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Chapter 8: Problem 23

A \(40000-\mathrm{kg}\) freight car is coasting at a speed of \(5.0 \mathrm{~m} / \mathrm{s}\) along a straight level track when it strikes a \(30000-\mathrm{kg}\) stationary freight car and couples to it. What will be their combined speed after impact?

Short Answer

Expert verified
The combined speed is approximately 2.857 m/s.

Step by step solution

01

Understand the Problem

We need to find the combined speed of two freight cars after they collide and couple together. The problem involves a conservation of momentum where an initially moving freight car contacts a stationary one.
02

Apply Conservation of Momentum

In this problem, no external forces act on the combined system of the freight cars in the direction of motion, so linear momentum is conserved. Let the initial momentum and final momentum be represented as:\[ m_1v_1 + m_2v_2 = (m_1 + m_2)v_f \]where \(m_1 = 40000\, \text{kg}\), \(v_1 = 5.0\, \text{m/s}\), \(m_2 = 30000\, \text{kg}\), and \(v_2 = 0\, \text{m/s}\). \(v_f\) is the final speed we need to find.
03

Substitute Known Values

Substitute the known values into the momentum conservation equation:\[ 40000 \times 5.0 + 30000 \times 0 = (40000 + 30000) v_f \]This simplifies to:\[ 200000 = 70000v_f \]
04

Solve for Final Speed

Rearrange the equation to solve for \(v_f\):\[ v_f = \frac{200000}{70000} \]\[ v_f = \frac{20}{7} \approx 2.857 \text{ m/s} \]

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Key Concepts

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

Linear Momentum
Linear momentum is a vital concept in physics, representing the quantity of motion an object possesses. It's the product of an object's mass and its velocity, and is mathematically expressed as \( p = mv \), where \( p \) is momentum, \( m \) is mass, and \( v \) is velocity.

Two main properties make linear momentum particularly useful:
  • Conservation: In the absence of external forces, the total momentum of a system remains constant.
  • Direction: Linear momentum is a vector quantity, meaning it has both magnitude and direction, crucial in fields like mechanics and collision analysis.

In the context of the freight car problem, the conservation of linear momentum helps us determine the final velocity of combined masses after a collision. This principle allows us to equate the initial momentum before collision with the final momentum, making it easier to solve for unknown quantities.
Collisions
Collisions occur when two or more objects exert forces on each other for a short duration. In physics, understanding collisions is critical for predicting outcomes in various everyday situations and scientific applications.

Collisions are categorized based on their energy and momentum conservation:
  • Elastic Collisions: Both kinetic energy and momentum are conserved.
  • Inelastic Collisions: Only momentum is conserved, while some kinetic energy is transformed into other forms of energy, like heat.
  • Perfectly Inelastic Collisions: The objects stick together post-collision, which is what happens in the freight car problem.

In this problem, the two freight cars sticking together after the collision is a classic example of a perfectly inelastic collision, where momentum conservation helps us find the final common velocity.
Freight Car Problems
Freight car problems involve computing various quantities like final velocities or forces when dealing with moving trains. Often modeled as perfectly inelastic collisions, these problems illustrate fundamental principles of momentum. Here's why they are essential:
  • Practical Applications: Railways are vital for transportation and logistics; understanding these problems aids in optimizing train operations.
  • Conceptual Clarity: Freight car problems provide a straightforward context for applying conservation of momentum concepts, especially in educational settings.
  • Safety Considerations: Predicting outcomes in collisions helps in designing safer transportation systems.

In the original problem, the calculation illustrates how mass and velocity changes after the coupling of two freight cars, enhancing understanding of real-world train coupling scenarios.

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Most popular questions from this chapter

Two balls of equal mass approach the coordinate origin, one moving downward along the \(y\) -axis at \(2.00 \mathrm{~m} / \mathrm{s}\) and the other moving to the right along the \(-x\) -axis at \(3.00 \mathrm{~m} / \mathrm{s}\). After they collide, one ball moves out to the right along the \(+x\) -axis at \(1.20 \mathrm{~m} / \mathrm{s}\). Find the scalar \(x\) and \(y\) velocity components of the other ball. This is a two-dimensional collision and momentum must be conserved independently in each perpendicular direction, \(x\) and \(y\). Take \(u p\) and to the right as positive. Accordingly, keeping in mind that before impact only one ball had an \(x\) -component of velocity, Or $$ \begin{aligned} \text { (Momentum before) }_{x} &=\text { (Momentum after) }_{x} \\ m(3.00 \mathrm{~m} / \mathrm{s})+0 &=m(1.20 \mathrm{~m} / \mathrm{s})+m v_{x} \end{aligned} $$ Here \(v_{\mathrm{x}}\) is the unknown \(x\) -component of velocity of the second ball acquired on impact. Since we know that the first ball lost some of its \(x\) -momentum, the second ball must have gained it. Moreover, (Momentum before) \(_{y}=(\text { Momentum after })_{y}\) or $$ 0+m(-2.00 \mathrm{~m} / \mathrm{s})=0+m v_{y} $$ Here \(v_{\mathrm{y}}\) is the \(y\) -component of velocity of the second ball. (Why the minus sign?) Solving each equation, after cancelling the mass we find that \(v_{x}=1.80 \mathrm{~m} / \mathrm{s}\) and \(v_{y}=-2.00 \mathrm{~m} / \mathrm{s}\)

As shown in Fig. \(8-1\), a 15 -g bullet is fired horizontally into a \(3.000-\mathrm{kg}\) block of wood suspended by a long cord. The bullet lodges in the block. Compute the speed of the bullet if the impact causes the block (and bullet) to swing \(10 \mathrm{~cm}\) above its initial level. Consider first the collision of block and bullet. During the collision, momentum is conserved, so Momentum just before \(=\) Momentum just after $$ (0.015 \mathrm{~kg}) v+0=(3.015 \mathrm{~kg}) V $$ where \(v\) is the speed of the bullet just prior to impact, and \(V\) is the speed of block and bullet just after impact. We have two unknowns in this equation. To find another equation, we can use the fact that the block swings \(10 \mathrm{~cm}\) high. If we let \(\mathrm{PE}_{\mathrm{G}}=0\) at the initial level of the block, energy conservation tells us that $$ \begin{array}{l} \text { KE just after collision }=\text { Final } \mathrm{PE}_{\mathrm{G}} \\ \qquad \frac{1}{2}(3.015 \mathrm{~kg}) V^{2}=(3.015 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.10 \mathrm{~m}) \end{array} $$ From this \(V=1.40 \mathrm{~m} / \mathrm{s}\). Substituting this combined speed into the previous equation leads to \(v=0.28 \mathrm{~km} / \mathrm{s}\) for the speed of the bullet. Notice that we cannot write the conservation of energy equation \(\frac{1}{2} m v^{2}=(m+M) g h\), where \(m=0.015 \mathrm{~kg}\) and \(M=3.000 \mathrm{~kg}\) because energy is lost (through friction) in the collision process.

A ball is dropped from a height \(h\) above a tile floor and rebounds to a height of \(0.65 h\). Find the coefficient of restitution between ball and floor. Assign floor quantities the subscript 1 , and ball quantities the subscript 2 . The initial and final velocities of the floor, \(u_{1}\) and \(v_{1}\), are zero. Therefore, $$ e=\frac{v_{2}-v_{1}}{u_{1}-u_{2}}=-\frac{v_{2}}{u_{2}} $$ Since we know both the drop and rebound heights \((h\) and \(0.65 h)\), we can write equations for the interchange of \(\mathrm{PE}_{\mathrm{G}}\) and \(\mathrm{KE}\) before and after the impact $$ m g h=\frac{1}{2} m u_{2}^{2} \quad \text { and } \quad m g(0.65 h)=\frac{1}{2} m v_{2}^{2} $$ Therefore, taking down as positive, \(u_{2}=\sqrt{2 g h}\) and \(v_{2}=-\sqrt{1.30 g h}\). Substitution leads to $$ e=\frac{\sqrt{1.30 g h}}{\sqrt{2 g h}}=\sqrt{0.65}=0.81 $$ Notice that the coefficient of restitution equals the square root of the final rebound height over the initial drop height.

A 90 -g ball moving at \(100 \mathrm{~cm} / \mathrm{s}\) collides head-on with a stationary 10 -g ball. Determine the speed of each after impact if \((a)\) they stick together, \((b)\) the collision is perfectly elastic, (c) the coefficient of restitution is \(0.90\).

A 6000 -kg truck traveling north at \(5.0 \mathrm{~m} / \mathrm{s}\) collides with a \(4000-\mathrm{kg}\) truck moving west at \(15 \mathrm{~m} / \mathrm{s}\). If the two trucks remain locked together after impact, with what speed and in what direction do they move immediately after the collision?
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