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FIVE READINGS

VIEWS: 4 PAGES: 44

									                                                                                    Chapter 6
                                    Momentum and Collisions

Quick Quizzes

1.   (d). We are given no information about the masses of the objects. If the masses are the
     same, the speeds must be the same (so that they have equal kinetic energies), and then
      p1  p2 . If the masses are not the same, the speeds will be different, as will the momenta,
     and either p1  p2 , or p1  p2 , depending on which particle has more mass. Without
     information about the masses, we cannot choose among these possibilities.
2.   (c). Because the momentum of the system (boy + raft) remains constant with zero
     magnitude, the raft moves towards the shore as the boy walks away from the shore.
3.   (c). The total momentum of the car-truck system is conserved. Hence, any change in
     momentum of the truck must be counterbalanced by an equal magnitude change of
     opposite sign in the momentum of the car.
4.   (a). The total momentum of the two-object system is zero before collision. To conserve
     momentum, the momentum of the combined object must be zero after the collision. Thus,
     the combined object must be at rest after the collision.
5.   (a) Perfectly inelastic. Any collision in which the two objects stick together afterwards is
     perfectly inelastic.
     (b) Inelastic. Both the Frisbee and the skater lose speed (and hence, kinetic energy) in this
     collision. Thus, the total kinetic energy of the system is not conserved.
     (c) Inelastic. The kinetic energy of the Frisbee is conserved. However, the skater loses
     speed (and hence, kinetic energy) in this collision. Thus, the total kinetic energy of the
     system is not conserved.
6.   (a). If all of the initial kinetic energy is transformed, then nothing is moving after the
     collision. Consequently, the final momentum of the system is necessarily zero. Because
     momentum of the system is conserved, the initial momentum of the system must be zero
     meaning that the two objects must have had equal magnitude momenta in opposite
     directions before the collision.




                                                                                                  191
192    CHAPTER 6



Answers to Even Numbered Conceptual Questions

 2.   The glass, concrete, and steel were part of a rigid structure that shattered upon impact of
      the airplanes with the towers and upon collapse of the buildings as the steel support
      structures weakened due to high temperatures of the burning fuel. The sheets of paper
      floating down were probably not in the vicinity of the direct impact, where they would
      have burned after being exposed to very high temperatures. The papers were most likely
      situated on desktops or open file cabinets and were blown out of the buildings as they
      collapsed.
 4.   No. Only in a precise head-on collision with equal and opposite momentum can both balls
      wind up at rest. Yes. In the second case, assuming equal masses for each ball, if Ball 2,
      originally at rest, is struck squarely by Ball 1, then Ball 2 takes off with the velocity of
      Ball 1. Then Ball 1 is at rest.
 6.   The skater gains the most momentum by catching and then throwing the Frisbee.
                                      p2
 8.   Kinetic energy can be written as   . Thus, even through the particles have the same
                                      2m
      kinetic energies their momenta may be different due to a difference in mass.
10.   The resulting collision is intermediate between an elastic and a completely inelastic
      collision. Some energy of motion is transformed as the pieces buckle, crumple, and heat
      up during the collision. Also, a small amount is lost as sound. The most kinetic energy is
      lost in a head-on collision, so the expectation of damage to the passengers is greatest.
12.   The less massive object loses the most kinetic energy in the collision.
14.   The superhero is at rest before the toss and the net momentum of the system is zero.
      When he tosses the piano, say toward the right, something must get an equal amount of
      momentum to the left to keep the momentum at zero. This something recoiling to the left
      must be the superhero. He cannot stay at rest.
16.   The passenger must undergo a certain momentum change in the collision. This means
      that a certain impulse must be exerted on the passenger by the steering wheel, the
      window, an air bag, or something. By increasing the time during which this momentum
      change occurs, the resulting force on the passenger can be decreased.
18.   A certain impulse is required to stop the egg. But, if the time during which the
      momentum change of the egg occurs is increased, the resulting force on the egg is
      reduced. The time is increased when the sheet billows out as the egg is brought to a stop.
      The force is reduced low enough so that the egg will not break.
                                                                        Momentum and Collisions   193


Answers to Even Numbered Problems

 2.   (a) 5.40 N·s                            (b) –27.0 J

 4.   (a) 0                                   (b)    1. kg  m s
                                                      1

 6.    7
      1. kN

 8.   1.  104 N upw ar
       91              d

10.   (a)    7. kg  m s w es w ar
              50             t d              (b)             t d
                                                     375 N eas w ar

12.   (a) 12.0 N·s                            (b)    6. m s
                                                      00                              (c)    00
                                                                                            4. m s

14.   260 N perpendicular to the wall

16.   (a)    6. kg  m s toward the pitcher
              3
      (b)    3.  103 N toward the pitcher
              2

18.   62 s

20.   (a)     49
             0. m s                           (b)    2.  10-2 m s
                                                      0

22.   vthrower  2. m s, vcatcher  2.  10 2 m s
                  48                 25

24.   (a) B exerts a horizontal force on A. (b) A exerts a horizontal force on B that is opposite in
      direction to the force B exerts on A. (c) The force on A is equal in magnitude to the force
      on B, but is oppositely directed. (d) Yes. The momentum of the system (the two skaters) is
      conserved because the net external force on the system is zero (neglecting friction). (e)
      2.22 m/s

26.   Fav  3.  103 N , no broken bones
             75

28.   5.  102 m s
       3

30.   143 m s

32.   (a)    20. m s Eas
               9        t                     (b)    8.  103 J into internal energy
                                                      68

34.   (a)    2. m s toward the right
              2                               (b) No

36.    40. cm s (10.0-g object),  10. cm s (15.0-g object)
          0                           0

38.   (a)     50
             2. m s                           (b)    3.  104 J
                                                      75
194    CHAPTER 6



40.   (a) 0, 1. m s
              50                              (b)   1. m s,1. m s
                                                      00     50
      (c) 1. m s,1. m s
            00      50

42.   (a)   12. m s at 14.9° N of E
              4                               (b) 7.20 %

44.   No, his speed was 41. m ih .
                          5

46.   40.5 g

48.   0.556 m

               4M 
50.   vm in          g
               m 

52.   0.960 m above the level of point B

54.   91 m s

56.   (a) 9. m s,  9. m s
            90           90                   (b)    16. m s, 3. m s
                                                        5       30
      (c) 13.9 m, 0.556 m

58.   0.980 m

60.   (a)   vred  0,vbl  3. m s
                       ue   00                (b) 0.212 m

                                     2
62.   (a)   vm  v0    2 ,v3m  v0            (b) 35.3°
                                     3

64.   (a) 90.0°                               (b)                                 ar 
                                                    3. m s  cue bal , 2. m s  t get
                                                     46            l     00

66.   0.31 m

68.   (a) See solution for diagrams.
      (b) From Newton’s third law, the forces have equal magnitudes and opposite directions.
      (c) pA   2M v 3 , pB   2M v 3 , pC  0
      (d) Kinetic energy is not conserved in this inelastic collision.

70.   33 m s, 2.  103 m s2
               9

72.   (a)    1
            1. m s at30°f om t pos tve x- s
                         r    he  ii     axi         (b) 0.32 or 32%

74.   (a)   7. m s
             1                                (b) 2.5 m
                                                                                            Momentum and Collisions   195


Problem Solutions

6.1   The velocity of the ball just before impact is found from vy  v0y  2ay y as
                                                                 2    2




                                                                     
              v1   v0y  2ayy   0  2 9. m s2  1. m    4. m s
                      2
                                             80         25         95

      and the rebound velocity with which it leaves the floor is

                                                                     
              v2   v2  2ayy   0  2 9. m s2  0. m    4. m s
                      f                     80         960        34

      The impulse given the ball by the floor is then

                               
              I Ft  m v  m v2  v1              
                     0. kg   4. m s  4. m s    1. N  s  1. N s upw ar
                        150       34         95           39         39          d



6.2   Assume the initial direction of the ball in the –x direction, away from the net.


      (a)                        0600 kg  40. m s  50. m s giving
            I p  m vf  vi   0.         0             0     
            I 5. kg  m s  5. N  s toward the net.
                40               40


      (b) W or  KE 
              k
                              1
                              2
                                 
                                m v2  vi
                                   f
                                        2
                                              
                         0. kg  40. m s   50. m s                     
                                                      2                   2
                           0600       0            0
                                                                               27. J
                                                                                      0
                                                  2


6.3   Use p  m v :

      (a)                     00    
            p  1.  1027 kg 5.  106 m s  8.  1021 kg  m s
                 67                           35                  
                           00   
      (b) p  1.  102 kg 3.  102 m s  4. kg  m s
               50                          50                 
      (c)   p 75. kg10. m s  750 kg  m s
                  0       0


                          98    
      (d) p 5.  1024 kg 2.  104 m s  1.  1029 kg  m s
              98                          78                  
196   CHAPTER 6



6.4   (a) Since the ball was thrown straight upward, it is at rest momentarily (v = 0) at its
          maximum height. Therefore, p 0 .

      (b) The maximum height is found from vy  v0y  2ay  y with vy  0 .
                                            2    2




                                                                                           2
                                                                                          v0y
               0  v0y  2  g y m ax
                    2
                                                              Thus,  y m ax 
                                                                                          2g


           We need the velocity at y 
                                         y                     m ax
                                                                         
                                                                               2
                                                                              v0y
                                                                                    , thus vy  v0y  2ay  y gives
                                                                                            2    2

                                                              2               4g

                                   v0y  v0y
                                     2      2
                                                        v0y 15 m s
               vy  v0y  2  g 
                2    2
                                            , or vy      
                                   4 g    2             2    2


           Therefore, p  m vy 
                                         0. kg15 m s 
                                           10
                                                                                   1. kg  m s upward.
                                                                                    1
                                                         2


6.5   (a) If pball  pbullet,


           then vball                 
                                         00   3
                        m bulletvbullet 3.  10 kg 1.  10 m s
                                                       50 3

                                                                31. m s
                                                                   0
                                                                                              
                           m ball                  145
                                                  0. kg

      (b) The kinetic energy of the bullet is

                                                                                                 
                                                                                                        2
                         1                  3.  10-3 kg 1.  103 m s
                                             00            50
               KEbullet  m bulletvbullet 
                                   2
                                                                       3.  103 J
                                                                         38
                         2                               2

                                                                 0. kg 31. m s  69. J
                                                                                                              2
                                                 1                 145      0
           while that of the baseball is KEball  m ballvball 
                                                         2
                                                                                       7
                                                 2                      2

           The bulethast l gerki i ener
                  l     he ar  netc    gy by a factor of 48.4.


6.6   From the impulse-momentum theorem,                                               
                                                                               Fav  t  p  m vf  m vi



      Thus, Fav 
                       
                     m vf  vi        55  10   -3
                                                             
                                                        kg 2.  102 f  0  1 m s 
                                                             0       ts                   
                                                                                      1. kN
                                                                                        7
                            t                         0020 s 0
                                                        0.                 281 f 
                                                                           3.   ts
                                                                                                Momentum and Collisions   197


6.7   If the diver starts from rest and drops vertically into the water, the velocity just before
      impact is found from

                    KE  PE    KE  PE 
                                    g
                                        f
                                                       g
                                                           i



                   1 2
                     m vi pact  0  0  m gh  vi pact  2gh
                        m                        m
                   2

      With the diver at rest after an impact time of t, the average force during impact is
                                    
                               m 0  vim pact     = m
                                                                                           directed upward 
                                                               2gh                m 2gh
      given by Fav =                                                 or Fav 
                                            t             t                       t

      Assuming a mass of 55 kg and an impact time of ~1.0 s, the magnitude of this average
      force is

                                55 kg 29. m
                                           80                    
                                                               s2 10 m   
                       Fav                                                   =770 N , or ~ 103 N
                                                  0
                                                 1. s


6.8   The speed just before impact is given by KE  PEg                                KE  PE 
                                                                                      f
                                                                                                    g
                                                                                                        i
                                                                                                            as

                   1 2
                     m vi pact  0  0  m gh , or vi pact  2gh
                        m                           m
                   2
      The time required for the stuntman to travel distance d as the mattresses bring him to
      rest is
                              d          d          2d       2d
                   t                                   
                             vav                 
                                   0  vim pact 2 vim pact   2gh

      Taking upward as positive, the impulse-momentum theorem gives the average net force
      exerted on the stuntman as he comes to rest as

                       p 0  m vim pact         
                                             m 2gh m gh  75. kg   9. m s   25. m 
                                                             0         80   2
                                                                                   0
       
       Fav
             net
                   
                       t
                          
                                t
                                          
                                            2d 2gh
                                                   
                                                     d
                                                        
                                                                        00
                                                                       1. m
                                                                                           1.  104 N
                                                                                              84



      or           F  av
                             net
                                    1.  104 N upw ar . But, this net upward force is the sum of an
                                      84              d

      upward force exerted by the mattresses and the downward gravitational force,
      Fnet   Fm attress  m g . Thus, the average upward force exerted by the mattresses is



                   Fm attress  Fnet  m g  1.  104 N   75. kg  9. m s2   1.  104 N
                                              84              0        80          91
198    CHAPTER 6



6.9               
       I Fav  t  p  m  v

       Thus, I  m v  70. kg 5. m s 0  364 kg  m s , and
                           0       20



                       I    364 kg  m s
               Fav                      438 kg  m s2
                       t       832
                              0. s


       or      Fav  438 N di ect f w ar
                            r ed or d



6.10   Choose toward the east as the positive direction.

       (a) The impulse delivered to the ball as it is caught is

                          I p  m v f  m vi  0  0. kg  15. m s   7. kg  m s
                                                       500        0           50

              or          I 7. kg  m s w es w ar
                              50             t d

       (b) The average force exerted by the ball on the receiver is the negative of the average
           force exerted by the receiver on the ball, or

                                                             7. kg  m s
               F av r c i r
                       e e ve
                                    
                                  Fav
                                          bal
                                            l
                                                
                                                     I
                                                     t
                                                         
                                                           
                                                                50
                                                                 0200 s 
                                                                0.
                                                                             375 N


               F av r c i r
                       e e ve
                                 375 N eas w ar
                                           t d


6.11   (a) The impulse equals the area under the F versus t graph. This area is the sum of the
           area of the rectangle plus the area of the triangle. Thus,

                                           1
               I  2. N   3. s 
                     0        0               2. N  2. s  8. N  s
                                                0       0       0
                                           2

                                     
       (b) I Fav  t  p  m vf  vi          
               8. N  s 1. kg vf  0, gi ng vf  5. m s
                0          5              vi         3
                                                                         Momentum and Collisions   199


       (c)                                
             I Fav  t  p  m vf  vi , s vf  vi 
                                             o
                                                          I
                                                          m

                                      8. N  s
                                        0
                 vf  2. m s
                        0                       3. m s
                                                  3
                                         5
                                       1. kg


6.12   (a) Impulse = area under curve = (two triangular areas of altitude 4.00 N and base
           2.00 s) + (one rectangular area of width 1.00 s and height of 4.00 N.)

                                4. N   2. s 
                                   00       00
             Thus,        I 2                     4. N 1. s  12. N  s
                                                        00     00       0
                              
                                      2         
                                                 


                                         o
       (b) I Fav  t  p  m vf  vi , s vf  vi 
                                                          I
                                                          m

                            12. N  s
                               0
                 vf  0               6. m s
                                         00
                              00
                             2. kg

                         I            12. N  s
                                         0
       (c)   vf  vi       2. m s
                               00                4. m s
                                                   00
                         m              00
                                       2. kg


6.13   (a) The impulse is the area under the curve between 0 and 3.0 s.

             This is:     I ( 0 N ) 3. s) 12 N  s
                              4.    ( 0

       (b) The area under the curve between 0 and 5.0 s is:

                                 0 ( 0
             I ( 0 N ) 3. s) ( 2. N ) 2. s) 8. N  s
                 4.    ( 0                      0


       (c)                                
             I Fav  t  p  m vf  vi , s vf  vi 
                                             o
                                                          I
                                                          m

                                      I      12 N  s
             at 3.0 s:    vf  vi       0           8. m s
                                                         0
                                      m       50
                                             1. kg

                                      I      8. N  s
                                              0
             at 5.0 s:    vf  vi       0           5. m s
                                                         3
                                      m       50
                                             1. kg
200    CHAPTER 6



               p                                 px                           py
6.14   Fav                so        Fav  x    t
                                                        and        Fav  y    t
               t


                         v  
                  m  vy
                                          m  vcos60.   vcos60. 
                                                      0           0
        Fav  y               
                                y
                           f        i
                                                                      0
                           t                           t

                  m  vx  f   vx i m  vsi               n60. 
                                                 n60.     vsi
                                                    0               0 
        Fav  x   t                 
                                                       t

                           n60.  2 3. kg10. m s si
                     2m vsi  0        00       0      n60. 
                                                          0
                                                             260 N
                          t                  200
                                            0. s

       Thus, Fav  260 N i t negatve x- r ton orper
                         n he     i    di ec i         c ar o he   l
                                                   pendi ul t t w al


                     x 2 x    21. m 
                                       20
6.15   (a)   t                            9.  102 s
                                                60
                     vav vf  vi 0  25. m s
                                        0

                     p m  v 1400 kg 25. m s
                                              0
       (b) Fav                            2
                                                     3.  105 N
                                                       65
                     t   t        9.  10 s
                                     60

                     v      0
                          25. m s                          1g   
       (c)   aav       
                          60   2
                     t 9.  10 s
                                   260 m s2  260 m s2 
                                                           80
                                                               2
                                                         9. m s 
                                                                    26. g
                                                                       6   

6.16   Choose the positive direction to be from the pitcher toward home plate.

       (a)                                        
             I Fav  t  p  m v f  vi   0. kg  22 m s   20 m s
                                               15                          

                        
             I Fav  t  6. kg  m s or
                             3                          6. kg  m s t ar t picher
                                                         3           ow d he t

                     I 6. kg  m s
                           3
       (b) Fav                 3
                                     3.  103 N
                                        2
                     t   2.  10 s
                           0


             or Fav  3.  103 N t ar t picher
                       2          ow d he t
                                                                                                 Momentum and Collisions   201


6.17   Choose eastward as the positive x-direction.


                 Fx av   
                             px m  vx 
                                          
                                              6          
                                             1.  103 kg 0  25 m s
                                                                             
                                                                        6.  103 N
                                                                           7
                             t      t                 0
                                                       6. s

       or       Fav  6.  103 N w es w ar
                       7             t d


6.18   We shall choose southward as the positive direction.

                                                     w   730 N
       The mass of the man is m                               74. kg . Then, from conservation of
                                                                   5
                                                     g 9. m s2
                                                         80
       momentum, we find

                 m m anvm an  m bookvbook  f   m m anvm an  m bookvbook i or

                74. kg v
                   5              m an    1. kg 5. m s  0  0 and vm an  8.  102 m s
                                             2        0                           1

       Therefore, the time required to travel the 5.0 m to shore is

                       x         0
                                5. m
                t                        62 s
                      vm an 8.  102 m s
                             1


6.19   Requiring that total momentum be conserved gives

                 m clubvclub  m ballvball f   m clubvclub  m ballvballi

       or        200 g 40 m s   46 g v            bal
                                                           l       200 g 55 m s  0


       and      vbal  65 m s
                   l




                                         w      30 N
6.20   (a) The mass of the rifle is m                  3. kg . We choose the direction of the
                                                           1
                                          g 9. m s2
                                               80
             bullet’s motion to be negative. Then, conservation of momentum gives

                m      v
                     r fe r fe
                      il il                        m
                                  m bulletvbullet
                                                     f
                                                                  v
                                                             r fe r fe
                                                              il il                      
                                                                          m bulletvbullet
                                                                                             i




                                                            
             or  3. kg vril  5.  103 kg  300 m s  0  0 and vril  0. m s
                   1        fe   0                                      fe   49
202    CHAPTER 6



                                                         730 N
       (b) The mass of the man plus rifle is m                 74. kg . We use the same
                                                                   5
                                                       9. m s2
                                                        80
                                             5.  103 kg 
                                               0
            approach as in (a), to find v                   300 m s  2.  102 m s
                                                                           0
                                             74. kg 
                                                   5       


6.21   The velocity of the girl relative to the ice, vG I , is vG I  vG P  vPI where
       vG P  vel t ofgi lr atve t pl
                ociy       r el i o ank, and
       vPI  vel t ofpl
                ociy      ank r atve t i Since we are given that
                                el i o ce.
       vG P  1. m s , this becomes
               50

               vG I  1. m s vPI
                       50                                                                   (1)

                                                                              m 
       (a) Conservation of momentum gives m G vG I  m P vPI  0 , or vPI    G  vG I    (2)
                                                                               mP 

                                          m 
            Then, Equation (1) becomes  1 G  vG I  1. m s
                                                        50
                                          mP 

                            50
                          1. m s
            or vG I                   1. m s
                                         15
                            45. kg 
                               0
                        1 
                            150 kg 
                                    

                                                  45. kg 
                                                     0
       (b) Then, using (2) above,        vPI             1. m s   0. m s
                                                              15          346
                                                  150 kg 
                                                          

            or vPI  0.        r ed       ie o he r       i
                      346 m s di ect oppos t t t gi ls m oton
                                                                           Momentum and Collisions   203


6.22   Consider the thrower first, with velocity after the throw of vthrower . Applying
       conservation of momentum yields

                                         0450 kg  30. m s   65. kg  0.
                65. kg  vthrow er   0.
                   0                                  0            0       0450 kg  2. m s
                                                                                       50


       or      vthrower  2. m s
                           48

       Now, consider the (catcher + ball), with velocity of vcatcher after the catch. From
       momentum conservation,

                           0450 kg  vcatcher   0.
                60. kg  0.
                   0                               0450 kg  30. m s   60. kg   0
                                                                0            0


       or              vcatcher  2.  102 m s
                                   25


6.23   The ratio of the kinetic energy of the Earth to that of the ball is

                                             2
               KEE 1 m E vE  m E   vE 
                          2
                   2                                                                               (1)
               KEb 1 m bvb  m b   vb 
                    2
                          2
                                  

       From conservation of momentum,

                                                           vE    m
               pf  pi  0 , giving m EvE  m bvb  0 or        b
                                                           vb    mE

                                                            2
                                 KEE  m E   m b   mb
       Equation (1) then becomes           m   m
                                 KEb  m b     E     E



                                                 KEE m b   1 kg
       Using order of magnitude numbers,                ~ 25   ~ 1025
                                                 KEb m E 10 kg


6.24   (a) B exerts a horizontal force on A.

       (b) A exerts a force on B that is opposite in direction to the force B exerts on A.

       (c) The force on A is equal in magnitude to the force on B, but is oppositely directed.

       (d) Yes. The momentum of the system (the two skaters) is conserved because the net
           external force on the system is zero (neglecting friction).
204    CHAPTER 6



       (e)    p 
                x s tm
                   ys e
                             px  A   px B  0  m A  vA  0  m B  vB  0  0


                       m             mB   
             or vA    B  vB             2. m s   2. m s
                                                  00          22
                       mA            900
                                     0. m B 


             vA  2. m s i t di ecton oppos t t vB
                   22    n he r i          ie o


6.25   Consider a system consisting of arrow and target from the instant just before impact
       until the instant after the arrow emerges from the target. No external horizontal forces
       act on the system, so total horizontal momentum must be conserved, or

                         m ava  m tvt f   m ava  m tvti

                                     m a  va i  m t  vti  m t  vt f
       Thus,             va  f 
                                                      ma


                                
                                      22. g 35. m s   300 g 2. m s  0 
                                         5        0                    50
                                                                                                      67
                                                                                                     1. m s
                                                                   5
                                                                 22. g

6.26   For each skater, the impulse-momentum theorem gives

                         p m  v  75. kg 5. m s
                                        0       00
                Fav                                  3.  103 N
                                                          75
                         t   t            100
                                           0. s

       Since Fav  4500 N , there are no broken bones


6.27   (a) If M is the mass of a single car, conservation of momentum gives

                 3M  vf  M  3. m s   2M  1. m s , or vf  1. m s
                                 00                20                80

       (b) The kinetic energy lost is KElost  KEi  KE f , or

                            1              1                                          1
                              M  3. m s   2M                  1. m s              3M    1. m s
                                         2                                    2                             2
                KElost            00                                20                           80
                            2              2                                          2

             With M  2.  104 kg , this yields KEl t  2.  104 J
                       00                         os     16
                                                                                 Momentum and Collisions   205


6.28   Let us apply conservation of energy to the block from the time just after the bullet has
       passed through until it reaches maximum height in order to find its speed V just after
       the collision.

               1 2             1                    1
                 m vi  m gyi  m v2  m gyf becomes m V 2  0  0  m gyf
                                   f
               2               2                    2

       or     V     2gyf                    
                                  2 9. m s2  0. m   1. m s
                                     80        120      53

       Now use conservation of momentum from before until just after the collision in order to
       find the initial speed of the bullet, v.

              7.  10
                0      3
                              
                            kg v  0  1. kg1. m s  7.  103 kg  200 m s
                                         5      53        0                    
       from which     v  5.  102 m s
                           3


6.29   Let M = mass of ball, m = mass of bullet, v = velocity of bullet, and V = the initial
       velocity of the ball-bullet combination. Then, using conservation of momentum from
       just before to just after collision gives

                                             m 
              M    m  V  m v  0 or V  
                                            M m
                                                 
                                                   v


       Now, we use conservation of mechanical energy from just after the collision until the
       ball reaches maximum height to find

                                                                                           2
                                      1                             V2   1  m  2
              0   M  m  ghm ax     M  m  V 2  0 or hm ax               v
                                      2                             2g 2g  M  m 

       With the data values provided, this becomes

                                                           2
                            1         030
                                      0. kg     
               hm ax                                           200 m s   2
                                                                                 57 m
                              2 
                                         
                       2 9. m s  0. kg  0. kg 
                          80       15      030 
206    CHAPTER 6



6.30   First, we will find the horizontal speed, v0x , of the block and embedded bullet just after
       impact. After this instant, the block-bullet combination is a projectile, and we find the
                                                     1 2
       time to reach the floor by use of y  v0yt ay t , which becomes
                                                     2


               1. m  0 
                 00
                               1
                               2
                                   
                                 9. m s2 t , giving
                                   80      2
                                                               t = 0.452 s


                     x 2. m
                         00
       Thus, v0x            4. m s
                                43
                         452
                      t 0. s

       Now use conservation of momentum for the collision, with vb = speed of incoming
       bullet:

               8.  10
                 00     3
                                                      
                             kg vb  0  258  10-3 kg  4. m s , so
                                                          43

               vb  143 m s                        (about 320 mph)


6.31   When Gayle jumps on the sled, conservation of momentum gives

                50. kg  5. kg v   50. kg 4. m s  0 , or v
                   0       00          2 0       00                     2    3. m s
                                                                               64

       After Gayle and the sled glide down 5.00 m, conservation of mechanical energy gives

               1
               2
                  55. kg v3  0  1 55. kg 3. m s   55. kg 9. m s2  5. m 
                     0      2

                                    2
                                         0       64
                                                       2
                                                              0      80       00    
       so      v3  10. m s
                      5

       After her Brother jumps on, conservation of momentum yields

                55. kg  30. kg v   55. kg10. m s  0 , and v
                   0        0          4  0       50                        4    6. m s
                                                                                   82

       After all slide an additional 10.0 m down, conservation of mechanical energy gives the
       final speed as

               1
               2
                 85. kg v5  0  1 85. kg 6. m s   85. kg 9. m s2 10. m 
                    0      2

                                   2
                                        0       82
                                                      2
                                                             0      80       0      

       or      v5  15. m s
                      6
                                                                                          Momentum and Collisions   207


6.32   (a) Conservation of momentum gives m T vfT  m cvfc  m T viT  m cvic , or


               vfT 
                                     
                        m T viT  m c vic  vfc   
                                   mT

                         9000 kg 20. m s  1200 kg   25.  18. 
                                      0                       0     0                   m s
                                                                                            
                    
                                                       9000 kg

               vf  20. m s Eas
                T     9        t

                                 1        1     2     1        1        
       (b) KElost  KEi  KE f   m cvic  m T viT    m cv2c  m T v2T 
                                       2
                                                              f         f
                                 2        2         2          2        

                  
                      1
                      2
                             
                         m c vic  v2c  m T viT  v2T 
                              2
                                    f        2
                                                   f         
                      1
                        1200 kg   625  324  m      2
                                                               s2    9000 kg   400  438.   m
                                                                                             2         2
                                                                                                           s2  
                      2                                                                                        


            KEl t  8.  103 J w hi becom es i er ener
              os     68       ,   ch         nt nal   gy

            Note: If 20.9 m/s were used to determine the energy lost instead of 20.9333, the
            answer would be very different. We keep extra digits in all intermediate answers
            until the problem is complete.


6.33   First, we use conservation of mechanical energy to find the speed of the block and
       embedded bullet just after impact:

                                                                    1                       1
                KE  PEs f   KE  PEsi becom es 2 m  M V 2  0  0  2 kx2

       and yields         V
                                   kx2
                                       
                                                  150 N   m         0. m 
                                                                        800
                                                                                2

                                                                                     29. m s
                                                                                        3
                                  m M                 0.  0.  kg
                                                         0120 100

       Now, employ conservation of momentum to find the speed of the bullet just before
       impact: m v  M  0   m  M  V ,


                  m M               0. kg 
                                          112
       or      v 
                   m            V  
                                        0120 kg 
                                                    29. m s  273 m s
                                                       3
                                       0.       
208    CHAPTER 6



6.34   (a) Using conservation of momentum,  p afer   pbef e , gives
                                                  t            or



              4.  10  3.  kg v   4. kg 5. m s  10 kg 3. m s   3. kg  4. m s
              0           0             0       0                  0           0         0


             Therefore,               v  2. m s , or 2. m s t ar t r ght
                                            2           2      ow d he i

       (b)    N o . For example, if the 10-kg and 3.0-kg mass were to stick together first, they
             would move with a speed given by solving

                13 kg v  10 kg 3. m s   3. kg  4. m s , or v
                             1        0           0         0                1     1. m s
                                                                                      38

             Then when this 13 kg combined mass collides with the 4.0 kg mass, we have

                17 kg v  13 kg1. m s   4. kg 5. m s , and v  2. m
                                     38          0       0                  2           s

             just as in part (a).


6.35   (a) From conservation of momentum,

                 5. g v
                   00         1f    10. g v2 f   5. g 20. cm s  0
                                        0              00      0                                        (1)

             Also for an elastic, head-on, collision, we have v1i  v1 f  v2i  v2 f , which becomes
             20. cm s v1 f  v2 f .
               0                                                                                        (2)

             Solving (1) and (2) simultaneously yields

                v1 f   6. cm s , and v2 f  13. cm s
                          67                    3


       (b) KEi  KE1i  KE2i 
                                        1
                                        2
                                           
                                          5.  103 kg  0.
                                           00              
                                                          200 m s  0  1.  104 J
                                                                  2
                                                                          00


                       1          1
                                                                    
                                                               2
             KE2 f      m 2v2 f  10.  103 kg 13.  10-2 m s  8.  105 J, so
                             2
                                     0             3               89
                       2          2

                 KE2 f       8.  105 J
                              89
                                         0.
                                            889
                 KEi         1.  104 J
                              00
                                                                            Momentum and Collisions   209


6.36   Using conservation of momentum gives

               10. g v
                  0      1f    15. g v2 f  10. g 20. cm s  15. g 30. cm s
                                   0              0       0            0        0                     (1)

       For elastic, head on collisions, v1i  v1 f  v2i  v2 f which becomes

               20. cm s v1 f   30. cm s+v2 f
                 0                  0                                                                 (2)


       Solving (1) and (2) simultaneously gives v1 f   40. cm s ,
                                                           0


       and v2 f  10. cm s
                    0


6.37   Conservation of momentum gives

                25. g v
                   0     1f    10. g v2 f   25. g 20. cm s  10. g15. cm s
                                   0               0       0            0      0                      (1)

       For head-on, elastic collisions, we know that v1i  v1 f  v2i  v2 f .

       Thus, 20. cm s v1 f  15. cm s+v2 f
               0                0                                                                     (2)

       Solving (1) and (2) simultaneously yields

               v1 f  17. cm s , and v2 f  22. cm s
                        1                     1
210    CHAPTER 6



6.38   (a) The internal forces exerted by the actor do not change total momentum.




            From conservation of momentum

                 4m  vi   3m   2. m s +m  4. m s
                                      00           00

                        6. m s 4. m s
                         00      00
                vi                      50
                                      = 2. m s
                              4

                                    1                                   1
                                        3m   2. m s  m  4. m s    4m   2. m s
                                                       2             2                     2
       (b) W actor  K f  K i                  00            00                   50
                                    2                                  2


            W actor 
                         2.  10
                           50      4
                                       kg       12.  16.  25.  m s   2
                                                                                 3.  104 J
                              2                   0      0     0                 75



6.39   We assume equal firing speeds v and equal forces F required for the two bullets to push
       wood fibers apart. These forces are directed opposite to the bullets displacements
       through the fibers.

       When the block is held in the vise, W net  KE f  KEi gives


                                      
                F 8.  102 m cos180  0 
                   00
                                                           1
                                                           2
                                                              
                                                             7.  103 kg v2
                                                              00                

       or
                1
                2
                  00                00    
                  7.  103 kg v2  8.  102 m F                                            (1)


       When a second 7.00-g bullet is fired into the block, now on a frictionless surface,
       conservation of momentum yields

                                                            7.  103 
                              00                        
                  1. kg vf  7.  103 kg v  0 , or vf  
                   014
                                                              00
                                                            1. 014   
                                                                         v                     (2)
                                                                                        Momentum and Collisions   211


       Also, applying the work-energy theorem to the second impact,


                Fdcos180 
                                1
                                2
                                    014
                                                2
                                                   00 
                                  1. kg v2f  1 7.  103 kg v2                                                (3)


       Substituting (2) into (3), we obtain

                                                      2
                                7.  103  2 1
                      1
                 Fd  1. kg 
                      2
                         014
                                  00
                                1. 014   
                                             v  7.  103 kg v2
                                                2
                                                  00                               

                     1                      7.  103 
       or
                     2
                        00                  
                F d  7.  103 kg v2   1 
                                       
                                                00
                                                  014 
                                                 1.      
                                                                                                                  (4)


       Finally, substituting (1) into (4) gives

                                           7.  103 
                      
                F d  8.  102 m F  1 
                        00                00
                                               014 
                                              1.
                                                                  94    2
                                                       , or d  7.  10 m  7. cm
                                                                              94



6.40   First, consider conservation of momentum and write

                m 1v1i  m 2v2i  m 1v1 f  m 2v2 f

       Since m 1  m 2 , this becomes v1i  v2i  v1 f  v2 f .                                                   (1)


                                                                              
       For an elastic head-on collision, we also have v1i  v2i   v1 f  v2 f ,           
       which may be written as                            v1i  v2i  v1 f  v2 f                                 (2)

       Subtracting Equation (2) from (1) gives                        v2 f  v1i                                  (3)

       Adding Equations (1) and (2) yields                            v1 f  v2i                                  (4)

       Equations (3) and (4) show us that, under the conditions of equal mass objects striking
       one another in a head-on elastic collision, the two objects simply exchange velocities.
       Thus, we may write the results of the various collisions as

       (a)   v1f  0 , v2 f  1. m s
                               50

       (b) v1 f   1. m s , v2 f  1. m s
                     00              50

       (c)   v1 f  1. m s , v2 f  1. m s
                     00              50
212    CHAPTER 6



6.41   Choose the +x-axis to be eastward and the +y-axis northward.

       (a) First, we conserve momentum in the x direction to find


               185 kg V cos   90 kg 5. m s , or V cos   185  5. m s
                                            0                    
                                                                 
                                                                    90
                                                                        0
                                                                       
                                                                                                     (1)


           Conservation of momentum in the y direction gives


               185 kg V si    95 kg 3. m s , or V si    185  3. m s
                           n                0              n     
                                                                 
                                                                    95
                                                                        0
                                                                       
                                                                                                     (2)


                                                               95 3. 
                                                                      0
           Divide equation (2) by (1) to obtain t  
                                                 an                       , and   32
                                                               90 5. 
                                                                      0

           Then, either (1) or (2) gives V = 2. m s, which rounds to V  2. m s
                                              88                          9

       (b) KElost  KEi  KE f

                       1
                         
                            90 kg 5. m s   95 kg 3. m s  185 kg 2. m s 
                                           2                   2                   2
                                    0                   0                   88
                       2                                                             

                    7.  102 J conver ed i o i er ener
                      9               t nt nt nal      gy


6.42   Choose the +x-axis to be eastward and the +y-axis northward.

       (a) Conserving momentum in the x direction gives

               0  10. kg v2x   8. kg15. m s  0 , or v2x  12. m s
                      0              00      0                       0

           Momentum conservation in the y direction yields

                8. kg  4. m s  10. kg v
                  00        00          0           2y    0  0 , or v2y  3. m s
                                                                              20


           After collision, v2  v2x  v2y  154 m s 12. m s
                                  2     2
                                                        4


                                   v2y          3. 
                                                     20
           and             t 1 
                              an         tan1         14.  . Thus, the final velocity of the
                                                             9
                                   v2x          12. 
                                                      0
           10.0-kg mass is v2  12. m s at14.  N ofE
                                  4         9
                                                                         Momentum and Collisions   213


             KElost KEi  KE f      KE f
       (b)                     1
             KEi      KEi           KEi


                   1 
                            00       00        
                         8.  4.  2  10.  154 2 
                                                0        0.
                                                             0720
                                 8. 15.   0
                                    00      0
                                              2
                                                        
                                                       

             or 7.
                 20%    of the original kinetic energy is lost in the collision.


6.43   Choose the +x-axis to be eastward and the +y-axis northward.

       If vi is the initial northward speed of the 3000-kg car, conservation of momentum in the
       y direction gives

               0   3000 kg vi   5000 kg  5. m s s n40.  , or vi  5. m s
                                               22        i   0              59

       Observe that knowledge of the initial speed of the 2000-kg car was unnecessary for this
       solution.


6.44   We use conservation of momentum for both northward and eastward components.

       For the eastward direction: M 13. m s  2 M V f cos55. 
                                        0                     0

                                                    n55. 
       For the northward direction: M v2i  2M V f si  0

       Divide the northward equation by the eastward equation to find:

               v2i  13. m s t 55. 
                        0       an 0

                                 237 m ih  
                                2.
                   13. m s 
                        0                     t 55.   41. m ih
                                                an 0       5
                    
                               1 m s   

       Thus, the driver of the north bound car was untruthful.
214    CHAPTER 6



6.45   Choose the x-axis to be along the original line of motion.

       (a) From conservation of momentum in the x direction,

               m  5. m s  0  m  4. m s cos30.   m v2 f cos
                    00                33          0

           or v2 f cos  1. m s
                           25                                                                    (1)

           Conservation of momentum in the y direction gives

               0  m  4. m s si 30.   m v2 f si  , or v2 f si    2. m s
                        33      n 0               n              n        16                     (2)

                                               2.16
           Dividing (2) by (1) gives t  
                                      an              1. and   60. 
                                                         73           0
                                               1.25

           Then, either (1) or (2) gives v2 f  2. m s , so the final velocity of the second ball is
                                                 50
            v2 f  2. m s at- 0
                    50       60.


       (b) KEi 
                    1 2
                    2
                                 1
                      m v1i  0  m  5. m s  m 12. m 2 s2
                                 2
                                       00
                                             2
                                                   5          
                  1 2 1 2
            KE f  m v1 f  m v2 f
                  2        2

                    1              1
                      m  4. m s  m  2. m s  m 12. m             s2 
                                 2             2
                          33            50            5           2

                    2              2

           Since KE f  KEi , this is an el tc coli i
                                          as i    ls on


6.46   The recoil speed of the subject plus pallet after a heartbeat is

                     x 6.  105 m
                          00
               V                   3.  104 m s
                                       75
                     t      160
                           0. s

       From conservation of momentum, m v  M V  0 0, so the mass of blood leaving the
       heart is

                     V              3.  104 m s
                                        75
               m  M     54. kg 
                               0                               2
                                                      4.  10 kg= 40. g
                                                         05           5
                      v             0. 500 m s 
                                                                                        Momentum and Collisions   215


6.47   From the Impulse-momentum theorem:

                                                                           F  t
                          
                 I F  t  p  m v f  vi               v f  vi 
                                                                             m

       (a) For the first 1.5-s interval, vi  0 and F  3. N
                                                          0


                              vf  0
                                           3. N 1. s 
                                              0      5
                                                                     9. m s
            so                                                         0
                                                   50
                                                  0. kg

       (b) For the next 3.0-s interval, vi   9. m s and F   4. N
                                                0                0


            giving            v f  9. m s
                                     0
                                                      4. N   3. s 
                                                         0        0
                                                                            15 m s
                                                             50
                                                            0. kg

6.48   First, we use conservation of mechanical energy to find the speed of m 1 at B just before
                            1
       collision. This gives m 1 v1  0  0  m 1ghi,
                                  2

                            2

       or        v1 
                  2
                           2 ghi                          
                                         2 9. m s2  5. m   9. m s
                                            80        00       90

       Next, we apply conservation of momentum and knowledge of elastic collisions to find
       the velocity of m 1 at B just after collision.

       From conservation of momentum, with the second object initially at rest,


       we have m 1v1 f  m 2v2 f  m 1v1i  0 , or v2 f 
                                                                    m1
                                                                    m2
                                                                       
                                                                       v1i  v1 f                                (1)


       For head-on elastic collisions, v1 f  v1i  v2 f  v2i . Since v2i  0 in this case, this becomes
       v2 f  v1 f  v1i and combining this with (1) above we obtain


                         m  m2         5.  10. 
                                            00    0
                 v1 f   1       v1i   5.  10.   9. m s   3. m s
                         m1  m2        00     0
                                                         90          30


       Finally, use conservation of mechanical energy for m 1 after the collision to find the
                                                          1
       maximum rebound height. This gives 0  m 1ghm ax  m 1v1 f  0
                                                                 2

                                                          2


       or        hm ax 
                            2
                           v1 f
                                  
                                      3. m s
                                         30
                                                    2

                                                         0. m
                                                           556
                           2g        
                                   2 9. m s2
                                      80            
216    CHAPTER 6



6.49   Choose the positive direction to be the direction of the truck’s initial velocity.

       Apply conservation of momentum to find the velocity of the combined vehicles after
       collision:

                4 000 kg  800 kg  V   4 000 kg  8. m s   800 kg  8. m s
                                                         00                     00

       which yields V   5. m s
                           33


                                                                             
       Use the impulse-momentum theorem, I Fav  t  p  m vf  vi , to find the   
       magnitude of the average force exerted on each driver during the collision.

       Truck Driver:

               Fav 
                       m vf  vi
                                   t uck
                                    r
                                           
                                                80. kg 5. m
                                                   0      33     s  8. m s
                                                                      00
                                                                               1.  103 N
                                                                                 78
                            t                             120
                                                          0. s

       Car Driver:
                       m vf  vi              80. kg 5. m
                                                 0      33      s   8. m s
                                                                        00
               Fav                car
                                                                                  8.  103 N
                                                                                     89
                           t                              120
                                                          0. s


6.50   If the pendulum bob barely swings through a complete circle, it arrives at the top of the
       arc (having risen a vertical distance of 2 ) with essentially zero velocity.

       From conservation of mechanical energy, we find the minimum velocity of the bob at

                                     
       the bottom of the arc as KE  PEg
                                         bot om
                                            t
                                                         t
                                                           
                                                 KE  PEg , or M V 2  0  M g 2  . This
                                                           op
                                                                1
                                                                2
                                                                     
       gives V  2 g as the needed velocity of the bob just after the collision.

       Conserving momentum through the collision then gives the minimum initial velocity of
       the bullet as

                 v
               m M 2 g
                 2                      m v  0 , or v    4M
                                                                 m
                                                                     g
                                                                               Momentum and Collisions         217


6.51   Note that the initial velocity of the target particle is zero (that is, v2i  0 ).

       From conservation of momentum, m 1v1 f  m 2v2 f  m 1v1i  0                                            (1)

       For head-on elastic collisions, v1 f  v1i  v2 f  0                                                    (2)

       Solving (1) and (2) simultaneously yields the final velocities as

                          m  m2                  2m 1 
                  v1 f   1       v1i and v2 f   m  m  v1i
                          m1  m2                 1    2



       (a) If m 1  2. g,m 2  1. g,and v1i  8. m s, then
                     0          0              0

                           8                32
                  v1 f      m s and v2 f     m s
                           3                 3

       (b) If m 1  2. g,m 2  10 g,and v1i  8. m s , we find
                     0                         0

                             16                8
                  v1 f        m s and v2 f    m s
                              3                3

       (c) The final kinetic energy of the 2.0 g particle in each case is:

                                                                       2
                                                          8   
            Case (a): KE1 f 
                                 1
                                 2
                                       2
                                            
                                            1
                                   m 1v1 f  2.  103 kg  m s  7.  103 J
                                            2
                                              0            
                                                          3   
                                                                    1


                                                                           2
                                                      16    
                             1
                             2
                                  2    1
            Case (b): KE1 f  m 1v1 f  2.  103 kg  
                                       2
                                         0           3
                                                           
                                                          m s  2.  102 J
                                                             
                                                                  8


            Since the incident kinetic energy is the same in cases (a) and (b), we observe that
              he nci         tcl os         e netc
             t i dentpar i e l es m or ki i ener i cas (  gy n     e a)



6.52   Use conservation of mechanical energy, KE  PEg               KE  PE 
                                                                   B
                                                                                g
                                                                                     A
                                                                                         , to find the speed of
                                                                                         1 2
       the bead at point B just before it collides with the ball. This gives               m v1i  0  0  m gyA ,
                                                                                         2
       or v1i       2gyA                 
                                2 9. m s2 1. m   5. m s
                                   80       50       42
218    CHAPTER 6


       Conservation of momentum during the collision gives

                0. kg v
                  400         1f     0. kg v2 f   0. kg 5. m s  0
                                        600             400     42

       or      v1 f  1. v2 f  5. m s
                       50        42                                                           (1)

       For a head-on elastic collision, we have v2 f  v2i  v1 f  v1i , which gives

               v2 f  v1 f  5. m s
                              42                                                              (2)

       Solving (1) and (2) simultaneously, the velocities just after collision are

               v1 f   1. m s and v2 f  4. m s
                         08                34

       Now, we use conservation of the mechanical energy of the ball after collision to find the
       maximum height the ball will reach. This gives

                             1
               0  M gym ax  M v2 f  0 , or ym ax 
                                 2
                                                       2
                                                      v2 f
                                                           
                                                              4. m s  0. m
                                                                34
                                                                          960
                                                                              2


                             2                        2 g 2 9. m s2 
                                                                 80


6.53   We first find the speed of the diver when he reaches the water by using
       vy  v0  2ay  y . This becomes
        2    2




                                           
               vy  0  2 9. m s2  3. m  , and yields vy   59 m s
                2
                            80         0

       The negative sign indicates the downward direction.

       Next, we use the impulse-momentum theorem to find the resistive force exerted by the
       water as the diver comes to rest.

                                              
               I Fnet  t  p  m vf  vi , or


                Fwater  784 N   2. s   80 kg 0   
                                     0
                                                                      
                                                                59 m s  , which gives
                                                                       

                                 80 59 
               Fwater  784 N          N = 1.  10 N
                                               1     3
                                                                  upw ard
                                 2 
                                                                                   Momentum and Collisions   219


6.54




       Using the work-energy theorem from immediately after impact to the end gives:

                W net  Ffriction scos180  KEend  KEafter

                                                         1
               or,            k  M  m  g s 0 
                                                          M  m  V 2 and V    2k gs
                                                         2

       Then, using conservation of momentum from immediately before to immediately after
       impact gives m v  0   M  m  V , or


                   M m    M m                             112 g 
                v 
                    m V  
                              m 
                                                      2k gs  
                                                                12. g 
                                                                   0 
                                                                                    
                                                                           2 0.  9. m s2 7. m
                                                                               650 80        5          

                v 91 m s


6.55   (a) Using conservation of momentum,

                 60. kg  120 kg v   60. kg 4. m s  0 ,
                    0                     f0       00


               or                        vf  1. m s
                                               33

       (b)                        80 
             Fy  n   60. kg 9. m s2  0
                           0                      
             so the normal force is n  588 N

             and f  kn   0.  588 N   235 N
                 k            400

       (c) Apply the impulse-momentum theorem to the person:

                                         
                I Fav  t  p  m vf  vi     

             so t
                        
                      m vf  vi      60. kg1. m
                                          0      33          s 4. m s
                                                                 00
                                                                             0. s
                                                                               681
                            fk                       235 N
220    CHAPTER 6



                                             
       (d) pperson  m vf  vi   60. kg1. m s 4. m s   160 N  s
                                      0      33      00


                                         
             pcart  M vf  0  120 kg1. m s 0   160 N  s
                                           33

                                     vf  vi 
       (e)   xperson  vav  t             t
                                     2 

                       1. m s 4. m s
                         33      00
                                      0. s  1. m
                                           681     82
                             2       

                                  vf  0
                                                
                                                    1. m s
                                                     33
       (f)                    
             xcart  vav  t           t           0. s  0. m
                                                               681     454
                                  2                 2  


       (g) KEperson 
                             1
                             2
                                      
                               m v2  vi
                                  f
                                       2
                                                      
                              60. kg  1. m s 2  4. m s 2  
                                 0
                                               00  
                                         33                                   427 J
                                      2

                                     120 kg  1. m s 2  0  107 J
       (h) KEcart 
                        1
                        2
                             f    
                          M v2  0 
                                        2       
                                                 33    
       (i)   Equal friction forces act through different distances on person and cart to do
             different amounts of work on them. This is a perfectly inelastic collision in which
             the total work on both person and cart together is –320 J, which becomes +320 J of
             internal energy.


6.56   (a) Let v1i and v2i be the velocities of m 1 and m 2 just before the collision. Then
           conservation of energy gives:

                v1i   v2i              2gh                    
                                                          2 9. m s2  5. m   9. m s
                                                             80        00       90

       (b) From conservation of momentum:

                 2.  v
                   00       1f     4.  v2 f   2.  9. m s   4.   9. m s
                                      00            00 90             00      90


             or  2.  v1 f   4.  v2 f  19. m s
                   00            00            8                                               (1)

             For an elastic head-on collision, v1 f  v1i  v2 f  v2i , giving

                 v1 f  9. m s  v2 f  9. m s, or v2 f  v1 f  19. m s
                         90              90                        8                           (2)
                                                                                             Momentum and Collisions   221


           Solving (1) and (2) simultaneously gives

                v1 f   16. m s , and v2 f  3. m s
                           5                   30

       (c) Applying conservation of energy to each block after the collision gives:


                  h1 f 
                            2
                           v1 f
                                      
                                         16. m s
                                             5
                                                        2

                                                                 13. m
                                                                    9
                               2g          
                                          2 9. m s2
                                             80         

           and h2 f 
                                2
                               v2 f
                                      
                                            3. m s
                                              30
                                                        2

                                                                 0. m
                                                                   556
                               2g          
                                           2 9. m s2
                                              80            

6.57   (a) Use conservation of mechanical energy to find the speed of m 1 just before collision.
           This gives

                v1i       2 gh1                               
                                               2 9. m s2  2. m   7. m s
                                                  80        50       00

           Apply conservation of momentum from just before to just after the collision:

                 0. kg v
                   500                1f    1. kg v2 f   0. kg 7. m s  0
                                               00              500     00

           or                  v1 f  2v2 f  7. m s
                                               00                                                                      (1)

           For a head-on elastic collision, v1 f  v1i  v2 f  v2i

           which becomes v1 f  v2 f   7. m s
                                          00                                                                           (2)

           Solving (1) and (2) simultaneously yields

                v1 f   2. m s , and v2 f  4. m s
                          33                  67

       (b) Apply conservation of mechanical energy to m 1 after the collision to find


                h1 
                 
                         2
                        v1 f
                               
                                  2. m s
                                     33
                                                    2

                                                         0. m
                                                           277            (rebound height)
                        2g            
                                  2 9. m s2
                                     80             
222    CHAPTER 6



                             1 2
       (c) From y  v0yt     ay t , with v0y  0 , the time for m 2 to reach the floor after it flies
                             2
           horizontally off the table is found to be

                     2 y       2 2. m
                                       00        0. s
                t                            2
                                                    639
                      ay           9.
                                  - 80 m s

           The horizontal distance traveled in this time is

                x  v0xt  4. m s  0. s  2. m
                              67        639     98

       (d) After the 0.500 kg mass comes back down the incline, it flies off the table with a
           horizontal velocity of 2.33 m/s. The time of the flight to the floor is 0.639 s as found
           above and the horizontal distance traveled is

                x  v0xt  2. m s  0. s  1. m
                              33        639     49


6.58   Use conservation of mechanical energy to find the velocity, v, of Tarzan just as he
       reaches Jane. This gives v         2ghi              
                                                    2 9. m s2  3. m   7. m s
                                                       80        00       67

       Now, use conservation of momentum to find the velocity, V, of Tarzan + Jane just after
       the collision. This becomes  M  m  V  M v  0 , or


                    M       80. kg 
                                 0
                V      v           7. m s  4. m s
                                          67        38
                   M m     140 kg 
                                      

       Finally, use conservation of mechanical energy from just after he picks her up to the end
       of their swing to determine the maximum height, H, reached. This yields


                H 
                    V2
                       
                          4. m s  0. m
                            38
                                       2

                                      980
                              
                    2 g 2 9. m s2
                             80            

6.59   (a) The momentum of the system is initially zero and remains constant throughout the
           motion. Therefore, when m 1 leaves the wedge, we must have m 2vw edge  m 1vblock  0 ,
           or

                           m               0. 
                                               500
                vwedge    1  vblock   
                            m2             3. 
                                                00 
                                                      4. m s   0. m s
                                                        00          667
                                                                         Momentum and Collisions   223


       (b) Using conservation of energy as the block slides down the wedge, we have
             KE  PE    KE  PE 
                      g
                          i
                                          g
                                              f
                                                  or


                               1            1
                0  m 1gh       m 1vblock  m 2vwedge  0
                                     2           2

                               2            2

                           1  2        m2 2 
           Thus, h          vblock    vw edge 
                          2g           m1       

                              1                 3. 
                                                    00               2
                                 2 
                                      4. m s              667 m s   0. m
                                                         0.
                                             2
                                      00                                  952
                          19. m s 
                            6                    0. 
                                                   500                


6.60   (a) Let m be the mass of each cart. Then, if v0 is the initial velocity of the red cart,
           applying conservation of momentum to the collision gives

                m vb  m vr  m v0  0 , or vb  vr  v0                                           (1)

           where vb and vr are the velocities of the blue and red carts after collision.

           In a head-on elastic collision, we have v2 f  v2i  v1 f  v1i which reduces to

                vb  vr  v0                                                                       (2)

           Solving (1) and (2) simultaneously gives vr  0 , and vb  3. m s
                                                                       00

       (b) Using conservation of mechanical energy for the blue cart-spring system,
            KE  PEs f   KE  PEsi becomes

                   1     1 2
                0  kx2  m vb  0
                   2     2

                                 m              250
                                               0. kg
           or             x
                                 k
                                   vb 
                                              50. N m
                                                 0
                                                       3. m s  0. m
                                                         00        212
224    CHAPTER 6



6.61   (a) Use conservation of the component of
           momentum in the horizontal direction
           from just before to just after the cannon
           firing.

                p    p 
                   x f          x i   gives


               m shell vshell cos45.   m cannonvrecoil  0 , or
                                    0


                            m        
               vrecoil    shell  vshellcos45. 
                                                0
                            m cannon 

                         200 kg 
                                 125 m s cos45.    3. m s
                                                    0        54
                         5000 kg 


       (b) Use conservation of mechanical energy for the cannon-spring system from right
           after the cannon is fired to the instant when the cannon comes to rest.

                KE  PE   g    PEs      KE  PE
                                       f
                                                       g    PEs      i



                      1 2      1
               0  0  kxm ax  m cannon vrecoil  0  0
                                          2

                      2        2


               xm ax 
                                    2
                          m cannon vrecoil
                                           
                                                5000 kg -3. m s
                                                             54
                                                                           2

                                                                                1. m
                                                                                  77
                                 k                2.  104 N m
                                                   00

       (c)                                          
             F ax  kxm ax  2.  104 N m 1. m   3.  104 N
              m               00            77       54

       (d) No. The rail exerts a vertical external force (the normal force) on the cannon and
           prevents it from recoiling vertically. Momentum is not conserved in the vertical
           direction. The spring does not have time to stretch during the cannon firing. Thus,
           no external horizontal force is exerted on the system (cannon plus shell) from just
           before to just after firing. Momentum is conserved in the horizontal direction
           during this interval.
                                                                         Momentum and Collisions     225


6.62   Conservation of the x-component of momentum gives

                                                            2
                 3m  v2x  0  m v0   3m  v0 , or v2x  v0                                     (1)
                                                            3

       Likewise, conservation of the y-component of momentum gives

                 m v1y   3m  v2y  0 , and v1y  3v2y                                            (2)

       Since the collision is elastic,  KE f   KEi , or

                1 2 1
                2        2
                                 
                  m v1y   3m  v2x  v2y  m v0   3m  v0
                                  2     2

                                            2
                                               
                                            1 2 1
                                                   2
                                                            2
                                                                                                     (3)


       Substituting (1) and (2) into (3) yields

                        4 2    2                       2
                9v2y  3 v0  v2y   4v0 , or v2y  v0
                  2                      2
                        9                              3

       (a) The particle of mass m has final speed v1y  3 v2y  v0     2

            and the particle of mass 3m moves at

                                     4 2 2 2        2
                v2  v2x  v2y 
                      2     2
                                       v0  v0  v0
                                     9     9        3

                     v2y          1
       (b)   t 1 
                an         tan1      35.3°
                     v2x          2
                                      


6.63   Let particle 1 be the neutron and particle 2 be the carbon nucleus. Then, we are given
       that m 2  12m 1 .

       (a) From conservation of momentum m 2v2 f  m 1v1 f  m 1v1i  0 . Since m 2  12m 1 , this
            reduces to 12v2 f  v1 f  v1i .                                                         (1)

            For a head-on elastic collision, v2 f  v2i  v1 f  v1i
226    CHAPTER 6


           Since v2i  0 , this becomes v2 f  v1 f  v1i                                              (2)

           Solve (1) and (2) simultaneously to find

                          11                 2
               v1 f        v1i, and v2 f  v1i
                          13                13

                                                                    1
           The initial kinetic energy of the neutron is KE1i         m 1v1i , and the final kinetic
                                                                          2

                                                                    2
           energy of the carbon nucleus is

                          1          1          4 2  48  1    2    48
               KE2 f       m 2v2 f  12m 1  
                                2
                                                    v1i   m 1v1i     KE
                          2          2          169  169  2      169 1i

                                                            KE2 f        48
           The fraction of kinetic energy transferred is                    0.
                                                                               28
                                                             KE1i       169

       (b) If KE1i  1.  1013 J, then
                      6


               KE2 f 
                           48
                          169
                              KE1i 
                                      48
                                     169
                                        1.  1013 J  4.  1014 J
                                          6             5


           The remaining energy 1.  1013 J  4.  1014 J 1.  1013 J stays with the
                                 6              5             1
           neutron.


6.64   Choose the positive x-axis in the direction of the initial velocity of the cue ball. Let vi be
       the initial speed of the cue ball, vc be the final speed of the cue ball, vT be the final
       speed of the target, and  be the angle the target’s final velocity makes with the x-axis.

       Conservation of momentum in the x-direction gives

               m vT cos  m vc cos30.   0  m vi, or vT c   vi  vc c
                                     0                      os            os30. 
                                                                              0                        (1)

       From conservation of momentum in the y-direction,

               m vT s n  m vc s n30.   0  0 , or vT s n  vc s n30. 
                     i           i   0                    i         i   0                              (2)

       Since this is an elastic collision, kinetic energy is conserved, giving

               1 2 1 2 1 2
                 m vT  m vc  m vi , or vT  vi  vc
                                          2    2    2
                                                                                                       (3)
               2       2      2
                                                                              Momentum and Collisions   227


       (b) To solve, square equations (1) and (2). Then add the results to obtain
           vT  vi  2vivc cos30.   vc . Substitute this into equation (3) and simplify to find
            2    2
                                0      2




               vc  vicos30.    4. m s cos30.   3. m s
                           0        00          0      46


            Then, equation (3) yields vT  vi  vc , or
                                            2    2




                       4. m s   3. m s
                                   2              2
               vT       00          46                2. m s
                                                         00

       (a) With the results found above, equation (2) gives

                      v               3. 
                                          46
               s n   c  s n30.   
                i            i   0            s n30.   0. , or   60. 
                                               i   0      866          0
                       vT             2. 
                                          00

            Thus, the angle between the velocity vectors after collision is

                 60.   30.   90. 
                     0       0       0



                                                            f     k  m g
6.65   The deceleration of the incident block is a         k
                                                                            k g
                                                            m        m

       Therefore, v2  v0  2a x gives the speed of the incident block just before collision as
                        2



       v  v0  2 kgd
            2




       Conservation of momentum from just before to just after collision gives

               m v1   2m  v2  m v , or 2v2  v1  v                                                 (1)

       where v1 and v2 are the speeds of the two blocks just after collision.

       Since this is a head-on elastic collision, v2 f  v2i  v1 f  v1i

       which becomes             v2  v1  v .                                                          (2)

                                                       2    2 2
       Adding equations (1) and (2) yields v2           v   v0  2 k gd
                                                       3    3
228    CHAPTER 6


       Note that the mass canceled in the calculation of the deceleration above. Thus, the
       second block will have the same deceleration after collision as the incident block had
       before. Then, v2  vi  2a x with vf  0 gives the stopping distance for the second
                      f
                           2


       block as 0  v2  2   k g D , or
                     2




                     2                          2
                                              2v0 4d
                D
                    v2
                        
                          2
                   2k g 9k g
                                2
                                          
                               v0  2k gd       
                                             9k g 9
                                                         

6.66   Observe from Figure P6.66, the platform exerts a 0.60-kN to support the weight of the
       standing athlete prior to t 0. s. From this, we determine the mass of the athlete:
                                      00
                         60
                   w 0. kN          600 N
              m                          61 kg
                   g       g       9. m s2
                                    8

       For the interval t 0. st t 1. s, we subtract
                              00 o      0
       the 0.60-kN used to counterbalance the weight
       to get the net upward force exerted on the
       athlete by the platform during the jump. The
       result is shown in the force-versus-time graph
       at the right. The net impulse imparted to the
       athlete is given by the area under this graph.
       Note that this area can be broken into two
       triangular areas plus a rectangular area.

       The net upward impulse is then:

                     1                  1
                I      0. s100 N    0. s 300 N    0. N 100 N   150 N  s
                          50                50                 50
                     2                  2

       The upward velocity vi of the athlete as he lifts off of the platform (at t 1. s) is found
                                                                                     0
       from
                                                     I 150 N  s
             I p  m vi  m v0  m vi  0  vi                   2. m s
                                                                        5
                                                    m      61 kg
       The height of the jump can then be found from v2  vi  2ayy (with vf  0 ) to be
                                                      f
                                                           2



                                     0   2. m s
                                                     2
                       v2  vi
                        f
                             2
                                            5
                y                                      0. m
                                                            31
                        2ay           
                                     2 9. m s2
                                         8           

6.67                    ai houl be deni . Immediately prior to impact, the total
       (a) The owner’s cl m s d       ed
            momentum of the two-car system had a northward component and an eastward
            component. Thus, after impact, the wreckage moved in a northeasterly direction
            and could not possibly have damaged the owner’s property on the southeast
            corner.
                                                                                                    Momentum and Collisions      229


       (b) From conservation of momentum:

             p 
               x atr
                  fe
                            px bef re 
                                    o
                                                   m 1  m 2  vx  m 1  v1ix  m 2 v2ix

                           m 1  v1i x  m 2  v2i x         1300 kg 30. km h +0  16. km
                                                                            0
             or vx                                                                       3                   h
                                   m1  m2                       1300 kg  1100 kg


             p 
               y
                    afe
                     tr
                             
                           py
                                  be or
                                    f e
                                                  m 1  m 2  vy  m 1  v1i y  m 2  v2i y

                           m 1  v1i y  m 2  v2i y         0  1100 kg 20. km h
                                                                                0
             or vy                                                                                9. km h
                                                                                                      17
                                   m1  m2                        1300 kg  1100 kg

             Thus, the velocity of the wreckage immediately after impact is

                                                     vy 
                v  vx  vy  18. km h and   t 1    t 1  0.   29. 
                     2    2
                                7               an         an     564     4
                                                     vx 

             or v  18. km h at29. nor h ofeas ,cons s entw ih par (
                      7          4°   t       t     it      t     t a)

6.68   (a)




       (b) From Newton’s third law, the force FBA exerted by B on A is at each instant equal in
             magnitude and opposite in direction to the force FA B exerted by A on B.

       (c) There are no horizontal external forces acting on System C which consists of both
           blocks. The forces FBA and FA B are internal forces exerted on one part of System C
           by another part of System C.

                                              p
             Thus,             Fexternal            C
                                                          0        p  0
                                                  t                      C




             This gives                   p   p   p   p 
                                              C   f        C i      A i       B i
                                                                                      or           M    2M  V  M  v  0
             so the velocity of the combined blocks after collision is                                      V  v 3
230    CHAPTER 6


            The change in momentum of A is then:


                p  p
                    A
                           p 
                           A    f    A i
                                                           v 
                                            M V  M v  M   v   2M v 3
                                                           3 

            and the change in momentum for B is:

                                                             v
                p  p
                    B
                           p 
                          B f        B i
                                            2M V  0  2M     2M v 3
                                                            3

                                                                         2

                                                             3M  
                                                                       v                   1
       (d) KE   KEC  f   KEA    KEB   i 
                                               
                                                        1
                                                                          2 M v  0   M v
                                                                            
                                                                              1    2
                                                                                       
                                                                                                2
                                                        2
                                                                      3                   3

            Thus, ki i ener i notcons ved i t s i as i coli i
                   netc    gy s      er   n hi nel tc ls on


6.69   (a) The speed vi of both balls just before the basketball reaches the ground may be
            found from vy  v0y  2ay y as
                        2    2




                                                   80   
                        vi  v0y  2ayy  0  2 9. m s2  1. m s  4. m s
                              2
                                                              20        85   
       (b) Immediately after the basketball rebounds from the floor, it and the tennis ball meet
           in an elastic collision. The velocities of the two ball just before collision are:

               For the tennis ball:                 v1i  vi   4. m s
                                                                   85
               For the basketball:                  v2i  vi   4. m s
                                                                   85

            We determine the velocity of the tennis ball immediately after this elastic collision
            as follows:

            Momentum conservation gives

                 57. g v
                    0      1f     590 g v2 f   57. g  4. m s   590 g  4. m s
                                                      0        85                    85


            which reduces to                        v2 f  4. m s 9.  102 v1f
                                                            38      66                                     (1)

            El i C oli on  v1 f  v2 f    v1i  v2i     4. m s 4. m s   9. m s
             astc    lsi                                          85      85          70
            and substituting from Equation (1) gives                    1 9.  10  v
                                                                             66      2
                                                                                          1f    9. m s 4. m s
                                                                                                  70      38
                          1
                        14. m s
            or v1 f              12. m s
                                      8
                          1.10
                                                                             Momentum and Collisions   231


            The vertical displacement of the tennis ball during its rebound is given by
            vy  v0y  2ay y as
             2    2




                                     0  12. m s
                                                     2
                      vy  v0y
                       2    2
                                            8
               y                                       8. m
                                                            41
                        2ay              
                                     2 9. m s2
                                         80          

6.70   Ignoring the force of gravity during the brief collision time, we use the conservation of
       momentum to obtain:

                0. kg v
                  45       bf     60 kg vpf   0. kg 25 m s   60 kg 4. m s
                                                    45                           0


       or             8              
               vpf  3. m s 7.  103 vbf
                              5                                                                       (1)


                                                                
       Also, el tc coli i  vbf  vpf   vbi  vpi    25 m s 4. m s
              as i    ls on                                         0

       or      vbf  29 m s vpf                                                                       (2)

       Substituting Equation (1) into Equation (2) yields
                     29 m s 3. m s
                               8
               vbf                    33 m s
                       1 7.  103
                           5

       The average acceleration of the ball during the collision is
                    vbf  vbi 33 m s  25 m s
              aav                      3
                                                   2.  103 m s2
                                                      9
                       t         20  10 s


6.71   Use conservation of mechanical energy from the instant an ice cube is released until it
       reaches the end of the frictionless track to find its speed as it leaves the track:

               1
               2                                                               
                   m v2  m gyf  0  m gyi  v  2g yi  yf  2 9. m s2 1. m   5. m s
                                                                  80       50       42


       The initial velocity components of the ice cube for the projectile phase of its trip are
       then:

               v0x   5. m s cos40.   4. m s and v0y   5. m s si 40.   3. m s
                        42          0      15                 42       n 0       49

       At the highest point on its trajectory, vy  0 and the velocity of the ice cube just before
       impact with the wall is vi  v0x  4. m s. The velocity it rebounds from the wall with
                                           15
       is then vf   vi 2    4. m s 2  2. m s
                                  15           08
232    CHAPTER 6


       From the impulse-momentum theorem, we find

                                                                
                      I Fav t p  5.  103 kg 2. m s 4. m s  3.  102 kg  m s
                                       00           08       15        12

       If the cubes strike the wall, one after the other, at the rate of 10.0 cubes per second, the
       collision of each cube with the wall has a duration of t 0. sand the average force
                                                                         100
       the wall exerts on the cubes is

                              I 3.  102 kg  m s
                                   12
                  Fav                              0. N or 0. N t ar t l t
                                                        312     312  ow d he ef
                              t       100
                                      0. s

       Thus, the reaction force the ice cubes exert on the wall is Fav                       on
                                                                                                  l
                                                                                               w al
                                                                                                       0. N t t r ght
                                                                                                         312  o he i



6.72   (a)




             The situations just before and just after the collision are shown above. Conserving
             momentum in both the x and y directions gives

             p   p 
                  y
                       f
                               y
                                   i
                                        m 1v1 f si 53  m 2v2 f si   0 or
                                                  n                n                  m 2v2 f si   m 1v1 f si 53
                                                                                               n              n           (1)


             p  x f        px i  m 1v1f cos53  m 2v2 f cos  m 1v1i  0

             or                    m 2v2 f cos  m 1v1i-m 1v1 f cos53                                                   (2)

             Dividing Equation (1) by Equation (2) yields


             t 
              an
                               v1 f s n53
                                     i
                                                 
                                                           1. m s si 
                                                             0        n53
                                                                                  0. or
                                                                                    57                   30
                             v1i-v1 f cos53          2. m s  1. m s cos53
                                                        0          0


             Equation (1) then gives:                   v2 f 
                                                                 m 1v1 f si 
                                                                          n53
                                                                                
                                                                                    0. kg1. m s si  
                                                                                      20      0       n53
                                                                                                                       1
                                                                                                                      1. m s
                                                                     m 2 si 
                                                                          n              0. kg si 
                                                                                           30     n30
                                                                                                    Momentum and Collisions        233


       (b) The fraction of the incident kinetic energy lost in this collision is

                                                               0. kg1. m s   0. kg1. m s
                                                                                       2                                 2
            KE KEi  KE f      KE f                      1
                                                          2      20     0            30    1    1
                                                                                                2
                           1       1
                                                                         0. kg 2. m s
                                                                                                             2
            KEi   KEi           KEi                                        20
                                                                           1
                                                                           2       0

            KE
                 0. or 32%
                   32
            KEi


6.73   (a) First, we use conservation of mechanical energy from the instant after the shot
           embeds itself in m 3 until the combined system comes to rest (with spring
           compressed). This gives  KE  PEsi   KE  PEs f , or                       1
                                                                                           2    m 2  m 3  V 2  0  0  1 kx2
                                                                                                                           2   f

           and the velocity of the combined system just after impact is


                 V
                             kx2
                               f
                                    
                                           4 500 N   m         0. m 
                                                                   500
                                                                           2

                                                                                0.
                                                                                  375 m s
                         m2 m3             10. kg  7 990 kg
                                              0

           Taking toward the right as positive, and using conservation of momentum from
           just before to just after the collision between m 2 and m 3 , we find
            m 2v2  0   m 2  m 3  V


                      m           7 990 kg 
           or v2   1 3  V   1
                      m2           10. kg 
                                                  0. m s   300 m s 300 m s t t l t
                                                     375                         o he ef
                                        0    

       (b) Now consider the “collision” between the shot and the cannon. Applying
           conservation of momentum to this event yields m 2v2  m 1v1   m 1  m 2   0

                                   m           10. kg 
                                                    0
           or     vrecoil  v1    2  v2              300 m s  3. m s to the ri
                                                                          75            ght
                                    m1         800 kg 
                                                         

       (c) Applying the work-energy theorem to the sliding cannon gives

                                                            3. m s
                                                  2                                                 2
                                  m 1v12     m 1 v1           75
             fd  0  m v or d 
                         1     2
                                                                      1. m
                                                                          20
              k          2   1 1
                                  2f  k               
                                           2 k m 1 g 2 0.  9. m s
                                                          600 80    2
                                                                                                        

6.74   (a) Apply conservation of momentum in the vertical direction to the squid-water
           system from the instant before to the instant after the water is ejected. This gives

                                                          m           0. kg 
                                                                          30
            m svs  m w vw   m s  m w   0 or vs    w  vw             20 m s  7. m s
                                                                                               1
                                                           ms         0. kg 
                                                                          85 
234   CHAPTER 6


      (b) Apply conservation of mechanical energy to the squid from the instant after the
          water is ejected until the squid reaches maximum height to find:


             0  m sgyf  m v  m gyi or y  s 
                         1     2             v2   7. m s  2. m
                                                    1
                                                             2

                                                              5
                         2   s s
                                                    
                                             2g 2 9. m s2
                                                     8           

								
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