CHAPTER 7 IMPULSE AND MOMENTUM

CHAPTER 7 IMPULSE AND MOMENTUM PROBLEMS 1. SSM REASONING The ipulse that the olleyball player applies to the ball can be ound ro the ipulse-oentu theore, Equation 7.4. Two orces act on the olleyball while it s being spiked: an aerage orce F exerted by the player, and the weight o the ball. As in Exaple 1, we will assue that F is uch greater than the weight o the ball, so the weight can be neglected. Thus, the net aerage orce F is equal to F. SOLUTION Fro Equation 7.4, the ipulse that the player applies to the olleyball is F t 0 Ipulse Final Initial oentu oentu 0 ( ) (0.35 kg) ( 1 /s) (+4.0 /s) 8.7 kg /s The inus sign indicates that the direction o the ipulse is the sae as that o the inal elocity o the ball.. REASONING AND SOLUTION According to the ipulse-oentu theore, Equation 7.4, F t, where F is the net aerage orce acting on the person. Taking 0 the direction o otion (downward) as the negatie direction and soling or the net aerage orce F, we obtain 0 6.0 kg 1.10 /s ( 5.50 /s) F t 1.65 s +165 N The plus sign indicates that the orce acts upward. 3. REASONING a. The change in oentu o the ball is the inal oentu inus the initial oentu, both o which can be deterined. b. According to the ipulse-oentu theore, F t = 0, the net aerage orce F applied to the ball is equal to the change ( 0 ) in the ball s oentu, diided by the tie t o ipact. In this situation the tee upon which the ball is placed supports its weight, so the net aerage orce is FF, the aerage orce that the club applies to the ball.

Chapter 7 Probles 83 SOLUTION a. The change p in the ball s oentu is p 0 0 0.045 kg 38 /s 0 /s 1.7 kg /s b. Soling the ipulse-oentu theore or the aerage orce gies 0.045 kg38 /s 0 /s 0 F = 570 N 3 t 3.0 10 s 4. REASONING During the collision, the bat exerts an ipulse on the ball. The ipulse is the product o the aerage orce that the bat exerts and the tie o contact. According to the ipulse-oentu theore, the ipulse is also equal to the change in the oentu o the ball. We will use these two relations to deterine the aerage orce exerted by the bat on the ball. SOLUTION The ipulse J is gien by Equation 7.1 as J = F t, where F is the aerage orce that the bat exerts on the ball and t is the tie o contact. According to the ipulseoentu theore, Equation 7.4, the net aerage ipulse F t is equal to the change in the ball s oentu; F t 0. Since we are ignoring the weight o the ball, the bat s orce is the net orce, so FF. Substituting this alue or the net aerage orce into the ipulse-oentu equation and soling or the aerage orce gies 0.149 kg45.6 /s 0.149 kg40. /s 0 F 11 600 N 3 t 1.1010 s where the positie direction or the elocity has been chosen as the direction o the incoing ball. 5. SSM REASONING AND SOLUTION The ipulse J is gien directly by Equation 7.1: 3 J F t (+1400 N) 7.910 s 11 Ns The plus sign indicates that the direction o the ipulse is the sae as that o. 6. REASONING The ipulse that the roo o the car applies to the hailstones can be ound ro the ipulse-oentu theore, Equation 7.4. Two orces act on the hailstones, the aerage orce F exerted by the roo, and the weight o the hailstones. Since it is assued

84 IMPULSE AND MOMENTUM that F is uch greater than the weight o the hailstones, the net aerage orce F is equal to F. SOLUTION Fro Equation 7.4, the ipulse that the roo applies to the hailstones is: F t 0 0 Ipulse Final Initial oentu oentu Soling or F (with up taken to be the positie direction) gies F ( 0) (0.060 kg/s) ( 15 /s) ( 15 /s) = +1.8 N t This is the aerage orce exerted on the hailstones by the roo o the car. The positie sign indicates that this orce points upward. Fro Newton's third law, the aerage orce exerted by the hailstones on the roo is equal in agnitude and opposite in direction to this orce. Thereore, Force on roo = The negatie sign indicates that this orce. 7. REASONING The ipulse that the wall exerts on the skater can be ound ro the ipulse-oentu theore, Equation 7.4. The aerage orce F exerted on the skater by the wall is the only orce exerted on her in the horizontal direction, so it is the net orce; F = F. SOLUTION Fro Equation 7.4, the aerage orce exerted on the skater by the wall is 46 kg 1. /s 46 kg0 /s 0 F 69 N t 0.80 s Fro Newton's third law, the aerage orce exerted on the wall by the skater is equal in agnitude and opposite in direction to this orce. Thereore, Force exerted on wall = 69 N The plus sign indicates that this orce points opposite to the elocity o the skater. 8. REASONING We will apply the ipulse oentu theore as gien in Equation 7.4 to sole this proble. Fro this theore we know that, or a gien change in oentu,

Chapter 7 Probles 85 greater orces are associated with shorter tie interals. Thereore, we expect that the orce in the sti-legged case will be greater than in the knees-bent case. SOLUTION a. Assuing that upward is the positie direction, we ind ro the ipulseoentu theore that 75 kg0 /s 75 kg 6.4 /s 5 0 F.4 10 N t 3.0 10 s b. Again using the ipulse-oentu theore, we ind that 75 kg0 /s 75 kg 6.4 /s 3 0 F 4.810 N t 0.10 s c. The net aerage orce acting on the an is F FGround W, where F Ground is the aerage upward orce exerted on the an by the ground and W is the downward-acting weight o the an. It ollows, then, that F F W. Since the weight is W = g, we hae Sti legged F F W Ground Knees bent F F W Ground Ground 5 5.410 N 75 kg 9.80 /s.4 10 N 3 3 4.810 N 75 kg 9.80 /s 5.510 N 9. SSM WWW REASONING The ipulse applied to the gol ball by the loor can be ound ro Equation 7.4, the ipulse-oentu theore: Ft. Two 0 orces act on the gol ball, the aerage orce F exerted by the loor, and the weight o the gol ball. Since F is uch greater than the weight o the gol ball, the net aerage orce F is equal to F. Only the ertical coponent o the ball's oentu changes during ipact with the loor. In order to use Equation 7.4 directly, we ust irst ind the ertical coponents o the initial and inal elocities. We begin, then, by inding these elocity coponents. SOLUTION The igures below show the initial and inal elocities o the gol ball.

86 IMPULSE AND MOMENTUM I we take up as the positie direction, then the ertical coponents o the initial and inal elocities are, respectiely, 0y 0 cos 30.0 and y cos 30.0. Then, ro Equation 7.4 the ipulse is F t ( y 0y ) ( cos 30.0 ) ( 0cos 30.0 ) Since 45 / s, the ipulse applied to the gol ball by the loor is 0 F t cos 30.0 (0.047 kg)(45 /s)(cos 30.0 ) 3.7 N s 0 10. REASONING This solution can be diided into two parts. First, the student alls reely ro rest and attains a certain elocity (which is unknown) just beore hitting the ground. This ipact elocity depends on the height H ro which the student alls, and we will use an equation o kineatics ro Chapter 3 to relate the height to the ipact elocity. Second, the student then collides with the ground and coes to rest. We will eploy the ipulse-oentu theore to deterine the ipact elocity ro a knowledge o the aerage ipact orce and the tie o ipact, both o which are known. SOLUTION Since the student alls reely ro rest, we know that 0y = 0 /s, and a y = 9.8 /s. The acceleration is negatie, because the downward direction is taken to be the negatie direction. The displaceent y o the student is y = H, where H is the height; the inus sign indicates that the student alls downward. These ariables are related by Equation 3.6b ro the equations o kineatics as y 0y y 0 /s y a y a H where y is the elocity o the student just beore hitting the ground. Soling this expression or the height yields y H (1) a At this point we do not know the ipact elocity y. Howeer, it can be deterined by exaining the collision o the student with the ground. Just beore the collision, the elocity y

Chapter 7 Probles 87 o the student is y. Since the student coes to rest, the inal elocity is = 0 /s. The aerage orce exerted on the student by the ground is F = +18 000 N, and the tie o collision is t = 0.010 s. The ipulse-oentu theore, Equation 7.4, relates these ariables: Ft 0 /s () 0 0 0 Now 0 is the ipact elocity o the student, which was labeled as y in Equation (1); thereore, 0 = y. Thus, Equation () becoes and substituting the result into Equation (1) yields Ft Ft F t 18 000 N 0.010 s y H a a 9.80 /s 63 kg y y. Soling this equation or y 0.4 11. REASONING AND SOLUTION According to the ipulse-oentu theore (Equation 7.4) t ( ) F (1) 0 Conseration o echanical energy can be used to relate the elocities to the heights. I the loor is used to deine the zero leel or the heights, we hae gh 0 1 B where h 0 is the height o the ball when it is dropped and B is the speed o the ball just beore it strikes the ground. Soling or B gies Siilarly, gh () B 0 gh 1 A where h is the axiu height o the ball when it rebounds and A is the speed o the ball just ater it rebounds ro the ground. Soling or A gies A gh (3) Substituting equations () and (3) into equation (1), where 0 = B, and = A gies (taking "upward" as the positie direction)

88 IMPULSE AND MOMENTUM Ft g h h0 = (0.500 kg) 9.80 /s 0.700 1.0 = +4.8 N s Since the ipulse is positie, it is directed. 1. REASONING This is a proble in ector addition, and we will use the coponent ethod or ector addition. Using this ethod, we will add the coponents o the indiidual oenta in the direction due north to obtain the coponent o the ector su in the direction due north. We will obtain the coponent o the ector su in the direction due east in a siilar ashion ro the indiidual coponents in that direction. For each jogger the oentu is the ass ties the elocity. SOLUTION Assuing that the directions north and east are positie, the coponents o the joggers oenta are as shown in the ollowing table: Direction due east b gb g 85 kg jogger 85 kg. 0 / s 170 kg / s b gb g 55 kg jogger 55 kg 3. 0 / s cos3 140 kg / s Direction due north 0 kg/s b gb g 55 kg 3. 0 / s sin 3 87 kg / s Total 310 kg /s 87 kg/s Using the Pythagorean theore, we ind that the agnitude o the total oentu is b g b g 310 kg / s 87 kg / s 3 kg / s The total oentu ector points north o east by an angle, which is gien by F HG IKJ 87 kg / s tan 1 16 310 kg / s 13. REASONING During the tie interal t, a ass o water strikes the turbine blade. The incoing water has a oentu 0 and that o the outgoing water is. In order to change the oentu o the water, an ipulse F t is applied to it by the stationary turbine blade. Now Ft F t, since only the orce o the blade is assued to act on

Chapter 7 Probles 89 the water in the horizontal direction. These ariables are related by the ipulse-oentu theore, F t = 0, which can be soled to ind the aerage orce F exerted on the water by the blade. SOLUTION Soling the ipulse-oentu theore or the aerage orce gies The ratio / t aerage orce is F t t = 0 0 is the ass o water per second that strikes the blade, or 30.0 kg/s, so the F 30.0 kg/s 16.0 /s 16.0 /s 960 N 0 t The agnitude o the aerage orce is 960 N. 14. REASONING AND SOLUTION The excess weight o the truck is due to the orce exerted on the truck by the sand. Newton's third law requires that this orce be equal in agnitude to the orce exerted on the sand by the truck. In tie t, a ass o sand alls into the truck bed and coes to rest. The ipulse is ( 0) Ft ( 0) so F t The sand gains a speed 0 in alling a height h so 0 gh 9.80 /s.00 6.6 /s The elocity o the sand just beore it hits the truck is 0 = 6.6 /s, where the downward direction is taken to be the negatie direction. The inal elocity o the sand is = 0 /s. Thus, the aerage orce exerted on the sand is F ( ) 55.0 kg/s0 /s 6.6 /s 0 344 N t 15. REASONING AND SOLUTION The oentu is zero beore the beat. Conseration o oentu requires that it is also zero ater the beat; thus so that 0 = p p + b b

90 IMPULSE AND MOMENTUM p = ( b / p ) b = (0.050 kg/85 kg)(0.5 /s) = 16. REASONING The su o the external orces acting on the swier/rat syste is zero, because the weight o the swier and rat is balanced by a corresponding noral orce and riction is negligible. The swier and rat constitute an isolated syste, so the principle o conseration o linear oentu applies. We will use this principle to ind the recoil elocity o the rat. SOLUTION As the swier runs o the rat, the total linear oentu o the swier/rat syste is consered: s s r r Total oentu ater swier runs o rat 0 Total oentu beore swier starts running where s and s are the ass and inal elocity o the swier, and r and r are the ass and inal elocity o the rat. Soling or r gies r r 55 kg4.6 /s s s 1. /s 10 kg 17. SSM REASONING AND SOLUTION Bonzo and Ender constitute an isolated syste. Thereore, the principle o conseration o linear oentu holds. Since both ebers o the syste are initially stationary, the initial linear oentu is zero, and ust reain zero during and ater any interaction between Bonzo and Ender. a. Since the total oentu o the syste is consered, and the total oentu o the syste is zero, Bonzo and Ender ust hae equal but opposite linear oenta. ust hae the larger ass, since. b. Conseration o linear oentu gies 0. Soling or the Bonzo Bonzo Ender Ender ratio o the asses, we hae Bonzo Ender Ender.5 / s 1.5 / s Bonzo 18. REASONING Let be Al s ass, which eans that Jo s ass is 168 kg. Since riction is negligible and since the downward-acting weight o each person is balanced by the upward-acting noral orce ro the ice, the net external orce acting on the two-person 1.7

Chapter 7 Probles 91 syste is zero. Thereore, the syste is isolated, and the conseration o linear oentu applies. The initial total oentu ust be equal to the inal total oentu. SOLUTION Applying the principle o conseration o linear oentu and assuing that the direction in which Al oes is the positie direction, we ind b gb gb g b gb gb g 0 / s 168 kg 0 / s 0. 90 / s 168 kg 1. / s Initial total oentu Final total oentu Soling this equation or and suppressing the units in the interests o clarity, we ind b gb gb g b g 0 0. 90 168 1. 1. b168gb 1. g 96 kg 0. 90 1. 19. REASONING During the breakup the linear oentu o the syste is consered, since the orce causing the breakup is an internal orce. We will assue that the +x axis is along the original line o otion (beore the breakup), and the +y axis is perpendicular to this line and points upward. We will apply the conseration o linear oentu twice, once or the oentu coponents along the x axis and again or the oentu coponents along the y axis. SOLUTION The ass o each piece o the rocket ater breakup is, and so the ass o the rocket beore breakup is. Applying the conseration o oentu theore along the original line o otion (the x axis) gies cos 30.0 cos 60.0 or cos 30.0 cos 60.0 (1) 1 0 1 0 P, x P 0, x Applying the conseration o oentu along the y axis gies sin 30.0 1 sin 30.0 sin 60.0 0 or sin 60.0 1 P P 0, y, y () a. To ind the speed 1 o the irst piece, we substitute the alue or ro Equation () into Equation (1). The result is sin 30.0 cos 30.0 cos 60.0 sin 60.0 1 1 0 Soling or 1 and setting 0 = 45.0 /s yields

9 IMPULSE AND MOMENTUM 1 0 45.0 /s sin 30.0 sin 30.0 cos 30.0 cos 60.0 cos 30.0 cos 60.0 sin 60.0 sin 60.0 77.9 /s b. The speed o the second piece can be ound by substituting 1 = 77.9 /s into Equation (): sin 30.0 1 77.9 /ssin30.0 45.0 /s sin 60.0 sin 60.0 0. REASONING During the tie that the skaters are pushing against each other, the su o the external orces acting on the two-skater syste is zero, because the weight o each skater is balanced by a corresponding noral orce and riction is negligible. The skaters constitute an isolated syste, so the principle o conseration o linear oentu applies. We will use this principle to ind an expression or the ratio o the skater s asses in ters o their recoil elocities. We will then obtain expressions or the recoil elocities by noting that each skater, ater pushing o, coes to rest in a certain distance. The recoil elocity, acceleration, and distance are related by Equation.9 o the equations o kineatics. SOLUTION While the skaters are pushing against each other, the total linear oentu o the two-skater syste is consered: 1 1 Total oentu ater pushing 0 Total oentu beore pushing Soling this expression or the ratio o the asses gies 1 1 (1) For each skater the (initial) recoil elocity, inal elocity, acceleration a, and displaceent x are related by Equation.9 o the equations o kineatics: Soling or the recoil elocity gies ax. ax. I we assue that skater 1 recoils in the positie direction and skater recoils in the negatie direction, the recoil elocities are Skater 1 a x 1 1 1 1 Skater a x

Chapter 7 Probles 93 Substituting these expressions into Equation (1) gies a x a x 1 a x a x 1 1 1 1 1 1 () Since the skaters coe to rest, their inal elocities are zero, so 1 = = 0 /s. We also know that their accelerations hae the sae agnitudes. This eans that a = a 1, where the inus sign denotes that the acceleration o skater is opposite that o skater 1, since they are oing in opposite directions and are both slowing down. Finally, we are gien that skater 1 glides twice as ar as skater. Thus, the displaceent o skater 1 is related to that o skater by x 1 = x, where, the inus sign denotes that the skaters oe in opposite directions. Substituting these alues into Equation () yields a x 1 0 /s 0 /s 1 0.707 1 a x 1 1. SSM REASONING No net external orce acts on the plate parallel to the loor; thereore, the coponent o the oentu o the plate that is parallel to the loor is consered as the plate breaks and lies apart. Initially, the total oentu parallel to the loor is zero. Ater the collision with the loor, the coponent o the total oentu parallel to the loor ust reain zero. The drawing in the text shows the pieces in the plane parallel to the loor just ater the collision. Clearly, the linear oentu in the plane parallel to the loor has two coponents; thereore the linear oentu o the plate ust be consered in each o these two utually perpendicular directions. Using the drawing in the text, with the positie directions taken to be up and to the right, we hae x direction (sin 5.0 ) + (cos 45.0 ) = 0 1 1 y direction (cos 5.0 ) + (sin 45.0 ) = 0 1 1 3 3 These equations can be soled siultaneously or the asses 1 and. SOLUTION Using the alues gien in the drawing or the elocities ater the plate breaks, we hae, suppressing units, 1. 7 1. 7 0 (1) 1. 7 1. 7 3. 99 0 () 1

94 IMPULSE AND MOMENTUM Subtracting (1) ro () gies 1 1.00 kg. Substituting this alue into either (1) or () then yields 1.00 kg.. REASONING For the syste consisting o the eale character, the gun and the bullet, the su o the external orces is zero, because the weight o each object is balanced by a corresponding upward (noral) orce, and we are ignoring riction. The eale character, the gun and the bullet, then, constitute an isolated syste, and the principle o conseration o linear oentu applies. SOLUTION a. The total oentu o the syste beore the gun is ired is zero, since all parts o the syste are at rest. Moentu conseration requires that the total oentu reains zero ater the gun has been ired. 0 1 1 Total oentu ater gun is ired Total oentu beore gun is ired where the subscripts 1 and reer to the woan (plus gun) and the bullet, respectiely. Soling or 1, the recoil elocity o the woan (plus gun), gies 1 (0.010 kg)(70 / s) 51 kg 1 0.14 / s b. Repeating the calculation or the situation in which the woan shoots a blank cartridge, we hae 4 (5.0 10 kg)(70 / s) 3 7.1 10 / s 1 51 kg 1 In both cases, the inus sign eans that the bullet and the woan oe in opposite directions when the gun is ired. The total oentu o the syste reains zero, because oentu is a ector quantity, and the oenta o the bullet and the woan hae equal agnitudes, but opposite directions. 3. SSM WWW REASONING The cannon and the shell constitute the syste. Since no external orce hinders the otion o the syste ater the cannon is unbolted, conseration o linear oentu applies in that case. I we assue that the burning gun powder iparts the sae kinetic energy to the syste in each case, we hae suicient inoration to deelop a atheatical description o this situation, and sole it or the elocity o the shell ired by the loose cannon. SOLUTION For the case where the cannon is unbolted, oentu conseration gies

Chapter 7 Probles 95 0 (1) 1 1 Total oentu ater shell is ired Initial oentu o syste where the subscripts "1" and "" reer to the cannon and shell, respectiely. In both cases, the burning gun power iparts the sae kinetic energy to the syste. When the cannon is bolted to the ground, only the shell oes and the kinetic energy iparted to the syste is 1 1 7 shellshell KE (85.0 kg)(551 /s) 1.9 10 J The kinetic energy iparted to the syste when the cannon is unbolted has the sae alue and can be written using the sae notation as equation (1): 1 1 1 1 KE () Soling equation (1) or 1, the elocity o the cannon ater the shell is ired, and substituting the resulting expression into Equation () gies Soling equation (3) or gies 1 1 KE (3) KE (1.910 J) 85 kg (85 kg) 1 1 3 1 5.8010 kg 7 +547 /s 4. REASONING AND SOLUTION Since no net external orce acts in the horizontal direction, the total horizontal oentu o the syste is consered regardless o which direction the ass is thrown. The oentu o the syste beore the ass is thrown o the wagon is W A, where W and A are the ass and elocity o the wagon, respectiely. When ten percent o the wagon s ass is thrown orward, the wagon is brought to a halt so its inal oentu is zero. The oentu o the ass thrown orward is 0.1 W ( A + M ), where M is the elocity o the ass relatie to the wagon and ( A + M ) is the elocity o the ass relatie to the ground. The conseration o oentu gies

96 IMPULSE AND MOMENTUM Soling or M yields (1) When the direction in which the ass is thrown is reersed, the elocity o the ass relatie to the ground is now ( A M ). The oentu o the ass and wagon is, thereore, where B is the elocity o the wagon. The conseration o oentu gies () Substituting equation (1) into equation () and soling or B / A gies B A. 5. REASONING AND SOLUTION The collision is an inelastic one, with the total linear oentu being consered: 1 1 = ( 1 + )V The ass o the receier is 11 115 kg 4.5 /s 115 kg 84 kg V.6 /s 1 6. REASONING Since the collision is an elastic collision, both the linear oentu and kinetic energy o the two-ehicle syste are consered. The inal elocities o the car and an are gien in ters o the initial elocity o the car by Equations 7.8a and 7.8b. SOLUTION a. The inal elocity 1 o the car is gien by Equation 7.8a as 1 1 01 1 where 1 and are, respectiely, the asses o the car and an, and 01 is the initial elocity o the car. Thus,

Chapter 7 Probles 97 1 715 kg 1055 kg.5 /s 0.43 /s 715 kg 1055 kg b. The inal elocity o the an is gien by Equation 7.8b: 715 kg 1 01.5 /s 1.8 /s 715 kg 1055 kg 1 7. SSM REASONING Since all o the collisions are elastic, the total echanical energy o the ball is consered. Howeer, since graity aects its ertical otion, its linear oentu is not consered. I is the axiu height o the ball on its inal bounce, conseration o energy gies SOLUTION In order to use this expression, we ust obtain the alues or the elocities and. The initial elocity has only a horizontal coponent,.the inal elocity also has only a horizontal coponent since the ball is at the top o its trajectory,. No orces act in the horizontal direction so the oentu o the ball in this direction is consered, hence. Thereore, h h0 3.00 8. REASONING a. During the collision between the bullet and the wooden block, linear oentu is consered, since no net external orce acts on the bullet and the block. The weight o each is balanced by the tension in the suspension wire, and the orces that the bullet and block exert on each other are internal orces. This conseration law will allow us to ind the speed o the bullet/block syste iediately ater the collision. b. Just ater the collision, the bullet/block rise up, ultiately reaching a inal height h beore coing to a oentary rest. During this phase, the tension in the wire (a nonconseratie orce) does no work, since it acts perpendicular to the otion. Thus, the work done by nonconseratie orces is zero, and the total echanical energy o the syste is consered. An application o this conseration law will enable us to deterine the height h.

98 IMPULSE AND MOMENTUM SOLUTION a. The principle o conseration o linear oentu states that the total oentu ater the collision is equal to that beore the collision. bullet block bullet 0,bullet block 0,block Moentu ater collision Moentu beore collision Soling this equation or the speed o the bullet/block syste just ater the collision gies bullet 0,bullet block 0,block bullet block 0.0050 kg45 /s 0.15 kg0 /s 0.0050 kg + 0.15 kg 4.89 /s b. Just ater the collision, the total echanical energy o the syste is all kinetic energy, since we take the zero-leel or the graitational potential energy to be at the initial height o the block. As the bullet/block syste rises, kinetic energy is conerted into potential energy. At the highest point, the total echanical energy is all graitational potential energy. Since the total echanical energy is consered, we hae gh 1 bullet block bullet block Total echanical energy at the top o the swing, all potential Soling this expression or the height h gies 4.89 /s Total echanical energy at the botto o the swing, all kinetic 1 1 h 1. g 9.80 /s 9. REASONING The elocity o the second ball just ater the collision can be ound ro Equation 7.8b (see Exaple 7). In order to use Equation 7.8b, howeer, we ust know the elocity o the irst ball just beore it strikes the second ball. Since we know the ipulse deliered to the irst ball by the pool stick, we can use the ipulse-oentu theore (Equation 7.4) to ind the elocity o the irst ball just beore the collision. SOLUTION According to the ipulse-oentu theore,, and setting /s and soling or, we ind that the elocity o the irst ball ater it is struck by the pool stick and just beore it hits the second ball is

Chapter 7 Probles 99 Ft N s 1. 50 9. 09 0.165 kg / s Substituting alues into Equation 7.8b (with ), we hae 30. REASONING The net external orce acting on the two-puck syste is zero (the weight o each ball is balanced by an upward noral orce, and we are ignoring riction due to the layer o air on the hockey table). Thereore, the two pucks constitute an isolated syste, and the principle o conseration o linear oentu applies. SOLUTION Conseration o linear oentu requires that the total oentu is the sae beore and ater the collision. Since linear oentu is a ector, the x and y coponents ust be consered separately. Using the drawing in the text, oentu conseration in the x direction yields while oentu conseration in the y direction yields Soling equation () or B, we ind that (1) () (3) Substituting equation (3) into Equation (1) leads to a. Soling or gies b. Fro equation (3), we ind that

300 IMPULSE AND MOMENTUM 31. SSM WWW REASONING The syste consists o the two balls. The total linear oentu o the two-ball syste is consered because the net external orce acting on it is zero. The principle o conseration o linear oentu applies whether or not the collision is elastic. 0 1 1 1 01 Total oentu ater collision Total oentu beore collision When the collision is elastic, the kinetic energy is also consered during the collision 1 1 1 0 1 1 1 01 Total kinetic energy ater collision Total kinetic energy beore collision SOLUTION a. The inal elocities or an elastic collision are deterined by siultaneously soling the aboe equations or the inal elocities. The procedure is discussed in Exaple 7 in the text, and leads to Equations 7.8a and 7.8b. According to Equation 7.8: 1 F HG I K J 1 and 1 01 1 F HG 1 Let the initial direction o otion o the 5.00-kg ball deine the positie direction. Substituting the alues gien in the text, these equations gie 5.00- kg ball 7.50- kg ball 1 F HG F HG 5.00 kg 7. 50 kg 5.00 kg 7. 50 kg (5.00 kg) 5.00 kg 7. 50 kg I KJ I KJ I K J 01 (.00 / s) = 0.400 / s (.00 / s) = +1.60 / s The signs indicate that, ater the collision, the 5.00-kg ball reerses its direction o otion, while the 7.50-kg ball oes in the direction in which the 5.00-kg ball was initially oing. b. When the collision is copletely inelastic, the balls stick together, giing a coposite body o ass which oes with a elocity. The stateent o conseration o linear oentu then becoes

Chapter 7 Probles 301 ( ) 01 1 1 0 Total oentu ater collision Total oentu beore collision The inal elocity o the two balls ater the collision is, thereore, 1 01 (5.00 kg)(.00 / s) 5.00 kg 7.50 kg +0.800 / s 1 3. REASONING AND SOLUTION The conseration o oentu gies Thereore, 33. REASONING AND SOLUTION The total linear oentu o the two-car syste is consered because no net external orce acts on the syste during the collision. We are ignoring riction during the collision, and the weights o the cars are balanced by the noral orces exerted by the ground. Moentu conseration gies ( ) 1 1 01 0 Total oentu ater collision Total oentu ater collision where = 0 /s since the 1900-kg car is stationary beore the collision. 0 a. Soling or, we ind that the elocity o the two cars just ater the collision is 100 kg +17 /s + 1900 kg0 /s 1 01 0 1 100 kg + 1900 kg +8.9 /s The plus sign indicates that the elocity o the two cars just ater the collision is in the sae direction as the direction o the elocity o the 100-kg car beore the collision. b. According to the ipulse-oentu theore, Equation 7.4, we hae Ft ( ) ( ) Ipulse due to riction 1 inal 1 ater Final oentu Total oentu when cars coe just ater collision to a halt

30 IMPULSE AND MOMENTUM where inal = 0 /s since the cars coe to a halt, and 8.9 /s. Thereore, we ater hae 4 Ft 100 kg + 1900 kg 0 /s 100 kg + 1900 kg 8.9 /s = 3.6 10 N s The inus sign indicates that the ipulse due to riction acts opposite to the direction o otion o the locked, two-car syste. This is reasonable since the elocity o the cars is decreasing in agnitude as the cars skid to a halt. c. Using the sae notation as in part (b) aboe, we hae ro the equations o kineatics (Equation.9) that ax inal ater where 0 /s inal and 8.9 /s ater. Fro Newton's second law we hae that k 1 a /, where k is the orce o kinetic riction that acts on the cars as they skid to a halt. Thereore, 0 x or x = k 1 1 k According to Equation 4.8, F k k N, where k is the coeicient o kinetic riction and F N is the agnitude o the noral orce that acts on the two-car syste. There are only two ertical orces that act on the syste; they are the upward noral orce F N and the weight ( 1 + )g o the cars. Taking upward as the positie direction and applying Newton's second law in the ertical direction, we hae F ( ) g ( ) a 0 N 1 1 y, or F ( ) g. Thereore,, and we hae N 1 ( ) (8.9 /s) x = 5.9 1 ( ) g g (0.68)(9.80 /s ) k 1 k 34. REASONING AND SOLUTION Moentu is consered in the horizontal direction during the "collision." Let the coal be object 1 and the car be object. Then

Chapter 7 Probles 303 The direction o the inal elocity is. 35. SSM REASONING The two skaters constitute the syste. Since the net external orce acting on the syste is zero, the total linear oentu o the syste is consered. In the x direction (the east/west direction), conseration o linear oentu gies P P, or ( ) cos 1 1 01 Note that since the skaters hold onto each other, they oe away with a coon elocity. In the y direction, P P, or ( ) sin 1 0 y 0 y These equations can be soled siultaneously to obtain both the angle and the elocity. SOLUTION a. Diision o the equations aboe gies F H G I L K J tan N M 1 0 1 tan 1 01 (70.0 kg)(7.00 / s) (50.0 kg)(3.00 / s) O QP b. Solution o the irst o the oentu equations gies 73.0 (50.0 kg)(3.00 / s) 1 01 ( ) cos (50.0 kg. 0 kg)(cos 73.0 ) 1 70 4.8 / s x 0 x 36. REASONING The ratio o the kinetic energy o the hydrogen ato ater the collision to that o the electron beore the collision is KE KE 1 hydrogen, ater collision H,H 1 electron, beore collison e 0,e where is the inal speed o the hydrogen ato, and,h 0,e is the initial speed o the electron. The ratio H / e o the asses is known. Since the electron and the stationary hydrogen ato experience an elastic head-on collision, we can eploy Equation 7.8b to deterine how is related to,h 0,e.

304 IMPULSE AND MOMENTUM SOLUTION According to Equation 7.8b, the inal speed,h o the hydrogen ato ater the collision is related to the initial speed 0,e o the electron by e,h 0,e e H Substituting this expression into the ratio o the kinetic energies gies KE KE e 1 H hydrogen, ater collision H,H e H 1 electron, beore collison e 0,e e The right hand side o this equation can be algebraically rearranged to gie KEhydrogen, ater collision H 3 1837.175 10 KE electron, beore collison e H 11837 1 e 37. REASONING AND SOLUTION a. The total oentu o the sled and person is consered since no external orce acts in the horizontal direction. ( p + s ) = p 0p gies p 0p 60.0 kg3.80 /s 60.0 kg 1.0 kg p s 3.17 /s The direction o is the sae as the direction o 0p. b. The rictional orce acting on the sled is k = µ k F N = µ k g. Newton's second law then gies the acceleration to be a = k / = µ k g The acceleration is constant so a = /(x) ro the equations o kineatics. Equating these two expressions and soling yields 3.17 /s k 0.0171 gx 9.80 /s 30.0

Chapter 7 Probles 305 38. REASONING AND SOLUTION During the elastic collision both the total kinetic energy and the total oentu o the balls are consered. The conseration o oentu gies 1 + = 01 + 0 where ball 1 has 01 = + 7.0 /s and ball has 0 = 4.0 /s. Hence, 1 + = 01 + 0 = 3.0 /s The conseration o kinetic energy gies or (1/) 1 + (1/) = (1/)01 + (1/)0 1 + = 01 + 0 = 65 /s Soling the irst equation or 1 and substituting into the second gies (6.0 /s) 56 /s = 0 The quadratic orula yields two solutions, = 7.0 /s and = 4.0 /s. The irst equation now gies 1 = 4.0 /s and 1 = + 7.0 /s. Hence, the 7.0-/s-ball has a inal elocity o The 4.0-/s-ball has a inal elocity o, opposite its original direction., opposite its original direction. 39. SSM REASONING The two balls constitute the syste. The tension in the wire is the only nonconseratie orce that acts on the ball. The tension does no work since it is perpendicular to the displaceent o the ball. Since J, the principle o conseration o echanical energy holds and can be used to ind the speed o the 1.50-kg ball just beore the collision. Moentu is consered during the collision, so the principle o conseration o oentu can be used to ind the elocities o both balls just ater the collision. Once the collision has occurred, energy conseration can be used to deterine how high each ball rises. SOLUTION a. Applying the principle o energy conseration to the 1.50-kg ball, we hae 1 1 gh gh 0 0 E E 0

306 IMPULSE AND MOMENTUM I we easure the heights ro the lowest point in the swing, aboe sipliies to 1 1 gh 0 0 Soling or, we hae, and the expression gh ( 5. 00 / s) (9.80 / s )(0.300 ) 5.56 / s 0 0 b. I we assue that the collision is elastic, then the elocities o both balls just ater the collision can be obtained ro Equations 7.8a and 7.8b: 1 F HG I K J 1 and 1 01 1 F HG 1 Since corresponds to the speed o the 1.50-kg ball just beore the collision, it is equal to the quantity calculated in part (a). With the gien alues o 1 1.50 kg and 4. 60 kg, and the alue o 01 = 5.56 / s obtained in part (a), Equations 7.8a and 7.8b yield the ollowing alues: 1 =.83 / s and = +.73 / s The inus sign in 1 indicates that the irst ball reerses its direction as a result o the collision. c. I we apply the conseration o echanical energy to either ball ater the collision we hae 1 1 gh gh 0 0 E E where 0 is the speed o the ball just ater the collision, and h is the inal height to which the ball rises. For either ball, h 0 = 0, and when either ball has reached its axiu height, = 0 /s. Thereore, the expression o energy conseration reduces to 0 I K J 01 1 gh or h 0 0 g Thus, the heights to which each ball rises ater the collision are 1.50- kg ball h 0 (.83 / s) g (9.80 / s ) 0.409

Chapter 7 Probles 307 4.60- kg ball h 0 (.73 / s) g (9.80 / s ) 0.380 40. REASONING AND SOLUTION Initially, the ball has a total echanical energy gien by E 0 = gh 0. Ater one bounce it reaches the top o its trajectory with an energy o E 1 = 0.900 E 0 = 0.900 gh 0 Ater two bounces it has energy E = (0.900) gh 0 Ater N bounces it has a reaining energy o E N = (0.900) N gh 0 In order to just reach the sill the ball ust hae E N = gh where h =.44. Hence, Taking the log o both sides gies (0.900) N gh 0 = gh or (0.900) N = h/h 0 Then N log(0.900) = log(h/h 0 ).44 log 6.10 N = 8.7 log (0.900) The ball can ake and still reach the sill. 41. SSM REASONING AND SOLUTION The location o the center o ass or a two-body syste is gien by Equation 7.10: x c x x 1 1 1 where the subscripts "1" and "" reer to the earth and the oon, respectiely. For conenience, we will let the center o the earth be coincident with the origin so that x 1 0 and x d, the center-to-center distance between the earth and the oon. Direct calculation then gies

308 IMPULSE AND MOMENTUM d 8 (7.3510 kg)(3.85 10 ) 6 xc 4.6710 4 1 5.9810 kg 7.3510 kg 4. REASONING AND SOLUTION The elocity o the center o ass o a syste is gien by Equation 7.11. Using the data and the results obtained in Exaple 5, we obtain the ollowing: a. The elocity o the center o ass o the two-car syste beore the collision is 1 01 0 c beore 1 3 3 (6510 kg)( + 0.80 /s) (910 kg)( +1. /s) 3 3 6510 kg 910 kg +1.0 /s b. The elocity o the center o ass o the two-car syste ater the collision is 1 c ater 1 +1.0 /s c. The answer in part (b) as the coon elocity. Since the cars are coupled together, eery point o the two-car syste, including the center o ass, ust oe with the sae elocity. 43. REASONING AND SOLUTION Equation 7.10 gies the center o ass o this two-ato syste as I we take the origin at the center o the carbon ato, then, and we hae 44. REASONING AND SOLUTION The drawing below shows the relatie positions and distances o the indiidual atos that ake up the nitric acid olecule.

Chapter 7 Probles 309 x Since the olecules are distributed in a plane, we ust speciy both the x and y coponents o the center o ass. Generalizing Equation 7.10 to account or all ie atos, we ind that the x coordinate o the center o ass is gien by x c x x x x 3 H H O O1 N N O O H N O where the subscripts on the asses reer to the nae o the ato. The distances x H, x O1, x N, and x O reer to the x distances o the atos ro the coordinate origin. The distance x O1 represents the location o the oxygen ato that is collinear with the hydrogen ato and the nitrogen ato, while x O reers to the x coponent o the location o the oxygen atos that are o the x axis. For conenience, we take the origin o coordinates at the location o the hydrogen ato, so x H 0, x x, x x x, and, O1 A N A B where the distances x, x, and d A B are identiied in the drawing aboe. Thus, we hae x c Fro the drawing we hae Substituting alues, we obtain H (0) O xa N ( xa xb) O ( xa xb d cos 65.0 ) 3 H N O 10 11 d cos 65.0 1.10 cos 65.0 = 5.16 10 7 10 7 10 (6.610 kg)(1.00 10 ) (3.310 kg)(.4110 ) 7 10 (6.6 10 kg)(.93 10 ) 10.8 10 1.67 10 kg 3.310 kg 3(6.610 kg) c 7 7 7

310 IMPULSE AND MOMENTUM Using siilar reasoning, we deduce that the y coordinate o the center o ass o the olecule is gien by y c y y y y y 3 H H O O1 N N O O aboe O O below H N O where y O aboe is the y coordinate o the oxygen ato located aboe the H O N line and y O below is the y coordinate o the oxygen ato located below the H O N line. Howeer, since the coordinate origin was chosen at the location o the hydrogen ato, the hydrogen ato, one o the oxygen atos, and the nitrogen ato are on the x axis and hae zero y coordinates, so that y y y 0. Furtherore, since the reaining two H O1 N oxygen atos are located syetrically on either side o the H O N line, we see that y y. Thus, the y coordinate o the center o ass o the nitric acid olecule O aboe O below is y c 0. Thereore, the center o ass lies along the x axis at x c = 45. SSM WWW REASONING The syste consists o the luberjack and the log. For this syste, the su o the external orces is zero. This is because the weight o the syste is balanced by the corresponding noral orce (proided by the buoyant orce o the water) and the water is assued to be rictionless. The luberjack and the log, then, constitute an isolated syste, and the principle o conseration o linear oentu holds. SOLUTION a. The total linear oentu o the syste beore the luberjack begins to oe is zero, since all parts o the syste are at rest. Moentu conseration requires that the total oentu reains zero during the otion o the luberjack. 1 1 Total oentu just beore the jup 0 Initial oentu Here the subscripts "1" and "" reer to the irst log and luberjack, respectiely. Let the direction o otion o the luberjack be the positie direction. Then, soling or 1 gies 1 (98 kg)(+3.6 / s) 30 kg 1 1.5 / s The inus sign indicates that the irst log recoils as the luberjack jups o.

Chapter 7 Probles 311 b. Now the syste is coposed o the luberjack, just beore he lands on the second log, and the second log. Graity acts on the syste, but or the short tie under consideration while the luberjack lands, the eects o graity in changing the linear oentu o the syste are negligible. Thereore, to a ery good approxiation, we can say that the linear oentu o the syste is ery nearly consered. In this case, the initial oentu is not zero as it was in part (a); rather the initial oentu o the syste is the oentu o the luberjack just beore he lands on the second log. Thereore, 1 1 1 01 0 Total oentu just ater luberjack lands Initial oentu In this expression, the subscripts "1" and "" now represent the second log and luberjack, respectiely. Since the second log is initially at rest, 01 0. Furtherore, since the luberjack and the second log oe with a coon elocity,. The stateent 1 o oentu conseration then becoes Soling or, we hae 1 0 0 (98 kg)(+3.6 / s) 30 kg 98 kg 1 + 1.1 / s The positie sign indicates that the syste oes in the sae direction as the original direction o the luberjack's otion. 46. REASONING AND SOLUTION The oentu o the spaceship is transerred to the rocket, so ( s + r ) s = r r, and the rocket's elocity is ( 6 s r ) s (4.0 10 kg 1300 kg) 30 /s 5 7.110 /s 1300 kg r r 47. SSM REASONING Batan and the boat with the criinal constitute the syste. Graity acts on this syste as an external orce; howeer, graity acts ertically, and we are concerned only with the horizontal otion o the syste. I we neglect air resistance and riction, there are no external orces that act horizontally; thereore, the total linear oentu in the horizontal direction is consered. When Batan collides with the boat, the horizontal coponent o his elocity is zero, so the stateent o conseration o linear oentu in the horizontal direction can be written as ( ) 0 1 1 01 Total horizontal oentu ater collision Total horizontal oentu beore collision

31 IMPULSE AND MOMENTUM Here, is the ass o the boat, and is the ass o Batan. This expression can be soled or, the elocity o the boat ater Batan lands in it. SOLUTION Soling or gies The plus sign indicates that the boat continues to oe in its initial direction o otion; it does not recoil. 48. REASONING AND SOLUTION According to the ipulse-oentu theore, Equation 7.4, the ipulse o the net aerage orce is equal to the change in the oentu o the car: F t 0 Since the initial elocity o the car is 0 = 0 /s, the inal oentu o the car is F t 680 N7. s 4900 kg /s 49. REASONING Since riction is negligible and since the downward-acting weight o each person is balanced by the upward-acting noral orce ro the sidewalk, the net external orce acting on the two-person syste is zero. Thereore, the syste is isolated, and the conseration o linear oentu applies. The initial total oentu ust be equal to the inal total oentu. SOLUTION Applying the principle o conseration o linear oentu and assuing that the direction in which Kein is oing is the positie direction, we ind b87 kg g kg 0 / s 87 kg kg.4 / s Kein b gb gb gb g Initial total oentu Kein b87 kg kggb.4 / sg 87 kg Final total oentu 3. 0 / s 50. REASONING AND SOLUTION a. According to Equation 7.4, the ipulse-oentu theore, t F. 0 Since the only horizontal orce exerted on the puck is the orce F exerted by the goalie,

Chapter 7 Probles 313 FF. Since the goalie catches the puck, 0 /s. Soling or the aerage orce exerted on the puck, we hae 0.17 kg 0 /s +65 /s 3 ( 0) F.10 N t 3 5.010 s By Newton s third law, the orce exerted on the goalie by the puck is equal in agnitude and opposite in direction to the orce exerted on the puck by the goalie. Thus, the aerage orce exerted on the goalie is 3. 10 N. b. I, instead o catching the puck, the goalie slaps it with his stick and returns the puck straight back to the player with a elocity o 65 /s, then the aerage orce exerted on the puck by the goalie is b g ( ) 0.17 kg ( 65 / s) (+65 / s) 0 F 3 t 5.0 10 s 4.4 10 3 N The aerage orce exerted on the goalie by the puck is thus 3 4.4 10 N. The answer in part (b) is twice that in part (a). This is consistent with the conclusion o Conceptual Exaple 3. The change in the oentu o the puck is greater when the puck rebounds ro the stick. Thus, the puck exerts a greater ipulse, and hence a greater orce, on the goalie. 51. SSM REASONING The two-stage rocket constitutes the syste. The orces that act to cause the separation during the explosion are, thereore, orces that are internal to the syste. Since no external orces act on this syste, it is isolated and the principle o conseration o linear oentu applies: ( ) 1 1 1 0 Total oentu ater separation Total oentu beore separation where the subscripts "1" and "" reer to the lower and upper stages, respectiely. This expression can be soled or 1. SOLUTION Soling or 1 gies

314 IMPULSE AND MOMENTUM 1 ( ) 1 1 0 400 kg 100 kg ( 4900 / s) (100 kg)(5700 / s) 400 kg +4500 / s Since 1 is positie. 5. REASONING AND SOLUTION The elocity o the center o ass is gien by Equation 7.11, a. When the asses are equal,, and we hae b. When the ass o the ball oing at 9.70 /s is twice the ass o the other ball, we hae 1 / =, and the elocity o the center o ass is 53. SSM REASONING AND SOLUTION The coet piece and Jupiter constitute an isolated syste, since no external orces act on the. Thereore, the head-on collision obeys the conseration o linear oentu: ( ) coet Jupiter coet coet Jupiter Jupiter Total oentu ater collision Total oentu beore collision where is the coon elocity o the coet piece and Jupiter ater the collision. We assue initially that Jupiter is oing in the +x direction so Jupiter 1. 310 4 / s. The coet piece ust be oing in the opposite direction so coet 6.010 4 / s. The inal elocity o Jupiter and the coet piece can be written as where is Jupiter the change in elocity o Jupiter due to the collision. Substituting this expression into the conseration o oentu equation gies ( )( ) coet Jupiter Jupiter coet coet Jupiter Jupiter

Chapter 7 Probles 315 Multiplying out the let side o this equation, algebraically canceling the ter ro both sides o the equation, and soling or yields Jupiter Jupiter ( coet coet Jupiter) coet Jupiter 1 4 4 (4.0 10 kg) ( 6.0 10 / s 1.310 / s) 1.510 1 7 4.0 10 kg 1. 9 10 kg 10 / s The change in Jupiter's speed is 1. 510 10 / s. 54. REASONING AND SOLUTION The conseration o oentu law applied in the horizontal direction gies c c + s s cos 30.0 = 0 so that c c s s cos 30.0 5.00 kg 8.00 /s cos 30.0 0.330 /s 105 kg The agnitude o the elocity is. The inus sign indicates that the direction is opposite the horizontal elocity coponent o the stone. 55. REASONING We will diide the proble into two parts: (a) the otion o the reely alling block ater it is dropped ro the building and beore it collides with the bullet, and (b) the collision o the block with the bullet. During the alling phase we will use an equation o kineatics that describes the elocity o the block as a unction o tie (which is unknown). During the collision with the bullet, the external orce o graity acts on the syste. This orce changes the oentu o the syste by a negligibly sall aount since the collision occurs oer an extreely short tie interal. Thus, to a good approxiation, the su o the external orces acting on the syste during the collision is negligible, so the linear oentu o the syste is consered. The principle o conseration o linear oentu can be used to proide a relation between the oenta o the syste beore and ater the collision. This relation will enable us to ind a alue or the tie it takes or the bullet/block to reach the top o the building. SOLUTION Falling ro rest ( 0, block = 0 /s), the block attains a inal elocity block just beore colliding with the bullet. This elocity is gien by Equation.4 as

316 IMPULSE AND MOMENTUM at block 0, block Final elocity o Initial elocity o block just beore block at top o bullet hits it building where a is the acceleration due to graity (a = 9.8 /s ) and t is the tie o all. The upward direction is assued to be positie. Thereore, the inal elocity o the alling block is block at (1) During the collision with the bullet, the total linear oentu o the bullet/block syste is consered, so we hae that bullet block bullet bullet block block Total linear oentu Total linear oentu ater collision beore collision () Here is the inal elocity o the bullet/block syste ater the collision, and bullet and block are the initial elocities o the bullet and block just beore the collision. We note that the bullet/block syste reerses direction, rises, and coes to a oentary halt at the top o the building. This eans that, the inal elocity o the bullet/block syste ater the collision ust hae the sae agnitude as block, the elocity o the alling block just beore the bullet hits it. Since the two elocities hae opposite directions, it ollows that = block. Substituting this relation and Equation (1) into Equation () gies Soling or the tie, we ind that at at bullet block bullet bullet block 0.015 kg810 /s bullet bullet t 0.34 s a bullet block 9.80 /s 0.015 kg 1.8 kg 56. REASONING The x- and y-coordinates o the center o ass can be ound by applying Equation 7.10 to the three atos in the sulur dioxide olecule. The x-coordinate o the center o ass uses the x-coordinates o the centers o the atos, and the y-coordinate o the center o ass uses the y-coordinates o the centers o the atos. Oxygen y Oxygen 0.143 n 60.0 60.0 0.143 n x Sulur