A lumberjack (mass = 98 kg) is standing at rest on one end of a floating log (mass = 270 kg) that is also at rest. The lumberjack runs to the other end of the log, attaining a velocity of +2.9 m/s relative to the shore, and then hops onto an identical floating log that is initially at rest. Neglect any friction and resistance between the logs and the water.

Answer 1

momentum is the same, zero, at all times
assume he runs in -x direction
as he leaves log 1
98*2.9 = 270 vlog1
so
vlog1 = 1.05 m/s in + x direction

momentum of log 1 = 270*1.05 = 283.5
therefore the momentum of log 2 with the logger now on board must be -283.5
(98+270)Vlog2 = -283.5

Answer 2

-1.30m/s