A lumberjack (mass = 103 kg) is standing at rest on one end of a floating log (mass = 255 kg) that is also at rest. The lumberjack runs to the other end of the log, attaining a velocity of + 2.93 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.

(a) - 1.18 m/s

(b) 0.84 m/s

Explanation:

(a)

The total linear momentum before the lumberjack begins to move is zero because all parts of the system are at res

From the law of conservation of momentum

m1v1+m2v2=0 hence m1v1=-m2v2 where m1 is mass of lumberjack, v1 is velocity of lumberjeck, m2 is mass of floating log, v2 is velocity of the floating log.

Substituting M1 for 103 Kg, V1 for 2.93 m/s, M2 for 255 Kg into the above equation we obtain

103Kg*2.93 m/s=-255Kg*V2

V2 = - (103 kg*2.93 m/s) / 255=-1.183490196 m/s

Hence V2=-1.18 m/s

(b)

For the second log

V (M1+M2) = m1v1 where V is the common velocity

V (103 Kg+255 Kg) = 103 Kg*2.93 m/s

V = (103 Kg*2.93 m/s) / (103 Kg+255 Kg) = 0.842988827 m/s

V=0.84 m/s