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A railroad car of mass \(m\) and initial speed \(v\) collides with and becomes coupled with the two identical cars. Compute the final speed \(v^{\prime}\) of the group of three cars and the fractional loss \(n\) of energy if \((a)\) the initial separation distance \(d=0\) (that is, the two stationary cars are initially coupled together with no slack in the coupling) and \((b)\) the distance \(d \neq 0\) so that the cars are uncoupled and slightly separated. Neglect rolling resistance.

Short Answer

Expert verified
The final speed of the cars is \( v^{\prime} = \frac{v}{3} \), with a fractional energy loss of \( \frac{2}{3} \).

Step by step solution

01

Analyze the System Before Collision

We have a railroad car with mass \( m \) moving with an initial speed \( v \). It collides with two identical, stationary cars which are initially at rest (in part (a)). The two stationary cars each also have a mass \( m \).
02

Apply Conservation of Momentum

According to the conservation of momentum, the initial momentum must equal the final momentum of the system. The initial momentum is \( m \times v \). After the collision, the three cars move together with mass \( 3m \) and speed \( v^{\prime} \). Therefore, we write: \[ m \cdot v = 3m \cdot v^{\prime} \]
03

Solve for Final Speed

From the momentum equation in Step 2, we solve for \( v^{\prime} \) by dividing both sides by \( 3m \):\[ v^{\prime} = \frac{v}{3} \]
04

Calculate Initial and Final Kinetic Energies

The initial kinetic energy is only that of the moving car: \[ KE_{initial} = \frac{1}{2} m v^2 \]The final kinetic energy, for the combined three cars moving at \( v^{\prime} \), is: \[ KE_{final} = \frac{1}{2} (3m) (v^{\prime})^2 \]Substituting \( v^{\prime} = \frac{v}{3} \), we get:\[ KE_{final} = \frac{1}{2} (3m) \left( \frac{v}{3} \right)^2 = \frac{1}{2} (3m) \cdot \frac{v^2}{9} = \frac{1}{6} m v^2 \]
05

Compute Fractional Loss of Energy

The fractional loss of energy \( n \) is given by:\[ n = \frac{KE_{initial} - KE_{final}}{KE_{initial}} \]Substituting the kinetic energies, we have:\[ n = \frac{\frac{1}{2} m v^2 - \frac{1}{6} m v^2}{\frac{1}{2} m v^2} = \frac{3 - 1}{3} = \frac{2}{3} \]
06

Consider the Second Situation (Distance \(d \neq 0\))

In this scenario, the only difference is that the car impacts each of the stationary cars one after the other. However, momentum is still conserved in the same way, and the final speed of the three cars is \( v^{\prime} = \frac{v}{3} \), just like before.There is no additional energy lost solely due to the initial separation, so the fractional energy loss remains the same.

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Key Concepts

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

Conservation of Momentum
In collision mechanics, one of the most fundamental principles is the conservation of momentum. This principle states that the total momentum of a closed system remains constant before and after an event, like a collision. For students imagining a high-speed train coming to meet some stationary cars, it's crucial to understand how momentum works. Given that momentum is a product of mass and velocity, when an object moves, it carries momentum. In the exercise, we had a single moving railroad car with mass \( m \) and velocity \( v \). Once it collided and coupled with two stationary cars of identical mass, the system of three cars now needed to share this momentum evenly.
  • Before the collision, the total momentum was \( m \times v \).
  • After the collision, since the three cars move together, the total momentum needs to equate to \( 3m \times v' \), ensuring balance and conservation.
  • From the principle of conservation of momentum, we derived the equation: \[ m \cdot v = 3m \cdot v' \]
  • Solving for \( v' \), the final velocity of the system becomes \( \frac{v}{3} \).
Understanding this helps in visualizing how momentum transfers within a system and remains consistent. It's not lost or gained, just reallocated among the involved masses.
Kinetic Energy
Kinetic energy is a measure of the energy an object possesses due to its motion. Similarly to momentum, kinetic energy depends on both mass and velocity, but it incorporates velocity in a squared form, expressed as \( KE = \frac{1}{2}mv^2 \). When evaluating collisions, especially inelastic ones like this exercise involves, kinetic energy provides insight into what happens to the energy within the system.
  • Initially, only the moving car has kinetic energy, calculated as: \[ KE_{initial} = \frac{1}{2} m v^2 \]
  • After the collision, with the three cars moving at the reduced speed \( v' \), the combined kinetic energy is: \[ KE_{final} = \frac{1}{2} (3m) \left(\frac{v}{3}\right)^2 = \frac{1}{6} m v^2 \]
  • This demonstrates a reduction of kinetic energy, showing us that it's partially transformed into other forms like sound or heat during the collision.
Kinetic energy, unlike momentum, isn't conserved in inelastic collisions. Instead, we witness the characteristic loss of kinetic energy, as shown in this calculation where we found a final value much less than the initial. Recognizing this distinction is crucial when analyzing real-world collisions.
Inelastic Collisions
Understanding inelastic collisions is vital, especially when objects stick together after collision, as observed in this exercise. An inelastic collision is characterized by a loss of kinetic energy, as opposed to perfectly elastic collisions where both momentum and kinetic energy are conserved.
  • Here, the railroad cars remain coupled, making it a perfectly inelastic collision.
  • The immediate result is a combined mass moving at a unified speed \( v' = \frac{v}{3} \), showcasing momentum conservation but kinetic energy loss.
  • The fractional loss of kinetic energy offers a quantitative perspective. The fractional energy loss \( n \) helps measure this, calculated as:\[ n = \frac{\frac{1}{2} m v^2 - \frac{1}{6} m v^2}{\frac{1}{2} m v^2} = \frac{2}{3} \]
  • The result \( n = \frac{2}{3} \) indicates significant energy conversion to non-mechanical forms.
Students should keep inelastic collisions in mind when considering the aftermath of a collision, as energy dissipation is a key characteristic. Visualizing how kinetic energy diminishes can aid in understanding physical processes beyond just numerical answers. This knowledge helps create a holistic view of energy conservation laws in practical scenarios.

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Most popular questions from this chapter

The 0.5 -kg cylinder \(A\) is released from rest from the position shown and drops the distance \(h_{1}=\) \(0.6 \mathrm{m} .\) It then collides with the 0.4 -kg block \(B ;\) the coefficient of restitution is \(e=0.8 .\) Determine the maximum downward displacement \(h_{2}\) of block \(B\) Neglect all friction and assume that block \(B\) is initially held in place by a hidden mechanism until the collision begins. The two springs of modulus \(k=500 \mathrm{N} / \mathrm{m}\) are initially unstretched, and the distance \(d=0.8 \mathrm{m}\).

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