# Week 8: Chapter 9. Linear Momentum. Newton Law and Momentum. Linear Momentum, cont. Conservation of Linear Momentum. Conservation of Momentum, 2

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1 Lnear omentum Week 8: Chapter 9 Lnear omentum and Collsons The lnear momentum of a partcle, or an object that can be modeled as a partcle, of mass m movng wth a velocty v s defned to be the product of the mass and velocty: p mv The terms momentum and lnear momentum wll be used nterchangeably n the text Lnear omentum, cont Lnear momentum s a vector quantty Its drecton s the same as the drecton of the velocty The dmensons of momentum are L/T The SI unts of momentum are kg m / s omentum can be expressed n component form: Newton Law and omentum Newton s Second Law can be used to relate the momentum of a partcle to the resultant force actng on t dv dmv dp Fma m dt dt dt wth constant mass p x = m v x p y = m v y p z = m v z Conservaton of Lnear omentum Whenever two or more partcles n an solated system nteract, the total momentum of the system remans constant The momentum of the system s conserved, not necessarly the momentum of an ndvdual partcle Ths also tells us that the total momentum of an solated system equals ts ntal momentum Conservaton of omentum, 2 Conservaton of momentum can be expressed mathematcally n varous ways p total = p 1 + p 2 = constant p 1 + p 2= p 1f + p2f In component form, the total momenta n each drecton are ndependently conserved p x = p fx p y = p fy p z = p fz Conservaton of momentum can be appled to systems wth any number of partcles Ths law s the mathematcal representaton of the momentum verson of the solated system model 1

2 Conservaton of omentum, Archer Example The archer s standng on a frctonless surface (ce) Approaches: Newton s Second Law no, no nformaton about F or a Energy approach no, no nformaton about work or energy omentum yes Archer Example, 2 Conceptualze The arrow s fred one way and the archer recols n the opposte drecton Categorze omentum Let the system be the archer wth bow (partcle 1) and the arrow (partcle 2) There are no external forces n the x-drecton, so t s solated n terms of momentum n the x-drecton Analyze Total momentum before releasng the arrow s 0 Archer Example, 3 Analyze, cont. The total momentum after releasng the arrow s p1 f p2f 0 Fnalze The fnal velocty of the archer s negatve Indcates he moves n a drecton opposte the arrow Archer has much hgher mass than arrow, so velocty s much lower Impulse and omentum d From Newton s Second Law, F p dt Solvng for dp gves dp F dt Integratng to fnd the change n momentum over some tme nterval t f p p p Fdt I f t The ntegral s called the mpulse,, of the I force actng on an object over t Impulse-omentum Theorem Ths equaton expresses the mpulsemomentum theorem: The mpulse of the force actng on a partcle equals the change n the momentum of the partcle p I Ths s equvalent to Newton s Second Law ore About Impulse Impulse s a vector quantty The magntude of the mpulse s equal to the area under the force-tme curve The force may vary wth tme Dmensons of mpulse are L / T Impulse s not a property of the partcle, but a measure of the change n momentum of the partcle 2

3 Impulse, Fnal The mpulse can also be found by usng the tme averaged force I Ft Ths would gve the same mpulse as the tme-varyng force does Impulse-omentum: Crash Test Example Categorze Assume force exerted by wall s large compared wth other forces Gravtatonal and normal forces are perpendcular and so do not effect the horzontal momentum Can apply mpulse approxmaton Collsons Example 1 Collsons may be the result of drect contact The mpulsve forces may vary n tme n complcated ways Ths force s nternal to the system Observe the varatons n the actve fgure omentum s conserved Collsons Example 2 The collson need not nclude physcal contact between the objects There are stll forces between the partcles Ths type of collson can be analyzed n the same way as those that nclude physcal contact Types of Collsons In an elastc collson, momentum and knetc energy are conserved Perfectly elastc collsons occur on a mcroscopc level In macroscopc collsons, only approxmately elastc collsons actually occur Generally some energy s lost to deformaton, sound, etc. In an nelastc collson, knetc energy s not conserved, although momentum s stll conserved If the objects stck together after the collson, t s a perfectly nelastc collson Collsons, cont In an nelastc collson, some knetc energy s lost, but the objects do not stck together Elastc and perfectly nelastc collsons are lmtng cases, most actual collsons fall n between these two types omentum s conserved n all collsons 3

4 Perfectly Inelastc Collsons Snce the objects stck together, they share the same velocty after the collson mv m v m m v f Clcker Queston In a perfectly nelastc one-dmensonal collson between two movng objects, what condton alone s necessary so that the fnal knetc energy of the system s zero after the collson? A. It s not possble B. The objects must have momenta wth the same magntude but opposte drectons. C. The objects must have the same mass. D. The objects must have the same velocty. E. The objects must have the same speed, wth velocty vectors n opposte drectons. Elastc Collsons Both momentum and knetc energy are conserved m1v1 m2v2 m1v1 f m2v2f m1v1 m2v m1v1 f m2v2f 2 2 Elastc Collsons, cont Typcally, there are two unknowns to solve for and so you need two equatons The knetc energy equaton can be dffcult to use Wth some algebrac manpulaton, a dfferent equaton can be used v 1 v 2 = v 1f + v 2f Ths equaton, along wth conservaton of momentum, can be used to solve for the two unknowns It can only be used wth a one-dmensonal, elastc collson between two objects Elastc Collsons, fnal Example of some specal cases m 1 = m 2 the partcles exchange veloctes When a very heavy partcle colldes head-on wth a very lght one ntally at rest, the heavy partcle contnues n moton unaltered and the lght partcle rebounds wth a speed of about twce the ntal speed of the heavy partcle When a very lght partcle colldes head-on wth a very heavy partcle ntally at rest, the lght partcle has ts velocty reversed and the heavy partcle remans approxmately at rest Collson Example Ballstc Pendulum Conceptualze Observe dagram Categorze Isolated system of projectle and block Perfectly nelastc collson the bullet s embedded n the block of wood omentum equaton wll have two unknowns Use conservaton of energy from the pendulum to fnd the velocty just after the collson Then you can fnd the speed of the bullet 4

5 Two-Dmensonal Collsons The momentum s conserved n all drectons Use subscrpts for Identfyng the object Indcatng ntal or fnal values The velocty components If the collson s elastc, use conservaton of knetc energy as a second equaton Remember, the smpler equaton can only be used for onedmensonal stuatons Two-Dmensonal Collson, example Partcle 1 s movng at velocty v 1 and partcle 2 s at rest In the x-drecton, the ntal momentum s m 1 v 1 In the y-drecton, the ntal momentum s 0 Clcker Queston What s the drecton of moton of v 1 m 2 after the collson? A. up-left B. up-rght C. down-left D. down-rght E. Rght only Two-Dmensonal Collson, example cont After the collson, the momentum n the x-drecton s m 1 v 1f cos m 2 v 2f cos After the collson, the momentum n the y-drecton s m 1 v 1f sn m 2 v 2f sn If the collson s elastc, apply the knetc energy equaton Ths s an example of a glancng collson Two-Dmensonal Collson Example Conceptualze See pcture Choose East to be the postve x-drecton and North to be the postve y- drecton Categorze Ignore frcton odel the cars as partcles The collson s perfectly nelastc The cars stck together The Center of ass There s a specal pont n a system or object, called the center of mass, that moves as f all of the mass of the system s concentrated at that pont The system wll move as f an external force were appled to a sngle partcle of mass located at the center of mass s the total mass of the system 5

6 Center of ass, Coordnates The coordnates of the center of mass are mx xc my y C mz zc s the total mass of the system Use the actve fgure to observe effect of dfferent masses and postons Center of ass, Extended Object Smlar analyss can be done for an extended object Consder the extended object as a system contanng a large number of partcles Snce partcle separaton s very small, t can be consdered to have a constant mass dstrbuton Center of ass, poston The center of mass n three dmensons can be located by ts poston vector, r C For a system of partcles, 1 rc mr r s the poston of the th partcle, defned by r x ˆ y ˆjzkˆ For an extended object, 1 rc dm r Fndng Center of ass, Irregularly Shaped Object Suspend the object from one pont The suspend from another pont The ntersecton of the resultng lnes s the center of mass Center of Gravty Each small mass element of an extended object s acted upon by the gravtatonal force The net effect of all these forces s equvalent to the effect of a sngle force g actng through a pont called the center of gravty If g s constant over the mass dstrbuton, the center of gravty concdes wth the center of mass Center of ass, Rod Conceptualze Fnd the center of mass of a rod of mass and length L The locaton s on the x- axs (or y C = z C = 0) Categorze Analyss problem Analyze Use equaton for x cm x C = L / 2 6

7 Velocty and omentum of a System of Partcles The velocty of the center of mass of a system of partcles s drc 1 C m dt v v The momentum can be expressed as v C m v p p tot The total lnear momentum of the system equals the total mass multpled by the velocty of the center of mass Acceleraton of the Center of ass The acceleraton of the center of mass can be found by dfferentatng the velocty wth respect to tme dvc 1 ac m dt a Forces In a System of Partcles The acceleraton can be related to a force ac F If we sum over all the nternal forces, they cancel n pars and the net force on the system s caused only by the external forces Newton s Second Law for a System of Partcles Snce the only forces are external, the net external force equals the total mass of the system multpled by the acceleraton of the center of mass: Fext ac The center of mass of a system of partcles of combned mass moves lke an equvalent partcle of mass would move under the nfluence of the net external force on the system Impulse and omentum of a System of Partcles The mpulse mparted to the system by external forces s I F dt d v p ext C tot The total lnear momentum of a system of partcles s conserved f no net external force s actng on the system v p constant when F 0 C tot ext oton of the Center of ass, Example A projectle s fred nto the ar and suddenly explodes Wth no exploson, the projectle would follow the dotted lne After the exploson, the center of mass of the fragments stll follows the dotted lne, the same parabolc path the projectle would have followed wth no exploson Use the actve fgure to observe a varety of explosons 7

8 Deformable Systems To analyze the moton of a deformable system, use Conservaton of Energy and the Impulse-omentum Theorem Esystem T K U 0 I p F dt mv tot ext If the force s constant, the ntegral can be easly evaluated Deformable System (Sprng) Example Conceptualze See fgure Push on left block, t moves to rght, sprng compresses At any gven tme, the blocks are generally movng wth dfferent veloctes The blocks oscllate back and forth wth respect to the center of mass Sprng Example, cont Categorze Non solated system Work s beng done on t by the appled force It s a deformable system The appled force s constant, so the acceleraton of the center of mass s constant odel as a partcle under constant acceleraton Analyze Apply mpulse-momentum Sprng Example, fnal Analyze, cont. Fnd energes Fnalze Answers do not depend on sprng length, sprng constant, or tme nterval Solve for v cm Rocket Propulson The operaton of a rocket depends upon the law of conservaton of lnear momentum as appled to a system of partcles, where the system s the rocket plus ts ejected fuel Rocket Propulson, 2 The ntal mass of the rocket plus all ts fuel s + m at tme t and speed v The ntal momentum of the system s p = ( + m) v 8

9 Rocket Propulson, 3 At some tme t + t, the rocket s mass has been reduced to and an amount of fuel, m has been ejected The rocket s speed has ncreased by v Rocket Propulson, 4 Because the gases are gven some momentum when they are ejected out of the engne, the rocket receves a compensatng momentum n the opposte drecton Therefore, the rocket s accelerated as a result of the push from the exhaust gases In free space, the center of mass of the system (rocket plus expelled gases) moves unformly, ndependent of the propulson process Rocket Propulson, 5 The basc equaton for rocket propulson s vf v veln f The ncrease n rocket speed s proportonal to the speed of the escape gases (v e ) So, the exhaust speed should be very hgh The ncrease n rocket speed s also proportonal to the natural log of the rato / f So, the rato should be as hgh as possble, meanng the mass of the rocket should be as small as possble and t should carry as much fuel as possble Thrust The thrust on the rocket s the force exerted on t by the ejected exhaust gases dv d thrust ve dt dt The thrust ncreases as the exhaust speed ncreases The thrust ncreases as the rate of change of mass ncreases The rate of change of the mass s called the burn rate 9

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