A B B A. the speed of the bat doesn t change significantly during the collision. Then the velocity of the baseball after being hit is v

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1 CHPTER 7: Linear oentu nswers to Questions. For oentu to be consered, the syste under analysis ust be closed not hae any forces on it fro outside the syste. coasting car has air friction and road friction on it, for eaple, which are outside forces and thus reduce the oentu of the car. If the ground and the air were considered part of the syste, and their elocities analyzed, then the oentu of the entire syste would be consered, but not necessarily the oentu of any single coponent, like the car.. Consider this proble as a ery light object hitting and sticking to a ery heay object. The large object sall object cobination (Earth + juper) would hae soe oentu after the collision, but due to the ery large ass of the Earth, the elocity of the cobination is so sall that it is not easurable. Thus the juper lands on the Earth, and nothing ore happens. 3. When you release an inflated but untied balloon at rest, the gas inside the balloon (at high pressure) rushes out the open end of the balloon. That escaping gas and the balloon for a closed syste, and so the oentu of the syste is consered. The balloon and reaining gas acquires a oentu equal and opposite to the oentu of the escaping gas, and so oe in the opposite direction to the escaping gas. 4. If the rich an would hae faced away fro the shore and thrown the bag of coins directly away fro the shore, he would hae acquired a elocity towards the shore by conseration of oentu. Since the ice is frictionless, he would slide all the way to the shore. 5. When a rocket epels gas in a gien direction, it puts a force on that gas. The oentu of the gasrocket syste stays constant, and so if the gas is pushed to the left, the rocket will be pushed to the right due to Newton s 3 rd law. So the rocket ust carry soe kind of aterial to be ejected (usually ehaust fro soe kind of engine) to change direction. 6. The air bag greatly increases the aount of tie oer which the stopping force acts on the drier. If a hard object like a steering wheel or windshield is what stops the drier, then a large force is eerted oer a ery short tie. If a soft object like an air bag stops the drier, then a uch saller force is eerted oer a uch longer tie. For instance, if the air bag is able to increase the tie of stopping by a factor of 0, then the aerage force on the person will be decreased by a factor of 0. This greatly reduces the possibility of serious injury or death. 7. Cruple zones are siilar to air bags in that they increase the tie of interaction during a collision, and therefore lower the aerage force required for the change in oentu that the car undergoes in the collision. 8. Fro Eq. 7-7 for a -D elastic collision,. Let represent the bat, and let represent the ball. The positie direction will be the (assued horizontal) direction that the bat is oing when the ball is hit. We assue the batter can swing the bat with equal strength in either case, so that is the sae in both pitching situations. ecause the bat is so uch heaier than the ball, we assue that the speed of the bat doesn t change significantly during the collision. Then the elocity of the baseball after being hit is. If 0, the ball tossed up into the air by the batter, then the ball oes away with twice the speed of the 005 Pearson Education, Inc., Upper Saddle Rier, NJ. ll rights resered. This aterial is protected under all copyright laws as they currently eist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the 64

2 Giancoli Physics: Principles with pplications, 6 th Edition bat. ut if 0, the pitched ball situation, we see that the agnitude of, and so the ball oes away with greater speed. If, for eaple, the pitching speed of the ball was about twice the speed at which the batter could swing the bat, then we would hae 4. Thus the ball has greater speed after being struck, and thus it is easier to hit a hoe run. This is siilar to the graitational slingshot effect discussed in proble The ipulse is the product of the force and the tie duration that the force is applied. So the ipulse fro a sall force applied oer a long tie can be larger than the ipulse applied by a large force oer a sall tie. 0. The oentu of an object can be epressed in ters of its kinetic energy, as follows. p KE. Thus if two objects hae the sae kinetic energy, then the one with ore ass has the greater oentu.. Consider two objects, each with the sae agnitude of oentu, oing in opposite directions. They hae a total oentu of 0. If they collide and hae a totally inelastic collision, in which they stick together, then their final coon speed ust be 0 so that oentu is consered. ut since they are not oing after the collision, they hae no kinetic energy, and so all of their kinetic energy has been lost.. The turbine blades should be designed so that the water rebounds. If the water rebounds, that eans that a larger oentu change for the water has occurred than if it just cae to a stop. nd if there is a larger oentu change for the water, there will also be a larger oentu change for the blades, aking the spin faster. 3. (a) The downward coponent of the oentu is unchanged. The horizontal coponent of oentu changes fro rightward to leftward. Thus the change in oentu is to the left in the picture. (b) Since the force on the wall is opposite that on the ball, the force on the wall is to the right. 4. (a) The oentu of the ball is not consered during any part of the process, because there is an eternal force acting on the ball at all ties the force of graity. nd there is an upward force on the ball during the collision. So considering the ball as the syste, there are always eternal forces on it, and so its oentu is not consered. (b) With this definition of the syste, all of the forces are internal, and so the oentu of the Earth-ball syste is consered during the entire process. (c) The answer here is the sae as for part (b). 5. In order to aintain balance, your ust be located directly aboe your feet. If you hae a heay load in your ars, your will be out in front of your body and not aboe your feet. So you lean backwards to get your directly aboe your feet. Otherwise, you would fall oer forwards. 6. The - length of pipe is unifor it has the sae density throughout, and so its is at its geoetric center, which is its idpoint. The ar and leg are not unifor they are ore dense where there is uscle, priarily in the parts that are closest to the body. Thus the of the ar or leg is closer the body than the geoetric center. The is located closer to the ore assie part of the ar or leg. 005 Pearson Education, Inc., Upper Saddle Rier, NJ. ll rights resered. This aterial is protected under all copyright laws as they currently eist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the 65

3 Chapter 7 Linear oentu 7. When you are lying flat on the floor, your is inside of the olue of your body. When you sit up on the floor with your legs etended, your is outside of the olue of your body. 8. The engine does not directly accelerate the car. The engine puts a force on the driing wheels, aking the rotate. The wheels then push backwards on the roadway as they spin. The Newton s 3 rd law reaction to this force is the forward-pushing of the roadway on the wheels, which accelerates the car. So it is the (eternal) road surface that accelerates the car. 9. The otion of the center of ass of the rocket will follow the original parabolic path, both before and after eplosion. Each indiidual piece of the rocket will follow a separate path after the eplosion, but since the eplosion was internal to the syste (consisting of the rocket), the center of ass of all the eploded pieces will follow the original path. Solutions to Probles. p 0.08 kg 8.4 s 0.4kg s. Fro Newton s second law, p F t. For a constant ass object, p. Equate the two epressions for p. t Ft F. If the skier oes to the right, then the speed will decrease, because the friction force is to the left. Ft 5 N0 s 7.7 s 65 kg The skier loses 7.7 s of speed. 3. Choose the direction fro the batter to the pitcher to be the positie direction. Calculate the aerage force fro the change in oentu of the ball. p Ft 5.0 s 39.0 s F t s kg N, towards the pitcher 3 4. The throwing of the package is a oentu-consering action, if the water resistance is ignored. Let represent the boat and child together, and let represent the package. Choose the direction that the package is thrown as the positie direction. pply conseration of oentu, with the initial elocity of both objects being 0. p p initial final 6.40 kg0.0 s 0.90 s 6.0 kg 45.0 kg The boat and child oe in the opposite direction as the thrown package. 005 Pearson Education, Inc., Upper Saddle Rier, NJ. ll rights resered. This aterial is protected under all copyright laws as they currently eist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the 66

4 Giancoli Physics: Principles with pplications, 6 th Edition 5. The force on the gas can be found fro its change in oentu. p 4 7 F 4.00 s500 kg s N downward t t t The force on the rocket is the Newton s 3 rd law pair (equal and opposite) to the force on the gas, and so is N upward. 6. The tackle will be analyzed as a one-diensional oentu consering situation. Let represent the halfback, and represent the tackling cornerback. p p initial final 95 kg 4. s 85 kg5.5 s 4.8 s 95 kg 85 kg 7. Consider the horizontal otion of the objects. The oentu in the horizontal direction will be consered. Let represent the car, and represent the load. The positie direction is the direction of the original otion of the car. p p initial final, 600 kg8.0 s 0.6 s,600 kg 5350 kg 8. Consider the otion in one diension, with the positie direction being the direction of otion of the first car. Let represent the first car, and represent the second car. oentu will be consered in the collision. Note that 0. p p initial final 9300 kg5.0 s 6.0 s 6.0 s kg 9. The force stopping the wind is eerted by the person, so the force on the person would be equal in agnitude and opposite in direction to the force stopping the wind. Calculate the force fro Eq. 7-, in agnitude only. 40 kg s s wind kg s 00 k h wind 7.8 s t 3.6 k h p wind wind wind wind F F on on wind 30 kg s7.8 s person wind t t t 833 N 8 0 N The typical aiu frictional force is F g s.0 70 kg 9.8 s 690 N, and so we fr see that F F on fr the wind is literally strong enough to blow a person off his feet. person 0. oentu will be consered in the horizontal direction. Let represent the car, and represent the snow. For the horizontal otion, 0 and. oentu conseration gies the following. 005 Pearson Education, Inc., Upper Saddle Rier, NJ. ll rights resered. This aterial is protected under all copyright laws as they currently eist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the 67

5 Chapter 7 Linear oentu p p initial final 3800 kg 8.60 s 7.94 s 7.9 s 3.50 kg 3800 kg 90.0 in in. Consider the otion in one diension, with the positie direction being the direction of otion of the original nucleus. Let represent the alpha particle, with a ass of 4 u, and represent the new nucleus, with a ass of 8 u. oentu conseration gies the following. p p initial final u40 s 8 u350 s s 4.0 u Note that the asses do not hae to be conerted to kg, since all asses are in the sae units, and a ratio of asses is what is significant.. Consider the otion in one diension with the positie direction being the direction of otion of the bullet. Let represent the bullet, and represent the block. Since there is no net force outside of the block-bullet syste (like frictions with the table), the oentu of the block and bullet cobination is consered. Note that 0. p p initial final 0.03 kg 30 s 70 s 0.69 s.0 kg 3. (a) Consider the otion in one diension with the positie direction being the direction of otion before the separation. Let represent the upper stage (that oes away faster) and represent the lower stage. It is gien that,, and. oentu rel conseration gies the following. p p initial final rel rel 975 kg5800 s 975 kg00 s 975 kg s, away fro Earth s.00 s s, away fro Earth rel (b) The change in KE had to be supplied by the eplosion. KE KE KE f i 487.5kg 6900 s 4700 s 975 kg5800 s J 005 Pearson Education, Inc., Upper Saddle Rier, NJ. ll rights resered. This aterial is protected under all copyright laws as they currently eist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the 68

6 Giancoli Physics: Principles with pplications, 6 th Edition 4. To alter the course by 35.0 o, a elocity perpendicular to the original elocity ust be added. Call the direction of the added elocity,, the positie direction. Fro the diagra, we see that tan. The oentu in add orig the perpendicular direction will be consered, considering that the gases are gien perpendicular oentu in the opposite direction of. The gas is epelled in the opposite direction to. gas gas gas rocket gas add before after add, and so a negatie alue is used for add p p 0 o 5 s tan 35 o rocket add.400 kg gas add gas 380 kg 5 s tan s add orig add final 5. (a) The ipulse is the change in oentu. The direction of trael of the struck ball is the positie direction. p 4.50 kg 45 s 0.0 kg s (b) The aerage force is the ipulse diided by the interaction tie. p.0 kg s F 5.80 N 3 t 3.50 s 6. (a) The ipulse gien to the nail is the opposite of the ipulse gien to the haer. This is the change in oentu. Call the direction of the initial elocity of the haer the positie direction. p p kg 8.5 s kg s nail haer i f (b) The aerage force is the ipulse diided by the tie of contact. F p.00 kg s t 8.00 s ag N 7. The ipulse gien the ball is the change in the ball s oentu. Fro the syetry of the proble, the ertical oentu of the ball does not change, and so there is no ertical ipulse. Call the direction WY fro the wall the positie direction for oentu perpendicular to the wall. o o o p sin 45 sin 45 sin 45 final initial o k 5 s sin 45.kg s, to the left 8. (a) The aerage force on the car is the ipulse (change in oentu) diided by the tie of interaction. The positie direction is the direction of the car s initial elocity. s 0 50 k h 3.6 k h p 5 5 F 500 kg.3890 N.40 N t t 0.5 s 005 Pearson Education, Inc., Upper Saddle Rier, NJ. ll rights resered. This aterial is protected under all copyright laws as they currently eist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the 69

7 Chapter 7 Linear oentu (b) The deceleration is found fro Newton s nd law. 5 F N F a a 93 s 500 kg 9. Call east the positie direction. (a) p original fullback original fullback 95 kg 4.0 s kg s (b) The ipulse on the fullback is the change in the fullback s oentu. p fullback final final 95 kg0 4.0 s kg s fullback fullback (c) The ipulse on the tackler is the opposite of the ipulse on the fullback, so kg s (d) The aerage force on the tackler is the ipulse on the tackler diided by the tie of interaction. p 3.80 kg s F t 0.75 s 5. 0 N 0. (a) The ipulse gien the ball is the area under the F s. t graph. pproiate the area as a triangle of height 50 N, and width 0.0 sec. p 50 N 0.0 s.5n s (b) The elocity can be found fro the change in oentu. Call the positie direction the direction of the ball s trael after being sered. p.5n s p i 0 - s f f i 6.00 kg. Find the elocity upon reaching the ground fro energy conseration. ssue that all of the initial potential energy at the aiu height h is conerted into kinetic energy. Take down to be the a positie direction, so the elocity at the ground is positie. gh gh a ground ground a When contacting the ground, the ipulse on the person causes a change in oentu. That relationship is used to find the tie of the stopping interaction. The force of the ground acting on the person is negatie since it acts in the upward direction. ground Ft 0 ground t F We assue that the stopping force is so large that we call it the total force on the person we ignore graity for the stopping otion. The aerage acceleration of the person during stopping a F is used with Eq. -b to find the displaceent during stopping, h. stop y y t at h h ground ground 0 0 stop ground stop F F F Fh F F F F g ground ground ground gh a stop h a 005 Pearson Education, Inc., Upper Saddle Rier, NJ. ll rights resered. This aterial is protected under all copyright laws as they currently eist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the 70

8 Giancoli Physics: Principles with pplications, 6 th Edition We assue that the person lands with both feet striking the ground siultaneously, so the stopping force is diided between both legs. Thus the critical aerage stopping force is twice the breaking strength of a single leg. h Fh F h kg9.8 s 70 0 N stop break stop a g g 69. Let represent the kg ball, and represent the 0.0-kg ball. We hae 3.30 s and 0. Use Eq. 7-7 to obtain a relationship between the elocities. Substitute this relationship into the oentu conseration equation for the collision. 0.0 kg 3.30 s.0 s east kg 3.30 s.0 s 4.40 s east 3. Let represent the kg puck, and let represent the kg puck. The initial direction of puck is the positie direction. We hae 3.00 s and 0. Use Eq. 7-7 to obtain a relationship between the elocities. Substitute this relationship into the oentu conseration equation for the collision kg 3.00 s.00 s.00 s west.350 kg 3.00 s.00 s.00 s east 4. Let represent the ball oing at.00 /s, and call that direction the positie direction. Let represent the ball oing at 3.00 /s in the opposite direction. So.00 s and 3.00 s. Use Eq. 7-7 to obtain a relationship between the elocities s Substitute this relationship into the oentu conseration equation for the collision, noting that..00 s 5.00 s 6.00 s 3.00 s 5.00 s.00 s The two balls hae echanged elocities. This will always be true for -D elastic collisions of objects of equal ass. 005 Pearson Education, Inc., Upper Saddle Rier, NJ. ll rights resered. This aterial is protected under all copyright laws as they currently eist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the 7

9 Chapter 7 Linear oentu 5. Let represent the kg tennis ball, and let represent the kg ball. The initial direction of the balls is the positie direction. We hae.50 s and.5 s. Use Eq. 7-7 to obtain a relationship between the elocities..35 s Substitute this relationship into the oentu conseration equation for the collision..35 s.35 s kg.50 s kg.5 s.35 s 0.50 kg 0.88 s.35 s.3 s oth balls oe in the direction of the tennis ball s initial otion. 6. Let represent the oing softball, and let represent the ball initially at rest. The initial direction of the softball is the positie direction. We hae 8.5 s, 0, and 3.7 s. (a) Use Eq. 7-7 to obtain a relationship between the elocities. 8.5 s s 4.8 s (b) Use oentu conseration to sole for the ass of the target ball. 0.0 kg 0.56 kg 4.8 s 8.5 s 3.7 s 7. Let the original direction of the cars be the positie direction. We hae 4.50 s and 3.70 s (a) Use Eq. 7-7 to obtain a relationship between the elocities s Substitute this relationship into the oentu conseration equation for the collision s 0.80 s 450 kg4.50 s 550 kg.90 s 0.80 s 4.4 s (b) Calculate p p p for each car kg s 4.00 kg s 000 kg p 450 kg 3.6 s 4.50 s kg s p 550 kg 4.4 s 3.70 s kg s The two changes are equal and opposite because oentu was consered. 3.6 s 005 Pearson Education, Inc., Upper Saddle Rier, NJ. ll rights resered. This aterial is protected under all copyright laws as they currently eist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the 7

10 Giancoli Physics: Principles with pplications, 6 th Edition 8. (a) oentu will be consered in one diension. Call the direction of the first ball the positie direction. Let represent the first ball, and represent the second ball. We hae 0 and. Use Eq. 7-7 to obtain a relationship between the elocities. Substitute this relationship into the oentu conseration equation for the collision. p p initial final kg kg (b) The fraction of the kinetic energy gien to the second ball is as follows. KE KE 9. Let represent the cube of ass, and represent the cube of ass. Find the speed of iediately before the collision,, by using energy conseration. gh gh 9.8 s s Use Eq. 7-7 for elastic collisions to obtain a relationship between the elocities in the collision. We hae 0 and. Substitute this relationship into the oentu conseration equation for the collision s 3.3 s 4 3 gh 9.8 s Each ass is oing horizontally initially after the collision, and so each has a ertical elocity of 0 as they start to fall. Use constant acceleration Eq. -b with down as positie and the table top as the ertical origin to find the tie of fall. y y t at H 0 0 gt t H g 0 0 Each cube then traels a horiztonal distance found by t. gh H t hh g 4 gh H 8 8 t hh g 30. (a) Use Eq. 7-7, along with 0, to obtain a relationship between the elocities. Substitute this relationship into the oentu conseration equation for the collision. Substitute this result into the result of Eq Pearson Education, Inc., Upper Saddle Rier, NJ. ll rights resered. This aterial is protected under all copyright laws as they currently eist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the 73

11 Chapter 7 Linear oentu (b) If, then approiate 0 0 The result is ; 0. n eaple of this is a ball bouncing off of the floor. The assie floor has no speed after the collision, and the elocity of the ball is reersed (if dissipatie forces are not present). (c) If, then approiate 0. The result is ;. n eaple of this would be a golf club hitting a golf ball. The speed of the club iediately after the collision is essentially the sae as its speed before the collision, and the golf ball takes off with twice the speed of the club. (d) If, then set. 0 The result is 0 ;. n eaple of this is one billiard ball aking a head-on collision with another. The first ball stops, and the second ball takes off with the sae speed that the first one had. 3. Fro the analysis in Eaple 7-0, the initial projectile speed is gien by gh. Copare the two speeds with the sae asses. gh h h 5. h.6 gh h 3. Fro the analysis in the Eaple 7-0, we know that gh h 0.08 kg30 s 9.8 s g 0.08 kg 3.6kg Fro the diagra we see that L L h L - h h L L L h Pearson Education, Inc., Upper Saddle Rier, NJ. ll rights resered. This aterial is protected under all copyright laws as they currently eist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the 74

12 Giancoli Physics: Principles with pplications, 6 th Edition 33. (a) In eaple 7-0,. i i KEi and KE f. The speeds are related by KE KE KE f i KE KE (b) For the gien alues, 380 g Thus 96% of the energy is lost. 394 g 34. Use conseration of oentu in one diension, since the particles will separate and trael in opposite directions. Call the direction of the heaier particle s otion the positie direction. Let represent the heaier particle, and represent the lighter particle. We hae.5, and. 0 p p 0 The negatie sign indicates direction. initial final 3 Since there was no echanical energy before the eplosion, the kinetic energy of the particles after the eplosion ust equal the energy added. Thus.5 E KE KE KE 5 5 added KE E 7500 J 4500 J KE E KE 7500 J 4500 J 3000 J added 5 added KE J KE J 35 Use conseration of oentu in one diension. Call the direction of the sports car s elocity the positie direction. Let represent the sports car, and represent the SUV. We hae 0 and. Sole for. p p 0 initial final The kinetic energy that the cars hae iediately after the collision is lost due to negatie work done by friction. The work done by friction can also be calculated using the definition of work. We assue the cars are on a leel surface, so that the noral force is equal to the weight. The distance the cars slide forward is. Equate the two epressions for the work done by friction, sole for, and use that to find. after 0 W KE KE fr final initial collision W F cos80 g o fr fr k g g k 005 Pearson Education, Inc., Upper Saddle Rier, NJ. ll rights resered. This aterial is protected under all copyright laws as they currently eist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the 75 k

13 Chapter 7 Linear oentu 90 kg 300 kg g s.8 k 90 kg 3.9 s 3 s 36. Consider conseration of energy during the rising and falling of the ball, between contacts with the floor. The graitational potential energy at the top of a path will be equal to the kinetic energy at the start and the end of each rising-falling cycle. Thus gh for any particular bounce cycle. Thus for an interaction with the floor, the ratio of the energies before and after the bounce is KE gh.0 after We assue that each bounce will further reduce the energy to KE gh.50 before 80% of its pre-bounce aount. The nuber of bounces to lose 90% of the energy can be epressed as follows. n log n 0.3 log 0.8 Thus after bounces, ore than 90% of the energy is lost. s an alternate ethod, after each bounce, 80% of the aailable energy is left. So after bounce, 80% of the original energy is left. fter the second bounce, only 80% of 80%, or 64% of the aailable energy is left. fter the third bounce, 5 %. fter the fourth bounce, 4%. fter the fifth bounce, 33 %. fter the sith bounce, 6%. fter the seenth bounce, %. fter the eight bounce, 7%. fter the ninth bounce, 3%. fter the tenth bounce, %. fter the eleenth bounce, 9% is left. So again, it takes bounces. 37. (a) For a perfectly elastic collision, Eq. 7-7 says coefficient of restitution definition. e.. Substitute that into the For a copletely inelastic collision,. Substitute that into the coefficient of restitution definition. e 0. (b) Let represent the falling object, and represent the heay steel plate. The speeds of the steel plate are 0 and 0. Thus e. Consider energy conseration during the falling or rising path. The potential energy of body at height h is transfored into kinetic energy just before it collides with the plate. Choose down to be the positie direction. gh gh The kinetic energy of body iediately after the collision is transfored into potential energy as it rises. lso, since it is oing upwards, it has a negatie elocity. gh gh Substitute the epressions for the elocities into the definition of the coefficient of restitution. gh e e h h gh 005 Pearson Education, Inc., Upper Saddle Rier, NJ. ll rights resered. This aterial is protected under all copyright laws as they currently eist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the 76

14 Giancoli Physics: Principles with pplications, 6 th Edition 38. Let represent the ore assie piece, and the less assie piece. Thus 3. In the eplosion, oentu is consered. We hae 0. p p 0 3 initial final 3 For each block, the kinetic energy gained during the eplosion is lost to negatie work done by friction on the block. W KE KE fr f i ut work is also calculated in ters of the force doing the work and the distance traeled. o W F cos80 F g fr fr k N k Equate the two work epressions, sole for the distance traeled, and find the ratio of distances. g 3 k g k g 9 nd so heay light 9 k g k 39. In each case, use oentu conseration. Let represent the 5.0-kg object, and let represent the 0.0-kg object. We hae 5.5 s and 4.0 s. (a) In this case,. 5.0 kg5.5 s 0.0 kg4.0 s 5.0 kg (b) In this case, use Eq. 7-7 to find a relationship between the elocities. 5.0 kg5.5 s 0.0 kg4.0 s.7 s 005 Pearson Education, Inc., Upper Saddle Rier, NJ. ll rights resered. This aterial is protected under all copyright laws as they currently eist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the kg 5.5 s 4.0 s. s 7.4 s (c) In this case, kg5.5 s 0.0 kg4.0 s 0.0 kg 4.3 s. s To check for reasonableness, first note the final directions of otion. has stopped, and has gone in the opposite direction. This is reasonable. Secondly, calculate the change in kinetic energy. KE 0.0 kg 5.0 kg 5.5 s 0.0 kg 4.0 s 0 J Since the syste has lost kinetic energy and the directions are possible, this interaction is reasonable.

15 Chapter 7 Linear oentu (d) In this case, kg5.5 s 0.0 kg4.0 s 5.0 kg.8 s This answer is not reasonable because it has oing in its original direction while has stopped. Thus has soehow passed through. If has stopped, should hae rebounded in the negatie direction. (e) In this case, 4.0 s. 5.0 kg5.5 s 4.0 s 0.0 kg4.0 s 0.3 s 0.0 kg The directions are reasonable, in that each object rebounds. Howeer, the speed of both objects is larger than its speed in the perfectly elastic case (b). Thus the syste has gained kinetic energy, and unless there is soe other source adding energy, this is not reasonable. 40. Use this diagra for the oenta after the decay. Since there was no oentu before the decay, the three oenta shown ust add to 0 in both the and y directions. p p p p nucleus neutrino nucleus y electron p p p p p nucleus nucleus nucleus y neutrino electron 3 pnucleus y p electron 3 pnucleus pneutrino kg s kg s.080 kg s kg s tan tan tan kg s o The second nucleus oentu is 50 o fro the oentu of the electron. p neutrino p electron p nucleus 4. Consider the diagra for the oenta of the eagles. oentu will be consered in both the and y directions. p p y y y 4.3 kg 7.8 s 5.6 kg 0. s y 4.3 kg 5.6 kg 6.7 s p p p 005 Pearson Education, Inc., Upper Saddle Rier, NJ. ll rights resered. This aterial is protected under all copyright laws as they currently eist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the 78

16 Giancoli Physics: Principles with pplications, 6 th Edition 5.6 kg0. s 4.3 kg7.8 s y o tan tan tan tan 60 rel. to eagle 4. (a) p : cos cos p : 0 sin sin y (b) Sole the equation for cos and the y equation for find the angle fro the tangent function. sin sin sin tan cos cos cos sin, and then sin.0 s sin 30.0 tan tan 33.0 o o o cos.80 s.0 scos30.0 With the alue of the angle, sole the y equation for the elocity. o sin kg.0 ssin s o sin kg sin Call the final direction of the joined objects the positie ais. diagra of the collision is shown. oentu will be consered in both the and y directions. Note that and 3. p : sin sin 0 sin sin y p : cos cos 3 cos cos 3 cos cos cos cos 70.5 o o 44. Write oentu conseration in the and y directions, and KE conseration. Note that both asses are the sae. We allow to hae both and y coponents. p : p : y y y KE : Substitute the results fro the oentu equations into the KE equation. y y y y 0 0 or 0 y y y Since we are gien that 0, we ust hae 0. This eans that the final direction of is the direction. Put this result into the oentu equations to find the final speeds. 3.7 s.0 s 005 Pearson Education, Inc., Upper Saddle Rier, NJ. ll rights resered. This aterial is protected under all copyright laws as they currently eist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the 79 y

17 Chapter 7 Linear oentu 45. Let represent the incoing neon ato, and represent the target ato. oentu diagra of the collision looks like the first figure. The figure can be re-drawn as a triangle, the second figure, since. Write the law of sines for this triangle, relating each final oentu agnitude to the initial oentu agnitude. sin sin sin sin sin sin sin sin The collision is elastic, so write the KE conseration equation, and substitute o o o the results fro aboe. lso note that sin sin sin sin o sin 0.0 u sin u o sin sin sin 74.4 sin Use Eq. 7-9a, etended to three particles..00 kg 0.50 kg kg 0.75 C C.00 kg.50 kg.0 kg 0.44 C o 55.6 o Choose the carbon ato as the origin of coordinates. u0 6 u C C O O 6.50 C O u 6 u fro the C ato. 48. Find the relatie to the front of the car. car car front front back back car front back 050 kg kg kg kg 70.0 kg kg Consider this diagra of the cars on the raft. Notice that the origin of coordinates is located at the of the raft. Reference all distances to that location. 00 kg9 00 kg9 00 kg kg 6800 kg y 00 kg9 00 kg9 00 kg9 300 kg 6800 kg.04 y 005 Pearson Education, Inc., Upper Saddle Rier, NJ. ll rights resered. This aterial is protected under all copyright laws as they currently eist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the 80

18 Giancoli Physics: Principles with pplications, 6 th Edition 50. y the syetry of the proble, since the centers of the cubes are along a straight line, the ertical coordinate will be 0, and the depth coordinate will be 0. The only coordinate to calculate is the one along the straight line joining the centers. The ass of each cube will be the olue ties the density, and so l, l, 3l. easuring fro the left edge of the sallest block, the locations of the s of the indiidual cubes are l, l, 4.5l. Use Eq. 7-9a to calculate the of the syste l 8l 7l l l l l l l 3.8 l fro the left edge of the sallest cube 5. Let each crate hae a ass. top iew of the pallet is shown, with the total ass of each stack listed. Take the origin to be the back left corner of the pallet. 5 l 3 3l 5l.l 0 7 l 3l 5l y 0.9l 0 y 3 5. Consider the following. We start with a full circle of radius R, with its at the origin. Then we draw a circle of radius R, with its at the coordinates 0.80 R,0. The full circle can now be labeled as a gray part and a white part. The y coordinate of the of the entire circle, the of the gray part, and the of the white part are all at y 0 by the syetry of the syste. The coordinate of the entire circle is at 0, and can be calculated by equation. gray gray gray white white total total white white total white white white white gray total white total white gray gray white white total. Rearrange this This is functionally the sae as treating the white part of the figure as a hole of negatie ass. The ass of each part can be found by ultiplying the area of the part ties the unifor density of the plate. R white white 0.80R 0.80R 0.7R gray R R 3 total white 53. Take the upper leg, lower leg, and foot all together. Note that Table 7- gies the relatie ass of OTH legs and feet, so a factor of / is needed. ssue a person of ass 70 kg kg kg Pearson Education, Inc., Upper Saddle Rier, NJ. ll rights resered. This aterial is protected under all copyright laws as they currently eist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the 8

19 Chapter 7 Linear oentu 54. With the shoulder as the origin of coordinates for easuring the center of ass, we hae the following relatie locations fro Table 7- for the ar coponents, as percentages of the height. Down is positie upper lower hand ar ar To find the, we can also use relatie ass percentages. Since the epression includes the total ass in the denoinator, there is no need to diide all asses by to find single coponent asses. Siply use the relatie ass percentages gien in the table. upper upper lower lower hand hand ar ar ar ar upper lower hand ar ar % of the person's height along the line fro the shoulder to the hand 55. Take the shoulder to be the origin of coordinates. We assue that the ar is held with the upper ar parallel to the floor and the lower ar and hand etended upward. easure horizontally fro the shoulder, and y ertically. Since the epression includes the total ass in the denoinator, there is no need to diide all asses by to find single coponent asses. Siply use the relatie ass percentages gien in the table. y upper upper lower lower hand hand ar ar ar ar upper lower hand ar ar y y y upper upper lower lower hand hand ar ar ar ar upper lower hand ar ar Conert the distance percentages to actual distance using the person s height. y 4.0% 55 c.7 c 4.9% 55 c 7.6 c 56. See the diagra of the person. The head, trunk, and neck are all lined up so that their s are on the torso s edian line. Call down the positie y direction. The y distances of the of each body part fro the edian line, in ters of percentage of full height, are shown below, followed by the percentage each body part is of the full body ass. trunk and neck upper legs lower legs feet head upper ars lower ars hands On edian line: head (h): 0 ; 6.9% body ass Trunk & neck (t n): 0 ; 46.% body ass Fro shoulder hinge point: upper ars (u a): = 9.5 ; 6.6% body ass lower ars (l a): = 5.9 ; 4.% body ass hands (ha): = 38. ;.7% body ass 005 Pearson Education, Inc., Upper Saddle Rier, NJ. ll rights resered. This aterial is protected under all copyright laws as they currently eist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the 8

20 Giancoli Physics: Principles with pplications, 6 th Edition Fro hip hinge point: upper legs (u l): = 9.6 ;.5% body ass lower legs (l l): = 33.9 ; 9.6% body ass feet (f): 5..8 = 50.3 ; 3.4% body ass Using this data, calculate the ertical location of the. y y y y y y y y h h t n t n u a u a l a l a ha ha u l u l l l l l f f y full body Thus the center of ass is 9.4% of the full body height below the torso s edian line. For a person of height.7, this is about 6 c. That is ost likely slightly outside the body. 57. (a) Find the relatie to the center of the Earth. E E E kg kg fro the center of the Earth This is actually inside the olue of the Earth, since kg kg R E (b) It is this Earth oon location that actually traces out the orbit as discussed in chapter 5. The Earth and oon will orbit about this location in (approiately) circular orbits. The otion of the oon, for eaple, around the Sun would then be a su of two otions: i) the otion of the oon about the Earth oon ; and ii) the otion of the Earth oon about the Sun. To an eternal obserer, the oon s otion would appear to be a sall radius, higher frequency circular otion (otion about the Earth oon ) cobined with a large radius, lower frequency circular otion (otion about the Sun). 58. (a) easure all distances fro the original position of the woan. W W 55 kg0 80 kg fro the woan 35 kg W (b) Since there is no force eternal to the an-woan syste, the will not oe, relatie to the original position of the woan. The woan s distance will no longer be 0, and the an s distance has changed to 7.5. W W 55 kg W 80 kg kg W W kg 80 kg kg W (c) When the an collides with the woan, he will be at the original location of the center of ass final initial He has oed 4. fro his original position. 005 Pearson Education, Inc., Upper Saddle Rier, NJ. ll rights resered. This aterial is protected under all copyright laws as they currently eist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the 83

21 Chapter 7 Linear oentu 59. The point that will follow a parabolic trajectory is the center of ass. Find the relatie to the botto of the allet. Each part of the haer (handle and head) can be treated as a point ass located at the of the respectie piece. So the of the handle is.0 c fro the botto of the handle, and the of the head is 8.0 c fro the botto of the handle. handle handle head head kg4.0 c.00 kg8.0 c 4.8 c.50 kg handle head Note that this is inside the head of the allet. 60. The of the syste will follow the sae path regardless of the way the ass splits, and so will still be d fro the launch point when the parts land. ssue that the eplosion is designed so that still is stopped in idair and falls straight down. I (a) (b) d 3 d 3 I I II II I I II II 7 d d II I II I 3 d 3d I I II II II II II II d 5d II 4 4 I II II 6. Call the origin of coordinates the of the balloon, gondola, and person at rest. Since the is at rest, the total oentu of the syste relatie to the ground is 0. The an clibing the rope cannot change the total oentu of the syste, and so the ust stay at rest. Call the upward direction positie. Then the elocity of the an with respect to the balloon is. Call the elocity of the balloon with respect to the ground. Then the elocity of the an with respect to the G ground is. pply Eq G G 0, upward G G G G G If the passenger stops, the balloon also stops, and the of the syste reains at rest. 6. To find the aerage force, diide the change in oentu by the tie oer which the oentu changes. Choose the direction to be the opposite of the baseball s incoing direction. The elocity with which the ball is oing after hitting the bat can be found fro conseration of energy, and knowing the height the ball rises. KE PE after gy initial final collision gy 9.80 s s The aerage force can be calculated fro the change in oentu and the tie of contact. p 0.45 kg s 3 F 3.60 N 3 t t.40 s p y y y 0.45 kg33.0 s 0 3 F 3.40 N y 3 t t.40 s F 3 y o F F F 5.00 N tan 43 y F y 005 Pearson Education, Inc., Upper Saddle Rier, NJ. ll rights resered. This aterial is protected under all copyright laws as they currently eist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the 84

22 Giancoli Physics: Principles with pplications, 6 th Edition 63. oentu will be consered in two diensions. The fuel was ejected in the y direction as seen fro the ground, and so the fuel had no -coponent of elocity. 3 p : 0 rocket 0 rocket fuel fuel 3 rocket 0 p : 0 y fuel fuel rocket fuel y 3 rocket 0 3 rocket y y In an elastic collision between two objects of equal ass, with the target object initially stationary, the angle between the final elocities of the objects is 90 o. Here is a proof of that fact. oentu conseration as a ector relationship says. Kinetic energy conseration says. The ector equation resulting fro oentu conseration can be illustrated by the second diagra. pply the law of cosines to that triangle of ectors, and then equate the two epressions for. cos Equating the two epressions for gies cos cos 0 90 o For this specific circustance, see the third diagra. We assue that the target ball is hit correctly so that it goes in the pocket. Find fro.0 o the geoetry of the left triangle: tan 30. Find fro o the geoetry of the right triangle: tan 60. Since the 3.0 balls will separate at a 90 o angle, if the target ball goes in the pocket, this does appear to be a good possibility of a scratch shot. 65. (a) The oentu of the astronaut space capsule cobination will be consered since the only forces are internal to that syste. Let represent the astronaut and represent the space capsule, and let the direction the astronaut oes be the positie direction. Due to the choice of reference frae, 0. We also hae.50 s. p p 0 initial final kg.50 s 0.94 s 800 kg The negatie sign indicates that the space capsule is oing in the opposite direction to the astronaut. (b) The aerage force on the astronaut is the astronaut s change in oentu, diided by the tie of interaction. p 40 kg.50 s 0 F 8.80 N t t 0.40 s Pearson Education, Inc., Upper Saddle Rier, NJ. ll rights resered. This aterial is protected under all copyright laws as they currently eist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the 85

23 Chapter 7 Linear oentu 66. Since the only forces on the astronauts are internal to the -astronaut syste, their will not change. Call the location the origin of coordinates. That is also the original location of the two astronauts. 60 kg 80 kg kg Their distance apart is Let represent the incoing ball, and represent the target ball. We hae 0 and. Use Eq. 7-7 to obtain a relationship between the elocities. 3 4 Substitute this relationship into the oentu conseration equation for the collision. p p 3 5 initial final We assue that all otion is along a single direction. The distance of sliding can be related to the change in the kinetic energy of a car, as follows. W KE W F cos80 F g o fr f i fr fr k N k g k f i For post-collision sliding, 0 and is the speed iediately after the collision,. Use this f i relationship to find the speed of each car iediately after the collision. Car : g g s s k k Car : g g s s k k During the collision, oentu is consered in one diension. Note that 0. p p initial final 900 kg4.55 s 00 kg8.78 s 900 kg For pre-collision sliding, again apply the friction energy relationship, with speed when the brakes were first applied. 5.4 s f and i 4 is the g g 5.4 s s 5 k i i k i h 8.68 s 64 i h s 69. ecause all of the collisions are perfectly elastic, no energy is lost in the collisions. With each collision, the horizontal elocity is constant, and the ertical elocity reerses direction. So, after each collision, the ball rises again to the sae height fro which it dropped. Thus, after fie bounces, the bounce height will be 4.00, the sae as the starting height. 70. This is a ballistic pendulu of sorts, siilar to Eaple 7-0 in the tetbook. There is no difference in the fact that the block and bullet are oing ertically instead of horizontally. The collision is still totally inelastic and conseres oentu, and the energy is still consered in the 005 Pearson Education, Inc., Upper Saddle Rier, NJ. ll rights resered. This aterial is protected under all copyright laws as they currently eist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the 86

24 Giancoli Physics: Principles with pplications, 6 th Edition rising of the block and ebedded bullet after the collision. So we siply quote the equation fro that eaple. gh kg50 s h 5.47 g 9.80 s kg.40 kg 7. This is a ballistic pendulu of sorts, siilar to Eaple 7-0 in the tetbook. oentu is. The kinetic energy present consered in the totally inelastic collision, and so iediately after the collision is lost due to negatie work being done by friction. W KE W F cos80 F g o fr f i after fr fr k N k collision g g k f i k Use this epression for in the oentu equation in order to sole for. g k 0.05 kg.35 kg 0.05 kg g s s k 7. Calculate the relatie to the 60-kg person s seat, at one end of the boat. See the first diagra. Don t forget to include the boat s ass. C C C 60 kg 0 80 kg.6 75 kg kg Now, when the passengers echange positions, the boat will oe soe distance d as shown, but the will not oe. We easure the location of the fro the sae place as before, but now the boat has oed relatie to that origin. C C.7 d 0.4 C 75 kg d 80 kg.6 d 60 kg 3. d 5 d kg 30 kg 5 kg 5 kg Thus the boat will oe 0. towards the initial position of the 75 kg person. 73. (a) The eteor striking and coing to rest in the Earth is a totally inelastic collision. Let represent the Earth and represent the eteor. Use the frae of reference in which the Earth is at rest before the collision, and so 0. Write oentu conseration for the collision. 60 kg 80 kg 75 kg 75 kg 80 kg 60 kg 005 Pearson Education, Inc., Upper Saddle Rier, NJ. ll rights resered. This aterial is protected under all copyright laws as they currently eist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the 87 d

25 Chapter 7 Linear oentu kg s.50 s kg.0 0 kg (b) The fraction of the eteor s KE transferred to the Earth is the final KE of the Earth diided by the initial KE of the eteor. KE final 4 3 Earth 6.00 kg.50 s KE initial.00 kg.50 s eteor (c) The Earth s change in KE can be calculated directly. 4 3 KE KE KE kg.50 s 0.9 J Earth final initial Earth Earth 74. oentu will be consered in one diension in the eplosion. Let represent the fragent with the larger KE. p p 0 initial final. KE KE The fragent with the larger KE energy has half the ass of the other fragent. 75. (a) The force is linear, with a aiu force of 580 N at 0 seconds, and a iniu force of 40 N at 3 illiseconds. (b) The ipulse gien the bullet is the area under the F s. t graph. The area is trapezoidal. 580 N 40 N 3 Ipulse 3.00 s 0.93N s (c) The ipulse gien the bullet is the change in oentu of the bullet. The starting speed of the bullet is 0. Ipulse 0.93 N s 3 Ipulse p kg 0 s 76. For the swinging balls, their elocity at the botto of the swing and the height to which they rise are related by conseration of energy. If the zero of graitational potential energy is taken to be the lowest point of the swing, then the kinetic energy at the low point is equal to the potential energy at the highest point of the swing, where the speed is zero. Thus we hae gh for any swinging ball, and so the relationship between speed and height is (a) Calculate the speed of the lighter ball at the botto of its swing. o botto botto gh. gh 9.8 s cos60.75 s.7 s (b) ssue that the collision is elastic, and use the results of proble 30. Take the direction that ball is oing just before the collision as the positie direction. Newtons illiseconds 005 Pearson Education, Inc., Upper Saddle Rier, NJ. ll rights resered. This aterial is protected under all copyright laws as they currently eist. No portion of this aterial ay be reproduced, in any for or by any eans, without perission in writing fro the 88