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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: PEaray = - where A is a constant, m, and m2 are the masses of the two objects, and ris the distance between them. Moreover, the new particle has an additional interaction with the heavy particle through the following force expression 1 qQ Fnew 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 KE1+ KE21 + PEgravf * Uelasticf + Unewf = KE1j + KE2j + 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 KE14+ + Unewf" + Unewi (Equation 1) For all energies, we know the following KE= mv² Am m2 PEgrav = r Ueastic =kx Unew = (1/ where in we have m1 = m, m2 = M, q1 = q and q2 = Q By substituting all these to Equation 1 and then simplifying results to = sqrt m ) - · (1/x ) ) +

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: PEaray = - where A is a constant, m, and m2 are the masses of the two objects, and ris the distance between them. Moreover, the new particle has an additional interaction with the heavy particle through the following force expression 1 qQ Fnew 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 KE1+ KE21 + PEgravf * Uelasticf + Unewf = KE1j + KE2j + 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 KE14+ + Unewf" + Unewi (Equation 1) For all energies, we know the following KE= mv² Am m2 PEgrav = r Ueastic =kx Unew = (1/ where in we have m1 = m, m2 = M, q1 = q and q2 = Q By substituting all these to Equation 1 and then simplifying results to = sqrt m ) - · (1/x ) ) +

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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QUESTION 4 Problem One newly discovered light particle has a mass of m and property g. 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: PEaray = where A is a constant, m1 and m, are the masses of the two objects, and ris the distance between them. r Moreover, the new particle has an additional interaction with the heavy particle through the following force expression 1 Fnew = qQ 4περ r 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 KE1f + KE21 + PEgravf * Uelasticf * Unewf = KE11 + KE2i * 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 KE16+ + Unewf = + Unewi (Equation 1) For all energies, we know the following KE = Am,m2 PEgray = - 1 Uelastic = kx² Unew = (1/ /(r where in we have m1 = m, m2 = M, 91 = q and g2 = Q By substituting all these to Equation 1 and then simplifying results to = sqrt( 2 + ( ( Q m ) - V V - (1/x ) ) + Take note that capital letters have different meaning than small letter variables/constants.
Transcribed Image Text:QUESTION 4 Problem One newly discovered light particle has a mass of m and property g. 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: PEaray = where A is a constant, m1 and m, are the masses of the two objects, and ris the distance between them. r Moreover, the new particle has an additional interaction with the heavy particle through the following force expression 1 Fnew = qQ 4περ r 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 KE1f + KE21 + PEgravf * Uelasticf * Unewf = KE11 + KE2i * 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 KE16+ + Unewf = + Unewi (Equation 1) For all energies, we know the following KE = Am,m2 PEgray = - 1 Uelastic = kx² Unew = (1/ /(r where in we have m1 = m, m2 = M, 91 = q and g2 = Q By substituting all these to Equation 1 and then simplifying results to = sqrt( 2 + ( ( Q m ) - V V - (1/x ) ) + Take note that capital letters have different meaning than small letter variables/constants.
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