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Chapter 8: Problem 17

Is it possible for two masses to undergo a collision such that the system of two masses has more kinetic energy than the two separate masses had? Explain.

Short Answer

Expert verified
Answer: No, it is not possible for two masses to undergo a collision and have more kinetic energy afterward than the two separate masses had before the collision. Both elastic and inelastic collisions cannot result in a system with an increase in total kinetic energy.

Step by step solution

01

Understand types of collisions and conservation of energy

In physics, collisions can be categorized into two types: elastic and inelastic. In an elastic collision, both kinetic energy and momentum are conserved. In an inelastic collision, only momentum is conserved, while kinetic energy is not. The principle of conservation of energy states that the total energy of an isolated system remains constant, meaning that energy cannot be created nor destroyed, only transformed.
02

Analyze elastic collisions

In an elastic collision, both momentum and kinetic energy are conserved. This means that the total kinetic energy of the system (the two colliding masses) before and after the collision remains the same. Thus, for elastic collisions, it is not possible for the system of two masses to have more kinetic energy after the collision than the two separate masses had initially.
03

Analyze inelastic collisions

In an inelastic collision, only momentum is conserved, while kinetic energy is not. This means that some of the initial kinetic energy will be transformed into other forms of energy (e.g., internal, potential, or sound energy) during the collision. Consequently, the total kinetic energy of the system after the collision will always be less than the total kinetic energy of the two separate masses before the collision. As a result, it is not possible for the system of two masses to have more kinetic energy after an inelastic collision than the two separate masses had initially.
04

Conclusion

After analyzing both types of collisions, it becomes evident that it is not possible for two masses to undergo a collision and have more kinetic energy afterward than the two separate masses had before the collision. Both elastic and inelastic collisions cannot result in a system with an increase in total kinetic energy.

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Most popular questions from this chapter

When a bismuth-208 nucleus at rest decays, thallium-204 is produced, along with an alpha particle (helium- 4 nucleus). The mass numbers of bismuth-208, thallium- 204 , and helium- 4 are 208,204 , and 4 , respectively. (The mass number represents the total number of protons and neutrons in the nucleus.) The kinetic energy of the thallium nucleus is a) equal to that of the alpha particle. b) less than that of the alpha particle. c) greater than that of the alpha particle.

A toy car of mass \(2.0 \mathrm{~kg}\) is stationary, and a child rolls a toy truck of mass 3.5 kg straight toward it with a speed of \(4.0 \mathrm{~m} / \mathrm{s}\) a) What is the velocity of the center of mass of the system consisting of the two toys? b) What are the velocities of the truck and the car with respect to the center of mass of the system consisting of the two toys?

The density of a \(1.00-\mathrm{m}\) long rod can be described by the linear density function \(\lambda(x)=\) \(100 \cdot \mathrm{g} / \mathrm{m}+10.0 x \mathrm{~g} / \mathrm{m}^{2}\) One end of the rod is positioned at \(x=0\) and the other at \(x=1.00 \mathrm{~m} .\) Determine (a) the total mass of the rod, and (b) the center-of-mass coordinate.

Many nuclear collisions studied in laboratories are analyzed in a frame of reference relative to the laboratory. A proton, with a mass of \(1.6605 \cdot 10^{-27} \mathrm{~kg}\) and traveling at a speed of \(70.0 \%\) of the speed of light, \(c\), collides with a tin\(116\left({ }^{116} \mathrm{Sn}\right)\) nucleus with a mass of \(1.9096 \cdot 10^{-25} \mathrm{~kg} .\) What is the speed of the center of mass with respect to the laboratory frame? Answer in terms of \(c\), the speed of light.

A thin rectangular plate of uniform area density \(\sigma_{1}=1.05 \mathrm{~kg} / \mathrm{m}^{2}\) has a length \(a=0.600 \mathrm{~m}\) and a width \(b=0.250 \mathrm{~m} .\) The lower left corner is placed at the origin, \((x, y)=(0,0) .\) A circular hole of radius \(r=0.048 \mathrm{~m}\) with center at \((x, y)=(0.068 \mathrm{~m}, 0.068 \mathrm{~m})\) is cut in the plate. The hole is plugged with a disk of the same radius that is composed of another material of uniform area density \(\sigma_{2}=5.32 \mathrm{~kg} / \mathrm{m}^{2}\) What is the distance from the origin of the resulting plate's center of mass?
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