An 88kg fisherman jumps from a dock into a 131kg rowboat at rest on the west side of the dock. If the velocity of the fisherman is 4.7m/s to the west as he leaves the dock, what is the final velocity of the fisherman and the boat. East is positive.

Answer 1

Apply conservation of momentum. If east is positive, the final momentum and velocity will be negative.

Answer 2

To find the momentum of the fisherman as he jumps, we use mass times velocity (mv). This leads us to 88kg x 4.7 m/s, giving us 413.6 kg m/s. Since this is how much momentum we start with, it will also be how much we end with. We still use mass times velocity (mv), but this time the boat is included in the mass (m). 88 kg + 131 kg = 219 kg, so now we solve for the velocity: 219 kg x v = 413.6 kg m/s. After dividing 413.6 kg m/s by 219 kg, we get the final velocity of 1.88 m/s. Since this is to the west, this value will be negative, giving us a FINAL ANSWER of -1.88 m/s.