11. an 85 kg fisherman in a 290 kg boat throws a package of bait, 15 kg,
horizontally toward the right with a speed of 3 m/s. assuming the boat
is at rest before the package is thrown, find the velocity of the boat
after the bait is thrown.

The velocity of the boat is, V = 0.12 m/s  

Explanation:

Given data,

The mass of the boat, M = 290 kg

The mass of the fisherman, m = 85 kg

The mass of the bait, m' = 15 kg

The velocity of the bait, v' = 3 m/s

Since the bodies were at rest initially, the net momentum of the three bodies is equal to zero.

According to the law of conservation of momentum the equation becomes,

                                     V(M + m) + m'v' = 0

                                     V = - m'v'/ (M + m)

Substituting the values,

                                            V = -0.12 m/s

The negative sign indicates the direction of the boat is opposite to the direction of the bait thrown.

Hence, the boat's velocity is, V = 0.12 m/s                                  

grav potential energy