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Chapter 9B -- Conservation of Momentum

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Chapter 9B -- Conservation of Momentum ChapterChapter 9B9B -- ConservationConservation ofof MomentumMomentum AAA PowerPointPowerPointPowerPoint PresentationPresentationPresentation bybyby PaulPaulPaul E.E.E. Tippens,Tippens,Tippens, EmeritusEmeritusEmeritus ProfessorProfessorProfessor SouthernSouthernSouthern PolytechnicPolytechnicPolytechnic StateStateState UniversityUniversityUniversity © 2007 Momentum is conserved in this rocket launch. The velocity of the rocket and its payload is determined by the mass and velocity of the expelled gases. Photo: NASA NASA Objectives:Objectives: AfterAfter completingcompleting thisthis module,module, youyou shouldshould bebe ableable to:to: •• StateState thethe lawlaw ofof conservationconservation ofof momentummomentum andand applyapply itit toto thethe solutionsolution ofof problems.problems. •• DistinguishDistinguish byby definitiondefinition andand exampleexample betweenbetween elasticelastic andand inelasticinelastic collisions.collisions. •• PredictPredict thethe velocitiesvelocities ofof twotwo collidingcolliding bodiesbodies whenwhen givengiven thethe coefficientscoefficients ofof restitution,restitution, masses,masses, andand initialinitial velocities.velocities. AA CollisionCollision ofof TwoTwo MassesMasses When two masses m1 and m2 collide, we will use the symbol u to describe velocities before collision. BeforeBefore u1 u2 m1 m2 The symbol v will describe velocities after collision. v AfterAfter v 1 m 2 m1 2 AA CollisionCollision ofof TwoTwo BlocksBlocks BeforeBefore u1 u2 m1 m2 Collision “u”= Before “v” = After m1 m2B AfterAfter v1 m v2 m1 2 ConservationConservation ofof EnergyEnergy u u1 2 m1 m2 The kinetic energy beforebefore colliding is equal to the kinetic energy afterafter colliding plus the energy lostlost in the collision. 112222 11 22mu11 mu 2 2 22 mv 11 mv 22 Loss ExampleExample 1.1. AA 22--kgkg massmass movingmoving atat 44 m/sm/s collidescollides withwith aa 11--kgkg massmass initiallyinitially atat rest.rest. AfterAfter thethe collision,collision, thethe 22--kgkg massmass movesmoves atat 11 m/sm/s andand thethe 11--kgkg massmass movesmoves atat 33 m/sm/s.. WhatWhat energyenergy waswas lostlost inin thethe collision?collision? ItIt’’ss importantimportant toto drawdraw andand labellabel aa sketchsketch withwith appropriateappropriate symbolssymbols andand givengiven information.information. u = 0 u1 = 4 m/s 2 v1 = 1 m/s v2 = 2 m/s m1 m2 m1 m2 m1 = 2 kg m1 = 1 kg m1 = 2 kg m1 = 1 kg BEFOREBEFORE AFTERAFTER ExampleExample 11 (Continued).(Continued). WhatWhat energyenergy waswas lostlost inin thethe collision?collision? EnergyEnergy isis conserved.conserved. u = 0 u1 = 4 m/s 2 v1 = 1 m/s v2 = 2 m/s m1 m2 m1 m2 m1 = 2 kg m1 = 1 kg m1 = 2 kg m1 = 1 kg 1122 1 2 BEFORE:BEFORE: 22mu11 mu 2 2 2(2 kg)(4 m/s) 0 16 J 1122 1 21 2 AFTERAFTER22mv11 mv 2 2 2(2 kg)(1 m/s) 2 (1 kg)(2 m/s) 3 J Energy Conservation: K(Before) = K(After) + Loss Loss = 16 J – 3 J EnergyEnergy LossLoss == 1515 JJ ImpulseImpulse andand MomentumMomentum u uA B A B Impulse = p --FF tt FF tt Ft= mv–mv A B B f o Opposite but Equal F t vA v A B B FB t= -FA t mB vB -mB uB = -(mA vA -mA uA ) Simplifying: m v + m v = m u + m u mAA vAA + mBB vBB = mAA uAA + mBB uBB ConservationConservation ofof MomentumMomentum The total momentum AFTER a collision is equal to the total momentum BEFORE. m v + m v = m u + m u mAA vAA + mBB vBB = mAA uAA + mBB uBB u uB A A B Recall that the total energy is also conserved: --FF tt FF tt A B B KineticKinetic Energy:Energy: KK == ½½mvmv2 vA v B KK ++ KK == KK ++ KK ++ LossLoss A B KA0A0 + KB0B0 = KAfAf + KBfBf + Loss ExampleExample 2:2: AA 22--kgkg blockblock AA andand aa 11--kgkg blockblock BB areare pushedpushed togethertogether againstagainst aa springspring andand tiedtied withwith aa cord.cord. WhenWhen thethe cordcord breaks,breaks, thethe 11--kgkg blockblock movesmoves toto thethe rightright atat 88 m/sm/s.. WhatWhat isis thethe velocityvelocity ofof thethe 22 kgkg block?block? TheThe initialinitial velocitiesvelocities areare zerozero,, soso thatthat thethe totaltotal A B momentummomentum beforebefore releaserelease isis zero.zero. 0 0 mmA vvA ++ mmB vvB == mmA uuA ++ mmB uuB mB vB mA vA = - mB vB vA = - mA ExampleExample 22 (Continued)(Continued) 2 kg vA2 8 m/s 1 kg A B A B 0 0 mmA vvA ++ mmB vvB == mmA uuA ++ mmB uuB mB vB mA vA = - mB vB vA = - mA (1 kg)(8 m/s) v = - vvA == --44 m/sm/s A (2 kg) A ExampleExample 22 (Cont.):(Cont.): IgnoringIgnoring friction,friction, howhow muchmuch energyenergy waswas releasedreleased byby thethe spring?spring? 2 kg 4 m/s 8 m/s 1 kg A B A B 2 2 2 Cons. of E: ½½kxkx == ½½mmA vvA ++ ½½mmB vvB ½½kxkx2 == ½½(2(2 kg)(4kg)(4 m/s)m/s)2 ++ ½½(1(1 kg)(8kg)(8 m/s)m/s)2 ½½kxkx2 == 1616 JJ ++ 3232 JJ == 4848 JJ ½½kxkx22 == 4848 JJ ElasticElastic oror Inelastic?Inelastic? An elastic collision loses In an inelastic collision, no energy. The deform- energy is lost and the ation on collision is fully deformation may be restored. permanent. (Click it.) CompletelyCompletely InelasticInelastic CollisionsCollisions CollisionsCollisions wherewhere twotwo objectsobjects stickstick togethertogether andand havehave aa commoncommon velocityvelocity afterafter impact.impact. Before After ExampleExample 3:3: AA 6060--kgkg footballfootball playerplayer standsstands onon aa frictionlessfrictionless lakelake ofof ice.ice. HeHe catchescatches aa 22--kgkg footballfootball andand thenthen movesmoves atat 4040 cm/scm/s.. WhatWhat waswas thethe initialinitial velocityvelocity ofof thethe football?football? A Given: uB = 0; mA = 2 kg; mB = 60 kg; B vA = vB = vC vC = 0.4 m/s 0 Momentum: mmA vvA ++ mmB vvB == mmA uuA ++ mmB uuB Inelastic collision: (m(mA ++ mmB )v)vC == mmA uuA (2 kg + 60 kg)(0.4 m/s) = (2 kg)uA u == 12.412.4 m/sm/s uuAA = 12.4 m/s ExampleExample 33 (Cont.):(Cont.): HowHow muchmuch energyenergy waswas lostlost inin catchingcatching thethe football?football? 0 1122 1 2 22muAA mu BB 2()Loss m A m B v C ½(2 kg)(12.4 m/s)2 = ½(62 kg)(0.4 m/s)2 + Loss 154 J = 4.96 J + Loss LossLossLoss === 149149149 JJJ 97% of the energy is lost in the collision!! General:General: CompletelyCompletely InelasticInelastic Collisions where two objects stick together and have a common velocity vC after impact. ConservationConservation ofof Momentum:Momentum: ()mmvmumuA Bc AA BB ConservationConservation ofof Energy:Energy: 1122 1 2 22muAA mu BB 2() m A m B v c Loss ExampleExample 4.4. AnAn 8787--kgkg skaterskater BB collidescollides withwith aa 2222--kgkg skaterskater AA initiallyinitially atat restrest onon ice.ice. TheyThey movemove togethertogether afterafter thethe collisioncollision atat 2.42.4 m/sm/s.. FindFind thethe velocityvelocity ofof thethe skaterskater BB beforebefore thethe collision.collision. Common speed after vvB == vvA == vvC == 2.42.4 m/sm/s colliding: 2.4 m/s. uuA = 0 uuB = ? muA ABBABC mu() m m v 87 kg (87 kg)u = (87 kg + 22 kg)(2.4 m/s) BB uB AA 22 kg (87(87 kg)kg)uuB =262=262 kgkg m/sm/s uB = 3.01 m/s ExampleExample 5:5: AA 5050 gg bulletbullet strikesstrikes aa 11--kgkg block,block, passespasses allall thethe wayway through,through, thenthen lodgeslodges intointo thethe 22 kgkg block.block. Afterward,Afterward, thethe 11 kgkg blockblock movesmoves atat 11 m/sm/s andand thethe 22 kgkg blockblock movesmoves atat 22 m/sm/s.. WhatWhat waswas thethe entranceentrance velocityvelocity ofof thethe bullet?bullet? 2 kg uA = ? 1 kg 1 m/s 2 m/s 1 kg 2 kg A C B Find entrance velocity of 50 g 1 kg 2 kg bullet: mA = 0.05 kg; uA = ? 1 m/s 2 m/s Momentum After = 1 kg 2 kg Momentum Before 0 0 mA uA + mB uB + mCu C = mB vB + (mA +mC) vAC (0.05 kg)uuA =(1 kg)(1 m/s)+(2.052.05 kgkg)(2 m/s) (0.05 kg) uuA =(5.1 kg m/s) u = 102 m/s uAA = 102 m/s CompletelyCompletely ElasticElastic CollisionsCollisions Collisions where two objects collide in such a way that zero energy is lost in the process. APPROXIMATIONS!APPROXIMATIONS! VelocityVelocity inin ElasticElastic CollisionsCollisions uA uB A B 1. Zero energy lost. 2. Masses do not change. vA vB A B 3. Momentum conserved. Equal but opposite impulses (F t) means that: (Relative v After) = - (Relative v Before) For elastic collisions: v -v = - (u -u ) vAA -vBB = - (uAA -uBB ) ExampleExample 6:6: AA 22--kgkg ballball movingmoving toto thethe rightright atat 11 m/sm/s strikesstrikes aa 44--kgkg ballball movingmoving leftleft atat 33 m/sm/s.. WhatWhat areare thethe velocitiesvelocities afterafter impact,impact, assumingassuming completecomplete elasticity?elasticity? 1 m/s 3 m/s vvA -- vvB == -- ((uuA -- uuB) ) A B vv -- vv == uu -- uu 1 kg 2 kg A B B A vvA vvB A B vvA -- vvB = (-3 m/s) - (1 m/s) From conservation of energy (relative v): v -v = -4 m/s vAA -vBB = -4 m/s ExampleExample 66 (Continued)(Continued) 1 m/s 3 m/s Energy:Energy: vvA -- vvB == -- 44 m/sm/s A B Momentum also conserved: vv 1 kg 2 kg vv A A B B mmA vvA ++ mmB vvB == mmA uuA ++ mmB uuB (1 kg)vvA +(2 kg)vvB =(1 kg)(1 m/s)+(2 kg)(-3 m/s) Two independent vvA ++ 2v2vB = -5 m/s equations to solve: vvA -- vvB == -- 44 m/sm/s ExampleExample 66 (Continued)(Continued) 1 m/s 3 m/s vA + 2vB = -5 m/s A B vvA -- vvB == -- 44 m/sm/s vv 1 kg 2 kg vv A A B B Subtract: 0 + 3vv B2 == -- 11 m/sm/s vvB == -- 0.3330.333 m/sm/s B vv A2 -- ((--0.3330.333 m/s)m/s) == -- 44 m/sm/s Substitution: v = -3.67 m/s vAA = -3.67 m/s vvA -- vvB == -- 44 m/sm/s Example 7.
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