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anonymous

  • one year ago

A firecracker stuck into a 150 g apple explodes and sends five apple fragments in different directions. The vector sum of momenta 1, 2, 3, and 4 is found from a video of the event to have components p_x = -1.20 kg * m/s, p_y = -0.80 kg * m/s (with no other component). The mass of the fifth fragment is 0.050 kg. What is its velocity right after the explosion? 2.1 m/s , 42 m/s , 18 m/s , or 29 m/s ?? not sure which it would be!

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

  2. anonymous
    • one year ago
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    ok!

  3. Michele_Laino
    • one year ago
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    in other words, we have to write these two components of the momentum of the fifth fragment: \[\Large \begin{gathered} {P_x} = M{V_x} \hfill \\ {P_y} = M{V_y} \hfill \\ \end{gathered} \]

  4. anonymous
    • one year ago
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    ok! what do we plug in? :O

  5. Michele_Laino
    • one year ago
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    where V_x and V_y are the components of the velocity of the fifth fragment, whereas M is the mass of the fifth fragment

  6. anonymous
    • one year ago
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    ohh okay!

  7. Michele_Laino
    • one year ago
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    so, we can write these two equations: \[\Large \begin{gathered} M{V_x} - 1.2 = 0 \hfill \\ M{V_y} - 0.8 = 0 \hfill \\ \end{gathered} \]

  8. anonymous
    • one year ago
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    ok!

  9. Michele_Laino
    • one year ago
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    since before the explosion, the object was at rest

  10. Michele_Laino
    • one year ago
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    then we get: \[\Large \begin{gathered} {V_x} = \frac{{1.2}}{M} = ... \hfill \\ \\ {V_y} = \frac{{0.8}}{M} = ... \hfill \\ \end{gathered} \]

  11. anonymous
    • one year ago
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    ok! how do we solve that? :/

  12. Michele_Laino
    • one year ago
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    it is simple you have to replace M with 0.15 Kg

  13. Michele_Laino
    • one year ago
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    you should get two values: V_x and V_y respectively

  14. anonymous
    • one year ago
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    ohh okay! so we get V_x=8 and V_y = 5.33333 ?

  15. Michele_Laino
    • one year ago
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    ok!

  16. Michele_Laino
    • one year ago
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    now, the requested speed V is given by the subsequent formula: \[\Large V = \sqrt {V_x^2 + V_y^2} = \sqrt {{8^2} + {{5.33}^2}} = ...\]

  17. anonymous
    • one year ago
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    so we get this? 9.6129 ?

  18. Michele_Laino
    • one year ago
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    please wait, I have made an error

  19. Michele_Laino
    • one year ago
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    you have to replace M with 0.05 Kg, not with 0.15 Kg

  20. Michele_Laino
    • one year ago
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    \[\Large \begin{gathered} {V_x} = \frac{{1.2}}{M} = \frac{{1.2}}{{0.05}}... \hfill \\ \hfill \\ {V_y} = \frac{{0.8}}{M} = \frac{{0.8}}{{0.05}}... \hfill \\ \end{gathered} \]

  21. anonymous
    • one year ago
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    ohh okay! so we get thiS? v_x=24 and V_y=16 so then we put that into sq rt and get this? 28.85? so our answer is choice D? 29 m/s?

  22. Michele_Laino
    • one year ago
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    yes! correct! Sorry for my error

  23. Michele_Laino
    • one year ago
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    \[\Large V = \sqrt {V_x^2 + V_y^2} = \sqrt {{{24}^2} + {{16}^2}} = ...\]

  24. anonymous
    • one year ago
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    yay!! and it is okay! :) thank you!

  25. Michele_Laino
    • one year ago
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    \[V = \sqrt {V_x^2 + V_y^2} = \sqrt {{{24}^2} + {{16}^2}} \cong 29m/\sec \]

  26. Michele_Laino
    • one year ago
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    thank you!! :)

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