Problem 38 - University Physics with Modern Physics

Question

An automobile with a mass of \(1450 \mathrm{~kg}\) is parked on a moving flatbed railcar; the flatbed is \(1.5 \mathrm{~m}\) above the ground. The railcar has a mass of \(38,500 \mathrm{~kg}\) and is moving to the right at a constant speed of \(8.7 \mathrm{~m} / \mathrm{s}\) on a frictionless rail. The automobile then accelerates to the left, leaving the railcar at a speed of \(22 \mathrm{~m} / \mathrm{s}\) with respect to the ground. When the automobile lands, what is the distance \(D\) between it and the left end of the railcar? See the figure.

Step-by-Step Solution

Answer

Answer: To find the distance between the automobile and the left end of the railcar, follow the solution steps provided and perform the calculations using the given values: mass of railcar = 38500 kg, speed of railcar = 8.7 m/s, mass of automobile = 1450 kg, relative speed of automobile = 22 m/s, height of railcar = 1.5 m. Using these values in the equations, you can calculate the distance between the automobile and the left end of the railcar when the automobile lands on the ground.

Step 1: 1. Find the initial total momentum

Since the railcar and automobile are moving together initially, the total momentum of the system can be calculated by adding their individual momenta: total momentum = momentum of railcar + momentum of automobile total momentum = (mass of railcar × speed of railcar) + (mass of automobile × speed of automobile) Given: mass of railcar = 38500 kg, speed of railcar = 8.7 m/s, mass of automobile = 1450 kg, and initially, speed of automobile = 8.7 m/s (same as the railcar) total momentum = (38500 kg × 8.7 m/s) + (1450 kg × 8.7 m/s)

Step 2: 2. Find the final momentum of the automobile

The automobile leaves the railcar with a speed of 22 m/s relative to the ground. Since it initially had the same speed as the railcar, find the final speed: final speed of automobile relative to the ground = 8.7 m/s - 22 m/s Next, calculate the final momentum of the automobile: final momentum of automobile = mass of automobile × final speed of automobile

Step 3: 3. Determine the final speed of the railcar

Due to the conservation of momentum, the total momentum of the system will remain the same before and after the automobile leaves the railcar. Therefore, we can set up an equation: total initial momentum = final momentum of automobile + final momentum of railcar We know the total initial momentum and the final momentum of the automobile. We can use this to solve for the final speed of the railcar. Final speed of the railcar can be represented as velocity Vr. Finall momentum of railcar = mass of railcar × Vr

Step 4: 4. Calculate the time taken for the automobile to land

The railcar is 1.5 m above the ground. As the automobile leaves the railcar horizontally, the vertical motion is free fall. We can use the equation: distance = (1/2) × (acceleration) × (time^2) Given: distance = 1.5 m, acceleration due to gravity = 9.8 m/s² We can solve for the time taken for the automobile to land.

Step 5: 5. Calculate the horizontal distance travelled by the automobile and railcar

Using the time calculated in the previous step, determine the horizontal distance travelled by the automobile and the railcar: distance travelled by automobile = final speed of automobile × time taken to land distance travelled by railcar = final speed of railcar × time taken to land

Step 6: 6. Find the distance between the automobile and the left end of the railcar

Finally, to find the distance D between the automobile and the left end of the railcar, subtract the horizontal distance travelled by the railcar from the horizontal distance travelled by the automobile: distance D = distance travelled by automobile - distance travelled by railcar By calculating the values for each step and plugging them into the equations, we can find the distance D between the automobile and the left end of the railcar when it lands on the ground.

Key Concepts

Momentum Calculation

Momentum is a vector quantity that represents the product of an object's mass and its velocity. It's described by the equation:
\( p = mv \)
Where \( p \) is momentum, \( m \) is mass, and \( v \) is velocity. In the context of our exercise, the momentum calculation helps us understand how the automotive and the railcar interact with each other. Since the railcar and automobile are initially moving together, their total system momentum is just the sum of their individual momenta.
By calculating the initial total momentum, considering the railcar's large mass and shared speed, we have a baseline to understand the system's behavior when the automobile starts accelerating in the opposite direction. This understanding is crucial because of the law of conservation of momentum, which states that in a closed system with no external forces, the total momentum is constant. This principle allows us to predict subsequent motions once the automobile leaves the railcar. Keeping the calculations simple but accurate is key in providing clear understanding to students. Make sure to accurately use the given values in the calculations and double-check units for consistency.

Relative Speed

Relative speed is the velocity of one object as observed from another moving object. It's essential for understanding how objects move in reference to each other. In the provided exercise, calculating the relative speed is important because the automobile's initial speed is given in relation to the ground, but we need to understand its speed in relation to the moving railcar.
When the automobile accelerates in the opposite direction, we observe its relative speed by subtracting the railcar's speed from the automobile's final ground speed. This calculation tells us how quickly the automobile is moving away from the railcar once it leaves it. Without a correct understanding of relative speed, it becomes challenging to visualize and solve problems where objects are moving in different directions and at different speeds. To keep concepts clear, we emphasize the difference between relative speed and the actual speeds of objects as measured from a stationary point. Introducing real-life examples, like two trains moving at different speeds on parallel tracks, can help make this abstract concept more tangible.

Free Fall Motion

Free fall motion occurs when gravity is the only force acting upon an object. It's an essential part of physics, particularly mechanics, and it's beautifully illustrated in this exercise when the automobile leaves the horizontal plane of the railcar and falls to the ground. There are two key components to remember when dealing with free fall: every object accelerates downwards at the same rate due to gravity (ignoring air resistance), and the object's horizontal motion is independent of its vertical fall.
Using the equation for free fall distance \( d = \frac{1}{2}gt^2 \) allows us to calculate the time it takes for the automobile to land. This calculation is vital as it enables us to use that time to figure out how far the railcar will travel while the automobile is falling. To convey this concept with clarity, it's useful to remind students that even though the automobile is moving horizontally, its vertical motion is entirely governed by gravity. This separation of horizontal and vertical movements is a pillar in solving projectile motion problems and is essential in our current scenario to determine the point of landing in relation to the railcar's new position.