A lumberjack (mass = 98 kg) is standing at rest on one end of a floating log (mass = 220 kg) that is also at rest. The lumberjack runs to the other end of the log, attaining a velocity of +3.5 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.
Determine the velocity of the second log if the lumberjack comes to rest on it.
To determine the velocity of the second log when the lumberjack comes to rest on it, we can use the concept of conservation of momentum. In an isolated system (where no external forces act), the total momentum before an event is equal to the total momentum after the event.
Let's denote the initial velocity of the log with mass 220 kg as v1 and the final velocity of the log with mass 98 kg as v2. We need to find v2.
Step 1: Calculate the initial momentum
The initial momentum of the system is the sum of the momentum of the lumberjack and the momentum of the first log:
Momentum_initial = momentum_lumberjack_initial + momentum_log1_initial
The momentum of an object is given by the product of its mass and velocity. Since both the lumberjack and the first log are initially at rest, their initial momenta are zero:
Momentum_initial = 0 + 0 = 0
Step 2: Calculate the final momentum
The final momentum of the system is the sum of the momentum of the lumberjack and the momentum of the second log:
Momentum_final = momentum_lumberjack_final + momentum_log2_final
Since the lumberjack comes to rest on the second log, the final momentum of the lumberjack is zero:
Momentum_final = 0 + momentum_log2_final
Step 3: Apply conservation of momentum
According to the conservation of momentum, the total momentum before the event is equal to the total momentum after the event. Therefore, we have:
Momentum_initial = Momentum_final
0 = 0 + momentum_log2_final
momentum_log2_final = 0
From this equation, we can conclude that the final momentum of the second log is zero. In other words, the second log comes to rest when the lumberjack lands on it. Therefore, the velocity of the second log is zero.
To solve this problem, we can use the principle of conservation of momentum.
The initial momentum of the system (lumberjack + log) is zero since both are at rest. The final momentum of the system should also be zero when the lumberjack comes to rest on the second log.
Let's label the initial log as Log 1, the lumberjack as Lumberjack, and the second log as Log 2.
1. Determine the momentum of Log 1 and Lumberjack before the lumberjack starts running:
- The momentum of Log 1 is 0 kg*m/s since it is at rest.
- The momentum of Lumberjack is 0 kg*m/s since they are also at rest.
2. Determine the momentum of Log 1 and Lumberjack when the lumberjack reaches the other end of the log:
- The momentum of Log 1 remains 0 kg*m/s since there is no external force acting on the log once the lumberjack starts moving.
- The momentum of Lumberjack can be calculated using the formula: momentum = mass * velocity.
Given that the mass of the lumberjack is 98 kg and the velocity is +3.5 m/s (relative to the shore), the momentum of the lumberjack is (98 kg) * (3.5 m/s) = 343 kg*m/s.
3. Determine the momentum of Log 2 and the lumberjack when the lumberjack comes to rest on the second log:
- The total momentum of the system should be zero when the lumberjack comes to rest. So, the momentum of Log 2 and Lumberjack combined should be zero.
- Let's label the velocity of Log 2 as v2. The momentum of Log 2 can be calculated using the formula: momentum = mass * velocity.
Therefore, the momentum of Log 2 is (220 kg) * (v2).
- The momentum of Lumberjack is zero since they come to rest on Log 2.
Using the principle of conservation of momentum:
Initial momentum = Final momentum
(0 kg*m/s) + (0 kg*m/s) = (220 kg) * (v2) + 0 kg*m/s
Therefore, we have:
0 = (220 kg) * (v2)
To have zero momentum, the velocity of Log 2 must be zero.
Hence, the velocity of the second log when the lumberjack comes to rest on it is zero m/s.