Marine Institute School of Maritime Studies

PHYSICS 1100 (Student Notes and Course Package)

(Instructor: Dave Woolridge)

The Fisheries and Marine Institute Of

Memorial University of Newfoundland

Revised June 21, 2005

Physics 1100 Formulas

)x = xfinal - xinitial And )t = tfinal - tinitial

speed = total distance / total time

v =)x /)t

vf = v0 + a)t

vf2 = v0

2 + 2a)x

)x = v0)t + ½ a)t2

g = -9.81 m/s2

v = (vf + vi) / 2

Fnet = ma = EF = EFwith - EFagainst

Fk = :kFN And Fs = :sFN

Fc = mv2 / r = EFin - EFout

Fg = mg = G(m1m2) / r2

G = 6.67 x 10-11 (N*m2)/kg2

KE = ½ mv2

PEg = mgh

Wnc =(F*d)cos2

Wnet = )KE =EWnc

KEi +PEgi + Wnc = KEf +PEgf

P = W/)t = (F*d) / )t = F*v

p = mv

Fnet )t = )p

J = (F×d )sin2 = F×l

Table of Contents

1. Introduction 1

2. Constant Velocity 9

3. Constant Acceleration 17

4. Vector Addition 2-D 27

5. 2-D Motion 31a. Projectile Motion 31b. Relative Velocity 34

6. Introduction to Force 39

7. Newton’s Laws of Motion 51

8. Systems of Bodies 55

9. Circular Motion and Universal Gravitation 57

10. Work and Energy 61

11. Momentum 77

12. Torque 83

13. Appendix A: Metric details 87

14. Appendix B: Student Evaluation and tips 88

15. Appendix C: Text readings and practice problems 90

16. Appendix D: Practice Labs 92a. Sheet 1 92b. Sheet 2 95c. Sheet 3 98d. Sample Final Exam 102

1

Physics 1100Instructor Dave WoolridgeCorridor # W2045 Office A

Physics Lab E1303Office Phone 778 0405

E-mail Dave.Woolridge@beacon.mi.mun.caWeb http://www.mi.mun.ca/~dwoolrid/

Lab Demonstrator Ann HarrisOffice Phone 778 0638

2

P Physics is the study of< the interactions between energy and matter.

< matter, interactions and change

What is Physics?General definition

3

The Branches of Physics.

Quantum MechanicsParticleRelativity

Modern

ThermodynamicsElectrcityWavesOptics

DynamicsKinematics

Mechanics

Classical

Physics

4

P Physics is the fundamental science.

< Chemistry – depends on the interactions of matter which depend on

the exchange of energy.

< Biology – dependant on chemistry, therefore on physics

< Earth science – based directly on chemistry and physics.

Physics and Other Sciences

5

P Facts< Discrete bits of knowledge that many experts

deem to be true.< Example:

P Models < simplified representations of reality including

analogies, pictures and equations used totranslate theories.

< Example:

Types of Knowledge

6

P Theories < the broad notions we have about the physical

world which can encompass many models< Example:

P Laws < simple, universal rules.< theories do not mature into laws over time< Example:

Types of Knowledge

7

P Principles< laws in the making, their validity has not been fully

tested.< Example:

Types of Knowledge

8

P A physical quantity is anything in the realworld that we can measure on an “objectivescale”.

Physical QuantitiesDefinition

9

P System International (SI)< A.K.A. as the metric system or metre, kilogram,

second (MKS) system< Base ten< Details on prefixs in Appendix A

P English or Imperial, mainly used in the US

Measuring Scales forPhysical Quantities

10

P Base< One of seven stand alone physical quantity

– Mass (kg), length (m), time (s), amount (mol), luminousintensity (lux), charge ( C) and temperature (K)

P Derived< Any quantity that is the result of the combination

of two or more base quantities– Force (kg*m/s2), velocity (m/s), electric current ( C/s),

etc.

Physical QuantitiesThe Principle Classification System

11

P Mass< A universally conserved property of matter that is

roughly the size of matter or resistance to changein motion.

P Time< A measure of temporial space or a progression

(sequencing) of occurances (events)

P Length< A measure of physical space

Important Base Quantitiesfor Mechanics

12

P Scalar< Some quantity with magnatude (size) alone.

– Example:

P Vector< Some quantity with magnatude and direction.

– Example:

Physical QuantitiesA Secondary Classification System

13

P We can use an arrow as a “model” of anyvector quantity.< The length of the arrow represents the size or

magnitude of the quantity.< The arrow head points in the direction the quantity

points.

The Vector Model

14

P Let vector ‘A’, ‘B’ and ‘C’ be vectors ofrepresenting the physical quantity of force.< Which force is “strongest” and how do we know?< Compare the “strengths” of all three forces.< In 1 dimension, we can use ‘+’ and ‘-’ to show the

only two directions.

The Vector Model1 dimension

Scale: 1 cm = ___ N

B = ___ N C = ___ N

A = ___ N

15

P When < two or more vectors are added together, the

answer is often called the “resultant vector” or just“resultant”.– You can only add vector of the same type or quantity.

< a vector is multiplied with a scalar, the answer is avector quantity that points in the same direction asthe original vector.– The quantity (type of vector) may or may not be changed

as a result.

Vector Algebra

16

P When < two or more vectors are multiplied or divided the

answer is sometimes – a vector quantity at 90E to the plane formed by the

original vectors. The quantity (type of vector) is changedas a result. (Cross Product)

– a scalar quantity. The quantity is changed as a result.(Dot product)

– Unlike quantities may be multiplied together.

Vector Algebra

17

P When vectors of the same quantity (type) liein the same line, then we may add themtogether using simple math to get a new,“resultant” vector.< What is the the resultant of A+B = R? The

answer should also indicate the direction.

Vector Addition Part 11 dimension

R = ___ N

Scale: 1 cm = ___ NA = ___ N

B = ___ N

18

P When vectors of the same quantity (type) liein the same line but opposite directions, westill add them using simple math butremember the signs are different.< What is the the resultant of C+B = R? The

answer should also indicate the direction.

Vector Addition Part 11 dimension

Scale: 1 cm = ___ NB = ___ N

C = ___ NR = ___ N

19

P Subtracting vectors is very similar to addingthem, lets examine the A-B = R< First the ‘-’ before the B, tells us to point B in the

opposite direction.< Then we add this “new B” to the A, as we did

before, A+(-B) = R.

Vector Subtraction Part 11 dimension

Scale: 1 cm = ___ N

B = ___ N

A = ___ N

-B = ___ N R = ___ N

20

P Significant figures< The number of ‘certain’ digits in a measurement< The last significant figure is the first estimated

digit.– Measuring scales– Estimate to the nearest 10th of a division.

P Insignificant figures< Zeros left of the first non-zero digit < Numbers right of the first estimated digit

Significant FiguresMeasurement errors

21

P Zeros left of the first nonzero digit areinsignificant< 0.0001209 Four sig. figs.

P Zeros right of the last nonzero digit areinsignificant, if there is no decimal point.< 209000 Three sig. figs.

P Zeros right of the decimal point aresignificant up to the first estimated digit< 209000.0 Seven sig. figs.

Significant Figures

22

Significant Figure RulesAddition and subtraction

P The answer has the samenumber of decimal places as theoperand with the fewest decimalplaces

P Determine the perimeter.

P A = 3.189 cmP B = 4.56 cmP C = 5.564469 cm B

CA

23

Significant Figure RulesMultiplication and division

P The answer has the same number of significant figures asthe operand with the fewest signficicant figures.

P Estimate the area.

P A = 3.189 cmP B = 4.56 cmP C = 5.564469 cm

B

CA

24

P The last significant figure rounds up if thenext digit is 6 or greater.

P Odd numbers round up if followed by 5.P Even numbers do not round if followed by 5.

Rounding

Target Quantity = Conversion Factor * Source QuantitiyT.Q. = C.F. * S.Q.

source to target

miles to kilometers

1 Target UnitConversion Factor (CF ) = Equivilant Source Units

1.0CF 0.6215

x

kmmile

=

T .Q . = C .F . * S .Q .

C o n v e r t t h e f o l lo w i n g( 1 ) 4 5 m g t o k g

1 k gT .Q . ( k g ) = * 4 5 m g = 0 .0 4 5 k g1 0 0 0 m g

( 2 ) 6 8 5 m t o m i le s1 m i l eT .Q . ( m i le s ) = * 6 8 5 m = 0 .4 2 6 m i le

1 6 0 9 m

25

P To convert within and between systems ofmeasure, we multiply a source quantity withan appropriate “conversion factor”. Theresult will be the target quantity.

Unit ConversionsHow to

26

P A conversion factor is a fracton of 1 unit ofthe target quantity over the equivilant amountof units in the source quantity.

Unit ConversionsCreating Conversion Factors

27

Unit ConversionsExamples

28

P Convert the following quantities< 78 days, 3 hours and 23 minutes to seconds < 100 k/h to m/s< 100 k/h to mph

Unit ConversionsSelf Practice

29

P Examine one approach that scientists use tocreate new knowledge.

< Modelling

Scientific ProcessObjective

30

P Find a situation that needs explaining.< Ask “how (why) does ...”or “what if …” questions.

P Develop a list of factors (variables)< that may relate to the situation being studied.

P Eliminate factors < cannot be measured < do not logically link to the situation

Situational Analysis Start with a problem

Fuel (l)Dist (km)23.8179.728.5221.032.0246.042.2290.436.3303.749.5357.347.8359.0

Fuel (l)Dist (km)23.8179.728.5221.032.0246.042.2290.436.3303.749.5357.347.8359.0

31

P Independent variable (x)< a factor, with a value set and altered by the you

P Dependent variable (y)< a factor monitored for variations in value, caused

by changing the independent variable

P Control variables < potential independent variables that have fixed

values during an experiment

Factors to Variables

32

P I tested my van’s fuel consumption. I tookseveral highway trips with the engine at anear constant 3600 rpm and measured thedistance travelled and the amount of gas Iput in the tank at each fill up. Ensuring that Ialways filled the tank the same way.

P List potential factors and their variable type.

Sample Experiment#1

33

Tabulate data (x) on the left (y) on the right

Data Collection

34

P Data Analysis< plotting graphs and determining critical values

P Data Interpretation< writing specific and general physical equations< Interpreting results

Model MakingThe real work begins.

35

Plot the points

Data AnalysisGraphing

36

Draw a “curve” thattouches the mostdata points, leavingmissed points evenlydivided by andequidistant from theline. Do notconnect the dots!

Data AnalysisLine of best fit

2 1

2 1

rise y y ysloperun x x x

∆ −= = =

∆ −

y mx b= +

37

Not always a straightline

Does not have topass through theorigin

Data AnalysisLine of best fit

38

Determine slope and y-intercept

Data AnalysisGraphing Software

Built in featuresquickly determineand display the slopeand y-intercept.

Slope = 0.1325 l/kmy-int = 0.0 l

39

P A Greek letter )< symbolizes a change in some quantity.< )d = (d - d0 ) = (dfinal - dinitial ) = (d2 - d1 )

P ‘Xn’ denotes the nth instantaneous value ofvariable X

Delta ()) Notation

40

Data AnalysisManual calculation

Using the coordinates ofany two points on yourline (not collected datapoints) compute theslope with the rise overrun method.

Reads the y-interceptdirectly off the graph.

y1

y2

x2x1

41

P Determine the base mathematical model

P Translate mathematical model into a specificphysical equation. (SPE)< Only good for your data

P Translate SPE into a general physicalequation. (GPE)< Good for anyone conducting similar tests

Data Interpretation3 steps to a model

42

P Identify the trend shown in the line of best fit.– linear thus it follows

P Where< Y is dependent variable< X is independent variable< M is slope of the line< B is y-intercept

Step 1Determine base the mathematical model

(0.1325) 0.0y x= +

(0.1325) 0.0F d= +

0FF d Fd

∆= +∆

2 1

2 1

rise y y ysloperun x x x

yy mx b x bx

∆ −= = =

∆ −∆

= + = +∆

43

P Replace the m and b with the experimentalvalues of the slope and y-intercept.

P Replace the y and x terms with the symbolsfor the appropriate variables.

P This is the SPE

Step 2The Specific Physical Equation (SPE)

44

P The GPE is simply the SPE with the slopeand y-intercept converted to symbols.

P Where < )F / )d is the rate of change in fuel per change in

distance travelled< F0 is the initial fuel consumed, when distance

travelled is zero km

Step 3Convert SPE to General Physical Equation (GPE)

45

P Once equations are made one has to< Determine the physical situations where it does

and does not apply.< Judge their confidence in the model outside the

data range.< Judge their confidence in the model between data

points.

Model Ramifications

46

P The final step is to summarize the model.< Note thesituation where it can and cannot be used< Note the shapes of graphs and their meaningful

features such as slopes, y-intercepts, areas, etc.< Note the general physical equation and

understand the meanings of all its terms.

P This summary contains all the tools forproblem solving.

Conclusions

47

Summary of Equations

f ix x x∆ = − f it t t∆ = −

1

Kinematics 01Constant Speed and Velocity

2

PDefine basic concepts of motion< reference systems, distance, displacement, time,

interval, speed and velocityPClassify these concepts as either vector or

scalarPDevelop and apply a simple model of motionPSolve problems by graphical & analytical

means

Objectives

3

PMechanics is the study of motion

PKinematics< The study of HOW things move.

– Constant speed– Constantly changing speed

PDynamics< The study of WHY things move.

– Force, Momentum, Energy

Classical MechanicsWhat it is.

4

PFrame of reference / coordinate system< An arbitrary rigid system by which position is

measured– You choose the zero mark and positive direction then

stick with it.– Measuring tapes

PPosition (xn)< A location on the frame of reference relative to an

arbitrary zero point< A vector quantity measured in metres

Concepts of KinematicsDefinitions

5

PDistance (d)< The length of a path travelled< A scalar quantity measured in metres

PDisplacement ()x)< The straight line distance between start and end

position, i.e., the change in position< A vector quantity measured in metres

Concepts of KinematicsDefinitions

)xd

6

PClock time (tn)< The current value of time, read from a clock< A scalar quantity measured in seconds

P Interval ()t)< The difference between successive clock times,

i.e., time elasped< A scalar quantity measured in seconds

Concepts of KinematicsDefinitions

distance distancespeed

elasped time interval= =

displacementvelocity = interval

avgx vt

∆= =∆

7

P In your own words answer these questions.< What does means to drive at a steady / constant

100 kilometers per hour?

< Is such cruising on the highway truly constantspeed motion?

Constant SpeedInitial thoughts

8

PConstant speed (v)< Equal distances (m) travelled over equal interval

(s). < A scalar quantity measured in m/s

Constant SpeedDefinition

9

PConstant average velocity (vavg.)< Equal displacement (m) travelled over equal

interval (s).< Equal changes in position with equal changes in

clock time.< A vector quantity measured in m/s

Constant Velocity

10

PWhen we talk about constant speed orvelocity, we really mean to say “constantaverage speed” or “constant averagevelocity”.

< These values do not necessarily reflect truecurrent (instantaneous) speed or velocity

< The word average does not suggest mathematicalaverage, i.e., ‘mean’ value

WARNING!

11

PUsing a constant velocity toy, we willexamine the relationship between. < Position and time< Velocity and time

P Independent variable (x-axis)< Time

PDependent variable (y-axis)< Velocity and position

Kinematics LabPart 1

CV 2005

12

Using these graphs and worksheet 1, learnabout the graphical representation of constantvelocity and develop the first equation ofmotion.

Graphical Analysis

Constant Velocity Motion Graphical Analysis

Worksheet 1: Constant velocity graphical analysis.

1. SKETCH the line of best fit on the position-time and velocity-time graphs. Use the lineof best fit to answer the following questions.

Position versus time graph

2. What is the y-intercept of the position-time graph and what does it tell us about thetractor’s motion?

3. What is the tractor’s position at....?a. 0.6 seconds

b. 2.8 seconds

4. Calculate the slope of the position-time graph. What does the slope tell us about thetractor’s motion?

5. Write the specific and general physical equations for this position-time graph.

SPE : GPE :

Velocity versus time graph

6. What is the y-intercept of the velocity-time graph and what does it tell us about thetractor’s motion?

7. What is the tractor’s velocity at....?a. 0.6 seconds

b. 2.8 seconds

8. Calculate the slope of the velocity-time graph. What does the slope tell us about thetractor’s motion?

9. Calculate the area under the velocity-time graph form 0.6 to 2.8 seconds. What does thisarea tell us about the tractor’s motion? How can we verify this with the position-timegraph?

0

d i s ta n c e t ra v e lle ds p e e dt im e

f i

f i

f

x x xt t t

dt

x x v tx v t

∆ = −∆ = −

= =∆

= + ∆

∆ = ∆

13

PGenerally < ‘+’

– Vectors that point to the right, East, vertically up or North

< ‘-’– Vectors that point to the left, West, vertically down or

South

PCan be arbitrarily assigned.

Sign Convention

14

PConstant Velocity / Uniform Linear Velocity< Used ONLY when

– the displacement is uniform with time.– velocity or speed appears constant.

PWe are not certain about the model’saccuracy < between the data points < beyond the data points

Limits of the Constant VelocityModel

15

Equation Summary

Do Examples 1 - 3

Constant Speed and Constant Velocity

1. A car leaves St. John’s headed for Gander, a total distance of 320 km andreaches Clarenville, a distance of 188 km, in 1 hour and 42 minutes.

a. Assuming the speed remains constant how long, in hours, does the wholetrip form St. John’ to Gander take?

2. An air force reconnaissance drone flies North at 56.0 km/h for 40 minutes, thenSouth at 42.3 km/h for a distance of 30.0 km.

a. Where is its final position, relative to its start point?b. How many minutes does the drone fly?c. What is its average speed in km/h and m/s?d. What is its average velocity in km/h and m/s?e. Sketch the x-t and v-t graphs for this trip.

3. Answer the following based on the given x-t graph.

a. Determine the average velocity(s) of objects A, B and C for each phase ofmotion they undergo. Note the interval for the phases.

b. Determine the initial positions of each object.c. Do any of the objects share the same speed, if so when?d. Do any of the objects share the same velocity, if so when?e. Generally what to the interceptions of the lines at tell us about the objects?f. Specifically what is happening at interception points 1, 2, 3, 4 and 5?g. What is significant about point ‘z’ for car A?h. Sketch the corresponding velocity time graph.

a c c e l e r a t i o n = v at

∆=

∆

1

Kinematics 02Constant Linear Acceleration

2

PExpand the list of base conceptsPDevelop a second model of motionPApply graphing and analytical problem

solving techniques

Objectives

3

P Instantaneous velocity (vn)< An object’s velocity at a specific clock time< The average velocity for an infinitesimal interval< Slope of the tangent on a curved x-t graph< Vector< m/s

Concepts of KinematicsDefinitions

4

PConstant average acceleration (a)< equal changes in velocity with equal changes in

clock time.< vector quantity< m/s/s or m/s2

Constant (Linear) Acceleration

5

PUsing an inclined plane and cart < we will develop the relationship between

– Velocity and time– Mean speed theorem

– Position and time< Afterwards we will state the relationship between

– Accleration and time – Velocity and position

Kinematics LabPart 2

CA 2005

6

Using these graphs and worksheet 2, learnabout the graphical representation of constantacceleration and develop the remainingequations of motion.

Graphical Analysis

Constant Acceleration Motion Graphs

Worksheet 2: Constant acceleration graphical analysis.

1. SKETCH the line of best fit on the velocity-time and position-time graphs. Use the lineof best fit to answer the following questions.

Velocity versus time graph

2. What is the y-intercept of the velocity-time graph and what does it tell us about the cart’smotion?

3. What is the cart’s velocity at....?a. 1.0 seconds

b. 3.2 seconds

4. Calculate the slope of the velocity-time graph. What does the slope tell us about thetractor’s motion?

5. Write the specific and general physical equations for this velocity-time graph.

SPE : GPE :

6. Calculate the area under the velocity-time graph form 1.0 to 3.2 seconds. What does thisarea tell us about the cart’s motion? How can we verify this with the position-timegraph?

7. What would be the average constant speed that would cover the same distance in thesame time as our accelerating cart did in the 3.6 seconds of travel it had? This speed iscalled the “Mean Speed.” Sketch a line to show the mean speed on the velocity-timegraph.

Constant Acceleration Motion Graphs

Position versus time graph

8. What is the y-intercept of the position-time graph and what does it tell us about the cart’smotion?

9. Calculate the slope of a tangent line at ....?

a. 1.0 seconds

b. and 3.2 seconds

c. What does these slopes tell us about the cart’s motion?

10. Calculate the cart’s acceleration.

11. Write the specific and general physical equations for this position-time graph.

SPE : GPE :

v v a xf2

02 2= + ∆

7

PSlope = 0, < constant acceleration

PY-intercept < Average acceleration.

PArea< The change in velocity

for that interval

Acceleration vs time

8

Velocity vs position

The resulting equation is:

9

PConstant Acceleration or Uniform LinearAcceleration< Used when the change in velocity is constant with

time

< A= 0, results in a constant velocity equation

< We do not know if this model is accurate betweenthe data points or beyond the data points

Ramifications

10

PWhen we talk about ACCELERATION weare concerned with CHANGING VELOCITY.

< In physics acceleration does not just meanspeeding up

< )v means a change in – speed or– direction of travel.

WARNING!

11

PAcceleration < Commonly means speeding up.< The v and a vectors point in the same direction.

PDeceleration < Commonly means slowing down.< The v and a vectors point in opposite directions.

Speeding Up & Slowing Down

12

PYou need to know how to relate graphshapes to ........< Speeding up

< Slowing down

< Constant velocity

Graphical Analysis

2

0

2 20

0

12

2

2

o

f

f

f

x v t a t

v v a t

v v a x

v vv

x v t

∆ = ∆ + ∆

= + ∆

= + ∆

+=

∆ = ∆

13

PGalileo < First to deduce free fall was accelerated motion< Ramp experiment

PAir resistance < A speed dependent retarding force that limits

speed.– Often ignored in this course

< The maximum (constant) speed of an objectthrough the atmosphere is called terminal velocity.

Free-Fall

14

PVertical motion affected by gravity only.< No resistive or driving forces

– Including air resistance

PConstant acceleration< a = g = -9.81 m/s2

< (-) shows that g points towards the centre of theEarth.

Free-Fall

15

Equation Summary

Constant Linear Acceleration and Free Fall

1. A 90-ton supertanker, at sea, is on a collision course with a barrier reef. At 10:45a.m. the vessel has a velocity of 20.0 knots. At 11:05 a.m. the ship has a velocityof 17.5 knots. 1 knot is 1 nautical mile per hour.

a. The reef is 30.0 nautical miles away. Will the tanker stop in time or runaground?

b. Plot the x-t, v-t and a-t graphs of this motion.

2. A Volkswagen Jetta travelling at 11.2 m/s can attain a velocity of 27.7 m/s over adisplacement of 75.0 m.

a. What is its acceleration?b. How much time elapses as the car accelerates?c. Plot the x-t, v-t and a-t graphs of this motion.

3. In the 100 metre sprint, Donovan Bailey burst out of the starting blocks to reach atop speed of 12.2 m/s in the first 14.0 m. He then maintains this terminal velocityto the finish line. Terminal velocity is the highest speed he can attain because ofare resistance.

a. How long does it take for him to reach the finish line?b. What is Bailey's average velocity?c. Plot the x-t, v-t and a-t graphs of this motion.

4. Answer the following questions based on the given velocity-time graph. Showworkings or state your rational for your answers.

a. What is the objects initial velocity?b. What is this object’s “acceleration” during the intervals:

i. 0 - 2 secondsii. 2 - 5 secondsiii. 5 - 10 seconds

c. What is this object’s “displacement” during the intervals:i. 0 - 2 secondsii. 2 - 5 secondsiii. 5 - 8.33 secondsiv. 8.33 - 10 seconds

d. What is the object’s initial position?e. What is the object’s overall average velocity?f. Accurately plot the corresponding position-time graph on the axis provided

and assume the initial position is 0 metres.

5. A car travelling a constant 15.0 m/s on the outer ring road drives past a truckstopped at a merge lane with the road. The moment the car passes the parkedtruck, the truck beings to accelerate at a constant 3.50 m/s2.

a. How long does it take for the truck to catch up to the car? b. How far has the truck travelled?c. How fast is the truck travelling then?d. Sketch the x-t and v-t graphs.

6. An egg is dropped from a 12.5-m high roof top.

a. How long does it take to reach the ground? b. What is its impact velocity?

7. A pop up ball, hit in a soft ball game, takes 4.26 seconds to return to the level itwas struck from. What was it’s initial velocity?

8. A Sea King helicopter flies straight up at 15.0 m/s. The tail wheel vibrates free ofthe chopper. This incident happens 200 metres above sea level.

a. When does the wheel strike the ocean? b. What is its impact velocity?

1

Vector Mathematics2 Dimensions

2

PStudy vector addition< Graphical< Analytical

PStudy vector subtraction

Objectives

3

PTechniques< Graphical

– Scale drawings with geometry sets– Tip to tail– Parallelogram

< Analytical– Components– Law of sines and cosines

Vector Addition

4

PSet a scalePDraw the x-y axisPDraw ANY vector from the origin< Use scale and ruler to decide length< Use a protractor to get the angle< Draw a miniature (parallel) axis on the tip of the

vector.PPick another vector and repeat the above

step starting at the last mini axis

Tip to Tail

5

PConnect the primary origin to the tip of thelast vector.< Called the resultant< Arrow head points away from the primary origin< Use ruler and scale to determine magnitude< Use protractor to determine direction

Tip to Tail

6

Practice1

x

y

BC

P Scale 1 cm = 1mP A = 7 m @ 60EP B = 11m @ 210EP C = 5 m @ 120E A

Resultant

Resultant = 9.83 m @ 150E

R A B= +

2 2

1tan ( )

R A BBA

θ −

= +

=

7

PSet a scalePDraw the x-y axisPDraw all vectors with tails at the origin< Use scale and ruler to decide length< Use a protractor to get the angle

PPick two vector and use set squares to slidecopies (creating a parallelogram)

PConnect origin to opposite corner< Temporary resultant

Parallelogram

8

PRepeat with last resultant and any othervector.

PUse ruler, scale and protractor to get the sizeand direction of the last resultant.

Parallelogram

9

Practice2

x

y

B

C

P Scale 1 cm = 1mP A = 7 m @ 60EP B = 11m @ 210EP C = 5 m @ 120E

A

1st resultant

Resultant

Resultant = 9.83 m @ 150E

10

PAlways draw a rough tip to tail sketch!

P Inline vectors< Simple algebra

POrthogonal vectors< Pythagorean’s Theorem< Arctan

Simple Analytical

2A

BResultant

A B

Resultant

11

PComponents are projections of a vector (A)on the principle axis

PX-component< Ax = Acos2

PY-component< Ay = Asin2

P2 is the standard angle

Components

2

A

Ax

AyAy

12

PAny number of vectorsPTabulate for clarityPUse standard anglePSketch a labeled the tip to tail solution

Component MethodPoints

2 2

2 28.625 4.8929.916

R x y

RR m

= +

= − +=

1

1

tan ( )

4.892tan ( )8.625

29.561 150.4

yx

θ

θ

θ

−

−

=

=−

= − =

2 2 2

2 2 2

2 cos7 11 2(7)(11)cos30

36.6 6.052

R A B ABR

R m

θ= + −

= + −

= =

1 1

1 1

sin sin sin6.052 12.105

sin sin307sin ( ) sin ( ) 35.3

12.105 12.10511sin ( ) sin ( ) 65.3 114.7

12.105 12.105

R A B

R

A

B

θ α β

θ

α

β

− −

− −

= =

= =

= = =

= = = =

R A B= + R A B= −1 2

Except when the anglebetween A and B is 90

R R≠

°

13

Component Method

Vector X-compt Y-compt

7 m @ 60E11m @ 210E

5 m @ 120E

Resultant

Ax = Acos2 Ay = Asin2

3.500

-9.526-2.500

-8.625

6.062

-5.5004.330

4.892

x

y

R

2

14

PTwo vectors at a timePSometimes quick and handyPDraw a tip to tail triangle< Determine the angle between the known sides

Law of Sines and Cosines

P A = 7 m @ 60EP B = 11m @ 210E

15

Law of Sines and CosinesBest when adding only 2 vectors

AB

Resultant

30E$ "

P A = 7 m @ 60EP B = 11m @ 210EP 2 = 30E

Law of Cosines forsize of R

16

Law of Sines and Cosines

Law of Sines for sizeof other angles

P A = 7 m @ 60EP B = 11m @ 210EP 2 = 30E

AB

Resultant

30E$ "

17

PA negative sign in front of any vector meanwe must< Add 180E to the standard angle i.e., make the

affected vector point in the opposite direction< Then apply regular vector addition techniques

Vector Subtraction

A

BR1A

-BR2

1

Kinematics 03Projectile Motion and Relative Velocity

2

PDefine key features of motion in 2-D PApply vector addition to solve 2-D motion

problems< Relative velocity< Projectile motion

Objectives

3

PMotion in two dimensions can be separatedinto component motions entirely in onedimension or the other.

< These components are completely independent ofeach other.

< Both motions are constrained to the same time oftravel.

Principle of 2-D Motion

4

PAssume

< that acceleration due to gravity is constant– Free fall in the vertical

< no air resistance– Constant velocity in the horizontal

Projectile MotionGeneral

5

PThe path a projectile follows through 2dimensional space.

Trajectory

BallisticTrajectory

RealTrajectory

6

PUsing the basketball movie and appropriatemotion graphs, we will determine the type ofmotion in the < Horizontal plane< Vertical plane

Projectile Motion LabGraphical Analysis

Basketball

Projectile Motion Graphical Analysis Worksheet 3

1. In the space below make a sketch of the horizontal position and velocity time graphs. What type of motion occurs in the horizontal plane?

2. In the space below make a sketch of the vertical position and velocity time graphs. Whattype of motion occurs in the horizontal plane?

7

PHorizontal launch< Launch angle is zero

PLanding on same level as launch site< has symmetry

PLanding above launch sitePLanding below launch site

– Launch angle may be above or below the horizontal

Common Problem Scenarios

8

PThe usual first step is to determine the x andy components of the launch velocity withtrigonometry.

PNote< V0x can be used through out a problem< V0y can only be use at the launch point

Velocity Components

2

v0x = v0cos(2)

v0y = v0ssin(2)v0

9

PSketch a diagram of the trajectory < Label the x,y co-ordinates as the origin (0,0)

– This is the launch point in all problems

< The launch should also start at time zero

< List the 7 variables at each point of interest– Variables (x, y, t, vx, vy, v, 2)– Points of interest (origin, max height, impact point and

question specified)

Problem Solving Strategy

10

PA table anchors the solution. The term v0 is thelaunch velocity and 2 is the launch angle. The onlyfeature shared by both motions is the time of flight.

Organize Your Work

Horizontal (constant velocity) Vertical (free fall)

v0x = v0cos2

)x = v0x )t

v0y = v0sin2

)y = v0y )t + ½g)t2

vy = v0y + g)tvy

2 = v0y2 + 2g)y

)t (time of flight)

Do Practice 1 - 4

11 12

13

PThe velocity of one moving object asmeasured from another moving object (orreference frame)

PThe rate of change of separationdisplacement between the moving object andthe moving reference

Relative Velocity

14

PAll relative velocity problems are solved asvector addition problems.

PKey solving tools< Vector equations

– pay attention to subscripts< Sketch of the tip to tail solution< Selecting appropriate vector addition methods< Putting yourself in at different reference points

Relative Velocity

15

PA plane has an air speed of 600 km/h. Wedenote this velocity as ‘Vpa ‘ when solvingequations.

< ‘Vpa ‘ reads as the velocity of “plane” relative to the“air.”

< The first subscript ‘p’ is the object while thesecond ‘a’ is the frame of reference.

Relative VelocitySubscripts

16

PThe relative velocity equation is the vectorsum of all velocities.< the inner subscripts are identical

– They cancel each other out< outer subscripts match the ones in the answer

– The resultant, i.e., what remains

Relative Velocity Equations

Vpe = Vpa + Vae

17

PCase 1:< A moving object relative to a moving surface or

medium.

PCase 2:< A moving object relative to another moving object.

Relative VelocityTypes of problems

18

PYou walk on a horizontal covayer belt at 2.5m/s, relative to the belt. The belt moves at aconstant 3.0 m/s, relative to the ground.

< What is your velocity relative to the ground?

1-D CaseCase 1

19

PWrite the relative velocity equation

sketch the tip-to-tail diagram

PVyg = Vyb + Vbg = 2.5 + 3.0 = 5.5 m/s

1-D Solution

Vyg = 5.5

Vyb = 2.5 Vbg = 3.0

Vyg = Vyb + Vbg

20

PA police car chases a speeding truck. Thetruck travels East at 100 kph and the policecar travels at 125 kph. All velocities arerelative to the ground.

< What is the velocity of the police relative to thetruck?

< What is the velocity of the truck relative to thepolice?

< How long does it take to catch the speeder?

1-D CaseCase 2

21

PWrite the relative velocity equation

< where Vgt = -Vtg

Psketch the tip-to-tail diagram

PVpt = Vpg + Vgt = 125 + (-100) = 25 km/hr

1-D Solution

Vpt = 25

Vpg = 125

Vgt = -100

Vpt = Vpg + Vgt

Projectile Motion

1. Jim runs straight out off a diving tower at 1.60 m/s. If the tower is 10.0 m abovethe pool, determine:

a. Jim's impact velocityb. The time of flight.c. The (horizontal) range.

2. A golf ball is driven straight down a level fairway. If the launch velocity is 20.0m/s at 20° above the horizontal then what is its:

a. Maximum height?b. Time of flight?c. Range?d. Impact velocity?

3. A deck hand on the MV Caribou throws a “heaving line” to a dock worker. Thedeck hand is 15.0 metres above the dock and gives the line an initial velocity of4.0 m/s @ 65° above horizontal. FYI the “heaving line” is a light line used to getthe heavier mooring rope from a ship to a dock or mooring dolphin.

a. What is the flight time for the heaving line?b. Where must the dock work stand in order to catch it?c. Assuming the dock worker lets the line hit the dock, what is the line’s

impact velocity?

4. A motorcycle dare devil drives off a 63.0 ° ramp. The biker attains a maximumheight of 14.5 m from the top of the ramp. He returns to the ground 4.50seconds after leaving the edge of the ramp. What was his:

a. Initial velocity?b. Horizontal range?c. Impact velocity?d. Vertical displacement?

Relative Velocity

1. My boat has a maximum still water speed of 5.0 m/s. If I attempt to go straightacross a river 200 metre wide with a 3.2 m/s current (w.r.t. the shore), then:

a. What is my boat's velocity relative to the shore?b. How long will it take to cross the river? c. Which way should I point my boat so I go straight across the river? d. How long does it take to cross the river now?

2. A Puma helicopter flies to the Hibernia platform. The chopper flies at 35 m/s 30°E of S, w.r.t. still air and the wind blows at 16 m/s SW, w.r.t. the Earth.

a. What is the velocity of the helicopter, relative to the Earth? b. What direction must the pilot fly to remain on course the 30° E of S and

what would be the new velocity?

3. Your are on a ferry travelling at 10 mph @18°S of W and a tanker is moving 25mph 5° N of W.

a. What is the velocity of the tanker relative to your ferry? b. What is your ferry’s velocity relative to the tanker?

w.r.t = with respect to

1

Dynamics 01Introduction to Force

2

PDefine forcePList the attributes of forcePDefine net forcePList common forces in naturePLook at the role of net force on the motion of

a single body

Objectives

3

PThe study of why things move (in the waysthey do)

PRole of force in moving

PRole of energy in moving

Dynamics

4

PPush or pull

PContact < Tension, friction

PField < Gravity, electric, magnetic

Applied Force

5

PAn interaction between two bodies

PObjects do not possess force< Forces exist as long as the bodies interact< They are not things that can be exchanged

between objects

Applied Force

6

PMeasured with spring scalesPDerrived, vector quantityPUnits< Newton (N), Pound (lb)< 1 N is a very small amount of force

– 4.45 N = 1 lb

Applied Force

7

PShort range< Strong Nuclear (attractive)< Weak nuclear (attractive)

PLong range< Electromagnetic (attractive and repulsive)< Gravitational (attractive)

Main Types of ForceAll are field forces

8

PMeasure the strength and direction of anapplied force on some test matter at differentpoints in space.

Mapping Field Forces

9

PGrand Unification Theory< It is believed that gravitational, electromagnetic

and nuclear forces are really the same thing. Modern physicists are trying to prove this notion.

PAll the common forces (contact or not) usedin mechanics problems are fundamentally ofone of the principal types!

Main Types of Forces

10

PGravity (weight)

PNormal Force

PTension

PFriction

Common Forces

11

PAn attractive pull of the earth on all objects.< Always points towards the centre of the Earth

– On diagrams this means towards the bottom of thepage.

< Fg = m(g) = mg– ‘m’ = mass in kilograms,– ‘g’ = the acceleration of gravity, 9.81 m/s2

Force of Gravity AKA True Weight or Weight

Fg Fg Fg

12

PMass (m)< is a measure of an object’s resistance to

acceleration< Universally conserved

PWeight or the force of gravity (Fg)< is the attractive interaction (force) between the

Earth and objects< Varies depending on location and environment.

Mass vs WeightThey are not the same

13

PStrictly speaking this is an attractive forcebetween any pair of objects.< Where the Earth is so massive we tend not to

notice the other gravitational forces around us. Hence it is safe to ignore them for purpose ofcalculations.

< We will explore this more with Newton’s Law ofUniversal Gravitation.

Force of GravityWarning

14

PA reactionary force of a surface to an appliedforce.< Always points away from, but perpendicular to the

surface.< This is the force we often precieve as our weight.< FN

Normal ForceApparent Weight

FN

FN

FN

15

PA pulling force of a string on an object – always points away from the object and along the

string– FT

Tension

16

PResistive force applied to objects that aresliding along a surface.< Always points against the direction of travel.< Independent of speed and contact area.< Depends only on surface roughness, aka the

coefficient of friction and normal force.< Fk = :k FN

Kinetic (Sliding) Friction

FkFk

Fk FkMotion

Motion Motion

17

PResistive force applied to objects that areattempting to slide along a surface.< Always points against the intended direction of

travel.< Independent of contact area.< Depends only on surface roughness, aka the

coefficient of friction and normal force.< Fs (maximum) = :s FN

Static (Sliding) Friction

18

< Greater than kinetic< Static friction must be overcome before kinetic

friction takes effect.– For any problem either static or kinitic friction may exist

but they do not exist simultaneously.< Static friction is only as large as it needs to be to

prevent motion.

Static (Sliding) Friction

net x right left 1 2 3

net y up down 1 2 3

2 2net net x net y

net y

net x

F F FF F F

F F F

Farctan( )

F

x x x

y y y

F F FF F F

θ

= Σ −Σ = + −

= Σ −Σ = − −

= +

=

19

PAll applied forces are vectors. < Add together by vector addition techniques. < The resultant is called the NET FORCE.< Only two possible results

– Fnet = 0– Fnet … 0

PThe value of the net force determines anobject’s type of motion!

– constant velocity – constant acceleration

Net Force

20

PThis is a tail to tail diagramof 3 forces on an object.

PThe object is the dot in thecentre.

PThe arrows represent theforces on the object, theirlength and directionrelative to the real force.

Free Body DiagramsFBD

F2

F1

F3

21

PThese forces can bearranged in a scaled, tip totail diagram.

PThe arrow connecting theobject to the last free tip isthe resultant or “Net Force.”

PUse the scale and protractorto determine the size anddirection of the Net Force.

Vector Sum of ForcesTip to tail

F2

F1

F3

Fnet

22

PWhen the forces are not on an axis we canredraw the FBD by breaking the forces intocomponets along both axis.

Vector Component DiagramsVCD

F2x

2F3x

F1y

F2y

F3y

F1xF1y = F1sin2F1

F1x = F1cos2

23

PApply the component solution, i.e., sum upthe forces in each axis, then add theseanswers with pythagorean and finally usearctan to get an angle.

Vector Sum of ForcesVCD

F2x

F3x

F1y

F2y

F3y

F1x

24

PView the first video clip and sketch the freebody diagram for the cart.

Net Force and MotionDemonstration 1

Net Force01, 2005

Dynamics and Motion Graphical Analysis

Worksheet 4 Part a

1. Is there a non zero or zero net force on this cart? How do you know?

2. What type of motion does the cart display? How do you know?

3. Write a rule that relates the net force to the type of motion.

4. Comment on the direction of the net force and the direction of the ______________.

25

PView the second video clip and sketch thefree body diagram for the cart, AFTER thehanging mass stops falling.

Net Force and MotionDemonstration 2

Net Force02, 2005

26

PWhy do you need a constant force from a carmotor to maintain constant velocity on the highway?< Explain

Real World Paradox

Dynamics and Motion Graphical Analysis

Worksheet 4 Part 2

1. Do the graphs, where the hanging mass is falling, still show us the same rule as before?

2. What is the net force on the cart after the hanging mass stops falling? How do we know?

3. What type of motion does the cart display, after the hanging mass stops falling? How doyou know?

4. Write a rule that relates the net force, after the hanging mass stops falling, to the type ofmotion.

1

w here k = 1

N ew ton 's S econd Law o f M otion

net

net

ne t ne t

net

a F am

Fam

F Fa km m

F ma

∝ ∝

∝

= =

=

F ma Fma F Fnet

with against

= == −

ΣΣ Σ

1

Dynamics 02Single Body Dynamics and Newton’s Laws of Motion

2

PPresent Newton’s laws of motionPCompare and contrast mass and weightPSolve dynamics problems using these laws

and common forces.

Objectives

3

PAll objects tend to move with constantvelocity, or stay at rest, unless acted upon byan external net force.< Remember that constant velocity means in a

straight line.

Newton’s 1st LawInertia

4

P Inertia is < A property of matter< The resistance to acceleration.

PMass is a measure of an object’s inertia< Kg

P Inertia is NOT a resistance to motion!PWeight is NOT a measure of inertia!

Mass and Inertia

5

PThe acceleration of an object is proportionalto the net force on the object and inverselyproportional to its mass.

Newton’s 2nd LawNet force

a

m

a

F

6

PCombining the results of Newton’s 2nd lawand the vector nature of applied force resultsin:

Newton’s 2nd LawNet force

A on B B on AF F= −

g

k k N

s s N

net w ith aga inst

F m gF FF FF m a F F

µµ

=

=== = Σ − Σ

7

Sketch the FBD of the forces on these objects.

1. A skater gliding across a smooth arena (CV)2. Car during a long, level highway trip (CV)3. Ball thrown straight up, after it has left the hand4. Projectile mid flight5. Fighter jet taking off on a level runway6. A car sliding down a steep hill, starting from rest with the park brake on.

Newton’s 1st and 2nd Law

8

PWhat conditions are needed to produce anapplied force?

Newton’s 3rd LawInteractions

9

PFor every action there is an equal andoppositely directed reaction, on the agentcausing the initial action.

< Forces act in pairs< There is one force in-between objects acting in

two directions simultaneously

– Important for determining normal force and tension

Newton’s 3rd LawInteractions

10

PThis law is true regardless of the< Masses of the objects< State of motion< And whether or not the objects are alive or not.

Newton’s 3rd LawInteractions

11

List some examples of interaction pairs.

1. When walking your foot pushes on the ground, theground reacts by pushing on you with the samestrength.

2.3.4.

Newton’s 3rd Law

Do Examples 1 - 6

12

Equation Summary

Single Body Dynamics

1. A barge (m = 1200 kg) is pulled up an English canal by a draft horse. The horseapplies 450.0 N at 20° S of E. Initially the barge accelerates at 0.30 m/s2 directlyup stream, that is to say due East.

a. What is the size and direction of the applied force from the rudder?

2. A tow truck accelerates a 2000 kg towed car at 1.50 m/s2 on a level street wherethe friction on the car is 6150 N.

a. What is the net force on the car?b. What is the horizontal applied force on the car?c. What is the tension in the towing arm if the arm is at 60° above the

horizontal?d. Determine the normal force on the car.e. Assuming the car’s park break was left engaged, determine the coefficient

of kinetic friction.

3. A typical European river barge is moored to a quay. The two mooring lines aretaught and practically horizontal. The effective force of the current is 5.0 kN at23.0° to the quay, downstream. The bow line has an angle of 11.1° downstreamand the stern line has an angle of 70.0° upstream. All angles are measure fromthe quay.

a. Find the tension in each mooring line.

4. An 8.0 kg block slides down a vertical wall with an acceleration of 2.00 m/s2. Thecoefficient of kinetic friction is 0.850.

a. What is the minimum applied force on this block (size and direction)?

5. A 5.00 kg block rests on a 40.0° inclined plane, µs = 1.192 and µk = 0.70.

a. What is the force of static friction on the block?b. If the block was given a nudge down the hill to overcome static friction,

what would be its acceleration after that nudge?c. What is the largest inclination of this surface that has the block just on the

verge of spontaneously sliding?

6. A block rests at the bottom of a 3.00 metre ramp inclined at 42.5°. The block isthen pulled up the ramp with a rope angled at 20.0° to the surface of the ramp. Given m = 2.15 kg, µk = 0.176 and FT = 23.1 N,

a. What is the speed of the block at the top of the ramp?

1

Dynamics 03Systems of Bodies

2

PSolve dynamics problems using Newton’slaws and common forces on systems ofconnected bodies.

Objectives

3

PSimultaneous equations< One force equation for each object< Common ‘a’ for all objects< Substitute for connecting forces

PSystems approach< General outcome of all simultaneous equation

solutions

Systems of BodiesTactics

4

PTreat the system as a single object where< m total = Em

PDetermine the vector sum of external forceswhen looking for acceleration< Ignore all internal (linking) forces

PDetermine the strengths of internal forces

Systems of BodiesSystems approach

Do Examples 1 - 3

Systems of bodies

1. A 2-wagon luggage train is pulled by a single tractor. The last wagon has a massof 1000 kg, the lead wagon has a mass of 700 kg. The lead wagon’s park brakeis engaged (µk is 0.45).

a. If the tractor applies 8000 N of force to the lead wagon, what is theacceleration of the system and the force in the connecting couplings?

2. Two masses (3.0 and 6.0 kg) are connected by a single string that runs over africtionless pulley hung from the ceiling (a device called an Atwood machine).

a. What is the acceleration of the system and the tension in the rope?

3. A 12.0 kg mass on a 30° slope is tied to a 9.0 kg mass which hangs freely from africtionless pulley at the top of the ramp.

a. Which way would the system attempt to move?b. Given :s is 0.62, will the system move?c. If it spontaneously moves or is nudged in the direction it would like to

move and :k is 0.18, what is the acceleration of the system?d. What is the minimum breaking strength needed for the connecting rope?

F ma F manet net c c= = =

2

cvar

=2

c c in outmvF ma F Fr

= = = Σ − Σ

1

Circular Motion 01

2

PDefine centripetal acceleration and forcePDescribe cyclic motionPState Newton’s Law of Universal GravitationPStudy satellite motion and weightlessness

Objectives

3

PConstant speed motion of an object around acircular path.

PThe object’s instantaneous velocity vector< always perpendicular to the radius of curvature.< because of uniform change in direction, thus

changes in velocity, uniform circular motion isaccelerated motion!

Uniform Circular Motion

4

PSince the object accelerates it must have anet force acting on it. This net force is calledcentripetal force and it points towards thecentre of the curved path.< Centripetal force is not an applied force!

PCentripetal acceleration is a centre seekingacceleration caused by the net force.

Centripetal Force (Fc)

5

PCentripetal acceleration is the rate at whichthe velocity vector changes direction.

Pac depends < speed< radius of the circular path.

Centripetal Acceleration

6

PBecause the motion type is constantacceleration, Newton’s 2nd Law applies toobjects in circular motion

PFwith can be replaced with Fin and Fagainst canbe replaced with Fout

Circular MotionAnd the dynamics equation

Fc

v ac

# o f round trip s

2

tT

d rvt T

π

∆=

= =∆

7

PCentre fleeing force

PReaction (Newton’s 3rd Law) to centripetalforce on the agent causing the object to turn.

PNEVER acts on the object traveling throughthe turn!

Centrifugal ForceFictitious force

8

PFor example, if you swing a ball on a string ina horizontal circular path the force of thestring on the ball is the centripetal force(towards centre) and the force of the stringpulling on your hand is the centrifugalforce (away from centre).

Centrifugal ForceFictitious force

9

Path ProblemsForce Identification

CCW

P Horizontal planeP Identify the

horizontal forcesP Which way does

the object travelwhen the redstring breaks?

10

PYou do not feel the force of gravity on you. Instead you feel the normal force of asurface as it reacts to your weight.< Apparent weight = FN

PWhen you (or anything) feels weightless,then gravity is the only force that acts uponyou!

Apparent Weight

11

PPeriodic motion is motion that reoccursaround a fixed point with regularity< Cyclic motion< Vibrations< Oscillations

PUniform circular motion is periodic, cyclicmotion

Periodic Motion

12

PPeriod (T) is the time for one< round trip, < revolution< cycle

PUniform circular speed< d = circumference

Periodic Motion

2 ( )* 260

d r RPM rvt T

π π= = =∆

F m mr

F G m mr

G x

g

g

N mkg

∝

=

= −

1 22

1 22

116 67 10 2

2. * 2

2

2

2

1 22

11 *

# of round trips2 (2 )

60

6.67 10

c

net c c

g

N mkg

tT

d r RPM rvt Tvar

mvF F mar

m mF mg Gr

G x

π π

−

∆=

= = =∆

=

= = =

= =

=

13

PRPM a.k.a. revolution per minute

PConverting RPM to velocity in m/s

RPM and Circular Speed

14

PAlways sketch the FBD and VCD of theobject .

PThen treat it as a regular dynamics problemwhere the acceleration is simply thecentripetal acceleration.

Problem Solving Tip

15

PThree main types of problems are:

< Horizontal Surface– e.g., record player turntable, car on a turn

< Vertical Drum– e.g., washing machine, amusement park ride

< Vertical Loop– e.g., ball on string, roller coaster

Common Problems

16

PHorizontal Surface< Fs or FTx provides the net centripetal force (Fc). < FN and or FTy balances Fg

PVertical Drum< FN provides the Fc< Fs balances Fg, i.e., prevents sliding down the wall

of the drumPVertical Loop< Fg, FT and/or FN interact to provide Fc

Common ProblemsDetails

20

PThe force of gravity between any two objectsis proportional to the product of the massesof each object and inversely proportional tothe square of the separation distance.

Newton’s Law of UniversalGravitation

21

Equation Summary

Circular Motion and Universal Gravitation

1. A highway engineer designs a flat curve for a highway. She knows that :s is 0.45on wet pavement and expects drivers to try to do 120 kph on the curve.

a. What must the radius of the curve be such that drivers will not slide off? b. Will this work for all cars and trucks?

2. The Rotor-Ride is a vertical cylinder with a removable floor. People stand on theinside of the cylinder, next to the wall. Then the 5.0 metre diameter cylinder isrotated at 30 rpm. At this point the floor drops out and the people are held up bystatic friction from the wall.

a. What is the minimum coefficient of static friction? b. Does it matter if an adult or small child is on this ride?

3. A 1400 kg propeller blade rotates in a vertical plane at 300 RPM. The effectiveradius of rotation is 2.0 metres.

a. What are the maximum and minimum forces felt by a single retaining bolt?

4. A 70 kg passenger in a roller coaster cart is travelling around the top portion of avertical loop with radius 18 m. The person feels an apparent weight of only halfhis true weight.

a. Calculate the speed of the cart, when it rides on thei. Outside of the curve. ii. Inside of the of the curve.

b. Calculate the speed of the cart, if the rider feels weightless.i. Does it matter which side of the track the car is on?

5. Use the Universal Law of Gravitation to determine the weight of a 150 kgastronaut

a. on Earth, b. on Mars c. and the acceleration of gravity on Mars.

i. Me = 5.976 x 1024 kg , Re = 6.378 x 106 mii. Mm = 6.418 x 1023 kg , Rm = 3.367 x 106 m

6. The 2400 kg Mars Explorer will orbit Mars at 1500 km above the surface.

a. What is the force of gravity on the space craft?b. What is the acceleration of gravity at that altitude?c. How fast must the Mars Explorer travel to maintain this orbit? d. What would be the speed if the orbit was reduced to 500 km?

212

KE mv=

1

Work and Energy 01

2

PDefine kinetic energy, potential energy andwork

PState and use the work energy theoremPDiscuss and use elastic potential energyPState and use the law of conservation of

energyPDefine and use power

Objectives

3

PList several places where we find water.

PFrom this list, what are the two situations inwhich water can be found?

EnergyThe water cycle analogy

4

PEnergy can only be in< Storage

– Water in lakes, oceans and glasses are contained andrepresents energy.

– Link: stored water represents energy, as both are keptin “sinks.”

< Or Transition– When water is not stored in a sink, it is flowing from one

sink to another via rivers, vapor or rain.– Link: flowing water represents work, which is the

transfer / transformation of energy from one sink toanother.

Our Energy Model

5

PMassless fluid like substance that an objectmay possess.

PDefined by what it does or can do and not bywhat it is.

EnergyQuasi Definition

6

PKinetic Energy < The energy that an object possesses due to its

motion.< Scalar< Depends on mass (sink) and speed squared< Measured in Joules

– 1J = 1 kg * 1 m2/s2 = 1N * 1m

Kinetic Energy (KE)

co sW Fd θ=

cosnet netW F d θ=

1 2 3 ...net nW W W W W= + + + +

n e tF m a=

220

220

"a" can be substituted for by re-writing the kinematic equation

2

( )2

v v a x

v vax

= + ∆

−=

∆

220

220

220

( )2

1 12 2

1 12 2

net

net

net f i

v vF mx

F x mv mv

W mv mv KE KE

−=

∆

∆ = −

= − = −

7

PWork (common) is the exertion of “effortwhich may or may not accomplish sometask.”

PWork (physics) is the exertion of force on anobject that causes motion over a distance. < Work is something done by forces to objects.

That is force transfers energy form one object (orfield) to another.

Work

8

PWhen calculating work, only consider thecomponent of force acting along the directionof motion.

< where – F is the force (N), – d is the displacement (m)– 2 is the angle between the displacement and the force

vectors< Work is a scalar quantity measure in joules.

– 1 J = 1N*1m

Determining Work done byApplied Force

Analytical

9

Determining WorkGraphical

F

d

F

d

F

dConstantforce

Constantlyvarying force

Varying force

Area under the line equals the work done.

10

PCan be found two ways

< Determine the Fnet, via vector addition of allapplied forces, then multiply it with thedisplacement.

< Determine the work done by each applied force and then add up all the individual works.

Determining Net Work

11

Relating Work and Energy

View this video on horizontal motion. Relatethe net work to the kinetic energy displayedby the cart.

Work -Energy, 2005

12

Work and Energy

Newton’s 2nd Law combined with the kinematicequations shows changes in speed.

Work-Energy Theorem

Note: The KE - x graph is the directly related to the v2 - x graph, albeit the slopes are not thesame

Worksheet 5 Part A

You have just viewed a video where a cart was accelerated horizontally by a falling mass.

1. Sketch the free body diagram and write the Net Force equation for the cart. Assuming nofriction determine the size of the net force.

2. Using the F vs x graph determine the net work done from x1 = 0.013 m to x2 = 0.267 m. comment on how the Net Work was calculated.

3. Determine the cart’s kinetic energy at:a. x1 = 0.013 m, KE1 =

b. x2 = 0.267 m, KE2 =

4. What happened to the cart’s kinetic energy as it moved from x1 to x2? Quantify.

5. Write a statement that relates the cart’s Kinetic Energy, as it moves, to the Net Workdone to the cart. Give consideration to the direction of the net force and the direction oftravel. This is called the Work-Energy Theorem.

6. What does the slope of the KE-x graph show?

2 21 102 2netW mv mv KE= − = ∆

* *W E F dP F vt t t

∆= = = =∆ ∆ ∆

13

PWhen the net work on an object is not zero,the object’s kinetic energy changes.

PThe relationship between work and kineticenergy called the Work-Energy theorem

Work-Energy Theorem

14

PWork is the mechanism that transfersenergy from one sink to another

PPositive work< Adding energy< Applying force with the motion < Causes a gain in speed.

PNegative work< Removing energy < Applying force against the motion.< Causes a loss in speed.

Positive & Negative Work

15

P If the net force is zero then zero work is doneto an object.

PTypical cases of zero work < An applied force cannot move the object at all

– Pushing against a wall

< An applied force acts perpendicular to thedisplacement vector– The normal force on a car travelling along a road

Zero Work

16

PEnergy is “the ability to do work”.

< When object ‘A’ has energy, it can apply force to object ‘B’, causing ‘B’ to move some distance.

– A swinging hammer has KE which is used to createforce on a nail and drive it some distance into a board. The hammer comes to a stop, transferring all its KE tothe nail.

Work Energy Interchange

17

PPower is the rate at which work is done.

< W is work (J),< )t is the time interval (s)< P is the power in Watts (W)< 1W = 1 J/s

P1 hp = 746 W

Power

Do Examples 1 - 3

18

Potential Energy

View this video on vertical motion. Using thegraphs and the Work-Energy Theorem,attempt to fully explain what goes on as anobject is raised in this way and then dropped.

VerticalLift, 2005

Vertical Work

Worksheet 5 Part B

You have just viewed a video where a basketball was lifted up, then dropped.

1. When lifted from the floor, a. Did the person raising the ball do work on it? Justify your answer.

b. If your answer to question 1a is “Yes”, theni. How can we determine the amount of work done? Determine how much

work was done between the heights of 0.085 m and 0.441 m.

ii. What should happen to the ball’s motion?

iii. Did you see what you expected and how can you tell?

iv. Account for discrepancies, if any, between what was expected and whatwas shown? A free body diagram of the lifting the ball may be helpful.

2. When dropped from the new height, a. Sketch a free body diagram of the falling ball. Explain what happened to the

ball’s motion with respect to the Work-Energy Theorem.

b. Where did the acquired energy come from?

gPE mgh=

19

PPotential Energy < Is stored energy as a result of position or

configuration.< It is stored in fields between points of matter.< This energy can be recovered and transformed

back into KE

Potential Energy

20

PPEg can be calculated by the formula:

< where – m is the object’s mass– g is the acceleration of gravity– h is the height that the object was raised

P It is, < stored in gravitational fields, a result of position.< Scalar< Measured in Joules

Gravitational Potential Energy

Do Example 4

21

Energy Analysis of a ClosedSystem

View this video on a video where a basketballdropped from some height. We will analysethe ball's total energy and its distributionfrom the point of the first impact with the floorto the next.

Conservationof Energy2005

Law of Conservation of Energy

Worksheet 5 Part C

You have just viewed a video where a basketball was dropped from a height and allowed tobounce several times. We will analyse the ball’s total energy and its distribution from the pointof the first impact with the floor to the second. This is a closed system because objects are notallowed to enter or leave the system until all the analysis is completed. Our system is rathersimple and consists of the basketball and the Earth.

1. List the forces acting on the ball between bounces.

2. Fill out this table with data from the graph. Add the KE and PE to get the total energy. On the axis sketch the total energy with respect to the height.

Height (m) KE (J) PE (J) Total E (J)

0.630

0.791

0.832

0.782

0.604

3. Write a statement, both English and Mathematical, about the total energy in thisdemonstration. Your instructor will then provide some less obvious details once youhave made your first statement. This will become the Law of Conservation of Energy.

total A A NC B BE PE KE W PE KE= + + = +

212

cosg

net net

netnet

nctotal A gA B gBA B

KE mvPE mghW F xW F x W KE

F xE WP F vt t t

E KE PE W KE PE

θ

→

=

=

= ∆= ∆ = Σ = ∆

∆∆= = = =

∆ ∆ ∆= + + = +

25

PEnergy transferred by “non-conservativeforce” is neither created or destroyed butcannot be recovered by reversing theprocess.

PThe amount of energy transfer is pathdependent!

PSuch forces include< Tension, friction, muscles, i.e, the “contact forces”

Work Done by Non-ConservativeForces

26

PWNC is the net work done by the non-conservative forces over the distance frompoint A to B.

PWhen non-conservative (contact) forces acton an object, the object’s total mechanicalenergy is not maintained. Instead it willeither increase (if the NC force does positivework) or decrease (if the NC force doesnegative work).

Non-Conservative Systems

27

Non-Conservative Systems

Do Examples 5 - 6

28

Equation Summary

Where Wnc/A6B = 0 for conservative systems

Work and Energy

1. Mark is a 73.0 kg speed skater training in Calgary. He can go from rest to 6.0m/s in a distance of 4.3 metres.

a. What is the applied force on Mark as he accelerates?b. How much horse power does he generate, if it takes him 1.43 seconds to

travel that distance?

2. The propeller on a 600 kg speed boat pushes the water with a constant force of2.0 kN directed at 15.0° below horizontal. The boat travels 300.0 metres acrossa smooth lake. Beware of Newton’s 3rd law.

a. How much work is done by the propeller, gravity and buoyant (normal)force? What is the net work of all these forces?

b. After all that work the boat to go from rest to 30.0 m/s. What is the network done by all forms of friction?

c. What is the effective force of friction on the boat?

3. A car that was travelling along a horizontal street skids to a halt, leaving 60.0metre skid marks. The coefficient of kinetic friction is 0.60 between the pavementand car tires. What was the initial speed of the car?

4. When you are moving you push a 30 kg trolley, with frictionless bearings, up a3.0 metre long ramp, at 18.0°, to the back of a truck.

a. How much work do you do to the trolley, if you walk up the ramp with aconstant velocity?

b. What was the work done by gravity?c. What was the net work done?

If the trolley were to roll back down the ramp, starting from rest,

d. What would be its acceleration? Use Dynamicse. What would be its speed at the bottom of the ramp? Use Kinematicsf. What would be impact its speed, if it were dropped off the back of the

truck? Hint: find the height of the ramp.g. How would the answers from ‘c’ to ‘f’ change if the mass of the trolley

doubled?

5. A roller coaster car is raised up to a height of 15.0 m above the ground andreleased with an initial speed of 10.0 m/s.

a. What is its speed at the bottom of a valley that is 6.0 metres above theground?

b. What is the maximum height the hill on the other side of the valley,assuming the cars will be at the verge of stopping when they are at thatpoint and the track is frictionless?

6. A roller 500 kg coaster car is raised up to a height of 15.0 m above the groundand released with an initial speed of 10.0 m/s.

a. What is its speed at the bottom of a valley that is 6.0 metres above theground, given 1000 N of friction act upon the car and the total length oftrack is 12.0 metres?

b. What is the maximum height the hill on the other side of the valley,assuming the cars will be at the verge of stopping when they are at thatpoint?

p mv=

0 0f fp p p mv mv∆ = − = −

1

Linear Momentum 01

2

PDefine momentum and impulsePState Netwon’s three laws of motion in terms

of momentum.PSolve dynamics problems with momentumPStudy centre of mass

Objectives

3

PMomentum, the drive that continues motioneven in the absence of a net force. It is thequality / durability of motion. Like energy,objects possess and can transfer it.< Depends on the product of mass and velocity.

< Vector, same direction as velocity< Units kg*m/s

Momentum (p)

4

PAll objects will move with constantmomentum, or stay at rest, unless actedupon by an external net force.

Newton’s 1st Law

5

P Impulse is the change in an object’smomentum.

< Vector< Same units as momentum

Impulse ()p)

6

PAn object’s impulse is proportional to the netforce on the object and the interval (elaspedtime) over which the net force is applied.

P Impulse has the same direction of the netforce.

Newton’s 2nd Law

netF ma=0

0

"a" can be substituted for by re-writing the kinematic equation

( )f

f

v v a tv v

at

= + ∆

−=

∆0

0

( )

= Impulse

fnet

net f f i

net

v vF m

tF t mv mv p p pF t p

−=

∆∆ = − = − = ∆

∆ = ∆

7

Deriving the Impulse form of the2nd Law

Newton’s 2nd Law, combined with the kinematicequations, can be expressed in terms ofimpulse.

8

Determining ImpulseGraphs

Effectiveforce

F

tConstantforce

Varying force

Do example 1

Area under the graph is the impulse or changein momentum for that interval.

F

t

9

Impulse Demonstration

View this video on horizontal motion. Relatethe impulse to the area under the graph.

Impulse,2005

Worksheet

1. Given the mass of the cart is _____ kg, use the velocity-time graph to determine thecart’s momentum before and after the collision with the force sensor.

2. What is the impulse of this collision?

3. How long does the collision last?

4. How does the area under the force-time graph compare with the impulse from question2?

5. What is the average force of the collision?

1 2' '

1 1 1 1 2 2 2 2' '

1 1 2 2 1 1 2 2' '

1 2 1 2

( ) ( )( )

net net

before after

F t F tm v m v m v m vm v m v m v m vp p p pp p

∆ = − ∆

− = − −

+ = +

+ = +Σ = Σ

2

' '1 2 1v v v v− = −

1 1 2 2

1 2

......

n ncm

n

m x m x m xxm m m+ + +

=+ + +

10

PFor every action there is an equal andopposite reaction on a different object.

PFor every impulse there is an equal andopposite impulse on a different object.

Conservation of MomentumNewton’s 3rd law

11

Conservation of MomentumNewton’s 3rd law

For an isolated system, i.e., no external netforce:

12

PMomentum is conserved in all directions.

P In multi dimension problems, one must findthe total momentum in each dimensionbefore and after the collision using thecomponents of the momentum vector.

Conservation of Momentum2D and 3D

13

PElastic< KE is conserved< Objects rebound and exchange velocities

P Inelastic< KE is lost< Objects stick together< Objects rebound do not completely exchange

velocity

Types of Collision

14

PMomentum is always conserved even whenenergy is not.

PWhen the collision between two objects iselastic, then

Collision and Conservation Laws

Do examples 2 - 4

15

PA system of many objects may be treated asa single object with all the mass located atthe centre of mass.

PWhere xcm is the displacement from anarbitrary position to the system centre ofmass.

Centre of MassLinear systems

1 1 2 2

1 2

......

n ncm

n

m v m v m vvm m m+ + +

=+ + +

p mvp F t mv mvp p

x m x m x m xm

v m v m v m vm

before after

cmn n

cmn n

== = −

=

=+ + +

=+ + +

∆ ∆Σ Σ

Σ

Σ

0

1 1 2 2

1 1 2 2

...

...

16

PUniform objects such as baseball have acentre of mass at the geometric centre of theobject

PNon-uniform objects have a centre of massat usually close to the concentration of mass< Fly rod

Centre of Mass3D objects

17

PSystem velocity may be determined < by taking the sum of the component momentums

and dividing by the system mass

System Velocity

18

Equation Summary

Momentum

1. A 0.50 kg baseball initially has 156.25 J of K.E. The ball is struck with a bat andrebounds with a velocity of 33.3 m/s.

a. What is the initial and rebound momentum of the ball?b. What is the average force on the ball if the contact time is 4.0 milliseconds

(ms)? c. What is the average force on the bat?

2. A 4.0 tonne truck travelling at 6.0 m/s, on a level road, collides with a 1500 kgparked car.

a. Determine the after collision velocity if the truck and car stick together.b. How much energy is lost?

3. Jane shoots her 1.40-kg “45" at the Rod and Gun club. The 20-g bullets have amuzzle velocity of 244 m/s. What is the recoil velocity of the gun?

4. A red billiard ball with a mass of 300-g travels at 0.75 m/s in a straight line on alevel pool table. It strikes resting, 400-g, blue ball. After the collision the blueball has a velocity of 0.40 m/s at 23.0° w.r.t. the red ball’s initial line of travel.

a. What is the red ball’s final velocity?b. How much energy is lost after the collision?

1

Torque and Equilibrium 01

2

PLook at the effect of where a force is appliedto an object, on its motion.

PDefine torque, line of action and moment armPState conditions for equilibriumPState types of stability and balance

Objectives

3

PTranslational motion is motion that takes youfrom point ‘A’ to point ‘B’

PHere we consider the forces acting on thecentre of mass.< Treat the object as if all its mass is compressed

into one spot.< FBD< VCD

Translational Dynamics

4

PNot all forces will cause translation, somecause rotation around a pivot. Rotationalmotion rotates an object around a fixed pointin space.

PHere we consider not only the size of theforces acting on the an object but also wherethese forces are applied.< Treat the object as a ruler.< Rigid body diagrams (RBD)

Rotational Dynamics

5

PPivot point< An arbitrary point in space around which an object

could rotate.< Aka

– Axis and Fulcrum

PLine of Action< A line that is parallel to the applied force and

drawn through the contact point of the force on abody.

Definitions

6

PLever Arm (d)< The distance along the rigid body between the

pivot and the applied force.

PMoment Arm (l)< A line that connects the pivot point to the applied

force, such that it is perpendicular to the line ofaction.

Definitions

s i nF d

F l

τ θ

τ

=

=

sinFdτ θ=F lτ =

7

PTorque is to rotation as force is to linearmotion. If one applies a continuous nettorque then the object will gain rotationalspeed, i.e., undergo angular acceleration.

Torque (J)

8

PTorque is the product of, < the perpendicular component of force and the

lever arm, its displacement to the point of rotation,

< Or the force and the moment arm.

Torque (J)

9

Pivot pointF

2 Fz= Fsin2

d

10

Pivot point

Moment arm (l)

22

Line of action

F

11

PTorque is < A vector< Measured in N×m

– Not the same as a joule

PForces on the pivot point produce no torque,because the lever or moment arms are zerometres long.< d = 0m

Torque (J)

12

PThe net force must equal zero in alldirections!

< Constant velocity or rest in all dimensions.

1st Condition of Equilibrium

cw c cwτ τΣ = Σ

s in

d o w n u p

le f t r ig h t

c cw cw

F d F lF FF F

τ θ

τ τ

= =Σ = Σ

Σ = Σ

Σ = Σ

13

PThe net torque must equal zero

< No rotation in any direction around any pivot point.

2nd Condition of Equilibrium

14

PCentre of gravity refers to the natural balancepoint of an object, i.e., where the force ofgravity appears to be concentrated.

< For simple “uniform” objects this is dead center inthe object.

Centre of Gravity

15

PStable< Balanced< Centre of gravity is below point of suspension< Object returns to rest position

– Pendulum

PNeutral< Object is balanced no matter where it is.

Balance and Stability

16

PUnstable< Unbalanced< Centre of gravity is above point of suspension< Object finds new rest position

– Tower

PStability depends on location of C of G andsize of base.

Balance and Stability

Do examples 1 - 4

17

Equation Summary

Torque and Rotational Equilibrium

1. You can use a uniform plank as a dead pan balance. This is done by placing thefulcrum at the centre of mass, such that the unloaded plank is initially balanced

a. With a 2.0 kg mass at 20.0 cm from the fulcrum, determine the mass of anobject 34.0 cm from the other side of the fulcrum, that keeps the plankbalanced.

2. A 1.5 m, uniform bar, with the fulcrum 30.0 cm in from one end, is in rotationalequilibrium, when a 1400-g mass is 10.0 cm in from the short end and a 100-gmass is 45-cm in from the long end.

a. What is the mass of the bar?b. What is the normal force on the bar?

3. A lawyer hangs a uniform 2.7 kg sign with a wire secured 17.0 cm in from the farend at an angle 55.0°. His sign is 0.85 mlong.

a. What is the FT and the FW exerted onthe sign?

b. What is the :s between the wall andsign?

4. Rose is 50 kg and 170 cm tall. Her centre of mass is 68.0 below the top of herhead. During a push up her hands are under her shoulders, 25 cm from the topof her head.

a. Determine the normal force on her hands and feet when her body ishorizontal.

Appendix A: Metric System and Base Quantities

The Seven Base Quantities1. Mass (kg)2. Length (m)3. Time (s)4. Amount (mol)5. Luminous intensity (lux)6. Electric charge ( C)7. Temperature (K)

Measured Quantity Metric Unit Some Other Units DefinitionMass m inch, foot, kilometer A universally conserved property of matter

that shows roughly the size of matter orresistance to acceleration

Length kg slug, dyne A measure of physical space

Time s minute, hour A measure of temporial space or aprogression or sequencing of events

Note: Mass and Time are scalar, base quantities where Length is a base quantity which may beeither scalar or vector.

Prefix Symbol Power of 10 MeaningTera T 1E+12 one trillion timesGiga G 1E+09 one billion timesMega M 1E+06 one million timesKilo k 1E+03 one thousand times

Hecto h 1E+02 one hundred timesDeka da 1E+01 ten times

1E+00 BASE UNITDeci d 1E-01 one tenth ofCenti c 1E-02 one hundredth ofMilli m 1E-03 one thousandth of

Micro : 1E-06 one millionth ofNano n 1E-09 one billionth ofPico p 1E-12 one trillionth of

Appendix B: Student Evaluation and Tips

Student Evaluation

50%, final exam25%, 2 term tests10%, 6 - 8 short quizzes and or assignments15%, practical and written lab exam

Course outline, up to the first test

1. Introduction to Physics2. Constant velocity (ULV) and kinetic energy in 1D3. Constant acceleration (ULA) in 1 D4. Vector addition5. Relative velocity6. Projectile motion

Course outline, from first to second test

1. Newton’s Lawsa. Single bodiesb. Systems of bodies

2. Circular Motion3. Work and Energy (Intro)

Course outline, from second test to final exam

1. Work and Energy (Conclusion)2. Momentum3. Torque

My expectations of students

1. You must read the selected text sections before class! See Appendix C.2. Complete assigned text book problems before the class moves onto the next

topic. See Appendix C.3. Do not hesitate to seek my help on any course related problem.

a. Drop by my office or make an appointmentb. Use e-mail c. Call me

Tips and suggestions

1. Form a buddy system or study groups. Your classmates may be able to explainthings more clearly than I.

2. Focus on skills and task execution but not final answers. Showing the strategy ofa solution is far more important than its outcome.

3. Allocate your time. Read the text every night and attempt a minimum of 3problems form the text every night. If you are unsuccessful with the problemssee me the following day for clarification.

Appendix C: Text References and Suggests Practice Problems

Textbook References James Walker, 2nd ed

Basic Concepts CH.1Read sect. 1.1 - 1.7Problems 5, 7, 8, 12, 13, 14, 16, 20, 23, 24, 27

Constant velocity CH.2Read sect. 2.1, 2.2, 2.3Problems 1, 2, 3, 5, 7, 9, 10, 12, 15, 17, 18, 19, 21, 23, 27

Acceleration CH.2Read sect. 2.4, 2.5, 2.6, 2.7Problems 28, 29, 31, 32, 33, 35, 37, 39, 40, 41, 49, 53, 55, 61, 62, 64, 67, 68, 71, 74, 75, 77 79, 86, 89, 93

Vector Addition / Relative Velocity CH.3Read sect. 3.1, 3.2, 3.3, 3.5, 3.6Problems 1, 2, 4, 14, 15, 16, 20, 41, 43, 44, 45, 47, 48, 58, 59

Projectile Motion CH.4Read sect. 4.1, 4.2, 4.3, 4.4, 4.5Problems 7, 9, 11, 12, 15, 19, 23, 25, 27, 31, 32, 34, 35, 37, 40, 57, 63

Forces and Dynamics CH. 5 & 6 Read sect. ALLProblems CH 5: 1, 3, 5, 6, 9, 14, 17, 19, 20, 23, 25, 29, 31, 34, 37, 38, 42, 43,

CH 6: 1, 2, 3, 7, 8, 9, 11, 15, 18, 21, 26, 28, 32, 34CH 6: (Systems) 36, 39, 37, 40

Circular Motion CH.6Read sect. 6.5Problems 43, 44, 47, 48, 49, 50, 52, 69, 88

Universal Law of Gravitation CH.12Read sect. 12.1, 12.2Problems: 1, 2, 4, 8, 9, 14, 19, 30

Work & Energy CH. 7 and CH. 8Read sect. 7.1, 7.2, 7.4 and 8.1, 8.2, 8.3, 8.4Problems CH 7: 1, 2, 4, 6, 7, 8, 9, 15, 16, 17, 25, 35, 39, 40, 48, 53, 54, 77

CH 8: 6, 9, 12, 13, 14, 15, 18, 19, 21, 26, 27, 28, 29, 30, 31, 32, 35, 37, 50, 60, 61, 63, 68

Momentum CH. 9Read sect. 9.1, 9.2, 9.3, 9.4, 9.5, 9.6, 9.7Problems 3, 8, 9, 11, 12, 17, 18, 22, 23, 25, 27, 28, 31, 32

Torque CH. 11Read sect. 11.1, 11.2, 11.3, 11.4Problems 1, 2, 3, 4, 6, 23, 27, 28, 30

Textbook References Cutnell & Johnson, 5th ed

Constant velocity CH.2Read sect. 2.1, 2.2Problems 1, 3, 5, 6, 7, 9, 10

Acceleration CH.2Read sect. 2.3 2.4, 2.5, 2.6, 2.7Problems 12, 13, 16, 20, 25, 26, 32, 35, 37, 38, 40, 42, 46, 53, 54

Projectile Motion CH.3Read sect. 3.1, 3.2, 3.3Problems 13, 16, 17, 19, 21, 22, 23, 27, 28, 30, 34, 35

Relative Velocity CH.3Read sect. 3.4Problems 47, 49, 50, 51, 52, 53, 54, 55, 56

Forces and Dynamics CH. 4Read sect. ALLProblems (No acceleration) 10, 34, 39, 46, 47, 50, 51, 52, 53, 58

(Acceleration) 1, 2, 7, 11, 12, 31, 35, 36*, 37, 38, 40 69, 72, 76, 79(Systems) 67, 70, 78, 84, 85

Circular Motion CH.5Read sect. 5.1, 5.2, 5.3, 5.7, , 5.5 (optional)Problems 1, 6, 8, 12, 14, 15, 18, 36, 39, 41

Work & Energy CH 6Read sect. ALL Problems 1, 3, 7, 11, 12, 18, 20, 25, 27, 29, 32, 34, 41, 46, 48, 49, 52, 55, 58, 60, 64, 67, 77, 78.

Momentum CH.7Read sect. 7.1, 7.2, 7.3Problems 1, 3, 5, 7, 8, 15, 17, 18, 19, 26,

Torque CH. 9Read sect. 9.1, 9.2Problems 2, 4, 5, 7, 15, 16, 20, 23, 24, 66, 68

Appendix D: Practice Assignments

Physics 1100 Practice Sheet 1

1. El Guerrouj ran 1-mile in 3 minutes, 43.13 seconds. What is his speed in m/s?

2. You can determine the speed of your car using the kilometer marks on the sideof the highway and a watch. How many seconds should elapse between twoconsecutive marks if your are driving at 100 km/h?

3. A motorist drives the 150 km distance between two cities in 2.5 hours, but makesthe return trip in 2.0 hours.

a. What is her average speed and velocity for (a) each half of the round tripand (b) the total trip?

b. Sketch the position-time and velocity-time graphs for this trip.

4. A confused physics student walks to the Marine Institute. He travels 650 mSouth in 5 minutes and realizes he has left his lab book home. Instantly he goesback home in 4 minutes and gets the book. He then returns to school, 840 mSouth of his apartment, in 8 minutes.

a. What was his overall average speed and average velocity (m/min)? b. Sketch the position-time and velocity-time graphs for this trip.

5. A 27 pound meteorite struck a car leaving a dent 22 cm deep in the trunk. If themeteorite struck the car with a speed of 550m/s, what was the magnitude of itsdeceleration?

6. Using the following v-t graph answer the following questions.

a. What is the object’s accelerationduring each distinct phase oftravel?

b. What is the object’sdisplacement during eachdistinct phase of travel?

c. What is the object’s averagevelocity?

d. Accurately plot the object’s x-tgraph given its initial position is5.0 metres.

7. Describes the motion of this object in a few sentences?

8. A volcano shoots out blobs of molten lava called lava bombs, from ground level. A geologist observing the eruption uses a stopwatch to time the flight of aparticular lava bomb that is projected straight upward. If the time for it to rise andfall back to the ground is 4.75 seconds, what is its initial speed?

9. A stone is thrown vertically upward from the edge of a 25.6 metre tall building,with an initial velocity of 16.7 m/s. The stone just misses the building on the waydown and strikes the street below. Determine:

a. the total time of flight b. velocity of the stone (impact velocity) just before it strikes the ground.

10. A drunk driver is driving his car at a constant 15.0 m/s, when he sees a dog inthe middle of the road. His reaction time (time to hit the brake) is a drastic 2.50seconds. Once the brakes are applied, the car decelerate at a constant 6.50m/s2. Consider the whole motion from the time he sees the dog until he stops:

a. How much time is spent in motion?b. How far does he travel?c. Sketch the corresponding position-time and velocity-time graphs. Plot and

label all critical points. However, you only need to sketch an appropriateline of best fit between them.

11. Deborah passes a ghost car travelling at 100 km/h. She travels at a constant125 km/h. When the cop begins the chase, she is already 40 metres away. Thecop will accelerate at a steady 15.0 m/s2.

a. When and where does he catch up with Deborah?b. What is the cop’s velocity when he catches her?c. Sketch the x-t and v-t graphs of this motion.

12. On a river that flows from West to East, a boat tries to head due North. Becauseof the river’s 4 m/s current the boat ends up drifting at 30° E of N. What is thespeed of the boat relative to the water? What angle must the sailor point theboat to make the desired crossing?

13. Many trans-Atlantic passenger jets were re-routed to St. John’s airport on September 11th.British Air flight 905, approached the airport witha velocity (relative to the ground) of 240 km/hr @30° N of W. Also, approaching St. John’s was,United Airlines flight 643, which was traveling ata velocity (relative to ground) of 310 km/hr @10° E of N. Assume both planes are in levelflight and flying at the same altitude.

Determine the velocity of “British Air flight 905”relative to “United Air flight 643.” Include a tip-totail diagram, which must be clearly and properlylabeled. Be sure to show all necessary workingsto find the magnitude and direction of thevelocity.

14. A person is playing a game of darts. He stands 2.6 m away from the dart boardand throws his dart with a velocity of 8.0 m/s @ 15° above horizontal. The dartfollows a projectile path and just after passing its maximum height, it sticks intothe bull’s-eye of the board.

a. Sketch an appropriate diagram.b. What are the initial velocity components of the dart?c. What is the total flight time of the dart?d. What is the height of the bull’s-eye above the point of release of the dart?e. What is the max height of the dart’s flight above its point of release?f. What are the components of the velocity of the dart as it hits the board?g. What is the dart’s impact velocity and angle?

Physics 1100 Practice Sheet 2

1. A hunter on an ATV pulls a 500 kg moose carcase out of horizontal bog. TheATV produces a steady horizontal 5000 N pull on the moose and moves at aconstant 5 km/h. Calculate the coefficient of kinetic friction of between the bogand moose?

2. A 4.0 kg block is pulled along a horizontal surface. The tension in the pullingstring is 7.0 N at 35.0° below the horizontal.a. What is the normal force exerted on the block?b. If the coefficient of kinetic friction is 0.15, then what is the acceleration?

3. A 10.0 kg wagon is pulled along a level floor by a 30 N force directed at 70°above the horizontal. The wagon’s horizontal acceleration is 0.52 m/s2 .Determine the coefficient of rolling friction. Hint: rolling friction is mathematicallythe same as sliding friction.

4. A 48.0 kg person decides to measure their weight on an elevator acceleratingdownward at 1.25 m/s2. What does the Newton scale read?

5. Bart fires a 250-g crossbow bolt (arrow) upward at an angle of 53° (this is theslope of the surface). Initially the bolt is at rest and it leaves the crossbow with avelocity of 70 m/s. The crossbow is 60.0 cm long and has a :k of 0.47. What isthe average applied force needed to launch the bolt?

6. A trawl is pulled up the 75° ramp (:s = 0.50 and :k = 0.30) on the back of afishing boat. The cables pulling the net up the ramp are parallel to the surfaceand effectively apply 15000 N of tension. If the net is raised with a constantvelocity, then what is the mass of the trawl and fish?

7. The diagram shows a pulley system that cangently lowers or raises the boat out of the water.If this system starts at rest and the truck’s parkbrake is on, determine:

a. The system velocity after 3.0 seconds.b. The tension in the connecting string.

8. What is the acceleration of this system and thetension in the connecting wires? Assume it is inmotion.

M1 = 12.0 kgM2 = 9.0 kgM3 = 8.0 kg:k1 = 0.10:k2 = 0.05

9. Answer the following for this system:a. Which way the system would try to

move?b. Does the resting system

spontaneously moves?c. If, for whatever reason, it were

moving in its desired direction, whatis the acceleration of the system?

d. What is the tension in theconnecting string?

M1 = 15.0 kgM2 = 10.0 kg:s2 = 0.10:k1 = 0.03

10. A jet fighter makes a horizontal arc of 3.0 km radius at 827 m/s. Calculate thecentripetal force on the 72 kg pilot.

11. A Ferris wheel travelling at a uniform speed 5.0 m/s makes you feel weightless atthe top of the ride. What is the diameter of the wheel? What is your apparentweight at the bottom if your mass is 50.0 kg?

12. In the Barrels of Fun ride at an amusement park people are spun around ahorizontal circle at a steady speed. The floor drops out of the ride. However, thepeople do not fall because of the static friction between them and the verticalwalls.

Given the ride has a diameter of 6.0 metres and spins at 90 rpm, determine thecoefficient of static friction.

13. What is the normal force on a 100 kg astronaut standing on a level plain onVenus, if the mass of Venus is 4.87 x 1024 kg and the diameter is 6.65 x 106 m?

14. What is the orbital speed of a satellite that has a stable orbit 200 km above theocean?

Physics 1100 Practice Sheet 3

1. A 2.0 kg block, initially at rest, slides from the top of a ramp to the bottom. Theramp is 1.5 metres long and the top is 0.75 m above the bottom. Determine thespeed of the block at the bottom of the ramp, if on the way down it experiences1.25 N of sliding friction.

2. George of the Jungle swings from a 10 metre long vine. Initially the vine ishorizontal.

a. What is George’s speed at the bottom of the swing?b. What is the minimum strength that the vine should withstand, if George

has a mass of 70 kg? Include a proper Free Body Diagram.c. At the bottom of the swing he lets go of the vine. Given that it is another

6.0 metres to the ground, what is his impact speed? (Solve with theprinciples of work and energy)

d. What is the angle of his impact velocity?

3. Martin, who is 47.0 kg, is out one winter afternoon, tobogganing on a hill. Thehill is sloped at 40 ° and the coefficient of kinetic friction between the snow andhis toboggan is 0.15. When he is at the top of this 20.0 metre long hill, hepushes himself to over come static friction. The result of the push is that he hasan initial speed of 1.5 m/s.

a. What is his speed at the bottom of the hill? b. How far will he slide on the level ground at the bottom of the hill before

stopping?

4. In 1996 a string of 60 box cars (60000 tonnes) rolled down a 6.0 km hill andcollided with a freight train (75000 tonnes) travelling up that same hill. Before thecollision the box cars had a velocity of 33.3 m/s East and the train had a velocityof 16.7 m/s West.

a. If the train and cars couple together without derailing, what would be thevelocity after the collision?

b. During the collision which object experienced the greatest impulse?

5. A 2.5 gram pellet travels from rest to 171 m/s in a 48 cm rifle barrel. What is therecoil velocity of the 5.5 kg rifle? Determine the impulse force.

6. A 180 kg raft with two swimmers 60 and 80 kg each. The two swimmerssimultaneously dive off opposite ends at 3.0 m/s each. What is the resultantvelocity of the raft?

7. A 35-g blob of putty moves towards a 450-g block of wood resting on a levelsurface with a :k of 0.815. The putty and sticks to the block of wood. After theimpulse of the collision ends, the putty and block combination slide a distance of25.0 cm before stopping.

a. Determine these values for the putty / block combination.i. K.E. at Pt. C _________ii. Fk from Pt. B to Pt. C _________iii. Work done by Fk, as the system goes from Pt. B to Pt. C

___________iv. Determine the velocity of the putty / block at Pt. B, using the

principles of work and energy.b. Determine the values of each of the following.

i. The momentum of the putty / block combination at Pt. B.____________

ii. The momentum of the block at Pt. A. ___________iii. The momentum of the putty at Pt. A. ___________iv. The velocity of the blob of putty before the impact with the wood.

c. The average force of the collision if the impulse lasts 60 milliseconds.

8. Jack (50 kg) and Jill (37 kg) are on a uniform 2.5 metre see-saw. If Jack is 0.50cm from the central pivot where must Jill sit just to keep the see-saw balanced? What is the normal force on the pivot?

9. This is a Dodge Dakota pickup truck. From wheel to wheel it measures 3.33metres and 5.46 metres from bumper to bumper. The front wheels are 0.90metres behind the front bumper. The centre of gravity, of this 1847.0 kg truck, is2.00 metres behind the front bumper. What is the normal force exerted on eachaxle?

10. Determine the tension in the cable and the force on the pivot for this simpleuniform boom.

Sample Final Exam Version 1

1. An archer stands on top of a 60 m tall castle wall. He fires an arrow straight upinto the air, with an initial velocity of 15.0 m/s. a. What is the maximum height of the arrow relative to the ground? (3 points)b. What is the velocity of the arrow as it hits the ground? (3 points)c. What is the time of flight? (2 points)d. Sketch the position-time and velocity-time graphs for this motion.(2 points)

2. Two jets are landing at, approaching, Pearson International on two differentrunways. One jet, a Jumbo 747, has a velocity of 150 km/h SW w.r.t. theground. The other, a Lear jet, has a velocity of 105 km/h at 20° W of N, w.r.t theground.a. Sketch and label an appropriate tip to tail solution of is the velocity of the

Lear jet w.r.t the Jumbo 747. (3 points)

b. What is the velocity of the Lear jet w.r.t the Jumbo 747? (7 points)

3. A fireman directs the fire hose to spray water onto a second story blaze. Individual water droplets have an initial velocity of 40 m/s at 55° above thehorizontal.a. What are the components of the initial velocity? (2 points)b. What is the maximum height a droplet may attain, relative to the end of the

hose? (2 points)c. Given the water strikes the building at a height of 10.0 metre above the

end of the hose, what is the vertical velocity of a water droplet, just beforehitting the building? Assume the droplet has already passed its maximumheight. (3 points)

d. What is the impact velocity of the droplet just before hitting the building? Include the angle. (3 points)

4. A 1350 kg pickup truck has gone over a steep bank and into a pond. Therecover crew attaches a cable to the truck and pull it out of the water. The bankis 40.0° with a µk of 0.32. Given the acceleration of the truck is 1.5 m/s2 up thehill do the following:a. Sketch and label both the free-body and vector component diagrams.(7 points)

b. Determine the net force on the truck. (1 point)c. Determine the tension in the cable. (2 points)

5. For the system below determine the acceleration and tension in all connectingstrings. Be sure to sketch and label all free-body and vector componentdiagrams. (10 points)

6. Jane swings from a vine on a jungle tree. The vine is 4.0 metres long. Initiallythe vine is horizontal and Jane is about to step off a branch.a. Sketch and label both the free-body / vector component diagram when the

vine is vertical. (2 points)b. What is her speed when the vine is vertical? (4 points)c. What is the tension in the vine given that Jane has a mass of 50 kg?(4 points)

7. The 600 kg cars of a roller coaster cross the top of a 12.0 metre high hill withsome unknown velocity. The next hill crest is 40 metres tall and the cars at thatpoint are on the verge of stopping. The track length between the hill tops is 152metres and an average frictional force of 800 N acts on the cars.a. What is the initial speed of the cars? (10 points)

8. A 70 kg hunter with a 5.0 kg rifle and a 20 kg canoe float motionless in a stillpond. The hunter shoots a moose such that the rifle is initially horizontal. a. Given the bullet has a mass of 25 grams and the recoil velocity of the

hunter, canoe and gun is 0.25 m/s, what is the bullet’s velocity? (4 points)b. What is the impulse on the bullet? (3 points)c. If it takes 0.6 ms for the bullet to leave the rifle, what is the recoil force on

the bullet? (3 points)

9. A 6.0 metre long, uniform draw bridge is raised to 14° above horizontal. Thecable that raises the bridge into this position is horizontal and is under 500 kN oftension.

a. Sketch and label the rigid body diagram. (2 points)b. What is the mass of the bridge? (4 points)c. What is the force of the pivot onto the bridge? (4 points)