Note: Take East as the positive direction. A(n) 81 kg fisherman jumps from a dock into a 128 kg rowboat at rest on the West side of the dock. If the velocity of the fisherman is 4.5 m/s to the West as he leaves the dock, what is the final velocity of the fisherman and the boat? Answer in units of m/s.

Respuesta :

Answer:

final velocity of fisherman will be 1.744 m /sec in west direction

Explanation:

We have given mass of the fisherman [tex]m_1=81kg[/tex]

Velocity of fisherman [tex]v_=-4.5m/sec[/tex] ( As east direction is positive direction )

Mass of the rowboat [tex]m_2=128kg[/tex]

As the rowboat is at rest so [tex]v_2=0m/sec[/tex]

Now according to conservation of momentum

[tex]m_1v_1+m_2v_2=(m_1+m_2)v[/tex]

[tex]81\times -4.5+128\times 0=(81+128)\times v[/tex]

[tex]v=-1.744m/sec[/tex]

So final velocity of fisherman will be 1.744 m /sec in west direction

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