Linear Momentum and Collisions

Transcription

1 chpter 9 Liner Momentum nd Collisions 9.1 Liner Momentum 9. Anlysis Model: Isolted ystem (Momentum) 9.3 Anlysis Model: Nonisolted ystem (Momentum) 9.4 Collisions in One Dimension 9.5 Collisions in Two Dimensions 9.6 The Center of Mss 9.7 ystems of Mny Prticles 9.8 Deformle ystems 9.9 Rocket Propulsion Consider wht hppens when cue ll on pool tle strikes the rest of the lls, s in the photogrph t the right. The white cue ll comes in from the top of the photogrph nd strikes the colored lls. The colored lls, initilly t rest, leve the site with vrious velocities. Becuse the verge force exerted on prticulr ll On pool tle, cue ll collides with collection of other lls. Even though the ensuing process is complicted, it is possile to during the collision is lrge (resulting in lrge ccelertion), the ll cn chieve lrge velocity very rpidly this chpter. (Dvid Leh/tone/Getty Imges) nlyze it with principles involving momentum, which we study in over the very short time intervl tht the ll is in contct with nother ll. Although the force nd ccelertion of given ll re relted y Newton s second lw, they vry in time, mking for complicted sitution! One of the min ojectives of this chpter is to enle you to understnd nd nlyze such events in simple wy. First, we introduce the concept of momentum, which is useful for descriing ojects in motion. The momentum of n oject is relted to oth its mss nd its velocity. The concept of momentum leds us to second conservtion lw, tht of conservtion of momentum. In turn, we identify new momentum versions of nlysis models for isolted 34

2 9.1 Liner Momentum 35 nd nonisolted system. These models re especilly useful for treting prolems tht involve collisions etween ojects nd for nlyzing rocket propulsion. This chpter lso introduces the concept of the center of mss of system of prticles. We find tht the motion of system of prticles cn e descried y the motion of one representtive prticle locted t the center of mss. 9.1 Liner Momentum In Chpter 8, we studied situtions tht re difficult to nlyze with Newton s lws. We were le to solve prolems involving these situtions y identifying system nd pplying conservtion principle, conservtion of energy. Let us consider nother sitution nd see if we cn solve it with the models we hve developed so fr: A 60-kg rcher stnds t rest on frictionless ice nd fires 0.50-kg rrow horizontlly t 50 m/s. With wht velocity does the rcher move cross the ice fter firing the rrow? From Newton s third lw, we know tht the force tht the ow exerts on the rrow is mtched y force in the opposite direction on the ow (nd the rcher). This force cuses the rcher to slide ckwrd on the ice with the speed requested in the prolem. We cnnot determine this speed using motion models such s the prticle under constnt ccelertion ecuse we don t hve ny informtion out the ccelertion of the rcher. We cnnot use force models such s the prticle under net force ecuse we don t know nything out forces in this sitution. Energy models re of no help ecuse we know nothing out the work done in pulling the owstring ck or the elstic potentil energy in the system relted to the tut owstring. Despite our inility to solve the rcher prolem using models lerned so fr, this prolem is very simple to solve if we introduce new quntity tht descries motion, liner momentum. To generte this new quntity, consider n isolted system of two prticles (Fig. 9.1) with msses m 1 nd m moving with velocities v 1 nd v t n instnt of time. Becuse the system is isolted, the only force on one prticle is tht from the other prticle. If force from prticle 1 (for exmple, grvittionl force) cts on prticle, there must e second force equl in mgnitude ut opposite in direction tht prticle exerts on prticle 1. Tht is, the forces on the prticles form Newton s third lw ction rection pir, nd F1 5F 1. We cn express this condition s F1 1 F1 5 0 Let us further nlyze this sitution y incorporting Newton s second lw. At the instnt shown in Figure 9.1, the intercting prticles in the system hve ccelertions corresponding to the forces on them. Therefore, replcing the force on ech prticle with m for the prticle gives m m 5 0 Now we replce ech ccelertion with its definition from Eqution 4.5: m 1 F 1 m v 1 F 1 v Figure 9.1 Two prticles interct with ech other. According to Newton s third lw, we must hve F 1 5F 1. m 1 dv 1 dt 1 m dv dt If the msses m 1 nd m re constnt, we cn ring them inside the derivtive opertion, which gives d1m 1 v 1 1 d1m v 5 0 dt dt d dt 1m 1v 1 1 m v 5 0 (9.1) 5 0

3 36 CHAPTER 9 Liner Momentum nd Collisions Notice tht the derivtive of the sum m 1 v 1 1 m v with respect to time is zero. Consequently, this sum must e constnt. We lern from this discussion tht the quntity mv for prticle is importnt in tht the sum of these quntities for n isolted system of prticles is conserved. We cll this quntity liner momentum: Definition of liner momentum of prticle The liner momentum of prticle or n oject tht cn e modeled s prticle of mss m moving with velocity v is defined to e the product of the mss nd velocity of the prticle: p ; mv (9.) Liner momentum is vector quntity ecuse it equls the product of sclr quntity m nd vector quntity v. Its direction is long v, it hs dimensions ML/T, nd its I unit is kg? m/s. If prticle is moving in n ritrry direction, p hs three components, nd Eqution 9. is equivlent to the component equtions p x 5 mv x p y 5 mv y p z 5 mv z As you cn see from its definition, the concept of momentum 1 provides quntittive distinction etween hevy nd light prticles moving t the sme velocity. For exmple, the momentum of owling ll is much greter thn tht of tennis ll moving t the sme speed. Newton clled the product mv quntity of motion; this term is perhps more grphic description thn our present-dy word momentum, which comes from the Ltin word for movement. We hve seen nother quntity, kinetic energy, tht is comintion of mss nd speed. It would e legitimte question to sk why we need nother quntity, momentum, sed on mss nd velocity. There re cler differences etween kinetic energy nd momentum. First, kinetic energy is sclr, wheres momentum is vector. Consider system of two equl-mss prticles heding towrd ech other long line with equl speeds. There is kinetic energy ssocited with this system ecuse memers of the system re moving. Becuse of the vector nture of momentum, however, the momentum of this system is zero. A second mjor difference is tht kinetic energy cn trnsform to other types of energy, such s potentil energy or internl energy. There is only one type of liner momentum, so we see no such trnsformtions when using momentum pproch to prolem. These differences re sufficient to mke models sed on momentum seprte from those sed on energy, providing n independent tool to use in solving prolems. Using Newton s second lw of motion, we cn relte the liner momentum of prticle to the resultnt force cting on the prticle. We strt with Newton s second lw nd sustitute the definition of ccelertion: F 5 m 5 m dv dt In Newton s second lw, the mss m is ssumed to e constnt. Therefore, we cn ring m inside the derivtive opertion to give us Newton s second lw for prticle F 5 d1mv dt 5 dp dt (9.3) This eqution shows tht the time rte of chnge of the liner momentum of prticle is equl to the net force cting on the prticle. This lterntive form of Newton s second lw is the form in which Newton presented the lw, nd it is ctully more generl thn the form introduced in Chpter 5. In ddition to situtions in which the velocity vector vries with time, we cn use 1 In this chpter, the terms momentum nd liner momentum hve the sme mening. Lter, in Chpter 11, we shll use the term ngulr momentum for different quntity when deling with rottionl motion.

4 9. Anlysis Model: Isolted ystem (Momentum) 37 Eqution 9.3 to study phenomen in which the mss chnges. For exmple, the mss of rocket chnges s fuel is urned nd ejected from the rocket. We cnnot use g F 5 m to nlyze rocket propulsion; we must use momentum pproch, s we will show in ection 9.9. Quick Quiz 9.1 Two ojects hve equl kinetic energies. How do the mgnitudes of their moment compre? () p 1, p () p 1 5 p (c) p 1. p (d) not enough informtion to tell Quick Quiz 9. Your physicl eduction techer throws sell to you t certin speed nd you ctch it. The techer is next going to throw you medicine ll whose mss is ten times the mss of the sell. You re given the following choices: You cn hve the medicine ll thrown with () the sme speed s the sell, () the sme momentum, or (c) the sme kinetic energy. Rnk these choices from esiest to hrdest to ctch. 9. Anlysis Model: Isolted ystem (Momentum) Using the definition of momentum, Eqution 9.1 cn e written d dt 1 p 1 1 p 5 0 Becuse the time derivtive of the totl momentum ptot 5 p1 1 p is zero, we conclude tht the totl momentum of the isolted system of the two prticles in Figure 9.1 must remin constnt: ptot 5 constnt (9.4) or, equivlently, p1i 1 pi 5 p1f 1 pf (9.5) Pitfll Prevention 9.1 Momentum of n Isolted ystem Is Conserved Although the momentum of n isolted system is conserved, the momentum of one prticle within n isolted system is not necessrily conserved ecuse other prticles in the system my e intercting with it. Avoid pplying conservtion of momentum to single prticle. where p 1i nd p i re the initil vlues nd p 1f nd p f re the finl vlues of the moment for the two prticles for the time intervl during which the prticles interct. Eqution 9.5 in component form demonstrtes tht the totl moment in the x, y, nd z directions re ll independently conserved: p 1ix 1 p ix 5 p 1fx 1 p fx p 1iy 1 p iy 5 p 1fy 1 p fy p 1iz 1 p iz 5 p 1fz 1 p fz (9.6) Eqution 9.5 is the mthemticl sttement of new nlysis model, the isolted system (momentum). It cn e extended to ny numer of prticles in n isolted system s we show in ection 9.7. We studied the energy version of the isolted system model in Chpter 8 nd now we hve momentum version. In generl, Eqution 9.5 cn e stted in words s follows: Whenever two or more prticles in n isolted system interct, the totl momentum of the system remins constnt. The momentum version of the isolted system model This sttement tells us tht the totl momentum of n isolted system t ll times equls its initil momentum. Notice tht we hve mde no sttement concerning the type of forces cting on the prticles of the system. Furthermore, we hve not specified whether the forces re conservtive or nonconservtive. We hve lso not indicted whether or not the forces re constnt. The only requirement is tht the forces must e internl to the system. This single requirement should give you hint out the power of this new model.

5 38 CHAPTER 9 Liner Momentum nd Collisions Exmple 9.1 The Archer Let us consider the sitution proposed t the eginning of ection 9.1. A 60-kg rcher stnds t rest on frictionless ice nd fires 0.50-kg rrow horizontlly t 50 m/s (Fig. 9.). With wht velocity does the rcher move cross the ice fter firing the rrow? OLUTION Conceptulize You my hve conceptulized this prolem lredy when it ws introduced t the eginning of ection 9.1. Imgine the rrow eing fired one wy nd the rcher recoiling in the opposite direction. Ctegorize As discussed in ection 9.1, we cnnot solve this prolem with models sed on motion, force, or energy. Nonetheless, we cn solve this prolem very esily with n pproch involving momentum. Let us tke the system to consist of the rcher (including the ow) nd the rrow. The system is not isolted ecuse the grvittionl force nd the norml force from the ice ct on the system. These forces, however, re verticl nd perpendiculr to the motion of the system. Therefore, there re no externl forces in the horizontl direction, nd we cn pply the isolted system (momentum) model in terms of momentum components in this direction. Figure 9. (Exmple 9.1) An rcher fires n rrow horizontlly to the right. Becuse he is stnding on frictionless ice, he will egin to slide to the left cross the ice. Anlyze The totl horizontl momentum of the system efore the rrow is fired is zero ecuse nothing in the system is moving. Therefore, the totl horizontl momentum of the system fter the rrow is fired must lso e zero. We choose the direction of firing of the rrow s the positive x direction. Identifying the rcher s prticle 1 nd the rrow s prticle, we hve m kg, m kg, nd v f 5 50 i^ m/s. Using the isolted system (momentum) model, set the finl momentum of the system equl to the initil vlue of zero: olve this eqution for v 1f nd sustitute numericl vlues: m 1 v 1f 1 m v f 5 0 1f v 5 m 0.50 kg v f 5 m 1 60 kg 150 i^ m/s i^ m/s Finlize The negtive sign for v 1f indictes tht the rcher is moving to the left in Figure 9. fter the rrow is fired, in the direction opposite the direction of motion of the rrow, in ccordnce with Newton s third lw. Becuse the rcher is much more mssive thn the rrow, his ccelertion nd consequent velocity re much smller thn the ccelertion nd velocity of the rrow. Notice tht this prolem sounds very simple, ut we could not solve it with models sed on motion, force, or energy. Our new momentum model, however, shows us tht it not only sounds simple, it is simple! WHAT IF? Wht if the rrow were fired in direction tht mkes n ngle u with the horizontl? How will tht chnge the recoil velocity of the rcher? Answer The recoil velocity should decrese in mgnitude ecuse only component of the velocity of the rrow is in the x direction. Conservtion of momentum in the x direction gives m 1 v 1f 1 m v f cos u 5 0 leding to v 1f 5 m v m f cos u 1 For u 5 0, cos u 5 1 nd the finl velocity of the rcher reduces to the vlue when the rrow is fired horizontlly. For nonzero vlues of u, the cosine function is less thn 1 nd the recoil velocity is less thn the vlue clculted for u 5 0. If u 5 908, then cos u 5 0 nd v 1f 5 0, so there is no recoil velocity. In this cse, the rcher is simply pushed downwrd hrder ginst the ice s the rrow is fired.

6 9.3 Anlysis Model: Nonisolted ystem (Momentum) 39 Exmple 9. Cn We Relly Ignore the Kinetic Energy of the Erth? In ection 7.6, we climed tht we cn ignore the kinetic energy of the Erth when considering the energy of system consisting of the Erth nd dropped ll. Verify this clim. OLUTION Conceptulize Imgine dropping ll t the surfce of the Erth. From your point of view, the ll flls while the Erth remins sttionry. By Newton s third lw, however, the Erth experiences n upwrd force nd therefore n upwrd ccelertion while the ll flls. In the clcultion elow, we will show tht this motion is extremely smll nd cn e ignored. Ctegorize We identify the system s the ll nd the Erth. We ssume there re no forces on the system from outer spce, so the system is isolted. Let s use the momentum version of the isolted system model. Anlyze We egin y setting up rtio of the kinetic energy of the Erth to tht of the ll. We identify v E nd v s the speeds of the Erth nd the ll, respectively, fter the ll hs fllen through some distnce. Use the definition of kinetic energy to set up this rtio: Apply the isolted system (momentum) model: the initil momentum of the system is zero, so set the finl momentum equl to zero: olve the eqution for the rtio of speeds: ustitute this expression for v E /v in Eqution (1): ustitute order-of-mgnitude numers for the msses: (1) K 1 E m E v E 5 1 K m v 5 m E v E m v p i 5 p f 0 5 m v 1 m E v E v E v 5 m m E K E 5 m E m 5 m K m m E m E K E 5 m, 1 kg K m E 10 5 kg, 105 Finlize The kinetic energy of the Erth is very smll frction of the kinetic energy of the ll, so we re justified in ignoring it in the kinetic energy of the system. 9.3 Anlysis Model: Nonisolted ystem (Momentum) According to Eqution 9.3, the momentum of prticle chnges if net force cts on the prticle. The sme cn e sid out net force pplied to system s we will show explicitly in ection 9.7: the momentum of system chnges if net force from the environment cts on the system. This my sound similr to our discussion of energy in Chpter 8: the energy of system chnges if energy crosses the oundry of the system to or from the environment. In this section, we consider nonisolted system. For energy considertions, system is nonisolted if energy trnsfers cross the oundry of the system y ny of the mens listed in ection 8.1. For momentum considertions, system is nonisolted if net force cts on the system for time intervl. In this cse, we cn imgine momentum eing trnsferred to the system from the environment y mens of the net force. Knowing the chnge in momentum cused y force is useful in solving some types of prolems. To uild etter understnding of this importnt concept, let us ssume net force g F cts on prticle nd this force my vry with time. According to Newton s second lw, g F 5 dp /dt, or dp 5 F dt (9.7)

7 40 CHAPTER 9 Liner Momentum nd Collisions We cn integrte this expression to find the chnge in the momentum of prticle when the force cts over some time intervl. If the momentum of the prticle chnges from p i t time t i to p f t time t f, integrting Eqution 9.7 gives t Dp 5 pf f pi 5 3 F dt (9.8) t i To evlute the integrl, we need to know how the net force vries with time. The quntity on the right side of this eqution is vector clled the impulse of the net force g F cting on prticle over the time intervl Dt 5 t f t i : Impulse of force t f I ; 3 F dt (9.9) t i From its definition, we see tht impulse I is vector quntity hving mgnitude equl to the re under the force time curve s descried in Figure 9.3. It is ssumed the force vries in time in the generl mnner shown in the figure nd is nonzero in the time intervl Dt 5 t f t i. The direction of the impulse vector is the sme s the direction of the chnge in momentum. Impulse hs the dimensions of momentum, tht is, ML/T. Impulse is not property of prticle; rther, it is mesure of the degree to which n externl force chnges the prticle s momentum. Comining Equtions 9.8 nd 9.9 gives us n importnt sttement known s the impulse momentum theorem: Dvid Woods/Terr/Coris Impulse momentum theorem for prticle Air gs in utomoiles hve sved countless lives in ccidents. The ir g increses the time intervl during which the pssenger is rought to rest, therey decresing the force on (nd resultnt injury to) the pssenger. The chnge in the momentum of prticle is equl to the impulse of the net force cting on the prticle: Dp 5 I (9.10) This sttement is equivlent to Newton s second lw. When we sy tht n impulse is given to prticle, we men tht momentum is trnsferred from n externl gent to tht prticle. Eqution 9.10 is identicl in form to the conservtion of energy eqution, Eqution 8.1, nd its full expnsion, Eqution 8.. Eqution 9.10 is the most generl sttement of the principle of conservtion of momentum nd is clled the conservtion of momentum eqution. In the cse of momentum pproch, isolted systems tend to pper in prolems more often thn nonisolted systems, so, in prctice, the conservtion of momentum eqution is often identified s the specil cse of Eqution 9.5. The left side of Eqution 9.10 represents the chnge in the momentum of the system, which in this cse is single prticle. The right side is mesure of how much momentum crosses the oundry of the system due to the net force eing pplied to the system. Eqution 9.10 is the mthemticl sttement of new nlysis model, the nonisolted system (momentum) model. Although this eqution is similr in form to Eqution 8.1, there re severl differences in its ppliction to prolems. First, Eqution 9.10 is vector eqution, wheres Eqution 8.1 is sclr eqution. Therefore, directions re importnt for Eqution econd, there is only one type of momentum nd therefore only one wy to store momentum in system. In contrst, s we see from Eqution 8., there re three wys to store energy in system: kinetic, potentil, nd internl. Third, there is only one wy to trnsfer momentum into system: y the ppliction of force on the system over time intervl. Eqution 8. shows six wys we hve identified s trnsferring energy into system. Therefore, there is no expnsion of Eqution 9.10 nlogous to Eqution 8.. Becuse the net force imprting n impulse to prticle cn generlly vry in time, it is convenient to define time-verged net force: 1 F vg ; 1 t f Dt 3 F dt (9.11) t i Here we re integrting force with respect to time. Compre this strtegy with our efforts in Chpter 7, where we integrted force with respect to position to find the work done y the force.

8 9.3 Anlysis Model: Nonisolted ystem (Momentum) 41 where Dt 5 t f t i. (This eqution is n ppliction of the men vlue theorem of clculus.) Therefore, we cn express Eqution 9.9 s I 5 1 F vg Dt (9.1) This time-verged force, shown in Figure 9.3, cn e interpreted s the constnt force tht would give to the prticle in the time intervl Dt the sme impulse tht the time-vrying force gives over this sme intervl. In principle, if g F is known s function of time, the impulse cn e clculted from Eqution 9.9. The clcultion ecomes especilly simple if the force cting on the prticle is constnt. In this cse, 1 g F vg 5 g F, where g F is the constnt net force, nd Eqution 9.1 ecomes I 5 F Dt (9.13) In mny physicl situtions, we shll use wht is clled the impulse pproximtion, in which we ssume one of the forces exerted on prticle cts for short time ut is much greter thn ny other force present. In this cse, the net force g F in Eqution 9.9 is replced with single force F to find the impulse on the prticle. This pproximtion is especilly useful in treting collisions in which the durtion of the collision is very short. When this pproximtion is mde, the single force is referred to s n impulsive force. For exmple, when sell is struck with t, the time of the collision is out 0.01 s nd the verge force tht the t exerts on the ll in this time is typiclly severl thousnd newtons. Becuse this contct force is much greter thn the mgnitude of the grvittionl force, the impulse pproximtion justifies our ignoring the grvittionl forces exerted on the ll nd t. When we use this pproximtion, it is importnt to rememer tht pi nd pf represent the moment immeditely efore nd fter the collision, respectively. Therefore, in ny sitution in which it is proper to use the impulse pproximtion, the prticle moves very little during the collision. Quick Quiz 9.3 Two ojects re t rest on frictionless surfce. Oject 1 hs greter mss thn oject. (i) When constnt force is pplied to oject 1, it ccelertes through distnce d in stright line. The force is removed from oject 1 nd is pplied to oject. At the moment when oject hs ccelerted through the sme distnce d, which sttements re true? () p 1, p () p 1 5 p (c) p 1. p (d) K 1, K (e) K 1 5 K (f) K 1. K (ii) When force is pplied to oject 1, it ccelertes for time intervl Dt. The force is removed from oject 1 nd is pplied to oject. From the sme list of choices, which sttements re true fter oject hs ccelerted for the sme time intervl Dt? ( F ) vg The impulse imprted to the prticle y the force is the re under the curve. F F t i The time-verged net force gives the sme impulse to prticle s does the timevrying force in (). t i Figure 9.3 () A net force cting on prticle my vry in time. () The vlue of the constnt force (o F ) vg (horizontl dshed line) is chosen so tht the re (o F ) vg Dt of the rectngle is the sme s the re under the curve in (). t f t f t t Quick Quiz 9.4 Rnk n utomoile dshord, set elt, nd ir g in terms of () the impulse nd () the verge force ech delivers to frontset pssenger during collision, from gretest to lest. Exmple 9.3 How Good Are the Bumpers? In prticulr crsh test, cr of mss kg collides with wll s shown in Figure 9.4 (pge 4). The initil nd finl velocities of the cr re v i i^ m/s nd v f 5.60 i^ m/s, respectively. If the collision lsts s, find the impulse cused y the collision nd the verge net force exerted on the cr. OLUTION Conceptulize The collision time is short, so we cn imgine the cr eing rought to rest very rpidly nd then moving ck in the opposite direction with reduced speed. continued

9 4 CHAPTER 9 Liner Momentum nd Collisions 9.3 cont. Ctegorize Let us ssume the net force exerted on the cr y the wll nd friction from the ground is lrge compred with other forces on the cr (such s ir resistnce). Furthermore, the grvittionl force nd the norml force exerted y the rod on the cr re perpendiculr to the motion nd therefore do not ffect the horizontl momentum. Therefore, we ctegorize the prolem s one in which we cn pply the impulse pproximtion in the horizontl direction. We lso see tht the cr s momentum chnges due to n impulse from the environment. Therefore, we cn pply the nonisolted system (momentum) model. Before 15.0 m/s After +.60 m/s Volvo Cr Corportion Figure 9.4 (Exmple 9.3) () This cr s momentum chnges s result of its collision with the wll. () In crsh test, much of the cr s initil kinetic energy is trnsformed into energy ssocited with the dmge to the cr. Anlyze Evlute the initil nd finl moment of the cr: Use Eqution 9.10 to find the impulse on the cr: pi 5 mv i kg115.0 i^ m/s i^ kg? m/s pf 5 mv f kg1.60 i^ m/s i^ kg? m/s I 5Dp 5 pf pi i^ kg? m/s i^ kg? m/s i^ kg? m/s Use Eqution 9.1 to evlute the verge net force exerted on the cr: 1 F vg 5 I Dt i^ kg? m/s s i^ N Finlize The net force found ove is comintion of the norml force on the cr from the wll nd ny friction force etween the tires nd the ground s the front of the cr crumples. If the rkes re not operting while the crsh occurs nd the crumpling metl does not interfere with the free rottion of the tires, this friction force could e reltively smll due to the freely rotting wheels. Notice tht the signs of the velocities in this exmple indicte the reversl of directions. Wht would the mthemtics e descriing if oth the initil nd finl velocities hd the sme sign? WHAT IF? Wht if the cr did not reound from the wll? uppose the finl velocity of the cr is zero nd the time intervl of the collision remins t s. Would tht represent lrger or smller net force on the cr? Answer In the originl sitution in which the cr reounds, the net force on the cr does two things during the time intervl: (1) it stops the cr, nd () it cuses the cr to move wy from the wll t.60 m/s fter the collision. If the cr does not reound, the net force is only doing the first of these steps stopping the cr which requires smller force. Mthemticlly, in the cse of the cr tht does not reound, the impulse is I 5Dp 5 pf pi i^ kg? m/s i^ kg? m/s The verge net force exerted on the cr is 1 F vg 5 I Dt i^ kg? m/s i^ N s which is indeed smller thn the previously clculted vlue, s ws rgued conceptully. 9.4 Collisions in One Dimension In this section, we use the isolted system (momentum) model to descrie wht hppens when two prticles collide. The term collision represents n event during which two prticles come close to ech other nd interct y mens of forces. The interction forces re ssumed to e much greter thn ny externl forces present, so we cn use the impulse pproximtion. A collision my involve physicl contct etween two mcroscopic ojects s descried in Active Figure 9.5, ut the notion of wht is ment y collision must e generlized ecuse physicl contct on sumicroscopic scle is ill-

10 9.4 Collisions in One Dimension 43 defined nd hence meningless. To understnd this concept, consider collision on n tomic scle (Active Fig. 9.5) such s the collision of proton with n lph prticle (the nucleus of helium tom). Becuse the prticles re oth positively chrged, they repel ech other due to the strong electrosttic force etween them t close seprtions nd never come into physicl contct. When two prticles of msses m 1 nd m collide s shown in Active Figure 9.5, the impulsive forces my vry in time in complicted wys, such s tht shown in Figure 9.3. Regrdless of the complexity of the time ehvior of the impulsive force, however, this force is internl to the system of two prticles. Therefore, the two prticles form n isolted system nd the momentum of the system must e conserved in ny collision. In contrst, the totl kinetic energy of the system of prticles my or my not e conserved, depending on the type of collision. In fct, collisions re ctegorized s eing either elstic or inelstic depending on whether or not kinetic energy is conserved. An elstic collision etween two ojects is one in which the totl kinetic energy (s well s totl momentum) of the system is the sme efore nd fter the collision. Collisions etween certin ojects in the mcroscopic world, such s illird lls, re only pproximtely elstic ecuse some deformtion nd loss of kinetic energy tke plce. For exmple, you cn her illird ll collision, so you know tht some of the energy is eing trnsferred wy from the system y sound. An elstic collision must e perfectly silent! Truly elstic collisions occur etween tomic nd sutomic prticles. These collisions re descried y the isolted system model for oth energy nd momentum. Furthermore, there must e no trnsformtion of kinetic energy into other types of energy within the system. An inelstic collision is one in which the totl kinetic energy of the system is not the sme efore nd fter the collision (even though the momentum of the system is conserved). Inelstic collisions re of two types. When the ojects stick together fter they collide, s hppens when meteorite collides with the Erth, the collision is clled perfectly inelstic. When the colliding ojects do not stick together ut some kinetic energy is trnsformed or trnsferred wy, s in the cse of ruer ll colliding with hrd surfce, the collision is clled inelstic (with no modifying dver). When the ruer ll collides with the hrd surfce, some of the ll s kinetic energy is trnsformed when the ll is deformed while it is in contct with the surfce. Inelstic collisions re descried y the momentum version of the isolted system model. The system could e isolted for energy, with kinetic energy trnsformed to potentil or internl energy. If the system is nonisolted, there could e energy leving the system y some mens. In this ltter cse, there could lso e some trnsformtion of energy within the system. In either of these cses, the kinetic energy of the system chnges. In the reminder of this section, we investigte the mthemticl detils for collisions in one dimension nd consider the two extreme cses, perfectly inelstic nd elstic collisions. Perfectly Inelstic Collisions Consider two prticles of msses m 1 nd m moving with initil velocities v 1i nd v i long the sme stright line s shown in Active Figure 9.6. The two prticles collide hed-on, stick together, nd then move with some common velocity v f fter the collision. Becuse the momentum of n isolted system is conserved in ny collision, we cn sy tht the totl momentum efore the collision equls the totl momentum of the composite system fter the collision: m 1 v 1i 1 m v i 5 1m 1 1 m v f (9.14) olving for the finl velocity gives f v 5 m 1v 1i 1 m v i (9.15) m 1 1 m F 1 p m 1 m ++ 4He F 1 ACTIVE FIGURE 9.5 () The collision etween two ojects s the result of direct contct. () The collision etween two chrged prticles. Pitfll Prevention 9. Inelstic Collisions Generlly, inelstic collisions re hrd to nlyze without dditionl informtion. Lck of this informtion ppers in the mthemticl representtion s hving more unknowns thn equtions. Before the collision, the prticles move seprtely. m 1 v 1i v i m 1 m v f ACTIVE FIGURE 9.6 m After the collision, the prticles move together. chemtic representtion of perfectly inelstic hed-on collision etween two prticles.

11 44 CHAPTER 9 Liner Momentum nd Collisions Before the collision, the prticles move seprtely. v1i v1f vi m 1 m After the collision, the prticles continue to move seprtely with new velocities. ACTIVE FIGURE 9.7 vf chemtic representtion of n elstic hed-on collision etween two prticles. Elstic Collisions Consider two prticles of msses m 1 nd m moving with initil velocities v 1i nd v i long the sme stright line s shown in Active Figure 9.7. The two prticles collide hed-on nd then leve the collision site with different velocities, v 1f nd v f. In n elstic collision, oth the momentum nd kinetic energy of the system re conserved. Therefore, considering velocities long the horizontl direction in Active Figure 9.7, we hve m 1 v 1i 1 m v i 5 m 1 v 1f 1 m v f (9.16) 1 m 1 v 1i 1 1 m v i 5 1 m 1 v 1f 1 1 m v f (9.17) Becuse ll velocities in Active Figure 9.7 re either to the left or the right, they cn e represented y the corresponding speeds long with lgeric signs indicting direction. We shll indicte v s positive if prticle moves to the right nd negtive if it moves to the left. In typicl prolem involving elstic collisions, there re two unknown quntities, nd Equtions 9.16 nd 9.17 cn e solved simultneously to find them. An lterntive pproch, however one tht involves little mthemticl mnipultion of Eqution 9.17 often simplifies this process. To see how, let us cncel the fctor 1 in Eqution 9.17 nd rewrite it s m 1 (v 1i v 1f ) 5 m (v f v i ) Fctoring oth sides of this eqution gives m 1 (v 1i v 1f ) (v 1i 1 v 1f ) 5 m (v f v i )(v f 1 v i ) (9.18) Pitfll Prevention 9.3 Not Generl Eqution Eqution 9.0 cn only e used in very specific sitution, onedimensionl, elstic collision etween two ojects. The generl concept is conservtion of momentum (nd conservtion of kinetic energy if the collision is elstic) for n isolted system. Next, let us seprte the terms contining m 1 nd m in Eqution 9.16 to otin m 1 (v 1i v 1f ) 5 m (v f v i ) (9.19) To otin our finl result, we divide Eqution 9.18 y Eqution 9.19 nd otin v 1i 1 v 1f 5 v f 1 v i v 1i v i 5 (v 1f v f ) (9.0) This eqution, in comintion with Eqution 9.16, cn e used to solve prolems deling with elstic collisions. This pir of equtions (Eqs nd 9.0) is esier to hndle thn the pir of Equtions 9.16 nd 9.17 ecuse there re no qudrtic terms like there re in Eqution According to Eqution 9.0, the reltive velocity of the two prticles efore the collision, v 1i v i, equls the negtive of their reltive velocity fter the collision, (v 1f v f ). uppose the msses nd initil velocities of oth prticles re known. Equtions 9.16 nd 9.0 cn e solved for the finl velocities in terms of the initil velocities ecuse there re two equtions nd two unknowns: v 1f 5 m 1 m m v 1i 1 v i (9.1) m 1 1 m m 1 1 m m 1 v f 5 v m 1 1 m 1i 1 m m 1 v m 1 1 m i (9.) It is importnt to use the pproprite signs for v 1i nd v i in Equtions 9.1 nd 9.. Let us consider some specil cses. If m 1 5 m, Equtions 9.1 nd 9. show tht v 1f 5 v i nd v f 5 v 1i, which mens tht the prticles exchnge velocities if they hve equl msses. Tht is pproximtely wht one oserves in hed-on illird ll collisions: the cue ll stops nd the struck ll moves wy from the collision with the sme velocity the cue ll hd.

12 9.4 Collisions in One Dimension 45 If prticle is initilly t rest, then v i 5 0, nd Equtions 9.1 nd 9. ecome v 1f 5 m 1 m m 1 1 m v 1i (9.3) m 1 v f 5 v m 1 1 m 1i (9.4) Elstic collision: prticle initilly t rest If m 1 is much greter thn m nd v i 5 0, we see from Equtions 9.3 nd 9.4 tht v 1f < v 1i nd v f < v 1i. Tht is, when very hevy prticle collides hed-on with very light one tht is initilly t rest, the hevy prticle continues its motion unltered fter the collision nd the light prticle reounds with speed equl to out twice the initil speed of the hevy prticle. An exmple of such collision is tht of moving hevy tom, such s urnium, striking light tom, such s hydrogen. If m is much greter thn m 1 nd prticle is initilly t rest, then v 1f < v 1i nd v f < 0. Tht is, when very light prticle collides hed-on with very hevy prticle tht is initilly t rest, the light prticle hs its velocity reversed nd the hevy one remins pproximtely t rest. Quick Quiz 9.5 In perfectly inelstic one-dimensionl collision etween two moving ojects, wht condition lone is necessry so tht the finl kinetic energy of the system is zero fter the collision? () The ojects must hve initil moment with the sme mgnitude ut opposite directions. () The ojects must hve the sme mss. (c) The ojects must hve the sme initil velocity. (d) The ojects must hve the sme initil speed, with velocity vectors in opposite directions. Quick Quiz 9.6 A tle-tennis ll is thrown t sttionry owling ll. The tle-tennis ll mkes one-dimensionl elstic collision nd ounces ck long the sme line. Compred with the owling ll fter the collision, does the tle-tennis ll hve () lrger mgnitude of momentum nd more kinetic energy, () smller mgnitude of momentum nd more kinetic energy, (c) lrger mgnitude of momentum nd less kinetic energy, (d) smller mgnitude of momentum nd less kinetic energy, or (e) the sme mgnitude of momentum nd the sme kinetic energy? Prolem-olving trtegy ONE-DIMENIONAL COLLIION You should use the following pproch when solving collision prolems in one dimension: 1. Conceptulize. Imgine the collision occurring in your mind. Drw simple digrms of the prticles efore nd fter the collision nd include pproprite velocity vectors. At first, you my hve to guess t the directions of the finl velocity vectors.. Ctegorize. Is the system of prticles isolted? If so, ctegorize the collision s elstic, inelstic, or perfectly inelstic. 3. Anlyze. et up the pproprite mthemticl representtion for the prolem. If the collision is perfectly inelstic, use Eqution If the collision is elstic, use Equtions 9.16 nd 9.0. If the collision is inelstic, use Eqution To find the finl velocities in this cse, you will need some dditionl informtion. 4. Finlize. Once you hve determined your result, check to see if your nswers re consistent with the mentl nd pictoril representtions nd tht your results re relistic.

13 46 CHAPTER 9 Liner Momentum nd Collisions Exmple 9.4 The Executive tress Reliever An ingenious device tht illustrtes conservtion of momentum nd kinetic energy is shown in Figure 9.8. It consists of five identicl hrd lls supported y strings of equl lengths. When ll 1 is pulled out nd relesed, fter the lmost-elstic collision etween it nd ll, ll 1 stops nd ll 5 moves out s shown in Figure 9.8. If lls 1 nd re pulled out nd relesed, they stop fter the collision nd lls 4 nd 5 swing out, nd so forth. Is it ever possile tht when ll 1 is relesed, it stops fter the collision nd lls 4 nd 5 will swing out on the opposite side nd trvel with hlf the speed of ll 1 s in Figure 9.8c? OLUTION Conceptulize With the help of Figure 9.8c, imgine one ll coming in from the left nd two lls exiting the collision on the right. Tht is the phenomenon we wnt to test to see if it could ever hppen. Ctegorize Becuse of the very short time intervl etween the rrivl of the ll from the left nd the deprture of the ll(s) from the right, we cn use the impulse pproximtion to ignore the grvittionl forces on the lls nd ctegorize the system of five lls s isolted in terms of oth momentum nd energy. Becuse the lls re hrd, we cn ctegorize the collisions etween them s elstic for purposes of clcultion. Anlyze The momentum of the system efore the collision is mv, where m is the mss of ll 1 nd v is its speed immeditely efore the collision. After the collision, we imgine tht ll 1 stops nd lls 4 nd 5 swing out, ech moving with speed v/. The totl momentum of the system fter the collision would e m(v/) 1 m(v/) 5 mv. Therefore, the momentum of the system is conserved. The kinetic energy of the system immeditely efore the collision is K i 5 1 mv nd tht fter the collision is K f 5 1 m1v/ 1 1 m1v/ 5 1 4mv. Tht shows tht the kinetic energy of the system is not conserved, which is inconsistent with our ssumption tht the collisions re elstic. Finlize Our nlysis shows tht it is not possile for lls 4 nd 5 to swing out when only ll 1 is relesed. The only wy to conserve oth momentum nd kinetic energy of the system is for one ll to move out when one ll is relesed, two lls to move out when two re relesed, nd so on. WHAT IF? Consider wht would hppen if lls 4 nd 5 re glued together. Now wht hppens when ll 1 is pulled out nd relesed? Answer In this sitution, lls 4 nd 5 must move together s single oject fter the collision. We hve rgued tht oth momentum nd energy of the system cnnot e conserved in this cse. We ssumed, however, ll 1 stopped fter striking ll. Wht if we do not mke this ssumption? Consider the conservtion equtions with the ssumption tht ll 1 moves fter the collision. For conservtion of momentum, p i 5 p f mv 1i 5 mv 1f 1 mv 4,5 where v 4,5 refers to the finl speed of the ll 4 ll 5 comintion. Conservtion of kinetic energy gives us K i 5 K f 1 mv 1i 5 1 mv 1f 1 1 1mv 4,5 v v c This cn hppen This cnnot hppen Figure 9.8 (Exmple 9.4) () An executive stress reliever. () If one ll swings down, we see one ll swing out t the other end. (c) Is it possile for one ll to swing down nd two lls to leve the other end with hlf the speed of the first ll? In () nd (c), the velocity vectors shown represent those of the lls immeditely efore nd immeditely fter the collision. Cengge Lerning/Chrles D. Winters v/ v

14 9.4 Collisions in One Dimension cont. Comining these equtions gives v 4,5 5 3v 1i v 1f 5 1 3v 1i Therefore, lls 4 nd 5 move together s one oject fter the collision while ll 1 ounces ck from the collision with one third of its originl speed. Exmple 9.5 Crry Collision Insurnce! An kg cr stopped t trffic light is struck from the rer y 900-kg cr. The two crs ecome entngled, moving long the sme pth s tht of the originlly moving cr. If the smller cr were moving t 0.0 m/s efore the collision, wht is the velocity of the entngled crs fter the collision? OLUTION Conceptulize This kind of collision is esily visulized, nd one cn predict tht fter the collision oth crs will e moving in the sme direction s tht of the initilly moving cr. Becuse the initilly moving cr hs only hlf the mss of the sttionry cr, we expect the finl velocity of the crs to e reltively smll. Ctegorize We identify the system of two crs s isolted in terms of momentum in the horizontl direction nd pply the impulse pproximtion during the short time intervl of the collision. The phrse ecome entngled tells us to ctegorize the collision s perfectly inelstic. Anlyze The mgnitude of the totl momentum of the system efore the collision is equl to tht of the smller cr ecuse the lrger cr is initilly t rest. et the initil momentum of the system equl to the finl momentum of the system: p i 5 p f m 1 v i 5 (m 1 1 m )v f olve for v f nd sustitute numericl vlues: v f 5 m 1v i 1900 kg10.0 m/s m/s m 1 1 m 900 kg kg Finlize Becuse the finl velocity is positive, the direction of the finl velocity of the comintion is the sme s the velocity of the initilly moving cr s predicted. The speed of the comintion is lso much lower thn the initil speed of the moving cr. WHAT IF? uppose we reverse the msses of the crs. Wht if sttionry 900-kg cr is struck y moving kg cr? Is the finl speed the sme s efore? Answer Intuitively, we cn guess tht the finl speed of the comintion is higher thn 6.67 m/s if the initilly moving cr is the more mssive cr. Mthemticlly, tht should e the cse ecuse the system hs lrger momentum if the initilly moving cr is the more mssive one. olving for the new finl velocity, we find v f 5 which is two times greter thn the previous finl velocity. m 1v i kg10.0 m/s m/s m 1 1 m kg kg Exmple 9.6 The Bllistic Pendulum The llistic pendulum (Fig. 9.9, pge 48) is n pprtus used to mesure the speed of fst-moving projectile such s ullet. A projectile of mss m 1 is fired into lrge lock of wood of mss m suspended from some light wires. The projectile emeds in the lock, nd the entire system swings through height h. How cn we determine the speed of the projectile from mesurement of h? continued

15 48 CHAPTER 9 Liner Momentum nd Collisions 9.6 cont. v1a vb m 1 m m 1 m h Cengge Lerning/Chrles D. Winters Figure 9.9 (Exmple 9.6) () Digrm of llistic pendulum. Notice tht v 1A is the velocity of the projectile immeditely efore the collision nd v B is the velocity of the projectile lock system immeditely fter the perfectly inelstic collision. () Multiflsh photogrph of llistic pendulum used in the lortory. OLUTION Conceptulize Figure 9.9 helps conceptulize the sitution. Run the nimtion in your mind: the projectile enters the pendulum, which swings up to some height t which it momentrily comes to rest. Ctegorize The projectile nd the lock form n isolted system in terms of momentum if we identify configurtion A s immeditely efore the collision nd configurtion B s immeditely fter the collision. Becuse the projectile imeds in the lock, we cn ctegorize the collision etween them s perfectly inelstic. Anlyze To nlyze the collision, we use Eqution 9.15, which gives the speed of the system immeditely fter the collision when we ssume the impulse pproximtion. Noting tht v A 5 0, solve Eqution 9.15 for v B : (1) v B 5 m 1v 1A m 1 1 m Ctegorize For the process during which the projectile lock comintion swings upwrd to height h (ending t configurtion we ll cll C), we focus on different system, tht of the projectile, the lock, nd the Erth. We ctegorize this prt of the prolem s one involving n isolted system for energy with no nonconservtive forces cting. Anlyze Write n expression for the totl kinetic energy of the system immeditely fter the collision: ustitute the vlue of v B from Eqution (1) into Eqution (): () K B 5 1 1m 1 1 m v B K B 5 m 1 v 1A 1m 1 1 m This kinetic energy of the system immeditely fter the collision is less thn the initil kinetic energy of the projectile s is expected in n inelstic collision. We define the grvittionl potentil energy of the system for configurtion B to e zero. Therefore, U B 5 0, wheres U C 5 (m 1 1 m )gh. Apply the conservtion of mechnicl energy principle to the system: K B 1 U B 5 K C 1 U C m 1 v 1A 1m 1 1 m m 1 1 m gh olve for v 1A : v 1A 5 m 1 1 m "gh m 1

16 9.4 Collisions in One Dimension cont. Finlize We hd to solve this prolem in two steps. Ech step involved different system nd different nlysis model: isolted system (momentum) for the first step nd isolted system (energy) for the second. Becuse the collision ws ssumed to e perfectly inelstic, some mechnicl energy ws trnsformed to internl energy during the collision. Therefore, it would hve een incorrect to pply the isolted system (energy) model to the entire process y equting the initil kinetic energy of the incoming projectile with the finl grvittionl potentil energy of the projectile lock Erth comintion. Exmple 9.7 A Two-Body Collision with pring A lock of mss m kg initilly moving to the right with speed of 4.00 m/s on frictionless, horizontl trck collides with light spring ttched to second lock of mss m 5.10 kg initilly moving to the left with speed of.50 m/s s shown in Figure The spring constnt is 600 N/m. (A) Find the velocities of the two locks fter the collision. v1i 4.00î m/s vi.50î m/s v1f 3.00î m/s vf k k m 1 m m 1 m x OLUTION Figure 9.10 (Exmple 9.7) A moving lock pproches second moving lock tht is ttched to spring. Conceptulize With the help of Figure 9.10, run n nimtion of the collision in your mind. Figure 9.10 shows n instnt during the collision when the spring is compressed. Eventully, lock 1 nd the spring will gin seprte, so the system will look like Figure 9.10 gin ut with different velocity vectors for the two locks. Ctegorize Becuse the spring force is conservtive, kinetic energy in the system of two locks nd the spring is not trnsformed to internl energy during the compression of the spring. Ignoring ny sound mde when the lock hits the spring, we cn ctegorize the collision s eing elstic nd the system s eing isolted for oth energy nd momentum. Anlyze Becuse momentum of the system is conserved, pply Eqution 9.16: Becuse the collision is elstic, pply Eqution 9.0: Multiply Eqution () y m 1 : Add Equtions (1) nd (3): olve for v f : ustitute numericl vlues: olve Eqution () for v 1f nd sustitute numericl vlues: (1) m 1 v 1i 1 m v i 5 m 1 v 1f 1 m v f () v 1i v i 5 (v 1f v f ) (3) m 1 v 1i m 1 v i 5 m 1 v 1f 1 m 1 v f m 1 v 1i 1 (m m 1 )v i 5 (m 1 1 m )v f v f 5 m 1v 1i 1 1m m 1 v i m 1 1 m kg14.00 m/s kg 1.60 kg1.50 m/s v f m/s 1.60 kg 1.10 kg v 1f 5 v f v 1i 1 v i m/s 4.00 m/s 1 (.50 m/s) m/s (B) Determine the velocity of lock during the collision, t the instnt lock 1 is moving to the right with velocity of m/s s in Figure OLUTION Conceptulize Focus your ttention now on Figure 9.10, which represents the finl configurtion of the system for the time intervl of interest. continued

17 50 CHAPTER 9 Liner Momentum nd Collisions 9.7 cont. Ctegorize Becuse the momentum nd mechnicl energy of the system of two locks nd the spring re conserved throughout the collision, the collision cn e ctegorized s elstic for ny finl instnt of time. Let us now choose the finl instnt to e when lock 1 is moving with velocity of m/s. Anlyze Apply Eqution 9.16: olve for v f : ustitute numericl vlues: m 1 v 1i 1 m v i 5 m 1 v 1f 1 m v f v f 5 m 1v 1i 1 m v i m 1 v 1f m kg14.00 m/s kg1.50 m/s kg13.00 m/s v f 5.10 kg m/s Finlize The negtive vlue for v f mens tht lock is still moving to the left t the instnt we re considering. (C) Determine the distnce the spring is compressed t tht instnt. OLUTION Conceptulize Once gin, focus on the configurtion of the system shown in Figure Ctegorize For the system of the spring nd two locks, no friction or other nonconservtive forces ct within the system. Therefore, we ctegorize the system s isolted in terms of energy with no nonconservtive forces cting. The system lso remins isolted in terms of momentum. Anlyze We choose the initil configurtion of the system to e tht existing immeditely efore lock 1 strikes the spring nd the finl configurtion to e tht when lock 1 is moving to the right t 3.00 m/s. Write conservtion of mechnicl energy eqution for the system: Evlute the energies, recognizing tht two ojects in the system hve kinetic energy nd tht the potentil energy is elstic: ustitute the known vlues nd the result of prt (B): olve for x: K i 1 U i 5 K f 1 U f 1 m 1 v 1i 1 1 m v i m 1 v 1f 1 1 m v f 1 1 kx kg14.00 m/s kg1.50 m/s kg13.00 m/s kg11.74 m/s N/mx x m Finlize This nswer is not the mximum compression of the spring ecuse the two locks re still moving towrd ech other t the instnt shown in Figure Cn you determine the mximum compression of the spring? 9.5 Collisions in Two Dimensions In ection 9., we showed tht the momentum of system of two prticles is conserved when the system is isolted. For ny collision of two prticles, this result implies tht the momentum in ech of the directions x, y, nd z is conserved. An importnt suset of collisions tkes plce in plne. The gme of illirds is fmilir exmple involving multiple collisions of ojects moving on two-dimensionl surfce. For such two-dimensionl collisions, we otin two component equtions for conservtion of momentum: m 1 v 1ix 1 m v ix 5 m 1 v 1fx 1 m v fx m 1 v 1iy 1 m v iy 5 m 1 v 1fy 1 m v fy

18 9.5 Collisions in Two Dimensions 51 where the three suscripts on the velocity components in these equtions represent, respectively, the identifiction of the oject (1, ), initil nd finl vlues (i, f ), nd the velocity component (x, y). Let us consider specific two-dimensionl prolem in which prticle 1 of mss m 1 collides with prticle of mss m initilly t rest s in Active Figure After the collision (Active Fig. 9.11), prticle 1 moves t n ngle u with respect to the horizontl nd prticle moves t n ngle f with respect to the horizontl. This event is clled glncing collision. Applying the lw of conservtion of momentum in component form nd noting tht the initil y component of the momentum of the two-prticle system is zero gives m 1 v 1i 5 m 1 v 1f cos u 1 m v f cos f (9.5) 0 5 m 1 v 1f sin u m v f sin f (9.6) where the minus sign in Eqution 9.6 is included ecuse fter the collision prticle hs y component of velocity tht is downwrd. (The symols v in these prticulr equtions re speeds, not velocity components. The direction of the component vector is indicted explicitly with plus or minus signs.) We now hve two independent equtions. As long s no more thn two of the seven quntities in Equtions 9.5 nd 9.6 re unknown, we cn solve the prolem. If the collision is elstic, we cn lso use Eqution 9.17 (conservtion of kinetic energy) with v i 5 0: 1 m 1 v 1i 5 1 m 1 v 1f 1 1 m v f (9.7) Knowing the initil speed of prticle 1 nd oth msses, we re left with four unknowns (v 1f, v f, u, nd f). Becuse we hve only three equtions, one of the four remining quntities must e given to determine the motion fter the elstic collision from conservtion principles lone. If the collision is inelstic, kinetic energy is not conserved nd Eqution 9.7 does not pply. Prolem-olving trtegy TWO-DIMENIONAL COLLIION The following procedure is recommended when deling with prolems involving collisions etween two prticles in two dimensions. 1. Conceptulize. Imgine the collisions occurring nd predict the pproximte directions in which the prticles will move fter the collision. et up coordinte system nd define your velocities in terms of tht system. It is convenient to hve the x xis coincide with one of the initil velocities. ketch the coordinte system, drw nd lel ll velocity vectors, nd include ll the given informtion.. Ctegorize. Is the system of prticles truly isolted? If so, ctegorize the collision s elstic, inelstic, or perfectly inelstic. 3. Anlyze. Write expressions for the x nd y components of the momentum of ech oject efore nd fter the collision. Rememer to include the pproprite signs for the components of the velocity vectors nd py creful ttention to signs throughout the clcultion. Write expressions for the totl momentum in the x direction efore nd fter the collision nd equte the two. Repet this procedure for the totl momentum in the y direction. Proceed to solve the momentum equtions for the unknown quntities. If the collision is inelstic, kinetic energy is not conserved nd dditionl informtion is proly required. If the collision is perfectly inelstic, the finl velocities of the two ojects re equl. If the collision is elstic, kinetic energy is conserved nd you cn equte the totl kinetic energy of the system efore the collision to the totl kinetic energy fter the collision, providing n dditionl reltionship etween the velocity mgnitudes. Before the collision v 1i m 1 m After the collision v 1f sin θ θ φ vf sinφ ACTIVE FIGURE 9.11 An elstic, glncing collision etween two prticles. v 1f v 1f cos θ v f cos φ v f Pitfll Prevention 9.4 Don t Use Eqution 9.0 Eqution 9.0, relting the initil nd finl reltive velocities of two colliding ojects, is only vlid for one-dimensionl elstic collisions. Do not use this eqution when nlyzing two-dimensionl collisions.

19 5 CHAPTER 9 Liner Momentum nd Collisions 4. Finlize. Once you hve determined your result, check to see if your nswers re consistent with the mentl nd pictoril representtions nd tht your results re relistic. Exmple 9.8 Collision t n Intersection A kg cr trveling est with speed of 5.0 m/s collides t n intersection with 500-kg truck trveling north t speed of 0.0 m/s s shown in Figure 9.1. Find the direction nd mgnitude of the velocity of the wreckge fter the collision, ssuming the vehicles stick together fter the collision. y v f OLUTION Conceptulize Figure 9.1 should help you conceptulize the sitution efore nd fter the collision. Let us choose est to e long the positive x direction nd north to e long the positive y direction. 5.0i ˆ m/s u x 0.0j ˆ m/s Ctegorize Becuse we consider moments immeditely efore nd immeditely fter the collision s defining our time intervl, we ignore the smll effect tht friction would hve on the wheels of the vehicles nd model the system of two vehicles s isolted in terms of momentum. We lso ignore the vehicles sizes nd model them s prticles. The collision is perfectly inelstic ecuse the cr nd the truck stick together fter the collision. Figure 9.1 (Exmple 9.8) An estound cr colliding with northound truck. Anlyze Before the collision, the only oject hving momentum in the x direction is the cr. Therefore, the mgnitude of the totl initil momentum of the system (cr plus truck) in the x direction is tht of only the cr. imilrly, the totl initil momentum of the system in the y direction is tht of the truck. After the collision, let us ssume the wreckge moves t n ngle u with respect to the x xis with speed v f. Equte the initil nd finl moment of the system in the x direction: Equte the initil nd finl moment of the system in the y direction: o p xi 5 o p xf o p yi 5 o p yf (1) m 1 v 1i 5 (m 1 1 m )v f cos u () m v i 5 (m 1 1 m )v f sin u Divide Eqution () y Eqution (1): olve for u nd sustitute numericl vlues: Use Eqution () to find the vlue of v f nd sustitute numericl vlues: m v i m 1 v 1i 5 sin u cos u 5 tn u u5tn 1 m v i kg10.0 m/s 5 tn 1 c m 1 v 1i kg15.0 m/s d v f 5 m v i 1m 1 1 m sin u kg10.0 m/s m/s kg kg sin 53.1 Finlize Notice tht the ngle u is qulittively in greement with Figure 9.1. Also notice tht the finl speed of the comintion is less thn the initil speeds of the two crs. This result is consistent with the kinetic energy of the system eing reduced in n inelstic collision. It might help if you drw the momentum vectors of ech vehicle efore the collision nd the two vehicles together fter the collision. Exmple 9.9 Proton Proton Collision A proton collides elsticlly with nother proton tht is initilly t rest. The incoming proton hs n initil speed of m/s nd mkes glncing collision with the second proton s in Active Figure (At close seprtions, the protons exert repulsive electrosttic force on ech other.) After the collision, one proton moves off t n ngle of to the originl direction of motion nd the second deflects t n ngle of f to the sme xis. Find the finl speeds of the two protons nd the ngle f.

20 9.6 The Center of Mss cont. OLUTION Conceptulize This collision is like tht shown in Active Figure 9.11, which will help you conceptulize the ehvior of the system. We define the x xis to e long the direction of the velocity vector of the initilly moving proton. Ctegorize The pir of protons form n isolted system. Both momentum nd kinetic energy of the system re conserved in this glncing elstic collision. Anlyze Using the isolted system model for oth momentum nd energy for two- dimensionl elstic collision, set up the mthemticl representtion with Equtions 9.5 through 9.7: Rerrnge Equtions (1) nd (): (1) v 1f cos u 1 v f cos f 5 v 1i () v 1f sin u v f sin f 5 0 (3) v 1f 1 v f 5 v 1i v f cos f 5 v 1i v 1f cos u v f sin f 5 v 1f sin u qure these two equtions nd dd them: v f cos f 1 v f sin f 5 v 1i v 1i v 1f cos u 1 v 1f cos u 1 v 1f sin u Incorporte tht the sum of the squres of sine nd cosine for ny ngle is equl to 1: (4) v f 5 v 1i v 1i v 1f cos u 1 v 1f ustitute Eqution (4) into Eqution (3): v 1f 1 (v 1i v 1i v 1f cos u 1 v 1f ) 5 v 1i (5) v 1f v 1i v 1f cos u 5 0 One possile solution of Eqution (5) is v 1f 5 0, which corresponds to hed-on, one-dimensionl collision in which the first proton stops nd the second continues with the sme speed in the sme direction. Tht is not the solution we wnt. Divide oth sides of Eqution (5) y v 1f nd solve for the remining fctor of v 1f : v 1f 5 v 1i cos u 5 ( m/s) cos m/s Use Eqution (3) to find v f : v f 5 "v 1i v 1f 5 " m/s m/s m/s Use Eqution () to find f: () f5sin 1 v 1f sin u 5 sin 1 B m/s sin 37.0 R v f m/s Finlize It is interesting tht u 1 f This result is not ccidentl. Whenever two ojects of equl mss collide elsticlly in glncing collision nd one of them is initilly t rest, their finl velocities re perpendiculr to ech other. 9.6 The Center of Mss In this section, we descrie the overll motion of system in terms of specil point clled the center of mss of the system. The system cn e either group of prticles, such s collection of toms in continer, or n extended oject, such s gymnst leping through the ir. We shll see tht the trnsltionl motion of the center of mss of the system is the sme s if ll the mss of the system were concentrted t tht point. Tht is, the system moves s if the net externl force were pplied to single prticle locted t the center of mss. This ehvior is independent of other motion, such s rottion or virtion of the system or deformtion of the system (for instnce, when gymnst folds her ody). This model, the prticle model, ws introduced in Chpter. Consider system consisting of pir of prticles tht hve different msses nd re connected y light, rigid rod (Active Fig on pge 54). The position of the center of mss of system cn e descried s eing the verge position of the

21 54 CHAPTER 9 Liner Momentum nd Collisions c The system rottes clockwise when force is pplied ove the center of mss. CM The system rottes counterclockwise when force is pplied elow the center of mss. CM The system moves in the direction of the force without rotting when force is pplied t the center of mss. CM ACTIVE FIGURE 9.13 A force is pplied to system of two prticles of unequl mss connected y light, rigid rod. system s mss. The center of mss of the system is locted somewhere on the line joining the two prticles nd is closer to the prticle hving the lrger mss. If single force is pplied t point on the rod ove the center of mss, the system rottes clockwise (see Active Fig. 9.13). If the force is pplied t point on the rod elow the center of mss, the system rottes counterclockwise (see Active Fig. 9.13). If the force is pplied t the center of mss, the system moves in the direction of the force without rotting (see Active Fig. 9.13c). The center of mss of n oject cn e locted with this procedure. The center of mss of the pir of prticles descried in Active Figure 9.14 is locted on the x xis nd lies somewhere etween the prticles. Its x coordinte is given y x CM ; m 1x 1 1 m x (9.8) m 1 1 m For exmple, if x 1 5 0, x 5 d, nd m 5 m 1, we find tht x CM 5 3d. Tht is, the center of mss lies closer to the more mssive prticle. If the two msses re equl, the center of mss lies midwy etween the prticles. We cn extend this concept to system of mny prticles with msses m i in three dimensions. The x coordinte of the center of mss of n prticles is defined to e x CM ; m 1x 1 1 m x 1 m 3 x 3 1 c 1 m n x n m 1 1 m 1 m 3 1 c 1 m n 5 i m i x i i m i 5 i m i x i M 5 1 M i m i x i (9.9) where x i is the x coordinte of the ith prticle nd the totl mss is M ; o i m i where the sum runs over ll n prticles. The y nd z coordintes of the center of mss re similrly defined y the equtions y CM ; 1 M m i y i nd z CM ; 1 i M m i z i (9.30) i The center of mss cn e locted in three dimensions y its position vector r CM. The components of this vector re x CM, y CM, nd z CM, defined in Equtions 9.9 nd Therefore, CM r 5 x CM i^ 1 y CM j^ 1 z CM k^ 5 1 M i m i x i i^ 1 1 M i m i y i j^ 1 1 M i m i z i k^ r CM ; 1 M i m i r i (9.31) where r i is the position vector of the ith prticle, defined y i r ; x i i^ 1 y i j^ 1 z i k^ Although locting the center of mss for n extended oject is somewht more cumersome thn locting the center of mss of system of prticles, the sic ides we hve discussed still pply. Think of n extended oject s system contining lrge numer of smll mss elements such s the cue in Figure Becuse the seprtion etween elements is very smll, the oject cn e considered to hve continuous mss distriution. By dividing the oject into elements of mss Dm i with coordintes x i, y i, z i, we see tht the x coordinte of the center of mss is pproximtely y m 1 x CM x 1 x CM m x x CM < 1 M i x i Dm i ACTIVE FIGURE 9.14 The center of mss of two prticles of unequl mss on the x xis is locted t x CM, point etween the prticles, closer to the one hving the lrger mss. with similr expressions for y CM nd z CM. If we let the numer of elements n pproch infinity, the size of ech element pproches zero nd x CM is given precisely. In this limit, we replce the sum y n integrl nd Dm i y the differentil element dm: x CM 5 lim Dm i 0 1 M i x i Dm i 5 1 M 3 x dm (9.3)

22 9.6 The Center of Mss 55 Likewise, for y CM nd z CM we otin y CM 5 1 M 3 y dm nd z CM 5 1 M 3 z dm (9.33) We cn express the vector position of the center of mss of n extended oject in the form r CM 5 1 M 3 r dm (9.34) which is equivlent to the three expressions given y Equtions 9.3 nd The center of mss of ny symmetric oject of uniform density lies on n xis of symmetry nd on ny plne of symmetry. For exmple, the center of mss of uniform rod lies in the rod, midwy etween its ends. The center of mss of sphere or cue lies t its geometric center. Becuse n extended oject is continuous distriution of mss, ech smll mss element is cted upon y the grvittionl force. The net effect of ll these forces is equivlent to the effect of single force Mg cting through specil point, clled the center of grvity. If g is constnt over the mss distriution, the center of grvity coincides with the center of mss. If n extended oject is pivoted t its center of grvity, it lnces in ny orienttion. The center of grvity of n irregulrly shped oject such s wrench cn e determined y suspending the oject first from one point nd then from nother. In Figure 9.16, wrench is hung from point A nd verticl line AB (which cn e estlished with plum o) is drwn when the wrench hs stopped swinging. The wrench is then hung from point C, nd second verticl line CD is drwn. The center of grvity is hlfwy through the thickness of the wrench, under the intersection of these two lines. In generl, if the wrench is hung freely from ny point, the verticl line through this point must pss through the center of grvity. Quick Quiz 9.7 A sell t of uniform density is cut t the loction of its center of mss s shown in Figure Which piece hs the smller mss? () the piece on the right () the piece on the left (c) oth pieces hve the sme mss (d) impossile to determine Figure 9.17 (Quick Quiz 9.7) A sell t cut t the loction of its center of mss. z An extended oject cn e considered to e distriution of smll elements of mss m i. y r i r CM m i CM Figure 9.15 The center of mss is locted t the vector position r CM, which hs coordintes x CM, y CM, nd z CM. The wrench is hung freely first from point A nd then from point C. The intersection of the two lines AB nd CD loctes the center of grvity. Figure 9.16 An experimentl technique for determining the center of grvity of wrench. A B A C D B x Exmple 9.10 The Center of Mss of Three Prticles A system consists of three prticles locted s shown in Figure Find the center of mss of the system. The msses of the prticles re m 1 5 m kg nd m kg. y (m) 3 m 3 OLUTION Conceptulize Figure 9.18 shows the three msses. Your intuition should tell you tht the center of mss is locted somewhere in the region etween the lue prticle nd the pir of tn prticles s shown in the figure. Figure 9.18 (Exmple 9.10) Two prticles re locted on the x xis, nd single prticle is locted on the y xis s shown. The vector indictes the loction of the system s center of mss. 1 0 r CM m 1 m 1 x (m) 3 continued

23 56 CHAPTER 9 Liner Momentum nd Collisions 9.10 cont. Ctegorize We ctegorize this exmple s sustitution prolem ecuse we will e using the equtions for the center of mss developed in this section. Use the defining equtions for the coordintes of the center of mss nd notice tht z CM 5 0: x CM 5 1 M m i x i 5 m 1x 1 1 m x 1 m 3 x 3 i m 1 1 m 1 m kg11.0 m kg1.0 m kg kg kg 1.0 kg 3.0 kg? m m 4.0 kg y CM 5 1 M m i y i 5 m 1y 1 1 m y 1 m 3 y 3 i m 1 1 m 1 m kg kg kg1.0 m kg 4.0 kg? m m 4.0 kg Write the position vector of the center of mss: CM r ; x CM i^ 1 y CM j^ i^ j^ m Exmple 9.11 The Center of Mss of Rod (A) how tht the center of mss of rod of mss M nd length L lies midwy etween its ends, ssuming the rod hs uniform mss per unit length. y dm = l dx OLUTION Conceptulize y CM 5 z CM 5 0. The rod is shown ligned long the x xis in Figure 9.19, so x L dx x Ctegorize We ctegorize this exmple s n nlysis prolem ecuse we need to divide the rod into smll mss elements to perform the integrtion in Eqution 9.3. Figure 9.19 (Exmple 9.11) The geometry used to find the center of mss of uniform rod. Anlyze The mss per unit length (this quntity is clled the liner mss density) cn e written s l 5 M/L for the uniform rod. If the rod is divided into elements of length dx, the mss of ech element is dm 5 l dx. Use Eqution 9.3 to find n expression for x CM : ustitute l 5 M/L: x CM 5 1 M 3 x dm 5 1 L M 3 xl dx 5 l M x CM 5 L M M L 5 1 L 0 x L ` 0 5 ll M One cn lso use symmetry rguments to otin the sme result. (B) uppose rod is nonuniform such tht its mss per unit length vries linerly with x ccording to the expression l 5 x, where is constnt. Find the x coordinte of the center of mss s frction of L. OLUTION Conceptulize Becuse the mss per unit length is not constnt in this cse ut is proportionl to x, elements of the rod to the right re more mssive thn elements ner the left end of the rod. Ctegorize This prolem is ctegorized similrly to prt (A), with the dded twist tht the liner mss density is not constnt. Anlyze In this cse, we replce dm in Eqution 9.3 y l dx, where l 5 x.

24 9.6 The Center of Mss cont. Use Eqution 9.3 to find n expression for x CM : x CM 5 1 M 3 x dm 5 1 M 3 L 0 xl dx 5 1 M 3 L 0 xx dx 5 M 3 L 0 x dx 5 L3 3M Find the totl mss of the rod: L L M 5 3 dm 5 3 l dx x dx 5 L ustitute M into the expression for x CM : x CM 5 L3 3L / 5 3L Finlize Notice tht the center of mss in prt (B) is frther to the right thn tht in prt (A). Tht result is resonle ecuse the elements of the rod ecome more mssive s one moves to the right long the rod in prt (B). Exmple 9.1 The Center of Mss of Right Tringle You hve een sked to hng metl sign from single verticl string. The sign hs the tringulr shpe shown in Figure 9.0. The ottom of the sign is to e prllel to the ground. At wht distnce from the left end of the sign should you ttch the support string? OLUTION Conceptulize Figure 9.0 shows the sign hnging from the string. The string must e ttched t point directly ove the center of grvity of the sign, which is the sme s its center of mss ecuse it is in uniform grvittionl field. Ctegorize As in the cse of Exmple 9.11, we ctegorize this exmple s n nlysis prolem ecuse it is necessry to identify infinitesiml mss elements of the sign to perform the integrtion in Eqution 9.3. Anlyze We ssume the tringulr sign hs uniform density nd totl mss M. Becuse the sign is continuous distriution of mss, we must use the integrl expression in Eqution 9.3 to find the x coordinte of the center of mss. We divide the tringle into nrrow strips of width dx nd height y s shown in Figure 9.0, where y is the height of the hypotenuse of the tringle ove the x xis for given vlue of x. The mss of ech strip is the product of the volume of the strip nd the density r of the mteril from which the sign is mde: dm 5 ryt dx, where t is the thickness of the metl sign. The density of the mteril is the totl mss of the sign divided y its totl volume (re of the tringle times thickness). Evlute dm: Use Eqution 9.3 to find the x coordinte of the center of mss: dm 5ryt dx 5 M My 1 yt dx 5 t dx (1) x CM 5 1 M 3 x dm 5 1 M 3 O y x Joe s Cheese hop c dm y dx Figure 9.0 (Exmple 9.1) () A tringulr sign to e hung from single string. () Geometric construction for locting the center of mss. 0 x My dx 5 3 To proceed further nd evlute the integrl, we must express y in terms of x. The line representing the hypotenuse of the tringle in Figure 9.0 hs slope of / nd psses through the origin, so the eqution of this line is y 5 (/)x. 0 xy dx x continued

25 58 CHAPTER 9 Liner Momentum nd Collisions 9.1 cont. ustitute for y in Eqution (1): x CM 5 3 x xdx x dx 5 c x 3 3 d 0 Therefore, the string must e ttched to the sign t distnce two-thirds of the length of the ottom edge from the left end. Finlize This nswer is identicl to tht in prt (B) of Exmple For the tringulr sign, the liner increse in height y with position x mens tht elements in the sign increse in mss linerly long the x xis, just like the liner increse in mss density in Exmple We could lso find the y coordinte of the center of mss of the sign, ut tht is not needed to determine where the string should e ttched. You might try cutting right tringle out of crdord nd hnging it from string so tht the long se is horizontl. Does the string need to e ttched t 3? ystems of Mny Prticles Consider system of two or more prticles for which we hve identified the center of mss. We cn egin to understnd the physicl significnce nd utility of the center of mss concept y tking the time derivtive of the position vector for the center of mss given y Eqution From ection 4.1, we know tht the time derivtive of position vector is y definition the velocity vector. Assuming M remins constnt for system of prticles tht is, no prticles enter or leve the system we otin the following expression for the velocity of the center of mss of the system: Velocity of the center of mss of system of prticles Totl momentum of system of prticles Accelertion of the center of mss of system of prticles CM v 5 d r CM 5 1 dt M m d r i i 5 1 i dt M m i v i (9.35) i where v i is the velocity of the ith prticle. Rerrnging Eqution 9.35 gives Mv CM 5 i m i v i 5 i p i 5 p tot (9.36) Therefore, the totl liner momentum of the system equls the totl mss multiplied y the velocity of the center of mss. In other words, the totl liner momentum of the system is equl to tht of single prticle of mss M moving with velocity v CM. Differentiting Eqution 9.35 with respect to time, we otin the ccelertion of the center of mss of the system: CM 5 dv CM 5 1 dt M m dv i i 5 1 i dt M m i i (9.37) i Rerrnging this expression nd using Newton s second lw gives M CM 5 i m i i 5 i F i (9.38) Newton s second lw for system of prticles where F i is the net force on prticle i. The forces on ny prticle in the system my include oth externl forces (from outside the system) nd internl forces (from within the system). By Newton s third lw, however, the internl force exerted y prticle 1 on prticle, for exmple, is equl in mgnitude nd opposite in direction to the internl force exerted y prticle on prticle 1. Therefore, when we sum over ll internl force vectors in Eqution 9.38, they cncel in pirs nd we find tht the net force on the system is cused only y externl forces. We cn then write Eqution 9.38 in the form F ext 5 M CM (9.39)

26 9.7 ystems of Mny Prticles 59 Tht is, the net externl force on system of prticles equls the totl mss of the system multiplied y the ccelertion of the center of mss. Compring Eqution 9.39 with Newton s second lw for single prticle, we see tht the prticle model we hve used in severl chpters cn e descried in terms of the center of mss: The center of mss of system of prticles hving comined mss M moves like n equivlent prticle of mss M would move under the influence of the net externl force on the system. Let us integrte Eqution 9.39 over finite time intervl: 3 F ext dt 5 3 M CM dt 5 3 M dv CM dt Notice tht this eqution cn e written s dt 5 M 3 dv CM 5 M Dv CM Dp tot 5 I (9.40) where I is the impulse imprted to the system y externl forces nd p tot is the momentum of the system. Eqution 9.40 is the generliztion of the impulse momentum theorem for prticle (Eq. 9.10) to system of mny prticles. It is lso the mthemticl representtion of the nonisolted system (momentum) model for system of mny prticles. Finlly, if the net externl force on system is zero, it follows from Eqution 9.39 tht Impulse momentum theorem for system of prticles so M CM 5 M dv CM dt 5 0 Mv CM 5 p tot 5 constnt 1when F ext 5 0 (9.41) Tht is, the totl liner momentum of system of prticles is conserved if no net externl force is cting on the system. It follows tht for n isolted system of prticles, oth the totl momentum nd the velocity of the center of mss re constnt in time. This sttement is generliztion of the isolted system (momentum) model for mny-prticle system. uppose the center of mss of n isolted system consisting of two or more memers is t rest. The center of mss of the system remins t rest if there is no net force on the system. For exmple, consider system of swimmer stnding on rft, with the system initilly t rest. When the swimmer dives horizontlly off the rft, the rft moves in the direction opposite tht of the swimmer nd the center of mss of the system remins t rest (if we neglect friction etween rft nd wter). Furthermore, the liner momentum of the diver is equl in mgnitude to tht of the rft, ut opposite in direction. Quick Quiz 9.8 A cruise ship is moving t constnt speed through the wter. The vctioners on the ship re eger to rrive t their next destintion. They decide to try to speed up the cruise ship y gthering t the ow (the front) nd running together towrd the stern (the ck) of the ship. (i) While they re running towrd the stern, is the speed of the ship () higher thn it ws efore, () unchnged, (c) lower thn it ws efore, or (d) impossile to determine? (ii) The vctioners stop running when they rech the stern of the ship. After they hve ll stopped running, is the speed of the ship () higher thn it ws efore they strted running, () unchnged from wht it ws efore they strted running, (c) lower thn it ws efore they strted running, or (d) impossile to determine?

27 60 CHAPTER 9 Liner Momentum nd Collisions Conceptul Exmple 9.13 Exploding Projectile A projectile fired into the ir suddenly explodes into severl frgments (Fig. 9.1). (A) Wht cn e sid out the motion of the center of mss of the system mde up of ll the frgments fter the explosion? OLUTION Neglecting ir resistnce, the only externl force on the projectile is the grvittionl force. Therefore, if the projectile did not explode, it would continue to move long the prolic pth indicted y the dshed line in Figure 9.1. Becuse the forces cused y the explosion re internl, they do not ffect the motion of the center of mss of the system (the frgments). Therefore, fter the explosion, the center of mss of the frgments follows the sme prolic pth the projectile would hve followed if no explosion hd occurred. (B) If the projectile did not explode, it would lnd t distnce R from its lunch point. uppose the projectile explodes nd splits into two pieces of equl mss. One piece lnds t distnce R from the lunch point. Where does the other piece lnd? R FIGURE 9.1 (Conceptul Exmple 9.13) When projectile explodes into severl frgments, the center of mss of the system mde up of ll the frgments follows the sme prolic pth the projectile would hve tken hd there een no explosion. OLUTION As discussed in prt (A), the center of mss of the two-piece system lnds t distnce R from the lunch point. One of the pieces lnds t frther distnce R from the lnding point (or distnce R from the lunch point), to the right in Figure 9.1. Becuse the two pieces hve the sme mss, the other piece must lnd distnce R to the left of the lnding point in Figure 9.1, which plces this piece right ck t the lunch point! Exmple 9.14 The Exploding Rocket A rocket is fired verticlly upwrd. At the instnt it reches n ltitude of m nd speed of v i m/s, it explodes into three frgments hving equl mss. One frgment moves upwrd with speed of v m/s following the explosion. The second frgment hs speed of v 5 40 m/s nd is moving est right fter the explosion. Wht is the velocity of the third frgment immeditely fter the explosion? OLUTION Conceptulize Picture the explosion in your mind, with one piece going upwrd nd second piece moving horizontlly towrd the est. Do you hve n intuitive feeling out the direction in which the third piece moves? Ctegorize This exmple is two-dimensionl prolem ecuse we hve two frgments moving in perpendiculr directions fter the explosion s well s third frgment moving in n unknown direction in the plne defined y the velocity vectors of the other two frgments. We ssume the time intervl of the explosion is very short, so we use the impulse pproximtion in which we ignore the grvittionl force nd ir resistnce. Becuse the forces of the explosion re internl to the system (the rocket), the system is modeled s isolted in terms of momentum. Therefore, the totl momentum p i of the rocket immeditely efore the explosion must equl the totl momentum p f of the frgments immeditely fter the explosion. Anlyze Becuse the three frgments hve equl mss, the mss of ech frgment is M/3, where M is the totl mss of the rocket. We will let v 3 represent the unknown velocity of the third frgment. Using the isolted system (momentum) model, equte the initil nd finl moment of the system nd express the moment in terms of msses nd velocities: pi 5 pf Mv i 5 M 3 v 1 1 M 3 v 1 M 3 v 3 olve for v 3: v 3 5 3v i v 1 v

28 9.8 Deformle ystems cont. ustitute the numericl vlues: v j^ m/s 1450 j^ m/s 140 i^ m/s i^ j^ m/s Finlize Notice tht this event is the reverse of perfectly inelstic collision. There is one oject efore the collision nd three ojects fterwrd. Imgine running movie of the event ckwrd: the three ojects would come together nd ecome single oject. In perfectly inelstic collision, the kinetic energy of the system decreses. If you were to clculte the kinetic energy efore nd fter the event in this exmple, you would find tht the kinetic energy of the system increses. (Try it!) This increse in kinetic energy comes from the potentil energy stored in whtever fuel exploded to cuse the rekup of the rocket. 9.8 Deformle ystems o fr in our discussion of mechnics, we hve nlyzed the motion of prticles or nondeformle systems tht cn e modeled s prticles. The discussion in ection 9.7 cn e pplied to n nlysis of the motion of deformle systems. For exmple, suppose you stnd on skteord nd push off wll, setting yourself in motion wy from the wll. How would we descrie this event? The force from the wll on your hnds moves through no displcement; the force is lwys locted t the interfce etween the wll nd your hnds. Therefore, the force does no work on the system, which is you nd your skteord. Pushing off the wll, however, does indeed result in chnge in the kinetic energy of the system. If you try to use the work kinetic energy theorem, W 5 DK, to descrie this event, you will notice tht the left side of the eqution is zero ut the right side is not zero. The work kinetic energy theorem is not vlid for this event nd is often not vlid for systems tht re deformle. Your ody hs deformed during this event: your rms were ent efore the event, nd they strightened out while you pushed off the wll. To nlyze the motion of deformle systems, we ppel to Eqution 8., the conservtion of energy eqution, nd Eqution 9.40, the impulse momentum theorem. For the exmple of you pushing off the wll on your skteord, identifying the system s you nd the skteord, Eqution 8. gives DE system 5 o T DK 1 DU 5 0 where DK is the chnge in kinetic energy due to the incresed speed of the system nd DU is the decrese in potentil energy stored in the ody from previous mels. This eqution tells us tht the system trnsformed potentil energy into kinetic energy y virtue of the musculr exertion necessry to push off the wll. Notice tht the system is isolted in terms of energy ut nonisolted in terms of momentum. Applying Eqution 9.40 to the system in this sitution gives us Dp tot 5 I m Dv 5 3 F wll dt where F wll is the force exerted y the wll on your hnds, m is the mss of you nd the skteord, nd Dv is the chnge in the velocity of the system during the event. To evlute the right side of this eqution, we would need to know how the force from the wll vries in time. In generl, this process might e complicted. In the cse of constnt forces, or well-ehved forces, however, the integrl on the right side of the eqution cn e evluted.

29 6 CHAPTER 9 Liner Momentum nd Collisions Exmple 9.15 Pushing on pring 3 As shown in Figure 9., two locks re t rest on frictionless, level tle. Both locks hve the sme mss m, nd they re connected y spring of negligile mss. The seprtion distnce of the locks when the spring is relxed is L. During time intervl Dt, constnt force of mgnitude F is pplied horizontlly to the left lock, moving it through distnce x 1 s shown in Figure 9.. During this time intervl, the right lock moves through distnce x. At the end of this time intervl, the force F is removed. (A) Find the resulting speed v CM of the center of mss of the system. m x 1 F m L m m x OLUTION Conceptulize Imgine wht hppens s you push on the left lock. It egins to move to the right in Figure 9., nd the spring egins to compress. As result, the spring pushes to the right on the right lock, which egins to move to the right. At ny given time, the locks re generlly moving with different velocities. As the center of mss of the system moves to the right fter the force is removed, the two locks oscillte ck nd forth with respect to the center of mss. Figure 9. (Exmple 9.15) () Two locks of equl mss re connected y spring. () The left lock is pushed with constnt force of mgnitude F nd moves distnce x 1 during some time intervl. During this sme time intervl, the right lock moves through distnce x. Ctegorize We pply three nlysis models in this prolem: the deformle system of two locks nd spring is modeled s nonisolted system in terms of energy ecuse work is eing done on it y the pplied force. It is lso modeled s nonisolted system in terms of momentum ecuse of the force cting on the system during time intervl. Becuse the pplied force on the system is constnt, the ccelertion of its center of mss is constnt nd the center of mss is modeled s prticle under constnt ccelertion. Anlyze Using the nonisolted system (momentum) model, we pply the impulse momentum theorem to the system of two locks, recognizing tht the force F is constnt during the time intervl Dt while the force is pplied. Write Eqution 9.40 for the system: During the time intervl Dt, the center of mss of the system moves distnce 1 1x 1 1 x. Use this fct to express the time intervl in terms of v CM,vg : Becuse the center of mss is modeled s prticle under constnt ccelertion, the verge velocity of the center of mss is the verge of the initil velocity, which is zero, nd the finl velocity v CM: (1) F Dt 5 1m1v CM 0 5 mv CM Dt 5 Dt 5 1 1x 1 1 x v CM,vg 1 1x 1 1 x v CM 5 1x 1 1 x v CM ustitute this expression into Eqution (1): F 1x 1 1 x v CM 5 mv CM olve for v CM : v CM 5 F 1x 1 1 x Å m (B) Find the totl energy of the system ssocited with virtion reltive to its center of mss fter the force F is removed. OLUTION Anlyze The virtionl energy is ll the energy of the system other thn the kinetic energy ssocited with trnsltionl motion of the center of mss. To find the virtionl energy, we pply the conservtion of energy eqution. The kinetic energy of the system cn e expressed s K 5 K CM 1 K vi, where K vi is the kinetic energy of the locks reltive to 3 Exmple 9.15 ws inspired in prt y C. E. Mungn, A primer on work energy reltionships for introductory physics, The Physics Techer 43:10, 005.

30 9.9 Rocket Propulsion cont. the center of mss due to their virtion. The potentil energy of the system is U vi, which is the potentil energy stored in the spring when the seprtion of the locks is some vlue other thn L. From the nonisolted system (energy) model, express Eqution 8. for this system: Express Eqution () in n lternte form, noting tht K vi 1 U vi 5 E vi : The initil vlues of the kinetic energy of the center of mss nd the virtionl energy of the system re zero. Use this fct nd sustitute for the work done on the system y the force F: olve for the virtionl energy nd use the result from prt (A): () DK CM 1 DK vi 1 DU vi 5 W DK CM 1 DE vi 5 W K CM 1 E vi 5 W 5 Fx 1 E vi 5 Fx 1 K CM 5 Fx 1 1 1mv CM 5 F 1x 1 x Finlize Neither of the two nswers in this exmple depends on the spring length, the spring constnt, or the time intervl. Notice lso tht the mgnitude x 1 of the displcement of the point of ppliction of the pplied force is different from the mgnitude 1 1x 1 1 x of the displcement of the center of mss of the system. This difference reminds us tht the displcement in the definition of work (Eq. 7.1) is tht of the point of ppliction of the force. 9.9 Rocket Propulsion When ordinry vehicles such s crs re propelled, the driving force for the motion is friction. In the cse of the cr, the driving force is the force exerted y the rod on the cr. We cn model the cr s nonisolted system in terms of momentum. An impulse is pplied to the cr from the rodwy, nd the result is chnge in the momentum of the cr s descried y Eqution A rocket moving in spce, however, hs no rod to push ginst. The rocket is n isolted system in terms of momentum. Therefore, the source of the propulsion of rocket must e something other thn n externl force. The opertion of rocket depends on the lw of conservtion of liner momentum s pplied to n isolted system, where the system is the rocket plus its ejected fuel. Rocket propulsion cn e understood y first considering our rcher stnding on frictionless ice in Exmple 9.1. Imgine tht the rcher fires severl rrows horizontlly. For ech rrow fired, the rcher receives compensting momentum in the opposite direction. As more rrows re fired, the rcher moves fster nd fster cross the ice. In ddition to this nlysis in terms of momentum, we cn lso understnd this phenomenon in terms of Newton s second nd third lws. Every time the ow pushes n rrow forwrd, the rrow pushes the ow (nd the rcher) ckwrd, nd these forces result in n ccelertion of the rcher. In similr mnner, s rocket moves in free spce, its liner momentum chnges when some of its mss is ejected in the form of exhust gses. Becuse the gses re given momentum when they re ejected out of the engine, the rocket receives compensting momentum in the opposite direction. Therefore, the rocket is ccelerted s result of the push, or thrust, from the exhust gses. In free spce, the center of mss of the system (rocket plus expelled gses) moves uniformly, independent of the propulsion process. 4 4 The rocket nd the rcher represent cses of the reverse of perfectly inelstic collision: momentum is conserved, ut the kinetic energy of the rocket exhust gs system increses (t the expense of chemicl potentil energy in the fuel), s does the kinetic energy of the rcher rrow system (t the expense of potentil energy from the rcher s previous mels). The force from nitrogen-propelled hnd-controlled device llows n stronut to move out freely in spce without restrictive tethers, using the thrust force from the expelled nitrogen. Courtesy of NAA

31 64 CHAPTER 9 Liner Momentum nd Collisions m M m v pi (M m) v M v v Figure 9.3 Rocket propulsion. () The initil mss of the rocket plus ll its fuel is M 1 Dm t time t, nd its speed is v. () At time t 1 Dt, the rocket s mss hs een reduced to M nd n mount of fuel Dm hs een ejected. The rocket s speed increses y n mount Dv. Expression for rocket propulsion uppose t some time t the mgnitude of the momentum of rocket plus its fuel is (M 1 Dm)v, where v is the speed of the rocket reltive to the Erth (Fig. 9.3). Over short time intervl Dt, the rocket ejects fuel of mss Dm. At the end of this intervl, the rocket s mss is M nd its speed is v 1 Dv, where Dv is the chnge in speed of the rocket (Fig. 9.3). If the fuel is ejected with speed v e reltive to the rocket (the suscript e stnds for exhust, nd v e is usully clled the exhust speed), the velocity of the fuel reltive to the Erth is v v e. Becuse the system of the rocket nd the ejected fuel is isolted, we cn equte the totl initil momentum of the system to the totl finl momentum nd otin 1M 1Dmv 5 M1v 1Dv 1Dm1v v e implifying this expression gives M Dv 5 v e Dm If we now tke the limit s Dt goes to zero, we let Dv dv nd Dm dm. Furthermore, the increse in the exhust mss dm corresponds to n equl decrese in the rocket mss, so dm 5 dm. Notice tht dm is negtive ecuse it represents decrese in mss, so dm is positive numer. Using this fct gives M dv 5 v e dm 5 v e dm (9.4) Now divide the eqution y M nd integrte, tking the initil mss of the rocket plus fuel to e M i nd the finl mss of the rocket plus its remining fuel to e M f. The result is v f M f dm 3 dv 5v e 3 v i M M i v f v i 5 v e ln M i M f (9.43) which is the sic expression for rocket propulsion. First, Eqution 9.43 tells us tht the increse in rocket speed is proportionl to the exhust speed v e of the ejected gses. Therefore, the exhust speed should e very high. econd, the increse in rocket speed is proportionl to the nturl logrithm of the rtio M i /M f. Therefore, this rtio should e s lrge s possile; tht is, the mss of the rocket without its fuel should e s smll s possile nd the rocket should crry s much fuel s possile. The thrust on the rocket is the force exerted on it y the ejected exhust gses. We otin the following expression for the thrust from Newton s second lw nd Eqution 9.4: Thrust 5 M dv dt 5 `v dm e dt ` (9.44) This expression shows tht the thrust increses s the exhust speed increses nd s the rte of chnge of mss (clled the urn rte) increses. Exmple 9.16 Fighting Fire Two firefighters must pply totl force of 600 N to stedy hose tht is dischrging wter t the rte of L/min. Estimte the speed of the wter s it exits the nozzle. OLUTION Conceptulize As the wter leves the hose, it cts in wy similr to the gses eing ejected from rocket engine. As result, force (thrust) cts on the firefighters in direction opposite the direction of motion of the wter. In this cse, we wnt the end of the hose to e modeled s prticle in equilirium rther thn to ccelerte s in the cse of the rocket. Consequently, the firefighters must pply force of mgnitude equl to the thrust in the opposite direction to keep the end of the hose sttionry.

32 ummry cont. Ctegorize This exmple is sustitution prolem in which we use given vlues in n eqution derived in this section. The wter exits t L/min, which is 60 L/s. Knowing tht 1 L of wter hs mss of 1 kg, we estimte tht out 60 kg of wter leves the nozzle ech second. Use Eqution 9.44 for the thrust: Thrust 5 `v e dm dt ` ustitute the known vlues: 600 N 5 0 v e 160 kg/s 0 olve for the exhust speed: v e 5 10 m/s Exmple 9.17 A Rocket in pce A rocket moving in spce, fr from ll other ojects, hs speed of m/s reltive to the Erth. Its engines re turned on, nd fuel is ejected in direction opposite the rocket s motion t speed of m/s reltive to the rocket. (A) Wht is the speed of the rocket reltive to the erth once the rocket s mss is reduced to hlf its mss efore ignition? OLUTION Conceptulize Figure 9.3 shows the sitution in this prolem. From the discussion in this section nd scenes from science fiction movies, we cn esily imgine the rocket ccelerting to higher speed s the engine opertes. Ctegorize This prolem is sustitution prolem in which we use given vlues in the equtions derived in this section. olve Eqution 9.43 for the finl velocity nd sustitute the known vlues: v f 5 v i 1 v e ln M i M f m/s m/sln 0.50M i m/s M i (B) Wht is the thrust on the rocket if it urns fuel t the rte of 50 kg/s? OLUTION Use Eqution 9.44 nd the result from prt (A), noting tht dm/dt 5 50 kg/s: Thrust 5 `v dm e dt ` m/s150 kg/s N Definitions ummry The liner momentum p of prticle of mss m moving with velocity v is p ; mv (9.) The impulse imprted to prticle y net force g F is equl to the time integrl of the force: t f I ; 3 F dt (9.9) t i continued

33 66 CHAPTER 9 Liner Momentum nd Collisions An inelstic collision is one for which the totl kinetic energy of the system of colliding prticles is not conserved. A perfectly inelstic collision is one in which the colliding prticles stick together fter the collision. An elstic collision is one in which the kinetic energy of the system is conserved. The position vector of the center of mss of system of prticles is defined s r CM ; 1 M m i r i (9.31) i where M 5 i m i is the totl mss of the system nd r i is the position vector of the ith prticle. Concepts nd Principles The position vector of the center of mss of n extended oject cn e otined from the integrl expression r CM 5 1 M 3 r dm (9.34) The velocity of the center of mss for system of prticles is v CM 5 1 M m i v i (9.35) i The totl momentum of system of prticles equls the totl mss multiplied y the velocity of the center of mss. Newton s second lw pplied to system of prticles is F ext 5 M CM (9.39) where CM is the ccelertion of the center of mss nd the sum is over ll externl forces. The center of mss moves like n imginry prticle of mss M under the influence of the resultnt externl force on the system. Anlysis Models for Prolem olving ystem oundry Impulse ystem oundry Momentum Momentum The chnge in the totl momentum of the system is equl to the totl impulse on the system. Nonisolted ystem (Momentum). If system intercts with its environment in the sense tht there is n externl force on the system, the ehvior of the system is descried y the impulse momentum theorem: Dp tot 5 I (9.40) With no impulse on the system, the totl momentum of the system is constnt. Isolted ystem (Momentum). The principle of conservtion of liner momentum indictes tht the totl momentum of n isolted system (no externl forces) is conserved regrdless of the nture of the forces etween the memers of the system: Mv CM 5 p tot 5 constnt 1when F ext 5 0 (9.41) In the cse of two-prticle system, this principle cn e expressed s p1i 1 pi 5 p1f 1 pf (9.5) The system my e isolted in terms of momentum ut nonisolted in terms of energy, s in the cse of inelstic collisions.

34 Ojective Questions 67 Ojective Questions 1. A cr of mss m trveling t speed v crshes into the rer of truck of mss m tht is t rest nd in neutrl t n intersection. If the collision is perfectly inelstic, wht is the speed of the comined cr nd truck fter the collision? () v () v/ (c) v/3 (d) v (e) None of those nswers is correct.. A hed-on, elstic collision occurs etween two illird lls of equl mss. If red ll is trveling to the right with speed v nd lue ll is trveling to the left with speed 3v efore the collision, wht sttement is true concerning their velocities susequent to the collision? Neglect ny effects of spin. () The red ll trvels to the left with speed v, while the lue ll trvels to the right with speed 3v. () The red ll trvels to the left with speed v, while the lue ll continues to move to the left with speed v. (c) The red ll trvels to the left with speed 3v, while the lue ll trvels to the right with speed v. (d) Their finl velocities cnnot e determined ecuse momentum is not conserved in the collision. (e) The velocities cnnot e determined without knowing the mss of ech ll. 3. A 3-kg oject moving to the right on frictionless, horizontl surfce with speed of m/s collides hed-on nd sticks to -kg oject tht is initilly moving to the left with speed of 4 m/s. After the collision, which sttement is true? () The kinetic energy of the system is 0 J. () The momentum of the system is 14 kg? m/s. (c) The kinetic energy of the system is greter thn 5 J ut less thn 0 J. (d) The momentum of the system is kg? m/s. (e) The momentum of the system is less thn the momentum of the system efore the collision. 4. A -kg oject moving to the right with speed of 4 m/s mkes hed-on, elstic collision with 1-kg oject tht is initilly t rest. The velocity of the 1-kg oject fter the collision is () greter thn 4 m/s, () less thn 4 m/s, (c) equl to 4 m/s, (d) zero, or (e) impossile to sy sed on the informtion provided. 5. A 5-kg crt moving to the right with speed of 6 m/s collides with concrete wll nd reounds with speed of m/s. Wht is the chnge in momentum of the crt? () 0 () 40 kg? m/s (c) 40 kg? m/s (d) 30 kg? m/s (e) 10 kg? m/s 6. A 57.0-g tennis ll is trveling stright t plyer t 1.0 m/s. The plyer volleys the ll stright ck t 5.0 m/s. If the ll remins in contct with the rcket for s, wht verge force cts on the ll? ().6 N () 3.5 N (c) 43.7 N (d) 7.1 N (e) 10 N 7. The momentum of n oject is incresed y fctor of 4 in mgnitude. By wht fctor is its kinetic energy chnged? () 16 () 8 (c) 4 (d) (e) 1 8. The kinetic energy of n oject is incresed y fctor of 4. By wht fctor is the mgnitude of its momentum chnged? () 16 () 8 (c) 4 (d) (e) 1 9. If two prticles hve equl moment, re their kinetic energies equl? () yes, lwys () no, never (c) no, except when their speeds re the sme (d) yes, s long s they move long prllel lines denotes nswer ville in tudent olutions Mnul/tudy Guide 10. If two prticles hve equl kinetic energies, re their moment equl? () yes, lwys () no, never (c) yes, s long s their msses re equl (d) yes, if oth their msses nd directions of motion re the sme (e) yes, s long s they move long prllel lines 11. A 10.0-g ullet is fired into 00-g lock of wood t rest on horizontl surfce. After impct, the lock slides 8.00 m efore coming to rest. If the coefficient of friction etween the lock nd the surfce is 0.400, wht is the speed of the ullet efore impct? () 106 m/s () 166 m/s (c) 6 m/s (d) 86 m/s (e) none of those nswers is correct 1. Two prticles of different mss strt from rest. The sme net force cts on oth of them s they move over equl distnces. How do their finl kinetic energies compre? () The prticle of lrger mss hs more kinetic energy. () The prticle of smller mss hs more kinetic energy. (c) The prticles hve equl kinetic energies. (d) Either prticle might hve more kinetic energy. 13. Two prticles of different mss strt from rest. The sme net force cts on oth of them s they move over equl distnces. How do the mgnitudes of their finl moment compre? () The prticle of lrger mss hs more momentum. () The prticle of smller mss hs more momentum. (c) The prticles hve equl moment. (d) Either prticle might hve more momentum. 14. A sketll is tossed up into the ir, flls freely, nd ounces from the wooden floor. From the moment fter the plyer releses it until the ll reches the top of its ounce, wht is the smllest system for which momentum is conserved? () the ll () the ll plus plyer (c) the ll plus floor (d) the ll plus the Erth (e) momentum is not conserved for ny system 15. A mssive trctor is rolling down country rod. In perfectly inelstic collision, smll sports cr runs into the mchine from ehind. (i) Which vehicle experiences chnge in momentum of lrger mgnitude? () The cr does. () The trctor does. (c) Their momentum chnges re the sme size. (d) It could e either vehicle. (ii) Which vehicle experiences lrger chnge in kinetic energy? () The cr does. () The trctor does. (c) Their kinetic energy chnges re the sme size. (d) It could e either vehicle. 16. A ll is suspended y string tht is tied to fixed point ove wooden lock stnding on end. L u The ll is pulled ck s shown in Figure OQ9.16 nd relesed. m In tril A, the ll reounds elsticlly from the lock. In tril B, two-sided tpe cuses the ll to stick to the lock. In which cse is the ll more likely to knock the lock over? () It is more likely Figure OQ9.16 in tril A. () It is more likely in tril B. (c) It mkes no difference. (d) It could e either cse, depending on other fctors. 17. You re stnding on sucer-shped sled t rest in the middle of frictionless ice rink. Your l prtner throws

35 68 CHAPTER 9 Liner Momentum nd Collisions you hevy Frisee. You tke different ctions in successive experimentl trils. Rnk the following situtions ccording to your finl speed from lrgest to smllest. If your finl speed is the sme in two cses, give them equl rnk. () You ctch the Frisee nd hold onto it. () You ctch the Frisee nd throw it ck to your prtner. (c) You ole the ctch, just touching the Frisee so tht it continues in its originl direction more slowly. (d) You ctch the Frisee nd throw it so tht it moves verticlly upwrd ove your hed. (e) You ctch the Frisee nd set it down so tht it remins t rest on the ice. 18. A oxcr t ril yrd is set into motion t the top of hump. The cr rolls down quietly nd without friction Conceptul Questions 1. Does lrger net force exerted on n oject lwys produce lrger chnge in the momentum of the oject compred with smller net force? Explin.. Does lrger net force lwys produce lrger chnge in kinetic energy thn smller net force? Explin. 3. A om, initilly t rest, explodes into severl pieces. () Is liner momentum of the system (the om efore the explosion, the pieces fter the explosion) conserved? Explin. () Is kinetic energy of the system conserved? Explin. 4. While in motion, pitched sell crries kinetic energy nd momentum. () Cn we sy tht it crries force tht it cn exert on ny oject it strikes? () Cn the sell deliver more kinetic energy to the t nd tter thn the ll crries initilly? (c) Cn the sell deliver to the t nd tter more momentum thn the ll crries initilly? Explin ech of your nswers. 5. You re stnding perfectly still nd then tke step forwrd. Before the step, your momentum ws zero, ut fterwrd you hve some momentum. Is the principle of conservtion of momentum violted in this cse? Explin your nswer. 6. A shrpshooter fires rifle while stnding with the utt of the gun ginst her shoulder. If the forwrd momentum of ullet is the sme s the ckwrd momentum of the gun, why isn t it s dngerous to e hit y the gun s y the ullet? 7. Two students hold lrge ed sheet verticlly etween them. A third student, who hppens to e the str pitcher on the school sell tem, throws rw egg t the center onto stright, level trck where it couples with fltcr of smller mss, originlly t rest, so tht the two crs then roll together without friction. Consider the two crs s system from the moment of relese of the oxcr until oth re rolling together. Answer the following questions yes or no. () Is mechnicl energy of the system conserved? () Is momentum of the system conserved? Next, consider only the process of the oxcr gining speed s it rolls down the hump. For the oxcr nd the Erth s system, (c) is mechnicl energy conserved? (d) Is momentum conserved? Finlly, consider the two crs s system s the oxcr is slowing down in the coupling process. (e) Is mechnicl energy of this system conserved? (f) Is momentum of this system conserved? denotes nswer ville in tudent olutions Mnul/tudy Guide of the sheet. Explin why the egg does not rek when it hits the sheet, regrdless of its initil speed. 8. A juggler juggles three lls in continuous cycle. Any one ll is in contct with one of his hnds for one fifth of the time. () Descrie the motion of the center of mss of the three lls. () Wht verge force does the juggler exert on one ll while he is touching it? 9. () Does the center of mss of rocket in free spce ccelerte? Explin. () Cn the speed of rocket exceed the exhust speed of the fuel? Explin. 10. On the suject of the following positions, stte your own view nd rgue to support it. () The est theory of motion is tht force cuses ccelertion. () The true mesure of force s effectiveness is the work it does, nd the est theory of motion is tht work done on n oject chnges its energy. (c) The true mesure of force s effect is impulse, nd the est theory of motion is tht impulse imprted to n oject chnges its momentum. 11. An irg in n utomoile infltes when collision occurs, which protects the pssenger from serious injury (see the photo on pge 40). Why does the irg soften the low? Discuss the physics involved in this drmtic photogrph. 1. In golf, novice plyers re often dvised to e sure to follow through with their swing. Why does this dvice mke the ll trvel longer distnce? If shot is tken ner the green, very little follow-through is required. Why? 13. An open ox slides cross frictionless, icy surfce of frozen lke. Wht hppens to the speed of the ox s wter from rin shower flls verticlly downwrd into the ox? Explin. Prolems The prolems found in this chpter my e ssigned online in Enhnced WeAssign 1. denotes strightforwrd prolem;. denotes intermedite prolem; 3. denotes chllenging prolem 1. full solution ville in the tudent olutions Mnul/tudy Guide 1. denotes prolems most often ssigned in Enhnced WeAssign; these provide students with trgeted feedck nd either Mster It tutoril or Wtch It solution video. denotes sking for quntittive nd conceptul resoning denotes symolic resoning prolem denotes Mster It tutoril ville in Enhnced WeAssign denotes guided prolem shded denotes pired prolems tht develop resoning with symols nd numericl vlues

36 Prolems 69 ection 9.1 Liner Momentum 1. A prticle of mss m moves with momentum of mgnitude p. () how tht the kinetic energy of the prticle is K 5 p /m. () Express the mgnitude of the prticle s momentum in terms of its kinetic energy nd mss.. An oject hs kinetic energy of 75 J nd momentum of mgnitude 5.0 kg? m/s. Find the speed nd mss of the oject. 3. At one instnt, 17.5-kg sled is moving over horizontl surfce of snow t 3.50 m/s. After 8.75 s hs elpsed, the sled stops. Use momentum pproch to find the verge friction force cting on the sled while it ws moving. 4. A sell pproches home plte t speed of 45.0 m/s, moving horizontlly just efore eing hit y t. The tter hits pop-up such tht fter hitting the t, the sell is moving t 55.0 m/s stright up. The ll hs mss of 145 g nd is in contct with the t for.00 ms. Wht is the verge vector force the ll exerts on the t during their interction? ection 9. Anlysis Model: Isolted ystem (Momentum) 5. A 65.0-kg oy nd his 40.0-kg sister, oth wering roller ldes, fce ech other t rest. The girl pushes the oy hrd, sending him ckwrd with velocity.90 m/s towrd the west. Ignore friction. () Descrie the susequent motion of the girl. () How much potentil energy in the girl s ody is converted into mechnicl energy of the oy girl system? (c) Is the momentum of the oy girl system conserved in the pushing-prt process? If so, explin how tht is possile considering (d) there re lrge forces cting nd (e) there is no motion eforehnd nd plenty of motion fterwrd. 6. A 45.0-kg girl is stnding on 150-kg plnk. Both re originlly t rest on frozen lke tht constitutes frictionless, flt surfce. The girl egins to wlk long the plnk t constnt velocity of 1.50 i^ m/s reltive to the plnk. () Wht is the velocity of the plnk reltive to the ice surfce? () Wht is the girl s velocity reltive to the ice surfce? 7. A girl of mss m g is stnding on plnk of mss m p. Both re originlly t rest on frozen lke tht constitutes frictionless, flt surfce. The girl egins to wlk long the plnk t constnt velocity v gp to the right reltive to the plnk. (The suscript gp denotes the girl reltive to plnk.) () Wht is the velocity v pi of the plnk reltive to the surfce of the ice? () Wht is the girl s velocity v gi reltive to the ice surfce? 8. When you jump stright up s high s you cn, wht is the order of mgnitude of the mximum recoil speed tht you give to the Erth? Model the Erth s perfectly solid oject. In your solution, stte the physicl quntities you tke s dt nd the vlues you mesure or estimte for them. 9. Two locks of msses m nd 3m re plced on frictionless, horizontl surfce. A light spring is ttched to the more mssive lock, nd the locks re pushed together with the spring etween them (Fig. P9.9). A cord initilly holding the locks together is urned; fter tht hppens, the lock of mss 3m moves to the right with speed of.00 m/s. () Wht is the m 3m velocity of the lock of mss m? () Find the system s originl elstic potentil energy, tking m kg. (c) Is the originl energy in the spring or in the v Before.00 m/s cord? (d) Explin your nswer to prt (c). (e) Is the momentum m 3m of the system conserved in the After ursting-prt process? Explin how tht is possile considering (f) there re lrge forces cting Figure P9.9 nd (g) there is no motion eforehnd nd plenty of motion fterwrd? ection 9.3 Anlysis Model: Nonisolted ystem (Momentum) 10. A mn clims tht he cn hold onto 1.0-kg child in hed-on collision s long s he hs his set elt on. Consider this mn in collision in which he is in one of two identicl crs ech trveling towrd the other t 60.0 mi/h reltive to the ground. The cr in which he rides is rought to rest in 0.10 s. () Find the mgnitude of the verge force needed to hold onto the child. () Bsed on your result to prt (), is the mn s clim vlid? (c) Wht does the nswer to this prolem sy out lws requiring the use of proper sfety devices such s set elts nd specil toddler sets? 11. An estimted force time curve for sell struck y t is shown in Figure P9.11. From this curve, determine () the mgnitude of the impulse delivered to the ll nd () the verge force exerted on the ll. F (N) F mx = N t (ms) Figure P Review. After kg ruer ll is dropped from height of 1.75 m, it ounces off concrete floor nd reounds to height of 1.50 m. () Determine the mgnitude nd direction of the impulse delivered to the ll y the floor. () Estimte the time the ll is in contct with the floor nd use this estimte to clculte the verge force the floor exerts on the ll. 13. A glider of mss m is free to slide long horizontl ir trck. It is pushed ginst luncher t one end of the trck. Model the luncher s light spring of force constnt k compressed y distnce x. The glider is relesed from rest. () how tht the glider ttins speed of v 5 x(k/m) 1/. () how tht the mgnitude of the impulse imprted to the glider is given y the expression I 5 x(km) 1/. (c) Is more work done on crt with lrge or smll mss? 14. A tennis plyer receives shot with the ll ( kg) trveling horizontlly t 50.0 m/s nd returns the shot with the ll trveling horizontlly t 40.0 m/s in the opposite direction. () Wht is the impulse delivered to the ll y the tennis rcquet? () Wht work does the rcquet do on the ll?

37 70 CHAPTER 9 Liner Momentum nd Collisions F (N) The mgnitude of the net force exerted in the x direction on.50-kg prticle vries in time s shown in Figure P9.15. Find () the impulse of the force over the 5.00-s time intervl, () the finl velocity the prticle ttins if it is originlly t rest, (c) its finl velocity if its originl velocity is.00 i^ m/s, nd (d) the verge force exerted on the prticle for the time intervl etween 0 nd 5.00 s. 16. Review. A force pltform is tool used to nlyze the performnce of thletes y mesuring the verticl force the thlete exerts on the ground s function of time. trting from rest, 65.0-kg thlete jumps down onto the pltform from height of m. While she is in contct with the pltform during the time intervl 0, t, s, the force she exerts on it is descried y the function F t t where F is in newtons nd t is in seconds. () Wht impulse did the thlete receive from the pltform? () With wht speed did she rech the pltform? (c) With wht speed did she leve it? (d) To wht height did she jump upon leving the pltform? 17. Wter flls without splshing t rte of 0.50 L/s from height of.60 m into kg ucket on scle. If the ucket is originlly empty, wht does the scle red in newtons 3.00 s fter wter strts to ccumulte in it? ection 9.4 Collisions in One Dimension 18. A 1 00-kg cr trveling initilly t v Ci m/s in n esterly direction crshes into the ck of kg truck moving in the sme direction t v Ti m/s (Fig. P9.18). The velocity of the cr immeditely fter the collision is v Cf m/s to the est. () Wht is the velocity of the truck immeditely fter the collision? () Wht is the chnge in mechnicl energy of the cr truck system in the collision? (c) Account for this chnge in mechnicl energy. v Ci v Ti Before t (s) Figure P9.15 After Figure P A 10.0-g ullet is fired into sttionry lock of wood hving mss m kg. The ullet imeds into the lock. The speed of the ullet-plus-wood comintion immeditely fter the collision is m/s. Wht ws the originl speed of the ullet? 0. A cr of mss m moving t speed v 1 collides nd couples with the ck of truck of mss m moving initilly in the sme direction s the cr t lower speed v. () Wht is the speed v f of the two vehicles immeditely fter the collision? () Wht is the chnge in kinetic energy of the cr truck system in the collision? v Cf v Tf 1. A neutron in nucler rector mkes n elstic, hedon collision with the nucleus of cron tom initilly t rest. () Wht frction of the neutron s kinetic energy is trnsferred to the cron nucleus? () The initil kinetic energy of the neutron is J. Find its finl kinetic energy nd the kinetic energy of the cron nucleus fter the collision. (The mss of the cron nucleus is nerly 1.0 times the mss of the neutron.). A tennis ll of mss m t is held just ove sketll of mss m, s shown in Figure P9.. With their centers verticlly ligned, oth re relesed from rest t the sme moment so tht the ottom of the sketll flls freely through height h nd strikes the floor. Figure P9. Assume n elstic collision with the ground instntneously reverses the velocity of the sketll while the tennis ll is still moving down ecuse the lls hve seprted it while flling. Next, the two lls meet in n elstic collision. () To wht height does the tennis ll reound? () How do you ccount for the height in () eing lrger thn h? Does tht seem like violtion of conservtion of energy? 3. A 1.0-g wd of sticky cly is hurled horizontlly t 100-g wooden lock initilly t rest on horizontl surfce. The cly sticks to the lock. After impct, the lock slides 7.50 m efore coming to rest. If the coefficient of friction etween the lock nd the surfce is 0.650, wht ws the speed of the cly immeditely efore impct? 4. A wd of sticky cly of mss m is hurled horizontlly t wooden lock of mss M initilly t rest on horizontl surfce. The cly sticks to the lock. After impct, the lock slides distnce d efore coming to rest. If the coefficient of friction etween the lock nd the surfce is m, wht ws the speed of the cly immeditely efore impct? 5. () Three crts of msses m kg, m kg, nd m kg move on frictionless, horizontl trck with speeds of v m/s to the right, v m/s to the right, nd v m/s to the left s shown in Figure P9.5. Velcro couplers mke the crts stick together fter colliding. Find the finl velocity of the trin of three crts. () Wht If? Does your nswer in prt () require tht ll the crts collide nd stick together t the sme moment? Wht if they collide in different order? v 1 v v 3 m1 m Figure P As shown in Figure P9.6, ullet of mss m nd speed v psses completely through pendulum o of mss M. The ullet emerges with speed of v/. The pendulum o is suspended y stiff rod (not string) of length, nd negligile mss. Wht is m3 m v M Figure P9.6 v/

38 Prolems 71 the minimum vlue of v such tht the pendulum o will rely swing through complete verticl circle? 7. Two locks re free to slide long the frictionless, wooden trck shown in Figure P9.7. The lock of mss m kg is relesed from the position shown, t height h m ove the flt prt of the trck. Protruding from its front end is the north pole of strong mgnet, which repels the north pole of n identicl mgnet emedded in the ck end of the lock of mss m kg, initilly t rest. The two locks never touch. Clculte the mximum height to which m 1 rises fter the elstic collision. h m 1 m Figure P9.7 ection 9.5 Collisions in Two Dimensions 8. Two utomoiles of equl mss pproch n intersection. One vehicle is trveling with speed 13.0 m/s towrd the est, nd the other is trveling north with speed v i. Neither driver sees the other. The vehicles collide in the intersection nd stick together, leving prllel skid mrks t n ngle of north of est. The speed limit for oth rods is 35 mi/h, nd the driver of the northwrd-moving vehicle clims he ws within the speed limit when the collision occurred. Is he telling the truth? Explin your resoning. 9. An oject of mss 3.00 kg, moving with n initil velocity of 5.00 i^ m/s, collides with nd sticks to n oject of mss.00 kg with n initil velocity of 3.00 j^ m/s. Find the finl velocity of the composite oject. 30. Two shuffleord disks of equl mss, one ornge nd the other yellow, re involved in n elstic, glncing collision. The yellow disk is initilly t rest nd is struck y the ornge disk moving with speed of 5.00 m/s. After the collision, the ornge disk moves long direction tht mkes n ngle of with its initil direction of motion. The velocities of the two disks re perpendiculr fter the collision. Determine the finl speed of ech disk. 31. Two shuffleord disks of equl mss, one ornge nd the other yellow, re involved in n elstic, glncing collision. The yellow disk is initilly t rest nd is struck y the ornge disk moving with speed v i. After the collision, the ornge disk moves long direction tht mkes n ngle u with its initil direction of motion. The velocities of the two disks re perpendiculr fter the collision. Determine the finl speed of ech disk. 3. A 90.0-kg fullck running est with speed of 5.00 m/s is tckled y 95.0-kg opponent running north with speed of 3.00 m/s. () Explin why the successful tckle constitutes perfectly inelstic collision. () Clculte the velocity of the plyers immeditely fter the tckle. (c) Determine the mechnicl energy tht disppers s result of the collision. Account for the missing energy. 33. A illird ll moving t 5.00 m/s strikes sttionry ll of the sme mss. After the collision, the first ll moves t 4.33 m/s t n ngle of with respect to the originl line of motion. Assuming n elstic collision (nd ignoring friction nd rottionl motion), find the struck ll s velocity fter the collision. 34. The mss of the lue puck in Figure P9.34 is 0.0% greter thn the mss of the green 30.0 puck. Before colliding, the 30.0 pucks pproch ech other with moment of equl mgnitudes nd opposite directions, nd the green puck hs n initil speed Figure P9.34 of 10.0 m/s. Find the speeds the pucks hve fter the collision if hlf the kinetic energy of the system ecomes internl energy during the collision. 35. An unstle tomic nucleus of mss kg initilly t rest disintegrtes into three prticles. One of the prticles, of mss kg, moves in the y direction with speed of m/s. Another prticle, of mss kg, moves in the x direction with speed of m/s. Find () the velocity of the third prticle nd () the totl kinetic energy increse in the process. ection 9.6 The Center of Mss 36. The mss of the Erth is kg, nd the mss of the Moon is kg. The distnce of seprtion, mesured etween their centers, is m. Locte the center of mss of the Erth Moon system s mesured from the center of the Erth. 37. Four ojects re situted long the y xis s follows:.00-kg oject is t m, 3.00-kg oject is t 1.50 m,.50-kg oject is t the origin, nd 4.00-kg oject is t m. Where is the center of mss of these ojects? 38. A uniform piece of sheet metl is shped s shown in Figure P9.38. Compute the x nd y coordintes of the center of mss of the piece. y (cm) 39. Explorers in the jungle find n ncient monument in the x (cm) shpe of lrge isosceles tringle s shown in Figure Figure P9.38 P9.39. The monument is mde from tens of thousnds of smll stone locks of density kg/m 3. The monument is 15.7 m high nd 64.8 m wide t its se nd is everywhere 3.60 m thick from front to ck. Before the monument ws uilt mny yers go, ll the stone locks ly on the ground. How much work did lorers do on the locks to put them in position while 15.7 m 64.8 m Figure P m

39 7 CHAPTER 9 Liner Momentum nd Collisions uilding the entire monument? Note: The grvittionl potentil energy of n oject Erth system is given y U g 5 Mgy CM, where M is the totl mss of the oject nd y CM is the elevtion of its center of mss ove the chosen reference level. 40. A rod of length 30.0 cm hs liner density (mss per length) given y l x where x is the distnce from one end, mesured in meters, nd l is in grms/meter. () Wht is the mss of the rod? () How fr from the x 5 0 end is its center of mss? ection 9.7 ystems of Mny Prticles 41. A.00-kg prticle hs velocity 1.00 i^ 3.00 j^ m/s, nd 3.00-kg prticle hs velocity i^ j^ m/s. Find () the velocity of the center of mss nd () the totl momentum of the system. 4. The vector position of 3.50-g prticle moving in the xy plne vries in time ccording to r i^ 1 3 j^ t 1 j^t, where t is in seconds nd r is in centimeters. At the sme time, the vector position of 5.50-g prticle vries s r 5 3 i^ i^t 6 j^t. At t 5.50 s, determine () the vector position of the center of mss, () the liner momentum of the system, (c) the velocity of the center of mss, (d) the ccelertion of the center of mss, nd (e) the net force exerted on the two-prticle system. 43. Romeo (77.0 kg) entertins Juliet (55.0 kg) y plying his guitr from the rer of their ot t rest in still wter,.70 m wy from Juliet, who is in the front of the ot. After the serende, Juliet crefully moves to the rer of the ot (wy from shore) to plnt kiss on Romeo s cheek. How fr does the 80.0-kg ot move towrd the shore it is fcing? 44. A ll of mss 0.00 kg with velocity of 1.50 i^ m/s meets ll of mss kg with velocity of i^ m/s in hed-on, elstic collision. () Find their velocities fter the collision. () Find the velocity of their center of mss efore nd fter the collision. ection 9.8 Deformle ystems 45. For technology project, student hs uilt vehicle, of totl mss 6.00 kg, tht moves itself. As shown in Figure P9.45, it runs on four light wheels. A reel is ttched to one of the xles, nd cord originlly wound on the reel goes up over pulley ttched to the vehicle to support n elevted lod. After the vehicle is relesed from rest, the lod descends very slowly, unwinding the cord to turn the xle nd mke the vehicle Figure P9.45 move forwrd (to the left in Fig. P9.45). Friction is negligile in the pulley nd xle erings. The wheels do not slip on the floor. The reel hs een constructed with conicl shpe so tht the lod descends t constnt low speed while the vehicle moves horizon- tlly cross the floor with constnt ccelertion, reching finl velocity of 3.00 i^ m/s. () Does the floor imprt impulse to the vehicle? If so, how much? () Does the floor do work on the vehicle? If so, how much? (c) Does it mke sense to sy tht the finl momentum of the vehicle cme from the floor? If not, where did it come from? (d) Does it mke sense to sy tht the finl kinetic energy of the vehicle cme from the floor? If not, where did it come from? (e) Cn we sy tht one prticulr force cuses the forwrd ccelertion of the vehicle? Wht does cuse it? 46. Figure P9.46 shows n overhed view of the initil configurtion of two pucks of mss m on frictionless ice. The pucks re tied together with string of length, nd negligile mss. At time t 5 0, constnt force of mgnitude F egins to pull to the right on the center point of the string. At time t, the moving pucks strike ech other nd stick together. At this time, the force hs moved through distnce d, nd the pucks hve ttined speed v (Fig. P9.46). () Wht is v in terms of F, d,,, nd m? () How much of the energy trnsferred into the system y work done y the force hs een trnsformed to internl energy?, m m t 0 F d CM d Figure P9.46 v t t 47. A prticle is suspended from post on top of crt y light string of length L s shown in Figure P9.47. The crt nd prticle re initilly moving to the right t constnt speed v i, with the string verticl. The crt suddenly comes to rest when it runs into nd sticks to umper s shown in Figure P9.47. The suspended prticle swings through n ngle u. () how tht the originl speed of the crt cn e computed from v i 5!gL11 cos u. () If the umper is still exerting horizontl force on the crt when the hnging prticle is t its mximum ngle forwrd from the verticl, t wht moment does the umper stop exerting horizontl force? v i L Figure P A 60.0-kg person ends his knees nd then jumps stright up. After his feet leve the floor, his motion is unffected y ir resistnce nd his center of mss rises y F u

40 Prolems 73 mximum of 15.0 cm. Model the floor s completely solid nd motionless. () Does the floor imprt impulse to the person? () Does the floor do work on the person? (c) With wht momentum does the person leve the floor? (d) Does it mke sense to sy tht this momentum cme from the floor? Explin. (e) With wht kinetic energy does the person leve the floor? (f) Does it mke sense to sy tht this energy cme from the floor? Explin. ection 9.9 Rocket Propulsion 49. A model rocket engine hs n verge thrust of 5.6 N. It hs n initil mss of 5.5 g, which includes fuel mss of 1.7 g. The durtion of its urn is 1.90 s. () Wht is the verge exhust speed of the engine? () This engine is plced in rocket ody of mss 53.5 g. Wht is the finl velocity of the rocket if it were to e fired from rest in outer spce y n stronut on spcewlk? Assume the fuel urns t constnt rte. 50. Review. The first stge of turn V spce vehicle consumed fuel nd oxidizer t the rte of kg/s with n exhust speed of m/s. () Clculte the thrust produced y this engine. () Find the ccelertion the vehicle hd just s it lifted off the lunch pd on the Erth, tking the vehicle s initil mss s kg. 51. A rocket for use in deep spce is to e cple of oosting totl lod (pylod plus rocket frme nd engine) of 3.00 metric tons to speed of m/s. () It hs n engine nd fuel designed to produce n exhust speed of 000 m/s. How much fuel plus oxidizer is required? () If different fuel nd engine design could give n exhust speed of m/s, wht mount of fuel nd oxidizer would e required for the sme tsk? (c) Noting tht the exhust speed in prt () is.50 times higher thn tht in prt (), explin why the required fuel mss is not simply smller y fctor of A rocket hs totl mss M i kg, including M f kg of fuel nd oxidizer. In interstellr spce, it strts from rest t the position x 5 0, turns on its engine t time t 5 0, nd puts out exhust with reltive speed v e m/s t the constnt rte k 5.50 kg/s. The fuel will lst for urn time of T 5 M f /k kg/(.5 kg/s) 5 13 s. () how tht during the urn the velocity of the rocket s function of time is given y Additionl Prolems 53. A ll of mss m is thrown stright up into the ir with n initil speed v i. Find the momentum of the ll () t its mximum height nd () hlfwy to its mximum height. 54. An mteur skter of mss M is trpped in the middle of n ice rink nd is unle to return to the side where there is no ice. Every motion she mkes cuses her to slip on the ice nd remin in the sme spot. he decides to try to return to sfety y throwing her gloves of mss m in the direction opposite the sfe side. () he throws the gloves s hrd s she cn, nd they leve her hnd with horizontl velocity v gloves. Explin whether or not she moves. If she does move, clculte her velocity v girl reltive to the Erth fter she throws the gloves. () Discuss her motion from the point of view of the forces cting on her. 55. A 3.00-kg steel ll strikes y wll with speed of 10.0 m/s t n ngle of u with the surfce. It ounces off with x u the sme speed nd ngle (Fig. P9.55). If the ll is u in contct with the wll for 0.00 s, wht is the verge force exerted y Figure P9.55 the wll on the ll? 56. () Figure P9.56 shows three points in the opertion of the llistic pendulum discussed in Exmple 9.6 (nd shown in Fig. 9.9). The projectile pproches the pendulum in Figure P9.56. Figure P9.56 shows the sitution just fter the projectile is cptured in the pendulum. In Figure P9.56c, the pendulum rm hs swung upwrd nd come to rest t height h ove its initil position. Prove tht the rtio of the kinetic energy of the projectile pendulum system immeditely fter the collision to the kinetic energy immeditely efore is m 1 /(m 1 1 m ). () Wht is the rtio of the momentum of the system immeditely fter the collision to the momentum immeditely efore? (c) A student elieves tht such lrge decrese in mechnicl energy must e ccompnied y t lest smll decrese in momentum. How would you convince this student of the truth? v 1t 5v e ln1 kt M i () Mke grph of the velocity of the rocket s function of time for times running from 0 to 13 s. (c) how tht the ccelertion of the rocket is 1t 5 kv e M i kt (d) Grph the ccelertion s function of time. (e) how tht the position of the rocket is x1t 5 v e M i k t ln 1 kt M i 1 v e t (f) Grph the position during the urn s function of time. v i m 1 m Figure P9.56 Prolems 56 nd 70. () A metl ll moves towrd the pendulum. () The ll is cptured y the pendulum. (c) The ll pendulum comintion swings up through height h efore coming to rest. v f 57. A 1.5-kg wooden lock rests on tle over lrge hole s in Figure P9.57 on pge 74. A 5.00-g ullet with n initil velocity v i is fired upwrd into the ottom of the lock nd remins in the lock fter the collision. The c h

41 74 CHAPTER 9 Liner Momentum nd Collisions lock nd ullet rise to mximum height of.0 cm. () Descrie how you would find the initil velocity of the ullet using ides you hve lerned in this chpter. () Clculte the initil velocity of the ullet from the informtion provided. 58. A wooden lock of mss M rests on tle over lrge hole s in Figure A ullet of mss m with n initil velocity of v i is fired upwrd into the ottom of the lock nd remins in the lock fter the collision. The lock nd ullet rise to mximum height of h. () Descrie how you would find the initil velocity of the ullet using ides you hve lerned in this chpter. () Find n expression for the initil velocity of the ullet. 59. Two gliders re set in motion on horizontl ir trck. A spring of force constnt k is ttched to the ck end of the second glider. As shown in Figure P9.59, the first glider, of mss m 1, moves to the right with speed v 1, nd the second glider, of mss m, moves more slowly to the right with speed v. When m 1 collides with the spring ttched to m, the spring compresses y distnce x mx, nd the gliders then move prt gin. In terms of v 1, v, m 1, m, nd k, find () the speed v t mximum compression, () the mximum compression x mx, nd (c) the velocity of ech glider fter m 1 hs lost contct with the spring. v 1 m 1 k Figure P9.59 m M v i Figure P9.57 Prolems 57 nd Pursued y ferocious wolves, you re in sleigh with no horses, gliding without friction cross n ice-covered lke. You tke n ction descried y the equtions 170 kg17.50 m/s i^ kg1v 1f i^ kg1v f i^ v 1f 1 v f m/s () Complete the sttement of the prolem, giving the dt nd identifying the unknowns. () Find the vlues of v 1f nd v f. (c) Find the mount of energy tht hs een trnsformed from potentil energy stored in your ody to kinetic energy of the system. 61. Two locks of msses m kg nd m kg re relesed from rest t height of h m on frictionless trck s shown in Figure P9.61. When they meet on the v m level portion of the trck, they undergo hed-on, elstic collision. Determine the mximum heights to which m 1 nd m rise on the curved portion of the trck fter the collision. 6. Why is the following sitution impossile? An stronut, together with the equipment he crries, hs mss of 150 kg. He is tking spce wlk outside his spcecrft, which is drifting through spce with constnt velocity. The stronut ccidentlly pushes ginst the spcecrft nd egins moving wy t 0.0 m/s, reltive to the spcecrft, without tether. To return, he tkes equipment off his spce suit nd throws it in the direction wy from the spcecrft. Becuse of his ulky spce suit, he cn throw equipment t mximum speed of 5.00 m/s reltive to himself. After throwing enough equipment, he strts moving ck to the spcecrft nd is le to gr onto it nd clim inside. 63. A kg lue ed slides on frictionless, curved wire, strting from rest t point in h Figure P9.63, where h m. At point, the lue ed collides elsticlly with kg Figure P9.63 green ed t rest. Find the mximum height the green ed rises s it moves up the wire. 64. Review. A metl cnnonll of mss m rests next to tree t the very edge of cliff 36.0 m ove the surfce of the ocen. In n effort to knock the cnnonll off the cliff, some children tie one end of rope round stone of mss 80.0 kg nd the other end to tree lim just ove the cnnonll. They tighten the rope so tht the stone just clers the ground nd hngs next to the cnnonll. The children mnge to swing the stone ck until it is held t rest 1.80 m ove the ground. The children relese the stone, which then swings down nd mkes hed-on, elstic collision with the cnnonll, projecting it horizontlly off the cliff. The cnnonll lnds in the ocen horizontl distnce R wy from its initil position. () Find the horizontl component R of the cnnonll s displcement s it depends on m. () Wht is the mximum possile vlue for R, nd (c) to wht vlue of m does it correspond? (d) For the stone cnnonll Erth system, is mechnicl energy conserved throughout the process? Is this principle sufficient to solve the entire prolem? Explin. (e) Wht if? how tht R does not depend on the vlue of the grvittionl ccelertion. Is this result remrkle? tte how one might mke sense of it. 65. Review. A ullet of mss m is fired into lock of mss M initilly t rest t the edge of frictionless tle of height h (Fig. P9.65). The ullet remins in the lock, nd fter m 1 m m M h h h d Figure P9.61 Figure P9.65

42 Prolems 75 impct the lock lnds distnce d from the ottom of the tle. Determine the initil speed of the ullet. 66. A smll lock of mss m kg is relesed from rest t the top of frictionless, curve-shped wedge of mss m kg, which sits on frictionless, horizontl surfce s shown in Figure P9.66. When the lock leves the wedge, its velocity is mesured to e 4.00 m/s to the right s shown in Figure P9.66. () Wht is the velocity of the wedge fter the lock reches the horizontl surfce? () Wht is the height h of the wedge? m 1 following verge dt: h cm, projectile mss m g, nd pendulum mss m 5 63 g. () Determine the initil speed v 1A of the projectile. () The second prt of her experiment is to otin v 1A y firing the sme projectile horizontlly (with the pendulum removed from the pth) nd mesuring its finl horizontl position x nd distnce of fll y (Fig. P9.70). Wht numericl vlue does she otin for v 1A sed on her mesured vlues of x 5 57 cm nd y cm? (c) Wht fctors might ccount for the difference in this vlue compred with tht otined in prt ()? v 1A h m v m v f y Figure P A kg sphere moving with velocity given y 1.00 i^ 3.00 j^ k^ m/s strikes nother sphere of mss 1.50 kg moving with n initil velocity of i^ 1.00 j^ 3.00k^ m/s. () The velocity of the kg sphere fter the collision is i^ j^ 8.00k^ m/s. Find the finl velocity of the 1.50-kg sphere nd identify the kind of collision (elstic, inelstic, or perfectly inelstic). () Now ssume the velocity of the kg sphere fter the collision is (0.50 i^ j^.00k^ ) m/s. Find the finl velocity of the 1.50-kg sphere nd identify the kind of collision. (c) Wht If? Tke the velocity of the kg sphere fter the collision s i^ j^ 1 k^ m/s. Find the vlue of nd the velocity of the 1.50-kg sphere fter n elstic collision. 68. A 75.0-kg firefighter slides down pole while constnt friction force of 300 N retrds her motion. A horizontl 0.0-kg pltform is supported y spring t the ottom of the pole to cushion the fll. The firefighter strts from rest 4.00 m ove the pltform, nd the spring constnt is N/m. Find () the firefighter s speed just efore she collides with the pltform nd () the mximum distnce the spring is compressed. Assume the friction force cts during the entire motion. 69. George of the Jungle, with mss m, swings on light vine hnging from sttionry tree rnch. A second vine of equl length hngs from the sme point, nd gorill of lrger mss M swings in the opposite direction on it. Both vines re horizontl when the primtes strt from rest t the sme moment. George nd the gorill meet t the lowest point of their swings. Ech is frid tht one vine will rek, so they gr ech other nd hng on. They swing upwrd together, reching point where the vines mke n ngle of with the verticl. Find the vlue of the rtio m/m. 70. Review. A student performs llistic pendulum experiment using n pprtus similr to tht discussed in Exmple 9.6 nd shown in Figure P9.56. he otins the x Figure P Review. A light spring of force constnt 3.85 N/m is compressed y 8.00 cm nd held etween 0.50-kg lock on the left nd kg lock on the right. Both locks re t rest on horizontl surfce. The locks re relesed simultneously so tht the spring tends to push them prt. Find the mximum velocity ech lock ttins if the coefficient of kinetic friction etween ech lock nd the surfce is () 0, () 0.100, nd (c) Assume the coefficient of sttic friction is greter thn the coefficient of kinetic friction in every cse. 7. Consider s system the un with the Erth in circulr orit round it. Find the mgnitude of the chnge in the velocity of the un reltive to the center of mss of the system over six-month period. Ignore the influence of other celestil ojects. You my otin the necessry stronomicl dt from the endppers of the ook. 73. A 5.00-g ullet moving with n initil speed of v i m/s is fired into nd psses through v i 1.00-kg lock s shown d v f in Figure P9.73. The lock, initilly t rest on frictionless, horizontl surfce, is connected to spring with force constnt 900 N/m. The Figure P9.73 lock moves d cm to the right fter impct efore eing rought to rest y the spring. Find () the speed t which the ullet emerges from the lock nd () the mount of initil kinetic energy of the ullet tht is converted into internl energy in the ullet lock system during the collision. 74. Review. There re (one cn sy) three coequl theories of motion for single prticle: Newton s second lw, stting tht the totl force on the prticle cuses its ccelertion; the work kinetic energy theorem, stting tht the totl work on the prticle cuses its chnge in kinetic

43 76 CHAPTER 9 Liner Momentum nd Collisions energy; nd the impulse momentum theorem, stting tht the totl impulse on the prticle cuses its chnge in momentum. In this prolem, you compre predictions of the three theories in one prticulr cse. A 3.00-kg oject hs velocity 7.00 j^ m/s. Then, constnt net force 1.0 i^ N cts on the oject for 5.00 s. () Clculte the oject s finl velocity, using the impulse momentum theorem. () Clculte its ccelertion from 5 1 v f v i /Dt. (c) Clculte its ccelertion from 5 g F/m. (d) Find the oject s vector displcement from Dr 5 v it 1 1 t. (e) Find the work done on the oject from W 5 F?Dr. (f) Find the finl 1 kinetic energy from mv f 5 1 mv f? v f. (g) Find the finl kinetic energy from 1 mv i 1 W. (h) tte the result of compring the nswers to prts () nd (c), nd the nswers to prts (f) nd (g). Chllenge Prolems 75. Two prticles with msses m nd 3m re moving towrd ech other long the x xis with the sme initil speeds v i. Prticle m is trveling to the left, nd prticle 3m is trveling to the right. They undergo n elstic glncing collision such tht prticle m is moving in the negtive y direction fter the collision t right ngle from its initil direction. () Find the finl speeds of the two prticles in terms of v i. () Wht is the ngle u t which the prticle 3m is scttered? 76. In the 1968 Olympic gmes, University of Oregon jumper Dick Fosury introduced new technique of high jumping clled the Fosury flop. It contriuted to rising the world record y out 30 cm nd is currently used y nerly every world-clss jumper. In this technique, the jumper goes over the r fce up while rching her ck s much s possile s shown in Figure P9.76. This ction plces her center of mss outside her ody, elow her ck. As her ody goes over the r, her center of mss psses elow the r. Becuse given energy input implies certin elevtion for her center of mss, the ction of rching her ck mens tht her ody is higher thn if her ck were stright. As model, consider the jumper s thin uniform rod of length L. When the rod is stright, its center of mss is t its center. Now end the rod in circulr rc so tht it sutends n ngle of t the center of the rc s shown in Figure P9.76. In this configurtion, how fr outside the rod is the center of mss? Mrk Ddswell/Getty Imges Figure P On horizontl ir trck, glider of mss m crries G-shped post. The post supports smll dense sphere, u lso of mss m, hnging just ove the top of the glider on cord of length L. The glider nd sphere re initilly t rest with the cord verticl. (Figure P9.47 shows crt nd sphere similrly connected.) A constnt horizontl force of mgnitude F is pplied to the glider, moving it through displcement x 1 ; then the force is removed. During the time intervl when the force is pplied, the sphere moves through displcement with horizontl component x. () Find the horizontl component of the velocity of the center of mss of the glider sphere system when the force is removed. () After the force is removed, the glider continues to move on the trck nd the sphere swings ck nd forth, oth without friction. Find n expression for the lrgest ngle the cord mkes with the verticl. 78. nd from sttionry hopper flls onto moving conveyor elt t the rte of 5.00 kg/s s shown in Figure P9.78. The conveyor elt is supported y frictionless rollers nd moves t constnt speed of v m/s under the ction of constnt horizontl externl force F ext supplied y the motor tht drives the elt. Find () the snd s rte of chnge of momentum in the horizontl direction, () the force of friction exerted y the elt on the snd, (c) the externl force F ext, (d) the work done y F ext in 1 s, nd (e) the kinetic energy cquired y the flling snd ech second due to the chnge in its horizontl motion. (f) Why re the nswers to prts (d) nd (e) different? v Figure P9.78 F ext 79. Review. A chin of length L nd totl mss M is relesed from rest with its lower end just touching the top of tle s shown in Figure P9.79. Find the force exerted y the tle on the chin fter the chin hs fllen through distnce x s shown in Figure P9.79. (Assume ech link comes to rest the instnt it reches the tle.) L Figure P9.79 x L x