Note: Take East as the positive direction. A(n) 81 kg fisherman jumps from a dock into a 136 kg rowboat at rest on the West side of the dock. If the velocity of the fisherman is 3.4 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

The final velocity of the fisherman and the boat = -1.27 m/s

Explanation:

Mass of fisherman, M₁ = 81 kg

Mas of Rowboat, M₂ = 136 kg

Initial velocity of the fisherman, v₁ = -3.4 m/s (Westwards)

Initial velocity of the rowboat, v₂ = 0 m/s ( It is at rest)

The final velocity of the fisherman and the boat, v = ?

Using the principle of momentum conservation

M₁v₁ + M₂v₂ = (M₁ + M₂) v

-(81*3.4) + (136*0) = (81 + 136) v

-275.4 = 217v

v = -275.4/217

v = -1.27 m/s