Question

In 1896 in Waco, Texas, William George Crush, owner of the K-T (or "Katy") Railroad, parked two locomotives at opposite ends of a 6.4 -km-long track, fired them up, tied their throttles open, and then allowed them to crash head- on at full speed in front of 30,000 spectators. Hundreds of people were hurt by flying debris; several were killed. Assuming that each locomotive weighed \(1.2 \cdot 10^{6} \mathrm{~N}\) and its acceleration along the track was a constant \(0.26 \mathrm{~m} / \mathrm{s}^{2},\) what was the total kinetic energy of the two locomotives just before the collision?

Short answer

Answer: The total kinetic energy of the two locomotives just before the collision was approximately \(4.06 \cdot 10^8 \mathrm{~J}\).

Step-by-step solution

Find the final velocity of each locomotive

To do this, we'll use the formula: \(v^2 = u^2 + 2as\), where \(v\) is the final velocity, \(u\) is the initial velocity (which is 0 since they start from rest), \(a\) is the acceleration, and \(s\) is the distance travelled. Therefore, \(v^2 = 0^2 + 2(0.26 \mathrm{~m/s^2})(6400 \mathrm{~m})\).

Solve for final velocity

By working out the equation from step 1, we have \(v^2 = 2(0.26 \mathrm{~m/s^2})(6400 \mathrm{~m}) = 3328\), so the final velocity of each locomotive is \(v = \sqrt{3328} \approx 57.70 \mathrm{~m/s}\).

Calculate the kinetic energy of each locomotive

We can now use the formula for kinetic energy: \(\text{KE} = \frac{1}{2}mv^2\), where \(\text{KE}\) is kinetic energy, \(m\) is mass, and \(v\) is velocity. We are given the weight (force due to gravity) of each locomotive, which is \(1.2 \cdot 10^{6} \mathrm{~N}\). To find mass, we can use \(w = mg\), so we have \(m = \frac{w}{g}\), where \(g \approx 9.8 \mathrm{~m/s^2}\). Therefore, \(m = \frac{1.2 \cdot 10^{6} \mathrm{~N}}{9.8\mathrm{~m/s^2}} \approx 1.22449 \cdot 10^5 \mathrm{~kg}\).

Calculate the kinetic energy

Now that we have the mass and velocity, we can calculate the kinetic energy of one locomotive: \(\text{KE} = \frac{1}{2}(1.22449 \cdot 10^5 \mathrm{~kg})(57.70 \mathrm{~m/s})^2 \approx 2.03 \cdot 10^8 \mathrm{~J}\).

Calculate the total kinetic energy for both locomotives

Since the two locomotives have the same mass and velocity, their kinetic energies are equal. Therefore, the total kinetic energy of both locomotives before the collision can be found by summing their individual kinetic energies: \(\text{Total KE} = 2 \times 2.03 \cdot 10^8 \mathrm{J} \approx 4.06 \cdot 10^8 \mathrm{~J}\). So, the total kinetic energy of the two locomotives just before the collision was approximately \(4.06 \cdot 10^8 \mathrm{~J}\).

Key concepts

Physics Problem Solving

There's something truly fascinating about the methodical approach required to solve physics problems. It's like unraveling a puzzle where each piece is governed by the universal laws of nature. For students embarking on these challenges, understanding the problem is the first crucial step. In this case, our problem revolved around a historical event where two locomotives were set on a collision course.

It's imperative to identify what's asked – here, the total kinetic energy just before the locomotives crash. We begin by establishing known variables and constants, like the locomotives starting from rest, the distance covered, and the acceleration. As part of our exercise improvement advice, it's also helpful to layout a strategy by choosing the right physics principles and equations that apply to the phenomenon under investigation before diving into calculations. This structured approach not only simplifies complex problems but also helps students learn how to apply theoretical knowledge to real-world scenarios.

Kinematics Equations

One might think of kinematics as the choreographer of motion in physics. The equations of kinematics guide us in analyzing objects' movements without considering the forces causing the motion. For students to grasp the intricacies of these equations, it's crucial to break down the movement into its components; in this particular problem, we're interested in the final velocity of the locomotives.

By using the equation v² = u² + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled, we can solve for v. This equation is pivotal as it relates velocity, acceleration, and displacement, allowing us to determine the crucial variable 'velocity.' As a tip for students, always ensure to include the correct measurement units and double-check calculations for accuracy. Kinematics isn't just about finding an answer; it's understanding the story of motion told through numbers and equations.

Energy Conservation

The principle of energy conservation is a cornerstone in physics, providing a foundational understanding that in an isolated system, energy can neither be created nor destroyed – only transformed. When we talk about kinetic energy in our locomotive problem, we refer to the energy an object possesses due to its motion.

The formula KE = \(\frac{1}{2}mv^2\), where KE represents the kinetic energy, m the mass, and v the velocity, gives us the tool to quantify this dynamic state. Understanding that the kinetic energy of the locomotives is proportional to the square of their velocity instills the concept that even small increases in speed can result in significant energy changes. This relationship is especially important for students to understand, as it often leads to profound implications in real-world applications, such as safety considerations in vehicle design and energy efficiency projects. Teaching this exercise, we'd emphasize the relevance of energy calculations in everyday life, enhancing the tangibility of physics for learners.