ProblemOne 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 lightparticle 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,m2Consider a new expression for gravitation potential energy as: PEgraywhere A is a constant, m1 and m2 are the masses of the two objects, and ris the distance between them.%3DMoreover, the new particle has an additional interaction with the heavy particle through the following force expression1Fnew =qQ4πεο r2where En 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 asKE1f + KE21 + PEgravf + Uelasticf+ Unewf = KE1i + KE21 + PEgravi +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+ UnewiKE16 ++ Unewf =+ Unewi(Equation 1) For all energies, we know the followingKE-글meAm,m2PEgrav =%3DUnew = (1/(rwhere in we havem1 = m, m2 = M, q1 = q and q2 = QBy substituting all these to Equation 1 and then simplifying results to= sgrt( v2 + ( (m) -DC CV.) - (1/x) ) +Take note that capital letters have different meaning than small letter variables/constants.

Name: ProblemOne newly discovered light particle has a mass of m and proper
Uploaded: 2024-03-05T11:56:34.000Z
Channel: Bartleby
Description: ProblemOne 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 lightparticle is xi distance from the heavy partic

Energy is a scalar quantity that means it only has magnitude but no direction to it. Energy comes in…

Science

Physics

PE grav r Uelastic= 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 = sqrt( 2 ( ( m)- V ) - (1/x ) ) + Take note that capital letters have different meaning than small letter variables/constants.

PE grav r Uelastic= 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 = sqrt( 2 ( ( m)- V ) - (1/x ) ) + Take note that capital letters have different meaning than small letter variables/constants.

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)...

See similar textbooks

Related questions

Q: Problem 3 Give all answers for this problem to 2 significant figures. Convert the following vectors…

A: 3. (a) Convert the following vectors from Cartesian to polar form. r=x2+y2=42+-62=7.2 The direction…

Q: A sphere of radius r has surface area A = 4πr2 and volumeV = (4/3)πr3. If the radius of sphere 2 is…

A: a)The ratio of the areas, A2/A1

Q: a) Write down the metric and the inverse metric, and use the definitionrpµv = 1/2gpσ(δµgvσ + δvgµσ -…

A: The objective of the question is to find the metric and inverse metric for a sphere of radius a, and…

Q: vector A is (2m) Long and is 60 above the x-axis vector B is (am) · Long and is 60 below the X. axis…

A: Since you have asked multiple question, we will solve the first question for you. If youwant any…

Q: For the graph of vx versus t shown, estimate the following: i) What is the slope of the tangent line…

A: The above problem can be solved in two different ways. The first method using the two-point formula…

Q: Where a>b the distance between the points (a, c) and (b, c) is (a-b) units. true or false

A: Given that a > b. We have to find distance between points (a, c) and (b, c). Take c as origin.

Q: Suppose you first walk A = 12.0 m in a direction 20.0 degrees west of north and then B = 20.0 m in a…

Q: Two vectors lying in the x-y plane are defined by their magnitudes and direction angles, as shown in…

Q: 10. Change of basis Let us now construct the vector (matrix representation) for a function g(x) =…

A: Given : g(x) = -2x+1

Q: Call the starting point (xo, yo) = (0,0). To find (x, y) for the ending point, use triangle trig on…

A: In the given solution we have arrived at a final equation which connects the distance, velocity and…

Q: dv B (а) = Ar +2, where A and B are constants, v and r are variables. At r = R1, v = dr Vo. At r =…

A: Given dvdr=Ar+Br Where A and B are constant So,…

Q: ds and a = dt dv Suppose quantity s is a length and quantity t is a time. Suppose the quantities v…

Q: Determine the following for a point with the rectangular coordinates (1.060, 2.930) m. (a) the polar…

A: Rectangular coordinates of a point = (1.06, 2.93) m to find : polar coordinates of this point

Q: Your formula is correct, but number is not. You forgot to multiply t=24s in step 1, at the line ....…

A: Apply the impulse-momentum equation to determine the final velocity of the car.

Q: Problem 3: A 500-kg dragster accelerates from rest to a final speed of 110 m/sec in 400 m (about a…

Q: What are the (a) x and (b) y components of the vector that must be added to the following three…

Q: What angle does the tangent to the curve y(x) = sin? (x/5) +17sin(x/5) at x = 5™ make with the…

Q: Vector R is defined as the vector sum of vectors A and D (R = À + D) shown below. What are the…

Q: (Figure 1) shows the standard way of measuring the angle. θ is measured to the vector from the x…

Concept explainers

Displacement, Velocity and Acceleration

In classical mechanics, kinematics deals with the motion of a particle. It deals only with the position, velocity, acceleration, and displacement of a particle. It has no concern about the source of motion.

Linear Displacement

The term "displacement" refers to when something shifts away from its original "location," and "linear" refers to a straight line. As a result, “Linear Displacement” can be described as the movement of an object in a straight line along a single axis, for example, from side to side or up and down. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Linear displacement is usually measured in millimeters or inches and may be positive or negative.

Topic Video

Question

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: PEgray
where A is a constant, m1 and m2 are the masses of the two objects, and ris the distance between them.
%3D
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 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 + KE21 + PEgravf + Uelasticf+ Unewf = KE1i + KE21 + PEgravi +
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
+ Unewi
KE16 +
+ Unewf =
+ Unewi
(Equation 1)

For all energies, we know the following
KE-글me
Am,m2
PEgrav =
%3D
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( v
2 + ( (
m) -
DC C
V.
) - (1/x
) ) +
Take note that capital letters have different meaning than small letter variables/constants.

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

SEE SOLUTION Check out a sample Q&A here

Step 1 Introduction

VIEW

Step 2 Solution

VIEW

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

Similar questions

Draw direction vectors in unit cells for the following cubic directions: a) [1 -2 3] b) [1 0 -3]
Consider two vectors A and B such that Ã × B – B (Ã· B Find the angle between A and B if ß = 6. O Report your numerical answer below, assuming three significant figures.
Compute the Force using Newtons Law. mass = 2.29, à = (-) Rows: 1 Columns: 1
a) Write down the metric and the inverse metric, and use the definitionrpµv = 1/2gpσ(δµgvσ + δvgµσ - δσgµv) = rpvµto reproduce the results written above for rθ ØØ and rØ θØ.