physics ib summer packet for hl + sl / 2017- 2018 · physics ib summer packet for ... the first and...

AmericanInternationalSchool/W–CairoPreparedbyDr.NasimSarira

PhysicsIBsummerPacketfor

HL+SL/2017-2018AllstudentstakingIB–Physicsareadvisedtoread,practiceandresearchtheselectedtopicsthatIincludedinthissummerpacket.

Physics.

Regards,

Dr.NasimSarira

[email protected]


Suggestedtopics:

2.Mechanics.

Introductiontotopic1:

• Scientificnotation:d=5.0⋅103m,sometimesd=5.0x103mord=5.0*103m• Prefixes,abbreviationsandvalues:

tera=T=1012,giga=G=109,mega=M=106,kilo=k=103,hecto=h=102,deca=da=101,deci=d=10-1,centi=c=10-2,milli=m=10-3,micro=µ=10-6,nano=n=10-9,pico=p=10-12,femto=f=10-15


Uncertainties and errors Before we learn the necessary rules governing propagation of uncertainties, note that we have been talking about experimental uncertainties rather than errors. To say that a result A = 10.5 ± 0.5 has an experimental error of ± 0.5 would imply that a reading of 10.0 or 11.0 would be in error since it does not agree with the "true" value of 10.5. For this reason, we will try to use "uncertainty" rather than ‘error."

1. Precision and accuracy A second word of caution is in order here. We should distinguish between the precision in our measurements and the accuracy in these measurements. The precision of a measurement I s a description of the size of the uncertainty relative to the size of the measured quantity. For example, if the speed of light (3 x 108 m/s) is measured to within 10 m/s, this would be a measurement of extremely high precision (0.000003%), since [10 /(3 x 108)] x 100% = 0.000003%. If the speed of a car (29m/s) were measured to within 10m/s, the precision would be relatively low (34%), since (10 /29) x 100% = 34%. On the other hand, accuracy refers to the extent to which our measured value agrees with the actual, or known, value. Returning to the first of the examples above. Suppose we measured the speed of light to be: c = (2.5005 ± 0.0002) x 108 m/s. Our measurement would not be accurate, compared to the known value of 3 x 108 m/sec, even though it is quite precise (.008 %) since (0.0002 /2.5005) x 100% = 0.008%. If however, we measured: c = (2.5 ± 0.8) x 108 m/s our measurement would be accurate, albeit of extremely low precision (32%) since (0.8 /2.5) x 100% = 32%. In this learning module, we will be concerned with ways to describe only the precision of experimental results.

2. ABSOLUTE UNCERTAINTY AND RELATIVE UNCERTAINTY As you have seen, we write our experimental results in the general form A ± Δ A. Here, Δ A is the uncertainty, or more specifically the absolute uncertainty, in A. We can also define a relative uncertainty: the relative uncertainty in a number is the absolute uncertainty (Δ A) divided by the number (A).

Number = A Absolute uncertainty = Δ A Relative Uncertainty = Δ A/A

3. For example, if we have a velocity v = 4.0 ± 0.2 m/sec, this equation follows the general form A ± Δ A. So, Δ A or 0.2 m/sec is the absolute uncertainty, and Δ A/A or 0.2/4.0 = 0.05 or 5% is the relative uncertainty. Then we can also write our velocity as v = 4.0 m/s ± .05 relative uncertainty, or v = 4.0 m/sec ± 5%.


Note that if would not be correct to write v = (4.0 m/sec ± 0.05) m/sec because the relative uncertainty (0.05) must not have units associated with it. The number appearing after the ±€sign is the absolute uncertainty if is has units associated with it. It is a relative uncertainty if the number has a percent sign or nothing after it. For example:

5.00g±0.0200.020isrelativeuncertainty5.00g±2.0%2.0%isrelativeuncertainty5.00g±0.10g0.10gisabsoluteuncertainty(5.00±0.10)g0.10gisabsoluteuncertainty

Exercise: decide for each one whether the uncertainty is expressed in relative or absolute form.

1. 5.2 g ± 0.1 g 2. 22lbs ± 5% 3. 4.09 cm ± 0.01 4. (50.0 ± 0.1)ml 5. 2.40 g ± 0.02 6. 112.1 mg ± 2% 7. 5 x 10-2 g ± 0.01 8. (7 ± 1)cm3 9. (5.00 ± 0.01) x 102 g 10. 2.7 x 103 m/sec ± 0.02

Ify=a±bthenΔy=Δa+Δb

Ify=ab/cthenΔy/y=Δa/a+Δb/b+Δc/c

Typesofphysicalquantities

Scalar quantity: has only magnitude, ex. time, mass, distance,temperature

Vector quantity: has both direction and magnitude, ex. force,velocity


Graphs

Lineargraphs

Inreports,refertographsasgraphordiagramnrso-and-so.Remembertoindicateunitsonthescales.Thezerocanbesuppressedifonlyhighvaluesareused:

Slopeorgradient:unitofverticalaxisdividedwithunitofhorizontal(here:thegradientistheaccelerationa,anditsunitis(ms-1)/s=ms-2.Slopeorgradient(m)andintercept(c)forstraightlineisgivenbyy=mx+c(herewegetv=at+u)

Area under line or curve: units of axes multiplied (here the area under the graph is thedisplacements;itsunitis(ms-1)*s=mTransformingnon-lineargraphstolinear

If thegraph isnot linearfromthestart, then itcanbemade linearbyplottingamanipulatedvariableononeorbothoftheaxes.Asanexample,takes=ut+½at2forUAMwhichbecomess=½at2whenu=0.Assumethata=2ms-2so½a=1ms-2,thenwecanexpressthedisplacementasafunctionoftimewithdatapoints,(timeinsec,displacementinm)=(t,s):(0,0),(1,1),(2,4),(3,9)etc.Thisgraphisnotlinear,butifweinsteadplot(t2,s)weget(0,0),(1,1),(4,4),(9,9)etc.whichisalineargraphwiththegradient1so½a=1ms-2anda=2ms-2.

Insteadwecouldhaveplotted(t,√s)giving(0,0),(1,1),(2,2),(3,3)whichalsoislinearandhasthegradient1givinga=2ms-2.


Insimilarwayswecantransformothermathematicalfeaturesinaformula:

Ifwe(foranidealgas)havePV=nRT=>P=nRT/V=k/Vforconstantn,RandTwegetastraightlinebyplottingPasafunctionof1/V

Forlogarithmicgraphs,seethesectionaboutradioactivedecayinnuclearphysicslater

Fittingalinetodatapoints:oneline,notpiecesfrompointtopoint.Asmanypointsabovethelineasbelow.

Fittingalineargraph("best-fit")toexperimentaldatapoints

Ifweareworkingwithexperimentalvaluesthatdoor"should"followastraightline(eitherastheyareoraftersomemathematicalmanipulationlikesquaringthemortakingthesquarerootofthem),thentheymaynotexactlylieonastraightline,butwecanfitalinetothembydrawingone line that approximately follows them (possibly disregarding "outliers", individual valueswhichareverydifferent from theothersandmaybe causedbymistakes in theexperimentalwork),leavingabouthalfofthedatapointsbelowandabovetheline.

Worksheet1canbeusedaspractice.


2.MECHANICS

2.1.Mechanics-thefoundationofphysics

Thefirstandmostimportantpartoftheareasofphysicsismechanics,whichformsabasisforotherparts tobepresented later. Thequantitiesusedherewill reappear inmanyplacer-therearemanytypesofforces,butallfollowthelawsofNewton.Velocityisnotonlyaquantitytobestudiedforitsownsake,butforexamplethevelocityofanelectriccharge affects how it reacts tomagnetic fields. Themechanics studied in this topic isclassical mechanics, developedmainly in the time period 1600-1800. Amore precisemodern theory of mechanics, involving Einstein's theory of relativity and quantummechanics can be learned later. Formost technical applications - including advancedtechnologylikesendingaspacecrafttotheplanetMars-thisclassicalmechanicsisstillsufficient.

2.2.Distanceanddisplacement

We start the physics course with mechanics which deals with questions like wheresomething is,how fast and inwhatdirection itmoves,how itsmotion changes,whatcauses it, and some consequences of the answers to these questions. All this fits theuniversalcharacterofphysics.Anexampleofthisisthequantityspeed(describedlater):acarmaydriveataspeed,ananimalmayrunorflyataspeed,bloodcanflowthroughyourveinsataspeed,adistantstarorgalaxymaymovetowardsorawayfromusatsomespeed.

Beforewegettothequantityspeed,weneedtodescribesomethingmorefundamental:wheresomethingis.Inphysics,therearetwowaystotellhowfarsomethinghasmovedorhowfarfromacertainpointitis:distance=howyouwentmeasuredalongthepathyouactuallytook.Thetripmeterinacarmeasuresdistance.Sincetheroadcanbecurveditisdifficulttosaywhatdirectionyoutook,anddistanceisthenascalar.Commonsymbol:s

displacement=howfaritisfromwhereyoustartedtowhereyoustoppedinastraightline.Ifyoulookatthemapmaybeyoucanfindoutthatthetownyoudrovetois25kmtothenorthwestofwhereyoustarted.Thisisavector,whichalsooftenhasthesymbols. If only two directions are possible, it is convenient to used positive values fordisplacementsinonedirectionandnegativeintheopposite.

Ex.Thetrainmoved500mforwardsandthen200mbackwards.

*Ifwecalltheforwardsdirectionpositive,thetotaldisplacementis500m+(-200m)=500m-200m=300m.

*Ifwecallthebackwardsdirectionpositive,wehave-500m+200m=-300m.


Note:

· inbothcaseswecouldadd thedisplacements,with their signs.Thesameformulacouldhavebeenused:stotal=s1+s2

·theanswersaredifferentalthoughtheyrepresentthesamemotionintherealworld.Theymustbeinterpretedusingthechosendefinitionofwhichdirectionthatispositive.

2.3.Speedandvelocity

Thiscanalsobedescribedeitherwithorwithoutadirection:

speed=distance/time Thisisascalar.Unitms-1

velocity=displacement/time Thisisavector,sameunit.

(notethat1ms-1=3.6kmh-1)

Thesameformulaisusedforboth(vforvelocityorspeed,sfordistanceordisplacement,tfortime):

v=Δs/Δt [DBp.4]

ThesymbolΔstandsforthechangeinsomething=thedifferencebetweenwhat it isnowandwhatitwasbefore.Inmanysituationsitcanbedropped-forexamplethetimeforsomethingtohappenisthedifferencebetweenwhattheclockshowedafteritandwhenitstarted,butifwestartedastopwatchfromzerowhentheeventstarted,thenthereadingonthestopwatchwhentheeventisoverequalsthetimeittook.Wethenoftenusetheformulaintheformv=s/t

Ifthevelocityisconstant(bothmagnitudeanddirection!),wehavewhatiscalleduniformmotion,UM.


Framesofreferenceandrelativevelocity

Example:AboatAmoveswith5ms-1downstream,anotherboatBwith5ms-1upstreaminariverflowing2ms-1relativetotheshore.Bothmoveat5ms-1inthe"river'sframeofreference",buttheirspeedsinthe"shoreframeofreference"(ortheirspeedsrelativetotheshore)are3ms-1and7ms-1.

2.4.Acceleration

Ifthevelocitychanges(magnitudeand/ordirection),wehaveanacceleration.Wewillfirstfocusoncaseswheresomethingmovesalongastraightline,butwherethespeed=themagnitudeofthevelocitychanges.Weusethesesymbols:

u=initialvelocity

v=finalvelocity

t=timetochangevelocityfromutov

a=acceleration

Δv=v-u=changeinvelocity

Thedefinitionofaccelerationisthen

a=Δv/Δt [DBp.4]

wherewecanwriteΔv/Δt=(v-u)/t(assumingthatt=thetimeittookforthevelocitytochangefromutov).Accelerationisavectoranditsunitisms-2(whichmeansm/s2).Theformulaisoftenwritteninthisformaftersolvingforv:

·a=(v-u)/t multiplybothsideswithtso

·at=(v-u)=v-u move-utotheleftside,lettingitchangesign

·at+u=v orasbelow:

v=u+at [DBp.5]

Iftheaccelerationisconstant,wehaveuniformlyacceleratedmotion,UAM.


·Nearearth,allthingsfalldownwithagravityaccelerationg=9.81ms-2ifwedonotthinkofairresistance.

• ForUMwewouldhavea=0andv=u+atwouldbecomev=u;thevelocityisconstant

2.5.GraphsofUMandUAM

UM:

·Thegraphofvelocityasafunctionoftime(velocityony-axis,timeonx-axis)isahorizontalstraightline(thevelocityisconstant).Ifanobjecthastraveledforthetimetwiththevelocityv,thedisplacement(howfaritasmoved)isgivenbyv=s/t=>s=vt.Thisisthearea(rectangle)underthegraph.

·Thegraphofdisplacementasafunctionoftimeisastraightlinewhichissteeperthehigherthevelocityis.Comparethistothegraphsofy=x,y=2x,y=3xetcwherethegraphy=kxissteeperthehigherkis.Herewehaves=vtwithsinsteadofy,vinsteadofkandtinsteadofx.Thevelocityisnowthegradient(slope)oftheline.ThismeansthatyoutakeanytwopointsAandBonthecurveandfindhowmuchhigherBisthanA,thendivideitbyhowmuchfurthertotheleftBisthanA.

UAM:

·Thegraphofvasafunctionoftisnowarisingstraightlinestartingfromu(initialvelocity)onthevelocityaxis.Duringthetimetitreachesthelevelv(finalvelocity).Thedistancetraveledisstilltheareaunderthisgraph-nowatrapeze(likeatriangleontopof a rectangle). This area canbe foundbyadding theareasof the rectangle and thetriangleorbyfindingthemeanoraveragevelocityvmwhichis(u+v)/2.Sinces=vmtwethenget:


s=[(u+v)/2]t [DBp.5]

·Thegraphofdisplacementsasafunctionoftimeisnownotastraightlinebutacurvebendingupwards(gettingsteeperandsteeper-thegradientorslopeisstill=thevelocity,butsincethisincreasesallthetime,wewouldneedtodrawa"helpline"(calledtangent)andfindthegradient=slopeofthisbychoosingtwopointsonit)

m05b

Othertypesofmotion(neitherUMorUAM)

Ifthevelocityisnotconstant(notUM)andtheaccelerationnotconstant(notUM)isstilltruethatthetravelleddisplacementistheareaunderthev-tcurve(whichmaybefoundwithgeometry,numericalapproximationsonacomputer,orothermethods)andthatthevelocityatacertaintimeistheslopeofthes-tcurve.

m05c


Instantaneousandaveragevalues

Ifonequantity is thegradient (slope)ofanother (e.g. velocity fromdisplacementoraccelerationfromvelocity)wecangraphicallyfindeitheranaverageoraninstantaneousvalue.Theaveragevalue is the change in thevertical coordinate / the change in thehorizontalcoordinate.The instantaneousvalue isthe"average"valueforan infinitelysmallchangeinthehorizontalcoordinate.

m05d

2.6.The4equationsofuniformlyacceleratedmotion=UAM

Wealreadyhave

v=u+at [DBp.5]

Wecannow

·replacevin2)byu+atandgetvm=(u+u+at)/2=(2u+at)/2

·simplifyvm=(2u+at)/2=2u/2+at/2=u+½at

·togets=vmtmultiplywithtandhavet(u+½at)=ut+½at2

sowehave:

s=ut+½at2

Anotherpossibilityisto

·solve1)fortwhichgivest=(v-u)/a

·replacetin2)withthis,sos=vmt=vm(v-u)/a


·usefrom2)thatvm=(u+v)/2togets=(u+v)(v-u)/2a

·let(u+v)(v-u)=(v+u)(v-u)=vv-vu+uv-uu=v2-vu+vu-u2=v2-u2

whichallgivesusthat

·s=(v2-u2)/2awhichwemultiplywith2atogetv2-u2=2as

·andfinallyv2=u2+2asso:

v2=u2+2as

Notethattheequationsarevalidonlyforconstantacceleration!

2.3.Forceandmass

Force (avectorquantity)

forexample

· deformation (stretching, bending, compressing, other). It ismeasuredwith aforcemeter (dynamometer, newtonmeter) containing a spiral metal spring which isextended(stretchedout)morethegreatertheforceis.Unit:1newton=1N.

·acceleration=changeinvelocitypertime.

Resultantforce (resultant,totalforce,netforce,sumofallforces)

Oftenseveralforcesactonthesameobject.Ifyouholdsomethinginyourhand,thereisaforceofgravitypullingitdownandaforcefromyourhandupwardswhichmaybalanceoutthedownwardsforcesotheresultantiszero.Thiscanbehandledbychoosingonedirectionaspositive(ex.up)andgivingtheforcessignsaccordingly.

Example:

Forceofgravity=Fg=-5.0NForceofhand=Fh=5.0N

Resultant=Fg+Fh=-5.0N+5.0N=0

Newton's3lawsforforces:

NewtonI:

Iftheresultantforceonanobjectiszero,itsvelocitywillbeconstant.

Thiscanmeaneitheroftwopossibilities:

·theobjectisatrestandwillremainsoaslongastheresultantiszero(liketheobjectinyourhand).


·theobjecthassomevelocityandwillkeepit(bothdirectionandmagnitude)aslongastheresultantiszero.Example:Acarcomestoacurvewheretheroadisextremelyslipperybecauseofice.Thedriverwouldliketoeitherslowdownorchangedirection,butbecauseoftheicenoforcecanbeappliedtoithorizontally,soitcontinuesoutintotheforestwhereforcesfromtreesitcollideswithslowsitdown.(Thiswasinthehorizontaldimension-intheverticaldimensionthereisaforceofgravitydownwhichisbalancedoutbytheforcefromthehardiceintheroadkeepingitfromsinkingintoit).

Afree-bodydiagram=sketchofanobjectshowingtheforcesactingonitusingarrowswitha lengthproportional to themagnitude (if known). Forces (asother vectors) canusing trigonometry be resolved into components in two dimensions perpendicular toeachother,andthecomponentsaddedseparately.Theresultantforce/resultant/totalforce/netforcecanbefoundusingPythagoras.

Translational equilibrium = a situation where the net force in all dimensions is zero.Example:anobjectslidingdownaslopeatconstantspeed,whenthecomponentoftheforceofgravitydowntheslopeandtheforce(ex.friction)balanceout,andthesameistrueforthenormalforce(perpendiculartosurface)andthecomponentoftheforceofgravity perpendicular to the surface (draw diagram, choose labels, resolve intocomponents!).

NewtonII

IfthereisaresultantforceF,thentherewillbeachangeinvelocity=acceleration

whichisgreaterthegreaterFis,butsmallerthegreaterthemassoftheobjectis.

a=F/m

Alargerenginegivingalargernetforcewillincreaseacceleration

Alargermasswilldecreaseit.

F=ma [DBp.5]


Thismeansthattheunit1N=1kgms-2.Massisascalar,butaccelerationisavector,sotheforceisalsoavector.

NewtonIII

IfAactsonBwiththeforceFthenBactsbackonAwith-F

(-FisaforceofthesamemagnitudebutoppositedirectiontoF).Examples:

·Ariflefiresabulletandactswithaforceonitacceleratingitforwards,butthebulletactsbackontheriflesoitrecoils

·Arocketengineinaspaceshipthrowsoutgasesactingwiththem,andthenthegasesactbackontherocketwithaforceforwards(notethattherocketdoesnot "push against the air" to drive it forwards, itworks out in emptyspace).

Massandweight

· massisapropertyofanobjectwhichithaswhereeverwetakeit-a100kgastronautisa100kgastronauthereoronthemoon

·weightistheforceofgravityactingonsomething-onthemoonwheretheforceofgravityisweaker,theweightinnewtonsislower.

Theforceofgravityis

Fg=mg

whereg=thegravityconstantorgravityacceleration=9.81ms-2onearth,1.6ms-2onthemoon.


·Inertialmass=F/a(whereFisresultantforce,regardlessofwhatkindofforcethis - forceofgravity, forceofhand, forceof rocketengine,electricalforcesorother).

· Gravitationalmass = F/g (near earth) the property of an objectwhichdetermineshowlargetheforceofgravityontheobjectis.

Thereisbasicallyno"good"reasonwhytheinertialandgravitationalmassesshouldbethe same - why the quantity which says how much force of any kind is needed toaccelerateanobjectshouldbethesameastheonewhichsayshowstrongoneparticularforce(gravity)is.Forotherthethreeotherfundamentalforces(electromagnetic,strongandweaknuclearforce)thestrengthoftheforceisdeterminedbyotherquantities(ex.electriccharge).

2.8.Work,energy,power

Workandenergy

m09a

IftheforceoracomponentFsof it is inthedirectionofitsdisplacement,thework(ascalar)doneis

W=(Fss=)Fscosθ [DBp.5]


withtheunit1joule=1J=1Nm.Theamountofworkdoneistheenergy(sameunit)convertedfromoneformtoanother.

Inavelocity-timediagramthedisplacementistheareaunderthegraphsinces=vtforUM,forothertypesofmotiontheareaisnotarectanglebutstillequaltos.Similarly,inagraphofFsasafunctionofs,theareaunderthegraph-rectangleorother-istheworkW.

Kineticenergy(liike-energia,rörelseenergi)

·ifacarisacceleratedfromrestbytheconstanthorizontalforceFthentheworkdoneisW=F.s=m.a.s;hereθ=0

·fromtheequationforUAMv2=u2+2aswenowgetv2=2asandthena=v2/2s

·insertingthisinW=masgivesW=½mv2whichis"stored"inthemovingcar,so

Ek=½mv2 [DBp.5]

Gravitationalpotentialenergy(potentiaalinenenergia,lägesenergi)

·ifanobjectfallsfromtheheighththeforceofgravitydoesaworkW=Fs=mgs=mghonit:

Ep=mgh [DBp.5]

Thesesumoftheseisthetotalmechanicalenergy,whichisconstant(thatis,conserved)unlessenergyislosttodoworkagainstfriction,airresistanceorother.

Power

P(=E/torW/t)=work/time=Fv [DBp.5]

unit1watt=1W=1Js-1.Poweristheamountofworkdoneorenergytransformedfromoneformtoanotherpertime;itcanbecalledtherateofworking."TherateofX"means"howmuchXpertime".Notethatforanobjectmovingataconstantspeedvthepower


P=W/t=Fs/t=FvwhereFisnottheresultantforcebuttheforcekeepingitinmotiondespitefriction,airresistanceetc.Notetheolderunit1horsepower=ca735W.

Efficiency

eorη=Eout/EinorPout/Pin[notinDBbutasimilardefinitionisgiveninthermalphysics,DBp.6]

whereEinistheworkorenergysuppliedandEoutthatwhichisconvertedtosomething"useful".What this isdependson thepurposeof thedevice; for a lightbulbwhereacertainamountofelectricenergyissupplied,theusefulenergyisthatconvertedtolightandtheenergyconvertedtoheatwasted.Forabreadtoaster,itistheopposite.Powercanbeusedinsteadofworkorenergysincethetimetiscanceled:Pout/Pin=(Eout/t)/(Ein/t)=Eout/Ein

2.9.Friction

Friction

Theforceoffrictioniscausedbyinteractionbetweenatomsinthematerialofasurfaceandinanobjectincontactwithit.Fortheforceoffrictionwehave

Ffr=µkNandFfr<or=µsN [DBp.5]

µ=positivefrictioncoefficient,withoutunit,whichcanbe

· kinetic (indexk)ordynamicorslidingformovingobject (forceoppositetovelocity)

·orstatic(indexs)forobjectatrest(forceoppositetonetforcetryingtosetitinmotion).Inthiscasethevalueissuchthattheforceoffrictionbalancesanynetforce trying to set the object inmotion until somemaximumvalue,when theobject"jumps"intomotionandtheforceoffrictiontheniskinetic(withaconstantcoefficientsomewhatsmallerthanthemaximumvalueofthestaticone)


N=normalforce,theforcewithwhichthesurfaceispressingtowardstheobject(onahorizontalsurfaceN=-FGso itcanbereplacedbytheforceofgravity inacalculationwhereonlymagnitudesareinvolved.

Alternatively:Weusedifferentpositive-negativedirectionsinthehorizontalandverticaldimensions.ThismeansthatNorFN(whichisintheverticaldimension,balancingouttheforceofgravityGorFG)maybegivenadifferentsignwhenusedtocalculatetheforceoffrictionastheexpressionµNsinceµ isalwayspositiveandtheforceoffrictioncanbeeitherpositiveornegativedependingourchoiceofdirections.Theforceoffrictionis,inprinciple,notaffectedbytheareaoftheobjectwhichisincontactwiththesurface.

m08a

For an object on an incline (slope) itmust be noted that the normal force is not theopposite of the force of gravity, but of the component of the force of gravityperpendiculartotheslope.


m08b

Foramovingobject,Ffrisintheoppositedirectiontothevelocity.Forastaticobject,itisintheoppositedirectiontotheresultantofallotherforcesactingonit.

2.10.Springs

Linearsprings

Ifaspringisextended(pulledout)orcompressed(pushedin)adisplacementxitactswithaforceaccordingto

F=(-)ks [DBp.5]

Aforcewhichfollowsthistypeofaformulaiscalledaharmonicforce.

m10a


wherek=springconstant,unitNm-1(higherthestrongerthespringis);

theminus sign shows that the force of the spring is in the opposite direction to thedisplacementsfromtheequilibriumposition

Elasticpotentialenergy

Whenaspringisextendedorcompressed,workisdoneonitwhichcanbestoredinitasanelasticpotentialenergy.Sincetheforceneededtoovercometheforceofthespringisnotconstantbutincreaseslinearlytheworkdone=theareaundertheforcegraph=½*thebase*theheight=½*x*F=½*x*kx=

Eelas=½kx2 [DBp.5]

m10b

2.11.*Simpleharmonicmotion

Massonspring

Itcanbeshownthatforamassmoscillatingonaspringwiththespringconstantk,thetimeperiodTfortheoscillationsfollowtheformula:

T=2π√(k/m)


Simplependulum

Inasimilarwayitcanbeshownthatforamassm(sometimescalledthependulum"bob")swingingattheendofanassumedlymasslesspendulumofthelengthlhasthetimeperiod

T=2π√(l/g) [notinDB]

2.12.Momentumandimpulse

(Linear)momentum

avectorquantity,unit1kgms-1,isdefinedas:

p=mv [DBp.5]

Ifwedefinemomentump=mvwecanalsowriteNIIasF=Δp/t(meaning"netforceistherateofchangeinthemomentum")sinceinitialmomentum=mu,finalmomentum=mvandchangeinmomentumpertime=(mv-mu)/t=m(v-u)/t=ma=F.Notethatmomentum = Fi. 'liikemäärä', Sw. 'rörelsemängd'. Fi. '(voiman) momentti' or'vääntömomentti'andSw.'(kraft)moment'or'vridmoment'all=torque,aquantitytobepresentedlater.

Note:hereFistheresultantforce

F=Δp/Δt [DBp.5]

WhentwoobjectsAandBcollideorotherwiseinteractforthetimetandnoexternalforceisacting(e.g.theforceoffrictioncanisneglected),thetotalmomentisconserved(thesamebeforeandafterthecollision)since

·NIII:AactsonBwithFsoBactsonAwith-F

·noexternalforces,sothesearetheresultantforcesonAandB

·NIIforA:-F=maA=m(vA-uA)/t=(mvA-muA)/t=ΔpA/t

·NIIforB:F=maB=m(vB-uB)/t=(mvB-muB)/t=ΔpB/t

·thereforeΔpA/t=-ΔpB/tandΔptotal=ΔpA+ΔpB=0

·nochangeintotalmomentummeansitisthesamebeforeandafter


m11a

Incalculationsforproblemswithtwoobjectscolliding,themostusefulformofthisis

m1u1+m2u2=m1v1+m2v2 [notinDB]

wheretheformulaisadaptedaccordingtothesituation,e.g.:

·ifobject2wasatrestbeforethecollisionthenu2=0andthetermm2u2dropped

·iftheobjectsstaytogetherafterthecollision,thenv1=v2=vandm1v1+m2v2=(m1+m2)v

·onedirectionischosenpositive,andthevelocitiesgivenpositiveornegativevaluesaccordingly.Ifavelocityiscalculated,thesignshowsitsdirection

Since momentum is a vector we can have collisions in two dimensions where themomentums and/or the velocities are split up into components in two perpendiculardimensions.Thesearethenbothconservedm1u1X+m2u2X=m1v1X+m2v2Xandm1u1Y+m2u2Y=m1v1Y+m2v2Y).Thecomponentsofthemomentumarefoundusingtrigonometrylikeforvelocities.


Anotherusefulrelationisthefollowing:Sincep=mv=>p2=m2v2=>p2/2m=½mv2so:

Ek=p2/2m [DBp.5]

Impulse

I=FΔt=Δp [DBp.5]

(unit1kgms-1=1Ns)whereFistheresultantforceactingonanobject,tthetimeduringwhichtheforceacts(canbeaveryshorttimeforacollision). Iftheforceactingisnotconstant,theonlywaytofindtheimpulseandwiththatthechangeinmomentumistofindtheareaunderthegraphofFasafunctiont.Ifwefindtheimpulsefromthegraph,thenI=Δp=m(v-u).

Elasticcollisions

Inanelasticcollision,e.g.twohardbilliardballscollidingandbouncingapart,thetotalkineticenergyisalsoconserved.

Example:AbilliardballAwiththemassmandvelocityuAcollideselasticallywithanotheridenticalballBatrest.Whatwillhappen?

Conservationofmomentum: muA+muB=mvA+mvB

=> muA=mvA+mvB

=> uA=vA+vB


Conservationofkineticenergy: ½muA2+½muB2=½mvA2+½mvB2

=> ½muA2=½mvA2+½mvB2

=> uA2=vA2+vB2

=> (vA+vB)2=vA2+vB2

=> vA2+vB2+2vAvB=vA2+vB2

=> 2vAvB=0

whichispossibleonlyifvBorvAis=0.ThefirstwouldrequirethatBisaffectedbyaforcewithoutanychangeinvelocity(impossible)sothelatteristrue.

Inelasticcollision

Ifsomekineticenergyislost,onlymomentumisconserved(ifnoexternalforcesact).Wemustassumethatacollisionisinelasticunlessotherinformationisgiven.Inacompletelyinelasticcollision,allkineticenergyislost(liketwoidenticalcarscollidingheadwiththesamespeedatformingawreckatresttogether.Sincemomentumisavector,thetotalisconserved-itiszerobothbeforeandafter!).

physics ib summer packet for hl + sl / 2017- 2018 · physics ib summer packet for ... the first and...

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