A 72-kg fisherman in a 114-kg boat throws a package of mass m = 15 kg horizontally toward the right with a speed of vi = 5.0 m/s. Neglecting water resistance, and assuming the boat is at rest before the package is thrown, find the velocity of the boat after the package is thrown.

We are going to use law of conservation of Momentum.

Momentum of package thrown = Momentum of the man and the ship

Mass of Package * Velocity of Package = Mass of man and ship * velocity of both.

15*5 = (72 + 114) * v

75 = 186*v

186v = 75

v = 75/186

v ≈ 0.403 m/s

The velocity of the boat and the man after the package is thrown is ≈ 0.403 m/s toward the left.