86 考试练习 ... 86

TOPIC 3
WAVES AND PARTICLE NATURE OF LIGHT
主题 3 光的波粒性质

3A BASIC WAVES 88 3a 基本波 88
1 WAVE BASICS 90 1 波基本知识 90
2 WAVE TYPES 93 2 种波形 93
THINKING BIGGER: 想得更远
EARTHQUAKE 地震
96
EXAM PRACTICE … 98 考试练习 ... 98
3B THE BEHAVIOUR 3b 行为
OF WAVES … 100 波浪...... 100
1 WAVE PHASE AND 1 个波相和
SUPERPOSITION … 102 叠加...... 102
2 STATIONARY WAVES … 105
2 个静止波 ... 105

3 DIFFRACTION … 108 3 衍射 ... 108
4 WAVE INTERFERENCE … 111
4 波干扰...... 111

THINKING BIGGER 大处着眼
THE MARIMBA … 114 马林巴琴...... 114
EXAM PRACTICE … 116 考试练习 ... 116
3C MORE WAVE 3C 更多波浪
PROPERTIES OF LIGHT … 118
光的特性...... 118

1 REFRACTION … 120 1 次折射 ... 120
2 TOTAL INTERNAL 共 2 个内部
REFLECTION … 122 反思 ... 122
3 POLARISATION … 124 3 极化 ... 124
THINKING BIGGER: 想得更远
GLASS FORENSICS … 126 玻璃取证 ... 126
EXAM PRACTICE … 128 考试练习 ... 128

3D QUANTUM PHYSICS 3D 量子物理学

1 WAVE-PARTICLE 1 波粒
DUALITY 双重性
2 THE PHOTOELECTRIC EFFECT
2 光电效应

3 ELECTRON DIFFRACTION AND INTERFERENCE
3 电子衍射和干涉

4 ATOMIC ELECTRON ENERGIES
4 原子电子能量
THINKING BIGGER: 想得更远
SOLAR CELLS TO POWER 太阳能电池供电
THE USA? 美国?
EXAM PRACTICE 考试练习
130 132 135 138
TOPIC 4 ELECTRIC CIRCUITS
专题 4 电路

4A ELECTRICAL QUANTITIES 4a 电气量

1 ELECTRIC CURRENT 1 电流
2 ELECTRICAL ENERGY TRANSFER
2 电能传输

3 CURRENT AND VOLTAGE RELATIONSHIPS
3 电流和电压关系

156
4 RESISTIVITY … 159 4 电阻率 ... 159
5 CONDUCTION AND 5 传导和
RESISTANCE … 161 阻力...... 161
6 SEMICONDUCTORS … 164 6 半导体...... 164
THINKING BIGGER: 想得更远
SHOCKING STUFF … 168 令人震惊的东西...... 168
EXAM PRACTICE … 170 考试练习 ... 170
4B COMPLETE 4B 已完成
ELECTRICAL CIRCUITS … 172
电路...... 172

1 SERIES AND PARALLEL 1 串联和并联
CIRCUITS … 174 电路...... 174
2 ELECTRICAL CIRCUIT 2 个电路
RULES … 178 规则 ... 178
3 POTENTIAL DIVIDERS … 180
3 个潜在分隔符 ... 180

4 EMF AND INTERNAL 4 EMF 和内部
RESISTANCE … 184 阻力...... 184
5 POWER IN ELECTRIC 5 电力
CIRCUITS … 186 电路...... 186
THINKING BIGGER: 想得更远
CURIOUS VOLTAGE DROP … 190
好奇的电压降 ... 190

EXAM PRACTICE … 192 考试练习 ... 192
MATHS SKILLS … 194 数学技能 ... 194
PREPARING FOR YOUR 准备您的
EXAMS … 200 考试 ... 200
SAMPLE EXAM ANSWERS … 202
考试答案示例 ... 202

COMMAND WORDS … 205 命令词...... 205
GLOSSARY … 207 术语表 ... 207
INDEX … 211 索引 ... 211

ABOUT THIS BOOK 关于本书

This book is written for students following the Pearson Edexcel International Advanced Subsidiary (IAS) Physics specification. This book covers the full IAS course and the first year of the International A Level (IAL) course.
本书是为学习培生爱德思国际高级中等(IAS)物理课程的学生编写的。本书涵盖整个 IAS 课程和国际 A Level (IAL) 课程的第一年。
The book contains full coverage of IAS units (or exam papers) 1 and 2. Each unit in the specification has two topic areas. The topics in this book, and their contents, fully match the specification. You can refer to the Assessment Overview on page X X XX for further information. Students can prepare for the written Practical Skills Paper (unit 3) by using the IAL Physics Lab Book (see page viii of this book).
本书涵盖了国际会计准则单元(或试卷)1 和 2 的全部内容。规范中的每个单元都有两个主题领域。本书中的题目及其内容完全符合考试说明。有关详细信息,请参阅 X X XX 页上的 "评估概述"。学生可以使用IAL物理实验手册(参见本书第viii页)准备实践技能书面试卷(第3单元)。
Each Topic is divided into chapters and sections to break the content down into manageable chunks. Each section features a mix of learning and activities.
每个主题都分为章节,将内容分解成易于管理的小块。每个部分都包含学习和活动。
Learning objectives 学习目标
Each chapter starts with a list of key assessment objectives.
每章开头都列出了主要评估目标。

1 A 1 VELOCITY AND ACGELERATION
1 a 1 速度和加速度
Mowemenn is fundamental to the denctioning of our univesse. Whetber you are rurring to catch a
Mowemenn 是我们学校的根本。如果您想赶上一个
RATE OF MOVEMENT 运动速度
One of the simplest thans we can measure is how tas an abject is moving You can cakculate an arestain distance
我们可以测量的最简单的方法之一就是物体移动的速度。
Howews the calculstion for powed will only tell you how fast an nbigect is moring Often it is also
功率的计算方法只能告诉您晶体的移动速度,但它还能告诉您晶体的移动速度。
OR v = s t v = Δ s M v = s t v = Δ s M quad^(v=(s)/(t))v=(Delta s)/(M)\quad{ }^{v=\frac{s}{t}} \mathrm{v}=\frac{\Delta s}{M}  v = s t v = Δ s M v = s t v = Δ s M quad^(v=(s)/(t))v=(Delta s)/(M)\quad{ }^{v=\frac{s}{t}} \mathrm{v}=\frac{\Delta s}{M}
the messurement is reseredet to os a a colalar quasunity The
将测量值重新设置为柱状准线。

scalar. aistarce is a scalar and depplacement is a vectoc.
aistarce 是一个标量,而 depplacement 是一个向量。

Specification reference The exact specification references covered in the section are provided.
规范参考 本节所涉及的确切规范参考均已提供。

Subject vocabulary 主题词汇
Key terms are highlighted in blue in the text. Clear definitions are provided at the end of each section for easy reference, and are also collated in a glossary at the back of the book.
关键术语在正文中以蓝色标出。每节末尾都提供了清晰的定义,以方便读者参考,同时还在书后的词汇表中进行了整理。

Exam hints 考试提示

Tips on how to answer examstyle questions and guidance for exam preparation. Worked examples also show you how to work through questions and set out calculations.
提示如何回答考试类型的问题,并提供备考指导。工作示例还向您展示了如何处理问题和进行计算。

Checkpoint 检查点

Questions at the end of each section check understanding of the key learning points in each chapter.
每节末尾的问题可检查对每章学习要点的理解程度。

These help you focus your learning and avoid common errors.
这有助于您集中精力学习,避免常见错误。

Did you know? 你知道吗?
Interesting facts help you to remember the key concepts.
有趣的事实有助于您记住关键概念。
Your learning, chapter by chapter, is always put in context:
您所学到的每一章内容,都会与实际情况相结合:
  • Links to other areas of Physics include previous knowledge that is built on in the topic, and future learning that you will cover later in your course.
    与物理学其他领域的联系包括在本专题中建立起来的先前知识,以及您将在课程后期学习的未来知识。
  • A checklist details maths knowledge required. If you need to practise these skills, you can use the Maths Skills reference at the back of the book as a starting point.
    核对表详细列出了所需的数学知识。如果您需要练习这些技能,可以将书后的数学技能参考作为起点。

TOPIC 1 MECHANICS 专题 1 力学

1A MOTION 1A 动议

MathS sKILL FOR THIS CHAPTER
本章数学任务

Unit of messuremem ieg are nemoan n n nn
信息管理系统的单位是 nemoan n n nn

1A THINKING BIGGER 1a 想得更远
THE BATTLE OF AGRA 阿格拉战役

Thinking Bigger 大处着眼

At the end of each topic, there is an opportunity to read and work with real-life research and writing about science. The activities help you to read real-life material that’s relevant to your course, analyse how scientists write, think critically and consider how different aspects of your learning piece together.
在每个主题的最后,都有机会阅读和处理现实生活中的科学研究和写作。这些活动有助于您阅读与课程相关的真实材料,分析科学家是如何写作的,进行批判性思考,并考虑如何将学习的不同方面结合起来。

Skills 技能

These sections will help you develop transferable skills, which are highly valued in further study and the workplace.
这些课程将帮助你培养可迁移的技能,这些技能在继续深造和工作中都非常重要。

Exam Practice 考试练习

Exam-style questions at the end of each chapter are tailored to the Pearson Edexcel specification to allow for practice and development of exam writing technique. They also allow for practice responding to the command words used in the exams (see the command words glossary at the back of this book).
每章末尾的考试题型均根据培生爱德思考试说明量身定制,便于练习和提高考试写作技巧。此外,还可以练习如何应对考试中使用的指令词(参见本书后面的指令词词汇表)。

1A EXAM PRACTICE 1a 考试练习

PRACTICAL SKILLS 实用技能

Practical work is central to the study of physics. The Pearson Edexcel International Advanced Subsidiary (IAS) Physics specification includes eight Core Practicals that link theoretical knowledge and understanding to practical scenarios.
实践是物理学习的核心。培生爱德思国际高级中等教育(IAS)物理课程规范包括八门核心实践课程,将理论知识和理解与实际情景联系起来。
Your knowledge and understanding of practical skills and activities will be assessed in all examination papers for the IAS Level Physics qualification.
IAS 物理水平资格证书的所有试卷都将评估您对实践技能和活动的认识和理解。
  • Papers 1 and 2 will include questions based on practical activities, including novel scenarios.
    试卷 1 和 2 将包括基于实践活动的问题,包括新颖的情景。
  • Paper 3 will test your ability to plan practical work, including risk management and selection of apparatus.
    试卷 3 将测试你计划实际工作的能力,包括风险管理和仪器选择。
In order to develop practical skills, you should carry out a range of practical experiments related to the topics covered in your course. Further suggestions in addition to the Core Practicals are included below.
为了培养实践技能,您应该进行一系列与课程主题相关的实践实验。以下是除核心实践之外的其他建议。
STUDENT BOOK TOPIC 学生用书主题 IAS CORE PRACTICALS IAS 核心实践
 专题 1 机械
TOPIC 1
MECHANICS
TOPIC 1 MECHANICS| TOPIC 1 | | :--- | | MECHANICS |
CP1

测定自由落体的加速度
Determine the acceleration of a freely-
falling object
Determine the acceleration of a freely- falling object| Determine the acceleration of a freely- | | :--- | | falling object |
 专题 2 材料
TOPIC 2
MATERIALS
TOPIC 2 MATERIALS| TOPIC 2 | | :--- | | MATERIALS |
CP2

使用落球法确定液体的粘度
Use a falling-ball method to determine
the viscosity of a liquid
Use a falling-ball method to determine the viscosity of a liquid| Use a falling-ball method to determine | | :--- | | the viscosity of a liquid |
CP3

测定材料的杨氏模量
Determine the Young modulus of
a material
Determine the Young modulus of a material| Determine the Young modulus of | | :--- | | a material |

专题 3 光的波和粒子本质
TOPIC 3
WAVES AND THE
PARTICLE NATURE
OF LIGHT
TOPIC 3 WAVES AND THE PARTICLE NATURE OF LIGHT| TOPIC 3 | | :--- | | WAVES AND THE | | PARTICLE NATURE | | OF LIGHT |
CP4

使用双光束示波器、信号发生器、扬声器和麦克风测定空气中的声速
Determine the speed of sound in air
using a two-beam oscilloscope, signal
generator, speaker and microphone
Determine the speed of sound in air using a two-beam oscilloscope, signal generator, speaker and microphone| Determine the speed of sound in air | | :--- | | using a two-beam oscilloscope, signal | | generator, speaker and microphone |
CP5

研究长度、张力和单位长度质量对振动弦或振动线频率的影响
Investigate the effects of length,
tension and mass per unit length on the
frequency of a vibrating string or wire
Investigate the effects of length, tension and mass per unit length on the frequency of a vibrating string or wire| Investigate the effects of length, | | :--- | | tension and mass per unit length on the | | frequency of a vibrating string or wire |
CP6

利用衍射光栅确定激光或其他光源发出的光的波长
Determine the wavelength of light from
a laser or other light source using a
diffraction grating
Determine the wavelength of light from a laser or other light source using a diffraction grating| Determine the wavelength of light from | | :--- | | a laser or other light source using a | | diffraction grating |
 专题 4 电路
TOPIC 4
ELECTRIC CIRCUITS
TOPIC 4 ELECTRIC CIRCUITS| TOPIC 4 | | :--- | | ELECTRIC CIRCUITS |
CP7

测定材料的电阻率
Determine the electrical resistivity of
a material
Determine the electrical resistivity of a material| Determine the electrical resistivity of | | :--- | | a material |
CP8

确定电池的电导率和内阻
Determine the e.m.f. and internal
resistance of an electrical cell
Determine the e.m.f. and internal resistance of an electrical cell| Determine the e.m.f. and internal | | :--- | | resistance of an electrical cell |
STUDENT BOOK TOPIC IAS CORE PRACTICALS "TOPIC 1 MECHANICS" CP1 "Determine the acceleration of a freely- falling object" "TOPIC 2 MATERIALS" CP2 "Use a falling-ball method to determine the viscosity of a liquid" CP3 "Determine the Young modulus of a material" "TOPIC 3 WAVES AND THE PARTICLE NATURE OF LIGHT" CP4 "Determine the speed of sound in air using a two-beam oscilloscope, signal generator, speaker and microphone" CP5 "Investigate the effects of length, tension and mass per unit length on the frequency of a vibrating string or wire" CP6 "Determine the wavelength of light from a laser or other light source using a diffraction grating" "TOPIC 4 ELECTRIC CIRCUITS" CP7 "Determine the electrical resistivity of a material" CP8 "Determine the e.m.f. and internal resistance of an electrical cell"| STUDENT BOOK TOPIC | IAS CORE PRACTICALS | | | :---: | :---: | :---: | | TOPIC 1 <br> MECHANICS | CP1 | Determine the acceleration of a freely- <br> falling object | | TOPIC 2 <br> MATERIALS | CP2 | Use a falling-ball method to determine <br> the viscosity of a liquid | | | CP3 | Determine the Young modulus of <br> a material | | TOPIC 3 <br> WAVES AND THE <br> PARTICLE NATURE <br> OF LIGHT | CP4 | Determine the speed of sound in air <br> using a two-beam oscilloscope, signal <br> generator, speaker and microphone | | | CP5 | Investigate the effects of length, <br> tension and mass per unit length on the <br> frequency of a vibrating string or wire | | | CP6 | Determine the wavelength of light from <br> a laser or other light source using a <br> diffraction grating | | TOPIC 4 <br> ELECTRIC CIRCUITS | CP7 | Determine the electrical resistivity of <br> a material | | | CP8 | Determine the e.m.f. and internal <br> resistance of an electrical cell |

UNIT 1 (TOPICS 1 AND 2) MECHANICS AND MATERIALS
第 1 单元(课题 1 和 2)力学与材料

Possible further practicals include:
可能的进一步实践包括
  • Strobe photography or the use of a video camera to analyse projectile motion
    频闪摄影或使用摄像机分析射弹运动
  • Determine the centre of gravity of an irregular rod
    确定不规则杆的重心
  • Investigate the conservation of momentum using light gates and air track
    利用光门和空气轨道研究动量守恒问题
  • Hooke’s law and the Young modulus experiments for a variety of materials
    各种材料的胡克定律和杨氏模量实验

UNIT 2 (TOPICS 3 AND 4)
第 2 单元(专题 3 和 4)

WAVES AND ELECTRICITY 电波
Possible further practicals include:
可能的进一步实践包括
  • Estimating power output of an electric motor
    估算电动机的输出功率
  • Using a digital voltmeter to investigate the output of a potential divider and investigating current/ voltage graphs for a filament bulb, thermistor and diode
    使用数字电压表调查电位分压器的输出,调查灯泡、热敏电阻和二极管的电流/电压图
  • Determining the refractive index of solids and liquids, demonstrating progressive and stationary waves on a slinky
    测定固体和液体的折射率,演示滑条上的渐变波和静止波

In the Student Book, the Core Practical specification and Lab Book references are supplied in the relevant sections.
在学生用书的相关章节中,提供了核心实践规范和实验用书参考资料。

Practical Skills 实用技能

Practical skills boxes explain techniques used in the Core Practicals, and also detail useful skills and knowledge gained in other related investigations.
实践技能框解释了核心实践中使用的技术,还详细介绍了在其他相关调查中获得的有用技能和知识。
This Student Book is accompanied by a Lab Book, which includes instructions and writing frames for the Core Practicals for students to record their results and reflect on their work.
本学生用书附有一本实验用书,其中包括核心实践活动的说明和书写框架,供学生记录结果和反思自己的工作。

Practical skills checklists, practice questions and answers are also provided.
还提供了实用技能检查表、练习题和答案。

The Lab Book records can be used as preparation and revision for the Practical Skills Paper.
实验手册记录可用于实践技能试卷的准备和复习。

ASSESSMENT OVERVIEW 评估概述

The following tables give an overview of the assessment for Pearson Edexcel International Advanced Subsidiary course in Physics. You should study this information closely to help ensure that you are fully prepared for this course and know exactly what to expect in each part of the examination. More information about this qualification, and about the question types in the different papers, can be found on page 200 of this book.
下表概述了培生爱德思国际高级中等物理课程的评估情况。您应仔细研究这些信息,以帮助确保您为本课程做好充分准备,并确切了解考试各部分的要求。有关本资格证书以及不同试卷题型的更多信息,请参阅本书第 200 页。
PAPER / UNIT 1 第 1 单元
 国际航空的发展
PERGENTAGE
OF IAS
PERGENTAGE OF IAS| PERGENTAGE | | :--- | | OF IAS |
 国家
PERGENTAGE
OF IAL
PERGENTAGE OF IAL| PERGENTAGE | | :--- | | OF IAL |
MARK TIME AVAILABILITY

机械与材料书面试卷试卷代码WPH11/01由PearsonEdexcel外部命题和阅卷单级入学
MECHANICS AND MATERIALS
Written exam paper
Paper code
WPH11/01
Externally set and marked by Pearson
Edexcel
Single tier of entry
MECHANICS AND MATERIALS Written exam paper Paper code WPH11/01 Externally set and marked by Pearson Edexcel Single tier of entry| MECHANICS AND MATERIALS | | :--- | | Written exam paper | | Paper code | | WPH11/01 | | Externally set and marked by Pearson | | Edexcel | | Single tier of entry |
40 % 40 % 40%40 \% 20 % 20 % 20%20 \% 80
 1 小时 30 分钟
1 hour
30 minutes
1 hour 30 minutes| 1 hour | | :--- | | 30 minutes |

一月、六月和十月首次评估 :2019 年 1 月
January, June and October
First assessment : January 2019
January, June and October First assessment : January 2019| January, June and October | | :--- | | First assessment : January 2019 |
PAPER / UNIT 2 第 2 单元 PERCENTAGE
 本金百分比
PERCENTAGE
OF IAL
PERCENTAGE OF IAL| PERCENTAGE | | :--- | | OF IAL |
MARK TIME AVAILABILITY

《波浪与电力》书面考试试卷试卷代码WPH12/01 由 PearsonEdexcel 外部命题和阅卷单级考试
WAVES AND ELECTRICITY
Written exam paper
Paper code
WPH12/01
Externally set and marked by Pearson
Edexcel
Single tier of entry
WAVES AND ELECTRICITY Written exam paper Paper code WPH12/01 Externally set and marked by Pearson Edexcel Single tier of entry| WAVES AND ELECTRICITY | | :--- | | Written exam paper | | Paper code | | WPH12/01 | | Externally set and marked by Pearson | | Edexcel | | Single tier of entry |
40 % 40 % 40%40 \% 20 % 20 % 20%20 \% 80
 1 小时 30 分钟
1 hour
30 minutes
1 hour 30 minutes| 1 hour | | :--- | | 30 minutes |

2019 年 1 月、6 月和 10 月首次评估 6 月
January, June and October
First assessment June 2019
January, June and October First assessment June 2019| January, June and October | | :--- | | First assessment June 2019 |
PAPER / UNIT 3 第 3 单元 PERCENTAGE
 本金百分比
PERCENTAGE
OF IAL
PERCENTAGE OF IAL| PERCENTAGE | | :--- | | OF IAL |
MARK TIME AVAILABILITY

物理实践技能 1书面考试试卷代码WPH13/01由 PearsonEdexcel 外部命题和阅卷单级入学
PRACTICAL SKILLS IN PHYSICS 1
Written examination
Paper code
WPH13/01
Externally set and marked by Pearson
Edexcel
Single tier of entry
PRACTICAL SKILLS IN PHYSICS 1 Written examination Paper code WPH13/01 Externally set and marked by Pearson Edexcel Single tier of entry| PRACTICAL SKILLS IN PHYSICS 1 | | :--- | | Written examination | | Paper code | | WPH13/01 | | Externally set and marked by Pearson | | Edexcel | | Single tier of entry |
20 % 20 % 20%20 \% 10 % 10 % 10%10 \% 50
 1 小时 20 分钟
1 hour
20 minutes
1 hour 20 minutes| 1 hour | | :--- | | 20 minutes |

一月、六月和十月首次评估 :2019 年 6 月
January, June and October
First assessment : June 2019
January, June and October First assessment : June 2019| January, June and October | | :--- | | First assessment : June 2019 |
PAPER / UNIT 1 "PERGENTAGE OF IAS" "PERGENTAGE OF IAL" MARK TIME AVAILABILITY "MECHANICS AND MATERIALS Written exam paper Paper code WPH11/01 Externally set and marked by Pearson Edexcel Single tier of entry" 40% 20% 80 "1 hour 30 minutes" "January, June and October First assessment : January 2019" PAPER / UNIT 2 PERCENTAGE "PERCENTAGE OF IAL" MARK TIME AVAILABILITY "WAVES AND ELECTRICITY Written exam paper Paper code WPH12/01 Externally set and marked by Pearson Edexcel Single tier of entry" 40% 20% 80 "1 hour 30 minutes" "January, June and October First assessment June 2019" PAPER / UNIT 3 PERCENTAGE "PERCENTAGE OF IAL" MARK TIME AVAILABILITY "PRACTICAL SKILLS IN PHYSICS 1 Written examination Paper code WPH13/01 Externally set and marked by Pearson Edexcel Single tier of entry" 20% 10% 50 "1 hour 20 minutes" "January, June and October First assessment : June 2019"| PAPER / UNIT 1 | PERGENTAGE <br> OF IAS | PERGENTAGE <br> OF IAL | MARK | TIME | AVAILABILITY | | :---: | :---: | :---: | :---: | :---: | :---: | | MECHANICS AND MATERIALS <br> Written exam paper <br> Paper code <br> WPH11/01 <br> Externally set and marked by Pearson <br> Edexcel <br> Single tier of entry | $40 \%$ | $20 \%$ | 80 | 1 hour <br> 30 minutes | January, June and October <br> First assessment : January 2019 | | PAPER / UNIT 2 | PERCENTAGE | PERCENTAGE <br> OF IAL | MARK | TIME | AVAILABILITY | | WAVES AND ELECTRICITY <br> Written exam paper <br> Paper code <br> WPH12/01 <br> Externally set and marked by Pearson <br> Edexcel <br> Single tier of entry | $40 \%$ | $20 \%$ | 80 | 1 hour <br> 30 minutes | January, June and October <br> First assessment June 2019 | | PAPER / UNIT 3 | PERCENTAGE | PERCENTAGE <br> OF IAL | MARK | TIME | AVAILABILITY | | PRACTICAL SKILLS IN PHYSICS 1 <br> Written examination <br> Paper code <br> WPH13/01 <br> Externally set and marked by Pearson <br> Edexcel <br> Single tier of entry | $20 \%$ | $10 \%$ | 50 | 1 hour <br> 20 minutes | January, June and October <br> First assessment : June 2019 |

ASSESSMENT OBJECTIVES AND WEIGHTINGS
评估目标和权重

ASSESSMENT OBJECTIVE 评估目标 DESCRIPTION % IN IAS 国际会计准则百分比 % IN IA2 % IN IAL
A01 Demonstrate knowledge and understanding of science
展示对科学的认识和理解
34 36 34 36 34-3634-36 29 31 29 31 29-3129-31 32 34 32 34 32-3432-34
A02

(b) 分析和评估科学信息,做出判断并得出结论。
(a) Application of knowledge and understanding of science in
familiar and unfamiliar contexts.
(b) Analysis and evaluation of scientific information to make
judgments and reach conclusions.
(a) Application of knowledge and understanding of science in familiar and unfamiliar contexts. (b) Analysis and evaluation of scientific information to make judgments and reach conclusions.| (a) Application of knowledge and understanding of science in | | :--- | | familiar and unfamiliar contexts. | | (b) Analysis and evaluation of scientific information to make | | judgments and reach conclusions. |
34 36 34 36 34-3634-36 33 36 33 36 33-3633-36 34 36 34 36 34-3634-36
A03

科学实验技能,包括分析和评估数据和方法
Experimental skills in science, including analysis and evaluation
of data and methods
Experimental skills in science, including analysis and evaluation of data and methods| Experimental skills in science, including analysis and evaluation | | :--- | | of data and methods |
20 14 16 14 16 14-1614-16 11 14 11 14 11-1411-14
ASSESSMENT OBJECTIVE DESCRIPTION % IN IAS % IN IA2 % IN IAL A01 Demonstrate knowledge and understanding of science 34-36 29-31 32-34 A02 "(a) Application of knowledge and understanding of science in familiar and unfamiliar contexts. (b) Analysis and evaluation of scientific information to make judgments and reach conclusions." 34-36 33-36 34-36 A03 "Experimental skills in science, including analysis and evaluation of data and methods" 20 14-16 11-14| ASSESSMENT OBJECTIVE | DESCRIPTION | % IN IAS | % IN IA2 | % IN IAL | | :---: | :--- | :---: | :---: | :---: | | A01 | Demonstrate knowledge and understanding of science | $34-36$ | $29-31$ | $32-34$ | | A02 | (a) Application of knowledge and understanding of science in <br> familiar and unfamiliar contexts. <br> (b) Analysis and evaluation of scientific information to make <br> judgments and reach conclusions. | $34-36$ | $33-36$ | $34-36$ | | A03 | Experimental skills in science, including analysis and evaluation <br> of data and methods | 20 | $14-16$ | $11-14$ |

RELATIONSHIP OF ASSESSMENT OBJECTIVES TO UNITS
评估目标与单元的关系

UNIT NUMBER 单位编号 ASSESSMENT OBJECTIVE 评估目标
A01 A02 (a) A02 (b) A03
UNIT 1 单元 1 17 18 17 18 17-1817-18 17 18 17 18 17-1817-18 4.5 5.5 4.5 5.5 4.5-5.54.5-5.5 0.0
UNIT 2 单元 2 17 18 17 18 17-1817-18 17 18 17 18 17-1817-18 4.5 5.5 4.5 5.5 4.5-5.54.5-5.5 0.0
UNIT 3 单元 3 0.0 0.0 0.0 20

国际高级子公司合计
TOTAL FOR INTERNATIONAL
ADVANCED SUBSIDIARY
TOTAL FOR INTERNATIONAL ADVANCED SUBSIDIARY| TOTAL FOR INTERNATIONAL | | :--- | | ADVANCED SUBSIDIARY |
34 36 34 36 34-3634-36 34 36 34 36 34-3634-36 9 11 9 11 9-119-11 20
UNIT NUMBER ASSESSMENT OBJECTIVE A01 A02 (a) A02 (b) A03 UNIT 1 17-18 17-18 4.5-5.5 0.0 UNIT 2 17-18 17-18 4.5-5.5 0.0 UNIT 3 0.0 0.0 0.0 20 "TOTAL FOR INTERNATIONAL ADVANCED SUBSIDIARY" 34-36 34-36 9-11 20| UNIT NUMBER | ASSESSMENT OBJECTIVE | | | | | :---: | :---: | :---: | :---: | :---: | | | A01 | A02 (a) | A02 (b) | A03 | | UNIT 1 | $17-18$ | $17-18$ | $4.5-5.5$ | 0.0 | | UNIT 2 | $17-18$ | $17-18$ | $4.5-5.5$ | 0.0 | | UNIT 3 | 0.0 | 0.0 | 0.0 | 20 | | TOTAL FOR INTERNATIONAL <br> ADVANCED SUBSIDIARY | $34-36$ | $34-36$ | $9-11$ | 20 |

WORKING AS A PHYSICIST 物理学家

Throughout your study of physics, you will develop knowledge and understanding of what it means to work scientifically. You will develop confidence in key scientific skills, such as handling and controlling quantities and units and making estimates. You will also learn about the ways in which the scientific community functions and how society as a whole uses scientific ideas.
在整个物理学习过程中,你将掌握并理解科学工作的含义。你将培养对关键科学技能的信心,如处理和控制数量和单位以及进行估算。您还将了解科学界的运作方式以及整个社会如何利用科学思想。

At the end of each chapter in this book, there is a section called Thinking Bigger. These sections are based broadly on the content of the chapter just completed, but they will also draw on your previous learning from earlier in the course or from your previous studies and point towards future learning and less familiar contexts. The Thinking Bigger sections will also help you to develop transferable skills. By working through these sections, you will:
本书每章末尾都有一节名为 "更广阔的思考 "的内容。这些章节大致基于刚刚完成的章节内容,但也会借鉴你在课程早期或以前学习的知识,并指向未来的学习和不那么熟悉的情境。深入思考 "部分还将帮助你发展可迁移的技能。通过学习这些章节,您将
  • read real-life scientific writing in a variety of contexts and aimed at different audiences
    阅读各种背景下针对不同受众的真实科学文章
  • develop an understanding of how the professional scientific community functions
    了解专业科学界的运作方式
  • learn to think critically about the nature of what you have read
    学会批判性地思考所读内容的性质
  • understand the issues, problems and challenges that may be raised
    了解可能提出的问题、难题和挑战
  • gain practice in communicating information and ideas in an appropriate scientific way
    练习以适当的科学方式交流信息和观点
  • apply your knowledge and understanding to unfamiliar contexts.
    将所学知识和理解运用到陌生的环境中。
You will also gain scientific skills through the hands-on practical work that forms an essential part of your course. As well as understanding the experimental methods of the practicals, it is important that you develop the skills necessary to plan experiments and analyse and evaluate data. Not only are these very important scientific skills, but they will be assessed in your examinations.
您还将通过实践操作获得科学技能,这也是课程的重要组成部分。除了了解实践的实验方法外,培养规划实验、分析和评估数据的必要技能也非常重要。这些不仅是非常重要的科学技能,而且将在考试中进行评估。

MATHS SKILLS FOR PHYSICISTS
物理学家的数学技能

  • Recognise and make use of appropriate units in calculations (e.g. knowing the difference between base and derived units)
    在计算中认识并使用适当的单位(如知道基数单位和导数单位的区别)
  • Estimate results (e.g. estimating the speed of waves on the sea)
    估算结果(如估算海浪的速度)
  • Make order of magnitude calculations (e.g. estimating approximately what an answer should be before you start calculating, including using standard form)
    进行数量级计算(例如,在开始计算前估计出答案的大致值,包括使用标准格式)。
  • Use algebra to rearrange and solve equations (e.g. finding the landing point of a projectile)
    使用代数重新排列和求解方程(如寻找弹丸的落点)
  • Recognise the importance of the straight line graph as an analysis tool for the verification and development of physical laws by experimentation (e.g. choosing appropriate variables to plot to generate a straight line graph with experimental data)
    认识到直线图作为分析工具对通过实验验证和发展物理规律的重要性(例如, 选择适当的变量绘制实验数据以生成直线图)
  • Determine the slope and intercept of a linear graph (e.g. finding acceleration from a velocity-time graph)
    确定线性图的斜率和截距(例如,从速度-时间图中找出加速度)
  • Calculate the area under the line on a graph (e.g. finding the energy stored in a stretched wire)
    计算图表中直线下的面积(例如,计算拉伸电线中储存的能量)
  • Use geometry and trigonometry (e.g. finding components of vectors)
    使用几何和三角法(如求矢量的分量)

What prior knowledge do I need?
我需要哪些先验知识?

  • Understanding and knowledge of physical facts, terminology, concepts, principles and practical techniques
    了解和掌握物理事实、术语、概念、原理和实用技术
  • Applying the concepts and principles of physics, including the applications of physics, to different contexts
    将物理学的概念和原理,包括物理学的应用,应用到不同的情境中
  • Appreciating the practical nature of physics and developing experimental and investigative skills based on the use of correct and safe laboratory techniques
    在使用正确和安全的实验室技术的基础上,了解物理的实用性并发展实验和探究技能
  • Communicating scientific methods, conclusions and arguments using technical and mathematical language
    使用技术和数学语言交流科学方法、结论和论据
  • Consideration of the implications, including benefits and risks, of scientific and technological developments
    审议科技发展的影响,包括益处和风险
  • Understanding how society uses scientific knowledge to make decisions about the implementation of technological developments
    了解社会如何利用科学知识做出实施技术发展的决策

    Understanding of how scientific ideas change over time, and the systems in place to validate these changes
    了解科学思想如何随着时间的推移而变化,以及验证这些变化的现有系统

What will I study later?
我以后要学什么?

  • Knowledge and understanding of further physical facts and terminology, deeper concepts, principles and more complex practical techniques
    掌握和理解更多物理事实和术语、更深层次的概念、原理和更复杂的实用技术
  • Practical skills and techniques for some key physics experiments
    一些关键物理实验的实用技能和技巧
  • How to communicate information and ideas in appropriate ways using appropriate terminology
    如何使用适当的术语,以适当的方式交流信息和观点
  • The implications of science and their associated benefits and risks
    科学的影响及其相关的益处和风险
  • The role of the scientific community in validating new knowledge and ensuring integrity
    科学界在验证新知识和确保完整性方面的作用
  • The ways in which society uses science to inform decision making
    社会利用科学为决策提供信息的方式

1 STANDARD UNITS IN PHYSICS
物理 1 个标准单位

LEARNING OBJECTIVES 学习目标

  • Understand the distinction between base and derived quantities.
    理解基量和导出量之间的区别。

    ■ Understand the idea of a fixed system of units, and explain the SI system.
    理解固定单位制的概念,并解释国际单位制。

BASE AND DERIVED QUANTITIES
基量和导出量

A fig A The international standard kilogram, officially known as the International Prototype Kilogram, is made from a mixture of platinum and iridium and is held at the Bureau International des Poids et Mesures in Paris. All other masses are defined by comparing with this metal cylinder.
A 图 A 国际标准千克的正式名称是 "国际千克原型",由铂和铱的混合物制成,存放在巴黎国际计量局。所有其他质量都是通过与这个金属圆筒进行比较来确定的。

Some measurements we make are of fundamental qualities of things in the universe. For example, the length of a pencil is a fundamental property of the object. Compare this with the pencil’s speed if you drop it. To give a value to the speed, we have to consider a distance moved, and the rate of motion over that distance - we also need to measure time and then do a calculation. You can see that there is a fundamental difference between the types of quantity that are length and speed. We call the length a base unit, whilst the speed is a derived unit. At present, the international scientific community uses seven base units, and from these all other units are derived. Some derived units have their own names. For example, the derived unit of force should be kg m s 2 kg m s 2 kgms^(-2)\mathrm{kg} \mathrm{m} \mathrm{s}^{-2}, but this has been named the newton (N). Other derived units do not get their own name, and we just list the base units that went together in deriving the quantity. For example, speed is measured in m s 1 m s 1 ms^(-1)\mathrm{m} \mathrm{s}^{-1}.
我们所做的一些测量是对宇宙中事物基本特性的测量。例如,铅笔的长度是物体的基本属性。将其与铅笔掉落时的速度进行比较。要给速度取一个值,我们必须考虑移动的距离,以及在这个距离上的运动速度--我们还需要测量时间,然后进行计算。可以看出,长度和速度这两种量之间有着本质的区别。我们称长度为基本单位,而速度则是派生单位。目前,国际科学界使用七种基本单位,所有其他单位都是从这些单位派生出来的。有些派生单位有自己的名称。例如,力的导出单位应该是 kg m s 2 kg m s 2 kgms^(-2)\mathrm{kg} \mathrm{m} \mathrm{s}^{-2} ,但它已被命名为牛顿(N)。其他派生单位则没有自己的名称,我们只是列出了推导出该量的基本单位。例如,速度的单位是 m s 1 m s 1 ms^(-1)\mathrm{m} \mathrm{s}^{-1}
BASIC QUANTITY 基本数量 UNIT NAME 单位名称 UNIT SYMBOL 单位符号
mass 质量 kilogram 公斤 kg 公斤
time 时间 second 附议 s
length 长度 metre 平方米 m
electric current 电流 ampere 安培 A
temperature 温度 kelvin 开尔文 K
amount of substance 物质的量 mole 鼹鼠 mol 
light intensity 光强 candela 蜡烛 cd
BASIC QUANTITY UNIT NAME UNIT SYMBOL mass kilogram kg time second s length metre m electric current ampere A temperature kelvin K amount of substance mole mol light intensity candela cd| BASIC QUANTITY | UNIT NAME | UNIT SYMBOL | | :--- | :--- | :---: | | mass | kilogram | kg | | time | second | s | | length | metre | m | | electric current | ampere | A | | temperature | kelvin | K | | amount of substance | mole | mol | | light intensity | candela | cd |
table A A A\mathbf{A} The base units.
A A A\mathbf{A} 基本单位。

The choice of which quantities are the base ones is somewhat a matter of choice. The scientists who meet to decide on the standard unit system have chosen these seven. You might think that electric current is not a fundamental property, as it is the rate of movement of charge. So it could be derived from measuring charge and time. However, scientists had to pick what was fundamental and they chose current. This means that electric charge is a derived quantity found by multiplying current passing for a given time.
选择哪些量作为基本量在某种程度上是一个选择问题。开会决定标准单位制的科学家们选择了这七个量。你可能会认为电流不是基本性质,因为它是电荷的运动速度。因此,它可以通过测量电荷和时间得出。然而,科学家们必须选择什么是基本性质,他们选择了电流。这就意味着,电荷是通过乘以一定时间内通过的电流而得到的一个导出量。

SI UNITS SI 单位

For each of the base units, a meeting is held every four or six years of the General Conference on Weights and Measures, under the authority of the Bureau International des Poids et Mesures in Paris. At this meeting, they either alter the definition, or agree to continue with the current definition. As we learn more and more about the universe, these definitions are gradually moving towards the fundamental constants of nature.
对于每个基本单位,在巴黎国际计量局的领导下,计量大会每四年或六年举行一次会议。在这次会议上,他们要么修改定义,要么同意继续沿用当前的定义。随着我们对宇宙的了解越来越多,这些定义也逐渐趋向于自然界的基本常数。
A fig B A standard metre, made to be exactly the length that light could travel in 1 / 299792458 1 / 299792458 1//2997924581 / 299792458 of a second.
A 无花果 B 标准米,其长度正好是光在 1 / 299792458 1 / 299792458 1//2997924581 / 299792458 秒内所能传播的长度。
The current definition of each of the seven base units is listed below:
七个基本单位的现行定义如下:
  • The kilogram is the unit of mass; it is equal to the mass of the International Prototype Kilogram, as in fig A A A\mathbf{A}.
    千克是质量单位;它等于国际千克原型的质量,如图 A A A\mathbf{A} 所示。
  • The second is the duration of 9192631770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium-133 atom.
    其次是与铯-133 原子基态两个超频级之间的转变相对应的辐射持续时间 9192631770 个周期。
  • The metre is the length of the path travelled by light in vacuum
    米是光在真空中所走路径的长度
  • during a time interval of 1 299792458 1 299792458 (1)/(299792458)\frac{1}{299792458} of a second (see fig B).
    1 299792458 1 299792458 (1)/(299792458)\frac{1}{299792458} 秒的时间间隔内(见图 B)。
  • The ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed 1 m apart in vacuum, would produce between these conductors a force equal to 2 × 10 7 2 × 10 7 2xx10^(-7)2 \times 10^{-7} newton per metre of length.
    安培是这样一个恒定电流:如果在两根无限长、圆形截面可忽略不计、相距 1 米的平行直导体中保持恒定电流,并将其放置在真空中,则在这两根导体之间产生的力等于 2 × 10 7 2 × 10 7 2xx10^(-7)2 \times 10^{-7} 每米长度上的牛顿。
  • The kelvin, unit of thermodynamic temperature, is the fraction 1 273.16 1 273.16 (1)/(273.16)\frac{1}{273.16} of the thermodynamic temperature of the triple point of water.
    开尔文是热力学温度单位,是水的三相点热力学温度的 1 273.16 1 273.16 (1)/(273.16)\frac{1}{273.16} 分数。
  • The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 0.012 kg of carbon-12. (When the mole is used, the elementary entities must be specified and may be atoms, molecules, ions, electrons, other particles or specified groups of such particles.)
    摩尔是一个系统的物质的量,它包含的基本实体的数量与 0.012 千克碳-12 中原子的数量相等。(在使用摩尔时,必须指明基本实体,可以是原子、分子、离子、电子、其他微粒或这些微粒的特定组)。

EXAM HINT 考试提示

Table A has the complete list of SI base units. You will not be asked questions about the candela in the exam.
表 A 列出了国际单位制基本单位的完整列表。考试中不会问到有关坎德拉的问题。

LEARNING TIP 学习提示

‘Metrology’ is the study of the science of measurement, and ‘metrics’ refers to ways of standardising measuring techniques.
计量学 "是对测量科学的研究,而 "度量衡 "指的是将测量技术标准化的方法。

DERIVED UNITS 衍生单位

In table B you will see many of the derived units that we will study in this book, but this is only a list of those that have their own name.
在表 B 中,您将看到我们将在本书中学习的许多派生单位,但这只是有自己名称的派生单位的列表。
 导出数量
DERIVED
QUANTITY
DERIVED QUANTITY| DERIVED | | :--- | | QUANTITY |
UNIT NAME 单位名称 UNIT SYMBOL 单位符号
 基本单位当量
BASE UNITS
EQUIVALENT
BASE UNITS EQUIVALENT| BASE UNITS | | :---: | | EQUIVALENT |
force 强逼 newton 牛顿 N kg m s 2 kg m s 2 kgms^(-2)\mathrm{kg} \mathrm{m} \mathrm{s}^{-2}
energy (work) 能量 joule 焦耳 J kg m 2 s 2 kg m 2 s 2 kgm^(2)s^(-2)\mathrm{kg} \mathrm{m}^{2} \mathrm{~s}^{-2}
power 权力 watt 功率 W W WW kg m 2 s 3 kg m 2 s 3 kgm^(2)s^(-3)\mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-3}
frequency 频率 hertz 赫兹 Hz 赫兹 s 1 s 1 s^(-1)\mathrm{s}^{-1}
charge 收费 coulomb 寇仑 C As 由于
voltage 电压 volt 伏特 V V VV kg m 2 s 3 A 1 kg m 2 s 3 A 1 kgm^(2)s^(-3)A^(-1)\mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-3} \mathrm{~A}^{-1}
resistance 阻力 ohm 欧姆 Ω Ω Omega\Omega kg m 2 s 3 A 2 kg m 2 s 3 A 2 kgm^(2)s^(-3)A^(-2)\mathrm{kg} \mathrm{m}^{2} \mathrm{~s}^{-3} \mathrm{~A}^{-2}
"DERIVED QUANTITY" UNIT NAME UNIT SYMBOL "BASE UNITS EQUIVALENT" force newton N kgms^(-2) energy (work) joule J kgm^(2)s^(-2) power watt W kgm^(2)s^(-3) frequency hertz Hz s^(-1) charge coulomb C As voltage volt V kgm^(2)s^(-3)A^(-1) resistance ohm Omega kgm^(2)s^(-3)A^(-2)| DERIVED <br> QUANTITY | UNIT NAME | UNIT SYMBOL | BASE UNITS <br> EQUIVALENT | | :--- | :---: | :---: | :---: | | force | newton | N | $\mathrm{kg} \mathrm{m} \mathrm{s}^{-2}$ | | energy (work) | joule | J | $\mathrm{kg} \mathrm{m}^{2} \mathrm{~s}^{-2}$ | | power | watt | $W$ | $\mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-3}$ | | frequency | hertz | Hz | $\mathrm{s}^{-1}$ | | charge | coulomb | C | As | | voltage | volt | $V$ | $\mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-3} \mathrm{~A}^{-1}$ | | resistance | ohm | $\Omega$ | $\mathrm{kg} \mathrm{m}^{2} \mathrm{~s}^{-3} \mathrm{~A}^{-2}$ |
table B Some well known derived units.
表 B 一些著名的衍生单位

POWER PREFIXES 电源前缀

Sometimes the values we have to work with for some quantities mean that the numbers involved are extremely large or small. For example, the average distance from the Earth to the sun, measured in metres, is 150000000000 m . Scientists have made an easier system for writing such large values by adding a prefix to the unit which tells us that it has been multiplied by a very large or very small amount. In the Earth orbit example, the distance is equivalent to 150 billion metres, and the prefix giga- means multiply by a billion. So the Earth-sun distance becomes 150 gigametres, or 150 Gm .
有时,我们必须使用某些量的数值,这意味着所涉及的数字非常大或非常小。例如,地球到太阳的平均距离是 150000000000 米。科学家们为书写这样大的数值提供了一种更简便的方法,即在单位前加上一个前缀,告诉我们该单位乘以一个非常大或非常小的数。在地球轨道的例子中,距离相当于 1500 亿米,前缀 giga- 表示乘以 10 亿。因此,地日距离就变成了 150 千兆米,或 150 Gm。
FACTOR NAME SYMBOL FACTOR NAME SYMBOL
10 1 10 1 10^(1)10^{1} deca-  da 10 1 10 1 10^(-1)10^{-1} deci-  d
10 2 10 2 10^(2)10^{2} hecto- 全株 h 10 2 10 2 10^(-2)10^{-2} centi-  c
10 3 10 3 10^(3)10^{3} kilo- 千克 k 10 3 10 3 10^(-3)10^{-3} milli-  m
10 6 10 6 10^(6)10^{6} mega-  M 10 6 10 6 10^(-6)10^{-6} micro- 微型 μ μ mu\mathrm{\mu}
10 9 10 9 10^(9)10^{9} giga- 十亿 G 10 9 10 9 10^(-9)10^{-9} nano- 纳米 n
10 12 10 12 10^(12)10^{12} tera-  T 10 12 10 12 10^(-12)10^{-12} pico-  P
10 15 10 15 10^(15)10^{15} peta-  P 10 15 10 15 10^(-15)10^{-15} femto- 女性 f
10 18 10 18 10^(18)10^{18} exa-  E 10 18 10 18 10^(-18)10^{-18} atto- 阿图 a
10 21 10 21 10^(21)10^{21} zetta- Z 10 21 10 21 10^(-21)10^{-21} zepto- z
10 24 10 24 10^(24)10^{24} yotta- Y 10 24 10 24 10^(-24)10^{-24} yocto- 溜溜球 y
FACTOR NAME SYMBOL FACTOR NAME SYMBOL 10^(1) deca- da 10^(-1) deci- d 10^(2) hecto- h 10^(-2) centi- c 10^(3) kilo- k 10^(-3) milli- m 10^(6) mega- M 10^(-6) micro- mu 10^(9) giga- G 10^(-9) nano- n 10^(12) tera- T 10^(-12) pico- P 10^(15) peta- P 10^(-15) femto- f 10^(18) exa- E 10^(-18) atto- a 10^(21) zetta- Z 10^(-21) zepto- z 10^(24) yotta- Y 10^(-24) yocto- y| FACTOR | NAME | SYMBOL | FACTOR | NAME | SYMBOL | | :---: | :---: | :---: | :---: | :---: | :---: | | $10^{1}$ | deca- | da | $10^{-1}$ | deci- | d | | $10^{2}$ | hecto- | h | $10^{-2}$ | centi- | c | | $10^{3}$ | kilo- | k | $10^{-3}$ | milli- | m | | $10^{6}$ | mega- | M | $10^{-6}$ | micro- | $\mathrm{\mu}$ | | $10^{9}$ | giga- | G | $10^{-9}$ | nano- | n | | $10^{12}$ | tera- | T | $10^{-12}$ | pico- | P | | $10^{15}$ | peta- | P | $10^{-15}$ | femto- | f | | $10^{18}$ | exa- | E | $10^{-18}$ | atto- | a | | $10^{21}$ | zetta- | Z | $10^{-21}$ | zepto- | z | | $10^{24}$ | yotta- | Y | $10^{-24}$ | yocto- | y |
table C Prefixes used with SI units.
表 C 与国际单位制单位一起使用的前缀。

CHECKPOINT 检查点

SKILLS PROBLEM SOLVING 解决问题的技能

  1. Refer to table B and answer the following questions:
    请参考表 B 并回答下列问题:

    (a) Pick any quantity that you have studied before and explain how its base unit equivalent is shown.
    (a) 任选一个你以前学过的量,并解释其基本单位等量如何表示。

    (b) All of the derived quantity units are named after scientists. Compare their names and abbreviations. What do you notice?
    (b) 所有衍生量单位都以科学家的名字命名。比较它们的名称和缩写。您发现了什么?
  2. Write the following in standard form:
    用标准格式写出下面的内容:

    (a) 9.2 GW (a) 9.2 千兆瓦
    © 6400 km © 6400 公里
    (b) 43 mm (b) 43 毫米
    (d) 44 ns. (d) 44 ns。
  3. Write the following using an appropriate prefix and unit symbol:
    用适当的前缀和单位符号写出下面的内容:

    (a) 3600000 joules (a) 3600000 焦耳
    © 10 millionths of an ampere
    10 百万分之一安培

    (b) 31536000 seconds (b) 31536000 秒
    (d) 105000 hertz. (d) 105000 赫兹。

2 ESTIMATION 2 估算

LEARNING OBJECTIVES 学习目标

Estimate values for physical quantities. Use your estimates to solve problems.
估算物理量的数值。利用估算值解决问题。

ORDER OF MAGNITUDE 数量级

In physics, it can be very helpful to be able to make approximate estimates of values to within an order of magnitude. This means that the power of ten of your estimate is the same as the true value. For example, you are the same height as the ceiling in your classroom, if we consider the order of magnitude. The ceiling may be twice your height, but it would need to be ten times bigger to reach the next order of magnitude.
在物理学中,能够对一个数量级内的数值进行近似估计是非常有帮助的。这意味着您估计值的十次方与真实值相同。例如,如果我们考虑数量级,你的身高与教室的天花板相同。天花板可能是你身高的两倍,但要达到下一个数量级,天花板需要大十倍。

This is made clearer if we express all values in standard form and then compare the power of ten. You are likely to be a thousand times taller than an ant, so we would say you are three orders of magnitude larger.
如果我们用标准形式表示所有数值,然后比较十的幂次,这一点就更清楚了。你可能比一只蚂蚁高一千倍,所以我们可以说你比蚂蚁大三个数量级。

typical ant height: 1.7 mm = 1.7 × 10 3 m 1.7 mm = 1.7 × 10 3 m 1.7mm=1.7 xx10^(-3)m1.7 \mathrm{~mm}=1.7 \times 10^{-3} \mathrm{~m}
典型的蚂蚁高度: 1.7 mm = 1.7 × 10 3 m 1.7 mm = 1.7 × 10 3 m 1.7mm=1.7 xx10^(-3)m1.7 \mathrm{~mm}=1.7 \times 10^{-3} \mathrm{~m}

typical human height: 1.7 m = 1.7 × 10 m 1.7 m = 1.7 × 10 m 1.7m=1.7 xx10^(@)m1.7 \mathrm{~m}=1.7 \times 10^{\circ} \mathrm{m}
典型的人类身高: 1.7 m = 1.7 × 10 m 1.7 m = 1.7 × 10 m 1.7m=1.7 xx10^(@)m1.7 \mathrm{~m}=1.7 \times 10^{\circ} \mathrm{m}

fig A We are three orders of magnitude taller than an ant.
图 A 我们比蚂蚁高三个数量级。

In many situations, physicists are not interested in specific answers, as circumstances can vary slightly and then the specific answer is incorrect. An order of magnitude answer will always be correct, unless you change the initial conditions by more than an order of magnitude. So a physicist could easily answer the question ‘What is the fastest speed of a car?’ because we don’t really want to know the exact true value. To give an exact answer would depend on knowing the model of car, and the weather and road conditions, and this answer would only be correct for that car on that day. By estimating important quantities, like a typical mass for cars, we can get an approximate - order of magnitude - answer. The reason for doing so would be that it allows us to develop ideas as possible or impossible. Also it helps us focus on developing the ideas along lines that will eventually be feasible when we get to developing a specific solution. This reduces time and money wasted by following ideas that are impossible. It also helps us quickly notice any miscalculations in an answer to a question. If we used an equation to calculate the answer to the fastest speed of a particular car in particular conditions, and the answer came out as 300000000 metres per second (the speed of light), we would immediately know that the answer is wrong, and re-check the calculation.
在很多情况下,物理学家对具体的答案并不感兴趣,因为情况可能稍有不同,那么具体的答案就是不正确的。数量级答案总是正确的,除非初始条件的变化超过一个数量级。因此,物理学家可以很容易地回答 "汽车的最快速度是多少 "这个问题,因为我们并不想知道确切的真实值。要给出准确的答案,就必须知道汽车的型号、天气和路况,而且这个答案只对当天的那辆车正确。通过估计重要的量,比如汽车的典型质量,我们可以得到一个近似的数量级的答案。这样做的原因是,它可以让我们提出可能或不可能的想法。同时,这也有助于我们在开发具体解决方案时,专注于按照最终可行的思路来开发想法。这就减少了因遵循不可能的想法而浪费的时间和金钱。它还能帮助我们迅速发现问题答案中的计算错误。如果我们用一个等式来计算某辆汽车在特定条件下的最快速度,而得出的答案是每秒 300000000 米(光速),我们就会立即知道答案是错误的,并重新检查计算结果。
 数量级
ORDER OF MAGNITUDE
SCALE
ORDER OF MAGNITUDE SCALE| ORDER OF MAGNITUDE | | :--- | | SCALE |
TYPICAL OBJECT 典型物体
1 × 10 13 m 1 × 10 13 m 1xx10^(13)m1 \times 10^{13} \mathrm{~m} size of the solar system
太阳系大小
1 × 10 11 m 1 × 10 11 m 1xx10^(11)m1 \times 10^{11} \mathrm{~m} size of Earth's orbit around the sun
地球围绕太阳的轨道大小
1 × 10 8 m 1 × 10 8 m 1xx10^(8)m1 \times 10^{8} \mathrm{~m} size of Moon's orbit around Earth
月径
1 × 10 4 m 1 × 10 4 m 1xx10^(4)m1 \times 10^{4} \mathrm{~m} diameter of Manchester 曼彻斯特直径
1 × 10 0 m 1 × 10 0 m 1xx10^(0)m1 \times 10^{0} \mathrm{~m} human height 人高
1 × 10 3 m 1 × 10 3 m 1xx10^(-3)m1 \times 10^{-3} \mathrm{~m} ant height 蚁高
1 × 10 5 m 1 × 10 5 m 1xx10^(-5)m1 \times 10^{-5} \mathrm{~m} biological cell diameter 生物细胞直径
1 × 10 8 m 1 × 10 8 m 1xx10^(-8)m1 \times 10^{-8} \mathrm{~m} wavelength of ultraviolet light
紫外线波长
1 × 10 10 m 1 × 10 10 m 1xx10^(-10)m1 \times 10^{-10} \mathrm{~m} diameter of an atom 原子直径
1 × 10 14 m 1 × 10 14 m 1xx10^(-14)m1 \times 10^{-14} \mathrm{~m} diameter of an atomic nucleus
原子核直径
"ORDER OF MAGNITUDE SCALE" TYPICAL OBJECT 1xx10^(13)m size of the solar system 1xx10^(11)m size of Earth's orbit around the sun 1xx10^(8)m size of Moon's orbit around Earth 1xx10^(4)m diameter of Manchester 1xx10^(0)m human height 1xx10^(-3)m ant height 1xx10^(-5)m biological cell diameter 1xx10^(-8)m wavelength of ultraviolet light 1xx10^(-10)m diameter of an atom 1xx10^(-14)m diameter of an atomic nucleus| ORDER OF MAGNITUDE <br> SCALE | TYPICAL OBJECT | | :--- | :--- | | $1 \times 10^{13} \mathrm{~m}$ | size of the solar system | | $1 \times 10^{11} \mathrm{~m}$ | size of Earth's orbit around the sun | | $1 \times 10^{8} \mathrm{~m}$ | size of Moon's orbit around Earth | | $1 \times 10^{4} \mathrm{~m}$ | diameter of Manchester | | $1 \times 10^{0} \mathrm{~m}$ | human height | | $1 \times 10^{-3} \mathrm{~m}$ | ant height | | $1 \times 10^{-5} \mathrm{~m}$ | biological cell diameter | | $1 \times 10^{-8} \mathrm{~m}$ | wavelength of ultraviolet light | | $1 \times 10^{-10} \mathrm{~m}$ | diameter of an atom | | $1 \times 10^{-14} \mathrm{~m}$ | diameter of an atomic nucleus |
table A Examples of object scales changing with powers of ten.
表 A 物体比例随 10 的幂次变化的示例。

FERMI QUESTIONS 费米问题
A fig B Enrico Fermi was one of the developers of both nuclear reactors and nuclear bombs, along with other work on particle physics, quantum physics and statistical mechanics. He was awarded the 1938 Nobel Prize for Physics for the discovery of new radioactive elements and induced radioactivity.
恩里科-费米是核反应堆和核弹的开发者之一,同时还从事粒子物理学、量子物理学和统计力学方面的工作。他因发现新的放射性元素和诱导放射性而获得 1938 年诺贝尔物理学奖。

Enrico Fermi was an Italian physicist who lived from 1901 to 1954. He was a pioneer of estimation. What have become known as Fermi questions are seemingly specific questions, to which
恩里科-费米是意大利物理学家,1901 年至 1954 年在世。他是估算学的先驱。被称为费米问题的是一些看似具体的问题,其中

only an order of magnitude answer is expected. It is common for the question to appear very difficult, as we do not have enough information to work out the answer. One of Fermi’s most interesting thought experiments was a consideration of whether or not alien life exists. Over a lunch with other scientists in 1950, Fermi surprised the group by asking ‘Where is everybody?’ referring to extraterrestrials. There seems to be no evidence of the existence of alien life. That is still as true today as it was in 1950. However, when Fermi made an estimation of what would be necessary for an extraterrestrial civilisation to travel to visit us, his estimate came out at a much shorter amount of time than the age of our galaxy.
只要求给出一个数量级的答案。由于我们没有足够的信息来计算出答案,因此问题看起来非常困难是很常见的。费米最有趣的思想实验之一是思考外星生命是否存在。1950 年,费米在与其他科学家共进午餐时,惊讶地问道:"大家都去哪儿了?"他指的是外星人。似乎没有证据表明外星生命的存在。今天的情况与 1950 年一样。然而,当费米估算外星文明前往拜访我们所需的时间时,他估算出的时间比银河系的年龄要短得多。

Δ Δ Delta\Delta fig C The Fermi Paradox: even the most conservative estimates of the requirements of exploring the galaxy mean that aliens should reach Earth within ten million years of their life beginning. If they existed, they would be here.
Δ Δ Delta\Delta fig C 费米悖论:即使是对探索银河系的要求最保守的估计,也意味着外星人应该在其生命开始后的一千万年内到达地球。如果他们存在,他们就会在这里。

You need to work out what steps are needed to make an estimation. First, think about what steps you would take to reach an answer, if you could have any information you wanted. Then, when the necessary data is not all available, make an estimate for the missing numbers. Making sensible assumptions is the key to solving Fermi questions.
您需要制定出估算所需的步骤。首先,想一想如果你能得到任何想要的信息,你会采取哪些步骤来得出答案。然后,在所需数据不全的情况下,对缺失的数据进行估算。做出合理的假设是解决费米问题的关键。

WORKED EXAMPLE 工作范例

Probably the most famous example of a Fermi question was this challenge to a class:
费米问题最有名的例子可能就是向全班同学提出的这个挑战:

‘How many piano tuners are there in Chicago?’
芝加哥有多少钢琴调音师?

The only piece of information he provided was that the population of Chicago was 3 million.
他提供的唯一信息是芝加哥人口为 300 万。
Step 1: How many pianos in Chicago?
第一步:芝加哥有多少架钢琴?

If each household is 4 people, then there are:
如果每个家庭有 4 人,那么就有

3000000 4 = 750000 3000000 4 = 750000 (3000000)/(4)=750000\frac{3000000}{4}=750000 households  3000000 4 = 750000 3000000 4 = 750000 (3000000)/(4)=750000\frac{3000000}{4}=750000 住户
If one household in ten owns a piano, then there are:
如果每十户家庭中就有一户拥有钢琴,那么就有

750000 10 = 75000 750000 10 = 75000 (750000)/(10)=75000\frac{750000}{10}=75000 pianos  750000 10 = 75000 750000 10 = 75000 (750000)/(10)=75000\frac{750000}{10}=75000 钢琴
Step 2: How many pianos per piano tuner?
步骤 2:每个钢琴调音师负责多少架钢琴?

Assume each piano needs tuning once a year. Further assume a piano tuner works 200 days a year, and can service 4 pianos a day. Each tuner can service: 200 × 4 = 800 200 × 4 = 800 200 xx4=800200 \times 4=800 pianos.
假设每架钢琴每年需要调音一次。再假设一名钢琴调音师每年工作 200 天,每天可为 4 架钢琴调音。每个调音师可以为 200 × 4 = 800 200 × 4 = 800 200 xx4=800200 \times 4=800 钢琴。
Step 3: How many tuners?
步骤 3:多少个调谐器?

Each piano tuner works with 800 pianos, and there are 75000 pianos in total. So there are: 75000 800 = 94 75000 800 = 94 (75000)/(800)=94\frac{75000}{800}=94 piano tuners.
每个钢琴调音师负责 800 台钢琴,总共有 75000 台钢琴。因此有 75000 800 = 94 75000 800 = 94 (75000)/(800)=94\frac{75000}{800}=94 钢琴调音师。

Your answer to Fermi would be ‘There are 100 piano tuners in Chicago’. This is not expected to be the exactly correct answer, but it will be correct to order of magnitude. We would not expect to find that Chicago has only 10 piano tuners, and it would be very surprising if there were 1000.
你给费米的答案是 "芝加哥有 100 个钢琴调音师"。这不可能是完全正确的答案,但在数量级上是正确的。我们不会想到芝加哥只有 10 个钢琴调音师,如果有 1000 个,那就太令人吃惊了。

CHECKPOINT 检查点

SKILLS ADAPTIVE LEARNING 技能适应性学习

  1. Give an order of magnitude estimate for the following quantities:
    估算下列数量的数量级:

    (a) the height of a giraffe
    (a) 长颈鹿的身高

    (b) the mass of an apple
    (b) 苹果的质量

    © the reaction time of a human
    人的反应时间

    (d) the diameter of a planet
    (d) 行星的直径

    (e) the temperature in this room.
    (e) 该房间的温度。
  2. Answer the following Fermi questions, showing all the steps and the assumptions and estimates you make.
    请回答下列费米问题,并说明所有步骤以及你所做的假设和估计。

    (a) How many tennis balls would fit into a soccer stadium?
    (a) 一个足球场能容纳多少个网球?

    (b) How many atoms are there in your body?
    (b) 你体内有多少原子?

    © How many drops of water are there in a swimming pool?
    © 游泳池里有多少滴水?

    (d) In your lifetime, how much money will you make in total?
    (d) 在你的一生中,你总共能赚多少钱?

    (e) How many Fermi questions could Enrico Fermi have answered whilst flying from Rome to New York?
    (e) 在从罗马飞往纽约的途中,恩里科-费米可以回答多少个费米问题?

TOPIC 1 MECHANICS 专题 1 力学

1A MOTION 1A 动议

How can we calculate how fast a plane is flying, in what direction it is going and how long it will take to reach a certain destination? If you were a pilot, how would you know what force to make the engines produce and where to direct that force so your plane moves to your destination?
我们如何计算飞机的飞行速度、飞行方向以及到达某个目的地所需的时间?如果你是一名飞行员,你如何知道该让发动机产生多大的力,以及该把力引向何处,从而让飞机飞向目的地?

An incredible number of intricate calculations need to be done to enable a successful flight, and the basis for all of them is simple mechanics.
要实现成功的飞行,需要进行大量复杂的计算,而所有计算的基础都是简单的力学。
This chapter explains the multiple movements of objects. It looks at how movement can be described and recorded, and then moves on to explaining why movement happens. It covers velocity and acceleration, including how to calculate these in different situations.
本章解释物体的多种运动。本章首先介绍如何描述和记录运动,然后解释运动发生的原因。它涵盖了速度和加速度,包括如何在不同情况下计算速度和加速度。

We only consider objects moving at speeds that could be encountered in everyday life. At these speeds (much less than the speed of light) Sir Isaac Newton succinctly described three laws of motion. With knowledge of basic geometry, we can identify aspects of movement in each dimension.
我们只考虑以日常生活中可能遇到的速度运动的物体。在这种速度下(远小于光速),艾萨克-牛顿爵士简明扼要地描述了三条运动定律。利用基本几何知识,我们可以确定每个维度中运动的各个方面。

Newton’s laws of motion have been constantly under test by scientists ever since he published them in 1687. Within the constraints established by Einstein in the early twentieth century, Newton’s laws have always correctly described the relationships between data collected. You may have a chance to confirm Newton’s laws in experiments of your own. With modern ICT recording of data, the reliability of such experiments is now much improved over traditional methods.
牛顿的运动定律自 1687 年发表以来,一直不断接受科学家的检验。在爱因斯坦于二十世纪初建立的约束条件下,牛顿定律始终正确地描述了所收集数据之间的关系。你也许有机会在自己的实验中证实牛顿定律。通过现代信息和通信技术记录数据,这类实验的可靠性比传统方法有了很大提高。

MATHS SKILLS FOR THIS CHAPTER
本章的数学技能

  • Units of measurement (e.g. the newton, N N NN )
    测量单位(如牛顿、 N N NN )
  • Using Pythagoras’ theorem, and the angle sum of a triangle (e.g. finding a resultant vector)
    使用勾股定理和三角形的角和(如求结果向量)
  • Using sin, cos and tan in physical problems (e.g. resolving vectors)
    在物理问题中使用 sin、cos 和 tan(如解决矢量问题)
  • Using angles in regular 2D structures (e.g. interpreting force diagrams to solve problems)
    在规则的二维结构中使用角(例如,解释力图来解决问题)
  • Changing the subject of an equation (e.g. re-arranging the kinematics equations)
    改变方程式的主题(例如重新排列运动学方程式)
  • Substituting numerical values into algebraic equations (e.g. calculating the acceleration)
    将数值代入代数方程(如计算加速度)
  • Plotting two variables from experimental or other data, understanding that y = m x + c y = m x + c y=mx+cy=m x+c represents a linear relationship and determining the slope of a linear graph (e.g. verifying Newton’s second law experimentally)
    根据实验数据或其他数据绘制两个变量的图形,理解 y = m x + c y = m x + c y=mx+cy=m x+c 代表线性关系,并确定线性图形的斜率(例如,通过实验验证牛顿第二定律)。
  • Estimating, by graphical methods as appropriate, the area between a curve and the x x xx-axis and realising the physical significance of the area that has been determined (e.g. using a speed-time graph)
    通过适当的图形方法估算曲线与 x x xx 轴之间的面积,并认识到所确定的面积的物理意义(例如使用速度-时间图形)

1A 1 VELOCITY AND AGGELERATION
1a 1 速度和加速度

LEARNING OBJECTIVES 学习目标

Explain the distinction between scalar and vector quantities.
解释标量和矢量的区别。

Distinguish between speed and velocity and define acceleration.
区分速度和速率,定义加速度。
  • Calculate values using equations for velocity and acceleration.
    使用速度和加速度方程计算数值。

Δ Δ Delta\Delta fig A A A\mathbf{A} These runners are accelerating to a high speed.
Δ Δ Delta\Delta A A A\mathbf{A} 这些选手正在高速加速。

LEARNING TIP 学习提示

The upper case Greek letter delta, Δ Δ Delta\Delta, is used in mathematics to indicate a change in a quantity. For example, Δ s Δ s Delta s\Delta s means the change in the displacement of an object, to be used here to calculate the velocity of the object.
希腊文大写字母 delta, Δ Δ Delta\Delta ,在数学中用来表示一个量的变化。例如, Δ s Δ s Delta s\Delta s 表示物体位移的变化,这里用来计算物体的速度。

DID YOU KNOW? 您知道吗?

The froghopper, a 6 mm long insect, can accelerate at 4000 m s 1 4000 m s 1 4000ms^(-1)4000 \mathrm{~m} \mathrm{~s}^{-1}.
蛙蛙是一种 6 毫米长的昆虫,能以 4000 m s 1 4000 m s 1 4000ms^(-1)4000 \mathrm{~m} \mathrm{~s}^{-1} 的速度加速。

LEARNING TIP 学习提示

Vector notation means that vectors are written in bold type to distinguish them from scalar variables.
矢量符号是指用粗体字书写矢量,以区别于标量变量。

Movement is fundamental to the functioning of our universe. Whether you are running to catch a bus or want to calculate the speed required for a rocket to travel to Mars or the kinetic energy of an electron in an X-ray machine, you need to be able to work out how fast things are moving.
运动是宇宙运行的基础。无论你是跑着去赶公共汽车,还是想计算火箭飞往火星所需的速度或 X 射线机器中电子的动能,你都需要能够计算出事物运动的速度。

RATE OF MOVEMENT 运动速度

One of the simplest things we can measure is how fast an object is moving. You can calculate an object’s speed if you know the amount of time taken to move a certain distance:
我们可以测量的最简单的事情之一就是物体移动的速度。如果知道物体移动一定距离所需的时间,就可以计算出物体的速度:
speed ( m s 1 ) = distance ( m ) time ( s ) v = d t  speed  m s 1 =  distance  ( m ) time ( s ) v = d t {:[" speed "(ms^(-1))=(" distance "(m))/(time(s))],[v=(d)/(t)]:}\begin{aligned} \text { speed }\left(\mathrm{m} \mathrm{~s}^{-1}\right) & =\frac{\text { distance }(\mathrm{m})}{\operatorname{time}(\mathrm{s})} \\ v & =\frac{d}{t} \end{aligned}
However, the calculation for speed will only tell you how fast an object is moving. Often it is also vitally important to know in what direction this movement is taking the object. When you include the direction in the information about the rate of movement of an object, this is then known as the velocity. So, the velocity is the rate of change of displacement, where the distance in a particular direction is called the ‘displacement’.
然而,速度计算只能告诉您物体移动的速度。通常,知道物体运动的方向也是非常重要的。如果在物体运动速度的信息中包含方向,这就是速度。因此,速度就是位移的变化率,其中特定方向上的距离称为 "位移"。
velocity ( m s 1 ) = displacement ( m ) time ( s ) v = s t OR v = Δ s Δ t  velocity  m s 1 =  displacement  ( m )  time  ( s ) v = s t  OR  v = Δ s Δ t {:[" velocity "(ms^(-1))=(" displacement "(m))/(" time "(s))],[v=(s)/(t)],[" OR "],[v=(Delta s)/(Delta t)]:}\begin{aligned} & \text { velocity }\left(\mathrm{m} \mathrm{~s}^{-1}\right)=\frac{\text { displacement }(\mathrm{m})}{\text { time }(\mathrm{s})} \\ & v=\frac{s}{t} \\ & \text { OR } \\ & v=\frac{\Delta s}{\Delta t} \end{aligned}
A fig B The displacement due north is only 75 m , whilst the actual distance this athlete has run is 300 m . So the velocity due north is much less than the actual speed.
A 图 B 正北方向的位移只有 75 米,而这名运动员实际跑了 300 米。因此,正北方向的速度远小于实际速度。
A quantity for which the direction must be stated is known as a vector. If direction is not important, the measurement is referred to as a scalar quantity. Therefore, velocity is a vector and speed is a scalar; distance is a scalar and displacement is a vector.
必须说明方向的量称为矢量。如果方向并不重要,测量结果就被称为标量。因此,速度是矢量,速度是标量;距离是标量,位移是矢量。
Scalar and vector quantities are not limited to measurements related to movement. Every measured quantity can be classified to include the direction (vector, e.g. force) or as being sufficiently stated by its magnitude only (scalar, e.g. mass).
标量和矢量并不局限于与运动有关的测量。每个测量量都可分为包括方向(矢量,如力)或仅由其大小充分说明(标量,如质量)的量。

AVERAGE AND INSTANTANEOUS SPEED
平均速度和瞬时速度

In most journeys, it is unlikely that speed will remain constant throughout. As part of his training programme, an athlete in fig A wants to keep a record of his speed for all races. From rest, before the starting gun starts the race, he accelerates to a top speed. However, the race timing will be made from start to finish, and so it is most useful to calculate an average speed over the whole race. Average speed is calculated by dividing the total distance for a journey by the total time for the journey. Thus it averages out the slower and faster parts of the journey, and even includes stops.
在大多数旅程中,速度不可能始终保持不变。作为训练计划的一部分,图 A 中的运动员希望记录下自己所有比赛的速度。在比赛发令枪响之前,他从静止状态加速到最高速度。不过,比赛计时将从起点到终点,因此计算整场比赛的平均速度最为有用。平均速度的计算方法是用总路程除以总时间。因此,它平均了路程中较慢和较快的部分,甚至包括停顿。

Instantaneous speed can be an important quantity, and we will look at how to measure it in the next topic.
瞬时速度是一个重要的量,我们将在下一个主题中探讨如何测量它。
A fig C Most speed checks look at instantaneous speed, but CCTV allows police to monitor average speed over a long distance.
A 无花果 C 大多数车速检查都是检查瞬时车速,但闭路电视可以让警方监测长距离的平均车速。

ACCELERATION 加速度

Acceleration is defined as the rate of change of velocity. Therefore, it must include the direction in which the speed is changing, and so acceleration is a vector quantity. The equation defining acceleration is:
加速度的定义是速度的变化率。因此,它必须包括速度变化的方向,所以加速度是一个矢量。定义加速度的方程为
acceleration ( m s 2 ) = change in velocity ( m s 1 ) time taken to change the velocity ( s )  acceleration  m s 2 =  change in velocity  m s 1  time taken to change the velocity  ( s ) " acceleration "(ms^(-2))=(" change in velocity "(ms^(-1)))/(" time taken to change the velocity "(s))\text { acceleration }\left(\mathrm{m} \mathrm{~s}^{-2}\right)=\frac{\text { change in velocity }\left(\mathrm{m} \mathrm{~s}^{-1}\right)}{\text { time taken to change the velocity }(\mathrm{s})}
OR 
a = v u t a = v u t a=(v-u)/(t)a=\frac{v-u}{t}
a = Δ v Δ t a = Δ v Δ t a=(Delta v)/(Delta t)a=\frac{\Delta v}{\Delta t}
where u u u\boldsymbol{u} is the initial velocity and v v v\boldsymbol{v} is the final velocity.
其中, u u u\boldsymbol{u} 为初速度, v v v\boldsymbol{v} 为末速度。

The vector nature of acceleration is very important. One of the consequences is that if an object changes only the direction of its velocity, it is accelerating, while remaining at a constant speed. Similarly, deceleration represents a negative change in velocity, and so could be stated as a negative acceleration.
加速度的矢量性质非常重要。其后果之一是,如果一个物体只改变了速度的方向,那么它就是在加速,而速度则保持不变。同样,减速代表速度的负变化,因此可以说是负加速度。

EXAM HINT 考试提示

Whilst accelerations can (very briefly) be extraordinarily high, like that for the electron in question 3(b), no speed or velocity can ever be greater than the speed of light, which is 3 × 10 8 m s 1 3 × 10 8 m s 1 3xx10^(8)ms^(-1)3 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}. If you calculate a speed that is higher than this, check your calculation again as it must be wrong.
虽然加速度可以(非常短暂地)非常高,如问题 3(b) 中电子的加速度,但任何速度或速率都不可能超过光速,即 3 × 10 8 m s 1 3 × 10 8 m s 1 3xx10^(8)ms^(-1)3 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1} 。如果你计算出的速度高于这个速度,请再次检查你的计算,因为它一定是错误的。

CHECKPOINT 检查点

SKILLS PROBLEM SOLVING 解决问题的技能
  1. The athlete in fig B B B\mathbf{B} has taken 36 seconds from the start to reach the 300 m mark as shown. Calculate:
    如图所示,图 B B B\mathbf{B} 中的运动员从起跑到跑到 300 米处用了 36 秒。计算一下

    (a) his average speed during this 36 seconds
    (a) 他在这 36 秒内的平均速度

    (b) his average velocity due north during this 36 seconds
    (b) 他在这 36 秒内正北方向的平均速度

    © his average velocity due east during this 36 seconds.
    在这 36 秒内,他向东的平均速度。
  2. A driver in a car travelling at about 40.2 km h 1 40.2 km h 1 40.2kmh^(-1)40.2 \mathrm{~km} \mathrm{~h}^{-1} sees a cat run onto the road ahead.
    一辆汽车以大约 40.2 km h 1 40.2 km h 1 40.2kmh^(-1)40.2 \mathrm{~km} \mathrm{~h}^{-1} 的速度行驶,司机看到一只猫跑到前面的路上。

    (a) Convert 40.2 km h 1 40.2 km h 1 40.2kmh^(-1)40.2 \mathrm{~km} \mathrm{~h}^{-1} into a speed in m s 1 m s 1 ms^(-1)\mathrm{m} \mathrm{s}^{-1}.
    (a) 将 40.2 km h 1 40.2 km h 1 40.2kmh^(-1)40.2 \mathrm{~km} \mathrm{~h}^{-1} 转换为 m s 1 m s 1 ms^(-1)\mathrm{m} \mathrm{s}^{-1} 中的速度。

    (b) The car travels 16.5 m whilst the driver is reacting to the danger. What is his reaction time?
    (b) 当司机对危险做出反应时,汽车行驶了 16.5 米。他的反应时间是多少?

    © The car comes to a stop in 2.5 s . What is its deceleration?
    汽车在 2.5 秒后停下,它的减速度是多少?
  3. An electron in an X X XX-ray machine is accelerated from rest to half the speed of light in 1.7 × 10 15 s 1.7 × 10 15 s 1.7 xx10^(-15)s1.7 \times 10^{-15} \mathrm{~s}. Calculate:
    X X XX 射线机中的电子在 1.7 × 10 15 s 1.7 × 10 15 s 1.7 xx10^(-15)s1.7 \times 10^{-15} \mathrm{~s} 内从静止加速到光速的一半。计算:

    (a) the speed the electron reaches in m s 1 m s 1 ms^(-1)\mathrm{m} \mathrm{s}^{-1}
    (a) 电子在 m s 1 m s 1 ms^(-1)\mathrm{m} \mathrm{s}^{-1} 中达到的速度

    (b) the acceleration the electron experiences.
    (b) 电子经历的加速度。

SUBJECT VOCABULARY 主题词汇

speed the rate of change of distance:
速度是距离的变化率:
speed ( m s 1 ) = distance ( m ) time ( s ) v = d t speed m s 1 = distance ( m ) time ( s ) v = d t {:[speed(ms^(-1))=(distance(m))/(time(s))],[v=(d)/(t)]:}\begin{aligned} \operatorname{speed}\left(\mathrm{m} \mathrm{~s}^{-1}\right) & =\frac{\operatorname{distance}(\mathrm{m})}{\operatorname{time}(\mathrm{s})} \\ v & =\frac{d}{t} \end{aligned}
velocity the rate of change of displacement:
速度 即位移的变化率:
velocity ( m s 1 ) = displacement ( m ) time ( s ) v = s t OR v = Δ s Δ t  velocity  m s 1 =  displacement  ( m )  time  ( s ) v = s t  OR  v = Δ s Δ t {:[" velocity "(ms^(-1))=(" displacement "(m))/(" time "(s))],[v=(s)/(t)" OR "v=(Delta s)/(Delta t)]:}\begin{aligned} & \text { velocity }\left(\mathrm{m} \mathrm{~s}^{-1}\right)=\frac{\text { displacement }(\mathrm{m})}{\text { time }(\mathrm{s})} \\ & v=\frac{s}{t} \text { OR } v=\frac{\Delta s}{\Delta t} \end{aligned}
displacement the vector measurement of distance in a certain direction
位移 测量某一方向距离的矢量

vector a quantity that must have both magnitude and direction scalar a quantity that has magnitude only average speed speed for a whole journey, calculated by dividing the total distance for a journey by the total time for the journey:
矢量 一个既有大小又有方向的量 标量 一个只有大小的量 整个行程的平均速度,用行程总距离除以行程总时间计算得出:
average speed ( m s 1 ) = total distance ( m ) total time ( s )  average speed  m s 1 =  total distance  ( m )  total time  ( s ) " average speed "(ms^(-1))=(" total distance "(m))/(" total time "(s))\text { average speed }\left(\mathrm{m} \mathrm{~s}^{-1}\right)=\frac{\text { total distance }(\mathrm{m})}{\text { total time }(\mathrm{s})}
instantaneous speed the speed at any particular instant in time on a journey, which can be found from the gradient of the tangent to a distance-time graph at that time acceleration the vector defined as the rate of change of velocity:
瞬时速度 指在旅程中任何特定时刻的速度,可根据当时距离-时间图形切线的梯度求得 加速度 定义为速度变化率的矢量:
acceleration ( m s 2 ) = change in velocity ( m s 1 ) time taken to change the velocity ( s )  acceleration  m s 2 =  change in velocity  m s 1  time taken to change the velocity  ( s ) " acceleration "(ms^(-2))=(" change in velocity "(ms^(-1)))/(" time taken to change the velocity "(s))\text { acceleration }\left(\mathrm{m} \mathrm{~s}^{-2}\right)=\frac{\text { change in velocity }\left(\mathrm{m} \mathrm{~s}^{-1}\right)}{\text { time taken to change the velocity }(\mathrm{s})}
a = v u t OR a = Δ v Δ t a = v u t  OR  a = Δ v Δ t a=(v-u)/(t)" OR "a=(Delta v)/(Delta t)a=\frac{v-u}{t} \text { OR } a=\frac{\Delta v}{\Delta t}

LEARNING OBJECTIVES 学习目标

  • Interpret displacement-time graphs, velocity-time graphs and acceleration-time graphs.
    解读位移-时间图、速度-时间图和加速度-时间图。
  • Make calculations from these graphs.
    根据这些图表进行计算。
  • Understand the graphical representations of accelerated motion.
    理解加速运动的图形表示。
One of the best ways to understand the movements of an object whilst on a journey is to plot a graph of the position of the object over time. Such a graph is known as a displacement-time graph. A velocity-time graph will also provide detail about the movements involved. A velocity-time graph can be produced from direct measurements of the velocity or generated from calculations made using the displacement-time graph.
了解物体在旅途中运动情况的最佳方法之一,就是绘制物体位置随时间变化的曲线图。这种图被称为位移-时间图。速度-时间图还能提供有关运动的详细信息。速度-时间图可以通过直接测量速度绘制,也可以通过使用位移-时间图进行计算生成。

DISPLACEMENT-TIME GRAPHS 位移-时间图

If we imagine a boat trip on a river, we could monitor the location of the boat over the hour that it travels for and plot the displacementtime graph for these movements. Depending on what information we want the graph to provide, it is often simpler to draw a distancetime graph in which the direction of movement is ignored.
如果我们想象在河上乘船旅行,我们可以监测船在一小时内的位置,并绘制这些运动的位移时间图。根据我们希望图表提供的信息,绘制忽略运动方向的距离时间图通常更为简单。

The graphs shown in fig A are examples of plotting position against time, and show how a distance-time graph cannot decrease with time. A displacement-time graph could have parts of it in the negative portions of the y y yy-axis, if the movement went in the opposite direction at some points in time.
图 A 所示的图形是绘制位置与时间关系图的示例,说明了距离-时间图不可能随时间的推移而减小。如果运动在某些时间点的方向相反,那么位移-时间曲线图的某些部分就可能位于 y y yy 轴的负部分。
The simplest thing we could find from these graphs is how far an object has moved in a certain time. For example, in fig A, both the graphs show that in the first 15 minutes the boat moved 150 m . Looking at the time from 40 to 48 minutes, both show that the boat travelled 120 m , but the displacement-time graph is in the negative region of the y y yy-axis, showing the boat was moving down river from the starting point - the opposite direction to the places it had been in the first 40 minutes.
从这些图表中,我们可以发现最简单的事情,那就是物体在一定时间内移动了多远。例如,在图 A 中,两个图都显示在前 15 分钟内,小船移动了 150 米。再看从 40 分钟到 48 分钟的时间,两个图都显示小船行驶了 120 米,但位移-时间图在 y y yy 轴的负区域,显示小船从起点沿河向下移动 - 与前 40 分钟的位置方向相反。
During the period from 20 to 25 minutes, both graphs have a flat line at a constant value, that shows no change in the distance or displacement. This means the boat was not moving - a flat line on a distance-time ( d t ) ( d t ) (d-t)(d-t) graph means the object is stationary. From 20 to 25 minutes on the velocity-time ( v t ) ( v t ) (v-t)(\boldsymbol{v}-t) graph of this journey (see fig B) the line would be at a velocity of 0 m s 1 0 m s 1 0ms^(-1)0 \mathrm{~m} \mathrm{~s}^{-1}.
在 20 至 25 分钟期间,两幅图上都有一条恒定值的平线,表明距离或位移没有变化。这意味着小船没有移动--距离-时间 ( d t ) ( d t ) (d-t)(d-t) 图上的平线意味着物体是静止的。在这段旅程的速度-时间 ( v t ) ( v t ) (v-t)(\boldsymbol{v}-t) 图上(见图 B),从 20 分钟到 25 分钟,这条线的速度为 0 m s 1 0 m s 1 0ms^(-1)0 \mathrm{~m} \mathrm{~s}^{-1}

SPEED AND VELOCITY FROM d t d t d-td-t GRAPHS
d t d t d-td-t 图形中的速度和速度值

The gradient of the d t d t d-td-t graphs in fig A A A\mathbf{A} will tell us how fast the boat was moving. Gradient is found from the ratio of changes in the y y yy-axis divided by the corresponding change on the x x xx-axis, so:
A A A\mathbf{A} d t d t d-td-t 图的梯度将告诉我们船行驶的速度有多快。梯度是由 y y yy 轴上的变化除以 x x xx 轴上相应变化的比值求得的,因此:

for a distance-time graph:
为距离-时间图:
gradient = distance ( m ) time ( s ) = speed ( m s 1 ) v = d t  gradient  = distance ( m ) time ( s ) = speed m s 1 v = d t {:[" gradient "=(distance(m))/(time(s))=speed(ms^(-1))],[v=(d)/(t)]:}\begin{aligned} \text { gradient } & =\frac{\operatorname{distance}(\mathrm{m})}{\operatorname{time}(\mathrm{s})}=\operatorname{speed}\left(\mathrm{m} \mathrm{~s}^{-1}\right) \\ v & =\frac{d}{t} \end{aligned}
for a displacement-time graph:
的位移时间图:
gradient = displacement ( m ) time ( s ) = velocity ( m s 1 ) v = Δ s Δ t  gradient  =  displacement  ( m )  time  ( s ) =  velocity  m s 1 v = Δ s Δ t {:[" gradient "=(" displacement "(m))/(" time "(s))=" velocity "(ms^(-1))],[v=(Delta s)/(Delta t)]:}\begin{aligned} \text { gradient } & =\frac{\text { displacement }(\mathrm{m})}{\text { time }(\mathrm{s})}=\text { velocity }\left(\mathrm{m} \mathrm{~s}^{-1}\right) \\ v & =\frac{\Delta s}{\Delta t} \end{aligned}
fig A A comparison of the displacement-time graph of the boat trip up and down a river with its corresponding distance-time graph.
图 A 船在河上航行的位移-时间图与相应的距离-时间图的比较。
For example, the first 15 minutes of the boat trip in fig A represents a time of 900 seconds. In this time, the boat travelled 150 m . Its velocity is:
例如,图 A 中小船航行的前 15 分钟为 900 秒。在这段时间内,小船行驶了 150 米。其速度为
v = Δ s Δ t = 150 900 = 0.167 m s 1 up river v = Δ s Δ t = 150 900 = 0.167 m s 1  up river  v=(Delta s)/(Delta t)=(150)/(900)=0.167ms^(-1)" up river "v=\frac{\Delta s}{\Delta t}=\frac{150}{900}=0.167 \mathrm{~m} \mathrm{~s}^{-1} \text { up river }

VELOCITY-TIME GRAPHS 速度-时间图

A velocity-time graph will show the velocity of an object over time. We calculated that the velocity of the boat on the river was 0.167 m s 1 0.167 m s 1 0.167ms^(-1)0.167 \mathrm{~m} \mathrm{~s}^{-1} up river for the first 15 minutes of the journey. Looking at the graph in fig B, you can see that the line is constant at + 0.167 m s 1 + 0.167 m s 1 +0.167ms^(-1)+0.167 \mathrm{~m} \mathrm{~s}^{-1} for the first 15 minutes.
速度-时间图将显示物体随时间变化的速度。我们计算出,小船在河上航行的前 15 分钟的速度是 0.167 m s 1 0.167 m s 1 0.167ms^(-1)0.167 \mathrm{~m} \mathrm{~s}^{-1} 。观察图 B 中的曲线图,可以发现在最初的 15 分钟内,该直线恒定在 + 0.167 m s 1 + 0.167 m s 1 +0.167ms^(-1)+0.167 \mathrm{~m} \mathrm{~s}^{-1} 处。

Also notice that the velocity axis includes negative values, so that the difference between travelling up river (positive y y yy-axis values) and down river (negative y y yy-axis values) can be represented.
还要注意的是,速度轴包含负值,因此可以表示逆流而上( y y yy 轴的正值)和顺流而下( y y yy 轴的负值)之间的差异。

fig B 
Velocity-time graph of the boat trip.
船行速度-时间曲线图。

ACCELERATION FROM v t v t v-tv-t GRAPHS
来自 v t v t v-tv-t 图形的加速度

Acceleration is defined as the rate of change in velocity. In order to calculate the gradient of the line on a v t v t v-t\boldsymbol{v}-t graph, we must divide a change in velocity by the corresponding time difference. This exactly matches with the equation for acceleration:
加速度的定义是速度的变化率。为了计算 v t v t v-t\boldsymbol{v}-t 图形上直线的梯度,我们必须用速度变化除以相应的时间差。这与加速度方程完全吻合:
gradient = Δ v Δ t = v u t = acceleration  gradient  = Δ v Δ t = v u t =  acceleration  " gradient "=(Delta v)/(Delta t)=(v-u)/(t)=" acceleration "\text { gradient }=\frac{\Delta v}{\Delta t}=\frac{v-u}{t}=\text { acceleration }
For example, between 15 and 20 minutes on the graphs, the boat slows evenly to a stop. The acceleration here can be calculated as the gradient:
例如,在曲线图上的 15 至 20 分钟之间,船会均匀地减速至停止。这里的加速度可以计算为梯度:
gradient = Δ v Δ t = v u t = 0 0.167 5 × 60 = 0.167 300 = 0.0006 m s 2  gradient  = Δ v Δ t = v u t = 0 0.167 5 × 60 = 0.167 300 = 0.0006 m s 2 " gradient "=(Delta v)/(Delta t)=(v-u)/(t)=(0-0.167)/(5xx60)=(-0.167)/(300)=-0.0006ms^(-2)\text { gradient }=\frac{\Delta v}{\Delta t}=\frac{v-u}{t}=\frac{0-0.167}{5 \times 60}=\frac{-0.167}{300}=-0.0006 \mathrm{~m} \mathrm{~s}^{-2}
So the acceleration is: a = 0.6 × 10 3 m s 2 a = 0.6 × 10 3 m s 2 a=-0.6 xx10^(-3)ms^(-2)\boldsymbol{a}=-0.6 \times 10^{-3} \mathrm{~m} \mathrm{~s}^{-2}.
因此,加速度为 a = 0.6 × 10 3 m s 2 a = 0.6 × 10 3 m s 2 a=-0.6 xx10^(-3)ms^(-2)\boldsymbol{a}=-0.6 \times 10^{-3} \mathrm{~m} \mathrm{~s}^{-2}

DISTANCE TRAVELLED FROM v t v t v-tv-t GRAPHS
v t v t v-tv-t 地形图的距离

Speed is defined as the rate of change in distance:
速度的定义是距离的变化率:
v = d t d = v × t v = d t d = v × t {:[v=(d)/(t)],[:.quad d=v xx t]:}\begin{aligned} & v=\frac{d}{t} \\ & \therefore \quad d=v \times t \end{aligned}
As the axes on the v t v t v-t\boldsymbol{v}-t graph represent velocity and time, an area on the graph represents the multiplication of velocity × × xx\times time, which gives distance. So to find the distance travelled from a v t v t v-t\boldsymbol{v}-t graph, find the area between the line and the x x xx-axis.
由于 v t v t v-t\boldsymbol{v}-t 图形上的坐标轴代表速度和时间,因此图形上的面积代表速度 × × xx\times 时间的乘积,即距离。因此,要从 v t v t v-t\boldsymbol{v}-t 图形中求得路程,请求得直线与 x x xx 轴之间的面积。

Δ Δ Delta\Delta fig C In the first 15 minutes ( 900 seconds) the distance travelled by the boat moving at 0.167 m s 1 0.167 m s 1 0.167ms^(-1)0.167 \mathrm{~m} \mathrm{~s}^{-1} is given by the area between the line and the x x xx-axis: d = v × t = 0.167 × 900 = 150 m d = v × t = 0.167 × 900 = 150 m d=v xx t=0.167 xx900=150md=\boldsymbol{v} \times t=0.167 \times 900=150 \mathrm{~m}.
Δ Δ Delta\Delta 图 C 在最初的 15 分钟(900 秒)内,小船以 0.167 m s 1 0.167 m s 1 0.167ms^(-1)0.167 \mathrm{~m} \mathrm{~s}^{-1} 的速度行驶的距离由直线与 x x xx 轴之间的面积给出: d = v × t = 0.167 × 900 = 150 m d = v × t = 0.167 × 900 = 150 m d=v xx t=0.167 xx900=150md=\boldsymbol{v} \times t=0.167 \times 900=150 \mathrm{~m}
If we are only interested in finding the distance moved, this also works for a negative velocity. You find the area from the line up to the time axis. This idea will still work for a changing velocity. Find the area under the line and you have found the distance travelled. For example, from 0 to 20 minutes, the area under the line, all the way down to the x x xx-axis, is a trapezium, so we need to find that area. To calculate the whole distance travelled in the journey for the first 40 minutes, we would have to find the areas under the four separate stages ( 0 15 0 15 0-150-15 minutes; 15 20 15 20 15-2015-20 minutes; 20 25 20 25 20-2520-25 minutes; and 25-40 minutes) and then add these four answers together.
如果我们只想找出移动的距离,这也适用于负速度。你可以找到从直线到时间轴的面积。对于速度变化的情况,这个方法仍然有效。找到这条直线下的面积,就找到了移动的距离。例如,从 0 到 20 分钟,从直线一直到 x x xx 轴的面积是一个梯形,因此我们需要找到这个面积。要计算出前 40 分钟内的全部路程,我们必须分别求出四个阶段( 0 15 0 15 0-150-15 分钟; 15 20 15 20 15-2015-20 分钟; 20 25 20 25 20-2520-25 分钟;25-40 分钟)下的面积,然后将这四个答案相加。

PRACTICAL SKILLS 实用技能

Finding the acceleration due to gravity by multiflash photography
通过多闪摄影测量重力加速度

Using a multiflash photography technique, or a video recording that can be played back frame by frame, we can observe the falling motion of a small object such as a marble (see fig D). We need to know the time between frames.
利用多闪摄影技术或可逐帧回放的视频录像,我们可以观察到弹珠等小物体的下落运动(见图 D)。我们需要知道帧与帧之间的时间。

From each image of the falling object, measure the distance it has fallen from the scale in the picture A carefully drawn distance-time graph will show a curve as the object accelerates. From this curve, take regular measurements of the gradient by drawing tangents to the curve. These gradients show the instantaneous speed at each point on the curve.
从下落物体的每幅图像中,测量它从图片中的刻度处下落的距离 仔细绘制的距离-时间曲线图将显示出物体加速时的曲线。根据这条曲线,通过画出曲线的切线,定期测量梯度。这些梯度显示了曲线上每一点的瞬时速度。

fig D Multiflash photography allows us to capture the accelerating movement of an object falling under gravity.
无花果 D 多闪摄影使我们能够捕捉到物体在重力作用下下落的加速运动。

Plotting these speeds on a velocity-time graph should show a straight line, as the acceleration due to gravity is a constant value. The gradient of the line on this v v v\boldsymbol{v} - t t tt graph will be the acceleration due to gravity, g g gg.
将这些速度绘制在速度-时间图上,应显示一条直线,因为重力加速度是一个恒定值。 v v v\boldsymbol{v} - t t tt 图上直线的梯度就是重力加速度 g g gg
Safety Note: Persons with medical conditions such as epilepsy or migraine may be adversely affected by multiflash photography.
安全提示:患有癫痫或偏头痛等疾病的人可能会受到多重闪光摄影的不利影响。

ACCELERATION-TIME GRAPHS 加速度-时间图

Acceleration-time graphs show how the acceleration of an object changes over time. In many instances the acceleration is zero or a constant value, in which case an acceleration-time ( a t ) ( a t ) (a-t)(\boldsymbol{a}-t) graph is likely to be of relatively little interest. For example, the object falling in our investigation above will be accelerated by gravity throughout. Assuming it is relatively small, air resistance will be minimal, and the a t a t a-t\boldsymbol{a}-t graph of its motion would be a horizontal line at a = 9.81 ms 2 a = 9.81 ms 2 a=-9.81ms^(-2)\boldsymbol{a}=-9.81 \mathrm{~ms}^{-2}. Compare this with your results to see how realistic it is to ignore air resistance.
加速度-时间曲线图显示了物体的加速度随时间的变化情况。在许多情况下,加速度为零或恒定值,在这种情况下,加速度-时间 ( a t ) ( a t ) (a-t)(\boldsymbol{a}-t) 图的意义可能相对较小。例如,在我们上面的探究中,下落的物体自始至终都会受到重力加速度的作用。假定它相对较小,空气阻力将是最小的,其运动的 a t a t a-t\boldsymbol{a}-t 图将是 a = 9.81 ms 2 a = 9.81 ms 2 a=-9.81ms^(-2)\boldsymbol{a}=-9.81 \mathrm{~ms}^{-2} 处的一条水平线。将这一结果与你的结果进行比较,看看忽略空气阻力有多现实。
For a larger object falling for a long period, such as a skydiver, then the acceleration will change over time as the air resistance increases with speed.
对于长时间下落的较大物体,如跳伞者,由于空气阻力随速度增加而增大,因此加速度会随时间变化。

fig E Acceleration-time graph for a skydiver.
图 E 跳伞者的加速度-时间图

The weight of a skydiver is constant, so the resultant force will be decreasing throughout, which means that the acceleration will also reduce (see Section 1A.5). The curve would look like that in fig E. See Section 2A. 4 for more details on falling objects and terminal velocity.
跳伞者的体重是恒定的,因此所产生的力自始至终都在减小,这意味着加速度也会减小(见第 1A.5 节)。曲线如图 E 所示。4 节,了解有关坠落物体和末端速度的更多详情。

EXAM HINT 考试提示

Remember that the gradient of a distance-time graph represents speed or velocity. So if the line is curved, the changing gradient indicates a changing speed, which you can describe as the same as the changes in gradient.
请记住,距离-时间图形的梯度代表速度或速率。因此,如果直线是弯曲的,梯度的变化表示速度的变化,你可以将其描述为梯度的变化。

CHECKPOINT 检查点

SKILLS 技能
INTERPRETATION 释义
(OF GRAPHICAL DATA) (图表数据)
  1. Describe in as much detail as you can, including calculated values, what happens in the bicycle journey shown on the d t d t d-td-t graph in fig F F F\mathbf{F}.
    请尽可能详细地描述图 F F F\mathbf{F} d t d t d-td-t 所示的自行车行程,包括计算值。

fig F F F\mathbf{F} Distance-time graph of a bike journey.
F F F\mathbf{F} 自行车旅行的距离-时间图。

2. Describe in as much detail as you can, including calculated values, what happens in the car journey shown on the v v v\boldsymbol{v} - t t tt graph in fig G G GG.
2.请尽可能详细地描述图 G G GG v v v\boldsymbol{v} - t t tt 图所示的汽车行程中发生的情况,包括计算值。

fig G Velocity-time graph of a car journey.
图 G 汽车行驶的速度-时间曲线图

3. From fig B, calculate the distance travelled by the boat from 40 to 60 minutes.
3.根据图 B,计算小船从 40 分钟到 60 分钟行驶的距离。

SUBJECT VOCABULARY 主题词汇

displacement-time graph a graph showing the positions visited on a journey, with displacement on the y y yy-axis and time on the x x xx-axis. velocity-time graph a graph showing the velocities on a journey, with velocity on the y y yy-axis and time on the x x xx-axis.
位移-时间曲线图 显示行程中访问位置的曲线图,位移在 y y yy 轴上,时间在 x x xx 轴上。 速度-时间曲线图 显示行程中速度的曲线图,速度在 y y yy 轴上,时间在 x x xx 轴上。

gradient the slope of a line or surface
坡度

LEARNING OBJECTIVES
- Add two or more vectors by drawing.
- Add two perpendicular vectors by calculation.
学习目标 - 通过绘图添加两个或更多向量。 - 通过计算将两个垂直向量相加。

Forces are vectors. This means that measuring their magnitude is important, but equally important is knowing the direction in which they act. In order to calculate the overall effect of multiple forces acting on the same object, we can use vector addition to work out the resultant force. This resultant force can be considered as a single force that has the same effect as all the individual forces combined.
力是矢量。这意味着测量它们的大小很重要,但同样重要的是知道它们的作用方向。为了计算作用在同一物体上的多个力的整体效果,我们可以使用矢量加法计算出结果力。这个结果力可以看作是一个单独的力,它的作用效果与所有单独力的总和相同。

ADDING FORCES IN THE SAME LINE
同线加力

If two or more forces are acting along the same line, then combining them is simply a case of adding or subtracting their magnitudes depending on their directions.
如果两个或两个以上的力沿同一条直线作用,那么将它们结合起来只需根据它们的方向相加或相减即可。

Δ f i g A Δ f i g A Delta figA\Delta \boldsymbol{f i g} \mathbf{A} Adding forces in the same line requires a consideration of their comparative directions.
Δ f i g A Δ f i g A Delta figA\Delta \boldsymbol{f i g} \mathbf{A} 在同一直线上添加力时,需要考虑它们的相对方向。

ADDING PERPENDICULAR FORCES
增加垂直力

The effect on an object of two forces that are acting at right angles (perpendicular) to each other will be the vector sum of their individual effects. We need to add the sizes with consideration for the directions in order to find the resultant.
两个相互成直角(垂直)作用的力对物体的影响是它们各自影响的矢量和。我们需要在考虑方向的情况下将大小相加,以求得结果。
A fig B These two rugby players are each putting a force on their opponent. The forces are at right angles, so the overall effect would be to move him in a third direction, which we could calculate.
A 无花果 B 这两名橄榄球运动员分别给对手施加了一个力。这两个力成直角,因此总的效果是使他向第三个方向移动,我们可以计算出这个结果。

MAGNITUDE OF THE RESULTANT FORCE
结果力大小

To calculate the resultant magnitude of two perpendicular forces, we can draw them, one after the other, as the two sides of a right-angled triangle and use Pythagoras’ theorem to calculate the size of the hypotenuse.
要计算两个垂直力的结果大小,我们可以把它们一个接一个地画成直角三角形的两条边,然后使用毕达哥拉斯定理计算斜边的大小。

Δ Δ Delta\Delta fig C C C\mathbf{C} The resultant force here is calculated using Pythagoras’ theorem: F = ( 70 2 + 110 2 ) = 130 N F = 70 2 + 110 2 = 130 N F=sqrt((70^(2)+110^(2)))=130NF=\sqrt{\left(70^{2}+110^{2}\right)}=130 \mathrm{~N}
Δ Δ Delta\Delta fig C C C\mathbf{C} 这里的结果力是用勾股定理计算出来的: F = ( 70 2 + 110 2 ) = 130 N F = 70 2 + 110 2 = 130 N F=sqrt((70^(2)+110^(2)))=130NF=\sqrt{\left(70^{2}+110^{2}\right)}=130 \mathrm{~N}

DIRECTION OF THE RESULTANT FORCE
结果力的方向

As forces are vectors, when we find a resultant force it must have both magnitude and direction. For perpendicular forces (vectors), trigonometry will determine the direction.
由于力是矢量,因此当我们找到一个结果力时,它必须同时具有大小和方向。对于垂直力(矢量),三角函数将确定其方向。

Δ Δ Delta\Delta fig D D D\mathbf{D} The resultant force here is at an angle up from the horizontal of: θ = tan 1 ( 70 110 ) = 32 θ = tan 1 70 110 = 32 theta=tan^(-1)((70)/(110))=32^(@)\theta=\tan ^{-1}\left(\frac{70}{110}\right)=32^{\circ}
Δ Δ Delta\Delta D D D\mathbf{D} 这里的结果力与水平面成向上的角度: θ = tan 1 ( 70 110 ) = 32 θ = tan 1 70 110 = 32 theta=tan^(-1)((70)/(110))=32^(@)\theta=\tan ^{-1}\left(\frac{70}{110}\right)=32^{\circ}

EXAM HINT 考试提示

Always take care to state where the angle for a vector’s direction is measured. For example, in fig D D DD, the angle should be stated as 32 32 32^(@)32^{\circ} up from the horizontal. This is most easily expressed on a diagram of the situation, where you draw in the angle.
一定要注意说明矢量方向角度的测量位置。例如,在图 D D DD 中,应将角度表述为从水平线向上的 32 32 32^(@)32^{\circ} 。这最容易在示意图上表达,在示意图上画出角度。

ADDING TWO NON-PERPENDICULAR FORCES
两个非垂直力相加

The geometry of perpendicular vectors makes the calculation of the resultant simple. We can find the resultant of any two vectors by drawing one after the other, and then the resultant will be the third side of the triangle from the start of the first one to the end of the second one. A scale drawing of the vector triangle will allow measurement of the size and direction of the resultant.
垂直向量的几何性质使得结果的计算非常简单。我们可以通过一个接一个地画出任意两个矢量的结果,然后结果就是从第一个矢量的起点到第二个矢量的终点的三角形的第三边。绘制矢量三角形的比例图可以测量结果的大小和方向。
A fig E The resultant force here can be found by scale drawing the two forces, and then measurement of the resultant on the drawing using a ruler and a protractor.
A fig E 这里的结果力可以通过按比例绘制两个力的图,然后用直尺和量角器在图上测量结果力。

70 110 N 70 110 N 70 obrace(110(N))70 \overbrace{110 \mathrm{~N}}

A fig G Free-body force diagram of a rugby player (red circle). The forces from the tacklers are marked on as force arrows.
A 图 G 橄榄球运动员(红圈)的自由体受力图。来自攻方的力被标记为力箭头。

THE PARALLELOGRAM RULE 平行四边形法则

There is another method for finding the resultant of two non-perpendicular forces (or vectors) by scale drawing, which can be easier to use. This is called the parallelogram rule. Draw the two vectors to scale - at the correct angle and scaled so their length represents the magnitude - starting from the same point. Then draw the same two vectors again parallel to the original ones, so that they form a parallelogram, as shown in fig F F F\mathbf{F}. The resultant force (or vector) will be the diagonal across the parallelogram from the starting point.
还有另一种通过比例尺作图求两个不垂直的力(或矢量)的结果的方法,这种方法可能更容易使用。这就是平行四边形法则。从同一点开始,按比例绘制两个矢量--以正确的角度和比例使它们的长度代表大小。然后再画出与原来平行的两个矢量,使它们构成一个平行四边形,如图 F F F\mathbf{F} 所示。从起点出发,结果力(或矢量)将是平行四边形的对角线。

fig F Finding the resultant vector using the parallelogram rule.
F 利用平行四边形法则求结果向量。

EXAM HINT 考试提示

The vector addition rules shown on these pages work for all vectors, not just forces. They are useful only for co-planar vectors, which means vectors that are in the same plane. If we have more than two vectors that are in more than one plane, add two vectors together first, in their plane, and then add the resultant to the next vector using these rules again. Keep doing this until all the vectors have been added in.
这些页面上显示的矢量加法规则适用于所有矢量,而不仅仅是力。它们只对共平面向量有用,即在同一平面内的向量。如果我们有两个以上的矢量在一个以上的平面内,首先在它们的平面内将两个矢量相加,然后再使用这些规则将结果加到下一个矢量上。一直这样做,直到所有向量都相加为止。

FREE-BODY FORCE DIAGRAMS 自由体力图

If we clarify what forces are acting on an object, it can be simpler to calculate how it will move. To do this, we usually draw a free-body force diagram, which has the object isolated, and all the forces that act on it drawn in at the points where they act. Forces acting on other objects, and those other objects, are not drawn. For example, fig G could be said to be a free-body force diagram of the rugby player being tackled in fig B, and this would lead us to draw fig C C C\mathbf{C} and fig D D D\mathbf{D} to make our resultant calculations.
如果我们明确了作用在物体上的力,那么计算物体如何运动就会变得简单。为此,我们通常绘制自由体受力图,将物体孤立起来,并在作用点上画出作用在物体上的所有力。作用在其他物体和这些其他物体上的力则不画出来。例如,图 G 可以说是图 B 中被扑倒的橄榄球运动员的自由体受力图,这将导致我们绘制图 C C C\mathbf{C} 和图 D D D\mathbf{D} 以进行结果计算。

CHECKPONT

  1. Work out the resultant force on a toy car if it has the following forces acting on it:
    如果玩具车受到以下力的作用,计算出玩具车受到的合力:
  • rubber band motor driving forwards 8.4 N
    橡皮筋电机向前驱动 8.4 N
  • air resistance 0.5 N 空气阻力 0.5 N
  • friction 5.8 N 摩擦 5.8 N
  • child’s hand pushing forward 10 N .
    孩子的手向前推 10 N .
  1. As a small plane accelerates to take off, the lift force on it is 6000 N vertically upwards, whilst the thrust is 2800 N horizontally forwards. What is the resultant of these forces on the plane?
    一架小型飞机加速起飞时,垂直向上的升力为 6000 N,而水平向前的推力为 2800 N。飞机受到的这些力的结果是什么?
  2. Draw a free-body force diagram of yourself sitting on your chair.
    画出自己坐在椅子上的自由体受力图。
  3. (a) Draw the scale diagram of fig E, and work out what the resultant force would be.
    (a) 画出图 E 的比例图,并计算出结果力是多少。

    (b) Use the parallelogram rule, as in fig F F F\mathbf{F}, to check your answer to part (a).
    (b) 如图 F F F\mathbf{F} 所示,使用平行四边形法则检查 (a) 部分的答案。
  4. In order to try and recover a car stuck in a muddy field, two tractors pull on it. The first acts at an angle of 20 20 20^(@)20^{\circ} left of the forwards direction with a force of 2250 N . The second acts 15 15 15^(@)15^{\circ} to the right of the forwards direction with a force of 2000 N . Draw a scale diagram of the situation and find the resultant force on the stuck car.
    为了把一辆陷在泥泞田地里的汽车拖出来,两辆拖拉机对其进行牵引。第一辆拖拉机向前进方向左侧 20 20 20^(@)20^{\circ} 倾斜,作用力为 2250 N。第二辆拖拉机以 15 15 15^(@)15^{\circ} 的角度向前进方向的右侧拉车,力为 2000 N。画出该情况的比例图,并求出卡住的汽车所受的合力。

SUBJECT VOCABULARY 主题词汇

resultant force the total force (vector sum) acting on a body when all the forces are added together accounting for their directions
作用在物体上的所有力相加后的总力(矢量总和),同时考虑到力的方向

free-body force diagram diagram showing an object isolated, and all the forces that act on it are drawn in at the points where they act, using arrows to represent the forces
自由体受力示意图 显示一个孤立的物体,所有作用在该物体上的力都画在其作用点上,并用箭头表示这些力

LEARNING OBJECTIVES 学习目标

  • Calculate the moment of a force.
    计算力矩。
Apply the principle of moments.
应用时刻原则。

Find the centre of gravity of an object.
找出物体的重心。

Forces on an object could act so that the object does not start to move along, but instead rotates about a fixed pivot. If the object is fixed so that it cannot rotate, it will bend.
物体上的力可能会使物体不开始移动,而是围绕一个固定的枢轴旋转。如果物体被固定,无法旋转,那么它就会弯曲。

THE MOMENT OF A FORCE
力矩

A fig A A force acts on a beam fixed at a point. The moment of a force causes rotation or, in this case, bending.
一个力作用在固定于一点的梁上。力矩会导致旋转,在本例中则会导致弯曲。

The tendency to cause rotation is called the moment of a force. It is calculated from:
导致旋转的趋势称为力矩。计算方法如下

moment ( Nm ) = ( Nm ) = (Nm)=(\mathrm{Nm})= force ( N ) × ( N ) × (N)xx(\mathrm{N}) \times perpendicular distance from the pivot to the line of action of the force ( m ) ( m ) (m)(m)
力矩 ( Nm ) = ( Nm ) = (Nm)=(\mathrm{Nm})= ( N ) × ( N ) × (N)xx(\mathrm{N}) \times 从支点到力的作用线的垂直距离 ( m ) ( m ) (m)(m)

moment = F x = F x =Fx=\boldsymbol{F} x 时刻 = F x = F x =Fx=\boldsymbol{F} x

A fig B The calculation of moment only considers the perpendicular distance between the line of action of the force and the axis of rotation, through the pivot point. When free to rotate, a body will turn in the direction of any net moment.
A fig B 力矩的计算只考虑力的作用线与旋转轴之间通过支点的垂直距离。当自由转动时,物体将沿任何净力矩的方向转动。

PRINCIPLE OF MOMENTS 力矩原理


fig C Balanced moments create an equilibrium situation.
图 C 平衡力矩创造了一种平衡状态。

If we add up all the forces acting on an object and the resultant force, accounting for their directions, is zero, then the object will be in equilibrium. Therefore it will remain stationary or, if it is already moving, it will carry on moving at the same velocity. The object could keep a constant velocity, but if the moments on it are not also balanced, it could be made to start rotating. The principle of moments tells us that if the total of all the moments trying to turn an object clockwise is equal to the total of all moments trying to turn an object anticlockwise, then it will be in rotational equilibrium. This means it will either remain stationary, or if it is already rotating it will continue at the same speed in the same direction.
如果我们把作用在一个物体上的所有力相加,并且考虑到它们的方向,结果力为零,那么这个物体将处于平衡状态。因此,它将保持静止,或者,如果它已经在运动,它将以相同的速度继续运动。物体可以保持匀速运动,但如果它受到的力矩不平衡,就会开始旋转。力矩原理告诉我们,如果试图使物体顺时针旋转的所有力矩的总和等于试图使物体逆时针旋转的所有力矩的总和,那么物体就会处于旋转平衡状态。这意味着它要么保持静止,要么如果它已经在旋转,就会以相同的速度沿相同的方向继续旋转。
A fig D As the metre-long beam is balanced, the sum of all the clockwise moments must equal the sum of all the anticlockwise moments.
D 由于一米长的横梁是平衡的,所有顺时针方向的力矩之和必须等于所有逆时针方向的力矩之和。

LEARNING TIP 学习提示

The clockwise moments and the anticlockwise moments must all be taken about the same pivot point.
顺时针力矩和逆时针力矩都必须取自同一个支点。

EXAM HINT 考试提示

Note that the steps and layout of the solution in this worked example are suitable for moments questions in the exam.
请注意,本工作示例中的解题步骤和布局适合考试中的片刻问题。

WORKED EXAMPLE 工作范例

In fig D, we can work out the weight of the beam if we know all the other weights and distances. The beam is uniform, so its weight will act from its centre. The length of the beam is 100 cm . So if x 1 = 20 cm x 1 = 20 cm x_(1)=20cmx_{1}=20 \mathrm{~cm}, then x 0 x 0 x_(0)x_{0} must be 30 cm , and x 2 = 80 cm x 2 = 80 cm x_(2)=80cmx_{2}=80 \mathrm{~cm}. The dinosaur ( W 1 ) W 1 (W_(1))\left(W_{1}\right) weighs 5.8 N and the toy car’s weight ( W 2 ) W 2 (W_(2))\left(W_{2}\right) is 0.95 N . In equilibrium, principle of moments: sum of anticlockwise moments = sum of clockwise moments
在图 D 中,如果我们知道所有其他重量和距离,就可以算出横梁的重量。横梁是均匀的,因此它的重量将从中心开始作用。横梁的长度为 100 厘米。因此,如果 x 1 = 20 cm x 1 = 20 cm x_(1)=20cmx_{1}=20 \mathrm{~cm} ,那么 x 0 x 0 x_(0)x_{0} 一定是 30 厘米, x 2 = 80 cm x 2 = 80 cm x_(2)=80cmx_{2}=80 \mathrm{~cm} 。恐龙 ( W 1 ) W 1 (W_(1))\left(W_{1}\right) 重 5.8 N,玩具车的重量 ( W 2 ) W 2 (W_(2))\left(W_{2}\right) 为 0.95 N。 在平衡状态下,力矩原理:逆时针力矩之和 = 顺时针力矩之和
W 1 x 1 = W 0 x 0 + W 2 x 2 5.8 × 0.20 = W 0 × 0.30 + 0.95 × 0.80 W 0 = 1.16 ( 0.76 ) 0.30 W 0 = 1.3 N W 1 x 1 = W 0 x 0 + W 2 x 2 5.8 × 0.20 = W 0 × 0.30 + 0.95 × 0.80 W 0 = 1.16 ( 0.76 ) 0.30 W 0 = 1.3 N {:[W_(1)x_(1)=W_(0)x_(0)+W_(2)x_(2)],[5.8 xx0.20=W_(0)xx0.30+0.95 xx0.80],[:.quadW_(0)=(1.16-(0.76))/(0.30)],[W_(0)=1.3N]:}\begin{aligned} W_{1} x_{1} & =W_{0} x_{0}+W_{2} x_{2} \\ 5.8 \times 0.20 & =W_{0} \times 0.30+0.95 \times 0.80 \\ \therefore \quad W_{0} & =\frac{1.16-(0.76)}{0.30} \\ W_{0} & =1.3 \mathrm{~N} \end{aligned}

CENTRE OF GRAVITY 重心

The weight of an object is caused by the gravitational attraction between the Earth and each particle contained within the object. The sum of all these tiny weight forces appears to act from a single point for any object, and this point is called the centre of gravity. For a symmetrical object, we can calculate the position of its centre of gravity, as it must lie on every line of symmetry. The point of intersection of all lines of symmetry will be the centre of gravity. Fig E illustrates this with two-dimensional shapes, but the idea can be extended into three dimensions. For example, the centre of gravity of a sphere is at the sphere’s centre. together and this answer used to calculate some sort of combined moment.
物体的重量是由地球与物体内每个微粒之间的引力造成的。对于任何物体来说,所有这些微小重力的总和似乎都是从一个点开始作用的,这个点就叫做重心。对于对称物体,我们可以计算出其重心的位置,因为它必须位于每条对称线上。所有对称线的交点就是重心。图 E 展示的是二维形状,但这一想法可以扩展到三维空间。例如,球体的重心位于球心。

LEARNING TIP You can consider the terms 'centre of gravity' and 'centre of mass' to mean the same thing. They are exactly the same for objects that are small compared
LEARNING TIP You can consider the terms 'centre of gravity' and 'centre of mass' to mean the same thing. They are exactly the same for objects that are small compared
LEARNING TIP You can consider the terms 'centre of gravity' and 'centre of mass' to mean the same thing. They are exactly the same for objects that are small compared
LEARNING TIP You can consider the terms 'centre of gravity' and 'centre of mass' to mean the same thing. They are exactly the same for objects that are small compared
LEARNING TIP You can consider the terms 'centre of gravity' and 'centre of mass' to mean the same thing. They are exactly the same for objects that are small compared
LEARNING TIP You can consider the terms 'centre of gravity' and 'centre of mass' to mean the same thing. They are exactly the same for objects that are small compared to the size of the Earth.
学习提示 你可以认为 "重心 "和 "质量中心 "是同一个意思。对于体积小的物体来说,这两个词的含义完全相同。 学习提示 你可以认为 "重心 "和 "质量中心 "是同一个意思。对于体积小的物体来说,这两个词的含义完全相同。 学习提示 你可以认为 "重心 "和 "质量中心 "是同一个意思。对于体积小的物体来说,这两个词的含义完全相同。 学习提示 你可以认为 "重心 "和 "质量中心 "是同一个意思。对于体积小的物体来说,这两个词的含义完全相同。 学习提示 你可以认为 "重心 "和 "质量中心 "是同一个意思。对于体积小的物体来说,这两个词的含义完全相同。 学习提示 你可以认为 "重心 "和 "质量中心 "是同一个意思。对于与地球大小相比很小的物体来说,这两个词的意思完全相同。

In order to calculate the sum of the moments in either direction, each individual moment must be calculated first and these individual moments then added together. The weights and/or
为了计算任一方向的力矩总和,必须先计算每个单独的力矩,然后将这些单独的力矩相加。权重和/或

distances cannot be added
距离不能增加
  • fig E E E\mathbf{E} The centre of gravity of a symmetrical object lies at the intersection of all lines of symmetry.
    E E E\mathbf{E} 对称物体的重心位于所有对称线的交点上。

IRREGULAR OBJECTS 不规则物体

The centre of gravity of an irregularly shaped object will still follow the rule that it is the point at which its weight appears to act on the object. A Bunsen burner, for example, has a heavy base. As such, the centre of gravity is low down near that concentration of mass, as there will be a greater attraction by the Earth’s gravity to this large mass.
不规则形状物体的重心仍然遵循这样的规则,即重心是物体重量的作用点。例如,本生灯的底座很重。因此,重心在靠近质量集中点的低处,因为地球引力对这个大质量物体的吸引力更大。
  • fig F Balancing a broom on its centre of gravity.
    F 用重心平衡扫帚。

fig G inding the centre of gravity of an irregular rod (broom).
为不规则杆(扫帚)确定重心。

PRACTICAL SKILLS 实用技能

Finding the centre of mass of an irregular rod
找到不规则杆的质心

In this investigation, we use the principle of moments to find the centre of gravity of a broom. It is not easy to estimate/determine the (location of the) centre of gravity by looking at a broom because it is not a symmetrical object. With the extra mass at the brush head end, the centre of gravity will be nearer that end. If you can balance the broom on the edge of a thick metal ruler, then the centre of gravity must lie above the ruler edge. As the perpendicular distance from the line of action to the weight is zero, the moment is zero so the broom sits in equilibrium.
在本探究中,我们利用力矩原理来找到扫帚的重心。要通过观察扫帚来估计/确定重心(位置)并不容易,因为它不是一个对称的物体。由于刷头一端有额外的质量,所以重心会靠近刷头一端。如果您能在厚金属尺的边缘上平衡扫帚,那么重心一定在尺边的上方。由于从作用线到重物的垂直距离为零,力矩为零,因此扫帚处于平衡状态。

You will probably find it difficult to balance the broom exactly, so you can use an alternative method. First you measure the mass of the broom ( M ) ( M ) (M)(M) using a digital balance. Then you use a set of hanging masses (of mass m m mm ) to balance the broom more in the middle of the handle, as in fig G . When the broom is balanced, you measure the distance ( d ) ( d ) (d)(d) from the hanging masses to the pivot. You calculate the distance ( x ) ( x ) (x)(x) from the pivot to the centre of gravity of the broom using the principle of moments:
您可能会发现很难精确地平衡扫帚,因此您可以使用另一种方法。首先,用数字天平测量扫帚的质量 ( M ) ( M ) (M)(M) 。然后使用一组悬挂物(质量为 m m mm )使扫帚在手柄中部保持平衡,如图 G 所示。扫帚平衡后,测量悬挂物到枢轴的距离 ( d ) ( d ) (d)(d) 。利用力矩原理计算从支点到扫帚重心的距离 ( x ) ( x ) (x)(x)

clockwise moment = anticlockwise moment
顺时针力矩 = 逆时针力矩
m g × d = M g × x x = m d M m g × d = M g × x x = m d M {:[mg xx d=Mg xx x],[x=(md)/(M)]:}\begin{aligned} m g \times d & =M g \times x \\ x & =\frac{m d}{M} \end{aligned}
Note: Do not get into the habit of using only the mass in moments calculations, as the definition is force times distance. It just happens that in this case g g g\boldsymbol{g} cancels on each side.
注意:在计算力矩时,不要养成只使用质量的习惯,因为力矩的定义是力乘以距离。在这种情况下, g g g\boldsymbol{g} 恰好在两边抵消。
Safety Note: Securely clamp the base of the stand to the bench. Do not use a very heavy broom as you would need a heavier counterweight which could fall and cause injury.
安全提示:将支架底座牢牢夹在工作台上。不要使用非常重的扫帚,因为您需要更重的配重,这样可能会掉下来造成伤害。

CHECKPOINT 检查点

  1. What is the moment of a 252 N force acting on a solid object at a perpendicular distance of 1.74 m from an axis of rotation of the object?
    在距离物体旋转轴 1.74 米的垂直距离上,一个 252 牛顿的力作用在一个实心物体上,其力矩是多少?
  2. A child and his father are playing on a seesaw, see fig H. They are exactly balanced when the boy (mass 46 kg ) sits at the end of the seesaw, 2.75 m from the pivot. If his father weighs 824 N , how far is he from the pivot?
    当男孩(质量为 46 千克)坐在跷跷板的末端,距离支点 2.75 米时,他们正好平衡。如果他的父亲重 824 N,他离支点有多远?

fig H H
3. The weight of the exercise book in the left-hand picture in fig B B BB causes a rotation so it moves towards the second position. Explain why it does not continue rotating but comes to rest in the position of the second picture.
3.图 B B BB 中左侧图片中练习本的重量引起了旋转,因此它向第二个位置移动。请解释为什么它没有继续旋转,而是停在了第二幅图的位置上。

4. If the same set-up as shown in fig D was used again, but the toy car was replaced with a banana weighing 1.4 N , find out where the banana would have to be positioned for the beam to balance calculate the new x 3 x 3 x_(3)x_{3}.
4.如果再次使用图 D 所示的相同装置,但将玩具车换成重 1.4 N 的香蕉,请找出香蕉必须放在什么位置才能使横梁平衡,计算出新的 x 3 x 3 x_(3)x_{3}

SUBJECT VOCABULARY 主题词汇

equilibrium the situation for a body where there is zero resultant force and zero resultant moment. It will have zero acceleration
平衡是指一个物体的结果力为零,结果力矩为零。加速度为零

principle of moments a body will be in equilibrium if the sum of clockwise moments acting on it is equal to the sum of the anticlockwise moments
力矩原理 如果作用在物体上的顺时针力矩之和等于逆时针力矩之和,则该物体处于平衡状态。

centre of gravity the point through which the weight of an object appears to act
重心 物体重量的作用点

1A 5 NEWTON'S LAWS OF MOTION
1A 5 牛顿运动定律

LEARNING OBJECTIVES 学习目标

Recall Newton’s laws of motion and use them to explain the acceleration of objects.
回顾牛顿运动定律,并用它们解释物体的加速度。

Make calculations using Newton’s second law of motion.
利用牛顿第二运动定律进行计算。

Identify pairs of forces involved in Newton’s third law of motion.
识别牛顿第三运动定律中涉及的力对。
Sir Isaac Newton was an exceptional thinker and scientist. His influence over science in the West is still enormous, despite the fact that he lived from 1642 to 1727 . He was a professor at Cambridge University, a member of the British Parliament, and a president of the respected scientific organisation, the Royal Society, in London. Probably his most famous contribution to science was the development of three simple laws governing the movement of objects subject to forces.
艾萨克-牛顿爵士是一位杰出的思想家和科学家。尽管他从 1642 年生活到 1727 年,但他对西方科学的影响仍然巨大。他曾任剑桥大学教授、英国议会议员和伦敦备受尊敬的科学组织皇家学会主席。他对科学最著名的贡献可能是提出了三条简单的定律,用以指导受力物体的运动。

Δ Δ Delta\Delta fig A A portrait painting of Sir Isaac Newton
Δ Δ Delta\Delta 图 A 艾萨克-牛顿爵士的肖像画

NEWTON'S FIRST LAW OF MOTION
牛顿第一运动定律

If an object is stationary there needs to be a resultant force on it to make it move. We saw how to calculate resultant forces in Section 1A.3. If the object is already moving then it will continue at the same speed in the same direction unless a resultant force acts on it. If there is no resultant force on an object - either because there is zero force acting or all the forces balance out then the object’s motion is not changed.
如果一个物体静止不动,就需要有一个作用在它上面的结果力来使它运动。我们在第 1A.3 节中学习了如何计算结果力。如果物体已经在运动,那么它将以相同的速度沿相同的方向继续运动,除非有一个作用在它身上的结果力。如果物体上没有结果力--要么是作用力为零,要么是所有力平衡,那么物体的运动就不会改变。

NEWTON'S SECOND LAW OF MOTION
牛顿第二运动定律

This law tells us how much an object’s motion will be changed by a resultant force. For an object with constant mass, it is usually written mathematically:
这一定律告诉我们,一个物体的运动会因一个结果力而发生多大的改变。对于质量不变的物体,通常用数学方法来表示:
Σ F = ma resultant force ( N ) = mass ( kg ) × acceleration ( m s 2 ) Σ F =  ma   resultant force  ( N ) =  mass  ( kg ) ×  acceleration  m s 2 {:[Sigma F=" ma "],[" resultant force "(N)=" mass "(kg)xx" acceleration "(ms^(-2))]:}\begin{aligned} \Sigma F & =\text { ma } \\ \text { resultant force }(\mathrm{N}) & =\text { mass }(\mathrm{kg}) \times \text { acceleration }\left(\mathrm{m} \mathrm{~s}^{-2}\right) \end{aligned}
For example, this relationship allows us to calculate the acceleration due to gravity ( g ) ( g ) (g)(g) if we measure the force ( F ) ( F ) (F)(F) accelerating a mass ( m ) ( m ) (m)(m) downwards.
例如,如果我们测量了加速质量 ( m ) ( m ) (m)(m) 向下运动的力 ( F ) ( F ) (F)(F) ,就可以通过这种关系计算出重力加速度 ( g ) ( g ) (g)(g)
F = m a = m g g = F m F = m a = m g g = F m {:[F=ma=mg],[:.quad g=(F)/(m)]:}\begin{aligned} & F=m a=m g \\ \therefore \quad g & =\frac{F}{m} \end{aligned}
A fig B A stationary object will not move unless it is acted upon by a resultant force.
A 无花果 B 静止的物体除非受到结果力的作用,否则不会移动。
PRACTICAL SKILLS 实用技能
You can use the set-up shown in fig C , or a similar experiment on an air track. This set-up measures the acceleration for various values of the resultant force that acts on the trolley whilst you keep its mass constant (table A). By plotting a graph of acceleration against resultant force, a straight line will show that acceleration is proportional to the resultant force. You could also plot a graph for varying masses of trolley whilst you keep the resultant force constant (table B).
您可以使用图 C 所示的装置,或在空气轨道上进行类似的实验。在保持小车质量不变的情况下,测量作用在小车上的不同结果力值的加速度(表 A)。通过绘制加速度与结果力的关系图,一条直线将显示加速度与结果力成正比。您还可以在保持结果力不变的情况下,绘制不同质量手推车的曲线图(表 B)。
Safety Note: Place heavy trolleys and runways where they cannot slide or fall off benches and cause injuries. Place air track ‘blowers’ on the floor. Secure the hose so that it cannot come loose and blow dirt and dust into eyes.
安全提示:将重型手推车和跑道放置在不会滑动或掉落的地方,以免造成伤害。将空气轨道 "鼓风机 "放在地板上。固定好软管,防止其松脱并将污物和灰尘吹入眼睛。
FORCE / N / N //N/ \mathbf{N} ACCELERATION / m s 2 / m s 2 //ms^(-2)/ \mathrm{m} \mathrm{s}^{-2} 加速度 / m s 2 / m s 2 //ms^(-2)/ \mathrm{m} \mathrm{s}^{-2}
0.1 0.20
0.2 0.40
0.3 0.60
0.4 0.80
0.5 1.00
0.6 1.20
FORCE //N ACCELERATION //ms^(-2) 0.1 0.20 0.2 0.40 0.3 0.60 0.4 0.80 0.5 1.00 0.6 1.20| FORCE $/ \mathbf{N}$ | ACCELERATION $/ \mathrm{m} \mathrm{s}^{-2}$ | | :---: | :---: | | 0.1 | 0.20 | | 0.2 | 0.40 | | 0.3 | 0.60 | | 0.4 | 0.80 | | 0.5 | 1.00 | | 0.6 | 1.20 |
table A Values of acceleration for different forces acting on a trolley.
表 A 作用在小车上的不同力的加速度值。
MASS / kg / kg //kg/ \mathrm{kg} 质量 / kg / kg //kg/ \mathrm{kg} ACCELERATION / m s 2 / m s 2 //ms^(-2)/ \mathrm{m} \mathrm{s}^{-2} 加速度 / m s 2 / m s 2 //ms^(-2)/ \mathrm{m} \mathrm{s}^{-2}
0.5 1.00
0.6 0.83
0.7 0.71
0.8 0.63
0.9 0.55
1.0 0.50
MASS //kg ACCELERATION //ms^(-2) 0.5 1.00 0.6 0.83 0.7 0.71 0.8 0.63 0.9 0.55 1.0 0.50| MASS $/ \mathrm{kg}$ | ACCELERATION $/ \mathrm{m} \mathrm{s}^{-2}$ | | :---: | :---: | | 0.5 | 1.00 | | 0.6 | 0.83 | | 0.7 | 0.71 | | 0.8 | 0.63 | | 0.9 | 0.55 | | 1.0 | 0.50 |
table B Values of acceleration resulting from an applied force of 0.5 N when the mass of the trolley is varied.
表 B 0.5 N 的外力在小车质量变化时产生的加速度值。
A fig D Graph of results from the first investigation into Newton’s second law. Acceleration is directly proportional to force.
A 图 D 牛顿第二定律首次探究的结果图。加速度与力成正比。
A fig E Graph of results from the second investigation into Newton’s second law. The x x xx-axis represents the derived data of ’ 1 /mass’, so the straight best-fit line shows that acceleration is inversely proportional to mass. Note that there is no need for a graph to start at the origin: choose axes scales that will best show the pattern in the data by making the points fill the graph paper.
图 E 牛顿第二定律第二次探究的结果图。 x x xx 轴表示 "1 /质量 "的推导数据,因此最佳拟合直线表示加速度与质量成反比。请注意,图表不需要从原点开始:选择最能显示数据模式的坐标轴刻度,使点填满图表纸。
Experimental verification of Newton’s second law is well established. The investigation shown in fig C C C\mathbf{C} demonstrates that: a = F m a = F m a=(F)/(m)\boldsymbol{a}=\frac{\boldsymbol{F}}{\boldsymbol{m}}
牛顿第二定律的实验验证是公认的。图 C C C\mathbf{C} 所示的研究证明了这一点: a = F m a = F m a=(F)/(m)\boldsymbol{a}=\frac{\boldsymbol{F}}{\boldsymbol{m}}

LEARNING TIP 学习提示

Straight-line graphs 直线图

Physicists always try to arrange their experimental data into graphs that produce a straight best-fit line. This proves a linear relationship between the experimental variables, and can also give us numerical information about the quantities involved.
物理学家总是试图将他们的实验数据绘制成能够产生最佳拟合直线的图表。这证明了实验变量之间的线性关系,也能为我们提供相关量的数字信息。

The equation for a straight line is:
直线的方程是
y = m x + c y = m x + c y=mx+cy=m x+c
m m mm ’ in the equation is the gradient of the straight line, and ’ c c cc ’ is the value on the y y yy-axis where the line crosses it, known as the y y yy-intercept. If we plot experimental data on a graph and get a straight best-fit line, then this proves the quantities we plotted on x x xx and y y yy have a linear relationship. It is referred to as ‘directly proportional’ (or simply ‘proportional’) only if the line also passes through the origin, meaning that c = 0 c = 0 c=0c=0. The graphs in fig D and fig E demonstrate experimental verification of Newton’s second law. In each case, the third variable in the equation was kept constant as a control variable. For example, in fig D D DD the straight best-fit line, showing that y x y x y prop xy \propto x, proves that a F a F a prop Fa \propto F, and the gradient of the best-fit line would represent the reciprocal of the mass that was accelerated and was kept constant throughout, as a control variable. In this example c = 0 c = 0 c=0c=0, which means that the proportional relationships are simple:
等式中的' m m mm '是直线的梯度,' c c cc '是直线与 y y yy 轴交叉处的值,称为 y y yy 截距。如果我们在图表上绘制实验数据并得到一条最佳拟合直线,那么这就证明我们在 x x xx y y yy 上绘制的量具有线性关系。只有当直线也通过原点,即 c = 0 c = 0 c=0c=0 时,才称为 "成正比"(或简称 "成正比")。图 D 和图 E 中的图形展示了牛顿第二定律的实验验证。在每种情况下,方程中的第三个变量作为控制变量保持不变。例如,在图 D D DD 中,最佳拟合直线显示 y x y x y prop xy \propto x ,证明了 a F a F a prop Fa \propto F ,最佳拟合直线的梯度代表被加速质量的倒数,作为控制变量始终保持不变。在这个例子中, c = 0 c = 0 c=0c=0 ,这意味着比例关系很简单:
y = m x a = F m y = m x a = F m {:[y=mx],[a=(F)/(m)]:}\begin{aligned} & y=m x \\ & a=\frac{F}{m} \end{aligned}

LEARNING TIP 学习提示

Identifying Newton’s third law of force pairs can confuse people. To find the two forces, remember they must always act on different objects, and must always have the same cause. In fig F F FF, it is the mutual repulsion of electrons in the atoms of the football and the boot that cause the action and reaction.
牛顿第三力定律对力的识别可能会让人感到困惑。要找到这两个力,请记住它们必须总是作用在不同的物体上,而且必须总是有相同的原因。在图 F F FF 中,是足球和靴子原子中电子的相互排斥造成了作用力和反作用力。

As both the above equations represent the graph in fig D D DD, it follows that the gradient, m m mm, equals the reciprocal of the mass, 1 / m 1 / m 1//m1 / \mathrm{m}. It is just coincidence that the symbol ’ m m mm ’ for gradient, and ’ m m mm ’ for mass are the same letter in this example.
由于上述两个等式都表示图 D D DD 中的图形,因此梯度 m m mm 等于质量 1 / m 1 / m 1//m1 / \mathrm{m} 的倒数。在这个例子中,表示梯度的符号" m m mm "和表示质量的符号" m m mm "是同一个字母,这只是巧合。

Note that graphs in physics are causal relationships. In fig D, the acceleration is caused by the force. It is very rare in physics that a graph would represent a statistical correlation, and so phrases such as ‘positive correlation’ do not correctly describe graphs of physics experiments.
请注意,物理学中的图形是因果关系。在图 D 中,加速度是由力引起的。在物理学中,图形表示统计相关性的情况非常罕见,因此 "正相关 "等短语并不能正确描述物理实验的图形。

Similarly, it is very rare that a graph of a physics experiment would be correctly drawn if the points are joined ‘dot-to-dot’. In most cases a best-fit line should be drawn, as has been done in fig E E E\mathbf{E}.
同样,如果把物理实验的点 "点对点 "连接起来,就能正确绘制出物理实验的图表,这种情况非常罕见。在大多数情况下,应该绘制一条最佳拟合线,如图 E E E\mathbf{E} 所示。

NEWTON'S THIRD LAW OF MOTION
牛顿第三运动定律

‘When an object A causes a force on another object B, then object B causes an equal force in the opposite direction to act upon object A’. For example, when a skateboarder pushes off from a wall, they exert a force on the wall with their hand. At the same time, the wall exerts a force on the skateboarder’s hand. This equal and opposite reaction force is what they can feel with the sense of touch, and as the skateboard has very low friction, the wall’s push on them causes acceleration away from the wall. As the wall is connected to the Earth, the Earth and wall combination will accelerate in the opposite direction. The Earth has such a large mass that its acceleration can’t be noticed. That’s Newton’s second law again: acceleration is inversely proportional to mass; huge mass means tiny acceleration.
当一个物体 A 对另一个物体 B 产生一个力时,物体 B 就会对物体 A 产生一个方向相反的相等的力"。例如,当滑板运动员从墙上推下时,他们的手对墙施加了一个力。与此同时,墙壁也对滑板者的手施加了一个力。由于滑板的摩擦力非常小,墙壁对他们的推力会导致他们加速离开墙壁。由于墙壁与地球相连,地球和墙壁的组合将向相反的方向加速。地球的质量很大,所以它的加速度不会被注意到。这又是牛顿第二定律:加速度与质量成反比;质量大意味着加速度小。

fig F F F\mathbf{F} When the boot puts a force on the football, the football causes an equal and opposite force on the boot. The footballer can feel the kick because they feel the reaction force from the ball on their toe.
F F F\mathbf{F} 当靴子对足球施加一个力时,足球也会对靴子施加一个相等且相反的力。足球运动员能感觉到踢球,因为他们的脚趾能感觉到球产生的反作用力。

CHECKPOINT 检查点

  1. In terms of Newton’s laws of motion:
    根据牛顿运动定律:

    (a) Explain why this book will sit stationary on a table.
    (a) 解释为什么这本书会静止地放在桌子上。

    (b) Describe and explain what will happen if your hands then put an upwards force on the book that is greater than its weight.
    (b) 描述并解释如果你的手向上施加的力大于书的重量,会发生什么情况。

    © Explain why you feel the book when your hands put that upwards force on it.
    解释为什么当你的手向上用力时会感觉到书的存在。
  2. (a) Calculate the gradient of the best-fit line on the graph in fig D D DD and thus work out the mass of the trolley that was accelerated in the first investigation.
    (a) 计算图 D D DD 中图形上最佳拟合线的梯度,从而算出在第一次探究中被加速的小车的质量。

    (b) State what quantity the gradient of the line on the graph in fig E represents. Calculate the value of that quantity.
    (b) 说出图 E 中曲线的梯度代表什么量。计算该量的值。
  3. Calculate the acceleration in each of the following cases:
    计算以下每种情况下的加速度:

    (a) A mass of 12.0 kg experiences a resultant force of 785 N .
    (a) 质量为 12.0 千克的物体受到 785 牛的作用力。

    (b) A force of 22.2 N acts on a 3.1 kg mass.
    (b) 22.2 N 的力作用在 3.1 kg 的质量上。

    © A 2.0 kg bunch of bananas is dropped. The bunch weighs 19.6 N .
    © 一束 2.0 千克的香蕉掉落。这串香蕉重 19.6 N。

    (d) During a tackle, two footballers kick a stationary ball at the same time, with forces acting in opposite directions. One kick has a force of 210 N , the other has a force of 287 N . The mass of the football is 430 g .
    (d) 在一次擒抱中,两名足球运动员同时踢一个静止的球,力的作用方向相反。一个踢球的力是 210 N,另一个踢球的力是 287 N。足球的质量为 430 克。

SUBJECT VOCABULARY 主题词汇

Newton’s first law of motion an object will remain at rest, or in a state of uniform motion, until acted upon by a resultant force
牛顿第一运动定律 一个物体将保持静止或匀速运动状态,直到受到一个结果力的作用

Newton’s second law of motion if an object’s mass is constant, the resultant force needed to cause an acceleration is given by the equation:
牛顿第二运动定律 如果物体的质量不变,则产生加速度所需的合力由等式给出:
Σ F = m a Σ F = m a Sigma F=ma\Sigma F=m a
Newton’s third law of motion for every action, there is an equal and opposite reaction
牛顿运动第三定律 每个作用力都有一个相等和相反的反作用力

1 A 1 A 1A1 A
6 KINEMATICS EQUATIONS
1 A 1 A 1A1 A 6 运动方程

LEARNING OBJECTIVES 学习目标

Recall the simple kinematics equations.
回顾一下简单的运动学方程。
  • Calculate unknown variables using the kinematics equations.
    利用运动学方程计算未知变量。
Kinematics is the study of the movement of objects. We can use equations to find out details about the motion of objects accelerating in one dimension.
运动学是研究物体运动的学科。我们可以利用方程来找出物体在一维加速运动的细节。

fig A Kinematics is the study of the description of the motion of objects. The equations could be used to make calculations about the stone falling through the air, and separately about its motion through the water. The acceleration in air and in water will be different as the resultant force acting in each will be different.
运动学是一门描述物体运动的学科。这些方程可以用来计算石头在空气中的下落,也可以分别计算石头在水中的运动。在空气中和在水中的加速度会有所不同,因为作用在两者身上的合力不同。

ZERO ACCELERATION 零加速度

If an object has no resultant force acting on it then it does not accelerate. This is uniform motion. In this situation, calculations on its motion are very easy, as they simply involve the basic velocity equation:
如果一个物体上没有作用力,那么它就不会加速。这就是匀速运动。在这种情况下,计算物体的运动非常简单,只需计算基本的速度方程即可:
v = s t v = s t v=(s)/(t)v=\frac{s}{t}
The velocity is the same at the beginning and end of the motion, and if we need to find the displacement travelled, it is a simple case of multiplying velocity by time:
运动开始和结束时的速度是相同的,如果我们需要计算所经过的位移,只需将速度乘以时间即可:
s = v × t s = v × t s=v xx t\mathbf{s}=\boldsymbol{v} \times t

CONSTANT ACCELERATION 恒加速度

There are equations that allow us to work out the motion of an object that is moving with a constant acceleration. For these kinematics equations, the first step is to define the five variables used:
通过一些方程,我们可以计算出以恒定加速度运动的物体的运动情况。对于这些运动学方程,第一步是定义所使用的五个变量:

s s s\boldsymbol{s} - displacement ( m )
s s s\boldsymbol{s} - 位移 ( 米 )

u u uu - initial velocity ( ms 1 ) ms 1 (ms^(-1))\left(\mathrm{ms}^{-1}\right)
u u uu - 初始速度 ( ms 1 ) ms 1 (ms^(-1))\left(\mathrm{ms}^{-1}\right)

v v vv - final velocity ( m s 1 ) m s 1 (ms^(-1))\left(\mathrm{m} \mathrm{s}^{-1}\right)
v v vv - 最终速度 ( m s 1 ) m s 1 (ms^(-1))\left(\mathrm{m} \mathrm{s}^{-1}\right)

a a a-\boldsymbol{a}- acceleration ( m s 2 ) m s 2 (ms^(-2))\left(\mathrm{m} \mathrm{s}^{-2}\right)
a a a-\boldsymbol{a}- 加速度 ( m s 2 ) m s 2 (ms^(-2))\left(\mathrm{m} \mathrm{s}^{-2}\right)

t t tt - time (s)
t t tt - 时间(秒)

Each equation uses four of the variables, which means that if we know the values of any three variables, we can find out the other two.
每个等式使用四个变量,也就是说,如果我们知道任意三个变量的值,就可以求出另外两个变量的值。

ACCELERATION REDEFINED 重新定义加速度

By re-arranging the equation that defined acceleration, we come to the usual expression of the first kinematics equation:
通过重新排列定义加速度的方程,我们可以得出第一个运动学方程的常规表达式:
v = u + a t v = u + a t v=u+atv=u+a t
For example, if a stone is dropped off a cliff (see fig B) and takes three seconds to hit the ground, what is its speed when it does hit the ground?
例如,如果一块石头从悬崖上掉下来(见图 B),需要 3 秒钟才能落地,那么它落地时的速度是多少?

Identify the three things we know:
确定我们知道的三件事:
  • falling under gravity, so a = g = 9.81 m s 2 a = g = 9.81 m s 2 a=g=9.81ms^(-2)\boldsymbol{a}=\boldsymbol{g}=9.81 \mathrm{~m} \mathrm{~s}^{-2}
    在重力作用下下落,所以 a = g = 9.81 m s 2 a = g = 9.81 m s 2 a=g=9.81ms^(-2)\boldsymbol{a}=\boldsymbol{g}=9.81 \mathrm{~m} \mathrm{~s}^{-2}

    (constant acceleration, so the kinematics equations can be used)
    (加速度恒定,因此可以使用运动学方程)
  • starts at rest, so u = 0 m s 1 u = 0 m s 1 u=0ms^(-1)\boldsymbol{u}=0 \mathrm{~m} \mathrm{~s}^{-1}
    从静止开始,所以 u = 0 m s 1 u = 0 m s 1 u=0ms^(-1)\boldsymbol{u}=0 \mathrm{~m} \mathrm{~s}^{-1}
  • time to fall t = 3 s t = 3 s t=3st=3 \mathrm{~s}
v = u + a t = 0 + 9.81 × 3 = 29.43 v = 29.4 m s 1 v = u + a t = 0 + 9.81 × 3 = 29.43 v = 29.4 m s 1 {:[v=u+at=0+9.81 xx3=29.43],[v=29.4ms^(-1)]:}\begin{aligned} & v=u+a t=0+9.81 \times 3=29.43 \\ & v=29.4 \mathrm{~m} \mathrm{~s}^{-1} \end{aligned}

EXAM HINT 考试提示

Often the acceleration will not be clearly stated, but the object is falling under gravity, so:
加速度通常不会明确指出,但物体是在重力作用下下落的,所以:
a = g = 9.81 m s 2 a = g = 9.81 m s 2 a=g=9.81ms^(-2)a=g=9.81 \mathrm{~m} \mathrm{~s}^{-2}

EXAM HINT 考试提示

Often the initial velocity will not be explicitly stated, but the object starts ‘at rest’. This means it is stationary at the beginning, so:
通常不会明确说明初速度,但物体开始时是 "静止 "的。这意味着它在开始时是静止的,因此:
u = 0 m s 1 u = 0 m s 1 u=0ms^(-1)\boldsymbol{u}=0 \mathrm{~m} \mathrm{~s}^{-1}

DISTANCE FROM AVERAGE SPEED
平均速度距离

As the kinematics equations only work with uniform acceleration, the average speed during any acceleration will be halfway from the initial velocity to the final velocity. Therefore the distance travelled is the average speed multiplied by the time:
由于运动学方程只适用于匀加速运动,因此任何加速过程中的平均速度都是从初速度到最终速度的一半。因此,行驶距离就是平均速度乘以时间:
s = ( u + v ) 2 × t s = ( u + v ) 2 × t s=((u+v))/(2)xx t\boldsymbol{s}=\frac{(\boldsymbol{u}+\boldsymbol{v})}{2} \times t
For example, for the same stone dropping off the cliff as in the previous example, we could work out how high the cliff is.
例如,对于上一个例子中从悬崖上掉下的同一块石头,我们可以计算出悬崖有多高。
Identify the three things we know:
确定我们知道的三件事:
  • final velocity came to be v = 29.4 m s 1 v = 29.4 m s 1 v=29.4ms^(-1)\boldsymbol{v}=29.4 \mathrm{~m} \mathrm{~s}^{-1}
    最终速度达到 v = 29.4 m s 1 v = 29.4 m s 1 v=29.4ms^(-1)\boldsymbol{v}=29.4 \mathrm{~m} \mathrm{~s}^{-1}
  • starts at rest, so u = 0 m s 1 u = 0 m s 1 u=0ms^(-1)\boldsymbol{u}=0 \mathrm{~m} \mathrm{~s}^{-1}
    从静止开始,所以 u = 0 m s 1 u = 0 m s 1 u=0ms^(-1)\boldsymbol{u}=0 \mathrm{~m} \mathrm{~s}^{-1}
  • time to fall t = 3 s t = 3 s t=3st=3 \mathrm{~s}
s = ( u + v ) 2 × t = ( 0 + 29.4 ) 2 × 3 = 44.1 s = 44.1 m s = ( u + v ) 2 × t = ( 0 + 29.4 ) 2 × 3 = 44.1 s = 44.1 m {:[s=((u+v))/(2)xx t=((0+29.4))/(2)xx3=44.1],[s=44.1m]:}\begin{aligned} & s=\frac{(u+v)}{2} \times t=\frac{(0+29.4)}{2} \times 3=44.1 \\ & s=44.1 \mathrm{~m} \end{aligned}

fig B Calculations about falling objects are very common.
B 关于坠落物体的计算非常常见。

EXAM HINT 考试提示

The kinematics equations are only valid if there is a constant acceleration. If the acceleration is changing they cannot be used.
运动学方程只有在加速度恒定的情况下才有效。如果加速度在变化,则无法使用。

COMBINING EQUATIONS A 将方程 a

We can combine the equations
我们可以把这两个方程结合起来
v = u + a t v = u + a t v=u+atv=u+a t
and 
s = ( u + v ) 2 × t s = ( u + v ) 2 × t s=((u+v))/(2)xx ts=\frac{(u+v)}{2} \times t
By substituting the first of these equations into the second, we get the combination equation:
将第一个等式代入第二个等式,就得到了组合等式:
s = ( u + ( u + a t ) ) 2 × t = ( 2 u t + a t 2 ) 2 s = u t + 1 2 a t 2 s = ( u + ( u + a t ) ) 2 × t = 2 u t + a t 2 2 s = u t + 1 2 a t 2 {:[s=((u+(u+at)))/(2)xx t=((2ut+at^(2)))/(2)],[s=ut+(1)/(2)at^(2)]:}\begin{aligned} & s=\frac{(u+(u+a t))}{2} \times t=\frac{\left(2 u t+a t^{2}\right)}{2} \\ & s=u t+\frac{1}{2} a t^{2} \end{aligned}
We can use this equation to check again how high the stone was dropped from:
我们可以利用这个等式再次检验石头是从多高的地方掉下来的:
  • falling under gravity, so a = g = 9.81 m s 2 a = g = 9.81 m s 2 a=g=9.81ms^(-2)\boldsymbol{a}=\boldsymbol{g}=9.81 \mathrm{~m} \mathrm{~s}^{-2}
    在重力作用下下落,所以 a = g = 9.81 m s 2 a = g = 9.81 m s 2 a=g=9.81ms^(-2)\boldsymbol{a}=\boldsymbol{g}=9.81 \mathrm{~m} \mathrm{~s}^{-2}
  • starts at rest, so u = 0 m s 1 u = 0 m s 1 u=0ms^(-1)\boldsymbol{u}=0 \mathrm{~m} \mathrm{~s}^{-1}
    从静止开始,所以 u = 0 m s 1 u = 0 m s 1 u=0ms^(-1)\boldsymbol{u}=0 \mathrm{~m} \mathrm{~s}^{-1}
  • time to fall t = 3 s t = 3 s t=3st=3 \mathrm{~s}
s = u t + 1 2 a t 2 = ( 0 × 3 ) + ( 1 2 × 9.81 × 3 2 ) = 44.1 s = 44.1 m s = u t + 1 2 a t 2 = ( 0 × 3 ) + 1 2 × 9.81 × 3 2 = 44.1 s = 44.1 m {:[s=ut+(1)/(2)at^(2)=(0xx3)+((1)/(2)xx9.81 xx3^(2))=44.1],[s=44.1m]:}\begin{aligned} & s=u t+\frac{1}{2} a t^{2}=(0 \times 3)+\left(\frac{1}{2} \times 9.81 \times 3^{2}\right)=44.1 \\ & s=44.1 \mathrm{~m} \end{aligned}
Notice that the answer must come out the same, as we are calculating for the same cliff. This highlights the fact that we can use whichever equation is most appropriate for the information given.
请注意,答案必须是相同的,因为我们计算的是同一个悬崖。这说明我们可以根据所给信息使用最合适的方程。

COMBINING EQUATIONS B 将方程 b

s = ( u + v ) 2 × t t = 2 s ( u + v ) and v = u + a t s = ( u + v ) 2 × t t = 2 s ( u + v )  and  v = u + a t {:[quad s=((u+v))/(2)xx t],[:.quad t=(2s)/((u+v))],[" and "quad v=u+at]:}\begin{aligned} & \quad s=\frac{(u+v)}{2} \times t \\ & \therefore \quad t=\frac{2 s}{(u+v)} \\ & \text { and } \quad v=u+a t \end{aligned}
By substituting the first of these equations into the second, we get the combination equation:
将第一个等式代入第二个等式,就得到了组合等式:
v = u + a × 2 s ( u + v ) v ( u + v ) = u ( u + v ) + 2 a s v u + v 2 = u 2 + u v + 2 a s v u = u v so subtract from each side v 2 = u 2 + 2 a s v = u + a × 2 s ( u + v ) v ( u + v ) = u ( u + v ) + 2 a s v u + v 2 = u 2 + u v + 2 a s v u = u v  so subtract from each side  v 2 = u 2 + 2 a s {:[v=u+a xx(2s)/((u+v))],[:.quad v(u+v)=u(u+v)+2as],[:.quad vu+v^(2)=u^(2)+uv+2as quad vu=uv" so subtract from each side "],[v^(2)=u^(2)+2as]:}\begin{aligned} v & =u+a \times \frac{2 s}{(u+v)} \\ \therefore \quad v(u+v) & =u(u+v)+2 a s \\ \therefore \quad v u+v^{2} & =u^{2}+u v+2 a s \quad v u=u v \text { so subtract from each side } \\ v^{2} & =u^{2}+2 a s \end{aligned}
Check again what the stone’s final velocity would be:
再看看石头的最终速度是多少:
  • falling under gravity, so a = g = 9.81 ms 2 a = g = 9.81 ms 2 a=g=9.81ms^(-2)\boldsymbol{a}=\boldsymbol{g}=9.81 \mathrm{~ms}^{-2}
    在重力作用下下落,所以 a = g = 9.81 ms 2 a = g = 9.81 ms 2 a=g=9.81ms^(-2)\boldsymbol{a}=\boldsymbol{g}=9.81 \mathrm{~ms}^{-2}
  • starts at rest, so u = 0 m s 1 u = 0 m s 1 u=0ms^(-1)\boldsymbol{u}=0 \mathrm{~m} \mathrm{~s}^{-1}
    从静止开始,所以 u = 0 m s 1 u = 0 m s 1 u=0ms^(-1)\boldsymbol{u}=0 \mathrm{~m} \mathrm{~s}^{-1}
  • height to fall s = 44.1 m s = 44.1 m s=44.1ms=44.1 \mathrm{~m} 坠落高度 s = 44.1 m s = 44.1 m s=44.1ms=44.1 \mathrm{~m}
v 2 = u 2 + 2 a s = 0 2 + ( 2 × 9.81 × 44.1 ) = 865 v = 865 = 29.4 v = 29.4 m s 1 v 2 = u 2 + 2 a s = 0 2 + ( 2 × 9.81 × 44.1 ) = 865 v = 865 = 29.4 v = 29.4 m s 1 {:[v^(2)=u^(2)+2as=0^(2)+(2xx9.81 xx44.1)=865],[:.quad v=sqrt865=29.4],[v=29.4ms^(-1)]:}\begin{aligned} v^{2} & =u^{2}+2 a s=0^{2}+(2 \times 9.81 \times 44.1)=865 \\ \therefore \quad v & =\sqrt{865}=29.4 \\ v & =29.4 \mathrm{~m} \mathrm{~s}^{-1} \end{aligned}
Notice that the answer must come out the same as previously calculated, and this again highlights that there are many ways to reach the answer.
注意,得出的答案必须与之前计算的相同,这再次强调了得出答案的方法有很多种。
KINEMATICS EQUATION 运动学方程 QUANTITY NOT USED 未使用数量
v = u + a t v = u + a t v=u+at\boldsymbol{v}=\boldsymbol{u}+\boldsymbol{a t} distance 距离
s = ( u + v ) 2 × t s = ( u + v ) 2 × t s=((u+v))/(2)xx t\boldsymbol{s}=\frac{(\boldsymbol{u}+\boldsymbol{v})}{2} \times t acceleration 加速度
s = u t + 1 2 a t 2 s = u t + 1 2 a t 2 s=ut+(1)/(2)at^(2)\boldsymbol{s}=\boldsymbol{u} t+\frac{1}{2} \boldsymbol{a} t^{2} final velocity 最终速度
v 2 = u 2 + 2 a v 2 = u 2 + 2 a v^(2)=u^(2)+2a\boldsymbol{v}^{2}=\boldsymbol{u}^{2}+2 \boldsymbol{a} time 时间
KINEMATICS EQUATION QUANTITY NOT USED v=u+at distance s=((u+v))/(2)xx t acceleration s=ut+(1)/(2)at^(2) final velocity v^(2)=u^(2)+2a time| KINEMATICS EQUATION | QUANTITY NOT USED | | :---: | :--- | | $\boldsymbol{v}=\boldsymbol{u}+\boldsymbol{a t}$ | distance | | $\boldsymbol{s}=\frac{(\boldsymbol{u}+\boldsymbol{v})}{2} \times t$ | acceleration | | $\boldsymbol{s}=\boldsymbol{u} t+\frac{1}{2} \boldsymbol{a} t^{2}$ | final velocity | | $\boldsymbol{v}^{2}=\boldsymbol{u}^{2}+2 \boldsymbol{a}$ | time |
table A Each of the kinematics equations is useful, depending on the information we are given. If you know three quantities, you can always find a fourth by identifying which equation links those four quantities and rearranging that equation to find the unknown.
表 A 根据所给的信息,每个运动学方程都很有用。如果你知道三个量,你总是可以通过确定哪个方程将这四个量联系起来,并重新排列该方程来找到未知量,从而找到第四个量。

PRACTICAL SKILLS 实用技能

Finding the acceleration due to gravity by freefall
计算自由落体的重力加速度

A system for timing the fall of an object under gravity can allow us to measure the acceleration due to gravity. In this experiment, we measure the time taken by a falling object to drop under gravity from a certain height, and then alter the height and measure again.
利用重力作用下物体下落计时系统可以测量重力加速度。在本实验中,我们测量一个下落物体在重力作用下从一定高度落下所需的时间,然后改变高度并再次测量。

fig C C C\mathbf{C} The freefall time of an object from different heights allows us to find the acceleration due to gravity, g g g\boldsymbol{g}.
根据物体从不同高度自由下落的时间,我们可以求出重力加速度 g g g\boldsymbol{g}

If we vary the height from which the object falls, the time taken to land will vary. The kinematics equations tell us that:
如果我们改变物体下落的高度,落地所需的时间也会不同。运动方程告诉我们:
s = u t + 1 2 a t 2 s = u t + 1 2 a t 2 s=ut+(1)/(2)at^(2)s=u t+\frac{1}{2} a t^{2}
As it always starts from rest, u = 0 u = 0 u=0\boldsymbol{u}=0 throughout; the acceleration is that caused by gravity, g g gg; and the distance involved is the height from which it is released, h h h\boldsymbol{h}. Thus:
因为它总是从静止开始, u = 0 u = 0 u=0\boldsymbol{u}=0 自始至终;加速度是由重力引起的, g g gg ;涉及的距离是它被释放的高度, h h h\boldsymbol{h} 。因此:
h = 1 2 g t 2 t 2 = 2 h g h = 1 2 g t 2 t 2 = 2 h g {:[h=(1)/(2)gt^(2)],[:.quadt^(2)=(2h)/(g)]:}\begin{aligned} & h=\frac{1}{2} g \boldsymbol{t}^{2} \\ & \therefore \quad t^{2}=\frac{2 h}{g} \end{aligned}
Compare this equation with the equation for a straight-line graph:
将此等式与直线图的等式进行比较:
y = m x + c y = m x + c y=mx+cy=m x+c
We plot a graph of h h h\boldsymbol{h} on the x x xx-axis and t 2 t 2 t^(2)t^{2} on the y y yy-axis to give a straight best-fit line. The gradient of the line on this graph will be 2 / g 2 / g 2//g2 / g, from which we can find g g gg.
我们在 x x xx 轴上绘制 h h h\boldsymbol{h} 图形,在 y y yy 轴上绘制 t 2 t 2 t^(2)t^{2} 图形,从而得到一条最佳拟合直线。该图上直线的梯度为 2 / g 2 / g 2//g2 / g ,由此我们可以求出 g g gg
We could find a value for g g g\boldsymbol{g} by taking a single measurement from this experiment and using the equation to calculate it:
我们可以通过这次实验中的一次测量,利用方程计算出 g g g\boldsymbol{g} 的值:
g = 2 h t 2 g = 2 h t 2 g=(2h)/(t^(2))g=\frac{2 h}{t^{2}}
However, a single measurement in any experiment is subject to uncertainty from both random and systematic errors. We can reduce such uncertainties significantly by taking many readings and plotting a graph, which leads to much more reliable conclusions.
然而,任何实验中的单次测量都会受到随机误差和系统误差带来的不确定性的影响。我们可以通过多次读数和绘制图表来大大减少这种不确定性,从而得出更可靠的结论。
Safety Note: Secure the tall stand holding the solenoid so that it cannot topple over.
安全提示:固定好螺线管的高脚支架,使其不会倾倒。

CHECKPOINT 检查点

  1. What is the final velocity of a bike that starts at 4 m s 1 4 m s 1 4ms^(-1)4 \mathrm{~m} \mathrm{~s}^{-1} and has zero acceleration act on it for 10 seconds?
    一辆自行车从 4 m s 1 4 m s 1 4ms^(-1)4 \mathrm{~m} \mathrm{~s}^{-1} 开始,在 10 秒内加速度为零,其最终速度是多少?
  2. How far will the bike in question 1 travel in the 10 -second time period?
    问题 1 中的自行车在 10 秒钟内能行驶多远?
  3. Calculate the acceleration in each of the following cases:
    计算以下每种情况下的加速度:

    (a) v = 22 m s 1 ; u = 8 m s 1 ; t = 2.6 s v = 22 m s 1 ; u = 8 m s 1 ; t = 2.6 s v=22ms^(-1);u=8ms^(-1);t=2.6s\boldsymbol{v}=22 \mathrm{~m} \mathrm{~s}^{-1} ; \boldsymbol{u}=8 \mathrm{~m} \mathrm{~s}^{-1} ; t=2.6 \mathrm{~s}
    (b) a ball starts at rest and after 30 m its velocity has reached 4.8 m s 1 4.8 m s 1 4.8ms^(-1)4.8 \mathrm{~m} \mathrm{~s}^{-1}
    (b) 小球从静止开始,经过 30 米后速度达到 4.8 m s 1 4.8 m s 1 4.8ms^(-1)4.8 \mathrm{~m} \mathrm{~s}^{-1}

    © in 15 s , a train moves 100 m from rest.
    在 15 秒内,一列火车从静止开始行驶了 100 米。
  4. The bird in fig A drops the stone from a height of 88 m above the water surface. Initially, the stone has zero vertical velocity. How long will it take the stone to reach the surface of the pond? Assume air resistance is negligible.
    图 A 中的小鸟将石块从距离水面 88 米的高度落下。最初,石块的垂直速度为零。石块到达池塘水面需要多长时间?假设空气阻力可以忽略不计。
  5. If the stone in fig A enters the water at 41.6 m s 1 41.6 m s 1 41.6ms^(-1)41.6 \mathrm{~m} \mathrm{~s}^{-1}, and takes 0.6 s to travel the 3 metres to the bottom of the pond, what is its average acceleration in the pond water?
    如果图 A 中的石块在 41.6 m s 1 41.6 m s 1 41.6ms^(-1)41.6 \mathrm{~m} \mathrm{~s}^{-1} 处进入水中,用 0.6 秒的时间经过 3 米的距离到达池塘底部,那么它在池塘水中的平均加速度是多少?

SUBJECT VOCABULARY 主题词汇

kinematics the study of the description of the motion of objects uniform motion motion when there is no acceleration:
运动学 研究描述物体在没有加速度时的匀速运动:
velocity ( m s 1 ) = displacement ( m ) time ( s ) v = s t  velocity  m s 1 =  displacement  ( m ) time ( s ) v = s t {:[" velocity "(ms^(-1))=(" displacement "(m))/(time(s))],[v=(s)/(t)]:}\begin{aligned} \text { velocity }\left(\mathrm{m} \mathrm{~s}^{-1}\right) & =\frac{\text { displacement }(\mathrm{m})}{\operatorname{time}(\mathrm{s})} \\ v & =\frac{s}{t} \end{aligned}

1A 7 RESOLVING VECTORS 1a 7 解析向量

LEARNING OBJECTIVES 学习目标

  • Explain that any vector can be split into two components at right angles to each other.
    解释一下,任何矢量都可以分成互成直角的两个分量。
  • Calculate the values of the component vectors in any such right-angled pair (resolution).
    计算任何这样的直角对中的分量向量值(分辨率)。
We have seen that vectors that act perpendicular to each other can be combined to produce a single vector that has the same effect as the two separate vectors. If we start with a single vector, we can find a pair of vectors at right angles to each other that would combine to give our single original vector. This reverse process is called resolution or resolving vectors. The resolved pair of vectors will both start at the same point as the original single vector.
我们已经看到,作用相互垂直的矢量可以合并成一个矢量,其效果与两个独立的矢量相同。如果我们从一个单一的矢量开始,我们可以找到一对互成直角的矢量,它们组合起来就会产生我们原来的单一矢量。这一逆向过程称为解析或解析向量。解析后的一对矢量的起点都与原始单矢量的起点相同。

Δ Δ Delta\Delta fig A The original velocity vector of 13 ms 1 13 ms 1 13ms^(-1)13 \mathrm{~ms}^{-1} has been resolved into a horizontal and a vertical velocity vector, which would separately be the same overall effect. An object moving right at 12 ms 1 12 ms 1 12ms^(-1)12 \mathrm{~ms}^{-1} and up at 5 m s 1 5 m s 1 5ms^(-1)5 \mathrm{~m} \mathrm{~s}^{-1} will move 13 metres each second at an angle of 22.6 22.6 22.6^(@)22.6^{\circ} up and right from the horizontal.
Δ Δ Delta\Delta 图 A 原来的速度矢量 13 ms 1 13 ms 1 13ms^(-1)13 \mathrm{~ms}^{-1} 被分解成水平速度矢量和垂直速度矢量,它们分别产生相同的整体效果。一个物体以 12 ms 1 12 ms 1 12ms^(-1)12 \mathrm{~ms}^{-1} 的速度向右运动,以 5 m s 1 5 m s 1 5ms^(-1)5 \mathrm{~m} \mathrm{~s}^{-1} 的速度向上运动,每秒将以 22.6 22.6 22.6^(@)22.6^{\circ} 的角度向上和向右运动 13 米。

RESOLVING VECTORS CALCULATIONS
解析向量计算

In order to resolve a vector into a pair at right angles, we must know all the details of the original vector. This means we must know its size and direction. The direction is most commonly given as an angle to either the vertical or the horizontal. This is useful, as we most commonly want to split the vector up into a horizontal and vertical pair.
为了将一个向量解析成一对直角,我们必须知道原始向量的所有细节。这意味着我们必须知道它的大小和方向。方向通常是与垂直方向或水平方向的夹角。这一点很有用,因为我们通常希望将矢量分成水平和垂直两对。
A fig B Resolving a velocity vector into vertical and horizontal components.
A 图 B 将速度矢量分解为垂直分量和水平分量。

In the example of fig B, a basketball is thrown into the air at an angle of 40 40 40^(@)40^{\circ} to the vertical. In order to find out if it will go high enough to reach the hoop, kinematics calculations could be done on its vertical motion. However, this can only be done if we can isolate the vertical component of the basketball’s motion. Similarly, we could find out if it will travel far enough horizontally to reach the hoop if we know how fast it is moving horizontally.
在图 B 的示例中,篮球以 40 40 40^(@)40^{\circ} 的垂直角度被抛向空中。为了确定篮球是否能够飞到足够高的地方,我们可以对篮球的垂直运动进行运动学计算。但是,只有当我们能够分离出篮球运动的垂直分量时,才能进行计算。同样,如果我们知道篮球水平运动的速度,我们就能知道它是否能水平运动到足够远的地方到达篮圈。

The basketball’s velocity must be resolved into horizontal and vertical components. This will require some trigonometrical calculations.
篮球的速度必须分解为水平和垂直两个部分。这需要进行一些三角计算。

THE VERTICAL COMPONENT OF VELOCITY
速度的垂直分量

Redrawing the components in fig B to show how they add up to produce the 4.2 m s 1 4.2 m s 1 4.2ms^(-1)4.2 \mathrm{~m} \mathrm{~s}^{-1} velocity vector, shows again that they form a right-angled triangle, as in fig C C C\mathbf{C}. This means we can use the relationship:
重新绘制图 B 中的分量,以显示它们如何相加产生 4.2 m s 1 4.2 m s 1 4.2ms^(-1)4.2 \mathrm{~m} \mathrm{~s}^{-1} 速度矢量,再次显示它们构成了一个直角三角形,如图 C C C\mathbf{C} 所示。这意味着我们可以使用以下关系:
cos 40 = v y 4.2 v y = 4.2 × cos 40 = 4.2 × 0.766 v y = 3.2 m s 1 cos 40 = v y 4.2 v y = 4.2 × cos 40 = 4.2 × 0.766 v y = 3.2 m s 1 {:[cos 40^(@)=(v_(y))/(4.2)],[:.quadv_(y)=4.2 xx cos 40^(@)],[=4.2 xx0.766],[:.quadv_(y)=3.2ms^(-1)]:}\begin{aligned} & \cos 40^{\circ}=\frac{v_{y}}{4.2} \\ \therefore \quad v_{y} & =4.2 \times \cos 40^{\circ} \\ & =4.2 \times 0.766 \\ \therefore \quad v_{y} & =3.2 \mathrm{~m} \mathrm{~s}^{-1} \end{aligned}

Δ Δ Delta\Delta fig C C C\mathbf{C} Finding the components of velocity.
Δ Δ Delta\Delta C C C\mathbf{C} 求速度分量。

THE HORIZONTAL COMPONENT OF VELOCITY
速度的水平分量

Similarly, for the horizontal component, we can use the relationship:
同样,对于水平分量,我们也可以使用这种关系:
sin 40 = v x 4.2 v x = 4.2 × sin 40 = 4.2 × 0.643 v x = 2.7 m s 1 sin 40 = v x 4.2 v x = 4.2 × sin 40 = 4.2 × 0.643 v x = 2.7 m s 1 {:[,sin 40^(@)=(v_(x))/(4.2)],[:.,v_(x)=4.2 xx sin 40^(@)=4.2 xx0.643],[:.,v_(x)=2.7ms^(-1)]:}\begin{array}{ll} & \sin 40^{\circ}=\frac{v_{\mathrm{x}}}{4.2} \\ \therefore & v_{\mathrm{x}}=4.2 \times \sin 40^{\circ}=4.2 \times 0.643 \\ \therefore & v_{\mathrm{x}}=2.7 \mathrm{~m} \mathrm{~s}^{-1} \end{array}

LEARNING TIP 学习提示

LEARNING TIP 学习提示

Resolving vectors could also be achieved using a scale diagram. Using this method in rough work may help you to get an idea as to what the answer should be in order to check that your calculations are about right.
也可以使用比例图来解决矢量问题。在粗略的作业中使用这种方法可以帮助你对答案有一个大致的了解,以便检查你的计算是否正确。

ALTERNATIVE RESOLUTION ANGLES
其他解决角度

If we know the velocity of an object, we have seen that we can resolve this into a pair of velocity vectors at right angles to each other. The choice of direction of the right-angle pair, though, is arbitrary, and can be chosen to suit a given situation.
如果我们知道一个物体的速度,我们已经看到,我们可以将其分解为一对互成直角的速度矢量。不过,这对直角矢量的方向选择是任意的,可以根据特定情况进行选择。

Imagine a submarine descending underwater close to an angled seabed.
想象一下,一艘潜艇在水下靠近倾斜的海床。
A fig E Resolving submarine velocity vectors helps to avoid collision.
A 图 E 解析潜艇速度矢量有助于避免碰撞。

The velocity could be resolved into a pair of vectors, horizontal and vertical.
速度可以分解成一对矢量,即水平矢量和垂直矢量。

Δ Δ Delta\Delta
fig F F F\mathbf{F} Submarine velocity resolved into horizontal and vertical components.
F F F\mathbf{F} 水下速度分为水平和垂直两部分。

However, the submarine commander is likely to be most interested in knowing how quickly the submarine is approaching the seabed. This could be found by resolving the velocity of the submarine into a component parallel with the seabed, and the right-angled pair with that will be a component perpendicular to the seabed. It is this v perp v perp  v_("perp ")\boldsymbol{v}_{\text {perp }} that will tell the submarine commander how quickly he is approaching the seabed.
不过,潜艇指挥官可能最想知道潜艇接近海床的速度。这可以通过将潜艇的速度分解为与海床平行的分量,以及与之成直角对的垂直于海床的分量来实现。正是这个 v perp v perp  v_("perp ")\boldsymbol{v}_{\text {perp }} 可以告诉潜艇指挥官他正在以多快的速度接近海底。
A fig G Submarine velocity resolved into components parallel and perpendicular to the seabed.
A 图 G 水下速度分为平行和垂直于海底的两个部分。

LEARNING TIP 学习提示

The vectors produced by resolution can be at any angle with respect to the original vector, as long as they are perpendicular to each other and their resultant equals the original vector. In order to allow for every pair to add up to the original vector, their magnitudes will vary as the angle changes.
解析所产生的矢量相对于原始矢量的角度可以是任意的,只要它们相互垂直,其结果等于原始矢量即可。为了让每一对矢量相加都等于原始矢量,它们的大小会随着角度的变化而变化。

CHECKPOINT 检查点

SKILLS INTERPRETATION 技能解释

  1. (a) On graph paper, draw a velocity vector for a stone fired from a catapult at 45 45 45^(@)45^{\circ} to the horizontal. Your arrow should be 10 cm long, representing a velocity of 10 m s 1 10 m s 1 10ms^(-1)10 \mathrm{~m} \mathrm{~s}^{-1}. Draw onto your diagram the horizontal and vertical components that would make up the overall velocity. Use a ruler to measure the size of the horizontal and vertical components, and convert these lengths into metres per second using the same scaling.
    (a) 在图画纸上,画出从弹射器上以 45 45 45^(@)45^{\circ} 向水平发射的石块的速度矢量。箭头应为 10 厘米长,代表速度 10 m s 1 10 m s 1 10ms^(-1)10 \mathrm{~m} \mathrm{~s}^{-1} 。在图中画出组成总速度的水平和垂直分量。用尺子测量水平和垂直分量的大小,并用同样的比例将这些长度换算成每秒米数。

    (b) Find the horizontal and vertical velocity components for this catapult stone by calculation, and compare with your answers from part (a).
    (b) 通过计算找出弹射石的水平和垂直速度分量,并与 (a) 部分的答案进行比较。
  2. A javelin is thrown at 16 m s 1 16 m s 1 16ms^(-1)16 \mathrm{~m} \mathrm{~s}^{-1} at an angle of 35 35 35^(@)35^{\circ} up from the horizontal. Calculate the horizontal and vertical components of the javelin’s motion.
    标枪以 16 m s 1 16 m s 1 16ms^(-1)16 \mathrm{~m} \mathrm{~s}^{-1} 扔出,与水平面成向上 35 35 35^(@)35^{\circ} 角。请计算标枪运动的水平分量和垂直分量。
  3. A ladder is leant against a wall, at an angle of 28 28 28^(@)28^{\circ} to the wall. The 440 N force from the floor acts along the length of the ladder. Calculate the horizontal and vertical components of the force from the floor that act on the bottom of the ladder.
    梯子靠在墙上,与墙壁成 28 28 28^(@)28^{\circ} 角。来自地面的 440 N 力沿梯子的长度方向作用。请计算作用在梯子底部的来自地面的力的水平分量和垂直分量。
  4. A plane is flying at 240 m s 1 240 m s 1 240ms^(-1)240 \mathrm{~m} \mathrm{~s}^{-1}, on a bearing of 125 125 125^(@)125^{\circ} from due north. Calculate its velocity component due south, and its velocity component due east.
    一架飞机以 240 m s 1 240 m s 1 240ms^(-1)240 \mathrm{~m} \mathrm{~s}^{-1} 的速度从正北方向飞向 125 125 125^(@)125^{\circ} 的方位。请计算其正南方向的速度分量和正东方向的速度分量。

SUBJECT VOCABULARY 主题词汇

resolution or resolving vectors the determination of a pair of vectors, at right angles to each other, that sums to give the single vector they were resolved from
分解或解析矢量 确定一对互成直角的矢量,其总和就是由它们分解而来的单一矢量

catapult a device that can throw objects at high speed
弹射器

1A 8 PROJECTILES 1a 8 枚射弹

LEARNING OBJECTIVES 学习目标

■Apply kinematics equations to moving objects. Apply the independence of horizontal and vertical motion to objects moving freely under gravity.
将运动学方程应用于运动物体。将水平运动和垂直运动的独立性应用于在重力作用下自由运动的物体。
  • Combine horizontal and vertical motion to calculate the movements of projectiles.
    结合水平运动和垂直运动来计算弹丸的运动。
Objects thrown or fired through the air generally follow projectile motion. Here we are going to combine ideas from the various earlier sections in order to solve questions about projectiles. Resolving vectors showed that the actions in each of two perpendicular directions are wholly independent. This means we can use Newton’s laws of motion and the kinematics equations separately for the horizontal and vertical motions of the same object. This will allow us to calculate its overall motion in two dimensions.
在空中投掷或发射的物体一般都会做弹丸运动。在这里,我们将综合前面各节的观点来解决有关射弹的问题。解决矢量问题表明,两个垂直方向上的运动是完全独立的。这意味着我们可以对同一物体的水平运动和垂直运动分别使用牛顿运动定律和运动学方程。这样,我们就可以计算出物体在两个维度上的整体运动。
We only consider the motion after the force projecting an object has finished - for example, after a cannonball has left the cannon. Air resistance is ignored in these calculations, so the only force acting will be the object’s weight. Thus, all vertical motion will follow kinematics equations, with gravity as the acceleration. There will be no horizontal force at all, which means no acceleration and therefore v = s / t v = s / t v=s//t\boldsymbol{v}=\boldsymbol{s} / t.
我们只考虑物体受力结束后的运动,例如,炮弹离开大炮后的运动。在这些计算中,空气阻力被忽略,因此唯一的作用力就是物体的重量。因此,所有垂直运动都将遵循运动学方程,以重力作为加速度。完全没有水平力,这意味着没有加速度,因此 v = s / t v = s / t v=s//t\boldsymbol{v}=\boldsymbol{s} / t

LEARNING TIP 学习提示

Vectors acting at right angles to each other act independently. Their combined effect can be calculated using vector addition, but they can also be considered to act separately and at the same time. This would cause independent effects, which themselves could then be combined to see an overall effect.
互成直角的矢量是独立作用的。它们的综合效应可以通过向量加法计算出来,但也可以认为它们是同时单独作用的。这将产生独立的效果,然后将这些效果结合起来,就能看到整体效果。

LEARNING TIP 学习提示

Physics is holistic: you can apply ideas from one area of physics to other areas of physics. Remember that physics is a set of rules that attempts to explain everything in the universe. Every rule should therefore be universally applicable.
物理学是整体性的:您可以将物理学某一领域的观点应用到物理学的其他领域。请记住,物理学是一套试图解释宇宙万物的规则。因此,每一条规则都应该是普遍适用的。

HORIZONTAL THROWS 水平投掷

If an object is thrown horizontally, it will start off with zero vertical velocity. However, gravity will act on it so that its motion will curve downwards in a parabola shape, like the stone in fig A A A\mathbf{A}.
如果一个物体被水平抛出,那么它一开始的垂直速度为零。然而,重力会对其产生作用,使其运动曲线呈抛物线形状向下弯曲,就像图 A A A\mathbf{A} 中的石头一样。
In the example of fig A , a stone is kicked horizontally off a cliff with a velocity of 8.2 m s 1 8.2 m s 1 8.2ms^(-1)8.2 \mathrm{~m} \mathrm{~s}^{-1}. How much time is the stone in flight? How far does it get away from the cliff by the time it lands?
在图 A 的示例中,一块石头被水平踢下悬崖,速度为 8.2 m s 1 8.2 m s 1 8.2ms^(-1)8.2 \mathrm{~m} \mathrm{~s}^{-1} 。石块飞行了多长时间?它落地时离悬崖有多远?

Horizontal and vertical motions are totally independent. Here the vertical component of velocity is initially zero, but the stone accelerates under gravity. Uniform acceleration means the kinematics equations can be used.
水平运动和垂直运动是完全独立的。在这里,速度的垂直分量最初为零,但石头在重力作用下会加速。匀加速度意味着可以使用运动学方程。

fig A Vertical acceleration on a horizontally moving stone.
图 A 水平移动的石头的垂直加速度

The time to hit the beach, t t tt, will be the same as if the stone was simply dropped. We know u = 0 m s 1 ; a = 9.81 m s 2 u = 0 m s 1 ; a = 9.81 m s 2 u=0ms^(-1);a=-9.81ms^(-2)\boldsymbol{u}=0 \mathrm{~m} \mathrm{~s}^{-1} ; \boldsymbol{a}=-9.81 \mathrm{~m} \mathrm{~s}^{-2}; and the height fallen, s = 60 m s = 60 m s=-60ms=-60 \mathrm{~m}.
落到沙滩上的时间 t t tt 将与石块直接掉落的时间相同。我们知道 u = 0 m s 1 ; a = 9.81 m s 2 u = 0 m s 1 ; a = 9.81 m s 2 u=0ms^(-1);a=-9.81ms^(-2)\boldsymbol{u}=0 \mathrm{~m} \mathrm{~s}^{-1} ; \boldsymbol{a}=-9.81 \mathrm{~m} \mathrm{~s}^{-2} ;以及落下的高度 s = 60 m s = 60 m s=-60ms=-60 \mathrm{~m}
s = u t + 1 2 a t 2 u = 0 u t = 0 s = 1 2 a t 2 t = 2 s a = ( 2 × 60 ) 9.81 t = 3.5 s s = u t + 1 2 a t 2 u = 0 u t = 0 s = 1 2 a t 2 t = 2 s a = ( 2 × 60 ) 9.81 t = 3.5 s {:[,,s,=ut+(1)/(2)at^(2)],[,u,=0:.ut=0],[:.quad,s,=(1)/(2)at^(2)],[:.quad,t,=sqrt((2s)/(a))=sqrt(((2xx-60))/(-9.81))],[,t,=3.5s]:}\begin{array}{rlrl} & & s & =u t+\frac{1}{2} a t^{2} \\ & u & =0 \therefore u t=0 \\ \therefore \quad & s & =\frac{1}{2} a t^{2} \\ \therefore \quad & t & =\sqrt{\frac{2 s}{a}}=\sqrt{\frac{(2 \times-60)}{-9.81}} \\ & t & =3.5 \mathrm{~s} \end{array}
Horizontally, there is no accelerating force once the stone is in flight, so it has a constant speed. Thus, to find the distance travelled horizontally, d d d\boldsymbol{d} :
在水平方向上,石头飞行时没有加速力,所以它的速度是恒定的。因此,要计算水平飞行的距离, d d d\boldsymbol{d}
v = d t d = v × t d = 8.2 × 3.5 d = 28.7 m v = d t d = v × t d = 8.2 × 3.5 d = 28.7 m {:[:.quad v=(d)/(t)],[:.quad d=v xx t],[d=8.2 xx3.5],[d=28.7m]:}\begin{aligned} \therefore \quad \boldsymbol{v} & =\frac{\boldsymbol{d}}{\boldsymbol{t}} \\ \therefore \quad \boldsymbol{d} & =\boldsymbol{v} \times t \\ & \boldsymbol{d}=8.2 \times 3.5 \\ \boldsymbol{d} & =28.7 \mathrm{~m} \end{aligned}

RECOMBINING VELOCITY COMPONENTS
重组速度成分

In the example of the stone kicked from the cliff to the beach, we might also want to calculate the final velocity of the stone on landing. This means adding vertical and horizontal components into their resultant.
在石头从悬崖踢向海滩的例子中,我们可能还想计算石头着陆时的最终速度。这意味着将垂直分量和水平分量加到它们的结果中。
A fig B The stone’s final velocity is the resultant of its vertical and horizontal components.
A 无花果 B 石头的最终速度是其垂直和水平分量的结果。
In the example of fig A , what is the velocity of the stone when it hits the beach?
在图 A 的例子中,石块落到沙滩上时的速度是多少?

The horizontal velocity was given as 8.2 m s 1 8.2 m s 1 8.2ms^(-1)8.2 \mathrm{~m} \mathrm{~s}^{-1}. To calculate the vertical velocity: u = 0 ms 1 ; a = 9.81 m s 2 u = 0 ms 1 ; a = 9.81 m s 2 u=0ms^(-1);a=-9.81ms^(-2)\boldsymbol{u}=0 \mathrm{~ms}^{-1} ; \boldsymbol{a}=-9.81 \mathrm{~m} \mathrm{~s}^{-2}; and s = 60 m s = 60 m s=-60m\boldsymbol{s}=-60 \mathrm{~m}.
水平速度为 8.2 m s 1 8.2 m s 1 8.2ms^(-1)8.2 \mathrm{~m} \mathrm{~s}^{-1} 。要计算垂直速度,需要 u = 0 ms 1 ; a = 9.81 m s 2 u = 0 ms 1 ; a = 9.81 m s 2 u=0ms^(-1);a=-9.81ms^(-2)\boldsymbol{u}=0 \mathrm{~ms}^{-1} ; \boldsymbol{a}=-9.81 \mathrm{~m} \mathrm{~s}^{-2} ;以及 s = 60 m s = 60 m s=-60m\boldsymbol{s}=-60 \mathrm{~m}
v vertical 2 = u 2 + 2 as = 0 2 + ( 2 × 9.81 × 60 ) = 1177.2 v vertical = 1177.2 = 34.3 v vertical = 34.3 m s 1 v vertical  2 = u 2 + 2  as  = 0 2 + ( 2 × 9.81 × 60 ) = 1177.2 v vertical  = 1177.2 = 34.3 v vertical  = 34.3 m s 1 {:[,,v_("vertical ")^(2),=u^(2)+2" as "=0^(2)+(2xx-9.81 xx-60)=1177.2],[:.quadv_("vertical "),=sqrt1177.2=-34.3],[,v_("vertical "),=-34.3ms^(-1)]:}\begin{array}{rlrl} & & \boldsymbol{v}_{\text {vertical }}^{2} & =\boldsymbol{u}^{2}+2 \text { as }=0^{2}+(2 \times-9.81 \times-60)=1177.2 \\ \therefore \quad v_{\text {vertical }} & =\sqrt{1177.2}=-34.3 \\ & v_{\text {vertical }} & =-34.3 \mathrm{~m} \mathrm{~s}^{-1} \end{array}
Pythagoras’ theorem gives the magnitude of the final velocity:
勾股定理给出了最终速度的大小:
v = ( 8.2 2 + 34.3 2 ) v = 35.3 m s 1 v = 8.2 2 + 34.3 2 v = 35.3 m s 1 {:[v=sqrt((8.2^(2)+34.3^(2)))],[:.quad v=35.3ms^(-1)]:}\begin{aligned} & v=\sqrt{\left(8.2^{2}+34.3^{2}\right)} \\ & \therefore \quad v=35.3 \mathrm{~m} \mathrm{~s}^{-1} \end{aligned}
Trigonometry will give the angle at which the stone is flying on impact with the beach:
三角函数将给出石头撞击沙滩时的飞行角度:
tan θ = 34.3 8.2 θ = 77 tan θ = 34.3 8.2 θ = 77 {:[tan theta=(34.3)/(8.2)],[:.theta=77^(@)]:}\begin{aligned} & & \tan \theta & =\frac{34.3}{8.2} \\ & \therefore & \theta & =77^{\circ} \end{aligned}
The stone’s velocity when it lands on the beach is 35.3 metres per second at an angle of 77 77 77^(@)77^{\circ} down from the horizontal.
石块落在沙滩上时的速度为每秒 35.3 米,与水平面成 77 77 77^(@)77^{\circ} 向下的角度。

EXAM HINT 考试提示

We use the usual approach that up is positive throughout this section. You can choose any direction to be negative, but you must refer to that direction as negative at every step of the calculation. If you are not consistent, your calculation will be incorrect.
在本节中,我们采用的通常方法是向上为正。您可以选择任何方向为负,但必须在计算的每一步都将该方向称为负。如果不一致,计算结果将是错误的。

VERTICAL THROWS 垂直投掷

Imagine throwing a ball to a friend. The ball goes up as well as forwards. One common idea in these calculations is that an object thrown with a vertical upwards component of motion will have a symmetrical trajectory. At the highest point, the vertical velocity is momentarily zero. Getting to this point will take half of the time of the whole flight.
想象一下把球扔给朋友的情景。球既会向上飞,也会向前飞。在这些计算中,一个常见的想法是,一个垂直向上运动的物体会有一个对称的轨迹。在最高点,垂直速度瞬间为零。到达这一点将花费整个飞行时间的一半。

fig C Considering only the vertical component of velocity.
图 C 仅考虑速度的垂直分量。

In the example of f i g C f i g C figC\mathbf{f i g} \mathbf{C}, a ball is thrown with a vertical velocity component of 5.5 m s 1 5.5 m s 1 5.5ms^(-1)5.5 \mathrm{~m} \mathrm{~s}^{-1}. How much time is it in flight? How high does it get?
f i g C f i g C figC\mathbf{f i g} \mathbf{C} 的例子中,一个球被抛出时的垂直速度分量为 5.5 m s 1 5.5 m s 1 5.5ms^(-1)5.5 \mathrm{~m} \mathrm{~s}^{-1} 。它飞行了多长时间?球飞多高?

These questions would have the same answer if a person threw the ball vertically up and caught it again themselves. This again highlights the independence of horizontal and vertical motions. It may be that an initial velocity at an angle is quoted, so that we need to resolve the velocity vector into its horizontal and vertical components in order to know that here v vertical = 5.5 m s 1 v vertical  = 5.5 m s 1 v_("vertical ")=5.5ms^(-1)\boldsymbol{v}_{\text {vertical }}=5.5 \mathrm{~m} \mathrm{~s}^{-1}.
如果一个人把球垂直抛起,然后自己再接住,这些问题的答案是一样的。这再次强调了水平运动和垂直运动的独立性。这里可能引用了一个角度的初速度,因此我们需要将速度矢量分解为水平和垂直分量,才能知道 v vertical = 5.5 m s 1 v vertical  = 5.5 m s 1 v_("vertical ")=5.5ms^(-1)\boldsymbol{v}_{\text {vertical }}=5.5 \mathrm{~m} \mathrm{~s}^{-1}

Consider the second question first: uniform acceleration under gravity means a = 9.81 m s 2 a = 9.81 m s 2 a=-9.81ms^(-2)\boldsymbol{a}=-9.81 \mathrm{~m} \mathrm{~s}^{-2} and the kinematics equations can be used. We know u = 5.5 m s 1 u = 5.5 m s 1 u=5.5ms^(-1)\boldsymbol{u}=5.5 \mathrm{~m} \mathrm{~s}^{-1}; at the top of the path, v = 0 m s 1 v = 0 m s 1 v=0ms^(-1)\boldsymbol{v}=0 \mathrm{~m} \mathrm{~s}^{-1}; and we want to find the height, s s s\boldsymbol{s}.
先考虑第二个问题:重力下的匀加速度表示 a = 9.81 m s 2 a = 9.81 m s 2 a=-9.81ms^(-2)\boldsymbol{a}=-9.81 \mathrm{~m} \mathrm{~s}^{-2} ,可以使用运动学方程。我们知道 u = 5.5 m s 1 u = 5.5 m s 1 u=5.5ms^(-1)\boldsymbol{u}=5.5 \mathrm{~m} \mathrm{~s}^{-1} ;在路径顶端, v = 0 m s 1 v = 0 m s 1 v=0ms^(-1)\boldsymbol{v}=0 \mathrm{~m} \mathrm{~s}^{-1} ;我们想找到高度, s s s\boldsymbol{s}
v 2 = u 2 + 2 a s s = v 2 u 2 2 a = 0 2 ( 5.5 ) 2 2 × 9.81 s = 1.54 m v 2 = u 2 + 2 a s s = v 2 u 2 2 a = 0 2 ( 5.5 ) 2 2 × 9.81 s = 1.54 m {:[v^(2)=u^(2)+2as],[:.quad s=(v^(2)-u^(2))/(2a)=(0^(2)-(5.5)^(2))/(2xx-9.81)],[s=1.54m]:}\begin{aligned} v^{2} & =u^{2}+2 a s \\ \therefore \quad s & =\frac{v^{2}-u^{2}}{2 a}=\frac{0^{2}-(5.5)^{2}}{2 \times-9.81} \\ s & =1.54 \mathrm{~m} \end{aligned}
Note that 1.54 metres is actually the height the ball reaches above the point of release at which it left the hand - the point where its initial speed was 5.5 m s 1 5.5 m s 1 5.5ms^(-1)5.5 \mathrm{~m} \mathrm{~s}^{-1} - but this is often ignored in projectiles calculations.
请注意,1.54 米实际上是球离开手的释放点以上的高度,也就是其初始速度为 5.5 m s 1 5.5 m s 1 5.5ms^(-1)5.5 \mathrm{~m} \mathrm{~s}^{-1} 的点,但在弹丸计算中通常会忽略这一点。

The time of flight for the ball will be just the time taken to rise and fall vertically. We find the time to reach the highest point, and then double that value. We know u = 5.5 m s 1 u = 5.5 m s 1 u=5.5ms^(-1)\boldsymbol{u}=5.5 \mathrm{~m} \mathrm{~s}^{-1}; at the top of the path, v = 0 m s 1 v = 0 m s 1 v=0ms^(-1)\boldsymbol{v}=0 \mathrm{~m} \mathrm{~s}^{-1}; and we want to find the time, t t tt.
小球的飞行时间就是垂直上升和下降的时间。我们找到到达最高点的时间,然后将该值加倍。我们知道 u = 5.5 m s 1 u = 5.5 m s 1 u=5.5ms^(-1)\boldsymbol{u}=5.5 \mathrm{~m} \mathrm{~s}^{-1} ;在路径的顶点, v = 0 m s 1 v = 0 m s 1 v=0ms^(-1)\boldsymbol{v}=0 \mathrm{~m} \mathrm{~s}^{-1} ;我们要找出时间, t t tt
v = u + a t t = v u a = 0 ( 5.5 ) 9.81 t = 0.56 s v = u + a t t = v u a = 0 ( 5.5 ) 9.81 t = 0.56 s {:[:.quad v=u+at],[t=(v-u)/(a)=(0-(5.5))/(-9.81)],[t=0.56s]:}\begin{aligned} \therefore \quad v & =u+a t \\ t & =\frac{v-u}{a}=\frac{0-(5.5)}{-9.81} \\ t & =0.56 \mathrm{~s} \end{aligned}
So the overall time of flight will be 0.56 seconds doubled: total time = 1.12 s = 1.12 s =1.12s=1.12 \mathrm{~s}
因此,总飞行时间将增加一倍,即 0.56 秒: = 1.12 s = 1.12 s =1.12s=1.12 \mathrm{~s} 总时间

EXAM HINT 考试提示

For virtually all projectile motion calculations, we assume that there is no air resistance, so the only force acting is gravity, vertically downwards.
在几乎所有的弹丸运动计算中,我们都假设没有空气阻力,因此唯一的作用力就是垂直向下的重力。

CHECKPOINT 检查点

SKILLS PROBLEM SOLVING 解决问题的技能

  1. A boy throws a ball vertically at a velocity of 4.8 m s 1 4.8 m s 1 4.8ms^(-1)4.8 \mathrm{~m} \mathrm{~s}^{-1}.
    一个男孩以 4.8 m s 1 4.8 m s 1 4.8ms^(-1)4.8 \mathrm{~m} \mathrm{~s}^{-1} 的速度垂直抛出一个球。

    (a) How long is it before he catches it again?
    (a) 多久后他会再次抓住它?

    (b) What will be the ball’s greatest height above the point of release?
    (b) 球离释放点的最大高度是多少?
  2. The boy in question 1 now throws his ball horizontally out of a high window with a velocity of 3.1 m s 1 3.1 m s 1 3.1ms^(-1)3.1 \mathrm{~m} \mathrm{~s}^{-1}.
    问题 1 中的男孩现在以 3.1 m s 1 3.1 m s 1 3.1ms^(-1)3.1 \mathrm{~m} \mathrm{~s}^{-1} 的速度将球水平抛出高窗。

    (a) How long will it take to reach the ground 18 m below?
    (a) 到达 18 米以下的地面需要多长时间?

    (b) How far away, horizontally, should his friend stand in order to catch the ball?
    (b) 他的朋友应该站在水平方向多远的地方才能接住球?
  3. A basketball is thrown with a velocity of 6.0 m s 1 6.0 m s 1 6.0ms^(-1)6.0 \mathrm{~m} \mathrm{~s}^{-1} at an angle of 40 40 40^(@)40^{\circ} to the vertical, towards the hoop.
    一个篮球以 6.0 m s 1 6.0 m s 1 6.0ms^(-1)6.0 \mathrm{~m} \mathrm{~s}^{-1} 的速度、与垂直方向成 40 40 40^(@)40^{\circ} 的角度被抛向篮圈。

    (a) If the hoop is 0.90 m above the point of release, will the ball rise high enough to go in the hoop?
    (a) 如果吊环距离释放点 0.90 米,球会上升到足以进入吊环的高度吗?

    (b) If the centre of the hoop is 3.00 m away, horizontally, from the point of release, explain whether or not you believe this throw will score in the hoop. Support your explanation with calculations.
    (b) 如果篮圈中心距离投掷点水平方向 3.00 米,请解释您是否认为这次投掷能进篮圈。请用计算支持你的解释。

SUBJECT VOCABULARY 主题词汇

projectile a moving object on which the only force of significance acting is gravity. The trajectory is thus pre-determined by its initial velocity
弹丸是一个运动物体,其上唯一的作用力是重力。因此,弹道是由其初始速度预先决定的

1A
THINKING BIGGER
1A 大处着眼

THE BATTLE OF AGRA 阿格拉战役

CRITICAL THINKING, PROBLEM SOLVING, ANALYSIS, INTERPRETATION, ADAPTIVE LEARNING, PRODUCTIVITY, ETHICS, COMMUNICATION, ASSERTIVE COMMUNICATION
批判性思维、解决问题、分析、解释、适应性学习、生产率、道德、交流、自信交流

Agra Fort was built in the 11th Century, although the present structure was built in 1573. In this activity, you need to imagine attacking the fort using a cannon that fires a cannonball as a projectile.
阿格拉堡始建于 11 世纪,不过现在的建筑是 1573 年建造的。在这项活动中,您需要想象用大炮发射炮弹攻击要塞。

STUDENT ESSAY 学生论文

  • fig A Agra Fort is now an UNESCO World Heritage Site.
    阿格拉堡现已成为联合国教科文组织的世界遗产。
In this section, I will use some basic mechanics to answer a question: could the Mughal Empire artillery really have attacked Agra Fort in the way described previously? The nineteenthcentury source material suggests that the fortress was under siege by the Mughals for three months and ‘battered by artillery’. However, the current walls bear little in the way of obvious battle scars.
在本节中,我将利用一些基本的力学原理来回答一个问题:莫卧儿帝国的大炮真的会以之前描述的方式攻击阿格拉堡吗?十九世纪的原始资料显示,莫卧儿帝国曾围攻阿格拉堡三个月,并 "炮火连天"。然而,现在的城墙上几乎没有明显的战痕。
Looking at fig B, the question that needs to be answered here is:
从图 B 可以看出,这里需要回答的问题是

‘How high up the front wall of the fortress will the cannonball hit?’ This height is marked on fig B B B\mathbf{B} as ’ H H HH '.
'炮弹会击中堡垒前墙多高?这个高度在图 B B B\mathbf{B} 中标记为' H H HH '。
A fig B The trajectory of a cannonball fired towards Agra Fort. We assume the cannonball leaves the cannon at ground level.
A 图 B 向阿格拉堡发射的炮弹轨迹。我们假设炮弹从地面离开大炮。
In addition to the layout shown in fig B, we need information about the initial velocity of the cannonball. The cannon explosion could act for 0.05 s to accelerate the cannonball (mass = 12 kg = 12 kg =12kg=12 \mathrm{~kg} ) with a force of 9300 N . It causes the cannonball to leave the cannon at an angle of 45 45 45^(@)45^{\circ} to the horizontal.
除了图 B 所示的布局外,我们还需要炮弹初速度的信息。加农炮爆炸可以在 0.05 秒内以 9300 N 的力加速炮弹(质量 = 12 kg = 12 kg =12kg=12 \mathrm{~kg} ),使炮弹以 45 45 45^(@)45^{\circ} 的角度离开加农炮。

Steps to the answer 找到答案的步骤

We can work out what calculations are required to solve this problem, by working back from the answer we want to find. The fundamental idea is that the parabola trajectory would be symmetrical if the flight was not interrupted by crashing into the fortress wall.
我们可以从想要找到的答案出发,推导出解决这个问题所需的计算。基本原理是,如果飞行没有因为撞上堡垒墙而中断,抛物线轨迹将是对称的。

1 To find the height up the wall from the ground, we will need to work out how far down from the cannonball’s maximum height it falls:
1 要想知道从地面到墙上的高度,我们需要计算出炮弹从最大高度落下的距离:
H = h max h H = h max h H=h_(max)-hH=h_{\max }-h
2 To find h h hh, we need to know the time of flight, t total t total  t_("total ")t_{\text {total }} so we can divide this into a time to reach h max h max h_(max)h_{\max }, and time left to fall height h h hh. We will use vertical gravitational acceleration to calculate the vertical drop in that remaining time:
2 要找到 h h hh ,我们需要知道飞行时间 t total t total  t_("total ")t_{\text {total }} ,这样我们就可以将其分为到达 h max h max h_(max)h_{\max } 的时间和下降高度 h h hh 的剩余时间。我们将使用垂直重力加速度来计算剩余时间内的垂直下落高度:
t total = s v horizontal t total  = s v horizontal  t_("total ")=(s)/(v_("horizontal "))t_{\text {total }}=\frac{s}{v_{\text {horizontal }}}
From fig B, we can see that s = 150 m s = 150 m s=150ms=150 \mathrm{~m}.
从图 B 可以看出, s = 150 m s = 150 m s=150ms=150 \mathrm{~m}

3 v horizontal 3 v horizontal  3v_("horizontal ")3 v_{\text {horizontal }} can be found by resolving the velocity to give the horizontal component:
3 v horizontal 3 v horizontal  3v_("horizontal ")3 v_{\text {horizontal }} 可以通过解析速度得到水平分量:
v horizontal = v total × cos 45 v horizontal  = v total  × cos 45 v_("horizontal ")=v_("total ")xx cos 45^(@)v_{\text {horizontal }}=v_{\text {total }} \times \cos 45^{\circ}
4 The overall velocity will come from the cannon’s acceleration of the cannonball:
4 总速度来自大炮对炮弹的加速度:
v = u + a t v = u + a t v=u+atv=u+a t
where u = 0 m s 1 u = 0 m s 1 u=0ms^(-1)u=0 \mathrm{~m} \mathrm{~s}^{-1}, and the question tells us that the explosion acts for 0.05 seconds.
其中 u = 0 m s 1 u = 0 m s 1 u=0ms^(-1)u=0 \mathrm{~m} \mathrm{~s}^{-1} ,问题告诉我们爆炸作用时间为 0.05 秒。

5 Newton’s second law of motion gives us the acceleration caused by the sling:
5 根据牛顿第二运动定律,我们可以得出吊索所产生的加速度:
a = F m a = F m a=(F)/(m)a=\frac{F}{m}
Calculate the answer by reversing these steps:
反过来计算答案:

The acceleration caused by the explosion:
爆炸产生的加速度:
a = F m a = a = F m a = {:[a=(F)/(m)],[a=]:}\begin{aligned} & a=\frac{F}{m} \\ & a= \end{aligned}

SCIENCE COMMUNICATION 科学传播

1 The extract opposite is a draft for a university essay about the Mughal siege of 1857. Consider the extract and comment on the type of writing being used. For example, think about whether this is a scientist reporting the results of their experiments, a scientific review of data, a newspaper or a magazine-style article for a specific audience. Try and answer the following questions:
1 对面的摘录是一篇关于 1857 年莫卧儿王朝围城的大学论文草稿。请仔细阅读摘要,并对所使用的写作类型发表评论。例如,想一想这是否是科学家在报告他们的实验结果,是否是对数据的科学评论,是否是一篇面向特定读者的报纸或杂志文章。尝试回答下列问题:

(a) How can you tell that the author is doubtful about the historical source material?
(a) 如何判断作者对史料来源的怀疑?

(b) What is the purpose of this mathematical analysis, for its inclusion in this essay?
(b) 将这一数学分析纳入本文的目的是什么?

PHYSICS IN DETAIL 物理详解

Now we will look at the physics in detail. You may need to combine concepts from different areas of physics to work out the answers.
现在我们来详细了解物理知识。您可能需要结合不同物理领域的概念来计算答案。

2 Complete the calculation steps, in reverse as suggested, in order to find out the answer, H H HH :
2 按照建议反向完成计算步骤,求出答案 H H HH

(a) the acceleration caused by the explosion
(a) 爆炸造成的加速度

(b) overall velocity that the cannonball is projected from the cannon
(b) 炮弹从大炮射出的总速度

© horizontal and vertical components of the velocity
速度的水平和垂直分量

(d) time of flight found from the horizontal travel
(d) 根据水平行程计算出的飞行时间

(e) time to reach maximum height using vertical motion
(e) 利用垂直运动达到最大高度的时间

(f) remaining flight time from maximum height
(f) 从最高高度算起的剩余飞行时间

(g) height fallen from the maximum in the remaining flight time
(g) 在剩余飞行时间内从最大高度下降的高度

(h) final answer, H H HH.
(h) 最后答案, H H HH

3 State two assumptions that have been made in these calculations.
3 说明计算中的两个假设。

4 Calculate what difference there would be in the answer if the cannon was loaded with different cannonballs of masses 10 kg and 14 kg . Note from fig B that the fortress walls are 24 m high. Comment on these answers.
4 计算如果大炮装载质量为 10 千克和 14 千克的不同炮弹,答案会有什么不同。从图 B 中可以看出,堡垒墙高 24 米。对这些答案进行评论。

5 If the available supply of cannonballs offered very variable masses, how might the Mughals be able to overcome the problems shown in question 4.
5 如果现有炮弹供应的质量变化很大,莫卧儿人如何才能克服问题 4 中显示的问题。

ACTIVITY 活动

Imagine the writer of this essay is a friend of yours, and he has come to you for help with the calculations as he is not an experienced scientist. His section ‘Steps to the answer’ was taken from a research source about a different fortress under siege. Write an email to Claus to explain the calculations required in each step.
想象一下,这篇文章的作者是你的一位朋友,由于他不是一位经验丰富的科学家,他来向你寻求计算方面的帮助。他的 "得出答案的步骤 "部分摘自一份关于被围困的不同要塞的研究资料。给克劳斯写一封电子邮件,解释每个步骤所需的计算。

INTERPRETATION NOTE 解释说明

Once you have answered the calculation questions below, decide whether you think the Mughal siege happened as the author suggests.
回答完下面的计算问题后,请判断您是否认为莫卧儿王朝的围城战正如作者所说的那样发生了。

THINKING BIGGER TIP 高瞻远瞩

Inside a cannon, an explosion exerts a force on the cannonball to fire it out of the cannon.
在加农炮内,爆炸对炮弹产生作用力,将其发射出去。

INTERPRETATION NOTE 解释说明

You can assume that the writer understands mathematics, and is generally intelligent - a student who could have done A Level physics but preferred arts subjects.
你可以假定作者懂数学,而且一般都很聪明--他本可以完成 A 级物理课程,但更喜欢文科科目。

1A
EXAM PRACTICE
1A 考试练习

1 Quantities can be scalar or vector. Select the row of the table that correctly states a scalar quantity and a vector quantity.
1 量可以是标量或矢量。请选择表格中正确表述标量和矢量的一行。
Scalar quantity 标量 Vector quantity 矢量数量
A acceleration 加速度 mass 质量
B mass 质量 weight 重量
C speed 速度 distance 距离
D weight 重量 speed 速度
Scalar quantity Vector quantity A acceleration mass B mass weight C speed distance D weight speed | | Scalar quantity | Vector quantity | | :--- | :--- | :--- | | A | acceleration | mass | | B | mass | weight | | C | speed | distance | | D | weight | speed | | | | |
(Total for Question 1 = 1 mark)
(问题 1 的总分 = 1 分)

2 How is the kinetic energy, E k E k E_(k)E_{\mathrm{k}}, of a car related to its speed, v v vv ?
2 汽车的动能 E k E k E_(k)E_{\mathrm{k}} 与速度 v v vv 有什么关系?

A E k v E k v E_(k)vE_{\mathrm{k}} v
B E k v 2 E k v 2 E_(k)v^(2)E_{\mathrm{k}} v^{2}
C E k v E k v E_(k)sqrtvE_{\mathrm{k}} \sqrt{v}
D E k 1 v E k 1 v E_(k)(1)/(v)E_{\mathrm{k}} \frac{1}{v}
(Total for Question 2 = 1 2 = 1 2=12=1 mark)
(问题 2 = 1 2 = 1 2=12=1 分值总计)

3 The unit of force is the newton. One newton is equivalent to:
3 力的单位是牛顿。一牛顿相当于

A 0.1 kg A 0.1 千克
B 1 kg m s 1 1 kg m s 1 1kgms^(-1)1 \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}
C 1 kg m s 2 1 kg m s 2 1kgms^(-2)1 \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-2}
D 1 ms 2 1 ms 2 1ms^(-2)1 \mathrm{~ms}^{-2}
(Total for Question 3 = 1 mark)
(问题 3 的总分 = 1 分)

4 A ball is thrown vertically upwards at a speed of 11.0 m s 1 11.0 m s 1 11.0ms^(-1)11.0 \mathrm{~m} \mathrm{~s}^{-1}. What is the maximum height it reaches?
4 将一个球以 11.0 m s 1 11.0 m s 1 11.0ms^(-1)11.0 \mathrm{~m} \mathrm{~s}^{-1} 的速度垂直向上抛出。它达到的最大高度是多少?

A 0.561 m A 0.561 米
B 1.12 m B 1.12 米
C 6.17 m C 6.17 米
D 12.3 m D 12.3 米
(Total for Question 4 = 1 4 = 1 4=14=1 mark)
(问题 4 = 1 4 = 1 4=14=1 分值总计)

5 Calculate the moment exerted on the nut by the spanner shown in the diagram.
5 计算图中所示扳手对螺母施加的力矩。
A 2.4 Nm A 2.4 牛米
B 4.2 Nm B 4.2 牛米
C 4.8 Nm C 4.8 牛米
D 420 Nm D 420 牛米
6 (a) What is meant by a vector quantity?
6 (a) 何谓矢量?

(b) A car is driven around a bend at a constant speed. Explain what happens to its velocity.
(b) 一辆汽车以恒定的速度驶过一个弯道。请解释它的速度会发生什么变化。

(Total for Question 6 = 3 6 = 3 6=36=3 marks)
(问题 6 = 3 6 = 3 6=36=3 总分)

7 You are asked to determine the acceleration of free fall at the surface of the Earth, g g gg, using a free fall method in the laboratory.
7 要求你在实验室使用自由落体法测定地球表面自由落体的加速度 g g gg

(a) Describe the apparatus you would use, the measurements you would take and explain how you would use them to determine g g gg.
(a) 描述你将使用的仪器,你将进行的测量,并解释你将如何使用它们来确定 g g gg

(b) Give one precaution you would take to ensure the accuracy of your measurements.
(b) 给出一个你会采取的预防措施,以确保测量的准确性。

(Total for Question 7 = 7 marks)
(第 7 题总分=7 分)

8 The graph shows how displacement varies with time for an object that starts from rest with constant acceleration.
8 该图显示了一个物体从静止开始以恒定加速度运动时,位移随时间的变化情况。

(a) Use the distance-time graph to determine the speed of the object at a time of 4.0 s .
(a) 利用距离-时间图确定物体在 4.0 秒时的速度。

(b) Calculate the acceleration.
(b) 计算加速度。

(Total for Question 8 = 5 8 = 5 8=58=5 marks)
(问题 8 = 5 8 = 5 8=58=5 总分)

9 The photograph shows a sequence of images of a bouncing tennis ball.
9 照片显示了一个弹跳的网球的一系列图像。
A student plots the following graph and claims that it shows the vertical motion of the ball in the photograph.
一名学生绘制了下图,并声称该图显示了照片中球的垂直运动。

(a) Without carrying out any calculations, describe how the following can be found from the graph
(a) 在不进行任何计算的情况下,描述如何从图中发现以下内容

(i) the vertical distance travelled by the ball between 0.5 s and 1.0 s
(i) 小球在 0.5 秒至 1.0 秒之间的垂直距离

(ii) the acceleration at Y .
(ii) Y 处的加速度。

(b) The graph contains several errors in its representation of the motion of the ball Explain two of these errors.
(b) 该图在表示小球的运动时有几处错误 请解释其中的两处错误。

(Total for Question 9 = 6 9 = 6 9=69=6 marks)
(问题 9 = 6 9 = 6 9=69=6 总分)

10 There has been a proposal to build a train tunnel underneath the Atlantic Ocean from England to America. The suggestion is that in the future the trip of 5000 km could take as little as one hour.
10 有人提议在大西洋海底建造一条从英国通往美国的火车隧道。该建议认为,未来 5000 公里的行程只需一个小时。

Assume that half the time is spent accelerating uniformly and the other half is spent decelerating uniformly with the same magnitude as the acceleration.
假设一半时间用于匀加速,另一半时间用于匀减速,减速幅度与加速度相同。

(a) Show that the acceleration would be about 2 ms 2 2 ms 2 2ms^(-2)2 \mathrm{~ms}^{-2}.
(a) 证明加速度约为 2 ms 2 2 ms 2 2ms^(-2)2 \mathrm{~ms}^{-2}

(b) Calculate the maximum speed.
(b) 计算最大速度。

© Calculate the resultant force required to decelerate the train.
计算火车减速所需的合力。

mass of train = 4.5 × 10 5 kg = 4.5 × 10 5 kg =4.5 xx10^(5)kg=4.5 \times 10^{5} \mathrm{~kg} 列车质量 = 4.5 × 10 5 kg = 4.5 × 10 5 kg =4.5 xx10^(5)kg=4.5 \times 10^{5} \mathrm{~kg}
(Total for Question 10 = 6 10 = 6 10=610=6 marks)
(问题 10 = 6 10 = 6 10=610=6 总分)

11 During a lesson on Newton’s laws of motion, a student says, ‘We don’t really need to bother with Newton’s first law because it is included in his second law’. State Newton’s first two laws of motion and explain how Newton’s second law includes the first law.
11 在一节关于牛顿运动定律的课上,一名学生说:"我们其实不需要费心牛顿第一定律,因为它已经包含在牛顿第二定律中了"。请说明牛顿的前两个运动定律,并解释牛顿第二定律是如何包含第一定律的。

(Total for Question 11 = 5 11 = 5 11=511=5 marks)
(问题 11 = 5 11 = 5 11=511=5 总分)

12 The diagram shows an arrangement used to launch a light foam rocket at a school science competition.
12 图中显示的是在学校科学竞赛中发射轻质泡沫火箭时使用的排列方式。
The rocket is launched at the level of one end of a long table and lands at the other end at the same level. The students measure the horizontal distance travelled by the rocket and the time of flight.
火箭在长桌一端的水平面发射,在另一端的水平面着陆。学生测量火箭飞行的水平距离和时间。

(a) The rocket travels 1.88 m in a time of 0.88 s .
(a) 火箭在 0.88 秒内飞行了 1.88 米。

(i) Show that the horizontal component of the initial velocity of the rocket is about 2 m s 1 2 m s 1 2ms^(-1)2 \mathrm{~m} \mathrm{~s}^{-1}.
(i) 证明火箭初速的水平分量约为 2 m s 1 2 m s 1 2ms^(-1)2 \mathrm{~m} \mathrm{~s}^{-1}

(ii) Show that the vertical component of the initial velocity of the rocket is about 4 m s 1 4 m s 1 4ms^(-1)4 \mathrm{~m} \mathrm{~s}^{-1}.
(ii) 证明火箭初速的垂直分量约为 4 m s 1 4 m s 1 4ms^(-1)4 \mathrm{~m} \mathrm{~s}^{-1}

(iii) Calculate the initial velocity of the rocket.
(iii) 计算火箭的初速度。

(b) The students obtained their data by filming the flight. When they checked the maximum height reached by the rocket they found it was less than the height predicted using this velocity.
(b) 学生们通过拍摄飞行获得数据。当他们检查火箭达到的最大高度时,发现它小于用这一速度预测的高度。

(i) Suggest why the maximum height reached was less than predicted.
(i) 说明为什么达到的最大高度低于预测值。

(ii) Give two advantages of filming the flight to obtain the data.
(ii) 说明拍摄飞行以获取数据的两个优势。

(Total for Question 12 = 11 marks)
(第 12 题共计 11 分)

TOPIC 1 MECHANICS 专题 1 力学

1B ENERGY 1B 能源

Chapter 1A finished with a discussion of the motion of a projectile cannonball’s flight. An alternative way of considering changes in movements of objects affected by gravity is to follow what happens to their energy. An object held up has an amount of gravitational potential energy as a result of its position. This energy can be transferred to kinetic, or movement, energy if the object falls. Humans have evolved over millions of years to avoid situations in which they might fall a great height, as such falls are generally dangerous. The kinetic energy that humans have gained from falling could be transferred to other stores of energy in their bodies, through large forces, and cause injuries.
第 1A 章最后讨论了抛射体炮弹的飞行运动。考虑受重力影响的物体运动变化的另一种方法是关注其能量的变化。被托起的物体因其位置而具有一定量的重力势能。如果物体下落,这种能量就会转化为动能或运动能。人类经过数百万年的进化,已经能够避免可能从高处坠落的情况,因为这种坠落通常是危险的。人类从坠落中获得的动能可以通过巨大的力量转移到体内的其他能量储存中,从而造成伤害。

The effect of gravity on the movement of an object should be considered in relation to the energy a body may possess or transfer. There are equations for calculating kinetic energy and gravitational potential energy, and the transfer of energy when a force is used to cause the transfer. These formulae and Newton’s laws can be used together to work out everything we might want to know about the movement of any everyday object in any everyday situation.
重力对物体运动的影响应与物体可能拥有或传递的能量联系起来考虑。有一些公式可以计算动能和重力势能,以及在力的作用下能量的转移。这些公式和牛顿定律一起使用,可以计算出我们想知道的任何日常物体在任何日常情况下的运动情况。

Whilst it is difficult for scientists to describe or identify the exact nature of energy, the equations that describe energy relationships have always worked correctly.
虽然科学家们很难描述或确定能量的确切性质,但描述能量关系的方程一直在正确地运行。

MATHS SKILLS FOR THIS CHAPTER
本章的数学技能

  • Units of measurement (e.g. the joule, J)
    测量单位(如焦耳,J)
  • Changing the subject of an equation (e.g. finding the velocity of a falling object)
    改变等式的主题(如求下落物体的速度)
  • Substituting numerical values into algebraic equations (e.g. calculating the power used)
    将数值代入代数方程(如计算所用功率)
  • Plotting two variables from experimental or other data, understanding that y = m x + c y = m x + c y=mx+cy=m x+c represents a linear relationship and determining the slope of a linear graph (e.g. finding the acceleration due to gravity experimentally)
    根据实验数据或其他数据绘制两个变量的图形,理解 y = m x + c y = m x + c y=mx+cy=m x+c 表示线性关系,并确定线性图形的斜率(例如,通过实验求出重力加速度)。
  • Using angles in regular 2D and 3D structures (e.g. finding the angle with which to calculate work done)
    在规则的二维和三维结构中使用角度(例如,找到计算做功的角度)
  • Using sin, cos and tan in physical problems (e.g. calculating the work done by a force acting at an angle)
    在物理问题中使用 sin、cos 和 tan(例如,计算作用在某一角度的力所做的功)

1B 1 GRAVITATIONAL POTENTIAL AND KINETIC ENERGIES
1B 1 重力势能和动能

LEARNING OBJECTIVES 学习目标

Calculate transfers of gravitational potential energy near the Earth’s surface.
计算地球表面附近重力势能的转移。

Calculate the kinetic energy of a body.
计算物体的动能。
  • Calculate exchanges between gravitational potential and kinetic energies, based on energy conservation.
    根据能量守恒,计算重力势能和动能之间的交换。

Δ Δ Delta\Delta fig A The gravitational potential energy transferred to kinetic energy for a falling coconut can be a significant hazard in tropical countries.
Δ Δ Delta\Delta 图 A 在热带国家,椰子下落时转化为动能的重力势能可能是一个重大危险。

DID YOU KNOW? 您知道吗?

The strength of the Earth’s gravitational field at the top of the Eiffel Tower is less than a hundredth of a per cent smaller than at the foot of the Eiffel Tower.
埃菲尔铁塔顶端的地球引力场强度比埃菲尔铁塔脚下的引力场强度小不到百分之一。
A fig B A jumbo jet plane has a lot of kinetic energy.
A 无花果 B 巨型喷气式飞机有很大的动能。

Gravitational potential energy ( E grav ) E grav  (E_("grav "))\left(E_{\text {grav }}\right) is the energy an object has by virtue of its position in a gravitational field. Kinetic energy ( E k ) E k (E_(k))\left(E_{k}\right) is the energy an object has by virtue of its movement. As objects rise or fall, gravitational potential energy can be transferred to kinetic energy and kinetic energy can be transferred to gravitational potential energy.
重力势能 ( E grav ) E grav  (E_("grav "))\left(E_{\text {grav }}\right) 是物体因其在重力场中的位置而具有的能量。动能 ( E k ) E k (E_(k))\left(E_{k}\right) 是物体因运动而具有的能量。当物体上升或下降时,重力势能可以转化为动能,动能也可以转化为重力势能。

GRAVITATIONAL POTENTIAL ENERGY
引力势能

Gravitational potential energy (gpe) can be calculated using the equation:
重力势能(gpe)可用公式计算:
gpe ( J ) = mass ( kg ) × gravitational field strength ( N kg 1 ) × height ( m ) E grav = m g h gpe ( J ) =  mass  ( kg ) ×  gravitational field strength  N kg 1 ×  height  ( m ) E grav  = m g h {:[gpe(J)=" mass "(kg)xx" gravitational field strength "(Nkg^(-1))xx" height "(m)],[E_("grav ")=mgh]:}\begin{aligned} \operatorname{gpe}(\mathrm{J}) & =\text { mass }(\mathrm{kg}) \times \text { gravitational field strength }\left(\mathrm{N} \mathrm{~kg}^{-1}\right) \times \text { height }(\mathrm{m}) \\ E_{\text {grav }} & =m g h \end{aligned}
Usually, the gpe is considered as a change caused by a change in height, for example, the change in gpe when you lift an object onto a shelf. This alters the equation slightly to consider transfers to or from gpe:
通常,gpe 被认为是由高度变化引起的变化,例如,当你把一个物体举到架子上时,gpe 会发生变化。这就稍微改变了等式,以考虑 gpe 之间的转移:
Δ E grav = m g Δ h Δ E grav  = m g Δ h DeltaE_("grav ")=mg Delta h\Delta E_{\text {grav }}=m g \Delta h
A brick of mass 2.2 kg is lifted vertically through a height of 1.24 m . The gpe gained is calculated as follows:
将一块质量为 2.2 千克的砖块垂直举过 1.24 米的高度。获得的 gpe 计算如下:
Δ E grav = m g Δ h Δ E grav = 2.2 × 9.81 × 1.24 Δ E grav = 26.8 J Δ E grav  = m g Δ h Δ E grav  = 2.2 × 9.81 × 1.24 Δ E grav  = 26.8 J {:[DeltaE_("grav ")=mg Delta h],[DeltaE_("grav ")=2.2 xx9.81 xx1.24],[DeltaE_("grav ")=26.8J]:}\begin{aligned} & \Delta E_{\text {grav }}=m g \Delta h \\ & \Delta E_{\text {grav }}=2.2 \times 9.81 \times 1.24 \\ & \Delta E_{\text {grav }}=26.8 \mathrm{~J} \end{aligned}
Writing the formula this way suggests that the gravitational field strength is a fixed value. The gravitational field strength is a measure of the pull of gravity by a planet at a distance from its centre. This is not actually constant, as the strength of the gravitational field experienced by a mass is inversely proportional to the square of the distance from the planet’s centre. However, close to the Earth’s surface, over small scales, such as the heights that humans deal with in everyday life, it is an acceptably close approximation to say g g gg is fixed at 9.81 Nkg 1 9.81 Nkg 1 9.81Nkg^(-1)9.81 \mathrm{Nkg}^{-1}.
这样写公式意味着引力场强度是一个固定值。引力场强度是一个星球在距离其中心一定距离时所受引力的量度。这实际上并不是恒定的,因为质量所经历的引力场强度与距离行星中心的距离的平方成反比。不过,在接近地球表面的小范围内,例如人类在日常生活中接触到的高度,说 g g gg 固定在 9.81 Nkg 1 9.81 Nkg 1 9.81Nkg^(-1)9.81 \mathrm{Nkg}^{-1} 是一个可以接受的近似值。

KINETIC ENERGY 动能

Kinetic energy can be calculated using the equation:
动能可以用公式计算:
kinetic energy ( J ) = 1 2 × mass ( kg ) × ( speed ) 2 ( m 2 s 2 ) E k = 1 2 × m × v 2  kinetic energy  ( J ) = 1 2 ×  mass  ( kg ) × (  speed  ) 2 m 2 s 2 E k = 1 2 × m × v 2 {:[" kinetic energy "(J)=(1)/(2)xx" mass "(kg)xx(" speed ")^(2)(m^(2)s^(-2))],[E_(k)=(1)/(2)xx m xxv^(2)]:}\begin{aligned} \text { kinetic energy }(\mathrm{J}) & =\frac{1}{2} \times \text { mass }(\mathrm{kg}) \times(\text { speed })^{2}\left(\mathrm{~m}^{2} \mathrm{~s}^{-2}\right) \\ E_{k} & =\frac{1}{2} \times m \times v^{2} \end{aligned}
For example, a large jumbo jet plane might have a cruising speed of 900 km h 1 900 km h 1 900kmh^(-1)900 \mathrm{~km} \mathrm{~h}^{-1} and a flight mass of 400 tonnes. What would its kinetic energy be?
例如,一架大型喷气式飞机的巡航速度为 900 km h 1 900 km h 1 900kmh^(-1)900 \mathrm{~km} \mathrm{~h}^{-1} ,飞行质量为 400 吨。它的动能是多少?

First convert into SI units:
首先转换成国际单位制单位:
v = 900 km h 1 = 900000 mh 1 = 9 × 10 5 60 × 60 = 250 m s 1 m = 400 × 1000 = 4 × 10 5 kg E k = 1 2 × m × v 2 E k = 1 2 × 4 × 10 5 × 250 2 E k = 1.25 × 10 10 J = 12.5 GJ v = 900 km h 1 = 900000 mh 1 = 9 × 10 5 60 × 60 = 250 m s 1 m = 400 × 1000 = 4 × 10 5 kg E k = 1 2 × m × v 2 E k = 1 2 × 4 × 10 5 × 250 2 E k = 1.25 × 10 10 J = 12.5 GJ {:[v=900kmh^(-1)=900000mh^(-1)=(9xx10^(5))/(60 xx60)=250ms^(-1)],[m=400 xx1000=4xx10^(5)kg],[E_(k)=(1)/(2)xx m xxv^(2)],[:.quadE_(k)=(1)/(2)xx4xx10^(5)xx250^(2)],[E_(k)=1.25 xx10^(10)J=12.5GJ]:}\begin{aligned} v & =900 \mathrm{~km} \mathrm{~h}^{-1}=900000 \mathrm{mh}^{-1}=\frac{9 \times 10^{5}}{60 \times 60}=250 \mathrm{~m} \mathrm{~s}^{-1} \\ m & =400 \times 1000=4 \times 10^{5} \mathrm{~kg} \\ E_{k} & =\frac{1}{2} \times m \times v^{2} \\ \therefore \quad E_{k} & =\frac{1}{2} \times 4 \times 10^{5} \times 250^{2} \\ E_{k} & =1.25 \times 10^{10} \mathrm{~J}=12.5 \mathrm{GJ} \end{aligned}

TRANSFER BETWEEN E grav E grav  E_("grav ")E_{\text {grav }} AND E K E K E_(K)E_{K}
E grav E grav  E_("grav ")E_{\text {grav }} E K E K E_(K)E_{K} 之间转移

The principle of conservation of energy tells us that we can never lose any energy or gain energy out of nowhere. In any energy transfer, we must have the same total energy before and after the transfer. Gravitational potential energy can be transferred to kinetic energy if an object falls to a lower height. Alternatively, an object thrown upwards will slow down as its kinetic energy is transferred to gpe. In either case:
能量守恒原理告诉我们,我们永远不会凭空失去任何能量或获得能量。在任何能量转移中,能量转移前后的总能量必须相同。如果物体下落到较低的高度,重力势能可以转移为动能。或者,向上抛出的物体在其动能转移到重力势能时会减速。无论哪种情况:
Δ E grav = m g Δ h = 1 2 m v 2 = E k Δ E grav  = m g Δ h = 1 2 m v 2 = E k DeltaE_("grav ")=mg Delta h=(1)/(2)mv^(2)=E_(k)\Delta E_{\text {grav }}=m g \Delta h=\frac{1}{2} m v^{2}=E_{k}
Depending on the situation, it can often be useful to divide out the mass that appears on both sides of this equation. This allows a convenient calculation of how fast an object will be travelling after falling a certain distance from rest:
根据具体情况,通常可以将等式两边的质量相除。这样就可以方便地计算出物体从静止下落一定距离后的速度:
v = 2 g Δ h v = 2 g Δ h v=sqrt(2g Delta h)v=\sqrt{2 g \Delta h}
or how high an object could rise if projected upwards at a certain speed:
或一个物体以一定速度向上投射时能上升多高:
Δ h = v 2 2 g Δ h = v 2 2 g Delta h=(v^(2))/(2g)\Delta h=\frac{v^{2}}{2 g}

LEARNING TIP 学习提示

The fact that mass divides out to give the relationship v = 2 g h v = 2 g h v=sqrt(2gh)v=\sqrt{2 g h} confirms Galileo’s idea that objects will all fall to the ground at the same rate regardless of their mass.
质量相除得出 v = 2 g h v = 2 g h v=sqrt(2gh)v=\sqrt{2 g h} 的关系这一事实证明了伽利略的观点,即无论物体的质量如何,都会以相同的速度坠落地面。

However, remember that all of the relationships shown in this section assume that no energy is lost through friction or air resistance, and that this can be an important factor when some objects fall.
不过,请记住,本节中显示的所有关系都假定没有通过摩擦或空气阻力损失能量,而这可能是某些物体下落时的一个重要因素。

fig C The Burj Khalifa tower in Dubai.
图 C 迪拜的哈利法塔。

For example, how fast would a coin hit the ground if it were dropped from the top of Burj Khalifa tower in Dubai, which is 830 m tall?
例如,如果一枚硬币从 830 米高的迪拜哈利法塔塔顶投下,它落地的速度会有多快?
v = 2 g Δ h v = 2 ( 9.81 × 830 ) = 16285 v = 128 m s 1 v = 2 g Δ h v = 2 ( 9.81 × 830 ) = 16285 v = 128 m s 1 {:[v=sqrt(2g Delta h)],[v=sqrt(2(9.81 xx830))=sqrt16285],[v=128ms^(-1)]:}\begin{aligned} & v=\sqrt{2 g \Delta h} \\ & v=\sqrt{2(9.81 \times 830)}=\sqrt{16285} \\ & v=128 \mathrm{~m} \mathrm{~s}^{-1} \end{aligned}
Another example: how high would water from a fountain rise if it were ejected vertically upwards from a spout at 13.5 m s 1 13.5 m s 1 13.5ms^(-1)13.5 \mathrm{~m} \mathrm{~s}^{-1} ?
另一个例子:如果喷泉的水从 13.5 m s 1 13.5 m s 1 13.5ms^(-1)13.5 \mathrm{~m} \mathrm{~s}^{-1} 处的喷口垂直向上喷出,喷泉的水会上升多高?
Δ h = v 2 2 g Δ h = 13.5 2 2 ( 9.81 ) = 182.25 19.62 h = 9.29 m Δ h = v 2 2 g Δ h = 13.5 2 2 ( 9.81 ) = 182.25 19.62 h = 9.29 m {:[Delta h=(v^(2))/(2g)],[Delta h=(13.5^(2))/(2(9.81))=(182.25)/(19.62)],[h=9.29m]:}\begin{aligned} \Delta h & =\frac{v^{2}}{2 g} \\ \Delta h & =\frac{13.5^{2}}{2(9.81)}=\frac{182.25}{19.62} \\ h & =9.29 \mathrm{~m} \end{aligned}
A fig D Fountain designers need to be able to calculate gpe and kinetic energy.
无花果D型喷泉设计人员需要能够计算gpe和动能。

PRAGTICAL SKILLS 实用技能

Finding g g gg from energy conservation
从节能中寻找 g g gg

There are a number of different experimental methods for finding the gravitational field strength. One example of these relies on the transfer of gravitational potential energy to kinetic energy. In this experiment, we measure the velocity that has been reached by a falling object after it has fallen under gravity from a certain height, and then alter the height and measure the velocity again.
有许多不同的实验方法可以找到引力场强度。其中一个例子是将重力势能转化为动能。在这个实验中,我们测量一个下落物体在重力作用下从一定高度落下后所达到的速度,然后改变高度并再次测量速度。

If we vary the height from which the object falls, the gravitational potential energy is different at each height. This gpe will all be transferred to kinetic energy.
如果我们改变物体下落的高度,在每个高度上的重力势能都是不同的。这些重力势能将全部转化为动能。
m g Δ h = 1 2 m v 2 m g Δ h = 1 2 m v 2 mg Delta h=(1)/(2)mv^(2)m g \Delta h=\frac{1}{2} m v^{2}
As we saw previously, this equation can be simplified to give:
如前所述,这个等式可以简化为
v 2 = 2 g Δ h v 2 = 2 g Δ h v^(2)=2g Delta hv^{2}=2 g \Delta h
Compare this equation with the equation for a straight-line graph: y = m x + c y = m x + c y=mx+cy=m x+c
将此方程与直线图形方程进行比较: y = m x + c y = m x + c y=mx+cy=m x+c

If we plot a graph of Δ h Δ h Delta h\Delta h on the x x xx-axis and v 2 v 2 v^(2)v^{2} on the y y yy-axis, we will get a straight best-fit line. The gradient of the line on this graph will be twice the gravitational field strength, 2 g 2 g 2g2 g, from which we can find g g gg.
如果我们在 x x xx 轴上绘制 Δ h Δ h Delta h\Delta h 的曲线图,在 y y yy 轴上绘制 v 2 v 2 v^(2)v^{2} 的曲线图,我们将得到一条最佳拟合直线。该图上直线的梯度将是引力场强度 2 g 2 g 2g2 g 的两倍,由此我们可以求出 g g gg

We could find a value for g g gg by taking a single measurement from this experiment and using the equation to calculate it:
我们可以通过这次实验中的一次测量,利用方程计算出 g g gg 的值:
g = v 2 2 Δ h g = v 2 2 Δ h g=(v^(2))/(2Delta h)\boldsymbol{g}=\frac{v^{2}}{2 \Delta h}
However, as mentioned in Section 1A.6, a single measurement in any experiment is prone to uncertainty from both random and systematic errors. The reliability of the conclusions is significantly improved with multiple readings and graphical analysis.
然而,正如第 1A.6 节所述,任何实验中的单次测量都可能因随机和系统误差而产生不确定性。通过多次读数和图形分析,可以大大提高结论的可靠性。

/_\\triangle fig E E E\mathbf{E} The freefall velocity of an object from different heights allows us to find the gravitational field strength, g g gg.
/_\\triangle E E E\mathbf{E} 通过物体从不同高度自由落体的速度,我们可以求出引力场强度 g g gg
Safety Note: Secure the tall stand so that it cannot topple over. The object must be positioned so that it cannot cause injury as it falls.
安全提示:固定好高脚支架,使其不会倾倒。物体的位置必须确保其在倒下时不会造成伤害。

SKILLS 技能

CREATIVITY 创造力

CHECKPOINT 检查点

  1. Estimate the speed at which a coconut from the tree in fig A would hit the sand.
    估计图 A 中树上的椰子砸到沙子上的速度。
  2. How fast would a fountain need to squirt its water upwards to reach a height of 15 m ?
    喷泉需要以多快的速度向上喷水才能达到 15 米的高度?
  3. How fast would a snowboarder be moving if he slid down a slope dropping a vertical height of 45 m ?
    如果一名滑雪板运动员从垂直高度为 45 米的斜坡上滑下,他的移动速度有多快?
  4. How high will a 48 kg trampolinist rise if he leaves the trampoline at a speed of 6.1 m s 1 6.1 m s 1 6.1ms^(-1)6.1 \mathrm{~m} \mathrm{~s}^{-1} ?
    如果一个体重 48 千克的蹦床运动员以 6.1 m s 1 6.1 m s 1 6.1ms^(-1)6.1 \mathrm{~m} \mathrm{~s}^{-1} 的速度离开蹦床,他将上升多高?
  5. What assumption must you make in order to answer all of the above questions?
    要回答上述所有问题,你必须做出什么假设?

SUBJECT VOCABULARY 主题词汇

gravitational potential energy ( E grav ) E grav  (E_("grav "))\left(E_{\text {grav }}\right) the energy an object stores by virtue of its position in a gravitational field:
重力势能 ( E grav ) E grav  (E_("grav "))\left(E_{\text {grav }}\right) 物体因其在重力场中的位置而储存的能量:
gpe ( J ) = mass ( kg ) × gravitational field strength ( N kg 1 ) × height ( m ) E grav = m g OR Δ E grav = m g Δ h  gpe  ( J ) =  mass  ( kg ) ×  gravitational field strength  N kg 1 ×  height  ( m ) E grav  = m g OR Δ E grav  = m g Δ h {:[" gpe "(J)=" mass "(kg)xx" gravitational field strength "(Nkg^(-1))xx" height "(m)],[E_("grav ")=mgORDeltaE_("grav ")=mg Delta h]:}\begin{aligned} \text { gpe }(J) & =\text { mass }(\mathrm{kg}) \times \text { gravitational field strength }\left(\mathrm{N} \mathrm{~kg}^{-1}\right) \times \text { height }(\mathrm{m}) \\ E_{\text {grav }} & =m g \mathrm{OR} \Delta E_{\text {grav }}=m g \Delta h \end{aligned}
kinetic energy ( E k ) E k (E_(k))\left(E_{k}\right) the energy an object stores by virtue of its movement:
动能 ( E k ) E k (E_(k))\left(E_{k}\right) 物体因运动而储存的能量:

kinetic energy ( J ) = 1 2 × ( J ) = 1 2 × (J)=(1)/(2)xx(\mathrm{J})=\frac{1}{2} \times mass ( kg ) × ( speed ) 2 ( m 2 s 2 ) ( kg ) × (  speed  ) 2 m 2 s 2 (kg)xx(" speed ")^(2)(m^(2)s^(-2))(\mathrm{kg}) \times(\text { speed })^{2}\left(\mathrm{~m}^{2} \mathrm{~s}^{-2}\right)
动能 ( J ) = 1 2 × ( J ) = 1 2 × (J)=(1)/(2)xx(\mathrm{J})=\frac{1}{2} \times 质量 ( kg ) × ( speed ) 2 ( m 2 s 2 ) ( kg ) × (  speed  ) 2 m 2 s 2 (kg)xx(" speed ")^(2)(m^(2)s^(-2))(\mathrm{kg}) \times(\text { speed })^{2}\left(\mathrm{~m}^{2} \mathrm{~s}^{-2}\right)
E k = 1 2 × m × v 2 E k = 1 2 × m × v 2 E_(k)=(1)/(2)xx m xxv^(2)E_{k}=\frac{1}{2} \times m \times v^{2}

LEARNING OBJECTIVES 学习目标

Calculate energy transferred as work done, including when the force is not along the line of motion.
计算作为功传递的能量,包括当力不沿运动线时。

Calculate the power of an energy transfer.
计算能量传递的功率。

Explain efficiency and be able to calculate.
解释效率并能够计算。

We often make assumptions that allow simplification of calculations in physics. In general, these assumptions make little difference as they are chosen to ignore effects which have a very small impact on the actual real world answers. An example of this is with the transfer of gravitational potential energy to kinetic energy, where the effects of air resistance are ignored.
在物理学中,我们经常会做出一些可以简化计算的假设。一般来说,这些假设的作用不大,因为它们是为了忽略对实际答案影响很小的效应。重力势能向动能的转移就是一个例子,其中空气阻力的影响被忽略了。
It is important to remember that the principle of conservation of energy insists that no energy can be lost in any scenario. Even if we were to consider the loss of kinetic energy to heating of the air through air resistance, the total amount of energy would be constant; it would simply have transferred to different stores.
重要的是要记住,能量守恒原则坚持认为在任何情况下都不能损失能量。即使我们考虑到空气阻力加热空气所损失的动能,能量的总量也是不变的,只是转移到了不同的存储空间。

WORK 工作

In physics, the phrase ‘doing work’ has a specific meaning to do with energy use. The amount of work done means the amount of energy transferred, so work is measured in joules.
在物理学中,"做功 "一词的特定含义与能量的使用有关。做功的量意味着转移的能量,因此功以焦耳为单位。
In general terms, we can express any energy transfer as work done. For example, a 15 W light bulb working for 10 seconds transfers 150 J of electrical energy as heat and light - it does 150 J of work. In any situation where we know how to calculate the energy before and after, we can calculate the energy transferred and thus the work done.
一般来说,我们可以用做功来表示任何能量的传递。例如,一个 15 W 的灯泡工作 10 秒钟,以热和光的形式传递了 150 J 的电能--它做了 150 J 的功。在任何情况下,只要我们知道如何计算前后的能量,就可以计算出传递的能量,从而计算出做的功。

FORCING WORK 强迫工作

If energy is transferred mechanically by means of a force, then the amount of work done can be calculated simply:
如果能量是通过力进行机械传递的,那么所做的功就可以简单计算出来:
work done ( J ) = force ( N ) × distance moved in the direction of the force ( m ) Δ W = F Δ s  work done  ( J ) =  force  ( N ) ×  distance moved in the direction of the force  ( m ) Δ W = F Δ s {:[" work done "(J)=" force "(N)xx" distance moved in the direction of the force "(m)],[Delta W=F Delta s]:}\begin{aligned} \text { work done }(\mathrm{J}) & =\text { force }(\mathrm{N}) \times \text { distance moved in the direction of the force }(\mathrm{m}) \\ \Delta W & =F \Delta s \end{aligned}

fig B ‘Work’ on a building site.
图 B 建筑工地上的 "工作"。

In the example of fig B, a brick of mass 2.2 kg is lifted vertically against its weight through a height of 1.24 m . The work done is:
在图 B 的示例中,一块质量为 2.2 千克的砖块被垂直举过 1.24 米的高度。所做的功为
Δ W = F Δ s Δ W = m g × Δ s W = 2.2 × 9.81 × 1.24 W = 26.8 J Δ W = F Δ s Δ W = m g × Δ s W = 2.2 × 9.81 × 1.24 W = 26.8 J {:[Delta W=F Delta s],[Delta W=mg xx Delta s],[:.quad W=2.2 xx9.81 xx1.24],[W=26.8J]:}\begin{aligned} \Delta W & =F \Delta s \\ \Delta W & =m g \times \Delta s \\ \therefore \quad W & =2.2 \times 9.81 \times 1.24 \\ W & =26.8 \mathrm{~J} \end{aligned}
Note that this is the same amount of energy as we calculated for the gravitational potential energy of this same brick undergoing the same lift in Section 1B.1.
请注意,这与我们在第 1B.1 节中计算出的同一块砖的重力势能相同。

/_\\triangle fig A Work is done by transferring gravitational potential energy to a heavy object.
/_\\triangle fig A 功是通过向重物传递重力势能来完成的。