In order to be able to calculate an answer, we must assume that the boat, with her on it, was motionless in the water until she jumped off of it.
I'll make that assumption, and then I'll go ahead and answer the question that I just invented:
-- Before she jumped off of the boat, she had no momentum and the boat had no momentum.
-- The SUM of (her momentum) + (the boat's momentum) was zero.
-- Momentum is conserved, so the SUM of (her momentum) + (the boat's momentum) has to still be zero after she jumps off of the boat.
-- Momentum = (mass) · (speed in some direction)
The boat's momentum after the jump = (42kg) · (1.5 m/s) that way ==>
The boat's momentum after the jump = 63 kg-m/s that way ==>
Her momentum after the jump has to be 63 kg-m/s this way <==
Her momentum = 63 kg-m/s = (59 kg) · (her speed this way <== )
Divide each side by (59 kg):
Her speed this way <== = (63 kg-m/s <==) / (59 kg)
Her speed = (63/59) · (m/s this way <== )
Her speed = 1.07 m/s opposite to the direction the boat is moving.
= = = = = = = = = =
A casual but striking observation:
Our 'student' is carrying 17 kg more mass than the boat she sails !
The mind boggles at the implied zaftigkeit.
