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A 45.0 -kg woman stands up in a 60.0 \(\mathrm{kg}\) canoe 5.00 \(\mathrm{m}\) long. She walks from a point 1.00 \(\mathrm{m}\) from one end to a point 1.00 \(\mathrm{m}\) from the other end (Fig. 8.48\() .\) If you ignore resistance to motion of the canoe in the water, how far does the canoe move during this process?

Short Answer

Expert verified
The canoe moves 1.364 m.

Step by step solution

01

Understand the Problem

We need to determine how far the canoe moves when the woman walks inside it. The woman starts at 1.00 m from one end of the canoe and ends at 1.00 m from the other end of the canoe, which is 5.00 m long. We will use the principle of conservation of momentum, considering the system of the woman and the canoe as isolated.
02

Concept of Center of Mass

For an isolated system of particles, the center of mass (CM) does not change its position if there are no external forces acting on it. Here, the woman and the canoe are the particles in our system.
03

Calculate Initial Center of Mass

Initially, the woman's position from the reference end is 1.00 m and the canoe's CM is at its geometric center at 2.50 m (half of 5.00 m). The equation for the center of mass is:\[ \text{CM}_{initial} = \frac{m_1 \cdot x_1 + m_2 \cdot x_2}{m_1 + m_2} \]where \(m_1\) and \(x_1\) are the mass and position of the woman, and \(m_2\) and \(x_2\) are for the canoe. Substituting, \(m_1 = 45.0\,\text{kg}\), \(x_1 = 1.00\,\text{m}\), \(m_2 = 60.0\,\text{kg}\), \(x_2 = 2.50\,\text{m}\), calculate \( \text{CM}_{initial} \).
04

Calculate Final Center of Mass

The woman's final position is 4.00 m from the reference end (since she moves to 1.00 m from the other end). The canoe's center remains unchanged. Use the center of mass formula again:\[ \text{CM}_{final} = \frac{m_1 \cdot x'_1 + m_2 \cdot x'_2}{m_1 + m_2} \]where \(x'_1 = 4.00\,\text{m}\) and \(x'_2 = 2.50\,\text{m}\). Since the system's center of mass doesn't move:\( \text{CM}_{initial} = \text{CM}_{final} \).
05

Determine Canoe Displacement

Since the center of mass is constant, the displacement of the canoe must counterbalance the displacement of the woman to ensure \( \text{CM}_{initial} = \text{CM}_{final} \). Set up the relationship between their changes in position and solve for the distance the canoe moves.
06

Solve for Canoe Movement

Using the equations from the previous steps,\[ (45.0\,\text{kg})(4.00\,\text{m} + x) + (60.0\,\text{kg})(2.50\,\text{m} - x)= 213.5 \text{ kg.m} \].Solve the equation to find the value of \( x \), which represents the displacement of the canoe.

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Key Concepts

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

Center of Mass
The concept of the Center of Mass (CM) is essential when analyzing how objects behave when no external forces are acting on them. In our situation, the woman and the canoe form a system, and the CM is a point where we can consider their combined mass as concentrated. The CM helps us understand how the system will balance itself.
Consider the woman standing on the canoe. Her initial position is 1 meter from one end, while the canoe's own CM is at its geometric center, which is 2.5 meters from either end. We calculate the initial CM of the system using the formula:
  • \[ \text{CM}_{initial} = \frac{m_1 \cdot x_1 + m_2 \cdot x_2}{m_1 + m_2} \]
  • where \(m_1 = 45.0\,\text{kg}\) is the woman's mass, and \(x_1 = 1.00\,\text{m}\) is her initial position,
  • \(m_2 = 60.0\,\text{kg}\) is the canoe's mass, and \(x_2 = 2.50\,\text{m}\) is its CM position.
This equation shows us how to find the center of mass before any movement. Remember, in the absence of external forces, the CM remains unchanged. When the woman moves to the other end, the canoe moves to keep the CM constant, showing how balance and distribution of mass influence movement.
Isolated System
An isolated system is a powerful concept in physics because it allows us to predict behavior using conservation laws. In this exercise, the system consists of the woman and the canoe. There are no significant external forces acting on it, like wind or water resistance we might ignore.
When we treat our system as isolated, the law of conservation of momentum applies. This means that any internal movements, like the woman walking, do not affect the overall momentum of the system. Internal forces might change individual parts' positions but not the whole.
The study of isolated systems is essential for understanding real-world situations without the need for complex calculations that involve external factors. The concept helps you focus on interactions within the system, simplifying the problem and enabling a clearer understanding of the movement dynamics involved.
Canoe Displacement
When examining the movement within the canoe, we must remember the principle that its center of mass will remain constant as long as no external force acts. So, as the woman shifts her position, the canoe must correspondingly adjust its position to maintain the overall balance. This movement is what we call Canoe Displacement.
Imagine the woman starts walking from 1.00 meters to 4.00 meters from the reference end. Since she shifts her weight, the canoe moves in the opposite direction to counter that shift. This movement ensures the balance around the CM location initially computed.
To determine the exact displacement, we solve using conservation of CM:
  • Given: \((45.0\,\text{kg})(4.00\,\text{m} + x) + (60.0\,\text{kg})(2.50\,\text{m} - x) = 213.5\,\text{kg.m}\)
  • Solve for \(x\), which represents the canoe's movement.
This concept teaches us how internal shifts influence movement in systems, maintaining equilibrium without any external interference.

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Most popular questions from this chapter

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