Physical Sciences Grade 12 Quanta

Physical Sciences Grade 12 Quanta Miemie Pretorius

ALL RIGHTS RESERVED ©Copyright Copyright subsists in this work. No part of this work may be reproduced in any form or by any means without the author’s written permission. Any unauthorised reproduction of this work will constitute a copyright infringement and render the doer liable under both civil and criminal law.

Whilst every effort has been made to ensure that the information published in this work is accurate, the editors, publishers, printers and author take no responsibility for any loss or damage suffered by any person as a result of the reliance upon the information contained therein.

All the answers are the original work of the author and have not been taken from the official memoranda of the Departments of Education.

ISBN 978-0-620-35264-2 Cover page and graphic design: Maximum Exposure, Somerset West.

DVD recordings: University of Stellenbosch, Telematic Services Animations: Egbert Westra from Enjoythecompany If you have any query, recommendation or discussion on the contents of this book, please feel free to contact the author at: www.quantabooks.co.za 021 852 5598 or 082 852 1826 or info@quantabooks.co.za or mimi.pretorius@gmail.com or 086 6596 220 (fax) Other books in this series by the author: Fisiese Wetenskappe Graad 12 – Kwanta (KABV) Physical Sciences Grade 11 – Quanta (CAPS) Fisiese Wetenskappe Graad 11 – Kwanta (KABV) Physical Sciences Grade 10 – Quanta (CAPS) Fisiese Wetenskappe Graad 10 – Kwanta (KABV)

CONTENTS MECHANICS Momentum: Definition Momentum of an Object Changing Direction Newton’s Second Law, Change in Momentum, Impulse Law of Conservation of Momentum Momentum: Elastic and Inelastic Collisions Vertical Projectile Motion: Gravitational Acceleration Vertical Projectile Motion: Signs of Displacement, Velocity, Acceleration Vertical Projectile Motion: Position-Time and Displacement-Time Graphs Vertical Projectile Motion: Velocity-Time Graphs Vertical Projectile Motion: Acceleration-Time Graphs Vertical Projectile Motion: Equations of Motion

1 1 2 6 10 12 14 14 16 19 20

MATTER AND MATERIALS Organic Chemistry: Terminology Organic Chemistry: Alkanes Organic Chemistry: Alkenes Organic Chemistry: Alkynes Organic Chemistry: Polymerisation Organic Chemistry: Haloalkanes Organic Chemistry: Alcohols Organic Chemistry: Carboxylic acids Organic Chemistry: Esters Organic Chemistry: Aldehydes Organic Chemistry: Ketones Organic Chemistry: Physical Properties of Organic Compounds Organic Chemistry: Mixed Questions on Organic Chemistry

26 27 31 37 38 40 43 46 47 51 52 52 58

MECHANICS Work and Energy Power Gravitational Potential Energy Kinetic Energy Mechanical Energy

67 70 71 73 75

WAVES, SOUND AND LIGHT Doppler Effect

82

CHEMICAL CHANGE Energy Change during Chemical Reactions Rate of Chemical Reactions Catalysts Collision Theory Chemical Equilibrium Factors Influencing Chemical Equilibrium The Common Ion Effect and Solutions The Equilibrium Constant Graphs of Amount Versus Time Graphs of Rate Versus Time Microscopic Representation Acids and Bases: Equilibrium Constant for Water pH Hydrolysis Titration, Neutralisation, Indicators

87 90 104 107 110 112 121 122 134 141 143 145 146 150 153

ELECTRICITY AND MAGNETISM Electrodynamics: Electrical Machines: Generators Electrodynamics: Electrical Machines: Motors Electrodynamics: Alternating Current (AC)

159 163 166

OPTICAL PHENOMENA AND PROPERTIES OF MATERIALS Photoelectric Effect Emission and Absorption Spectra

169 175

CHEMICAL CHANGE Electrochemical Cells Electrochemical Cells: Writing Equations for Redox Reactions Electrochemical Cells: Voltaic/Galvanic Cells Electrochemical Cells: Electrolytic Cells Electrochemical Cells: Spontaneous and Non-Spontaneous Reactions Electrochemical Cells: Recovery of Aluminium Electrochemical Cells: Production of Chlorine (Chlor-Alkali Industry)

178 179 182 193 197 200 202

CHEMICAL SYSTEMS The Fertilizer Industry (N, P, K)

207

Grade 12 Paper Paper 1 (Physics)

Paper 2 (Chemistry)

Content Mechanics Waves, Sound and Light Electricity and Magnetism Matter and Materials Chemical Change Chemical Systems Matter and Materials

Grade 12 Examination Physics from grade 11 Chemistry from grade 11 Newton’s Laws and forces Intermolecular forces Electrostatics Mol and Stoichiometry Electric Circuits Energy and Change

Marks 63 17 55 15 84 18 48

Total 150

150

CONTENTS: MEMORANDUM MECHANICS Exercise 1: Momentum: Definition Exercise 2: Momentum of an Object Changing Direction Exercise 3: Newton’s Second Law, Change in Momentum, Impulse Exercise 4: Law of Conservation of Momentum Exercise 5: Momentum: Elastic an Inelastic Collisions Exercise 6: Vertical Projectile Motion: Gravitational Acceleration Exercise 7: Vertical Projectile Motion: Position-Time and Displacement-Time Graphs Exercise 8: Vertical Projectile Motion: Velocity-Time Graphs Exercise 9: Vertical Projectile Motion: Acceleration-Time Graphs Exercise 10: Vertical Projectile Motion: Equations of Motion

215 215 215 217 219 221 221 222 223 223

MATTER AND MATERIALS Exercise 11: Organic Chemistry: Terminology Exercise 12: Organic Chemistry: Alkanes Exercise 13: Organic Chemistry: Alkenes Exercise 14: Organic Chemistry: Alkynes Exercise 15: Organic Chemistry: Haloalkanes Exercise 16: Organic Chemistry: Alcohols Exercise 17: Organic Chemistry: Carboxylic Acids Exercise 18: Organic Chemistry: Esters Exercise 19: Organic Chemistry: Aldehydes Exercise 20: Organic Chemistry: Ketones Exercise 21: Organic Chemistry: Physical Properties of Organic Compounds Exercise 22: Organic Chemistry: Mixed Questions on Organic Chemistry

228 228 229 232 232 233 235 235 237 237 238 240

MECHANICS Exercise 23: Work and Energy Exercise 24: Power Exercise 25: Gravitational Potential Energy Exercise 26: Kinetic Energy Exercise 27: Mechanical Energy

244 245 246 246 247

WAVES, SOUND AND LIGHT Exercise 28: Doppler Effect

252

CHEMICAL CHANGE Exercise 29: Energy Change during Chemical Reactions Exercise 30: Rate of Chemical Reactions Exercise 31: Catalysts Exercise 32: Collision Theory Exercise 33: Chemical Equilibrium Exercise 34: Factors Influencing Chemical Equilibrium Exercise 35: The Common Ion Effect and Solutions Exercise 36: The Equilibrium Constant Exercise 37: Graphs of Amount Versus Time Exercise 38: Graphs of Rate Versus Time Exercise 39: Microscopic Representation Exercise 40: Acids and Bases: Equilibrium Constant for Water Exercise 41: pH Exercise 42: Hydrolysis Exercise 43: Titration, Neutralisation, Indicators

254 255 259 260 261 261 265 266 276 279 280 280 281 284 285

ELECTRICITY AND MAGNETISM Exercise 44: Electrodynamics: Electrical Machines: Generators Exercise 45: Electrodynamics: Electrical Machines: Motors Exercise 46: Electrodynamics: Alternating Current (AC)

288 289 290

OPTICAL PHENOMENA AND PROPERTIES OF MATERIALS Exercise 47: Photoelectric Effect Exercise 48: Emission and Absorption Spectra

292 295

CHEMICAL CHANGE Exercise 49: Electrochemical Cells Exercise 50: Electrochemical Cells: Writing Equations for Redox Reactions Exercise 51: Electrochemical Cells: Voltaic/Galvanic Cells Exercise 52: Electrochemical Cells: Electrolytic Cells Exercise 53: Electrochemical Cells: Spontaneous and Non-Spontaneous Reactions Exercise 54: Electrochemical Cells: Recovery of Aluminium Exercise 55: Electrochemical Cells: Production of Chlorine (Chlor-Alkali Industry)

295 295 296 302 303 305 305

CHEMICAL SYSTEMS Exercise 56: The Fertilizer Industry (N, P, K)

307

SKILLS FOR PHYSICAL SCIENCES LEARNERS How to Solve Problems in Physical Sciences As you read through the question, list all the quantities that are given, and very important, the quantity which is asked.

Look for an equation under the relevant section on your data sheet, containing the quantities given and asked.

Substitute the given values into the equation, but first ensure that your units are correct. Usually you have to use SI units, but in some instances you have to use other units, e.g. in n = m/M, the mass has to be in gram and not in the SI unit of kilogram!

In all problems on vectors, e.g. graphs, equations of motion, work, Newton, momentum, etc., start by determining which direction is +. Keep this in mind in ALL your answers and calculations.

If you see a problem contains quantities like displacement, time, acceleration, initial velocity or final velocity, you know to use equations of motion.

If a problem is about explaining if or why an object will move or will not move, it is about the net force – if it is zero, Newton I kicks in and the object will either remain at rest or continue moving at constant velocity. If the net force is NOT zero, Newton II kicks in and the object accelerates.

If it is about the force that objects exert on one another – be it electrostatic, gravitational, etc., Newton III kicks in and those two forces are equal and opposite, whatever the two objects are doing or whatever they look like.

If two objects are moving towards or away from each other and they collide or explode, and there is no friction or another external force you use law of conservation of momentum.

Remember that gravitational acceleration, g, is always down, whether the object is moving upwards, downwards, or at the highest point.

If a moving object changes height under the influence of its weight only (conservative force only), no friction, you use the law of conservation of mechanical energy.

If an object moves under the influence of a non-conservative force (e.g. friction), work done by the non-conservative force can be calculated using Wnc = Ep k.

If a moving object changes height and an external force, e.g. friction, is present, you use the workenergy theorem. Chances are almost 100% that you will get a question on work-energy theorem.

Please note that for objects moving on a horizontal plane you can also use the work-energy theorem, but not the law of conservation of mechanical energy.

Example 1: The diagram below shows a track, ABC. The curved section, AB, is frictionless. The rough horizontal section, BC, is 8 m long.

An object of mass 10 kg is released from point A which is 4 m above the ground. It slides down the track and comes to rest at point C.

From B to C its height, and therefore its potential energy (Ep), remain the same, while friction, a nonconservative force, is acting on it: Wnc = Ep because Ep remains the same.

k

can therefore be used between B and C, with Ep = 0,

OR: Friction is the net force acting on it between B and C and hence the work-energy theorem can be used: W net = Ek Whether we use W nc = Ep

k,

with Ep = 0 or W net = Ek, we need to know the change in kinetic

2 2 k = Ekf – Eki = ½ mvf – ½ mvi

We know vf = 0 and m = 10 kg, but we need to find vi, which is the velocity at B, where frictional force starts, but which also happens to be at the bottom of AB. STEP 1:

Calculate the velocity of the object at B: MEA = MEB (Ep + Ek)A = (Ep + Ek)B mghA + 0 = 0 + ½ mvB2 (10)(9,8)(4) + 0 = 0 + ½ (10)vB2 vB2 = 78,4

STEP 2:

Use the velocity calculated in STEP 1 to calculate the frictional force, f.

W net OR Use W nc = Ep k k Wf

= Ekf – Eki

Then W f

k

p

=0

f x cos180 = 0 – ½ mv B2 f(8)( –1) = –½ (10)(78,4) f = 49 N Example 2: A motor pulls a crate of mass 300 kg with a constant force by means of a light inextensible rope running over a light frictionless pulley as shown below. The coefficient of kinetic friction between the crate and the surface of the inclined plane is 0,19. Calculate the magnitude of the frictional force acting between the crate and the surface of the inclined plane.

(3) rope motor

A

300 kg 25o

4m

8m B C Using ENERGY PRINCIPLES only, calculate the magnitude of the frictional force exerted on the object as it moves along BC. (6) How to solve the problem: First look at the bigger picture: From A to B it changes height under the influence of its weight only, hence law of conservation of mechanical energy can be used for AB.

STEP 1: STEP 2:

Write down an expression from the data sheet to obtain kinetic friction fk: fk = µkN Use the equation in STEP 1 to calculate fk fk = µkN = µk w NOTE N

Procedure: Resolve w into rectangular components along the inclined plane.

wcos25 25o

wsin25 25o

Example 3: A 5 kg block is released from rest from a 5 kg height of 5 m and slides down a frictionless incline to point P as shown in R the diagram. It then moves along a frictionless horizontal portion PQ and 5m finally moves up a second rough inclined 3m plane. It comes to a stop at point R which P Q is 3 m above the horizontal.

The frictional force, which is a non-conservative force, between the surface and the block is 18 N. 3.1 Using ENERGY PRINCIPLES only, calculate the speed of the block at point P. (4) 3.2 Explain why the kinetic energy at point P is the same as that at point Q. (2) 3.3 ) of the slope QR. (7) How to solve the problem: First look at the bigger picture: From rest to P it changes height under influence of its weight only, no friction. This qualifies for law of conservation of mechanical energy.

From P to Q no net force and therefore no change in net energy.

From Q to R it changes height and experiences friction, i.e. its kinetic energy and potential energy change under the influence of a conservative and non-conservative force. Work-energy theorem can be used or equation for non-conservative force, W nc p k.

3.1

ME0 = MEP (Ep + Ek)0 = (Ep + Ek)P mgh0 + ½ mv02 = 0 + ½ mvP2 (5)(9,8)(5) + 0 = 0 + ½ (5)vP2 vP = 9,90 m·s–1

3.2

From P to Q no net force and therefore no change in net energy – Ep and Ek remain the same.

3.3

Work done by non-conservative force, friction: Wnc = Ep k (on data sheet) Wf = (Ep at R – Ep at Q ) + (Ek at R – Ek at Q ) f x cos180 = (mghR – 0) + (0 – ½ mvR2) (We take cos 180 because friction acts in direction opposite to motion – angle between direction of motion and friction is 180º) 18 x (–1) = (5)(9,8)(3) – ½ (5)(9,90)2 x = 5,45 m

OR

From work-energy theorem: Wnet k (Net work W net = work done by friction + work done by gravity = W f + W G and change in kinetic k = Ek at R – Ek at Q) Wf + W G = EkR – EkQ (Work done by friction = f· x·cos 180 and work done by gravity = component of gravitational force along p f x cos180 x cos180 = 0 – ½ mvQ2 x) (18) x (–1) + (5)(9,8)(3/ x)( x) (–1) = – ½ (5)(9,90)2 –18 x – 147 = –245,03 x = 5,45 m

· x·cos 180) N

mg cos

Scientific Notation Very large or very small numbers are preferably written in the scientific notation, e.g. a charge of 0,0000000326 C can be written as 3,26 x 10–8 C or a pressure of 350 625 Pa can be written as 3,51 x 105 Pa. Notice that in the scientific notation the number is written as a product of two numbers – the first number is between 1 and 10 and the second as 10n.

The first number is usually rounded off to two decimal places.

Please ensure that you know how to multiply and divide by numbers with exponents and how to use your calculator when dealing with exponents.

Free Body Diagrams The object is shown as a dot Vectors are shown as arrows. The head shows direction.

The length shows the approximate size of the vector. The tails must all touch the dot.

Note: Marks are lost for missing arrow heads, tails not touching the dot.

Marks can also be lost if relative sizes of arrows are incorrect.

All answers to be rounded off to two decimal places.

Conversion of Units

T2

T1

Fg

The 7 basic SI units: Mass (m) is measured in kilogram (kg) Time (t) is measured in second (s) Length ( ) is measured in metre (m) Electric current (I) is measured in ampere (A) Amount of substance (n) is measured in mole (mol) Temperature (T) is measured in kelvin (K) Luminous intensity is measured in candela (cd) All answers should be given in SI units, unless otherwise specified.

Mass is often given in gram.

In order to convert it to kg, it is useful to know that ‘kilo’ means ‘1 000’ or 103.

Therefore 1 kg = 103 g and 1 g = 10–3 kg.

Time is often given in hours (h). Remember that 1 h is 60 minutes and 1 minute is 60 seconds.

Hence 1 h = 3 600 s or 3,6 x 103 s and 1 s = 1/(3,6 x 103) h Length is sometimes given in millimetre (mm) or centimetre (cm) or decimetre (dm) or kilometre (km).

In order to convert from one to the other, it is useful to understand the following diagram: For each space that you move to the right, you x10–1 x10–1 x10–1 x10–1 x10–1 x10–1 10 or x10–1, e.g. 5 mm = 5 x 10–1 cm = 5 x 10–2 dm mm cm dm m Dm Hm km = 5 x 10–3 m = 5 x 10–6 km, etc.

For each space that you move to the left, you x10 x10 x10 x10 x10 x10 x10, e.g. 5 m = 5 x 10 dm = 5 x 102 cm = 5 x 103 mm, etc.

If the conversion is for area, which is x10–2 x10–2 x10–2 x10–2 x10–2 x10–2 measured in m2, then for each space that you 2 –2 move to the right, you 10 or x10 , e.g. mm2 cm2 dm2 m2 Dm2 Hm2 km2 5 cm2 = 5 x 10–2 dm2 = 5 x 10–4 m2, etc.

For each space you move to the left, you x102 x102 x102 x102 x102 x102 x102, e.g. 6 km2 = 6 x 106 m2, = 6 x 108 dm2 = 6 x 1012 mm2, etc. x10–3 x10–3 x10–3 x10–3 x10–3 x10–3 If the conversion is for volume, which is 3 measured in m , then for each space that you mm3 cm3 dm3 m3 Dm3 Hm3 km3 move to the right, you 103 or x10–3, e.g.

250 cm3 = 250 x 10–3 dm3 = 250 10–6 m3, etc. x103 x103 x103 x103 x103 x103 For each space you move to the left, you 3 3 9 3 15 3 x10 , e.g. 1 km = 1 x 10 m = 1 x 10 cm , etc.

Electric current is sometimes indicated as micro ampere ( A) or milli ampere (mA).

‘Micro’ means one millionth, or 10–6 and ‘milli’ means one thousandth, or 10–3.

The scientific temperature scale that is used in Science is the kelvin scale, but the celsius scale is popular for everyday use.

In order to convert from celsius to kelvin and vice versa, the following equation can be used: T = t + 273, where T is the temperature in kelvin and t is the temperature in celsius.

What is ‘Rate’?

Rate is the change of a quantity per second.

E.g. power is rate at which work is done or the amount of work done per second, i.e. (work done)/second.

Current is rate at which charge flows or the amount of charge flowing past a point per second, i.e. (amount of charge)/second.

Directly and Inversely Proportional If we say that quantity A is directly proportional to quantity B, it means that if A doubles, B also doubles; if A increases 10 times, B also increases 10 times, if A is halved, B is halved, etc.

This is written as A B, which we read as “A is directly proportional to B”.

To change this into a mathematical equation, we write it as A = kB, where k is a constant.

This means that it we measure the distance covered by different people for the same time, say 10 s, then the greater the speed of the person, the greater the distance covered by the person.

D v, which can be written as D = kv, where the constant k is the time – the equation becomes The graph of two quantities that are directly proportional, is always a straight line passing through the origin.

If we say that P is inversely proportional to Q, it means that if P increases 2 times, Q decreases 2 times; if P increases 10 times, Q decreases 10 times, etc.

This is written as P 1/Q, which we read as “P is inversely proportional to Q”.

To change this into a mathematical equation, we write it as P = k/Q, where k is a constant and PQ = k.

The graph for two quantities that are inversely proportional is a hyperbola.

Graph of A and B, which are Graph of P and Q, which are directly proportional inversely proportional Q

B

A

P

Ratios

In Physical Sciences we are often required to divide a quantity into ratios.

Suppose three workers, X, Y and Z divide the profit of a business in a ratio of 1:3:5.

This means that for every R1 that X earns, Y earns R3 and Z earns R5, i.e. every R(1 + 3 + 5) = R9 is split up into R1, R3 and R5.

Hence X earns 1/9 of the profit, Y earns 3/9 of the profit and Z earns 5/9 of the profit.

Now it easy to calculate each one’s share for any profit, e.g. if the profit is R1 800, X earns 1/9(R1 800) = R200; Y earns 3/9(R1 800) = R600 and Z earns 5/9(R1 800) = R1 000 Example: A potential difference of 24 V divides in a ratio of 2:2:4 across resistors P, Q and R. Share of potential difference for P: 2/(2 + 2 + 4) = 2/8 2/8(24 V) = 6 V Share of potential difference for Q: 2/(2 + 2 + 4) = 2/8 2/8(24 V) = 6 V Share of potential difference for R: 4/(2 + 2 + 4) = 4/8 4/8(24 V) = 12 V

Practical Investigations A practical investigation involves the relationship between two quantities.

All other quantities have to be kept constant.

The investigative question should always relate the two quantities investigated, e.g. suppose we investigate the relationship between temperature and volume, then the question can be “What is the relationship between temperature and volume?” or “Does volume increase with temperature?”, etc.

The hypothesis should be an ‘answer’ to the investigative question, e.g. “Volume is directly proportional to temperature” or “Volume increases as temperature increases”, etc.

The hypothesis may even be a wrong statement, but it has to ‘answer’ the investigative question.

The results should be recorded in a table with suitable headings that contain units, e.g. the column that contains the temperature should have the heading “Temperature (°C)” and the column containing the volume should have the heading “Volume (cm3)”.

The independent variable is the quantity that you change – suppose you decide you are going to measure the volume at 0 °C, 10 °C, 20 °C, then temperature is the independent variable.

The dependent variable is the quantity that changes because of the change in the one that you control – in this case, the volume.

The constant variables are those quantities that you will have to keep the same in order to have a fair investigation, i.e. you cannot have anything that will influence your readings except your two variables.

E.g. you will have to work with the same substance – you cannot use hydrogen in one measurement and then iron in the next. You will also have to keep the mass of the substance constant – you cannot take 10 g of hydrogen in one measurement and 50 g of hydrogen in the next. You will also have to keep the pressure constant – one measurement cannot be taken at sea level and the other at high altitudes, etc.

The graph should always have the two axes labelled properly, similar to the table, including units and then also a suitable scale.

The independent variable should always be on the horizontal or X-axis and the dependent variable on the vertical or Y-axis.

The conclusion should state the relationship between the two variables.

Precautions have to be taken to exclude dangers and factors which could cause the investigation to fail.

A quantitative analysis involves measurements, e.g. when you determine the relationship between the volume of a gas and temperature, you have to measure temperature and volume.

A qualitative analysis involves a study, e.g. when you determine which materials conduct electricity; you only change the substances and observe in which cases the light bulb glows – no measurement required.

Graphs: Gradient and Area In both Chemistry and Physics you often will come across graphs which you will have to interpret.

At this stage there are two important skills to master that come in very useful: Look at the gradient of the graph and the area included by the graph and the axes.

In any graph the . This means, the gradient will be v whatever is on the Y-axis divided by whatever is on the X-axis.

Whenever you get a question based on a graph, see if you can find an equation on the data sheet that contains the two quantities represented by the two axes. Rewrite the equation in such a way that the expression on the right hand side contains the two axes-quantities. If they happen to t be in the form of then whatever is on the left-hand side of the equation = gradient.

Example Now look for an equation that contains the two variables, v and t, and see if you can rewrite the equation in such a way that v and t are on one side of the equation – the rest of the equation then represents the gradient. In our example we find on the da Whatever the gradient does, acceleration a does the same – the first part of the graph has a constant and negative gradient, hence the acceleration a will be constant and negative. In the second part of the graph the gradient is constant and positive; hence the acceleration is constant and positive.

In any graph the area included by the graph and the axes is calculated as length times width.

In our example it is v x t. Again see if you can find an equation containing the two variables represented by the axes. Rearrange them in such a way that they are alone on one side of the equation. In our – area under a velocity/time graph represents displacement.

Models in Science Models are used to represent something.

It tries to make a difficult concept understandable.

SUMMARY OF DEFINITIONS & STATEMENTS OF LAWS FOR MATRIC FINAL PHYSICAL SCIENCE EXAMINATIONS (Taken from Examination Guidelines) PHYSICS Newton's laws and application of Newton's laws The normal force, N, is the force or the component of a force which a surface exerts on an object with which it is in contact, and which is perpendicular to the surface.

The frictional force, f, is the force that opposes the motion of an object and which acts parallel to the surface.

The static frictional force, fs, is the force that opposes the tendency of motion of a stationary object relative to a surface.

The kinetic frictional force, fk, is the force that opposes the motion of a moving object relative to a surface.

Newton's first law of motion: A body will remain in its state of rest or motion at constant velocity unless a non-zero resultant/net force acts on it.

Newton's second law of motion: When a resultant/net force acts on an object, the object will accelerate in the direction of the force at an acceleration directly proportional to the force and inversely proportional to the mass of the object.

Newton's third law of motion: When one body exerts a force on a second body, the second body simultaneously exerts a force of equal magnitude in the opposite direction on the first body.

Newton's Law of Universal Gravitation Newton's Law of Universal Gravitation: Each body in the universe attracts every other body with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.

Weight is the gravitational force the Earth exerts on any object on or near its surface.

Momentum Momentum is the product of an object's mass and its velocity.

Newton's second law of motion in terms of momentum: The resultant/net force acting on an object is equal to the rate of change of momentum of the object in the direction of the resultant/net force.

Impulse is the product of the resultant/net force acting on an object and the time the resultant/net force acts on the object.

A closed/an isolated system (in Physics), is a system on which the resultant/net external force is zero.

The Principle of conservation of linear momentum: The total linear momentum of a closed system remains constant (is conserved).

Vertical Projectile Motion in One Dimension A projectile is an object upon which the only force acting is the force of gravity.

Work, Energy and Power

The work done

the magnitude of the force, the magn the angle between the force and the displacement.

The work-energy theorem: The net work done on an object is equal to the change in the object's kinetic energy OR the work done on an object by a resultant/net force is equal to the change in the object's kinetic energy.

A conservative force is a force for which the work done in moving an object between two points is independent of the path taken.

(Examples are gravitational force, the elastic force in a spring and electrostatic forces (coulomb forces).

A non-conservative force is a force for which the work done in moving an object between two points depends on the path taken.

(Examples are frictional force, air resistance, tension in a chord, etc.)

The principle of conservation of mechanical energy: The total mechanical energy (sum of gravitational potential energy and kinetic energy) in an isolated system remains constant.

Power is the rate at which work is done or energy is expended.

Doppler Effect The Doppler Effect with sound, is the change in frequency (or pitch) of the sound detected by a listener because the sound source and the listener have different velocities relative to the medium of sound propagation.

Electrostatics Coulomb's law: The magnitude of the electrostatic force exerted by one point charge (Q1) on another point charge (Q2) is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance (r) between their centres.

An electric field is a region of space in which an electric charge experiences an electrostatic force.

The direction of the electric field at a point is the direction that a positive test charge would move if placed at that point.

The electric field strength at a point is the electrostatic force experienced per unit positive charge placed at that point.

Electric Circuits Ohm's law: The potential difference across a conductor is directly proportional to the current in the conductor at constant temperature.

Power is the rate at which work is done Alternating Current The RMS voltage, Vrms, is equivalent to the direct current (DC) voltage that will produce the same heating effect.

The RMS current, Irms, is equivalent to the direct current (DC) current that will produce the same heating effect.

Photo-electric effect The photoelectric effect is the process whereby electrons are ejected from a metal surface when light of suitable frequency is incident on that surface.

Threshold frequency, f0, is the minimum frequency of light needed to emit electrons from a certain metal surface.

Work function, W 0, is the minimum energy that an electron in the metal needs to be emitted from the metal surface.

Emission and absorption spectra An atomic absorption spectrum is formed when certain frequencies of electromagnetic radiation that passes through a medium, e.g. a cold gas, are absorbed.

An atomic emission spectrum is formed when certain frequencies of electromagnetic radiation are emitted due to an atom's electrons making a transition from a high-energy state to a lower energy state.

CHEMISTRY Intermolecular forces and interatomic forces (chemical bonds) Boiling point is the temperature at which the vapour pressure of a substance equals atmospheric pressure.

Melting point is the temperature at which the solid and liquid phases of a substance are at equilibrium.

Vapour pressure is the pressure exerted by a vapour at equilibrium with its liquid in a closed system.

Organic Molecules Organic molecules are molecules containing carbon atoms.

Molecular formula: is a chemical formula that indicates the type of atoms and the correct number of each in a molecule. Example: C4H8O A structural formula of a compound shows which atoms are attached to which within the molecule.

Atoms are represented by their chemical symbols and lines are used to represent ALL the bonds that hold the atoms together.

Condensed structural formula is a notation that shows the way in which atoms are bonded together in the molecule, but DOES NOT SHOW ALL the bond lines.

Example: CH3CH2COCH3 Hydrocarbons are organic compounds that consist of hydrogen and carbon only.

Homologous series is a series of organic compounds that can be described by the same general formula OR in which one member differs from the next with a CH2 group.

Saturated compounds are compounds in which there are no multiple bonds between C atoms in their hydrocarbon chains. Example: Alkanes Unsaturated compounds are compounds with one or more multiple bonds between C atoms in their hydrocarbon chains. Examples: Alkenes and Alkynes Functional group is a bond or an atom or a group of atoms that determine(s) the physical and chemical properties of a group of organic compounds.

Structural isomers are organic molecules with the same molecular formula, but different structural formulae.

Chain isomers are organic molecules with the same molecular formula, but different types of chains.

Positional isomers are organic molecules with the same molecular formula, but different positions of the side chain, substituents or functional groups on the parent chain.

Functional isomers are organic molecules with the same molecular formula, but different functional groups.

Organic Chemical Reactions The following are addition reactions of alkenes: Hydrohalogenation: The addition of a hydrogen halide to an alkene.

Halogenation: The reaction of a halogen (Br2 2) with a compound.

Hydration: The addition of water to a compound. (Do NOT confuse with hydrolysis!)

Hydrogenation: The addition of hydrogen to an alkene.

The following are elimination reactions: Dehydrohalogenation of haloalkanes: The elimination of hydrogen and a halogen from a haloalkane.

Dehydration of alcohols: Elimination of water from an alcohol.

Cracking of alkanes: The chemical process in which longer chain hydrocarbon molecules are broken down to shorter more useful molecules.

The following are substitution reactions: Hydrolysis: The reaction of a compound with water (Do NOT confuse with hydration!)

Halogenation of alkanes: The reaction of a halogen (Br2 2) with an alkane compound Plastics and Polymers Macromolecule: A molecule that consists of a large number of atoms.

Polymer: A large molecule composed of smaller monomer units covalently bonded to each other in a repeating pattern.

Monomer: Small organic molecules that can be covalently bonded to each other in a repeating pattern.

Polymerisation: A chemical reaction in which monomer molecules join to form a polymer.

Addition polymerisation: A reaction in which small molecules join to form very large molecules by adding on double bonds.

Addition polymer: A polymer formed when monomers (usually containing a double bond) combine through an addition reaction.

Condensation polymerisation: Molecules of two monomers with different functional groups undergo condensation reactions with the loss of small molecules, usually water.

Condensation polymer: A polymer formed by two monomers with different functional groups that are linked together in a condensation reaction in which a small molecule, usually water, is lost.

Energy changes in reactions related to bond energy changes Heat of reaction Exothermic reactions are reactions that release energy.

Endothermic reactions are reactions that absorb energy.

Activation energy is the minimum energy needed for a reaction to take place.

Activated complex is the unstable transition state from reactants to products in a chemical reaction.

Reaction rate is the change in concentration of reactants or products per unit time.

Collision theory is a model that explains reaction rate as the result of particles colliding with a certain minimum energy.

A positive catalyst is a substance that increases the rate of a chemical reaction without itself undergoing a permanent change.

Chemical Equilibrium An open system continuously interacts with its environment, while a closed system is isolated from its surroundings.

A reversible reaction: A reaction is reversible when products can be converted back to reactants.

Chemical equilibrium is a dynamic equilibrium when the rate of the forward reaction equals the rate of the reverse reaction.

Le Chatelier's principle: When the equilibrium in a closed system is disturbed, the system will re-instate a new equilibrium by favouring the reaction that will oppose the disturbance.

Acid-base reactions Arrhenius theory:

+

+

An acid is a substance that produces hydrogen ions (H )/hydronium ions (H3O ) when it dissolves in water.

A base is a substance that produces hydroxide ions (OH–) when it dissolves in water.

Lowry-Brønsted theory: An acid is a proton (H+ ion) donor.

A base is a proton (H+ ion) acceptor.

Strong acids ionise completely in water to form a high concentration of H3O+ ions.

Examples of strong acids are hydrochloric acid, sulphuric acid and nitric acid.

Weak acids ionise incompletely in water to form a low concentration of H3O+ ions.

Examples of weak acids are ethanoic acid and oxalic acid.

Strong bases dissociate completely in water to form a high concentration of OH– ions.

Examples of strong bases are sodium hydroxide and potassium hydroxide.

Weak bases dissociate/ionise incompletely in water to form a low concentration of OH– ions.

Examples of weak bases are ammonia, potassium carbonate, calcium carbonate and sodium hydrogen carbonate.

Concentrated acids contain a large amount (number of moles) of acid in proportion to the volume of water.

Concentrated bases contain a large amount (number of moles) of base in proportion to the volume of water.

Dilute acids contain a small amount (number of moles) of acid in proportion to the volume of water.

Dilute bases contain a small amount (number of moles) of base in proportion to the volume of water.

A substance that can act as either acid or base is amphiprotic or is an ampholyte.

Water is a good example of an ampholyte.

215

216

10

D. A momentum/time graph is similar to a velocity/time graph, because p = mv, and with m constant, p v. In the exercise on projectile motion we do velocity/time graphs.

Exercise 1:

C. According to Newton III, they experience forces equal in magnitude. But because of the smaller mass of the car, its acceleration is greater (Newton II), i.e. its change in velocity is greater.

11

A. From F t = p

2

B. p

12

3

C. From p = mv, with m constant, p v.

But from vf = vi + a t, with vi = 0 and a = g = constant, vf t.

Therefore p v t, and if time increases by factor 3, so does the momentum.

C. Impulse = F t, i.e. the product of F and t. In any graph, the product of the two quantities on the axes, gives the area under the graph. In this case Fnet = 100 and t = from 1 to 10 = 9.

13

C. F t = p

1

v, therefore when v doubles, p also doubles.

4

A. p

v and magnitude of v is maximum at A.

5

D. From Newton’s Second Law constant force means constant acceleration. In a velocity/time graph the gradient is determined by a, which is constant. Therefore a v/t graph would be a straight line (constant gradient) and the p/t graph would look the same.

Exercise 2: D. The initial +20 m s–1 implies that the direction towards the wall is +. p = pf – pi = m(vf – vi) Therefore –4 = 0,1(vf – 20) therefore vf = –20, thus 20 m s–1, away from wall

2

A.

3

C. Take towards the wall as positive: p = pf – pi = m(–v) – m(2v) = –3mv. 3mv away from the wall.

4

D. See no. 2.

5

C. Take west as +.

6

A. Change in momentum has the same units as momentum.

7

C. Take toward the wall as +.

8

D. Take down +.

p = pf – pi = (–3) – (+4) = –7, i.e. 7 kg m s–1, to the left

p = pf – pi = m(vf – vi) = 0,1(–5 – 10) = –1,5 kg m s–1. Therefore 1,5 m s–1 east.

p = pf – pi = (–mv) – (+mv) = –2mv. Magnitude is 2mv.

p = pf – pi = –mv – (+2mv) = –3mv, magnitude 3mv

Net force (Fnet = p/ t) Impulse (F t = p) Impulse (F t = p)

1.2 1.4

2

D. F t (impulse) = p, therefore F = p/ t (rate of change in momentum)

3

16.1 16.2

p = F t = 3 000 x 0,0005 = 1,5 kg m s–1, in the direction of the force. p = mvf – mvi = m(vf – vi) 1,5 = 0,06(vf – 0) vf = 25 m s–1, in direction of the force The change in momentum with or without the crumple zone will be the same.

But with the crumple zone the change in momentum takes place over a longer period.

From F t = p it follows that if the change in momentum remains the same and the time increases, F decreases, therefore the force experienced by the passengers decreases.

18.1 Sketch! Take towards the wall as +.

F = p/ t = m(vf – vi)/ t = 0,1(– 8 – 10)/0,01 = –180 N 180 N away from wall.

18.2 F = m v/ t; m and t remain the same, therefore F v and hence ball with greater change in velocity experiences greater force. The force exerted on the putty (where v = 0 – 10 = –10) is smaller than the force exerted on the ball (where v = –8 – 10 = –18).

19.1 It is only the car that is stopped by the wall, therefore force is only on car: F = m(vf – vi)/ t = (850)(0 – 20)/0,1 = –170 000 N, i.e. magnitude 170 000 N. 19.2 19.2.1 Less.

19.2.2 Airbag deforms, so it takes longer to bring dummy to rest. p remains the same, thus F t = p remains the same, thus for longer time, F is smaller.

20.1 Net force = rate of change of momentum 20.2 F = m(vf – vi)/ t = 60(0 – 14)/0,25 = –3 360 N Force exerted by seat belt on dummy is 3 360 N in a direction opposite to motion.

Exercise 3: 1.1 1.3 1.5

D. Fnet = p/ t

15.1 The shell moves to the right, therefore if momentum is conserved, the cannon has to move to the left.

The braking force acts against direction of motion to produce deceleration, therefore to the right.

15.2 Take right +. Cannon: F t = p, therefore t = p/F = m(vf – vi)/F = 500[0 – (–1)]/180 = 2,8 s

17

1

14

Newton’s Second Law of Motion.

Impulse (F t = p)

21

From Newton’s Third Law both cars will experience the same magnitude force during collision and they remain in contact for the same time. Therefore F t = p = m(vf – vi), will be the same for both, which implies that the vehicle with the greater mass experiences a smaller change in velocity and hence causes less injury for its occupants.

A. F t (impulse) = p, therefore F = p/ t, and hence if F is constant, then p/ t is also constant.

22

Unit of impulse F t: N s = (kg m s–2) s = kg m s–1 which is the same as the unit for momentum.

4

A

5

C. Fnet = p/ t = m(vf – vi)/ t = m(0 – v)/t = – mv/t

6

D. Take left as +. For ball: F t = p = mvf – mvi = m(vf – vi) = 1[1,5 – (–2)] = 3,5 Therefore 3,5 kg m s–1 to the left

7

C. p = F t From Newton’s Third Law they experience forces equal in magnitude but opposite, t is the same, therefore p must be equal in magnitude, but opposite in direction.

23.1 Dimension of area = F t, i.e. impulse = change in momentum.

23.2 F t = p = m(vf – vi) But F t = area = ½(0,5)(150) = 37,5, therefore 37,5 = 0,15(v – 0) Thus v = 250 m s–1, in direction of motion of hockey stick.

23.3 If m, vi and vf remain the same, F t = m(vf – vi) = area also remains the same.

Because it is a softer ball, t is longer (ball takes longer to deform due to its softness it deforms more) therefore base t is longer and thus F smaller, therefore height of triangle decreases.

s–1

= (N m) m

s–1

= [(kg m

s–2)

m] m

s–1

=

Magnitude is mv/t

kg m3 s–3

Impulse: N s or kg m

24

Take east as +: For m: F t = m v = m(vf – vi) = m(1,5 – 3) = –1,5m N s For 2m: F t = 2m(1,75 – 1) = 1,5m N s

25

Take down as +: On ball: F t = m(vf – vi) = 0,1(–4 – 5) = –0,9 Therefore impulse on ball is 0,9 N s upwards

s–1

8

C. J m

9

C. According to Newton’s Third Law, the force they exert on one another is the same in magnitude and the time they are in contact is same, hence their impulse, F t, also has to be the same in magnitude.