Ch_9Word - Chapter 9 CHAPTER 9 Linear Momentum and...

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Chapter  9 CHAPTER 9 - Linear Momentum and Collisions 1. We find the force on the expelled  gases from F  = ? p /? t  = (? m /? t ) v  = (1200 kg/s)(50,000 m/s)  = 6.0 × 10 7  N. An equal, but opposite, force will be exerted  on the rocket:        6.0 × 10 7  N, up . 2. For the momentum   p  = 4.8 t 2 i  – 8.0 j  – 8.9 t k , we find the force from F  = d p /dt  =        9.6 t i  – 8.9 k , in SI units . 3. ( a ) p  =  mv  = (0.030 kg)(12 m/s)  =         0.36 kg     m/s . ( b ) The force, opposite the direction of the velocity, changes the momentum: F  = ? p /? t ; – 2.0 × 10 –2  N = ( p 2  – 0.36 kg     m/s)/(12  s), which gives  p 2  =        0.12 kg     m/s . 4. The change in momentum  is ?  =  p 2  –  p 1  =  mv j  –  mv i   = (0.145 kg)(30 m/s) j  – (0.145 kg)(30 m/s) i  =       – (4.35 kg     m/s) i  + (4.35 kg     m/s) j . 5. The force changes the momentum: F  = (26 N) i  – (12 N/s 2 ) t 2 j  = d p /d t . Because  F  is a variable force, we integrate to find the momentum  change: 6. If  M  is the initial mass of the rocket and   m 2  is the mass of the  expelled  gases, the final mass of the rocket is  m 1  =  M  –  m 2 .   Because the gas is expelled perpendicular  to the rocket in the  rocket’s frame, it will still have the initial forward  velocity,  so the velocity of the rocket in the original direction will not  change.  We find the  y -component  of the rocket’s velocity after  firing from v 1  =  v 0  tan  θ  = (120 m/s)  tan 23.0 °  = 50.9 m/s. Using the coordinate system  shown, for momentum   conservation  in the  y -direction we have 0 + 0 =  m 1 v 1  –  m 2 v 2   ,   or ( M  –  m 2 ) v 1  =  m 2 v 2 ; (4200 kg –  m 2 )(50.9 m/s)  =  m 2 (2200 m/s), which gives  m 2  =        95 kg . Page 1 # ,# #h ��� �� # ,# #h ��� ��
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Chapter  9 7. ( a ) We choose downward  as positive.  For the fall we have y  =  y 0  +  v 0 t 1  +  ! at 1 2 ; h  = 0 + 0 +  ! gt 1 2 , which gives  t 1  = (2 h / g ) 1/2 . To reach the same height on the rebound,  the upward  motion must be a reversal of the downward   motion.  Thus the time to rise will be the same, so the total time is t t otal  = 2 t 1  = 2(2 h / g ) 1/2  =        (8 h / g ) 1/2 . ( b ) We find the speed  from v  =  v 0  +  at 1  = 0 +  g (2 h / g ) 1/2  =        (2 gh ) 1/2 . ( c ) To reach the same height on the rebound,  the upward  speed  at the floor must be the same as the  speed  striking the floor.  Thus the change in momentum  is ? p  =  m (–  v ) –  mv  = – 2 m (2 gh ) 1/2  =  =          – (8 m 2 gh ) 1/2  (up) .
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