A lumberjack (mass = 98.3 kg) is standing at rest on one end of a floating log (mass = 258 kg) that is also at rest. The lumberjack runs to the other end of the log, attaining a velocity of +3.69 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.
To solve this problem, we can use the principle of conservation of momentum, which states that the total momentum before an event is equal to the total momentum after the event, as long as no external forces act on the system.
Let's break down the problem step by step:
(a) What is the velocity of the first log just before the lumberjack jumps off?
We need to calculate the velocity of the first log just before the lumberjack jumps off. Since no external forces act on the system, the total momentum of the system (lumberjack + first log) is conserved.
The momentum of an object is given by the product of its mass and velocity: momentum = mass × velocity.
Let's denote the velocity of the first log just before the lumberjack jumps off as v1 and the velocity of the lumberjack as v_lumberjack.
Using the conservation of momentum principle:
(mass of lumberjack × v_lumberjack) + (mass of first log × 0) = (mass of lumberjack × 0) + (mass of first log × v1)
Since the lumberjack starts at rest and jumps off, his final velocity is 0 m/s (relative to the first log). Therefore, we can write:
(mass of lumberjack × 0) + (mass of first log × 0) = (mass of lumberjack × v_lumberjack) + (mass of first log × v1)
Simplifying the equation:
0 = (mass of lumberjack × v_lumberjack) + (mass of first log × v1)
Rearranging the equation:
v1 = -(mass of lumberjack × v_lumberjack) / mass of first log
Substituting the given values:
v1 = -(98.3 kg × 3.69 m/s) / 258 kg
Calculating:
v1 ≈ -1.401 m/s
So, the velocity of the first log just before the lumberjack jumps off is approximately -1.401 m/s (negative because it's in the opposite direction to the lumberjack's velocity).
(b) Determine the velocity of the second log if the lumberjack comes to rest relative to the second log.
Since the lumberjack comes to rest on the second log, his final velocity relative to the second log is 0 m/s.
We can use the conservation of momentum principle again to find the velocity of the second log.
Let's denote the velocity of the second log as v2.
(mass of lumberjack × 0) + (mass of second log × 0) = (mass of lumberjack × 0) + (mass of second log × v2)
Simplifying the equation:
0 = (mass of second log × v2)
Therefore, the velocity of the second log is 0 m/s.
So, the velocity of the second log (relative to the shore) is 0 m/s.