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A 10,000 kg railroad car is rolling at \(2.00 \mathrm{m} / \mathrm{s}\) when \(\mathrm{a} 4000 \mathrm{kg}\) load of gravel is suddenly dropped in. What is the car's speed just after the gravel is loaded?

Short Answer

Expert verified
The car's speed just after the gravel is loaded is \(1.43 m/s\).

Step by step solution

01

Identify Initial Momentum

The first step is to calculate the initial momentum of the car before the gravel is dropped in. Momentum (\(p\)) is a vector quantity which is the product of the mass and velocity of an object. Thus, the initial momentum of the car can be calculated using the formula \(p_{initial} = m_{car} \cdot v_{car}\), where \(m_{car}\) is the mass of the car and \(v_{car}\) is its velocity. Substituting the given values, we get \(p_{initial} = 10,000kg \cdot 2.00m/s = 20,000kgâ‹…m/s\).
02

Calculate Final Velocity

We can solve the equation in step 2 for the final velocity \(v_{final}\) of the car and the gravel. Rearranging the equation, we get \(v_{final}= \frac{p_{final}}{m_{car} + m_{gravel}}\). Substituting the known values leads to \(v_{final}= \frac{20,000kgâ‹…m/s}{10,000kg + 4000kg} = 1.43 m/s\).

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Key Concepts

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

Understanding Momentum
Momentum is a fundamental concept in physics, encapsulating the quantity of motion an object has. It's defined as the product of an object's mass and velocity and is given by the equation \( p = mv \). It's a vector quantity, meaning it has both magnitude and direction.

For instance, in the case of a moving railroad car, its momentum can be intuitively understood as the 'oomph' it carries due to its motion. The heavier the car or the faster it's moving, the more momentum it has. In mathematical terms, if the car has a mass of 10,000 kg and is traveling at 2.00 m/s, its momentum is calculated as \( p = 10,000 \text{ kg} \times 2.00 \text{ m/s} = 20,000 \text{ kg} \cdot \text{m/s} \).

Students often benefit from visualizing momentum with analogies, such as thinking of it as a moving truck; the more massive the truck or the faster it goes, the harder it is to stop, representing greater momentum. Additionally, understanding momentum helps in grasping other physics concepts like collisions, where the total momentum before and after the event remains constant, assuming no external forces act on the system.
Determining Final Velocity
Calculating final velocity is crucial in problems involving changes in mass or force. It's a step that requires careful attention to the conservation of momentum. When an object or a system of objects undergoes a change, the final velocity can be different from the initial velocity, particularly if mass is added or removed, or if an external force is applied.

To calculate the final velocity, the formula \( v_{final} = \frac{p_{final}}{m_{total}} \) is commonly used. In this context, \( p_{final} \) is the total momentum after the event has occurred, and \( m_{total} \) is the combined mass of the system at that point. As shown in the solution for the railroad car problem, after dropping in 4000 kg of gravel, the final velocity is inferred from the preservation of momentum and is found by dividing the initial momentum by the total mass of the car plus the gravel.
Solving Momentum Conservation Problems
When tackling momentum conservation problems, it's essential to recognize that the total momentum of a closed system remains constant if no external forces are at play. This principle is grounded in the conservation of momentum law, which underpins many physics problems.

To solve such problems, one typically follows a structured approach: First, calculate the initial momentum of the system, as done with the railroad car's momentum before the gravel is added. Then, apply the law of conservation of momentum, which states that initial momentum equals final momentum. It's crucial to remember that any added mass, like the gravel, must be factored into the total system mass. By applying the correct formulas and principles, these three essential steps can be followed to determine the final velocity or any other unknown in a momentum conservation problem.

It's often helpful for students to draw a diagram before starting calculations to visualize the system's before and after states. It not only aids in understanding but also minimizes the risk of oversight in complex problems.

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Most popular questions from this chapter

A billiard ball of mass \(0.28 \mathrm{kg}\) hits a second, identical ball at a speed of \(7.2 \mathrm{m} / \mathrm{s}\) and comes to rest as the second ball flies off. The collision takes \(250 \mu\) s. What is the average force on the first ball? The second ball?

A water pipe in a building delivers 1000 liters (with mass \(1000 \mathrm{kg}\) ) of water per second. The water is moving through the pipe at \(1.4 \mathrm{m} / \mathrm{s}\). The pipe has a \(90^{\circ}\) bend, and the pipe will require a supporting structure, called a thrust block, at the bend, as in Figure \(\mathrm{P} 9.70\). We can use the ideas of momentum and impulse to understand why. Each second, \(1000 \mathrm{kg}\) of water moving at \(v_{x}=1.4 \mathrm{m} / \mathrm{s}\) changes direction to move at \(v_{y}=1.4 \mathrm{m} / \mathrm{s}\) a. What are the magnitude and direction of the change in momentum of the \(1000 \mathrm{kg}\) of water? b. What are the magnitude and direction of the necessary impulse? c. This impulse takes place over 1.0 s. What is the necessary force?

A \(20 \mathrm{g}\) ball of clay traveling east at \(2.0 \mathrm{m} / \mathrm{s}\) collides with a \(30 \mathrm{g}\) ball of clay traveling \(30^{\circ}\) south of west at \(1.0 \mathrm{m} / \mathrm{s}\). What are the speed and direction of the resulting \(50 \mathrm{g}\) blob of clay?

Two ice skaters, with masses of \(75 \mathrm{kg}\) and \(55 \mathrm{kg},\) stand facing each other on a \(15-\mathrm{m}\) -wide frozen river. The skaters push off against each other, glide backward straight toward the river's edges, and reach the edges at exactly the same time. How far did the 75 kg skater glide?

\(\mathrm{A} 300 \mathrm{g}\) bird flying along at \(6.0 \mathrm{m} / \mathrm{s}\) sees a \(10 \mathrm{g}\) insect heading straight toward it with a speed of \(30 \mathrm{m} / \mathrm{s}\). The bird opens its mouth wide and enjoys a nice lunch. What is the bird's speed immediately after swallowing?

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