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Answer :
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.
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.
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Rewritten by : Barada
Using conservation of momentum, the velocity of the boat after the package is thrown is calculated to be 0.556 m/s opposite to the direction of the throw.
The question pertains to conservation of momentum in a system consisting of a fisherman, a boat, and a package. Initially, the system is at rest, meaning the total momentum is zero. When the fisherman throws the package, the system must still adhere to the principle of conservation of momentum, meaning the momentum of the boat and fisherman system after the throw equals the negative momentum of the package.
To calculate the velocity of the boat (including the fisherman) after the package is thrown, we use the formula for conservation of momentum: (m1 + m2) * v1 = m3 * v3, where m1 is the mass of the boat, m2 is the mass of the fisherman, v1 is the final velocity of the boat and fisherman, m3 is the mass of the package, and v3 is the velocity of the package relative to the boat and fisherman before it was thrown.
Plugging in the values, (114 kg + 72 kg) * v1 = 15 kg * 5.0 m/s. Solving for v1, we find that the velocity of the boat after the package is thrown is 0.556 m/s in the opposite direction of the package's throw.
