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)

21 SECTION 9.5 The Center of Mass 271 For example, f x 1 0, x 2 d, and m 2 2m 1, we fnd that x CM 2 That s, the center of mass les closer to the more massve partcle. If the two masses are equal, the cen- 3 d. ter of mass les mdway between the partcles. We can extend ths concept to a system of many partcles wth masses m n three dmensons. The x coordnate of the center of mass of n partcles s defned to be CM x CM m 1x 1 m 2 x 2 m 3 x 3 m n x n m 1 m 2 m 3 m n m x m m x M (9.28) (a) where x s the x coordnate of the th partcle. For convenence, we express the total mass as M m where the sum runs over all n partcles. The y and z coordnates of CM the center of mass are smlarly defned by the equatons (b) y CM m y M and z CM m z M (9.29) The center of mass can also be located by ts poston vector r CM. The Cartesan coordnates of ths vector are x CM, y CM, and z CM, defned n Equatons 9.28 and Therefore, r CM x CM î y CM ĵ z CM kˆ where r s the poston vector of the th partcle, defned by (9.30) Although locatng the center of mass for an extended object s somewhat more cumbersome than locatng the center of mass of a system of partcles, the basc deas we have dscussed stll apply. We can thnk of an extended object as a system contanng a large number of partcles (Fg. 9.18). The partcle separaton s very small, and so the object can be consdered to have a contnuous mass dstrbuton. By dvdng the object nto elements of mass m wth coordnates x, y, z, we see that the x coordnate of the center of mass s approxmately wth smlar expressons for y CM and z CM. If we let the number of elements n approach nfnty, then x CM s gven precsely. In ths lmt, we replace the sum by an ntegral and m by the dfferental element dm: Lkewse, for y CM and z CM we obtan x CM r CM x CM r x î y ĵ z kˆ lm m : 0 m x î m y ĵ m z kˆ m r x m x m y CM 1 M y dm and z CM 1 M z dm (9.31) (9.32) We can express the vector poston of the center of mass of an extended object n the form M M M M 1 M x dm CM (c) Actve Fgure 9.16 Two partcles of unequal mass are connected by a lght, rgd rod. (a) The system rotates clockwse when a force s appled between the less massve partcle and the center of mass. (b) The system rotates counterclockwse when a force s appled between the more massve partcle and the center of mass. (c) The system moves n the drecton of the force wthout rotatng when a force s appled at the center of mass. At the Actve Fgures lnk at you can choose the pont at whch to apply the force. y m 1 x CM x 1 x 2 CM m 2 Actve Fgure 9.17 The center of mass of two partcles of unequal mass on the x axs s located at x CM, a pont between the partcles, closer to the one havng the larger mass. At the Actve Fgures lnk at you can adjust the masses and postons of the partcles to see the effect on the locaton of the center of mass. x

22 272 CHAPTER 9 Lnear Momentum and Collsons y r CM 1 M r dm (9.33) m r z Fgure 9.18 An extended object can be consdered to be a dstrbuton of small elements of mass m. The center of mass s located at the vector poston r CM, whch has coordnates x CM, y CM, and z CM. A r CM B CM C x C whch s equvalent to the three expressons gven by Equatons 9.31 and The center of mass of any symmetrc object les on an axs of symmetry and on any plane of symmetry. 4 For example, the center of mass of a unform rod les n the rod, mdway between ts ends. The center of mass of a sphere or a cube les at ts geometrc center. The center of mass of an rregularly shaped object such as a wrench can be determned by suspendng the object frst from one pont and then from another. In Fgure 9.19, a wrench s hung from pont A, and a vertcal lne AB (whch can be establshed wth a plumb bob) s drawn when the wrench has stopped swngng. The wrench s then hung from pont C, and a second vertcal lne CD s drawn. The center of mass s halfway through the thckness of the wrench, under the ntersecton of these two lnes. In general, f the wrench s hung freely from any pont, the vertcal lne through ths pont must pass through the center of mass. Because an extended object s a contnuous dstrbuton of mass, each small mass element s acted upon by the gravtatonal force. The net effect of all these forces s equvalent to the effect of a sngle force Mg actng through a specal pont, called the center of gravty. If g s constant over the mass dstrbuton, then the center of gravty concdes wth the center of mass. If an extended object s pvoted at ts center of gravty, t balances n any orentaton. Center of mass A B Quck Quz 9.10 A baseball bat s cut at the locaton of ts center of mass as shown n Fgure The pece wth the smaller mass s (a) the pece on the rght (b) the pece on the left (c) Both peces have the same mass. (d) mpossble to determne. D Fgure 9.19 An expermental technque for determnng the center of mass of a wrench. The wrench s hung freely frst from pont A and then from pont C. The ntersecton of the two lnes AB and CD locates the center of mass. Fgure 9.20 (Quck Quz 9.10) A baseball bat cut at the locaton of ts center of mass. Example 9.13 The Center of Mass of Three Partcles A system conssts of three partcles located as shown n Fgure 9.21a. Fnd the center of mass of the system. Soluton We set up the problem by labelng the masses of the partcles as shown n the fgure, wth m 1 m kg and m kg. Usng the defnng equatons for the coordnates of the center of mass and notng that z CM 0, we obtan x CM m x M m 1x 1 m 2 x 2 m 3 x 3 m 1 m 2 m 3 (1.0 kg)(1.0 m) (1.0 kg)(2.0 m) (2.0 kg)(0) 1.0 kg 1.0 kg 2.0 kg 3.0 kgm 4.0 kg 0.75 m 4 Ths statement s vald only for objects that have a unform mass per unt volume.

23 SECTION 9.5 The Center of Mass 273 y(m) 3 y CM m y M m 1y 1 m 2 y 2 m 3 y 3 m 1 m 2 m 3 (1.0 kg)(0) (1.0 kg)(0) (2.0 kg)(2.0 m) 4.0 kg 2 m kgm 4.0 kg 1.0 m 1 The poston vector to the center of mass measured from the orgn s therefore r CM r CM x CM î y CM ĵ (0.75î 1.0ĵ) m m 1 m (a) 3 x(m) We can verfy ths result graphcally by addng together m 1 r 1 m 2 r 2 m 3 r 3 and dvdng the vector sum by M, the total mass. Ths s shown n Fgure 9.21b. Mr CM m3 r 3 r CM m 1 r 1 m 2 r 2 (b) Fgure 9.21 (Example 9.13) (a) Two 1.0-kg partcles are located on the x axs and a sngle 2.0-kg partcle s located on the y axs as shown. The vector ndcates the locaton of the system s center of mass. (b) The vector sum of m r and the resultng vector for r CM. Example 9.14 The Center of Mass of a Rod (A) Show that the center of mass of a rod of mass M and length L les mdway between ts ends, assumng the rod has a unform mass per unt length. Soluton The rod s shown algned along the x axs n Fgure 9.22, so that y CM z CM 0. Furthermore, f we call the mass per unt length (ths quantty s called the lnear mass densty), then M/L for the unform rod we assume here. If we dvde the rod nto elements of length dx, then the mass of each element s dm dx. Equaton 9.31 gves x CM 1 M x dm 1 M L Because M/L, ths reduces to x CM 0 x dx L2 2M M L One can also use symmetry arguments to obtan the same result. (B) Suppose a rod s nonunform such that ts mass per unt length vares lnearly wth x accordng to the expresson M L 2 x 2 2 L 0 L 2 2M x, where s a constant. Fnd the x coordnate of the center of mass as a fracton of L. Soluton In ths case, we replace dm by dx, where s not constant. Therefore, x CM s x CM 1 M x dm 1 M L L x 2 dx M y 0 L 0 L 3 3M x dx 1 L x x dx M dm = ldx x O x dx Fgure 9.22 (Example 9.14) The geometry used to fnd the center of mass of a unform rod. 0

24 274 CHAPTER 9 Lnear Momentum and Collsons We can elmnate by notng that the total mass of the rod s related to through the relatonshp M dm L 0 dx L 0 x dx L 2 2 Substtutng ths nto the expresson for x CM gves x CM L 3 3L 2 /2 2 3 L Example 9.15 The Center of Mass of a Rght Trangle You have been asked to hang a metal sgn from a sngle vertcal wre. The sgn has the trangular shape shown n Fgure 9.23a. The bottom of the sgn s to be parallel to the ground. At what dstance from the left end of the sgn should you attach the support wre? Soluton The wre must be attached at a pont drectly above the center of gravty of the sgn, whch s the same as ts center of mass because t s n a unform gravtatonal feld. We assume that the trangular sgn has a unform densty and total mass M. Because the sgn s a contnuous dstrbuton of mass, we must use the ntegral expresson n Equaton 9.31 to fnd the x coordnate of the center of mass. We dvde the trangle nto narrow strps of wdth dx and heght y as shown n Fgure 9.23b, where y s the heght of the hypotenuse of the trangle above the x axs for a gven value of x. The mass of each strp s the product of the volume of the strp and the densty of the materal from whch the sgn s made: dm yt dx, where t s the thckness of the metal sgn. The densty of the materal s the total mass of the sgn dvded by ts total volume (area of the trangle tmes thckness), so dm Usng Equaton 9.31 to fnd the x coordnate of the center of mass gves x CM 1 M x dm 1 M a yt dx M 1 2 abt yt dx 2My ab 0 x 2My ab dx dx 2 a xy dx ab 0 To proceed further and evaluate the ntegral, we must express y n terms of x. The lne representng the hypotenuse of the trangle n Fgure 9.23b has a slope of b/a and passes through the orgn, so the equaton of ths lne s y (b/a)x. Wth ths substtuton for y n the ntegral, we have x CM 2 ab a 2 3 a 0 x b a x dx 2 a 2 a 0 x 2 dx 2 a 2 x 3 3 a 0 Thus, the wre must be attached to the sgn at a dstance two thrds of the length of the bottom edge from the left end. We could also fnd the y coordnate of the center of mass of the sgn, but ths s not needed n order to determne where the wre should be attached. (a) Fgure 9.23 (Example 9.15) (a) A trangular sgn to be hung from a sngle wre. (b) Geometrc constructon for locatng the center of mass. O y x c a dm (b) y dx b x 9.6 Moton of a System of Partcles We can begn to understand the physcal sgnfcance and utlty of the center of mass concept by takng the tme dervatve of the poston vector gven by Equaton From Secton 4.1 we know that the tme dervatve of a poston vector s by defnton a velocty. Assumng M remans constant for a system of partcles, that s, no partcles enter or leave the system, we obtan the followng expresson for the velocty of the center of mass of the system: Velocty of the center of mass Total momentum of a system of partcles v CM dr CM dt 1 M where v s the velocty of the th partcle. Rearrangng Equaton 9.34 gves m dr dt Mv CM m v p p tot m v M (9.34) (9.35)

25 SECTION 9.6 Moton of a System of Partcles 275 Therefore, we conclude that the total lnear momentum of the system equals the total mass multpled by the velocty of the center of mass. In other words, the total lnear momentum of the system s equal to that of a sngle partcle of mass M movng wth a velocty v CM. If we now dfferentate Equaton 9.34 wth respect to tme, we obtan the acceleraton of the center of mass of the system: a CM dv CM dt 1 M m dv dt 1 M m a Rearrangng ths expresson and usng Newton s second law, we obtan (9.36) Acceleraton of the center of mass Ma CM m a F (9.37) where F s the net force on partcle. The forces on any partcle n the system may nclude both external forces (from outsde the system) and nternal forces (from wthn the system). However, by Newton s thrd law, the nternal force exerted by partcle 1 on partcle 2, for example, s equal n magntude and opposte n drecton to the nternal force exerted by partcle 2 on partcle 1. Thus, when we sum over all nternal forces n Equaton 9.37, they cancel n pars and we fnd that the net force on the system s caused only by external forces. Thus, we can wrte Equaton 9.37 n the form F ext Ma CM (9.38) That s, the net external force on a system of partcles equals the total mass of the system multpled by the acceleraton of the center of mass. If we compare ths wth Newton s second law for a sngle partcle, we see that the partcle model that we have used for several chapters can be descrbed n terms of the center of mass: Newton s second law for a system of partcles The center of mass of a system of partcles of combned mass M moves lke an equvalent partcle of mass M would move under the nfluence of the net external force on the system. Fnally, we see that f the net external force s zero, then from Equaton 9.38 t follows that Ma CM M d v CM dt 0 so that M v CM p tot constant (when F ext 0) (9.39) That s, the total lnear momentum of a system of partcles s conserved f no net external force s actng on the system. It follows that for an solated system of partcles, both the total momentum and the velocty of the center of mass are constant n tme, as shown n Fgure Ths s a generalzaton to a many-partcle system of the law of conservaton of momentum dscussed n Secton 9.1 for a two-partcle system. Suppose an solated system consstng of two or more members s at rest. The center of mass of such a system remans at rest unless acted upon by an external force. For example, consder a system made up of a swmmer standng on a raft, wth the system ntally at rest. When the swmmer dves horzontally off the raft, the raft moves n the drecton opposte to that of the swmmer and the center of mass of the system remans at rest (f we neglect frcton between raft and water). Furthermore, the lnear momentum of the dver s equal n magntude to that of the raft, but opposte n drecton. As another example, suppose an unstable atom ntally at rest suddenly breaks up nto two fragments of masses M 1 and M 2, wth veloctes v 1 and v 2, respectvely. Because the total momentum of the system before the breakup s zero, the total momentum of the

26 276 CHAPTER 9 Lnear Momentum and Collsons Rchard Megna/Fundamental Photographs Fgure 9.24 Multflash photograph showng an overhead vew of a wrench movng on a horzontal surface. The whte dots are located at the center of mass of the wrench and show that the center of mass moves n a straght lne as the wrench rotates. system after the breakup must also be zero. Therefore, M 1 v 1 M 2 v 2 0. If the velocty of one of the fragments s known, the recol velocty of the other fragment can be calculated. Quck Quz 9.11 The vacatoners on a cruse shp are eager to arrve at ther next destnaton. They decde to try to speed up the cruse shp by gatherng at the bow (the front) and runnng all at once toward the stern (the back) of the shp. Whle they are runnng toward the stern, the speed of the shp s (a) hgher than t was before (b) unchanged (c) lower than t was before (d) mpossble to determne. Quck Quz 9.12 The vacatoners n Quck Quz 9.11 stop runnng when they reach the stern of the shp. After they have all stopped runnng, the speed of the shp s (a) hgher than t was before they started runnng (b) unchanged from what t was before they started runnng (c) lower than t was before they started runnng (d) mpossble to determne. Conceptual Example 9.16 The Sldng Bear Suppose you tranqulze a polar bear on a smooth glacer as part of a research effort. How mght you estmate the bear s mass usng a measurng tape, a rope, and knowledge of your own mass? Soluton Te one end of the rope around the bear, and then lay out the tape measure on the ce wth one end at the bear s orgnal poston, as shown n Fgure Grab hold of the free end of the rope and poston yourself as x p CM x b Fgure 9.25 (Conceptual Example 9.16) The center of mass of an solated system remans at rest unless acted on by an external force. How can you determne the mass of the polar bear?

27 SECTION 9.7 Rocket Propulson 277 shown, notng your locaton. Take off your spked shoes, and pull on the rope hand over hand. Both you and the bear wll slde over the ce untl you meet. From the tape, observe how far you slde, x p, and how far the bear sldes, x b. The pont where you meet the bear s the fxed locaton of the center of mass of the system (bear plus you), and so you can determne the mass of the bear from m b x b m p x p. (Unfortunately, you cannot return to your spked shoes and so you are n bg trouble f the bear wakes up!) Conceptual Example 9.17 Explodng Projectle A projectle fred nto the ar suddenly explodes nto several fragments (Fg. 9.26). What can be sad about the moton of the center of mass of the system made up of all the fragments after the exploson? Soluton Neglectng ar resstance, the only external force on the projectle s the gravtatonal force. Thus, f the projectle dd not explode, t would contnue to move along the parabolc path ndcated by the dashed lne n Fgure Because the forces caused by the exploson are nternal, they do not affect the moton of the center of mass of the system (the fragments). Thus, after the exploson, the center of mass of the fragments follows the same parabolc path the projectle would have followed f there had been no exploson. Fgure 9.26 (Conceptual Example 9.17) When a projectle explodes nto several fragments, the center of mass of the system made up of all the fragments follows the same parabolc path the projectle would have taken had there been no exploson. Example 9.18 The Explodng Rocket A rocket s fred vertcally upward. At the nstant t reaches an alttude of m and a speed of 300 m/s, t explodes nto three fragments havng equal mass. One fragment contnues to move upward wth a speed of 450 m/s followng the exploson. The second fragment has a speed of 240 m/s and s movng east rght after the exploson. What s the velocty of the thrd fragment rght after the exploson? Soluton Let us call the total mass of the rocket M; hence, the mass of each fragment s M/3. Because the forces of the exploson are nternal to the system and cannot affect ts total momentum, the total momentum p of the rocket just before the exploson must equal the total momentum p f of the fragments rght after the exploson. Before the exploson, p M v M(300 ĵ m/s) After the exploson, p f M 3 (240î m/s) M 3 (450ĵ m/s) M 3 where v f s the unknown velocty of the thrd fragment. Equatng these two expressons (because p p f ) gves M 3 v f M 3 (240 î m/s) M 3 v f M(300 ĵ m/s) (450 ĵ m/s) ( 240î 450ĵ) m/s What does the sum of the momentum vectors for all the fragments look lke? v f 9.7 Rocket Propulson When ordnary vehcles such as cars and locomotves are propelled, the drvng force for the moton s frcton. In the case of the car, the drvng force s the force exerted by the road on the car. A locomotve pushes aganst the tracks; hence, the drvng force s the force exerted by the tracks on the locomotve. However, a rocket movng n space has no road or tracks to push aganst. Therefore, the source of the propulson of a rocket must be somethng other than frcton. Fgure 9.27 s a dramatc photograph of a spacecraft at lftoff. The operaton of a rocket depends upon the law of conservaton of lnear momentum as appled to a system of partcles, where the system s the rocket plus ts ejected fuel.

28 278 CHAPTER 9 Lnear Momentum and Collsons Courtesy of NASA Fgure 9.27 At lftoff, enormous thrust s generated by the space shuttle s lqud-fuel engnes, aded by the two sold-fuel boosters. Ths photograph shows the lftoff of the space shuttle Columba, whch was lost n a tragc accdent durng ts landng attempt on February 1, 2003 (shortly before ths volume went to press). m M + m p = (M + m)v (a) (b) M v v + v Fgure 9.28 Rocket propulson. (a) The ntal mass of the rocket plus all ts fuel s M m at a tme t, and ts speed s v. (b) At a tme t t, the rocket s mass has been reduced to M and an amount of fuel m has been ejected. The rocket s speed ncreases by an amount v. Rocket propulson can be understood by frst consderng a mechancal system consstng of a machne gun mounted on a cart on wheels. As the gun s fred, each bullet receves a momentum m v n some drecton, where v s measured wth respect to a statonary Earth frame. The momentum of the system made up of cart, gun, and bullets must be conserved. Hence, for each bullet fred, the gun and cart must receve a compensatng momentum n the opposte drecton. That s, the reacton force exerted by the bullet on the gun accelerates the cart and gun, and the cart moves n the drecton opposte that of the bullets. If n s the number of bullets fred each second, then the average force exerted on the gun s F nm v. In a smlar manner, as a rocket moves n free space, ts lnear momentum changes when some of ts mass s released n the form of ejected gases. Because the gases are gven momentum when they are ejected out of the engne, the rocket receves a compensatng momentum n the opposte drecton. Therefore, the rocket s accelerated as a result of the push, or thrust, from the exhaust gases. In free space, the center of mass of the system (rocket plus expelled gases) moves unformly, ndependent of the propulson process. 5 Suppose that at some tme t, the magntude of the momentum of a rocket plus ts fuel s (M m)v, where v s the speed of the rocket relatve to the Earth (Fg. 9.28a). Over a short tme nterval t, the rocket ejects fuel of mass m, and so at the end of ths nterval the rocket s speed s v v, where v s the change n speed of the rocket (Fg. 9.28b). If the fuel s ejected wth a speed v e relatve to the rocket (the subscrpt e stands for exhaust, and v e s usually called the exhaust speed), the velocty of the fuel relatve to a statonary frame of reference s v v e. Thus, f we equate the total ntal mo- 5 It s nterestng to note that the rocket and machne gun represent cases of the reverse of a perfectly nelastc collson: momentum s conserved, but the knetc energy of the system ncreases (at the expense of chemcal potental energy n the fuel).

29 SECTION 9.7 Rocket Propulson 279 mentum of the system to the total fnal momentum, we obtan (M m)v M(v v) m(v v e ) where M represents the mass of the rocket and ts remanng fuel after an amount of fuel havng mass m has been ejected. Smplfyng ths expresson gves We also could have arrved at ths result by consderng the system n the center-ofmass frame of reference, whch s a frame havng the same velocty as the center of mass of the system. In ths frame, the total momentum of the system s zero; therefore, f the rocket gans a momentum M v by ejectng some fuel, the exhausted fuel obtans a momentum v e m n the opposte drecton, so that M v v e m 0. If we now take the lmt as t goes to zero, we let v : dv and m : dm. Furthermore, the ncrease n the exhaust mass dm corresponds to an equal decrease n the rocket mass, so that dm dm. Note that dm s negatve because t represents a decrease n mass, so dm s a postve number. Usng ths fact, we obtan (9.40) We dvde the equaton by M and ntegrate, takng the ntal mass of the rocket plus fuel to be M and the fnal mass of the rocket plus ts remanng fuel to be M f. Ths gves (9.41) Ths s the basc expresson for rocket propulson. Frst, t tells us that the ncrease n rocket speed s proportonal to the exhaust speed v e of the ejected gases. Therefore, the exhaust speed should be very hgh. Second, the ncrease n rocket speed s proportonal to the natural logarthm of the rato M /M f. Therefore, ths rato should be as large as possble, whch means that the mass of the rocket wthout ts fuel should be as small as possble and the rocket should carry as much fuel as possble. The thrust on the rocket s the force exerted on t by the ejected exhaust gases. We can obtan an expresson for the thrust from Equaton 9.40: Thrust M M v v e m M dv v e dm v e dm v f dv v e v Mf M v f v v e ln M M f dv dt v e dm dt (9.42) Ths expresson shows us that the thrust ncreases as the exhaust speed ncreases and as the rate of change of mass (called the burn rate) ncreases. dm M Courtesy of NASA The force from a ntrogenpropelled hand-controlled devce allows an astronaut to move about freely n space wthout restrctve tethers, usng the thrust force from the expelled ntrogen. Expresson for rocket propulson Example 9.19 A Rocket n Space A rocket movng n free space has a speed of m/s relatve to the Earth. Its engnes are turned on, and fuel s ejected n a drecton opposte the rocket s moton at a speed of m/s relatve to the rocket. (A) What s the speed of the rocket relatve to the Earth once the rocket s mass s reduced to half ts mass before gnton? Soluton We can guess that the speed we are lookng for must be greater than the orgnal speed because the rocket s acceleratng. Applyng Equaton 9.41, we obtan v f v v e ln M M f m/s ( m/s)ln M 0.5 M m/s (B) What s the thrust on the rocket f t burns fuel at the rate of 50 kg/s? Soluton Usng Equaton 9.42, Thrust v e dm dt ( m/s)(50 kg/s) N

30 280 CHAPTER 9 Lnear Momentum and Collsons Example 9.20 Fghtng a Fre Two frefghters must apply a total force of 600 N to steady a hose that s dschargng water at the rate of L/mn. Estmate the speed of the water as t exts the nozzle. Soluton The water s extng at L/mn, whch s 60 L/s. Knowng that 1 L of water has a mass of 1 kg, we estmate that about 60 kg of water leaves the nozzle every second. As the water leaves the hose, t exerts on the hose a thrust that must be counteracted by the 600-N force exerted by the frefghters. So, applyng Equaton 9.42 gves Thrust v e dm dt 600 N v e (60 kg/s) v e 10 m/s Frefghtng s dangerous work. If the nozzle should slp from ther hands, the movement of the hose due to the thrust t receves from the rapdly extng water could njure the frefghters. Take a practce test for ths chapter by clckng on the Practce Test lnk at SUMMARY The lnear momentum p of a partcle of mass m movng wth a velocty v s p mv (9.2) The law of conservaton of lnear momentum ndcates that the total momentum of an solated system s conserved. If two partcles form an solated system, the momentum of the system s conserved regardless of the nature of the force between them. Therefore, the total momentum of the system at all tmes equals ts ntal total momentum, or p 1 p 2 p 1f p 2f (9.5) The mpulse mparted to a partcle by a force F s equal to the change n the momentum of the partcle: I t f F dt p t (9.8, 9.9) Ths s known as the mpulse momentum theorem. Impulsve forces are often very strong compared wth other forces on the system and usually act for a very short tme, as n the case of collsons. When two partcles collde, the total momentum of the solated system before the collson always equals the total momentum after the collson, regardless of the nature of the collson. An nelastc collson s one for whch the total knetc energy of the system s not conserved. A perfectly nelastc collson s one n whch the colldng bodes stck together after the collson. An elastc collson s one n whch the knetc energy of the system s conserved. In a two- or three-dmensonal collson, the components of momentum of an solated system n each of the drectons (x, y, and z) are conserved ndependently. The poston vector of the center of mass of a system of partcles s defned as r CM m r M (9.30) where M m s the total mass of the system and r s the poston vector of the th partcle.

31 Questons 281 The poston vector of the center of mass of an extended object can be obtaned from the ntegral expresson r CM 1 M r dm (9.33) The velocty of the center of mass for a system of partcles s v CM m v M (9.34) The total momentum of a system of partcles equals the total mass multpled by the velocty of the center of mass. Newton s second law appled to a system of partcles s F ext Ma CM (9.38) where a CM s the acceleraton of the center of mass and the sum s over all external forces. The center of mass moves lke an magnary partcle of mass M under the nfluence of the resultant external force on the system. QUESTIONS 1. Does a large force always produce a larger mpulse on an object than a smaller force does? Explan. 2. If the speed of a partcle s doubled, by what factor s ts momentum changed? By what factor s ts knetc energy changed? 3. If two partcles have equal knetc energes, are ther momenta necessarly equal? Explan. 4. Whle n moton, a ptched baseball carres knetc energy and momentum. (a) Can we say that t carres a force that t can exert on any object t strkes? (b) Can the baseball delver more knetc energy to the object t strkes than the ball carres ntally? (c) Can the baseball delver to the object t strkes more momentum than the ball carres ntally? Explan your answers. 5. An solated system s ntally at rest. Is t possble for parts of the system to be n moton at some later tme? If so, explan how ths mght occur. 6. If two objects collde and one s ntally at rest, s t possble for both to be at rest after the collson? Is t possble for one to be at rest after the collson? Explan. 7. Explan how lnear momentum s conserved when a ball bounces from a floor. 8. A bomb, ntally at rest, explodes nto several peces. (a) Is lnear momentum of the system conserved? (b) Is knetc energy of the system conserved? Explan. 9. A ball of clay s thrown aganst a brck wall. The clay stops and stcks to the wall. Is the prncple of conservaton of momentum volated n ths example? 10. You are standng perfectly stll, and then you take a step forward. Before the step your momentum was zero, but afterward you have some momentum. Is the prncple of conservaton of momentum volated n ths case? 11. When a ball rolls down an nclne, ts lnear momentum ncreases. Is the prncple of conservaton of momentum volated n ths process? 12. Consder a perfectly nelastc collson between a car and a large truck. Whch vehcle experences a larger change n knetc energy as a result of the collson? 13. A sharpshooter fres a rfle whle standng wth the butt of the gun aganst hs shoulder. If the forward momentum of a bullet s the same as the backward momentum of the gun, why sn t t as dangerous to be ht by the gun as by the bullet? 14. A pole-vaulter falls from a heght of 6.0 m onto a foam rubber pad. Can you calculate hs speed just before he reaches the pad? Can you calculate the force exerted on hm by the pad? Explan. 15. Frefghters must apply large forces to hold a fre hose steady (Fg. Q9.15). What factors related to the projecton of the water determne the magntude of the force needed to keep the end of the fre hose statonary? Bll Stormont/The Stock Market Fgure Q A large bed sheet s held vertcally by two students. A thrd student, who happens to be the star ptcher on the baseball team, throws a raw egg at the sheet. Explan why the egg does not break when t hts the sheet, regardless of ts ntal speed. (If you try ths demonstraton, make sure the ptcher hts the sheet near ts center, and do not allow the egg to fall on the floor after beng caught.)

32 282 CHAPTER 9 Lnear Momentum and Collsons 17. A skater s standng stll on a frctonless ce rnk. Her frend throws a Frsbee straght at her. In whch of the followng cases s the largest momentum transferred to the skater? (a) The skater catches the Frsbee and holds onto t. (b) The skater catches the Frsbee momentarly, but then drops t vertcally downward. (c) The skater catches the Frsbee, holds t momentarly, and throws t back to her frend. 18. In an elastc collson between two partcles, does the knetc energy of each partcle change as a result of the collson? 19. Three balls are thrown nto the ar smultaneously. What s the acceleraton of ther center of mass whle they are n moton? 20. A person balances a meter stck n a horzontal poston on the extended ndex fngers of her rght and left hands. She slowly brngs the two fngers together. The stck remans balanced and the two fngers always meet at the 50-cm mark regardless of ther orgnal postons. (Try t!) Explan. 21. NASA often uses the gravty of a planet to slngshot a probe on ts way to a more dstant planet. The nteracton of the planet and the spacecraft s a collson n whch the objects do not touch. How can the probe have ts speed ncreased n ths manner? 22. The Moon revolves around the Earth. Model ts orbt as crcular. Is the Moon s lnear momentum conserved? Is ts knetc energy conserved? 23. A raw egg dropped to the floor breaks upon mpact. However, a raw egg dropped onto a thck foam rubber cushon from a heght of about 1 m rebounds wthout breakng. Why s ths possble? If you try ths experment, be sure to catch the egg after ts frst bounce. 24. Can the center of mass of an object be located at a poston at whch there s no mass? If so, gve examples. 25. A juggler juggles three balls n a contnuous cycle. Any one ball s n contact wth hs hands for one ffth of the tme. Descrbe the moton of the center of mass of the three balls. What average force does the juggler exert on one ball whle he s touchng t? 26. Does the center of mass of a rocket n free space accelerate? Explan. Can the speed of a rocket exceed the exhaust speed of the fuel? Explan. 27. Early n the twenteth century, Robert Goddard proposed sendng a rocket to the moon. Crtcs objected that n a vacuum, such as exsts between the Earth and the Moon, the gases emtted by the rocket would have nothng to push aganst to propel the rocket. Accordng to Scentfc Amercan (January 1975), Goddard placed a gun n a vacuum and fred a blank cartrdge from t. (A blank cartrdge contans no bullet and fres only the waddng and the hot gases produced by the burnng gunpowder.) What happened when the gun was fred? 28. Explan how you could use a balloon to demonstrate the mechansm responsble for rocket propulson. 29. On the subject of the followng postons, state your own vew and argue to support t. (a) The best theory of moton s that force causes acceleraton. (b) The true measure of a force s effectveness s the work t does, and the best theory of moton s that work done on an object changes ts energy. (c) The true measure of a force s effect s mpulse, and the best theory of moton s that mpulse njected nto an object changes ts momentum. PROBLEMS 1, 2, 3 = straghtforward, ntermedate, challengng = full soluton avalable n the Student Solutons Manual and Study Gude = coached soluton wth hnts avalable at = computer useful n solvng problem = pared numercal and symbolc problems Secton 9.1 Lnear Momentum and ts Conservaton 1. A 3.00-kg partcle has a velocty of (3.00î 4.00ĵ) m/s. (a) Fnd ts x and y components of momentum. (b) Fnd the magntude and drecton of ts momentum. 2. A kg ball s thrown straght up nto the ar wth an ntal speed of 15.0 m/s. Fnd the momentum of the ball (a) at ts maxmum heght and (b) halfway up to ts maxmum heght. 3. How fast can you set the Earth movng? In partcular, when you jump straght up as hgh as you can, what s the order of magntude of the maxmum recol speed that you gve to the Earth? Model the Earth as a perfectly sold object. In your soluton, state the physcal quanttes you take as data and the values you measure or estmate for them. 4. Two blocks of masses M and 3M are placed on a horzontal, frctonless surface. A lght sprng s attached to one M v 3M M Before (a) 2.00 m/s 3M After (b) Fgure P9.4

33 Problems 283 of them, and the blocks are pushed together wth the sprng between them (Fg. P9.4). A cord ntally holdng the blocks together s burned; after ths, the block of mass 3M moves to the rght wth a speed of 2.00 m/s. (a) What s the speed of the block of mass M? (b) Fnd the orgnal elastc potental energy n the sprng f M kg. 5. (a) A partcle of mass m moves wth momentum p. Show that the knetc energy of the partcle s K p 2 /2m. (b) Express the magntude of the partcle s momentum n terms of ts knetc energy and mass. Secton 9.2 Impulse and Momentum 6. A frend clams that, as long as he has hs seatbelt on, he can hold on to a 12.0-kg chld n a 60.0 m/h head-on collson wth a brck wall n whch the car passenger compartment comes to a stop n s. Show that the volent force durng the collson wll tear the chld from hs arms. A chld should always be n a toddler seat secured wth a seat belt n the back seat of a car. 7. An estmated force tme curve for a baseball struck by a bat s shown n Fgure P9.7. From ths curve, determne (a) the mpulse delvered to the ball, (b) the average force exerted on the ball, and (c) the peak force exerted on the ball. 10. A tenns player receves a shot wth the ball ( kg) travelng horzontally at 50.0 m/s and returns the shot wth the ball travelng horzontally at 40.0 m/s n the opposte drecton. (a) What s the mpulse delvered to the ball by the racquet? (b) What work does the racquet do on the ball? 11. In a slow-ptch softball game, a kg softball crosses the plate at 15.0 m/s at an angle of 45.0 below the horzontal. The batter hts the ball toward center feld, gvng t a velocty of 40.0 m/s at 30.0 above the horzontal. (a) Determne the mpulse delvered to the ball. (b) If the force on the ball ncreases lnearly for 4.00 ms, holds constant for 20.0 ms, and then decreases to zero lnearly n another 4.00 ms, what s the maxmum force on the ball? 12. A professonal dver performs a dve from a platform 10 m above the water surface. Estmate the order of magntude of the average mpact force she experences n her collson wth the water. State the quanttes you take as data and ther values. 13. A garden hose s held as shown n Fgure P9.13. The hose s orgnally full of motonless water. What addtonal force s necessary to hold the nozzle statonary after the water flow s turned on, f the dscharge rate s kg/s wth a speed of 25.0 m/s? F(N) F = N t(ms) Fgure P9.7 Fgure P A ball of mass kg s dropped from rest from a heght of 1.25 m. It rebounds from the floor to reach a heght of m. What mpulse was gven to the ball by the floor? 9. A 3.00-kg steel ball strkes a wall wth a speed of 10.0 m/s at an angle of 60.0 wth the surface. It bounces off wth the same speed and angle (Fg. P9.9). If the ball s n contact wth the wall for s, what s the average force exerted by the wall on the ball? 14. A glder of mass m s free to slde along a horzontal ar track. It s pushed aganst a launcher at one end of the track. Model the launcher as a lght sprng of force constant k compressed by a dstance x. The glder s released from rest. (a) Show that the glder attans a speed of v x(k/m) 1/2. (b) Does a glder of large or of small mass attan a greater speed? (c) Show that the mpulse mparted to the glder s gven by the expresson x(km) 1/2. (d) Is a greater mpulse njected nto a large or a small mass? (e) Is more work done on a large or a small mass? y Fgure P9.9 x Secton 9.3 Collsons n One Dmenson 15. Hgh-speed stroboscopc photographs show that the head of a golf club of mass 200 g s travelng at 55.0 m/s just before t strkes a 46.0-g golf ball at rest on a tee. After the collson, the club head travels (n the same drecton) at 40.0 m/s. Fnd the speed of the golf ball just after mpact. 16. An archer shoots an arrow toward a target that s sldng toward her wth a speed of 2.50 m/s on a smooth, slppery

34 284 CHAPTER 9 Lnear Momentum and Collsons 17. surface. The 22.5-g arrow s shot wth a speed of 35.0 m/s and passes through the 300-g target, whch s stopped by the mpact. What s the speed of the arrow after passng through the target? A 10.0-g bullet s fred nto a statonary block of wood (m 5.00 kg). The relatve moton of the bullet stops nsde the block. The speed of the bullet-plus-wood combnaton mmedately after the collson s m/s. What was the orgnal speed of the bullet? 18. A ralroad car of mass kg s movng wth a speed of 4.00 m/s. It colldes and couples wth three other coupled ralroad cars, each of the same mass as the sngle car and movng n the same drecton wth an ntal speed of 2.00 m/s. (a) What s the speed of the four cars after the collson? (b) How much mechancal energy s lost n the collson? 19. Four ralroad cars, each of mass kg, are coupled together and coastng along horzontal tracks at speed v toward the south. A very strong but foolsh move actor, rdng on the second car, uncouples the front car and gves t a bg push, ncreasng ts speed to 4.00 m/s southward. The remanng three cars contnue movng south, now at 2.00 m/s. (a) Fnd the ntal speed of the cars. (b) How much work dd the actor do? (c) State the relatonshp between the process descrbed here and the process n Problem Two blocks are free to slde along the frctonless wooden track ABC shown n Fgure P9.20. The block of mass m kg s released from A. Protrudng from ts front end s the north pole of a strong magnet, repellng the north pole of an dentcal magnet embedded n the back end of the block of mass m kg, ntally at rest. The two blocks never touch. Calculate the maxmum heght to whch m 1 rses after the elastc collson. A m forces, of course, s false. Newton s thrd law tells us that both objects experence forces of the same magntude. The truck suffers less damage because t s made of stronger metal. But what about the two drvers? Do they experence the same forces? To answer ths queston, suppose that each vehcle s ntally movng at 8.00 m/s and that they undergo a perfectly nelastc head-on collson. Each drver has mass 80.0 kg. Includng the drvers, the total vehcle masses are 800 kg for the car and kg for the truck. If the collson tme s s, what force does the seatbelt exert on each drver? A neutron n a nuclear reactor makes an elastc headon collson wth the nucleus of a carbon atom ntally at rest. (a) What fracton of the neutron s knetc energy s transferred to the carbon nucleus? (b) If the ntal knetc energy of the neutron s J, fnd ts fnal knetc energy and the knetc energy of the carbon nucleus after the collson. (The mass of the carbon nucleus s nearly 12.0 tmes the mass of the neutron.) 24. As shown n Fgure P9.24, a bullet of mass m and speed v passes completely through a pendulum bob of mass M. The bullet emerges wth a speed of v/2. The pendulum bob s suspended by a stff rod of length and neglgble mass. What s the mnmum value of v such that the pendulum bob wll barely swng through a complete vertcal crcle? m M v v/2 Fgure P m B m 2 Fgure P9.20 A 45.0-kg grl s standng on a plank that has a mass of 150 kg. The plank, orgnally at rest, s free to slde on a frozen lake, whch s a flat, frctonless supportng surface. The grl begns to walk along the plank at a constant speed of 1.50 m/s relatve to the plank. (a) What s her speed relatve to the ce surface? (b) What s the speed of the plank relatve to the ce surface? 22. Most of us know ntutvely that n a head-on collson between a large dump truck and a subcompact car, you are better off beng n the truck than n the car. Why s ths? Many people magne that the collson force exerted on the car s much greater than that experenced by the truck. To substantate ths vew, they pont out that the car s crushed, whereas the truck s only dented. Ths dea of unequal C 25. A 12.0-g wad of stcky clay s hurled horzontally at a 100-g wooden block ntally at rest on a horzontal surface. The clay stcks to the block. After mpact, the block sldes 7.50 m before comng to rest. If the coeffcent of frcton between the block and the surface s 0.650, what was the speed of the clay mmedately before mpact? 26. A 7.00-g bullet, when fred from a gun nto a 1.00-kg block of wood held n a vse, penetrates the block to a depth of 8.00 cm. What If? Ths block of wood s placed on a frctonless horzontal surface, and a second 7.00-g bullet s fred from the gun nto the block. To what depth wll the bullet penetrate the block n ths case? 27. (a) Three carts of masses 4.00 kg, 10.0 kg, and 3.00 kg move on a frctonless horzontal track wth speeds of 5.00 m/s, 3.00 m/s, and 4.00 m/s, as shown n Fgure P9.27. Velcro couplers make the carts stck together after colldng. Fnd the fnal velocty of the tran of three carts. (b) What If? Does your answer requre that all the carts collde and stck together at the same tme? What f they collde n a dfferent order?

35 Problems 285 Secton m/s 3.00 m/s 4.00 m/s 4.00 kg 10.0 kg 3.00 kg Fgure P9.27 Two-Dmensonal Collsons 28. A 90.0-kg fullback runnng east wth a speed of 5.00 m/s s tackled by a 95.0-kg opponent runnng north wth a speed of 3.00 m/s. If the collson s perfectly nelastc, (a) calculate the speed and drecton of the players just after the tackle and (b) determne the mechancal energy lost as a result of the collson. Account for the mssng energy. 29. Two shuffleboard dsks of equal mass, one orange and the other yellow, are nvolved n an elastc, glancng collson. The yellow dsk s ntally at rest and s struck by the orange dsk movng wth a speed of 5.00 m/s. After the collson, the orange dsk moves along a drecton that makes an angle of 37.0 wth ts ntal drecton of moton. The veloctes of the two dsks are perpendcular after the collson. Determne the fnal speed of each dsk. 30. Two shuffleboard dsks of equal mass, one orange and the other yellow, are nvolved n an elastc, glancng collson. The yellow dsk s ntally at rest and s struck by the orange dsk movng wth a speed v. After the collson, the orange dsk moves along a drecton that makes an angle wth ts ntal drecton of moton. The veloctes of the two dsks are perpendcular after the collson. Determne the fnal speed of each dsk. 31. The mass of the blue puck n Fgure P9.31 s 20.0% greater than the mass of the green one. Before colldng, the pucks approach each other wth momenta of equal magntudes and opposte drectons, and the green puck has an ntal speed of 10.0 m/s. Fnd the speeds of the pucks after the collson f half the knetc energy s lost durng the collson. 33. A bllard ball movng at 5.00 m/s strkes a statonary ball of the same mass. After the collson, the frst ball moves, at 4.33 m/s, at an angle of 30.0 wth respect to the orgnal lne of moton. Assumng an elastc collson (and gnorng frcton and rotatonal moton), fnd the struck ball s velocty after the collson. 34. A proton, movng wth a velocty of v î, colldes elastcally wth another proton that s ntally at rest. If the two protons have equal speeds after the collson, fnd (a) the speed of each proton after the collson n terms of v and (b) the drecton of the velocty vectors after the collson. 35. An object of mass 3.00 kg, movng wth an ntal velocty of 5.00î m/s, colldes wth and stcks to an object of mass 2.00 kg wth an ntal velocty of 3.00ĵ m/s. Fnd the fnal velocty of the composte object. 36. Two partcles wth masses m and 3m are movng toward each other along the x axs wth the same ntal speeds v. Partcle m s travelng to the left, whle partcle 3m s travelng to the rght. They undergo an elastc glancng collson such that partcle m s movng downward after the collson at rght angles from ts ntal drecton. (a) Fnd the fnal speeds of the two partcles. (b) What s the angle at whch the partcle 3m s scattered? 37. An unstable atomc nucleus of mass kg ntally at rest dsntegrates nto three partcles. One of the partcles, of mass kg, moves along the y axs wth a speed of m/s. Another partcle, of mass kg, moves along the x axs wth a speed of m/s. Fnd (a) the velocty of the thrd partcle and (b) the total knetc energy ncrease n the process. Secton 9.5 The Center of Mass 38. Four objects are stuated along the y axs as follows: a 2.00 kg object s at 3.00 m, a 3.00-kg object s at 2.50 m, a 2.50-kg object s at the orgn, and a 4.00-kg object s at m. Where s the center of mass of these objects? 39. A water molecule conssts of an oxygen atom wth two hydrogen atoms bound to t (Fg. P9.39). The angle between the two bonds s 106. If the bonds are nm long, where s the center of mass of the molecule? H nm Fgure P Two automobles of equal mass approach an ntersecton. One vehcle s travelng wth velocty 13.0 m/s toward the east, and the other s travelng north wth speed v 2. Nether drver sees the other. The vehcles collde n the ntersecton and stck together, leavng parallel skd marks at an angle of 55.0 north of east. The speed lmt for both roads s 35 m/h, and the drver of the northward-movng vehcle clams he was wthn the speed lmt when the collson occurred. Is he tellng the truth? O nm Fgure P9.39 H

36 286 CHAPTER 9 Lnear Momentum and Collsons 40. The mass of the Earth s kg, and the mass of the Moon s kg. The dstance of separaton, measured between ther centers, s m. Locate the center of mass of the Earth Moon system as measured from the center of the Earth. 41. A unform pece of sheet steel s shaped as n Fgure P9.41. Compute the x and y coordnates of the center of mass of the pece. y(cm) Fgure P9.41 x(cm) where x s the dstance from one end, measured n meters. (a) What s the mass of the rod? (b) How far from the x 0 end s ts center of mass? 44. In the 1968 Olympc Games, Unversty of Oregon jumper Dck Fosbury ntroduced a new technque of hgh jumpng called the Fosbury flop. It contrbuted to rasng the world record by about 30 cm and s presently used by nearly every world-class jumper. In ths technque, the jumper goes over the bar face up whle archng hs back as much as possble, as n Fgure P9.44a. Ths acton places hs center of mass outsde hs body, below hs back. As hs body goes over the bar, hs center of mass passes below the bar. Because a gven energy nput mples a certan elevaton for hs center of mass, the acton of archng hs back means hs body s hgher than f hs back were straght. As a model, consder the jumper as a thn unform rod of length L. When the rod s straght, ts center of mass s at ts center. Now bend the rod n a crcular arc so that t subtends an angle of 90.0 at the center of the arc, as shown n Fgure P9.44b. In ths confguraton, how far outsde the rod s the center of mass? 42. (a) Consder an extended object whose dfferent portons have dfferent elevatons. Assume the free-fall acceleraton s unform over the object. Prove that the gravtatonal potental energy of the object Earth system s gven by U g Mgy CM where M s the total mass of the object and y CM s the elevaton of ts center of mass above the chosen reference level. (b) Calculate the gravtatonal potental energy assocated wth a ramp constructed on level ground wth stone wth densty kg/m 3 and everywhere 3.60 m wde. In a sde vew, the ramp appears as a rght trangle wth heght 15.7 m at the top end and base 64.8 m (Fgure P9.42). Eye Ubqutous/CORBIS (a) 90 (b) Fgure P9.44 Fgure P A rod of length 30.0 cm has lnear densty (mass-perlength) gven by 50.0 g/m 20.0x g/m 2, Secton 9.6 Moton of a System of Partcles 45. A 2.00-kg partcle has a velocty (2.00î 3.00ĵ) m/s, and a 3.00-kg partcle has a velocty (1.00î 6.00ĵ) m/s. Fnd (a) the velocty of the center of mass and (b) the total momentum of the system. 46. Consder a system of two partcles n the xy plane: m kg s at the locaton r 1 (1.00î 2.00ĵ) m and has a velocty of (3.00î ĵ) m/s; m kg s at r 2 ( 4.00î 3.00ĵ) m and has velocty (3.00î 2.00ĵ) m/s.

37 Problems (a) Plot these partcles on a grd or graph paper. Draw ther poston vectors and show ther veloctes. (b) Fnd the poston of the center of mass of the system and mark t on the grd. (c) Determne the velocty of the center of mass and also show t on the dagram. (d) What s the total lnear momentum of the system? Romeo (77.0 kg) entertans Julet (55.0 kg) by playng hs gutar from the rear of ther boat at rest n stll water, 2.70 m away from Julet, who s n the front of the boat. After the serenade, Julet carefully moves to the rear of the boat (away from shore) to plant a kss on Romeo s cheek. How far does the 80.0-kg boat move toward the shore t s facng? 48. A ball of mass kg has a velocty of 150î m/s; a ball of mass kg has a velocty of 0.400î m/s. They meet n a head-on elastc collson. (a) Fnd ther veloctes after the collson. (b) Fnd the velocty of ther center of mass before and after the collson. Secton Rocket Propulson The frst stage of a Saturn V space vehcle consumed fuel and oxdzer at the rate of kg/s, wth an exhaust speed of m/s. (a) Calculate the thrust produced by these engnes. (b) Fnd the acceleraton of the vehcle just as t lfted off the launch pad on the Earth f the vehcle s ntal mass was kg. Note: You must nclude the gravtatonal force to solve part (b). 50. Model rocket engnes are szed by thrust, thrust duraton, and total mpulse, among other characterstcs. A sze C5 model rocket engne has an average thrust of 5.26 N, a fuel mass of 12.7 g, and an ntal mass of 25.5 g. The duraton of ts burn s 1.90 s. (a) What s the average exhaust speed of the engne? (b) If ths engne s placed n a rocket body of mass 53.5 g, what s the fnal velocty of the rocket f t s fred n outer space? Assume the fuel burns at a constant rate. 51. A rocket for use n deep space s to be capable of boostng a total load (payload plus rocket frame and engne) of 3.00 metrc tons to a speed of m/s. (a) It has an engne and fuel desgned to produce an exhaust speed of m/s. How much fuel plus oxdzer s requred? (b) If a dfferent fuel and engne desgn could gve an exhaust speed of m/s, what amount of fuel and oxdzer would be requred for the same task? 52. Rocket Scence. A rocket has total mass M 360 kg, ncludng 330 kg of fuel and oxdzer. In nterstellar space t starts from rest, turns on ts engne at tme t 0, and puts out exhaust wth relatve speed v e m/s at the constant rate k 2.50 kg/s. The fuel wll last for an actual burn tme of 330 kg/(2.5 kg/s) 132 s, but defne a projected depleton tme as T p M /k 144 s. (Ths would be the burn tme f the rocket could use ts payload and fuel tanks as fuel, and even the walls of the combuston chamber.) (a) Show that durng the burn the velocty of the rocket s gven as a functon of tme by v(t) v e ln[1 (t/t p )] (b) Make a graph of the velocty of the rocket as a functon of tme for tmes runnng from 0 to 132 s. (c) Show that the acceleraton of the rocket s (d) Graph the acceleraton as a functon of tme. (e) Show that the poston of the rocket s (f) Graph the poston durng the burn. 53. An orbtng spacecraft s descrbed not as a zero-g, but rather as a mcrogravty envronment for ts occupants and for on-board experments. Astronauts experence slght lurches due to the motons of equpment and other astronauts, and due to ventng of materals from the craft. Assume that a kg spacecraft undergoes an acceleraton of 2.50 g m/s 2 due to a leak from one of ts hydraulc control systems. The flud s known to escape wth a speed of 70.0 m/s nto the vacuum of space. How much flud wll be lost n 1 h f the leak s not stopped? Addtonal Problems a(t) v e /(T p t) x(t) v e (T p t)ln [1 (t/t p )] v e t 54. Two glders are set n moton on an ar track. A sprng of force constant k s attached to the near sde of one glder. The frst glder, of mass m 1, has velocty v 1, and the second glder, of mass m 2, moves more slowly, wth velocty v 2, as n Fgure P9.54. When m 1 colldes wth the sprng attached to m 2 and compresses the sprng to ts maxmum compresson x max, the velocty of the glders s v. In terms of v 1, v 2, m 1, m 2, and k, fnd (a) the velocty v at maxmum compresson, (b) the maxmum compresson x max, and (c) the velocty of each glder after m 1 has lost contact wth the sprng. v 1 m 1 k v 2 Fgure P Revew problem. A 60.0-kg person runnng at an ntal speed of 4.00 m/s jumps onto a 120-kg cart ntally at rest (Fgure P9.55). The person sldes on the cart s top surface and fnally comes to rest relatve to the cart. The coeffcent of knetc frcton between the person and the cart s Frcton between the cart and ground can be neglected. (a) Fnd the fnal velocty of the person and cart relatve to the ground. (b) Fnd the frcton force actng on the person whle he s sldng across the top surface of the cart. (c) How long does the frcton force act on the person? (d) Fnd the change n momentum of the person and the change n momentum of the cart. (e) Determne the dsplacement of the person relatve to the ground whle he s sldng on the cart. (f) Determne the dsplacement of the cart relatve to the ground whle the person s sldng. (g) Fnd the change n m 2

38 288 CHAPTER 9 Lnear Momentum and Collsons knetc energy of the person. (h) Fnd the change n knetc energy of the cart. () Explan why the answers to (g) and (h) dffer. (What knd of collson s ths, and what accounts for the loss of mechancal energy?) 60.0 kg 4.00 m/s 120 kg 2.00ˆk) m/s, fnd the fnal velocty of the 1.50-kg sphere and dentfy the knd of collson. (c) What If? If the velocty of the kg sphere after the collson s ( 1.00î 3.00ĵ aˆk) m/s, fnd the value of a and the velocty of the 1.50-kg sphere after an elastc collson. 60. A small block of mass m kg s released from rest at the top of a curve-shaped frctonless wedge of mass m kg, whch sts on a frctonless horzontal surface as n Fgure P9.60a. When the block leaves the wedge, ts velocty s measured to be 4.00 m/s to the rght, as n Fgure P9.60b. (a) What s the velocty of the wedge after the block reaches the horzontal surface? (b) What s the heght h of the wedge? Fgure P9.55 m A golf ball (m 46.0 g) s struck wth a force that makes an angle of 45.0 wth the horzontal. The ball lands 200 m away on a flat farway. If the golf club and ball are n contact for 7.00 ms, what s the average force of mpact? (Neglect ar resstance.) 57. An 80.0-kg astronaut s workng on the engnes of hs shp, whch s drftng through space wth a constant velocty. The astronaut, wshng to get a better vew of the Unverse, pushes aganst the shp and much later fnds hmself 30.0 m behnd the shp. Wthout a thruster, the only way to return to the shp s to throw hs kg wrench drectly away from the shp. If he throws the wrench wth a speed of 20.0 m/s relatve to the shp, how long does t take the astronaut to reach the shp? 58. A bullet of mass m s fred nto a block of mass M ntally at rest at the edge of a frctonless table of heght h (Fg. P9.58). The bullet remans n the block, and after mpact the block lands a dstance d from the bottom of the table. Determne the ntal speed of the bullet. h m M Fgure P A kg sphere movng wth a velocty (2.00î 3.00ĵ 1.00ˆk) m/s strkes another sphere of mass 1.50 kg movng wth a velocty ( 1.00î 2.00ĵ 3.00ˆk) m/s. (a) If the velocty of the kg sphere after the collson s ( 1.00î 3.00ĵ 8.00ˆk) m/s, fnd the fnal velocty of the 1.50-kg sphere and dentfy the knd of collson (elastc, nelastc, or perfectly nelastc). (b) If the velocty of the kg sphere after the collson s ( 0.250î 0.750ĵ d h m 2 (a) v 2 Fgure P9.60 m 2 (b) 4.00 m/s 61. A bucket of mass m and volume V s attached to a lght cart, completely coverng ts top surface. The cart s gven a quck push along a straght, horzontal, smooth road. It s ranng, so as the cart cruses along wthout frcton, the bucket gradually flls wth water. By the tme the bucket s full, ts speed s v. (a) What was the ntal speed v of the cart? Let represent the densty of water. (b) What If? Assume that when the bucket s half full, t develops a slow leak at the bottom, so that the level of the water remans constant thereafter. Descrbe qualtatvely what happens to the speed of the cart after the leak develops. 62. A 75.0-kg frefghter sldes down a pole whle a constant frcton force of 300 N retards her moton. A horzontal 20.0-kg platform s supported by a sprng at the bottom of the pole to cushon the fall. The frefghter starts from rest 4.00 m above the platform, and the sprng constant s N/m. Fnd (a) the frefghter s speed just before she colldes wth the platform and (b) the maxmum dstance the sprng s compressed. (Assume the frcton force acts durng the entre moton.) 63. George of the Jungle, wth mass m, swngs on a lght vne hangng from a statonary tree branch. A second vne of equal length hangs from the same pont, and a gorlla of larger mass M swngs n the opposte drecton on t. Both vnes are horzontal when the prmates start from rest at the same moment. George and the gorlla meet at the lowest pont of ther swngs. Each s afrad that one vne wll break, so they grab each other and hang on. They swng upward together, reachng a pont where the vnes make an angle of 35.0 wth the vertcal. (a) Fnd the value of the rato m/m. (b) What If? Try ths at home. Te a small magnet and a steel screw to opposte ends of a strng. Hold the cen-

39 Problems 289 ter of the strng fxed to represent the tree branch, and reproduce a model of the motons of George and the gorlla. What changes n your analyss wll make t apply to ths stuaton? What If? Assume the magnet s strong, so that t notceably attracts the screw over a dstance of a few centmeters. Then the screw wll be movng faster just before t stcks to the magnet. Does ths make a dfference? 64. A cannon s rgdly attached to a carrage, whch can move along horzontal rals but s connected to a post by a large sprng, ntally unstretched and wth force constant k N/m, as n Fgure P9.64. The cannon fres a 200-kg projectle at a velocty of 125 m/s drected 45.0 above the horzontal. (a) If the mass of the cannon and ts carrage s kg, fnd the recol speed of the cannon. (b) Determne the maxmum extenson of the sprng. (c) Fnd the maxmum force the sprng exerts on the carrage. (d) Consder the system consstng of the cannon, carrage, and shell. Is the momentum of ths system conserved durng the frng? Why or why not? through the relaton v 1A What numercal value does she obtan for v 1A based on her measured values of x 257 cm and y 85.3 cm? What factors mght account for the dfference n ths value compared to that obtaned n part (a)? 66. Small ce cubes, each of mass 5.00 g, slde down a frctonless track n a steady stream, as shown n Fgure P9.66. Startng from rest, each cube moves down through a net vertcal dstance of 1.50 m and leaves the bottom end of the track at an angle of 40.0 above the horzontal. At the hghest pont of ts subsequent trajectory, the cube strkes a vertcal wall and rebounds wth half the speed t had upon mpact. If 10.0 cubes strke the wall per second, what average force s exerted on the wall? x 2y/g m 40.0 Fgure P A student performs a ballstc pendulum experment usng an apparatus smlar to that shown n Fgure 9.11b. She obtans the followng average data: h 8.68 cm, m g, and m g. The symbols refer to the quanttes n Fgure 9.11a. (a) Determne the ntal speed v 1A of the projectle. (b) The second part of her experment s to obtan v 1A by frng the same projectle horzontally (wth the pendulum removed from the path), by measurng ts fnal horzontal poston x and dstance of fall y (Fg. P9.65). Show that the ntal speed of the projectle s related to x and y Fgure P A 5.00-g bullet movng wth an ntal speed of 400 m/s s fred nto and passes through a 1.00-kg block, as n Fgure P9.67. The block, ntally at rest on a frctonless, horzontal surface, s connected to a sprng wth force constant 900 N/m. If the block moves 5.00 cm to the rght after mpact, fnd (a) the speed at whch the bullet emerges from the block and (b) the mechancal energy converted nto nternal energy n the collson. 400 m/s v 1A 5.00 cm v y Fgure P9.67 x Fgure P Consder as a system the Sun wth the Earth n a crcular orbt around t. Fnd the magntude of the change n the velocty of the Sun relatve to the center of mass of the

40 290 CHAPTER 9 Lnear Momentum and Collsons system over a perod of 6 months. Neglect the nfluence of other celestal objects. You may obtan the necessary astronomcal data from the endpapers of the book. 69. Revew problem. There are (one can say) three coequal theores of moton: Newton s second law, statng that the total force on an object causes ts acceleraton; the work knetc energy theorem, statng that the total work on an object causes ts change n knetc energy; and the mpulse momentum theorem, statng that the total mpulse on an object causes ts change n momentum. In ths problem, you compare predctons of the three theores n one partcular case. A 3.00-kg object has velocty 7.00ĵ m/s. Then, a total force 12.0î N acts on the object for 5.00 s. (a) Calculate the object s fnal velocty, usng the mpulse momentum theorem. (b) Calculate ts acceleraton from a (v f v )/t. (c) Calculate ts acceleraton from a F/m. (d) Fnd the object s vector dsplacement from r v t 1. 2 at 2 (e) Fnd the work done on the object from W Fr. 1 (f) Fnd the fnal knetc energy from 1 (g) Fnd the fnal knetc energy from 2 mv 2 mv 2 f mv f v f. W. 70. A rocket has total mass M 360 kg, ncludng 330 kg of fuel and oxdzer. In nterstellar space t starts from rest. Its engne s turned on at tme t 0, and t puts out exhaust wth relatve speed v e m/s at the constant rate 2.50 kg/s. The burn lasts untl the fuel runs out, at tme 330 kg/(2.5 kg/s) 132 s. Set up and carry out a computer analyss of the moton accordng to Euler s method. Fnd (a) the fnal velocty of the rocket and (b) the dstance t travels durng the burn. 71. A chan of length L and total mass M s released from rest wth ts lower end just touchng the top of a table, as n Fgure P9.71a. Fnd the force exerted by the table on the chan after the chan has fallen through a dstance x, as n Fgure P9.71b. (Assume each lnk comes to rest the nstant t reaches the table.) (a) L Fgure P9.71 (b) x L x 72. Sand from a statonary hopper falls onto a movng conveyor belt at the rate of 5.00 kg/s as n Fgure P9.72. The conveyor belt s supported by frctonless rollers and moves at a constant speed of m/s under the acton of a constant horzontal external force F ext suppled by the motor that drves the belt. Fnd (a) the sand s rate of change of momentum n the horzontal drecton, (b) the force of frcton exerted by the belt on the sand, (c) the external force F ext, (d) the work done by F ext n 1 s, and (e) the knetc energy acqured by the fallng sand each second due to the change n ts horzontal moton. (f ) Why are the answers to (d) and (e) dfferent? 73. A golf club conssts of a shaft connected to a club head. The golf club can be modeled as a unform rod of length and mass m 1 extendng radally from the surface of a sphere of radus R and mass m 2. Fnd the locaton of the club s center of mass, measured from the center of the club head. Answers to Quck Quzzes 9.1 (d). Two dentcal objects (m 1 m 2 ) travelng at the same speed (v 1 v 2 ) have the same knetc energes and the same magntudes of momentum. It also s possble, however, for partcular combnatons of masses and veloctes to satsfy K 1 K 2 but not p 1 p 2. For example, a 1-kg object movng at 2 m/s has the same knetc energy as a 4-kg object movng at 1 m/s, but the two clearly do not have the same momenta. Because we have no nformaton about masses and speeds, we cannot choose among (a), (b), or (c). 9.2 (b), (c), (a). The slower the ball, the easer t s to catch. If the momentum of the medcne ball s the same as the momentum of the baseball, the speed of the medcne ball must be 1/10 the speed of the baseball because the medcne ball has 10 tmes the mass. If the knetc energes are the same, the speed of the medcne ball must be 1/ 10 the speed of the baseball because of the squared speed term n the equaton for K. The medcne ball s hardest to catch when t has the same speed as the baseball. 9.3 (c). The ball and the Earth exert forces on each other, so nether s an solated system. We must nclude both n the system so that the nteracton force s nternal to the system. 9.4 (c). From Equaton 9.4, f p 1 p 2 constant, then t follows that p 1 p 2 0 and p 1 p 2. Whle the change n momentum s the same, the change n the velocty s a lot larger for the car! 9.5 (c) and (e). Object 2 has a greater acceleraton because of ts smaller mass. Therefore, t takes less tme to travel the dstance d. Even though the force appled to objects 1 and 2 s the same, the change n momentum s less for object 2 because t s smaller. The work W Fd done on m/s F ext Fgure P9.72

41 Problems 291 both objects s the same because both F and d are the same n the two cases. Therefore, K 1 K (b) and (d). The same mpulse s appled to both objects, so they experence the same change n momentum. Object 2 has a larger acceleraton due to ts smaller mass. Thus, the dstance that object 2 covers n the tme nterval t s larger than that for object 1. As a result, more work s done on object 2 and K 2 K (a) All three are the same. Because the passenger s brought from the car s ntal speed to a full stop, the change n momentum (equal to the mpulse) s the same regardless of what stops the passenger. (b) Dashboard, seatbelt, arbag. The dashboard stops the passenger very quckly n a front-end collson, resultng n a very large force. The seatbelt takes somewhat more tme, so the force s smaller. Used along wth the seatbelt, the arbag can extend the passenger s stoppng tme further, notably for hs head, whch would otherwse snap forward. 9.8 (a). If all of the ntal knetc energy s transformed, then nothng s movng after the collson. Consequently, the fnal momentum of the system s necessarly zero and, therefore, the ntal momentum of the system must be zero. Whle (b) and (d) together would satsfy the condtons, nether one alone does. 9.9 (b). Because momentum of the two-ball system s conserved, p T 0 p Tf p B. Because the table-tenns ball bounces back from the much more massve bowlng ball wth approxmately the same speed, p Tf p T. As a consequence, p B 2p T. Knetc energy can be expressed as K p 2 /2m. Because of the much larger mass of the bowlng ball, ts knetc energy s much smaller than that of the table-tenns ball (b). The pece wth the handle wll have less mass than the pece made up of the end of the bat. To see why ths s so, take the orgn of coordnates as the center of mass before the bat was cut. Replace each cut pece by a small sphere located at the center of mass for each pece. The sphere representng the handle pece s farther from the orgn, but the product of less mass and greater dstance balances the product of greater mass and less dstance for the end pece: 9.11 (a). Ths s the same effect as the swmmer dvng off the raft that we just dscussed. The vessel passengers system s solated. If the passengers all start runnng one way, the speed of the vessel ncreases (a small amount!) the other way (b). Once they stop runnng, the momentum of the system s the same as t was before they started runnng you cannot change the momentum of an solated system by means of nternal forces. In case you are thnkng that the passengers could do ths over and over to take advantage of the speed ncrease whle they are runnng, remember that they wll slow the shp down every tme they return to the bow!