2.00 kg 4.00 kg 3.00 kg m. y com. (2.00 kg)(0.500 m) 4.00 kg m 3.00 kg m m m kg 4.00 kg 3.00 kg m.

Transcription

1 Chapter 9. We use Eq. 9-5 to sole or ( x, y ). (a) The x coordnate o the system s center o mass s: x com x m x m (.00 kg)(.0 m) 4.00 kg m.00 kg x mx m m m.00 kg 4.00 kg.00 kg m. Solng the equaton yelds x =.50 m. (b) The y coordnate o the system s center o mass s: y com y m y m y m y (.00 kg)(0.500 m) 4.00 kg m.00 kg m m m.00 kg 4.00 kg.00 kg m. Solng the equaton yelds y =.4 m.. Our notaton s as ollows: x = 0 and y = 0 are the coordnates o the m =.0 kg partcle; x =.0 m and y =.0 m are the coordnates o the m = 4.0 kg partcle; and x =.0 m and y =.0 m are the coordnates o the m = 8.0 kg partcle. (a) The x coordnate o the center o mass s x m x m x m x kg.0 m 8.0 kg.0 m. m. com m m m.0 kg 4.0 kg 8.0 kg (b) The y coordnate o the center o mass s y m y m y m y kg.0 m 8.0 kg.0 m. m. com m m m.0 kg 4.0 kg 8.0 kg (c) As the mass o m, the topmost partcle, s ncreased, the center o mass shts toward that partcle. As we approach the lmt where m s nntely more masse than the others, the center o mass becomes nntesmally close to the poston o m.. We use Eq. 9-5 to locate the coordnates. 44

2 45 (a) By symmetry x com = d / = ( cm)/ = 6.5 cm. The negate alue s due to our choce o the orgn. (b) We nd y com as y com m y m y V y V y m m V V com, a com, a com, a a cm, a a a a cm / 7.85 g/cm cm /.7 g/cm 7.85 g/cm.7 g/cm 8. cm. (c) Agan by symmetry, we hae z com = (.8 cm)/ =.4 cm. 4. We wll reer to the arrangement as a table. We locate the coordnate orgn at the let end o the tabletop (as shown n Fg. 9-7). Wth +x rghtward and +y upward, then the center o mass o the rght leg s at (x,y) = (+L, L/), the center o mass o the let leg s at (x,y) = (0, L/), and the center o mass o the tabletop s at (x,y) = (L/, 0). (a) The x coordnate o the (whole table) center o mass s x com 0 / M L M M L L. M M M Wth L = cm, we hae x com = ( cm)/ = cm. (b) The y coordnate o the (whole table) center o mass s / / 0 M L M L M L ycom, M M M 5 or y com = ( cm)/5 = 4.4 cm. From the coordnates, we see that the whole table center o mass s a small dstance 4.4 cm drectly below the mddle o the tabletop. 5. Snce the plate s unorm, we can splt t up nto three rectangular peces, wth the mass o each pece beng proportonal to ts area and ts center o mass beng at ts geometrc center. We ll reer to the large 5 cm 0 cm pece (shown to the let o the y axs n Fg. 9-8) as secton ; t has 6.6% o the total area and ts center o mass s at (x,y ) = (5.0 cm,.5 cm). The top 0 cm 5 cm pece (secton, n the rst quadrant) has 8.% o the total area; ts center o mass s at (x,y ) = (0 cm,.5 cm). The bottom 0 cm x 0 cm pece (secton ) also has 8.% o the total area; ts center o mass s at (x,y ) = (5 cm, 5 cm). (a) The x coordnate o the center o mass or the plate s

3 46 CHAPTER 9 x com = (0.66)x + (0.8)x + (0.8)x = 0.45 cm. (b) The y coordnate o the center o mass or the plate s y com = (0.66)y + (0.8)y + (0.8)y =.0 cm. 6. The centers o mass (wth centmeters understood) or each o the e sdes are as ollows: ( x, y, z ) (0, 0, 0) or the sde n the yz plane ( x, y, z ) (0, 0, 0) or the sde n the xz plane ( x, y, z ) (0, 0, 0) or the sde n the xy plane ( x, y, z ) (40, 0, 0) or the remanng sde parallel to sde ( x, y, z ) (0, 40, 0) or the remanng sde parallel to sde Recognzng that all sdes hae the same mass m, we plug these nto Eq. 9-5 to obtan the results (the rst two beng expected based on the symmetry o the problem). (a) The x coordnate o the center o mass s x com mx mx mx mx mx 5m cm 5 (b) The y coordnate o the center o mass s y com my my my my my 5m cm 5 (c) The z coordnate o the center o mass s z com mz mz mz mz mz 5m cm 5 7. (a) By symmetry the center o mass s located on the axs o symmetry o the molecule the y axs. Thereore x com = 0. (b) To nd y com, we note that m H y com = m N (y N y com ), where y N s the dstance rom the ntrogen atom to the plane contanng the three hydrogen atoms: Thus, y N 0.40 m 9.40 m.80 0 m.

4 47 y mny m N com. 0 m mn mh where Appendx F has been used to nd the masses. 8. (a) Snce the can s unorm, ts center o mass s at ts geometrcal center, a dstance H/ aboe ts base. The center o mass o the soda alone s at ts geometrcal center, a dstance x/ aboe the base o the can. When the can s ull ths s H/. Thus the center o mass o the can and the soda t contans s a dstance b g b g M H / m H h / M m H aboe the base, on the cylnder axs. Wth H = cm, we obtan h = 6.0 cm. (b) We now consder the can alone. The center o mass s H/ = 6.0 cm aboe the base, on the cylnder axs. (c) As x decreases the center o mass o the soda n the can at rst drops, then rses to H/ = 6.0 cm agan. (d) When the top surace o the soda s a dstance x aboe the base o the can, the mass o the soda n the can s m p = m(x/h), where m s the mass when the can s ull (x = H). The center o mass o the soda alone s a dstance x/ aboe the base o the can. Hence b g b g b g b gb g p b g b g M H / mp x / M H / m x / H x / h M m M mx / H MH mx. MH mx We nd the lowest poston o the center o mass o the can and soda by settng the derate o h wth respect to x equal to 0 and solng or x. The derate s dh dx mx MH mx m m x MmHx MmH MH mx MH mx MH mx b The soluton to m x + MmHx MmH = 0 s g c h b g b g F HG MH m x. m M The poste root s used snce x must be poste. Next, we substtute the expresson ound or x nto h = (MH + mx )/(MH + mx). Ater some algebrac manpulaton we obtan I KJ.

5 48 CHAPTER 9 HM m ( cm)(0.4 kg) 0.54 kg h 4. cm. m M 0.54 kg 0.4 kg 9. We use the constant-acceleraton equatons o Table - (wth +y downward and the orgn at the release pont), Eq. 9-5 or y com and Eq. 9-7 or com. (a) The locaton o the rst stone (o mass m ) at t = 00 0 s s y = (/)gt = (/)(9.8 m/s ) (00 0 s) = 0.44 m, and the locaton o the second stone (o mass m = m ) at t = 00 0 s s y = (/)gt = (/)(9.8 m/s )(00 0 s 00 0 s) = 0.0 m. Thus, the center o mass s at y com m y m y m m m b0. 44 mg mb0. 0 mg 0. 8 m. m m (b) The speed o the rst stone at tme t s = gt, whle that o the second stone s = g(t 00 0 s). Thus, the center-o-mass speed at t = 00 0 s s com 9.8 m/s 00 0 s 9.8 m/s 00 0 m s 000 s m m m m m m m. m/s. 0. We use the constant-acceleraton equatons o Table - (wth the orgn at the trac lght), Eq. 9-5 or x com and Eq. 9-7 or com. At t =.0 s, the locaton o the automoble (o mass m ) s x at 4. 0 m/ s 0. s 8 m, c hb g whle that o the truck (o mass m ) s x = t = (8.0 m/s)(.0s) = 4 m. The speed o the automoble then s at 4.0 m/s.0 s m/s, whle the speed o the truck remans = 8.0 m/s. (a) The locaton o ther center o mass s

6 49 x com m x m x m m b000 kggb8 mg b000 kggb4 mg m. 000 kg 000 kg (b) The speed o the center o mass s m m kg m / s kg m / s b000 gb g b000 gb8. 0 g com 9. m / s. m m 000 kg 000 kg. The mplcaton n the problem regardng 0 s that the ole and the nut start at rest. Although we could proceed by analyzng the orces on each object, we preer to approach ths usng Eq The total orce on the nut-ole system s F ˆ ˆ o Fn ( j) N. Thus, Eq. 9-4 becomes ( ˆ ˆj) N Ma where M =.0 kg. Thus, acom ( ˆ ˆ j) m/s. Each component s constant, so we apply the equatons dscussed n Chapters and 4 and obtan com ˆ ˆ com ( 4.0 m) (4.0 m)j rcom a t when t = 4.0 s. It s perhaps nstructe to work through ths problem the long way (separate analyss or the ole and the nut and then applcaton o Eq. 9-5) snce t helps to pont out the computatonal adantage o Eq Snce the center o mass o the two-skater system does not moe, both skaters wll end up at the center o mass o the system. Let the center o mass be a dstance x rom the 40-kg skater, then 65 kg 0 m x 40 kg x x 6. m. b gb g b g Thus the 40-kg skater wll moe by 6. m.. THINK A shell explodes nto two segments at the top o ts trajectory. Knowng the moton o one segment allows us to analyze the moton o the other usng the momentum conseraton prncple. EXPRESS We need to nd the coordnates o the pont where the shell explodes and the elocty o the ragment that does not all straght down. The coordnate orgn s at the rng pont, the +x axs s rghtward, and the +y drecton s upward. The y component o the elocty s gen by = 0 y gt and ths s zero at tme t = 0 y /g = ( 0 /g) sn 0, where 0 s the ntal speed and 0 s the rng angle. The coordnates o the hghest pont on the trajectory are

7 40 CHAPTER 9 and 0 m/s 0 x 0 xt 0t cos0 sn0 cos0 sn 60 cos m g 9.8 m/s b g 0 0 m / s y 0 yt gt sn 0 sn m. g 9. 8 m / s Snce no horzontal orces act, the horzontal component o the momentum s consered. In addton, snce one ragment has a elocty o zero ater the exploson, the momentum o the other equals the momentum o the shell beore the exploson. At the hghest pont the elocty o the shell s 0 cos 0, n the poste x drecton. Let M be the mass o the shell and let V 0 be the elocty o the ragment. Then M 0 cos 0 = MV 0 /, snce the mass o the ragment s M/. Ths means V cos 0 m/s cos60 0 m/s. Ths normaton s used n the orm o ntal condtons or a projectle moton problem to determne where the ragment lands. ANALYZE Resettng our clock, we now analyze a projectle launched horzontally at tme t = 0 wth a speed o 0 m/s rom a locaton hang coordnates x 0 = 7.7 m, y 0 = 5. m. Its y coordnate s gen by y y gt, and when t lands ths s zero. The 0 tme o landng s t y 0 / g and the x coordnate o the landng pont s y 5. m 0 x x0 V0t x0 V m 0 m / s 5 m. g 9. 8 m / s b g b g LEARN In the absence o exploson, the shell wth a mass M would hae landed at 0 m/s 0 R x0 sn 0 sn[(60 )] 5. m g 9.8 m/s whch s shorter than x 5 m ound aboe. Ths makes sense because the broken ragment, hang a smaller mass but greater horzontal speed, can trael much arther than the orgnal shell. 4. (a) The phrase (n the problem statement) such that t [partcle ] always stays drectly aboe partcle durng the lght means that the shadow (as a lght were drectly aboe the partcles shnng down on them) o partcle concdes wth the poston o partcle, at each moment. We say, n ths case, that they are ertcally

8 4 algned. Because o that algnment, x = = 0.0 m/s. Because the ntal alue o s gen as 0.0 m/s, then (usng the Pythagorean theorem) we must hae = 00 m/s y x or the ntal alue o the y component o partcle s elocty. Equaton -6 (or conseraton o energy) readly yelds y max = 00/9.6 = 5. m. Thus, we obtan H max = m y max /m total = (.00 g)(5. m)/(8.00 g) = 5.74 m. (b) Snce both partcles hae the same horzontal elocty, and partcle s ertcal component o elocty anshes at that hghest pont, then the center o mass elocty then s smply (0.0 m/s) ˆ (as one can ery usng Eq. 9-7). (c) Only partcle experences any acceleraton (the ree all acceleraton downward), so Eq. 9-8 (or Eq. 9-9) leads to a com = m g /m total = (.00 g)(9.8 m/s )/(8.00 g) =.68 m/s or the magntude o the downward acceleraton o the center o mass o ths system. Thus, a (.68 m/s ) ˆj. com 5. (a) The net orce on the system (o total mass m + m ) s m g. Thus, Newton s second law leads to a = g(m /( m + m )) = 0.4g. For block, ths acceleraton s to the rght (the ^ drecton), and or block ths s an acceleraton downward (the j^ drecton). Thereore, Eq. 9-8 ges a com = m a + m a m + m = (0.6)(0.4g^ ) + (0.4)( 0.4gj^ ) = (.5 ^.57 j^ ) m/s. (b) Integratng Eq. 4-6, we obtan com = (.5 ^.57j^ ) t (wth SI unts understood), snce t started at rest. We note that the rato o the y- component to the x-component (or the elocty ector) does not change wth tme, and t s that rato whch determnes the angle o the elocty ector (by Eq. -6), and thus the drecton o moton or the center o mass o the system. (c) The last sentence o our answer or part (b) mples that the path o the center-o-mass s a straght lne. (d) Equaton -6 leads to = 4º. The path o the center o mass s thereore straght, at downward angle 4.

9 4 CHAPTER 9 6. We denote the mass o Rcardo as M R and that o Carmelta as M C. Let the center o mass o the two-person system (assumed to be closer to Rcardo) be a dstance x rom the mddle o the canoe o length L and mass m. Then M R (L/ x) = mx + M C (L/ + x). Now, ater they swtch postons, the center o the canoe has moed a dstance x rom ts ntal poston. Thereore, x = 40 cm/ = 0.0 m, whch we substtute nto the aboe equaton to sole or M C :. 0 MRbL / xg mx b80gb 0. 0g b0gb0. 0g MC 58 kg. L / x 0. / There s no net horzontal orce on the dog-boat system, so ther center o mass does not moe. Thereore by Eq. 9-6, b g whch mples Mx 0 mb x com b md x, d m d xb xd mb. Now we express the geometrcal condton that relate to the boat the dog has moed a dstance d =.4 m: x x d b whch accounts or the act that the dog moes one way and the boat moes the other. We substtute or x b rom aboe: md bxdg xd d m b d whch leads to d.4 m x d.9 m. m / m (4.5/8) d b The dog s thereore.9 m closer to the shore than ntally (where t was D = 6. m rom t). Thus, t s now D x d = 4. m rom the shore. 8. The magntude o the ball s momentum change s p m (0.70 kg) (5.0 m/s) (.0 m/s) 4.9 kg m/s. 9. (a) The change n knetc energy s

10 4 00 kg 5 km/h 4 km/h K m m kg km/h 0 m/km h/600 s J. (b) The magntude o the change n elocty s 4 km/h 5 km/h 65.4 km/h so the magntude o the change n momentum s F000 m / km b H G I 00 kg gb65. 4 km / hg K J p m 600 s / h (c) The ector p ponts at an angle south o east, where F HG I F KJ H G I K J 4 km / h tan tan 9. 5 km / h kg m / s. 0. We ner rom the graph that the horzontal component o momentum p x s 4.0 kg m/s. Also, ts ntal magntude o momentum p o s 6.0 kg m/s. Thus, cos o = p x p o o = 48.. We use coordnates wth +x horzontally toward the ptcher and +y upward. Angles are measured counterclockwse rom the +x axs. Mass, elocty, and momentum unts are SI. Thus, the ntal momentum can be wrtten p 0 b4. 55g n magntude-angle notaton. (a) In magntude-angle notaton, the momentum change s ( ) (4.5 5 ) = (5.0 4 ) (ecently done wth a ector-capable calculator n polar mode). The magntude o the momentum change s thereore 5.0 kg m/s. (b) The momentum change s (6.0 0 ) (4.5 5 ) = (0 5 ). Thus, the magntude o the momentum change s 0 kg m/s.. (a) Snce the orce o mpact on the ball s n the y drecton, p x s consered: p p m sn m sn. x x

11 44 CHAPTER 9 Wth = 0.0, we nd = 0.0. (b) The momentum change s p m cos ˆj m cos ˆj 0.65 kg.00 m/s cos 0 ˆj ( 0.57 kg m/s)j. ˆ. We estmate hs mass n the neghborhood o 70 kg and compute the upward orce F o the water rom Newton s second law: F mg ma, where we hae chosen +y upward, so that a > 0 (the acceleraton s upward snce t represents a deceleraton o hs downward moton through the water). Hs speed when he arres at the surace o the water s ound ether rom Eq. -6 or rom energy conseraton: gh, where h m, and snce the deceleraton a reduces the speed to zero oer a dstance d = 0.0 m we also obtan ad. We use these obseratons n the ollowng. Equatng our two expressons or leads to a = gh/d. Our orce equaton, then, leads to F HG F F mg m g h h mg H G I d K J d whch yelds F kg. Snce we are not at all certan o hs mass, we express ths as a guessed-at range (n kn) 5 < F < 0. Snce F mg, the mpulse J due to the net orce (whle he s n contact wth the water) s oerwhelmngly caused by the upward orce o the water: F dt J to a good approxmaton. Thus, by Eq. 9-9, z Fdt p p 0 md gh (the mnus sgn wth the ntal elocty s due to the act that downward s the negate (70 kg) 9.8 m/s m. 0 kg m s. drecton), whch yelds as a range we estmate.00 kg m s F dt.0 kg m s. I K J z Expressng ths 4. We choose +y upward, whch mples a > 0 (the acceleraton s upward snce t represents a deceleraton o hs downward moton through the snow). (a) The maxmum deceleraton a max o the paratrooper (o mass m and ntal speed = 56 m/s) s ound rom Newton s second law

12 45 F mg ma snow where we requre F snow =. 0 5 N. Usng Eq. -5 = a max d, we nd the mnmum depth o snow or the man to sure: max 85kg56m s m d. m. 5 amax Fsnow mg.0 N (b) Hs short trp through the snow noles a change n momentum p p p 0 85kg 56m s 4.80 kg m s, or p 4.80 kg m s. The negate alue o the ntal elocty s due to the act that downward s the negate drecton. By the mpulse-momentum theorem, ths equals the mpulse due to the net orce F snow mg, but snce Fsnow mg we can approxmate ths as the mpulse on hm just rom the snow. 5. We choose +y upward, whch means 5m s and 0m s. Durng the collson, we make the reasonable approxmaton that the net orce on the ball s equal to F ag, the aerage orce exerted by the loor up on the ball. (a) Usng the mpulse momentum theorem (Eq. 9-) we nd (b) From Eq. 9-5, we obtan F J m m b gb g b gb g kg m s. J 4. 0 t ag N. 6. (a) By energy conseraton, the speed o the ctm when he alls to the loor s m mgh gh (9.8 m/s )(0.50 m). m/s. Thus, the magntude o the mpulse s J p m m (70 kg)(. m/s). 0 N s. (b) Wth duraton o t 0.08 s or the collson, the aerage orce s F ag J.0 Ns t 0.08 s.7 0 N.

13 46 CHAPTER 9 7. THINK The elocty o the ball s changng because o the external orce appled. Impulse-lnear momentum theorem s noled. EXPRESS The ntal drecton o moton s n the +x drecton. The magntude o the aerage orce F ag s gen by J.4 Ns Fag.00 N. t.700 s The orce s n the negate drecton. Usng the lnear momentum-mpulse theorem stated n Eq. 9-, we hae Fag t J p m( ). where m s the mass, the ntal elocty, and the nal elocty o the ball. The equaton can be used to sole or. ANALYZE (a) Usng the aboe expresson, we nd m 0.40 kg 4 m s 00 N 7 0 s Fag t 67 m s. m 0.40 kg The nal speed o the ball s 67 m/s. (b) The negate sgn n ndcates that the elocty s n the x drecton, whch s opposte to the ntal drecton o trael. (c) From the aboe, the aerage magntude o the orce s Fag.0 0 N. (d) The drecton o the mpulse on the ball s x, same as the appled orce. LEARN In ector notaton, Fag t J p m( ), whch ges J Fag t m m Snce J or F ag s n the opposte drecton o, the elocty o the ball wll decrease under the appled orce. The ball rst moes n the +x-drecton, but then slows down and comes to a stop, and then reerses ts drecton o trael. 8. (a) The magntude o the mpulse s J p m m (0.70 kg)( m/s) 9. kg m/s 9. N s.

14 47 (b) Wth duraton o t s or the collson, the aerage orce s F J 9. Ns t 5.00 s ag.8 0 N. 9. We choose the poste drecton n the drecton o rebound so that 0 and 0. Snce they hae the same speed, we wrte ths as and. Thereore, the change n momentum or each bullet o mass m s p m m. Consequently, the total change n momentum or the 00 bullets (each mnute) P 00p 00 m. The aerage orce s then F ag 00 0 kg500 m s mn60s mn P 5 N. t 0. (a) By Eq. 9-0, mpulse can be determned rom the area under the F(t) cure. Keepng n mnd that the area o a trangle s (base)(heght), we nd the mpulse n ths case s.00 Ns. (b) By denton (o the aerage o uncton, n the calculus sense) the aerage orce must be the result o part (a) dded by the tme (0.00 s). Thus, the aerage orce s ound to be 00 N. (c) Consder ten hts. Thnkng o ten hts as 0 F(t) trangles, our total tme nteral s 0(0.050 s) = 0.50 s, and the total area s 0(.0 Ns ). We thus obtan an aerage orce o 0/0.50 = 0.0 N. One could consder 5 hts, 7 hts, and so on, and stll arre at ths same answer.. (a) By energy conseraton, the speed o the passenger when the eleator hts the loor s m mgh gh (9.8 m/s )(6 m) 6.6 m/s. Thus, the magntude o the mpulse s J p m m (90 kg)(6.6 m/s).9 0 N s. (b) Wth duraton o t s or the collson, the aerage orce s F J.90 Ns ag t 5.00 s N.

15 48 CHAPTER 9 (c) I the passenger were to jump upward wth a speed o downward elocty would be and the magntude o the mpulse becomes 6.6 m/s 7.0 m/s 9.6 m/s, 7.0 m/s, then the resultng J p m m (90 kg)(9.6 m/s).76 0 N s. (d) The correspondng aerage orce would be J.760 Ns F ag t 5.00 s N.. (a) By the mpulse-momentum theorem (Eq. 9-) the change n momentum must equal the area under the F(t) cure. Usng the acts that the area o a trangle s (base)(heght), and that o a rectangle s (heght)(wdth), we nd the momentum at t = 4 s to be (0 kg. m/s)^. (b) Smlarly (but keepng n mnd that areas beneath the axs are counted negately) we nd the momentum at t = 7 s s (8 kg. m/s)^. (c) At t = 9 s, we obtan = (6.0 m/s)^.. We use coordnates wth +x rghtward and +y upward, wth the usual conentons or measurng the angles (so that the ntal angle becomes = 5 ). Usng SI unts and magntude-angle notaton (ecent to work wth when usng a ector-capable calculator), the change n momentum s J p p p (a) The magntude o the mpulse s J p 5.86 kgm/s 5.86 N s. (b) The drecton o J s 59.8 measured counterclockwse rom the +x axs. (c) Equaton 9-5 leads to 5.86 Ns J F t F.000 s ag 5.86 N s ag.9 0 N. We note that ths orce s ery much larger than the weght o the ball, whch justes our (mplct) assumpton that graty played no sgncant role n the collson.

16 49 (d) The drecton o Fag s the same as J, 59.8 measured counterclockwse rom the +x axs. 4. (a) Choosng upward as the poste drecton, the momentum change o the oot s p 0 moot (0.00 kg) (.50 m s)= N s. (b) Usng Eq. 9-5 and now treatng downward as the poste drecton, we hae J F t m g t ag lzard (0.090 kg)(9.80 m/s )(0.60 s) 0.59 N s. (c) Push s what prodes the prmary support. 5. We choose our poste drecton n the drecton o the rebound z (so the ball s ntal elocty s negate-alued). We ealuate the ntegral J F dt by addng the approprate areas (o a trangle, a rectangle, and another trangle) shown n the graph (but wth the t conerted to seconds). Wth m = kg and = 4 m/s, we apply the mpulse-momentum theorem: wall F dt m m F dt F dt F dt m m Fmax 0.00s Fmax 0.00s Fmax 0.00s m whch yelds Fmax 0.004s 0.058kg 4 m s = N. 6. (a) Perormng the ntegral (rom tme a to tme b) ndcated n Eq. 9-0, we obtan b ( t ) dt ( ba) ( b a ) a n SI unts. I b =.5 s and a = 0.50 s, ths ges 7.7 Ns. (b) Ths ntegral (the mpulse) relates to the change o momentum n Eq. 9-. We note that the orce s zero at t =.00 s. Ealuatng the aboe expresson or a = 0 and b =.00 ges an answer o 6.0 kg m/s. 7. THINK We re gen the orce as a uncton o tme, and asked to calculate the correspondng mpulse, the aerage orce and the maxmum orce. EXPRESS Snce the moton s one-dmensonal, we work wth the magntudes o the ector quanttes. The mpulse J due to a orce Ft () exerted on a body s

17 40 CHAPTER 9 t () t ag, J F t dt F t where F ag s the aerage orce and t t t. To nd the tme at whch the maxmum orce occurs, we set the derate o F wth respect to tme equal to zero, and sole or t. ANALYZE (a) We take the orce to be n the poste drecton, at least or earler tmes. Then the mpulse s J Fdt (6.00 ) t (.00 ) t dt (6.0 0 ) t (.0 0 ) t 9.0 N s. (b) Usng J = F ag t, we nd the aerage orce to be F J 9.0 Ns t.0 0 s ag N. (c) Derentatng Ft () wth respect to t and settng t to zero, we hae df d dt dt (6.0 0 ) t (.0 0 ) t (6.0 0 ) (4.0 0 ) t 0, whch can be soled to ge t =.5 0 s. At that tme the orce s c hc h c hc h 6 9 F max N. (d) Snce t starts rom rest, the ball acqures momentum equal to the mpulse rom the kck. Let m be the mass o the ball and ts speed as t leaes the oot. The speed o the ball mmedately ater t loses contact wth the player s oot s p J 9.0 Ns m m 0.45 kg 0 m/s. LEARN The orce as uncton o tme s shown below. The area under the cure s the mpulse J. From the plot, we readly see that Ft () s a maxmum at t s, wth Fmax 4500 N.

18 4 8. From Fg. 9-54, +y corresponds to the drecton o the rebound (drectly away rom the wall) and +x toward the rght. Usng unt-ector notaton, the ball s ntal and nal eloctes are cosˆ snˆj 5. ˆ.0ˆj respectely (wth SI unts understood). cos ˆ snˆj 5. ˆ.0ˆj (a) Wth m = 0.0 kg, the mpulse-momentum theorem (Eq. 9-) yelds ˆ ˆ J m m 0.0 kg (.0 m/s j) (.8 N s)j. (b) Usng Eq. 9-5, the orce on the ball by the wall s J t ( )j ˆ (80 N) ˆj. By Newton s thrd law, the orce on the wall by the ball s ( 80 N)j ˆ (that s, ts magntude s 80 N and ts drecton s drectly nto the wall, or down n the ew proded by Fg. 9-54). 9. THINK Ths problem deals wth momentum conseraton. Snce no external orces wth horzontal components act on the man-stone system and the ertcal orces sum to zero, the total momentum o the system s consered. EXPRESS Snce the man and the stone are ntally at rest, the total momentum s zero both beore and ater the stone s kcked. Let m s be the mass o the stone and s be ts elocty ater t s kcked. Also, let m m be the mass o the man and m be hs elocty ater he kcks the stone. Then, by momentum conseraton, m m m. s s s m m 0 m s mm ANALYZE We take the axs to be poste n the drecton o moton o the stone. Then or m m.0 0 m/s. ms kg 4.0 m/s.0 0 s m/s m 9 kg m LEARN The negate sgn n m ndcates that the man moes n the drecton opposte to the moton o the stone. Note that hs speed s much smaller (by a actor o ms / m m) compared to the speed o the stone. 40. Our notaton s as ollows: the mass o the motor s M; the mass o the module s m; the ntal speed o the system s 0 ; the relate speed between the motor and the module

19 4 CHAPTER 9 s r ; and, the speed o the module relate to the Earth s ater the separaton. Conseraton o lnear momentum requres Thereore, 0 (M + m) 0 = m + M( r ). M m r b4 gb8 km / hg 400 km / h M m 4m m km / h. 4. (a) Wth SI unts understood, the elocty o block L (n the rame o reerence ndcated n the gure that goes wth the problem) s ( )^. Thus, momentum conseraton (or the exploson at t = 0) ges m L ( ) + (m C + m R ) = 0 whch leads to = m L m L + m C + m R = ( kg) 0 kg = 0.60 m/s. Next, at t = 0.80 s, momentum conseraton (or the second exploson) ges m C + m R ( + ) = (m C + m R ) = (8 kg)(0.60 m/s) = 4.8 kg m/s. Ths yelds = 0.5. Thus, the elocty o block C ater the second exploson s = (0.5 m/s)^. (b) Between t = 0 and t = 0.80 s, the block moes t = (0.60 m/s)(0.80 s) = 0.48 m. Between t = 0.80 s and t =.80 s, t moes an addtonal t = ( 0.5 m/s)(.00 s) = 0.0 m. Its net dsplacement snce t = 0 s thereore 0.48 m 0.0 m = 0.8 m. 4. Our notaton (and, mplctly, our choce o coordnate system) s as ollows: the mass o the orgnal body s m; ts ntal elocty s 0 ; the mass o the less masse pece s m ; ts elocty s 0 ; and, the mass o the more masse pece s m. We note that the condtons m = m (speced n the problem) and m + m = m generally assumed n classcal physcs (beore Ensten) lead us to conclude m m and m m. 4 4 Conseraton o lnear momentum requres m m m m î 0 m 4 0

20 4 whch leads to 4. The ncrease n the system s knetc energy s thereore 0 4 K m m m 0 m m m Wth ˆ ˆ 0 ( j) m/s, the ntal speed s 0 x0 y0 (9.5 m/s) (4.0 m/s) 0. m/s and the takeo angle o the athlete s y tan tan.8. x0 9.5 Usng Equaton 4-6, the range o the athlete wthout usng halteres s R sn (0. m/s) sn (.8 ) 7.75 m. g 9.8 m/s On the other hand, two halteres o mass m = 5.50 kg were thrown at the maxmum heght, then, by momentum conseraton, the subsequent speed o the athlete would be ( ) M m M m x M x x x M 0 0 Thus, the change n the x-component o the elocty s M m m (5.5 kg) x x x0 x0 x0 x0 (9.5 m/s).4 m/s. M M 78 kg The maxmum heght s attaned when 0 gt 0, or y y y m/s t 0.4s. g 9.8 m/s Thereore, the ncrease n range wth use o halteres s R ( ) t (.4 m/s)(0.4s) 0.55 m. x

21 44 CHAPTER We can thnk o the sldng-untl-stoppng as an example o knetc energy conertng nto thermal energy (see Eq. 8-9 and Eq. 6-, wth F N = mg). Ths leads to = gd beng true separately or each pece. Thus we can set up a rato: L R = L gd L R gd R = 5. But (by the conseraton o momentum) the rato o speeds must be nersely proportonal to the rato o masses (snce the ntal momentum beore the exploson was zero). Consequently, m R m L = 5 m R = 5 m L =.9 kg. Thereore, the total mass s m R + m L.4 kg. 45. THINK The mong body s an solated system wth no external orce actng on t. When t breaks up nto three peces, momentum remans consered, both n the x- and the y-drectons. EXPRESS Our notaton s as ollows: the mass o the orgnal body s M = 0.0 kg; ts ntal elocty s ˆ 0 (00 m/s) ; the mass o one ragment s m = 0.0 kg; ts elocty s ˆ (00 m/s)j; the mass o the second ragment s m = 4.0 kg; ts elocty s ˆ ( 500 m/s); and, the mass o the thrd ragment s m = 6.00 kg. Conseraton o lnear momentum requres M m m m. 0 The energy released n the exploson s equal to K, the change n knetc energy. ANALYZE (a) The aboe momentum-conseraton equaton leads to M0 m m m (0.0 kg)(00 m/s) ˆ (0.0 kg)(00 m/s) ˆj (4.0 kg)( 500 m/s) ˆ kg ˆ (.00 0 m/s) (0.670 m/s) ˆj The magntude o s (000 m/s) ( 67 m/s).0 0 m/s. It ponts at tan ( 67/000) = 9.48 (that s, at 9.5 measured clockwse rom the +x axs). (b) The energy released s K:

22 45 6 K K K m m m M0.0 J. LEARN The energy released n the exploson, o chemcal nature, s conerted nto the knetc energy o the ragments. 46. Our +x drecton s east and +y drecton s north. The lnear momenta or the two m =.0 kg parts are then p m m j where =.0 m/s, and e j e j p m m j m cos sn j x y where = 5.0 m/s and = 0. The combned lnear momentum o both parts s then P p p m ˆj m cos ˆ sn ˆj m cos ˆ m m sn ˆj.0 kg 5.0 m/s cos 0 ˆ.0 kg.0 m/s 5.0 m/s sn 0 ˆj 8.66 ˆˆj kgm/s. From conseraton o lnear momentum we know that ths s also the lnear momentum o the whole kt beore t splts. Thus the speed o the 4.0-kg kt s 8.66 kg m/s kg m/s P Px Py.5 m/s. M M 4.0 kg 47. Our notaton (and, mplctly, our choce o coordnate system) s as ollows: the mass o one pece s m = m; ts elocty s ˆ ( 0 m/s) ; the mass o the second pece s m = m; ts elocty s ˆ ( 0 m/s)j; and, the mass o the thrd pece s m = m. (a) Conseraton o lnear momentum requres ˆ ˆ m m m m 0 m 0 m 0j m 0 whch leads to ˆ ˆ (0 0j) m/s. Its magntude s 0 4 m/ s. (b) The drecton s 45 counterclockwse rom +x (n ths system where we hae m lyng o n the x drecton and m lyng o n the y drecton).

23 46 CHAPTER Ths problem noles both mechancal energy conseraton U K K, where U = 60 J, and momentum conseraton m m 0 where m = m. From the second equaton, we nd, whch n turn mples (snce and lkewse or ) b g. F K m m m K HG I K J F HG I K J (a) We substtute K = K nto the energy conseraton relaton and nd (b) And we obtan K = (0) = 40 J. U K K K U 0 J. 49. We reer to the dscusson n the textbook (see Sample Problem Conseraton o momentum, ballstc pendulum, whch uses the same notaton that we use here) or many o the mportant detals n the reasonng. Here we only present the prmary computatonal step (usng SI unts): m M m.00 gh (9.8)(0.).0 m / s (a) We choose +x along the ntal drecton o moton and apply momentum conseraton: mbullet mbullet mblock ( 5. g)(67 m / s) (5. g)(48 m / s) (700 g) whch yelds =.8 m/s. (b) It s a consequence o momentum conseraton that the elocty o the center o mass s unchanged by the collson. We choose to ealuate t beore the collson: com mbullet (5. g)(67 m/s) 4.96 m/s. m m 5. g 700 g bullet block 5. In solng ths problem, our +x drecton s to the rght (so all eloctes are postealued). (a) We apply momentum conseraton to relate the stuaton just beore the bullet strkes the second block to the stuaton where the bullet s embedded wthn the block.

24 47 (0.005 kg) (.805 kg)(.4 m/s) 7 m/s. (b) We apply momentum conseraton to relate the stuaton just beore the bullet strkes the rst block to the nstant t has passed through t (hang speed ound n part (a)). (0.005 kg) 0 (.0 kg)(0.60 m/s) ( kg)(7 m/s) whch yelds 0 = 97 m/s. 5. We thnk o ths as hang two parts: the rst s the collson tsel where the bullet passes through the block so quckly that the block has not had tme to moe through any dstance yet and then the subsequent leap o the block nto the ar (up to heght h measured rom ts ntal poston). The rst part noles momentum conseraton (wth +y upward): 00. kg 000m s 50. kg 00. kg 400m s b gb g b g b gb g whch yelds. m s. The second part noles ether the ree-all equatons rom Ch. (snce we are gnorng ar rcton) or smple energy conseraton rom Ch. 8. Choosng the latter approach, we hae whch ges the result h = 0.07 m. b gb g b gd kg. m s. kg. m s 5. Wth an ntal speed o, the ntal knetc energy o the car s K mc /. Ater a totally nelastc collson wth a moose o mass m m, by momentum conseraton, the speed o the combned system s m c mc ( mc mm ), mc mm wth nal knetc energy mc mc c m c m mc mm mc mm K ( m m ) ( m m ). (a) The percentage loss o knetc energy due to collson s K K K K mc mm 500 kg.%. K K K m m m m 000 kg 500 kg c m c m (b) I the collson were wth a camel o mass camel 00 kg, m then the percentage loss o knetc energy would be h

25 48 CHAPTER 9 m K m m K camel 00 kg c camel 000 kg 00 kg %. (c) As the anmal mass decreases, the percentage loss o knetc energy also decreases. 54. The total momentum mmedately beore the collson (wth +x upward) s p = (.0 kg)(0 m/s) + (.0 kg)( m/s) = 6 kg m/s. Ther momentum mmedately ater, when they consttute a combned mass o M = 5.0 kg, s p = (5.0 kg). By conseraton o momentum, then, we obtan = 7. m/s, whch becomes ther "ntal" elocty or ther subsequent ree-all moton. We can use Ch. methods or energy methods to analyze ths subsequent moton; we choose the latter. The leel o ther collson prodes the reerence (y = 0) poston or the gratatonal potental energy, and we obtan K 0 + U 0 = K + U M = 0 + Mgy max. Thus, wth 0 = 7. m/s, we nd y max =.6 m. 55. We choose +x n the drecton o (ntal) moton o the blocks, whch hae masses m = 5 kg and m = 0 kg. Where unts are not shown n the ollowng, SI unts are to be understood. (a) Momentum conseraton leads to m m m m 5 kg.0 m/s 0 kg.0 m/s (5 kg) 0 kg.5 m/s whch yelds.0 m/s. Thus, the speed o the 5.0 kg block mmedately ater the collson s. 0m s. (b) We nd the reducton n total knetc energy: K K.5 J. J. 5 kg m/s 0 kg m/s 5 kg m/s 0 kg.5 m/s (c) In ths new scenaro where 4. 0m s, momentum conseraton leads to 0. m s and we obtan K 40 J.

26 49 (d) The creaton o addtonal knetc energy s possble, say, some gunpowder were on the surace where the mpact occurred (ntally stored chemcal energy would then be contrbutng to the result). 56. (a) The magntude o the deceleraton o each o the cars s a = /m = k mg/m = k g. I a car stops n dstance d, then ts speed just ater mpact s obtaned rom Eq. -6: ad ad gd 0 snce 0 = 0 (ths could alternately hae been dered usng Eq. 8-). Thus, k A k gd A (0.)(9.8 m/s )(8. m) 4.6 m/s. (b) Smlarly, B k gdb (0.)(9.8 m/s )(6. m).9 m/s. (c) Let the speed o car B be just beore the mpact. Conseraton o lnear momentum ges m B = m A A + m B B, or ( maa mbb ) (00)(4.6) (400)(.9) 7.5 m / s. m 400 B (d) The conseraton o lnear momentum durng the mpact depends on the act that the only sgncant orce (durng mpact o duraton t) s the orce o contact between the bodes. In ths case, that mples that the orce o rcton exerted by the road on the cars s neglected durng the bre t. Ths neglect would ntroduce some error n the analyss. Related to ths s the assumpton we are makng that the transer o momentum occurs at one locaton, that the cars do not slde apprecably durng t, whch s certanly an approxmaton (though probably a good one). Another source o error s the applcaton o the rcton relaton Eq. 6- or the sldng porton o the problem (ater the mpact); rcton s a complex orce that Eq. 6- only partally descrbes. 57. (a) Let be the nal elocty o the ball-gun system. Snce the total momentum o the system s consered m = (m + M). Thereore, m (60 g)( m/s) 4.4 m/s m M 60 g + 40 g. (b) The ntal knetc energy s K m and the nal knetc energy s K m M m m M b g b g. The problem ndcates E th 0, so the derence K K must equal the energy U s stored n the sprng:

27 440 CHAPTER 9 U m s b m m M g F HG m m m M I K J m M. m M Consequently, the racton o the ntal knetc energy that becomes stored n the sprng s Us M K m M We thnk o ths as hang two parts: the rst s the collson tsel, where the blocks jon so quckly that the.0-kg block has not had tme to moe through any dstance yet, and then the subsequent moton o the.0 kg system as t compresses the sprng to the maxmum amount x m. The rst part noles momentum conseraton (wth +x rghtward): m = (m +m ) (. 0 kg)(4.0 m s) ( 0. kg) whch yelds. 7 m s. The second part noles mechancal energy conseraton: whch ges the result x m = 0. m. 0 (. kg) (.7 m s) (00 N m) x 59. As hnted n the problem statement, the elocty o the system as a whole, when the sprng reaches the maxmum compresson x m, satses m + m = (m + m ). The change n knetc energy o the system s thereore m ( m m ) K ( m m ) m m m m ( ) m m whch yelds K = 5 J. (Although t s not necessary to do so, stll t s worth notng mm that algebrac manpulaton o the aboe expresson leads to K d m m rel where rel = ). Conseraton o energy then requres K ( 5 J) kxm K xm = 0.5 m. k 0 N/m 60. (a) Let m A be the mass o the block on the let, A be ts ntal elocty, and A be ts nal elocty. Let m B be the mass o the block on the rght, B be ts ntal elocty, and B be ts nal elocty. The momentum o the two-block system s consered, so

28 44 and A m A A + m B B = m A A + m B B maa mbb mbb (.6 kg)(5.5 m/s) (.4 kg)(.5 m/s) (.4 kg)(4.9 m/s) ma.6 kg.9 m/s. (b) The block contnues gong to the rght ater the collson. (c) To see whether the collson s elastc, we compare the total knetc energy beore the collson wth the total knetc energy ater the collson. The total knetc energy beore s K maa mbb (.6 kg)(5.5 m/s) (.4 kg)(.5 m/s).7 J. The total knetc energy ater s K maa mbb (.6 kg)(.9 m/s) (.4 kg)(4.9 m/s).7 J. Snce K = K the collson s ound to be elastc. 6. THINK We hae a mong cart colldng wth a statonary cart. Snce the collson s elastc, the total knetc energy o the system remans unchanged. EXPRESS Let m be the mass o the cart that s orgnally mong, be ts elocty beore the collson, and be ts elocty ater the collson. Let m be the mass o the cart that s orgnally at rest and be ts elocty ater the collson. Conseraton o lnear momentum ges m m. m Smlarly, the total knetc energy s consered and we hae m m m. Solng or and, we obtan: m m, m m m m m m m The speed o the center o mass s com m m. ANALYZE (a) Wth m = 0.4 kg,. m/s and 0.66 m/s, we obtan

29 44 CHAPTER 9 m. m/s 0.66 m/s m (0.4 kg) kg kg.. m/s 0.66 m/s (b) The elocty o the second cart s: m (0.4 kg) (. m/s).9 m/s. 0.4 kg kg mm (c) From the aboe, we nd the speed o the center o mass to be com m m m m (0.4 kg)(. m/s) kg kg 0.9 m/s. LEARN In solng or, com alues or the ntal eloctes were used. Snce the system s solated wth no external orce actng on t, com remans the same ater the collson, so the same result s obtaned alues or the nal eloctes are used. That s, com m m m m (0.4 kg)(0.66 m/s) (0.099 kg)(.9 m/s) 0.4 kg kg 0.9 m/s. 6. (a) Let m be the mass o one sphere, be ts elocty beore the collson, and be ts elocty ater the collson. Let m be the mass o the other sphere, be ts elocty beore the collson, and be ts elocty ater the collson. Then, accordng to Eq. 9-75, m m m m m m m. Suppose sphere s orgnally traelng n the poste drecton and s at rest ater the collson. Sphere s orgnally traelng n the negate drecton. Replace wth, wth, and wth zero to obtan 0 = m m. Thus, m m / (00 g) / 00 g. (b) We use the eloctes beore the collson to compute the elocty o the center o mass: m 00 g.00 m s 00 g.00 m s m com.00 m/s. m m 00 g 00 g 6. (a) The center o mass elocty does not change n the absence o external orces. In ths collson, only orces o one block on the other (both beng part o the same system) are exerted, so the center o mass elocty s.00 m/s beore and ater the collson.

30 44 (b) We can nd the elocty o block beore the collson (when the elocty o block s known to be zero) usng Eq. 9-7: Now we use Eq to nd : (m + m ) com = m + 0 =.0 m/s. = m m + m = 6.00 m/s. 64. Frst, we nd the speed o the ball o mass m rght beore the collson (just as t reaches ts lowest pont o swng). Mechancal energy conseraton (wth h = m) leads to mgh m gh.7 m s. (a) We now treat the elastc collson usng Eq. 9-67: m m m m 0.5 kg.5 kg (.7 m/s).47 m/s 0.5 kg.5 kg whch means the nal speed o the ball s. 47 m s. (b) Fnally, we use Eq to nd the nal speed o the block: m (0.5 kg) (.7 m/s). m/s. m m 0.5 kg.5 kg 65. THINK We hae a mass colldng wth another statonary mass. Snce the collson s elastc, the total knetc energy o the system remans unchanged. EXPRESS Let m be the mass o the body that s orgnally mong, be ts elocty beore the collson, and be ts elocty ater the collson. Let m be the mass o the body that s orgnally at rest and be ts elocty ater the collson. Conseraton o lnear momentum ges m m m. Smlarly, the total knetc energy s consered and we hae m m m. m m The soluton to s gen by Eq. 9-67: m m. We sole or m to obtan

31 444 CHAPTER 9 The speed o the center o mass s m com m. m m m m. ANALYZE (a) gen that / 4, we nd the second mass to be /4 m m m m (.0 kg). kg. / (b) The speed o the center o mass s m m.0 kg4.0 m/s com mm.0 kg. kg.5 m s. LEARN The nal speed o the second mass s m (.0 kg) (4.0 m/s) 5.0 m/s..0 kg. kg mm Snce the system s solated wth no external orce actng on t, com remans the same ater the collson, so the same result s obtaned alues or the nal eloctes are used: m m (.0 kg)(.0 m/s) (. kg)(5.0 kg) com mm.0 kg. kg.5 m/s. 66. Usng Eq and Eq. 9-68, we hae ater the collson m m m 0.40m (4.0 m/s).7 m/s m m m 0.40m m m (4.0 m/s) 5.7 m/s. m m m 0.40m (a) Durng the (subsequent) sldng, the knetc energy o block K m s conerted nto thermal orm (E th = k m g d ). Solng or the sldng dstance d we obtan d = m 0 cm. (b) A ery smlar computaton (but wth subscrpt replacng subscrpt ) leads to block s sldng dstance d =. m. m.

32 We use Eq 9-67 and 9-68 to nd the eloctes o the partcles ater ther rst collson (at x = 0 and t = 0): m m 0.0 kg 0.40 kg (.0 m/s) 0.9 m/s 0.0 kg 0.40 kg m (0.0 kg) (.0 m/s).7 m/s. 0.0 kg 0.40 kg mm mm At a rate o moton o.7 m/s, x w = 40 cm (the dstance to the wall and back to x = 0) wll be traersed by partcle n 0.8 s. At t = 0.8 s, partcle s located at x = ( /7)(0.8) = cm, and partcle s ganng at a rate o (0/7) m/s letward; ths s ther relate elocty at that tme. Thus, ths gap o cm between them wll be closed ater an addtonal tme o (0. m)/(0/7 m/s) = 0.6 s has passed. At ths tme (t = = 0.98 s) the two partcles are at x = ( /7)(0.98) = 8 cm. 68. (a) I the collson s perectly elastc, then Eq apples = m m + m = m m + (.00)m gh = gh where we hae used the act (ound most easly rom energy conseraton) that the speed o block at the bottom o the rctonless ramp s gh (where h =.50 m). Next, or block s rough slde we use Eq. 8-7: m = E th = k d = k m g d where k = Solng or the sldng dstance d, we nd that m cancels out and we obtan d =. m. (b) In a completely nelastc collson, we apply Eq. 9-5: = m + m (where, as aboe, = gh ). Thus, n ths case we hae = gh /. Now, Eq. 8-7 (usng the total mass snce the blocks are now joned together) leads to a sldng dstance o d m (one-ourth o the part (a) answer). 69. (a) We use conseraton o mechancal energy to nd the speed o ether ball ater t has allen a dstance h. The ntal knetc energy s zero, the ntal gratatonal potental energy s M gh, the nal knetc energy s M, and the nal potental energy s zero. Thus Mgh M and gh. The collson o the ball o M wth the loor s an elastc collson o a lght object wth a statonary masse object. The elocty o the lght object reerses drecton wthout change n magntude. Ater the collson, the ball s m

33 446 CHAPTER 9 traelng upward wth a speed o gh. The ball o mass m s traelng downward wth the same speed. We use Eq to nd an expresson or the elocty o the ball o mass M ater the collson: M m m M m m M m M M m gh gh gh. M m M m M m M m M m For ths to be zero, m = M/. Wth M = 0.6 kg, we hae m = 0. kg. (b) We use the same equaton to nd the elocty o the ball o mass m ater the collson: m M M m gh M M m gh M m m M m gh whch becomes (upon substtutng M = m) m gh. We next use conseraton o mechancal energy to nd the heght h' to whch the ball rses. The ntal knetc energy m m, the ntal potental energy s zero, the nal knetc energy s zero, and the nal s potental energy s mgh'. Thus, Wth h =.8 m, we hae m mm mgh' h' 4h. g h 7. m. 70. We use Eqs. 9-67, 9-68, and 4- or the elastc collson and the subsequent projectle moton. We note that both pucks hae the same tme-o-all t (durng ther projectle motons). Thus, we hae m x = t where x = d and = m + m x = t where x = d and = m m m + m. Ddng the rst equaton by the second, we arre at d d = m m + m t m m m + m t. Ater cancelng, t, and d, and solng, we obtan m =.0 kg. 7. We apply the conseraton o lnear momentum to the x and y axes respectely.

34 447 m m cos m cos 0 m sn m sn. 5 We are gen.0 0 m/s, 64.0and 5.0. Thus, we are let wth two unknowns and two equatons, whch can be readly soled. (a) We sole or the nal alpha partcle speed usng the y-momentum equaton: sn 5.0 m sn m/s. m sn 4.00 sn 64.0 (b) Pluggng our result rom part (a) nto the x-momentum equaton produces the ntal alpha partcle speed: m cos m cos m cos cos m/s. 7. We orent our +x axs along the ntal drecton o moton, and specy angles n the standard way so = 90 or the partcle B, whch s assumed to scatter downward and > 0 or partcle A, whch presumably goes nto the rst quadrant. We apply the conseraton o lnear momentum to the x and y axes, respectely mbb mb B cos maa cos 0 m sn m sn B B A A (a) Settng B = and B, the y-momentum equaton yelds m m A A sn B and the x-momentum equaton yelds maa cos mb. Ddng these two equatons, we nd tan, whch yelds = 7. (b) We can ormally sole or A (usng the y-momentum equaton and the act that 5 ) A 5 mb m A

35 448 CHAPTER 9 but lackng numercal alues or and the mass rato, we cannot ully determne the nal speed o A. Note: substtutng cos 5, nto the x-momentum equaton leads to exactly ths same relaton (that s, no new normaton s obtaned that mght help us determne an answer). 7. Suppose the objects enter the collson along lnes that make the angles > 0 and > 0 wth the x axs, as shown n the dagram that ollows. Both hae the same mass m and the same ntal speed. We suppose that ater the collson the combned object moes n the poste x drecton wth speed V. Snce the y component o the total momentum o the twoobject system s consered, m sn m sn = 0. Ths means =. Snce the x component s consered, m cos = mv. We now use V to nd that cos. Ths means = 60. The angle between the ntal eloctes s (a) Conseraton o lnear momentum mples m m m m. A A B B A A B B Snce m A = m B = m =.0 kg, the masses dde out and we obtan (5 ˆ 0j) ˆ m/s ( 0ˆ 5j) ˆ m/s ( 5 ˆ 0ˆ B A B A j) m/s (0 ˆ5ˆj) m/s. (b) The nal and ntal knetc energes are c K m' A m' B (. 0) ( 5) J K m A mb (. 0) c5 0 ( 0) 5 h. 0 J. The change knetc energy s then K = J (that s, 500 J o the ntal knetc energy s lost). h

36 We orent our +x axs along the ntal drecton o moton, and specy angles n the standard way so = +60 or the proton (), whch s assumed to scatter nto the rst quadrant and = 0 or the target proton (), whch scatters nto the ourth quadrant (recall that the problem has told us that ths s perpendcular to ). We apply the conseraton o lnear momentum to the x and y axes, respectely. m m cos m cos 0 m sn m sn. We are gen = 500 m/s, whch prodes us wth two unknowns and two equatons, whch s sucent or solng. Snce m = m we can cancel the mass out o the equatons entrely. (a) Combnng the aboe equatons and solng or we obtan sn (500 m/s)sn(60 ) sn ( ) sn (90 ) 4 m/s. We used the dentty sn cos cos sn = sn ( ) n smplyng our nal expresson. (b) In a smlar manner, we nd sn (500 m/s)sn( 0 ) sn ( ) sn ( 90 ) 50 m/s. 76. We use Eq Then M 6090 kg rel ln 05 m/s (5 m/s) ln 08 m/s. M 600 kg 77. THINK The mass o the aster barge s ncreasng at a constant rate. Addtonal orce must be proded n order to mantan a constant speed. EXPRESS We consder what must happen to the coal that lands on the aster barge durng a tme nteral t. In that tme, a total o m o coal must experence a change o elocty (rom slow to ast), ast slow where rghtwards s consdered the poste drecton. The rate o change n momentum or the coal s thereore p ( m) m ( ast slow ) t t t

37 450 CHAPTER 9 whch, by Eq. 9-, must equal the orce exerted by the (aster) barge on the coal. The processes (the shoelng, the barge motons) are constant, so there s no ambguty n equatng p wth dp. Note that we gnore the transerse speed o the coal as t s t dt shoeled rom the slower barge to the aster one. ANALYZE (a) Wth ast 0 km/h 5.56 m/s, slow 0 km/h.78 m/s and the rate o mass change ( m/ t) 000 kg/mn (6.67 kg/s), the orce that must be appled to the aster barge s m Fast ( ast slow ) (6.67 kg/s)(5.56 m/s.78 m/s) 46. N t (b) The problem states that the rctonal orces actng on the barges does not depend on mass, so the loss o mass rom the slower barge does not aect ts moton (so no extra orce s requred as a result o the shoelng). LEARN The orce that must be appled to the aster barge n order to mantan a constant speed s equal to the rate o change o momentum o the coal. 78. We use Eq and smply wth = 0, =, and rel = u. M M M M e ln / u rel (a) I = u we obtan M M e. 7. (b) I = u we obtan M M e THINK As uel s consumed, both the mass and the speed o the rocket wll change. EXPRESS The thrust o the rocket s gen by T = R rel where R s the rate o uel consumpton and rel s the speed o the exhaust gas relate to the rocket. On the other hand, the mass o uel ejected s gen by M uel = R t, where t s the tme nteral o the burn. Thus, the mass o the rocket ater the burn s M = M M uel. ANALYZE (a) Gen that R = 480 kg/s and rel =.7 0 m/s, we nd the thrust to be 6 T R rel 480 kg s.70 m s.57 0 N.

38 45 (b) Wth the mass o uel ejected gen by M uel = Rt (480 kg/s)(50 s) =.00 5 kg, the nal mass o the rocket s M = M M uel = ( kg ) ( kg) = kg. (c) Snce the ntal speed s zero, the nal speed o the rocket s 5 M.550 kg rel ln.70 m/sln.08 0 m s. 5 M.50 kg LEARN The speed o the rocket contnues to rse as the uel s consumed. From the rst rocket equaton gen n Eq. 9-87, the thrust o the rocket s related to the acceleraton by T Ma. Usng ths equaton, we nd the ntal acceleraton to be a T M.550 kg N m/s. 80. The elocty o the object s dr d ( t) ˆ 700 ˆj 00 k ˆ (60 m/s). ˆ dt dt p m 50 kg 60 m/s ˆ ( kg m/s) ˆ. (a) The lnear momentum s 4 (b) The object s mong west (our î drecton). (c) Snce the alue o p does not change wth tme, the net orce exerted on the object s zero, by Eq We assume no external orces act on the system composed o the two parts o the last stage. Hence, the total momentum o the system s consered. Let m c be the mass o the rocket case and m p the mass o the payload. At rst they are traelng together wth elocty. Ater the clamp s released m c has elocty c and m p has elocty p. Conseraton o momentum yelds (m c + m p ) = m c c + m p p. (a) Ater the clamp s released the payload, hang the lesser mass, wll be traelng at the greater speed. We wrte p = c + rel, where rel s the relate elocty. When ths expresson s substtuted nto the conseraton o momentum condton, the result s Thereore, dm m m m m. c p c c p c p rel