rayceharvey9
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A(n) 83 kg fisherman jumps from a dock into a 139 kg rowboat at rest on the West side of the dock. If the velocity of the fisherman is 3.1 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 :

here we can say that there is no external force on fisherman and dock

so here we will use momentum conservation theory

As per momentum conservation

initial momentum of fisherman + boat = final momentum of fisherman + boat

[tex]m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}[/tex]

now we will have

[tex]83(3.1) + 139(0) = 83 v + 139 v[/tex]

[tex]257.3 = 222v[/tex]

[tex]v = 1.16 m/s[/tex]

so the speed of boat and fisherman will be 1.16 m/s

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