anonymous
  • anonymous
A canoeist on a lake throws a 0.145 kg baseball at 40.0 m/s to a friend in a different canoe. Each canoe with its contents has a mass of 120 kg, and is at rest initially. How fast is the first canoe moving after the pitch, assuming no friction with the water? A. 0.048 m/s B. 0.33 m/s C. 20 m/s D. 40 m/s ** :/
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Michele_Laino
  • Michele_Laino
here, since we can neglet any friction forces acting on our canoe, we have to apply the theorem of conservation of total momentum
anonymous
  • anonymous
ok!
Michele_Laino
  • Michele_Laino
here is the situation described in your problem: |dw:1433009286262:dw|

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anonymous
  • anonymous
ok!
Michele_Laino
  • Michele_Laino
the momentum of the ball is: \[\Large mv = 0.145 \times 40 = ...\]
Michele_Laino
  • Michele_Laino
the speed of the ball is referred to the water, and the water is at rest with respect to the earth
anonymous
  • anonymous
so 5.8?
anonymous
  • anonymous
5.8 = mv?
Michele_Laino
  • Michele_Laino
ok! that's right!
anonymous
  • anonymous
yay! what happens next?
Michele_Laino
  • Michele_Laino
the momentum gained, by the system canoeist+canoe is: \[\Large MV = 120 \times V\]
Michele_Laino
  • Michele_Laino
now both of those momentum have to be equal each other, so we can write: \[\Large 120 \times V = 5.8\]
anonymous
  • anonymous
so 5.8=120*V V=0.0483333...? so our solution is choice A?
Michele_Laino
  • Michele_Laino
correct!
anonymous
  • anonymous
yay!! thank you:)
Michele_Laino
  • Michele_Laino
thanks!! :)

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