A 74-kg fisherman in a 129-kg boat throws a package of mass
m = 15 kg
horizontally toward the right with a speed of
vi = 4.7 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.
magnitude m/s
direction
To find the velocity of the boat after the package is thrown, we can use the principle of conservation of momentum. According to this principle, the total momentum before the package is thrown should be equal to the total momentum after the package is thrown.
Before the package is thrown, only the fisherman is occupying the boat. Therefore, the total momentum before the package is thrown is given by:
momentum_before = mass_of_fisherman * velocity_of_fisherman
momentum_before = 74 kg * 0 m/s (since the boat is at rest)
After the package is thrown, the boat and the fisherman together will have a non-zero velocity due to the impulse from throwing the package. The total momentum after the package is thrown is given by:
momentum_after = (mass_of_fisherman + mass_of_boat + mass_of_package) * velocity_of_boat
momentum_after = (74 kg + 129 kg + 15 kg) * velocity_of_boat
Since momentum is conserved, we can set these two equations equal to each other and solve for the velocity of the boat:
mass_of_fisherman * velocity_of_fisherman = (mass_of_fisherman + mass_of_boat + mass_of_package) * velocity_of_boat
Plugging in the given values:
74 kg * 0 m/s = (74 kg + 129 kg + 15 kg) * velocity_of_boat
We can rearrange the equation to solve for velocity_of_boat:
velocity_of_boat = (74 kg * 0 m/s) / (74 kg + 129 kg + 15 kg)
Simplifying this expression:
velocity_of_boat = 0 m/s
Therefore, the velocity of the boat after the package is thrown is 0 m/s.