A 82-kg fisherman in a 112-kg boat throws a package of mass [tex]$m = 15 \, \text{kg}$[/tex] horizontally toward the right with a speed of [tex]$v_i = 4.8 \, \text{m/s}$[/tex] as in the figure below. 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.

To solve for the velocity of the boat after the fisherman throws the package, we can use the law of conservation of momentum, which states that the total momentum of a system remains constant if no external forces are acting on it. Before the package is thrown, the system (fisherman, boat, and package) is at rest, so the initial momentum is zero. After the package is thrown, the momentum of the package and the momentum of the boat (with fisherman) must add up to zero.

Let's denote the final velocity of the boat as vb. The momentum of the thrown package is given by its mass times its velocity, which is (15 kg)(4.8 m/s). The total mass of the boat and fisherman is the sum of their masses, (82 kg + 112 kg). Since momentum is conserved, the momentum of the boat and fisherman is equal in magnitude but opposite in direction to the momentum of the package.

The equation for conservation of momentum can be stated as:

mpackage × vi = (mboat + mfisherman) × vb

Plugging in the values we have:

(15 kg × 4.8 m/s) = (82 kg + 112 kg) × vb

Solving for vb:

vb = (15 kg × 4.8 m/s) / (82 kg + 112 kg)

vb = 72 kg·m/s / 194 kg

vb = 0.371 m/s

Therefore, the velocity of the boat after the package is thrown is 0.371 m/s to the left (opposite to the direction of the thrown package).