A 68-kg fisherman in a 136-kg boat throws a package of mass m = 15 kg horizontally toward the right with a speed of vi = 4.0 m/s 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.

www.webassign.net/sercp11/6-op-026-alt.png

A - magnitude ???????
B - direction - Towards the left

momentum is conserved

package mass * velocity = (boat + man) mass * velocity

as you have deduced, the velocities are in opposite directions

To find the velocity of the boat after the package is thrown, we can use the law of conservation of momentum. According to this law, the total momentum before and after an event remains constant, as long as no external forces are acting.

In this scenario, we have an isolated system consisting of the fisherman, the boat, and the package. Before the package is thrown, the fisherman and the boat are at rest, so their initial momentum is zero. Therefore, the initial momentum of the system is also zero.

After the package is thrown horizontally towards the right, the package exerts a forward momentum on the system. To conserve momentum, the fisherman and the boat must have an equal and opposite momentum in the backward direction after the package is thrown.

Let's denote the velocity of the boat after the package is thrown as V_f and the velocity of the package as v_p. We can then write the conservation of momentum equation as:

(boat mass) * (-V_f) + (package mass) * v_p = 0

Substituting the given values:
(136 kg) * (-V_f) + (15 kg) * (4.0 m/s) = 0

Rearranging the equation to solve for V_f:
V_f = (15 kg * 4.0 m/s) / 136 kg

Calculating the value:
V_f ≈ 0.441 m/s (rounded to three decimal places)

Therefore, the magnitude of the velocity of the boat after the package is thrown is approximately 0.441 m/s, and the direction of the velocity is towards the left.