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Multiple-Concept Example 7 presents a model for solving problems such as this one. A \(1055-\mathrm{kg}\) van, stopped at a traffic light, is hit directly in the rear by a \(715-\mathrm{kg}\) car traveling with a velocity of \(+2.25 \mathrm{m} / \mathrm{s}\) . Assume that the transmission of the van is in neutral, the brakes are not being applied, and the collision is elastic. What are the final velocities of (a) the car and (b) the van?

Short Answer

Expert verified
The final velocity of the car is approximately \(0.925 \text{ m/s}\), and the van is approximately \(1.325 \text{ m/s}\).

Step by step solution

01

Understand Elastic Collision

An elastic collision is a type of collision where both momentum and kinetic energy are conserved. For two objects undergoing a one-dimensional elastic collision, the equations to solve are derived from these conservation principles.
02

Identify the Variables

Let \( m_1 = 715 \text{ kg} \) be the mass of the car, \( v_{1i} = +2.25 \text{ m/s} \) be its initial velocity, \( m_2 = 1055 \text{ kg} \) be the mass of the van, and \( v_{2i} = 0 \text{ m/s} \) be its initial velocity as it's initially at rest. We need to find the final velocities \( v_{1f} \) and \( v_{2f} \) of the car and van respectively.
03

Write the Conservation of Momentum Equation

The conservation of momentum for the system can be expressed as:\[ m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \]Substituting the known values, we get:\[ 715 \times 2.25 + 1055 \times 0 = 715 v_{1f} + 1055 v_{2f} \]
04

Write the Conservation of Kinetic Energy Equation

The conservation of kinetic energy for an elastic collision is given by:\[ \frac{1}{2}m_1 v_{1i}^2 + \frac{1}{2}m_2 v_{2i}^2 = \frac{1}{2}m_1 v_{1f}^2 + \frac{1}{2}m_2 v_{2f}^2 \]Substituting the given values, we have:\[ \frac{1}{2} \times 715 \times (2.25)^2 + 0 = \frac{1}{2} \times 715 \times v_{1f}^2 + \frac{1}{2} \times 1055 \times v_{2f}^2 \]
05

Solve the Equations

From Step 3:\[ 1608.75 = 715 v_{1f} + 1055 v_{2f} \] From Step 4:\[ 1809.375 = 357.5 v_{1f}^2 + 527.5 v_{2f}^2 \]Solve these equations simultaneously to find \( v_{1f} \) and \( v_{2f} \). This can often involve substitution or using a system of equations solver.
06

Calculate the Final Velocities

Solving the equations simultaneously, we find:For the car (\( v_{1f} \)):\[ v_{1f} = 0.925 \text{ m/s} \ (approx.) \] For the van (\( v_{2f} \)):\[ v_{2f} = 1.325 \text{ m/s} \ (approx.) \]

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Key Concepts

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

Conservation of Momentum
In physics, the conservation of momentum is a fundamental principle that states the total momentum of a closed system remains constant, provided no external forces are acting on it. Momentum is defined as the product of an object's mass and its velocity.
The equation for conservation of momentum in a collision is expressed as follows:
  • \( m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \)
For the collision scenario involving the van and the car, we know the initial and final velocities of the objects. Both the car and the van contribute to the total system momentum. By applying this principle, we ensure that despite the impact between the vehicles, their combined momentum remains unchanged, allowing us to solve for unknown variables in the system.
Conservation of Kinetic Energy
The conservation of kinetic energy is crucial in an elastic collision, where no energy is lost to sound, heat, or deformation. In such collisions, the total kinetic energy before impact equals the total kinetic energy after impact.
This principle can be expressed using the following equation for two objects:
  • \( \frac{1}{2}m_1 v_{1i}^2 + \frac{1}{2}m_2 v_{2i}^2 = \frac{1}{2}m_1 v_{1f}^2 + \frac{1}{2}m_2 v_{2f}^2 \)
In our scenario, this equation implies that even though the car and van might change their velocities post-collision, the total kinetic energy between them remains constant. This conservation helps in predicting post-collision velocities when combined with other equations.
Collision Physics
Collision physics provides the framework to understand interactions between objects upon impact. It encompasses the types of collisions, including elastic and inelastic, and impacts on momentum and energy transfers.
Elastic collisions, such as the one between the van and car here, are characterized by both conservation of momentum and kinetic energy. This allows us to predict the behavior of colliding bodies — both in terms of direction and speed post-collision. Understanding these principles helps us apply physical laws to real-world situations, such as vehicle safety analysis and sports sciences.
Momentum Equations
Momentum equations are tools used to solve for unknowns in physics problems involving motion and collisions. They are derived from the principle of conservation of momentum.
Given two objects colliding:
  • The initial momentum: \( m_1 v_{1i} + m_2 v_{2i} \)
  • The final momentum: \( m_1 v_{1f} + m_2 v_{2f} \)
By setting these equal, we recognize any discrepancies ought to stem from external forces — absent in this context of ideal elastic collisions. Applying these equations allows us to solve simultaneously with kinetic energy equations to find final velocities — as seen with the van and car here, yielding \( v_{1f} = 0.925 \text{ m/s} \) and \( v_{2f} = 1.325 \text{ m/s} \). These steps showcase how momentum equations help bridge theoretical physics with practical problem-solving.

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Most popular questions from this chapter

ssm mmh Two friends, Al and Jo, have a combined mass of 168 \(\mathrm{kg}\) . At an ice skating rink they stand close together on skates, at rest and facing each other, with a compressed spring between them. The spring is kept from pushing them apart because they are holding each other. When they release their arms, Al moves off in one direction at a speed of 0.90 \(\mathrm{m} / \mathrm{s}\) , while Jo moves off in the opposite direction at a speed of 1.2 \(\mathrm{m} / \mathrm{s}\) . Assuming that friction is negligible, find Al's mass.

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