Question

A lumberjack (mass = 112 kg) is standing at rest on one end of a floating log (mass = 250 kg) that is also at rest. The lumberjack runs to the other end of the log, attaining a speed of 3.0 m/s relative to the shore, and then hops onto an identical floating log that is initially at rest. What is the speed of the first log relative to the second log when the lumberjack comes to rest with respect to the second log? Neglect any friction and resistance between the logs and the water.

The answer is supposed to be 2.3 m/s, however I am struggling to reach this answer.

Answer 1

lumberjack mass = m

lumberjack velocity = v

log mass = M

log velocity = V

using conservation of momentum
mv(i) + MV(i) = mv(f) +MV(f)
0 = 112(3) +250*V(f)
log velocity = V(f) = -1.344 m/s

For lumberjack final velocity
mv(i) + MV(i) = (m + M)v(f)
112(3) + 250(0) = (112+250)v(f)
v(f) = +0.928 m/s

speed of the first log relative to the second log when the lumberjack comes to rest with respect to the second log V_R= v(f) – V(f)

V_R = 0.928 + 1.344

V_R = 2.27 m/s

V_R= 2.3 m/s