A lumberjack (mass=98 kg) is standing at rest on one end of a floating that is a

Question

A lumberjack (mass=98 kg) is standing at rest on one end of a floating that is also at rest. The lumberjack runs to the other end of the log (mass=230 kg), attaining a velocity of +3.6 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 just before the lumberjack jumps off? (b) Determine the velocity of the second log if the lumberjack comes to rest on it.

Explanation / Answer

here no externel force and lumberjack and log constitue and isolate system

so here we can use linear momentum conservation

part a ) initial momentum = final momentum

initial momentum = m1u1 + m2u2

final momentum = m1v1 + m2v2

initially system at rest

0 = m1v1 + m2v2

m1 = mass of lumber jack

v1 = velocity of lumber jack

m2 = mass of log

v2 = velocity of log

v2 = -m1v1/ m2

v2 = -(98*3.6)/230 = - 1.53 m/s

here negative signs means that log recoils off as lumberjack jump off

part b )

apply the linear momentum conservation here

second log initally at rest

after hop lumberjack and secong log move with same velocity

m1u1 + m2u2 = m1v1 + m2v2

m1 = mass of lumber jack ; u1 = initial velocity = 3.6 m/s

m2 = mass of secong log = 230 kg ; u2 = initial velocity of log = 0

v1 = v2

v = m1u1 /(m1+m2)

v = 1.08 m/s

the system will move in same direction of lumberjack's original motion direction

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