Conservation of momentum and toboggan

[jaded18]

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)) ??

 

[learningphysics]

You're not doing momentum properly. What is the definition of momentum?

 

[jaded18]

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??

 

[jaded18]

becuase it is an inelastic collision and there is energy loss?

 

[jaded18]

nvm, ignore #4, i don't know what I am talking about

 

[jaded18]

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)

 

[learningphysics]

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)

yes, exactly.