A lumberjack (mass = 112 kg) is standing at rest on one end of a floating log (m

Question

A lumberjack (mass = 112 kg) is standing at rest on one end of a floating log (mass = 203 kg) that is also at rest. The lumberjack runs to the other end of the log, attaining a velocity of +2.16 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. (a) What is the velocity of the first log (again relative to the shore) just before the lumberjack jumps off? (b) Determine the velocity of the second log (again relative to the shore) if the lumberjack comes to rest relative to the second log.

Explanation / Answer

A) momentum will be conserved

final momentum = initial momentum =0

So m1v1 + m2v2 = 0

112×2.16 + 203 v2 = 0

v2 = -112×2.16/203 = -1.19 m/s

B) final momentum of second system= initial momentum

since lumberjack is at rest with respect to second log, both have same final velocity relative to shore

(203+112)V = 112×2.16

v =2.16×112/315 = 0.768 m/s

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