Momentum, Impulse,

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Chapter 8Momentum, Impulse,and CollisionsPowerPoint Lectures forUniversity Physics, Thirteenth Edition– Hugh D. Young and Roger A. FreedmanLectures by Wayne AndersonCopyright 2012 Pearson Education Inc.

Goals for Chapter 8 To learn the meaning of the momentum of a particleand how an impulse causes it to change To learn how to use the conservation of momentum To learn how to solve problems involving collisions To learn the definition of the center of mass of asystem and what determines how it moves To analyze situations, such as rocket propulsion, inwhich the mass of a moving body changesCopyright 2012 Pearson Education Inc.

Introduction In many situations, such as a bullet hitting a carrot, wecannot use Newton’s second law to solve problems becausewe know very little about the complicated forces involved. In this chapter, we shall introduce momentum and impulse,and the conservation of momentum, to solve such problems.Copyright 2012 Pearson Education Inc.

Impulse and MomentumCopyright 2012 Pearson Education Inc.

More About ImpulseImpulse is a vector quantityThe magnitude of theimpulse is equal to the areaunder the force-time curve The force may vary with timeDimension of impulse is N sImpulse is a measure of thechange in momentum of theparticleCopyright 2012 Pearson Education Inc.

Impulse and momentum The impulse of a force is theproduct of the force and thetime interval during which itacts. On a graph of Fx versus time,the impulse is equal to the areaunder the curve, as shown infigure to the right. Impulse-momentum theorem:The change in momentum of aparticle during a time intervalis equal to the impulse of thenet force acting on the particleduring that interval.Copyright 2012 Pearson Education Inc.

Momentum and Newton’s second law The momentum of aparticle is the productof its mass and itsvelocity: Newton’s second lawcan be written interms of momentumasCopyright 2012 Pearson Education Inc.p

Impulse-Momentum: Crash Test Example 1Copyright 2012 Pearson Education Inc.

Example 2A golf ball strikes a hard, smooth floor at an angleof 25 0 and rebounds at the same angle. Themass of the ball is 0.0047 kg, and its speed is 45m/s just before and after striking the floor.a) What are the average magnitude and thedirection of impulse on the ball from the wall?Δpx mvfx – mvix 0.0047 kg(45m/s·cos250 -45 m/s·cos250) 0Δpy mvfy – mviy 0.0047 kg(45m/s·sin250 – (-45 m/ssin250) 0.28NsJ 0.28 Nsj b) If the ball is in contact with the wall for 10ms, what are the magnitude and direction ofthe average force on the ball?FΔt Δp 0.28 NsjNs/0.01s 28 N jF Δp/ Δt 0.28

Kicking a soccer ballCopyright 2012 Pearson Education Inc.

Linear MomentumThe linear momentum of a particle of mass mmoving with a velocity v is defined to be theproduct of the mass and velocity:p mvLinear momentum is a vector quantity Its direction is the same as the direction of the velocityThe SI units of momentum are kg · m / sMomentum can be expressed in component form:px m vxCopyright 2012 Pearson Education Inc.py m vypz m vz

dv d mv dp F ma m dtdtdt

An isolated system The total momentum of a system of particles is the vector sumof the momenta of the individual particles. No external forces act on the isolated system consisting of thetwo astronauts shown below, so the total momentum of thissystem is conserved.Copyright 2012 Pearson Education Inc.

Conservation of momentum External forces (the normalforce and gravity) act on theskaters shown in Figure 8.9at the right, but their vectorsum is zero. Therefore thetotal momentum of theskaters is conserved. Conservation of momentum:If the vector sum of theexternal forces on a systemis zero, the total momentumof the system is constant.Copyright 2012 Pearson Education Inc.

Conservation of Linear MomentumWhenever two or more particles in an isolatedsystem interact, the total momentum of thesystem remains constant Law of conservation of linear momentum can beapplied to analyze such events as collisions,explosion and recoil Types of collisions : elastic collisioninelastic collisionCopyright 2012 Pearson Education Inc.

Types of CollisionsIn an elastic collision, momentum and kineticenergy are conserved elastic collisions occur on a microscopic level In macroscopic collisions, only approximately elasticcollisions actually occur–Generally some energy is lost to deformation, sound, etc.In an inelastic collision, kinetic energy is notconserved, although momentum is stillconserved If the objects stick together after the collision, it is a perfectlyinelastic collisionCopyright 2012 Pearson Education Inc.

Recoil of a rifleCopyright 2012 Pearson Education Inc.

Conservation of Momentum, Recoil Example 4Copyright 2012 Pearson Education Inc.

Objects colliding along a straight line Two gliders collide on an air track.0.5 kg 2.0 m/s 0.3 kg (-2.0m/s) -0.5kgvA2 0.3kg 2.0m/svA2 0.4 m/s to the leftCopyright 2012 Pearson Education Inc.

Elastic collisions In an elastic collision, thetotal kinetic energy of thesystem is the same after thecollision as before. Figure at the left illustratesan elastic collision betweenair track gliders.Copyright 2012 Pearson Education Inc.

Inelastic collisions In an inelastic collision, thetotal kinetic energy after thecollision is less than before thecollision. A collision in which the bodiesstick together is called acompletely inelastic collisionIn any collision in which theexternal forces can beneglected, the total momentumis conserved.Copyright 2012 Pearson Education Inc.

The ballistic pendulumCopyright 2012 Pearson Education Inc.

Some inelastic collisions Cars are intended to haveinelastic collisions so the carabsorbs as much energy aspossible.Copyright 2012 Pearson Education Inc.

elastic collisions(one dimension)All rights reserved JO

Elastic collisions The behavior of thecolliding objects isgreatly affected by theirrelative masses.Copyright 2012 Pearson Education Inc.

elastic collisions(one dimension)x only or y only0v1f -v1i0000v2f 0m1cntdm2All rights reserved JO

elastic collisions(one dimension)x only or y only0v1f v1i0v2f 2*v1i000m1m2All rights reserved JO

An elastic straight-line collisionTwo gliders on air track undergo elastic collision. Find thevelocity of each glider after the collision,Copyright 2012 Pearson Education Inc.

Two-Dimensional CollisionFor two-dimensional collision conservation ofmomentum has to be expressed by twocomponent equations:m1v1ix m2v2ix m1v1fx m2v2fxm1v1iy m2v2iy m1v1fy m2v2fyCopyright 2012 Pearson Education Inc.

A two-dimensional collisionTwo robots collide and go off atdifferent angles.Assume mA 20 kg, vA,I 2m/s,mB 12 kg,vA,f 1m/s at α 300Copyright 2012 Pearson Education Inc.

A two-dimensional perfectly inelastic collisionA 1500 kg car traveling east with a speed of 25m/s collides at an intersection with a 2500 kgvan traveling north at a speed of 20 m/s. Findthe direction and magnitude of the velocity ofthe wreckage after the collision, assuming thevehicles stick together after the collision.Ignore friction Model the carsas particles The collision isperfectly inelasticThe cars sticktogetherCopyright 2012 Pearson Education Inc.

The Center of MassThere is a special point in a system or object,called the center of mass, that moves as if allof the mass of the system is concentrated atthat pointThe system will move as if an external force wereapplied to a single particle of mass M located atthe center of mass M is the total mass of the systemCopyright 2012 Pearson Education Inc.

Center of Mass of System of Particles,CoordinatesThe coordinates of thecenter of mass arexCM y CM zCM m xi iiM mi y iiM mi ziiMM is the total mass of thesystemCopyright 2012 Pearson Education Inc.

ExampleFind the position of centerof mass of the systemconsisting of two particlesm1 2 kg and m2 10 kg ifx1 20 cmand x2 60 cmCopyright 2012 Pearson Education Inc.

Center of mass of a water moleculeFor simple model of water molecule, find the position of centerof mass if d 0.0957nm, mH 1u and mO 16 uCopyright 2012 Pearson Education Inc.

Center of Mass, positionThe center of mass in three dimensions can belocated by its position vector, rCM For a system of particles,rCM 1mi ri M iri xi ˆi y i ˆj zi kˆ For an extended object,rCMCopyright 2012 Pearson Education Inc.1 r dmM

Center of mass of symmetrical objects It is easy to find the centerof mass of a homogeneoussymmetric object, as shownin Figure 8.28 at the left.Copyright 2012 Pearson Education Inc.

Center of Mass, RodCopyright 2012 Pearson Education Inc.

Motion of a System of ParticlesAssume the total mass, M, of the system remainsconstantWe can describe the motion of the system interms of the velocity and acceleration of thecenter of mass of the systemWe can also describe the momentum of thesystem and Newton’s Second Law for thesystemCopyright 2012 Pearson Education Inc.

Velocity and Momentum of a System of ParticlesThe velocity of the center of mass of a system of particles isvCM drCM 1 mi vidtM iThe momentum can be expressed asMvCM mi vi pi ptotiiThe total linear momentum of the system equals the totalmass multiplied by the velocity of the center of massCopyright 2012 Pearson Education Inc.

External forces and center-of-mass motion When a body or collection of particles is acted upon byexternal forces, the center of mass moves as though all themass were concentrated there (see Figure 8.31 below).Copyright 2012 Pearson Education Inc.

Rocket PropulsionThe operation of a rocket depends upon the law ofconservation of linear momentum as applied to anisolated system, where the system is the rocket plusits ejected fuel.As the rocket moves in free space, its linearmomentum changes when some of its mass isejected in the form of exhaust gases. Because the gases are given momentum when they areejected out of the engine, the rocket receives acompensating momentum in the opposite direction. In free space, the center of mass of the system movesSection 9.9uniformly.Copyright 2012 Pearson Education Inc.

Rocket Propulsion, 2The initial mass of the rocket plus all itsfuel is M Dm at time ti and speed v.The initial momentum of the system ispi (M Dm)vAt some time t Dt, the rocket’s masshas been reduced to M and an amountof fuel, Dm has been ejected.The rocket’s speed has increased byDv.Copyright 2012 Pearson Education Inc.

Rocket Propulsion The rocket’s mass is MThe mass of the fuel, Δm, has beenejectedThe rocket’s speed has increased tov DvAll rights reserved JO

ʃdv ʃ -dmvf - vi ve*ln MiMfMi is the initial mass of the rocket plus fuelMf is theMfinalmassthe rocketplus Dvany fuelM(v Dv) remainingDm(v-v)Mv MDmv - Dmv Dmv ofinitial(M Dm)ve v e finalThe speed of the rocket is proportional to the exhaust speed0M DvDmveAll rights reserved JO

Rocket Propulsion, 3The basic equation for rocket propulsion is Mi v f v i v e ln M f The increase in rocket speed is proportional to the speed ofthe escape gases (ve). So, the exhaust speed should be very high.The increase in rocket speed is also proportional to the naturallogarithm of the ratio Mi/Mf. So, the ratio should be as high as possible, meaning the mass ofthe rocket should be as small as possible and it should carry asmuch fuel as possible.Section 9.9Copyright 2012 Pearson Education Inc.

ThrustThe thrust on the rocket is the force exerted on it by theejected exhaust gases.dvdMthrust M vedtdtThe thrust increases as the exhaust speed increases.The thrust increases as the rate of change of mass increases. The rate of change of the mass is called the burn rate.Section 9.9Copyright 2012 Pearson Education Inc.

Rocket propulsionCopyright 2012 Pearson Education Inc.

Impulse and momentum The impulse of a force is the product of the force and the time interval during which it acts. On a graph of F x versus time, the impulse is equal to the area under the curve, as shown in figure to the right. Impulse-momentum theorem: The change in momentum of a particle during a time interval

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