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Packages in an automobile parts supply house are transported to the loading dock by pushing them along on a roller track with very little friction. At the instant shown, packages B and C are at rest, and package A has a velocity of 2 m/s. Knowing that the coefficient of restitution between the packages is 0.3, determine (a) the velocity of package C after A hits B and B hits C, (b) the velocity of A after it hits B for the second time.

Short Answer

Expert verified
Package C's velocity after collision is approximately 0.6 m/s; A's velocity after its second hit is approximately 0.42 m/s.

Step by step solution

01

Understand the Problem

We have three packages, A, B, and C, aligned on a track. Initially, only package A is moving with a velocity of 2 m/s. The packages collide sequentially, first A with B, then B with C. We need to calculate the velocities of packages C after the collisions and A after its second collision with B. The coefficient of restitution (e) is 0.3.
02

Apply the Concept of Coefficient of Restitution

The coefficient of restitution, which is 0.3 in this scenario, will help us find the relative velocity of separation after collision compared to the relative velocity of approach before collision. For collisions between two packages m and n, the formula is:\[ e = \frac{v_n' - v_m'}{v_m - v_n} \]where \(v_m\) and \(v_n\) are velocities before the collision, and \(v_m'\) and \(v_n'\) are velocities after the collision.
03

Calculate the Velocity of Package B After Collision with A

Since B is initially at rest and A has a velocity of 2 m/s, using the restitution equation between A and B:\[ 0.3 = \frac{v_B' - v_A'}{2 - 0} \]Substituting \(v_A' = 0.3(0 - 2) + v_B'\), you resolve this to find:\[ v_A' = 0.3(v_B' - 2) + v_B' \]
04

Calculate the Velocity of Package C After Collision with B

Initially, B moves towards C after being hit by A. Let's use the coefficient of restitution between B and C:\[ 0.3 = \frac{v_C' - v_B'}{v_B - 0} \]Since B hits C with its velocity from the previous step, solve for \(v_C'\) using the simultaneous equations from B's previous velocity.
05

Calculate Velocity of A after it Hits B Again

After B hits C and makes them move, A, which was already slowed down from its first collision with B, will strike B again. We must revisit the velocity garnered by A after its initial collision with B and use conservation of energy or successive application of the coefficient of restitution for second-time calculations.

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Key Concepts

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

Collision Mechanics
Collision mechanics is the study of how objects interact with each other when they collide. In our problem, this involves packages A, B, and C on a roller track. Each package has an initial velocity, and when they collide, their velocities change based on specific physical rules. A key concept here is the coefficient of restitution. This is a value that measures how elastic or inelastic a collision is. With a coefficient of restitution of 0.3, our collisions are inelastic, meaning some kinetic energy is lost as heat or deformation. During a collision, the principle of conservation of momentum is also at play. This means that the total momentum before collision will be equal to the total momentum after collision. In simple terms, if two objects collide, the speed and direction they move in after the hit depends on their masses and velocities before they collided. However, for exact calculations in our problem, we use the given restitution coefficient to adjust the velocities post-collision.
Sequential Collisions
Sequential collisions are a series of impacts where one collision leads to another. In the given scenario, package A first collides with package B. This causes B to move and, because the track allows almost frictionless motion, B continues to move until it collides with package C. Such sequential behavior is common in systems with multiple bodies in motion. To solve problems like this, you must analyze each collision step-by-step. After calculating the result of the first collision, you proceed to the next, using the velocities obtained from the previous collisions as starting values for the new ones. This process continues until all collisions have been resolved. In our example, after package B has been hit and moves towards package C, it carries forward the energy and partly transfers it to package C upon contact.
Velocity Calculation
Velocity calculation in collision problems involves the use of formulas that relate the velocities of the objects before and after collision. The key formula here involves the coefficient of restitution, where: \[ e = \frac{v_n' - v_m'}{v_m - v_n} \]This helps us to find the velocities after a collision, given we know the coefficient of restitution and initial velocities. Here, velocity calculation becomes crucial after each impact since package movements occur in sequence. For example, in the initial collision between A and B, knowing A’s velocity and using the above formula allows us to solve for B's velocity. Similarly, once we know B's new velocity, we can calculate C's velocity after B hits C. These calculations allow us to understand how energy transfers across the packages in motion.
Roller Track Dynamics
Roller track dynamics refers to how objects behave when moving on a roller track. In our exercise, the packages move along such a track with very little friction, allowing for easy motion transfer upon collision. Key components of track dynamics include the initial push-off and the absence of friction. Here, friction is minimal, allowing packages to continue moving after the initial collision with minimal resistance. This affects how accurately we can predict the movements post-collision since energy lost to friction is negligible. This results in longer-lasting motion and more pronounced effects of each collision. Understanding roller track dynamics is essential when calculating the outcome of sequential collisions, as it provides a near-ideal environment to study the pure effects of kinetic energy transfer and the coefficient of restitution in collision scenarios.

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Most popular questions from this chapter

Collar \(A\) has a mass of \(3 \mathrm{kg}\) and is attached to a spring of constant \(1200 \mathrm{N} / \mathrm{m}\) and of undeformed length equal to \(0.5 \mathrm{m} .\) The system is set in motion with \(r=0.3 \mathrm{m}, v_{\theta}=2 \mathrm{m} / \mathrm{s},\) and \(v_{r}=0 .\) Neglecting the mass of the rod and the effect of friction, determine the radial and transverse components of the velocity of the collar when \(r=0.6 \mathrm{m} .\)

Rockfalls can cause major damage to roads and infrastructure. To design mitigation bridges and barriers, engineers use the coefficient of restitution to model the behavior of the rocks. Rock \(A\) falls a distance of \(20 \mathrm{m}\) before striking an incline with a slope of \(\alpha=40^{\circ}\). Knowing that the coefficient of restitution between rock \(A\) and the incline is \(0.2,\) determine the velocity of the rock after the impact.

A 5 -kg sphere is dropped from a height of \(y=2 \mathrm{m}\) to test newly designed spring floors used in gymnastics. The mass of the floor section is \(10 \mathrm{kg}\), and the effective stiffness of the floor is \(k=120 \mathrm{kN} / \mathrm{m}\). Knowing that the coefficient of restitution between the ball and the platform is \(0.6,\) detrmine \((a)\) the height \(h\) reached by the sphere after rebound, \((b)\) the maximum force in the springs.

A 0.25 -lb ball thrown with a horizontal velocity \(\mathrm{v}_{0}\) strikes a 1.5 -lb plate attached to a vertical wall at a height of \(36 \mathrm{in.}\) above the ground. It is observed that after rebounding, the ball hits the ground at a distance of \(24 \mathrm{in.}\) from the wall when the plate is rigidly attached to the wall (Fig. 1) and at a distance of 10 in. when a foam- rubber mat is placed between the plate and the wall (Fig. 2 ). Determine (a) the coefficient of restitution \(e\) between the ball and the plate, (b) the initial velocity \(v_{0}\) of the ball.

A spring is used to stop a \(50-\mathrm{kg}\) package that is moving down a \(20^{\circ}\) incline. The spring has a constant \(\mathrm{k}=30 \mathrm{kN} / \mathrm{m}\) and is held by cables so that it is initially compressed \(50 \mathrm{mm}\). Knowing that the velocity of the package is \(2 \mathrm{m} / \mathrm{s}\) when it is \(8 \mathrm{m}\) from the spring and neglect- ing friction, determine the maximum additional deformation of the spring in bringing the package to rest.

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