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anonymous

  • one year ago

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  1. Michele_Laino
    • one year ago
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    here we have to apply the total momentum conservation law

  2. Michele_Laino
    • one year ago
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    here is the situation described in your problem: |dw:1437247150772:dw|

  3. Michele_Laino
    • one year ago
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    if, after collision the two carts are attached together, then we can write this equation: \[\Large 2m{v_0} = \left( {m + 2m} \right)u\] where u is the final velocity

  4. Michele_Laino
    • one year ago
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    and v_0=5 m/sec

  5. Michele_Laino
    • one year ago
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    so after a simplification, we get: \[\Large u = \frac{2}{3}{v_0} = ...?\]

  6. Michele_Laino
    • one year ago
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    more explanation: the total momentum of our system before collision is: \[\Large 2m{v_0}\] whereas the total momentum of our system, after collision is: \[\Large \left( {m + 2m} \right)u\] those momentum have to be equal each other

  7. Michele_Laino
    • one year ago
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    |dw:1437247606502:dw|

  8. Michele_Laino
    • one year ago
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    that's right! u=3.33 m/sec

  9. Michele_Laino
    • one year ago
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    where are you confused?

  10. Michele_Laino
    • one year ago
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    thanks! :)

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