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

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.
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: PE Egrav where A is a constant, mj and m2 are the r 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 qQ new 4πεο 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 = KE1 + KE21 + 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
Transcribed Image Text:KE16 + + Unewf = + + + Unewi (Equation 1) + + For all energies, we know the following KE = Am¡m2 PEgrav r Uekstic = kx? Unew = (1/ /(r )) %3D where in we have m1 = m, m2 = M, q1 = q and q2 = Q %3D By substituting all these to Equation 1 and then simplifying results to 2 = sqrt( + ( ( m ) V V %3D ) - (1/x ) ) +

Question

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

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.

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: PE
Egrav
where A is a constant, mj and m2 are the
r
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
qQ
new
4πεο 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 = KE1 + KE21 + 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
PEgrav
r
Uekstic = kx?
Unew = (1/
/(r
))
%3D
where in we have
m1 = m, m2 = M, q1 = q and q2 = Q
%3D
By substituting all these to Equation 1 and then simplifying results to
2
= sqrt(
+ ( (
m )
V
V
%3D
) - (1/x
) ) +

Accepted Answer

The energy equation be expressed as,…