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anonymous

  • one year ago

A 6,600 kg train car moving at +2.0 m/s bumps into and locks together with one of mass 5,400 kg moving at -3.0 m/s. What is their final velocity? A. -0.25 m/s B. -1.0 m/s C. +0.50 m/s D. +2.5 m/s

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  1. Michele_Laino
    • one year ago
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    since the system composed by the two trains, is isolated along the direction of the motion, namely no external forces, along the direction of motion are acting on our trains, then we can write: \[\Large {M_1}{V_1} + {M_2}{V_2} = \left( {{M_1} + {M_2}} \right)V\]

  2. Michele_Laino
    • one year ago
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    V is the final velocity

  3. Michele_Laino
    • one year ago
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    namely: \[\Large V = \frac{{{M_1}{V_1} + {M_2}{V_2}}}{{{M_1} + {M_2}}}\]

  4. anonymous
    • one year ago
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    okay! what do i plug in?

  5. Michele_Laino
    • one year ago
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    here is the next step: \[\Large V = \frac{{{M_1}{V_1} + {M_2}{V_2}}}{{{M_1} + {M_2}}} = \frac{{\left( {6600 \times 2} \right) + \left\{ {5400 \times \left( { - 3} \right)} \right\}}}{{6600 + 5400}}\]

  6. anonymous
    • one year ago
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    i cannot see the last part you wrote!! i just see (6600*2) + [5400 ------------------- 6600+5400 what was the rest of it?

  7. Michele_Laino
    • one year ago
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    sorry, here is the formula: \[V = \frac{{{M_1}{V_1} + {M_2}{V_2}}}{{{M_1} + {M_2}}} = \frac{{\left( {6600 \times 2} \right) + \left\{ {5400 \times \left( { - 3} \right)} \right\}}}{{6600 + 5400}}\]

  8. anonymous
    • one year ago
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    ohh okay!! i see it now:) so we get -0.495 ?

  9. Michele_Laino
    • one year ago
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    I got a different result

  10. anonymous
    • one year ago
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    oh oops sorry i got -0.25 !! i entered something wrong earlier!

  11. Michele_Laino
    • one year ago
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    that's right!

  12. anonymous
    • one year ago
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    yay! so our solution is choice A?

  13. Michele_Laino
    • one year ago
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    yes!

  14. anonymous
    • one year ago
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    yay!! thanks!

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

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