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4 January, 17:30

A 300. kg motorboat is coasting toward the dock at 0.50 m/s. Isaac, whose mass is 62.0 kg, jumps off the front of the boat with a speed of 3.0 m/s relative to the boat. What is the velocity of the boat after Isaac jumps?

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  1. 4 January, 17:45
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    0.087 m/s

    -0.12 m/s

    Explanation:

    The momentum is restored

    (M₁ + M₂) V = M₁V₁ + M₂V₂ ... Eqn 1

    M₁ = Mass of the boat = 300 kg

    M₂ = Mass of Isaac = 62 kg

    V = collective velocity before Isaac jumped = 0.5 m/s

    V₁ = Velocity of the boat after Isaac jumped = ?

    V₂ = Velocity of Isaac after he jumped =

    1st Scenaro; If Isaac jumps forward, V₂ = 2.5 m/s

    2nd Scenaro; If Isaac jumps backward V₂ = 3.5 m/s

    Using equation for Eqn one first scenario = (300 + 62) 0.5 = 300V₁ + 62 (2.5) = 0.086 m/s

    Using equation for Eqn one second scenario = (300 + 62) 0.5 = 300V₁ + 62 (3.5) = - 0.12 m/s

    Note: The negative sign indicates the jump was made backwards
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