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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 Fnew 1 qQ 4TTE, ? 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 = 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 KE16+ + Unewf = + Unewi (Equation 1) For all energies, we know the following 1 -mv² KE = Am¡m2 PEgrav r Uelastic = kx? 1 kx² Unew = (1/ where in we have m1 = m, m2 = M, 91 = q and q2 = Q By substituting all these to Equation 1 and then simplifying results to = sqrtt v 2 + ( ( m ) - V Q ) - (1/x ) ) Take note that capital letters have different meaning than small letter variables/constants. +

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 Fnew 1 qQ 4TTE, ? 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 = 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 KE16+ + Unewf = + Unewi (Equation 1) For all energies, we know the following 1 -mv² KE = Am¡m2 PEgrav r Uelastic = kx? 1 kx² Unew = (1/ where in we have m1 = m, m2 = M, 91 = q and q2 = Q By substituting all these to Equation 1 and then simplifying results to = sqrtt v 2 + ( ( m ) - V Q ) - (1/x ) ) Take note that capital letters have different meaning than small letter variables/constants. +

Physics for Scientists and Engineers with Modern Physics
Physics for Scientists and Engineers with Modern Physics
10th Edition
ISBN:9781337553292
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter38: Relativity
Section: Chapter Questions
Problem 38P
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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? Amım2 Consider a new expression for gravitation potential energy as: PEgrav = É, where A is a constant, m, and m2 are the masses of the two r 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 qQ 4TE, r? 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 KE1F+ KE2F + PEgravf + Uelasticf + 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 KE16 + + Unewf = + + Unewi (Equation 1) + + For all energies, we know the following KE =mv Am,m2 PE grav Uelastic = kx? 2 Unew = (1/ /(r where in we have m1 = m, m2 = M, 91 = q and q2 = Q By substituting all these to Equation 1 and then simplifying results to = sqrt( v 2 + ( ( Q m ) - V ) - (1/x ) ) + Take note that capital letters have different meaning than small letter variables/constants.
Transcribed Image Text: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? Amım2 Consider a new expression for gravitation potential energy as: PEgrav = É, where A is a constant, m, and m2 are the masses of the two r 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 qQ 4TE, r? 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 KE1F+ KE2F + PEgravf + Uelasticf + 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 KE16 + + Unewf = + + Unewi (Equation 1) + + For all energies, we know the following KE =mv Am,m2 PE grav Uelastic = kx? 2 Unew = (1/ /(r where in we have m1 = m, m2 = M, 91 = q and q2 = Q By substituting all these to Equation 1 and then simplifying results to = sqrt( v 2 + ( ( Q m ) - V ) - (1/x ) ) + Take note that capital letters have different meaning than small letter variables/constants.
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