An 85.0 kg fisherman jumps from a dock at a speed of 4.30 m/s onto their 135.0 kg boat. If the boat was at rest to begin but moves with the

Question

An 85.0 kg fisherman jumps from a dock at a speed of 4.30 m/s onto their 135.0 kg boat. If the boat was at rest to begin but moves with the fisherman after he lands, what is the final speed of the boat and fisherman? (Hint: this is an inelastic “collision”)
m1v1 + m2v2 = (m1+m2)vf

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RuslanHeatt 3 years 2021-08-24T02:51:24+00:00 1 Answers 28 views 0

Answers ( )

  1. Dau
    0
    2021-08-24T02:52:42+00:00
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    Answer:

    Final speed of boat + man is 1.66 m/s

    Explanation:

    As we know that there is no friction on the system or there is no external force on this system

    So here we can use momentum conservation here

    mv = (m + M)v_f

    so we have

    m = 85 kg

    M = 135 kg

    v = 4.30 m/s

    now we have

    85 \times 4.30 = (85 + 135) v

    v = 1.66 m/s

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