Ramirezjasmine9
Answered

Discover answers to your most pressing questions at Westonci.ca, the ultimate Q&A platform that connects you with expert solutions. Join our platform to connect with experts ready to provide precise answers to your questions in different areas. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

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?

Sagot :

AL2006

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


Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.