/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 46 A 900 -lb space vehicle travelin... [FREE SOLUTION] | 91Ó°ÊÓ

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A 900 -lb space vehicle traveling with a velocity \(\mathbf{v}_{0}=(1500 \mathrm{ft} / \mathrm{s}) \mathrm{k}\) passes through the origin \(O .\) Explosive charges then separate the vehicle into three parts \(A, B\), and \(C,\) with masses of \(150 \mathrm{lb}, 300 \mathrm{lb}\), and \(450 \mathrm{lb}\), respectively. Knowing that shortly thereafter the positions of the three part are, respectively, \(A(250,250,2250), B(600,1300,\) \(3200),\) and \(C(-475,-950,1900),\) where the coordinates are expressed in feet, that the velocity of \(B\) is \(\mathbf{v}_{B}=(500 \mathrm{ft} / \mathrm{s}) \mathrm{i}+(1100 \mathrm{ft} / \mathrm{s}) \mathrm{j}+\) \((2100 \mathrm{ft} / \mathrm{s}) \mathrm{k},\) and that the \(x\) component of the velocity of \(C\) is \(-400 \mathrm{ft} / \mathrm{s}\), determine the velocity of part \(A\)

Short Answer

Expert verified
Velocity of part A is: \( (200 \mathrm{ft/s}, \ 0, \ -3300 \mathrm{ft/s}) \).

Step by step solution

01

Determine Initial Momentum

The initial momentum of the space vehicle is given by its total mass times its velocity. The total mass of the space vehicle is 900 lb, and its velocity is \( \mathbf{v}_{0} = (1500 \mathrm{ft/s}) \mathbf{k} \). Thus, the initial momentum \( \mathbf{p}_{0} \) is:\[ \mathbf{p}_{0} = 900 \times (0 \mathbf{i} + 0 \mathbf{j} + 1500 \mathbf{k}) = (0, 0, 135000) \text{ lbâ‹…ft/s}. \]
02

Determine Final Momentum of Parts A, B, and C

After the explosion, the vehicle is divided into three parts with known masses and velocities (or velocity components). We calculate the momentum for each part.- **Part B:** Given velocity \( \mathbf{v}_{B} = (500 \mathbf{i} + 1100 \mathbf{j} + 2100 \mathbf{k}) \mathrm{ft/s} \), its momentum \( \mathbf{p}_{B} \) is:\[ \mathbf{p}_{B} = 300 \times (500 \mathbf{i} + 1100 \mathbf{j} + 2100 \mathbf{k}) = (150000, 330000, 630000) \text{ lbâ‹…ft/s}. \]- **Part C:** \(-400 \mathrm{ft/s}\) for \(x\)-component; we assume unknown \(y\) and \(z\) components \(v_{Cy}\) and \(v_{Cz}\). Part C momentum:\[ \mathbf{p}_{C} = 450 \times (-400 \mathbf{i} + v_{Cy} \mathbf{j} + v_{Cz} \mathbf{k}) = (-180000, 450v_{Cy}, 450v_{Cz}) \text{ lbâ‹…ft/s}. \]
03

Apply Conservation of Momentum

The conservation of momentum states the initial total momentum equals the final total momentum:\[ \mathbf{p}_{0} = \mathbf{p}_{A} + \mathbf{p}_{B} + \mathbf{p}_{C}. \] Let's write this for each vector component (\(x, y, z\)):- **x-component:**\[ 0 = 150v_{Ax} + 150000 - 180000. \]- **y-component:**\[ 0 = 150v_{Ay} + 330000 + 450v_{Cy}. \]- **z-component:**\[ 135000 = 150v_{Az} + 630000 + 450v_{Cz}. \]
04

Solve Conservation of Momentum for x-component

Solve the momentum equation for the \(x\)-component.\[ 0 = 150v_{Ax} + 150000 - 180000 \Rightarrow 150v_{Ax} = 30000 \Rightarrow v_{Ax} = 200 \mathrm{ft/s}. \]
05

Use Momentum to Find Unknown Components of C's Velocity

Substitute known values and solve for the unknowns in y and z equations to help determine the conservation requirement:- **z-component:** \[ 135000 = 150v_{Az} + 630000 + 0 \Rightarrow 150v_{Az} = 135000 - 630000 \Rightarrow v_{Az} = -3300 \mathrm{ft/s}. \]

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Key Concepts

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

Vector Mechanics
Vector mechanics is a key tool used in analyzing systems where direction and magnitude play a crucial role. In problems involving momentum, like the explosion of a space vehicle, vectors help in breaking down the velocity and momentum into components in the x, y, and z directions. Each vector component corresponds to the direction of movement. Knowing these components, you can better analyze and predict the behavior of each part after the explosion.

For instance, when the original space vehicle explodes, each of the resultant parts moves in a specific direction. By examining the vectors, you can find how much force is directed along each axis. This insight is achieved by resolving the velocity or momentum vectors, such as finding the velocities needed for conservation of momentum. In this scenario, vector mechanics is invaluable for breaking down the initial and post-explosion movements of the vehicle into easily workable components.
Momentum Equations
Momentum equations form the backbone of calculations in systems involving motion and collision. The principle of conservation of momentum states that the total momentum of a closed system remains constant, provided no external forces are acting upon it. This principle can be expressed using momentum equations.

The initial momentum \( \mathbf{p}_0 \) of a system (like our space vehicle) is derived from multiplying its mass by its initial velocity. After the vehicle explodes, this momentum is distributed among the individual parts, each now having its own mass and velocity. For each piece, we calculate momentum separately. Using momentum equations, we can express these relationships mathematically:
  • For part A: \( \mathbf{p}_A = 150 \times (v_{Ax} \mathbf{i} + v_{Ay} \mathbf{j} + v_{Az} \mathbf{k}) \)
  • For part B: \( \mathbf{p}_B = 300 \times (v_{Bx} \mathbf{i} + v_{By} \mathbf{j} + v_{Bz} \mathbf{k}) \)
  • For part C: \( \mathbf{p}_C = 450 \times (v_{Cx} \mathbf{i} + v_{Cy} \mathbf{j} + v_{Cz} \mathbf{k}) \)
These equations allow for detailed calculations, helping you solve for unknown velocities or affirm that total momentum is conserved.
System of Particles
When analyzing this space vehicle's explosive separation, you deal with a system of particles. Each fragment—parts A, B, and C—is considered a particle within the system. A system of particles in mechanics involves dealing with the collective motion and interaction of multiple particles.

This analysis lets you apply the conservation of momentum principle for the entire system. The initial system, composed of the space vehicle, has its momentum. Post-explosion, you divide this momentum among the individual particles (A, B, and C), observing how each part interacts with the rest of the system.

The total mass and initial velocity give rise to the system's overall initial momentum. After the explosion, calculating the momentum of each part and ensuring they sum up to the original momentum helps affirm the conservation law. This step is crucial in gaining a clearer understanding of particle dynamics, allowing you to predict velocities and positions of each piece accurately.
Velocity Components
Velocity components are fundamental in dissecting how an object moves in different directions. In our example, the exploded parts of the space vehicle have specific velocity components along the \( x, y, \) and \( z \) axes. When a problem provides only partial velocity information, such as the x-component for part C, you must seek the y and z components using momentum conservation rules.

Each velocity component reflects how fast a particle moves in that direction. By isolating these components, you can gather a complete picture of the particle's actual velocity vector. For instance, the given velocities of part B are broken down as follows:
  • \( v_{Bx} = 500 \text{ ft/s} \)
  • \( v_{By} = 1100 \text{ ft/s} \)
  • \( v_{Bz} = 2100 \text{ ft/s} \)
Utilizing velocity components lets you cater solutions to scenarios where motion isn't along a single line but involves complex directional movement. It ensures all parts of the system after an explosion abide by the conservation principles and helps in resolving velocities that aren't straightforward, paving the way to analyze intricate physical systems more effectively.

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Most popular questions from this chapter

A wind turbine generator system having a diameter of \(82.5 \mathrm{m}\) produces \(1.5 \mathrm{MW}\) at a wind speed of \(12 \mathrm{m} / \mathrm{s}\). Determine the diameter of blade necessary to produce \(10 \mathrm{MW}\) of power assuming the efficiency is the same for both designs and \(\rho=1.21 \mathrm{kg} / \mathrm{m}^{3}\) for air.

A bullet is fired with a horizontal velocity of \(1500 \mathrm{ft} / \mathrm{s}\) through a \(6-\) -lb block \(A\) and becomes embedded in a \(4.95-\mathrm{lb}\) block \(B\). Knowing that blocks \(A\) and \(B\) start moving with velocities of \(5 \mathrm{ft} / \mathrm{s}\) and \(9 \mathrm{ft} / \mathrm{s}\), respectively, determine (a) the weight of the bullet, \((b)\) its velocity as it travels from block \(A\) to block \(B\)

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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.

The helicopter shown can produce a maximum downward air speed of \(80 \mathrm{ft}\) in a 30 -ft-diameter slipstream. Knowing that the weight of the helicopter and its crew is \(3500 \mathrm{lb}\) and assuming \(\gamma=0.076 \mathrm{lbft}^{3}\) for air, determine the maximum load that the helicopter can lift while hovering in midair.

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