CENTER OF MASS AND LINEAR MOMENTUM

Transcription

1 CHAPTER CENTER OF MASS 9 AND LINEAR MOMENTUM 9-1 WHAT IS PHYSICS? Ever mechanical engineer hired as an eper winess o reconsruc a raffic acciden uses phsics. Ever rainer who coaches a ballerina on how o leap uses phsics. Indeed, analzing complicaed moion of an sor requires simplificaion via an undersanding of phsics. In his chaper we discuss how he complicaed moion of a ssem of objecs, such as a car or a ballerina, can be simplified if we deermine a special poin of he ssem he cener of mass of ha ssem. Here is a quick eample. If ou oss a ball ino he air wihou much spin on he ball (Fig. 9-1a), is moion is simple i follows a parabolic pah, as we discussed in Chaper 4, and he ball can be reaed as a paricle. If, insead, ou flip a baseball ba ino he air (Fig. 9-1b), is moion is more complicaed. Because ever par of he ba moves differenl, along pahs of man differen shapes, ou canno represen he ba as a paricle. Insead, i is a ssem of paricles each of which follows is own pah hrough he air. However, he ba has one special poin he cener of mass ha does move in a simple parabolic pah. The oher pars of he ba move around he cener of mass. (To locae he cener of mass, balance he ba on an ousreched finger; he poin is above our finger, on he ba s cenral ais.) You canno make a career of flipping baseball bas ino he air, bu ou can make a career of advising long-jumpers or dancers on how o leap properl ino he air while eiher moving heir arms and legs or roaing heir orso. Your saring poin would be he person s cener of mass because of is simple moion. (a) 9-2 The Cener of Mass We define he cener of mass (com) of a ssem of paricles (such as a person) in order o predic he possible moion of he ssem. The cener of mass of a ssem of paricles is he poin ha moves as hough (1) all of he ssem s mass were concenraed here and (2) all eernal forces were applied here. In his secion we discuss how o deermine where he cener of mass of a ssem of paricles is locaed. We sar wih a ssem of onl a few paricles, and hen we consider a ssem of a grea man paricles (a solid bod, such as a baseball ba). Laer in he chaper, we discuss how he cener of mass of a ssem moves when eernal forces ac on he ssem. (b) Fig. 9-1 (a) A ball ossed ino he air follows a parabolic pah. (b) The cener of mass (black do) of a baseball ba flipped ino he air follows a parabolic pah, bu all oher poins of he ba follow more complicaed curved pahs. (a Richard Megna/Fundamenal Phoographs) 201

2 202 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM Ssems of Paricles Figure 9-2a shows wo paricles of masses m 1 and m 2 separaed b disance d.we have arbiraril chosen he origin of an ais o coincide wih he paricle of mass m 1.We define he posiion of he cener of mass (com) of his wo-paricle ssem o be com m 2 m 1 m 2 d. (9-1) Suppose, as an eample, ha m 2 0. Then here is onl one paricle, of mass m 1, and he cener of mass mus lie a he posiion of ha paricle; Eq. 9-1 duifull reduces o com 0. If m 1 0, here is again onl one paricle (of mass m 2 ), and we have, as we epec, com d.if m 1 m 2,he cener of mass should be halfwa beween he wo paricles; Eq. 9-1 reduces o com 1 2 d, again as we epec. Finall, Eq. 9-1 ells us ha if neiher m 1 nor m 2 is zero, com can have onl values ha lie beween zero and d;ha is, he cener of mass mus lie somewhere beween he wo paricles. Figure 9-2b shows a more generalized siuaion, in which he coordinae ssem has been shifed lefward.the posiion of he cener of mass is now defined as com m 1 1 m 2 2. (9-2) m 1 m 2 Noe ha if we pu 1 0, hen 2 becomes d and Eq. 9-2 reduces o Eq. 9-1, as i mus. Noe also ha in spie of he shif of he coordinae ssem, he cener of mass is sill he same disance from each paricle. We can rewrie Eq. 9-2 as com m 1 1 m 2 2 M (9-3) in which M is he oal mass of he ssem. (Here, M m 1 m 2.) We can eend his equaion o a more general siuaion in which n paricles are srung ou along he ais.then he oal mass is M m 1 m 2... m n,and he locaion ofhe cener of mass is com m 1 1 m 2 2 m 3 3 m n n M, 1 M n m i i. i 1 The subscrip i is an inde ha akes on all ineger values from 1 o n. (9-4) com This is he cener of mass of he wo-paricle ssem. com m 1 m 2 com d (a) m 1 m 2 com 1 d 2 (b) Shifing he ais does no change he relaive posiion of he com. Fig. 9-2 (a) Two paricles of masses m 1 and m 2 are separaed b disance d.the do labeled com shows he posiion of he cener of mass, calculaed from Eq (b) The same as (a) ecep ha he origin is locaed farher from he paricles.the posiion of he cener of mass is calculaed from Eq. 9-2.The locaion of he cener of mass wih respec o he paricles is he same in boh cases.

3 9-2 THE CENTER OF MASS 203 PART 1 If he paricles are disribued in hree dimensions, he cener of mass mus be idenified b hree coordinaes. B eension of Eq. 9-4, he are com 1 M n m i i, com 1 i 1 M n m i i, z com 1 i 1 M n m i z i. i 1 (9-5) We can also define he cener of mass wih he language of vecors. Firs recall ha he posiion of a paricle a coordinaes i, i,and z i is given b a posiion vecor r i i î i ĵ z i kˆ. (9-6) Here he inde idenifies he paricle, and î, ĵ, and ˆk are uni vecors poining, respecivel, in he posiive direcion of he,,and z aes. Similarl, he posiion of he cener of mass of a ssem of paricles is given b a posiion vecor r com com î com ĵ z com ˆk. (9-7) The hree scalar equaions of Eq. 9-5 can now be replaced b a single vecor equaion, r com 1 M n m i r i, i 1 (9-8) where again M is he oal mass of he ssem. You can check ha his equaion is correc b subsiuing Eqs. 9-6 and 9-7 ino i, and hen separaing ou he,,and z componens.the scalar relaions of Eq. 9-5 resul. Solid Bodies An ordinar objec, such as a baseball ba, conains so man paricles (aoms) ha we can bes rea i as a coninuous disribuion of maer. The paricles hen become differenial mass elemens dm, he sums of Eq.9-5 become inegrals, and he coordinaes of he cener of mass are defined as com 1 M dm, com 1 M dm, z com 1 M z dm, (9-9) where M is now he mass of he objec. Evaluaing hese inegrals for mos common objecs (such as a elevision se or a moose) would be difficul, so here we consider onl uniform objecs. Such objecs have uniform densi, or mass per uni volume; ha is, he densi r (Greek leer rho) is he same for an given elemen of an objec as for he whole objec. From Eq. 1-8, we can wrie dm dv M V, (9-10) where dv is he volume occupied b a mass elemen dm,and V is he oal volume of he objec. Subsiuing dm (M/V) dv from Eq ino Eq. 9-9 gives com 1 V dv, com 1 V dv, z com 1 V z dv. (9-11) You can bpass one or more of hese inegrals if an objec has a poin, a line, or a plane of smmer. The cener of mass of such an objec hen lies a ha poin, on ha line, or in ha plane. For eample, he cener of mass of a uniform sphere (which has a poin of smmer) is a he cener of he sphere (which is he poin of smmer). The cener of mass of a uniform cone (whose ais is a line of smmer) lies on he ais of he cone. The cener of mass of a banana

4 204 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM (which has a plane of smmer ha splis i ino wo equal pars) lies somewhere in he plane of simmer. The cener of mass of an objec need no lie wihin he objec. There is no dough a he com of a doughnu, and no iron a he com of a horseshoe. Sample Problem com of plae wih missing piece Figure 9-3a shows a uniform meal plae P of radius 2R from which a disk of radius R has been samped ou (removed) in an assembl line. The disk is shown in Fig. 9-3b. Using he coordinae ssem shown, locae he cener of mass com P of he remaining plae. KEY IDEAS (1) Le us roughl locae he cener of plae P b using smmer. We noe ha he plae is smmeric abou he ais (we ge he porion below ha ais b roaing he upper porion abou he ais). Thus, com P mus be on he ais. The plae (wih he disk removed) is no smmeric abou he ais. However, because here is somewha more mass on he righ of he ais, com P mus be somewha o he righ of ha ais. Thus, he locaion of com P should be roughl as indicaed in Fig. 9-3a.Our job here is o find he acual value of ha locaion. (2) Plae P is an eended solid bod, so in principle we can use Eqs o find he acual coordinaes of he cener of mass of plae P.Here we are simpl looking for he coordinaes of he cener of mass because he plae is hin and uniform. If i had an appreciable hickness, we would jus sa ha he cener of mass is midwa across he hickness. Sill, even neglecing he widh, using Eqs would be challenging because we would need a funcion for he shape of he plae wih is hole, and hen we would need o inegrae he funcion in wo dimensions. (3) Here is a much easier wa In working wih ceners of mass, we can assume ha he mass of a uniform objec (as we have here) is concenraed in a paricle a he objec s cener of mass.thus we can rea he objec as a paricle and avoid an wo-dimensional inegraion. Calculaions Firs, pu he samped-ou disk (call i disk S) back ino place (Fig. 9-3c) o form he original composie plae (call i plae C). Because of is circular smmer, he cener of mass com S for disk S is a he cener of S, a R (as shown). Similarl, he cener of mass com C for composie plae C is a he cener of C,a he origin (as shown). We hen have he following Cener Locaion Plae of Mass of com Mass P com P P? m P S com S S R m S C com C C 0 m C m S m P Assume ha mass m S of disk S is concenraed in a paricle a S R, andmass m P is concenraed in a paricle a P (Fig. 9-3d). Ne rea hese wo paricles as a woparicle ssem, using Eq. 9-2 o find heir cener of mass S P.We ge (9-12) Ne noe ha he combinaion of disk S and plae P is composie plae C. Thus, he posiion S P of com S P mus coincide wih he posiion C of com C,which is a he origin;so S P C 0. Subsiuing his ino Eq and solving for P,we ge (9-13) We can relae hese masses o he face areas of S and P b noing ha Then mass densi volume densi hickness area. m S m P Because he plae is uniform, he densiies and hicknesses are equal; we are lef wih m S m P S P m S S m P P m S m P. densi S densi P area S area P P S m S m P. hickness S hickness P R 2 (2R) 2 R Subsiuing his and S R ino Eq. 9-13, we have P 1 3 R. area S area C area S area S area P. (Answer) Addiional eamples, video, and pracice available a WilePLUS

5 9-2 THE CENTER OF MASS 205 PART 1 A 2R R com P (a) Plae P Assume he plae's mass is concenraed as a paricle a he plae's cener of mass. Disk S Here oo, assume he mass is concenraed as a paricle a he cener of mass. com S (b) Composie plae C = S + P Here oo. com C (c) Fig. 9-3 (a) Plae P is a meal plae of radius 2R,wih a circular hole of radius R. The cener of mass of P is a poin com P. (b) Disk S. (c) Disk S has been pu back ino place o form a composie plae C. The cener of mass com S of disk S and he cener of mass com C of plae C are shown. (d) The cener of mass com S P of he combinaion of S and P coincides wih com C, which is a 0. (d) com S Disk paricle com C com P Plae paricle The com of he composie plae is he same as he com of he wo pieces. Here are hose hree paricles.

6 206 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM Sample Problem com of hree paricles Three paricles of masses m kg, m kg, and m kg form an equilaeral riangle of edge lengh a 140 cm.where is he cener of mass of his ssem? KEY IDEA We are dealing wih paricles insead of an eended solid bod, so we can use Eq. 9-5 o locae heir cener of mass. The paricles are in he plane of he equilaeral riangle, so we need onl he firs wo equaions. Calculaions We can simplif he calculaions b choosing he and aes so ha one of he paricles is locaed a he com 0 0 a r com m 3 m 2 m 1 50 com a This is he posiion vecor r com for he com (i poins from he origin o he com). Fig. 9-4 Three paricles form an equilaeral riangle of edge lengh a.the cener of mass is locaed b he posiion vecor rcom. origin and he ais coincides wih one of he riangle s sides (Fig. 9-4). The hree paricles hen have he following coordinaes Paricle Mass (kg) (cm) (cm) The oal mass M of he ssem is 7.1 kg. From Eq. 9-5, he coordinaes of he cener of mass are com 83 cm and com (Answer) In Fig. 9-4, he cener of mass is locaed b he posiion vecor r, which has componens com and com. com 1 M 3 m i i m 1 1 m 2 2 m 3 3 i 1 M (1.2 kg)(0) (2.5 kg)(140 cm) (3.4 kg)(70 cm) 7.1 kg 1 M 3 m i i m 1 1 m 2 2 m 3 3 i 1 M (1.2 kg)(0) (2.5 kg)(0) (3.4 kg)(120 cm) 7.1 kg 58 cm. (Answer) Addiional eamples, video, and pracice available a WilePLUS CHECKPOINT 1 The figure shows a uniform square plae from which four idenical squares a he corners will be removed. (a) Where is he cener of mass of he plae originall? Where is i afer he removal of (b) square 1; (c) squares 1 and 2; (d) squares 1 and 3; (e) squares 1, 2, and 3; (f) all four squares? Answer in erms of quadrans, aes, or poins (wihou calculaion, of course). 9-3 Newon s Second Law for a Ssem of Paricles Now ha we know how o locae he cener of mass of a ssem of paricles, we discuss how eernal forces can move a cener of mass. Le us sar wih a simple ssem of wo billiard balls. If ou roll a cue ball a a second billiard ball ha is a res, ou epec ha he wo-ball ssem will coninue o have some forward moion afer impac. You would be surprised, for eample, if boh balls came back oward ou or if boh moved o he righ or o he lef. Wha coninues o move forward, is sead moion compleel unaffeced b he collision, is he cener of mass of he wo-ball ssem. If ou focus on his poin which is alwas halfwa beween hese bodies because he have ideni-

7 9-3 NEWTON S SECOND LAW FOR A SYSTEM OF PARTICLES 207 PART 1 cal masses ou can easil convince ourself b rial a a billiard able ha his is so. No maer wheher he collision is glancing, head-on, or somewhere in beween, he cener of mass coninues o move forward, as if he collision had never occurred. Le us look ino his cener-of-mass moion in more deail. To do so, we replace he pair of billiard balls wih an assemblage of n paricles of (possibl) differen masses. We are ineresed no in he individual moions of hese paricles bu onl in he moion of he cener of mass of he assemblage. Alhough he cener of mass is jus a poin, i moves like a paricle whose mass is equal o he oal mass of he ssem; we can assign a posiion, a veloci, and an acceleraion o i. We sae (and shall prove ne) ha he vecor equaion ha governs he moion of he cener of mass of such a ssem of paricles is (ssem of paricles). (9-14) This equaion is Newon s second law for he moion of he cener of mass of a ssem of paricles. Noe ha is form is he same as he form of he equaion (F ne ma ) for he moion of a single paricle. However, he hree quaniies ha appear in Eq mus be evaluaed wih some care 1. is he ne force of all eernal forces ha ac on he ssem. Forces on one par of he ssem from anoher par of he ssem (inernal forces) are no included in Eq M is he oal mass of he ssem. We assume ha no mass eners or leaves he ssem as i moves, so ha M remains consan. The ssem is said o be closed. 3. a com is he acceleraion of he cener of mass of he ssem. Equaion 9-14 gives no informaion abou he acceleraion of an oher poin of he ssem. F ne F ne Equaion 9-14 is equivalen o hree equaions involving he componens of and a along he hree coordinae aes. These equaions are com F ne Ma com The inernal forces of he eplosion canno change he pah of he com. Fig. 9-5 A fireworks rocke eplodes in fligh. In he absence of air drag, he cener of mass of he fragmens would coninue o follow he original parabolic pah, unil fragmens began o hi he ground. F ne, Ma com, F ne, Ma com, F ne, z Ma com, z. (9-15) Now we can go back and eamine he behavior of he billiard balls. Once he cue ball has begun o roll, no ne eernal force acs on he (wo-ball) ssem. Thus, because F ne 0, Eq ells us ha a com 0 also. Because acceleraion is he rae of change of veloci, we conclude ha he veloci of he cener of mass of he ssem of wo balls does no change. When he wo balls collide, he forces ha come ino pla are inernal forces, on one ball from he oher. Such forces do no conribue o he ne force F ne, which remains zero. Thus, he cener of mass of he ssem, which was moving forward before he collision, mus coninue o move forward afer he collision, wih he same speed and in he same direcion. Equaion 9-14 applies no onl o a ssem of paricles bu also o a solid bod, such as he ba of Fig. 9-1b.In ha case,m in Eq is he mass of he ba and F ne is he graviaional force on he ba. Equaion 9-14 hen ells us ha a com g. In oher words, he cener of mass of he ba moves as if he ba were a single paricle of mass M,wih force F g acing on i. Figure 9-5 shows anoher ineresing case. Suppose ha a a fireworks displa, a rocke is launched on a parabolic pah. A a cerain poin, i eplodes ino fragmens. If he eplosion had no occurred, he rocke would have coninued along he rajecor shown in he figure. The forces of he eplosion are inernal o he ssem (a firs he ssem is jus he rocke, and laer i is is fragmens); ha is, he are forces on pars of he ssem from oher pars. If we ignore air drag, he ne eernal force F ne acing on he ssem is he graviaional force on he ssem, regardless of wheher he rocke eplodes. Thus, from Eq. 9-14, he acceleraion a com of he cener of mass of he fragmens (while he are in fligh) remains equal o g. This means ha he cener of mass of he fragmens follows he same parabolic rajecor ha he rocke would have followed had i no eploded.

8 208 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM Pah of head Pah of cener of mass Fig. 9-6 A grand jeé. (Adaped from The Phsics of Dance, b Kenneh Laws, Schirmer Books, 1984.) When a balle dancer leaps across he sage in a grand jeé, she raises her arms and sreches her legs ou horizonall as soon as her fee leave he sage (Fig. 9-6). These acions shif her cener of mass upward hrough her bod.alhough he shifing cener of mass faihfull follows a parabolic pah across he sage, is movemen relaive o he bod decreases he heigh ha is aained b her head and orso, relaive o ha of a normal jump. The resul is ha he head and orso follow a nearl horizonal pah, giving an illusion ha he dancer is floaing. Proof of Equaion 9-14 Now le us prove his imporan equaion. From Eq. 9-8 we have, for a ssem of n paricles, Mr com m 1 r 1 m 2 r 2 m 3 r 3 m n r n, (9-16) in which M is he ssem s oal mass and r com is he vecor locaing he posiion of he ssem s cener of mass. Differeniaing Eq wih respec o ime gives Mv com m 1 v 1 m 2 v 2 m 3 v 3 m n v n. (9-17) Here v is he veloci of he ih paricle, and v com ( dr i ( dr i /d) com/d) is he veloci of he cener of mass. Differeniaing Eq wih respec o ime leads o Ma com m 1 a 1 m 2 a 2 m 3 a 3 m n a n. (9-18) Here a i ( dv i/d) is he acceleraion of he ih paricle, and a com ( dv com /d) is he acceleraion of he cener of mass. Alhough he cener of mass is jus a geomerical poin, i has a posiion, a veloci, and an acceleraion, as if i were a paricle. From Newon s second law, m i a i is equal o he resulan force F i ha acs on he ih paricle.thus, we can rewrie Eq as Ma com F 1 F2 F3 Fn. (9-19) Among he forces ha conribue o he righ side of Eq will be forces ha he paricles of he ssem eer on each oher (inernal forces) and forces eered on he paricles from ouside he ssem (eernal forces). B Newon s hird law, he inernal forces form hird-law force pairs and cancel ou in he sum ha appears on he righ side of Eq Wha remains is he vecor sum of all he eernal forces ha ac on he ssem. Equaion 9-19 hen reduces o Eq. 9-14, he relaion ha we se ou o prove.

9 9-3 NEWTON S SECOND LAW FOR A SYSTEM OF PARTICLES 209 PART 1 CHECKPOINT 2 Two skaers on fricionless ice hold opposie ends of a pole of negligible mass. An ais runs along i, wih he origin a he cener of mass of he wo-skaer ssem. One skaer, Fred, weighs wice as much as he oher skaer, Ehel. Where do he skaers mee if (a) Fred pulls hand over hand along he pole so as o draw himself o Ehel, (b) Ehel pulls hand over hand o draw herself o Fred, and (c) boh skaers pull hand over hand? Sample Problem Moion of he com of hree paricles The hree paricles in Fig. 9-7a are iniiall a res. Each eperiences an eernal force due o bodies ouside he hree-paricle ssem. The direcions are indicaed, and he magniudes are F N, F 2 12 N, and F 3 14 N. Wha is he acceleraion of he cener of mass of he ssem, and in wha direcion does i move? KEY IDEAS The posiion of he cener of mass is marked b a do in he figure.we can rea he cener of mass as if i were a real paricle, wih a mass equal o he ssem s oal mass M 16 kg. Fig. 9-7 (a) Three paricles, iniiall a res in he posiions shown, are aced on b he eernal forces shown.the cener of mass (com) of he ssem is marked. (b) The forces are now ransferred o he cener of mass of he ssem, which behaves like a paricle wih a mass M equal o he oal mass of he ssem.the ne eernal force F ne and he acceleraion acom of he cener of mass are shown. F 1 F kg 45 2 com 8.0 kg kg 2 F The com of he ssem 3 3 will move as if all he (a) mass were here and he ne force aced here. F 2 Fne 3 M = 16 kg 2 F 1 θ a com 1 com F (b) We can also rea he hree eernal forces as if he ac a he cener of mass (Fig. 9-7b). Calculaions We can now appl Newon s second law (F ne ma ) o he cener of mass, wriing or (9-20) so a com F 1 F2 F3. (9-21) M Equaion 9-20 ells us ha he acceleraion of he cener of mass is in he same direcion as he ne eernal force F ne on he ssem (Fig. 9-7b). Because he paricles are iniiall a res, he cener of mass mus also be a res. As he cener of mass hen begins o accelerae, i mus move off in he common direcion of a and F com ne. We can evaluae he righ side of Eq direcl on avecor-capable calculaor,or we can rewrie Eq.9-21 in componen form, find he componens of a com, and hen find a Along he ais, we have com. a com, F 1 F 2 F 3 M 6.0 N (12 N) cos N 16 kg Along he ais, we have F ne Ma com F 1 F2 F3 Ma com a com, F 1 F 2 F 3 M 0 (12 N) sin kg From hese componens, we find ha a a com 2(a com, ) 2 (a com, ) 2 has he magniude 1.16 m/s m/s 2 (Answer) and he angle (from he posiive direcion of he ais) an 1 a com, a com, m/s 2. com 27. a com 1.03 m/s 2. (Answer) Addiional eamples, video, and pracice available a WilePLUS

10 210 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM 9-4 Linear Momenum In his secion, we discuss onl a single paricle insead of a ssem of paricles, in order o define wo imporan quaniies. Then in Secion 9-5, we eend hose definiions o ssems of man paricles. The firs definiion concerns a familiar word momenum ha has several meanings in everda language bu onl a single precise meaning in phsics and engineering. The linear momenum of a paricle is a vecor quani p ha is defined as p mv (linear momenum of a paricle), (9-22) in which m is he mass of he paricle and v is is veloci. (The adjecive linear is ofen dropped, bu i serves o disinguish p from angular momenum, which is inroduced in Chaper 11 and which is associaed wih roaion.) Since m is alwas a posiive scalar quani, Eq ells us ha p and v have he same direcion. From Eq. 9-22, he SI uni for momenum is he kilogram-meer per second (kg m/s). Newon epressed his second law of moion in erms of momenum The ime rae of change of he momenum of a paricle is equal o he ne force acing on he paricle and is in he direcion of ha force. In equaion form his becomes F ne dp d. (9-23) In words, Eq sas ha he ne eernal force F ne on a paricle changes he paricle s linear momenum p. Conversel, he linear momenum can be changed onl b a ne eernal force. If here is no ne eernal force, p canno change. As we shall see in Secion 9-7, his las fac can be an eremel powerful ool in solving problems. Manipulaing Eq b subsiuing for p from Eq gives, for consan mass m, F ne dp d d d (mv ) m dv d ma. Thus, he relaions F and F ne dp /d ne ma are equivalen epressions of Newon s second law of moion for a paricle. CHECKPOINT 3 The figure gives he magniude p of he linear momenum versus ime for a paricle moving along an ais.a force direced along he ais acs on he paricle.(a) Rank he four regions indicaed according o he magniude of he force, greaes firs. (b) In which region is he paricle slowing? p

11 9-6 COLLISION AND IMPULSE 211 PART The Linear Momenum of a Ssem of Paricles Le s eend he definiion of linear momenum o a ssem of paricles. Consider a ssem of n paricles, each wih is own mass, veloci, and linear momenum. The paricles ma inerac wih each oher, and eernal forces ma ac on hem. The ssem as a whole has a oal linear momenum P, which is defined o be he vecor sum of he individual paricles linear momena. Thus, P p 1 p 2 p 3 p n m 1 v 1 m 2 v 2 m 3 v 3 m n v n. If we compare his equaion wih Eq. 9-17, we see ha (9-24) P Mv com (linear momenum, ssem of paricles), (9-25) which is anoher wa o define he linear momenum of a ssem of paricles The linear momenum of a ssem of paricles is equal o he produc of he oal mass M of he ssem and he veloci of he cener of mass. If we ake he ime derivaive of Eq. 9-25, we find dp d M dv com d Ma com. (9-26) Comparing Eqs and 9-26 allows us o wrie Newon s second law for a ssem of paricles in he equivalen form F ne dp d (ssem of paricles), (9-27) where F ne is he ne eernal force acing on he ssem. This equaion is he generalizaion of he single-paricle equaion F ne dp /d o a ssem of man paricles. In words, he equaion sas ha he ne eernal force F on a ssem of paricles changes he linear momenum P ne of he ssem. Conversel, he linear momenum can be changed onl b a ne eernal force. If here is no ne eernal force, canno change. P 9-6 Collision and Impulse The momenum of an paricle-like bod canno change unless a ne eernal force changes i. For eample, we could push on he bod o change is momenum. More dramaicall, we could arrange for he bod o collide wih a baseball ba. In such a collision (or crash), he eernal force on he bod is brief, has large magniude, and suddenl changes he bod s momenum. Collisions occur commonl in our world, bu before we ge o hem, we need o consider a simple collision in which a moving paricle-like bod (a projecile) collides wih some oher bod (a arge). Single Collision p Le he projecile be a ball and he arge be a ba.the collision is brief, and he ball eperiences a force ha is grea enough o slow, sop, or even reverse is moion. Figure 9-8 depics he collision a one insan. The ball eperiences a force F () ha The collision of a ball wih a ba collapses par of he ball. (Phoo b Harold E. Edgeron. The Harold and Esher Edgeron Famil Trus, coures of Palm Press, Inc.) Ba F () Ball Fig. 9-8 Force F () acs on a ball as he ball and a ba collide.

12 212 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM F avg F F i i The impulse in he collision is equal o he area under he curve. J (a) F() Fig. 9-9 (a) The curve shows he magniude of he ime-varing force F() ha acs on he ball in he collision of Fig. 9-8.The area under he curve is equal o he magniude of he impulse J on he ball in he colli- sion. (b) The heigh of he recangle represens he average force F avg acing on he ball over he ime inerval.the area wihin he recangle is equal o he area under he curve in (a) and hus is also equal o he magniude of he impulse J in he collision. f The average force gives he same area under he curve. (b) J f varies during he collision and changes he linear momenum of he ball. Tha change is relaed o he force b Newon s second law wrien in he form F dp /d. Thus, in ime inerval d,he change in he ball s momenum is (9-28) We can find he ne change in he ball s momenum due o he collision if we inegrae boh sides of Eq from a ime i jus before he collision o a ime f jus afer he collision (9-29) The lef side of his equaion gives us he change in momenum p f p i p. The righ side, which is a measure of boh he magniude and he duraion of he collision force, is called he impulse J of he collision (impulse defined). (9-30) Thus, he change in an objec s momenum is equal o he impulse on he objec p J This epression can also be wrien in he vecor form and in such componen forms as f i dp F () d. dp f F () d. i f J F () d i (linear momenum impulse heorem). (9-31) p f p i J (9-32) p J (9-33) f and p f p i F d. (9-34) If we have a funcion for F (), we can evaluae J (and hus he change in momenum) b inegraing he funcion. If we have a plo of F versus ime,we can evaluae J b finding he area beween he curve and he ais, such as in Fig. 9-9a.In man siuaions we do no know how he force varies wih imebu we do know he average magniude F avg of he force and he duraion ( f i ) of he collision.then we can wrie he magniude of he impulse as J F avg. (9-35) The average force is ploed versus ime as in Fig. 9-9b.The area under ha curve is equal o he area under he curve for he acual force F() in Fig. 9-9a because boh areas are equal o impulse magniude J. Insead of he ball, we could have focused on he ba in Fig A an insan, Newon s hird law ells us ha he force on he ba has he same magniude bu he opposie direcion as he force on he ball. From Eq. 9-30, his means ha he impulse on he ba has he same magniude bu he opposie direcion as he impulse on he ball. i p CHECKPOINT 4 A pararooper whose chue fails o open lands in snow; he is hur slighl. Had he landed on bare ground, he sopping ime would have been 10 imes shorer and he collision lehal. Does he presence of he snow increase, decrease, or leave unchanged he values of (a) he pararooper s change in momenum, (b) he impulse sopping he pararooper, and (c) he force sopping he pararooper?

13 9-6 COLLISION AND IMPULSE 213 PART 1 Series of Collisions Now le s consider he force on a bod when i undergoes a series of idenical, repeaed collisions. For eample, as a prank, we migh adjus one of hose machines ha fire ennis balls o fire hem a a rapid rae direcl a a wall. Each collision would produce a force on he wall, bu ha is no he force we are seeking. We wan he average force F avg on he wall during he bombardmen ha is, he average force during a large number of collisions. In Fig. 9-10, a sead sream of projecile bodies, wih idenical mass m and linear momena mv, moves along an ais and collides wih a arge bod ha is fied in place. Le n be he number of projeciles ha collide in a ime inerval. Because he moion is along onl he ais, we can use he componens of he momena along ha ais. Thus, each projecile has iniial momenum mv and undergoes a change p in linear momenum because of he collision. The oal change in linear momenum for n projeciles during inerval is n p. The resuling impulse J on he arge during is along he ais and has he same magniude of n p bu is in he opposie direcion. We can wrie his relaion in componen form as J n p, (9-36) where he minus sign indicaes ha J and p have opposie direcions. B rearranging Eq and subsiuing Eq. 9-36, we find he average force F avg acing on he arge during he collisions v Projeciles Targe Fig A sead sream of projeciles, wih idenical linear momena, collides wih aarge,which is fied in place.the average force F avg on he arge is o he righ and has a magniude ha depends on he rae a which he projeciles collide wih he arge or, equivalenl, he rae a which mass collides wih he arge. F avg J n p n m v. (9-37) This equaion gives us F avg in erms of n/, he rae a which he projeciles collide wih he arge, and v,he change in he veloci of hose projeciles. If he projeciles sop upon impac, hen in Eq we can subsiue, for v, v v f v i 0 v v, (9-38) where v i ( v) and v f ( 0) are he velociies before and afer he collision, respecivel. If, insead, he projeciles bounce (rebound) direcl backward from he arge wih no change in speed, hen v f v and we can subsiue v v f v i v v 2v. (9-39) In ime inerval, an amoun of mass m nm collides wih he arge. Wih his resul, we can rewrie Eq as F avg m (9-40) This equaion gives he average force F avg in erms of m/, he rae a which mass collides wih he arge. Here again we can subsiue for v from Eq or 9-39 depending on wha he projeciles do. v. CHECKPOINT 5 The figure shows an overhead view of a ball bouncing from a verical wall wihou an change in is speed. Consider he change p in he ball s linear momenum. (a) Is p posiive, negaive, or zero? (b) Is p posiive, negaive, or zero? (c) Wha is he direcion of p? θ θ

14 214 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM Sample Problem Two-dimensional impulse, race car wall collision Race car wall collision. Figure 9-11a is an overhead view of he pah aken b a race car driver as his car collides wih he racerack wall. Jus before he collision, he is raveling a speed v i 70 m/s along a sraigh line a 30 from he wall. Jus afer he collision, he is raveling a speed v f 50 m/s along a sraigh line a 10 from he wall. His mass m is 80 kg. (a) Wha is he impulse J on he driver due o he collision? KEY IDEAS We can rea he driver as a paricle-like bod and hus appl he phsics of his secion. However, we canno calculae J direcl from Eq because we do no know anhing abou he force F () on he driver during he collision.tha is, we do no have a funcion of F () or a plo for i and hus canno inegrae o find J. However, we can find J from he change in he driver s linear momenum p via Eq ( J p f p i). Calculaions Figure 9-11b shows he driver s momenum p i before he collision (a angle 30 from he posiive direcion) and his momenum p afer he collision (a angle 10 ). f From Eqs and 9-22 ( p mv ), we can wrie J p f p i mv f mv i m(v f v i). (9-41) We could evaluae he righ side of his equaion direcl on a vecor-capable calculaor because we know m is 80 kg, v is 50 m/s a 10, and f v is 70 m/s a 30. Insead, here we i evaluae Eq in componen form. componen Along he ais we have J m(v f v i ) (80 kg)[(50 m/s) cos( 10 ) (70 m/s) cos 30 ] 910 kg m/s. componen Along he ais, J m(v f v i ) (80 kg)[(50 m/s) sin( 10 ) (70 m/s) sin 30 ] 3495 kg m/s 3500 kg m/s. Impulse The impulse is hen J ( 910î 3500ĵ) kg m/s, which means he impulse magniude is J J 2 J kg m/s 3600 kg m/s. The angle of J is given b an 1 J (Answer) (Answer) which a calculaor evaluaes as Recall ha he phsicall correc resul of an inverse angen migh be he displaed answer plus 180. We can ell which is correc here b drawing he componens of J (Fig. 9-11c). We find ha u is acuall , which we can wrie as u 105. (Answer) (b) The collision lass for 14 ms. Wha is he magniude of he average force on he driver during he collision? KEY IDEA From Eq (J F avg ), he magniude F avg of he average force is he raio of he impulse magniude J o he duraion of he collision. Calculaions We have F avg J 3616 kg m/s s N N. (Answer) Using F ma wih m 80 kg, ou can show ha he magniude of he driver s average acceleraion during he collision is abou m/s 2 329g, which is faal. Surviving Mechanical engineers aemp o reduce he chances of a faali b designing and building racerack walls wih more give, so ha a collision lass longer. For eample, if he collision here lased 10 imes longer and he oher daa remained he same, he magniudes of he average force and average acceleraion would be 10 imes less and probabl survivable. J, Fig (a) Overhead view of he pah aken b a race car and is driver as he car slams ino he racerack wall. (b) The iniial momenum p i and final momen- um p of he driver. (c) The f impulse J on he driver during he collision. 30 Wall Pah 10 (a) The collision changes he momenum. (b) p i 30 p f J J J The impulse on he car is equal o he change in he momenum. (c) Addiional eamples, video, and pracice available a WilePLUS

15 9-7 CONSERVATION OF LINEAR MOMENTUM 215 PART Conservaion of Linear Momenum Suppose ha he ne eernal force F ne (and hus he ne impulse J ) acing on a ssem of paricles is zero (he ssem is isolaed) and ha no paricles leave or ener he ssem (he ssem is closed). Puing F in Eq hen ields dp ne 0 /d 0,or P consan (closed, isolaed ssem). (9-42) In words, If no ne eernal force acs on a ssem of paricles, he oal linear momenum he ssem canno change. P of This resul is called he law of conservaion of linear momenum. I can also be wrien as P i P f (closed, isolaed ssem). (9-43) In words, his equaion sas ha, for a closed, isolaed ssem, oal linear momenum a some iniial ime i oal linear momenum a some laer ime. f Cauion Momenum should no be confused wih energ. In he sample problems of his secion, momenum is conserved bu energ is definiel no. Equaions 9-42 and 9-43 are vecor equaions and, as such, each is equivalen o hree equaions corresponding o he conservaion of linear momenum in hree muuall perpendicular direcions as in, sa, an z coordinae ssem. Depending on he forces acing on a ssem, linear momenum migh be conserved in one or wo direcions bu no in all direcions. However, If he componen of he ne eernal force on a closed ssem is zero along an ais, hen he componen of he linear momenum of he ssem along ha ais canno change. As an eample, suppose ha ou oss a grapefrui across a room. During is fligh, he onl eernal force acing on he grapefrui (which we ake as he ssem) is he graviaional force F g, which is direced vericall downward. Thus, he verical componen of he linear momenum of he grapefrui changes, bu since no horizonal eernal force acs on he grapefrui, he horizonal componen of he linear momenum canno change. Noe ha we focus on he eernal forces acing on a closed ssem. Alhough inernal forces can change he linear momenum of porions of he ssem, he canno change he oal linear momenum of he enire ssem. The sample problems in his secion involve eplosions ha are eiher onedimensional (meaning ha he moions before and afer he eplosion are along a single ais) or wo-dimensional (meaning ha he are in a plane conaining wo aes). In he following secions we consider collisions. CHECKPOINT 6 An iniiall saionar device ling on a fricionless floor eplodes ino wo pieces, which hen slide across he floor.one piece slides in he posiive direcion of an ais. (a) Wha is he sum of he momena of he wo pieces afer he eplosion? (b) Can he second piece move a an angle o he ais? (c) Wha is he direcion of he momenum of he second piece?

16 216 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM Sample Problem One-dimensional eplosion, relaive veloci, space hauler One-dimensional eplosion Figure 9-12a shows a space hauler and cargo module, of oal mass M,raveling along an ais in deep space. The have an iniial veloci v i of magniude 2100 km/h relaive o he Sun. Wih a small eplosion, he hauler ejecs he cargo module, of mass 0.20M (Fig. 9-12b).The hauler hen ravels 500 km/h faser han he module along he ais; ha is, he relaive speed v rel beween he hauler and he module is 500 km/h.wha hen is he veloci v HS of he hauler relaive o he Sun? KEY IDEA Because he hauler module ssem is closed and isolaed, is oal linear momenum is conserved; ha is, Cargo module (a) P i P f, (9-44) The eplosive separaion can change he momenum of he pars bu no he momenum of he ssem. v i Hauler Fig (a) A space hauler, wih a cargo module, moving a iniial veloci v i. (b) The hauler has ejeced he cargo module. Now he velociies relaive o he Sun are v for he module and MS v for he hauler. HS v MS 0.20M 0.80M (b) v HS where he subscrips i and f refer o values before and afer he ejecion, respecivel. Calculaions Because he moion is along a single ais, we can wrie momena and velociies in erms of heir componens, using a sign o indicae direcion. Before he ejecion, we have P i Mv i. (9-45) Le v MS be he veloci of he ejeced module relaive o he Sun.The oal linear momenum of he ssem afer he ejecion is hen P f (0.20M)v MS (0.80M)v HS, (9-46) where he firs erm on he righ is he linear momenum of he module and he second erm is ha of he hauler. We do no know he veloci v MS of he module relaive o he Sun, bu we can relae i o he known velociies wih veloci of module relaive. o Sun In smbols, his gives us v HS v rel v MS (9-47) or v MS v HS v rel. Subsiuing his epression for v MS ino Eq. 9-46, and hen subsiuing Eqs and 9-46 ino Eq. 9-44, we find which gives us or veloci of hauler relaive o Sun veloci of Mv i 0.20M(v HS v rel ) 0.80Mv HS, v HS v i 0.20v rel, v HS 2100 km/h (0.20)(500 km/h) 2200 km/h. hauler relaive o module (Answer) Sample Problem Two-dimensional eplosion, momenum, coconu Two-dimensional eplosion A firecracker placed inside a coconu of mass M, iniiall a res on a fricionless floor, blows he coconu ino hree pieces ha slide across he floor. An overhead view is shown in Fig. 9-13a.Piece C,wih mass 0.30M,has final speed v fc 5.0 m/s. (a) Wha is he speed of piece B,wih mass 0.20M? KEY IDEA Firs we need o see wheher linear momenum is conserved. We noe ha (1) he coconu and is pieces form a closed ssem, (2) he eplosion forces are inernal o ha ssem, and (3) no ne eernal force acs on he ssem. Therefore, he linear momenum of he ssem is conserved. Calculaions To ge sared, we superimpose an coordinae ssem as shown in Fig. 9-13b,wih he negaive direcion of he ais coinciding wih he direcion of v fa. The ais is a 80 wih he direcion of v fc and 50 wih he direcion of v fb. Linear momenum is conserved separael along each ais. Le s use he ais and wrie P i P f, (9-48) where subscrip i refers o he iniial value (before he eplosion), and subscrip refers o he componen of or. P i P f

17 9-8 MOMENTUM AND KINETIC ENERGY IN COLLISIONS 217 PART 1 The componen P i of he iniial linear momenum is zero, because he coconu is iniiall a res. To ge an epression for P f,we find he componen of he final linear momenum of each piece, using he -componen version of Eq ( p mv ) p fa, 0, p fb, 0.20Mv fb, 0.20Mv fb sin 50, p fc, 0.30Mv fc, 0.30Mv fc sin 80. (Noe ha p fa, 0 because of our choice of aes.) Equaion 9-48 can now be wrien as v fa 100 A P i P f p fa, p fb, p fc,. The eplosive separaion can change he momenum of he pars bu no he momenum of he ssem. B v fb C v fc 130 (a) (b) Fig Three pieces of an eploded coconu move off in hree direcions along a fricionless floor. (a) An overhead view of he even. (b) The same wih a wo-dimensional ais ssem imposed. v fa A B v fb C 50 v fc 80 Then, wih v fc 5.0 m/s, we have Mv fb sin 50 (0.30M)(5.0 m/s) sin 80, from which we find v fb 9.64 m/s 9.6 m/s. (b) Wha is he speed of piece A? (Answer) Calculaions Because linear momenum is also conserved along he ais, we have P i P f, (9-49) where P i 0 because he coconu is iniiall a res. To ge P f,we find he componens of he final momena, using he fac ha piece A mus have a mass of 0.50M ( M 0.20M 0.30M) p fa, 0.50Mv fa, p fb, 0.20Mv fb, 0.20Mv fb cos 50, p fc, 0.30Mv fc, 0.30Mv fc cos 80. Equaion 9-49 can now be wrien as P i P f p fa, p fb, p fc,. Then, wih v fc 5.0 m/s and v fb 9.64 m/s, we have Mv fa 0.20M(9.64 m/s) cos 50 from which we find v fa 3.0 m/s. 0.30M(5.0 m/s) cos 80, (Answer) Addiional eamples, video, and pracice available a WilePLUS 9-8 Momenum and Kineic Energ in Collisions In Secion 9-6, we considered he collision of wo paricle-like bodies bu focused on onl one of he bodies a a ime. For he ne several secions we swich our focus o he ssem iself, wih he assumpion ha he ssem is closed and isolaed. In Secion 9-7, we discussed a rule abou such a ssem The oal linear momenum P of he ssem canno change because here is no ne eernal force o change i.this is a ver powerful rule because i can allow us o deermine he resuls of a collision wihou knowing he deails of he collision (such as how much damage is done). We shall also be ineresed in he oal kineic energ of a ssem of wo colliding bodies. If ha oal happens o be unchanged b he collision, hen he kineic energ of he ssem is conserved (i is he same before and afer he collision). Such a collision is called an elasic collision. In everda collisions of common bodies, such as wo cars or a ball and a ba, some energ is alwas ransferred from kineic energ o oher forms of energ, such as hermal energ or energ of sound. Thus, he kineic energ of he ssem is no conserved. Such a collision is called an inelasic collision. However, in some siuaions, we can approimae a collision of common bodies as elasic. Suppose ha ou drop a Superball ono a hard floor. If he collision

18 218 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM Bod 1 Bod 2 v 1i v 2i Before Afer Here is he generic seup for an inelasic collision. m 1 m 2 m 1 m 2 Fig Bodies 1 and 2 move along an ais, before and afer he have an inelasic collision. v 1f v 2f beween he ball and floor (or Earh) were elasic, he ball would lose no kineic energ because of he collision and would rebound o is original heigh. However, he acual rebound heigh is somewha shor, showing ha a leas some kineic energ is los in he collision and hus ha he collision is somewha inelasic. Sill, we migh choose o neglec ha small loss of kineic energ o approimae he collision as elasic. The inelasic collision of wo bodies alwas involves a loss in he kineic energ of he ssem. The greaes loss occurs if he bodies sick ogeher, in which case he collision is called a compleel inelasic collision. The collision of a baseball and a ba is inelasic. However, he collision of a we pu ball and a ba is compleel inelasic because he pu sicks o he ba. 9-9 Inelasic Collisions in One Dimension One-Dimensional Inelasic Collision Figure 9-14 shows wo bodies jus before and jus afer he have a onedimensional collision. The velociies before he collision (subscrip i) and afer he collision (subscrip f ) are indicaed.the wo bodies form our ssem, which is closed and isolaed. We can wrie he law of conservaion of linear momenum for his wo-bod ssem as which we can smbolize as oal momenum P i before he collision oal momenum P f, afer he collision p 1i p 2i p 1f p 2f (conservaion of linear momenum). (9-50) Because he moion is one-dimensional, we can drop he overhead arrows for vecors and use onl componens along he ais, indicaing direcion wih a sign. Thus, from p mv,we can rewrie Eq.9-50 as m 1 v 1i m 2 v 2i m 1 v 1f m 2 v 2f. (9-51) If we know values for, sa, he masses, he iniial velociies, and one of he final velociies, we can find he oher final veloci wih Eq Before Afer In a compleel inelasic collision, he bodies sick ogeher. v 1i m 1 Projecile m 2 Targe v 2i = 0 Fig A compleel inelasic collision beween wo bodies. Before he collision, he bod wih mass m 2 is a res and he bod wih mass m 1 moves direcl oward i.afer he collision, he suckogeher bodies move wih he same veloci V. V m 1 + m 2 One-Dimensional Compleel Inelasic Collision Figure 9-15 shows wo bodies before and afer he have a compleel inelasic collision (meaning he sick ogeher).the bod wih mass m 2 happens o be iniiall a res (v 2i 0). We can refer o ha bod as he arge and o he incoming bod as he projecile. Afer he collision, he suck-ogeher bodies move wih veloci V. For his siuaion, we can rewrie Eq as m 1 v 1i (m 1 m 2 )V (9-52) or V v 1i. (9-53) m 1 m 2 If we know values for, sa, he masses and he iniial veloci v 1i of he projecile, we can find he final veloci V wih Eq Noe ha V mus be less han v 1i because he mass raio m 1 /(m 1 m 2 ) mus be less han uni. Veloci of he Cener of Mass m 1 com In a closed, isolaed ssem, he veloci v of he cener of mass of he ssem canno be changed b a collision because, wih he ssem isolaed, here is no ne

19 9-9 INELASTIC COLLISIONS IN ONE DIMENSION 219 PART 1 The com of he wo bodies is beween hem and moves a a consan veloci. Here is he incoming projecile. v 1i v com v 2i = 0 m 1 m 2 Here is he saionar arge. Fig Some freezeframes of he wo-bod ssem in Fig. 9-15, which undergoes a compleel inelasic collision. The ssem s cener of mass is shown in each freeze-frame.the veloci v com of he cener of mass is unaffeced b he collision. Because he bodies sick ogeher afer he collision, heir common veloci V mus be equal o v com. Collision! The com moves a he same veloci even afer he bodies sick ogeher. m 1 + m 2 V = v com eernal force o change i. To ge an epression for v, le us reurn o he wobod ssem and one-dimensional collision of Fig From Eq (P Mv com),we can relae v o he oal linear momenum P com of ha wobod ssem b wriing P Mv com (m 1 m 2 )v com. (9-54) The oal linear momenum is conserved during he collision; so i is given b eiher side of Eq Le us use he lef side o wrie P com P p 1i p 2i. (9-55) Subsiuing his epression for P in Eq and solving for v give us com v com P m 1 m 2 p 1i p 2i. (9-56) m 1 m 2 The righ side of his equaion is a consan, and vcom has ha same consan value before and afer he collision. For eample, Fig shows, in a series of freeze-frames, he moion of he cener of mass for he compleel inelasic collision of Fig Bod 2 is he arge, and is iniial linear momenum in Eq is p 2i m 2 v 2i 0. Bod 1 is he projecile, and is iniial linear momenum in Eq is p1i m 1 v 1i. Noe ha as he series of freeze-frames progresses o and hen beond he collision, he cener of mass moves a a consan veloci o he righ. Afer he collision, he common final speed V of he bodies is equal o vcom because hen he cener of mass ravels wih he suck-ogeher bodies. CHECKPOINT 7 Bod 1 and bod 2 are in a compleel inelasic one-dimensional collision. Wha is heir final momenum if heir iniial momena are, respecivel,(a) 10kg m/s and 0; (b) 10 kg m/s and 4 kg m/s; (c) 10 kg m/s and 4kg m/s?

20 220 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM Sample Problem Conservaion of momenum, ballisic pendulum The ballisic pendulum was used o measure he speeds of bulles before elecronic iming devices were developed. The version shown in Fig consiss of a large block of wood of mass M 5.4 kg, hanging from wo long cords. A bulle of mass m 9.5 g is fired ino he block, coming quickl o res. The block bulle hen swing upward, heir cener of mass rising a verical disance h 6.3 cm before he pendulum comes momenaril o res a he end of is arc. Wha is he speed of he bulle jus prior o he collision? KEY IDEAS We can see ha he bulle s speed v mus deermine he rise heigh h.however,we canno use he conservaion of mechanical energ o relae hese wo quaniies because surel energ is ransferred from mechanical energ o oher forms (such as hermal energ and energ o break apar he wood) as he bulle peneraes he block. Neverheless, we can spli his complicaed moion ino wo seps ha we can separael analze (1) he bulle block collision and (2) he bulle block rise, during which mechanical energ is conserved. Reasoning sep 1 Because he collision wihin he bulle block ssem is so brief, we can make wo imporan assumpions (1) During he collision, he graviaional force on he block and he force on he block from he cords are sill balanced. Thus, during he collision, he ne eernal impulse on he bulle block ssem is zero. Therefore, he ssem is isolaed and is oal linear momenum is conserved (9-57) (2) The collision is one-dimensional in he sense ha he direcion of he bulle and block jus afer he collision is in he bulle s original direcion of moion. Because he collision is one-dimensional, he block is iniiall a res, and he bulle sicks in he block, we use Eq o epress he conservaion of linear momenum. Replacing he smbols here wih he corresponding smbols here, we have (9-58) Reasoning sep 2 As he bulle and block now swing up ogeher, he mechanical energ of he bulle block Earh ssem is conserved oal momenum before he collision oal momenum afer he collision. V m m M v. mechanical energ a boom mechanical energ. a op (9-59) (This mechanical energ is no changed b he force of he cords on he block, because ha force is alwas direced perpendicular o he block s direcion of ravel.) Le s ake he block s iniial level as our reference level of zero graviaional poenial energ. Then conservaion of mechanical energ means ha he ssem s kineic energ a he sar of he swing mus equal is graviaional poenial energ a he highes poin of he swing. Because he speed of he bulle and block a he sar of he swing is he speed V immediael afer he collision, we ma wrie his conservaion as (9-60) Combining seps Subsiuing for V from Eq leads o v m M m 630 m/s. 1 2 (m M)V 2 (m M)gh. 22gh (9-61) kg 5.4 kg kg 2(2)(9.8 m/s2 )(0.063 m) (Answer) The ballisic pendulum is a kind of ransformer, echanging he high speed of a ligh objec (he bulle) for he low and hus more easil measurable speed of a massive objec (he block). Fig bulles. There are wo evens here. The bulle collides wih he block. Then he bulle block ssem swings upward b heigh h. m v M A ballisic pendulum, used o measure he speeds of h Addiional eamples, video, and pracice available a WilePLUS

21 9-10 ELASTIC COLLISIONS IN ONE DIMENSION 221 PART Elasic Collisions in One Dimension As we discussed in Secion 9-8, everda collisions are inelasic bu we can approimae some of hem as being elasic; ha is, we can approimae ha he oal kineic energ of he colliding bodies is conserved and is no ransferred o oher forms of energ oal kineic energ before he collision oal kineic energ afer he collision. (9-62) This does no mean ha he kineic energ of each colliding bod canno change. Raher, i means his In an elasic collision, he kineic energ of each colliding bod ma change, bu he oal kineic energ of he ssem does no change. Before m 1 Projecile Afer v 1i m 2 Targe v 2i = 0 Here is he generic seup for an elasic collision wih a saionar arge. Fig Bod 1 moves along an ais before having an elasic collision wih bod 2, which is iniiall a res. Boh bodies move along ha ais afer he collision. v 1f m 1 m 2 v 2f For eample, he collision of a cue ball wih an objec ball in a game of pool can be approimaed as being an elasic collision. If he collision is head-on (he cue ball heads direcl oward he objec ball), he kineic energ of he cue ball can be ransferred almos enirel o he objec ball. (Sill, he fac ha he collision makes a sound means ha a leas a lile of he kineic energ is ransferred o he energ of he sound.) Saionar Targe Figure 9-18 shows wo bodies before and afer he have a one-dimensional collision, like a head-on collision beween pool balls. A projecile bod of mass m 1 and iniial veloci v 1i moves oward a arge bod of mass m 2 ha is iniiall a res (v 2i 0). Le s assume ha his wo-bod ssem is closed and isolaed. Then he ne linear momenum of he ssem is conserved, and from Eq we can wrie ha conservaion as m 1 v 1i m 1 v 1f m 2 v 2f (linear momenum). (9-63) If he collision is also elasic, hen he oal kineic energ is conserved and we can wrie ha conservaion as (kineic energ). (9-64) In each of hese equaions, he subscrip i idenifies he iniial velociies and he subscrip f he final velociies of he bodies. If we know he masses of he bodies and if we also know v 1i,he iniial veloci of bod 1,he onl unknown quaniies are v 1f and v 2f,he final velociies of he wo bodies.wih wo equaions a our disposal, we should be able o find hese wo unknowns. To do so, we rewrie Eq as and Eq as* 1 m 2 2 1v 1i 1 m 2 2 1v 1f 1 m 2 2 2v 2f m 1 (v 1i v 1f ) m 2 v 2f (9-65) m 1 (v 1i v 1f )(v 1i v 1f ) m 2 v 2 2f. Afer dividing Eq b Eq and doing some more algebra, we obain (9-66) v 1f m 1 m 2 v m 1 m 1i 2 (9-67) and v 2f 2m 1 v (9-68) m 1 m 1i. 2 *In his sep, we use he ideni a 2 b 2 (a b)(a b). I reduces he amoun of algebra needed o solve he simulaneous equaions Eqs and 9-66.

22 222 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM We noe from Eq ha v 2f is alwas posiive (he iniiall saionar arge bod wih mass m 2 alwas moves forward). From Eq we see ha v 1f ma be of eiher sign (he projecile bod wih mass m 1 moves forward if m 1 m 2 bu rebounds if m 1 m 2 ). Le us look a a few special siuaions. 1. Equal masses If m 1 m 2,Eqs.9-67 and 9-68 reduce o v 1f 0 and v 2f v 1i, which we migh call a pool plaer s resul. I predics ha afer a head-on collision of bodies wih equal masses, bod 1 (iniiall moving) sops dead in is racks and bod 2 (iniiall a res) akes off wih he iniial speed of bod 1. In head-on collisions, bodies of equal mass simpl echange velociies. This is rue even if bod 2 is no iniiall a res. 2. A massive arge In Fig. 9-18, a massive arge means ha m 2 m 1.For eample, we migh fire a golf ball a a saionar cannonball. Equaions 9-67 and 9-68 hen reduce o v 1f v 1i and v 2f 2m 1 m 2 v 1i. (9-69) This ells us ha bod 1 (he golf ball) simpl bounces back along is incoming pah, is speed esseniall unchanged. Iniiall saionar bod 2 (he cannonball) moves forward a a low speed, because he quani in parenheses in Eq is much less han uni.all his is wha we should epec. 3. A massive projecile This is he opposie case; ha is, m 1 m 2.This ime,we fire a cannonball a a saionar golf ball. Equaions 9-67 and 9-68 reduce o v 1f v 1i and v 2f 2v 1i. (9-70) Equaion 9-70 ells us ha bod 1 (he cannonball) simpl keeps on going, scarcel slowed b he collision. Bod 2 (he golf ball) charges ahead a wice he speed of he cannonball. You ma wonder Wh wice he speed? Recall he collision described b Eq. 9-69, in which he veloci of he inciden ligh bod (he golf ball) changed from v o v,a veloci change of 2v.The same change in veloci (bu now from zero o 2v) occurs in his eample also. Here is he generic seup for an elasic collision wih a moving arge. m 1 v 1i m 2 Fig Two bodies headed for a onedimensional elasic collision. v 2i Moving Targe Now ha we have eamined he elasic collision of a projecile and a saionar arge, le us eamine he siuaion in which boh bodies are moving before he undergo an elasic collision. For he siuaion of Fig. 9-19, he conservaion of linear momenum is wrien as and he conservaion of kineic energ is wrien as m 1 v 1i m 2 v 2i m 1 v 1f m 2 v 2f, (9-71) (9-72) To solve hese simulaneous equaions for v 1f and v 2f,we firs rewrie Eq.9-71 as and Eq as 1 2 m 2 1v 1i 1 2 m 2 2v 2i 1 2 m 2 1v 1f 1 2 m 2 2v 2f. m 1 (v 1i v 1f ) m 2 (v 2i v 2f ), (9-73) m 1 (v 1i v 1f )(v 1i v 1f ) m 2 (v 2i v 2f )(v 2i v 2f ). (9-74) Afer dividing Eq b Eq and doing some more algebra, we obain v 1f m 1 m 2 v m 1 m 1i 2m 2 v 2 m 1 m 2i 2 (9-75) and v 2f 2m 1 v (9-76) m 1 m 1i m 2 m 1 v 2 m 1 m 2i. 2

23 9-10 ELASTIC COLLISIONS IN ONE DIMENSION 223 PART 1 Noe ha he assignmen of subscrips 1 and 2 o he bodies is arbirar. If we echange hose subscrips in Fig and in Eqs and 9-76, we end up wih he same se of equaions. Noe also ha if we se v 2i 0, bod 2 becomes a saionar arge as in Fig. 9-18, and Eqs and 9-76 reduce o Eqs and 9-68, respecivel. CHECKPOINT 8 Wha is he final linear momenum of he arge in Fig if he iniial linear momenum of he projecile is 6 kg m/s and he final linear momenum of he projecile is (a) 2 kg m/s and (b) 2 kg m/s? (c) Wha is he final kineic energ of he arge if he iniial and final kineic energies of he projecile are, respecivel, 5 J and 2 J? Sample Problem Elasic collision, wo pendulums Two meal spheres, suspended b verical cords, iniiall jus ouch, as shown in Fig Sphere 1, wih mass m 1 30 g, is pulled o he lef o heigh h cm, and hen released from res. Afer swinging down, i undergoes an elasic collision wih sphere 2, whose mass m 2 75 g. Wha is he veloci v 1f of sphere 1 jus afer he collision? KEY IDEA We can spli his complicaed moion ino wo seps ha we can analze separael (1) he descen of sphere 1 (in which mechanical energ is conserved) and (2) he wo-sphere collision (in which momenum is also conserved). Sep 1 As sphere 1 swings down, he mechanical energ of he sphere Earh ssem is conserved. (The mechanical energ is no changed b he force of he cord on sphere 1 because ha force is alwas direced perpendicular o he sphere s direcion of ravel.) Calculaion Le s ake he lowes level as our reference level of zero graviaional poenial energ. Then he kineic energ of sphere 1 a he lowes level mus equal he graviaional poenial energ of he ssem when sphere 1 is a heigh h 1.Thus, which we solve for he speed v 1i of sphere 1 jus before he collision m/s. 1 m 2 2 1v 1i m 1 gh 1, v 1i 22gh 1 2(2)(9.8 m/s 2 )(0.080 m) Sep 2 Here we can make wo assumpions in addiion o he assumpion ha he collision is elasic. Firs, we can assume ha he collision is one-dimensional because he moions of he spheres are approimael horizonal from jus before he collision o jus afer i. Second, because he collision is so brief, we can assume ha he wo-sphere ssem is closed and isolaed.this means ha he oal linear momenum of he ssem is conserved. Calculaion Thus, we can use Eq o find he veloci of sphere 1 jus afer he collision v 1f m 1 m 2 v 1i m 1 m kg kg (1.252 m/s) kg kg m/s 0.54 m/s. (Answer) The minus sign ells us ha sphere 1 moves o he lef jus afer he collision. h 1 Ball 1 swings down and collides wih ball 2, which hen swings upward. If he collision is elasic, no mechanical energ is los. 1 2 m 1 Fig Two meal spheres suspended b cords jus ouch when he are a res. Sphere 1, wih mass m 1,is pulled o he lef o heigh h 1 and hen released. m 2 Addiional eamples, video, and pracice available a WilePLUS

24 224 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM A glancing collision ha conserves boh momenum and kineic energ. m 1 v 1i m 2 Fig An elasic collision beween wo bodies in which he collision is no head-on.the bod wih mass m 2 (he arge) is iniiall a res. θ 1 θ 2 v 1f v 2f 9-11 Collisions in Two Dimensions When wo bodies collide, he impulse beween hem deermines he direcions in which he hen ravel. In paricular, when he collision is no head-on, he bodies do no end up raveling along heir iniial ais. For such wo-dimensional collisions in a closed, isolaed ssem, he oal linear momenum mus sill be conserved P 1i P 2i P 1f P 2f. (9-77) If he collision is also elasic (a special case), hen he oal kineic energ is also conserved K 1i K 2i K 1f K 2f. (9-78) Equaion 9-77 is ofen more useful for analzing a wo-dimensional collision if we wrie i in erms of componens on an coordinae ssem. For eample, Fig shows a glancing collision (i is no head-on) beween a projecile bod and a arge bod iniiall a res. The impulses beween he bodies have sen he bodies off a angles u 1 and u 2 o he ais, along which he projecile iniiall raveled. In his siuaion we would rewrie Eq.9-77 for componens along he ais as and along he ais as m 1 v 1i m 1 v 1f cos u 1 m 2 v 2f cos u 2, (9-79) 0 m 1 v 1f sin 1 m 2 v 2f sin 2. (9-80) We can also wrie Eq (for he special case of an elasic collision) in erms of speeds 1 m 2 2 1v 1i 1 m 2 2 1v 1f 1 m 2 2 2v 2f (kineic energ). (9-81) Equaions 9-79 o 9-81 conain seven variables wo masses, m 1 and m 2 ;hree speeds, v 1i, v 1f,and v 2f ;and wo angles,u 1 and u 2.If we know an four of hese quaniies, we can solve he hree equaions for he remaining hree quaniies. CHECKPOINT 9 In Fig. 9-21, suppose ha he projecile has an iniial momenum of 6 kg m/s,a final componen of momenum of 4 kg m/s, and a final componen of momenum of 3 kg m/s. For he arge, wha hen are (a) he final componen of momenum and (b) he final componen of momenum? 9-12 Ssems wih Varing Mass A Rocke In he ssems we have deal wih so far, we have assumed ha he oal mass of he ssem remains consan. Someimes, as in a rocke, i does no. Mos of he mass of a rocke on is launching pad is fuel, all of which will evenuall be burned and ejeced from he nozzle of he rocke engine. We handle he variaion of he mass of he rocke as he rocke acceleraes b appling Newon s second law, no o he rocke alone bu o he rocke and is ejeced combusion producs aken ogeher. The mass of his ssem does no change as he rocke acceleraes. Finding he Acceleraion Assume ha we are a res relaive o an inerial reference frame, waching a rocke accelerae hrough deep space wih no graviaional or amospheric drag forces acing on i. For his one-dimensional moion, le M be he mass of he rocke and v is veloci a an arbirar ime (see Fig. 9-22a).

25 9-12 SYSTEMS WITH VARYING MASS A ROCKET 225 PART 1 Ssem boundar The ejecion of mass from he rocke's rear increases he rocke's speed. Ssem boundar M v dm U M + dm v + dv (a) (b) Fig (a) An acceleraing rocke of mass M a ime,as seen from an inerial reference frame. (b) The same bu a ime d.the ehaus producs released during inerval d are shown. Figure 9-22b shows how hings sand a ime inerval d laer. The rocke now has veloci v dv and mass M dm,where he change in mass dm is a negaive quani. The ehaus producs released b he rocke during inerval d have mass dm and veloci U relaive o our inerial reference frame. Our ssem consiss of he rocke and he ehaus producs released during inerval d.the ssem is closed and isolaed,so he linear momenum of he ssem mus be conserved during d;ha is, P i P f, (9-82) where he subscrips i and f indicae he values a he beginning and end of ime inerval d. We can rewrie Eq as Mv dm U (M dm)(v dv), (9-83) where he firs erm on he righ is he linear momenum of he ehaus producs released during inerval d and he second erm is he linear momenum of he rocke a he end of inerval d. We can simplif Eq b using he relaive speed v rel beween he rocke and he ehaus producs, which is relaed o he velociies relaive o he frame wih veloci of rocke. relaive o frame veloci of rocke relaive o producs veloci of producs relaive o frame In smbols, his means (v dv) v rel U, or U v dv v rel. (9-84) Subsiuing his resul for U ino Eq ields, wih a lile algebra, Dividing each side b d gives us dm v rel M dv. (9-85) dm d v rel M dv d. (9-86) We replace dm/d (he rae a which he rocke loses mass) b R, where R is he (posiive) mass rae of fuel consumpion, and we recognize ha dv/d is he acceleraion of he rocke. Wih hese changes, Eq becomes Rv rel Ma (firs rocke equaion). (9-87) Equaion 9-87 holds for he values a an given insan. Noe he lef side of Eq has he dimensions of force (kg/s m/s kg m/s 2 N) and depends onl on design characerisics of he rocke engine namel, he rae R a which i consumes fuel mass and he speed v rel wih which

26 226 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM ha mass is ejeced relaive o he rocke. We call his erm Rv rel he hrus of he rocke engine and represen i wih T.Newon s second law emerges clearl if we wrie Eq as T Ma,in which a is he acceleraion of he rocke a he ime ha is mass is M. Finding he Veloci How will he veloci of a rocke change as i consumes is fuel? From Eq we have Inegraing leads o v f v i dv v rel dm M. M dv v rel f dm M i M, in which M i is he iniial mass of he rocke and M f is final mass. Evaluaing he inegrals hen gives v f v i v rel ln M i M f (second rocke equaion) (9-88) for he increase in he speed of he rocke during he change in mass from M i o M f.(the smbol ln in Eq.9-88 means he naural logarihm.) We see here he advanage of mulisage rockes, in which M f is reduced b discarding successive sages when heir fuel is depleed. An ideal rocke would reach is desinaion wih onl is paload remaining. Sample Problem Rocke engine, hrus, acceleraion A rocke whose iniial mass M i is 850 kg consumes fuel a he rae R 2.3 kg/s. The speed v rel of he ehaus gases relaive o he rocke engine is 2800 m/s. Wha hrus does he rocke engine provide? KEY IDEA Thrus T is equal o he produc of he fuel consumpion rae R and he relaive speed v rel a which ehaus gases are epelled, as given b Eq Calculaion Here we find T Rv rel (2.3 kg/s)(2800 m/s) 6440 N 6400 N. (b) Wha is he iniial acceleraion of he rocke? KEY IDEA (Answer) We can relae he hrus T of a rocke o he magniude a of he resuling acceleraion wih T Ma, where M is he rocke s mass. However, M decreases and a increases as fuel is consumed. Because we wan he iniial value of a here, we mus use he inial value M i of he mass. Calculaion We find a T M i 6440 N 850 kg 7.6 m/s2. (Answer) To be launched from Earh s surface, a rocke mus have an iniial acceleraion greaer han g 9.8 m/s 2. Tha is, i mus be greaer han he graviaional acceleraion a he surface. Pu anoher wa, he hrus T of he rocke engine mus eceed he iniial graviaional force on he rocke, which here has he magniude M i g,which gives us (850 kg)(9.8 m/s 2 ) = 8330 N. Because he acceleraion or hrus requiremen is no me (here T 6400 N), our rocke could no be launched from Earh s surface b iself; i would require anoher, more powerful, rocke. Addiional eamples, video, and pracice available a WilePLUS

27 REVIEW & SUMMARY 227 PART 1 Cener of Mass The cener of mass of a ssem of n paricles is defined o be he poin whose coordinaes are given b com 1 M n m i i, com i 1 1 M n m i z i, i 1 (9-5) or r com 1 (9-8) M n m i r i, where M is he oal mass of he ssem. Newon s Second Law for a Ssem of Paricles The moion of he cener of mass of an ssem of paricles is governed b Newon s second law for a ssem of paricles, which is F ne Ma com. (9-14) Here F ne is he ne force of all he eernal forces acing on he ssem, M is he oal mass of he ssem, and a com is he acceleraion of he ssem s cener of mass. Linear Momenum and Newon s Second Law For a single paricle, we define a quani p called is linear momenum as p mv, (9-22) and can wrie Newon s second law in erms of his momenum For a ssem of paricles hese relaions become P Mv com (9-23) and F ne dp (9-25, 9-27) d. Collision and Impulse Appling Newon s second law in momenum form o a paricle-like bod involved in a collision leads o he impulse linear momenum heorem p f p i p J, (9-31,9-32) where p f p i p is he change in he bod s linear momenum, and J is he impulse due o he force F () eered on he bod b he oher bod in he collision f J F ne dp d. i F () d. (9-30) If F avg is he average magniude of F () during he collision and is he duraion of he collision, hen for one-dimensional moion J F avg. (9-35) When a sead sream of bodies, each wih mass m and speed v,collides wih a bod whose posiion is fied, he average force on he fied bod is F avg n p n m v, (9-37) where n/ is he rae a which he bodies collide wih he fied bod, and v is he change in veloci of each colliding bod. This average force can also be wrien as i 1 F avg m 1 M n m i i, z com i 1 v, (9-40) where m/ is he rae a which mass collides wih he fied bod. In Eqs and 9-40, v v if he bodies sop upon impac and v 2v if he bounce direcl backward wih no change in heir speed. Conservaion of Linear Momenum If a ssem is isolaed so ha no ne eernal force acs on i, he linear momenum P of he ssem remains consan P consan This can also be wrien as P i P f (closed, isolaed ssem). (9-42) (closed, isolaed ssem), (9-43) where he subscrips refer o he values of a some iniial ime and a a laer ime. Equaions 9-42 and 9-43 are equivalen saemens of he law of conservaion of linear momenum. Inelasic Collision in One Dimension In an inelasic collision of wo bodies, he kineic energ of he wo-bod ssem is no conserved. If he ssem is closed and isolaed, he oal linear momenum of he ssem mus be conserved, which we can wrie in vecor form as p 1i p 2i p 1f p 2f, (9-50) where subscrips i and f refer o values jus before and jus afer he collision, respecivel. If he moion of he bodies is along a single ais, he collision is one-dimensional and we can wrie Eq in erms of veloci componens along ha ais m 1 v 1i m 2 v 2i m 1 v 1f m 2 v 2f. (9-51) If he bodies sick ogeher, he collision is a compleel inelasic collision and he bodies have he same final veloci V (because he are suck ogeher). Moion of he Cener of Mass The cener of mass of a closed, isolaed ssem of wo colliding bodies is no affeced b a collision. In paricular, he veloci v com of he cener of mass canno be changed b he collision. Elasic Collisions in One Dimension An elasic collision is a special pe of collision in which he kineic energ of a ssem of colliding bodies is conserved. If he ssem is closed and isolaed, is linear momenum is also conserved. For a one-dimensional collision in which bod 2 is a arge and bod 1 is an incoming projecile, conservaion of kineic energ and linear momenum ield he following epressions for he velociies immediael afer he collision v 1f m 1 m 2 v m 1 m 1i 2 (9-67) and v 2f 2m 1 v (9-68) m 1 m 1i. 2 Collisions in Two Dimensions If wo bodies collide and heir moion is no along a single ais (he collision is no head-on), he collision is wo-dimensional. If he wo-bod ssem is closed and isolaed, he law of conservaion of momenum applies o he P

28 228 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM collision and can be wrien as P 1i P 2i P 1f P 2f. (9-77) rocke acceleraes a an insananeous rae given b Rv rel Ma (firs rocke equaion), (9-87) In componen form, he law gives wo equaions ha describe he collision (one equaion for each of he wo dimensions). If he collision is also elasic (a special case), he conservaion of kineic energ during he collision gives a hird equaion Variable-Mass Ssems K 1i K 2i K 1f K 2f. (9-78) In he absence of eernal forces a in which M is he rocke s insananeous mass (including unepended fuel), R is he fuel consumpion rae, and v rel is he fuel s ehaus speed relaive o he rocke. The erm Rv rel is he hrus of he rocke engine. For a rocke wih consan R and v rel,whose speed changes from v i o v f when is mass changes from M i o M f, v f v i v rel ln M i M f (second rocke equaion). (9-88) 1 Figure 9-23 shows an overhead view of hree paricles on which eernal forces ac. The magniudes 1 and direcions of he forces on wo of 3 5 N he paricles are indicaed. Wha are he magniude and direcion of he force acing on he hird paricle if 2 3 N he cener of mass of he hree-paricle ssem is (a) saionar, (b) moving Fig Quesion 1. a a consan veloci righward, and (c) acceleraing righward? 2 Figure 9-24 shows an overhead view of four paricles of equal mass sliding over a fricionless surface a consan veloci.the direcions of he velociies are indicaed; heir magniudes are equal. Consider pairing he paricles. Which pairs form a ssem wih a cener of mass ha (a) is saionar, (b) is saionar and a he origin, and (c) passes hrough he origin? c a 4 (m) Fig Quesion 2. b d (m) F 2F 0 4F (a) (b) (c) 8 Ν 8 Ν 6 Ν 3 Ν Ν 4 Ν 6 Ν 2 Ν 3 Ν 2 Ν 5 Ν 5 Ν 60 2 Ν 6 Ν (a) (b) (c) F 2F 0 Fig Quesion 4. 5 The free-bod diagrams in Fig give, from overhead views, he horizonal forces acing on hree boes of chocolaes as he boes move over a fricionless confecioner s couner. For each bo, is is linear momenum conserved along he ais and he ais? Fig Quesion 5. 6 Figure 9-28 shows four groups of hree or four idenical paricles ha move parallel o eiher he ais or he ais, a idenical speeds. Rank he groups according o cener-of-mass speed, greaes firs. F 3 Consider a bo ha eplodes ino wo pieces while moving wih a consan posiive veloci along an ais. If one piece, wih mass m 1,ends up wih posiive veloci v 1,hen he second piece,wih mass m 2,could end up wih (a) a posiive veloci v 2 (Fig.9-25a), (b) a negaive veloci v 2 (Fig. 9-25b), or (c) zero veloci (Fig. 9-25c). Rank hose hree possible resuls for he second piece according o he corresponding magniude of, greaes firs. v 1 (a) (b) v 2 v 1 v 2 v 1 v 1 (a) (b) Fig Quesion 3. (c) 4 Figure 9-26 shows graphs of force magniude versus ime for a bod involved in a collision. Rank he graphs according o he magniude of he impulse on he bod, greaes firs. (c) (d) Fig Quesion 6.

29 7 A block slides along a fricionless floor and ino a saionar second block wih he same mass. Figure 9-29 shows four choices for a graph of he kineic energies K of he blocks. (a) Deermine which represen phsicall impossible siuaions. Of he ohers, which bes represens (b) an elasic collision and (c) an inelasic collision? K K p (a) p (b) QUESTIONS p (c) 229 PART 1 p p p (a) (b) K K (d) (e) Fig Quesion 10. ( f ) (c) (d) Fig Quesion 7. 8 Figure 9-30 shows a snapsho of block 1 as i slides along an ais on a fricionless floor, before i undergoes an elasic collision wih saionar block 2. The figure also shows hree possible posiions of he cener of mass (com) of he wo-block ssem a he ime of he snapsho. (Poin B is halfwa beween he ceners of he wo blocks.) Is block 1 saionar, moving forward, or moving backward afer he collision if he com is locaed in he snapsho a (a) A,(b) B,and (c) C? 1 A B C 2 Fig Quesion 8. 9 Two bodies have undergone an elasic one-dimensional collision along an ais. Figure 9-31 is a graph of posiion versus ime for hose bodies and for heir cener of mass. (a) Were boh bodies iniiall moving, or was one iniiall saionar? Which line segmen corresponds o he moion of he cener of mass (b) before he collision and (c) afer he collision? (d) Is he mass of he bod ha was moving faser before he collision greaer han, less han, or equal o ha of he oher bod? Block 1 wih mass m 1 slides along an ais across a fricionless floor and hen undergoes an elasic collision wih a saionar block 2 wih mass m 2.Figure 9-33 shows a plo of posiion versus ime of block 1 unil he collision occurs a posiion c and ime c.in which of he leered regions on he graph will he plo be coninued (afer he collision) if (a) m 1 m 2 and (b) m 1 m 2? (c) Along which of he numbered dashed lines will he plo be coninued if m 1 m 2? c 1 c A D 5 2 B C Fig Quesion Figure 9-34 shows four graphs of posiion versus ime for wo bodies and heir cener of mass.the wo bodies form a closed, isolaed ssem and undergo a compleel inelasic, one-dimensional collision on an ais. In graph 1, are (a) he wo bodies and (b) he cener of mass moving in he posiive or negaive direcion of he ais? (c) Which graphs correspond o a phsicall impossible siuaion? Eplain Fig Quesion Figure 9-32A block on a horizonal floor is iniiall eiher saionar, sliding in he posiive direcion of an ais, or sliding in he negaive direcion of ha ais. Then he block eplodes ino wo pieces ha slide along he ais. Assume he block and he wo pieces form a closed, isolaed ssem. Si choices for a graph of he momena of he block and he pieces are given, all versus ime. Deermine which choices represen phsicall impossible siuaions and eplain wh. (1) (2) (3) (4) Fig Quesion 12.

30 230 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM SSM Tuoring problem available (a insrucor s discreion) in WilePLUS and WebAssign Worked-ou soluion available in Suden Soluions Manual WWW Worked-ou soluion is a Number of dos indicaes level of problem difficul ILW Ineracive soluion is a Addiional informaion available in The Fling Circus of Phsics and a flingcircusofphsics.com hp// sec. 9-2 The Cener of Mass 1 A 2.00 kg paricle has he coordinaes ( 1.20 m, m), and a 4.00 kg paricle has he coordinaes (0.600 m, m). Boh lie on a horizonal plane. A wha (a) and (b) coordinaes mus ou place a 3.00 kg paricle such ha he cener of mass of he hree-paricle ssem has he coordinaes ( m, m)? 2 Figure 9-35 shows a hreeparicle ssem, wih masses m 1 (m) 3.0 kg, m kg, and m m 3 s kg. The scales on he aes are se b s 2.0 m and s 2.0 m. m 2 Wha are (a) he coordinae and m 1 (b) he coordinae of he ssem s cener of mass? (c) If m 3 is (m) 0 s graduall increased, does he cener of mass of he ssem shif oward or awa from ha paricle, or does i remain Fig Problem 2. saionar? 3 Figure 9-36 shows a slab wih dimensions d cm, d cm, and d cm. Half he slab consiss of aluminum (densi 2.70 g/cm 3 ) and half consiss of iron (densi 7.85 g/cm 3 ).Wha are (a) he coordinae, (b) he coordinae, and (c) he z coordinae of he slab s cener of mass? 5 Wha are (a) he coordinae and (b) he coordinae of he cener of mass for he uniform plae shown in Fig if L 5.0 cm? 3L 4L 2L 2L 4L 2L Fig Problem 5. 6 Figure 9-39 shows a cubical bo ha has been consruced from uniform meal plae of negligible hickness. The bo is open a he op and has edge lengh L 40 cm. Find (a) he coordinae, (b) he coordinae, and (c) he z coordinae of he cener of mass of he bo. z L 2d 1 z d 3 Iron Midpoin Aluminum d 2 L O d 1 d 1 Fig Problem 6. Fig Problem 3. 4 In Fig. 9-37, hree uniform hin rods, each of lengh L 22 cm, form an invered U. The verical rods each have a mass of 14 g; he horizonal rod has a mass of 42 g.wha are (a) he coordinae and (b) he coordinae of he ssem s cener of mass? L L L Fig Problem 4. 7 ILW In he ammonia (NH 3 ) molecule of Fig. 9-40, hree hdrogen (H) aoms form an equilaeral riangle, wih he cener of he riangle a disance d m from each hdrogen aom. The nirogen (N) aom is a he ape of a pramid, wih he hree hdrogen aoms forming he base. The nirogen-o-hdrogen aomic mass raio is 13.9, and he nirogen-o-hdrogen disance is L m. Wha are he (a) and (b) coordinaes of he molecule s cener of mass? H H N L H d Fig Problem 7.

31 8 A uniform soda can of mass kg is 12.0 cm all and filled wih kg of soda (Fig. 9-41). Then small holes are drilled in he op and boom (wih negligible loss of meal) o drain he soda.wha is he heigh h of he com of he can and conens (a) iniiall and (b) afer he can loses all he soda? (c) Wha happens o h as he soda drains ou? (d) If is he heigh of he remaining soda a an given insan, find when he com reaches is lowes poin. sec. 9-3 Newon s Second Law for a Ssem of Paricles 9 ILW A sone is dropped a 0. A second sone, wih wice he mass of he firs, is dropped from he same poin a 100 ms. (a) How far below he release poin is he cener of mass of he wo sones a 300 ms? (Neiher sone has e reached he ground.) (b) How fas is he cener of mass of he wosone ssem moving a ha ime? 10 A 1000 kg auomobile is a res a a raffic signal. A he insan he ligh urns green, he auomobile sars o move wih a consan acceleraion of 4.0 m/s 2.A he same insan a 2000kg ruck, raveling a a consan speed of 8.0 m/s, overakes and passes he auomobile. (a) How far is he com of he auomobile ruck ssem from he raffic ligh a 3.0 s? (b) Wha is he speed of he com hen? 11 A big olive (m 0.50 kg) lies a he origin of an coordinae ssem, and a big Brazil nu (M 1.5 kg) lies a he poin (1.0, 2.0) m. A 0, a force F begins o ac on he olive, and a force F o (2.0î 3.0ĵ) N n ( 3.0î 2.0ĵ) N begins o ac on he nu. In uni-vecor noaion, wha is he displacemen of he cener of mass of he olive nu ssem a 4.0 s, wih respec o is posiion a 0? 12 Two skaers, one wih mass 65 kg and he oher wih mass 40 kg, sand on an ice rink holding a pole of lengh 10 m and negligible mass. Saring from he ends of he pole, he skaers pull hemselves along he pole unil he mee. How far does he 40 kg skaer move? 13 SSM A shell is sho wih an iniial veloci v0 of 20 m/s, a an angle of 0 60 wih he horizonal.a he op of he rajecor, he shell eplodes ino wo fragmens of equal mass (Fig. 9-42). One fragmen, whose speed immediael afer he eplosion is zero, falls vericall. How far from he gun does he oher fragmen land, assuming ha he errain is level and ha air drag is negligible? v 0 θ0 Eplosion Fig Problem 13. Fig Problem In Figure 9-43, wo paricles are launched from he origin of he coordinae ssem a ime 0. Paricle 1 of mass m g is sho direcl along he ais on a fricionless floor, wih consan speed 10.0 m/s. Paricle 2 of mass m g is sho wih a veloci of magniude 20.0 m/s, a an upward angle such ha i alwas sas PROBLEMS 231 direcl above paricle 1. (a) Wha is he maimum heigh H ma reached b he com of he wo-paricle ssem? In 2 uni-vecor noaion, wha are he (b) 1 veloci and (c) acceleraion of he com when he com reaches H ma? Fig Problem Figure 9-44 shows an arrangemen wih an air rack, in which a car is conneced b a cord o a hanging block. The car has mass m kg, and is cener is iniiall a coordinaes ( m, 0 m); he block has mass m kg, and is cener is iniiall a coordinaes (0, m).the mass of he cord and pulle are negligible.the car is released from res, and boh car and block move unil he car his he pulle. The fricion beween he car and he air rack and beween he pulle and is ale is negligible. (a) In uni-vecor noaion, wha is he acceleraion of he cener of mass of he car block ssem? (b) Wha is he veloci of he com as a funcion of ime? (c) Skech he pah aken b he com. (d) If he pah is curved, deermine wheher i bulges upward o he righ or downward o he lef, and if i is sraigh, find he angle beween i and he ais. 16 Ricardo, of mass 80 kg, and Carmelia, who is ligher, are enjoing Lake Merced a dusk in a 30 kg canoe. When he canoe is a res in he placid waer, he echange seas, which are 3.0 m apar and smmericall locaed wih respec o he canoe s cener. If he canoe moves 40 cm horizonall relaive o a pier pos, wha is Carmelia s mass? 17 In Fig. 9-45a, a 4.5kg dog sands on an 18 kg flaboa a disance D 6.1 m from he shore. I walks 2.4 m along he boa oward shore and hen sops. Assuming no fricion beween he boa and he waer, find how far he dog is hen from he shore. (Hin See Fig. 9-45b.) m 1 Fig Problem 15. sec. 9-5 The Linear Momenum of a Ssem of Paricles 18 A 0.70 kg ball moving horizonall Boa's displacemen d b (b) a 5.0 m/s srikes a verical wall and rebounds wih speed 2.0 Fig Problem 17. m/s. Wha is he magniude of he change in is linear momenum? 19 ILW A 2100 kg ruck raveling norh a 41 km/h urns eas and acceleraes o 51 km/h. (a) Wha is he change in he ruck s kineic energ? Wha are he (b) magniude and (c) direcion of he change in is momenum? 20 A ime 0, a ball is sruck a ground level and sen over level ground. The momenum p versus during he fligh is D (a) m 2 Dog's displacemen d d PART 1

32 232 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM given b Fig ( p kg m/s p 0 and p kg m/s). A wha iniial angle is he ball launched? (Hin find a soluion ha does no require ou o read he ime of he low poin of he plo.) 21 A 0.30 kg sofball has a veloci p of 15 m/s a an angle of 35 below he horizonal jus before making conac wih he ba.wha is he magniude of he change in momenum of Fig (s) Problem 20. he ball while in conac wih he ba if he ball leaves wih a veloci of (a) 20 m/s, vericall downward, and (b) 20 m/s, horizonall back oward he picher? 22 Figure 9-47 gives an overhead view of he pah aken b a kg cue ball as i bounces from a rail of a pool able. The ball s iniial speed is 2.00 m/s, and he angle u 1 is The bounce reverses he componen of he ball s veloci bu does no aler he componen. Wha are θ 1 θ 2 (a) angle u 2 and (b) he change in he ball s linear momenum in uni-vecor noaion? (The fac ha he ball rolls is irrelevan o he problem.) Fig Problem 22. sec. 9-6 Collision and Impulse 23 Unil his sevenies, Henri LaMohe (Fig. 9-48) ecied audiences b bell-flopping from a heigh of 12 m ino 30 cm of p (kg m/s) waer. Assuming ha he sops jus as he reaches he boom of he waer and esimaing his mass, find he magniude of he impulse on him from he waer. 24 In Februar 1955, a pararooper fell 370 m from an airplane wihou being able o open his chue bu happened o land in snow, suffering onl minor injuries. Assume ha his speed a impac was 56 m/s (erminal speed), ha his mass (including gear) was 85 kg, and ha he magniude of he force on him from he snow was a he survivable limi of N. Wha are (a) he minimum deph of snow ha would have sopped him safel and (b) he magniude of he impulse on him from he snow? 25 A 1.2 kg ball drops vericall ono a floor, hiing wih a speed of 25 m/s. I rebounds wih an iniial speed of 10 m/s. (a) Wha impulse acs on he ball during he conac? (b) If he ball is in conac wih he floor for s, wha is he magniude of he average force on he floor from he ball? 26 In a common bu dangerous prank, a chair is pulled awa as a person is moving downward o si on i, causing he vicim o land hard on he floor. Suppose he vicim falls b 0.50 m, he mass ha moves downward is 70 kg, and he collision on he floor lass s. Wha are he magniudes of he (a) impulse and (b) average force acing on he vicim from he floor during he collision? 27 SSM A force in he negaive direcion of an ais is applied for 27 ms o a 0.40 kg ball iniiall moving a 14 m/s in he posiive direcion of he ais. The force varies in magniude, and he impulse has magniude 32.4 N s. Wha are he ball s (a) speed and (b) direcion of ravel jus afer he force is applied? Wha are (c) he average magniude of he force and (d) he direcion of he impulse on he ball? 28 In ae-kwon-do, a hand is slammed down ono a arge a a speed of 13 m/s and comes o a sop during he 5.0 ms collision. Assume ha during he impac he hand is independen of he arm and has a mass of 0.70 kg. Wha are he magniudes of he (a) impulse and (b) average force on he hand from he arge? 29 Suppose a gangser spras Superman s ches wih 3 g bulles a he rae of 100 bulles/min, and he speed of each bulle is 500 m/s. Suppose oo ha he bulles rebound sraigh back wih no change in speed. Wha is he magniude of he average force on Superman s ches? 30 Two average forces. A sead sream of kg snowballs is sho perpendicularl ino a wall a a speed of 4.00 m/s. Each ball sicks o he wall. Figure 9-49 gives he magniude F of he force on he wall as a funcion of ime for wo of he snowball impacs. Impacs occur wih a repeiion ime inerval r 50.0 ms, las a duraion ime inerval d 10 ms, and produce isosceles riangles on he graph, wih each impac reaching a force maimum F ma 200 N. During each impac, wha are he magniudes of (a) he impulse and (b) he average force on he wall? (c) During a ime in- F F ma d d Fig Problem 23. Bell-flopping ino 30 cm of waer. r (George Long/ Spors Illusraed/ Time, Inc.) Fig Problem 30.

33 erval of man impacs, wha is he magniude of he average force on he wall? 31 Jumping up before he elevaor his. Afer he cable snaps and he safe ssem fails, an elevaor cab free-falls from a heigh of 36 m. During he collision a he boom of he elevaor shaf, a 90 kg passenger is sopped in 5.0 ms. (Assume ha neiher he passenger nor he cab rebounds.) Wha are he magniudes of he (a) impulse and (b) average force on he passenger during he collision? If he passenger were o jump upward wih a speed of 7.0 m/s relaive o he cab floor jus before he cab his he boom of he shaf, wha are he magniudes of he (c) impulse and (d) average force (assuming he same sopping ime)? 32 A 5.0 kg o car can move along an ais; Fig gives F of he force acing on he car, which begins a res a ime 0. The scale on he F ais is se b F s 5.0 N. In uni-vecor noaion, wha is p a (a) 4.0 s and (b) 7.0 s, and (c) wha is v a 9.0 s? F (N) Figure 9-51 shows a kg baseball jus before and jus afer i collides wih a ba. Jus Fig Problem 32. before, he ball has veloci v 1 of magniude 12.0 m/s and angle u Jus afer, i is raveling direcl upward wih veloci v 2 of magniude 10.0 m/s. The duraion of he collision is 2.00 ms. Wha are he (a) magniude and (b) direcion (relaive o he posiive 2 v direcion of he ais) of he impulse θ1 v 1 on he ball from he ba? Wha are he (c) magniude and (d) direcion of he average force on he ball from he ba? Fig Problem Basilisk lizards can run across he op of a waer surface (Fig. 9-52). Wih each sep, a lizard firs slaps is foo agains he waer and hen pushes i down ino he waer rapidl enough o form an air cavi around he op of he foo. To avoid having o pull he foo back up agains waer drag in order o complee he sep, he lizard wihdraws he foo before waer can flow ino he F s F s (s) PROBLEMS 36 A 0.25 kg puck is iniiall saionar on an ice surface wih negligible fricion. A ime 0, a horizonal force begins o move he puck.the force is given b F ( )î,wih F in newons and in seconds, and i acs unil is magniude is zero. (a) Wha is he magniude of he impulse on he puck from he force beween s and 1.25 s? (b) Wha is he change in momenum of he puck beween 0 and he insan a which F 0? 37 SSM A soccer plaer kicks a soccer ball of mass 0.45 kg ha is iniiall a res. The foo of he plaer is in conac wih he ball for s, and he force of he kick is given b F() [( ) ( ) 2 ] N 233 air cavi. If he lizard is no o sink, he average upward impulse on he lizard during his full acion of slap, downward push, and wihdrawal mus mach he downward impulse due o he graviaional force. Suppose he mass of a basilisk lizard is 90.0 g, he mass of each foo is 3.00 g, he speed of a foo as i slaps he waer is 1.50 m/s, and he ime for a single sep is s. (a) Wha is he magniude of he impulse on he lizard during he slap? (Assume his impulse is direcl upward.) (b) During he s duraion of a sep, wha is he downward impulse on he lizard due o he graviaional force? (c) Which acion, he slap or he push, provides he primar suppor for he lizard, or are he approimael equal in heir suppor? 35 Figure 9-53 shows an approimae plo of force magniude F versus ime during he collision of a 58 g Superball wih a wall. The iniial veloci of he ball is 34 m/s perpendicular o he wall; he ball rebounds direcl back wih approimael he same speed, also perpendicular o he wall. Wha is F ma,he maimum magniude of he force on he ball from he wall during he collision? F (N) F ma (ms) Fig Problem 35. for s, where is in seconds. Find he magniudes of (a) he impulse on he ball due o he kick, (b) he average force on he ball from he plaer s foo during he period of conac, (c) he maimum force on he ball from he plaer s foo during he period of conac, and (d) he ball s veloci immediael afer i loses conac wih he plaer s foo. 38 In he overhead view of Fig. 9-54, a 300 g ball wih a speed v of 6.0 m/s srikes a wall a an angle u of 30 and hen rebounds wih he 6 PART 1 v v θ θ Fig Problem 34. Lizard running across waer. (Sephen Dalon/Phoo Researchers) Fig Problem 38.

34 234 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM same speed and angle. I is in conac wih he wall for 10 ms. In univecor noaion, wha are (a) he impulse on he ball from he wall and (b) he average force on he wall from he ball? sec. 9-7 Conservaion of Linear Momenum 39 SSM A 91 kg man ling on a surface of negligible fricion shoves a 68 g sone awa from himself, giving i a speed of 4.0 m/s. Wha speed does he man acquire as a resul? 40 A space vehicle is raveling a 4300 km/h relaive o Earh when he ehaused rocke moor (mass 4m) is disengaged and sen backward wih a speed of 82 km/h relaive o he command module (mass m). Wha is he speed of he command module relaive o Earh jus afer he separaion? 41 Figure 9-55 shows a wo-ended rocke ha is iniiall saionar on a fricionless floor, wih is cener a he origin of an ais.the rocke consiss of a cenral block C (of mass M 6.00 kg) and blocks L and R (each of mass m 2.00 kg) on he lef and righ sides. Small eplosions can shoo eiher of he side blocks awa from block C and along he ais. Here is he sequence (1) A ime 0, block L is sho o he lef wih a speed of 3.00 m/s relaive o he veloci ha he eplosion gives he res of he rocke. (2) Ne, a ime 0.80 s, block R is sho o he righ wih a speed of 3.00 m/s relaive o he veloci ha block C hen has. A 2.80 s, wha are (a) he veloci of block C and (b) he posiion of is cener? Fig Resource) L C 0 R Fig Problem An objec, wih mass m and speed v relaive o an observer, eplodes ino wo pieces, one hree imes as massive as he oher; he eplosion akes place in deep space. The less massive piece sops relaive o he observer. How much kineic energ is added o he ssem during he eplosion, as measured in he observer s reference frame? 43 In he Olmpiad of 708 B.C., some ahlees compeing in he sanding long jump used handheld weighs called haleres o lenghen heir jumps (Fig. 9-56).The weighs were swung up in fron jus before lifoff and hen swung down and hrown backward during he fligh. Suppose a modern 78 kg long jumper similarl uses wo 5.50 kg haleres, hrowing hem horizonall o he rear a his maimum heigh such ha heir horizonal veloci is zero relaive o he ground. Le his lifoff veloci be v (9.5î 4.0ĵ) m/s Problem 43. (Réunion des Musées Naionau/Ar wih or wihou he haleres, and assume ha he lands a he lifoff level. Wha disance would he use of he haleres add o his range? 44 In Fig. 9-57, a saionar block eplodes ino wo pieces L and R ha slide across a fricionless floor and hen ino regions wih fricion, where he sop. Piece L,wih a mass of 2.0kg,encouners a coefficien of kineic fricion m L 0.40 and slides o a sop in disance d L 0.15 m. Piece R encouners a coefficien of kineic fricion m R 0.50 and slides o a sop in disance d R 0.25 m. Wha was he mass of he block? µ = 0 µ L µ R d L Fig Problem SSM WWW A 20.0 kg bod is moving hrough space in he posiive direcion of an ais wih a speed of 200 m/s when, due o an inernal eplosion, i breaks ino hree pars. One par, wih a mass of 10.0 kg, moves awa from he poin of eplosion wih aspeed of 100m/s in he posiive direcion. A second par, wih a mass of 4.00 kg, moves in he negaive direcion wih a speed of 500 m/s. (a) In uni-vecor noaion, wha is he veloci of he hird par? (b) How much energ is released in he eplosion? Ignore effecs due o he graviaional force. 46 A 4.0 kg mess ki sliding on a fricionless surface eplodes ino wo 2.0 kg pars 3.0 m/s, due norh, and 5.0 m/s, 30 norh of eas.wha is he original speed of he mess ki? 47 A vessel a res a he origin of an coordinae ssem eplodes ino hree pieces. Jus afer he eplosion, one piece, of mass m,moves wih veloci ( 30 m/s) î and a second piece, also of mass m, moves wih veloci ( 30 m/s) ĵ. The hird piece has mass 3m. Jus afer he eplosion, wha are he (a) magniude and (b) direcion of he veloci of he hird piece? 48 Paricle A and paricle B are held ogeher wih a compressed spring beween hem. When he are released, he spring pushes hem apar, and he hen fl off in opposie direcions, free of he spring. The mass of A is 2.00 imes he mass of B,and he energ sored in he spring was 60 J. Assume ha he spring has negligible mass and ha all is sored energ is ransferred o he paricles. Once ha ransfer is complee, wha are he kineic energies of (a) paricle A and (b) paricle B? sec. 9-9 Inelasic Collisions in One Dimension 49 A bulle of mass 10 g srikes a ballisic pendulum of mass 2.0 kg.the cener of mass of he pendulum rises a verical disance of 12 cm. Assuming ha he bulle remains embedded in he pendulum, calculae he bulle s iniial speed. 50 A 5.20 g bulle moving a 672 m/s srikes a 700 g wooden block a res on a fricionless surface. The bulle emerges, raveling in he same direcion wih is speed reduced o 428 m/s. (a) Wha is he resuling speed of he block? (b) Wha is he speed of he bulle block cener of mass? 51 In Fig. 9-58a, a 3.50g bulle is fired horizonall a wo blocks a res on a fricionless able. The bulle passes hrough block 1 (mass 1.20 kg) and embeds iself in block 2 (mass 1.80 kg). The blocks end up wih speeds v m/s and v m/s (Fig. 9-58b). Neglecing he maerial removed from block 1 b he d R

35 hallida_c09_ v2.qd Page 235 bulle, find he speed of he bulle as i (a) leaves and (b) eners block Fricionless PROBLEMS 235 ances d A 8.2 m and d B 6.1 m.wha are he speeds of (a) car A and (b) car B a he sar of he sliding, jus afer he collision? (c) Assuming ha linear momenum is conserved during he collision, find he speed of car B jus before he collision. (d) Eplain wh his assumpion ma be invalid. PART 1 (a) v 1 v 2 Before A v 0 B d A (b) Fig Problem In Fig. 9-59, a 10 g bulle moving direcl upward a 1000 m/s srikes and passes hrough he cener of mass of a 5.0 kg block iniiall a res. The bulle emerges from he block moving direcl upward a 400 m/s. To wha maimum heigh does he block hen rise above is iniial posiion? Bulle Fig Problem 52. Afer A B d B Fig Problem In Fig. 9-61, a ball of mass m 60 g is sho wih speed v i 22 m/s ino he barrel of a spring gun of mass M 240 g iniiall a res on a fricionless surface. The ball sicks in he barrel a he poin of maimum compression of he spring. Assume ha he increase in hermal energ due o fricion beween he ball and he barrel is negligible. (a) Wha is he speed of he spring gun afer he ball sops in he barrel? (b) Wha fracion of he iniial kineic energ of he ball is sored in he spring? M v i m 53 In Anchorage, collisions of a vehicle wih a moose are so common ha he are referred o wih he abbreviaion MVC. Suppose a 1000 kg car slides ino a saionar 500 kg moose on a ver slipper road, wih he moose being hrown hrough he windshield (a common MVC resul). (a) Wha percen of he original kineic energ is los in he collision o oher forms of energ? A similar danger occurs in Saudi Arabia because of camel vehicle collisions (CVC). (b) Wha percen of he original kineic energ is los if he car his a 300 kg camel? (c) Generall, does he percen loss increase or decrease if he animal mass decreases? 54 A compleel inelasic collision occurs beween wo balls of we pu ha move direcl oward each oher along a verical ais. Jus before he collision, one ball, of mass 3.0 kg, is moving upward a 20 m/s and he oher ball, of mass 2.0 kg, is moving downward a 12 m/s. How high do he combined wo balls of pu rise above he collision poin? (Neglec air drag.) 55 ILW A 5.0 kg block wih a speed of 3.0 m/s collides wih a 10 kg block ha has a speed of 2.0 m/s in he same direcion.afer he collision, he 10 kg block ravels in he original direcion wih a speed of 2.5 m/s. (a) Wha is he veloci of he 5.0 kg block immediael afer he collision? (b) B how much does he oal kineic energ of he ssem of wo blocks change because of he collision? (c) Suppose, insead, ha he 10 kg block ends up wih a speed of 4.0 m/s. Wha hen is he change in he oal kineic energ? (d) Accoun for he resul ou obained in (c). 56 In he before par of Fig. 9-60, car A (mass 1100 kg) is sopped a a raffic ligh when i is rear-ended b car B (mass 1400 kg). Boh cars hen slide wih locked wheels unil he fricional force from he slick road (wih a low m k of 0.13) sops hem, a dis- Fig Problem In Fig. 9-62, block 2 (mass 1.0 kg) is a res on a fricionless surface and ouching he end of an unsreched spring of spring consan 200 N/m. The oher end of he spring is fied o a wall. Block 1 (mass 2.0 kg), raveling a speed v m/s, collides wih block 2, and he wo blocks sick ogeher. When he blocks momenaril sop, b wha disance is he spring compressed? v Fig Problem ILW In Fig. 9-63, block 1 (mass 2.0 kg) is moving righward a 10 m/s and block 2 (mass 5.0 kg) is moving righward a 3.0 m/s. The surface is fricionless, and a spring wih a spring consan of 1120 N/m is fied o block 2. When he blocks collide, he compression of he spring is maimum a he insan he blocks have he same veloci. Find he maimum compression. 1 2 Fig Problem 59. sec Elasic Collisions in One Dimension 60 In Fig. 9-64, block A (mass 1.6 kg) slides ino block B (mass 2.4 kg), along a fricionless surface. The direcions of hree velociies before (i) and afer ( f ) he collision are indicaed; he corresponding

36 236 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM speeds are v Ai 5.5 m/s, v Bi 2.5 v Ai v Bi 68 In Fig. 9-67, block 1 of mass m 1 slides from res along a m/s, and v Bf 4.9 m/s. Wha are he fricionless ramp from heigh h 2.50 m and hen collides wih (a) speed and (b) direcion (lef or saionar block 2, which has mass m m 1.Afer he collision, righ) of veloci vaf? (c) Is he collision block 2 slides ino a region where he coefficien of kineic fricion elasic? m k is and comes o a sop in disance d wihin ha region. v 61 SSM A car wih mass 340 g Af =? v Bf Wha is he value of disance d if he collision is (a) elasic and (b) moving on a fricionless linear air compleel inelasic? rack a an iniial speed of 1.2 m/s 1 undergoes an elasic collision wih Fig an iniiall saionar car of unknown Problem 60. mass. Afer he collision, he firs car coninues in is origi- h Fricionless nal direcion a 0.66 m/s. (a) Wha is he mass of he second car? 2 µ k (b) Wha is is speed afer impac? (c) Wha is he speed of he wocar cener of mass? Fig Problem Two ianium spheres approach each oher head-on wih he same speed and collide elasicall. Afer he collision, one of he spheres, whose mass is 300 g, remains a res. (a) Wha is he mass of he oher sphere? (b) Wha is he speed of he wo-sphere cener of mass if he iniial speed of each sphere is 2.00 m/s? 63 Block 1 of mass m 1 slides along a fricionless floor and ino a one-dimensional elasic collision wih saionar block 2 of mass m 2 3m 1.Prior o he collision,he cener of mass of he woblock ssem had a speed of 3.00 m/s. Aferward, wha are he speeds of (a) he cener of mass and (b) block 2? 64 A seel ball of mass kg is fasened o a cord ha is 70.0 cm long and fied a he far end.the ball is hen released when he cord is horizonal (Fig. 9-65). A he boom of is pah, he ball srikes a 2.50 kg seel block iniiall a res on a fricionless surface. The collision is elasic. Find (a) he speed of he ball and (b) he speed of he block, boh jus afer he collision. 65 SSM A bod of mass 2.0 kg makes an elasic collision wih anoher bod a res and coninues o move in he original direcion bu wih one-fourh of is original speed. (a) Wha is he mass of he oher bod? (b) Wha is he speed of he wo-bod cener of mass if he iniial speed of he 2.0 kg bod was 4.0 m/s? 66 Block 1, wih mass m 1 and speed 4.0 m/s, slides along an ais on a fricionless floor and hen undergoes a one-dimensional elasic collision wih saionar block 2, wih mass m m 1.The wo blocks hen slide ino a region where he coefficien of kineic fricion is 0.50; here he sop. How far ino ha region do (a) block 1 and (b) block 2 slide? 67 In Fig. 9-66, paricle 1 of mass m kg slides righward along an ais on a fricionless floor wih a speed of 2.0 m/s. When i reaches 0, i undergoes a one-dimensional elasic collision wih saionar paricle 2 of mass m kg. When paricle 2 hen reaches a wall a w 70 cm, i bounces from he wall wih no loss of speed. A wha posiion on he ais does paricle 2 hen collide wih paricle 1? w (cm) Fig Problem 67. Fig Problem A small ball of mass m is aligned above a larger ball of mass M 0.63 kg (wih a sligh separaion, as wih he baseball and baskeball of Fig. 9-68a), and he wo are dropped simulaneousl from a heigh of h 1.8 m. (Assume he radius of each ball is negligible relaive o h.) (a) If he larger ball rebounds elasicall from he floor and hen he small ball rebounds elasicall from he larger ball, wha value of m resuls in he larger ball sopping when i collides wih he small ball? (b) Wha heigh does he small ball hen reach (Fig. 9-68b)? 70 In Fig. 9-69, puck 1 of mass m kg is sen sliding across a fricionless lab bench, o undergo a one-dimensional elasic collision wih saionar puck 2. Puck 2 hen slides off he bench and lands a disance d from he base of he bench. Puck 1 rebounds from he collision and slides off he opposie edge of he bench, landing a disance 2d from he base of he bench. Wha is he mass of puck 2? (Hin Be careful wih signs.) 2d (a) Before 1 2 Fig Problem 70. Baseball Baskeball d (b) Afer Fig Problem 69. sec Collisions in Two Dimensions 71 ILW In Fig. 9-21, projecile paricle 1 is an alpha paricle and arge paricle 2 is an ogen nucleus. The alpha paricle is scaered a angle u and he ogen nucleus recoils wih speed m/s and a angle u In aomic mass unis, he mass of he alpha paricle is 4.00 u and he mass of he ogen nucleus is 16.0 u.wha are he (a) final and (b) iniial speeds of he alpha paricle? 72 Ball B,moving in he posiive direcion of an ais a speed v,collides wih saionar ball A a he origin. A and B have differen masses. Afer he collision, B moves in he negaive direcion of he ais a speed v/2. (a) In wha direcion does A move? (b)

37 hallida_c09_ v2.qd Page 237 Show ha he speed of A canno be deermined from he given informaion. 73 Afer a compleel inelasic collision, wo objecs of he same mass and same iniial speed move awa ogeher a half heir iniial speed. Find he angle beween he iniial velociies of he objecs. 74 Two 2.0 kg bodies, A and B,collide.The velociies before he collision are v A (15î 30ĵ) m/s and v B ( 10î 5.0ĵ) m/s. Afer he collision, v A ( 5.0î 20ĵ) m/s. Wha are (a) he final veloci of B and (b) he change in he oal kineic energ (including sign)? 75 A projecile proon wih a speed of 500 m/s collides elasicall wih a arge proon iniiall a res. The wo proons hen move along perpendicular pahs, wih he projecile pah a 60 from he original direcion. Afer he collision, wha are he speeds of (a) he arge proon and (b) he projecile proon? sec Ssems wih Varing Mass A Rocke 76 A 6090 kg space probe moving nose-firs oward Jupier a 105 m/s relaive o he Sun fires is rocke engine, ejecing 80.0 kg of ehaus a a speed of 253 m/s relaive o he space probe.wha is he final veloci of he probe? 77 SSM In Fig. 9-70, wo long barges are moving in he same direcion in sill waer, one wih a speed of 10 km/h and he oher wih a speed of 20 km/h. While he are passing each oher, coal is shoveled from he slower o he faser one a a rae of 1000 kg/min. How much addiional force mus be provided b he driving engines of (a) he faser barge and (b) he slower barge if neiher is o change speed? Assume ha he shoveling is alwas perfecl sidewas and ha he fricional forces beween he barges and he waer do no depend on he mass of he barges. PROBLEMS 237 of which kg is fuel. The rocke engine is hen fired for 250 s while fuel is consumed a he rae of 480 kg/s. The speed of he ehaus producs relaive o he rocke is 3.27 km/s. (a) Wha is he rocke s hrus? Afer he 250 s firing, wha are (b) he mass and (c) he speed of he rocke? Addiional Problems 80 An objec is racked b a radar saion and deermined o have a posiion vecor given b r ( ) î 2700ĵ 300 kˆ, wih r in meers and in seconds. The radar saion s ais poins eas, is ais norh, and is z ais vericall up. If he objec is a 250 kg meeorological missile, wha are (a) is linear momenum, (b) is direcion of moion, and (c) he ne force on i? 81 The las sage of a rocke, which is raveling a a speed of 7600 m/s, consiss of wo pars ha are clamped ogeher a rocke case wih a mass of kg and a paload capsule wih a mass of kg.when he clamp is released, a compressed spring causes he wo pars o separae wih a relaive speed of m/s. Wha are he speeds of (a) he rocke case and (b) he paload afer he have separaed? Assume ha all velociies are along he same line. Find he oal kineic energ of he wo pars (c) before and (d) afer he separae. (e) Accoun for he difference. 82 Pancake collapse of a all building. In he secion of a all building shown in Fig. 9-71a, he infrasrucure of an given floor K mus suppor he weigh W of all higher floors. Normall he infrasrucure is consruced wih a safe facor s so ha i can wihsand an even greaer downward force of sw.if,however,he suppor columns beween K and L suddenl collapse and allow he higher floors o free-fall ogeher ono floor K (Fig. 9-71b), he force in he collision can eceed sw and, afer a brief pause, cause K o collapse ono floor J,which collapses on floor I,and so on unil he ground is reached. Assume ha he floors are separaed b d 4.0 m and have he same mass. Also assume ha when he floors above K free-fall ono K, he collision lass 1.5 ms.under hese simplified condiions, wha value mus he safe facor s eceed o preven pancake collapse of he building? PART 1 N M d L K J I Fig Problem 77. (a) (b) Fig Problem Consider a rocke ha is in deep space and a res relaive o an inerial reference frame. The rocke s engine is o be fired for a cerain inerval. Wha mus be he rocke s mass raio (raio of iniial o final mass) over ha inerval if he rocke s original speed relaive o he inerial frame is o be equal o (a) he ehaus speed (speed of he ehaus producs relaive o he rocke) and (b) 2.0 imes he ehaus speed? 79 SSM ILW A rocke ha is in deep space and iniiall a res relaive o an inerial reference frame has a mass of kg, 83 Relaive is an imporan word. In Fig. 9-72, block L of mass m L 1.00 kg and block R of mass m R kg are held in place wih a compressed spring beween hem. When he blocks are released, he spring sends hem sliding across a fricionless floor. (The spring has negligible mass and falls o he floor afer he L R Fig Problem 83.

38 238 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM blocks leave i.) (a) If he spring gives block L a release speed of 1.20 m/s relaive o he floor, how far does block R ravel in he ne s? (b) If, insead, he spring gives block L a release speed of 1.20 m/s relaive o he veloci ha he spring gives block R,how far does block R ravel in he ne s? 84 Figure 9-73 shows an overhead 1 A view of wo paricles sliding a consan 2 veloci over a fricionless surface. The B paricles have he same mass and he θ θ same iniial speed v 4.00 m/s, and θ θ he collide where heir pahs inersec. C An ais is arranged o bisec he angle 3 4 beween heir incoming pahs, such ha D u The region o he righ of he Fig Problem 84. collision is divided ino four leered secions b he ais and four numbered dashed lines. In wha region or along wha line do he paricles ravel if he collision is (a) compleel inelasic, (b) elasic, and (c) inelasic? Wha are heir final speeds if he collision is (d) compleel inelasic and (e) elasic? 85 Speed deamplifier. In Fig. 9-74, block 1 of mass m 1 slides along an ais on a fricionless floor a speed 4.00 m/s. Then i undergoes a one-dimensional elasic collision wih saionar block 2 of mass m m 1.Ne,block 2 undergoes a one-dimensional elasic collision wih saionar block 3 of mass m m 2.(a) Wha hen is he speed of block 3? Are (b) he speed, (c) he kineic energ, and (d) he momenum of block 3 greaer han, less han, or he same as he iniial values for block 1? Fig Problem Speed amplifier. In Fig. 9-75, block 1 of mass m 1 slides along an ais on a fricionless floor wih a speed of v 1i 4.00 m/s. Then i undergoes a one-dimensional elasic collision wih saionar block 2 of mass m m 1.Ne,block 2 undergoes a onedimensional elasic collision wih saionar block 3 of mass m m 2.(a) Wha hen is he speed of block 3? Are (b) he speed, (c) he kineic energ, and (d) he momenum of block 3 greaer han, less han, or he same as he iniial values for block 1? Fig Problem A ball having a mass of 150 g srikes a wall wih a speed of 5.2 m/s and rebounds wih onl 50% of is iniial kineic energ. (a) Wha is he speed of he ball immediael afer rebounding? (b) Wha is he magniude of he impulse on he wall from he ball? (c) If he ball is in conac wih he wall for 7.6 ms, wha is he magniude of he average force on he ball from he wall during his ime inerval? 88 A spacecraf is separaed ino wo pars b deonaing he eplosive bols ha hold hem ogeher. The masses of he pars are 1200 kg and 1800 kg; he magniude of he impulse on each par from he bols is 300 N s. Wih wha relaive speed do he wo pars separae because of he deonaion? 89 SSM A 1400 kg car moving a 5.3 m/s is iniiall raveling norh along he posiive direcion of a ais.afer compleing a 90 righ hand urn in 4.6 s, he inaenive operaor drives ino a ree, which sops he car in 350 ms. In uni-vecor noaion, wha is he impulse on he car (a) due o he urn and (b) due o he collision? Wha is he magniude of he average force ha acs on he car (c) during he urn and (d) during he collision? (e) Wha is he direcion of he average force during he urn? 90 ILW A cerain radioacive (paren) nucleus ransforms o a differen (daugher) nucleus b emiing an elecron and a neurino. The paren nucleus was a res a he origin of an coordinae ssem. The elecron moves awa from he origin wih linear momenum ( kg m/s) î; he neurino moves awa from he origin wih linear momenum ( kg m/s) ĵ. Wha are he (a) magniude and (b) direcion of he linear momenum of he daugher nucleus? (c) If he daugher nucleus has a mass of kg, wha is is kineic energ? 91 A 75 kg man rides on a 39 kg car moving a a veloci of 2.3 m/s. He jumps off wih zero horizonal veloci relaive o he ground. Wha is he resuling change in he car s veloci, including sign? 92 Two blocks of masses 1.0 kg and 3.0 kg are conneced b a spring and res on a fricionless surface. The are given velociies oward each oher such ha he 1.0 kg block ravels iniiall a 1.7 m/s oward he cener of mass, which remains a res. Wha is he iniial speed of he oher block? 93 SSM A railroad freigh car of mass kg collides wih a saionar caboose car. The couple ogeher, and 27.0% of he iniial kineic energ is ransferred o hermal energ, sound, vibraions, and so on. Find he mass of he caboose. 94 An old Chrsler wih mass 2400 kg is moving along a sraigh srech of road a 80 km/h. I is followed b a Ford wih mass 1600 kg moving a 60 km/h. How fas is he cener of mass of he wo cars moving? 95 SSM In he arrangemen of Fig. 9-21, billiard ball 1 moving a a speed of 2.2 m/s undergoes a glancing collision wih idenical billiard ball 2 ha is a res. Afer he collision, ball 2 moves a speed 1.1 m/s, a an angle of u 2 60.Wha are (a) he magniude and (b) he direcion of he veloci of ball 1 afer he collision? (c) Do he given daa sugges he collision is elasic or inelasic? 96 A rocke is moving awa from he solar ssem a a speed of m/s. I fires is engine, which ejecs ehaus wih a speed of m/s relaive o he rocke. The mass of he rocke a his ime is kg, and is acceleraion is 2.0 m/s 2.(a) Wha is he hrus of he engine? (b) A wha rae, in kilograms per second, is ehaus ejeced during he firing? 97 The hree balls in he overhead view of Fig are idenical. Balls 2 and 3 ouch each oher and are aligned perpendicular o he pah of ball 1. The veloci of ball 1 has magniude v 0 10 m/s and is direced a he conac poin of balls 1 and 2. Afer he collision, wha are he (a) speed and (b) direcion of he veloci of ball 2, he (c) speed and (d) direcion of he veloci of ball 3, and he (e) speed and (f) direcion of he veloci of ball 1? (Hin Wih fricion absen, each impulse is direced along he line connecing he ceners of he colliding balls, normal o he colliding surfaces.) 1 v 0 Fig Problem

39 A45678SF hallida_c09_ v2.qd Page A 0.15 kg ball his a wall wih a veloci of (5.00 m/s) î (6.50 m/s) ĵ (4.00 m/s) kˆ. I rebounds from he wall wih a veloci of (2.00 m/s) î (3.50 m/s) ĵ ( 3.20 m/s) kˆ. Wha are (a) he change in he ball s momenum, (b) he impulse on he ball, and (c) he impulse on he wall? 99 In Fig. 9-77, wo idenical conainers of sugar are conneced b a cord ha passes over a fricionless pulle. The cord and pulle have negligible mass, each conainer and is sugar ogeher have a mass of 500 g, he ceners of he conainers are separaed b 50 mm, and he conainers are held fied a he same heigh. Wha is he horizonal disance beween he cener of conainer 1 and he cener of mass of he wo-conainer ssem (a) iniiall and (b) afer 20 g of sugar is ransferred from conainer 1 o conainer 2? Afer he ransfer and 1 2 Fig Problem 99. afer he conainers are released, (c) in wha direcion and (d) a wha acceleraion magniude does he cener of mass move? 100 In a game of pool, he cue ball srikes anoher ball of he same mass and iniiall a res. Afer he collision, he cue ball moves a 3.50 m/s along a line making an angle of 22.0 wih he cue ball s original direcion of moion, and he second ball has a speed of 2.00 m/s. Find (a) he angle beween he direcion of moion of he second ball and he original direcion of moion of he cue ball and (b) he original speed of he cue ball. (c) Is kineic energ (of he ceners of mass, don consider he roaion) conserved? 101 In Fig. 9-78, a 3.2 kg bo of running shoes slides on a horizonal fricionless able and collides wih a 2.0 kg bo of balle slippers iniiall a res on he edge of he able, a heigh h 0.40 m. The speed of he 3.2 kg bo is 3.0 m/s jus before he collision. If he wo boes sick ogeher because of packing ape on heir sides, wha is heir kineic energ jus before he srike he floor? h Fig Problem In Fig. 9-79, an 80 kg man is on a ladder hanging from a balloon ha has a oal mass of 320 kg (including he baske passenger). The balloon is iniiall saionar relaive o he ground. If he man on he ladder begins o climb a 2.5 m/s relaive o he ladder, (a) in wha direcion and (b) a wha speed does he balloon move? (c) If he man hen sops climbing, wha is he speed of he balloon? 103 In Fig. 9-80, block 1 of mass m kg is a res on a long fricionless able ha is up agains a wall. Block 2 of mass m 2 is placed beween block 1 and he wall and sen sliding o he lef, oward block 1, wih consan speed v 2i.Find he value of m 2 for which boh blocks move wih he same Fig Problem 102. PROBLEMS 239 veloci afer block 2 has collided once wih block 1 and once wih he wall. Assume all collisions are elasic (he collision wih he wall does no change he speed of block 2). 1 v 2i Fig Problem The scrip for an acion movie calls for a small race car (of mass 1500 kg and lengh 3.0 m) o accelerae along a flaop boa (of mass 4000 kg and lengh 14 m), from one end of he boa o he oher, where he car will hen jump he gap beween he boa and a somewha lower dock. You are he echnical advisor for he movie. The boa will iniiall ouch he dock, as in Fig. 9-81; he boa can slide hrough he waer wihou significan resisance; boh he car and he boa can be approimaed as uniform in heir mass disribuion. Deermine wha he widh of he gap will be jus as he car is abou o make he jump. Dock Boa Fig Problem SSM A 3.0 kg objec moving a 8.0 m/s in he posiive direcion of an ais has a one-dimensional elasic collision wih an objec of mass M, iniiall a res.afer he collision he objec of mass M has a veloci of 6.0 m/s in he posiive direcion of he ais.wha is mass M? 106 A 2140 kg railroad flacar, which can move wih negligible fricion, is moionless ne o a plaform. A 242 kg sumo wresler runs a 5.3 m/s along he plaform (parallel o he rack) and hen jumps ono he flacar.wha is he speed of he flacar if he hen (a) sands on i, (b) runs a 5.3 m/s relaive o i in his original direcion, and (c) urns and runs a 5.3 m/s relaive o he flacar opposie his original direcion? 107 SSM A 6100 kg rocke is se for verical firing from he ground. If he ehaus speed is 1200 m/s, how much gas mus be ejeced each second if he hrus (a) is o equal he magniude of he graviaional force on he rocke and (b) is o give he rocke an iniial upward acceleraion of 21 m/s 2? 108 A kg module is aached o a kg shule craf, which moves a 1000 m/s relaive o he saionar main spaceship. Then a small eplosion sends he module backward wih speed m/s relaive o he new speed of he shule craf. As measured b someone on he main spaceship, b wha fracion did he kineic energ of he module and shule craf increase because of he eplosion? 109 SSM (a) How far is he cener of mass of he Earh Moon ssem from he cener of Earh? (Appendi C gives he masses of Earh and he Moon and he disance beween he wo.) (b) Wha percenage of Earh s radius is ha disance? 110 A 140 g ball wih speed 7.8 m/s srikes a wall perpendicularl and rebounds in he opposie direcion wih he same speed. The 2 PART 1

40 240 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM collision lass 3.80 ms. Wha are he magniudes of he (a) impulse and (b) average force on he wall from he ball? 111 SSM A rocke sled wih a mass of 2900 kg moves a 250 m/s on a se of rails. A a cerain poin, a scoop on he sled dips ino a rough of waer locaed beween he racks and scoops waer ino an emp ank on he sled. B appling he principle of conservaion of linear momenum, deermine he speed of he sled afer 920 kg of waer has been scooped up. Ignore an rearding force on he scoop. 112 SSM A pelle gun fires en 2.0 g pelles per second wih a speed of 500 m/s. The pelles are sopped b a rigid wall. Wha are (a) he magniude of he momenum of each pelle, (b) he kineic energ of each pelle, and (c) he magniude of he average force on he wall from he sream of pelles? (d) If each pelle is in conac wih he wall for 0.60 ms, wha is he magniude of he average force on he wall from each pelle during conac? (e) Wh is his average force so differen from he average force calculaed in (c)? 113 A railroad car moves under a grain elevaor a a consan speed of 3.20 m/s. Grain drops ino he car a he rae of 540 kg/min. Wha is he magniude of he force needed o keep he car moving a consan speed if fricion is negligible? 114 Figure 9-82 shows a uniform square plae of edge lengh 6d 6.0 m from which a square piece of edge lengh 2d has been removed.wha are (a) he coordinae and (b) he coordinae of he cener of mass of he remaining piece? ssem. (a) Wha is he speed of he cener of mass of P and Q when he separaion is 0.50 m? (b) A wha disance from P s original posiion do he paricles collide? 117 A collision occurs beween a 2.00 kg paricle raveling wih veloci v 1 ( 4.00 m/s)î ( 5.00 m/s)ĵ and a 4.00 kg paricle raveling wih veloci v 2 (6.00 m/s)î ( 2.00 m/s)ĵ.the collision connecs he wo paricles. Wha hen is heir veloci in (a) uni-vecor noaion and as a (b) magniude and (c) angle? 118 In he wo-sphere arrangemen of Fig. 9-20, assume ha sphere 1 has a mass of 50 g and an iniial heigh of h cm, and ha sphere 2 has a mass of 85 g. Afer sphere 1 is released and collides elasicall wih sphere 2, wha heigh is reached b (a) sphere 1 and (b) sphere 2? Afer he ne (elasic) collision, wha heigh is reached b (c) sphere 1 and (d) sphere 2? (Hin Do no use rounded-off values.) 119 In Fig. 9-83, block 1 slides along an ais on a fricionless floor wih a speed of 0.75 m/s. When i reaches saionar block 2, he wo blocks undergo an elasic collision. The following able gives he mass and lengh of he (uniform) blocks and also he locaions of heir ceners a ime 0.Where is he cener of mass of he wo-block ssem locaed (a) a 0, (b) when he wo blocks firs ouch, and (c) a 4.0 s? Block Mass (kg) Lengh (cm) Cener a m d 3d 3d SSM A ime 0, force F 1 ( 4.00î 5.00ĵ) N acs on an iniiall saionar paricle of mass kg and force F 2 (2.00î 4.00ĵ) N acs on an iniiall saionar paricle of mass kg. From ime 0 o 2.00 ms, wha are he (a) magniude and (b) angle (relaive o he posiive direcion of he ais) of he displacemen of he cener of mass of he woparicle ssem? (c) Wha is he kineic energ of he cener of mass a 2.00 ms? 116 Two paricles P and Q are released from res 1.0 m apar. P has a mass of 0.10 kg, and Q a mass of 0.30 kg. P and Q arac each oher wih a consan force of N. No eernal forces ac on he d d 3d 2d 2d Fig Problem m 0 Fig Problem A bod is raveling a 2.0 m/s along he posiive direcion of an ais; no ne force acs on he bod. An inernal eplosion separaes he bod ino wo pars, each of 4.0 kg, and increases he oal kineic energ b 16 J. The forward par coninues o move in he original direcion of moion. Wha are he speeds of (a) he rear par and (b) he forward par? 121 An elecron undergoes a one-dimensional elasic collision wih an iniiall saionar hdrogen aom. Wha percenage of he elecron s iniial kineic energ is ransferred o kineic energ of he hdrogen aom? (The mass of he hdrogen aom is 1840 imes he mass of he elecron.) 122 A man (weighing 915 N) sands on a long railroad flacar (weighing 2415 N) as i rolls a 18.2 m/s in he posiive direcion of an ais, wih negligible fricion. Then he man runs along he flacar in he negaive direcion a 4.00 m/s relaive o he flacar. Wha is he resuling increase in he speed of he flacar? Ask quesions ge answers, homework help

41 10-1 WHAT IS PHYSICS? As However, so far we have eamined onl he moion of ranslaion, in which an objec moves along a sraigh or curved line, as in Fig. 10-1a. We now urn o he moion of roaion, in which an objec urns abou an ais, as in Fig. 10-1b. You see roaion in nearl ever machine, ou use i ever ime ou open a beverage can wih a pull ab, and ou pa o eperience i ever ime ou go o an amusemen park. Roaion is he ke o man fun aciviies, such as hiing a long drive in golf (he ball needs o roae in order for he air o keep i alof longer) and hrowing a curveball in baseball (he ball needs o roae in order for he air o push i lef or righ). Roaion is also he ke o more serious maers, such as meal failure in aging airplanes. We begin our discussion of roaion b defining he variables for he moion, jus as we did for ranslaion in Chaper 2. As we shall see, he variables for roaion are analogous o hose for one-dimensional moion and, as in Chaper 2, an imporan special siuaion is where he acceleraion (here he roaional acceleraion) is consan. We shall also see ha Newon s second law can be wrien for roaional moion, bu we mus use a new quani called orque insead of jus force. Work and he work kineic energ heorem can also be applied o roaional moion, bu we mus use a new quani called roaional ineria insead of jus mass. In shor, much of wha we have discussed so far can be applied o roaional moion wih, perhaps, a few changes. we have discussed, one focus of phsics is moion The Roaional Variables ROTATION We wish o eamine he roaion of a rigid bod abou a fied ais. A rigid bod is a bod ha can roae wih all is pars locked ogeher and wihou an change in is shape. A fied ais means ha he roaion occurs abou an ais ha does no move. Thus, we shall no eamine an objec like he Sun, because he pars of he Sun (a ball of gas) are no locked ogeher. We also shall no eamine an objec like a bowling ball rolling along a lane, because he ball roaes abou a moving ais (he ball s moion is a miure of roaion and ranslaion). (a) CHAPTER 10 Fig Figure skaer Sasha Cohen in moion of (a) pure ranslaion in a fied direcion and (b) pure roaion abou a verical ais. (a Mike Segar/Reuers/Landov LLC; b Elsa/Ge Images, Inc.) (b) 241