Question

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.

Answer 1

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.