Physical Sciences
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Grade 12 • Examination Preparation
Physical SciencesSample
Electrostatics
Electric
Exercise
Exercise
Exercise
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 W nc = 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.
Example1:
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: W nc = Ep k can therefore be used between B and C, with Ep = 0, because Ep remains the same.
OR: Friction is the net force acting on it between B and C and hence the work-energy theorem can be used: Wnet = Ek
Whether we use W nc = Ep k, with Ep = 0 or W net = Ek, we need to know the change in kinetic k =Ekf –Eki = ½ mvf2 –½ mvi2
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.
Sample
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: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. Wnet k OR Use W nc = Ep k
Wf =Ekf –Eki Then W f k p = 0
f xcos180 = 0 –½ mv B 2
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)
STEP 1: Write down an expressionfrom the data sheetto obtain kinetic friction fk:fk = μkN
STEP 2: Use the equation in STEP 1 to calculate fk
fk = μkN = μk w
NOTE
Procedure:
Resolve w into rectangular components along the inclined plane.
25 wcos25 wsin25 25o N
Example 3:
A 5 kg block is released from rest from a height of 5 m and slides down a frictionless incline to point P as shown in the diagram. It then moves along a frictionless horizontal portion PQ and finally moves up a second rough inclined plane. It comes to a stop at point R which is 3 m above the horizontal.
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 350625 Pa can bewritten 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 frictional force, which is a non-conservative force, between the surface and the block is 18 N.
3.1Using ENERGY PRINCIPLES only, calculate the speed of the block at point P. (4)
3.2Explain why the kinetic energy at point Pis 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
3.1ME0 = 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
3.2From P to Q no net force and therefore no change in net energy–Ep and Ek remain the same.
3.3Work 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 xcos180= (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
ORFrom 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 · x·cos 180)
f xcos180 xcos180= 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
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.
The objectis shown as a dot
Free Body Diagrams
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.
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)
Conversion of Units
Electric current (I) is measured in ampere (A)
Temperature (T) is measured in kelvin (K)
Sample
Amount of substance (n) is measured in mole (mol)
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 ‘1000’ 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 = 3600s 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 10 or x10–1 , e.g. 5 mm = 5 x 10–1 cm = 5 x 10–2 dm
= 5 x 10–3 m = 5 x 10–6 km, etc.
For each space that you move to the left, you 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 measured in m2, then for each space that you move to the right, you 102 or x10–2 , e.g. 5 cm2 =5 x 10–2 dm2 = 5 x 10–4 m2, etc
For each space you move to the left, you x102, e.g. 6 km2 = 6 x 106 m2, = 6 x 108 dm2 = 6 x 1012 mm2, etc.
If the conversion is for volume, which is measured in m3, then for each space that you move to the right, you 103 or x10–3, e.g. 250 cm3 = 250 x 10–3 dm3 = 250 10–6 m3, etc.
For each space you move to the left, you x103, e.g. 1 km3 = 1 x 109 m3 = 1 x 1015 cm3, 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, thefollowing 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 persecond, 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 astraight 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
Ratios
In Physical Sciences we are often required to divide a quantity into ratios. Supposethree 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 R1800, X earns 1/9(R1800) = R200; Y earns 3/9(R1800) = R600 and Z earns 5/9(R1800) = R1000
Example: A potential differenceof 24 V divides in a ratio of 2:2:4 across resistors P, Q and R.
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)”.
Sample
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 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, similarto the table, including units and then also a suitablescale
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 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 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.
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 frictionalforce, 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 secondlaw 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 theuniverse 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.
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.
Sample
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.
An electric field is a region of space in which an electric charge experiences an electrostaticforce.
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 fieldstrength 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, areabsorbed.
An atomic emission spectrum is formed when certain frequencies of electromagnetic radiation are emitted due toan 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 followingare 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 followingare 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 followingare substitution reactions:
Hydrolysis: The reaction of a compound with water (Do NOT confuse with hydration!)
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
Sample
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.
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.
Hydrolysis is the reaction of a salt with water.
The equivalence point of a titration is the point atwhich the acid and the base havecompletely reacted with each other.
The endpoint of a titration is the point where the indicator changes colour.
The pH scale is a scale of numbers from 0 to 14 used to express the acidity or alkalinity of a solution.
Kw is the equilibrium constant for the ionisation of water or the ionic product of water or the ionisation constant of water, i.e. Kw = [H3O+][OH–] = 1 x 1014 at 298 K.
Auto-ionisation of water is the reaction of water with itself to form H3O+ ions and OH– ions.
Electrochemical Reactions
A Galvanic cell is a cell in which chemical energy is converted into electrical energy.
A galvanic (voltaic) cell has self-sustaining electrode reactions.
An Electrolytic cell is a cell in which electrical energy is converted into chemical energy.
Oxidation is a loss of electrons.
Reduction is again of electrons.
Oxidation is an increase in oxidation number.
Reduction is a decrease in oxidation number.
An oxidising agent is a substance that is reduced i.e. it gains electrons.
A reducing agent is a substance that is oxidized i.e. it loses electrons.
The anode is the electrode where oxidation takes place.
The cathode is the electrode where reduction takes place.
An electrolyte is a solution/liquid/dissolved substance that conducts electricity through the movement of ions.
Electrolysis is the chemical process in which electrical energy is converted to chemical energy OR the use of electrical energy to produce a chemical change.
The Fertiliser Industry (N, P, K)
Eutrophication is the process by which an ecosystem, e.g. a river or dam, becomes enriched with inorganic plant nutrients; especially phosphorus and nitrogen, resulting in excessive plantgrowth. As plantgrowth becomes excessive, the amount of dead and decaying plantmaterial increases rapidly.
Quantities, symbolsand units
The most common quantities, symbols and SI units used in introductory Physics are listed below. Aquantity should notbe confusedwith the units inwhich it is measured
Sample
MECHANICS
Different Kinds of Forces and Newton’s Laws
Please see that you do this section properly from the grade 11 book
Momentum: Definition
Momentum (p) of an object is the product of the mass (m) and the velocity (v) of that object. p = mv (On data sheet.)
The SI unit, in which momentum is measured, is kg m s–1 It is a vector quantity.
Its direction is the same as the direction of the velocity (i.e. the direction in which it is moving). If the mass of an object remains constant, then from p = mv, p v Therefore, whatever happens to velocity, the same will happen to momentum. For example, when velocity doubles, momentum doubles too; momentum/time sketch graphs have the same shape as velocity/time graphs; etc.
Exercise 1:
1 A stone is thrown vertically upwards and allowed to fall freely. (Ignore the effects of air resistance.) Which ONE of the following graphs is the best representation of its momentum p as a function of time? (2)
2 Which ONE of the following is doubled when the velocity of an object is doubled?
A Kinetic energy B Momentum C Acceleration D Inertia (2)
3 The momentum of an object which falls vertically from rest is p after a time t. Ignore air resistance. What will the momentum of the object be after a time 3t? (2)
A p B 2p C 3p D 9p (2)
4 A person dives from a high platform into a pool. At which ONE of the positions A, B, C or D will the magnitude of his momentum be a maximum?
A Position A B Position B C Position C D Position D (2)
5 Which ONE of the following momentum versus time graphs best represents the motion of an object that starts from rest and moves in a straight line under the influence of a constant force? (2)
Momentum of an Object Changing Direction
When a ball moves towards a wall, collides and moves back, a direction (e.g. towards the wall) must be chosen as positive. The change in momentum is then p = pf – pi = mvf – mv
2 A tennis ball strikes a wall horizontally with a momentum of 4 kg m s–1 and rebounds in the opposite direction with a momentum of 3 kg m s–1. Which ONE of the vector diagrams correctly shows the incident momentum, rebound momentum and change of momentum ( p) vectors for the ball (in kg m s–1)? (2)
Sample
Remember that one of these velocities will be positive and one negative!
Exercise 2:
1 A ball of mass 100 g, collides elastically with a wall at +20 m s–1. If the change in momentum of the ball is –4 kg m s–1, what is the velocity (in m s–1) at which the ball rebounds?
A +60,0 B +20,0 C –60,0 D –20,0 (2)
3 A squash ball of mass m hits the wall perpendicularly at a speed of 2v. The ball rebounds in the opposite direction at a speed of v. The magnitude of the change in the momentum of the ball is
(2)
4 A boy plays a snooker ball perpendicularly towards the side of a snooker table from which it bounces back. The vector marked p represents the momentum of the ball as it travels towards the side and the vector marked q represents the momentum of the ball after it has bounced back. Which vector represents the change in momentum of the ball as a result of the collision with the side of the table? (2)
5 A ball of mass 0,1 kg is thrown at a velocity of 10 m s–1, direction west, against a wall. The ball bounces back at 5 m s–1, direction east. The change of momentum the ball experiences during the collision with the wall is
A 1,5 kg m s–1; west. B 0,5 kg m s–1; west. C 1,5 kg m s–1; east. D 0,5 kg m s–1; east. (2)
6 Which ONE of the following combinations of base units represents the change in momentum?
7 A ball of mass m strikes a wall perpendicularly at a speed v. Immediately after the collision the ball moves in the opposite direction at the same speed v, as shown in the diagram. Which ONE of the following represents the magnitude of the change in momentum of the ball?
(2)
8 A cricket ball, mass m, falls vertically and strikes the ground with a speed of 2v. It rebounds vertically with a speed of v. What is the magnitude of the change in its momentum?
Newton’s
Second
Law of Motion, Change in Momentum and Impulse
Newton’s Second Law of Motion:
A net force on an object or system causes a change in momentum. This net (or resultant) force acting on the object or system is equal to the rate of change of momentum of the object or system in the direction of the resultant/net force.
(“rate of ...” implies “divided by t” or “per second”)
In symbols: Fnet = p/ t
From this equation it follows that the change in momentum is in the same direction as the force causing this change in momentum.
If the net force acting on the object or system is zero, there is no change in momentum (momentum is conserved).
Impulse (F t) is defined as the product of the force F and the contact time t of the applied force. It is a vector quantity with its direction in the direction of the net force producing the change in momentum.
The SI unit is the N s.
Impulse–momentum Theorem: Impulse equals the change in momentum F t = p = mvf – mvi (On data sheet.)
Please note: Force = rate of change of momentum, while impulse = change of momentum. When the impulse by B on A has to be calculated, the mass and change in velocity of A is used. The impulse by A on B is equal in magnitude, but opposite in direction to the impulse by B on A. This follows from Newton’s Third Law which states that the force exerted by B on A is equal but opposite to the force exerted by A on B. Furthermore, the time that the two forces in the force pair are applied, is the same, therefore for two colliding bodies F t as well as p, are the same in magnitude but opposite in direction
Exercise 3:
1 Give ONE word/term for
1.1 the rate of change of momentum. (1)
1.2 the law that states that the net (resultant) force is equal to the rate of change in momentum. (1)
1.3 change in momentum. (1)
1.4 the product of force and the time the force is applied. (1)
1.5 the physical quantity that is equivalent to the change in the momentum of a body. (1)
2 Which of the following can be represented by the rate of change of momentum?
A Mass x velocity B Impulse C Resultant force x displacement D Resultant force (2)
3 Which ONE of the following rates remains unchanged when an object experiences a constant resultant force?
A Rate of change of momentum of the object.
B Rate of change of displacement of the object.
C Rate at which the resultant force does work.
D Rate of change in kinetic energy of the object. (2)
4 Which ONE of the following physical quantities is the same as the rate of change of momentum?
A Resultant force B Work C Power D Impulse (2)
5 A car of mass m moves along a straight line with a velocity of magnitude v. The driver sees an obstruction and immediately applies the brakes. The car stops uniformly in t seconds from the moment that the brakes are applied. The car does not hit the obstruction. Which ONE of the following represents the magnitude of the average force exerted on the car during the braking period of t seconds?
A v/t B mv C mv/t D mvt (2)
6 A ball with a mass of 1 kg strikes a wall at right angles with a velocity of 2 m s–1. It bounces back with a velocity of 1,5 m s–1 The impulse of the wall on the ball (in kg m s–1) is (2)
A 0,5 to the right. B 0,5 to the left. C 3,5 to the right. D 3,5 to the left.
7 Two boys, Moloko and Thabiso, bump head–on into each other. They both have the same mass but Moloko is moving at twice the speed of Thabiso. How does the magnitude of the change in momentum of the two boys compare? The change in the momentum of Thabiso is A greater. B smaller. C the same. (2) 1,5
8 Which ONE of the following is not a unit of impulse? (2)
9 A trolley A with a mass of 8 kg collides head–on with a trolley B with a mass of 16 kg. The impulse of B on A during the collision
A is double the impulse by A on B.
B is four times greater than the impulse by A on B.
C has the same magnitude as the impulse by A on B.
D is half as great as the impulse by A on B. (2)
10 A truck and a car collide head–on. After the collision both vehicles move together to the right. The impulse of the truck on the car is to the
A right and bigger in magnitude than the impulse of the car on the truck.
B left and bigger in magnitude than the impulse of the car on the truck.
C right and equal in magnitude to the impulse of the car on the truck.
D left and equal in magnitude to the impulse af the car on the truck. (2)
11 Which ONE of the physical quantities is the same as the ‘change in momentum’?
12 The sketch graph may be used to calculate the impulse of a constant net force of 100 N that acts on an object over a period of time.
Which ONE of the following can be used to calculate the impulse (in kg m s–1) of the force for the time interval t = 1 s to t = 10 s?
A 100 x 1 B 100 x 10
C 100 x 9 D 10 x 9 (2)
13 Impulse is equal to the ...
A initial momentum of the body. B final momentum of a body.
C change in momentum of a body. D rate of change in momentum of a body. (2)
14 The net force acting on an object is equal to the ...
A mass of the object. B acceleration of the object.
C change in momentum of the object D rate of change in momentum of the object. (2)
15 A cannon of mass 500 kg fires a shell of mass 1 kg horizontally to the right at 500 m s–1 .
The initial recoil velocity of the cannon is 1 m s–1 to the left.
A constant force of 180 N is applied to brake the recoil of the cannon.
15.1 What is the direction of this force? Explain briefly. (3)
15.2 Calculate the time taken for the cannon to come to rest. (5)
16 A golf club exerts an average force of 3 kN on a ball of mass 0,06 kg. If the golf club is in contact with the golf ball for 5 x 10–4 s, calculate
16.1 the change in the momentum of the golf ball. (4)
16.2 the velocity of the golf ball as it leaves the club. (4)
17 The fronts of modern cars are deliberately designed in such a way that in case of a head–on collision, the front would crumple. Why is it desirable that the front of the car should crumple? (2)
18 A ball of mass 100 g strikes a wall horizontally at 10 m s–1 and rebounds at 8 m s–1 . It is in contact with the wall for 0,01 s.
18.1 Calculate the average force exerted by the wall on the ball. (7)
18.2 Consider a lump of putty also of mass 100 g that strikes the wall at 10 m s–1 and comes to rest in 0,01 s against the surface. Explain qualitatively whether the force exerted on the putty will be less, greater than or the same as the force exerted on the ball by the wall. Do no calculations. (4)
19 A dummy, mass 50 kg, is placed on the driver’s seat of a car during an experiment to investigate the effects of collisions. The car, mass 850 kg, starts from rest and accelerates towards a wall. The car collides with the wall at a speed of 20 m s–1. When the car hits the wall, the dummy carries on moving forward and collides with the windscreen.
19.1 When the car hits the wall, it comes to rest in 0,1 s.
Calculate the magnitude of the average force exerted by the wall on the car. (6)
19.2 The experiment is repeated with the car fitted with a large pillow (air bag) between the dummy and the dashboard.
19.2.1 Will the magnitude of the force exerted by the pillow (air bag) on the dummy be greater than, less than or the same as the magnitude of the force exerted by the dashboard (on the dummy)? (1)
19.2.2 Explain briefly, by using appropriate physics concepts. (3)
20 Akhona and Scelo had an argument about the importance of using a seat belt in a car. Scelo does not believe that it is necessary to use seat belts. Akhona takes Scelo to the Kyalami race track, where a safety test is performed on a car. During this test, a 60 kg crash-dummy is placed on the front seat, without a seat belt. At high speed, the brakes are applied very hard in an emergency situation. The dummy is then secured with a seat belt. The car is involved in a collision which brings it to rest from a speed of 14 m s–1 .
20.1 What is the relationship between change in momentum and resultant force? (2)
20.2 Calculate the force exerted by the seat belt on the dummy if the dummy comes to rest in 0,25 s. (6)
21 A traffic officer mentions the dangers of head-on collisions. He mentions that for cars involved in a head-on collision, the risk of injury for passengers in a heavier car would be less than for passengers in a lighter car. Use principles of Physics to explain why the statement made by the officer is correct. (3)
22 Show that impulse and momentum have the same units. (3)
23 During a game of hockey, a player strikes a stationary ball of mass 150 g.
The graph shows how the force on the ball varies with the time.
23.1 What does the area under this graph represent? (1)
23.2 Calculate the speed at which the ball leaves the hockey stick. (5)
23.3 The same player hits a practice ball of the same mass, but which is made from a softer material. The hit is such that the ball moves off with the same speed as before. How will the area, the height and the base of the triangle that forms the graph, compare with that for the original ball? (5)
24 A trolley of mass m travelling at 3 m s–1 due east collides with a second trolley of mass 2m travelling at 1 m s–1 in the same direction. After collision the velocity of the first trolley is 1,5 m s–1 due east. The velocity of the second trolley, after collision, is 1,75 m s–1 due east. Derive an expression in terms of m for the magnitude of the impulse experienced by each trolley during collision. (4)
25 A child drops a ball of mass 0,1 kg.
26.2 Calculate the impulse of the ball. (5)
26.3
Calculate the magnitude of the force exerted by the bat on the ball. (3)
27 A billboard advertises a car from a certain manufacturer. It says “Safety first! Both in one package! Airbag and crumple zone!”
Use your knowledge of momentum and impulse to justify how the safety features mentioned in the advertisement contribute to the safety of passengers. (3)
28 New cars have a crumple zone to help minimise injuries during accidents. In addition seat belts, air bags and padded interiors can reduce the chance of death or serious injury.
Sample
The ball strikes the ground with a velocity of 5 m s–1 and rebounds with a velocity of 4 m s–1
Determine the magnitude and direction of the impulse exerted on the ball by the ground. (4)
26 A cricket ball, mass 175 g, is thrown directly towards a player at a velocity of 12 m s–1
It is hit back in the opposite direction with a velocity of 30 m s–1 .
The ball is in contact with the bat for a period of 0,05 s.
26.1 Define, in words, the term impulse. (2)
28.1 Use principles in Physics to explain how air bags can reduce the chance of death or injury. (3)
28.2 In a crash test, a car of mass 1,2 x 103 kg collides with a wall and rebounds as illustrated. The initial and final velocities of the car are 12 m s–1 to the left and 2 m s–1 to the right respectively. The collision lasts 0,1 s.
Calculate the
28.2.1 impulse of the car during the accident (4)
28.2.2 average force exerted on the car. (3)
28.3 How will the magnitude of the force be affected if the time interval of the collision remains 0,1 s, but the car does not bounce off the wall?
Write down only INCREASES, DECREASES or REMAINS THE SAME. Explain your answer. (2)
29 The bounce of a cricket ball is tested before it is used. The standard test is to drop a ball from a certain height onto a hard surface and then measure how high it bounces. During such a test, a cricket ball of mass 0,15 kg is dropped from rest from a certain height and it strikes the floor at a speed of 6,2 m s–1. The ball bounces straight upwards at a velocity of 3,62 m s–1 to a height of 0,65 m. The effects of friction may be ignored.
29.1 Define the term impulse in words. (2)
29.2 Calculate the magnitude of the impulse of the net force applied to the ball during its collision with the floor. (3)
30 The roof of a tall building is 25 m above the ground A rigid ball of mass 0,3 kg falls freely when dropped from the roof. It strikes the concrete floor on the ground with velocity v1. It bounces to a maximum vertical height of 6 m. The ball was in contact with the floor for 0,9 s. Ignore the effects of friction.
30.1 Calculate the velocity v1 when the ball first hits the floor. (3)
30.2 Calculate the impulse of the ball as a result of the collision. (7)
30.3 Calculate the magnitude of the net force exerted on the ball. (3)
30.4 The rigid ball is now replaced with a softer ball of the same mass and volume as the rigid ball. It is then dropped from the same height onto the concrete floor. Will the ball reach the SAME, GREATER or LESSER height compared to the previous ball? Use principles of Physics to explain your answer. (3)
Law of Conservation of Momentum
Law of Conservation of Momentum:
In the absence of an external force acting on a system, total linear momentum is conserved OR The total linear momentum in a closed system remains constant in magnitude and direction. The Law of Conservation of Momentum is applied whenever there is a collision or explosion (but it has to be in a closed system).
The Law of Conservation of Momentum can be applied to closed systems only.
A system is a small part of the universe that we consider when solving a particular problem, e.g. all the bodies involved in the collision or explosion. Everything outside the system is called the environment. A closed/isolated system is a system on which no net external force (e.g. friction) is acting – only the internal forces, i.e. the forces which the colliding bodies exert on one another.
In short, all objects involved in the collision must be totally free to move after collision. Total momentum before collision/explosion = total momentum after collision/explosion.
7
m1vi1 + m2vi2 = m1vf1 + m2vf2 where vi1 and vi2 are the velocities of the objects (masses m1 and m2) before collision and vf1 and vf2 are the velocities of the objects after collision NB: When applying the Law of Conservation of Momentum, a direction has to be taken as positive, before starting the problem.
Exercise 4:
1 Two trolleys approach one another at the same speed. After collision, they remain in contact and at rest. Which ONE of the following is false?
A The total momentum remains constant.
B The two objects have the same mass.
C The momenta of the two objects were of the same magnitude before collision.
D The total kinetic energy remains constant. (2)
2 Two objects moving in opposite directions collide and come to rest immediately after the collision. This outcome is possible only if their
A masses are equal. B initial momenta are equal in magnitude.
C initial kinetic energies are equal. D initial velocities are equal in magnitude. (2)
3 Two trolleys are placed on a horizontal plane, as shown in the figure. The trolleys separate when a compressed spring between them is released. They hit buffers Y and Z simultaneously. What is the mass of X in kg? (Friction is negligible.)
A 0,2 B 0,3
C 0,6 D 0,9 (2)
4 A trolley is moving with velocity v across a horizontal table. A bag filled with sand and having the same mass as the trolley is dropped vertically onto it. The subsequent velocity of the trolley with the bag on top of it will be
A zero B v C ½ v D ¼ v (2)
5 Two trolleys are moving together at constant speed towards the right on a smooth horizontal surface when a compressed spring between them is suddenly released. After the ‘explosion’ the trolleys moved in opposite directions as indicated in the sketch. Which statement regarding the momentums of the trolleys after the ‘explosion’ is true?
A The total momentum of the system is zero.
B The momentum of the trolley moving to the right is greater in magnitude than the momentum of the trolley moving to the left.
C The momentum of the trolley moving to the left is greater in magnitude than the momentum of the trolley moving to the right.
D The momentum of the trolley moving to the left is equal in magnitude to the momentum of the trolley moving to the right. (2)
7 Trolley A has twice the mass of trolley B. In a head-on collision between trolley A and trolley B they both come to rest immediately after the collision. Which ONE of the following statements is TRUE?
A Trolley A has twice the velocity of trolley B before the collision.
B The two trolleys had the same velocity before the collision.
C Trolley B had twice the velocity of trolley A before the collision.
D Trolley A experienced a greater change in momentum than trolley B. (2)
Sample6 A projectile is fired vertically upwards from the ground. At the highest point of its motion, the projectile explodes and separates into two pieces of equal mass. If one of the pieces is projected vertically upwards after the explosion, the second piece will ... (2)
A drop to the ground at zero initial speed.
B be projected downwards at the same initial speed as the first piece.
C be projected upwards at the same initial speed as the first piece.
D be projected downwards at twice the initial speed as the first piece.
8 Two dynamics trolleys, with masses of 2 kg and 1 kg respectively, are moving at 3 m s–1, under frictionless conditions as shown in the figure. The compressed spring between the two trolleys is released and immediately after it has expanded fully, the velocity of the 2 kg trolley is found to be 1 m s–1 in the opposite direction (i.e. towards the left.)
8.1 Does the total momentum of the system change? (2)
8.2 Calculate the new speed of the 1 kg-trolley. (7)
9 A cannon of mass 500 kg fires a shell of mass 1 kg horizontally to the right at 500 m s–1 . What is the magnitude and direction of the initial recoil velocity of the cannon? (8)
10 A bullet of mass 10 g, travelling horizontally with a velocity of 300 m s–1, embeds itself in a block of wood of mass 190 g, which rests on a smooth horizontal surface. Calculate their combined velocity after impact. (6)
11 A 7 200 kg empty railway truck is stationary. A fertilizer firm loads 10 800 kg fertilizer into the truck. A second, identical, empty truck is moving at 10 m s–1 when it collides with the loaded truck.
11.1 If the empty truck stops completely immediately after collision, use a conservation law to calculate the velocity of the loaded truck immediately after the collision. (6)
11.2 Calculate the distance that the loaded truck moves after collision, if a constant frictional force of 24 kN acts on the truck. (6)
12 A trolley of mass m travelling at 3 m s–1 due east collides with a second trolley of mass 2m travelling at 1 m s–1 in the same direction.
12.1 Explain how the Law of Conservation of Momentum applies to this situation. (3)
12.2 If the velocity of the first trolley after the collision is 1,5 m s–1 due east, calculate the velocity of the second trolley after the collision. (7)
13 A child drops a ball of mass 0,1 kg.
The ball strikes the ground with a velocity of 5 m s–1 and rebounds with a velocity of 4 m s–1 Does the Law of Conservation of Momentum apply to this situation? Discuss. (3)
14 A boy with a mass of 45 kg, jumps with a horizontal speed of 5 m s–1 onto a stationary skateboard, with a mass of 5 kg, resting on the horizontal surface of a skateboard ramp. Show, by calculation, that the speed of the boy is 4,5 m s–1 immediately after he has jumped onto the skateboard. (6)
15 A bullet of mass 50 g travelling horizontally at 600 m s–1 strikes a stationary wooden block of mass 2 kg resting on a smooth horizontal surface. The bullet goes through the block and comes out the other side at 200 m s–1
15.1 Name and state the principle which can be applied to find the speed of the block after the bullet has passed through it. (4)
15.2 Calculate the speed of the block after the bullet has come out the other side. (6)
15.3 If the bullet takes 5 x 10–4 s to travel through the block, calculate the average acceleration of the block while the bullet is passing through it. (6)
Before explosion
After explosion
16 John, mass 47 kg, rides a skateboard, mass 3 kg, with a velocity of 1 m s–1, when his cat, mass 5 kg, drops from a tree into his arms.
Calculate the velocity of John and the cat as they move together on the skateboard. (5)
17 In a railway shunting yard, a locomotive, mass 4 000 kg, travelling due east at a velocity of 1,5 m s–1 , collides with a stationary goods wagon of mass 3 000 kg in an attempt to couple with it. The coupling fails and instead the goods wagon moves due east with a velocity of 2,8 m s–1
17.1 Calculate the magnitude and direction of the velocity of the locomotive immediately after collision.(6) 17.2 Name and state in words the law you used to answer the question. (4)
18 Hendrik, of mass 65 kg, on roller-skates, holds an iron bar of mass 5 kg in his hands. He is moving forward on a frictionless, horizontal track at a speed of 1 m s–1. In order to slow himself down, he throws the iron bar away from himself at a speed of 4 m s–1. Ignore the effects of all types of friction.
18.1 State, in words, the Principle of Conservation of Momentum (3)
18.2 In which direction should Hendrik throw the iron bar to get the maximum decrease in his velocity?(2)
18.3 Calculate the magnitude and direction of Hendrik’s velocity immediately after he has thrown the iron bar to get the maximum decrease in his velocity. (7)
19 The most common reasons for rear-end collisions are too short following distance, speeding and failing brakes. The sketch represents one such collision. Car A of mass 1 000 kg, stationary at a traffic light, is hit from behind by car B of mass 1 200 kg, travelling at 18 m s–1 . Immediately after the collision car A moves forward at 12 m s–1
19.1 Assume that linear momentum is conserved during this collision. Calculate the speed of car B immediately after the collision. (4)
19.2 Modern cars are designed to crumple partially on impact. Explain why the assumption made in the previous question may NOT be valid in this case. (2)
20 A bullet of mass 2 g is fired horizontally into a stationary wooden block of mass 1 kg. The bullet strikes the wooden block at a velocity of 490 m s–1 Ignore friction.
20.1 Name and state the law, in words, that can be used to calculate the velocity of the block-bullet system immediately after impact. (3)
20.2 Calculate the magnitude of the velocity of the block-bullet system immediately after impact. (5)
21 A railway truck A of mass 2 000 kg moves westwards with a velocity of 3 m s–1 It collides with a stationary truck B of mass 1 200 kg loaded with electronic equipment of mass 300 kg. The two trucks combined after the collision. Ignore the effects of friction.
21.4 The electronic equipment on the stationary truck is wrapped in bubble plastic (plastic filled with air bubbles). Use physics principles to explain why bubble plastic is preferred to ordinary plastic. (3)
22 A man of mass 87 kg on roller skates, moving horizontally at constant speed in a straight line, sees a boy of mass 22 kg standing directly in his path.
The man grabs the boy and they both continue in a straight line at 2,4 m s–1 .
Calculate the man’s speed just before he grabs the boy. Ignore the effects of friction. (4)
23 While travelling at 40 m s–1, a car of mass 1 000 kg, collides head-on with a truck of mass 5 000 kg moving at 20 m s–1. After the collision, the car and the truck move together. Ignore the effects of friction.
Sample
23.1 State the law of conservation of linear momentum in words. (2)
23.2 Calculate the velocity of the car immediately after collision. (6)
23.3 Research has shown that forces greater than 85 000 N during collisions may cause fatal injuries. The collision described above lasts for 0,5 s.
Determine, by means of a calculation, whether the collision above could result in fatal injury. (5)
Momentum: Elastic and Inelastic Collisions
Kinetic energy Ek = ½mv2 (on data sheet).
An elastic collision is one during which kinetic energy is conserved It is independent of whether or not momentum is conserved.
Thus, when the Law of Conservation of Momentum applies, the collision may be elastic or inelastic.
For an elastic collision:
total kinetic energy before collision = total kinetic energy after collision
During an inelastic collision, kinetic energy is usually lost as heat and/or sound.
During an explosion the kinetic energy of the system increases. (E.g. elastic potential energy of compressed spring between two trolleys is transformed into kinetic energy of the trolleys.)
If an object has momentum, v 0 and hence it must have kinetic energy, and vice versa.
Exercise 5:
1 Give ONE word/term for
1.1 the type of collision in which kinetic energy is conserved. (1) 1.2 a collision during which the kinetic energy changes. (1)
2 Two bodies approach each other head-on at equal speeds and collide. After colliding they stick together and remain stationary. Which ONE of the following statements is false?
A The total kinetic energy of the bodies did not change during the collision.
B The combined momentum of the two bodies was zero before collision.
C The two bodies must have been of equal mass.
D Before collision the momentum of the body was equal but opposite to the momentum of the other. (2)
3 A 1 kg trolley travelling at 2 m s–1 strikes an identical trolley which is stationary.
The trolleys remain coupled together after the collision. What change in the total momentum (i.e. of the system of two trolleys) and the total kinetic energy takes place here? (2)
Total momentum Total kinetic energy
A Stays the same Reduces
B Stays the same Stays the same
C Reduces Reduces
D Reduces Stays the same
21.1 Write down the magnitude and direction of the ‘reaction force’ to the weight of truck A. (3)
21.2 Calculate the velocity of truck B after the collision. (5)
21.3 Calculate the magnitude of the force that truck A exerts on truck B if the collision lasts for 0,5 s. (4)
4 Three separate, identical balls are in contact with each other in a straight line and are at rest on a smooth, horizontal surface.
Each of these balls has a mass m. Another ball, of the same size, but of mass 2m, moving at a velocity v, collides elastically, and in the same straight line with the three stationary balls. Which ONE of the following diagrams represents the situation immediately after the collision? (2)
5 A car has kinetic energy Ek when it is travelling at velocity v. Which ONE of the following expressions represents the magnitude of the car’s momentum?
A Ek/2v B Ekv/2 C 2v/Ek D 2Ek/v (2)
6 Two dynamics trolleys, with masses of 2 kg and 1 kg respectively, are moving at 3 m s–1, under frictionless conditions. The compressed spring between the two trolleys is released and immediately after it has expanded fully, the velocity of the 2 kg trolley is 1 m s–1 in the opposite direction (i.e. towards the left.)
6.1 Does the total momentum of the system change? (2)
6.2 Calculate the new speed of the 1 kg-trolley. (7)
6.3 Calculate the change in kinetic energy of the system. (7)
6.4 Explain why kinetic energy is not conserved during an event such as this. (4)
7 The velocity of a moving trolley of mass 1 kg is 3 m s–1
A block of mass 0,5 kg is dropped vertically on to the trolley. Immediately after the collision, the speed of the trolley and block is 2 m s–1
7.1 Show by calculation whether momentum is conserved in the collision. (5)
7.2 By way of calculation, show whether kinetic energy is conserved in the collision. (6)
8 A bullet with a mass of 20 g is fired horizontally at 500 m s–1 into a stationary wooden block with a mass of 4 kg. The block with the bullet embedded in it, then slides for 1,55 m across a horizontal surface before it comes to a stop.
8.1 Show by calculation that the velocity of the bullet and the block immediately after impact, is 2,5 m s–1 in the direction of motion. (5)
8.2 Calculate the average magnitude of the frictional force between the block and the surface when the block slides over it. (7)
8.3 Calculate the loss of kinetic energy of the system when the bullet strikes the block. What happened to the energy? (7)
9 A trolley of mass 3 kg collides with a compressed spring device (attached to a wall) while moving at a speed of 4 m s–1 (figure 1). At the instant the spring is completely released, the trolley rebounds with a speed of 6 m s–1 (figure 2) and comes to rest after 10 m.
9.1 Calculate the change in kinetic energy of the trolley. (5)
9.2 State whether the kinetic energy of the trolley increased or decreased. (2)
9.3 Explain what caused the change in the kinetic energy of the trolley. (2)
11 Sipho (60 kg) is chasing John (56 kg) in the hall of the school. John’s velocity is 3 m s–1 Sipho catches up with John, tackles him from behind and both of them continue moving together as a unit at 4,5 m s–1 on a slippery floor. Assume friction is negligible.
11.1 Determine Sipho’s velocity before he tackles John. (6)
11.2 Determine if the collision is elastic or inelastic. (5)
11.3 Name and state the law that was used to determine the velocity of above players. (3)
12 A man of mass 87 kg on roller skates, moving horizontally at constant speed in a straight line, sees a boy of mass 22 kg standing directly in his path.
Sample
10 Collisions occur daily on the roads in our country. In one of these collisions a car with mass 1 600 kg, moving at a speed of 30 m s–1 to the left, collides with a minibus, mass 3 000 kg, moving at 20
1 to the right. After collision the two vehicles move together as a unit in a straight line.
10.1 Calculate the velocity of the two vehicles after the collision. (6)
10.2 Do the necessary calculations to indicate that the collision was inelastic. (6)
The man grabs the boy and they both continue in a straight line at 2,4 m s–1
12.1 Calculate the man’s speed just before he grabs the boy. Ignore the effects of friction. (4)
12.2 Is the collision elastic? Use a calculation to support your answer. (6)
12.3 After grabbing the boy, they both continue at a velocity of 2,4 m s–1 along a straight line until they arrive at a loose gravel surface near the end of the path. They now move at a constant acceleration in a straight line through the loose gravel for 2 m before coming to rest.
Calculate the magnitude of the force exerted by the gravel surface on the man and the boy. (5)
Vertical Projectile Motion: Gravitational Acceleration
A projectile is an object upon which the only force acting is the force of gravity.
Near the earth’s surface, the gravitational acceleration g = 9,8 m s–2 (on data sheet), down, for all objects in the absence of air resistance and g does not depend on the mass of the object. All objects near a planet's surface will fall with the same acceleration, g, independent of the object's size, shape or mass in the absence of air resistance.
When an object is dropped or projected in any direction, with no air resistance, the only force acting on it after leaving your hand, will be its weight down and its acceleration will thus be the gravitational acceleration on that planet.
The object will only experience 'the force that drives it forward' while in contact with the body that projected it (e.g. while the hand is throwing the ball, while a bullet is inside a gun, while a ball is in contact with a bat/racket, etc. and not after it left that body).
NB! An object that is projected vertically upwards near the earth’s surface, moving under influence of its weight only, has a constant acceleration of 9,8 m s–2 down while it is moving upwards, at its highest point and while moving downwards.
Therefore, at its maximum height, its velocity is zero, but its acceleration is 9,8 m s–2 down.
Exercise 6:
1 Give ONE word/term for
1.1 the force that acts on a body in free fall. (1)
1.2 motion of an object near the surface of the earth under the influence of the earth's gravitational force alone. (1)
2 Two heavy metal spheres, P and Q, are dropped simultaneously from the same height above the floor. P is much smaller than Q. Ignore the effect of air resistance. Which ONE of the following is the correct observation of and possible reason for what happens? (2)
Observation Possible reason
A They reach the floor simultaneously Gravitational forces are equal on both
B P reaches the floor first Smaller objects always fall faster than bigger ones
C Q reaches the floor first The earth’s force of gravity is greater on the heavier object
D They reach the floor simultaneously The gravitational acceleration is the same for both
3 Two spheres of different mass are dropped simultaneously from the same point. Ignore air friction. When they are 1 m above the ground, they have the same
A momentum.
B velocity.
C gravitational potential energy. D kinetic energy. (2)
4 An astronaut stands on the surface of the moon and releases a spanner. What will happen to the spanner? The spanner will
A remain stationary at the height that the astronaut’s hand is above the moon’s surface.
B remain at the same height but drift slowly away.
C rise upwards away from the moon’s surface.
D fall to the moon’s surface. (2)
5 The sketch shows the path followed by a ball bouncing on the floor in the absence of air resistance.
Which ONE of the following sketches indicates correctly the force F acting on the ball when it is in position H?
6 The sketch shows the path of a cricket ball thrown by a player. Ignore air friction. Which arrow best indicates the direction of the force acting on the ball when it is at point P?
A B C D (2)
7 Mzi throws a marble upwards and catches it again on its way down. Which diagram best represents the force(s) on the marble at A and B respectively?
Air friction can be ignored.
8 A ball is thrown vertically upward and then returns to the ground. What is the acceleration of the ball at the top of its path?
A Zero B 9,8 m s–2; upward
C 9,8 m s–2; downward D Changes from 9,8 m s–2 upward to 9,8 m s–2 downward (2)
9 A ball is thrown vertically upwards and returns to the point of projection. Which statement about the acceleration at points X and Y is correct?
A The acceleration is downwards at both points.
B The acceleration is upwards at both points.
C The acceleration is downwards at X and upwards at Y.
D The acceleration is upwards at X and downwards at Y. (2)
10 A bullet fired vertically upwards, reaches a maximum height and falls back to the ground. Which ONE of the following statements is true with reference to the acceleration of the bullet during its motion, if air resistance is ignored? The acceleration
A is always downwards. B is first upwards then downwards.
C is first downwards then upwards. D decreases first and then increases. (2)
11 When an object is thrown upwards its velocity at maximum height is equal to
A zero. B greater than zero.
C less than zero. D equal to 9,8 m s–2. (2)
12 An object projected vertically upwards reaches its maximum height and returns to its original point of projection. Ignoring the effects of friction, the direction of the acceleration of the object during its motion is ...
A always vertically downwards.
B first vertically upwards and then vertically downwards.
C first vertically downwards and then vertically upwards.
D always vertically upwards. (2)
13 Which ONE of the following diagrams is the correct representation of the forces acting on a ball immediately after it has been thrown vertically upwards? The relative lengths of the vectors (arrows) indicate the magnitudes of the relevant forces. (2)
Vertical Projectile Motion: Signs (+ or –) of Displacement, Velocity, Acceleration
In graphs and equations of motion, it is of utmost importance to understand the significance of the SIGN (+ or –), of displacement, velocity and acceleration.
Whenever an object moves along a straight line in opposite directions (e.g. up and down), it is extremely important to pay careful attention to the signs (directions) of the displacement, velocity and acceleration, because they are vector quantities. Therefore a direction has to be selected as positive and the signs of displacement, velocity and acceleration have to be established accordingly. Consider the motion of an object along a vertical line. Up or down may be selected as positive and for this example we take up as positive and down as negative, with O as reference point.
The sign (direction) of the displacement or position is determined by drawing an arrow from the reference point to where the object is. If the arrow points upwards, in the positive direction, its displacement is + and if the arrow points downwards, in the negative direction, displacement is negative. Note that the sign of the displacement does not depend on the direction of motion – it depends only on its position relative to the reference point. Therefore its displacement is + while it is on the positive side of its reference point and negative while on the negative side of its reference point, irrespective of how it is moving.
The sign of the displacement therefore tells us where it is relative to its reference point.
The sign (direction) of the velocity is determined by the direction in which it is busy moving. If it is busy moving in the + direction (upwards in this case), velocity is + and while moving in the negative direction (downwards in this case), its velocity is negative, irrespective of where it is (irrespective of the sign of position).
The sign of the velocity thus tells us the direction in which it is busy moving.
The sign (direction) of acceleration is determined by the direction of the velocity and also whether it is speeding up or slowing down: If it is speeding up, the signs of velocity and acceleration will be the same (both positive or both negative).
If it slows down, the velocity and acceleration have different signs (one + and one –).
The sign of the acceleration (together with the sign of the velocity) therefore tells us whether the object speeds up or slows down.
Scalar quantities like distance and speed can only be positive, because they have no direction. Whenever you have to determine a vector quantity, the direction of the vector in the answer, e.g. displacement, may not be given as merely –5 m or 5 m in the negative direction, but the sign has to be interpreted, e.g. 5 m due east or 5 m left, etc.
When an object is dropped from a moving “vehicle” (e.g. a hand throwing, a rising balloon, a helicopter ascending, etc.), the initial velocity is the same as the velocity it had inside the moving “vehicle”.
Vertical Projectile Motion: Position-Time and Displacement-Time Graphs
A sketch graph is a graph where the labelled axes and the shape of the graph are of significance and not the numerical values.
Position is where the object is relative to the reference point and is indicated by y.
Displacement is the change in position and is indicated by .
Distance is the length of the path followed and can never be negative, because it is a scalar quantity without direction. Therefore a distance-time graph is always above the time axis.
Average speed = distance/time
Speed can be determined from a distance-time graph by calculating the gradient of the distance-
Average velocity during a time interval is the (total displacement during a time interval)/time
The velocity can be determined by calculating the gradient of the position-time or displacementtime graph. If the gradient is negative, it is busy moving in a negative direction and if the gradient is +, it is busy moving in a + direction.
During an interval of constant velocity, the position-time or displacement-time graph is a straight line (constant velocity means constant gradient).
During an interval of changing velocity, the position-time or displacement-time graph is a curve If velocity increases, the gradient increases and if the velocity decreases, the gradient decreases
The instantaneous velocity at any instant can be determined by calculating the gradient of the tangent to the graph at that point.
The average velocity between any two points is determined by drawing a straight line joining those two points on the graph and calculating the gradient of this line. For an example of the graph for a bouncing ball, see no. 2 and 5 below.
Exercise 7:
1 A ball of clay is dropped from a height of 45 m. It does not bounce back. Which ONE of the following graphs best represents the displacement (measured from the starting point) against the time t? Do not take air resistance into account. (2)
2 A ball is dropped, bounces and returns to its initial height. The best displacement-time graph of its motion is (2)
3 The position-time graph for an object that is thrown downwards is:
4 A stone is thrown vertically upwards and returns to the thrower’s hands after a while. Which ONE of the following position-versus-time graphs best represents the motion of the stone? (2)
7 A student is at the top of a building of height h. He throws a stone, X, upward with a speed v He then throws a second identical stone, Y, downward at the same speed v Redraw the following set of axes and sketch the graphs of position versus time for each of the stones X and Y. Use the ground as the point of zero position. (4)
8 The roof of a tall building is 25 m above the ground. A rigid ball of mass 0,3 kg falls freely when dropped from the roof. It strikes the concrete floor on the ground with velocity v1. It bounces to a maximum vertical height of 6 m.
Sample
5 A ball is released from rest from a certain height above the floor and bounces off the floor a number of times. The position-time graph represents the motion of the bouncing ball from the instant it is released from rest. Neglecting air resistance, which point (A, B, C or D) on the graph represents the position-time coordinates of the maximum height reached by the ball after the SECOND bounce? A B C D (2)
6 The position-time graph represents the motion of an object, which has been thrown vertically upwards.
6.1 Describe the motion of the object in terms of its velocity and acceleration. (6)
6.2 Draw neat sketch graphs of velocity-time and of acceleration-time of the object’s motion corresponding to the position-time graph. (8)
The ball was in contact with the floor for 0,9 s. Ignore the effects of friction. Using the ground as zero reference, draw a sketch graph of position (displacement) versus time for the motion of the ball from its original height until it reaches its second maximum height. Indicate the relevant position values on the y-axis. (4)
Vertical Projectile Motion: Velocity-Time Graphs
Speed is a scalar quantity and therefore speed-time graphs can have positive values only, i.e. be above the t-axis only.
For a speed-time graph the area between the graph and the time axis gives the distance covered The direction (sign) of the velocity is the direction in which the object is busy moving.
The gradient of a velocity-time graph gives the acceleration, which is 9,8 m s–2 for an object moving under the influence of its weight only.
If the direction (sign) of the acceleration is the same as the direction (sign) of the velocity, the object moves faster (accelerates) and if the directions (signs) of acceleration and velocity are different, the object slows down (decelerates).
For a velocity-time graph the area between the graph and the time axis gives the displacement Areas below the time-axis represent negative displacement and have negative values.
Distance can be obtained by calculating the area under the v-t graph, with all areas taken as positive, whether they lie above or below the time axis, because distance cannot be negative.
The velocity-time graph for a bouncing ball, dropped from a certain height, upward taken as positive:
Note that it starts moving downwards, in the negative direction and hence the velocity is negative. After touching the floor (at A) it starts moving upwards (velocity instantly becomes + at broken line), in the positive direction with velocity decreasing, hence the velocity is positive until it reaches the highest point (at B), where the velocity is zero. It then starts moving down again, with negative increasing velocity and touches the ground again at C. The gradient of the graph remains the same, because it represents gravitational acceleration. The period of each bounce decreases, because the ball loses energy and hence the height decreases.
Exercise 8:
1 A ball is thrown vertically upwards from a point and is then caught at the same height when it returns. Which ONE of the following velocity-time graphs is the best representation of this situation if the effect of air resistance is negligible? (2)
t v A t v B t v C t v D
2 The graph shows the variation of velocity with time of a body moving under the influence of the force of gravity.
Downward is taken as positive. What is the total displacement of the body, in m, in time 2t?
A 5t2 B 2vt
C vt D zero (2)
3 A man leans over the edge of a high building and throws a ball vertically upwards.
The following graph shows the velocity of the ball during the first 5 s of the motion:
After 5 s the ball will be
A back at the initial position.
B 20 m above the initial position, at rest.
C 25 m below the initial position, moving downwards.
D 30 m below the initial position, moving downwards. (2)
4 A solid rubber ball held above a horizontal table is dropped from rest. It strikes the table and bounces several times. The collisions are inelastic. Which ONE of the following graphs is the best representation of the variation of the velocity v of the ball with time t? Ignore air friction. (2)
5 A ball is released from rest from a certain height above the floor and bounces off the floor a number of times. Ignore the effects of air resistance.
Which ONE of the following velocity-time graphs best represents the motion of the ball? (2)
6 A crane is used on a building site to hoist bricks up in a steel bucket at a constant speed. When the bucket is 12 m above the ground and still going upwards, a brick is dislodged from the bucket and strikes the ground 2 s later.
Ignore the effects of air resistance.
Use the system of axes shown to draw sketch graphs of the motions of both the bucket and the brick on the same system of axes, from the moment the brick leaves the bucket until it hits the ground. Distinguish clearly between the graphs. (5)
7 Mzi throws a ball with a mass of 200 g vertically upward with an initial velocity of 15 m s–1. The ball falls back in her hand. The velocity-time graph represents the motion of the ball.
7.1 How long was the ball in the air? (1)
7.2 What is the velocity of the ball after 3 s? (2)
7.3 What will
7.3.1 the velocity and
7.3.2 the acceleration of the ball be at its maximum height? (3)
7.4 Use the graph to calculate the total distance the ball travelled in the air. (4)
7.5 Draw a sketch graph of distance-time for the movement of the ball in the air.
Show relevant values on both axes of the graph. (4)
Take gravitational acceleration as 10 m s–2 and answer the following questions:
8.1 What is the initial velocity u of the binoculars? (2)
8.2 What is the value of the slope of the graph? Explain. (4)
8.3 Which point on the path of the binoculars corresponds to time 0,8 s? (2)
8.4 Write an equation for the straight-line graph in terms of v and t. (3)
9 A boy stands at the edge of a high cliff. He throws a stone vertically upwards with an initial velocity of 10 m s–1. The stone strikes the ground at a point below the cliff after 3,5 s. The velocity-time graph was obtained from measurements made during the motion of the stone. Use the information on the graph to answer the following questions:
Sample
8 A hot-air balloon is rising vertically upwards with a constant velocity of 8 m s–1 when one of the passengers accidentally lets go of a pair of binoculars while leaning over the edge. The binoculars reach the ground after 6 seconds. The velocity-time graph represents the motion of the binoculars from the time the passenger lets go until they reach the ground.
9.1 Calculate the acceleration of the stone between times t = 2 s and t = 3 s. (3)
9.2 At which time(s) is the stone moving at 5 m s–1? (2)
9.3 After how many seconds does the stone reach its highest point? (1)
9.4 Determine the height of the cliff from which the stone was thrown. (4)
9.5 Using the top of the cliff as the initial position of the stone, sketch the position-time graph (displacement-time graph) for the motion of the stone from its highest point until it reaches the ground.
Only indicate relevant values on the x-axis. (3)
10 A lift at the side of a high building under construction is used to hoist building material to the construction team at the top. The platform on which the building material is transported, ascends vertically upward at a constant speed of 2,5 m s–1 when an iron bolt breaks off it and reaches the ground 4 s later. The velocity-time sketch graph represents the vertical motion of the falling bolt from the moment it breaks off. (Downward direction is +.)
Do not use the formal equations of motion in the following questions, but where applicable, show how you obtained your answer.
10.1 What is the velocity of the bolt at point p? (3)
10.2 The gradient of the graph is approximately 10. With what physical quantity relating to the motion of the bolt does this correspond?(3)
10.3 What is the significance of point q on the graph with respect to the motion of the bolt? (3)
10.4 What is the velocity of the bolt when it reaches the ground? (5)
10.5 At what height did the bolt break off the platform? (5)
11 A crane in the East London harbour lifts a crate, mass 300 kg, from the deck of a ship. It then moves the crate horizontally above the surface of the water and stops. The crate is now lifted vertically upwards at a constant speed of 2 m s–1
When the crate is 30 m above the surface of the water, while it is still moving upwards, the cable holding the crate snaps.
The velocity-time graph represents the motion of the crate from the moment the cable snaps until it hits the water.
Downward motion is taken as positive and the effects of air resistance can be ignored.
11.1 What does the gradient of this velocity-time graph represent?
State also, without doing any calculations, the magnitude of the gradient. (3)
11.2 Determine by using the graph, but not any equations of motion, the maximum height which the crate reaches above the position where the cable snapped. (4)
12 A man fires a projectile X vertically upwards at a velocity of 29,4 m s–1 from the EDGE of a cliff of height 100 m. After some time the projectile lands on the ground below the cliff. The velocity-time graph (NOT DRAWN TO SCALE) represents the motion of projectile X. (Ignore effects of friction.)
12.1 Use the graph to determine the time that the projectile takes to reach its maximum height. (A calculation is not required.) (1)
12.2 Calculate the maximum height that projectile X reaches above the ground. (4)
12.3 Sketch the position-time graph for projectile X for the period t = 0 s to t = 6 s.
USE THE EDGE OF THE CLIFF AS ZERO POSITION. Indicate the following on the graph:
The time when projectile X reaches its maximum height. The time when projectile X reaches the edge of the cliff. (4)
13 A ball is released from a certain height. The velocity-time graph represents the motion of the ball as it bounces vertically on a concrete floor. The interaction time of the ball with the floor is negligibly small and is ignored.
13.1 Describe the changes, if any, in velocity and acceleration of the ball from t = 0 s to t = 0,4 s. (4)
13.2 Without using the equations of motion, calculate the height from which the ball has been dropped initially. (4)
13.3 Copy the set of axes. Use the given velocity versus time graph for the motion of the ball to sketch the corresponding position-time graph for the time interval 0 s to 0,7 s. (3)
13.4 Is the first collision of the ball with the floor elastic or inelastic? Give a reason for your answer. (2)
Vertical Projectile Motion: Acceleration-Time Graphs
Acceleration is the change in velocity per time unit or the rate of change in velocity. f – vi
Acceleration is measured in m s–2
It is a vector quantity.
When an object moves faster (accelerates), the direction (sign) of the acceleration is the same as the direction (sign) of the velocity.
If the object slows down (decelerates), the directions (signs) of velocity and acceleration differ The area under the acceleration-time graph gives the change in velocity
NB! An object that is projected vertically upwards, moving under influence of its weight only, has a constant acceleration of 9,8 m s–2 down while it is moving upwards, at its highest point and while moving downwards. Therefore, if up is taken as positive, gravitational acceleration is negative and the graph is a straight horizontal line with an a-axis intercept of –9,8 m s–2
If down is taken as positive, gravitational acceleration is positive throughout the motion and the graph is a straight horizontal line with an a-axis intercept of +9,8 m s–2
In order to get a better feeling of what acceleration is:
Acceleration is the amount by which the velocity changes each second.
E.g. if the acceleration is 10 m s–2, it means that the velocity changes by 10 m s–1 each second, hence if an object starts moving from rest, the velocity after 1 second is 10 m s–1, after 2 seconds 20 m s–1, after 3 seconds 30 m s–1, etc.
The acceleration-time graph for a bouncing ball is a straight horizontal line (constant gravitational acceleration, 9,8 m s–2 downwards) interrupted by spikes when the ball hits the floor, because while the ball is in contact with the floor, it experiences the force exerted by the floor, upwards and hence for that brief moment the acceleration is upwards.
Exercise 9:
1 A ball is thrown vertically upwards. It takes 4 s to return to the thrower’s hand. Which ONE of the following graphs can be used to represent the acceleration-time graph for the motion? Take up as positive. (2)
Vertical Projectile Motion: Equations of Motion
NB! When an object moves up and down, a direction has to be selected as positive and the signs of a, v , vf and y have to be established accordingly, because they are vector quantities.
When an object is projected vertically upwards it has zero velocity at its greatest height, although acceleration will still be 9,8 m s–2, down, at that point.
Time symmetry: An object will take the same time to reach its highest point from projection as the time taken to fall back from the highest point to the point of projection.
The up and down motion can be considered as one motion, with v as the initial velocity at which it was projected, vf as the final velocity at which it strikes the surface, a = g = 9,8 m s–2 down and y the straight line drawn from reference point (start) to end, because y is displacement, which is the straight line drawn from reference point to end, irrespective of the path followed.
For an object falling freely (i.e. from rest under the influence of its weight only, no air resistance),
vi = 0 and a = g = 9,8 m s–2 down (near the earth’s surface), hence for the equations of motion:
vf = vi + a t becomes vf = g t
y = vi t + ½ a t2 becomes y = ½ g t2
vf2 = vi2 + 2 a y becomes vf2 = 2 g y
y = (vi + vf) t/2 becomes y = vf t/2
Furthermore, because g is constant, we can deduce from these equations that vf t y t2
vf2 y or vf y
Exercise 10:
1 Two identical bodies fall freely from rest from two different heights. Air resistance is negligible and the times taken to reach the ground are 1 s and 2 s respectively. The heights are in the ratio of
2 A golf ball dropped from a height h above the ground, strikes the ground at a particular speed. Ignoring air friction, from what height must the ball be dropped in order to strike the ground at double the speed?
2h B 3h C 4h D 5h (2)
3 Jonty and Paul stand on a balcony. They lean over the balcony rail and each throws a cricket ball. Jonty throws it vertically upwards and Paul throws it vertically downwards at the same speed If air resistance is negligible, how will the speeds of the two balls compare on reaching the ground?
• All the examination preparation for physics and chemistry you need in small, digestible, bite-sized ‘quanta’.
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• Includes formula sheets, and summaries of definitions and statements of law.
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