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Chapter 8: Problem 22

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 -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 1.13 m/s; (b) The change in kinetic energy is -1480 J.

Step by step solution

01

Understand the Conservation of Momentum

The law of conservation of momentum states that the total momentum of a system remains constant if no external forces act on it. For this collision, since there is no road friction, the total momentum before and after the collision remains the same.
02

Calculate Initial Momentum

Calculate the momentum of each car before the collision. \( p_1 = m_1 \times v_1 = 1750 \times 1.50 \) (to the right), and \( p_2 = m_2 \times (-v_2) = 1450 \times (-1.10) \) (to the left). Thus, the total initial momentum is \( p_{\text{initial}} = p_1 + p_2 \).
03

Calculate Final Momentum of the Heavier Car

Given that the speed of the heavier car after the collision is 0.250 m/s, calculate its final momentum: \( p_1' = m_1 \times v_1' = 1750 \times 0.250 \).
04

Set Up the Momentum Conservation Equation

Set the total initial momentum from Step 2 equal to the total final momentum: \[ p_{\text{initial}} = p_1' + p_2' \] where \( p_2' \) is the final momentum of the lighter car. Solve this equation for \( v_2' \), which is the speed of the lighter car after the collision.
05

Calculate the Speed of the Lighter Car (Part a)

Substitute the known values into the momentum equation: \[ 1750 \times 1.50 + 1450 \times (-1.10) = 1750 \times 0.250 + 1450 \times v_2' \]. Solve for \( v_2' \).
06

Calculating Initial Kinetic Energy

The kinetic energy of the cars before the collision is calculated using the formula: \[ KE_{\text{initial}} = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]. Substitute the known values to find \( KE_{\text{initial}} \).
07

Calculating Final Kinetic Energy

The kinetic energy after the collision is given by: \[ KE_{\text{final}} = \frac{1}{2} m_1 (v_1')^2 + \frac{1}{2} m_2 (v_2')^2 \]. Use the speed of the heavier car after the collision and the speed found for the lighter car.
08

Calculate the Change in Kinetic Energy (Part b)

The change in the kinetic energy of the system during the collision is: \[ \Delta KE = KE_{\text{final}} - KE_{\text{initial}} \]. Substitute the initial and final kinetic energies to find this change.

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Key Concepts

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

Conservation of Momentum
In collision physics, the law of conservation of momentum is fundamental. It states that if no external forces act on a system, the total momentum will stay constant throughout the process.
In our car collision scenario, we have two cars moving towards each other with no external forces like road friction interfering. The total momentum before and after the collision remains unchanged.
This principle allows us to set the sum of momenta before and after the event to be equal, essentially setting up our equation to solve for unknown velocities. This powerful law helps us understand not just car collisions, but also how objects interact in a closed system.
Kinetic Energy
Kinetic energy is all about the energy something has due to motion. In our collision case, the cars have kinetic energy because they are moving. When the cars collide, some of that energy might be transformed into other forms like sound and heat, especially if they crash hard without bouncing back.
To calculate the kinetic energy, we use the formula:\[KE = \frac{1}{2}mv^2\]where \(m\) is the mass and \(v\) is the velocity.
By comparing the kinetic energy before and after the collision, we can determine whether energy was transformed or stayed as kinetic. This comparison is crucial to understanding energy dynamics in collisions.
Elastic Collisions
Elastic collisions occur when the total kinetic energy before and after the collision remains constant. This means the objects bounce off each other without losing energy to sound, heat, or deformation.
For our scenario, the fact that the heavier car's speed is measured right after the collision lets us apply this concept of elastic collision, assuming minimal energy is lost. However, real-life collisions might not be perfectly elastic.
The assumption of elasticity simplifies understanding and solving collision problems, even though not always 100% applicable.
Momentum Equation
The momentum equation is the heart of solving collision problems. It involves setting initial and final total momenta equal to apply the law of conservation of momentum.
For the cars:\[p_{\text{initial}} = p_1 + p_2 \]\[p_{\text{final}} = p_1' + p_2'\]Here, \(p_1\) and \(p_2\) are initial momenta of the cars before collision, and \(p_1'\) and \(p_2'\) are after.
By using this equation, we solve for unknowns, such as post-collision speeds. The momentum equation is straightforward, yet essential, for understanding and analyzing interactions in physics effectively.

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Most popular questions from this chapter

Energy Sharing. An object with mass \(m,\) initially at rest, explodes into two fragments, one with mass \(m_{A}\) and the other with mass \(m_{B},\) where \(m_{A}+m_{B}=m .\) (a) If energy \(Q\) is released in the explosion, how much kinetic energy does each fragment have immediately after the explosion? (b) What percentage of the total energy released does each fragment get when one fragment has four times the mass of the other?

A 15.0 -kg fish swimming at 1.10 \(\mathrm{m} / \mathrm{s}\) suddenly gobbles up a 4.50 -kg fish that is initially stationary. Neglect any drag effects of the water. (a) Find the speed of the large fish just after it eats the small one. (b) How much mechanical energy was dissipated during this meal?

On a very muddy football field, a 110 -kg linebacker tackles an \(85-\) kg halfback. Immediately before the collision, the line-backer is slipping with a velocity of 8.8 \(\mathrm{m} / \mathrm{s}\) north and the halfback is sliding with a velocity of 7.2 \(\mathrm{m} / \mathrm{s}\) east. What is the velocity (magnitude and direction) at which the two players move together immediately after the collision?

A 1200 -kg station wagon is moving along a straight highway at 12.0 \(\mathrm{m} / \mathrm{s}\) . Another car, with mass 1800 \(\mathrm{kg}\) and speed \(20.0 \mathrm{m} / \mathrm{s},\) has its center of mass 40.0 \(\mathrm{m}\) ahead of the center of mass of the station wagon (Fig. E8.54). (a) Find the position of the center of mass of the system consisting of the two automobiles. (b) Find the magnitude of the total momentum of the system from the given data. (c) Find the speed of the center of mass of the system. (d) Find the total momentum of the system, using the speed of the center of mass. Compare your result with that of part (b).

A C6-5 model rocket engine has an impulse of 10.0 \(\mathrm{N} \cdot \mathrm{s}\) while burning 0.0125 \(\mathrm{kg}\) of propellant in 1.70 s. 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.
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