A 78.5-kg man is standing on a frictionless ice surface when he throws a 2.40-kg book horizontally at a speed of 11.3 m/s. With what speed d

Question

A 78.5-kg man is standing on a frictionless ice surface when he throws a 2.40-kg book horizontally at a speed of 11.3 m/s. With what speed does the man move across the ice as a result? Give your answer in m/s, and be sure to use three significant figures in your answer.

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Sigridomena 3 years 2021-08-28T13:22:24+00:00 2 Answers 11 views 0

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  1. Cherry
    0
    2021-08-28T13:23:46+00:00
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    Answer:

    The man moves across the ice with a speed of 0.345m/s.

    Explanation:

    From the conservation of linear momentum, we have that the total linear momentum before the book throw is equal to the total linear momentum just after it. Since the initial velocity of the system is zero (so the initial momentum is zero), we have that:

    m_mv_m+m_bv_b=0\\\\v_m=-\frac{m_b}{m_m} v_b

    Where m_m is the mass of the man, m_b is the mass of the book, and v_m and v_b are their velocities. Plugging in the given values, we can compute the speed of the man (ignoring the negative sign, because we care about the magnitude, not the direction):

    v_m=\frac{2.40kg}{78.5kg}(11.3m/s)=0.345m/s

    In words, the resulting speed of the man is 0.345m/s.

  2. Amity
    0
    2021-08-28T13:24:08+00:00
    Reply

    Answer:

    0.345 m/s²

    Explanation:

    From the law of conservation of momentum,

    The total momentum before the book was thrown = Total momentum after the book is thrown.

    Note: Since both the man and the book are stationary before the book was thrown, then the total momentum before the book was thrown = 0.

    0 = m’v’ + mv ……………….. Equation 1

    m’v’+mv = 0

    Where m’ = mass of the man, v’ = final velocity of the man after the throw, m = mass of the book, v = final velocity after the throw

    make v’ the subject of the equation

    v’ = -mv/m’…………… Equation 2

    Given: m = 2.40 kg, v = 11.3 m/s, m’ = 78.5 kg.

    Substitute into equation 2

    v’ = 2.4(11.3)/78.5

    v’ = 27.12/78.5

    v’ = 0.345 m/s²

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