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Chapter 14: Problem 2

Two identical 1350 -kg automobiles \(A\) and \(B\) are at rest with their brakes released when \(B\) is struck by a \(5400-\mathrm{kg}\) truck \(C\) that is moving to the left at \(8 \mathrm{km} / \mathrm{h}\). A second collision then occurs when \(B\) strikes \(A\). Assuming the first collision is perfectly plastic and the second collision is perfectly elastic, determine the velocities of the three vehicles just after the second collision.

Short Answer

Expert verified
After the second collision, A: -0.036 m/s, B: 0.018 m/s, C: -0.018 m/s.

Step by step solution

01

Analyze the First Collision

In the first collision, truck \( C \) strikes car \( B \). Since the collision is perfectly plastic, \( B \) and \( C \) move together after the collision. We apply the conservation of momentum here. \( C \) has mass 5400 kg and initially moves to the left at 8 km/h, while \( B \) is at rest. Using: \( m_C \cdot v_C + m_B \cdot v_B = (m_C + m_B) \cdot v_{BC} \).\[ 5400 \times (-8/3.6) + 0 = (5400 + 1350) \cdot v_{BC} \]\[ v_{BC} = \frac{-120}{6750} \;\text{m/s} = -0.018 \;\text{m/s} \] (approximately). This means \( B \) and \( C \) together move to the left at 0.018 m/s.
02

Analyze the Second Collision

The second collision is between cars \( B \) and \( A \) and is perfectly elastic. In an elastic collision, both momentum and kinetic energy are conserved. Initially, \( B \) is moving at -0.018 m/s and \( A \) is at rest. Applying conservation of momentum: \( m_B \cdot v_{BC} + m_A \cdot 0 = m_B \cdot v_B'+ m_A \cdot v_A' \) leads to \[ 1350\times(-0.018) = 1350\cdot v_B' + 1350\cdot v_A'\] which simplifies to \( v_B' + v_A' = -0.018 \;\text{m/s} \).
03

Apply Conservation of Kinetic Energy

For perfectly elastic conditions, kinetic energy is also conserved. Thus, \[ \frac{1}{2}\cdot 1350\cdot (-0.018)^2 = \frac{1}{2}\cdot 1350\cdot (v_B')^2 + \frac{1}{2}\cdot 1350\cdot (v_A')^2 \]\[ 0.2187 = 675(v_B')^2 + 675(v_A')^2 \]. Solving these gives us two equations: \( v_B' + v_A' = -0.018 \) and \( 675(v_B')^2 + 675(v_A')^2 = 0.2187 \).
04

Solve the Equations

Substitute one equation into the other. Use \( v_A' = -0.018 - v_B' \) and solve for \( v_B' \), then substitute back to find \( v_A' \). Solving yields: \( v_A' = -0.036 \;\text{m/s} \) and \( v_B' = 0.018 \;\text{m/s} \). Finally, vehicle \( C \) remains at velocity -0.018 m/s, as it did not participate in the second collision.

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Key Concepts

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

Conservation of Momentum
The principle of conservation of momentum is fundamental in understanding collisions and movements of systems in engineering mechanics. It states that the total momentum of a closed system remains constant if no external forces act upon it. The mathematical expression for momentum conservation is:
  • In a system of masses: \[ m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2' \] where \( m \) is mass and \( v \) is velocity.
This principle applies regardless of whether collisions are elastic or inelastic. It helps in predicting post-collision velocities of objects. In our exercise, we utilized momentum conservation during collisions to determine the final velocities of cars and the truck involved.
Perfectly Elastic Collision
A perfectly elastic collision is one where no kinetic energy is lost throughout the event. This means that both kinetic energy and momentum are conserved. In real-world scenarios, perfectly elastic collisions are idealized since some energy is usually transformed into sound, heat, or deformation.
  • For perfect elasticity:

    Both momentum and kinetic energy equations must hold:\[ m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2' \]\[ \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 = \frac{1}{2} m_1 (v_1')^2 + \frac{1}{2} m_2 (v_2')^2 \]
In the exercise, the collision between cars B and A is assumed perfectly elastic, allowing us to solve for their velocities using these conservation principles.
Perfectly Plastic Collision
In a perfectly plastic collision, the two colliding bodies stick together, effectively becoming a single mass post-collision. This type of plastic or inelastic collision does not conserve kinetic energy, though it does conserve momentum.
  • The equation for momentum conservation is:\[ m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_{combined} \]
Such collisions result in energy loss typically transformed into internal energy like heat or deformation. In our exercise, the collision between truck C and car B was considered perfectly plastic, leading them to move together as one unit initially.
Kinetic Energy Conservation
The conservation of kinetic energy is crucial in analyzing elastic collisions. After a perfectly elastic collision, the total kinetic energy before and after the collision remains unchanged. This contrasts with inelastic collisions, where kinetic energy is not conserved.
  • In equations, kinetic energy conservation is expressed as:\[ \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 = \frac{1}{2} m_1 (v_1')^2 + \frac{1}{2} m_2 (v_2')^2 \]
This principle guarantees that velocities after an elastic collision can be accurately calculated. In the exercise, this principle was vital in analyzing the outcome of the collision between vehicles A and B, providing insight into their final velocities.

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Most popular questions from this chapter

Three small spheres \(A, B\), and \(C,\) each of mass \(m,\) are connected to a small ring \(D\) of negligible mass by means of three inextensible, inelastic cords of length \(l\). The spheres can slide freely on a frictionless horizontal surface and are rotating initially at a speed \(v_{0}\) about ring \(D\) which is at rest. Suddenly the cord \(C D\) breaks. After the other two cords have again become taut, determine ( \(a\) ) the speed of ring \(D,(b)\) the relative speed at which spheres \(A\) and \(B\) rotate about \(D,\) (c) the fraction of the original energy of spheres \(A\) and \(B\) that is dissipated when cords \(A D\) and \(B D\) again became taut.

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Four small disks \(A, B, C,\) and \(D\) can slide freely on a frictionless horizontal surface. Disks \(B, C,\) and \(D\) are connected by light rods and are at rest in the position shown when disk \(B\) is struck squarely by disk \(A\) which is moving to the right with a velocity \(v_{0}=\left(38.5 \text { fits) i. The weights of the disks are } W_{A}=W_{B}=W_{C}=15\right.\) Ib, and \(W_{D}=30\) lb. Knowing that the velocities of the disks immediately after the impact are \(v_{A}=v_{B}=(8.25 \mathrm{ft} / \mathrm{s}) \mathrm{i}\), \(\mathrm{v}_{C}=v_{C} \mathrm{i},\) and \(v_{D}=v_{D} \mathrm{i}\), determine \((a)\) the speeds \(v_{C}\) and \(v_{D},(b)\) the fraction of the initial kinetic energy of the system which is dissipated during the collision.

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