A lumberjack (mass = 104 kg) is standing at rest on one end of a floating log (mass = 234 kg) that is also at rest. The lumberjack runs to the other end of the log, attaining a velocity of +3.18 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. (b) Determine the velocity of the second log (again relative to the shore) if the lumberjack comes to rest relative to the second log.

Question

A lumberjack (mass = 104 kg) is standing at rest on one end of a floating log (mass = 234 kg) that is also at rest. The lumberjack runs to the other end of the log, attaining a velocity of +3.18 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. (b) Determine the velocity of the second log (again relative to the shore) if the lumberjack comes to rest relative to the second log.

I know that it has to do with conservation of momentum, but I have confused myself with the formula and have no idea how to start working it out.

Answer 1

To solve this problem, you can apply the principle of conservation of momentum. According to this principle, the total momentum before an event is equal to the total momentum after the event when there are no external forces acting on the system.

Let's consider the different stages of the problem:

1. Initial situation: The lumberjack and the log are at rest, so their total momentum is initially zero.

2. When the lumberjack runs to the other end of the log, two things happen:
- The lumberjack gains some velocity relative to the shore.
- The log acquires an equal and opposite velocity relative to the shore to maintain momentum conservation.

3. When the lumberjack hops onto the second log, two things happen:
- The lumberjack comes to rest relative to the second log.
- The second log acquires a velocity relative to the shore.

To determine the velocity of the second log, let's consider the total momentum before and after the lumberjack jumps. The total momentum before the event is zero since both the lumberjack and the log are at rest. The total momentum after the event is the momentum of the second log.

Using the equation for conservation of momentum:

Total initial momentum = Total final momentum

0 = (104 kg)(3.18 m/s) + (234 kg)(v)

Here, v represents the velocity of the second log relative to the shore. Rearranging the equation:

- (104 kg)(3.18 m/s) = (234 kg)(v)

Solve for v:

v = - (104 kg)(3.18 m/s) / (234 kg)

Calculating this expression gives:

v ≈ -1.42 m/s

Therefore, the velocity of the second log, relative to the shore, is approximately -1.42 m/s. The negative sign indicates that the second log is moving in the opposite direction to the lumberjack's initial velocity.