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261_PartUniversity Physics Solution

261_PartUniversity Physics Solution - 8-20 Chapter 8...

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8-20 Chapter 8 E XECUTE : / / / A E A B B E = + v v v ! ! ! / 5.00 m/s B E v = + / 5.00 m/s A B v = − (minus since the mass is moving west relative to the car). This gives / 0; A E v = the mass is at rest relative to the earth after it is thrown backwards from the car. As in part (a), ( ) 1 2 2 . A B A A x B B x m m v m v m v + = + Now 2 0, A x v = so ( ) 1 2 . A B B B x m m v m v + = ( ) 2 1 200 kg 5.00 m/s 5.71 m/s. 175 kg A B B x B m m v v m + = = = The final velocity of the car is 5.71 m/s, east. E VALUATE : The thrower exerts a force in the -direction x so the mass exerts a force on him in the -direction x + and he and the car speed up. (c) S ET U P : Let A be the 25.0 kg mass and B be the car (mass 200 kg). B m = Figure 8.70b x P is conserved so ( ) 1 1 2 A A x B B x A B x m v m v m m v + = + . E XECUTE : ( ) 1 1 2 A A B B A B x m v m v m m v + = + . ( )( ) ( )( ) 1 1 2 200 kg 5.00 m/s 25.0 kg 6.00 m/s 3.78 m/s. 200 kg 25.0 kg B B A A x A B m v m v v m m = = = + + The final velocity of the car is 3.78 m/s, east. E VALUATE : The mass has negative x p so reduces the total x P of the system and the car slows down. 8.71. I DENTIFY : The horizontal component of the momentum of the sand plus railroad system is conserved. S ET U P : As the sand leaks out it retains its horizontal velocity of 15.0 m/s. E XECUTE : The horizontal component of the momentum of the sand doesn°t change when it leaks out so the speed of the railroad car doesn°t change; it remains 15.0 m/s. In Exercise 8.27 the rain is falling vertically and initially has no horizontal component of momentum. Its momentum changes as it lands in the freight car. Therefore, in order to conserve the horizontal momentum of the system the freight car must slow down. E VALUATE : The horizontal momentum of the sand does change when it strikes the ground, due to the force that is external to the system of sand plus railroad car. 8.72. I DENTIFY : Kinetic energy is 2 1 2 K mv = and the magnitude of the momentum is p mv = . The force and the time t it acts are related to the change in momentum whereas the force and distance d it acts are related to the change in kinetic energy. S ET U P : Assume the net forces are constant and let the forces and the motion be along the x axis. The impulse- momentum theorem then says Ft p = Δ and the work-energy theorem says Fd K = Δ . E XECUTE : (a) 2 4 1 N 2 (840 kg)(9.0 m/s) 3.40 10 J K = = × . 2 4 1 P 2 (1620 kg)(5.0 m/s) 2.02 10 J K = = × . The Nash has the greater kinetic energy and N P 1.68 K K = . (b) 3 N (840 kg)(9.0 m/s) 7.56 10 kg m/s p = = × . 3 P (1620 kg)(5.0 m/s) 8.10 10 kg m/s p = = × . The Packard has the greater magnitude of momentum and N P 0.933 p p = . (c) Since the cars stop the magnitude of the change in momentum equals the initial momentum. Since P N p p > , P N F F > and N N P P 0.933 F p F p = = . (d) Since the cars stop the magnitude of the change in kinetic energy equals the initial kinetic energy. Since N P K K > , N P F F > and N N P P 1.68 F K F K = = .
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