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? Consider a new expression for gravitation potential energy as: , where A is a constant, m1 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 where 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 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 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 v = sqrt( v2 + ( ( Q )/( m ) - )( ( ) - (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? Consider a new expression for gravitation potential energy as: , where A is a constant, m1 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 where 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 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 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 v = sqrt( v2 + ( ( Q )/( m ) - )( ( ) - (1/x) ) + ) Take note that capital letters have different meaning than small letter variables/constants.

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Question

Consider a new expression for gravitation potential energy as: , where A is a constant, m₁ 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

where 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

KE_1f + KE_2f+ PE_grav_f+ U_elasticf + U_newf = KE_1i + KE_2i + PE_gravi + + U_newi

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

KE_1f + + + + U_newf = + + + + U_newi (Equation 1)

For all energies, we know the following

U_new = (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

v = sqrt( v² + ( ( Q )/( m ) - )( ( ) - (1/x) ) + )

Take note that capital letters have different meaning than small letter variables/constants.