Chapter 9. Linear Momentum and Collisions

Transcription

1 Chapter 9 Lnear Momentum and Collsons CHAPTER OUTLINE 9.1 Lnear Momentum and Its Conservaton 9.2 Impulse and Momentum 9.3 Collsons n One Dmenson 9.4 Two-Dmensonal Collsons 9.5 The Center of Mass 9.6 Moton of a System of Partcles 9.7 Rocket Propulson A movng bowlng ball carres momentum, the topc of ths chapter. In the collson between the ball and the pns, momentum s transferred to the pns. (Mark Cooper/Corbs Stock Market)

2 Consder what happens when a bowlng ball strkes a pn, as n the openng photograph. The pn s gven a large velocty as a result of the collson; consequently, t fles away and hts other pns or s projected toward the backstop. Because the average force exerted on the pn durng the collson s large (resultng n a large acceleraton), the pn acheves the large velocty very rapdly and experences the force for a very short tme nterval. Accordng to Newton s thrd law, the pn exerts a reacton force on the ball that s equal n magntude and opposte n drecton to the force exerted by the ball on the pn. Ths reacton force causes the ball to accelerate, but because the ball s so much more massve than the pn, the ball s acceleraton s much less than the pn s acceleraton. Although F and a are large for the pn, they vary n tme a complcated stuaton! One of the man objectves of ths chapter s to enable you to understand and analyze such events n a smple way. Frst, we ntroduce the concept of momentum, whch s useful for descrbng objects n moton. Imagne that you have ntercepted a football and see two players from the opposng team approachng you as you run wth the ball. One of the players s the 180-lb quarterback who threw the ball; the other s a 300-lb lneman. Both of the players are runnng toward you at 5 m/s. However, because the two players have dfferent masses, ntutvely you know that you would rather collde wth the quarterback than wth the lneman. The momentum of an object s related to both ts mass and ts velocty. The concept of momentum leads us to a second conservaton law, that of conservaton of momentum. Ths law s especally useful for treatng problems that nvolve collsons between objects and for analyzng rocket propulson. In ths chapter we also ntroduce the concept of the center of mass of a system of partcles. We fnd that the moton of a system of partcles can be descrbed by the moton of one representatve partcle located at the center of mass. 9.1 Lnear Momentum and Its Conservaton In the precedng two chapters we studed stuatons that are complex to analyze wth Newton s laws. We were able to solve problems nvolvng these stuatons by applyng a conservaton prncple conservaton of energy. Consder another stuaton a 60-kg archer stands on frctonless ce and fres a 0.50-kg arrow horzontally at 50 m/s. From Newton s thrd law, we know that the force that the bow exerts on the arrow wll be matched by a force n the opposte drecton on the bow (and the archer). Ths wll cause the archer to begn to slde backward on the ce. But wth what speed? We cannot answer ths queston drectly usng ether Newton s second law or an energy approach there s not enough nformaton. Despte our nablty to solve the archer problem usng our technques learned so far, ths s a very smple problem to solve f we ntroduce a new quantty that descrbes moton, lnear momentum. Let us apply the General Problem-Solvng Strategy and conceptualze an solated system of two partcles (Fg. 9.1) wth masses m 1 and m 2 and movng wth veloctes v 1 and v 2 at an nstant of tme. Because the system s solated, the only force on 252

3 SECTION 9.1 Lnear Momentum and Its Conservaton 253 one partcle s that from the other partcle and we can categorze ths as a stuaton n whch Newton s laws wll be useful. If a force from partcle 1 (for example, a gravtatonal force) acts on partcle 2, then there must be a second force equal n magntude but opposte n drecton that partcle 2 exerts on partcle 1. That s, they form a Newton s thrd law acton reacton par, so that F 12 F 21. We can express ths condton as v 1 F 21 F 12 0 Let us further analyze ths stuaton by ncorporatng Newton s second law. Over some tme nterval, the nteractng partcles n the system wll accelerate. Thus, replacng each force wth ma gves m 1 a 1 m 2 a 2 0 Now we replace the acceleraton wth ts defnton from Equaton 4.5: m 1 d v 1 dt m 2 d v 2 dt If the masses m 1 and m 2 are constant, we can brng them nto the dervatves, whch gves 0 m 1 F 21 m 2 F 12 v 2 Fgure 9.1 Two partcles nteract wth each other. Accordng to Newton s thrd law, we must have F 12 F 21. d(m 1 v 1 ) dt d(m 2v 2 ) dt 0 d dt (m 1v 1 m 2 v 2 ) 0 (9.1) To fnalze ths dscusson, note that the dervatve of the sum m 1 v 1 m 2 v 2 wth respect to tme s zero. Consequently, ths sum must be constant. We learn from ths dscusson that the quantty m v for a partcle s mportant, n that the sum of these quanttes for an solated system s conserved. We call ths quantty lnear momentum: 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: Defnton of lnear momentum of a partcle p m v (9.2) Lnear momentum s a vector quantty because t equals the product of a scalar quantty m and a vector quantty v. Its drecton s along v, t has dmensons ML/T, and ts SI unt s kg m/s. If a partcle s movng n an arbtrary drecton, p must have three components, and Equaton 9.2 s equvalent to the component equatons p x mv x p y mv y p z mv z As you can see from ts defnton, the concept of momentum 1 provdes a quanttatve dstncton between heavy and lght partcles movng at the same velocty. For example, the momentum of a bowlng ball movng at 10 m/s s much greater than that of a tenns ball movng at the same speed. Newton called the product m v quantty of moton; ths s perhaps a more graphc descrpton than our present-day word momentum, whch comes from the Latn word for movement. Usng Newton s second law of moton, we can relate the lnear momentum of a partcle to the resultant force actng on the partcle. We start wth Newton s second law and substtute the defnton of acceleraton: F ma m d v dt 1 In ths chapter, the terms momentum and lnear momentum have the same meanng. Later, n Chapter 11, we shall use the term angular momentum when dealng wth rotatonal moton.

4 254 CHAPTER 9 Lnear Momentum and Collsons In Newton s second law, the mass m s assumed to be constant. Thus, we can brng m nsde the dervatve notaton to gve us Newton s second law for a partcle F d(mv) dt d p dt Ths shows that the tme rate of change of the lnear momentum of a partcle s equal to the net force actng on the partcle. Ths alternatve form of Newton s second law s the form n whch Newton presented the law and s actually more general than the form we ntroduced n Chapter 5. In addton to stuatons n whch the velocty vector vares wth tme, we can use Equaton 9.3 to study phenomena n whch the mass changes. For example, the mass of a rocket changes as fuel s burned and ejected from the rocket. We cannot use F ma to analyze rocket propulson; we must use Equaton 9.3, as we wll show n Secton 9.7. The real value of Equaton 9.3 as a tool for analyss, however, arses f we apply t to a system of two or more partcles. As we have seen, ths leads to a law of conservaton of momentum for an solated system. Just as the law of conservaton of energy s useful n solvng complex moton problems, the law of conservaton of momentum can greatly smplfy the analyss of other types of complcated moton. (9.3) Quck Quz 9.1 Two objects have equal knetc energes. How do the magntudes of ther momenta compare? (a) p 1 p 2 (b) p 1 p 2 (c) p 1 p 2 (d) not enough nformaton to tell. Quck Quz 9.2 Your physcal educaton teacher throws a baseball to you at a certan speed, and you catch t. The teacher s next gong to throw you a medcne ball whose mass s ten tmes the mass of the baseball. You are gven the followng choces: You can have the medcne ball thrown wth (a) the same speed as the baseball (b) the same momentum (c) the same knetc energy. Rank these choces from easest to hardest to catch. Usng the defnton of momentum, Equaton 9.1 can be wrtten d dt (p 1 p 2 ) 0 Because the tme dervatve of the total momentum p tot p 1 p 2 s zero, we conclude that the total momentum of the system must reman constant: p tot p 1 p 2 constant (9.4) PITFALL PREVENTION 9.1 Momentum of a System s Conserved Remember that the momentum of an solated system s conserved. The momentum of one partcle wthn an solated system s not necessarly conserved, because other partcles n the system may be nteractng wth t. Always apply conservaton of momentum to an solated system. or, equvalently, p 1 p 2 p 1f p 2f where p l and p 2 are the ntal values and p 1f and p 2f the fnal values of the momenta for the two partcles for the tme nterval durng whch the partcles nteract. Equaton 9.5 n component form demonstrates that the total momenta n the x, y, and z drectons are all ndependently conserved: p x p fx p y p fy p z p fz Ths result, known as the law of conservaton of lnear momentum, can be extended to any number of partcles n an solated system. It s consdered one of the most mportant laws of mechancs. We can state t as follows: (9.5) (9.6)

5 SECTION 9.1 Lnear Momentum and Its Conservaton 255 Whenever two or more partcles n an solated system nteract, the total momentum of the system remans constant. Conservaton of momentum Ths law tells us that the total momentum of an solated system at all tmes equals ts ntal momentum. Notce that we have made no statement concernng the nature of the forces actng on the partcles of the system. The only requrement s that the forces must be nternal to the system. Quck Quz 9.3 A ball s released and falls toward the ground wth no ar resstance. The solated system for whch momentum s conserved s (a) the ball (b) the Earth (c) the ball and the Earth (d) mpossble to determne. Quck Quz 9.4 A car and a large truck travelng at the same speed make a head-on collson and stck together. Whch vehcle experences the larger change n the magntude of momentum? (a) the car (b) the truck (c) The change n the magntude of momentum s the same for both. (d) mpossble to determne. Example 9.1 The Archer Interactve Let us consder the stuaton proposed at the begnnng of ths secton. A 60-kg archer stands at rest on frctonless ce and fres a 0.50-kg arrow horzontally at 50 m/s (Fg. 9.2). Wth what velocty does the archer move across the ce after frng the arrow? Soluton We cannot solve ths problem usng Newton s second law, F ma, because we have no nformaton about the force on the arrow or ts acceleraton. We cannot solve ths problem usng an energy approach because we do not know how much work s done n pullng the bow back or how much potental energy s stored n the bow. However, we can solve ths problem very easly wth conservaton of momentum. Let us take the system to consst of the archer (ncludng the bow) and the arrow. The system s not solated because the gravtatonal force and the normal force act on the system. However, these forces are vertcal and perpendcular to the moton of the system. Therefore, there are no external forces n the horzontal drecton, and we can consder the system to be solated n terms of momentum components n ths drecton. The total horzontal momentum of the system before the arrow s fred s zero (m 1 v 1 m 2 v 2 0), where the archer s partcle 1 and the arrow s partcle 2. Therefore, the total horzontal momentum after the arrow s fred must be zero; that s, the drecton of moton of the arrow, n accordance wth Newton s thrd law. Because the archer s much more massve than the arrow, hs acceleraton and consequent velocty are much smaller than the acceleraton and velocty of the arrow. What If? What f the arrow were shot n a drecton that makes an angle wth the horzontal? How wll ths change the recol velocty of the archer? Answer The recol velocty should decrease n magntude because only a component of the velocty s n the x drecton. m 1 v 1f m 2 v 2f 0 We choose the drecton of frng of the arrow as the postve x drecton. Wth m 1 60 kg, m kg, and v 2f 50î m/s, solvng for v 1f, we fnd the recol velocty of the archer to be v 1f m 2 m 1 v 2f 0.50 kg 60 kg (50î m/s) 0.42î m/s The negatve sgn for v 1f ndcates that the archer s movng to the left after the arrow s fred, n the drecton opposte Fgure 9.2 (Example 9.1) An archer fres an arrow horzontally to the rght. Because he s standng on frctonless ce, he wll begn to slde to the left across the ce.

6 256 CHAPTER 9 Lnear Momentum and Collsons If the arrow were shot straght up, for example, there would be no recol at all the archer would just be pressed down nto the ce because of the frng of the arrow. Only the x component of the momentum of the arrow should be used n a conservaton of momentum statement, because momentum s only conserved n the x drecton. In the y drecton, the normal force from the ce and the gravtatonal force are external nfluences on the system. Conservaton of momentum n the x drecton gves us m 1 v 1f m 2 v 2f cos 0 leadng to v 1f m 2 m 1 v 2f cos For 0, cos 1 and ths reduces to the value when the arrow s fred horzontally. For nonzero values of, the cosne functon s less than 1 and the recol velocty s less than the value calculated for 0. If 90, cos 0, and there s no recol velocty v 1f, as we argued conceptually. At the Interactve Worked Example lnk at you can change the mass of the archer and the mass and speed of the arrow. Example 9.2 Breakup of a Kaon at Rest One type of nuclear partcle, called the neutral kaon (K 0 ), breaks up nto a par of other partcles called pons ( and ) that are oppostely charged but equal n mass, as llustrated n Fgure 9.3. Assumng the kaon s ntally at rest, prove that the two pons must have momenta that are equal n magntude and opposte n drecton. p p An mportant pont to learn from ths problem s that even though t deals wth objects that are very dfferent from those n the precedng example, the physcs s dentcal: lnear momentum s conserved n an solated system. Soluton The breakup of the kaon can be wrtten K 0 9: Κ 0 Before decay (at rest) If we let p be the fnal momentum of the postve pon and p the fnal momentum of the negatve pon, the fnal momentum of the system consstng of the two pons can be wrtten p f p p Because the kaon s at rest before the breakup, we know that p 0. Because the momentum of the solated system (the kaon before the breakup, the two pons afterward) s conserved, p p f 0, so that p p 0, or p π π + After decay Fgure 9.3 (Example 9.2) A kaon at rest breaks up spontaneously nto a par of oppostely charged pons. The pons move apart wth momenta that are equal n magntude but opposte n drecton. p Impulse and Momentum Accordng to Equaton 9.3, the momentum of a partcle changes f a net force acts on the partcle. Knowng the change n momentum caused by a force s useful n solvng some types of problems. To buld a better understandng of ths mportant concept, let us assume that a sngle force F acts on a partcle and that ths force may vary wth tme. Accordng to Newton s second law, F d p/dt, or dp Fdt We can ntegrate 2 ths expresson to fnd the change n the momentum of a partcle when the force acts over some tme nterval. If the momentum of the partcle changes from p at tme t to p f at tme t f, ntegratng Equaton 9.7 gves (9.7) 2 Note that here we are ntegratng force wth respect to tme. Compare ths wth our efforts n Chapter 7, where we ntegrated force wth respect to poston to fnd the work done by the force.

7 SECTION 9.2 Impulse and Momentum 257 p p f p t f (9.8) To evaluate the ntegral, we need to know how the force vares wth tme. The quantty on the rght sde of ths equaton s called the mpulse of the force F actng on a partcle over the tme nterval t t f t. Impulse s a vector defned by t Fdt I t f F dt t (9.9) Impulse of a force Equaton 9.8 s an mportant statement known as the mpulse momentum theorem: 3 The mpulse of the force F actng on a partcle equals the change n the momentum of the partcle. Impulse momentum theorem Ths statement s equvalent to Newton s second law. From ths defnton, we see that mpulse s a vector quantty havng a magntude equal to the area under the force tme curve, as descrbed n Fgure 9.4a. In ths fgure, t s assumed that the force vares n tme n the general manner shown and s nonzero n the tme nterval t t f t. The drecton of the mpulse vector s the same as the drecton of the change n momentum. Impulse has the dmensons of momentum that s, ML/T. Note that mpulse s not a property of a partcle; rather, t s a measure of the degree to whch an external force changes the momentum of the partcle. Therefore, when we say that an mpulse s gven to a partcle, we mean that momentum s transferred from an external agent to that partcle. Because the force mpartng an mpulse can generally vary n tme, t s convenent to defne a tme-averaged force F 1 t f F dt t (9.10) where t t f t. (Ths s an applcaton of the mean value theorem of calculus.) Therefore, we can express Equaton 9.9 as t I F t (9.11) F F t (a) t f t F Courtesy of Saab Arbags n automobles have saved countless lves n accdents. The arbag ncreases the tme nterval durng whch the passenger s brought to rest, thereby decreasng the force on (and resultant njury to) the passenger. 3 Although we assumed that only a sngle force acts on the partcle, the mpulse momentum theorem s vald when several forces act; n ths case, we replace F n Equaton 9.8 wth F. t Area = F t (b) Fgure 9.4 (a) A force actng on a partcle may vary n tme. The mpulse mparted to the partcle by the force s the area under the force-versus-tme curve. (b) In the tme nterval t, the tme-averaged force (horzontal dashed lne) gves the same mpulse to a partcle as does the tme-varyng force descrbed n part (a). t f t

8 258 CHAPTER 9 Lnear Momentum and Collsons Ths tme-averaged force, shown n Fgure 9.4b, can be nterpreted as the constant force that would gve to the partcle n the tme nterval t the same mpulse that the tme-varyng force gves over ths same nterval. In prncple, f F s known as a functon of tme, the mpulse can be calculated from Equaton 9.9. The calculaton becomes especally smple f the force actng on the partcle s constant. In ths case, F F and Equaton 9.11 becomes I F t (9.12) In many physcal stuatons, we shall use what s called the mpulse approxmaton, n whch we assume that one of the forces exerted on a partcle acts for a short tme but s much greater than any other force present. Ths approxmaton s especally useful n treatng collsons n whch the duraton of the collson s very short. When ths approxmaton s made, we refer to the force as an mpulsve force. For example, when a baseball s struck wth a bat, the tme of the collson s about 0.01 s and the average force that the bat exerts on the ball n ths tme s typcally several thousand newtons. Because ths contact force s much greater than the magntude of the gravtatonal force, the mpulse approxmaton justfes our gnorng the gravtatonal forces exerted on the ball and bat. When we use ths approxmaton, t s mportant to remember that p and p f represent the momenta mmedately before and after the collson, respectvely. Therefore, n any stuaton n whch t s proper to use the mpulse approxmaton, the partcle moves very lttle durng the collson. Quck Quz 9.5 Two objects are at rest on a frctonless surface. Object 1 has a greater mass than object 2. When a constant force s appled to object 1, t accelerates through a dstance d. The force s removed from object 1 and s appled to object 2. At the moment when object 2 has accelerated through the same dstance d, whch statements are true? (a) p 1 p 2 (b) p 1 p 2 (c) p 1 p 2 (d) K 1 K 2 (e) K 1 K 2 (f ) K 1 K 2. Quck Quz 9.6 Two objects are at rest on a frctonless surface. Object 1 has a greater mass than object 2. When a force s appled to object 1, t accelerates for a tme nterval t. The force s removed from object 1 and s appled to object 2. After object 2 has accelerated for the same tme nterval t, whch statements are true? (a) p 1 p 2 (b) p 1 p 2 (c) p 1 p 2 (d) K 1 K 2 (e) K 1 K 2 (f ) K 1 K 2. Quck Quz 9.7 Rank an automoble dashboard, seatbelt, and arbag n terms of (a) the mpulse and (b) the average force they delver to a front-seat passenger durng a collson, from greatest to least. Example 9.3 Teeng Off A golf ball of mass 50 g s struck wth a club (Fg. 9.5). The force exerted by the club on the ball vares from zero, at the nstant before contact, up to some maxmum value and then back to zero when the ball leaves the club. Thus, the force tme curve s qualtatvely descrbed by Fgure 9.4. Assumng that the ball travels 200 m, estmate the magntude of the mpulse caused by the collson. Soluton Let us use to denote the poston of the ball when the club frst contacts t, to denote the poston of the ball when the club loses contact wth the ball, and to denote the poston of the ball upon landng. Neglectng ar resstance, we can use Equaton 4.14 for the range of a projectle: R x C v B 2 g sn 2B Let us assume that the launch angle B s 45, the angle that provdes the maxmum range for any gven launch velocty. Ths assumpton gves sn 2 B 1, and the launch velocty of the ball s v B Rg (200 m)(9.80 m/s 2 ) 44 m/s

9 SECTION 9.2 Impulse and Momentum 259 I p mv B mv A ( kg)(44 m/s) kgm/s Harold and Esther Edgerton Foundaton 2002, courtesy of Palm Press, Inc. Fgure 9.5 (Example 9.3) A golf ball beng struck by a club. Note the deformaton of the ball due to the large force from the club. Consderng ntal and fnal values of the ball s velocty for the tme nterval for the collson, v v A 0 and v f v B. Hence, the magntude of the mpulse mparted to the ball s What If? What f you were asked to fnd the average force on the ball durng the collson wth the club? Can you determne ths value? Answer Wth the nformaton gven n the problem, we cannot fnd the average force. Consderng Equaton 9.11, we would need to know the tme nterval of the collson n order to calculate the average force. If we assume that the tme nterval s 0.01 s as t was for the baseball n the dscusson after Equaton 9.12, we can estmate the magntude of the average force: F I t 2.2 kgm/s 0.01 s N where we have kept only one sgnfcant fgure due to our rough estmate of the tme nterval. Example 9.4 How Good Are the Bumpers? In a partcular crash test, a car of mass kg colldes wth a wall, as shown n Fgure 9.6. The ntal and fnal veloctes of the car are v 15.0î m/s and v f 2.60î m/s, respectvely. If the collson lasts for s, fnd the mpulse caused by the collson and the average force exerted on the car. Soluton Let us assume that the force exerted by the wall on the car s large compared wth other forces on the car so that we can apply the mpulse approxmaton. Furthermore, we note that the gravtatonal force and the normal force exerted by the road on the car are perpendcular to the moton and therefore do not affect the horzontal momentum. The ntal and fnal momenta of the car are p mv (1 500 kg)(15.0î m/s) î kgm/s p f m v f (1 500 kg)(2.60î m/s) î kgm/s Hence, the mpulse s equal to (b) Before 15.0 m/s After m/s (a) Tm Wrght/CORBIS Fgure 9.6 (Example 9.4) (a) Ths car s momentum changes as a result of ts collson wth the wall. (b) In a crash test, much of the car s ntal knetc energy s transformed nto energy assocated wth the damage to the car.

10 260 CHAPTER 9 Lnear Momentum and Collsons The average force exerted by the wall on the car s F p t I p p f p î kg m/s I ( î kgm/s) î kg m/s î kgm/s s In ths problem, note that the sgns of the veloctes ndcate the reversal of drectons. What would the mathematcs be descrbng f both the ntal and fnal veloctes had the same sgn? What If? What f the car dd not rebound from the wall? Suppose the fnal velocty of the car s zero and the tme nterval of the collson remans at s. Would ths represent a larger or a smaller force by the wall on the car? î N Answer In the orgnal stuaton n whch the car rebounds, the force by the wall on the car does two thngs n the tme nterval t (1) stops the car and (2) causes t to move away from the wall at 2.60 m/s after the collson. If the car does not rebound, the force s only dong the frst of these, stoppng the car. Ths wll requre a smaller force. Mathematcally, n the case of the car that does not rebound, the mpulse s I p p f p 0 ( î kgm/s) The average force exerted by the wall on the car s F p t î kgm/s î kgm/s s î N whch s ndeed smaller than the prevously calculated value, as we argued conceptually. F 21 F Collsons n One Dmenson p + m 1 m 2 (a) 4 (b) ++ He Fgure 9.7 (a) The collson between two objects as the result of drect contact. (b) The collson between two charged partcles. Elastc collson Inelastc collson In ths secton we use the law of conservaton of lnear momentum to descrbe what happens when two partcles collde. We use the term collson to represent an event durng whch two partcles come close to each other and nteract by means of forces. The tme nterval durng whch the veloctes of the partcles change from ntal to fnal values s assumed to be short. The nteracton forces are assumed to be much greater than any external forces present, so we can use the mpulse approxmaton. A collson may nvolve physcal contact between two macroscopc objects, as descrbed n Fgure 9.7a, but the noton of what we mean by collson must be generalzed because physcal contact on a submcroscopc scale s ll-defned and hence meanngless. To understand ths, consder a collson on an atomc scale (Fg. 9.7b), such as the collson of a proton wth an alpha partcle (the nucleus of a helum atom). Because the partcles are both postvely charged, they repel each other due to the strong electrostatc force between them at close separatons and never come nto physcal contact. When two partcles of masses m 1 and m 2 collde as shown n Fgure 9.7, the mpulsve forces may vary n tme n complcated ways, such as that shown n Fgure 9.4. Regardless of the complexty of the tme behavor of the force of nteracton, however, ths force s nternal to the system of two partcles. Thus, the two partcles form an solated system, and the momentum of the system must be conserved. Therefore, the total momentum of an solated system just before a collson equals the total momentum of the system just after the collson. In contrast, the total knetc energy of the system of partcles may or may not be conserved, dependng on the type of collson. In fact, whether or not knetc energy s conserved s used to classfy collsons as ether elastc or nelastc. An elastc collson between two objects s one n whch the total knetc energy (as well as total momentum) of the system s the same before and after the collson. Collsons between certan objects n the macroscopc world, such as bllard balls, are only approxmately elastc because some deformaton and loss of knetc energy take place. For example, you can hear a bllard ball collson, so you know that some of the energy s beng transferred away from the system by sound. An elastc collson must be perfectly slent! Truly elastc collsons occur between atomc and subatomc partcles. An nelastc collson s one n whch the total knetc energy of the system s not the same before and after the collson (even though the momentum of the system s conserved). Inelastc collsons are of two types. When the colldng objects stck together after the collson, as happens when a meteorte colldes wth the Earth,

11 SECTION 9.3 Collsons n One Dmenson 261 the collson s called perfectly nelastc. When the colldng objects do not stck together, but some knetc energy s lost, as n the case of a rubber ball colldng wth a hard surface, the collson s called nelastc (wth no modfyng adverb). When the rubber ball colldes wth the hard surface, some of the knetc energy of the ball s lost when the ball s deformed whle t s n contact wth the surface. In most collsons, the knetc energy of the system s not conserved because some of the energy s converted to nternal energy and some of t s transferred away by means of sound. Elastc and perfectly nelastc collsons are lmtng cases; most collsons fall somewhere between them. In the remander of ths secton, we treat collsons n one dmenson and consder the two extreme cases perfectly nelastc and elastc collsons. The mportant dstncton between these two types of collsons s that momentum of the system s conserved n all collsons, but knetc energy of the system s conserved only n elastc collsons. Perfectly Inelastc Collsons Consder two partcles of masses m 1 and m 2 movng wth ntal veloctes v 1 and v 2 along the same straght lne, as shown n Fgure 9.8. The two partcles collde head-on, stck together, and then move wth some common velocty v f after the collson. Because the momentum of an solated system s conserved n any collson, we can say that the total momentum before the collson equals the total momentum of the composte system after the collson: Solvng for the fnal velocty gves Elastc Collsons (9.13) (9.14) Consder two partcles of masses m 1 and m 2 movng wth ntal veloctes v 1 and v 2 along the same straght lne, as shown n Fgure 9.9. The two partcles collde head-on and then leave the collson ste wth dfferent veloctes, v 1f and v 2f. If the collson s elastc, both the momentum and knetc energy of the system are conserved. Therefore, consderng veloctes along the horzontal drecton n Fgure 9.9, we have (9.15) (9.16) Because all veloctes n Fgure 9.9 are ether to the left or the rght, they can be represented by the correspondng speeds along wth algebrac sgns ndcatng drecton. We shall ndcate v as postve f a partcle moves to the rght and negatve f t moves to the left. In a typcal problem nvolvng elastc collsons, there are two unknown quanttes, and Equatons 9.15 and 9.16 can be solved smultaneously to fnd these. An alternatve approach, however one that nvolves a lttle mathematcal manpulaton of Equaton 9.16 often smplfes ths process. To see how, let us cancel the factor 9.16 and rewrte t as and then factor both sdes: m 1 v 1 m 2 v 2 (m 1 m 2 )v f v f m 1v 1 m 2 v 2 m 1 m 2 m 1 v 1 m 2 v 2 m 1 v 1f m 2 v 2f 1 2 m 1v m 2v m 1v 1f m 2v 2f 2 m 1 (v 1 2 v 1f 2 ) m 2 (v 2f 2 v 2 2 ) m 1 (v 1 v 1f )(v 1 v 1f ) m 2 (v 2f v 2 )(v 2f v 2 ) n Equaton Next, let us separate the terms contanng m 1 and m 2 n Equaton 9.15 to obtan m 1 (v 1 v 1f ) m 2 (v 2f v 2 ) 1 2 (9.17) (9.18) PITFALL PREVENTION 9.2 Inelastc Collsons Generally, nelastc collsons are hard to analyze unless addtonal nformaton s provded. Ths appears n the mathematcal representaton as havng more unknowns than equatons. Before collson m 1 m 2 v 1 v 2 Before collson v 1 v 2 m 1 m 2 v 1f (a) After collson m 1 + m 2 (b) Actve Fgure 9.8 Schematc representaton of a perfectly nelastc head-on collson between two partcles: (a) before collson and (b) after collson. At the Actve Fgures lnk at you can adjust the masses and veloctes of the colldng objects to see the effect on the fnal velocty. (a) After collson v 2f (b) Actve Fgure 9.9 Schematc representaton of an elastc head-on collson between two partcles: (a) before collson and (b) after collson. At the Actve Fgures lnk at you can adjust the masses and veloctes of the colldng objects to see the effect on the fnal veloctes. v f

12 262 CHAPTER 9 Lnear Momentum and Collsons PITFALL PREVENTION 9.3 Not a General Equaton We have spent some effort on dervng Equaton 9.19, but remember that t can only be used n a very specfc stuaton a one-dmensonal, elastc collson between two objects. The general concept s conservaton of momentum (and conservaton of knetc energy f the collson s elastc) for an solated system. To obtan our fnal result, we dvde Equaton 9.17 by Equaton 9.18 and obtan v 1 v 1f v 2f v 2 v 1 v 2 (v 1f v 2f ) (9.19) Ths equaton, n combnaton wth Equaton 9.15, can be used to solve problems dealng wth elastc collsons. Accordng to Equaton 9.19, the relatve velocty of the two partcles before the collson, v 1 v 2, equals the negatve of ther relatve velocty after the collson, (v 1f v 2f ). Suppose that the masses and ntal veloctes of both partcles are known. Equatons 9.15 and 9.19 can be solved for the fnal veloctes n terms of the ntal veloctes because there are two equatons and two unknowns: v 1f m 1 m 2 m 1 m 2 v 1 2m 2 m 1 m 2 v 2 (9.20) v 2f 2m 1 m 1 m 2 v 1 m 2 m 1 m 1 m 2 v 2 (9.21) It s mportant to use the approprate sgns for v 1 and v 2 n Equatons 9.20 and For example, f partcle 2 s movng to the left ntally, then v 2 s negatve. Let us consder some specal cases. If m 1 m 2, then Equatons 9.20 and 9.21 show us that v 1f v 2 and v 2f v 1. That s, the partcles exchange veloctes f they have equal masses. Ths s approxmately what one observes n head-on bllard ball collsons the cue ball stops, and the struck ball moves away from the collson wth the same velocty that the cue ball had. If partcle 2 s ntally at rest, then v 2 0, and Equatons 9.20 and 9.21 become Elastc collson: partcle 2 ntally at rest v 1f m 1 m 2 m 1 m 2 v 1 v 2f 2m 1 m 1 m 2 v 1 (9.22) (9.23) PITFALL PREVENTION 9.4 Momentum and Knetc Energy n Collsons Momentum of an solated system s conserved n all collsons. Knetc energy of an solated system s conserved only n elastc collsons. Why? Because there are several types of energy nto whch knetc energy can transform, or be transferred out of the system (so that the system may not be solated n terms of energy durng the collson). However, there s only one type of momentum. If m 1 s much greater than m 2 and v 2 0, we see from Equatons 9.22 and 9.23 that v 1f v 1 and v 2f 2v 1. That s, when a very heavy partcle colldes head-on wth a very lght one that s ntally at rest, the heavy partcle contnues ts moton unaltered after the collson and the lght partcle rebounds wth a speed equal to about twce the ntal speed of the heavy partcle. An example of such a collson would be that of a movng heavy atom, such as uranum, strkng a lght atom, such as hydrogen. If m 2 s much greater than m 1 and partcle 2 s ntally at rest, then v 1f v 1 and v 2f 0. That s, when a very lght partcle colldes head-on wth a very heavy partcle that s ntally at rest, the lght partcle has ts velocty reversed and the heavy one remans approxmately at rest. Quck Quz 9.8 In a perfectly nelastc one-dmensonal collson between two objects, what condton alone s necessary so that all of the orgnal knetc energy of the system s gone after the collson? (a) The objects must have momenta wth the same magntude but opposte drectons. (b) The objects must have the same mass. (c) The objects must have the same velocty. (d) The objects must have the same speed, wth velocty vectors n opposte drectons. Quck Quz 9.9 A table-tenns ball s thrown at a statonary bowlng ball. The table-tenns ball makes a one-dmensonal elastc collson and bounces back along the same lne. After the collson, compared to the bowlng ball, the table-tenns ball has (a) a larger magntude of momentum and more knetc energy (b) a smaller

13 SECTION 9.3 Collsons n One Dmenson 263 magntude of momentum and more knetc energy (c) a larger magntude of momentum and less knetc energy (d) a smaller magntude of momentum and less knetc energy (e) the same magntude of momentum and the same knetc energy. Example 9.5 The Executve Stress Relever Interactve An ngenous devce that llustrates conservaton of momentum and knetc energy s shown n Fgure It conssts of fve dentcal hard balls supported by strngs of equal lengths. When ball 1 s pulled out and released, after the almost-elastc collson between t and ball 2, ball 5 moves out, as shown n Fgure 9.10b. If balls 1 and 2 are pulled out and released, balls 4 and 5 swng out, and so forth. Is t ever possble that when ball 1 s released, balls 4 and 5 wll swng out on the opposte sde and travel wth half the speed of ball 1, as n Fgure 9.10c? Soluton No, such movement can never occur f we assume the collsons are elastc. The momentum of the system before the collson s mv, where m s the mass of ball 1 and v s ts speed just before the collson. After the collson, we would have two balls, each of mass m movng wth speed v/2. The total momentum of the system after the collson would be m(v/2) m(v/2) mv. Thus, momentum of the system s conserved. However, the knetc energy just before the collson s K 1 and that after the collson s K f 1 2 m(v/2)2 1 2 mv 2 2 m(v/2)2 1 4 mv 2. Thus, knetc energy of the system s not conserved. The only way to have both momentum and knetc energy conserved s for one ball to move out when one ball s released, two balls to move out when two are released, and so on. What If? Consder what would happen f balls 4 and 5 are glued together so that they must move together. Now what happens when ball 1 s pulled out and released? Answer We are now forcng balls 4 and 5 to come out together. We have argued that we cannot conserve both momentum and energy n ths case. However, we assumed that ball 1 stopped after strkng ball 2. What f we do not make ths assumpton? Consder the conservaton equatons wth the assumpton that ball 1 moves after the collson. For conservaton of momentum, p p f mv 1 mv 1f 2mv 4,5f where v 4,5f refers to the fnal speed of the ball 4 ball 5 combnaton. Conservaton of knetc energy gves us K K f 1 2 mv mv 2 1f 1 2 (2m)v2 4,5f Combnng these equatons, we fnd v 4,5f 2 3 v 1 v 1f 1 3 v 1 Thus, balls 4 and 5 come out together and ball 1 bounces back from the collson wth one thrd of ts orgnal speed. 1 5 v Ths can happen. (b) v (a) v Can ths happen? (c) Fgure 9.10 (Example 9.5) An executve stress relever. v/2 At the Interactve Worked Example lnk at you can glue balls 4 and 5 together to see the stuaton dscussed above.

14 264 CHAPTER 9 Lnear Momentum and Collsons Example 9.6 Carry Collson Insurance! An kg car stopped at a traffc lght s struck from the rear by a 900-kg car, and the two become entangled, movng along the same path as that of the orgnally movng car. If the smaller car were movng at 20.0 m/s before the collson, what s the velocty of the entangled cars after the collson? Soluton The phrase become entangled tells us that ths s a perfectly nelastc collson. We can guess that the fnal speed s less than 20.0 m/s, the ntal speed of the smaller car. The total momentum of the system (the two cars) before the collson must equal the total momentum mmedately after the collson because momentum of an solated system s conserved n any type of collson. The magntude of the total momentum of the system before the collson s equal to that of the smaller car because the larger car s ntally at rest: p m 1 v (900 kg)(20.0 m/s) kgm/s After the collson, the magntude of the momentum of the entangled cars s p f (m 1 m 2 )v f (2 700 kg)v f Equatng the ntal and fnal momenta of the system and solvng for v f, the fnal velocty of the entangled cars, we have v f Because the fnal velocty s postve, the drecton of the fnal velocty s the same as the velocty of the ntally movng car. What If? Suppose we reverse the masses of the cars a statonary 900-kg car s struck by a movng kg car. Is the fnal speed the same as before? Answer Intutvely, we can guess that the fnal speed wll be hgher, based on common experences n drvng. Mathematcally, ths should be the case because the system has a larger momentum f the ntally movng car s the more massve one. Solvng for the new fnal velocty, we fnd v f p kgm/s m 1 m kg p m 1 m 2 whch s ndeed hgher than the prevous fnal velocty. (1 800 kg)(20.0 m/s) kg 6.67 m/s 13.3 m/s Example 9.7 The Ballstc Pendulum The ballstc pendulum (Fg. 9.11) s an apparatus used to measure the speed of a fast-movng projectle, such as a bullet. A bullet of mass m 1 s fred nto a large block of wood of mass m 2 suspended from some lght wres. The bullet embeds n the block, and the entre system swngs through a heght h. How can we determne the speed of the bullet from a measurement of h? Soluton Fgure 9.11a helps to conceptualze the stuaton. Let confguraton be the bullet and block before the collson, and confguraton be the bullet and block mmedately after colldng. The bullet and the block form an solated system, so we can categorze the collson between them as a conservaton of momentum problem. The collson s perfectly nelastc. To analyze the collson, we note that Equaton 9.14 gves the speed of the system rght after the collson when we assume the mpulse approxmaton. Notng that v 2A 0, Equaton 9.14 becomes m 1 + m 2 v 1A v B m 1 m 2 (a) h (1) v B m 1v 1A m 1 m 2 For the process durng whch the bullet block combnaton swngs upward to heght h (endng at confguraton ), we focus on a dfferent system the bullet, the block, and the Earth. Ths s an solated system for energy, so we categorze ths part of the moton as a conservaton of mechancal energy problem: K B U B K C U C We begn to analyze the problem by fndng the total knetc energy of the system rght after the collson: (2) K B 1 2 (m 1 m 2 )v B 2 Courtesy of Central Scentfc Company (b) Fgure 9.11 (Example 9.7) (a) Dagram of a ballstc pendulum. Note that v 1A s the velocty of the bullet just before the collson and v B s the velocty of the bullet-block system just after the perfectly nelastc collson. (b) Multflash photograph of a ballstc pendulum used n the laboratory.

15 SECTION 9.3 Collsons n One Dmenson 265 Substtutng the value of v B from Equaton (1) nto Equaton (2) gves K B m 1 2 v 1A 2 2(m 1 m 2 ) Ths knetc energy mmedately after the collson s less than the ntal knetc energy of the bullet, as expected n an nelastc collson. We defne the gravtatonal potental energy of the system for confguraton to be zero. Thus, U B 0 whle U C (m 1 m 2 )gh. Conservaton of energy now leads to Solvng for v 1A, we obtan v 1A m 1 m 2 m 1 2gh To fnalze ths problem, note that we had to solve ths problem n two steps. Each step nvolved a dfferent system and a dfferent conservaton prncple. Because the collson was assumed to be perfectly nelastc, some mechancal energy was converted to nternal energy. It would have been ncorrect to equate the ntal knetc energy of the ncomng bullet to the fnal gravtatonal potental energy of the bullet block Earth combnaton. m 1 2 v 1A 2 2(m 1 m 2 ) 0 0 (m 1 m 2 )gh Example 9.8 A Two-Body Collson wth a Sprng Interactve A block of mass m kg ntally movng to the rght wth a speed of 4.00 m/s on a frctonless horzontal track colldes wth a sprng attached to a second block of mass m kg ntally movng to the left wth a speed of 2.50 m/s, as shown n Fgure 9.12a. The sprng constant s 600 N/m. (A) Fnd the veloctes of the two blocks after the collson. Soluton Because the sprng force s conservatve, no knetc energy s converted to nternal energy durng the compresson of the sprng. Ignorng any sound made when the block hts the sprng, we can model the collson as beng elastc. Equaton 9.15 gves us Equaton 9.19 gves us m 1 v 1 m 2 v 2 m 1 v 1f m 2 v 2f (1.60 kg)(4.00 m/s) (2.10 kg)(2.50 m/s) (1.60 kg)v 1f (2.10 kg)v 2f (1) 1.15 kgm/s (1.60 kg)v 1f (2.10 kg)v 2f v 1 v 2 (v 1f v 2f ) 4.00 m/s (2.50 m/s) v 1f v 2f (2) 6.50 m/s v 1f v 2f Multplyng Equaton (2) by 1.60 kg gves us (3) 10.4 kgm/s (1.60 kg)v 1f (1.60 kg)v 2f Addng Equatons (1) and (3) allows us to fnd v 2f : kgm/s (3.70 kg)v 2f v 2f kgm/s 3.70 kg Now, Equaton (2) allows us to fnd v 1f : 6.50 m/s v 1f 3.12 m/s v 1f 3.38 m/s (B) Durng the collson, at the nstant block 1 s movng to the rght wth a velocty of 3.00 m/s, as n Fgure 9.12b, determne the velocty of block 2. Soluton Because the momentum of the system of two blocks s conserved throughout the collson for the system of two blocks, we have, for any nstant durng the collson, m 1 v 1 m 2 v 2 m 1 v 1f m 2 v 2f 3.12 m/s We choose the fnal nstant to be that at whch block 1 s movng wth a velocty of 3.00 m/s: v 1 = (4.00î) m/s v 2 = ( 2.50î) m/s v 1f = (3.00î) m/s v 2f m1 m 2 k k m m 2 1 (a) x Fgure 9.12 (Example 9.8) A movng block approaches a second movng block that s attached to a sprng. (b)

16 266 CHAPTER 9 Lnear Momentum and Collsons (1.60 kg)(4.00 m/s) (2.10 kg)(2.50 m/s) v 2f (1.60 kg)(3.00 m/s) (2.10 kg)v 2f 1.74 m/s The negatve value for v 2f means that block 2 s stll movng to the left at the nstant we are consderng. (C) Determne the dstance the sprng s compressed at that nstant. Soluton To determne the dstance that the sprng s compressed, shown as x n Fgure 9.12b, we can use the prncple of conservaton of mechancal energy for the system of the sprng and two blocks because no frcton or other nonconservatve forces are actng wthn the system. We choose the ntal confguraton of the system to be that exstng just before block 1 strkes the sprng and the fnal confguraton to be that when block 1 s movng to the rght at 3.00 m/s. Thus, we have K U K f U f 1 2 m 1v m 2v m 1v 1f m 2v 2f kx 2 Substtutng the gven values and the result to part (B) nto ths expresson gves (D) What s the maxmum compresson of the sprng durng the collson? Soluton The maxmum compresson would occur when the two blocks are movng wth the same velocty. The conservaton of momentum equaton for the system can be wrtten where the ntal nstant s just before the collson and the fnal nstant s when the blocks are movng wth the same velocty v f. Solvng for v f, v f m 1v 1 m 2 v 2 m 1 m m/s m 1 v 1 m 2 v 2 (m 1 m 2 )v f (1.60 kg)(4.00 m/s) (2.10 kg)(2.50 m/s) 1.60 kg 2.10 kg Now, we apply conservaton of mechancal energy between these two nstants as n part (C): K U K f U f 1 2 m 1v m 2v (m 1 m 2 )v f kx 2 Substtutng the gven values nto ths expresson gves x m x m At the Interactve Worked Example lnk at you can change the masses and speeds of the blocks and freeze the moton at the maxmum compresson of the sprng. Example 9.9 Slowng Down Neutrons by Collsons In a nuclear reactor, neutrons are produced when an atom splts n a process called fsson. These neutrons are movng at about 10 7 m/s and must be slowed down to about 10 3 m/s before they take part n another fsson event. They are slowed down by passng them through a sold or lqud materal called a moderator. The slowng-down process nvolves elastc collsons. Show that a neutron can lose most of ts knetc energy f t colldes elastcally wth a moderator contanng lght nucle, such as deuterum (n heavy water, D 2 O) or carbon (n graphte). Soluton Let us assume that the moderator nucleus of mass m m s at rest ntally and that a neutron of mass m n and ntal speed v n colldes wth t head-on. Because these are elastc collsons, both momentum and knetc energy of the neutron nucleus system are conserved. Therefore, Equatons 9.22 and 9.23 can be appled to the head-on collson of a neutron wth a moderator nucleus. We can represent ths process by a drawng such as Fgure 9.9 wth v 2 0. The ntal knetc energy of the neutron s K n 1 2 m nv n 2 After the collson, the neutron has knetc energy, 2 m 2 nv nf and we can substtute nto ths the value for v nf gven by 1 Equaton 9.22: K nf 1 2 m nv nf m n m n m m m n m m 2 v n 2 Therefore, the fracton f n of the ntal knetc energy possessed by the neutron after the collson s (1) f n K nf K n From ths result, we see that the fnal knetc energy of the neutron s small when m m s close to m n and zero when m n m m. We can use Equaton 9.23, whch gves the fnal speed of the partcle that was ntally at rest, to calculate the knetc energy of the moderator nucleus after the collson: K mf 1 2 m mv mf 2 Hence, the fracton f m of the ntal knetc energy transferred to the moderator nucleus s (2) f m Kmf K n m n m m m n m m 2 2m n 2 m m (m n m m ) 2 v n 2 4m n m m (m n m m ) 2

17 SECTION 9.4 Two-Dmensonal Collsons 267 Because the total knetc energy of the system s conserved, Equaton (2) can also be obtaned from Equaton (1) wth the condton that f n f m 1, so that f m 1 f n. Suppose that heavy water s used for the moderator. For collsons of the neutrons wth deuterum nucle n D 2 O (m m 2m n ), f n 1/9 and f m 8/9. That s, 89% of the neutron s knetc energy s transferred to the deuterum nucleus. In practce, the moderator effcency s reduced because head-on collsons are very unlkely. How do the results dffer when graphte ( 12 C, as found n pencl lead) s used as the moderator? 9.4 Two-Dmensonal Collsons In Secton 9.1, we showed that the momentum of a system of two partcles s conserved when the system s solated. For any collson of two partcles, ths result mples that the momentum n each of the drectons x, y, and z s conserved. An mportant subset of collsons takes place n a plane. The game of bllards s a famlar example nvolvng multple collsons of objects movng on a two-dmensonal surface. For such twodmensonal collsons, we obtan two component equatons for conservaton of momentum: m 1 v 1x m 2 v 2x m 1 v 1fx m 2 v 2fx m 1 v 1y m 2 v 2y m 1 v 1fy m 2 v 2fy where we use three subscrpts n these equatons to represent, respectvely, (1) the dentfcaton of the object, (2) ntal and fnal values, and (3) the velocty component. Let us consder a two-dmensonal problem n whch partcle 1 of mass m 1 colldes wth partcle 2 of mass m 2, where partcle 2 s ntally at rest, as n Fgure After the collson, partcle 1 moves at an angle wth respect to the horzontal and partcle 2 moves at an angle wth respect to the horzontal. Ths s called a glancng collson. Applyng the law of conservaton of momentum n component form and notng that the ntal y component of the momentum of the two-partcle system s zero, we obtan m 1 v 1 m 1 v 1f cos m 2 v 2f cos (9.24) PITFALL PREVENTION 9.5 Don t Use Equaton 9.19 Equaton 9.19, relatng the ntal and fnal relatve veloctes of two colldng objects, s only vald for one-dmensonal elastc collsons. Do not use ths equaton when analyzng two-dmensonal collsons. 0 m 1 v 1f sn m 2 v 2f sn (9.25) where the mnus sgn n Equaton 9.25 comes from the fact that after the collson, partcle 2 has a y component of velocty that s downward. We now have two ndependent equatons. As long as no more than two of the seven quanttes n Equatons 9.24 and 9.25 are unknown, we can solve the problem. If the collson s elastc, we can also use Equaton 9.16 (conservaton of knetc energy) wth v 2 0 to gve 1 2 m 1v m 1v1f m 2v 2f 2 (9.26) v 1f sn θ v 1f v 1 v 2f sn φ v 1f cos θ θ φ v 2f cos φ (a) Before the collson (b) After the collson Actve Fgure 9.13 An elastc glancng collson between two partcles. v 2f At the Actve Fgures lnk at you can adjust the speed and poston of the blue partcle and the masses of both partcles to see the effects.

18 268 CHAPTER 9 Lnear Momentum and Collsons Knowng the ntal speed of partcle 1 and both masses, we are left wth four unknowns (v 1f, v 2f,, and ). Because we have only three equatons, one of the four remanng quanttes must be gven f we are to determne the moton after the collson from conservaton prncples alone. If the collson s nelastc, knetc energy s not conserved and Equaton 9.26 does not apply. PROBLEM-SOLVING HINTS Two-Dmensonal Collsons The followng procedure s recommended when dealng wth problems nvolvng two-dmensonal collsons between two objects: Set up a coordnate system and defne your veloctes wth respect to that system. It s usually convenent to have the x axs concde wth one of the ntal veloctes. In your sketch of the coordnate system, draw and label all velocty vectors and nclude all the gven nformaton. Wrte expressons for the x and y components of the momentum of each object before and after the collson. Remember to nclude the approprate sgns for the components of the velocty vectors. Wrte expressons for the total momentum of the system n the x drecton before and after the collson and equate the two. Repeat ths procedure for the total momentum of the system n the y drecton. If the collson s nelastc, knetc energy of the system s not conserved, and addtonal nformaton s probably requred. If the collson s perfectly nelastc, the fnal veloctes of the two objects are equal. Solve the momentum equatons for the unknown quanttes. If the collson s elastc, knetc energy of the system s conserved, and you can equate the total knetc energy before the collson to the total knetc energy after the collson to obtan an addtonal relatonshp between the veloctes. Example 9.10 Collson at an Intersecton A kg car travelng east wth a speed of 25.0 m/s colldes at an ntersecton wth a kg van travelng north at a speed of 20.0 m/s, as shown n Fgure Fnd the drecton and magntude of the velocty of the wreckage after the collson, assumng that the vehcles undergo a perfectly nelastc collson (that s, they stck together). Soluton Let us choose east to be along the postve x drecton and north to be along the postve y drecton. Before the collson, the only object havng momentum n the x drecton s the car. Thus, the magntude of the total ntal momentum of the system (car plus van) n the x drecton s p x (1 500 kg)(25.0 m/s) kgm/s Let us assume that the wreckage moves at an angle and speed v f after the collson. The magntude of the total momentum n the x drecton after the collson s (25.0î) m/s y θ v f x (20.0ĵ) m/s Fgure 9.14 (Example 9.10) An eastbound car colldng wth a northbound van.

19 SECTION 9.4 Two-Dmensonal Collsons 269 p xf (4 000 kg)v f cos Because the total momentum n the x drecton s conserved, we can equate these two equatons to obtan (1) kgm/s (4 000 kg)v f cos Smlarly, the total ntal momentum of the system n the y drecton s that of the van, and the magntude of ths momentum s (2 500 kg)(20.0 m/s) kg m/s. Applyng conservaton of momentum to the y drecton, we have p y p yf (2) kgm/s (4 000 kg)v f sn If we dvde Equaton (2) by Equaton (1), we obtan sn tan cos 53.1 When ths angle s substtuted nto Equaton (2), the value of v f s v f kgm/s (4 000 kg)sn m/s It mght be nstructve for you to draw the momentum vectors of each vehcle before the collson and the two vehcles together after the collson. Example 9.11 Proton Proton Collson A proton colldes elastcally wth another proton that s ntally at rest. The ncomng proton has an ntal speed of m/s and makes a glancng collson wth the second proton, as n Fgure (At close separatons, the protons exert a repulsve electrostatc force on each other.) After the collson, one proton moves off at an angle of 37.0 to the orgnal drecton of moton, and the second deflects at an angle of to the same axs. Fnd the fnal speeds of the two protons and the angle. Soluton The par of protons s an solated system. Both momentum and knetc energy of the system are conserved n ths glancng elastc collson. Because m 1 m 2, 37.0, and we are gven that v m/s, Equatons 9.24, 9.25, and 9.26 become (1) v 1f cos 37 v 2f cos m/s (2) v 1f sn 37.0 v 2f sn 0 Substtutng nto Equaton (3) gves v 1f ( )v 1f v 1f 2 One possblty for the soluton of ths equaton s v 1f 0, whch corresponds to a head-on collson the frst proton stops and the second contnues wth the same speed n the same drecton. Ths s not what we want. The other possblty s From Equaton (3), v 1f 2 ( )v 1f (2v 1f )v 1f 0 2v 1f : v 1f m/s v 2f v 1f ( ) 2 (3) v 1f 2 v 2f 2 ( m/s) m/s m 2 /s 2 We rewrte Equatons (1) and (2) as follows: v 2f cos m/s v 1f cos 37.0 v 2f sn v 1f sn 37.0 Now we square these two equatons and add them: v 2 2f cos 2 v 2 2f sn m 2 /s 2 ( m/s)v 1f cos 37.0 v 2 1f cos v 2 1f sn and from Equaton (2), sn 1 v 1f sn v 2f sn1 ( ) sn It s nterestng to note that 90. Ths result s not accdental. Whenever two objects of equal mass collde elastcally n a glancng collson and one of them s ntally at rest, ther fnal veloctes are perpendcular to each other. The next example llustrates ths pont n more detal. v 2f ( )v 1f v 1f 2 Example 9.12 Bllard Ball Collson In a game of bllards, a player wshes to snk a target ball n the corner pocket, as shown n Fgure If the angle to the corner pocket s 35, at what angle s the cue ball deflected? Assume that frcton and rotatonal moton are unmportant and that the collson s elastc. Also assume that all bllard balls have the same mass m. Soluton Let ball 1 be the cue ball and ball 2 be the target ball. Because the target ball s ntally at rest, conservaton of knetc energy (Eq. 9.16) for the two-ball system gves 1 2 m 1v m 1v 2 1f 1 2 m 2 2v 2f But m 1 m 2 m, so that

20 270 CHAPTER 9 Lnear Momentum and Collsons (1) v 1 2 v 1f 2 v 2f 2 Applyng conservaton of momentum to the two-dmensonal collson gves (2) m 1 v 1 m 1 v 1f m 2 v 2f Note that because m 1 m 2 m, the masses also cancel n Equaton (2). If we square both sdes of Equaton (2) and use the defnton of the dot product of two vectors from Secton 7.3, we obtan v 1 2 (v 1f v 2f ) (v 1f v 2f ) v 1f 2 v 2f 2 2v 1f v 2f y v 2f Because the angle between v 1f and v 2f s 35, v 1f v 2f v 1f v 2f cos( 35 ), and so (3) v 1 2 v 1f 2 v 2f 2 2v 1f v 2f cos( 35) Subtractng Equaton (1) from Equaton (3) gves 0 2v 1f v 2f cos( 35) v 1 Cue ball θ 35 x cos( 35) or 55 Fgure 9.15 (Example 9.12) The cue ball (whte) strkes the number 4 ball (blue) and sends t toward the corner pocket. v 1f Ths result shows that whenever two equal masses undergo a glancng elastc collson and one of them s ntally at rest, they move n perpendcular drectons after the collson. The same physcs descrbes two very dfferent stuatons, protons n Example 9.11 and bllard balls n ths example. 9.5 The Center of Mass In ths secton we descrbe the overall moton of a mechancal system n terms of a specal pont called the center of mass of the system. The mechancal system can be ether a group of partcles, such as a collecton of atoms n a contaner, or an extended object, such as a gymnast leapng through the ar. We shall see that the center of mass of the system moves as f all the mass of the system were concentrated at that pont. Furthermore, f the resultant external force on the system s F ext and the total mass of the system s M, the center of mass moves wth an acceleraton gven by a F ext /M. That s, the system moves as f the resultant external force were appled to a sngle partcle of mass M located at the center of mass. Ths behavor s ndependent of other moton, such as rotaton or vbraton of the system. Ths s the partcle model that was ntroduced n Chapter 2. Consder a mechancal system consstng of a par of partcles that have dfferent masses and are connected by a lght, rgd rod (Fg. 9.16). The poston of the center of mass of a system can be descrbed as beng the average poston of the system s mass. The center of mass of the system s located somewhere on the lne jonng the two partcles and s closer to the partcle havng the larger mass. If a sngle force s appled at a pont on the rod somewhere between the center of mass and the less massve partcle, the system rotates clockwse (see Fg. 9.16a). If the force s appled at a pont on the rod somewhere between the center of mass and the more massve partcle, the system rotates counterclockwse (see Fg. 9.16b). If the force s appled at the center of mass, the system moves n the drecton of F wthout rotatng (see Fg. 9.16c). Thus, the center of mass can be located wth ths procedure. The center of mass of the par of partcles descrbed n Fgure 9.17 s located on the x axs and les somewhere between the partcles. Its x coordnate s gven by x CM m 1x 1 m 2 x 2 m 1 m 2 (9.27)