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Problem One newly discovered light particle has a mass of m and property q. Suppose it moves within the vicinity of an extremely heavy (fixed in place) particle with a property Q and mass M. When the light particle is xi distance from the heavy particle, it is moving directly away from the heavy particle with a speed of vi. a) What is the lighter particle's speed when it is xf away from the heavy particle? Am m2 Consider a new expression for gravitation potential energy as: PE -grav where A is a constant, m, and m2 are the masses of the two objects, and r is the distance between them. Moreover, the new particle has an additional interaction with the heavy particle through the following force expression 1 Fnew 4TtEO 2 where Eo is a constant that is read as epsilon subscript 0, q and Q are their new properties, r is the distance between the new particle and the heavy particle. Solution: We may solve this using two approaches. One involves the Newton's Laws and the other involving Work-Energy theorem. To avoid the complexity of vector solution. we will instead employ the Work-Energy theorem, more specifically, the Conservation of Energy Principle. Let us first name the lighter particle as object 1 and the heavy particle as object 2. Through work-energy theorem, we will take into account all of the energy of the two-charged particle system before and after traveling a certain distance as KE1F+ KE2F + PEgravf + Uelasticf + Unewf = KE + KE2 + PEgravi + + Unewi Since the heavy particle remains fixed, before and after the motion of the lighter particle. it does not have any velocity, moreover, there is no spring involved, so KE4+ - Unewf =

Problem One newly discovered light particle has a mass of m and property q. Suppose it moves within the vicinity of an extremely heavy (fixed in place) particle with a property Q and mass M. When the light particle is xi distance from the heavy particle, it is moving directly away from the heavy particle with a speed of vi. a) What is the lighter particle's speed when it is xf away from the heavy particle? Am m2 Consider a new expression for gravitation potential energy as: PE -grav where A is a constant, m, and m2 are the masses of the two objects, and r is the distance between them. Moreover, the new particle has an additional interaction with the heavy particle through the following force expression 1 Fnew 4TtEO 2 where Eo is a constant that is read as epsilon subscript 0, q and Q are their new properties, r is the distance between the new particle and the heavy particle. Solution: We may solve this using two approaches. One involves the Newton's Laws and the other involving Work-Energy theorem. To avoid the complexity of vector solution. we will instead employ the Work-Energy theorem, more specifically, the Conservation of Energy Principle. Let us first name the lighter particle as object 1 and the heavy particle as object 2. Through work-energy theorem, we will take into account all of the energy of the two-charged particle system before and after traveling a certain distance as KE1F+ KE2F + PEgravf + Uelasticf + Unewf = KE + KE2 + PEgravi + + Unewi Since the heavy particle remains fixed, before and after the motion of the lighter particle. it does not have any velocity, moreover, there is no spring involved, so KE4+ - Unewf =

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Since the heavy particle remains fixed, before and after the motion of the lighter particle, it does not have any velocity, moreover, there is no spring involved, so KE1 + +Unewf + Unewi (Equation 1) For all energies, we know the following KE= mv 2. Am,m2 PEgrav 1. U elastic /(r Unew = (1/ where in we have m1 = m, m2 = M, q1 = q and q, = Q By substituting all these to Equation 1 and then simplifying results to m) - 2 ( ( = sqrt ) (1/x Take note that capital letters have different meaning than small letter variables/constants.
Transcribed Image Text:Since the heavy particle remains fixed, before and after the motion of the lighter particle, it does not have any velocity, moreover, there is no spring involved, so KE1 + +Unewf + Unewi (Equation 1) For all energies, we know the following KE= mv 2. Am,m2 PEgrav 1. U elastic /(r Unew = (1/ where in we have m1 = m, m2 = M, q1 = q and q, = Q By substituting all these to Equation 1 and then simplifying results to m) - 2 ( ( = sqrt ) (1/x Take note that capital letters have different meaning than small letter variables/constants.
Problem One newly discovered light particle has a mass of m and property q. Suppose it moves within the vicinity of an extremely heavy (fixed in place) particle with a property Q and mass M. When the light particle is xi distance from the heavy particle, it is moving directly away from the heavy particle with a speed of vi. a) What is the lighter particle's speed when it is xf away from the heavy particle? Am,m2 where A is a constant, m, and m, are the masses of the two Consider a new expression for gravitation potential energy as: PE grav objects, and r is the distance between them. Moreover, the new particle has an additional interaction with the heavy particle through the following force expression Fnew 4TE0 where En is a constant that is read as epsilon subscript 0, q and Q are their new properties.r is the distance between the new particle and the heavy particle. Solution: We may solve this using two approaches. One involves the Newton's Laws and the other involving Work-Energy theorem. To avoid the complexity of vector solution. we will instead employ the Work-Energy theorem, more specifically, the Conservation of Energy Principle. Let us first name the lighter particle as object 1 and the heavy particle as object 2. Through work-energy theorem, we will take into account all of the energy of the two-charged particle system before and after traveling a certain distance as -Unewi KE1F+ KE2F + PEgravf Velasticf- Unewf- KE1it KE2 + PEgravi Since the heavy particle remains fixed, before and after the motion of the Ilighter partice, it does not have any velocity, moreover, there is no spring involved, so -Unewf KEf-
Transcribed Image Text:Problem One newly discovered light particle has a mass of m and property q. Suppose it moves within the vicinity of an extremely heavy (fixed in place) particle with a property Q and mass M. When the light particle is xi distance from the heavy particle, it is moving directly away from the heavy particle with a speed of vi. a) What is the lighter particle's speed when it is xf away from the heavy particle? Am,m2 where A is a constant, m, and m, are the masses of the two Consider a new expression for gravitation potential energy as: PE grav objects, and r is the distance between them. Moreover, the new particle has an additional interaction with the heavy particle through the following force expression Fnew 4TE0 where En is a constant that is read as epsilon subscript 0, q and Q are their new properties.r is the distance between the new particle and the heavy particle. Solution: We may solve this using two approaches. One involves the Newton's Laws and the other involving Work-Energy theorem. To avoid the complexity of vector solution. we will instead employ the Work-Energy theorem, more specifically, the Conservation of Energy Principle. Let us first name the lighter particle as object 1 and the heavy particle as object 2. Through work-energy theorem, we will take into account all of the energy of the two-charged particle system before and after traveling a certain distance as -Unewi KE1F+ KE2F + PEgravf Velasticf- Unewf- KE1it KE2 + PEgravi Since the heavy particle remains fixed, before and after the motion of the Ilighter partice, it does not have any velocity, moreover, there is no spring involved, so -Unewf KEf-
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