Question

A freight car, open at the top, weighing \(9.75\) metric tons, is coasting along a level track with negligible friction at \(1.36 \mathrm{~m} / \mathrm{s}\) when it begins to rain hard. The raindrops fall vertically with respect to the ground. What is the speed of the car when it has collected \(0.50\) metric tons of rain? What assumptions, if any, must you make to get your answer?

Short answer

The final speed of the freight car after collecting 0.50 metric tons of rain will be \(1.30 \, \mathrm{m/s}\). The calculations assume that raindrops fall vertically with respect to the ground, momentum is conserved and the friction in the system is negligible.

Step-by-step solution

Understand the Problem

We need to find the speed of the freight car after it has collected a certain weight of water by the rain. The rain falls vertically and system experiences no external force so total momentum should be conserved. An important point to keep in mind is that the weights given in the problem need to converted to masses by dividing with acceleration due to gravity. In SI units, acceleration due to gravity (g) is approximately \(9.8 \, \mathrm{m/s^2}\).

Apply the Principle of Conservation of Momentum

According to the conservation of momentum, the initial momentum of the system equal to the final momentum. The initial momentum is only the momentum of the car (mass_car * velocity_car). The final momentum when the rain has been collected is the mass of car plus the collected rain mass, both moving with the same final speed (velocity_final). Setting these two equal to each other: mass_car * velocity_car = (mass_car + mass_rain) * velocity_final. Solve this equation to get the final velocity.

Perform the Calculations

First convert the weights to mass by dividing with g: Mass of the car, mass_car = 9.75 metric tons / g = 9.75 * 10^3 kg / 9.8 m/s^2 = 995 kg (approximately). Mass of the rain, mass_rain = 0.50 metric tons / g = 0.50 * 10^3 kg / 9.8 m/s^2 = 51 kg (approximately). Substitute these values and the known initial velocity into the equation from Step 2:995 kg * 1.36 m/s = (995 kg + 51 kg) * v_final. Simplifying and solving the equation gives v_final = 1.30 m/s.

Key concepts

Momentum

Momentum is a key concept in physics that involves the movement of objects. It is defined as the product of an object's mass and its velocity. In our exercise, the freight car has an initial momentum due to its mass and velocity as it moves along the track. We calculate it using the formula:

• Initial Momentum = mass_car * velocity_car

When rain collects in the car, both the mass and the velocity change. Yet, because momentum is conserved, the system's total momentum remains the same before and after the rain falls. This principle allows us to solve for the final speed of the car after the rain has collected.
Conservation of momentum assumes no external net forces acting on the system. Here, we consider the track frictionless and the rain as additional mass falling vertically, which means no horizontal external force alters the momentum of the system.

Frictionless Motion

In physics, frictionless motion refers to moving without any resistance. In our exercise, we assume the freight car moves along a level track with negligible friction. This simplifies the problem as it allows us to apply the conservation of momentum.
Without friction, the only forces acting are those related to the changing masses when rain gets into the car. Friction usually opposes motion and would reduce the car's speed over time. However, in this idealized scenario, friction is absent. This means the only factor influencing the change in speed is the mass change due to the rain falling into the car.
This setup is crucial because it sets the conditions for using the formula given to determine the car's new speed. Whenever we assume frictionless motion, calculations can focus purely on changes in mass and velocity.

Vertical Motion

Vertical motion is essential to understanding how the rain affects the freight car. The rain falls straight down, or vertically, from the sky into the car. This aspect is important because it means that no horizontal force is being exerted on the car.
In our calculation of the final speed of the car, the vertical nature of the rain's motion helps us maintain conservation of momentum. Since the rain is not affecting the car's horizontal motion directly, it doesn't add to or take away from the existing momentum in that direction.
Thus, by focusing only on the mass change, we can apply the conservation laws to solve for the car's final speed without worrying about external forces in the horizontal direction.

Mass and Weight Conversion

Converting weight to mass is crucial in the exercise due to units and physical definitions. Weight is the force of gravity acting on an object, typically measured in newtons, while mass is the amount of matter in the object, measured in kilograms.
In the given problem, the weights provided for both the car and the rain are in metric tons, which must be converted to mass before any calculations. Since weight \( W = m imes g \), with \( g \approx 9.8 \, \mathrm{m/s^2} \), you convert weight to mass by dividing by \( g \).

• For the freight car: \( \text{mass}_{\text{car}} = \frac{9.75 \times 10^3 \text{ kg}}{9.8 \, \mathrm{m/s^2}} \approx 995 \, \text{kg} \)
• For the rain: \( \text{mass}_{\text{rain}} = \frac{0.50 \times 10^3 \text{ kg}}{9.8 \, \mathrm{m/s^2}} \approx 51 \, \text{kg} \)

This conversion ensures accurate calculations for momentum, as momentum depends directly on mass, not weight.