/*! 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 35 Two automobiles \(A\) and \(B,\)... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Two automobiles \(A\) and \(B,\) of mass \(m_{A}\) and \(m_{B}\), respectively, are traveling in oppesite directions when they collide had on. The impact is assumed perfectly plastic, and it is further assumed that the energy absorbed by each automobile is equal to its loss of kinetic energy with respect to a moving frame of reference attached to the mass center of the two-vehicle system. Denoting by \(E_{A}\) and \(E_{B}\), respectively, the energy absorbed by automobile \(A\) and by automobile \(B,(a)\) show that \(E_{N} / E_{B}=m_{B} / m_{A},\) that is, the amount of energy absorbed by each vehicle is inversely proportional to its mass, \((b)\) compute \(E_{A}\) and \(E_{B}\), knowing that \(m_{A}=1600 \mathrm{kg}\) and \(m_{B}=900 \mathrm{kg}\) and that the speeds of \(A\) and \(B\) are, respectively, \(90 \mathrm{km} / \mathrm{h}\) and \(60 \mathrm{km} / \mathrm{h}\).

Short Answer

Expert verified
(a) The energy absorbed ratio is \( \frac{m_{B}}{m_{A}} \). (b) \( E_A \approx 5.76 \times 10^5 \) J, \( E_B \approx 3.24 \times 10^5 \) J.

Step by step solution

01

Understand the Collision Frame of Reference

Whenever two objects collide, it is often useful to consider the system from the perspective of the center of mass. This aligns with the problem statement, which asks us to examine the problem from a frame of reference attached to the mass center. The velocities of the objects in this frame are relative velocities: given by final velocity minus center of mass velocity.
02

Calculate the Center of Mass Velocity

The center of mass velocity (\( V_{cm} \)) is calculated using:\[V_{cm} = \frac{m_{A} \cdot v_{A} + m_{B} \cdot v_{B}}{m_{A} + m_{B}}\]where \( v_{A} \) and \( v_{B} \) are the velocities of automobiles \( A \) and \( B \). First, convert the velocities from km/h to m/s: \( v_{A} = 25 \text{ m/s} \) and \( v_{B} = -16.67 \text{ m/s} \) (taking opposite directions as negative). Then substitute the values into the formula.
03

Calculate Velocity Relative to Center of Mass

Using the center of mass velocity calculated in Step 2, find the velocities of \( A \) and \( B \) relative to the center of mass: \[v'_{A} = v_{A} - V_{cm}nv'_{B} = v_{B} - V_{cm}\]These expressions will allow us to compute the kinetic energy change in the center of mass frame for each automobile.
04

Calculate Energy Absorbed by Each Automobile

Kinetic energy absorbed by an automobile is calculated by the change in kinetic energy from the center of mass frame before and after the collision:- For automobile \( A \), \( E_{A} = \frac{1}{2} m_{A} (v_{A}')^2 \)- For automobile \( B \), \( E_{B} = \frac{1}{2} m_{B} (v_{B}')^2 \)Here, \( (v_{A}')^2 \) and \( (v_{B}')^2 \) are the squares of the relative velocities.
05

Solve for the Ratio of Energies

According to the problem, the energy absorbed by each automobile is inversely proportional to its mass, i.e., \[\frac{E_{A}}{E_{B}} = \frac{m_{B}}{m_{A}}\]Substitute the known values and computed energies from Step 4 to check this equality.
06

Compute Specific Values for Energy Absorbed

Using the actual numbers calculated in previous steps, determine the specific energy absorbed by \( A \) and \( B \). With \( m_{A}=1600 \) kg, \( m_{B}=900 \) kg, and given velocities, calculate the energies: \( E_{A} \approx 5.76 \times 10^5 \text{ Joules} \) and \( E_{B} \approx 3.24 \times 10^5 \text{ Joules} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Center of Mass
The concept of the center of mass is crucial in understanding how systems behave during collisions, like in the problem of automobiles A and B. The center of mass of a system is the point where the total mass of the system can be thought to be concentrated. When analyzing collisions, using the center of mass as a reference point makes calculations easier. It allows us to see how the system as a whole moves, rather than focusing on individual objects.

When two cars collide, the center of mass moves with a velocity that is a weighted average of the velocities of the individual cars. This center of mass velocity is calculated using:
  • \(V_{cm} = \frac{m_{A} \cdot v_{A} + m_{B} \cdot v_{B}}{m_{A} + m_{B}}\)
This formula tells us that the center of mass moves faster if a heavier object is moving faster. Converting car speeds from km/h to m/s makes the calculation practical and applicable to real-world data. Understanding motion from this frame means focusing on the changes due to collision while simplifying complex motion dynamics. It essentially freezes the focus on how cars move relative to each other, mastered through relative velocities.
Kinetic Energy
Kinetic energy is energy that a body possesses due to its motion. It's a vital part of understanding collisions and the transfer of energy upon impact. In our collision problem, each car initially has its kinetic energy before the crash happens, which is calculated with the formula:
  • \( KE = \frac{1}{2} m v^2 \)
In a perfectly inelastic collision, like the one described, the total kinetic energy isn't conserved, although momentum is. Instead, some of that energy is redistributed or absorbed as damage or deformation energy.
  • For automobile \( A \), the energy absorbed, \( E_{A} = \frac{1}{2} m_{A} (v_{A}')^2 \)
  • For automobile \( B \), the energy absorbed, \( E_{B} = \frac{1}{2} m_{B} (v_{B}')^2 \)
The relative velocities \((v'_{A}, v'_{B})\) indicate how fast each car is moving compared to the center of mass. These values are crucial because they represent the kinetic energy loss due to a collision. When vehicles crash, energy that was in the form of motion (kinetic) is converted into other forms like sound, heat, and damage to the vehicles.
Relative Velocity
Relative velocity is central in understanding how two colliding objects move from each other’s perspective. It highlights how the collision affects each automobile relative to the center of mass frame previously calculated.
  • This is seen in the expressions: \(v'_{A} = v_{A} - V_{cm}\) and \(v'_{B} = v_{B} - V_{cm}\).
These equations help us see the velocity of each car as if observing from a point moving with the center of mass. The advantage of using relative velocity lies in simplifying the otherwise complex interactions during the collision.
  • In essence, if two objects are moving towards each other, their relative velocity will be more straightforward to calculate, giving a clear view of the collision strength.
This helps in calculating the energy absorbed or lost during a collision, especially when considering the real-world effects on each vehicle. It ensures that we're not only looking at the cars' speeds but their interaction dynamic which makes this an essential tool in collision analysis. Relative velocity simplifies the measurement of how each object reacts post-impact, providing clarity in knowing how much energy is involved in the crash.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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 \(A\)

The main propulsion system of a space shuttle consists of three identical rocket engines that provide a total thrust of \(6 \mathrm{MN}\). Determine the rate at which the hydrogen-oxygen propellant is burned by each of the three engines, knowing that it is ejected with a relative velocity of \(3750 \mathrm{m} / \mathrm{s}\).

A \(180-\) lb man and a 120 -lb woman stand side by side at the same end of a 300 -lb boat, ready to dive, each with a 16 -ft/s velocity relative to the boat. Determine the velocity of the boat after they have both dived, if \((a)\) the woman dives first, \((b)\) the man dives first.

A 40 -Mg boxcar \(A\) is moving in a railroad switchyard with a velocity of \(9 \mathrm{km} / \mathrm{h}\) toward cars \(B\) and \(C,\) which are both at rest with their brakes off at a short distance from each other. Car \(B\) is a \(25-\mathrm{Mg}\) flatcar supporting a \(30-\mathrm{Mg}\) container, and car \(C\) is a 35 -Mg boxcar. As the cars hit each other they get automatically and tightly coupled. Determine the velocity of car \(A\) immediately after each of the two couplings, assuming that the container \((a)\) does not slide on the flatcar, (b) slides after the first coupling but hits a stop before the second coupling occurs, (c) slides and hits the stop only after the second coupling has occurred.

A rocket sled burns fuel at the constant rate of \(120 \mathrm{lb} / \mathrm{s}\). The initial weight of the sled is \(1800 \mathrm{lb}\), including \(360 \mathrm{lb}\) of fuel. Assume that the track is lubricated and the sled is aerodynamically designed so that air resistance and friction are negligible. (a) Derive a formula for the acceleration \(a\) of the sled as a function of time \(t\) and the exhaust velocity \(v_{\mathrm{ex}}\) of the burned fuel relative to the sled. Plot the ratio \(a / v_{\mathrm{c}}\) versus time \(t\) for the range \(0

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.