/*! 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 18 When cars are equipped with flex... [FREE SOLUTION] | 91Ó°ÊÓ

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When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less damage. In one such accident, a \(1750-\mathrm{kg}\) car traveling to the right at 1.50 \(\mathrm{m} / \mathrm{s}\) collides with a \(1450-\mathrm{kg}\) car going to the left at 1.10 \(\mathrm{m} / \mathrm{s}\) . Measurements show that the heavier car's speed just after the collision was 0.250 \(\mathrm{m} / \mathrm{s}\) in its original direction. You can ignore any road friction during the collision. (a) What was the speed of the lighter car just after the collision? (b) Calculate the change in the combined kinetic energy of the two-car system during this collision.

Short Answer

Expert verified
(a) The speed of the lighter car after the collision is approximately 0.409 m/s. (b) The total kinetic energy decreased by about 2670 J.

Step by step solution

01

Identify Initial Conditions

We have two cars involved in a collision. Car 1 (the heavier car) has a mass of \( m_1 = 1750 \, \text{kg} \) and a velocity \( v_{1i} = 1.50 \, \text{m/s} \) to the right. Car 2 (the lighter car) has a mass of \( m_2 = 1450 \, \text{kg} \) and a velocity \( v_{2i} = -1.10 \, \text{m/s} \) (to the left, hence the negative sign). After the collision, car 1 has a velocity \( v_{1f} = 0.250 \, \text{m/s} \) in its original direction.
02

Apply Conservation of Momentum

The law of conservation of momentum states that the total momentum before the collision equals the total momentum after the collision. That is, \( m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \). Plug in the known values: \( (1750)(1.50) + (1450)(-1.10) = (1750)(0.250) + (1450)v_{2f} \).
03

Solve for Final Velocity of Car 2

Solving the equation from Step 2: \( 2625 - 1595 = 437.5 + 1450v_{2f} \). This simplifies to \( 1030 = 437.5 + 1450v_{2f} \), therefore \( 1450v_{2f} = 592.5 \). Solve for \( v_{2f} \): \( v_{2f} = \frac{592.5}{1450} \approx 0.409 \, \text{m/s} \).
04

Calculate Initial Kinetic Energy

The initial kinetic energy \( KE_i \) is the sum of the kinetic energies of both cars before the collision: \( KE_i = \frac{1}{2} m_1 v_{1i}^2 + \frac{1}{2} m_2 v_{2i}^2 \). Calculating, \( KE_i = \frac{1}{2}(1750)(1.50)^2 + \frac{1}{2}(1450)(1.10)^2 = 1968.75 + 877.25 = 2846 \, \text{J} \).
05

Calculate Final Kinetic Energy

The final kinetic energy \( KE_f \) is the sum of the kinetic energies of both cars after the collision: \( KE_f = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2 \). Calculating, \( KE_f = \frac{1}{2}(1750)(0.250)^2 + \frac{1}{2}(1450)(0.409)^2 = 54.6875 + 121.23005 \approx 175.91755 \, \text{J} \).
06

Determine Change in Kinetic Energy

The change in the total kinetic energy \( \Delta KE \) is \( KE_f - KE_i \). Thus, \( \Delta KE = 175.91755 - 2846 \approx -2670.08 \, \text{J} \). This indicates that the kinetic energy decreased by about 2670 J.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. It can be calculated using the formula: \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass and \( v \) is the velocity of the object.
In our car collision scenario, we first calculate the kinetic energy of both vehicles before and after the collision to understand how energy is distributed. Initially, car 1 and car 2 have kinetic energies of 1968.75 J and 877.25 J, respectively. After the collision, these values change to 54.6875 J for car 1 and 121.23005 J for car 2.
The significant decrease in the total kinetic energy from 2846 J to about 175.92 J suggests that much of the energy was transferred elsewhere, such as sound or heat, typical in inelastic collisions. This loss is essential to understanding real-world physics, where energy is conserved but often changes form.
Elastic Collision
An elastic collision is a collision in which both momentum and kinetic energy are conserved. When the cars collide, it’s essential to differentiate whether the collision is elastic or inelastic. In an elastic collision, the total kinetic energy before and after the impact would remain the same.
In our example, the total kinetic energy before the collision significantly decreased after the collision—about 2670 J was lost. This suggests that the collision was not perfectly elastic. Although the cars bounce off each other, energy is lost to other forms, classically seen in real-world, low-speed collisions.
Understanding the distinction between elastic and inelastic collisions helps predict post-collision conditions: in elastic collisions, both objects rebound with no kinetic energy loss; in inelastic ones, some energy typically converts to heat or deformation, as observed in this scenario.
Momentum Equation
The momentum equation is a fundamental principle in physics described by the law of conservation of momentum. It states that the total momentum in an isolated system remains constant if no external forces act upon it.
The equation is expressed as: \( m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \). This equation simply represents the total momentum before a collision equaling the total momentum after the collision.
In the given problem, applying this principle, we calculated the final velocity of the lighter car. With the initial velocities and the final velocity of the heavier car known, we solved for the unknown velocity and found it to be approximately 0.409 m/s. This constant exchange of momentum, irrespective of energy losses, illustrates the core concept underpinning many physical scenarios.

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Most popular questions from this chapter

A stationary object with mass \(m_{B}\) is struck head-on by an object with mass \(m_{A}\) that is moving initially at speed \(v_{0}\) , (a) If the collision is elastic, what percentage of the original energy does each object have after the collision? (b) What does your answer in part (a) give for the special cases \((1) m_{A}=m_{B}\) and \((i i) m_{A}=5 m_{B} ?(c)\) For what values, if any, of the mass ratio \(m_{A} / m_{B}\) is the original kinetic energy shared equally by the two objects after the collision?

A \(\mathrm{C} 6-5\) model rocket engine has an impolse of 10.0 \(\mathrm{N}\) . for 1.70 \(\mathrm{s}\) , while buming 0.0125 \(\mathrm{kg}\) of propellant. It has a maximum thrust of 13.3 \(\mathrm{N}\) . The initial mass of the engine plus propellant is 0.0258 \(\mathrm{kg}\) , (a) What fraction of the maximum thrust is the average thrust? (b) Calculate the relative speed of the exhaust gases, assuming it is constant. (c) Assuming that the relative speed of the exhaust gases is constant, find the final speed of the engine if it was attached to a very light frame and fired from rest in gravity-free outer space.

A 10.0 -g marble slides to the left with a velocity of magnitude 0.400 \(\mathrm{m} / \mathrm{s}\) on the frictionless, horizontal surface of an icy New York sidewalk. and has a head- on, elastic collision with a larger \(30.0-\mathrm{g}\) marblesliding to the right with a velocity of magnitude 0.200 \(\mathrm{m} / \mathrm{s}\) (Fig. 8.38\() .\) (a) Find the velocity of each marble (magnitude and direction) after the collision. (Since the collision is head-on, all the motion is along a line. (b) Calculate the change in momentum (that is, the momentum after the collision minus the momentum before the collision) for each marble. Compare the values you get for each marble. (c) Calculate the change in kinetic energy (that is, the kinetic energy after the collision minus the kinetic energy before the collision) for each marble. Compare the values you get for each marble.

Pluto's diameter is approximately \(2370 \mathrm{km},\) and the diameter of its satellite Charon is 1250 \(\mathrm{km}\) . Although the distance varies, they are often about \(19,700 \mathrm{km}\) apart, center-to-center. Assuming that both Pluto and Charon have the same composition and hence the same average density, find the location of the center of mass of this system relative to the center of Pluto.

Changing Mass. An open-topped freight car with mass \(24,000 \mathrm{kg}\) is coasting without friction along a level track. It is raining very hard, and the rain is falling vertically downward. Originally, the car is empty and moving with a speed of 4.00 \(\mathrm{m} / \mathrm{s}\) . What is the speed of the car after it has collected 3000 \(\mathrm{kg}\) of rainwater?

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