Edexcel AS Physics in 100 Pages PDF

Edexcel AS Physics in 100 Pages -----an easy-to-understand textbook & exam preparation guide Copyright ©2011 by Yajun Wei All rights reserved under international Copyright Conventions. No part of the text of this book may be reproduced in any form or by any electronic or mechanical means, including information storage and retrieval systems, without permission in writing by the publisher, except by reviewers or catalogues not limited to online for purpose of promotion. Front Cover photo by Fritz Diorico Back Cover photo by Yajun Wei Edexcel AS Physics in 100 Pages / Yajun Wei.--1st ed. ISBN 978-1-257-84056-4. Edexcel AS Physics in 100 Pages -----an easy-to-understand textbook & exam preparation guide Yajun Wei Edexcel AS Physics in 100 Pages Preface The purpose of this book is to offer a concise and easy-to-understand textbook and exam preparation guide of the Edexcel AS Physics. This book is slim, with only 100 pages! But it covers almost all that you need to learn to crack the Edexcel AS Physics Unit 1 and Unit 2 exams. It can be used as a textbook for your AS physics courses. It is an excellent revision guide as well. The principles and terms are explained in very simple words and sentences. They are illustrated in the students’ point of view, rather in the physicists or teachers’ point of view. There are plenty of carefully designed diagrams and examples to illustrate difficult concepts. This work is written based on my teaching experience. During my teaching, I always tried to find the ways of instruction that are easy for the students to understand. I decide to put down the contents in the ways that my students found easiest to understand and memorize. You will find it a bit easier to understand some difficult physical concepts with this book compared to some other books. This book is written based on the specifications published by Edexcel and my study on the past papers. It is excellent for exam preparation. Contents of particular importance are highlighted in gray background. The examiners frequently request you to write down these sentences or use these formulae for calculations. You will find a great match between highlighted contents in this book and the exam paper questions! Finally, good luck. ---------The author, Yajun Wei About the author The author is an A-level physics teacher. He has a B.S. in physics and M.S. in electrical engineering. He is interest in physics, engineering sciences and physics education. He began to teach A-level physics since 2007 and most of his students end up getting A in the exams. He also has experience in teaching International Baccalaureate physics and mathematics. He also do research in these fields and published some papers. 2 Edexcel AS Physics in 100 Pages Table of content Preface............................................................................................................................ 2 Table of content.............................................................................................................. 3 Chapter 0 Physical Quantities and Units....................................................................... 6 Chapter 1 Mechanics...................................................................................................... 8 1.1 Motion in one dimension................................................................................. 9 Speed, velocity, distance and displacement................................................... 9 Acceleration................................................................................................. 10 Accelerating uniformly along a straight line............................................... 10 Motion graphs.............................................................................................. 11 1.2 Combining and resolving vectors.................................................................. 13 Combining/adding vectors........................................................................... 13 Resolving/decomposing vectors.................................................................. 15 1.3 Motion and Force........................................................................................... 16 Newton’s First Law...................................................................................... 16 Newton’s Second Law................................................................................. 16 Newton’s Third Law.................................................................................... 17 Free-body diagram....................................................................................... 17 Centre of gravity & centre of mass.............................................................. 18 1.4 Motion in two dimensions—Projectile motion.............................................. 19 Projectile motion.......................................................................................... 19 Vertical projection........................................................................................ 20 Horizontal projection................................................................................... 20 1.5 Work and Energy............................................................................................ 23 Energy conservation..................................................................................... 23 Kinetic energy.............................................................................................. 24 Gravitational potential energy...................................................................... 24 Power........................................................................................................... 24 Efficiency..................................................................................................... 25 Chapter 2 Materials...................................................................................................... 27 2.1 Statics of Fluid............................................................................................... 28 Density......................................................................................................... 28 Upthrust........................................................................................................ 28 2.2 Dynamics of Fluid.......................................................................................... 32 Laminar and turbulent flows........................................................................ 32 Viscous drag................................................................................................. 33 Terminal velocity......................................................................................... 34 2.3 Strength of Solid Materials............................................................................ 37 Hooke’s Law................................................................................................ 37 Elastic strain energy..................................................................................... 37 Force ~ extension graph............................................................................... 38 3 Edexcel AS Physics in 100 Pages Stress, strain and Young Modulus................................................................ 39 Stress ~ strain graph..................................................................................... 40 Describing materials.................................................................................... 41 Chapter 3 Elementary of Waves................................................................................... 43 3.1 Introduction to Waves.................................................................................... 44 Classifying waves........................................................................................ 44 Quantities describing waves........................................................................ 47 Using graphs to describe waves................................................................... 48 3.2 Reflection....................................................................................................... 50 Law of reflection.......................................................................................... 50 Pulse echo detection--imaging..................................................................... 50 Pulse echo detection--echolocation.............................................................. 51 Doppler effect.............................................................................................. 51 3.3 Refraction....................................................................................................... 54 Law of refraction.......................................................................................... 54 Total internal reflection................................................................................ 55 3.4 Interference.................................................................................................... 57 Superposition............................................................................................... 57 Path difference and interference.................................................................. 57 Stationary wave............................................................................................ 59 3.5 Diffraction...................................................................................................... 62 3.6 Polarisation................................................................................................ 63 3.7 More about Electromagnetic Waves.......................................................... 66 Chapter 4 Direct Current Electricity............................................................................ 68 4.1 Electrical Quantities....................................................................................... 69 Electric charge and current.......................................................................... 69 Potential difference...................................................................................... 70 Electronvolt.................................................................................................. 71 Ohm’s Law and resistance........................................................................... 71 Resistivity.................................................................................................... 74 Thermistor and photoresistor....................................................................... 75 Transport equation....................................................................................... 76 4.2 Electrical Circuits........................................................................................... 78 Power and work in electrical circuit............................................................ 78 Series and parallel circuits........................................................................... 79 Potential divider........................................................................................... 81 EMF and Internal resistance........................................................................ 85 Chapter 5 Nature of Light............................................................................................ 89 5.1 Photoelectric Effect........................................................................................ 90 Investigating photoelectric effect by experiment......................................... 90 Inconsistence of the experiment results with wave theory of light.............. 93 Particle theory of light................................................................................. 93 Wave-particle duality................................................................................... 96 5.2 Spectra............................................................................................................ 97 4 Edexcel AS Physics in 100 Pages Bohr model of atom and energy levels........................................................97 Spectra..........................................................................................................98 5.3 Solar cells.....................................................................................................100 5 Edexcel AS Physics in 100 Pages Chapter 0 Physical Quantities and Units A physical quantity is a physical property that can be measured quantitatively (in numbers). Mass, pressure, density, temperature and forces are some examples of physical quantities. Physical quantities can be divided to base quantities and derived quantities. Base quantities are expressed in base units. There are 7 base quantities, length, mass, time, electric current, temperature, amount of substance and luminous intensity. All other quantities such as force, energy are derived quantities. Derived quantities are expressed in derived units. The following table summarizes all the 7 base quantities and their corresponding SI base units. All derived units can be derived from the 7 base units. Physical quantity SI unit Symbol for the unit length metre m mass kilogram kg time second s electric current ampere A temperature kelvin K amount of substance mole mol luminous intensity candela cd The seven base quantities and base units To deal with the AS physics tests, you only need to memorize the five highlighted base quantities and units as the other two are not be covered in this quantification. Example: (get back to this example after you finish the whole book) (a) Express the unit N in terms of base units only. (b) Express the unit of potential difference volt in terms of base units only. Solution: (a) F ma. Since the unit of mass m is kg and the unit of acceleration a if ms-2, the unit Force F is kgms-2. (b)Potential difference is given by W Fs mas V Q It It The unit of m is kg, unit of a is ms-2, unit of s is m, unit of I is A and unit of t is s. So the unit of potential difference is then given by 6 Edexcel AS Physics in 100 Pages kg ms 2 m V kgm 2 s 3 A1. A s Also, remember the following scale. It will make your life easier while doing unit conversions. 7 Edexcel AS Physics in 100 Pages Chapter 1 Mechanics 8 Edexcel AS Physics in 100 Pages 1.1 Motion in one dimension Speed, velocity, distance and displacement We use speed to describe how fast dose an object moves. In physics, speed is defined as the distance traveled in unit time (one second). That is dis tan ce traveled speed (1.1) time taken or s v (1.2) t Equation(1.1) and Equation(1.2) describe the same thing, but Equation(1.1) is called symbol equation and Equation(1.2) is called word equation. The most commonly used unit for speed is ms-1 and kmh-1. The conversion between these two units are given by 1kmh 1 1u1000m/3600s= 1/ 3.6 ms 1 and 1ms 1 (1/1000) km / (1/ 3600)h 3.6 kmh 1 For example, if a train travels 300 km in one and half hours, then its speed is 200 kmh-1, or 55.6 ms-1. In most cases, the speed is not constant. For the motion of the train, it may go as slow as 20km/h a few seconds after it departures, but may increase its speed to 300km/h a few minutes later. The speed at a particular instant is called instantaneous speed. But in the case of train discussed above, we are more interested in the average speed which is defined as the total distance traveled during a given time interval. Take a look at the Fig 1.1.1. A student walks from A to B along a curved path along a water pond (the blue line). Then the distance walked is the length of the path. The length of the direct line joining A and B is called the displacement. Fig 1.1.1 Distance and displacement 9 Edexcel AS Physics in 100 Pages But actually, this is only part of displacement; it is the size of displacement. Displacement is a quantity that has both size (also called magnitude) and direction. This kind of physical quantity is called a vector quantity. Distance is a quantity that has only size but no direction. This kind of physical quantity is called a scalar quantity. When an object is moving, it has a direction of its movement. The direction of velocity is the direction of movement. The size of the velocity is speed. Velocity is also a vector. We may say that the train is traveling at a velocity of 300km/h towards southeast or the train is traveling at a speed of 300km/m. Acceleration When an object increases its velocity, we say that it accelerates. Acceleration describes how fast velocity is increased. It is defined as the rate of change of velocity with time. That is final velocity initial velocity 'v vu acceleration or a (1.3) timetaken 't 't The SI unit of acceleration is ms-2. Acceleration is a vector quantity. Sometimes even though the speed doesn’t change, there is acceleration. An example is the uniform circular motion where the size of the velocity (speed) is constant but the direction of the velocity is changing all the time. If the velocity is decreasing, we may say that it decelerates. Then the acceleration is a negative value, which is also called deceleration. Accelerating uniformly along a straight line This is a motion where the acceleration is a constant and it is a very important type of motion. This type of motion can be described by five quantities and three equations: s ----displacement or distance t ---- time taken u ----initial velocity or speed v ---final velocity or speed a ---acceleration (for free fall motion under gravity, it is written as g and is equal to 9.81ms-2) v u at (1.4) 1 s ut at 2 (1.5) 2 v2 u 2 2as (1.6) Any problems of uniform acceleration motion can be solved by using the above three equations. A wise first step to solve problems during exam is to list the known 10 Edexcel AS Physics in 100 Pages quantities and the quantity you are asked to calculate. Then you will easily see which of the above equation(s) could be used to find the unknown. Example: A car accelerates uniformly from stop with a rate of 6.1 ms-2. Find the displacement and speed after 0.82 seconds. Solution: List the known quantities and unknown quantities as below: Known: u 0, t 0.82 s , a 6.1ms 2 Unknown: s , v 1 1 To find s , equation(1.5) can be used: s ut at 2 u 6.1u 0.822 2.05m 2 2 To find v , equation(1.4) should be used: v u at 6.1u 0.82 5.0ms 1 Example: A small stone is fired vertically upwards by a catapult with a speed of 10 ms-1, find the maximum height the stone can reach. Solution: At maximum height, its speed is zero. List the known quantities and unknown quantities as below: Known: u 10ms 1 , v 0, a g 9.81ms 2 Unknown: s , t To find the maximum height s with the knowledge of u , v and a , equation(1.6) v2 u 2 0 102 should be used: v 2 u 2 2as gives s 5.1m 2a 2 u 9.81 The minus sign indicates that the direction of displacement is opposite to the direction of the acceleration. The maximum height the stone can reach is 5.1m. Motion graphs Motion graphs show how displacement or velocity changes with time. It is a very useful and straightforward way of describing motion by graphs. In a displacement/distance--time graph, a straight line indicates that the object is moving at a constant velocity. If the graph is a curve, it means that the object is accelerating. The gradient of a displacement/distance—time graph is the change of displacement/distance over time, which is velocity/speed. In a velocity/speed--time graph, a straight line parallel to the time axis indicates that the object is moving at a constant velocity. If the graph is a straight line with a slope, it means that the object is accelerating uniformly. The gradient of a velocity/speed—time graph is the change of velocity/speed over time, which is 11 Edexcel AS Physics in 100 Pages acceleration. The area under the velocity/speed—time graph represents the displacement/distance moved. Example: Consider the motion of the object whose velocity-time graph is given in the diagram below. (a)What is the acceleration of the object between times t=0s and t=2s? (b)What is the average speed between times t=8s and t=12s? Solution: (a) The acceleration is equal to the gradient of the line between t=0s and t=2s, which is 'v v u 8 0 2 a ms 4ms 2 't 't 2 (b)The average speed should be calculated using total distance divided by time taken. Time taken is 12-8=4s. Total distance traveled during this time interval is equal to the shaded area under the v—t graph, as shown in the figure below. total displacement shaded area 2 u 8 1/ 2 u 4 8 u 2 28m Then the average speed=28m/4s=7ms-1. 12 Edexcel AS Physics in 100 Pages 1.2 Combining and resolving vectors We have already known that scalars have only magnitude while vectors have both magnitude and direction. In this section we will discuss vectors in more detail. Displacement, velocity, acceleration and force are four vector quantities we have already or will study in this book. All other quantities encountered in this book (therefore AS Unit1 and Unit2 tests) are scalars. Combining/adding vectors Mass is a scalar quantity, if you add masses of 5 kg and 3 kg, you simply get a total mass of 8 kg. But while combining vectors, you can't just simply add their magnitudes together. Directions need to be taken into account as well. If you want to find the resultant of two vectors, the parallelogram rule or the triangle rule should be used. Triangle rule: Fig 1.2.1 illustrates how to add/combine vectors. Draw the two vectors to be combined (added) tip-to tail and the resultant vector is the third side of the triangle pointing from the tail of the first vector (A) to the tip of the second vector (B). Fig 1.2.1 combining vectors using triangle rule Example: A student walks due north from point O to P for 3 miles and then walks 4 miles due east from P to Q, find the total displacement of the student OQ. Solution: As shown in the diagram below, the size of the total displacement is equal to the length of OQ, which is 3 2 4 2 =5 miles 4 The direction of the displacement is measure by the angle T. Since tan T , 3 T 53o. 13 Edexcel AS Physics in 100 Pages Parallelogram rule: Draw the two vectors to be combined tail-to-tail as two sides of a parallelogram, as shown in Fig 1.2.2. The resultant vector is simply the diagonal of the parallelogram. Fig 1.2.2 Combing vetors using parallelogram rule Example: Two forces F1 and F2 are acting on an object with a mass of 5kg, as shown in the graph below. Both forces are 10 N. Find the acceleration of the object. Solution: To find the acceleration, we need to find the resultant force first. Using geometrical knowledge, it is easy to find the diagonal of the parallelogram. It is 2 u10 u cos 30R 17.3N. So the size of the resultant force is 17.3N. The acceleration is given by a ¦F 17.3 N 3.46ms 2 m 5kg 14 Edexcel AS Physics in 100 Pages Resolving/decomposing vectors Vector decomposition is the process of representing a vector as the sum of two vectors perpendicular to each other. These two vectors are called the components. These two components have the same effect as the single vector to be decomposed. This process is also called resolving vectors. Example: A student is towing a small box on a table with a force of 10N at an angle of 40o to the horizontal. Find the horizontal and vertical components of the force. Solution: Draw the force diagram as follows The horizontal component of the force is Fhorizontal 10 N u cos 40o 7.7 N The vertical component of the force is Fvertical 10 N u sin 40o 6.4 N 15 Edexcel AS Physics in 100 Pages 1.3 Motion and Force Newton’s First Law Newton’s first law of motion states that: If the resultant force on an object is zero, it will stay at rest or move at constant velocity. When the resultant force on an object is zero and it stays at rest, then we say that the object is in stationary, or in equilibrium. The branch of physics that study bodies in equilibrium is called statics. The Newton’s first law is also called inertia law. Inertia describes any object’s resistance to change of its state of motion. Inertia is measured by mass. The larger the mass, the larger the inertia. Newton’s Second Law Newton’s second law of motion states that: Acceleration is proportional to the resultant force on it and inversely proportional to the mass. In mathematical form, it is 6F ma (1.7) where 6F is the resultant force, m is the mass of the object and a is the acceleration. Newton’s first law and second indicate that force is not the cause of motion, but it is the cause of change of motion, acceleration. Example: A driver is starting his car with a mass of 1000kg when the traffic light changes from red to green. (a)When the driving force is 500N, the car is still at rest, find the friction at this time. (b)When the driving force is increased to 6000N, the speed of the car increases from 0 to 10ms-1 in 2 seconds, what is the friction during this process? Solution: (a) Since the car is at rest, the resultant force should be zero. This means the driving force and the friction cancel each other. Then the friction is also 500N, with its direction opposite of the driving force. (b) The acceleration of the car is given by v u 10 0 a 5ms 1 t 2 If f is the friction, according to Newton’s second law, we have Fdriving force f ma. That is 6000 N f 1000kg u 5ms 2. Solving this gives f 1000 N 16 Edexcel AS Physics in 100 Pages Newton’s Third Law Newton’s third law of motion states that: If body A exerts a force on body B, then body B exerts a force of the same type on body A that is equal in size but opposite in direction. These two forces are called a Newton third law pair. Note that the Newton third law pair forces (action and reaction) are exerted on different bodies. Example: A man is standing still on the floor. List all the forces acting on him and find their Newton third law pair forces. Solution: Force Body the force Newton third Body the Type of the acting on law pair force Newton third forces law pair force acting on weight man Attraction earth Gravitational force on earth force from the car Push from the man Normal contact floor surface Normal contact ground force from the force feet of the man Free-body diagram A free-body diagram is a diagram that shows all forces of all types acting on a particular body that we are interested in. It is also called a free-body force diagram. The body of our interest is isolated. The free-body diagram dose not show forces acting on other objects. Fig 1.3.1 shows the free body diagram for a block on a ramp. Fig 1.3.1 Free body diagram for a block on a ramp 17 Edexcel AS Physics in 100 Pages Usually the first step to solve mechanics problems is to draw free body diagram. And then the forces are decomposed to two directions perpendicular to each other. Equations of Newton’s second law in each direction can be written down. Finally the equations can be solved to find the unknown. Centre of gravity & centre of mass The gravitational force from the earth acts on every point on an object. But we often draw the weight of an object as acting on a single point. This point is the mean location of the gravitational force acting on the object. It is called the centre of gravity. Centre of mass is the point at which all the object’s mass are considered to concentrate. This point is the mean location of all the mass of the object. The term center of mass is often used interchangeably with center of gravity. For a uniform object, centre of gravity and centre of mass are located in the middle of the object. 18 Edexcel AS Physics in 100 Pages 1.4 Motion in two dimensions—Projectile motion Projectile motion The object that is thrown out (projected) is called a projectile. The path this projectile follows is called trajectory. The trajectory of a projectile is a parabola (if projected vertically, it is a straight line). In a projectile motion, the motion can be decomposed into two directions, the horizontal direction and the vertical direction. In horizontal direction, there is no force acting on the object (air resistance is neglected) and thus the object will move with a constant velocity in this direction. vx ux u cos D (1.8) s vx t (u cos D )t (1.9) In vertical direction, the acceleration is g=9.8ms-2 as weight is the only force acting on the object. vy u y gt u sin D gt (1.10) 1 1 h u y t gt 2 (u sin D )t gt 2 (1.11) 2 2 Using the above four equations enables you to find the velocity and displacement of the projectile at any given time. Sometimes, it is useful to know how far a projectile can go if the initial velocity is given. This is called the range of the projection. Equation (1.9) should be used to find out the range. Let h=0 m in equation(1.11) to find out the time it takes for the projectile to land. 1 h (u sin D )t gt 2 0 (1.12) 2 So, 2u sin D t (1.13) g 19 Edexcel AS Physics in 100 Pages Substituting equation(1.13) into equation(1.9), the range of the projection R (horizontal displacement when it lands) is obtained as R ut cos D 2u sin D u cos D (1.14) g u 2 sin 2D g We can see that the range depends both on the magnitude and direction of the initial velocity. When D 45o , the maximum range is achieved. It is u 2 / g Vertical projection Vertical projection is a special case of a projectile motion when D 90o. This means the motion is only along the vertical direction. Horizontal projection Horizontal projection is another special case of a projectile motion when D 0o. This means that the initial velocity is completely in the horizontal direction. In this case, the object undergoes free fall in vertical direction and uniform motion in horizontal direction. Letting the download direction as the positive direction, the motion equations can be simplified as follows. In horizontal direction, it moves at uniform velocity. vx u (1.15) s vx t ut (1.16) In vertical direction, the acceleration is g=9.8ms-2 as weight is the only force acting on the object. vy gt (1.17) 1 2 h gt (1.18) 2 Example: A table tennis ball is thrown by a player with a height of 0.5 m from the table. The ball leaves the player’s hand horizontally with an initial velocity u. After bouncing once, it passes just over the net at the top of its bounce. The height of a table tennis 20 Edexcel AS Physics in 100 Pages net is approximately 15.3cm. The length of the table is 274cm. (a) How long dose it take the ball to reach the table. (b) How long does it take the ball to rise 15.3cm after bouncing? (c) Use your answers to parts (a) and (b) to calculate the initial horizontal velocity of the ball. Solution: (a)Known: h 0.5m. To find: t. 1 2 From equations(1.15)--(1.18) we can see that equation(1.18) h gt can be used 2 to find the t when the height is given. 2h 2 u 0.5 The time taken to fall is t 0.32 s. g 9.81 (b)Since the ball passes just over the net at the top of its bounce, its speed at this point is zero and it reached its maximum height. Consider the motion in vertical direction. Known: final velocity v y 0 , displacement hnet 0.153m , acceleration a g 9.81ms 2 Unkown: initial velocity u y , time t In the five quantities of motion, you have already known three, so you know you can find the other two using the three equations. Now you should feel happy. 1 List the three equations. v u at s ut at 2 and v 2 u 2 2as. 2 1 To find t, we need to use v u at or s ut at 2. But in either way, we need to 2 known the initial velocity. The initial velocity can be find using v2 u 2 2as. That is vy 2 u y 2 2 ghnet. Substituting the values into the above equation and solving for u y yields 21 Edexcel AS Physics in 100 Pages 02 u y 2 2 u 9.81u 0.153 uy 1.73ms 1 Now we can use v u at to find the time t. For our case, it is 0 1.73 9.81t and thus the time taken is t 5.67 s (c) Consider the motion in horizontal direction. The equation is s ut. So the initial velocity is given by 0.137 m u 0.023ms 1 0.32 s 5.67 s Comment: To solve projectile motions problems the first step is to resolve the motion into horizontal and vertical directions. Then for the motion in each direction, just use the three equations to find the unkown. It is quite simple! 22 Edexcel AS Physics in 100 Pages 1.5 Work and Energy Energy is the ability to perform work. There are many forms of energy. A few examples are kinetic energy, gravitational potential energy, elastic potential energy, electrical energy, solar energy, chemical energy. Energy conservation The principle of conservation of energy states that: Energy cannot be created or destroyed. This is a very important statement that you should memorize. Total energy in the universe is conserved, which means its amount doesn’t change. However, energy can be transferred from one object to another or be transformed from one form to another form. For example, if billiard ball A hits billiard ball B, then some kinetic energy of ball A is transferred to ball B, but the total energy of the two doesn’t change. A light bulb transforms electrical energy to light and heat energy. There are two ways to transfer energy, by heating and working. By heating an object, heat energy (also called thermal energy) is transferred to it. If there is a force acting on an object and there is a displacement along the direction of the force, the force dose work on the object and energy is transferred to the object. Fig 1.5.1 Work done by a force The work done is equal to the displacement multiply the force component along the direction of the displacement, that is W Fs cos T (1.19) The SI unit for work is joule (J). 1 J of work is done if a 1N force is applied on an object and the object moves 1metre along the direction of the force. 1J=1Nm. When the force is not a constant but changes with displacement, the work done can be estimated by the area under the force-displacement graph. Work done on an object is equal to the increase of its energy. When 1 J work is done on an object by a force, the object increases its energy by 1 J. The unit of energy is also J. 23 Edexcel AS Physics in 100 Pages Kinetic energy Kinetic energy of an object is the energy that it possesses due to its motion. The amount of kinetic energy that an object with mass m moving in speed v possesses is 1 2 Ek mv (1.20) 2 Example: An airplane has a mass of 1500 kg. Calculate the kinetic energy of the airplane when it is flying at a speed of 500miles/h. Solution: The speed of the plane is v 500miles / h (500 u1.64)km / h 820km / h 227.8ms 1 The kinetic energy is then 1 2 1 Ek mv u1500 u 227.82 3.89 u107 J 2 2 Gravitational potential energy When an object is lifted to a higher position, work is done against the gravitational force (weight). Thus, the gravitational potential energy is increased. The change of gravitational potential energy for an object is 'Egrav mg 'h (1.21) Example: A bucket of water together with the bucket has a total mass of 15kg. What is the change of gravitational potential energy of the bucket when it is raised from the 1st floor to the 2nd floor (the height difference is 3.0 metres)? Solution: The change of gravitational potential energy is 'Egrav mg 'h 15 u 9.81u 3.0 J 441J Power Power is the work done (or energy transferred) during unit time. That is W E P (1.22) t t The SI unit of power is Watt (W). 1W=1J/s. Another commonly used unit is Horsepower (HP). 1HP=750W. From the definition of power, we can also derive the following expression 24 Edexcel AS Physics in 100 Pages W Fs P Fv (1.23) t t This equation explains a phenomenon in our daily life. For a car to climb a steep slope, a bigger driving force is required. What the driver dose is to reduce the speed because by doing so, the driving force can be increased, as explained by equation(1.23). Example: What is the power of a car’s engine if it is moving at a steady speed of 30ms-1 and the driving force is 1.5kN? Solution: The power is P Fv 1.5kN u 30ms 1 45kW You might ask that since there is a force of 1.5kN exerted on the car, how could it move at a steady speed? This is because the force is not a resultant force. The 1.5kN driving force is cancelled out by the friction and air resistance. The resultant force is still zero. The power calculated here is the power of the driving force only. Efficiency Efficiency is defined as the useful energy output divided by total energy input. useful energy output efficiency u100% (1.24) total energy input This can be also written as useful energy output useful power output u time efficiency u100% u100% total energy input total power input u time (1.25) Cancelling the variable time in the above equation gives useful power output efficiency u100% (1.26) total power input Fig 1.5.2 Efficiency 25 Edexcel AS Physics in 100 Pages Example: The input power of a car engine is 100kW. Its maximum speed is 120km/h if the drag force on it is 1500N. What is the efficiency of the car engine in this case? Where is the lost energy? Solution: The total power input is Pin 100kW 105W The useful power output is Pout Fv 1500 u (120 u103 / 3600)W 50000W The efficiency of the car engine = Pin / Pout =50% Some energy is lost due to heating. 26 Edexcel AS Physics in 100 Pages Chapter 2 Materials 27 Edexcel AS Physics in 100 Pages 2.1 Statics of Fluid Fluid is any substance that can flow. For example, water, milk and air are all fluid. Fluid includes gases and liquids. Density Density is an important property of materials. It is defined as the mass per unit volume (one cubic metre). That is mass m density or U (2.1) volume V The above equation holds for all types of materials, including fluids and solids, pure substance and mixtures. The SI unit of density is kgm-3. Another commonly used unit is gcm-3. 1 kgm-3=1000 gcm-3. The density of water is 1000 kgm-3. This means that 1m3 water has a mass of 1000 kg. Example A bottle of beer contains 250ml beer. The mass of the beer (excluding the bottle) is 0.238 kg. What is the density of the beer? Solution: m 0.228kg The density of the beer is U 6 3 9.52 u104 kgm 3 V 250 u 10 m Example One kilogram of salt is poured into 49.8L of water and makes 50L salt water. Find the density of the salt water. Solution: The mass of the salt water is equal to the sum of the mass of salt and water, that is msalt water msalt mwater msalt U waterVwater 1kg 1000 u (49.7 u103 )kg 50.7 kg msalt water 50.7 kg The density of the salt water is U 1.014 u103 kgm 3 Vsalt water 50 u103 m3 Upthrust When an object is submerged in a fluid, it feels an upward force that is called upthrust. This is the force that keeps ships and boats floating. Greek scientist Archimedes found that the size of the upthrust force is equal to the weight of the fluid that has been displaced by the object. This is called the Archimedes’ Principle. It can be written in mathematical form as 28 Edexcel AS Physics in 100 Pages F mdisplaced fluid g U f Vg (2.2) Note that in the above equation U f is the density of the fluid. Do not mistake it as the density of the object! Example: A brick with a size of 20cm u10cm u 8cm has a density of 2000kgm-3. This brick was thrown by a naughty boy to a swimming pool and it lies in the bottom of the pool. Find the normal contact force exerted by the brick on the floor of the swimming pool. Solution: To find the normal contact force from the brick to the floor, we may find its Newton third law pair force, the normal contact force from the floor to the brick first. To do this, a free body force diagram is draw as follows. Fig 2.1.1 Free-body force diagram of the brick As the brick is at rest in the bottom, we have Upthrust+ Normal contact force form the floor= Weight U f Vg N UbrickVg Solving the equation for N gives N UbrickVg U f Vg ( Ubrick U f )Vg where V=20cm u 10cm u 8cm=1600cm3 1.6 u 103 m3 is the volume of the brick and Ubrick U f 2000 1000 1000kgm 3 is the density difference between the brick and water. Substituting these values into the above equation, we get N=15.7N. So the normal contact force exerted on the floor of the swimming pool will also be 15.7N, but with its direction pointing downwards. If you put a wood block into a water pool, it will float. If you put an iron block into a water pool, it will sink to the bottom of the pool. Whether an object released in a fluid will float or sink depends on the size of its weight and the upthrust. If the upthrust is smaller than its weight, then the object will sink. If the upthrust is just equal to its 29 Edexcel AS Physics in 100 Pages weight, then the object will be able to stay anywhere in the fluid, this state is called suspension. If the upthrust is larger than the object’s weight, then the object will move upwards. When it gets to the surface of the fluid, part of its body will get out of the fluid and thus reduce the volume submerged in the fluid. So, the upthrust on the object will decrease to a value that is equal to the weight of the object. This state is called floating. Fig 2.1.2 Sinking, suspension or floating Example: The hydrometer is a device that is used to determine the density of a liquid. It has a constant weight W. To balance the weight, you need to have a floating force with the same size. Thus, in liquids with different densities, the volumes (thus the depths) the hydrometer sink in the liquid are different. Fig 2.1.3 Hydrometer 30 Edexcel AS Physics in 100 Pages If a hydrometer is put into pure water, 30% of the total volume of the hydrometer floats above the water surface. If the hydrometer is put into salt water with a density of 1.1 gm-3, how much will it float above the surface of the salt water? Solution: According to Archimedes’ Principle, the volume of the displaced liquid (also the volume that the hydrometer sinks) is inversely proportional to the density of the liquid. So, if we assume that x% of the hydrometer will float above the surface of the salt water, then (1 30%)V 1.1gcm 3 (1 x%)V 1gcm 3 where V is the total volume of the hydrometer. Solving the above equation gives x%=36%. 31 Edexcel AS Physics in 100 Pages 2.2 Dynamics of Fluid Laminar and turbulent flows In the section dynamics of fluid, we will focus on the flow of fluid. Fluid flow includes laminar flow and turbulent flow, as shown in Fig 2.2.1. Table 2.2.1 summarizes the most important features of these two types of flows. Fig 2.2.1 Laminar and turbulent water flows Laminar flow Turbulent flow z Velocity may be different at different z Velocity at any given place places, but at any given place the changes with time. velocity must be constant. z There are sudden changes of speed. z Smooth, no sudden change of speed. There are eddies. z Slow (occurs at low speed). z Occurs at higher speed. z Associated with small resistance z Associated with big resistance Example: slow water flow along a smooth Example: air flow in a storm pipe Table 2.2.1 Laminar flow and turbulent flow 32 Edexcel AS Physics in 100 Pages Laminar flow is also called streamlined flow. The lines we use to represent the flow of laminar fluid flow is called streamlines, as shown in figure in Table 2.2.1 Objects moving in a turbulent flow experience a much bigger air resistance/drag force than moving in a laminar flow. Many vehicles are designed in streamlined shapes (plane, racing car) to reduce turbulent flow and thus reduce air resistance. Fig 2.2.2 Streamlined design of high speed trains Viscous drag When an object is moving in a fluid, it will experience a drag force (resistance) that resists the motion of the object. If the object is moving downwards, this drag force is partially due to the upthrust. Another important source of the drag force is the viscosity of the fluid. The drag force caused by viscosity is called viscous drag. Viscosity is an intrinsic property of fluid. The factor coefficient of viscosity K (or simply viscosity) describes how large the viscosity of the fluid is. Or, in even simpler words, coefficient of viscosity describes how “sticky” a fluid is. The flow rate (volume of fluid passing through in unit time) of a fluid is inversely proportional to viscosity. The viscous drag is larger while moving in a fluid with larger coefficient of viscosity. Honey obviously has a much larger viscosity than water. A ball moving in honey will experience a much larger resistance than moving in water. Viscosity, like density, is a property of material. Viscosity is dependent on temperature. In general, for gases, viscosity increases with temperature. For liquid, viscosity decreases with temperature. Table 2.2.2 lists viscosities for some common fluids. Fluid Temperature/°C Viscosity/Pa s Air 0 0.000017 20 0.000018 100 0.000022 Water 0 0.0018 20 0.0010 100 0.0003 Glycerine 20 1.5 30 0.63 Table 2.2.2 viscosities for some common fluids. 33 Edexcel AS Physics in 100 Pages The viscous drag on an arbitrary shaped object is difficult to calculate as it causes a lot of turbulent flow. Sir Stokes studied the viscous drag for a small sphere moving at low speed in a fluid. He found that in this case, the viscous drag is given by F 6SK rv (2.3) where K is the viscosity of the fluid(Pags), r is the radius(m) of the sphere and v is the speed(ms-1). This result is called Stokes’s law and this viscous drag is also called Stokes’ force in honor of Sir Stokes. You should be aware that Stokes’ law is only valid for small sphere moving at low speed. In this case the fluid flow is laminar flow. Fig 2.2.3 Streamlines around a sphere Example: Use Stokes’ Law to calculate the viscous drag on a ball bearing with a diameter of 5mm, falling at a speed of 2mms-1 through glycerine at room temperature (viscosity of glycerine at 20°C is 1.5 Pas). Solution: According to the Stokes’ Law, the viscous drag on the ball is given by 5 u103 F 6SK rv 6 u 3.14 u1.5 u u 2 u103 N 1.27 u104 N 2 Terminal velocity In Chapter 1 we have discussed the free fall motion of an object due to gravity, where we neglected the air resistance and upthrust. In many cases, these two forces are comparable to the weight of the object and can not be neglected. If upthrust and viscous drag (air resistance) are also taken into account, the object will move under the influence of three forces. The weight and upthrust are constant forces, but the viscous drag increases with the increase of speed. At the moment the object is released, the speed is still zero; the viscous drag is also zero. The object will accelerate due to the gravitational force and upthrust. However, when the speed increases, the viscous drag increases accordingly, causing the net force to decrease. Although the acceleration is 34 Edexcel AS Physics in 100 Pages decreasing, the speed is increasing because there is acceleration. This process continues until when the speed reach a certain value so that the viscous drag is large enough to balance the weight and upthrust. Then the net force becomes zero and so do the acceleration. The speed will not change anymore. This velocity is called the terminal velocity. The condition for terminal velocity is that the resultant force be 0. That is weight upthrust viscous drag (2.4) For a small sphere, the above equation can be written as ms g U f Vg 6SK rvterm (2.5) where ms , V , r and vterm are the mass, volume radius and terminal velocity of the sphere and U f and K are the density and viscosity of the fluid respectively. 4 3 4 Substituting V S r and ms U sV U s S r 3 ( U s is the density of the sphere) into 3 3 the above equation gives 4 4 U s S r 3 g U f S r 3 g 6SK rvterm (2.6) 3 3 Solving it for vterm gives 2r 2 g ( U s U f ) vterm (2.7) 9K We can see from the above expression that the terminal velocity is dependent on the radius of the sphere, the density difference between the sphere and the fluid, as well as the viscosity of the fluid. Example: A spherical meteorite, of radius 1m and made of pure iron whose density is 7800kgm-3, falls into the Pacific Ocean, what is the final velocity the meteorite will move at? Suggest the assumptions you have made while doing the calculation and comment your result. (Viscosity of water can be found in Table 2.2.2) Solution: The meteorite will reach the terminal velocity after a sufficient time of travel in the sea. 2r 2 g ( U s U f ) 2 u12 u 9.81u (7800 1000) vterm 1.48 u107 ms 1 9K 9 u 0.001 The assumptions we made include that the temperature of the water is 20 °C, the ocean is deep enough for the meteorite to reach its terminal velocity and the velocity of the meteorite is slow to use the Stokes’ formula. But actually from the result we see that this speed is too large to be true. The assumptions are not all valid. The speed of the 35 Edexcel AS Physics in 100 Pages meteorite gets very large and the Stokes’ law doesn’t hold anymore. 36 Edexcel AS Physics in 100 Pages 2.3 Strength of Solid Materials Hooke’s Law In 1676 Robert Hooke discovered the famous relationship between the extension (or compression) of a spring and the force it exerts. This relationship is therefore called Hooke’s law and it states that: The force F exerted by a spring is proportional to the extension (or compression) 'x. That is: F k 'x (2.8) where k is a constant called spring constant, Hooke’s constant or stiffness of the spring. The minus sign here only indicates that the force is in the opposite direction to the extension (or compression). You do not have to include the minus sign in your calculation. Fig 2.3.1 Hooke’s Law experiment. The slope of the graph is the stiffness. Elastic strain energy When a spring is stretched or compressed, energy is stored in the spring because it has potential to do work. This form of energy is called elastic potential energy or elastic strain energy. The amount of elastic strain energy stored in a spring is equal to the work done by the force exerted on it. Thus, it can be obtained by calculating the area under the force-extension graph, which is 1 1 Eel F 'x k 'x 2 (2.9) 2 2 Note that this equation is valid for springs and other elastic materials that obey Hooke’s law. Some elastic materials do not follow Hooke’s law, which means that the force is 37 Edexcel AS Physics in 100 Pages not proportional to extension. In this case, the elastic potential energy can still be determined by calculating the area under the force—extension graph. Fig 2.3.2 Elastic strain energy stored in a catapult Example: A spring is compressed by 0.2m when a 28 Newton force is exerted on it. Find the spring constant and elastic strain energy stored in the spring. Solution: F 28 N The spring constant is k 140 Nm 1. 'x 0.2m 1 1 The elastic strain energy stored is Eel k 'x 2 u140 u 0.22 J 2.8 J 2 2 Force ~ extension graph Some materials follow Hooke’s law up to a certain point. The point beyond which force is no longer proportional to extension is called the proportionality limit. The region between the origin and the proportionality limit is called the liner region. For spring and many materials, there is a short further region where the behavior is still elastic (they return to their original lengths when the force is removed). This region ends at the elastic limit. After the elastic limit, further force produces permanent deformation, which means that after removing the force they can’t return to their original lengths. This is also called the plastic deformation. 38 Edexcel AS Physics in 100 Pages Fig 2.3.3 Force—extension graph Stress, strain and Young Modulus Stiffness of a spring measures how hard it is to produce extension on a spring. It is dependent on the material that the spring is made of and the dimensions of the spring. We now want to introduce a quantity that measures how hard it is to produce extension for a certain material. In other words, we want this quantity to be a property of the material and to be independent on its geometrical dimensions. To do this, I firstly wish you to know the two terms stress and strain. Stress V is defined as the force divided by cross section area, that is force F stress or V (2.10) cross sec tion area A There are three types of stress, tensile stress (also called tensional stress), compressive stress (also called compressional stress) and shear stress. Fig 2.3.3 shows the difference between them. 39 Edexcel AS Physics in 100 Pages Fig 2.3.4 Different types of stresses A material may break if the stress caused by a tension or compression is too big. The tensile stress at which a material fails is called tensile strength. The compressive stress at which a material fails is called compressive strength of the material. Strain H is the relative extension, that is extension 'x strain or H (2.11) original length x Similar to the Hooke’s law which states that force is proportional to extension, in many materials, stress is proportional to strain. The proportionality constant is called Young Modulus. The letter E is used to denote Young Modulus. Written in mathematical form, it is V E (2.12) H Young Modulus is a quantity that is similar to the spring constant (stiffness) of a spring as it also measures how difficult it is to produce an extension. The difference is that Young Modulus is a property of material and is not dependent on the dimensions of a particular sample. Anything made from the same material have the same Young Modulus, regardless of their shape or length. We can see this from the definition of Young Modulus. The purpose of dividing the force by cross section area while defining stress and dividing extension by original length while defining strain is to eliminate the influence of the dimensions of the sample. Example: A wire fence is made of steel wire of diameter 2.5 mm. A force of 1500 N is applied to tension a single length of this wire. (a) Calculate the stress produced in the wire. (b) If the Young modulus of steel is 210 GPa, calculate the extension, produced in a 33 m length of this wire when it is tensioned. Solution: (a) The stress in the wire is given by F F 1500 V Pa 3.1u108 Pa A S (d / 2) 2 S u ( 2.5 u 103 /2) 2 V V (b) In the equation E , the extension 'x is the quantity we need to H 'x / x calculate, while all the other quantities are given. Substituting the values into the equation and solve for 'x gives the value of extension to be 0.048 m Stress ~ strain graph A stress—strain graph is a very convenient tool to describe characteristics of a solid 40 Edexcel AS Physics in 100 Pages material. Figure 2.3.5 shows such a graph for a metal. Fig 2.3.5 Stress—strain graph The stress strain graph is similar to the force extension graph. The gradient of the linear part of the stress-strain graph is equal to Young Modulus of the material. Some important points and regions on the graph you should be familiar with are: z Proportionality limit: before this point, stress v strain; after this point, stress—strain graph is no longer linear. z Elastic limit: before this point, the material behaviors elastically, which means that it will return to its original shape if the stress is removed; after this point, the material behaviors plastically, which means that it can’t return to its original shape if the stress is removed. z Yield point: after this point, it seems that the material gives up “fighting against” the stress. Even a very small increase of stress causes a big increase of strain. In other words, the gradient of the graph becomes much smaller after this point. z Linear region: the region before the proportionality limit. z Elastic region: the region before the elastic limit. z Plastic region: the region after the elastic limit. While solving exam problems, it is often difficult to locate the elastic limit on the stress strain graph. But do not worry about this, as long as you label the elastic limit somewhere between proportionality limit and yield point, you will get the mark. Describing materials Different materials behavior differently under external forces. You should be familiar with the following words that are used to describe materials. z Brittle materials break/shatter/snap with little or no plastic deformation. Eg.ceramics, biscuit z Malleable materials can be beaten into sheets and show a large plastic deformation under compression. Eg.iron, gold z Ductile materials can be pulled into wires or threads; these materials show plastic 41 Edexcel AS Physics in 100 Pages deformation before failure under tension. Eg.copper wire z Hard materials resist plastic deformation. Eg.diamond z Tough materials can withstand impact forces and absorb a lot of energy before breaking; large forces produce a moderate deformation. Eg. rubber, Kevlar(a type of material for fabricating bulletproof vest) Example: Boxes containing glass objects are often labeled “Be careful, fragile”. Which of the following words best matches the meaning of fragile? A. plastic B. hard C. brittle D. tough Solution: The answer should be C because brittle materials are easy to break or shatter. 42 Edexcel AS Physics in 100 Pages Chapter 3 Elementary of Waves 43 Edexcel AS Physics in 100 Pages 3.1 Introduction to Waves When you throw a stone into a pond, a water wave will be formed. This is because the stone causes the water near it to oscillate (move up and down periodically). This oscillation will influence the water in the surrounding area and lead them to oscillate as well. This process continues and the oscillation travels from the point that the stone enters the pond. So, a wave is some sort of disturbance travelling through space. Fig 3.1.1 Various kinds of waves Classifying waves Waves can be classified as mechanical waves and electromagnetic waves depending on whether it requires a medium for the waves to travel (propagate) in. Mechanical waves require a medium to propagate. For example, sound wave is mechanical wave and it can propagate in air, water or other media. It can’t propagate in vacuum. Electromagnetic waves require no medium to propagate. This means that they can travel in vacuum. Light is an example of electromagnetic wave. Another way to classify waves is according to the relationship between the direction of oscillation and the direction of propagation. If the direction of oscillation is perpendicular to the direction of propagation of the waves, they are called transverse waves. Figure 3.1.2(a) shows a transverse wave propagating along horizontal direction on a string. The wave may be produced by vibrating one end of a string vertically. All electromagnetic waves are transverse waves. Water waves are also transverse waves. If the direction of oscillation is parallel to the direction of propagation of the waves, they are called longitudinal waves. Figure 3.1.2(b) shows a longitudinal wave propagating along horizontal direction on a spring. This wave may be produced by vibrating one end of the spring horizontally. Sound wave is a longitudinal wave. 44 Edexcel AS Physics in 100 Pages (a) (b) (c) Fig 3.1.2 (a) A transverse waveform on a string (b)A longitudinal waveform on a spring (c) Displacement~time graph of point P on the string(spring). Waves that have an infinite length are called continuous waves. Waves that have a beginning and end are called wave trains. Waves with a very short length(less than a couple of wavelengths) are called pulses. Fig 3.1.3 illustrates these concepts. Actually, according to the definition of continuous wave, it has infinite length, which is not possible in reality. It is only a physics model. Usually, people regard long wave trains as continuous waves. 45 Edexcel AS Physics in 100 Pages Fig 3.1.3 Continuous wave, Wave train, and pulse Fig 3.1.2 and Fig 3.1.3 ‘view’ the waves sideway. If we look from above, the pictures would look like those shown in Fig 3.1.4. Fig 3.1.4 Plane and circular water waves in a pond viewed from above Using lines or circles to represent crests, we can draw diagrams as shown in Fig 3.1.5 to describe waves. In such a diagram, the lines (or circles) representing crests are referred to as wavefronts and the rays tell the direction of propagation of the waves. 46 Edexcel AS Physics in 100 Pages Fig 3.1.5 Wavefronts and rays to represent waves Quantities describing waves Four quantities are used to describe waves. Their definitions are Amplitude: the maximum displacement from the equilibrium position. Wavelength: distance between two adjacent peaks(for transverse waves) or distance between two adjacent compressions(for longitudinal waves). Frequency: number of complete oscillations per second. Period: time taken for one complete oscillation. Figure 3.1.2 illustrates theses definitions. If a certain point completes 5 oscillations in 1s, then its frequency is 5Hz. The time it takes to complete one oscillation is then 0.2s. So, its period is 0.2s. Period is the reciprocal of frequency. That is 1 T (3.1) f or fT 1 (3.2) Speed of the wave is the distance traveled by the wave during unit time. In a period (the time interval that the wave source finish one complete oscillation), the wave propagates a distance of a wavelength. So, O v Of (3.3) T The speed of all electromagnetic waves(including light) in vacuum is 3.0h108ms-1. You should memorize this speed. Points that are a whole wavelength apart oscillate in exactly the same way. We say that they oscillate in phase. Points that are a distance of half wavelength away oscillate in exactly the opposite way. We say that they oscillate in antiphase (or out of phase). 47 Edexcel AS Physics in 100 Pages Example: 13.725 kHz is one of the frequencies that BBC broadcasts its East Asian programs. What is the wavelength of these radio waves? Solution: According to the equation v O f , the wavelength is given by v 3.0 u108 O m 21858m f 13725 Using graphs to describe waves Displacement-distance graph Fig 3.1. (a) is a graph that shows the waveform on the string at a certain time. It’s like a picture of the string. In this graph, the horizontal axis represents the distance from the source of the wave, the vertical axis represents the displacement of each point of the string at this time. This graph is called a displacement-distance graph. You can know the wavelength and amplitude from such a graph. Two points that are one wavelength apart have a phase difference of 360o. If two points are x wavelengths apart, then we say the phase difference between these two points is 360 x degrees. Here, x can be either integer or fraction. Displacement-time graph Fig 3.1. (c) is a graph that shows the oscillation of a certain point on the string. It describes how the displacement of this point changes with time. In this graph, the horizontal axis represents the time and the vertical axis represents the displacement of this point at different times. This graph is called a displacement-time graph. You can know the period and amplitude from such a graph. Example: The diagram below shows a transverse wave travelling towards right at a speed of 2.0ms-1. (a) From the diagram, determine the amplitude and wavelength of the wave. (b) State whether point M is moving downwards or upwards at the moment this displacement-distance graph is obtained. (c) Sketch the displacement-time graph of point M to describe its motion during next period. 48 Edexcel AS Physics in 100 Pages (d) Mark a point on the waveform whose motion is exactly 180o out of phase with the motion of point M. Label this point N. Solution: (a)The amplitude is 20cm. The wave length is 0.4m. (b)Since the wave is travelling to the right. A point will duplicate the motion of its immediate left point. Point M’s immediate left point is in the upper of M, so M will follow that point and travels to the upper first. Thus, the direction of the movement of point M at this time is upwards. (c)To sketch the displacement-time graph of point M, we need to know the amplitude and period of its oscillation. The amplitude is 20cm, the period can be calculated O O 0.4m using v O f. The period is T 0.2 s. T v 2.0ms 1 Also, we need to know the initial displacement of point M and its direction of movement at the moment. The initial displacement is about 10cm, according to the displacement-distance graph given. The direction of movement at the moment is upwards, which was worked out in (b). Then the displacement-time graph of point M can be sketched as follows: (d)We have known that in a displacement-distance graph, if two points are x wavelengths apart, then the phase difference between these two points is 360 x degrees. Now the phase difference between M and N is 180 o. So, 360 x =180. x 1/ 2. Thus, N is any point that is 1/2 wavelength away from point M. 49 Edexcel AS Physics in 100 Pages 3.2 Reflection Law of reflection When a travelling wave reaches a boundary between two mediums, part of the wave will reflect back and part of the wave will transmit(go into the second medium). In wave reflection, the angle of incidence is equal to the angle of reflection. That is Ti T r , as shown in Figure 3.2.1. Fig 3.2.1 Reflection of wave Pulse echo detection--imaging For wave reflection, the greater the difference in density between the two materials, the stronger the reflection will be. This principle is used make imaging devices such as ultrasound scanning machine. Fig 3.2.2 Ultrasound scanning Ultrasound scanning technique is widely used in medical imaging. An ultrasound 50 Edexcel AS Physics in 100 Pages pulse is sent to the target, the pulse will partially reflect between the boundary of air and skin, the boundary between fat and muscle and the boundary between muscles and bones. Since these materials have different densities, the intensities of the reflected waves vary. From the intensities of the reflected waves, we can know the structure of the imaged objects. In imaging technology, the smallest level of detail that can be seen is called the resolution of the image. Resolution of ultrasound scanning can be improved by reducing the wavelength of the sound waves used. There are two rules defining resolution. The first one defines resolution simply as the wavelength of the sound wave used. The second one defines resolution as half of the length of the pulse. Pulse echo detection--echolocation Pulse echo technique is also used to detect the position of an object. Fig 3.2.3 Bats use echolocation to detect obstacles A pulse is sent to the object and the time interval between the moment the device sends out the pulse and the moment it receives the reflected pulse is recorded. If the velocity of the wave is known, the distance between the sender and the obstacle can be decided. This is called echolocation. Bat uses this method to ‘see’ its surrounding environment at night. Radars use this method to locate airplanes as well. Doppler effect When waves are emitted from a moving source or detected by a moving receiver, the detected frequency differs from the emitted frequency. The shift in frequency is proportional to the relative speed of the motion. This phenomenon is called the Doppler effect. When the wave source moves relatively away from the detector (receiver), the detector will ‘see’ a stretched wavelength (or lower frequency); when the wave source moves relatively towards the detector, the detector will ‘see’ a compressed wavelength (or higher frequency). 51 Edexcel AS Physics in 100 Pages Fig 3.2.4 Doppler effect (From wikipedia) A Doppler speedometer is a device that uses the Doppler effect to measure speed of moving vehicles. Its working principle is as follows: A radio pulse of known frequency is sent from the speedometer to the targeting vehicle. The pulse is reflected from the moving vehicle back to the speedometer. The shift of frequency of reflected wave is proportional to the speed of the vehicle. Fig 3.2.5 Doppler speedometer and traffic cameras Example: An air traffic control system sent out a radio pulse with frequency of 20GHz to an aircraft. After 0.003seconds the system received the pulse reflected back by the aircraft. The frequency of the received waves is now 20.00002GHz. (a) What is the distance between the air control system and the aircraft? (b) What other information about the aircraft can you get? Solution: (a) The distance travelled by the wave during 0.003s is s vt 3.0 u108 u 0.003m 9 u105 m 900km The radio waves travel to the aircraft and back for a total distance of 900km. So, the distance between the air control system and the aircraft is 450km. 52 Edexcel AS Physics in 100 Pages (b)Since we are given that the frequency of the reflected waves is increased. We can know that the wavelength is compressed due to Doppler effect. So, the aircraft is moving towards the control system according. Astronomers observed that the spectra from other galaxies far away from our galaxy are shifted towards the red end. This is called red shift. This phenomenon is due to Doppler effect. Red shift indicates that other galaxies are moving away from us. This means that our universe is expanding. Fig 3.2.6 Red shift Red shift dose not mean that the colour of the light from other galaxies is all red. It only means that the wavelengths of the light from other galaxies are stretched (become longer). It is named red shift because red light has the longest wavelength in the visible light range. 53 Edexcel AS Physics in 100 Pages 3.3 Refraction Law of refraction As shown in Fig 3.3.1, when a wave travels to the boundary of two mediums, part of the wave will go into the second medium and its direction of propagation will change. This is called refraction and the wave that gets into the second medium is called the refracted wave. Fig 3.3.1 Refraction of wave The direction of the propagation of the refracted wave is governed by the law of refraction (also called Snell’s Law). Fig 3.3.2 Snell’s law In symbol form, Snell’s Law is 54 Edexcel AS Physics in 100 Pages sin T1 n2 v1 (3.4) sin T 2 n1 v2 where: T1 and T 2 ----the angle of incidence and angle of refraction, n1 and n2 ---- the refractive index of medium 1 and medium 2; v1 and v2 ----speed of light in medium 1 and medium 2. Refractive index is a quantity that measures the ability of a medium to deflect the direction of propagation of light when it travels from vacuum to the medium. The value of refractive index is related with the density of the medium. Denser medium has higher refractive index. Vacuum has a refractive index of 1. The refractive index of air is 1.003. In most calculations, we regard the refractive index of air as 1, unless otherwise indicated. The refractive indexes of water and glass are 1.33 and 1.5 respectively. When light travels from one medium to another, its frequency doesn’t change, but its speed and wavelength change. The speed of light in vacuum is 3.0h10ms-1. In air, the speed of light is only slight slower than the speed in vacuum. So we usually take the speed of light in air as 3.0h10 ms-1 as well. In other media, its speed is much slower. Example: The refractive index of a type of glass for red light is 1.51. A ray of red light travels from air into the glass. If the incident angle of the ray is 61o, calculate the speed of the ray in the glass and the angle of refraction. Solution: sin T1 n2 v1 According to Snell’s law,. sin T 2 n1 v2 So, sin 61o 1.51 3.0 u108 ms 1 sin T 2 1 v2 So, the speed in glass is v2 1.99 u108 ms 1. and angle of refraction is 41.5o. Total internal reflection If a light ray passes from a denser to a less dense material and the incident angle is larger than a certain value called critical angle, refracted ray disappears and the entire 55 Edexcel AS Physics in 100 Pages incident light is reflected. This is called total internal reflection. The value of the critical angle C is decided by the equation n2 sin C (3.5) n1 We can see that critical angle is only decided by the relative refractive index between the two media. For example, the critical angle of glass-air interface is 42o. This means that if light travels from glass to air and its angle of incidence is larger than 42o, then all the light will be reflected back and no light will refract into air. Fig 3.3.3 Total internal reflection There are two conditions for total internal reflection to happen. 1. Light travels from a denser medium to a less dense medium ( n1 ! n2 ). 2. The incident angle is larger than the critical angle. Example: Calculate the critical angle of a water-air interface given the refractive index of water to be 1.33. Solution: The critical angle is calculated by using n2 1 sin C 0.752 n1 1.33 The critical angle C is 48.8o 56 Edexcel AS Physics in 100 Pages 3.4 Interference Superposition Where two waves meet, the total displacement at any point is the sum of the displacements that each individual wave would cause at that point. This is called the principle of superposition. For example, as shown in Fig 3.4.1, there are two sound generators S1 and S2 oscillating. They cause the air molecules in the surrounding area to oscillate as well. Consider the oscillation of air molecules at point P at a particular time t, if only S1 was in present and S2 dose not exist, the wave generated by S1 would cause a displacement of x1 at point P. If only S2 was in present and S1 dose not exist, the wave generated by S2 would cause a displacement of x2 at point P. Now both S1 and S2 are in present, the actual displacement of air molecules at point P at this time should be x1+ x2. Fig 3.4.1 An example of superposition Path difference and interference If two sets of waves have the same frequency and a constant phase difference, then we say that they are coherent. When two sets of coherent waves meet and superpose, the oscillation at any point will be the resultant vibration caused by both wave sources, we say that they interfere. And they can produce a stable interference pattern. Fig 3.4.2 Interference pattern of water waves in a ripple tank 57 Edexcel AS Physics in 100 Pages In some places, the displacements caused by the waves from the two sources are always the opposite and thus they cancel each other, so the vibration in these places are minima. We call this destructive interference. In some places, the displacements caused by the two sources are always the same and thus they add up, so the vibrations in these places are maxima. We call this constructive interference. Fig 3.4.2 shows an interference pattern of water waves. If the distance from source S1 to a point P is S1P and the distance from source S2 to the point P is S2P, then S1P-S2P is called the path difference. For example, in Fig 3.4.1, if the distance between P and S1 is 3m and the distance between P and S2 is 4m, then the path difference at point P is 1m. Consider the following case: S1 and S2 are two wave sources oscillating in phase (oscillate exactly in the same way). Waveforms at a particular instant are shown in Fig 3.4.3. A solid semicircle represents a peak and a dashed semicircle represents a trough of the waves. Fig 3.4.3 Path difference and interference O The path difference of point Q is S1Q S2 Q 2.5O 2O. We can see from the 2 figure that when the wave from source S1 causes a peak at point Q, the wave from source S2 causes a trough at point Q, so the resultant displacement at point Q is zero. The oscillation is at minima. Actually this result can be generalized. If the path difference is an odd number of half wavelength, then destructive interference(minima) takes place. The path difference of point P is S1P S2 P 2.5O 1.5O O. We can see from the figure that when the wave from source S1 causes a peak at point P, the wave from source S2 causes a peak at point P as well, so the resultant displacement at point P is two times of the displacement caused by each source. The oscillation is at maxima. Acturally, if the path difference is an even number of half wavelength, then constructive interference (maxima) takes place. During the exam, you might be asked to determine whether constructive or destructive interference takes at a point. To solve this kind of problems, you may 58 Edexcel AS Physics in 100 Pages follow these three simple steps: 1. Calculate or measure the path difference. O 2. Calculate the value of half wavelength. 2 O 3. Compare the path difference with half wavelength to see whether the path 2 O O difference is equal to an even number times of or an odd number times of : 2 2 O If the path difference= an even numbeUh , constructive interference(maxima); 2 O If the path difference= an odd numbeUh , destructive interference(minima); 2 Example: In a ripple tank, two dippers S1 and S2 vibrate in exactly the same way, producing coherent water waves with wavelength of 2cm. X and Y are two points in the ripple tank. Their distances from the two dippers are shown in the diagram below. Tell if constructive interference or destructive interference takes place at X and Y respectively. Solution: For point X, since its distances from the two dippers are the same, the path difference at point X is 0. Thus, constructive interference takes place at point X. For point Y, the path difference is Y S1- YS2= 16 162 122 16 20 4cm The half wavelength is O 2cm 1cm. 2 2 Since, 4cm 4 u1cm and 4 is an even number, constructive interference takes place at point Y. Stationary wave Waves can be classified as travelling waves and stationary waves. For a travelling 59 Edexcel AS Physics in 100 Pages wave, its peaks and troughs are travelling. Travelling wave is also called progressive wave. For a stationary wave, its peaks and troughs do not move. Stationary wave is also called standing wave. When two traveling waves with the same frequency and amplitude propagating in opposite direction meet, they interfere and can form a stationary wave. Fig 3.4.4 illustrates how a stationary wave is formed. Fig 3.4.4 Two travelling waves(red and blue) form a stationary wave(black) A stationary wave is basically an interference pattern. The minima are called nodes and maxima are called antinodes. Fig 3.4.5 shows the pattern of a stationary wave on a string fixed at both ends. The distance between two adjacent nodes (or antinodes) is equal to half wavelength. 60 Edexcel AS Physics in 100 Pages Fig 3.4.5 Stationary wave on a string Since in such a string, the ends must be nodes, the following condition exists in order to form a stable stationary wave on a string with length l.

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