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Chapter 7: Problem 19

ssm A lumberjack (mass \(=98 \mathrm{kg} )\) is standing at rest on one end of a floating log (mass \(=230 \mathrm{kg}\) ) that is also at rest. The lumberjack runs to the other end of the log, attaining a velocity of \(+3.6 \mathrm{m} / \mathrm{s}\) 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 just before the lumberjack jumps off? (b) Determine the velocity of the second log if the lumberjack comes to rest on it.

Short Answer

Expert verified
(a) The first log's velocity is approximately \(-1.53 \text{ m/s}\). (b) The second log's velocity is \(1.53 \text{ m/s}\).

Step by step solution

01

Apply Conservation of Momentum for First Log

Using the principle of conservation of momentum, we know that the total momentum before and after the lumberjack runs should be the same. Before running, the momentum is 0 because both the lumberjack and the log are at rest. After running, the equation is set up as follows: \[ m L v_L + m J v_J = 0 \] where \( m_J = 98 \text{ kg} \) is the mass of the lumberjack, \( v_J = +3.6 \text{ m/s} \) is the velocity of the lumberjack relative to the shore, \( m_L = 230 \text{ kg} \) is the mass of the log, and \( v_L \) is the velocity of the log we need to find.
02

Solve for the First Log's Velocity

Rearranging the equation from Step 1 for the velocity of the first log:\[ v_L = -\frac{m_J v_J}{m_L} \]Substituting the known values:\[ v_L = -\frac{98 \text{ kg} \times 3.6 \text{ m/s}}{230 \text{ kg}} \approx -1.53 \text{ m/s} \]This means the first log moves with a velocity of \(-1.53 \text{ m/s}\) just before the lumberjack jumps off.
03

Analyze the Lumberjack's Movement to the Second Log

When the lumberjack jumps off the first log and lands on the second log, the relative velocity situation changes. Initially, the second log is at rest.
04

Apply Conservation of Momentum for the Second Log

For the lumberjack and the second log system, apply conservation of momentum. Initially, the total momentum of the system is the momentum of the lumberjack alone, since the second log is at rest:\[ m_J v_J = m_J v'_{J} + m_L v'_{L} \]Here, \( v'_J = 0 \) (velocity of lumberjack after coming to rest on the second log) and \( v'_L \) is the velocity of the second log we need to find.
05

Solve for the Second Log's Velocity

Substitute values into the momentum equation for the second log:\[ 98 \text{ kg} \times 3.6 \text{ m/s} = 0 + 230 \text{ kg} \times v'_L \]Solve for \( v'_L \):\[ v'_L = \frac{98 \text{ kg} \times 3.6 \text{ m/s}}{230 \text{ kg}} \approx 1.53 \text{ m/s} \]Therefore, the second log moves with a velocity of \(1.53 \text{ m/s}\) in the direction the lumberjack was initially moving.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Lumberjack Physics Problem
In this exercise, we explore a classic example of momentum conservation using a lumberjack and logs on water. Imagine a thrilling scenario where a lumberjack needs to jump between two floating logs. This problem not only challenges one's physics understanding but also shows the elegance of physics in action. Let's break it down!

The situation begins with a lumberjack, initially at rest, standing on a log that is also at rest. The lumberjack's mass is 98 kg, and the log's mass is significantly heavier at 230 kg. As the lumberjack runs across the log to reach the other end, two primary actions regard momentum conservation. First, the momentum of the lumberjack affects the log, causing them to move in opposite directions once the lumberjack runs. Second, when the lumberjack hops onto another identical log, the relative motion needed for understanding momentum interactions is observed again.

Through this problem, students not only engage with theoretical physics but also relate these ideas to real-world scenarios involving motion and force. The lack of friction resistance makes it an ideal situation to perfectly observe the conservation laws at work.
Physics Problem Solving
Breaking down a physics problem into manageable steps is key. The lumberjack exercise is a perfect application of step-by-step problem-solving skills, primarily focusing on conservation laws.

First, one must identify the system and the forces acting upon it. Here, the system includes the lumberjack and the logs. Using the conservation of momentum principle, we calculate that the total momentum should remain the same before and after the lumberjack runs or jumps. This conservation principle serves as a constant guide throughout solving the problem.

Each step is essential:
  • Start by noting initial states: the initial rest state of the lumberjack and logs.
  • Apply momentum formulas to both logs independently and solve for unknown velocities.
  • Understand the relative motion, as the lumberjack's movement relates to the log, even affecting the direction of movement.
By working through these steps methodically, students learn how problem-solving in physics bridges abstract concepts with tangible results.
Log Velocity Calculation
Calculating the velocity of the logs involves applying mathematical equations stemming from conservation laws. Let's see how each velocity is determined.

When the lumberjack runs across the first log, both start from rest. Using the equation derived from conservation of momentum, \[ m_L v_L + m_J v_J = 0 \]we solve for the first log's velocity. With the lumberjack's speed at +3.6 m/s and weight differences, the result tells us the first log moves at -1.53 m/s in the opposite direction, as per the principle that momentum is maintained by counterbalancing movements.

Next, the lumberjack jumps onto the second log. Before the jump, the log was at rest. We apply the conservation principle again to find the second log's new velocity: \[ m_J v_J = m_J v'_J + m_L v'_L \]While the lumberjack comes to rest on the second log, the log adopts a velocity of 1.53 m/s, indicating its movement in the lumberjack's direction just before he stopped.

Understanding how to manipulate these equations equips students to tackle similar real-world applications of physics.

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