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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: PEgrav = where A is a constant, m1 and m2 are the masses of the two objects, and r is the distance r between them. Moreover, the new particle has an additional interaction with the heavy particle through the following force expression 1 qQ 4TEO 12 Fnew = where En is a constant that is read as epsilon subscript 0, g 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 + Velasticf + Unewf = KE1¡ + 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 KE1F+ - Unewf = + Unewi (Equation 1) For all energies, we know the following KE= Am¡m2 PEgrav= r 1 U elastic %3D 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 = sgrt( 2. m ) - V v ) - (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 Consider a new expression for gravitation potential energy as: PEgrav = where A is a constant, m1 and m2 are the masses of the two objects, and r is the distance r between them. Moreover, the new particle has an additional interaction with the heavy particle through the following force expression 1 qQ 4TEO 12 Fnew = where En is a constant that is read as epsilon subscript 0, g 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 + Velasticf + Unewf = KE1¡ + 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 KE1F+ - Unewf = + Unewi (Equation 1) For all energies, we know the following KE= Am¡m2 PEgrav= r 1 U elastic %3D 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 = sgrt( 2. m ) - V v ) - (1/x ) ) + Take note that capital letters have different meaning than small letter variables/constants.

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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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, m1 and m2 are the masses of the two objects, and r is the distance r between them. Moreover, the new particle has an additional interaction with the heavy particle through the following force expression 1 qQ Fnew 4TEO r2 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 = KE1i + KE2i + PEgravi + + Unewi %3D 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 KE1F+ Unewf = + + + + + Unewi (Equation 1) For all energies, we know the following 1 KE=mv? 2 Am,m2 PEgrav r 1 U elastic = Unew = (1/ /(r 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( 2 + ( ( V m ) - %3D V ) - (1/x ) ) + Take note that capital letters have different meaning than small letter variables/constants.
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 Consider a new expression for gravitation potential energy as: PE grav where A is a constant, m1 and m2 are the masses of the two objects, and r is the distance r between them. Moreover, the new particle has an additional interaction with the heavy particle through the following force expression 1 qQ Fnew 4TEO r2 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 = KE1i + KE2i + PEgravi + + Unewi %3D 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 KE1F+ Unewf = + + + + + Unewi (Equation 1) For all energies, we know the following 1 KE=mv? 2 Am,m2 PEgrav r 1 U elastic = Unew = (1/ /(r 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( 2 + ( ( V m ) - %3D V ) - (1/x ) ) + Take note that capital letters have different meaning than small letter variables/constants.
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