physics

A 300kg motorboat is turned off as it approaches a dock and it coasts in toward the dock at .5 m/s. Isaac, whose mass is 62kg, jumps off the front of the boat with a speed of 3m/s relative to the boat. what is the velocity of the boat after Isaac jumps?

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  1. The combined momentum of motorboat and Isaac remains the same. Choose a coordinate system based on land. The momentum conservation equation can then be written

    0.5 m/s*(Mb + Mi) = Mb*Vb + Mi*(Vi+0.5)

    Solve for the unknown velocity of the boat afterwards, Vb.

    0.5*362 = 300 Vb + 62*3.5
    300 Vb = 181 - 217 = -36
    Vb = -0.12 m/s (backwards)

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  2. boat is m1 = 300 kg
    Isaac is m2 = 62 kg
    towards dock is +
    away from dock is -
    original velocity of both is
    v = +0.50 m/s
    v2' = final velocity of Isaac
    v1' = final velocity of boat
    v2'-v1' = +3.0 m/s (because Isaac jumps off the front of the boat & is moving toward the dock; this will make v2' positive and v1' negative.
    (m1+m2)v = m1v1' + m2v2'
    v1' = 0.014 m/s away from the dock (boat) and v2' - 2.986 m/s towards the dock (Isaac)

    1. 👍
    2. 👎
  3. boat is m1 = 300 kg
    Isaac is m2 = 62 kg
    towards dock is +
    away from dock is -
    original velocity of both is
    v = +0.50 m/s
    v2' = final velocity of Isaac
    v1' = final velocity of boat
    v2'-v1' = +3.0 m/s (because Isaac jumps off the front of the boat & is moving toward the dock; this will make v2' positive and v1' negative.
    (m1+m2)v = m1v1' + m2v2'
    v1' = 0.014 m/s away from the dock (boat) and v2' = 2.986 m/s towards the dock (Isaac)

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