Mechanics question

  • Thread starter Cognizant
  • Start date
  • #1
3
0
I'm attempting to understand an issue that a mechanics student is having, and I was wondering if someone could break down this particle mechanics equation for me so I know where it fits in:

V(x)=-A|x|^(n)

The exact problem that they are dealing with is:

A particle moves towards [x = 0] under the influence of a potential V( x ) = -A|x|^(n) (assume n > 0). The particle has barely enough energy to reach x = 0. For what values of n will it reach x = 0 in a finite time?

Precisely, I need to know what values A, x, and n are supposed to represent. I'm not specifically a mechanics student, but these sort of things always spark my interest in a self-teaching direction.
 

Answers and Replies

  • #2
StatusX
Homework Helper
2,564
1
A is a (positive) constant, which is just there to show this is an abstract, generalized problem; in a real world situation you would experimentally determine constants like A so you could make actual numerical predictions. When you ask what n means, I assume you know about powers, and are asking for its physical significance? If so, there are some real potentials that can be approximated by certain powers of x, but in this case it's probaly just a textbook problem; for practice, nothing more. And x is the position of the particle.

To solve the problem, you would equate KE=E-V. Since it will just barely pass, I assume E=0, so KE=A|x|^n. Then use KE=1/2 m v^2, where v = dx/dt, and solve the resulting differential equation to get x(t), which can then be inspected to see if it crosses 0 in a finite time.
 
  • #3
3
0
You are correct. I understand the basic mathematical principles, but since I'm not familiar with these types of physics, I was not sure if the A and n were arbitrary (I had assumed x is a variable in this case, and representative of some aspect of the particle, but was not clear on which), or if they were a significant constant. V(x), as I would understand it, is simply the calculated potential of the particle in x position.

[tex]KE = \frac {1}{2}m (\frac {dx}{dt})^2 [/tex]

I'll work with it a bit, and see what I can come up with. Thanks for the starting point! :)
 

Related Threads on Mechanics question

  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
14
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
1
Views
786
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
3
Views
1K
Top