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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 is the final velocity of (a) the car and (b) the van?

Short Answer

Expert verified
The car's final velocity is -0.787 m/s and the van's is 1.46 m/s.

Step by step solution

01

Understand the Problem

We have a 715-kg car moving at a velocity of \(+2.25 \text{ m/s}\) that collides with a stationary 1055-kg van. The collision is elastic, which means both momentum and kinetic energy are conserved. We need to find the final velocities of both vehicles.
02

Apply Conservation of Momentum

In an elastic collision, the total momentum before and after the collision is the same. Use the formula for momentum: \[ m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2'\]where \(m_1 = 715 \text{ kg}\), \(v_1 = +2.25 \text{ m/s}\), \(m_2 = 1055 \text{ kg}\), \(v_2 = 0 \text{ m/s}\), \(v_1'\) and \(v_2'\) are the final velocities of the car and van, respectively.
03

Apply Conservation of Kinetic Energy

Since the collision is elastic, kinetic energy is conserved:\[\frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 = \frac{1}{2}m_1(v_1')^2 + \frac{1}{2}m_2(v_2')^2\]Substitute the known values into the equation to relate \(v_1'\) and \(v_2'\).
04

Solve the Equations

Use the system of equations derived from the conservation of momentum and kinetic energy:\[\begin{align*}715(2.25) &= 715v_1' + 1055v_2' \quad \text{(conservation of momentum)} \ \frac{1}{2}(715)(2.25)^2 &= \frac{1}{2}(715)(v_1')^2 + \frac{1}{2}(1055)(v_2')^2 \quad \text{(conservation of kinetic energy)}\end{align*}\]Solve the equations simultaneously to find \(v_1'\) and \(v_2'\).
05

Calculate the Final Velocities

Through solving the equations, we find:\[v_1' = \frac{2.25(715 - 1055)}{715 + 1055} = -0.787 \text{ m/s}\]and\[v_2' = \frac{2(2.25)}{715 + 1055} (1055) = 1.46 \text{ m/s}\].

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Key Concepts

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

Conservation of Momentum
In the realm of physics, the principle of conservation of momentum is quite fundamental. Momentum, which is the product of mass and velocity, remains constant in a closed system where no external forces are acting. This is especially true for elastic collisions, where two objects collide and bounce off, without any loss in the overall energy. In our example, a 715-kg car moving at a velocity of \(+2.25 \text{ m/s}\) strikes a stationary 1055-kg van. Just before the impact, the car's momentum is the product of its mass and velocity, while the van's momentum is zero since it is stationary. Using the conservation of momentum law, we equate the total momentum before and after the collision:
  • \( m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2' \)
Here,
  • \( m_1 = 715 \text{ kg} \)
  • \( v_1 = +2.25 \text{ m/s} \)
  • \( m_2 = 1055 \text{ kg} \)
  • \( v_2 = 0 \text{ m/s} \)
The equation demonstrates how the initial total momentum equals the final total momentum, helping us determine the velocities after collision.
Conservation of Kinetic Energy
Kinetic energy, the energy an object possesses due to its motion, follows the law of conservation in elastic collisions. This principle means that the total kinetic energy before the collision equals the total kinetic energy after it. It’s important to note that while energy can transform forms in inelastic collisions, in elastic collisions, the type of energy remains consistent. For the collision between our car and van, conservation of kinetic energy is expressed as:
  • \( \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 = \frac{1}{2}m_1(v_1')^2 + \frac{1}{2}m_2(v_2')^2 \)
Where:
  • \( m_1 = 715 \text{ kg} \)
  • \( v_1 = +2.25 \text{ m/s} \)
  • \( m_2 = 1055 \text{ kg} \)
  • \( v_2 = 0 \text{ m/s} \)
  • \( v_1' \) and \( v_2' \) are the final velocities we need to find.
This equation shows us how the distribution of kinetic energy plays out post-collision, allowing us to derive the velocities after the collision much like the conservation of momentum.
Collision Physics
Collision physics gives us insights into what happens when two or more bodies exert forces upon each other for a short amount of time. Elastic collisions are specific types where both momentum and kinetic energy are conserved. These types of collisions are ideal and mostly used for studying outcomes in a frictionless environment or where no energy is lost to sound, heat, or deformation. The intricacies of these collisions revolve around:
  • The interaction of different masses and velocities
  • The application of physical laws such as the conservation of momentum and kinetic energy
  • Resulting behaviors post-collision, like reversing or maintaining speed based on the masses' movements
In our car-van collision example, the calculated velocities post-impact—\( v_1' = -0.787 \text{ m/s} \) for the car and \( v_2' = 1.46 \text{ m/s} \) for the van—perfectly illustrate the transfer of energy and momentum. The car reverses direction, while the van moves forward, showcasing the balanced exchange facilitated by the laws governing elastic collisions.

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Most popular questions from this chapter

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