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Two swimmers \(A\) and \(B\), of weight 190 lb and 125 lb, respectively, are at diagonally opposite corners of a floating raft when they realize that the raft has broken away from its anchor. Swimmer \(A\) immediately starts walking toward \(B\) at a speed of \(2 \mathrm{ft} / \mathrm{s}\) relative to the raft. Knowing that the raft weighs \(300 \mathrm{lb}\), determine (a) the speed of the raft if \(B\) does not move, \((b)\) the speed with which \(B\) must walk toward \(A\) if the raft is not to move.

Short Answer

Expert verified
(a) Raft's speed is 1.27 ft/s. (b) B must walk at 3.04 ft/s towards A.

Step by step solution

01

Understanding the Problem

We have a system consisting of two swimmers and a raft. The principle of conservation of momentum applies here because there are no external horizontal forces acting on the system. Therefore, the center of mass of the entire system should not move (assuming it started stationary).
02

Define the System Mass and Initial Conditions

The total mass of the system is the sum of the weights of swimmer A, swimmer B, and the raft. Convert weights to mass by dividing by gravitational acceleration, but since g will be the same for all components, we don't need to put the actual mass but focus on weight ratios/context. Total weight is: 190 lb (A) + 125 lb (B) + 300 lb (raft) = 615 lb.
03

Apply Conservation of Momentum for Part (a)

For part (a), swimmer B remains stationary. In the horizontal direction, according to conservation of momentum, swimmer A walking towards B means the raft must move in the opposite direction to keep the center of mass stationary. Use the formula: \[ m_A v_A + m_{ ext{raft}} v_{ ext{raft}} = 0 \] Substituting known quantities, for swimmer A, \(m_A = 190/615\) and \(v_A = 2 \mathrm{ft/s}\), solve for \(v_{ ext{raft}}\).
04

Calculate Raft's Speed for Part (a)

- Substitute into conservation equation:\[ (\frac{190}{615}) \,\cdot\, 2 + \frac{300}{615} \,\cdot\, v_{ ext{raft}} = 0 \]- Rearrange to find \(v_{ ext{raft}}\): \[ v_{ ext{raft}} = - \frac{190 \,\cdot\, 2}{300} \]- Simplify to get the raft's speed:\[ v_{ ext{raft}} = -1.27 \mathrm{ft/s} \]
05

Apply Conservation of Momentum for Part (b)

Now consider Part (b): Swimmer B moves towards A, and the raft should not move, implying the system's center of mass does not change. Therefore, resolve conservation equation as:\[ m_A v_A + m_B v_B = 0 \] Here, swimmers A and B's velocities should counterbalance each other. Utilizing weights for mass: \(v_B\) for B will be solved knowing \(v_A = 2 \mathrm{ft/s}\).
06

Calculate the Required Speed of Swimmer B for Part (b)

- Set up momentum equation:\[ \frac{190}{615} \,\cdot\, 2 + \frac{125}{615} \,\cdot\, v_B = 0 \]- Solve for \(v_B\):\[ v_B = - \frac{190 \,\cdot\, 2}{125} \]- Simplify to determine the speed of B:\[ v_B = -3.04 \mathrm{ft/s} \]

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Key Concepts

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

Center of Mass
In physics, the center of mass is a crucial concept that helps us understand how objects behave when forces are applied. Imagine you have several people standing on a raft. Each person, along with the raft, contributes to the overall center of mass of the system. The center of mass is the point where the entire mass of the system can be considered to be concentrated.

For our swimmers and raft, we start by identifying the total weight, which includes both swimmers and the raft. This total weight is 615 lb. Although these forces are acting downward due to gravity, we focus on the horizontal movement where these weights create a balanced system in the absence of external horizontal forces.

When swimmer A moves towards B, the raft will move in the opposite direction if swimmer B remains still. This keeps the horizontal center of mass stationary, showing how the center of mass stays balanced unless acted upon by an external force.
System Dynamics
System dynamics in physics involves understanding how different objects within a system interact with each other. It’s all about how movements within the system affect the entire system’s behavior.

In the exercise with swimmers A and B and the raft, each participant contributes to the momentum of the system. Momentum is preserved because no external forces are affecting our horizontal plane. This means that if part of the system (like swimmer A) moves, other parts (like the raft) must adjust to maintain equilibrium.
  • The movement of swimmer A towards B causes the raft to move in the opposite direction if swimmer B does not move.
  • Conversely, if swimmer B moves towards A, the system should adjust such that the raft remains stationary.
This interplay showcases the interconnectedness of system elements. Understanding these dynamics helps predict how changes in part of the system affect the rest of it.
Physics Problem Solving
Physics problem solving often requires systematic application of physical laws. The key here is the law of conservation of momentum, stating that if no external forces act on a system, the total momentum remains constant.

To solve part (a) of our problem, where swimmer A moves towards B, we set swimmer B stationary. The challenge is calculating how the raft must move to keep the momentum conserved. Using the formula: \[m_A v_A + m_{\text{raft}} v_{\text{raft}} = 0\]We substitute the given values, solve, and find the raft's speed.

For part (b), swimmer B moves towards A, and we need to maintain the raft's position stationary. The setup becomes: \[ m_A v_A + m_B v_B = 0 \]Solving this reveals how fast swimmer B needs to move to counteract swimmer A's movement.

By breaking the problem into manageable steps, using known equations, and inserting given values, complex problems can be tackled elegantly. This process exemplifies successful physics problem-solving by analyzing systems, applying principles, and using algebraic manipulation.

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Most popular questions from this chapter

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