276_PartUniversity Physics Solution

276_PartUniversity Physics Solution - Momentum Impulse and...

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Momentum, Impulse, and Collisions 8-35 (b) For the wagon, 2 4 1 1 2 (300 kg)(10.12 m/s) 1.54 10 J K = = × . Assume that the two heroes drop from a small height, so their kinetic energy just before the wagon can be neglected compared to 1 K of the wagon. 2 4 1 2 2 (435 kg)(6.98 m/s) 1.06 10 J K = = × . The kinetic energy of the system decreases by 3 1 2 4.8 10 J K K = × . E VALUATE : The wagon slows down when the two heroes drop into it. The mass that is moving horizontally increases, so the speed decreases to maintain the same horizontal momentum. In the collision the vertical momentum is not conserved, because of the net external force due to the ground. 8.110. I DENTIFY : Gravity gives a downward external force of magnitude mg . The impulse of this force equals the change in momentum of the rocket. S ET U P : Let + y be upward. Consider an infinitesimal time interval dt . In Example 8.15, ex 2400 m/s v = and 0 120 s dm m dt = − . In Example 8.16, 0 /4 m m = after 90 s t = . E XECUTE : (a) The impulse-momentum theorem gives ex ( )( ) ( )( ) mgdt m dm v dv dm v v mv = + + + . This simplifies to ex mgdt mdv v dm = + and ex dv dm m v mg dt dt = − . (b) ex dv v dm a g dt m dt = = − . (c) At 0 t = , 2 2 ex 0 1 (2400 m/s) 9.80 m/s 10.2 m/s 120 s v dm a g m dt = − = − = . (d) ex v dv dm gdt m = − . Integrating gives 0 0 ex ln m v v v gt m = + . 0 0 v = and 2 (2400 m/s)ln4 (9.80 m/s )(90 s) 2445 m/s v = + = . E VALUATE : Both the initial acceleration in Example 8.15 and the final speed of the rocket in Example 8.16 are reduced by the presence of gravity. 8.111. I DENTIFY and S ET U P : Apply Eq. 8.40 to the single-stage rocket and to each stage of the two-stage rocket. (a) E XECUTE : ( ) 0 ex 0 ln / ; v v v m m = 0 0 v = so ( ) ex 0 ln / v v m m = The total initial mass of the rocket is 0 12,000 kg 1000 kg 13,000 kg. m = + = Of this, 9000 kg 700 kg 9700 kg + = is fuel, so the mass m left after all the fuel is burned is 13,000 kg 9700 kg 3300 kg. = ( ) ex ex ln 13,000 kg/3300 kg 1.37 v v v = = . (b) First stage: ( ) ex 0 ln / v v m m = 0 13,000 kg m = The first stage has 9000 kg of fuel, so the mass left after the first stage fuel has burned is 13,000 kg 9000 kg 4000 kg. = ( ) ex ex ln 13,000 kg/4000 kg 1.18 v v v = = . (c) Second stage: 0 1000 kg, m = 1000 kg 700 kg 300 kg m = = . ( ) ( ) 0 ex 0 ex ex ex ln / 1.18 ln 1000 kg/300 kg 2.38 v v v m m v v v = + = + = . (d) 7.00 km/s v = ( ) ex / 2.38 7.00 km/s / 2.38 2.94 km/s v v = = = . q E VALUATE : The two-stage rocket achieves a greater final speed because it jetisons the left-over mass of the first stage before the second-state fires and this reduces the final m and increases 0 / . m m 8.112. I DENTIFY : During an interval where the mass is constant the speed of the rocket is constant. During an interval where the mass is changing at a constant rate, the equations of Section 8.6 apply.
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