Question

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.

Answer 1

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

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

now we will have

$83(3.1) + 139(0) = 83 v + 139 v$

$257.3 = 222v$

$v = 1.16 m/s$

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