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Chapter 4: Problem 86

A coal car with an empty mass of \(25 \mathrm{Mg}\) is moving freely with a speed of \(1.2 \mathrm{m} / \mathrm{s}\) under a hopper which opens and releases coal into the moving car at the constant rate of \(4 \mathrm{Mg}\) per second. Determine the distance \(x\) moved by the car during the time that \(32 \mathrm{Mg}\) of coal are deposited in the car. Neglect any frictional resistance to rolling along the horizontal track.

Short Answer

Expert verified
The coal car moves approximately 6.91 meters.

Step by step solution

01

Calculate time taken for coal to fill

We need to find the time, \( t \), it takes for 32 Mg of coal to fill the car. Given that the constant rate is 4 Mg/s, use the formula: \[ t = \frac{\text{Total coal mass}}{\text{Rate}} = \frac{32\, \text{Mg}}{4\, \text{Mg/s}} = 8\, \text{s}. \] So, it takes 8 seconds to fill the coal.
02

Determine initial and final mass of the car

The initial mass of the car (empty) is 25 Mg. After receiving 32 Mg of coal, the final mass of the car becomes: \[ \text{Final mass} = 25\, \text{Mg} + 32\, \text{Mg} = 57\, \text{Mg}. \] Convert these masses to kilograms by multiplying by 1000: \( m_0 = 25000 \, \text{kg}, m_1 = 57000 \, \text{kg}. \)
03

Use conservation of momentum to find final velocity

According to the conservation of momentum principle, the initial momentum equals the final momentum: \( m_0 v_0 = m_1 v_1. \) Where \( v_0 \) is the initial velocity. Substitute known values: \[ 25000 \times 1.2 = 57000 \times v_1. \] Solve for \( v_1 \): \[ v_1 = \frac{25000 \times 1.2}{57000} = 0.5263 \text{ m/s}. \] The final velocity of the car is 0.5263 m/s.
04

Calculate the average velocity

The average velocity \( v_{avg} \) during the coal loading is the arithmetic mean of the initial and final velocities: \[ v_{avg} = \frac{v_0 + v_1}{2} = \frac{1.2 + 0.5263}{2} = 0.86315 \text{ m/s}. \]
05

Determine the distance moved by the car

The distance \( x \) moved by the car is given by \( x = v_{avg} \times t \). Substituting the known values: \[ x = 0.86315 \times 8 = 6.9052 \text{ m}. \] Therefore, the car moves approximately 6.91 meters.

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Key Concepts

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

Dynamics
Dynamics is a branch of mechanics that explores forces and their effects on motion. In our problem, dynamics help us understand how the speed of the coal car changes when it collects coal. By looking at different forces like gravity and the force of the moving coal, dynamics tell us how these contribute to the car鈥檚 movement. With the coal loading, the external forces are minimal since we neglect friction. This simplifies the analysis, keeping our focus on the car's changing momentum. This simplification makes it easier to apply conservation of momentum principles in this exercise.
Mechanics
Mechanics is the umbrella term for the study of motion and the forces that cause it. It encompasses both kinematics鈥攖he description of motion鈥攁nd dynamics鈥攖he causes of motion. In this exercise, we primarily focus on a mechanics principle called **momentum conservation**. The coal car, due to the incoming coal, has changing mass, but as there are no external horizontal forces, mechanics tells us the system鈥檚 momentum remains constant. By accounting for the initial mass and velocity, we can accurately deduce how the coal affects the car's velocity and the subsequent distance it travels along the track.
Velocity Calculation
Velocity is a measure of speed with a direction. Calculating it after the coal is loaded involves understanding how the mass and speed of the car change. Initially, the car's velocity is 1.2 m/s. As coal fills the car, its mass increases, causing an adjustment in velocity which we determine by applying the conservation of momentum.
  • Initial mass (\( m_0 \)): 25 Mg
  • Final mass (\( m_1 \)): 57 Mg
  • Initial velocity (\( v_0 \)): 1.2 m/s
Evaluating the final velocity (\( v_1 \)) using momentum conservation reveals it as approximately 0.5263 m/s. Understanding this shift helps in the calculation of average velocity and consequently, the distance moved.
Conservation Principles
The conservation principles, like conservation of momentum, are foundational to understanding mechanics problems. In momentum conservation, the total system momentum remains constant when there are no external forces. For the coal car exercise, the initial and final momentum (\( m_0 v_0 = m_1 v_1 \)) must be equal, as external forces are neglected. This principle not only aids in finding the final velocity but ensures the analysis upholds natural rules that govern motion. With both momentum and energy conservation principles forming the backbone of mechanics, they provide a reliable framework for solving various problems on motion and impact in physics.

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Most popular questions from this chapter

A small rocket-propelled vehicle weighs \(125 \mathrm{lb}\), in cluding 20 lb of fuel. Fuel is burned at the constant rate of 2 lb/sec with an exhaust velocity relative to the nozzle of \(400 \mathrm{ft} / \mathrm{sec}\). Upon ignition the vehicle is released from rest on the \(10^{\circ}\) incline. Calculate the maximum velocity \(v\) reached by the vehicle. Neglect all friction.

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