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Answer :
Momentum = (mass) x (speed)
-- Before he jumps, while he's just standing there on the dock thinking about it,
the total momentum of the man and the boat is zero, since nothing is moving.
-- Suddenly, he uses his leg muscles to push against the dock, and gives himself
(85kg) x (4.3 m/s) = 365.5 kg-m/s of momentum west.
-- When he lands in the boat and sticks to it, that same amount of momentum
has to account for the boat's motion as well as the man's motion.
Together, their mass is (135 + 85) = 220 kg.
So
(220 kg) x (their speed) = the same 365.5 kg-m/s of momentum west.
Divide each side by (220 kg):
their speed = (365 kg-m/s) / (220 kg) = 1.661 m/s west
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Final answer:
Based on the law of conservation of momentum, the final velocity of the boat and the fisherman together, following the fisherman's jump, is 1.66 m/s to the west.
Explanation:
This question is based on the law of conservation of momentum which states that the total momentum of a closed system is constant unless acted upon by an external force. Here, the fisherman jumping off the dock and the boat constitutes our system. According to the law of conservation of momentum, the momentum of the fisherman before jumping is equal to the collective momentum of the fisherman and the boat after the jump.
The initial momentum can be calculated as the product of the fisherman's mass and his velocity i.e., (85.0 kg) * (4.30 m/s) = 365.5 kg*m/s.
After the jump, the fisherman and the boat are moving together, so their total mass is 85.0 kg + 135 kg = 220.0 kg. Their final velocity (v') can be calculated by setting the initial and final momentum equal to each other and solving for v': 365.5 kg*m/s = (220.0 kg)*(v') => v' = 365.5 kg*m/s ÷ 220.0 kg = 1.66 m/s.
Therefore, the final velocity of the boat and the fisherman together is 1.66 m/s to the west (in the same direction as the fisherman was moving initially).
Learn more about Conservation of Momentum here:
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