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anonymous

  • one year ago

A 1,600 kg car traveling north at 10.0 m/s crashes into a 1,400 kg car traveling east at 15 m/s at an unexpectedly icy intersection. The cars lock together as they skid on the ice. What is their speed after the crash? 8.8 m/s , 12 m/s , 18 m/s , or 26 m/s ?

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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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    |dw:1433014089573:dw|

  3. anonymous
    • one year ago
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    ok:)

  4. Michele_Laino
    • one year ago
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    we have to find the momentum p_1 and the momentum p_2

  5. Michele_Laino
    • one year ago
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    so we have: \[\begin{gathered} {p_1} = 1600 \times 10 = ... \hfill \\ \hfill \\ {p_2} = 1400 \times 15 = ... \hfill \\ \end{gathered} \]

  6. anonymous
    • one year ago
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    okay so we get 16000 and 21000?

  7. Michele_Laino
    • one year ago
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    correct!

  8. anonymous
    • one year ago
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    yay!! what happens now?

  9. Michele_Laino
    • one year ago
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    now, after collision, the system car #1 + car#2 has two components of its momentum, namely P_1 and P_2 such that the subsequent condition holds: \[\Large \begin{gathered} {P_1} = \left( {{m_1} + {m_2}} \right){V_1} = {p_1} \hfill \\ \hfill \\ P2 = \left( {{m_1} + {m_2}} \right){V_2} = {p_2} \hfill \\ \end{gathered} \]

  10. Michele_Laino
    • one year ago
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    where m_1 and m_2 are the masses of the two cars respectively, and V_1 and V_2 are the components of the velocity of the system car #1 + car#2, namely: |dw:1433014582296:dw|

  11. anonymous
    • one year ago
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    yes :) how do we find the speed from that?

  12. Michele_Laino
    • one year ago
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    we have to divide the formulas above, by m_1+m_2, like this: \[\Large \begin{gathered} {V_1} = \frac{{{p_1}}}{{{m_1} + {m_2}}} \hfill \\ \hfill \\ {V_2} = \frac{{{p_2}}}{{{m_1} + {m_2}}} \hfill \\ \end{gathered} \]

  13. anonymous
    • one year ago
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    ok! what do we plug in? :/

  14. Michele_Laino
    • one year ago
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    it is simple: p_1=10,000, p_2=21,000, m_1=1,600 and m_2=1,400

  15. anonymous
    • one year ago
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    ok! so we get 7?

  16. Michele_Laino
    • one year ago
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    and the othe velocity?

  17. Michele_Laino
    • one year ago
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    other*

  18. anonymous
    • one year ago
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    it is 3.333 (oops sorry forgot to write this one earlier haha)

  19. Michele_Laino
    • one year ago
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    I got 5.33

  20. anonymous
    • one year ago
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    ohh yes sorry!! so from there what do we do?

  21. Michele_Laino
    • one year ago
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    ok! so the requested velocity, has the subsequent magnitude: \[V = \sqrt {V_1^2 + V_2^2} = \sqrt {{{5.33}^2} + {7^2}} = ...\]

  22. anonymous
    • one year ago
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    8.79? so our solution is 8.8 m/s?

  23. Michele_Laino
    • one year ago
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    yes! That's right!

  24. anonymous
    • one year ago
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    yay!! thank you!!

  25. Michele_Laino
    • one year ago
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    :):)

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