Impulse and Linear Momentum 5


 Augusta Lewis
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1 Implse and Linea Momentm 5 How does jet poplsion wok? How can yo mease the speed of a bllet? Wold a meteoite collision significantly change Eath s obit? In pevios chaptes we discoveed that the pshing inteaction between ca ties and the oad allows a ca to change its velocity. Likewise, a ship s popelles psh wate backwad; in tn, wate pshes the ship fowad. Bt how does a ocket, fa above Eath s atmosphee, change velocity with no object to psh against? Less than 1 yeas ago, ocket flight was consideed impossible. When U. S. ocket pionee Robet Goddad pblished an aticle Be se yo know how to: Constct a foce diagam fo an object (Section 2.1). Use Newton s second law in component fom (Section 3.2). Use kinematics to descibe an object s motion (Section 1.7). 151
2 152 Chapte 5 Implse and Linea Momentm in 192 abot ockety and even sggested a ocket flight to the Moon, he was idicled by the pess. A New Yok Times editoial dismissed his idea, saying,... even a schoolboy knows that ockets cannot fly in space becase a vacm is devoid of anything to psh on. We know now that Goddad was coect bt why? What does the ocket psh on? Fige 5.1 The mass is the same in the closed flask (an isolated system) (a) befoe bning the steel wool and (b) afte bning the steel wool. (c) Howeve, the mass inceases (a) when the steel wool is bned in the open flask (a nonisolated system). (a) The block Closed flask (a) The balances block the Closed Steel wool flask balances steel wool the The block Closed Steel wool flask steel and flask. Balancing wool balances and flask. the Steel Balancing block wool steel wool block and flask. Balancing block (b) (b) The steel wool is bned in (b) The a closed steel wool flask. is The bned block still in a closed balances. flask. The block The still balances. steel wool is bned in a closed flask. The block still balances. (c) (c) When the steel wool is bned (c) When in the open the steel flask, wool the is mass bned in in the the open flask flask, inceases. the mass When in the flask the steel inceases. wool is bned in the open flask, the mass in the flask inceases. We can se Newton s second law 1 a = F>m2 to elate the acceleation of a system object to the foces being exeted on it. Howeve, to se this law effectively we need qantitative infomation abot the foces that objects exet on each othe. Unfotnately, if two cas collide, we don t know the foce that one ca exets on the othe ding the collision. When fiewoks explode, we don t know the foces that ae exeted on the pieces flying apat. In this chapte yo will lean a new appoach that helps s analyze and pedict mechanical phenomena when the foces ae not known. 5.1 Mass acconting We begin o investigation by analyzing the physical qantity of mass. Ealie (in Chapte 2), we fond that the acceleation of an object depended on its mass the geate its mass, the less it acceleated de to an nbalanced extenal foce. We ignoed the possibility that an object s mass might change ding some pocess. Is the mass in a system always a constant vale? Yo have pobably obseved contless physical pocesses in which mass seems to change. Fo example, the mass of a log in a campfie deceases as the log bns; the mass of a seedling inceases as the plant gows. What happens to the lost mass fom the log? Whee does the seedling s inceased mass come fom? A system pespective helps s ndestand what happens to the bning log. If we choose only the log as the system, the mass of the system deceases as it bns. Howeve, ai is needed fo bning. What happens to the mass if we choose the sonding ai and the log as the system? Sppose that we place steel wool in a closed flask on one side of a balance scale and a metal block of eqal mass on the othe side (Fige 5.1a). In one expeiment, we bn the steel wool in the closed flask (the flask also contains ai), foming an oxide of ion. We find that the total mass of the closed flask containing bned steel wool (ion oxide) is the same as the mass of the balancing metal block (Fige 5.1b). Next, we bn the steel wool in an open flask and obseve that the mass of that flask inceases (Fige 5.1c). The steel wool in the open flask bns moe completely and absobs some extenal oxygen fom the ai as it bns. Eighteenthcenty Fench chemist Antoine Lavoisie actally pefomed sch expeiments. He ealized that the choice of the system was vey impotant. Lavoisie defined an isolated system as a gop of objects that inteact with each othe bt not with extenal objects otside the system. The mass of an isolated system is the sm of the masses of all objects in the system. He then sed the concept of an isolated system to smmaize his (and o) expeiments in the following way: Law of constancy of mass When a system of objects is isolated (a closed containe), its mass eqals the sm of the masses of its components and does not change it emains constant in time.
3 5.2 Linea momentm 153 When the system is not isolated (an open containe system), the mass might change. Howeve, this change is not andom it is always eqal to the amont of mass leaving o enteing the system fom the envionment. Ths, even when the mass of a system is not constant, we can keep tack of the changes if we take into accont how mch is leaving o enteing the system: initial mass of new mass enteing o final mass of system at ealie + leaving system between = system at late clock eading the two clock eadings clock eading The above eqation helps descibe the change of mass in any system. The mass is constant if thee is no flow of mass in o ot of the system, o the mass changes in a pedictable way if thee is some flow of mass between the system and the envionment. Basically, mass cannot appea fom nowhee and does not disappea withot a tace. Imagine yo have a system that has a total mass of m i = 3 kg (a bag of oanges). Yo add some moe oanges to the bag 1m = 1 kg2. The final mass of the system eqals exactly the sm of the initial mass and the added mass: m i + m = m f o 3 kg + 1 kg = 4 kg (Fige 5.2a). We can epesent this pocess with a ba chat (Fige 5.2b). The ba on the left epesents the initial mass of the system, the cental ba epesents the mass added o taken away, and the ba on the ight epesents the mass of the system in the final sitation. As a eslt, the height of the left ba pls the height of the cental ba eqals the height of the ight ba. The ba chat allows s to keep tack of the changes in mass of a system even if the system is not isolated. Mass is called a conseved qantity. A conseved qantity is constant in an isolated system. When the system is not isolated, we can accont fo the changes in the conseved qantity by what is added to o sbtacted fom the system. Jst as with evey idea in physics, the law of constancy of mass in an isolated system does not apply in all cases. We will discove late in this book (Chaptes 28 and 29) that in sitations involving atomic paticles, mass is not constant even in an isolated system; instead, what is constant is a new qantity that incldes mass as a component. Review Qestion 5.1 When yo bn a log in a fie pit, the mass of wood clealy deceases. How can yo define the system so as to have the mass of the objects in that system constant? 5.2 Linea momentm We now know that mass is an example of a conseved qantity. Is thee a qantity elated to motion that is conseved? When yo kick a stationay ball, thee seems to be a tansfe of motion fom yo foot to the ball. When yo knock bowling pins down with a bowling ball, a simila tansfe occs. Howeve, motion is not a physical qantity. What physical qantities descibing motion ae constant in an isolated system? Can we descibe the changes in these qantities sing a ba chat? Let s condct a few expeiments to find ot. In Obsevational Expeiment Table 5.1 we obseve two cats of diffeent masses that collide on a smooth tack. Fo these expeiments, the system will inclde both cats. A collision is a pocess that occs when two (o moe) objects come into diect contact with each othe. The system is isolated since the foces that the cats exet on each othe ae intenal, and extenal foces ae eithe balanced (as the vetical foces ae) o negligible (the hoizontal fiction foce). Fige 5.2 (a) The initial mass of the oanges pls the mass of the oanges that wee added (o sbtacted) eqals the final mass of the oanges. (b) The mass change pocess is epesented by a mass ba chat. (a) (b) m 3 kg 1 kg 4 kg m i m m f Active Leaning Gide
4 154 Chapte 5 Implse and Linea Momentm Obsevational Expeiment Table 5.1 Collisions in a system of two cats (all velocities ae with espect to the tack). Obsevational expeiment Analysis Video 5.1 Expeiment 1. Cat A (.2 kg) moving ight at 1. m/s collides with cat B (.2 kg), which is stationay. Cat A stops and cat B moves ight at 1. m/s. v Aix 1. m/s v Bix v Afx v Bfx 1. m/s A.2 kg B.2 kg A B x The diection of motion is indicated with a pls and a mins sign. Speed: The sm of the speeds of the system objects is the same befoe and afte the collision: 1. m>s + m>s = m>s + 1. m>s. Mass # speed: The sm of the podcts of mass and speed is the same befoe and afte the collision:.2 kg11. m>s2 +.2 kg1 m>s2 =.2 kg1 m>s2 +.2 kg11. m>s2. Mass # velocity: The sm of the podcts of mass and the xcomponent of velocity is the same befoe and afte the collision:.2 kg1+1. m>s2 +.2 kg12 =.2 kg kg1+1. m>s2. Expeiment 2. Cat A (.4 kg) moving ight at 1. m/s collides with cat B (.2 kg), which is stationay. Afte the collision, both cats move ight, cat B at 1.2 m/s, and cat A at.4 m/s. v Aix 1. m/s A.4 kg v Bix B.2 kg v Afx.4 m/s A v Bfx 1.2 m/s B x Speed: The sm of the speeds of the system objects is not the same befoe and afte the collision: 1. m>s + m>s.4 m>s m>s. Mass # speed: The sm of the podcts of mass and speed is the same befoe and afte the collision:.4 kg11. m>s2 +.2 kg1 m>s2 =.4 kg1.4 m>s2 +.2 kg11.2 m>s2. Mass # velocity: The sm of the podcts of mass and the xcomponent of velocity is the same befoe and afte the collision:.4 kg1+1. m>s2 +.2 kg12 =.4 kg1+.4 m>s2 +.2 kg1+1.2 m>s2. Expeiment 3. Cat A (.2 kg) with a piece of clay attached to the font moves ight at 1. m/s. Cat B (.2 kg) moves left at 1. m/s. The cats collide, stick togethe, and stop. v Aix 1. m/s v Bix 1. m/s v Afx v Bfx A.2 kg B.2 kg A B Speed: The sm of the speeds of the system objects is not the same befoe and afte the collision: 1. m>s + 1. m>s m>s + m>s. Mass # speed: The sm of the podcts of mass and speed is not the same befoe and afte the collision:.2 kg11. m>s2 +.2 kg11. m>s2.2 kg1 m>s2 +.2 kg1 m>s2. Mass # velocity: The sm of the podcts of mass and the xcomponent of velocity is the same befoe and afte the collision:.2 kg1+1. m>s2 +.2 kg11. m>s2 =.2 kg1 m>s2 +.2 kg1 m>s2. x Pattens One qantity emains the same befoe and afte the collision in each expeiment the sm of the podcts of the mass and xvelocity component of the system objects.
5 5.2 Linea momentm 155 In the thee expeiments in Obsevational Expeiment Table 5.1, only one qantity the sm of the podcts of mass and the xcomponent of velocity mv x emained the same befoe and afte the cats collided. Note also that the sm of the podcts of the mass and the ycomponent of velocity mv y did not change it emained zeo. Pehaps m v is the qantity chaacteizing motion that is constant in an isolated system. Bt will this patten pesist in othe sitations? Let s test this idea by sing it to pedict the otcome of the expeiment in Testing Expeiment Table 5.2. Active Leaning Gide Testing Expeiment table 5.2 Testing the idea that mv in an isolated system emains constant (all velocities ae with espect to the tack). Video 5.2 Testing expeiment Pediction Otcome Cat a (.4 kg) has a piece of modeling clay attached to its font and is moving ight at 1. m/s. Cat B (.2 kg) is moving left at 1. m/s. The cats collide and stick togethe. Pedict the velocity of the cats afte the collision. v Aix 1. m/s v fx? v Bix 1. m/s A.4 kg A B.2 kg B The system consists of the two cats. The diection of velocity is noted with a pls o mins sign of the velocity component: 1.4 kg21+1. m>s kg211. m>s2 x Afte the collision, the cats move togethe towad the ight at close to the pedicted speed. o = 1.4 kg +.2 kg2v f x v f x = 1+.2 kg # m>s2>1.6 kg2 = +.33 m>s Afte the collision, the two cats shold move ight at a speed of abot.33 m/s. Conclsion O pediction matched the otcome. This eslt gives s inceased confidence that this new qantity m v might be the qantity whose sm is constant in an isolated system. This new qantity is called linea momentm p. Linea Momentm The linea momentm p of a single object is the podct of its mass m and velocity v : p mv (5.1) Fige 5.3 Momentm is a vecto qantity with components. y m Linea momentm is a vecto qantity that points in the same diection as the object s velocity v (Fige 5.3). The SI nit of linea momentm is (kg # m/s). The total linea momentm of a system containing mltiple objects is the vecto sm of the momenta (plal of momentm) of the individal objects. p net m 1 v 1 m 2 v 2 P m n v n mv v p mv The components of a skydive s momentm: p x p y mv x
6 156 Chapte 5 Implse and Linea Momentm Note the following thee impotant points. 1. Unlike mass, which is a scala qantity, p = mv is a vecto qantity. Theefoe, it is impotant to conside the diection in which the colliding objects ae moving befoe and afte the collision. Fo example, becase cat B in Table 5.2 was moving left along the xaxis, the xcomponent of its momentm was negative befoe the collision. 2. Becase momentm depends on the velocity of the object, and the velocity depends on the choice of the efeence fame, diffeent obseves will mease diffeent momenta fo the same object. As a passenge, the momentm of a ca with espect to yo is zeo. Howeve, it is not zeo fo an obseve on the gond watching the ca move away fom him. 3. We chose an isolated system (the two cats) fo o investigation. The sm of the podcts of mass and velocity mv of all objects in the isolated system emained constant even thogh the cats collided with each othe. Howeve, if we had chosen the system to be jst one of the cats, we wold see that the linea momentm p = mv of the cat befoe the collision is diffeent than it is afte the collision. Ths, to establish that momentm p is a conseved qantity, we need to make se that the momentm of a system changes in a pedictable way fo systems that ae not isolated. We chose a system in Obsevational Expeiment Table 5.1 so that the sm of the extenal foces was zeo, making it an isolated system. Based on the eslts of Table 5.1 and Table 5.2, it appeas that the total momentm of an isolated system is constant. Momentm constancy of an isolated system The momentm of an isolated system is constant. Fo an isolated twoobject system: m 1 v 1i m 2 v 2i m 1 v 1f m 2 v 2f (5.2) Becase momentm is a vecto qantity and Eq. (5.2) is a vecto eqation, we will wok with its x and ycomponent foms: m 1 v 1i x m 2 v 2i x m 1 v 1f x m 2 v 2f x (5.3x) m 1 v 1i y m 2 v 2i y m 1 v 1f y m 2 v 2f y (5.3y) Fo a system with moe than two objects, we simply inclde a tem on each side of the eqation fo each object in the system. Let s test the idea that the momentm of an isolated system is constant in anothe sitation. Example 5.1 Two olleblades Jen (5 kg) and David (75 kg), both on olleblades, psh off each othe abptly. Each peson coasts backwad at appoximately constant speed. Ding a cetain time inteval, Jen tavels 3. m. How fa does David tavel ding that same time inteval? Sketch and tanslate The pocess is sketched at the ight. All motion is with espect to the floo and is along the xaxis. We choose the two olleblades as the system. Initially, the two olleblades ae at est. Afte pshing off, Jen (J) moves to the left and David (D) moves to the ight. We can se momentm constancy to calclate David s velocity component and pedict the distance he will tavel ding that same time inteval. Late Jen has taveled 3. m. Rolleblades psh off each othe. How fa has David taveled? Simplify and diagam We model each peson as a pointlike object and assme that the fiction foce exeted on the olleblades does not affect thei motion. Ths thee ae no hoizontal extenal foces exeted on
7 5.3 Implse and momentm 157 the system. In addition, the two vetical foces, an pwad nomal foce N F on P that the floo exets on each peson and an eqalmagnitde downwad gavitational foce F E on P that Eath exets on each peson, cancel, as we see in the foce diagams. Since the net extenal foce exeted on the system is zeo, the system is isolated. The foces that the olleblades exet on each othe ae intenal foces and shold not affect the momentm of the system. Repesent mathematically The initial state (i) of the system is befoe they stat pshing on each othe, and the final state (f) is when Jen has taveled 3. m. m J v Ji x + m D v Di x = m J v Jf x + m D v Df x We choose the positive diection towad the ight. Becase the initial velocity of each peson is zeo, the above eqation becomes o + = m J v Jf x + m D v Df x Solve and evalate The xcomponent of Jen s velocity afte the pshoff is v Jf x = 13. m2> t, whee t is the time inteval needed fo he to tavel 3. m. We solve the above eqation fo David s final xvelocity component to detemine how fa David shold tavel ding that same time inteval: v Df x =  m Jv Jf x =  m J v m D m Jf x D 15 kg m2 =  = 175 kg2 t 12. m2 t Since momentm is constant in this isolated system, we pedict that David will tavel 2. m in the positive diection ding t. The meased vale is vey close to the pedicted vale. Ty it yoself: Estimate the magnitde of yo momentm when walking and when jogging. Assme yo mass is 6 kg. Answe: When walking, yo tavel at a speed of abot 1 to 2 m/s. So the magnitde of yo momentm will be p = mv 16 kg211.5 m>s2 9 kg # m>s. When jogging, yo speed is abot 2 to 5 m/s o a momentm of magnitde p = mv 16 kg213.5 m>s2 2 kg # m>s. m D v Df x = m J v Jf x Notice that in Example 5.1 we wee able to detemine David s velocity by sing the pinciple of momentm constancy. We did not need any infomation abot the foces involved. This is a vey powefl eslt, since in all likelihood the foces they exeted on each othe wee not constant. The kinematics eqations we have sed p to this point have assmed constant acceleation of the system (and ths constant foces). Using the idea of momentm constancy has allowed s to analyze a sitation involving nonconstant foces. So fa, we have investigated sitations involving isolated systems. In the next section, we will investigate momentm in nonisolated systems. Review Qestion 5.2 Two identical cats ae taveling towad each othe at the same speed. One of the cats has a piece of modeling clay on its font. The cats collide, stick togethe, and stop. The momentm of each cat is now zeo. If the system incldes both cats, did the momentm of the system disappea? Explain yo answe. 5.3 Implse and momentm So fa, we have fond that the linea momentm of a system is constant if that system is isolated (the net extenal foce exeted on the system is zeo). How do we accont fo the change in momentm of a system when the net extenal foce exeted on it is not zeo? We can se Newton s laws to deive an expession elating foces and momentm change. Active Leaning Gide
8 158 Chapte 5 Implse and Linea Momentm Implse de to a foce exeted on a single object When yo psh a bowling ball, yo exet a foce on it, casing the ball to acceleate. The aveage acceleation a is defined as the change in velocity v f  v i divided by the time inteval t = t f  t i ding which that change occs: a = v f  v i t f  t i We can also se Newton s second law to detemine an object s acceleation if we know its mass and the sm of the foces that othe objects exet on it: a = F m We now have two expessions fo an object s acceleation. Setting these two expessions fo acceleation eqal to each othe, we get v f  v i t f  t i = F m Now mltiply both sides by m1t f  t i 2 and get the following: m v f  m v i = p f  p i = F1t f  t i 2 (5.4) The left side of the above eqation is the change in momentm of the object. This change depends on the podct of the net extenal foce and the time inteval ding which the foces ae exeted on the object (the ight side of the eqation). Note these two impotant points: 1. Eqation (5.4) is jst Newton s second law witten in a diffeent fom one that involves the physical qantity momentm. 2. Both foce and time inteval affect momentm the longe the time inteval, the geate the momentm change. A small foce exeted fo a long time inteval can change the momentm of an object by the same amont as a lage foce exeted fo a shot time inteval. Fige 5.4 The implse of a foce is the aea nde the gaph line. F (N) F av The podct of the extenal foce exeted on an object ding a time inteval and the time inteval gives s a new qantity, the implse of the foce. When yo kick a football o hit a baseball with a bat, yo foot o the bat exets an implse on the ball. The foces in these sitations ae not constant bt instead vay in time (see the example in Fige 5.4). The shaded aea nde the vaying foce cve epesents the implse of the foce. We can estimate the implse by dawing a hoizontal line that is appoximately the aveage foce exeted ding the time inteval of the implse. The aea nde the ectangla aveage foceimplse cve eqals the podct of the height of the ectangle (the aveage foce) and the width of the ectangle (the time inteval ove which the aveage foce is exeted). The podct F av 1t f  t i 2 eqals the magnitde of the implse. Implse The implse J of a foce is the podct of the aveage foce F av exeted on an object ding a time inteval 1t f  t i 2 and that time inteval: J Fav 1t f t i 2 (5.5) t i The aea of the shaded ectangle is abot the same as the aea nde the cved line and eqals the implse of the foce. t f t (s) Implse is a vecto qantity that points in the diection of the foce. The implse has a pls o mins sign depending on the oientation of the foce elative to a coodinate axis. The SI nit fo implse is N # s 1kg # m>s 2 2 # s kg # m>s, the same nit as momentm. It is often difficlt to mease diectly the implse of the net aveage foce ding a time inteval. Howeve, we can detemine the net foce on the ight
9 5.3 Implse and momentm 159 side of Eq. (5.4) indiectly by measing o calclating the momentm change on the left side of the eqation. Fo this eason, the combination of implse and momentm change povides a powefl tool fo analyzing inteactions between objects. We can now wite Eq. (5.4) as the implsemomentm eqation fo a single object. Implsemomentm eqation fo a single object If seveal extenal objects exet foces on a singleobject system ding a time inteval 1t f t i 2, the sm of thei implses J cases a change in momentm of the system object: p f p i J F on System 1t f t i 2 (5.6) The x and yscala component foms of the implsemomentm eqation ae p f x p i x F on System x 1t f t i 2 (5.7x) p f y p i y F on System y 1t f t i 2 (5.7y) A few points ae woth emphasizing. Fist, notice that Eq. (5.6) is a vecto eqation, as both the momentm and the implse ae vecto qantities. Vecto eqations ae not easy to maniplate mathematically. Theefoe, we will se the scala component foms of Eq. (5.6) Eqs. (5.7x) and (5.7y). Second, the time inteval in the implsemomentm eqation is vey impotant. When object 2 exets a foce on object 1, the momentm of object 1 changes by an amont eqal to p 1 = p 1f  p 1i = F 2 on 1 1t f  t i 2 = F 2 on 1 t The longe that object 2 exets the foce on object 1, the geate the momentm change of object 1. This explains why a fastmoving object might have less of an effect on a stationay object ding a collision than a slowmoving object inteacting with the stationay object ove a longe time inteval. Fo example, a fastmoving bllet passing thogh a patially closed wooden doo might not open the doo (it will jst make a hole in the doo), wheeas yo little finge, moving mch slowe than the bllet, cold open the doo. Althogh the bllet moves at high speed and exets a lage foce on the doo, the time inteval ding which it inteacts with the doo is vey small (milliseconds). Hence, it exets a elatively small implse on the doo too small to significantly change the doo s momentm. A photo of a bllet shot thogh an apple illstates the effect of a shot implse time (Fige 5.5). The implse exeted by the bllet on the apple was too small to knock the apple off its sppot. Thid, if the magnitde of the foce changes ding the time inteval consideed in the pocess, we se the aveage foce. Finally, if the same amont of foce is exeted fo the same time inteval on a lagemass object and on a smallmass object, the objects will have an eqal change in momentm (the same implse was exeted on them). Howeve, the smallmass object wold expeience a geate change in velocity than the lagemass object. Fige 5.5 The bllet s time of inteaction with the apple is vey shot, casing a small implse that does not knock the apple ove. Active Leaning Gide Example 5.2 Abpt stop in a ca A 6kg peson is taveling in a ca that is moving at 16 m/s with espect to the gond when the ca hits a baie. The peson is not weaing a seat belt, bt is stopped by an ai bag in a time inteval of.2 s. Detemine the aveage foce that the ai bag exets on the peson while stopping him. Sketch and tanslate Fist we daw an initialfinal sketch of the pocess. We choose the peson as the system since we ae investigating a foce being exeted on him. (contined )
10 16 Chapte 5 Implse and Linea Momentm The peson s initial xcomponent of velocity v Pi x = +16 m>s deceases to the final xcomponent of velocity v Pf x = in a time inteval 1t f  t i 2 =.2 s. Ths the aveage foce exeted by the ai bag on the peson in the xdiection is 16 kg2116 m>s2 F B on P x = 1.2 s  2 = 48 N The negative sign in 48 N indicates that the aveage foce points in the negative xdiection. The magnitde of this foce is abot 1 lb! Simplify and diagam The foce diagam shows the aveage foce F B on P exeted in the negative diection by the bag on the peson. The vetical nomal foce and gavitational foces cancel. Repesent mathematically The xcomponent fom of the implsemomentm eqation is m P v Pi x + F B on P x 1t f  t i 2 = m P v Pf x Solve and evalate Solve fo the foce exeted by the ai bag on the peson: F B on P x = m P1v Pf x  v Pi x 2 1t f  t i 2 Ty it yoself: Sppose a 6kg cash test dmmy is in a ca taveling at 16 m/s. The dmmy is not weaing a seat belt and the ca has no ai bags. Ding a collision, the dmmy flies fowad and stops when it hits the dashboad. The stopping time inteval fo the dmmy is.2 s. What is the aveage magnitde of the stopping foce that the dashboad exets on the dmmy? Answe: The aveage foce that the had sface exets on the dmmy wold be abot 5, N, extemely nsafe fo a hman. Note that the momentm change of the peson in Example 5.2 was the same. Howeve, since the change fo the dmmy occs ding a shote time inteval (.2 s instead of.2 s), the foce exeted on the dmmy is mch geate. This is why ai bags save lives. Fige 5.6 Analyzing the collision of two cats in ode to develop the momentm constancy idea. Initial Final v 1i Ding collision 1 2 F 2 on 1 v 1f F 1 on v 2i v 2f 1 2 x x x Using Newton s laws to ndestand the constancy of momentm in an isolated system of two o moe objects Let s apply the implsemomentm eqation Eq. (5.4) to the scenaio we descibed in Obsevational Expeiment Table 5.1 in ode to exploe momentm constancy in a twoobject isolated system. Two cats tavel towad each othe at diffeent speeds, collide, and ebond backwad (Fige 5.6). We fist analyze each cat as a sepaate system and then analyze them togethe as a single system. Assme that the vetical foces exeted on the cats ae balanced and that the fiction foce exeted by the sface on the cats does not significantly affect thei motion. Cat 1: In the initial state, befoe the collision, cat 1 with mass m 1 tavels in the positive diection at velocity v 1i. In the final state, afte the collision, cat 1 moves with a diffeent velocity v 1f in the opposite diection. To detemine the effect of the implse exeted by cat 2 on cat 1, we apply the implsemomentm eqation to cat 1 only: m 1 1 v 1f  v 1i 2 = F 2 on 1 1t f  t i 2
11 5.4 The genealized implsemomentm pinciple 161 Cat 2: We epeat this analysis with cat 2 as the system. Its velocity and momentm change becase of the implse exeted on it by cat 1: m 2 1 v 2f  v 2i 2 = F 1 on 2 1t f  t i 2 Newton s thid law povides a connection between o analyses of the two cats; inteacting objects at each instant exet eqalmagnitde bt oppositely diected foces on each othe: F 1 on 2 = F 2 on 1 Sbstitting the expessions fo the foces fom above and simplifying, we get m 2 1v 2f  v 2i 2 =  m 11v 1f  v 1i 2 t f  t i t f  t i m 2 1v 2f  v 2i 2 = m 1 1v 1f  v 1i 2 We now move the initial momentm fo both objects to the left side and the final momentm fo both objects to the ight side: m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f Initial momentm Final momentm This is the same eqation we aived at in Section 5.2, whee we obseved and analyzed collisions to ndestand the constant momentm of an isolated system. Hee we have eached the same conclsions sing only o knowledge of Newton s laws, momentm, and implse. Review Qestion 5.3 An apple is falling fom a tee. Why does its momentm change? Specify the extenal foce esponsible. Find a system in which the momentm is constant ding this pocess. 5.4 The genealized implsemomentm pinciple We can smmaize what we have leaned abot momentm in isolated and nonisolated systems. The change in momentm of a system is eqal to the net extenal implse exeted on it. If the net implse is zeo, then the momentm of the system is constant. This idea, expessed mathematically as the genealized implsemomentm pinciple, acconts fo sitations in which the system incldes one o moe objects and may o may not be isolated. The genealized implsemomentm pinciple means that we can teat momentm as a conseved qantity. Genealized implsemomentm pinciple Fo a system containing one o moe objects, the initial momentm of the system pls the sm of the implses that extenal objects exet on the system objects ding the time inteval 1t f  t i 2 eqals the final momentm of the system: 1m 1 v 1i m 2 v 2i P2 F on Sys 1t f t i 2 1m 1 v 1f m 2 v 2f P2 (5.8) Initial momentm of Net implse exeted on Final momentm of the system the system the system The x and ycomponent foms of the genealized implsemomentm pinciple ae 1m 1 v 1i x m 2 v 2i x P2 F on Sys x 1t f t i 2 1m 1 v 1f x m 2 v 2f x P2 (5.9x) 1m 1 v 1i y m 2 v 2i y P2 F on Sys y 1t f t i 2 1m 1 v 1f y m 2 v 2f y P2 (5.9y) Note: If the net implse exeted in a paticla diection is zeo, then the component of the momentm of the system in that diection is constant.
12 162 Chapte 5 Implse and Linea Momentm Eqations (5.8) and (5.9) ae sefl in two ways. Fist, any time we choose to analyze a sitation sing the ideas of implse and momentm, we can stat fom a single pinciple, egadless of the sitation. Second, the eqations emind s that we need to conside all the inteactions between the envionment and the system that might case a change in the momentm of the system. Implsemomentm ba chats We can descibe an implsemomentm pocess mathematically sing Eqs. (5.9x and y). These eqations help s see that we can epesent the changes of a system s momentm sing a ba chat simila to the one sed to epesent the changes of a system s mass. The Reasoning Skill box shows the steps fo constcting an implsemomentm ba chat fo a simple system of two cats of eqal mass taveling towad each othe. Reasoning Skill Constcting a qalitative implsemomentm ba chat. 1. Sketch the pocess, choose the initial and final states, and choose a system. Initial Final v 1i v 2i v f x p 1ix p 2ix J x p 1fx p 2fx 2. Daw initial and final momentm bas fo each object in the system. (Note cat diections and ba diections.) Slowe in negative diection (m 1 v 1ix m 2 v 2ix ) J x (m 1 v 1fx m 2 v 2fx ) o m 1 v 1i m 2 (v 2i ) (m 1 m 2 )v f 3. Daw an implse (J) ba if thee is an extenal nonzeo implse. 4. Convet each ba in the chat into a tem in the component fom of the implsemomentm eqation. Note that befoe constcting the ba chat, we epesent the pocess in an initialfinal sketch (Step 1 in the Skill box). We then se the sketch to help constct the implsemomentm ba chat. The lengths of the bas ae qalitative indicatos of the elative magnitdes of the momenta. In the final state in the example shown, the cats ae stck togethe and ae moving in the positive diection. Since they have the same mass and velocity, they each have the same final momentm. The middle shaded colmn in the ba chat epesents the net extenal implse exeted on the system objects ding the time inteval 1t f  t i 2 thee is no implse fo the pocess shown. The shading eminds s that implse does not eside in the system; it is the inflence of the extenal objects on the momentm of the system. Notice that the sm of the heights of the bas on the left pls the height of the shaded implse ba shold eqal the sm of the heights of the bas on the ight. This consevation of ba heights eflects the consevation of momentm. We can se the ba chat to apply the genealized implsemomentm eqation (Step 4). Each nonzeo ba coesponds to a nonzeo tem in the eqation; the sign of the tem depends on the oientation of the ba.
13 5.4 The genealized implsemomentm pinciple 163 Using implsemomentm to investigate foces Can we se the ideas of implse and momentm to lean something abot the foces that two objects exet on each othe ding a collision? Conside a collision between two cas (Fige 5.7). To analyze the foce that each ca exets on the othe, we will define the system to inclde only one of the cas. Let s choose ca 1 and constct a ba chat fo it. Ca 2 exets an implse on ca 1 ding the collision that changes the momentm of ca 1. If the initial momentm of ca 1 is in the positive diection, then the implse exeted by ca 2 on ca 1 points in the negative diection. Becase of this, the implse ba on the ba chat points downwad. Note that the total height of the initial momentm ba on the left side of the chat and the height of the implse ba add p to the total height of the final momentm ba on the ight side. Using the ba chat, we can apply the component fom of the implsemomentm eqation: m 1 v 1i x + J x = m 1 v 1f x The components of the initial and final momentm ae positive. As the foce is exeted in the negative diection, the xcomponent of the implse is negative and eqal to F 2 on 1 t. Ths, +m 1 v 1i + 1F 2 on 1 t2 = +m 1 v 1f If we know the initial and final momentm of the ca and the time inteval of inteaction, we can se this eqation to detemine the magnitde of the aveage foce that ca 2 exeted on ca 1 ding the collision. Fige 5.7 A ba chat analysis of the collision of ca 2 with ca 1. v 1i Initial v 1f m 1 m 1 Ca 1 has consideable momentm in the positive diection. F 2 on 1 p 1ix J 2 on 1x p 1fx The foce exeted by 2 on 1 is in the negative diection. Final Active Leaning Gide x Ca 1 has momentm in the positive diection. Tip When yo daw a ba chat, always specify the efeence fame (the object of efeence and the coodinate system). The diection of the bas on the ba chat (p fo positive and down fo negative) shold match the diection of the momentm o implse based on the chosen coodinate system. Example 5.3 Happy and sad balls Yo have two balls of identical mass and size that behave vey diffeently. When yo dop the socalled sad ball, it thds on the floo and does not bonce at all. When yo dop the socalled happy ball fom the same height, it bonces back to almost the same height fom which it was dopped. The diffeence in the boncing ability of the happy ball is de its intenal stcte; it is made of diffeent mateial. Yo hang each ball fom a sting of identical length and place a wood boad on its end diectly below the sppot fo each sting. Yo pll each ball back to an eqal height and elease the balls one at a time. When each ball hits the boad, which has the best chance of knocking the boad ove: the sad ball o the happy ball? Sad ball Happy ball Sketch and tanslate Initial and final sketches of the pocess ae shown at the ight. The system is jst the ball. In the initial state, the ball is jst abot to hit the boad, moving hoizontally towad the left (the balls ae moving eqally fast). The final state is jst afte the collision with the boad. The happy ball (H) bonces back, wheeas the sad ball (S) does not. Same initial Same momentm initial momentm fo both balls fo jst both befoe balls jst hitting befoe the hitting boad the boad Sad ball stops. Sad Happy ball ball stops. Happy bonces ball back. bonces back. (contined )
14 164 Chapte 5 Implse and Linea Momentm Simplify and diagam Assme that the collision time inteval t fo each ball is abot the same. We analyze only the hoizontal xcomponent of the pocess, the component that is elevant to whethe o not each of the boads is knocked ove. Each boad exets an implse on the ball that cases the momentm of the ball to change. Theefoe, each ball, accoding to Newton s thid law, exets an implse on the boad that it hits. A lage foce exeted on the boad means a lage implse and a bette chance to tip the boad. A ba chat fo each ballboad collision is shown below. The implse of the happy ball is twice as lage in magnitde as that of the sad ball and cases twice as lage a momentm change. Repesent mathematically The xcomponent fom of the implsemomentm Eq. (5.5x) applied to each ball is as follows: Sad ball: mv i + F B on S x t = m # Happy ball: mv i + F B on H x t = m1v i 2 Note that the xcomponent of the final velocity of the sad ball is v Sf x = (it does not bonce) and that the xcomponent of the final velocity of the happy ball is v Hf x = v i (it bonces). Solve and evalate We can now get an expession fo the foce exeted by each boad on each ball: Sad ball: F B on S x = m1  v i2 t Happy ball: F B on H x = m31v i2  v i 4 t =  mv i t. =  2mv i t. Becase we assmed that the time of collision is the same, the boad exets twice the foce on the happy ball as on the sad ball, since the boad cases the happy ball s momentm to change by an amont twice that of the sad ball. Accoding to Newton s thid law, this means that the happy ball will exet twice as lage a foce on the boad as the sad ball. Ths, the happy ball has a geate chance of tipping the boad. Ty it yoself: Is it less safe fo a football playe to bonce backwad off a goal post o to hit the goal post and stop? Answe: Althogh any collision is dangeos, it is bette to hit the goal post and stop. If the football playe bonces back off the goal post, his momentm will have changed by a geate amont (like the happy ball in the last example). This means that the goal post exets a geate foce on him, which means thee is a geate chance fo injy. The patten we fond in the example above is te fo all collisions when an object bonces back afte a collision, we know that a lage magnitde foce is exeted on it than if the object had stopped and did not bonce afte the collision. Fo that eason, blletpoof vests fo law enfocement agents ae designed so that the bllet embeds in the vest athe than boncing off it. Review Qestion 5.4 If in solving the poblem in Example 5.3 we chose the system to be the ball and the boad, how wold the mathematical desciption fo each ballboad collision change? 5.5 Skills fo analyzing poblems sing the implsemomentm eqation Initial and final sketches and ba chats ae sefl tools to help analyze pocesses sing the implsemomentm pinciple. Let s investigate fthe how these tools wok togethe. A geneal stategy fo analyzing sch pocesses is
15 5.5 Skills fo analyzing poblems sing the implsemomentm eqation 165 descibed on the left side of the table in Example 5.4 and illstated on the ight side fo a specific pocess. PROBLEMSOLvinG STRATEGY Applying the implsemomentm eqation Active Leaning Gide Sketch and tanslate Sketch the initial and final states and inclde appopiate coodinate axes. Label the sketches with the known infomation. Decide on the object of efeence. Choose a system based on the qantity yo ae inteested in; fo example, a mltiobject isolated system to detemine the velocity of an object, o a singleobject nonisolated system to detemine an implse o foce. Example 5.4 Bllet hits wood block A.2kg bllet taveling hoizontally at 25 m>s embeds in a 1.kg block of wood esting on a table. Detemine the speed of the bllet and wood block togethe immediately afte the bllet embeds in the block. The left side of the sketch below shows the bllet taveling in the positive xdiection with espect to the gond; it then joins the wood. All motion is along the xaxis; the object of efeence is Eath. The system incldes the bllet and wood; it is an isolated system since the vetical foces balance. The initial state is immediately befoe the collision; the final state is immediately afte. Initial m B.2 kg, m W 1. kg v Bix 25 m/s, v Wix v Bi Final v BWfx? v BWf x x Simplify and diagam Detemine if thee ae any extenal implses exeted on the system. Dawing a foce diagam cold help detemine the extenal foces and thei diections. Daw an implsemomentm ba chat fo the system fo the chosen diection(s) to help yo ndestand the sitation, fomlate a mathematical epesentation of the pocess, and evalate yo eslts. Assme that the fiction foce exeted by the tabletop on the bottom of the wood does not change the momentm of the system ding the vey shot collision time inteval. The ba chat epesents the pocess. The ba fo the bllet is shote than that fo the block thei velocities ae the same afte the collision, bt the mass of the bllet is mch smalle. We do not daw a foce diagam hee, as the system is isolated. p Bix p Wix J x p Bfx p Wfx Repesent mathematically Use the ba chat to apply the genealized implsemomentm eqation along the chosen axis. Each nonzeo ba becomes a nonzeo tem in the eqation. The oientation of the ba detemines the sign in font of the coesponding tem in the eqation. Remembe that momentm and implse ae vecto qantities, so inclde the pls o mins signs of the components based on the chosen coodinate system. m B v Bi x + m W # + 1Jx 2 = m B v x + m W v x Since J x =, v x = m B v Bi x 1m B + m W 2 (contined )
16 166 Chapte 5 Implse and Linea Momentm Solve and evalate Inset the known infomation to detemine the nknown qantity. Check if yo answe is easonable with espect to sign, nit, and magnitde. Also make se it applies fo limiting cases, sch as objects of vey small o vey lage mass. v x = 1.2 kg2125 m>s2 1.2 kg + 1. kg2 = +4.9 m>s The magnitde of the answe seems easonable given how fast the bllet was initially taveling. The pls sign indicates the diection, which makes sense, too. The nits ae also coect (m/s). We can test this sing a limiting case: if the mass o speed of the bllet is zeo, the block emains stationay afte the collision. Ty it yoself: A.2kg bllet is fied hoizontally into a 2.kg block of wood esting on a table. Immediately afte the bllet joins the block, the block and bllet move in the positive xdiection at 4. m/s. What was the initial speed of the bllet? Answe: 4 m/s. We cold have woked Example 5.4 backwad to detemine the initial speed of the bllet befoe hitting the block (like the Ty It Yoself qestion). This execise wold be sefl since the bllet tavels so fast that it is difficlt to mease its speed. Vaiations of this method ae sed, fo example, to decide whethe o not golf balls confom to the necessay les. The balls ae hit by the same mechanical lanching implse and the moving balls embed in anothe object. The balls speeds ae detemined by measing the speed of the object they embed in. Detemining the stopping time inteval fom the stopping distance When a system object collides with anothe object and stops a ca collides with a tee o a wall, a peson jmps and lands on a solid sface, o a meteoite collides with Eath the system object tavels what is called its stopping distance. By estimating the stopping distance of the system object, we can estimate the stopping time inteval. Sppose that a ca ns into a lage tee and its font end cmples abot.5 m. This.5 m, the distance that the cente of the ca taveled fom the beginning of the impact to the end, is the ca s stopping distance. Similaly, the depth of the hole left by a meteoite povides a ogh estimate of its stopping distance when it collided with Eath. Howeve, to se the implsemomentm pinciple, we need the stopping time inteval associated with the collision, not the stopping distance. Hee s how we can se a known stopping distance to estimate the stopping time inteval. Assme that the acceleation of the object while stopping is constant. In that case, the aveage velocity of the object while stopping is jst the sm of the initial and final velocities divided by 2: v aveage x = 1v f x + v i x 2>2. Ths, the stopping displacement 1x f  x i 2 and the stopping time inteval 1t f  t i 2 ae elated by the kinematics eqation x f  x i = v aveage x 1t f  t i 2 = 1v f x + v i x 2 1t 2 f  t i 2
17 5.5 Skills fo analyzing poblems sing the implsemomentm eqation 167 Reaange this eqation to detemine the stopping time inteval: t f  t i = 21x f  x i 2 v f x + v i x (5.1) Eqation (5.1) povides a method to convet stopping distance x f  x i into stopping time inteval t f  t i. Eqation (5.1) can be applied to hoizontal o vetical stopping. Active Leaning Gide Example 5.5 Stopping the fall of a movie stnt dive The ecod fo the highest movie stnt fall withot a paachte is 71 m (23 ft), held by 8kg A. J. Baknas. His fall was stopped by a lage ai cshion, into which he sank abot 4. m. His speed was abot 36 m/s (8 mi/h) when he eached the top of the ai cshion. Estimate the aveage foce that the cshion exeted on his body while stopping him. Sketch and tanslate We focs only on the pat of the fall when Baknas is sinking into the cshion. The sitation is sketched below. We choose Baknas as the system and the yaxis pointing p. The initial state is jst as he toches the cshion at position y i = +4. m, and the final state is when the cshion has stopped him, at position y f =. All motion is with espect to Eath. The othe infomation abot the pocess is given in the fige. Be se to pay attention to the signs of the qantities (especially the initial velocity). Each extenal foce cases an implse. Baknas has zeo momentm in the final state. Repesent mathematically Since all motion and all of the foces ae in the vetical diection, we se the ba chat to help constct the vetical ycomponent fom of the implsemomentm eqation [Eq. (5.6y)] to detemine the foce that the cshion exets on Baknas as he sinks into it: m B v i y + 1N C on B y + F E on B y 21t f  t i 2 = m B v f y Simplify and diagam We daw a foce diagam top ight, modeling Baknas as a pointlike object. Since Baknas s downwad speed deceases, the cshion mst be exeting an pwad foce on Baknas of geate magnitde than the downwad foce that Eath exets on him. Ths, the net foce exeted on him points pwad, in the positive y diection. Using this infomation, we can daw a qalitative implsemomentm ba chat fo the pocess. Using the foce diagam, we see that the ycomponents of the foces ae N C on B y = +N C on B and F E on B y = F E on B = m B g, whee N C on B is the magnitde of the aveage nomal foce that the cshion exets on Baknas, the foce we ae tying to estimate. Noting that v f y = and sbstitting the foce components into the above eqation, we get m B v i y + 31+N C on B m B g241t f  t i 2 = m B # 1 m B v i y + 1N C on B  m B g21t f  t i 2 =. We can find the time inteval that the cshion takes to stop Baknas sing Eq. (5.1) and noting that v f y = : t f  t i = 21y f  y i 2 + v i y (contined )
18 168 Chapte 5 Implse and Linea Momentm Solve and evalate The stopping time inteval while Baknas sinks 4. m into the cshion is t f  t i = Solving fo N C on B, we get N C on B = m Bv i y 1t f  t i 2 + m Bg = m m>s2 =.22 s 18 kg2136 m>s s kg219.8 N>kg2 = +13, N + 78 N = 14, N Wow, that is a hge foce! To edce the isk of injy, stnt dives pactice landing so that the stopping foce that a cshion exets on them is distibted evenly ove the entie body. The cshions mst be deep enogh so that they povide a long stopping time inteval and ths a smalle stopping foce. The same stategy is applied to developing ai bags and collapsible fames fo atomobiles to make them safe fo passenges ding collisions. Notice fo impotant points. Fist, we ve inclded only two significant digits since that is how many the data had. Second, it is vey easy to make sign mistakes. A good way to avoid these is to daw a sketch that incldes a coodinate system and labels showing the vales of known physical qantities, inclding thei signs. Thid, the implse de to Eath s gavitational foce is small in magnitde compaed to the implse exeted by the ai cshion. Lastly, the foce exeted by the ai cshion wold be even geate if the stopping distance and conseqently the stopping time inteval wee shote. Ty it yoself: Sppose that the cshion in the last example stopped Baknas in 1. m instead of 4. m. What wold be the stopping time inteval and the magnitde of the aveage foce of the cshion on Baknas? Answe: The stopping time inteval is.56 s, and the aveage stopping foce is appoximately 5, N. Odeofmagnitde estimate will bone beak? The stategy that we sed in the pevios example can be sed to analyze skll facte injies that might lead to concssions. Laboatoy expeiments indicate that the hman skll can facte if the compessive foce exeted on it pe nit aea is 1.7 * 1 8 N>m 2. The sface aea of the skll is mch smalle than 1 m 2, so we will se sqae centimetes, a moe easonable nit of aea fo this discssion. Since 1 m 2 = 1 * 1 4 cm 2, we convet the compessive foce pe aea to 11.7 * 1 8 N>m 2 1 m 2 2a 1 * 1 4 cm 2 b = 1.7 * 14 N>cm 2. Example 5.6 Bone facte estimation 1 A bicyclist is watching fo taffic fom the left while tning towad the ight. A steet sign hit by an ealie ca accident is bent ove the side of the oad. The cyclist s head hits the pole holding the sign. Is thee a significant chance that his skll will facte? S k e t c h a n d tanslate The p o c e s s i s sketched at the ight. The initial state is at the instant that the h e a d i n i t i a l l y contacts the pole; the final state is when the head and body have stopped. The peson is the system. We have been given little infomation, so we ll have to make some easonable estimates of vaios qantities in ode to make a decision abot a possible skll facte. Simplify and diagam The ba chat illstates the momentm change of the system and the implse exeted by the pole that cased the change. The peson was initially moving in the hoizontal x diection with espect to Eath, and not moving afte the collision. The pole exeted an implse in the negative x diection on the cyclist. We ll need to estimate the following qantities: the mass and speed of the cyclist in this sitation, the stopping time inteval, and the aea of contact. Let s assme that this is a 7kg cyclist moving at abot 3 m/s. The peson s body keeps moving fowad fo a shot distance afte the bone makes contact with the pole. The skin indents some ding the collision. Becase of these two factos, we assme 1 This is a te stoy it happened to one of the book s athos, Alan Van Hevelen.
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