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Chapter 8: Problem 22

When cars are equipped with flexible bumpers, they will bounce off cach other during low-specd 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. 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 eneryy of the two-car system during this collision.

Short Answer

Expert verified
a) The lighter car's speed will be 0.499 m/s to the left; b) The change in kinetic energy can be calculated using the formula from Step 4 and the previously found speeds and given masses and will result in the value.

Step by step solution

01

Apply Conservation of Momentum

The total momentum before the collision (\(p_{i}\)) equals the total momentum after the collision (\(p_{f}\)). The formula is \(p_{i} = p_{f}\). We have \(p_{i} = m_{1}v_{1i} + m_{2}v_{2i}\) and \(p_{f} = m_{1}v_{1f} + m_{2}v_{2f}\) where \(v_{1i}\) and \(v_{2i}\) are initial velocities (1.50 m/s and -1.10 m/s), \(m_{1}\) = 1750 kg and \(m_{2}\) = 1450 kg are the masses and \(v_{1f} = 0.250 m/s\) is the final velocity of the first car. Subtracting the knowns, and rearranging for \(v_{2f}\), gives \(v_{2f} = (m_{1}(v_{1i} - v_{1f}) + m_{2}v_{2i}) / m_{2}\)
02

Calculating speed of the lighter car post-collision

Substitute the known values into the equation above. This gives \(v_{2f} = ((1750 kg)(1.50 m/s - 0.250 m/s) + (1450 kg)(-1.10 m/s)) / 1450 kg\). Solving this gives \(v_{2f}\) = -0.499 m/s. The negative sign indicates that the car keeps moving to the left as originally.
03

Calculation of initial and final kinetic energy

The formula for kinetic energy (KE) is \(KE=0.5*m*v^2\). The initial kinetic energy is \(KE_{1i} = 0.5*m_{1}*(v_{1i})^2\) and \(KE_{2i} = 0.5*m_{2}*(v_{2i})^2\). Similarly, the final kinetic energy is \(KE_{1f} = 0.5*m_{1}*(v_{1f})^2\) and \(KE_{2f} = 0.5*m_{2}*(v_{2f})^2\). The total initial and final kinetic energy is then the sum of individual KEs before and after the collision.
04

Calculation of change in kinetic energy

The change in KE (Delta_KE) is \(Delta_KE = KE_i - KE_f\). Substitute the appropriate expressions from Step 3 to obtain \(Delta_KE = [(0.5*m_{1}*(v_{1i})^2 + 0.5*m_{2}*(v_{2i})^2)] - [(0.5*m_{1}*(v_{1f})^2 + 0.5*m_{2}*(v_{2f})^2)]\). Include the values already known.

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Key Concepts

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

Inelastic Collisions
An inelastic collision is one in which the participating bodies do not retain their shapes and kinetic energies after the impact. Unlike elastic collisions, where both the kinetic energy and momentum are conserved, in inelastic collisions, only momentum is conserved while some kinetic energy is transformed into other forms such as heat, sound, or deformation energy.

In the provided exercise, the collision between the two cars can be considered partially inelastic because the cars bounce off each other, indicating some energy is converted into other forms, but they do not stick together, which would be the case in a perfectly inelastic collision. To analyze such collisions, one uses the momentum conservation principle to predict the post-collision velocity of the involved objects, as done in the exercise.
Kinetic Energy
Kinetic energy (KE) is the energy that objects possess due to their motion. The formula for kinetic energy, given by KE = 0.5 * m * v2, reflects how it depends on both the mass (m) of the object and the square of its velocity (v).

In the context of collisions, kinetic energy is important because it helps us understand the dynamics of the process. It also enables us to determine how much energy is dissipated during inelastic collisions, such as sound or heat, by comparing the kinetic energy before and after the collision, as shown in the exercise solution.
Momentum Conservation Principle
The momentum conservation principle is a fundamental law in physics stating that if no external forces act on a system, the total momentum of the system remains constant. The principle is encapsulated in the equation pi = pf, where pi is the initial total momentum and pf is the final total momentum.

This principle is pivotal when solving collision problems in physics, as it allows us to calculate unknown velocities post-collision. During the collision between two cars, as depicted in the exercise, the forces the cars exert on each other are internal to the system, so the system's total momentum is conserved.
Physics of Collisions
The physics of collisions involves analyzing the interactions between objects when they come into contact. There are two primary types of collisions: elastic and inelastic. Elastic collisions are characterized by the conservation of both momentum and kinetic energy. Inelastic collisions, however, only conserve momentum, and some kinetic energy is lost to other forms of energy.

When studying collisions, factors like mass, velocity, and the nature of the collision (elastic or inelastic) are crucial. By using conservation laws, one can glean insightful information about the system's behavior before and after the collision. As outlined in the textbook exercise, by applying these principles, one can determine velocities and changes in kinetic energy, which are essential for understanding the outcomes of the collision.

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Most popular questions from this chapter

One \(110 \mathrm{~kg}\) football lineman is running to the right at \(2.75 \mathrm{~m} / \mathrm{s}\) while another 125 kg lineman is nunning directly toward him at \(2,60 \mathrm{~m} / \mathrm{s}\). What ure (a) the magnitude and direction of the net momentum of these two athletes, and (b) their total kinctic encrgy?

You are standing on a large shect of frictionless ice and holding a large rock. In orver to get off the ice. you throw the rock so it has velocity \(12.0 \mathrm{~m} / \mathrm{s}\) relutive to the eurth at an angle of \(35.0^{\circ}\) above the horizontal. If your mass is \(70.0 \mathrm{~kg}\) and the rock's mass is \(3.00 \mathrm{~kg}\), what is your speed after you throw the rock? (See Discussion Question \(Q 8.7 .)\)

Two ice skaters, Daniel (mass \(65.0 \mathrm{~kg}\) ) and Rebecca (mass \(45.0 \mathrm{~kg}\) ) are practicing. Danicl stops to tic his shoclace and, while at rest, is struck by Rebecca, who is moving at \(13.0 \mathrm{~m} / \mathrm{s}\) before she collides with him. After the collision, Rchecca has a velocity of magnitude \(8.00 \mathrm{~m} / \mathrm{s}\) at an angle of \(53.1^{\circ}\) from her initial direction, Both skaters move on the fric. tionless, horizontal surface of the rink. (a) What are the magnitude and direction of Danicl 's velocity after the collision? (b) What is the change in total kinetic energy of the two skaters as a result of the collision?

In Section 8.5 we calculated the center of mass by considering objects composed of a finite number of point masses or objects that, by symmetry, could be represented by a finite number of point masses. For a solid object whose mass distribution does not allow for a simple determination of the center of mass by symmetry, the sums of Eqs. (8.28) must be generalized to integrals $$ x_{\mathrm{cm}}=\frac{1}{M} \int x d m \quad y_{\mathrm{cm}}=\frac{1}{M} \int y d m $$ where \(x\) and \(y\) are the coordinates of the small picce of the object that has mass \(d m\). The integration is over the whole of the object. Consider a thin rod of length \(I_{4}\), mass \(M\), and cross-sectional area \(A\). I et the origin of the coordinates be at the left end of the rod and the positive \(x\) -axis lie along the rod. (a) If the density \(\rho=M / V\) of the object is uniform, perform the integration described above to show that the \(x\) -coordinate of the center of mass of the rod is at its gcometrical center. (b) If the density of the object varies linearly with \(x\) - that is, \(\rho=\alpha x,\) where \(\alpha\) is a positive constant - calculate the \(x\) -coordinate of the rod's center of mass.

A hunter on a frozen, essentially frictionless pond uses a rifle that shoots \(4.20 \mathrm{~g}\) bullets at \(965 \mathrm{~m} / \mathrm{s}\). The mass of the hunter (including his gun) is \(72.5 \mathrm{~kg}\). and the hunter holds tight to the gun after firing it. Find the recoil velocity of the hunter if he fires the rifle (a) horizontally and (b) at 56.0 " above the horizontal.
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