/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 106 A \(2140 \mathrm{~kg}\) railroad... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(2140 \mathrm{~kg}\) railroad flatcar, which can move with negligible friction, is motionless next to a platform. A \(242 \mathrm{~kg}\) sumo wrestler runs at \(5.3 \mathrm{~m} / \mathrm{s}\) along the platform (parallel to the track) and then jumps onto the flatcar. What is the speed of the flatcar if he then (a) stands on it, \((\mathrm{b})\) runs at \(5.3 \mathrm{~m} / \mathrm{s}\) relative to it in his original direction, and (c) turns and runs at \(5.3 \mathrm{~m} / \mathrm{s}\) relative to the flatcar opposite his original direction?

Short Answer

Expert verified
(a) 0.538 m/s, (b) 0.287 m/s, (c) 0.790 m/s.

Step by step solution

01

Understanding the Conservation of Momentum

The law of conservation of momentum states that the total momentum of a system remains constant if no external forces are acting on it. Here, the sumo wrestler and the flatcar are the system, and the track is frictionless. Thus, the total momentum before and after the wrestler jumps onto the flatcar remains the same.
02

Calculate Initial Momentum

Initially, the flatcar is motionless, so its momentum is zero. The wrestler has momentum given by the product of his mass and velocity: \[ p_{wrestler} = m_{wrestler} \cdot v_{wrestler} = 242 \times 5.3 \ = 1282.6\, \text{kg}\cdot\text{m/s}. \]
03

Calculate Final Momentum when Wrestler Stands on Flatcar

When the wrestler stands still on the flatcar, the combined mass of the flatcar and wrestler is \(2140 + 242 = 2382\, \text{kg}\). Using momentum conservation, \[ 1282.6 = 2382 \cdot v_{final} \] Solving for \(v_{final}\) gives \[ v_{final} = \frac{1282.6}{2382} \approx 0.538\, \text{m/s}. \]
04

Calculate Final Momentum when Wrestler Runs on Flatcar

If the wrestler runs at 5.3 m/s relative to the flatcar, his absolute speed is the speed of flatcar plus 5.3 m/s. Conservation of momentum gives: \[ 1282.6 = 2140 \cdot v_{flatcar} + 242(v_{flatcar} + 5.3). \] Simplifying and solving for \(v_{flatcar}\), we find: \[ v_{flatcar} = \frac{1282.6 - 242 \times 5.3}{2140 + 242} \approx 0.287\, \text{m/s}. \]
05

Calculate Final Momentum when Wrestler Runs Opposite to his Original Direction

If the wrestler runs at 5.3 m/s opposite to his original direction, his absolute speed becomes \(v_{flatcar} - 5.3\). Conservation of momentum gives: \[ 1282.6 = 2140 \cdot v_{flatcar} + 242(v_{flatcar} - 5.3). \] Solving for \(v_{flatcar}\), we obtain: \[ v_{flatcar} = \frac{1282.6 + 242 \times 5.3}{2140 + 242} \approx 0.790\, \text{m/s}. \]

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Key Concepts

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

Physics Problem Solving
Physics problem solving is a strategic approach that involves understanding a problem, identifying the principles involved, and applying mathematical calculations to find a solution. Here, we deal with a momentum conservation problem involving a sumo wrestler and a flatcar on a frictionless track.
The first step in tackling such problems is to identify the system, which in this case is the sumo wrestler and the flatcar. Since there are no external forces acting, we apply the law of conservation of momentum. This law states that the total momentum of a closed system remains unchanged unless acted upon by external forces.
Next, calculations are made by considering the initial momentum of the wrestler, which is the product of his mass and speed. As the flatcar is initially at rest, its initial momentum is zero. By summing the initial momentum of both objects, we establish a total system momentum, which will help in calculating the final momentum and eventual speed scenarios of the flatcar in different conditions.
It's important to break down the given information and work systematically through the problem to avoid unnecessary errors.
Kinematics
Kinematics, the branch of mechanics concerning objects in motion, is essential for understanding the movement of the wrestler and flatcar in this problem. We need to consider both velocity and mass to predict the subsequent motion after the wrestler jumps onto the flatcar.
Each scenario provides different kinematic effects. Initially, the wrestler moves with a specific velocity, adding his own momentum to the system. When he stands still on the flatcar, the new velocity of the combined mass can be derived, showing how the system slows due to the increased mass with unchanged momentum.
In other scenarios, where the wrestler runs on or against the direction of the flatcar, his relative speed to the flatcar modifies the overall velocity calculations. The combination of these kinematic principles, alongside momentum conservation, allows us to predict the flatcar's speed accurately in each situation.
Newton's Laws of Motion
Newton's laws of motion provide foundational concepts that help understand why the wrestler and flatcar behave as they do. Of particular importance here is Newton's third law, which tells us that for every action, there is an equal and opposite reaction. This means that as the wrestler exerts force by jumping onto the flatcar, he imparts momentum onto it.
Newton’s second law, which relates force, mass, and acceleration, helps us delineate how the wrestler’s actions affect the flatcar. By understanding these forces, we can deduce changes in velocity. The action of the wrestler changes the overall motion state of both him and the flatcar, effectively a direct application of his weight and velocity impacting the flatcar's motion.
These laws work together to explain the equilibrium of motion and forces, illustrating how different velocities evolve in the specified situations once the initial action of the wrestler jumping or running takes place.

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