Chapter 9 Linear Momentum and Collisions
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1 Chapter 9 Lnear Momentum and Collsons m = 3. kg r = ( ˆ ˆ j ) P9., r r (a) p m ( ˆ ˆj ) m s = = 9.. kg m s Thus, p x = 9. kg m s and p y =. kg m s (b) p px p y p y θ = tan = tan (.33) = 37 px = + = = 5. kg m s *P9.4 (a) For the system o two blocks Δ p =, or p = p Thereore, = M + ( 3M )(. m s) m Solng ges m = 6. m s (moton toward the let). kx = MM + 3 M 3M = 8.4 J (b) FIG. P9.4 (c) The orgnal energy s n the sprng. A orce had to be exerted oer a dstance to compress the sprng, transerrng energy nto t by work. The cord exerts orce, but oer no dstance. (d) System momentum s consered wth the alue zero. The orces on the two blocks are o equal magntude n opposte drectons. Ther mpulses add to zero. The nal momenta o the two blocks are o equal magntude n opposte drectons. *P9.6 From the mpulse-momentum theorem, requred to hold onto the chld s ( ) F Δ t =Δ p= m m, the aerage orce m kg 6 m h m s 3 F = = = 6.44 N ( Δt ).5 s.37 m h In tryng to hang onto the chld, he would hae to exert a orce o 6.44 kn (oer 4 lb) toward the back o the car, to slow down the chld s orward moton. He s not strong enough to exert so large a orce. I he were belted n and hs arms were rmly ted around the chld, the chld would exert ths sze orce on hm toward the ront o
2 the car. A person cannot saely exert or eel a orce o ths magntude and a saety dece should be used. P9.7 (a) I = Fdt = area under cure 3 I = (.5 s )( 8 N ) = 3.5 N s 3.5 N s (b) F = = 9. kn.5 3 s FIG. P9.7 P9.9 (c) From the graph, we see that F max = 8. kn r r Δ p = FΔt Δ p = m = m cos 6. m cos 6. = Δ p = m sn 6. sn 6. = msn 6. F ag y y y x = 3. kg. m s.866 = 5. kg m s Δpx 5. kg m s = = = 6 N Δt. s FIG. P9.9 P9. Assume the ntal drecton o the ball n the x drecton. r r r r I =Δ p = p p =.6 4. ˆ.6 5. ˆ = 5.4 î N s (a) Impulse, = = = 7. J (b) Work K K P9. A graph o the expresson or orce shows a parabola openng down, wth the alue zero at the begnnng and end o the.8 s nteral..8s.8s ( 9 N/ s 5 N/ s ) 3 t t I = Fdt = t t dt = 9 N/ s / 5 N/ s / 3 3 = 9 N/ s (.8 s) / 5 N/ s (.8 s) / 3 = 944 N s 963 N s = 98 N s The athlete mparts downward mpulse to the platorm, so the platorm mparts 98 N s o upward mpulse to her. (b) We could nd her mpact speed as a ree-all calculaton, but we choose to wrte t as a conseraton-o energy calculaton: mgy top = (/)m mpact mpact = ( g y top ) / = [(9.8 m/s ).6 m] / = 3.43 m/s down (c) Graty, as well as the platorm, mparts mpulse to her durng the nteracton wth the platorm. m + I platorm + mgt = m (65 kg)( 3.43 m/s) + 98 N s (65 kg)(9.8 m/s )(.8 s) = 65 kg.8s
3 up 3 N s + 98 N s 5 N s = 65 kg = 49 N s/65 kg = 3.83 m/s Note that the athlete s puttng a lot o eort nto jumpng and does not exert any orce "on hersel." The useulness o the orce platorm s to measure her eort by showng the orce she exerts on the loor. (d) Agan energy s consered n upward lght. (/)m takeo = mgytop y top = takeo / g = (3.83 m/s) /(9.8 m/s ) =.748 m P9.8 Energy s consered or the bob-earth system between bottom and top o swng. At the top the st rod s n compresson and the bob nearly at rest. K + U = K + U : Mb + = + Mg l = g 4l so = g l b Momentum o the bob-bullet system s consered n the collson: b FIG. P9.8 ( ) 4M m = m + M g l = gl m P9.9 Frst we nd, the speed o m at B beore collson. m = mgh = = 9.9 m s Now we use the text's analyss o one-dmensonal elastc collsons to nd, the speed o m at B just ater collson. m m = = ( 9.9 ) m s = 3.3 m s m + m 3 Now the 5-kg block bounces back up to ts hghest pont ater collson accordng to mgh = m ( 3.3) max h ( 3.3 m s) ( 9.8 m s ) = = max.556 m FIG. P9.9 *P9. (a) We assume that energy s consered n the all o the basketball and the tenns ball. Each reaches ts lowest pont wth a speed gen by ( K+ Ug) = ( K+ Ug) release + mgy = mb + b bottom = gy = 9.8 m s. m = 4.85 m s
4 (b) The two balls exert no orces on each other as they moe down. They collde wth each other ater the basketball has ts elocty reersed by the loor. We choose upward as poste. Momentum conseraton: 4.85 m/s 4.85 m/s ( 57 g )( 4.85 m s) + ( 59 g )( 4.85 m s) = ( 57 g ) + ( 59 g ) FIG. P9. To descrbe the elastc character o the collson, we use the relate elocty equaton 4.85 m s 4.85 m s = we sole by substtuton = 9.7 m s s ( ) 58 gm s = 57 g + 59 g 9.7 m s = ( 57 g ) + ( 59 g ) 5 7 gm s 83 m s = =.8 m s 647 Now the tenns ball-earth system keeps constant energy as the ball rses: g p 57 g.8 m s 57 g 9.8 m s 65 m s y = = 8.4 m 9.8 m s = P9. (a), (b) Let and be the x-components o elocty o the grl and the plank relate to the ce surace. Then we may say that g p s the elocty o the grl relate to the plank, so that g p =.5 () But also we must hae m g g + m p p =, snce total momentum o the grl-plank system s zero relate to the ce surace. Thereore 45. g +5 p =, or g = 3.33 p y Puttng ths nto the equaton () aboe ges FIG. P p p =.5 or p =.346 î m s (answer b) = =.5 ˆ m s (answer a) Then g
5 r r P9.4 (a) Usng conseraton o momentum, ( ) = ( ) p p, ges + + ( ) = ( + + ) 4. kg 5. m s kg 3. m s 3. kg 4. m s kg beore ater Thereore, =+.4 m s, or.4 m s toward the rght (b) No. For example, the -kg and 3.-kg mass were to stck together rst, they would moe wth a speed gen by solng ( 3 kg ) = ( kg )( 3. m s) + ( 3. kg )( 4. m s), or =+.38 m s Then when ths 3 kg combned mass colldes wth the 4. kg mass, we hae ( 7 kg ) = ( 3 kg )(.38 m s) + ( 4. kg )( 5. m s), and =+.4 m s just as n part (a). Couplng order makes no derence to the nal elocty. *P9.6 (a) Oer a ery short tme nteral, outsde orces hae no tme to mpart sgncant mpulse thus the nteracton s a collson. The opponent grabs the ullback and does not let go, so the two players moe together at the end o ther nteracton thus the collson s completely nelastc. (b) Frst, we consere momentum or the system o two ootball players n the x drecton (the drecton o trael o the ullback). ( 9. kg )( 5. m s) + = ( 85 kg ) V cosθ where θ s the angle between the drecton o the nal elocty V and the x axs. We nd V cosθ =.43 m s () Now consder conseraton o momentum o the system n the y drecton (the drecton o trael o the opponent). 95. kg 3. m s + = 85 kg V sn θ whch ges V snθ =.54 m s () Dde equaton () by () tanθ = =.633 From whch θ = 3.3 Then, ether () or () ges V =.88 m s 3 K = 9. kg 5. m s kg 3. m s =.55 J (c) K = ( 85 kg )(.88 m s ) = 7.67 J
6 Thus, the knetc energy lost s 786 J nto nternal energy. r r r P9.3 m + m = m + m : = ˆ ˆ r j 5. r = 3. ˆ. ˆ j m s P9.3 x-component o momentum or the system o the two objects: p x + p x = p x + p x : m + 3m = + 3m x y-component o momentum o the system: + = m y + 3m y by conseraton o energy o the system: + m + 3m = my + 3m x + y we hae also So the energy equaton becomes x = 3 y = 3 y 4 = 9 y 8 3 = y y or y = 3 (a) The object o mass m has nal speed y = 3 y = and the object o mass 3 m moes at x + y = x + y = 3 (b) θ = y tan x θ = 3 tan = P9.39 Ths object can be made by wrappng tape around a lght st unorm rod..3 m (a) M = λdx = 5. g m +.x g m dx.3 m M = 5.x g m+.x g m.3 m = 5.9 g xdm all mass (b) x CM = = M M.3 m λ xdx = M.3 m x CM = 5.9 g 5.x g m+ x 3 g m 3 5.x g m+.x g m dx.3 m =.53 m
7 P9.4 Take the orgn at the center o curature. We hae L = L πr, r = 4 π. An ncremental bt o the rod at angle θ rom the x axs has mass gen by dm rdθ = M Mr, dm = dθ where we hae used the denton o L L radan measure. Now y θ x y CM = M all mass = L π ydm= M 35 θ=45 ( L cosθ) Mr rsnθ L dθ = r L snθ dθ = 4L π + = 4 L π FIG. P9.4 The top o the bar s aboe the orgn by r = L, so the center o mass s below the π mddle o the bar by L π 4 L = π π π L =.63 5L. dm P9.5 (a) Thrust = e dt Thrust 3 4 =.6 m s.5 kg s = N = Thrust Mg= Ma 7 ( 6 6 ) = ( ) (b) F y : a a= 3. m s P9.5 (a) From the equaton or rocket propulson n the text, M M = eln = eln M M M kt k Now, M = M kt, so = eln = eln t M M M Wth the denton, ths becomes k () t = t e ln t = 5 m s ln 44 s (b) Wth e = 5 m s, and T p = 44 s,
8 ( m s) t s (m/s) t d d e ln T (c) p e at () = = = = e, or t t dt dt T T p p at ()= e T t p FIG. P9.5(b) t (s) (d) Wth e = 5 m s, and T p = 44 s, a= 5 m s 44 s t ( m s ) t s a a (m/s ) FIG. P9.5(d) 4 t (s) (e) () = + t = t = t t t dt x t dt eln dt e ln () x t t t t = e ln t x() t = e( t) ln + et t () Wth e = 5 m s=.5 km s, and T p = 44 s, t x =.5( 44 t) ln +.5t 44
9 x( km) t s x (km) FIG. P9.5() 4 t (s) P9.6 (a) Each prmate swngs down accordng to mgr = m MgR = M = gr The collson: m M m M M m = M + m + = + + cos35 M + m = M + m gr ) Swngng up: ( = gr( cos35 ) gr( cos35 )( M + m) = ( M m) gr.45m +.45m= M m.45m=.575m m M =.43 (b) No change s requred the orce s derent. The nature o the orces wthn the system o colldng objects does not aect the total momentum o the system. Wth strong magnetc attracton, the heaer object wll be mong somewhat aster and the lghter object aster stll. Ther extra knetc energy wll all be mmedately conerted nto extra nternal energy when the objects latch together. Momentum conseraton guarantees that none o the extra knetc energy remans ater the objects jon to make them swng hgher.
10 P9.67 (a) Fnd the speed when the bullet emerges rom the block by usng momentum conseraton: 4 m/s m = M V + m The block moes a dstance o 5. cm. Assume or an approxmaton that the block quckly reaches ts maxmum elocty, V, and the bullet kept gong wth a constant elocty,. The block then compresses the sprng and stops. 5. cm MV = kx ( 9 N m )( 5. m ) V = =.5 m s. kg m M V m = = = m s FIG. P ( 5. kg)( 4 m s) (. kg)(.5 m s) 3 5. kg 3 3 (b) Δ E= Δ K+Δ U = ( 5. kg)( m s) ( 5. kg)( 4 m s) + 9 N m 5. m Δ E = 374 J, or there s a mechancal energy loss o 374 J. P9.69 The orce exerted by the table s equal to the change n momentum o each o the lnks n the chan. By the calculus chan rule o derates, F dm = = = +m dp dm d dt dt dt dt We choose to account or the change n momentum o each lnk by hang t pass rom our area o nterest just beore t hts the table, so that FIG. P9.69 dm d and m = dt dt Snce the mass per unt length s unorm, we can express each lnk o length dx as hang a mass dm: M dm = dx L
11 The magntude o the orce on the allng chan s the orce that wll be necessary to stop each o the elements dm. dm M dx M F = = = dt L dt L Ater allng a dstance x, the square o the elocty o each lnk knematcs), hence = gx (rom Mgx F = L The lnks already on the table hae a total length x, and ther weght s supported by a orce : F Mgx F = L Hence, the total orce on the chan s F = F + F = total 3Mgx L That s, the total orce s three tmes the weght o the chan on the table at that nstant.
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