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1 January, 16:03

A swimmer of mass 64.38 kg is initially standing still at one end of a log of mass 237 kg which is floating at rest in water. He runs toward the other end of the log and dives off with a horizontal speed of 3.472 m/s relative to the water. What is the speed of the log relative to water after the swimmer jumps off

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  1. 1 January, 18:19
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    0.9432 m/s

    Explanation:

    We are given;

    Mass of swimmer; m_s = 64.38 kg

    Mass of log; m_l = 237 kg

    Velocity of swimmer; v_s = 3.472 m/s

    Now, if we consider the first log and the swimmer as our system, then the force between the swimmer and the log and the log and the swimmer are internal forces. Thus, there are no external forces and therefore momentum must be conserved.

    So;

    Initial momentum = final momentum

    m_l * v_l = m_s * v_s

    Where v_l is speed of the log relative to water

    Making v_l the subject, we have;

    v_l = (m_s * v_s) / m_l

    Plugging in the relevant values, we have;

    v_l = (64.38 * 3.472) / 237

    v_l = 0.9432 m/s
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