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# General Physics I PHY 113

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This 266 page Class Notes was uploaded by Miss Lucienne Hamill on Wednesday October 28, 2015. The Class Notes belongs to PHY 113 at Wake Forest University taught by Staff in Fall. Since its upload, it has received 15 views. For similar materials see /class/230732/phy-113-wake-forest-university in Physics 2 at Wake Forest University.

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Date Created: 10/28/15

Review of Mechanical Waves Oscillations I Traveling waves a Sound Waves Wave interactions Harmonic Oscillations When a restoring force acts on a mechanical system spring on a block gravity on a pendulum the resultant motion is a simple harmonic motion xt Acosot I Harmonic oscillators move with an amplitude frequency and period Frequency Period 1sHz s a 1 27r T f 27r f a Harmonic Oscillation Systems Springblock system Simple pendulum 1113K 6 6 cosat m L Energy of Harmonic Oscillators I The total mechanical energy of a harmonic oscillator stays constant 1 2 1 2 E 3194 2 E mvmax for the springblock system I This energy is transferred back and forth between a potential energy elastic gravitational and kinetic energy I The object undergoing harmonic motion will have maximum kinetic energy at the equilibrium position and maximum potential energy at the extremes of its motion Traveling Waves Traveling waves water waves sound waves waves on a string are disturbances that move In a medIum Can be longitudinal disturbance is parallel to wave motion or transverse disturbance is perpendicular to wave motion Waves that are transmitted from one medium to another are not inverted Waves that are reflected from an interface with a denser medium are inverted upon re ection Waves that are reflected from an interface with a lighter medium are not inverted upon re ection Sinusoidal Waves x1 Wavelength T Period A Amplitude 1 Frequency Hz 27 k 7 Wave number m1 27 a 7 2727 Angular frequency rads 1 T V 4 Speed ms v 1 String u l rA lylEx A m mgtl b LE1 2004 Thomsunlamoks Cole Sound Waves l Sound waves are longitudinal pressure waves where air is compressed or decompressed periodically Air displacement 5x t Smax C0500 wt Pressure variation APxt 2 AP sinkx wt 111 x I The intensity of a sound wave is the power per unit area of the wave 3 Doppler Effect I If either the source or the observer of a wave is in motion the apparent frequency of the wave will depend on the relative velocities of the two Moving observer Moving source Both moving f vvojf V I All signs are with source observer moving toward the other Wave Interactions When two or more waves pass through each other they neither impede or destroy the other s motion I The total wave at a given point in time and space is the algebraic sum of the waves at that point I If the this sum is zero the waves interfere destructively if the sum is a maximum they interfere constructively Sinusoidal Wave Interactions if gt 2Acos s1n ICC 01 A smkx 01 b y 2 2 092747fau2mf Maximum amplitude 15 7f37f2quot 17I Minimum amplitude xi xi Ar 2143 Constructive Interference xi Ar 214 1 Destructive Interference The amplitude of the resultant wave depends on the phase difference between the two If the waves come from the same source but travel different paths Standing Waves I If two waves travel in opposite directions their interference will create a standing wave yR 2A sinkx cosat I This wave will have maxima antinodes and minima nodes at stationary points in space Nodes Antinodes Standing Waves on Strings and Air Columns I Standing waves created on a string with both ends fixed will only support a discrete number of frequencies Natural frequencies I An air column with both ends open supports all the natural frequencies of the system I If one end is closed only odd multiples of the fundamental frequency is supported Static Equilibrium and Elasticity Equilibrium Conditions Center of Gravity Examples Elastic Properties of Solids Static Equilibrium Objects under static equilibrium do not move I do not rotate This means that If the total force acting on an object is zero and the total torque about an axis is zero then the total torque about any other axis is also zero Point of Action Matters Net force is zero The lines of action pass through But lines of action do not cross a common point no net torque net torque is not zero But the net force is not zero A Simplification Normally we would need to solve many equations with many unknowns However if we restrict ourselves to forces acting on the xy plane F2 I How About Gravity Gravitational force mg acts upon the center of gravity of an object For uniform g center of gravity is at the same position as the center of mass Example P125 y 40 cm 2251712 2 1712 8 3850171 T V 1 2251712 1712 L120 cm 4 miJyi 2 251712 9 1713 2 quot 1 1 6850171 M 2251712 1712 A o rcM 3850171 6850171 ProblemSolving Hints I Draw a free body diagram I Show and label all the external forces acting on the object I Indicate where the forces are applied I Establish a convenient coordinate system for forces Then apply condition 1 Net force must equal zero I Establish a convenient coordinate system for torque Then apply condition 2 Net torque must equal zero Example A 1kg rock is suspended by a massless string from one end of a 1m measuring stick What is the weight of the measuring stick if it is balanced by a support force at the 025m mark 2F Iq quotrockgn s ckg 0 Z T n39zmckg025m mmkg025m 0 quotIsn39ck quotTrack 2 m m rockg stlckg Example 122 mg 500 N F I 300 cm cm F ng 50 583N d 3 Example 124 anHngQZ mg Z P 2lusmg 2113 Elastic Properties of Solids I External forces can cause a deformation in an object length shape volume I Stress An external force per unit area causing the deformation I Strain A measure of the deformation I Strain is proportional to Stress for small deformations elastic modulus I Length Young s Modulus I Shape Shear Modulus I Volume Bulk Modulus Young s Modulus Change in Length km Mm Shear Modulus FAA39 SE 0 g 39 3h AssZ52 Bulk Modulus gmg Mf g M3 13 f W gm m AV V AWEVQ Example P1228 150x1010Nmz StressmaX 150 x 108 N m2 dmin 2500771 L 25 0077 tens11e stressmax 150 gtlt10b F z125x10 3i gtlt150gtlt108 736x104N tensile stress FA AL tensile stress 150gtlt108 tensile strain AL L Y quot mesaW 0250 Review I Static equilibrium requires a zero net force and torque as well as no linear or rotational motion 0 I Gravity acts upon the center of gravity The various deformations of objects are quantified by the moduli Length Young s mod Shape Shear mod Volume Bulk mod tensile stress F A V shear stress F A volume stress F A AP tensile strain AL Ll Shear strain Ax 7 volume strain AVl39l Al39 I Oscillatory Motion I Object attached to a spring I Simple harmonic motion I Energy of a simple harmonic oscillator I Simple harmonic motion and circular motion I The pendulum An Object Attached to a Spring When acceleration is grogortional Q and in the opposite direction of the displacement from equilibrium the object moves with Simple Harmonic Motion 121104 ThomsonBrooks Cale Equation of Motion Second order differential equation for the motion of the block The harmonic solution for the springblock system xt A cosat where a 2 JE Some Terminology Angular frequency Phase constant xt oosat Properties of Periodic Functions a The function is periodic with T xt xt T The maximum value is the amplitude Angular Frequency Frequency Period rads 1sHz s 1 27 a T I f 2 f a Simple Harmonic Motion A cosat 5 xt x111 ax l 10 4 w l m S 0 Am dx dt Properties of Simple Harmonic Motion Displacement velocity and acceleration are sinusoidal with the same frequency The frequency and period of motion are independent of the amplitude Velocity is 90 outof phase with displacement Acceleration is proportional to displacement but in the opposite direction Example P151O I A piston in a gasoline engine is in simple harmonic motion If the extremes of its position relative to its center point are 575 cm find the maximum velocity and acceleration of the piston when the engine is running at the rate of 3750 revmin 3750rev 60s f 625HZ a 279 275625Hz 3927rads Aa 00575m3927mds 226ms armax A02 OOS75m3927rcla s2 8867ms2 The BlockSpring System i 1 w 1 T 2 2 m Frequency is only dependent on the mass of the object and the force constant of the spring I71 T221227 a k MW ms r Cue T 126s w 1T1 x a 111 x a Example 153 x0 I A gt a f I I mamm Jim WKMMQI WWIWW xt A coswt 005171cos500t x0 005mcos 2 005m gt JE J 500rads m 0200 27 xt A cos wt 2 005mcos 500t r at a 20A 500md s005m 2 025m s vmax sin wt 2 025m ssin 500t wZA 500139ads 005m 125m s2 1T1 21X cos wt 2 125m s3cos 500t Energy of the Harmonic Oscillator Consider the blockspring system a If there is no friction total mechanical energy is conserved At any given time this energy is the sum of the kinetic energy of the block and the elastic potential energy of the spnng Their relative share of the total energy changes as the block moves back and forth Energy of the Harmonic Oscillator 1 E 231542 mvm x0 A cosat 5 V0 2 Aa sinat K mv2 Z gin1sz sinz W U 1062 2 gig12 0052wt E 2 K U constant Energy of the Harmonic Oscillator quot y WMlt5A 1MMWsv x t 0ltWltW 42r aMAMMWWWme I v 1 Example P1518 I A blockspring system oscillates with an amplitude of 370 cm The spring constant is 250 Nm and the mass of the block is 0700 kg I Determine the mechanical energy of the system Determine the frequency of oscillation Ifthe system starts oscillating at a point of maximum potential energy when will it have maximum kinetic energy a When is the next time it will have maximum potential energy E ch2 2 250Nm0037m2 0171 1 k 1 I 250 27 27 m 27 0700 The Simple Pendulum to 2004 ThomsonBrows Cale m Sin 9 mg cos 9 The tangential component of the gravitational force is a restoring force dis E mg sm 6 m c f For small 66 lt 10 6 gt The form as simple dr L harmonic motion 61mm cosar L 5 g gwg 6 62 g The Physical Pendulum 2 26 2 Iz nzgdsin621a 1 dt O For small 66lt 10 2 26 mgd 2 0 1 Si 9 ch2 1 6 mgd 6 Hmax cosat 1 3233 f 7 ma Example 157 If we had assumed a simple pendulum of mass M hanging from a string of length L 9 2004 ThomsonBrows Cale Simple Harmonic Motion and Uniform Circular Motion i I 39I 2004 Thomsunmmoks Cole X component xt A cos 9 A cos at y component yt A sin 49 A sin wt Damped Oscillations Suppose a nonconservative force friction retarding force acts upon the harmonic oscillator Review I Restoring forces can result in oscillatory motion xt ACOSat Displacement velocity and acceleration all a i a 5 oscillate with the same frequency m L Energy of a harmonic oscillator will remain constant Simple harmonic motion is a projection of Circular motion Resistive forces will dampen the oscillations Classical Mechanics Motion of objects Large objects much larger than an atom Slow objects much smaller than the speed of light Key concepts displacement velocity and acceleration a force energy momentum Physics and Measurement I Basic quantities of mechanics I SI units I Modeling and quantifying matter I Problem solving I Dimensional analysis I Units and conversion I Estimates and order of magnitude calculations I Significant figures Basic Quantities of Mechanics Length SI Unit meters m Standard is related to the speed of light I Mass SI Unit kilograms kg Standard is a specific Pt Ru cylinder in France I Time SI Unit seconds 5 Standard is a vibration of a Ce atom Range of Values Length m u The farthest quasar 14 x1026 The moon 15 x1011 Football field 91 x101 3 Typical cell 105 Atom 103910 Time s Age of universe 5 x1017 Age of earth 13 x1017 Mass kg I Universe 1052 You 63 X108 1 24 n One heartbeat 8 x 10 Earth 598 X10 Nuclear collision 103922 Human 102 Bacterium 1015 Hydrogen atom 167 x 1027 Prefixes for Powers of Ten Table 14 Pre xes for Powers of Ten Power Pre x Abbreviation HF HF chlu 7 21110 IO 13 13mm 10 pins 10 quot mum H mirm milli ccnli 10 10quot 10 10 1 7quot 10 m 10 yullu E 2004 ThomsonBrnoks Cole Structure of Matter Macroscopic objects I Molecules Atoms Electrons protons neutrons Quarks Measuring the Amount of Matter I Atomic number of protons I Specific to an element I Mass number protons of neutrons I Varies from isotope to isotope 14C I Atomic Mass the mass of a single atom of the element in atomic mass units I An average quantity I 1 atomic mass unit u 166 x 103927 kg I Aluminum 27 u I Lead 207 u I Mole the amount of substance that contains as many particles atoms molecules other particles as there are atoms in 12 g of the 120 isotope I One mole contains 6022 x 1023 particles Avogadro s number I In general the mass in 1 mole of any element is the element s atomic mass expressed in grams Mass Density I Mass per unit volume A derived quantity Densities of various p44 27 gcm3 i amer 2721 SUbstances pm 113 gCm3 amuph 20721 Substance p103 kgm3 39 7 Lead 113 Density is not only a measure of the amount of matter but also 3 Aluminum 27 measure of how tightly It IS Water 1 packed Air 00012 Example An Iron Block A small cube of iron is observed under a microscope a Find the mass of the block b Find the number of iron atoms in the block p 786 gcm3 L 10 um Molar mass 559 gmole m 786x10 12kg of atoms 2 847 gtlt10l3 Dimensional Analysis A dimension in physics is the physical nature of a quantity L M T 1 speed v L T a area A L2 It can be treated as an algebraic quantity Always do a dimensional analysis of a problem Newton39s law of universal gravitation is represented by the following equation Here F is the gravitational force M and m are masses and ris a length Force has the SI units kg msz What are the SI units of the proportionality constant G Units and Conversion It is very important to use the right units for the right physical quantity and keep them consistent throughout a problem Convert the 65 mph highway speed limit to SI units 65mph265 1609m 1h 1mm 22905 h lmi 60mm 603 s Estimation and Order of Magnitude Assign common sense typical numbers to unknown quantities Calculate the required quantity to an order of magnitude Do a reality check does the result make sense How much gas is used annually by all the automobiles in the US I Estimate amount of cars I Estimate miles driven per car I Estimate mpg Significant Figures I The number of reliably known digits in a measurement I When adding and subtracting use the smallest number of decimal places in any of the terms I When multiplying or dividing use the lowest number of significant figures in any of the terms Review I Length mass and time are the basic quantities of mechanics I We use SI units m kg s I The structure of matter goes from molecules to atoms to nuclear particles to subnuclear particles I The amount of matter can be density or moles I When solving problems use dimensional analysis estimation and order of magnitude reasoning to get an idea and to check you answers I Keep your units and significant figures straight Wave Motion I Propagation of a disturbance I Sinusoidal waves I Speed of waves I Reflection and Transmission Propagation of a Disturbance I Waves on a string I Stadium waves I Sound waves I Waves on a spring I Water waves I All require I a disturbance I a medium to be disturbed I some mechanism of influence Transverse Waves particle Stadium waves H if Waves on a string m a I The disturbed particle and the wave move perpendicular to each other Longitudinal Waves Sound waves Waves on a spring r z 4 Compressed 11 if we Stretched Stre ch id particle wave H The disturbed particle and the wave move parallel to each other Combined Wave Motion Water waves Velmjily 0f 1 pagati n The disturbed particle moves both parallel and perpendicular to the wave One Dimensional Wave The pulse is moving at a speed of v in the xdirection It creates a displacement of y y is a function ofX and t Right moving pulse Left moving pulse 4 quotquotquotquotjlt rtt 1 39 y f x Vt b PIIISC at filmquot I Example A wave pulse is moving as illustrated with uniform speed v along a rope Which of the graphs 1 4 below correctly shows the relation between the displacements of point P and time t Sinusoidal Waves Mx A m2fltxw D Q and time t Variation in position x Sinusoidal Waves x1 Wavelength f T Period A Amplitude 1 f 2 Frequency Hz Wave number m1 27yquot Angular frequency rads v 412 Speed ms l V A ly g g Ae mltl 09 b LE1 2004 Thomsunlamoks Cole Example P1613 A sinusoidal wave is described by y 025msin030x 40t where x and y are in meters and t is in seconds Determine a the amplitude b the angular frequency c the angular wave number d the wavelength e the wave speed f and the direction of motion Waves on a String Each element of the string oscillates vertically with simple harmonic motion yx t A sinkx wt V x f 0A cosloc or a x t a2A sinkx or T tension in the string u 2 mass per unit length of the string Reflection of Waves Incide 311 fitquot zfza ty Invidom PHI Rigid Support The Wave is m upOn reflection Loose SuPport m upOn reflectiOn Transmission of Waves When the boundary is neither totally rigid or loose the wave is partially reflected and partially transmitted Light to heavy string Reflected wave is inverted Transmitted wave is not inverted gt Ill39ll lll plll Roman1 misc 4 a ame Heavy to light string Reflected wave is not inverted Transmitted wave is not inverted lnrirlcnl 1 39r In General I Transmitted waves are not inverted I If the wave is traveling from a medium of high speed to a medium of low speed traveling from a lighter to a denser medium the reflected wave is inverted I If the wave is traveling from a denser low speed to a lighter medium high speed then the reflected wave is not inverted Review I Longitudinal and transverse waves I Traveling waves have amplitude and velocity I Sinusoidal traveling waves are periodic in time and space I Transmitted waves are not inverted I If a wave reflects off of a denser medium it gets inverted othenNise it does not Potential Energy I Potential Energy I Conservation of Mechanical Energy I Conservative and Nonconservative Forces I Energy Diagrams and Equilibrium Potential Energy I Think of it as stored energy in a system that can do work or change the system s kinetic energy Gravitational u Elastic Chemical electrical etc I When work gets done on a system its potential andor kinetic energy increases Gravitational Potential Energy A o W Fapp Ar mgj yl ya J mgyb ya Ugamgy Gravitational potential energy only depends on height y and not on lateral distance x Example P83 l v 0W 120 ms Eff 25 P The mass of water flow per second 111 pV 1gIquotc1113X300c111gtlt 506111X1206111S 18 X106g1quots 1800kgs The Change in potential energy per second of this mass as it falls down 5 m Power generated by this Change in energy P 025AUAz 22kW AUAZ mgh 1800gtlt98gtlt5 882X104JS Elastic Potential Energy The work done by the spring The potential energy U 1 3 stored in the spring 5 5 1399 2004 ThomsonBrooks Cola Example 00878 A mass m is bobbing up and down on a spnng Describe the various forms of energy of this system a At the highest point Gravitational and elastic potential energy b At the point where the kinetic energy is highest Kinetic energy gravitational potential energy c At the lowest point Gravitational and elastic potential energy Conservative Forces I Conservative Forces I The work done by conservative forces is independent of path II The work done over a closed path is zero I We can always associate a potential energy with them I We can only associate a potential energy with them I Gravity springs are examples Work done by a conservative force Relationship between a conservative force and potential energy Mechanical Energy and Energy Conservation l Mechanl Ene g Kine Erma 4 Potential Ener l In an isolated system mechanical energy is conserved when only conservative forces act upon the system Roller Coaster Example 0086 Three balls are thrown at different angles from the roof of a building with the same speed Describe the motion of the balls Projectile motion parabola 8 O V Rank their speed when they hit the ground All the same 1 2004 Thumsonr roaks Cale Example 83 mgL1 cos 6A msz VB 2gLil c056A 2 ZFr mg TB mv TB mgm2g1 c056A TB mg3 2c056A 2004 Thomsunf muks CD S Example 85 quot quot m 350 g AX 0120 m h 200 m 1 7 kAx mgh 71quot1l k 7 2 2ngh 953Nm 1 2 1 2 kAx mvB mgAx VB r 2 2 VB AxZE 2gAx 197mS m t 1 J 9 2001 lemsonl mcks Cole Nose Basher Nonconservative Forces I A nonconservative force will change the mechanical energy of the system I The total work done by the force depends on the path taken by the object I Example Friction A quot The amount of work is done against friction and depends on the path The change in the mechanical energy of the book is irreversibly converted other forms of energy heat Work Done by Nonconservative Forces I Work can be done by an applied force Lifting an object increases the potential energy of the object Work can be done by kinetic friction As an object moves on a surface with friction it will lose kinetic energy Example 810 AE mech AUS Ang AE 1118017 1 397 kh I1L h 2 g AE fkh Ltki111gh mech 1 ML kh g 2 u 1111 g Review Gravitational Elastic Potential Energy Potential Energy U m U 1k 2 g 83 S 2 x Mechanical Energy Kinetic Energy Potential Energy ME is conserved when only conservative forces act on a system spring gravity Example of a nonconservative force friction Fluid Mechanics Pressure I Pascal s Law Archimedes Principle Fluid Dynamics I Bernoulli s Equation Pressure Fluids apply a compressive force to submerged objects from all sides This means that the force is spread out over a surface area Elcuum Force per unit area F Pressure P E 1 Pa 1 Nmz If pressure varies over the area 61F PdA Example 0142 Both dams have the same height and width Which needs to be stronger an i a 1 e Example You hold a thumb tack between your index finger and thumb with a force of 10 N The needle has a point that is 01mm in radius whereas the flat end has a radius of 5 mm aWhat is the force experienced by your finger what is the force experienced by your thumb F 10N 1 32gtlt108Pa bYour thumb holds the pointy end What is the pressure on the thumb what is the pressure on your finger 2 13gtlt105Pa F 72395 gtlt 1073111 A F A Variation of Pressure with Depth Pressure exerted by a liquid increases with depth EP Pressure at sea level is taken to be 1 atmospheres atm latm 21013gtlt105 Pa Example 144 F Pressure varies with depth dF pg H ywdy Pascal s Law A change in the pressure applied to a fluid is transmitted to every point of the fluid and to the walls of the container Example 142 A1 7rd1 22 196cm2 A Adz2 217670177 F 13300N F2 13300N 752X105Pa 2 743m EF3A1213300 21478N P 74 A2 1767 A2 1767gtlt10 m Buoyant Forces Archimedes Principle Archimedes Principle The magnitude of the buoyant force on an object equals the weight of the fluid displaced by the object B mfg Z prfg Example 145 Weight in air 784 N Weight in water 684 N L B IN V Vc 2 3 1 102gtlt10 4m3 pfg lkgm 98ms r 784 98 K l02gtlt10 Totally Submerged Objects a is upward if pfgt pobj a is downward if pflt pobj 2F probjg poijobjg Floating Objects B Mong pk Vfg poijobJg prfg Example Consider an object that floats in water but sinks in oil When the object floats in water half of it is submerged If we slowly pour oil on top of the water so it completely covers the object the object 1 moves up 2 stays in the same place 3 moves down Fluid Dynamics I We now put the fluid in motion flow I Here are several assumptions about the fluid and its flow The flow is to be laminar steady not turbulent u The fluid is nonviscous negligible internal friction Think water not honey I The fluid is incompressible The flow is irrotational no angular momentum Equation of Continuity Point 2 The product of the velocity of flow and the area of the pipe remains constant A1121 2 A2122 constant Example A blood platelet drifts along with the flow of blood through an artery that is partially blocked by deposits As the platelet moves from the narrow region to the wider region its speed 1 increases 2 remains the same 3 decreases What about the pressure on the platelet Bernoulli s Equation Point 2 Using conservation of energy AE W Point 1 A961 P1 Ali x 5 1 2 I P E pv pgy constant Example 149 A1 7rd1 2V2 196cm2 A2 7ra 22 707cm2 szzm 196gtlt10 4X1510gtlt 60 6361113 V out Review F Pressure PE PEPO pgh Pascal s Law A change in the Archimedes39 Law The pressure applied to a fluid is magnitude of the buoyant force transmitted to every point of the on an object equals the weight fluid and to the walls of the of the fluid displaced by the container FIAZ F2141 ObleCt B mfg Equation of Continuity Am A2quot constant Fluid Dynamics 1 7 Bernoulli s Equation P Epvt pgy constant Uniform Circular Motion Motion in a circular path with constant radius r and constant speed v Velocity does change in direction Then there is an acceleration g Centripetal Acceleration From the similarity of the velocity and radial vector triangles lArl At Centripetal acceleration V2 27 directed towards the Q g T center 37 V Uniform Circular Motion and Newton s Second Law Centripetal acceleration is directed towards the center of the circle of motion ar ac It is always perpendicular to the velocity no change in speed The string is applying the force to keep the ball from going in a straight line When the String Breaks When the string breaks meaning no radial force is applied anymore the ball continues to move at constant velocity in a straight line along the tangent to circle at the break point Example 62 1 6t 6 Conical Pendulum Example 64 nlElmoks Cola fisumax 3quot n lac I I I What is the maximum speed of the car without skidding m 1500 kg rcurve m 3 0 5 7 1 y max uSgT 131ms Radial and Tangential Acceleration Path of particle g V g V 2 g The object changes speed and direction Acceleration has a radial and tangential component Nonuniform Circular Motion I Ifthere are radial and tangential components to acceleration the force must also have these components Resistive Forces I It is the interaction between the object and the medium it moves in a Air drag viscous forces in liquids I It is dependent to the speed of the object I The exact dependence is an empirical relationship Objects in liquids tiny objects in air D Air drag coefficient p Density of air A Area of object v velocity of object Large objects in air Terminal Velocity Occurs when the resistive force equals gravity Then the net force and acceleration is zero The object moves at constant speed VT Sky divers soap bubbles Example 613 Find the resistive force on a 90 mph fastball mbaseball 0145 kg Assume the baseball is dropped vertically and using the values in the book for vT A and p calculate the drag coefficient Then use this value to find R with v 90mph D 2mg pVT 2 A Review l Radial acceleration aC mvr2 I Radial force F maC Viscous drag will limit velocity Angular Momentum Vector Products I Torque Angular Momentum I Conservation of Angular Momentum Vector Cross Product Remember the definition of torque We now define a new vector Righthand rule relationship Vector or Cross Product Properties of Vector Product AXB BXA AgtltBC AgtltBAgtltC Examples A x B AXE ABXI2 2 X 2 3 x 112 712 EX A BX1y BAXIA 1gtlt 3 2x 22 712 AxBBgtltA AXBC BxAz AXB AXBXBX A C2 Torque EEXF T rFsin F 200i3oojiv T r 400i 5003 rer 17 4gtlt3 5gtlt21A 2Nml Angular Momentum Example P 1111 1 A o L3kg 05m gtlt 3ng5m SJ 75kgm2 M L L3kg L 175kgm s Example 114 L my I I ZZVR MVR L 2 ml 7712 MVR 2 rm 2 2 ml 7712 MVR dt mg 2 ml 7712 M Ra m a 2 1g ml 7712 M Angular Momentum of a Rotating Rigid Object For an object rotating about the zaxis L 2m E The angular momentum and velocity are along the zaxis Example 115 M 2 6kg R 126m w10revS L2 9 I sz2 6kg012m2 0035kgm2 L 0 035kgn12S Conservation of Angular Momentum The total angular momentum of a system is conserved if the net external torque acting on the system is zero L constant I The initial and final angular momentum of an isolated system is constant after an internal rearrangement L I Lf constant I For a rigid rotating object 110 I fa f constant Example P 1130 1m 2 3kg X 1177 3kg X 1177 6kgm IS III39iai 3 6075 6751ng s Li Li 2251 ad s 3 of 0750mds m 300kg 1 2100177 I r S f 3 0300177 00kg m2 Review Vector Product C 2 AB sin 6 Torque Angular Momentum 2e I Angular momentum is conserved Energy and Energy Transfer I Systems and Environments Work Done by a Constant Force Scalar Vector Product Work Done by a Variable Force Kinetic Energy A System and Its Environment Identify a system One or more objects I A region of space I Set a system boundary not necessarily a physical boundary I Everything else is the environment Work as defined by a physicist The product of the force acting on an object in the direction of its displacement and the magnitude of the displacement I s 1 l 6 FCOSQ 39 l No work is done in either case k 3 The walking waiter is W Z II donng some work Ar Work and Energy Transfer I If the applied force and displacement are parallel work done by the environment on the object is positive Energy is transferred from the environment to the system I If they are antiparallel negative work is done on the object and energy is transferred out of the system Example 71 500 N d3m What is the work done by a the janitor W Fd 0086 50NX3mcos30 130J b gravity c the normal force WGW O n More on Work I Work is a scalar quantity I Its units are 1 Nm 1 Joule J I It is the clot product scalar product of the force vector and the displacement vector 6 d Wchos6Fd The Dot Product of Two Vectors AogAmgl ABCABAC A Axi A Azk A B AxBx 1413 AZBZ B Bxi By 13213 A A AA AA AZAZ A2 Example 73 Ar 20i3ojm F 50320N 4m 2 arctan 2 563 20 4F 2 arctan j 218 50 W FAr cos563 218 160J W FAr 20503020160J Work Done by a Variable Force Area 2 AA Ax Consider force and displacement along the xaxis W lim iAW lim iFvAx ARC gt0 Av gt0 Work is equal to the area under the FX vs X graph 3 a grade 3 For more than one force acting upon a particle a K ncgmiw Hooke s Law Spring Constant Iquot i ni1it39 ix negzuiw Work Done by a Spring For arbitrary displacement between xi and xf a 1 J 1 megwxegwj f When an external force is applied to compress or expand the spring Yf 39 39j w calyx Yf xi i 71 WWEWEK Example 76 WorkKinetic Energy Theorem Consider a block moving under the influence of a net force A Z W F dx Tmadx W gt Km dx gt m dx mvdV Z dt dx all x y Ti 2 2 The total work Wg 2W W gt done by the force 72 1 r 2 lag 253 The Klnetlc Energy of the particle thatme eettemr eneane mxmmm me 2W 2 Kfzgg 2M Example 77 m6kg F12N Ax3m Work Due to Friction I When friction is involved work must be done against the kinetic frictional force I This transfers the kinetic energy of the particle to its internal energy Example 79 m60kg F12N uk015 Ax30m vf W 2 FM 2 12N3m 36J n mg 2 588N ukn 882N AK mn fd 882N3m 265J other 1 q E 77Vf AK ic on Z W meaX W FAxcos Kf d W uk mg F sin 9d 67 6 Fd cos 9 1 d6 uk cos6 sin 9 0 9 arctanuk 85o Power Power is the time rate of energy transfer or the rate at which work is done or the rate at which energy is transferred from on system to the other Instantaneous power Average power work done per time interval At Power 1 watt W 1 Js 1 kgm2s2 1 horsepower hp 746 W 1 kWh total energy delivered in an hour at a rate of 1 kW 103 W 3600 sec 36 X 106 J Example 712 Motor mam 1600 kg mpass 200 kg f 4000 N a v3mls T f Mg0 P T216gtlt104N Mm 777 el ev pass P 2 TV 2 648x104W b a 100 ms2 TfMgMa T234gtlt104N PTv234gtlt104v P 200d ThumsnnBmuks Cole Review Work done by a constant force Work done by a variable force Hooke s law Kinetic energy Work kinetic energy theorem Work done by friction Power Wdecos zFd Superposition and Standing Waves l Interactions of two waves in the same medium I Superposition Interference Standing Waves Strings Air columns Superposition of Waves If two or more traveling waves pass through each other at any given point the resultant wave is the algebraic sum of the wave functions of the waves This means that two waves can pass through each other without being destroyed or altered Examples v10cms tO6s Total wave 2004 ThomsonBrooks Cole Interference I The formation of a resultant wave by superposition is called interference I If the algebraic sum of two waves add up displacements in the same direction we have constructive interference If the amplitudes cancel each other out displacements are in opposite directions we have destructive Constructive Destructive interference interference interference Sinusoidal Waves l and y are idenlival A sinkx of y2 A sinkx 01 5 y y1 y2 Asinkx ol sinkx 01 15 y 2A m limawg 1 The resultant wave has the same frequency and wavelength of the interfering waves lts amplitude is dependent on the phase 39 difference between the two waves 02754752n7f Maximum amplitude to 150quot 7r37r2n 17r Minimum amplitude mmmmm Interference of Sound Waves When sound waves from a single source follow different paths to a receiver they can interfere constructively or destructively depending on their relative phase jsinkx 01 xi Ar 214 Constructive Interference 2 xi Ar 214 1 Destructive Interference Example 181 0 18 II 39a I I 7 115 m P0350111 L 1 P is the first minimum of sound intensity f Standing Waves I Iftwo waves traveling in opposite directions yl ASIIIUOC wt interfere I The result is a standing wave yR sm cosa n A standing wave is a an oscillating pattern with a stationary outline It has nodes and antinodes Standing Wave Patterns The nodes occur when the outline is zero A 31 2n11 4 4 j 4 Example 182 yl 400111sin30x 20t y2MOa in 0x2m yR x 2 236172 2 yR 2A sinkx 00502 24sin3 x 23 00522 46 00522 Xnode Xantinode Qn02 4 antinodes 2n 1cm ymax antinode yanzz nodemnax yR111aX 24Cn1 8C Standing Waves in a Fixed String I Since both ends of the string are fixed they have to be nodes This leads to quantization where only certain frequencies called the normal modes are allowed to oscillate I For a string of length L Example I A cello Astring vibrates in its first normal mode with a frequency of 220 Hz The vibrating segment is 690 cm long and has a mass of 190 g I Find the tension in the string Determine the frequency of vibration when the string vibrates in three segments fizzonzzi i 1 T 3 00019 069 T 2202069 2539N 2L Lz2069 00019069 f 2 3f 2 660Hz Notes and Their Fundamental Frequencies 31011 31111 H rm 3 D E auanhalg Resonance Assume an oscillatory system that has a series of natural frequencies Now assume an external periodic force that is applied to the system If the external force has a frequency that matches one of the natural frequencies of the system the amplitude of the resultant motion is greatly amplified 0 quotC5 ye 5 EL 2w Thomso Frequency of driving force Standing Waves in Air Columns I A standing wave pattern can be set up by sound waves that reflect back and forth in an instrument I The closed ends would be displacement nodes and pressure antinodes I The open ends would be displacement antinodes and pressure nodes Standing Waves in Air Columns First lu u mmlit Remind IIEII39IXIDIIlL Third harmonic a Open at boll ends 1 2004 ThomSDnElmoks Cole When both ends of a pipe are open the natural frequencies of oscillation are all the integer multiples of the fundamental frequency Standing Waves in Air Columns Firsl harmonic i 3 i quot 39l39llil d ll r l l0lli39 n Filthharmonic l l l r a r t 4 7 bl Clusul at um vml 11pm al the ullwr 2004 ThomsunJElmoks Cola When one end of a pipe is closed the natural frequencies of oscillation are all the odd multiples of the fundamental frequency Example 187 Lmin f L for the next two resonances 900 cm First resonance A4L 111 in 2 36cm Second resonance third harnu mic Third v 343m S Lfi Z1 036m 953H resonance fth harmonic Review I Waves can travel through each other without affecting the other I The total wave at a given point is the algebraic sum of all the waves at that point I This leads to constructive or destructive interference I Waves traveling in opposite directions create standing wave patterns Rotation of Rigid Objects About a Fixed Axis I Angular position velocity acceleration I Constant angular acceleration I Rotational kinetic energy I Moments of inertia I Torque I Work power and energy I Rolling motion Rotational Motion Arc length 3 Reference line measured in radians L5 2004 ThomsonBrooks Cole Angular Quantities Angular displacement Average angular speed Instantaneous angular speed Average angular 0f 0 acceleration Instantaneous angular acceleration A particles of a rigid object rotate at the same angular displacement speed and acceleration Angular Velocity Angular velocity a is a vector For rotation about a fixed axis the direction of the velocity is along the axis of rotation Use the right hand rule to determine direction Rotational Kinematics Linear 1 D Motion with constant linear acceleration a Rotational Motion with constant rotational acceleration a a 1 wiat f Q 61 ml wf t 1 6ft 2 61 alt ort2 2 52 of 2a 9f al Example 101 I 0s m 200 rads 1f 200s 9f 2 If of 0 0512rads35radSZX2s 9rads 6f 2 61 011 01lZ 2rads2s35radSZXZS2 11rad Angular and Linear Quantities Arc length 3 S g 37 Tangential speed 5 a ofa point P Tangential acceleration ofa point P 27 Rotational Energy A rigid object made up of a collection of i particles with mass m has a rotational kinetic energy of 1 2 Kagm it 2 where I is the moment of inertia 7 I 22 fgzm Calculating Moments of Inertia For an extended rigid object 1 lim ZIQZAmI Irzdm Am gt0 Example Uniform Solid Cylinder with mass M and an axis of rotation along the zaxis dV LdA L27zrdr I I prldV I pl L27Z7 dl 4 R R I pL27rIr3dr pL27r 0 The moment of inertia is dependent on the axis of 1 lMLZ rotation 3 l Moments of Inertia of Various Objects CM Moment of inertia about an axis of rotation through the center of mass Parallel Axis Theorem If CM is known the moment of inertia through a parallel axis of rotation a distance D away is 1 CM MD2 Long thin rod L Long thin with rotation axis rod with thrmlgh center 1 rotation axis through end Q 39 Q d K 00 x Q x o I Consider a rigid object about a pivot point I A force is applied along the line of action 1 FF Sin Fd I This force causes the object to rotate Torque and Angular Acceleration Example P1036 The combination of an applied force and a constant frictional force produces a constant total torque of 360 Nm on a wheel rotating about a fixed axis The applied force acts for 100 s During this time the angular speed of the wheel increases from 0 to 100 rads The applied force is then removed and the wheel comes to rest in 600 s Find a the moment of inertia of the wheel b the magnitude of the frictional torque c the total number of revolutions of the wheel Am 10 0rad s 7 Am 10 0rads 2 a 2 2 20167mds lmd S 1 Aid 603 Z Z39 Z39 lord 36kngX017radsz 6kgm 36Nm 36k 1112 l 3 l 3 a hads2 8 A6 2 Gama 2 ordAld 3501ad A6 350md 27 557I39ev Work Power and Energy Work in linear motion Work in rotational motion 2 w W 2 sew Work Kinetic Energy Theorem The workkinetic energy theorem for linear motion External work done on an object changes its kinetic energy and for rotational motion External rotational work done on an object changes its rotational kinetic energy Rolling Motion Pure rolling motion No slipping occurs Rolling Motion The linear speed of the center of mass Ik The linear acceleration of the center of mass While the angular velocity and acceleration of any point of the object is the same the linear speed changes Rolling Motion I gt Ii W M 11 21 M 7 7 M I 739 Ha 4 07 I h I urc mmlinn Ell l urc translation C Inmbinuliun ul ll unsluliun um I39Ulilllll xmmr Pure rolling motion is a superposition of pure translation and pure rotation motions Kineticenergy of K lMy2ij 2 rolling motion quot 2 2 CM Example P1055 h 3msin 25 2127171 1 l M gh 0 0 EMVCM r a 4m S R 003l9m 1254rads M w 02150 3 a I a Zgh VCM I W2gl27m 4ms 1122gtlt10 4kgm Review I Linear quantities have analogous angular counterparts I Torque is the tendency of a force to rotate an object I The total kinetic energy of a rotating object has to include its rotational kinetic energy I Pure rolling motion is the superposition of pure translational and pure rotational motion The Laws of Motion I The Concept of Force I Newton s First Law I Mass I Newton s Second Law I Gravitational Force and Weight I Newton s Third Law I Friction The Concept of Force I Eg Push pull throw an object gravity magnetic attraction I Only a force can cause a change in velocity I which results in acceleration I Force is a vector quantity Contact and Field Forces Contact forces Field forces act through physical contact act through empty space Measuring Force Force is measured as the elongation of the spring I The total net force is the vector sum of all forces acting on the spring lfthe net force is zero the object will be in equilibrium its acceleration will be zero and its velocity will be constant Newton s First Law I In the absence of an external force u an object in motion remains in motion with constant velocity u an object at rest remains at rest I This means acceleration is zero I The tendency of an object to resist any change in its velocity is called inertia Inertial Frame of Reference I A frame of reference that is not accelerating u Examples A stationary observer a car moving with a constant speed in a straight line I The first law is only valid in inertial reference frames Mass I It is an inherent property independent of the environment and measurement I It can be thought of as the resistance of an object to any changes in its velocity I Don t confuse mass with weight which is the gravitational force exerted on an object Newton s Second Law l The acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass Units of Force I The SI unit is 1 Newton N Equal to the force required to accelerate a 1 kg object to 1 msZ I 1 N 1 kgms2 Example 51 m 030 kg What is the acceleration of the puck av 4a 2 arctan quot a X L1 2004 Thomom39Bmoks Cole Gravity and Weight l Gravity is the attractive force every object feels towards Earth I The magnitude of the gravitational force is the weight F 2 m Gravitational Force g g Fg mg Weight Weight can Change from planet to planet but mass is constant Newton s Third Law If two objects interact the force F12 exerted by object 1 on object 2 is equal in magnitude and opposite in direction to the force F21 exerted by object 2 on object 1 Example Computer monitor at rest on a table The action and reaction forces act on different objects Free Body Diagrams Draw a diagram of all the forces acting on an object Take care of which forces act on which objects Example 54 I Conceptualize the problem diagram I Categorize the problem equilibrium I Isolate an object and identify the forces acting upon it I Establish a convenient coordinate system I Apply Newton s second law to each diagram I Solve the equations simultaneously I Check your answer The Normal Force Consider a person standing in an elevator that is accelerating upward The upward normal force N exerted by the elevator floor on the person is 1 larger than 2 identical to 3 smaller than the downward weight Wof the person Example 56 aa 2 Fx 2 mg 5111 6 max 2 n mg cos 6 2 may Example 59 W m mg bT T g n I V 39 E1 2004 Thomscnr mu ks Gala 1m 71722 Atwood machine Fnc on Friction is due to the interaction of two adjacent surfaces at the atomic or molecular level It is parallel to the surface of interaction and opposite the direction of motion or impending motion Fnc on Friction is proportional to the Static Friction normal force Friction is independent of fl m area us is generally larger than uk The coefficients do not depend on weight or speed and generally range between 0003 and 1 Kinetic Friction Example Friction I A person pulls a block across a rough horizontal surface at a constant speed by applying a force F The arrows correctly indicate the directions but not necessarily the magnitudes of the forces on the block Which of the following relations among the force magnitudes of W k N and F must be true kandNW gtkandNltW Example 512 mg sin 6 fS O A fs111ax Ills n mg005620 mgcosO Review I Only a force can cause acceleration I F ma Weight F mg 39 F12 21 I Static and kinetic friction is proportional to the normal force Motion in Two Dimensions I Position velocity and acceleration vectors 2D motion with constant acceleration Projectile motion Position Vector r is drawn from origin to the position of the object at a given time Displacement vector T Path of particle Velocity Average velocity A v E M Speed Instantaneous velocity Mdth Venime Me Ag gig Acceleration Average acceleration Acceleration can cause a change in the magnitude or the direction of the velocity of an object or both Two Dimensional Motion With Constant Acceleration I Use 1D equations but replace scalars with vectors I Use the superposition principle Break down each vector to its components Calculate each component independently ViVfm f 1 2 rfnVJ at Two Dimensional Motion With Constant Acceleration Ki 391 21 Example P 465 The coyote has skates that give a constant acceleration of 15 ms2 He starts from rest and catches up the roadrunner at the edge of the cliff 70 m later a What is the speed of the coyote when he 7 catches up With the roadrunner 458ms 1 2 xf x1 vxit 709 fo Vx at Of course the roadrunner makes a sharp turn right at the edge and the coyote flies off the cliff and maintains his horizontal acceleration If the cliff is 100 m above the floor of the canyon yf Z M H371 aytz l 2 1f xi v agt 1 b Where will the coyote land c What will its final velocity be Projectile Motion The object has an initial velocity vi and is launched at an angle 61 Two assumptions Freefall acceleration g is constant Air resistance is negligible Then the trajectory path of the projectile is a parabola The velocity changes magnitude and direction The acceleration in the ydirection is constant 9 The acceleration in the Xdirection is zero Projectile Motion We ll analyze projectile motion as a superposition of two independent motions Constant velocity motion in the Freefall motion in the vertical horizontal direction direction anSIiaHtZ Demo Bull s eye Target Y i Point of collision f9 2004 Thomsanleoks Cole Example 45 1 200 IllS w20m a3o h 45m building at C Xf YXVZ f 7 x Y 450 m a ma a m quot 0 I E 2004 ThomsunijakS CDIB 39r Review I Use the same basic equations as 1D but with vectors Solve each direction separately l Projectile motion is superposition of 1D motion with constant velocity and free fall motion Vectors I Vector vs Scalar Quantities I Coordinate Systems Properties of Vectors Vector Algebra I Components and Unit Vectors Scalars and Vectors I Scalar quantities only have a single value and an appropriate unit Distance 5 m time 4 sec mass 5 gr I Vector quantities have a magnitude and direction coupled with a unit Displacement 5 ft to left velocity 10 mph due north acceleration 98 ms2 downward B The curved line The path taken The red arrow Displacement vector A Coordinate Systems Proper solution of a problem requires a uniform point of reference and a system to identify location Displacement from point Q to point P Rectangular and Polar Coordinates Vectors I Indicated by an arrow I Tip points away from the starting point in the direction of the vector I Denoted by or I Magnitude is equal to the length of the arrow and denoted by or I Two vectors are equal if they have the same magnitude and direction Adding Vectors Draw vector A Draw vector B starting from the tip of A Draw the resultant vector R A B from the beginning ofA to the tip of B Rules of Vector Addition Cumulative law A B B A Associative law ABCABC Negative of a Vector and Vector Subtraction l A has the same magnitude as A but is in the opposite direction A lt A gt 0 I To subtract B from A just add the negative of B to A A BAbB Multiplying a Vector by a Scalar l lfthe scalar is positive the direction of the vector does not change but its magnitude is multiplied by the scalar value I If it is negative its direction is reversed and its magnitude multiplied 22 Example 32 l J 71 39vkm I HRH 2 M2 533 342m gg f m iampm 39c R Components of a Vector Component vectors of a vector u when added result in the vector are along the coordinate axis x y z Components ofa vector u are scalar values I are projections of the vector on the coordinate axis Unit Vectors A dimensionless vector with a magnitude of 1 Used only to specify direction i j k represent unit vectors in the x y and 2 directions respectively Using Unit Vectors R AB 132A 13 4FB REJEJ AB may y Ari 353 Example 36 M1911 if Cxy Cx 953km Cy 232m R251km 4R1113 Linear Momentum and Collisions I Linear Momentum I Conservation of Momentum I Collisions I Center of Mass Impulse Linear Momentum I The linear momentum of a particle with mass m and velocity v is d I Linear momentum is a vector quantity in the direction of velocity Conservation of Momentum l Linear momentum is conserved in an isolated system Changes in velocity collisions etc 2 p constant Example Skaters m1 100 kg v15mls m2 50 kg ptot p1 v2 pm 2 pi mlvli mzvzi 0 p1 mlvlf 100kg5m 500N pmt p1p2 0 p2 p1500Nsi Elastic and Inelastic Collisions I Perfectly inelastic collision Particles stick together and lose some energy deformation Elastic collision Particles bounce off each other no loss of energy I Inelastic collision u Particles don t stick together but lose some energy Energy is conserved ONLY in elastic collisions Momentum is conserved in both elastic and inelastic collisions Perfectly Inelastic Collisions Before collision Momentum is conserved pi mivii mzvzi pf ml 1712Vf Don t forget the vector nature of velocity Energy is not After collision I 1 397 7 K mlv1i mvi 2 k i O ll 7772 b Eloss Kf Ki a EDD1 ThumsnnIBmoks Cole Elastic Collisions Before collision After collision V11 V2139 1 Momentum is conserved so is Energy 1 2 2 1 2 2 K 2 mlvh mv7i K 2 mlvlf m7v7f 2 k k 2 L k kn I f Example 97 Perfectly inelastic collision can only use momentum conservation 1111le m1 m2 VB IMIVIA m1 m2 After collision we can use energy conservation KB UB KC UC KC O U C m1 m2 gz 1 7 KB 30111 m2 VB U3 0 7 1 31er m2 VB m v 1 1A VB 2g2 m1 m2 m1 m2 gz m m 1 2 11A 2g2 m1 Elastic Collisions in 1D For two masses m1 and m2 and initial velocities v1 and v2 Before collision After collision 0quot Example 98 v 1100 1115 viii 2501 ms Elastic Collision m v 17 V m v 1 1 Z Z 1 1 f l74ms I771V1i mzvm39 quotIlvlf quotIlvlf m2 vlf 2 338m S vzf 2 312m S Twodimensional Collisions Conservation of momentum still stands mlvh I l leZi mlvlf mzvzf Analyze each component separately 3 Before the collision Vl mlv mzv mlvlfv mzvva lix 2ix 111 sin 9 M gt1 111 2111 tilC056 Use conservation of kinetic cccccc energy If the colISIon IS elastic Vx l 4112 sin q mlv mzv mlvl mzvz ll 2 2M 2 2 a mlvh mzv2i 5 mlvlf mZv2f H b After the collision i Lek Example 911 397 v1 35 x 105 ms x125 6 37 4 739 Elastic collision p and E are conserved m1 m2 3 Before the collision V 1 vlix v2ix Vi v 21 v sinQ IllCOS 0 vlzy V2131 vlfv v2 i 2 2 2 2 v1 122 v1f v2f 2 112 COS I i 5 5 wm v vlf 28x 10 ms vzf 212gtlt 10 ms 3 w K lb After the collision Center of Mass I Consider a collection of particles or an extended object with a total mass M I We can consider the system as a single particle with the total mass M concentrated at a single point called the center of mass I If a net force is applied to the system you can apply Newton s second law to the center of mass Center of Mass 2W2 M 33 A group of particles yaw g M An extended object ym E y w 1 gm EEJW Motion of a System of Particles l Applying Newton s second law to a system of particles The center of mass of a system of particles of combined mass M moves like a single particle of equivalent mass under the influence of a net external force Example P 947 55 kg 80 kg 70 kg h H Xi 27 m lt xf gt lt XB gt Z 0 39 rCMj FCMJ VCM i VCM f Impulse l A change in momentum is caused by a net force acting upon a particle F2 0113th f Impulse is the net change in momentum due to F If F is constant over a period At

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