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Three coupled railroad cars roll along and couple with a fourth car, which is initially at rest. These four cars roll along and couple with a fifth car initially at rest. This process continues until the speed of the final collection of railroad cars is one-fifth the speed of the initial three railroad cars. All the cars are identical. Ignoring friction, how many cars are in the final collection?

Short Answer

Expert verified
There are 15 cars in the final collection.

Step by step solution

01

Define the Variables

Let's start by defining the variables. Let the mass of each car be \(m\) and the initial speed of the three moving cars be \(v_i\). The speed of the final collection of railcars is given as \(v_f = \frac{1}{5}v_i\). The initial three cars have a combined mass of \(3m\) and an initial momentum of \(p_i = 3mv_i\).
02

Understand the Conservation of Momentum

Since there are no external forces acting on the system , momentum is conserved. The momentum of the initial moving cars before they collide with the fourth car is equal to the total momentum after they couple. This means:\[ p_i = (3m + m)v_1 \] \where \(v_1\) is the speed after coupling with the fourth car.
03

Calculate the Speed After the First Collision

Using the conservation of momentum from Step 2, plug in the values:\[ 3mv_i = (3m + m)v_1 \] This simplifies to:\[ v_1 = \frac{3}{4}v_i \]So, after the first collision, the speed of the cars is \(\frac{3}{4}v_i\).
04

Continue the Conservation of Momentum for Each Coupling

Each time the rolling cars couple with an additional stationary car, apply the conservation of momentum again.For example, after coupling with the fifth car:\[ (4m)(\frac{3}{4}v_i) = (4m + m)v_2 \]Solving for \(v_2\), we have:\[ v_2 = \frac{3}{5}v_i \]Repeating this process, eventually you need:\[ v_n = \frac{1}{5}v_i \]
05

Generalize the Equation

Generalizing for n cars, each time an n-th car is added the momentum equation becomes:\[ (mv_i)(3) = (n \times m)v_n \]Solving for \(v_n\) simplifies to:\[ v_n = \frac{3}{n}v_i \] We need \(\frac{3}{n} = \frac{1}{5} \), thus \(n = 15\).
06

Verify the Number of Cars

Each step has shown the gradual reduction of speed, due to the incremental buildup of mass. Cross check, when \(n = 15\), the final velocity equation \(v_n = \frac{3}{15}v_i\) satisfies the given condition of \(v_f = \frac{1}{5}v_i\). This confirms our solution that the final collection comprises 15 cars.

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Key Concepts

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

Physics Problem Solving
To tackle physics problems effectively, it's crucial to follow a structured approach. Start by clearly defining all known variables and identifying what you're solving for. This gives you a solid understanding of what the problem entails.
In our railroad cars problem, defining variables like the mass of each car and initial speeds helped lay the groundwork. Breaking the problem down into manageable steps ensures nothing is overlooked. This technique not only simplifies complex sequences but also empowers you to connect physical principles to the problem at hand.
Moreover, always verify your solution. This means checking if your final result fits logically with the problem context. Such thoroughness ensures accuracy and deepens your understanding of the physics concepts involved.
Railroad Cars Collision
Collisions in physics, particularly those like the railroad cars collision, are fascinating because they illustrate real-world applications of momentum. When two or more objects collide and stick together, it's often referred to as a perfectly inelastic collision.
In this railroad cars problem, initially, three cars moving along a track collide and couple with a fourth stationary car. This first collision illustrates momentum transfer during a perfectly inelastic collision. Each additional coupling follows this pattern, gradually increasing the total mass while decreasing the speed.
These couplings are purely theoretical, ignoring real-world variables such as friction or energy loss. However, concepts like these guide our understanding of how momentum keeps the system in equilibrium during such interactions.
Momentum Conservation Equation
One of the most pivotal equations in collisions is the momentum conservation equation. In our exercise, this principle was crucial in determining the final scenario of the coupled railroad cars.
Momentum is a vector quantity, calculated by multiplying an object's mass and velocity. The conservation principle states that, in an isolated system with no external forces, the total momentum before an event, like a collision, is equal to the total momentum after.
In equation form, the equation is stated as \[ p_{ ext{initial}} = p_{ ext{final}} \]. For the railroad cars, this translated to repeatedly combining the moving system's momentum with that of a stationary car's mass. The problem's step-by-step solution highlighted this through each coupling, gradually reaching a velocity that was one-fifth of the original speed. Understanding this principle is essential for solving many physics problems involving moving bodies.

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Most popular questions from this chapter

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