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Conservation of Momentum difficulty

  1. Mar 19, 2006 #1
    I'm not sure if i did this problem correctly. Could someone check it over.

    A 730-N man stands in the middle of a frozen pond of a radius of 5.0 m. He is unable to get to the other side because of a lack of friction between his shoes and the ice. To overcome this difficulty, he throws his 1.2 kg physics textbook horizontally toward the north shore at a speed of 5.0 m/s. How long does it take him to reach the south shore?

    [tex] p_i = p_f [/tex]
    [tex] 0 = m_1v_1f + m_2v_2f [/tex]

    [tex] v_1f = - \frac{m_2}{m_1} * v_2f = - \frac{1.20 kg}{74.5 kg} * 5.0 m/s = -.081 m/s [/tex]

    [tex] t = \frac{x}{v} = \frac{5.0 m}{2.46 m/s} = 2.03 s [/tex]
  2. jcsd
  3. Mar 19, 2006 #2
    [tex] t = \frac{x}{v} = \frac{5.0 m}{2.46 m/s} = 2.03 s [/tex][/QUOTE]

    [tex] t = \frac{x}{v} = \frac{5.0 m}{.081 m/s} = 62 s [/tex][/QUOTE]
  4. Mar 19, 2006 #3


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    That is correct :smile:
  5. Mar 19, 2006 #4
    Thank you for checking my problem.
  6. Jul 9, 2009 #5
    Why is 5.0m/s the final speed of the book but not its initial speed? Wasn't it thrown at a speed of 5.0m/s?
    Last edited: Jul 10, 2009
  7. Jul 9, 2009 #6
    we assume that initially all the system is at rest
  8. Jul 10, 2009 #7
    so is it wrong to assume that the initial speed of the book is 5m/s?
  9. Jul 10, 2009 #8


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    Yeah, it is. In this case "initial" and "final" mean "before the throw" and "after the throw" respectively.

    Of course, there's technically no reason you couldn't pick "initial" and "final" to both be after the throw, but it wouldn't tell you anything useful.
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