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 r 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 Fnew 1 qQ 4περ 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
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 r 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 Fnew 1 qQ 4περ 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
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I am having troubles with this problem. Please help me find the answers needed for each box.
Transcribed Image Text:KE1 +
+ Unewf =
+
+
+
+ Unewi
(Equation 1)
For all energies, we know the following
1
KE =mv
Am,m2
PEgrav
r
Uelastic = 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
= sąrt(
2 + ( (
V
V
m ) -
) -
(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
r
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
Fnew
qQ
4TE, r2
where Eg 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; + 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
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