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anonymous

  • one year ago

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 ** :/

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  1. Michele_Laino
    • one year ago
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    here, since we can neglet any friction forces acting on our canoe, we have to apply the theorem of conservation of total momentum

  2. anonymous
    • one year ago
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    ok!

  3. Michele_Laino
    • one year ago
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    here is the situation described in your problem: |dw:1433009286262:dw|

  4. anonymous
    • one year ago
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    ok!

  5. Michele_Laino
    • one year ago
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    the momentum of the ball is: \[\Large mv = 0.145 \times 40 = ...\]

  6. Michele_Laino
    • one year ago
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    the speed of the ball is referred to the water, and the water is at rest with respect to the earth

  7. anonymous
    • one year ago
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    so 5.8?

  8. anonymous
    • one year ago
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    5.8 = mv?

  9. Michele_Laino
    • one year ago
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    ok! that's right!

  10. anonymous
    • one year ago
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    yay! what happens next?

  11. Michele_Laino
    • one year ago
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    the momentum gained, by the system canoeist+canoe is: \[\Large MV = 120 \times V\]

  12. Michele_Laino
    • one year ago
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    now both of those momentum have to be equal each other, so we can write: \[\Large 120 \times V = 5.8\]

  13. anonymous
    • one year ago
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    so 5.8=120*V V=0.0483333...? so our solution is choice A?

  14. Michele_Laino
    • one year ago
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    correct!

  15. anonymous
    • one year ago
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    yay!! thank you:)

  16. Michele_Laino
    • one year ago
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    thanks!! :)

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