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Physics201Final–Examples1.Idea:1‐dimensionalmotion(Chapter2)Anobjectstartsatheight(H)abovethegroundwithaninitialupwardspeed(Vi).Theobjectreachesapogeeafter4.0sec.andhitstheground12.0sec.afterreachingapogee.Calculate(H)and(Vi).Solution:Startwithupwardmotion.Vf −Vi = 0 −Vi = at = (−9.8)(4) = −39.2 So,Vi = 39.2 m s Howfarabove(H)istheapogee?Vf2 −Vi

2 = 2az = 0 − (39.2)2 = 2(−9.8)z

⇒ z =(39.2)2

(19.6)= 78.4meters

Nowstartatapogee.Objectfallsfromrest⇒ (H + 78.4) = 1

2 (9.8)(12)2 = 4.9 ×144 = 705.6

⇒ H = 627.2meters2.Idea:VectorsandTorque(Chapters3+10)Aforce

F = (5n.)x + (20n.)y isexertedonanobjectadistance

V = (4meters)z fromthepivot

pointoftheobject.Calculatethetorqueontheobject.Solution:Torque

τ =V ×F = [(4m.)z]× [(5n.)x × (20n.)y]

= (20 joule)y − (80 joule)x

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3.Idea:Motionintwodimensions(Chapter4)A(5.0)kgballisthrownwithaninitialspeedof80.0m/satanangle (θ) withrespecttothehorizontal.Theballlandsonanincline.Theinclineis 30° upwardfromhorizontal.Determinetheheight(H)abovethegroundatwhichithitstheincline.Extracredit:Determinetheangle (θ) atwhich(H)isamaximum.Solution:Let x and z denotehorizontalandverticaldirections.x :Vix = (80.0 m s )cosθ x(t) = (80.0 m s )cosθ(t) where(t)istimeinairafterballisthrown.z = Viz = (80.0 m s )sinθ z(t) = [(80.0 m s )(sinθ)(t)]− [( 12 (9.8 m s

2 )(t 2 )] =heightabovehorizontal.Wewantheightaboveincline.Whatisheightofinclineabovehorizontal?Itis[tan 30°](x(t)) Soheightofballabovehorizontalwhenballhitsinclineisattimessatisfying:[(80sinθ)(t)]− [4.9t 2 ] = [tan 30°](80cosθ)(t) OR[80sinθ]− [4.9t] = [tan 30°][80cosθ] ORt = ( 1

4.9 )[80sinθ − 80 tan 30°cosθ]

Forthistimetheheightis:z(t) = {[80sinθ]( 1

4.9 )[80sinθ − 80 tan 30°cosθ]} − ( 1

4.9 ){80sinθ − 80 tan 30°cosθ}2

= ( 14.9 )(6400){sinθ cosθ tan 30° − cos2θ}

where θ>30°

Extracredit:

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dzdθ

= 0 =64004.9

{cos2θ tan 30° − sin2θ tan 30° + 2cosθ sinθ}

Term in {} must equal zero.{}= 0 = {tan 30°cos2θ + sin2θ}tan2θ = − tan 30°⇒θ = 75°

4.Idea:Circularmotion(Chapter6)A4.0kgsphereisheldbyasteelwire.ThespheremovesinaverticalofradiusR=1.5metersataconstantspeedV = 5m s .Determinethetensioninthewireatthetop,bottom,andhorizontalpointsofthecircle.Solution:

Tension T = mg(V2

Rg+ cosθ)

where(θ )isanglewithrespecttobottom.

Bottom:θ = 0°⇒ T = mgV 2

Rg+1

⎛⎝⎜

⎞⎠⎟

= (4)(9.8) 251.5( ) 9.8( )

⎡

⎣⎢

⎤

⎦⎥ +1

⎛

⎝⎜⎞

⎠⎟

Top:θ = 180°⇒ T = 4( ) 9.8( ) 251.5( ) 9.8( )

⎡

⎣⎢

⎤

⎦⎥ −1

⎛

⎝⎜⎞

⎠⎟

Horizontal:θ = 90°⇒ T =4( ) 25( )1.5( )

5.Ideas:Energy&Friction(Chapters7+8)A(5.0kg)ballslidesdowna 30° inclinefor80m.,thenmoveshorizontallyfor60m.untilstopping.Itstartswithaspeedof11m s .Thecoefficientoffrictionontheinclineistwiceaslargeasonthehorizontalpart.Determinethecoefficientoffrictiononthehorizontalpart.Solution:Let µ( )bethecoefficientoffrictiononthehorizontalpart.Allkineticandpotentialenergyisdissipatedintofriction.Thismeans:12 5( ) 11( )2⎡⎣ ⎤⎦ + 5.0( ) 9.8( ) 40( )⎡⎣ ⎤⎦ = 2µN( ) 80( )⎡⎣ ⎤⎦ + µmg( ) 60( )⎡⎣ ⎤⎦

N =mg 32

So: 499 joule( ) = µ 6811+ 2940{ } = µ 9751{ } ⇒ µ ≅ 0.05

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6.Idea:Linearmomentum(Chapter9)Twomasseshitinanelasticcollision.MassAis(6.0kg)andhasvelocity

VA = 8m s( ) x + 4 m s( ) y + 12m s( ) z

MassBis(8.0kg)andhasvelocity

VB = 7m s( ) x − 5m s( ) y − 2m s( ) z

Determinethefinaltotallinearmomentumafterthiselasticcollision.Thereisnoforceactingonthetwomasses.Solution:Linearmomentumisconserved.Sothefinaltotallinearmomentumis:

104 kg ⋅msec.

⎛⎝⎜

⎞⎠⎟x − 16 kg ⋅m

sec.⎛⎝⎜

⎞⎠⎟y + 56 kg ⋅m

sec.⎛⎝⎜

⎞⎠⎟z

7.Idea:Rotation(Chapter10)Startingfromrest,a(8.0kg)pointmassslidesdownthefrictionlessinnersurfaceofa(2.5meter)radiushollowsphere.Calculatetheangularvelocityandangularaccelerationatthebottomofthespherewithrespecttothecenterofthesphere.Solution:Angularvelocity:Sincethereisnofriction,energyisconserved,sothisisaconversionfromgravitationalpotentialenergytokineticenergy.

mgR =12mv2 ⇒ v = 2Rg = 2 2.5( ) 9.8( ) ≅ 7.0 m s

Whatangularvelocitydoes V( ) correspondto?w = V R =

7.02.5

= 2.8 sec.−1

Whataboutangularacceleration?Atthebottomofthesphere,thereisnonetforceonthepointmass,sotheangularaccelerationiszero.8.Ideas:Angularmomentum(Chapters10+11) ParallelAxisTheoremThe xy ‐planeisthehorizontalplaneofafrictionlesssurface.A(4.0kg)pointmasswithaninitialvelocity

Vi = (8 m s )x movesalongaline y = +2.0meters.Thepointmassstrikesa(3.0

kg),4.0meterlongstickcenteredat(0,0)andlyingonthe y ‐axis.Thecollisioniselastic.Calculatethetranslationalspeedofthepointmass,thetranslationalspeedofthestick,andtheangularspeedofthestickafterthecollision.Solution:Whatareconservedquantities?Energy,linearmomentum,andangularmomentum.Let s( ) denote stick, p( ) point mass.

Linearmomentum:mpvpi = mpvpf + msvs

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Or: 32 − 4vpf = (3.0 kg)vs Angularmomentumwithrespectto(0,0): − 2.0m.( ) 4.0kg( ) 8m s( ) = − 2.0m.( ) 4.0kg( )vpf = Iw where I( ) ismomentofinertiaofstickand w( ) isangularvelocityofstick. I =

112

msLs2 =

112

3.0kg( ) 4.0m( )2 = 4kg ⋅m2

So:−64 = −8vpf + 4w Or:−16 = −2vpf + w

Energy:12mpvpi

2 =12mpvpf

2 +12Iw2 +

12msvs

2

128 j.=2vpf2 +1.5vs

2 + 2w2

vs =323

−43vpf

w = 2vpf −16

So: 128 = 2vpf2 +

3z323

−43vpf

⎡⎣⎢

⎤⎦⎥

2

+ 2 2vpf −16⎡⎣ ⎤⎦2

= 2vpf2 +

32322

9−2569vpf +

169vpf2⎡

⎣⎢

⎤

⎦⎥ + 2 4vpf

2 − 64vpf + 256⎡⎣ ⎤⎦

Or: 0 = 383vpf2 −

4( ) 128( )3

⎡⎣⎢

⎤⎦⎥vpf +

16643

Or: 0 = 38vpf2 − 512vpf +1664

Or: vpf =512 ±142

76= 4.9 or 8.6 m

s

vpf = 8.6 m s implesthatthestickmovesinthe− x directionaftercollision,whichis

unphysical.So vpf = 4.9 m s istheanswer.Usethistofind vs( ) and w( ). 9.Ideas:Fluidmechanics(Chapter14) Bernoulliequation Torricelli’sLaw BuoyancyImayuseanyoftheseideas,sothisproblemdoesnotlimitwhichideaImaychoose.Youaregivenahelium‐filledballoonwitha2.0meterlong,0.50kgmassuniformdensitystring.Themassoftheballoonisnegligible.Theballoonissphericalwitharadiusof0.40meters.Youreleasetheballoon.Itrisesaheight(H)abovethegroundandstops.Calculate(H).

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Solution:Thisisacompetitionbetweengravityandbuoyancy.Buoyancywouldcausetheballoontorise:Helium density=1.79 ×10−1 kg

m3 Air density=1.29 kg

m3 Sothedifferenceis:1.11kgm3

Volumeofballoonis:43Π 0.4( )3 = 2.56 ×10

−3 × 3.143

= 2.7 ×10−1m3

whichproducesabuoyancyforce: B = mg = 0.27( ) 9.8( ) = 2.6n. Thisisequaltotheforceofgravitywhentheballoonstops:

2.6n. = g H2.0

⎛⎝⎜

⎞⎠⎟0.5kg( ) ≅ 2.45H

⇒ H =2.62.45

≅ 1.1 meters

10.Ideas:Simpleharmonicoscillator(Chapter15) FrictionA(5.0kg)massmovestotherightwithaninitialspeedof 25.0 m s( ) .Itmovesadistanceof(12.0m.)onahorizontalsurfacewithacoefficientoffrictionµ = 0.08. Itthenhitsandcompressesaspringwithaspringconstant k = 40 n

m . Whilecompressingthespring,themassmovesonasmooth,frictionlesssurface.Calculatethespeedoftheobjectwhenitreachestheoriginallocationandthetimetoreachtheoriginallocation.Solution:Thisexamplehastwoideas:

• Friction–aconstantforce• Simpleharmonicoscillator

Whatisthetimetocompressandextendthespring?Themasscompressesandextendsthe

spring,whichisonehalfofoneperiod.Theperiod T( ) is:T =2πw.

w = k m = 40 / 5 = 8 = 2.82

T =6.28w

=6.282.82

≅ 2.2 sec.

Onehalfofthisis(1.1sec.).Whataboutthefrictionpart?Thefrictionisconstant.Thefrictionforceis:

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FFR = µmg = 0.08( ) m( ) g( ) = m( ) aFR( ) Sothefrictionaccelerationis:aFR = µg = 0.8 m s 2 Thisconstantaccelerationactsthroughadistanceof24.0m.(12.0m.x2).Thisisonedimensionalmotion:

24.0m.= 25.0 m s( ) t( ) − 12aFR( )t 2 = 25t − 0.4t 2

Or: 0.4t 2 − 25t + 24 = 0 Or: t 2 − 62.5t + 60 = 0

Or: t =62.5 ± 62.5( )2 − 4 60( )

2=62.5 ± 60.5

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=1.0sec.Sothetotaltimeis2.1sec.11.Ideas:Wavemotion(Chapter16)Thisfollowsexample16.3.TheideasIwillusearewavespeedbasedonmediumdensityandtension.Seetextfordetails.12. Ideas:Dopplereffect(Chapter17) SimpleharmonicmotionAspeakerismountedontheendofaspring.Themassofthespeakeris(6.0kg)andthespringconstantis k = 4 ×10+4 n

m . Thespringiscompressedby(1.4meters)andreleased.Thespeakeremitsafrequencyof620Hzwhenatrest.Theflooronwhichthespeakermovesishorizontalandfrictionless.Determinethelocationsatwhichthefrequencyis620Hz,andalsothemaximumandminimumfrequenciesandthelocationsatwhichthe(motionless)observerhearsthesefrequencies.Solution:620Hzismeasuredwhenthespeakerisatrestwithrespecttotheobserver.Thishappensatthemaximumcompressionandextensionlocations.Themaximumandminimumfrequenciesoccuratthespringequilibriumlocation,wherethespeakerspeedisthemaximum.Todeterminethespeed:

12mv2 =

12kx2

Or: v = kx2

m=

4 ×104 × 1.4( )26

= 1.33×104 ≅ 116m s

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Tofindthemaximumandminimumfrequencies,use:

′f =v + v0v − vs

⎛⎝⎜

⎞⎠⎟f

Or: fMAX =343

343−116⎛⎝⎜

⎞⎠⎟620( ) = 936Hz

Or: fMIN =343

3443+116⎛⎝⎜

⎞⎠⎟620( ) = 463Hz

EXTRACREDITIdea:Friction&EnergyA(5.0kg)ballslidesdowntheinsideofahollow,(3.2meter)radiussphere,startingfromthehorizontalpositionwithaninitialdownwardspeedof 20.0 m s( ) .Theballstopsatthebottom.Calculatethecoefficientoffriction.Solution:Kineticandpotentialenergydissipatedtofriction.Thus:125( ) 20( )2⎡

⎣⎢⎤⎦⎥+ 5( ) 9.8( ) 3.2( )⎡⎣ ⎤⎦ = FFR∫ ⋅dx

= − °∫π 2

µN

N = mgcosθ Righthandside= µmg

π 2

°

∫ Rcosθdθ = −µmgRsinθ π 20

= µmgR = µ 5( ) 9.8( ) 3.2( ) = 1157

⇒ µ =1157157

≅ 7.37