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 + Unewf = + Unewi (Equation 1) For all energies, we know the following KE=-m Am,m2 PEg Fgrav Uelastic = kx? Unew = (1/ where in we have m, = m, m2 = M, q1 = q and q2 = Q By substituting all these to Equation 1 and then simplifying results to = sąrt v 2 + ( ( m ) - ) - (1/x ) ) +
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 + Unewf = + Unewi (Equation 1) For all energies, we know the following KE=-m Am,m2 PEg Fgrav Uelastic = kx? Unew = (1/ where in we have m, = m, m2 = M, q1 = q and q2 = Q By substituting all these to Equation 1 and then simplifying results to = sąrt v 2 + ( ( m ) - ) - (1/x ) ) +
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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Question
![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
KE14+
+ Unewf =
+ Unewi
(Equation 1)
For all energies, we know the following
XF -m
KE =
Am,m2
PEgrav =
r
1
Uelastic =
kx²
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
= sgrt(
2 + ( (
Q
m ) -
V
v
) - (1/x
) ) +](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F33563087-dc2f-4ea2-82ec-799b0cb9393a%2Fddbf27f0-2a66-4470-baad-7f3354d3cc19%2Fv30pxx7_processed.png&w=3840&q=75)
Transcribed Image Text: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
KE14+
+ Unewf =
+ Unewi
(Equation 1)
For all energies, we know the following
XF -m
KE =
Am,m2
PEgrav =
r
1
Uelastic =
kx²
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
= sgrt(
2 + ( (
Q
m ) -
V
v
) - (1/x
) ) +
![QUESTION 4
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: PEaray = -
where A is a constant, m, and m, are the masses of the
r
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
Fnew =
qQ
4πεο r2
where En is a constant that is read as epsilon naught, 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
KE16 + KE2F + PEgravf + Velasticf + 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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F33563087-dc2f-4ea2-82ec-799b0cb9393a%2Fddbf27f0-2a66-4470-baad-7f3354d3cc19%2Fqw24xs_processed.png&w=3840&q=75)
Transcribed Image Text:QUESTION 4
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: PEaray = -
where A is a constant, m, and m, are the masses of the
r
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
Fnew =
qQ
4πεο r2
where En is a constant that is read as epsilon naught, 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
KE16 + KE2F + PEgravf + Velasticf + 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
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