An 85.0 kg fishermen jumps from a dock into an135 kg boat at rest. If the velocity of the fisherman as he leaves the dock is 4.30 m/s west, what is the final velocity of the boat and fisherman together?

Question

Westie

Physics

7 February, 11:52

An 85.0 kg fishermen jumps from a dock into an135 kg boat at rest. If the velocity of the fisherman as he leaves the dock is 4.30 m/s west, what is the final velocity of the boat and fisherman together?

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Noble · Answer 1

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