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Conservation of momentum and toboggan

  1. Oct 8, 2007 #1
    a runaway toboggan of mass 8.6 kg is moving horizontally at 23 km/h. as it passes under a tree 15 kg of snow drop onto it.
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    isn't the ans 13.88 if we use conservation of momentum equation (.5mv1 initial + 0.5mv2 initial) = (0.5mv1 final + 0.5 mv2final) --> (0.5 (8.6) (23^2)) = (0.5 (8.6+15)(v2^2)) ??
     
  2. jcsd
  3. Oct 8, 2007 #2

    learningphysics

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    You're not doing momentum properly. What is the definition of momentum?
     
  4. Oct 8, 2007 #3
    well, actually, i guess i had used the conservation statement for kinetic energy and not the conservation for momentum... and so i should use (mv 1 initial) (mv 2 initial) = (mv 1 final) (mv 1 initial) but why is it again that i cannot use what i used in #1??
     
  5. Oct 8, 2007 #4
    becuase it is an inelastic collision and there is energy loss?
     
  6. Oct 8, 2007 #5
    nvm, ignore #4, i don't know what im talking about
     
  7. Oct 8, 2007 #6
    oh haha, ok, i take that back. it IS an inelastic collision and so the equation i must use is (mv1 in) + (mv2 in) = (m1+m2)(Vfin)
     
  8. Oct 8, 2007 #7

    learningphysics

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    yes, exactly.
     
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