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anonymous

  • one year ago

An air-track glider of mass 0.25 kg moving at 0.60 m/s collides with and sticks to a glider of mass 0.50 kg at rest. How much kinetic energy is converted into other forms of energy as a direct result of this collision? A. 0.030 J B. 0.045 J C. 0.015 J D. 0.060 J ***not sure at all! :( Thank you!! :D

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  1. Michele_Laino
    • one year ago
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    here we have to apply the conservation of momentum, so we can write: \[\Large {m_1}{v_1} = \left( {{m_1} + {m_2}} \right)u\]

  2. anonymous
    • one year ago
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    okay!

  3. Michele_Laino
    • one year ago
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    where m_1= 0.25 Kg, m_2=0.50 Kg, and v_1=0.60 m/sec

  4. Michele_Laino
    • one year ago
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    u is the final speed of the system particle1+ particle2

  5. Michele_Laino
    • one year ago
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    we have to find what is u: \[\Large u = \frac{{{m_1}{v_1}}}{{{m_1} + {m_2}}} = ...m/\sec \]

  6. anonymous
    • one year ago
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    okay, so i plug in like this? (0.25*0.60)=(0.25+0.50)u so like this? 0.15=0.75u u=0.2 ?

  7. anonymous
    • one year ago
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    oh oops

  8. anonymous
    • one year ago
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    oh wait... it is 0.2, correct? the value of u?

  9. Michele_Laino
    • one year ago
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    that's right u=0.2 m/sec

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

  11. Michele_Laino
    • one year ago
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    Now the final kinetic energy is: \[\Large {E_f} = \frac{1}{2}\left( {{m_1} + {m_2}} \right){u^2} = ...Joules\]

  12. anonymous
    • one year ago
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    okay! so 1/2(0.25+0.50)0.2^2 = 0.015 J ? so the answer is 0.015 J? :O

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

  14. anonymous
    • one year ago
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    yay!! thank you:D

  15. Michele_Laino
    • one year ago
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    whereas the initial kinetic energy is: \[\Large {E_i} = \frac{1}{2}{m_1}v_1^2 = ...Joules\]

  16. Michele_Laino
    • one year ago
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    @iheartfood

  17. anonymous
    • one year ago
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    ahh so it would be 1/2(0.25+0.5^2)? so 0.25 is the initial kinetic energy?

  18. Michele_Laino
    • one year ago
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    I got 0.045 Joules

  19. Michele_Laino
    • one year ago
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    \[\Large {E_i} = \frac{1}{2}{m_1}v_1^2 = \frac{1}{2} \times 0.25 \times {\left( {0.6} \right)^2} = ...Joules\]

  20. anonymous
    • one year ago
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    ohhh yes 0.045 sorry, i plugged in the wrong thing! wait so for this problem, i now have to do this to find the answer? 0.045-0.015=0.030 J ? and that would be this problem's final solution? :O

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

  22. anonymous
    • one year ago
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    yay!! and so this problem is completed now? :O

  23. Michele_Laino
    • one year ago
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    yes! We have finished!

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

  25. Michele_Laino
    • one year ago
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    thank you!! :D

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