VCE Physics

Unit 2 Topic 1

Motion

View physics as a system of thinking about the world rather than information that can be dumped into your brain without integrating it into your own belief

systems.

Unit Outline• Identify parameters of motion as scalars or vectors• Analyse straight line motion under constant acceleration graphically, numerically and algebraically;• Analyse graphically non-uniform in a straight line;• Compare the effects of a force as defined by Aristotle, Galileo and Newton.• Describe the change in motion that result from the application of a force;• Model weight, w, as the force of gravity acting at the centre of mass point (approximated as the geometric centre)

of a body, w = mg• Model forces as vectors acting at the point of application (with magnitude and direction) labelling these forces

using the convention “force of … on…”• Apply Newton’s 3 laws of motion to a body on which a resultant vector force acts• Model forces as external actions through the centre of mass point of each body;• Apply the vector model of forces including vector addition, vector subtraction and components to readily

observable forces including weight, friction and reaction forces;• Explain movement in terms of the Newtonian model and some of its assumptions, including Newton’s 3 laws of

motion, forces act on point particles, and the ideal, frictionless world.• Apply the concept of work done by a constant force

work done = force multiplied by distance in the direction of the force work done = area under the force versus distance graph.

• Analyse Hooke’s Law for an ideal spring, F = -kΔx• Analyse energy transfers and transformations using an energy conservation model, including transfers between

– gravitational potential energy near the Earth’s surface, mgΔh ; and kinetic energy ½mv2 – potential energy in ideal springs ½ kx2 and kinetic energy ½mv2;

• Apply rate of energy transfer, power P = E/t• Apply the concept of momentum, p = mv;• Describe how action of a net force causes changes in momentum• Analyse impulse (momentum transfer) in an isolated system, for elastic collisions between objects moving in a

straight line.• Apply graphical, numerical and algebraic models to primary data collected during practical investigations of

motion, and to secondary data.• Identify and apply safe and responsible practices when investigating motion

Chapter 1

Introduction

1.0 An Ideal World1.0 An Ideal World

In the ideal world the In the ideal world the laws of motion apply laws of motion apply exactly, eg. objects exactly, eg. objects which are moving will which are moving will continue to move continue to move with the same speed with the same speed unless or until unless or until something occurs to something occurs to change this.change this.

To make life easier for To make life easier for Physics students Physics students situations or events situations or events which require which require mathematical analysis mathematical analysis are often described as are often described as occuring in an ideal, occuring in an ideal, frictionless world. frictionless world.

In the ideal world an object In the ideal world an object under the influence of Earth’s under the influence of Earth’s gravity will accelerate at 9.8 msgravity will accelerate at 9.8 ms-2 -2

throughout its journey never throughout its journey never reaching a terminal velocity.reaching a terminal velocity.

In the ideal world energy In the ideal world energy transformations are always transformations are always 100% efficient, so that the 100% efficient, so that the potential energy of a potential energy of a pendulum at the top of its pendulum at the top of its swing is all converted to swing is all converted to Kinetic Energy (motion Kinetic Energy (motion energy) at the bottom.energy) at the bottom.

In the ideal world perpetual motion In the ideal world perpetual motion machines are commonplace.machines are commonplace.

1.1 The S.I. System1.1 The S.I. SystemIn 1960, the “General Conference of Weights and Measures” , a Paris based In 1960, the “General Conference of Weights and Measures” , a Paris based international organisation, agreed that one set of units would be adopted international organisation, agreed that one set of units would be adopted world wide for the measurement of physical quantities. world wide for the measurement of physical quantities. This system is called the Systeme Internationale d’Units, or more simply the This system is called the Systeme Internationale d’Units, or more simply the S.I. System. S.I. System. The system is used and recognised worldwide and defines 7 fundamental The system is used and recognised worldwide and defines 7 fundamental units. units.

Physical QuantityPhysical Quantity S.I. UnitS.I. Unit SymbolSymbol

LengthLength metre m metre mMassMass kilogram kilogram kg kg

TimeTime second second s s

Electric CurrentElectric Current ampere ampere A A

TemperatureTemperature kelvin kelvin K KLuminous IntensityLuminous Intensity candela candela cd cd

Amount of SubstanceAmount of Substance mole mole mol mol

A derived unit is the force unit, the Newton, A derived unit is the force unit, the Newton, which is found from mass x length x 1/(time)which is found from mass x length x 1/(time)22

Thus the Newton has Thus the Newton has dimensions kg x m x sdimensions kg x m x s-2-2

All other units are derived from these 7 fundamentals.All other units are derived from these 7 fundamentals.

1.2 S.I. Definitions1.2 S.I. DefinitionsLength; metre [m]

It is the distance light travels, in a vacuum, in 1/299,792,458th of a second.

Luminous Intensity; candela [cd] It is the intensity of a source of light of a specified frequency, which gives a specified amount of power in a given direction.

Amount of Substance; mole [mol] It is the amount of substance that contains as many elementary units as there are atoms in 0.012 kg of 12C

Temperature; kelvin [K] It is 1/273.16th of the thermodynamic temperature of the triple point of water.

Current; ampere [A] It is that current which produces a force of 2 x 10-7 N between two parallel wires which are 1 metre apart in a vacuum.

Time; second [s] It is the length of time taken for 9,192,631,770 periods of vibration of the caesium-133 atom to occur.

Questions

Mass; kilogram [kg] It is the mass of a platinum-iridium cylinder kept at Sevres in France. It is now the only basic unit still defined in terms of a material object.

FundamentalsQUESTIONS1. Which of the following quantities have fundamental units and which have derived ?

Quantity (Unit) Fundamental DerivedPower (Watts)Distance (metre)Time (second)Force (Newton)Energy (Joule)Mass (kilogram)Electrical Resistance (ohms)Temperature (kelvin)Electric Current (amperes)

√√√

√√

√√

√√

Fundamentals2. From which of the fundamental units do the following derive their units ?

Quantity (Unit) Fundamental Units

e.g Force (Newton) Mass (kg), length (m), time (s)Acceleration (ms-2)

Momentum (kgms-1)

Impulse (Newton.second)

Velocity (ms-1)

Work (Joule) Note: W = F.d

Length (m), time (s)

Mass (kg), length (m), time (s)

Mass (kg), length (m), time (s)

Length (m), time (s)

Mass (kg), length (m), time (s)

Fundamentals

3. Show that 1 ms-1 = 3.6 kmh-1

1m x 1km x 3600s s 1000m 1h

Two relevant conversion factors are: 1 km = 1000 m, 1 h = 3600 s

so 1 ms-1 = 3.6 kmh-1

Which ones to use ?Easy, you want to end up with km on the top line and h on the bottom

These can be written as 1km1000m or 1000m

1km and 1h3600 s

3600s1hor

3.6

1.3 Position1.3 PositionIn order to specify theIn order to specify the position position of an object we first need to define an ORIGIN of an object we first need to define an ORIGIN or starting point from which measurements can be taken. or starting point from which measurements can be taken.

For example, on the number line, the point 0 is taken as the origin and all For example, on the number line, the point 0 is taken as the origin and all measurements are related to that point.measurements are related to that point.

-40 -35 -30 -25 -20 -15 -10 -5-40 -35 -30 -25 -20 -15 -10 -5 00 5 10 15 20 25 30 35 405 10 15 20 25 30 35 40

Numbers to the right of zero are labelled positiveNumbers to the right of zero are labelled positiveNumbers to the left of zero are labelled negativeNumbers to the left of zero are labelled negativeA number 40 is 40 units to the right of 0A number 40 is 40 units to the right of 0A number -25 is 25 units to the left of 0A number -25 is 25 units to the left of 0

Questions

Position4. What needs to be defined before the position of any object can be

specified ?

A zero point needs to be defined before the position of an object can be defined

5. (a) What distance has been covered when an object moves from position +150 m to position + 275 m ?

Change in position = final position – initial position = +275 – (+150) = + 125 m. Just writing 125 m is OK

(b) What distance has been covered when an object moves from position + 10 m to position -133.5 m ?

Change in position = final position – initial position = -133.5 – (+10) = - 143.5 m. Negative sign IS required

Chapter 2

Vectors & Scalars

2.0 Scalars and Vectors2.0 Scalars and VectorsBefore proceeding further we need to define two new quantities:Before proceeding further we need to define two new quantities:

SCALAR QUANTITIESSCALAR QUANTITIESThese are completely defined by These are completely defined by

•A Number andA Number and•A UnitA Unit

Examples of scalars are:Examples of scalars are:Temperature 17Temperature 1700, Mass 1.5 kg, Mass 1.5 kg

VECTOR QUANTITIESVECTOR QUANTITIESThese are completely defined byThese are completely defined by

•A NumberA Number•A Unit and A Unit and •A DirectionA Direction

Examples of vectors are:Examples of vectors are:Displacement 25 km West, Displacement 25 km West, Force 14 Newtons SouthForce 14 Newtons South

Vectors are usually represented by an ARROW, with the length of the arrow Vectors are usually represented by an ARROW, with the length of the arrow indicating the size of the quantity and the direction of the arrow the indicating the size of the quantity and the direction of the arrow the direction of the quantity.direction of the quantity.

This vector represents a This vector represents a Force of 4 N, acting North WestForce of 4 N, acting North West

NN

Questions

Vectors and Scalars6. Which of the following quantities are scalars and which vectors ?

Quantity Unit Scalar VectorDistance metreMomentum kgms-1 EastKinetic Energy jouleAcceleration ms-2 N45oEGravitational Field Strength Nkg-1

downwardsDisplacement metre

sidewaysAge yearsVelocity ms-1 WestTemperature oC

√√

√√√√

√√

√

2.1 Vector Addition & 2.1 Vector Addition & Subtraction Subtraction

Vectors can be at any angle to one another and still be added.Vectors can be at any angle to one another and still be added.This can be done in two ways:This can be done in two ways:

Draw accurate, scale vectors on graph paper and measure the size Draw accurate, scale vectors on graph paper and measure the size and direction of the result of the addition, called the “resultant vector”and direction of the result of the addition, called the “resultant vector”Draw sketch vectors and use trig and algebraic methods to calculate Draw sketch vectors and use trig and algebraic methods to calculate the size and direction of the resultant.the size and direction of the resultant.

ADDITIONADDITION

The tail of the second adds to The tail of the second adds to the head of the firstthe head of the first

The resultant is The resultant is drawn from the tail drawn from the tail of the first to the of the first to the

head of the secondhead of the second

SUBTRACTIONSUBTRACTION

To subtract, reverse the direction To subtract, reverse the direction of the negative vector then add.of the negative vector then add.

7.1 units East7.1 units East5.0 units5.0 units

NENE

++5.0 units5.0 units

SESE

==

5.0 units5.0 units NENE

__ 5.0 units5.0 unitsSESE

====5.0 units5.0 units

NENE5.0 units5.0 units

NWNW++ 7.1 units7.1 units

NorthNorth

2.2 Vector Components2.2 Vector ComponentsA single vector can be broken up into two or more parts called COMPONENTS.A single vector can be broken up into two or more parts called COMPONENTS.This process is useful when, for example, trying to find the vertical and This process is useful when, for example, trying to find the vertical and horizontal parts of a force which is accelerating a mass through the Earth’s horizontal parts of a force which is accelerating a mass through the Earth’s atmosphere.atmosphere.

F = 5 x 10F = 5 x 1066 N N

3030oo

F = 5 x 10

F = 5 x 10 66 N N

FFHH

FFVV

The Horizontal component of the force (FThe Horizontal component of the force (FHH) can be found using trig methods:) can be found using trig methods:FFHH = F cos 30 = F cos 30oo

= (5 x 10= (5 x 1066) ( 0.866)) ( 0.866) = 4.3 x 10= 4.3 x 106 6 NN

Similarly for the Vertical component (FSimilarly for the Vertical component (FVV), ), FFVV = F sin 30 = F sin 30oo

= (5 x10= (5 x1066)(0.5))(0.5) = 2.5 x 10= 2.5 x 1066 N N

At present, the total At present, the total force is directed at 30force is directed at 30oo above the horizontalabove the horizontal

FFHH and F and FVV are the COMPONENTS of are the COMPONENTS of the force F.the force F.

300

Questions

Vector Addition

7. What is the resultant force when 2 forces (6.0 N west and 4.0 N south) act on an object at the same time ?

θ = tan-1 4/6 = 33.7o

θ

Resultant Force = √(6)2 + (4)2 = 7.2 N

6 N west

4 N south

Vector Subtraction

8. Calculate the change in velocity of an object initially travelling at 8.5 ms-1 East whose final velocity was 8.5 ms-1 West. (remember Change in Velocity = Final Velocity – Initial Velocity)

- =8.5 ms-1 East8.5 ms-1 West

+ =8.5 ms-1 West

8.5 ms-1 West

17 ms-1 West

Vector Components9. An boy fires a stone from slingshot. The stone leaves with a velocity of 27 ms-1 at an angle 320 above the horizontal. Calculate the vertical and horizontal components of the stone’s velocity.

VH

Vv

θ = 32o

V = 27 ms-1

Vv = 27 sin 32o = 14.3 ms-1

VH = 27 Cos 32o = 22.9 ms-1

Vector addition/subtraction10. Calculate the acceleration of a car whose velocity changes from 16 ms-1 west to 21 ms-1 north in 1.5 seconds (acceleration = change in velocity/change in time) Acceleration is a vector quantity so a vector calculation is required to calculate it.

16 ms-1 West

Initial Velocity Final

Velocity21 ms-1 North

Change in Velocity = VF – Vi

- = + =

Resultant Velocity= √(21)2 + (16)2

= 26.4 ms-1

θ θ = tan-1 16/21 = 37.3o Acceleration = change in velocity/change in time

= 26.4/1.5 = 17.6 ms-2 at N37.3oW

Chapter 3

Kinematics

3.0 Distance & Displacement3.0 Distance & DisplacementDistance is a SCALAR quantity. It Distance is a SCALAR quantity. It has a Unit (metres) but no has a Unit (metres) but no Direction.Direction.Distance is best defined as “How far Distance is best defined as “How far you have travelled in your journey”you have travelled in your journey”

Displacement is a VECTOR quantityDisplacement is a VECTOR quantityHaving both a Unit (metres) and a Having both a Unit (metres) and a Direction.Direction.Displacement is best defined as Displacement is best defined as “How far from your starting point you “How far from your starting point you are at the end of your journey”are at the end of your journey”

The difference between these two quantities is easily illustrated with a The difference between these two quantities is easily illustrated with a simple example. You are sent on a message from home to tell the butcher simple example. You are sent on a message from home to tell the butcher his meat is off.his meat is off.

2 km2 km

At the end of the journey, Distance travelled = 2 + 2 = 4 km At the end of the journey, Distance travelled = 2 + 2 = 4 km

Positive DirectionPositive Direction

while Displacement = +2 + (-2) = 0 km while Displacement = +2 + (-2) = 0 km

At this point in the journey , Distance travelled = 2 km and Displacement = + 2 kmAt this point in the journey , Distance travelled = 2 km and Displacement = + 2 km

3.1 Speed & Velocity3.1 Speed & VelocityThese two terms are used interchangeably in the community but strictly These two terms are used interchangeably in the community but strictly speaking they speaking they areare different: different:

Speed is the time rate of change of Speed is the time rate of change of distance, i.e., distance, i.e.,

Speed = Speed = DistanceDistance TimeTime

Speed is a SCALAR QUANTITY, Speed is a SCALAR QUANTITY, having a unit (mshaving a unit (ms-1-1), but no ), but no direction.direction.Thus a speed would be:Thus a speed would be:100 kmh100 kmh-1 -1 or,or,27 ms27 ms-1-1

Velocity is the time rate of change Velocity is the time rate of change of displacement, i.e.,of displacement, i.e.,

Velocity = Velocity = Displacement Displacement Time Time

Velocity is a VECTOR QUANTITY, Velocity is a VECTOR QUANTITY, having a unit (mshaving a unit (ms-1-1) AND a direction.) AND a direction.Thus a velocity would be:Thus a velocity would be:100 kmh100 kmh-1-1 South or South or - 27 ms- 27 ms-1-1

3.2 Acceleration3.2 AccelerationAcceleration is defined as the time Acceleration is defined as the time rate of change of velocity, i.e.,rate of change of velocity, i.e.,

Acceleration = Acceleration = VelocityVelocityTimeTime

Acceleration is a VECTOR QUANTITY Acceleration is a VECTOR QUANTITY having both a unit (mshaving both a unit (ms-2-2) and a ) and a direction.direction.

There is no scalar measurement There is no scalar measurement of acceleration, so acceleration of acceleration, so acceleration MUST always be quoted with a MUST always be quoted with a direction.direction.

Typically, Acceleration means Typically, Acceleration means an increase in velocity over an increase in velocity over time, while Deceleration means time, while Deceleration means a decrease in velocity over time.a decrease in velocity over time.

aavv

When v and a are in the same direction,When v and a are in the same direction,the car accelerates and its velocity will the car accelerates and its velocity will increase over time.increase over time.

vvaa

When v and a are in the opposite When v and a are in the opposite direction, the car decelerates and direction, the car decelerates and its velocity will decrease over time.its velocity will decrease over time.

3.3 Instantaneous & Average 3.3 Instantaneous & Average VelocityVelocity

The term velocity can be The term velocity can be misleading, depending upon misleading, depending upon whether you are concerned with an whether you are concerned with an Instantaneous or an Average value.Instantaneous or an Average value.The best way to illustrate the The best way to illustrate the difference between the two is with difference between the two is with an example.an example.

You take a car journey out of a city to You take a car journey out of a city to your gran’s place in a country town 90 your gran’s place in a country town 90 km away. The journey takes you a total of km away. The journey takes you a total of 2 hours.2 hours.

The average velocity for this journey, The average velocity for this journey, vvAVAV = = Total DisplacementTotal Displacement = = 9090 = 45 kmh = 45 kmh-1-1

Total Time 2Total Time 2

However, your instantaneous velocity measured at a particular time during However, your instantaneous velocity measured at a particular time during the journey would have varied between 0 kmhthe journey would have varied between 0 kmh-1-1 when stopped at traffic when stopped at traffic lights, to, say 120 kmhlights, to, say 120 kmh-1 -1 when speeding along the freeway. when speeding along the freeway. Average and Instantaneous velocities are rarely the same.Average and Instantaneous velocities are rarely the same.Unless otherwise stated, all the problems you do in this section of the Unless otherwise stated, all the problems you do in this section of the course require you to use Instantaneous Velocities.course require you to use Instantaneous Velocities. Questions

Kinematics

11. A runner completes a 400 m race (once around the track) in 21 seconds what is: (a) her distance travelled (in km), (b) her displacement (in km), (c) her speed (in ms-1) and (d) her velocity (in ms-1) ?

(a) Distance = 0.4 km(b) Displacement = 0 km(c) Speed = distance/time = 400/21 = 19ms-1

(d) Velocity = displacement/time = 0/21 = 0 ms-1

Acceleration

12. A roller coaster, at the end of its journey, changes it’s velocity from 36 ms-1 to 0 ms-1 in 2.5 sec. Calculate the roller coaster’s acceleration.

a = change in velocity/change in time = (0 – 36)/2.5 = - 14.4 ms-2

Chapter 4Chapter 4

Motion by GraphsMotion by Graphs

4.0 Graphical Relationships4.0 Graphical RelationshipsIt is often useful and convenient It is often useful and convenient to represent information about to represent information about things like position, velocity, things like position, velocity, acceleration etc., using graphs.acceleration etc., using graphs.

Graphs “tell you a story”. Graphs “tell you a story”.

You need to develop the skills You need to develop the skills and abilities to “read the story”. and abilities to “read the story”. There are two basic types of graphs used There are two basic types of graphs used

in Physics:in Physics:(a)(a) Sketch GraphsSketch Graphs – give a “broad brush” – give a “broad brush”

picture of the general relationship picture of the general relationship between the two quantities graphed.between the two quantities graphed.

(b) (b) Numerical GraphsNumerical Graphs – give the exact – give the exact mathematical relationship between the mathematical relationship between the two quantities graphed and may be two quantities graphed and may be used to calculate or deduce numerical used to calculate or deduce numerical values.values.

4.1 Sketch Graphs4.1 Sketch Graphs

DistanceDistance

TimeTime

Sketch graphs have labelled axes Sketch graphs have labelled axes but no numerical values, they give a but no numerical values, they give a general broad brush relation general broad brush relation between the quantities.between the quantities.

The Story:The Story:As time passes, the As time passes, the distance of the distance of the object from its object from its starting point does starting point does not changenot change. . This is the graph of a This is the graph of a stationary objectstationary object

The Story:The Story:The object begins its The object begins its journey at the origin at t = 0. journey at the origin at t = 0. As time passes its As time passes its displacement increases at a displacement increases at a constant rate (slope is constant rate (slope is constant). So time rate of constant). So time rate of change of displacement change of displacement which equals velocity is which equals velocity is constant.constant.This is a graph of an object This is a graph of an object travelling at constant travelling at constant velocityvelocity

DisplacementDisplacement

TimeTime

VelocityVelocity

TimeTime

The Story:The Story:As time passes the As time passes the velocity remains velocity remains constant. constant. This is a graph of an This is a graph of an object travelling at object travelling at constant velocityconstant velocity

The Story:The Story:As time passes its As time passes its displacement gets displacement gets larger at an increasing larger at an increasing rate.rate. This is the graph of an This is the graph of an object moving with object moving with constant accelerationconstant acceleration

DisplacementDisplacement

TimeTime

Questions

Sketch Graphs Distance versus time graph. As

time passes displacement remains the same. This is the graph of a stationary object

Displacement versus time graph. As time passes its displacement is increasing in a uniform manner. This is a graph of an object travelling at constant velocity.

Distance

TimeTime

13 (a)

Displacement

TimeTime

(b)

Sketch GraphsVelocity

TimeTime

(c) Velocity versus time graph. As time passes the velocity of the object remains the same. This is a graph of an object travelling at constant velocity.

Displacement versus time graph. As time passes its displacement gets larger at an increasing rate. This is a graph of an accelerating object. (if the shape is parabolic the object is increasing its speed in a uniform fashion i.e. it has a constant acceleration)

Displacement

TimeTime

(d)

4.2 Exact Graphical Relationships4.2 Exact Graphical RelationshipsThe graphs you are required to interpret mathematically are those where The graphs you are required to interpret mathematically are those where distance or displacement, speed or velocity or acceleration are plotted distance or displacement, speed or velocity or acceleration are plotted against time.against time.

The information available from these graphs are summarised in the table The information available from these graphs are summarised in the table given below.given below.

Learn this table off by heart.Learn this table off by heart.Put it on any cheat sheet you are allowed to use.Put it on any cheat sheet you are allowed to use.

Graph TypeGraph Type

Distance orDistance orDisplacement Displacement

versusversusTimeTime

Speed or VelocitySpeed or VelocityversusversusTimeTime

AccelerationAccelerationversusversusTimeTime

Read directly fromRead directly fromthe Graphthe Graph

Obtain from Slope ofObtain from Slope ofthe Graphthe Graph

Obtain from Area Obtain from Area Under the GraphUnder the Graph

Distance orDistance orDisplacementDisplacement

Speed or Speed or VelocityVelocity

AccelerationAcceleration

Speed or Speed or VelocityVelocity

AccelerationAcceleration

No Useful No Useful InformationInformation

No Useful No Useful InformationInformation VelocityVelocity

Distance orDistance orDisplacementDisplacement

Questions

Graphical Interpretation14. Given below is the Distance vs Time graph for a cyclist riding along a straight path.

010

20 30

40

50

60

A B C D

Time (s)

10

20

Distance (a) In which section (A,B,C or D) is the cyclist stationary ? (b) In which section is the cyclist travelling at her slowest (but not zero) speed ? (c) What is her speed in part (b) above ? (d) What distance did she cover in the first 40 seconds of her journey ? (e) In which section(s) of the graph is her speed the greatest ? (f) What is her displacement from her starting point at t = 50 sec ?

(a) Stationary in section C(b) Section B (c) Travels 10 m in 20 s speed = 10/20 = 0.5 ms-1

(d) 20 m (read directly from graph)(e) Section D (travels 20 m in 10 s) speed = 2 ms-1

(f) Displacement at t = 50 s is 0 m (i.e., back at starting point)

Graphical Interpretation15. Shown below is the Velocity vs Time graph for a motorist travelling along a straight section of road.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 200

2

4

6

8

10

-2

-4

-6

-8

Time(s)

-10

Velocity (ms-1)(a) What is the motorist's displacement after 4.0 sec ?(b) What is the motorists acceleration during this 4.0 sec period ?(c) What distance has the motorist covered in the 20.0 sec of his journey ?(d) What is the motorist's displacement at t = 20.0 sec (e) What happens to the motorists velocity at t = 20.0 sec? Is this realistic ?(f) Sketch an acceleration vs time graph for this journey.

(a) Displacement = area under velocity time graph. Between t = 0 and t = 4 s. Area = ½ (10 x 4) = 20 m

(b) Acceleration = slope of velocity time graph = (10 – 0)/(4 – 0) = 2.5 ms-2

(c) Distance = area under graph (disregarding signs) Total area = ½(10 x 4) + (6 x 10) + ½(10 x 2) + ½(9 x 2) + (6 x 9) = 20 + 60 + 10 + 9 + 54 = 153 m

(d) Displacement = area under graph (taking signs into account) = ½(10 x 4) + (6 x 10) + ½(10 x 2) - ½(9 x 2) - (6 x 9) = 20 + 60 + 10 - 9 – 54 = 27 m

(e) Velocity falls from 9 ms-1 to zero in no time – no realistic, as it would require an infinite deceleration to achieve this.

Graphical Interpretation

15, continued

-4.54

10

12

14 20

To Infinitya

-5

2.5

(e)

t

Graphical Interpretation16. An object is fired vertically upward on a DISTANT PLANET. Shown below is the Velocity vs Time graph for the object. The time commences the instant the object leaves the launcher

Velocity (ms

-1)

Time (s)

30

-30

2 4 6 8 10 120

(a) What is the acceleration of the object ?(b) What is the maximum height attained by the object ?(c) How long does the object take to stop ?(d) How far above the ground is the object at time t = 10.0 sec ?

(a) Acceleration = slope of velocity time graph. Slope = (30 – 0)/(0 – 6) = -5.0 ms-2

(b) Displacement = area under velocity time graph = ½ (6 x 30) = 90 m(c) Stops at t = 6.0 sec(d) The rocket has risen to a height of 90 m in 6 sec. It then falls a distance of

½ (4 x 20) = 40 m, so it will be 90 – 40 = 50 m above the ground at t = 10 s

Chapter 5Chapter 5

The Equations of The Equations of MotionMotion

5.0 The Equations of Motion5.0 The Equations of MotionThe Equations of Motion are a set of equations The Equations of Motion are a set of equations linking displacement, velocity, acceleration and linking displacement, velocity, acceleration and time.time.They allow calculations of these quantities They allow calculations of these quantities without the need for graphical representations.without the need for graphical representations.

THESE EQUATION CAN ONLY BE USED IF THESE EQUATION CAN ONLY BE USED IF THE ACCELERATION IS CONSTANTTHE ACCELERATION IS CONSTANT

The 3 main equations are:The 3 main equations are: v = u + atv = u + atvv22 = u = u22 + 2as + 2ass = ut + s = ut + ½at½at22

Where,Where,u = initial velocity (msu = initial velocity (ms-1-1))v = final velocity (msv = final velocity (ms-1-1))a = acceleration (msa = acceleration (ms-2-2))s = displacement (m)s = displacement (m)t = time (s)t = time (s)

When using the equations, always list out the When using the equations, always list out the information given and note what you need to find, information given and note what you need to find, then choose the most appropriate equation.then choose the most appropriate equation.In some cases you also need to define a positive direction, up or down for In some cases you also need to define a positive direction, up or down for vertical motion, left or right for horizontal motion questionsvertical motion, left or right for horizontal motion questions

u =u =v =v =a =a =s =s =t =t =

+ve+ve

5.1 Motion Under Gravity5.1 Motion Under GravityObjects (close to the surface) Objects (close to the surface) falling through the Earth’s falling through the Earth’s gravitational field are subject gravitational field are subject to a constant acceleration of to a constant acceleration of 9.8 ms9.8 ms-2-2. . Since the acceleration is Since the acceleration is constant this motion can be constant this motion can be analysed by the equations of analysed by the equations of motion.motion.

You need to go through the You need to go through the same process of listing same process of listing informationinformation

u =u =v =v =a =a =s =s =t =t =

+ve+ve

The acceleration in this case is ALWAYS directed downward.Objects thrown or fired directly upwards would thus have their velocity and acceleration in opposite directions.

v = u + atv = u + atvv22 = u = u22 + 2as + 2ass = ut + s = ut + ½at½at22

The calculations using the equations of motion always ignore the effects of friction and air resistance

and deciding on a positive direction

Questions

Equations of Motion

17. A truck travels from rest for 10.0 sec with an acceleration of 3.0 ms-2. Calculate the truck's final velocity and total distance travelled.

List information: u = 0, v = ?, a = 3.0 ms-2, s = ?, t = 10 sfirstly find v, use v = u + at v = 0 + (3.0)(10) = 30 ms-1

then find x use s = ut + ½at2 (0)(10) +½(3.0)(10)2 = 150 m.

18. A ball rolling down an inclined plane from rest travels a distance of 20.0 m in 4.00 sec. Calculate its acceleration and its final speed

List information: u = 0, v = ?, a = ?, s = 20.0 m, t = 4.0 sFirstly find a, use s = ut + ½at2 20.0 = (0)(4.0) + ½a(4.0)2 a = 1.25 ms-2

The find v, use v = u + at v = 0 + (1.25)(4.0) 5.0 ms-1

Equations of Motion19. The speed of a freewheeling skateboard travelling on a level surface falls from 10.0 ms-1 to 5.00 ms-1 in moving a distance of 30.0 m. If the rate of slowdown is constant, how much further will the skateboard travel before coming to rest ?

List information u = 10 ms-1, v = 5.0 ms-1, a = ?, s = 30 m, t = ?Cannot get to answer in 1 step. First find accelerationUse v2 = u2 + 2as a = (v2 – u2)/2s a = - 1.25 ms-2

Now new information u = 5.0 ms-1, v = 0, a = -1.25ms-2, s = ?, t = ?Use v2 = u2 + 2as s = (v2 – u2)/2a x = 10 m

20. A bullet leaves the barrel of a gun aimed vertically upwards at 140 ms-1. How long will it take to reach its maximum height ? (Ignore air resistance and use g = 10 ms-2) .

List information (up is +ve) u = 140 ms-1 v = 0, a = -10 ms-2, s = ? t = ?Use v = u + at t = (v – u)/a = (0 – 140)/-10 = 14 s

Chapter 6Chapter 6

Forced Change Forced Change

6.0 What is a Force ?6.0 What is a Force ?

First, a force is an "interaction". You can compare a force to another common interaction - a conversation. A conversation is an interaction between 2 people involving the exchange of words (and ideas). Some things to notice about a conversation (or any interaction) are:To have a conversation, you need two people. One person can't have a conversation A conversation is something that happens between two people. It is not an independently existing "thing" (object), in the sense that a chair is an independently existing "thing".

In the definition, "(material) objects" means that both objects have to be made out of matter - atoms and molecules. They both have to be "things", in the sense that a chair is a "thing". A force is something that happens between 2 objects. It is not an independently existing "thing" (object) in the sense that a chair is an independently existing "thing".

"A force is an interaction between two material objects involving a push or a pull."How is this different from the usual textbook definition of a Force simply being a “push or a pull” ?

Forces are like conversations in that:To have a force, you have to have 2 objects - one object pushes, the other gets pushed.

Questions

Force21. A force is an interaction between 2 objects. Therefore a force can be likened to

A: Loving chocolate B: Fear of flyingC: Hatred of cigarettesD: Having an argument with your partner

22. Between which pair can a force NOT exist ?A: A book and a tableB: A person and a ghostC: A bicycle and a footpathD: A bug and a windscreen

6.1 What Kinds of Forces Exist ?6.1 What Kinds of Forces Exist ?

2. Field Forces are forces in which the two interacting objects are not in contact with each other, yet are able to exert a push or pull despite a physical separation. Examples of field forces include Gravitational Forces, Electrostatic Forces and Magnetic Forces

For simplicity sake, all forces (interactions) between objects can be placed into two broad categories:

1. Contact forces are types of forces in which the two interacting objects are physically contacting each other. Examples of contact forces include frictional forces, tensional forces, normal forces, air resistance forces, and applied forces.

Force is a quantity which is measured using the derived metric unit known as the Newton. One Newton (N) is the amount of force required to give a 1 kg mass an acceleration of 1 ms-2. So 1N = 1 kgms-2

 Force is a vector quantity, you must describe both the magnitude (size) and the direction.

Questions

Contact or Field Forces23. Classify the following as examples of either Contact or Field forces in action (or maybe both acting at the same time).

EXAMPLE CONTACT FORCE

FIELD FORCE

(a) A punch in the nose

(b) A parachutist free falling

(c) Bouncing a ball on the ground(d) A magnet attracting a nail

(e) Two positive charges repelling each other(f) Friction when dragging a refrigerator across the floor(g) A shotput after leaving the thrower’s hand

√

√√

√

√

√

√

√

√

6.2 What Do Forces Do ?6.2 What Do Forces Do ?Forces affect motion. They can:• Begin motion• Change motion• Stop motion• Have no effect

BEGINNING MOTION:A constant force (in the same direction as the motion) produces an ever increasing velocity.

STOPPING MOTION:A constant force (in the opposite direction to the motion) produces an ever decreasing velocity.

CHANGING MOTION:A constant force (at right angles to the motion) produces an ever changing direction of velocity.

NO EFFECT:A total applied force smaller than friction will not move the mass

FR

6.3 Where Forces Act6.3 Where Forces ActForces acting on objects Forces acting on objects must have a point of must have a point of application, a place application, a place where the force acts.where the force acts.

For Contact Forces the For Contact Forces the point of application is point of application is simply the point at simply the point at which the force initiator which the force initiator contacts the object.contacts the object.

For Field Forces, the only one applicable in movement being gravity, will act through the centre of mass of the object

Force of finger on cartonForce of carton on finger

Gravitational Force

C of M

Questions

Net Force24. A body is at rest. Does this necessarily mean that it has no force acting on it ? Justify your answer.

NO – A body will remain at rest if the NET FORCE acting is zero – it could have any number of forces acting on it. So long as these forces add to zero it will remain at rest.

25. Calculate the net force acting on the object in each of the situations shown.

1200 N 900 N 75 N95 N

250 N

250 N

150 N

450 N

(a) (b)

(c) (d)

300 N Left 20 N Left

0 N 300 N Down

6.4 Forces in 2 Dimensions6.4 Forces in 2 Dimensions

TimTim

TomTom

TamTam

Tim, Tom and Tam, the triplets, Tim, Tom and Tam, the triplets, are fighting over a teddy bear. are fighting over a teddy bear. Each exerts a different force.Each exerts a different force.What will be the net force on the What will be the net force on the bear ?bear ?

FFTIM TIM = 26 N= 26 NFFTOMTOM = 25 N = 25 N

FFTAMTAM = 18 N = 18 N

Add the vectors head to tail. Add the vectors head to tail.

TimTim

TomTom TamTam

FFRESULTANTRESULTANT

The resultant force is the vector joining the The resultant force is the vector joining the starting point to the finishing point starting point to the finishing point

A force diagram A force diagram shows each boy’s shows each boy’s contributioncontribution

The bear will then accelerate in the direction of the resultant forceThe bear will then accelerate in the direction of the resultant force

Forces can act in any direction and the total or resultant force is the Forces can act in any direction and the total or resultant force is the vector sum of all the forces acting.vector sum of all the forces acting.

Questions

Net Force26. Tim, Tom and Tam, the triplets, are fighting over a teddy bear. Each exerts a different force. A force diagram shows each boy’s contribution. What will be the net force on the bear ?

FFTIM TIM = 35= 35 NN

FFTOMTOM = 26 = 26 NN

FFTAMTAM = 12 = 12 NN

Resolve FTIM and FTOM

to give F = 9 N leftThen resolve this force with FTAM

12 N

9 N

Xθ

X = √(92 + 122) = 15 Nθ = sin-1 (9/12) = 48.60

Net force is 15 N directed at S48.6oW

6.5 Weight6.5 WeightWeight is the outcome of a gravitational field acting on a massWeight is the outcome of a gravitational field acting on a massWeight is a FORCE and is measured in Newtons. Weight is a FORCE and is measured in Newtons. Its direction is along the line joining the centres of the two bodies which, Its direction is along the line joining the centres of the two bodies which, between them, generate the Gravitational Field. between them, generate the Gravitational Field.

Near the surface of the Earth, each kilogram of mass is Near the surface of the Earth, each kilogram of mass is attracted toward the centre of the earth by a force of 9.8 N.attracted toward the centre of the earth by a force of 9.8 N.(Of course each kilogram of Earth is also attracted to the (Of course each kilogram of Earth is also attracted to the mass by the same force, Newton 3)mass by the same force, Newton 3)So, the Gravitational Field Strength near the Earth’s So, the Gravitational Field Strength near the Earth’s surface = 9.8 Nkgsurface = 9.8 Nkg-1-1

Weight and mass are NOT the same, but they are related Weight and mass are NOT the same, but they are related through the formula:through the formula:

W = mgW = mg Where: Where: W = Weight (N)W = Weight (N)m = mass (kg)m = mass (kg)g = Grav. Fieldg = Grav. Field Strength (NkgStrength (Nkg-1-1))

1 kg1 kg

1 kg 1 kg

1 kg1 kg

9.8 N9.8 N

Questions

Mass & Weight27. Fill in the blank spaces in the table based on a person whose mass on earth is 56 kg

Planet Mass on planet (kg)

Grav Field Strength (Nkg-

1)

Weight on planet (N)

Earth 56 9.81

Mercury 0.36

Venus 0.88

Jupiter 26.04

Saturn 11.19

Uranus 10.49

56

56

56

56

56

549.4

20.2

49.3

1458.2

626.6

587.4

6.6 Reaction Force6.6 Reaction ForceAll objects on and near the Earth’s All objects on and near the Earth’s surface are subject to the surface are subject to the gravitational force.gravitational force.Any object subject to a net or Any object subject to a net or resultant force will accelerate in the resultant force will accelerate in the direction of that force (Newton 2).direction of that force (Newton 2).Why then do objects placed on a Why then do objects placed on a table on the Earth’s surface remain table on the Earth’s surface remain stationary ?stationary ?

The Reaction Force only exists as a The Reaction Force only exists as a result of the action of the weight of result of the action of the weight of the vase acting on the table top and the vase acting on the table top and as such the reaction force does not as such the reaction force does not exist as an isolated force in its own exist as an isolated force in its own right.right.

Because there is no net or resultant Because there is no net or resultant force on the vase, it remains force on the vase, it remains stationary on the table stationary on the table Remove the table, the reaction force Remove the table, the reaction force disappears and the vase accelerates disappears and the vase accelerates under the action of W, until it under the action of W, until it encounters the floor and probably encounters the floor and probably smashes.smashes.

WW

RR

WW

There must be a force equal in size There must be a force equal in size and opposite in direction to cancel and opposite in direction to cancel out the gravitational force.out the gravitational force.There is such a force. It is called the There is such a force. It is called the REACTION or NORMAL FORCE.REACTION or NORMAL FORCE.

Note: W and R are NOT an action reaction pair. Why?Because when R disappears W does not.

Chapter 7Chapter 7

Centre of MassCentre of Mass

7.0 Centre of Mass7.0 Centre of MassIn order to deal with large objects it is useful to think of all the object’s mass In order to deal with large objects it is useful to think of all the object’s mass being concentrated at one point, this point being the Centre of Mass of the being concentrated at one point, this point being the Centre of Mass of the object.object.

Centre of MassCentre of Mass

For odd shaped For odd shaped objects such as a objects such as a boomerang, the boomerang, the Centre of Mass may Centre of Mass may fall outside the fall outside the perimeter of the perimeter of the object.object.

C of MC of M

For regularly shaped For regularly shaped objects eg. squares or objects eg. squares or rectangles, cubes or rectangles, cubes or spheres the Centre of Mass spheres the Centre of Mass of the object is in the of the object is in the geometric centre of the geometric centre of the objectobject

The C of M is the point around which the object it The C of M is the point around which the object it will spin if a torque or turning force is applied to will spin if a torque or turning force is applied to the object.the object.

7.1 Translation and Rotation7.1 Translation and Rotation

When a Force acts C through the Centre of Mass (of M) of an object or structure, it causes Translational Motion, ie. The object moves in the direction of the applied force according to Newton’s 2nd Law. (see slide 6.3)

A Force acting through the Centre of Mass causes Translational Motion only

When the force is applied to another part of the object or structure, a TORQUE or TWISTING FORCE or TURNING MOMENT is applied and Rotational as well as Translational motion occurs

A Force acting at a point other than the C of M will cause BOTH Translational AND Rotational Motion.

Chapter 8Chapter 8

Newton’s LawsNewton’s Laws

8.0 Aristotle to Newton8.0 Aristotle to Newton

Isaac Newton (1642 – 1727) was Isaac Newton (1642 – 1727) was the first to realise there WAS a the first to realise there WAS a universal set of laws which could universal set of laws which could describe the motion of ALL describe the motion of ALL bodies, BUT these laws had to be bodies, BUT these laws had to be modified for use within the modified for use within the friction riddled confines of the friction riddled confines of the Earth and its atmosphere.Earth and its atmosphere.

Attempts to explain the “causes Attempts to explain the “causes of motion” (a field of study of motion” (a field of study called dynamics), were first called dynamics), were first recorded in the time of the recorded in the time of the ancient Greek philosopher ancient Greek philosopher Aristotle (384 – 322 BC). Aristotle (384 – 322 BC).

However there were problems However there were problems with the theories which could with the theories which could not, for example, explain why not, for example, explain why falling objects tended to falling objects tended to increase their speed in the increase their speed in the absence of any visible force or absence of any visible force or why heavenly bodies behaved why heavenly bodies behaved differently than those on earth.differently than those on earth.

This seemed logical as This seemed logical as everyone could see that a everyone could see that a horse needed to apply a horse needed to apply a constant pull to haul a cart at constant pull to haul a cart at constant speed.constant speed.

It was Galileo (1564 – 1642) who It was Galileo (1564 – 1642) who was the first to define the was the first to define the property of matter we call property of matter we call INERTIA, (matter’s tendency to INERTIA, (matter’s tendency to resist changes in its motion), with resist changes in its motion), with his law which said “when no his law which said “when no force exists a body will stay at force exists a body will stay at rest or move with constant speedrest or move with constant speed

It was believed that constant It was believed that constant speed required a constant force.speed required a constant force.

Philosophers prior to Newton Philosophers prior to Newton believed a set of laws covering believed a set of laws covering motion on earth could be motion on earth could be developed, but they needed to be developed, but they needed to be modified to explain the motions of modified to explain the motions of heavenly bodies.heavenly bodies.Galileo

Newton

Aristotle

Questions

Aristotle to Newton

Scientist Statement

Newton Constant speed requires constant force

Aristotle Defined the property of matter called inertia

Galileo A universal set of laws applicable everywhere but must be modified for use on earth

8.1 Newton’s Laws8.1 Newton’s LawsNewton developed 3 laws which cover all aspects of Newton developed 3 laws which cover all aspects of motion (provided objects travel at speeds are well motion (provided objects travel at speeds are well below the speed of light). below the speed of light).

Law 1 Law 1 (The Law of Inertia)(The Law of Inertia)A body will remain at rest, or in a state of uniform A body will remain at rest, or in a state of uniform

motion, unless acted upon by a net external force.motion, unless acted upon by a net external force.

Law 2Law 2The acceleration of a body is directly proportional to The acceleration of a body is directly proportional to net force applied and inversely proportional to its net force applied and inversely proportional to its

mass. Mathematically, a = F/m more commonly mass. Mathematically, a = F/m more commonly written as written as F = maF = ma

Law 3Law 3 (Action Reaction Law)(Action Reaction Law)For every action there is an equal and For every action there is an equal and

opposite reaction.opposite reaction.

Motion at or near the speed of lightMotion at or near the speed of lightis explained by Albert Einstein’s is explained by Albert Einstein’s Theory of Special Relativity.Theory of Special Relativity.Newton, at age 26

8.2 Newton’s 18.2 Newton’s 1stst Law Law

Newton 1 deals with non Newton 1 deals with non accelerated motion.accelerated motion.It does not distinguish It does not distinguish between the states of between the states of “rest” and “uniform “rest” and “uniform motion” (constant motion” (constant velocity).velocity).As far as the law is As far as the law is concerned these are the concerned these are the same thing (state).same thing (state).

There is no experiment that There is no experiment that can be performed in an can be performed in an isolated windowless room isolated windowless room which can show whether the which can show whether the room is stationary or moving room is stationary or moving at constant velocity.at constant velocity.

Objects want to keep on doing what

they are doing

Newton’s 1Newton’s 1stst Law states: Law states:A body will remain at rest, or A body will remain at rest, or in a state of uniform motion, in a state of uniform motion, unless acted upon by a net unless acted upon by a net external force.external force.

If NO net external force exists

No Net Force means No

Acceleration

It requires an unbalanced force

to change the velocity of an

object

Another way of saying this is:Another way of saying this is:

Most importantly: Most importantly: Force is NOT needed to Force is NOT needed to keep an object in motion keep an object in motion

Is this how you understand the world works ? Is this how you understand the world works ?

8.3 Newton’s 28.3 Newton’s 2ndnd Law LawNewton’s 2Newton’s 2ndnd Law states: Law states:The acceleration of an object as The acceleration of an object as produced by a net force is produced by a net force is directly proportional to the directly proportional to the magnitude of the net force Fmagnitude of the net force FNETNET, , in the same direction as the net in the same direction as the net force, and inversely proportional force, and inversely proportional to the mass of the object.to the mass of the object.Mathematically, a = FMathematically, a = FNETNET/m more /m more commonly written as Fcommonly written as FNETNET = ma = ma

The Net Force on an object equals

the rate of change of its momentum

Using the Using the formula Fformula FNETNET = ma is only = ma is only valid for valid for situations situations where the where the mass remains mass remains constantconstant

Newton actually expressed his 2nd law in terms of momentum.

So, FNET = change in momentum = Δp = mΔv = ma change in time Δt Δt

Momentum (p) = mass x velocity

FFNET NET is the VECTOR SUM is the VECTOR SUM of all the forces acting of all the forces acting on an object.on an object.The acceleration and FThe acceleration and FNETNET are ALWAYS in the same are ALWAYS in the same direction.direction.

Newton 2 deals with accelerated Newton 2 deals with accelerated motion.motion.

8.4 Newton’s 38.4 Newton’s 3rdrd Law LawNewton's 1Newton's 1stst and 2 and 2ndnd Laws tell you Laws tell you what forces do. what forces do. Newton's 3Newton's 3rdrd Law tells you what Law tells you what forces are.forces are.

"opposite""opposite" means that the two forces means that the two forces always act in opposite directions - always act in opposite directions - exactly 180exactly 180oo apart. apart.

This statement is correct, This statement is correct, but terse and confusing. but terse and confusing. You need to understand You need to understand that it means:that it means:"action...reaction""action...reaction" means that means that forces always occur in pairs. forces always occur in pairs. Single, isolated forces never Single, isolated forces never happen.happen.

For example, first Suzie For example, first Suzie annoys Johnnie (action) then annoys Johnnie (action) then Johnny says "Mommy! Johnny says "Mommy! Suzie’s annoying me!" Suzie’s annoying me!" (reaction).(reaction).This isThis is NOT anNOT an example what is going on here! The action and reaction The action and reaction forces exist forces exist at the same time.at the same time.

"action " and "reaction " are unfortunate names for a couple of reasons :

For every action there is an equal

and opposite reaction

1. Either force in an interaction can be the "action" force or the "reaction" force.

"equal" means :Both forces are the same in magnitude.Both forces are the same in magnitude.Both forces exist at exactly the same time. Both forces exist at exactly the same time. They both start at exactly the same instant, They both start at exactly the same instant, and they both stop at exactly the same and they both stop at exactly the same instant.instant. They are equal in time.

2. People associate 2. People associate action/reaction with "first action/reaction with "first an action, then a reaction”an action, then a reaction”

Questions

Newton’s Laws29. At what speeds are Newton’s Laws applicable ?

At speeds way below the speed of light

30. Newton’s First Law:A: Does not distinguish between accelerated motion and constant velocity motionB: Does not distinguish between stationary objects and those moving with constant accelerationC: Does not distinguish between stationary objects and those moving with constant velocityD: None of the above

31. Newton’s Second Law:A: Implies that for a given force, large masses will accelerate faster than small massesB: Implies that for a given force, larger masses will accelerate slower than smaller massesC: Implies that for a given force, the acceleration produced is independent of massD: Implies that for a given force, no acceleration is produced irrespective of the mass.

Newton’s laws

32. Newton’s Third Law:A: Does not distinguish which force of a pair is the “action” force and which is the “reaction” force.B: Implies that both action and reaction forces begin and end at the same instantC: Implies that forces always exist in pairsD: All of the above.

33. Which of Newton’s Laws require that the vector sum of all the forces acting is needed before a calculation of acceleration can be made ?A: Newton’s 1st LawB: Newton’s 2nd LawC: Newton’s 3rd LawD: Newtons 1st and 2nd Laws

Newton’s 2nd Law

34. A car of mass 1250 kg is travelling at a constant speed of 78 kmh-1 (21.7 ms-1). It suffers a constant retarding force (from air resistance, friction etc) of 12,000 N

(a) What is the net force on the car when travelling at its constant speed of 78 kmh-1 ?

At constant velocity, acc = 0 thus ΣF = 0

(b) What driving force is supplied by the car’s engine when travelling at 78 kmh-1 ?

(c) If the car took 14.6 sec to reach 78 kmh-1 from rest , what was its acceleration (assumed constant) ?

At constant velocity ΣF = 0, so driving force = retarding force = 12,000 N

Use eqns of motionu = 0 ms-1 , v = 21.7 ms-1, a = ?, x = ?, t = 14.6 suse v = u + at -> 21.7 = 0 + 14.6(a) -> a = 1.49 ms-2

8.5 The Horse and Cart Problem8.5 The Horse and Cart ProblemIf the horse and cart exert equal and opposite forces on each other, how come the combination can move ?

An explanation hinges on a couple of An explanation hinges on a couple of simple points: (Lets assume no friction)simple points: (Lets assume no friction)1. An object accelerates (or not) because of 1. An object accelerates (or not) because of the forces that push or pull on it. (Newton 2)the forces that push or pull on it. (Newton 2)2. Only the forces that act on an object can 2. Only the forces that act on an object can cancel. Forces that act on different objects cancel. Forces that act on different objects don't cancel - after all, they affect the don't cancel - after all, they affect the motion of different objects! motion of different objects! Why does the cart accelerate?Why does the cart accelerate?

Looking at the cart alone, just one Looking at the cart alone, just one force is exerted on it, (Fforce is exerted on it, (FHCHC) - the ) - the force that the horse exerts on it. force that the horse exerts on it. The cart accelerates because the The cart accelerates because the horse pulls on it! horse pulls on it! The cart’s acceleration equals the The cart’s acceleration equals the net force on it divided by its massnet force on it divided by its mass

FCH FHC

FHR FRH

The cart pulls the horse backwards (FThe cart pulls the horse backwards (FCHCH),), and and the road pushes the horse forward (Fthe road pushes the horse forward (FRHRH).). The net force is the vector sum of these two The net force is the vector sum of these two forces. forces. The horse’s acceleration equals the net force The horse’s acceleration equals the net force on it divided by the its mass. on it divided by the its mass. There are 2 pairs of Newton 3 There are 2 pairs of Newton 3

forces in this situation:forces in this situation:FFHCHC and F and FCHCH

FFHRHR and F and FRHRH

If FNET on the horse is zero, what happens ?The obvious answer is the horse and cart are at rest. BUT, they could also be moving at constant speed ! Newton 1

Why does the horse accelerate?Why does the horse accelerate?There are 2 forces acting on the horse.There are 2 forces acting on the horse.

Questions

Newton’s 3rd Law35. Explain why, if a cart exerts an equal an opposite force on a horse as the horse exerts on the cart, the combination is able to move forward. It is the forces that act on the individual components (i.e., on the horse or cart individually) that decide whether each will move. If the friction of the road on the horse’s feet is larger than the force of the cart on the horse then the horse will accelerate. If the force of the horse on the cart is greater than the frictional force acting on the cart then the cart will accelerate.

36. A car mass 1500 kg is towing a trailer of mass 750 kg. The car/trailer combination accelerate at 3.4 ms-2. The trailer suffers a constant retarding force of 500 N, while the car suffers a constant retarding force of 1000 N.

Calculate the net force acting on the trailer.

Calculate the driving force supplied by the car’s engine.

The net force is that force that provides the acceleration. From Newton 2, ΣF = maΣF = (750)(3.4) = 2550 N (2.55 x 103 N)

The driving force must (i) overcome friction and (ii) provide extra force to accelerate the combination

ΣF = (1000 + 500) + ma = 1500 + (1500 + 750)(3.4) = 1500 + 7650 = 9150 N (9.15 x 103 N)

8.6 Momentum and Impulse8.6 Momentum and ImpulseNewton described Momentum as the “quality of motion”, a measure of the Newton described Momentum as the “quality of motion”, a measure of the ease or difficulty of changing the motion of an object.ease or difficulty of changing the motion of an object.Momentum is a vector quantity having both magnitude and direction.Momentum is a vector quantity having both magnitude and direction.Mathematically,Mathematically,

p = mvp = mvWhere,Where,p = momentum (kgmsp = momentum (kgms-1-1))m = mass (kg)m = mass (kg)v = velocity (msv = velocity (ms-1-1))

In order to change the momentum of an object a mechanism for that change is In order to change the momentum of an object a mechanism for that change is required. required. This mechanism of change is called Impulse.This mechanism of change is called Impulse.Mathematically,Mathematically,

I = FtI = FtWhere,Where,I = Impulse (N.s)I = Impulse (N.s)F = Force (N)F = Force (N)t = Time (s)t = Time (s)

The relationship between momentum and impulse can be derived from The relationship between momentum and impulse can be derived from Newton’s 2Newton’s 2ndnd Law: Law:

F = ma and a = v/t, so F = mv/t F = ma and a = v/t, so F = mv/t Rearranging we get:Rearranging we get:

Ft = mvFt = mvie. Impulse = Momentumie. Impulse = Momentum

8.7 Conservation of Momentum8.7 Conservation of MomentumThe concept of Momentum is particularly useful in analysing collisions.The concept of Momentum is particularly useful in analysing collisions.This is because of the Law of Conservation of Momentum which states:This is because of the Law of Conservation of Momentum which states:

IN AN ISOLATED SYSTEM, TOTAL MOMENTUM IS CONSERVED.IN AN ISOLATED SYSTEM, TOTAL MOMENTUM IS CONSERVED.The term “isolated system” means no external forces are acting in the situation The term “isolated system” means no external forces are acting in the situation under investigation.under investigation.In a crash situation, where the vehicle comes to a In a crash situation, where the vehicle comes to a halt after, say, hitting a tree, halt after, say, hitting a tree, both its velocity and momentum fall to zero. both its velocity and momentum fall to zero. The apparently “lost” momentum, has, in fact, been The apparently “lost” momentum, has, in fact, been transferred via the tree to the Earth. transferred via the tree to the Earth. Since the Earth has a huge mass (6 x 10Since the Earth has a huge mass (6 x 102424 kg). The kg). The change in its velocity is so small as to be negligible. change in its velocity is so small as to be negligible. In the crash mentioned, the momentum change is a In the crash mentioned, the momentum change is a fixed quantity so the Impulse (the product of F and t) is fixed quantity so the Impulse (the product of F and t) is also a fixed quantity.also a fixed quantity.However the individual values of F and t can vary as long as the multiply to However the individual values of F and t can vary as long as the multiply to give that fixed value.give that fixed value.If t, the time during which the crash occurs, can be lengthened, then the force If t, the time during which the crash occurs, can be lengthened, then the force which needs to be absorbed by the car and its occupants is reduced.which needs to be absorbed by the car and its occupants is reduced.Modern vehicles use this concept in crumple zones and air bags as both are Modern vehicles use this concept in crumple zones and air bags as both are designed to extend the time and so reduce the force.designed to extend the time and so reduce the force. Questions

Momentum37. A car (and its occupants) is of total mass of 2250 kg and is travelling at 50

kmh-1 . Approaching, head on, is a motorcycle (and rider) of total mass 350kg travelling at 180 kmh-1

(a) Which vehicle (car or bike) has the greater momentum ?

(b) They collide head on and stick together. What velocity will the “wreck” have immediately after collision ?

Firstly need to convert speeds to ms-1 50 kmh-1 = 13.9 ms-1 : 180 kmh-1 = 50.0 ms-1

p4wd = mv = (2250)(13.9) = 31275 kgms-1 pcycle = mv = (350)(50) = 17500 kgms-1 So 4WD has the greater momentum.

Assume 4WD is travelling to the right and motorcycle to the left.So Σp = p4wd - pcycle = 31275 – 17500 = 13775 kgms-1 to the right.If the vehicles stick together total mass = 2250 + 350 = 2600 kgSo Σp = mtotalv v = Σp/mtotal = 13775/2600 = 5.3 ms-1 to the right

Momentum and Impulse

38. While talking on a mobile phone a truck driver loses concentration and runs off the road and hits a tree. His speed goes from 20 ms-1 to 0 ms-1 in 0.7 sec. His truck has a mass of 42 tonnes (1 tonne = 1000 kg)

(a) Calculate his change in momentum

(b) Calculate the Impulse during the collision

(c) Calculate the force he will experience during the collision

Change in momentum = final mom – initial mom = mvfinal - mvinitial = m (vfinal – vinitial) = 4.2 x 104(0 – 20)= - 8.4 x 105 kgms-1 (negative sign can be omitted from answer as it only indicates direction of momentum change)

Since change in momentum = impulse. Impulse = 8.4 x 105 Ns

I = Ft F = I/t = (8.4 x 105)/0.7 = 1.2 x 106 N

Airbags/Crumple Zones39. Explain why, in a modern car equipped with seat belts and an air bag , he would likely survive the collision whereas in the past, with no such safety devices, he would most likely have been killed.

The change in momentum in any collision is a fixed value thus impulse is also fixed, but the individual values of F and t can vary as long as their product is the that fixed value. In modern vehicles seat belts and crumple zones are designed to increase to time it takes to stop thus necessarily reducing the force needed to be absorbed by the driver because Impulse = Ft. This reduced force will lead to reduced injuries. In the old days the driver would have been “stopped” be some hard object like a metal dashboard and his time to stop would have been much shorter and thus the force experienced would have been larger leading to more severe injury and likely death.

Chapter 9Chapter 9

Work, Energy & PowerWork, Energy & Power

9.0 Work9.0 WorkIn Physics, the term WORK is very In Physics, the term WORK is very strictly defined.strictly defined.When a When a forceforce moves an object through moves an object through a a distance,distance, workwork has been done. has been done.

Mathematically:Mathematically:W = F x dW = F x d

Where,Where,W = Work (Joules)W = Work (Joules)F = Force (N)F = Force (N)d = distance (m)d = distance (m)

If a force is applied and the object If a force is applied and the object does notdoes not move, NO WORK has been done.move, NO WORK has been done.

If the force applied is constant, If the force applied is constant, the work done can be calculated the work done can be calculated from the formula, W = F x dfrom the formula, W = F x dBut, if the force varies during the But, if the force varies during the course of doing the work, as in course of doing the work, as in compressing a spring, the work compressing a spring, the work must be calculated from the area must be calculated from the area under the force versus distance under the force versus distance graphgraph

ForceForce

DistanceDistance

Area = Work doneArea = Work done

Work is a SCALAR Work is a SCALAR quantity, meaning it quantity, meaning it has a magnitude but has a magnitude but no direction.no direction.

Questions

Work40. Calculate the work done on a refrigerator when a net force of 125 N acts over a distance of 4.5 m

Work = Fxd = (125) x (4.5) = 562.5 J (5.63 x 102 J)

41. The graph shows the force required to compress a spring

Distance Distance (cm)(cm)

Force Force (kN)(kN)

1.1.00

2.2.00

3.3.00

4.4.00

15015000

30030000

45045000

60060000

(a) Calculate the work done in compressing the spring by 3.0 cm.

(b) Calculate the further work required to compress the spring from 3.0 cm to 4.0 cm

Work done = area under graph = ½ base x height = ½ (3.0 x 10-2)(4.5 x 106) = 1.35 x 105 J

Further work to compress from 3.0 cm to 4.0 cm = area under graph between these two distances= area of trapesium = ½ (height(1) + height(2)) x base = ½ (4.5 x 106 + 6.0 x 106 )(1.0 x 10-2)= 5.25 x 104 J

9.1 Work and Energy9.1 Work and EnergyThe concept of WORK was The concept of WORK was developed BY PHYSICISTS as a developed BY PHYSICISTS as a means of quantifying and means of quantifying and measuring ENERGY.measuring ENERGY.

It is very easy to say what energy can do, but It is very easy to say what energy can do, but very difficult to define exactly what energy is. very difficult to define exactly what energy is.

The relation between work and The relation between work and energy is summarised by one energy is summarised by one simple but powerful statement:simple but powerful statement:

WORK DONE = ENERGY TRANSFERRED

If work has been done on an object, the amount of If work has been done on an object, the amount of energy it has MUST have increased.energy it has MUST have increased.By how much ?By how much ?By exactly the amount of work done on the object.By exactly the amount of work done on the object.

If an object has done some work, the amount of If an object has done some work, the amount of energy it has MUST have decreased.energy it has MUST have decreased.By how much ?By how much ?By exactly the amount of work done by the object.By exactly the amount of work done by the object. Questions

Work & Energy42. How much energy is stored in the spring in question 41 when it has been compressed by 2.0 cm

Work Done = ½ base x height = ½ (2.0 x 10-2)(3.0 x 106) = 3.0 x 104 J Since Work Done = Energy Transferred, the energy stored in the spring = 3.0 x 104 J

9.2 Kinetic Energy 9.2 Kinetic Energy Kinetic Energy is the energy possessed by moving objects.Kinetic Energy is the energy possessed by moving objects.It is called the “Energy Of Motion”.It is called the “Energy Of Motion”.Kinetic Energy is a SCALER quantity.Kinetic Energy is a SCALER quantity.

Mathematically:Mathematically:K.E. = ½mvK.E. = ½mv22

Where:Where:K.E. = Kinetic Energy (Joule)K.E. = Kinetic Energy (Joule) m = mass (kg)m = mass (kg) v = speed (msv = speed (ms-1-1))

Horse has K.E. Horse has K.E. due to its movementdue to its movement

Arrow has K.E. Arrow has K.E. due to its motiondue to its motion

Gears have K.E. Gears have K.E. due to their rotationdue to their rotation

9.3 Gravitational Potential 9.3 Gravitational Potential Energy Energy

Gravitational Potential Energy, often just called Potential Energy, is the Gravitational Potential Energy, often just called Potential Energy, is the energy possessed by an object due to its position.energy possessed by an object due to its position.It is called the “Energy of Position”It is called the “Energy of Position”Potential Energy is a SCALAR quantity.Potential Energy is a SCALAR quantity.Mathematically:Mathematically:

P.E. = mghP.E. = mgh

Where:Where:P.E. = Potential Energy (Joules)P.E. = Potential Energy (Joules) m = mass (kg)m = mass (kg) g = Grav. Field Strength (Nkgg = Grav. Field Strength (Nkg-1-1)) h = height (m)h = height (m)

Potential Energy needs a zero point for the measurement of the height, h.Potential Energy needs a zero point for the measurement of the height, h.The zero point is usually, but not always, the surface of the Earth.The zero point is usually, but not always, the surface of the Earth.The zero point needs to be known for the calculation to have meaning.The zero point needs to be known for the calculation to have meaning.

The man has P.E. due to hisThe man has P.E. due to his height above the groundheight above the ground Skier has P.E. while he is on the Skier has P.E. while he is on the

slope + KE due to his speedslope + KE due to his speed Questions

Kinetic & Potential Energy

43. A cyclist is riding her bike along a flat road. She and her bike have a mass of 105 kg. she is travelling at a constant speed of 15 ms-1.

(a) Calculate her Kinetic Energy

She accidentally rides over a 15 m cliff. (b) What is her potential energy at the top of the cliff ? (take g = 10 ms-2)

KE = ½ mv2 = ½ (105)(15)2 = 1.18 x 104 J

(c) If all the PE she had at the top of the cliff is converted to KE at the bottom, calculate her vertical speed just before she hits the ground.

PE = mgh = (105)(10)(15) = 1.58 x 104 J

PETOP = KEBOTTOM 1.58 x 104 = ½ (105) v2 v = 17.3 ms-1

9.4 Hooke’s Law9.4 Hooke’s LawDeveloped by English scientist Robert Hooke in 1676, the law states that the Restoring Force in an elastic material is directly proportional to its extension.

Mathematically:F = - kx

Where:F = Restoring Force (N)k = Spring Constant (Nm-1)x = Extension (m)

The spring constant (k) is a measure of the nature or quality of the elastic material. The higher its value the greater is the restoring force for a given extension.

The negative sign in the equation indicates that the restoring force and the extension are in opposite directions.

Questions

Hooke’s Law44. A spring of length 100 cm and spring constant 2.5 x 102 Nm-1 hangs vertically from a retort stand. A total mass of 15.6 kg is hung from the spring. Calculate the extent of the spring’s extension under this load. (Take g = 10 Nkg-1)

F = -kx: Force produced by hanging mass = mg = (15.6) (10) = 156 N. So Restoring Force = 156 N x = -F/k = -156/(2.5 x 102) = -0.62 m (could do this calculation without negative sign) by using applied force in Hooke’s Law equation)

9.5 Energy Transfers9.5 Energy TransfersENERGY CANNOT BE CREATED OR DESTROYED BUT ONLY ENERGY CANNOT BE CREATED OR DESTROYED BUT ONLY

TRANSFERRED FROM ONE FORM TO ANOTHER.TRANSFERRED FROM ONE FORM TO ANOTHER.When doing problems concerning energy and energy transfers, it is assumedWhen doing problems concerning energy and energy transfers, it is assumed that the transfers are 100% efficient, meaning no energy losses occur.that the transfers are 100% efficient, meaning no energy losses occur.

For instance, a roller coaster will have a large For instance, a roller coaster will have a large Gravitational Potential Energy component at its Gravitational Potential Energy component at its highest point most of which will have been converted highest point most of which will have been converted to Kinetic Energy at its lowest point and in to Kinetic Energy at its lowest point and in calculations a direct equality can be made between calculations a direct equality can be made between P.E. at the top and K.E at the bottomP.E. at the top and K.E at the bottomIn real life, these types of conversion processes are In real life, these types of conversion processes are never 100% efficient. never 100% efficient. Friction and the production of heat and sound mean Friction and the production of heat and sound mean significant losses occur.significant losses occur.

The efficiency of energy transfer processes can be calculated from: The efficiency of energy transfer processes can be calculated from: % Eff = % Eff = Energy OutEnergy Out x x 100100 Energy In 1Energy In 1

The Law of Conservation of Energy says:The Law of Conservation of Energy says:

9.6 Elastic Potential Energy9.6 Elastic Potential EnergyEnergy stored in springs is called elastic potential energy

• Elastic materials store energy when Elastic materials store energy when they are deformed and release that they are deformed and release that energy when they return to their energy when they return to their original conditionoriginal condition..

• The amount of energy stored can be The amount of energy stored can be found from the Elastic Potential found from the Elastic Potential Energy Formula: Energy Formula: EESS = = ½½kxkx22 where where EES S = Elastic P. E. (J ) = Elastic P. E. (J ) k = Spring Constant (Nm k = Spring Constant (Nm-1-1) ) x = extension and or x = extension and or compression (m) compression (m)

Area = Area = ½½Fx = Fx = ½½kxkx22

= Energy stored = Energy stored up to extensionup to extension xx

ForceForce

ExtensionExtension

Slope = Spring Slope = Spring constant (k)constant (k)

xx

FF

““REGULAR” ELASTIC BEHAVIOURREGULAR” ELASTIC BEHAVIOUR

The stored energy can be used to increase the kinetic energy of the “arrows” such as used in cross bows

Questions

Elastic Potential Energy

45. Calculate the amount of elastic potential energy stored in the extended spring in question 44.

EPE = ½ kx2 = ½ (2.5 x 102) (0.62)2 = 48.1 J

9.7 Energy Transfers & Heat9.7 Energy Transfers & Heat

The end result of the action of these frictional effects is The end result of the action of these frictional effects is the production of what is called “low grade, the production of what is called “low grade, unrecoverable heat”unrecoverable heat”

Large sources of this kind of heat are the Large sources of this kind of heat are the exhausts from fossil fuelled transport such exhausts from fossil fuelled transport such as cars, trucks and trains and from the as cars, trucks and trains and from the electricity generation industry.electricity generation industry.

This means the heat energy cannot This means the heat energy cannot be harnessed to do any useful be harnessed to do any useful work, such as boil water or run a work, such as boil water or run a pump.pump.

In our friction riddled world, no energy transfer process is 100% efficient.In our friction riddled world, no energy transfer process is 100% efficient.

9.8 Power9.8 PowerPower is the time rate of doing work, and since Power is the time rate of doing work, and since work and energy are equivalent is also the time work and energy are equivalent is also the time rate of energy transfer.rate of energy transfer.

Mathematically:Mathematically:P = W/t = E/tP = W/t = E/t

Where:Where: P = Power (W, Watts)P = Power (W, Watts)W = Work (J)W = Work (J) E = Energy (J)E = Energy (J) t = time (s) t = time (s)

SinceSinceP = W/t and W = F.d, we can say P = W/t and W = F.d, we can say P = F.d/t but d/t = v so P = F.d/t but d/t = v so P = F.v P = F.v

So, the power for a body moving at So, the power for a body moving at constant velocity can be found in a one constant velocity can be found in a one step calculation.step calculation.

Questions

Power46. A train of mass 1.5 x 104 kg is travelling at a constant speed of 70 kmh-1. If the engine is providing a driving force of 3.0 x 107 N , at what power is the engine operating ?70 kmh-1 = 19.4 ms-1 Power = F.v = (3.0 x 107)(19.4) = 5.82 x 108 W

A

BC D

E

F

47. A roller coaster moves through its journey from A to F. The coaster has no motor and is not powered after it leaves point A. Its total mass is 650 kg. The heights above ground of each portion of the track are given below. Positi

on

Height above

Ground (m)A 25.0B 5.0C 12.5D 12.5E 15.0F 7.5

Energy Transformations(a) Between which points is (i) the Kinetic Energy increasing, (ii) the Potential

Energy falling, (iii) the coaster accelerating

(d) Assuming no frictional losses, calculate the Kinetic Energy at point F

(i) AB, EF (ii) BC, DE (iii) All except CD(b) At which point is the force exerted on the coaster by the track at its greatest ?Calculate the Gravitational Potential Energy at Point A. (take g = 10 ms-2)

B (At this point the track must supply the reaction force to counteract the weight of the coaster AND the centripetal force to make the coaster go round the ‘corner”

PE = mgh = (650)(10)(25) = 1.63 x 105 J

(c) Calculate the Gravitational Potential Energy at Point A. (take g = 10 ms-2)

(e) Calculate the speed of the coaster at point F

Loss in PE between A and F = gain in KE between A and F PELOSS = mgh1 – mgh2 = (650)(10)(25 – 7.5) = 1.14 x 105 J

At point F, KE = 1.14 x 105 = ½ mv2 v = 18.7 ms-1

Ollie Leitl 2008