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anonymous

  • one year ago

Two hard steel balls collide in a perfectly elastic collision. The first ball has a mass of 1kg and is traveling at 5m/s east. The second ball has a mass of 2kg and is initially traveling at 7 m/s west along the same path so they collide head on. After the collision the second ball is traveling at 1m/s east. What is the velocity of the 1st ball after the collision? **How do I solve and show the work for this? :/ Thanks!!

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  1. Michele_Laino
    • one year ago
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    here we have to apply the law of conservation of total momentum and the law of conservation of kinetic energy, since the collision is elastic

  2. Michele_Laino
    • one year ago
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    the situation, before collision is: |dw:1433610623171:dw|

  3. anonymous
    • one year ago
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    ok!

  4. Michele_Laino
    • one year ago
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    so, total momentum is: \[\Large {m_1}{v_1} - {m_2}{v_2} = 1 \times 5 - 2 \times 7 = ...\]

  5. anonymous
    • one year ago
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    ok! so we get 5-14? = -9 ?

  6. Michele_Laino
    • one year ago
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    minus sign, means that the vector of the resultant momentum is oriented towards west

  7. anonymous
    • one year ago
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    ohh okay! so -9, meaning 9 m/s to the west is the velocity of the 1st ball after collision?

  8. Michele_Laino
    • one year ago
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    no, -9 is the magnitude and sign of the resultant momentum

  9. anonymous
    • one year ago
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    ohh okay! so it's like this? 5-14=-9 =resultant momentum ?

  10. Michele_Laino
    • one year ago
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    yes! now total kinetic energy is: \[\large KE = \frac{1}{2}{m_1}v_1^2 + \frac{1}{2}{m_2}v_2^2 = \frac{1}{2} \times 1 \times 25 + \frac{1}{2} \times 2 \times 49 = ...\]

  11. anonymous
    • one year ago
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    ok! so we get this? 61.5 ?

  12. Michele_Laino
    • one year ago
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    ok! correct!

  13. Michele_Laino
    • one year ago
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    now let's consider the situation after the collision

  14. anonymous
    • one year ago
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    ohhh so 61.5 is just the KE right? so we solve for velocity now? :/

  15. Michele_Laino
    • one year ago
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    |dw:1433611229406:dw|

  16. Michele_Laino
    • one year ago
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    no, we have to apply the conservation of total momentum first

  17. anonymous
    • one year ago
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    oh ok! how do we do that?

  18. Michele_Laino
    • one year ago
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    I call with u the speed of the first ball after collision

  19. anonymous
    • one year ago
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    ok!

  20. Michele_Laino
    • one year ago
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    so we can write: conservation of total momentum: \[\Large \begin{gathered} {m_1}u + {m_2}v{'_2} = - 9 \hfill \\ 1 \times u + 7 \times 1 = - 9 \hfill \\ \end{gathered} \]

  21. Michele_Laino
    • one year ago
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    what is u?

  22. anonymous
    • one year ago
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    -16?

  23. Michele_Laino
    • one year ago
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    in my reasoning i have considered positive velocity if the velocity is towards east, and negative velocity if that velocity is towards west

  24. anonymous
    • one year ago
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    oh! so u=-16?

  25. anonymous
    • one year ago
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    16 m/s to the west?

  26. Michele_Laino
    • one year ago
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    correct! that means the first ball, after the collision is goinf towards west

  27. Michele_Laino
    • one year ago
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    going*

  28. anonymous
    • one year ago
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    ahh so that is our solution? :O

  29. Michele_Laino
    • one year ago
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    yes!

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

  31. Michele_Laino
    • one year ago
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    :)

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