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Chapter 7: Problem 4

The value of the momentum for a system is the same at a later time as at an earlier time if there are no a) collisions between particles within the system. b) inelastic collisions between particles within the system. c) changes of momentum of individual particles within the system. d) internal forces acting between particles within the system. e) external forces acting on particles of the system.

Short Answer

Expert verified
a) Collisions within the system b) Inelastic collisions between particles c) No changes in the momentum of individual particles within the system d) Internal forces acting between particles e) External forces acting on particles of the system Answer: c) No changes in the momentum of individual particles within the system

Step by step solution

01

Understanding Conservation of Momentum Principle

The conservation of momentum states that the total momentum of a closed system remains constant if no external forces act on the system. In other words, the sum of the momenta of individual particles within the system remains the same over time if no external forces influence the particles.
02

Evaluating each option

Now, let's evaluate each given option: a) Collisions within the system might transfer momentum between particles, but the total momentum will remain constant, so this option is not right. b) Inelastic collisions between particles might cause energy dissipation as heat, but they still conserve the overall momentum of the system. Therefore, this option is incorrect as well. c) If there are no changes in the momentum of individual particles, the total momentum will definitely be conserved. This option seems correct, but we should evaluate other options as well. d) Internal forces acting between particles only redistribute the total momentum among the particles, they do not change the overall momentum of the system. Thus, this option is not correct. e) External forces acting on particles of the system can change the total momentum, but the question is about scenarios when the momentum remains constant. Hence, this option is not correct.
03

Choosing the correct option

Based on the evaluation of each option and principles of conservation of momentum, the correct answer is (c) changes of momentum of individual particles within the system.

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Most popular questions from this chapter

During an ice-skating extravaganza, Robin Hood on Ice, a 50.0 -kg archer is standing still on ice skates. Assume that the friction between the ice skates and the ice is negligible. The archer shoots a \(0.100-\mathrm{kg}\) arrow horizontally at a speed of \(95.0 \mathrm{~m} / \mathrm{s} .\) At what speed does the archer recoil?

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Rank the following objects from highest to lowest in terms of momentum and from highest to lowest in terms of energy. a) an asteroid with mass \(10^{6} \mathrm{~kg}\) and speed \(500 \mathrm{~m} / \mathrm{s}\) b) a high-speed train with a mass of \(180,000 \mathrm{~kg}\) and a speed of \(300 \mathrm{~km} / \mathrm{h}\) c) a 120 -kg linebacker with a speed of \(10 \mathrm{~m} / \mathrm{s}\) d) a 10 -kg cannonball with a speed of \(120 \mathrm{~m} / \mathrm{s}\) e) a proton with a mass of \(6 \cdot 10^{-27} \mathrm{~kg}\) and a speed of \(2 \cdot 10^{8} \mathrm{~m} / \mathrm{s}\).

A method for determining the chemical composition of a material is Rutherford backscattering (RBS), named for the scientist who first discovered that an atom contains a high-density positively charged nucleus, rather than having positive charge distributed uniformly throughout (see Chapter 39 ). In RBS, alpha particles are shot straight at a target material, and the energy of the alpha particles that bounce directly back is measured. An alpha particle has a mass of \(6.65 \cdot 10^{-27} \mathrm{~kg} .\) An alpha particle having an initial kinetic energy of \(2.00 \mathrm{MeV}\) collides elastically with atom X. If the backscattered alpha particle's kinetic energy is \(1.59 \mathrm{MeV}\), what is the mass of atom \(\mathrm{X}\) ? Assume that atom \(X\) is initially at rest. You will need to find the square root of an expression, which will result in two possible an- swers (if \(a=b^{2},\) then \(b=\pm \sqrt{a}\) ). Since you know that atom \(X\) is more massive than the alpha particle, you can choose the correct root accordingly. What element is atom X? (Check a periodic table of elements, where atomic mass is listed as the mass in grams of 1 mol of atoms, which is \(6.02 \cdot 10^{23}\) atoms.)
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