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# GENERAL PHYSICS 1 PHYS 1110

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This 10 page Class Notes was uploaded by Mrs. Peter Toy on Friday October 30, 2015. The Class Notes belongs to PHYS 1110 at University of Colorado at Boulder taught by Michael Dubson in Fall. Since its upload, it has received 11 views. For similar materials see /class/232086/phys-1110-university-of-colorado-at-boulder in Physics 2 at University of Colorado at Boulder.

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Date Created: 10/30/15

Vectorsl Vectors A vector is a mathematical object consisting of a magnitude size and a direction A vector can be represented graphically by an arrow direction of arrow direction of vector length of arrow magnitude of vector A vector quantity is written in bold A or with a little arrow overhead A A no arrow not bold magnitude of the vector positive number magnitudes are positive by de nition Examples of vector quantities position velocity acceleration force electric eld If two vectors have the same direction and the same magnitude then they are the same vector Vector magnitude direction not location y A Ay A In 2D we need 2 numbers to specify a vector A 9 I X 0 magnitude A and angle 6 H J 1 Ax 0 components AK and Ay more on components later Addition of Vectors gt1 4 cm Oi 1182009 University of Colorado at Boulder Vectors2 Vector addition is commutative 11 D D1 C B A Graphical addition quottiptotailquot or quottailtoheadquot method A B C Addition by quotparallelogram methodquot same result as tiptotail method l C B A Can add lots of vectors like steps in a treasure map quottake 20 steps east then 15 steps northwest then quot AT3E g A De nition of negative of vector same size opposite A direction A De nition of multiplication of a vector by a number b 3 b A c 2 c B M times as negative c v long as A i ips direction What about multiplication of a vector by a vector There are two different ways to de ne multiplication of two vectors 1 Dot product or scalar product 3D and 2 Cross product Ax B These will be de ned later 1182009 University of Colorado at Boulder Vectors3 Vector subtraction AiB A 7B quotsubstractquot means quotadd negative of Graphically 13 A7 A B B B B A D lzs DA 45 B H H4 H l DA7B isthesameas DBA issameas Blj A Components of a Vector YA A AX gt2 Ay y I 2 quotXhatquot is the unit vector explained below Ay X Ax A cos 9 Xcomponent quotprojection ofA onto XaXisquot 39 Ay A sin 6 ycomponent quotprojection ofA onto yaXisquot Ax X Think ofthe Ax as the quotshadowquot or quotprojectionquot of the l l l l l l light vector A cast onto the XaXis by a distant light source rays directly quotoverheadquot in the direction of y y gt A Components are numbers not vectors They do not have a direction but they do have a Sign a or 7 sign If the quotshadowquot onto the XaXis points in the X direction then Ax X is positive B BM L Bx Ax Here Ex is negative because the Xprojection is along the X direction By is positive because the yproj ection is along the y direction 1182009 University of Colorado at Boulder Vectors4 A cose AX 5 Ax Acose Ay AA sine Ky 5 Ay A sine Ax A A A2A2 tan6 y y x y A x Magnitude A l is positive always but AK and Ay can be or Example of vector math Ax 2 Ay 3 What is the magnitude A and the angle 0L with the XaXis A JAX2AYZ 2232 44 J13 2 36 tanoc 2 3 04 tan 1 563 2 2 University of Colorado at Boulder 1182009 Vectors5 Vector Addition by Components Proof by diagram 6 A CXA Cy A Similarly subtraction by components 13117123 7 D A7B x x x D A7B y y 3 Position Velocity and Acceleration Vectors Velocity is a vector quantity it has a magnitude called the speed and a direction which is the direction of motion Position is also a vector quantity Huh What do we mean by the magnitude and direction of position How can position have a direction In order to specify the position of something we must give its location in some coordinate system that is its location relative to some origin We de ne the position vector r as n X a y the vector which stretches from the origin of our coordinate system to the location of the object The X and y components of the position vector are simply the X and y ry coordinates of the position Notice that that the position y vector depends on the coordinate system that we have chosen location of object rxx If the object is moving the position vector is a lnction of time r rt Consider the position vector at two different times t1 and t2 separated by a short time interval At t2 7 t1 The position vector is initially r1 and a short time later it is r2 The change in position during the interval At is the vector Ar r1 7 r2 Notice that although r1 and r2 depend on the choice of the origin the change in position Ar r1 7 r is independent of choice of origin 1182009 University of Colorado at Boulder Vectors6 In 2D or 3D we de ne the velocity vector as v lim r AtAo At As At gets smaller and smaller 1 2 is getting closer and closer to 1 1 and Ar is becoming tangent to the path of the object Note that the velocity V is in the same direction as the in nitesimal Ar since the vector V is a positive number lAt times the vector Ar Therefore the velocity vector like the in nitesimal Ar is always tangent to the trajectory of the object A AT The vector equation v ilm A r has X and r The r J t are A0 A A A A A v 11m X v 11m y Any vector equatlon like A B C 1s shorthand X AtAo At y AtAo At notation for 2 or3 component equations AX 13K CX Ay By Cy AZ BZ CZ The change in velocity between two times t1 and t is AV V2 A V1 remember that change is n Av always nal m1nus 1n1t1al We de ne the acceleration vector as a iim As we A0 mentioned in the chapter on 1D motion the direction of the acceleration is the same as the direction of AV The direction of the acceleration is NOT the direction of the velocity it is the direction towards which the velocity is tending that is the direction of AV We will get more experience thinking about the velocity and acceleration vectors in the next few chapters 1 182009 University of Colorado at Boulder Pnysies mu FaH 2mm Lecture 15 27 Septeth 2004 Announcements Reading AssignmentfurWEd SeptZB Knight 7 2 7 3 Constant Acceleration with two dimensions cont Anetnerprebierntnatiiiustratestne appiieatien quEeri s iavvstu nee faH arid 2D mutiun istne surcaiied Hunter arid Ceeenut rnenkeyy prebiern A nunter aims nis gun giremiy ata eeeenut nanging in a tree a gistanee D away at a neigntH a eve tne gruund Just as tne gun isfirEd tne eeeenut beginstu faiifrum tnetree Assuming air resistanee is negiigibie WiH tne buHet stiii strike tne faHing eeeenut iuvv is a figure sneWing a pietenai representatiun ufthe prubiem arid SEImE gennitien ufvariabies quotBi VBxi D X YBi VByi Eire unsidering tne qu prubiem iet s Examine Wnat Weuiu nappen figure Tne furmuiasthatgive tne pesitien ufthe buiiet are tnese fur eenstant speed in eaen eernpenent l n W y m WMA L154 Physics 1110 Fall 2004 In fact let s take the initial bullet position as the origin so that we just have x3 VBJciAt yB vByiAt 39 Note that the condition that the bullet is aimed at the coconut tells is that the ratio of y to X components ofthe velocity are just the same as the coconut height to the distance from the start position These are both just given by the tangent of the angle 6 H vByi D vai tan6 Now we add forces but there is only one since we are neglecting drag and it is gravity of course This gives us a net force of value mg pointed down and therefore the acceleration is O in the Xdirection and g in the ydirection Thus our solution for the horizontal motion of the bullet is unchanged The vertical motion ofthe bullet and coconut are changed but both are just constant acceleration cases which we have solved several times before For the coconut we get xc D y H gAt2 while for the bullet we get x3 VBJCiAt yB vByiAt gAt2 Finally we can answer the question about the bullet hitting the coconut lfthey are to hit there should be a place where both coordinates X and y of the bullet and coconut are equal Since the coconut X position is constant we should look at the y position of the bullet when it has an X position equal to that of the coconut So we first find the time for the bullet to get from the origin to an X position of D At D VBxi and now substitute this time into the y position formula 1 2 yB VByi 73At 39 vai L152 Physms 1MB FaH 2mm NM mn tn get that 2 ys r t T ts ts the same sdmttdh as tdr the EDEDHUL sd yesthe buHEt htts the EDEDHUU The buHetfaHs by the same arhddht as the EDEDHU theth ts tt rhdwhg aerdss spaee m a edrved parabuhc trajemury A su hate that m th case the UmE at the mEIUEIH was deterrmned by a hunzunta cunstramt ratherthan a mantra CINE FmaHy we eah use the taetthat v lt Relative Velocity Th the past T have asserted thatddr resmts ddh t depend an the the t t a tw h ppehsttwe have dtrt t dt systems Whmh are m vmg reTatwe td eaeh dther7 Th drde dhderst hdthe rteet Hm H t n WE Ht t d rh wtu sucktu Euurdmate systemsthathave paraHe X y ah z axes The resths are true EVEN r we aHdwthe axes td be arhttrardy rutated but the aTgehra ahd thg are reaHy hurreh BUS erst we wtu edhstdertwd trarhes dt reterehee s and 5 The EIHQWS utthe twu trarhes have a rewatwe pusmun gWEn by the vectur in Whmh may depend uh UmE The hgdre hexdw shdws the Wu trarhes at same mstant m Mme rurh thts we can wrtte uur rhaster eduatmh check thts mmEdtatE y tor the ease ur the 5 system Nuvv We H tank at eases uf edhstaht reTatwe pustuth cunstant retatwe VE DEm and cunstant retatwe acceteratmn Mara Physics 1110 Fall 2004 Constant Relative Position In this case i is a constant Let s call it R0 and therefore we have 7 0 In orderto go further we ll take the derivative of both sides ofthe equation with respect to time Then we have d7 d 7 dt dt since the derivative of a constant R0 is zero By the way to take the derivative of a vector you just take the derivative of each component Since the derivative of position is the instantaneous velocity we can rewrite this new formula simple as v 3 We can proceed even further by taking the time derivative again which gives us a a39 where we used a d17dt So now we can make a more precise statement about what we mean about the physics not depending on the coordinate system If the reference systems are fixed relative to each other objects will have the same velocity and acceleration vectors independent of the frame of reference In the lingo of physics we say that the velocity and acceleration are invariant under a transformation from one frame to the other The last result about the accelerations is even more profound than it may appear We have argued that Newton s 2nd law gives the acceleration Forces do not depend on frames of reference you measure them with a scale and they are what they are So the invariance ofthe acceleration vectors tells us that our force analysis works in both frames without change Constant Relative Velocity In this case we have R R0 Vot When we take our derivatives we get v X70 and a a39 We now see that the velocity is frame dependent but the acceleration is not This is known as Galilean invariance To transform velocity between frames we just add the relative velocity of the frames Whenever you actually do this in a problem always check your method with a simple case so that you can guess the answers The simplest cases to guess are always the ones where an object is at rest in one of the frames We ll discuss this further in the next lecture L154

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