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# Notes, week 2 phys 114

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This 18 page Class Notes was uploaded by Shelby Logsdon on Wednesday January 21, 2015. The Class Notes belongs to phys 114 at University of Washington taught by kazumi tolich in Winter2015. Since its upload, it has received 142 views. For similar materials see general physics 1- mechanics in Physics at University of Washington.

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Date Created: 01/21/15

Phys 114 1515 Clicker registration httbscatalvstuweduwebdsurvevktolich254782 exam 1 jan 30 exam 2 feb 24 nal mar 17 230420 study center PAB AM018 lecture chapter 11 and 12 scienti c method is a careful application of experience and reason to answer questions introduced in the 16th century by galileo 5 steps to scienti c method 1 recognize a problem 2 form a hypothesis Educated suggestion that is experimentally veri able 3 predict consequences of hypothesis 4 perform experiments to test predictions if experiment disagrees w predictions must go back to step 2 5 formulate theory simplest rule that organizes hypothesis prediction and outcome only if experiment agrees w predictions Theory is a very strong statement means it has not been disproven w experiments Theory cannot be proven to be certainly true Theory can be proven to be wrong UNIT a standard relative to which a quantity such as length is measured Ex Inches pounds miles kilometers SI unts length in meters mass in kilograms time in seconds electric current in amps temp in kelvin amount in mol luminosity on candela Length distance between two points Mass quantitative measure of inertia Lecture chapter 13 18 Dimension physical nature of a quantity dimension of length L feet meters inches etc dimension of mass M kiograms tonne slug etc two quantities can only be added or subtracted if they have the same dimensions the terms on both sides of an equations must have the same dimensions Scienti c notation power of 10 100010x10x1010quot3 multiplication and division in multiplication exponents are added in division exponents are subtracted Conversion of units your employer sends you on a trip to a foreign country where the road signs give distances in kilometers and the automobile speedometers are calibrated in kilometers per hour If you drive 90 kmh how fast are you going in a meters per second 1hr3600s 1000m1km 90kmhr90kmhr1hr360051000m1km25ms b miles per hour 1mi1609km 90kmhr90kmhr1mile1609km56mihr a literL is the volume of a cube that is 10 cm by 10cm by 10cm if you drink 1 L of water how much volume in cubic centimeters and in cubic meters would it occupy in your stomach aquot310cmquot3 10quot3 cmquot3 convert to mquot3 10quot3cmquot310quot2 m1cmquot3 10quot3cmquot310quot6mquot31cmquot310quot3 mquot3 mary is driving in a straight line at 60mph nate is driving at a velocity 2500 micrometersms slower than mary in the same direction How fast is nate driving in ms express in scienti c notation Mary speed 60 milehr 1609 km mile 10quot3 m 1 km 1 hr3600 s2682 ms Nate speed Mary speed 2500 micrometers ms Mary speed 25x10quot3 mcmms10quot6mmcm1 ms10quot3s mary speed 25 ms 262 ms 25ms 243 ms 243x10 ms Order of magnitude calculations in doing rough calculations or order of magnitude calculations we sometimes round off a number to the nearest power of 10 the height of an ant might be 8x10quot4 m or approximately 10quot3 m the order of magnitude of an ant39s height is 10quot3 m Ex What thickness of rubber tread is worn off the tire of your automobile as it travels 1km 6mi Thickness of tread of a new tire 1 cm Distance to wear out new tire 100000 km 1cm of wear 100000km of travel 1x10quot7 m of wearkm Ex Estimate the number of grains of sand on a beach Size of beach 1 km long 100 m wide 1 m deep Diameter of grain of sand 1mm Volume of beach V Number of grains N Volume of single grain G VNXG G 43piRquot3 V 1X10A3 X 100m X 1m 4pi 5x10quot3mquot3 10quot14 Vector and scalar quantities A vector quantity has a magnitude and direction 0 Displacement m how far something moved in what direction 0 Velocity ms how fast something is moving in what direction 0 Acceleration msquot2 how fast velocity is changing in what direction A scalar quantity has only magnitude 0 Time 5 how long 0 Temperature k how hot something is 0 Mass kg how much stuff there is Lecture chapter 21 to 26 Coordinate system and position A coordinate system de nes the position of an object De ne origin and positive direction Displacement is the difference in the initial and nal positions Change in X x naxinitial Displacement is a vector quantity Total distance traveled is total length of travel Distance is a scalar quantity no negative Average velocity and average speed Average velocity is de ned to be xfxitfti Average speed of the trip is de ned to be total distancechange in time Instantaneous velocity and speed Instantaneous velocity in the x direction is de ned to be lim as change in time goes to zero of change in xchange in time Instantaneous speed is the magnitude of the velocity Instantaneous vs average speed Why is instantaneous speed more interesting to the highway patrol than average speed 0 Suppose you drive 100km in one hour your average speed would be 100kmhr o How likely is it that you travel 100kmhr the whole time Motion with constant velocity If the velocity is constant instantaneous at any time is equal to the aveage velocity For a particle with an initial position and a constant velocity the nal position as a function of time is given by X xinitia vt Average and instantaneous accelerations Acceleration is the rate of change of velocity or how quickly velocity is changing Average acceleration in the x direction is de ned to be vfvitfti Speed is absolute value of velocity When direction of acceleration and velocity are the same object is speeding up When direction of acceleration and velocity are opposite object is slowing down Acceleration and force Acceleration is caused by force Motion w constant acceleration For an object with an initial position xO initial velocity vO and a constant acceleration a o The velocity v as a function of time t is given by vv0 at o The position x as a function of time is given by xx0 v0t12atquot2 For an object with a initial velocity vO and a constant acceleration a since velocity varies linearly in time 0 During the period when an object accelerates from vO to its nal velocity v its average velocity is given by vav12v0at Ex A motor cycle is moving at v0300 ms when the rider applies the brakes giving the motorcycle a constant deceleration At t130 s after braking begins the speed decreases to v1150ms what distance does the motorcycle travel from the instant braking begins until it comes to rest t00 t130 s v0300ms v1150ms v20 v1 at vO a V1V0t1 1 v2quot2 vOquot22achange in x change in x v2quot2 vOquot22at2v1v0 300msquot22 30s1SO300 90 m Ex At t0 a ball initially at rest starts to roll down a ramp with constant acceleration Suppose it moves 1 foot between t0 and t1 sec how far does it move between t1 and t2 sec Xx0 v0t12atquot2 X00 VOt0 X12atquot2 1ft12a1squots A2ftsquot2 X2512atquot2122ftsquot225quot2 XZsx1s4ft1ft3 ft Lecture ch 27 Free fall The motion of an object falling freely under only the in uence of gravity Without air resistance objects of different weight fall with the same downward acceleration An object is in free fall from the time it is released until it lands whether it is dropped from rest thrown downward or thrown upward Acceleration due to gravity g The acceleration due to gravity is constant at g 981 msquot2 Free faing equations For a free faing object the acceleration is g de ning upward to be the positive x direction For a free falling object with an initial position x0 and initial velocity VO Ex A monkey drops from a tree limb to grab a piece of fruit on the ground 180 m below Neglect the effects of air resistance A How long does it take the monkey to reach the ground X XO VOt 1thquot2 X 0 V0 0 12gtquot2X0 Tquot22XOg T sqrt2X0g sqrt 2180m981mSquot2 T sqrt 606 s B How fast is the monkey moving just before it reaches the ground VV0gt V00 vgt v 981msquot2 6065 594 ms Speed 594ms Alternate solution d displacement VA2 VOquot2ng VO0 Vquot22gX0 V sqrt2981msquot2180m 594ms Speed 594ms Returning speed When a particle thrown upward with an initial velocity v0 returns to the initial height what is its velocity When it comes back it is falling down with the same speed Ex In a jump a springbok leaves the ground at a speed of VO 600 ms A What maximum height above the ground does the springbok attain ddispacement Vquot2V0quot22gd Vquot20 DV0quot22g xX0V0quot2Zg x00 x 600msquot22982msquot2 184 m B How long does it take the springbok to attain this height VV0gt V0 gtV0 TV0g 600mS981msquot2 6125 C What is the springboks velocity when it is x125m above the ground Vquot2V0quot22gd VsqrtV02gd sqrt600msquot22981msquot2125m 339ms D At what times is the springbok 125 m above the ground VV0gt tV0Vg Moving up t 600ms339ms 981msquot22555 Moving down t600ms339ms981msquot2958 5 Ex A tennis ball is hit straight up at V0 200 ms from the edge of a sheer cliff Ignore effects of air resistance A How fast is the ball moving when the ball passes the original height from which it was hit 20ms B If the cliff is 300 m high how long will it take the ball to reach the ground level X X0 V0t12gtquot2 12gtquot2V0txx0 Use quadratic formula Xb sqrtbquot24ac2a TV0l sqrtlvloquot2Zgxx0g T 524s choose pos time C What total distance did the ball travel Vquot2V0quot22gd at max height Vquot2 0 2gdV0quot2 dv0quot22g Total distance 2dx0 2V0quot22g X0 708 m Lecture 31 to 34 Scalars verses vectors A scalar is a number with units It can be positive negative or zero A vector is a quantity with a direction and a magnitude VECtOFS Vectors are expressed as arrows The arrow head points to the direction of the vector and the length of the arrow indicates magnitude The magnitude is always positive A variable for a vector usually has an arrow on top or is a boldface letter Vectors in same direction are parallel opposite direction are antiparallel If two vectors have same magnitude and point in the same direction they are equal Any vector can be broken down to its x and y components w trig Ax Acosth Ay Asinth A sqrtAxquot2 Ayquot2 Pythagorean theorem th tanquot1AyAx Direction angle th theta Be careful when calculating th You may need to add 180 degrees to get the correct angle Depending on situation visualize vector direction and check to see angle makes sense Direction of a vector can be de ned relative to the x axis or the y axis Ex nd the x and y components of a position vector r of magnitude r 75 m if its angle relative to the x axis is th 350 degreesa FX rCOS Eh 75m cos35 61 m ry rsinlth 75m sin35 43 m check r sqrt rxquot2 ryquot2 sqrt 61quot2 43 A2 75 m Vector addition Head to tail method from the tail of A to the head B Parallelogram method from the tails of A and B to the other corner of the parallelogram formed by the vectors Suppose you want to calculate C A B You can add each component separately Cx Ax Bx Cy Ay By Negative of a vector A negative of a vector is de ned as the vector that gives zero for the vector sum when added to the vector The vectors A and A have the same magnitude but points in the opposite directions Negative vector is opposite directions same magnitude vectors Vector subtraction A You can subtract a vector by adding its anti parallel vector with the same magnitude B From the head of B to the head of A Multiplying a vector by a scalar Keep the direction of the vector xed and multiply the magnitude of the vector by the scalar 3A A A A Unit vectors Xquot and Yquot are called unit vectors vector of length 1 in the directions of x and y respectively A vector can be expressed as A AXXquot AyYquot Vector addition and subtraction using unit vectors Vector addition can be calculated by A B AXBXXquot AyByYquot Subtracted by subtraction inside parentheses Ex given two vectors A 40xquot 30yquot and B 60xquot 80yquot nd magnitudes and directions of A AB 40xquot 30yquot 60xquot 80yquot 40xquot60xquot 30xquot80yquot 10xquot 5y AB sqrt100quot2 50quot2 112 th tanquot150100 27 degrees B BA C AB Ex you are driving up a long inclined road after 12 miles you noticed that signs say your elevation has increased by 530 feet A What is the angle of the road above the horizontal 12 miles 5260ft1 mi 6336ft th sinquot1 5306336 48 degrees B How far do you have to drive to gain an additional 150 feet of elevation r 150ft sin48degrees 1793 ft or 34 mi Lecture 35 11315 Position vectors Position is a location describd with the distance and direction from a reference point In 2 dimensional space it can be expressed as rxxquot yyA Displacement vectors Displacement is the difference between the nal and initial positions It is independent of the path taken In 2 dimensional space it can be expressed as change inr rf ri Rearranging the equation you can see that rfrichange inr Ex In its daily prowl of the neighborhood a cat makes a displacement of 120 m due north followed by a 72 m displacement due west A Find the magnitude and direction of the displacement required for the cat to return home Rf 72mxquot 120myquot sqrt 72mquot2 120mquot2 140 m th tanquot1 12072 59 degrees 5 of e B If instead the cat had rst prowled 72m west and then 120 m north how would this affect the displacement needed to bring it home R3 r1 r2 r3 displacement is not affected Average velocity Average velocity in two dimensions is de ned to be chxchtxquot chyllchltDYquot Dividing a vector with a scalar changes only the magnitude of the vector leaving the direction the same Instantaneous velocity Lim chrcht Magnitude of the velocity vector is the speed Ex if the cat in example 1 takes 45 minutes to complete the 120 m displacement and 17 minutes to complete the 72 m displacement what are the magnitude and direction of its average velocity during this 62 mnute time Vav chrcht chr1 chr2cht 120myquot 72mxquot62min 6051 min 0194msxquot 0323 ms yquot Vav sqrt 0194quot2 0323quot2 037 ms th tanquot1 0323 0194 121 degrees 31 degrees w of n Average acceleration chvxchtXquot chvyChtYquot Aav Ex you throw a ball upward with an initial speed of 45 ms when it returns to your hand 92 sec later it has the same speed in the downward direction What was te average acceleration vector of the ball V1 45 ms yquot V2 45ms yquot Aav chvcht V2V1cht 4545925 98 msquot2yquot Acceleration and constant speed An object can accelerate even if its speed stays the same if its direction of motion changes Instantaneous acceleration In two dimensions is de ned to be lim chvcht axXquot ayYquot The acceleration vector can point in direction other than the direction of motion 0 When a is perpendicular to v speed remains constant and direction changes 0 When a is parallel to v speed increases without changing direction of motion 0 When a is parallel to v speed decreases without changing direction of motion Ex a water molecule is shown schematically in the gure The distance from the center of the oxygen atom to the center of a hydrogen atom is 96 A and the angle between the hydrogen atoms is 1045 degrees Find the center to center distance between the hydrogen atoms 1A 10quot10 m A angstrom D2 96Asin1045s D 296Asin5225 15 A 15A10quot10 m 1 A 15 X lOAlO m Lecture chapter 36 Frame of reference A coordinate system x and y axis for which an observer is at rest with respect to the origin Position direction velocity acceleration etc are measured with respect to a particular coordinate system Let vab denote quotvelocity of a relative to bquot Reversing the subscripts of a velocity vector reverses the corresponding velocity vector Relative velocity A particle is moving vpb measured by observer B in frame of reference SB and SB is moving at vba with respect to the frame of reference Sa the velocity is measured by observer A va va VBA ordering of subscripts is important EX as you hurry to catch your ight at the local airport you encounter a moving walkway that is 85 m long and has a speed of 22 mx relative to the ground A If it takes you 68 s to cover 85 m when walking on the ground how long will it take you to cover the same distance on the walkway Vyw velocity of you w respect to walkway 85m68s 125ms ng velocity of walkway w respect to ground 22 ms Vyg velocity of you w respect to ground Vyg VyW ng 125ms 22 ms 345 ms T 85m345ms 25 s B How long would it take you to cover the 85 m length of the walkway if once you get on the walkway you immediately turn around and started walking in the opposite direction w a speed of 13ms relative to the walkway ng 22ms Vyw 13 ms Vyg VyW ng 22ms 13ms 9ms T 85 m 9ms 94 5 Relative velocity in 2D A Consider a man walking on top of a traveling car Consider two frames of reference one associated w the car and the other w the ground B The man is walking at Vpc with respect to the car The car is traveling at ch w respect to the ground C The man is moving w respect the ground at VpgVpCch Example 2 You are riding on a jet ski at an angle of 35 degrees upstream on a river owing with a speed of 28ms if your velocity relative to the ground is 95ms at an angle of 20 degrees upstream what is the speed of the jet ski relative to the water ng ijng ng 95ms sin20xquot 95ms cos20yquot 3249ms xquot 8927ms yquot ijngng ngng3249msxquot 8927msyquot 28ms xquot 605msxquot 893ms yquot ij sqrt 605quot2 893quot2 11 ms Example 3 A passenger on a stopped bus notices that rain is fallng vertically just outside the window When the bus moves w constant velocity the passenger observes 15 degree angle w respect ot vertical A What is the ratio fo the speed of the rain drops to the speed of the bus Vrgvel of rain w rt ground ng vel of bus wrt ground Vrb vel rain wrt bus v9 vgyA vbgvngA vrbvrg vgb vrg vbg vgyA vngquot 39ng XA 39 VrgyA vrb Vrbxquot Vrbxquot th tanquot1 vrbyvrbx tanquot1 vbvbg Tan255 vrngg 37 B Find the speed of the raindrops given that the bus is moving with a speed of 18 ms 2D constant acceleration motion Ex an object experiences a constant acceleration of 200msquot2 along the x axis for 27 s attaining a velocity of 160ms in a direction 45 from the x axis calculate intial velocity vector of the object A axXA ayyA ay0 A 200 msquot2 xquot VVxxquotVyyquot 160mscos45xquot 160mssin45yquot 1131ms xquot 1131ms yquot Vx V0xaxt V0xvx39axt VyV0yayt VyV0y Vo 167msxquot 113msyquot Ex initially a particle is moving at 410 ms at an angle of 335 above the horizontal Two seconds later its velocity is 605ms at an angle of 590 below horizontal What was the particles average acceleration during these 200 5 Vi 41cos335xquot 41sin335yquot 3419ms xquot 2263ms yquot Vf605c0559xquot 605sin59 yquot 3116 ms xquot 5816ms yquot Aav vt 3116xquot5816 yquot 3419Xquot 2263yquot 2 s 152msquot2 Xquot 372msquot2 yquot Projectile motion assumption The free fall acceleration is constant over the range of motion The free fall acceleration is directed downward The rotation of earth is ignored The effect of air resistance is negligible 0 This assumption is often not justi ed especially for high velocity projectiles Projectile motion In projectile motion velocities in the x and y directions are independent from each other Constant velocity motion in the x direction Constant acceleration motion in the y direction Projectile motion x component X component of a projectile is a constant velocity motion 39 0 Projectile motion y component 39 Ayt quotg vy2y v20y2ayy V025in2th3929y Zero launch angle x component For a projectile with a zero launch angle set th 0 39 0 Xt xo vocostht Y component V2yy29y Ex a diver runs horizontally off of a diving board with an initial speed of Va 175 ms if the diving board is h 300 m above the water what is the divers speed just before she enters the water Vxvo 175ms vy Vy22gy 2g h 2981msquot23 m Vy767ms V sqrt 175quot2 767quot2 787ms Ex a swimmer runs horizontally off a diving board w a speed of Va 262 ms and hits the water a horizontal distance of d 188 m from the end of the board D Vot t dVo Height Y yo12gt h 12 gtquot2 h 12gtquot2 12gdvoquot2 l2988 188262 253m Lecture 44 45 Jumping on a train If you jump straight up in a train traveling at a constant velocity you would land on the same spot on the train 0 In the reference frame on the train you just jumped straight up so there is no horizontal component of initial velocity 0 In the reference frame of the ground you are a projectile The horizontal component of your velocity is the same as tat of the train You and the train travel the same distance horizontally Ex A ball is shot from the ground into the air At a height of 91 m the velocity is observed to be v 76xquot 61yquot ms to what maximum height will the ball rise vf2vi229y y Vzi29 61m522981m52 19 m ymax yiy91m19m 11m Ex a tennis player serves at 236 ms and the ball leaves the racquet at 500 below the horizontal and 237m above the court surface Does the ball clear the net which is 120m away and 9 m high If so by how much Vi236ms Vix Vicosth Viy Visinth In X XfVixt X Xf39Xi Vixt D Vixt T dVix dVicosth In y Y1yiyiyt12gt2 yfyiVisinthdvicosth l2 gdvicosth2 237m 12mtan512981msz12m236cos5 004m No it will not clear the net by 86 cm y 9m 04 m 86m Path of a projectile The path of a projectile without air resistance is a parabolic trajectory Horizontal range The horizontal range horizontal distance a projectile travels when it comes back to the originlal height is given by R V02gsin2th For a xed v0 the maximum range is obtained when sin2th1 or th 45 Ex A projectile is red with an initial speed vo 300 ms from the level ground at a target on the ground a distance R 200 m awary Find the two projectile angles R Vizg sin2th th l2 sin391Rgv2 th l2 sin391 200981300 629 387 less than 45 45 387 8370 The maximum height A projectile reaches its maximum hight when the vertical component of its velocity is zero For a given initial velocity and launch angle the maximum height is geven by ymax vozsin2th2g for a xed theta the greater initial velocity yields the greater maximum hiehgt for a xed initial velocity theta 90 yeilds the greatest height

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