Problem 48
Question
A 1600 -kg (empty) truck rolls with a speed of \(2.5 \mathrm{~m} / \mathrm{s}\) under a loading bin, and a mass of \(3500 \mathrm{~kg}\) is deposited into the truck. What is the truck's speed immediately after loading?
Step-by-Step Solution
Verified Answer
The truck's speed after loading is approximately \(0.7843 \, \mathrm{m/s}\).
1Step 1: Identifying the System
In this problem, we consider the truck and the mass added to it as one single system. Before the loading, the truck has a velocity of \(2.5 \, \mathrm{m/s}\) and the mass of the truck is \(1600 \, \mathrm{kg}\).
2Step 2: Understanding Conservation of Momentum
The principle of conservation of momentum states that the total momentum of a closed system before the event (loading) must be equal to the total momentum of the system after the event, assuming no external forces are acting.
3Step 3: Calculating Initial Momentum
Calculate the initial momentum of the truck before loading: \[ p_{ ext{initial}} = m_{ ext{truck}} \times v_{ ext{truck}} = 1600 \, \mathrm{kg} \times 2.5 \, \mathrm{m/s} = 4000 \, \mathrm{kg} \cdot \mathrm{m/s} \]
4Step 4: Calculating Final Mass After Loading
After loading, the total mass of the truck and the additional load is:\[ m_{ ext{final}} = 1600 \, \mathrm{kg} + 3500 \, \mathrm{kg} = 5100 \, \mathrm{kg} \]
5Step 5: Calculating Final Velocity Using Conservation of Momentum
By the conservation of momentum, the initial momentum equals the final momentum. Let \(v_{ ext{final}}\) be the final velocity of the truck. Therefore:\[ p_{ ext{initial}} = p_{ ext{final}} \]\[ 1600 \, \mathrm{kg} \cdot 2.5 \, \mathrm{m/s} = 5100 \, \mathrm{kg} \cdot v_{ ext{final}} \]Solving for \(v_{ ext{final}}\):\[ v_{ ext{final}} = \frac{4000 \, \mathrm{kg} \cdot \mathrm{m/s}}{5100 \, \mathrm{kg}} \approx 0.7843 \, \mathrm{m/s} \]
6Step 6: Finalize the Result
The speed of the truck immediately after loading is approximately \(0.7843 \, \mathrm{m/s}\). This value reflects the reduced speed due to the increased mass after loading, adhering to the law of conservation of momentum.
Key Concepts
Physics Problem SolvingMomentum CalculationVelocity Determination
Physics Problem Solving
When tackling physics problems, it's important to first understand the scenario and identify key elements. Breaking the problem down step-by-step can make it more manageable and clear. In our example exercise, we initially need to identify the system, which includes both the truck and the additional load.
It's crucial to determine the parameters for the system, such as masses and velocities. This forms a foundation for further calculations. We also need to recognize the physical principles at play, like the conservation of momentum.
Once the system and governing laws are understood, planning the solution involves setting up equations based on these principles. By approaching the problem logically and systematically, even complex phenomena, like momentum transfer, can become straightforward. This step-by-step method not only aids in solving physics problems but also enhances comprehension of fundamental concepts.
It's crucial to determine the parameters for the system, such as masses and velocities. This forms a foundation for further calculations. We also need to recognize the physical principles at play, like the conservation of momentum.
Once the system and governing laws are understood, planning the solution involves setting up equations based on these principles. By approaching the problem logically and systematically, even complex phenomena, like momentum transfer, can become straightforward. This step-by-step method not only aids in solving physics problems but also enhances comprehension of fundamental concepts.
Momentum Calculation
Momentum is a fundamental concept in physics, describing the quantity of motion an object possesses. It is defined as the product of an object's mass and velocity. In our exercise, the initial momentum of the truck is calculated using this principle.
We know the truck's mass (1600 \, \text{kg}) and its velocity (2.5 \, \text{m/s}), so we compute the initial momentum like this: \[ p_{\text{initial}} = m_{\text{truck}} \times v_{\text{truck}} = 1600 \, \text{kg} \times 2.5 \, \text{m/s} = 4000 \, \text{kg} \cdot \text{m/s} \]This step highlights how quantities can be calculated using known values. Understanding the momentum calculation is vital. It allows us to predict how changes in mass and velocity affect an object's motion. It also lays the groundwork for the conservation of momentum principle.
We know the truck's mass (1600 \, \text{kg}) and its velocity (2.5 \, \text{m/s}), so we compute the initial momentum like this: \[ p_{\text{initial}} = m_{\text{truck}} \times v_{\text{truck}} = 1600 \, \text{kg} \times 2.5 \, \text{m/s} = 4000 \, \text{kg} \cdot \text{m/s} \]This step highlights how quantities can be calculated using known values. Understanding the momentum calculation is vital. It allows us to predict how changes in mass and velocity affect an object's motion. It also lays the groundwork for the conservation of momentum principle.
Velocity Determination
To determine velocity after an event like loading, we use the conservation of momentum. This principle states that in the absence of external forces, the system's momentum before and after the event remains constant.
For the truck, post-loading, the total system mass becomes 5100 \, \text{kg}. The initial momentum was 4000 \, \text{kg} \cdot \text{m/s}. By setting the initial momentum equal to the final momentum, we solve for the truck's final velocity: \[ v_{\text{final}} = \frac{4000 \, \text{kg} \cdot \text{m/s}}{5100 \, \text{kg}} \approx 0.7843 \, \text{m/s} \]This subsequent velocity is lower due to the truck's increased mass, illustrating how mass affects motion. Reliability on conservation laws, like momentum, is key in accurately determining unknowns such as velocity in an isolated system. This practice demonstrates the utility of fundamental physics principles in real-world applications.
For the truck, post-loading, the total system mass becomes 5100 \, \text{kg}. The initial momentum was 4000 \, \text{kg} \cdot \text{m/s}. By setting the initial momentum equal to the final momentum, we solve for the truck's final velocity: \[ v_{\text{final}} = \frac{4000 \, \text{kg} \cdot \text{m/s}}{5100 \, \text{kg}} \approx 0.7843 \, \text{m/s} \]This subsequent velocity is lower due to the truck's increased mass, illustrating how mass affects motion. Reliability on conservation laws, like momentum, is key in accurately determining unknowns such as velocity in an isolated system. This practice demonstrates the utility of fundamental physics principles in real-world applications.
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