Using energy considerations to analyse particle motion

AI Thread Summary
For a particle to cross the origin, it must possess sufficient kinetic energy to overcome the potential energy at that point, specifically at U(0) = A. The equations presented show the relationship between potential energy and kinetic energy at any position x=a, indicating that the total energy must satisfy the condition A + Ae^{-\alpha a^2} ≤ (1/2)(mv^2_a - ka^2). There is confusion regarding whether to use KE - PE or KE + PE in energy considerations, with the consensus that total energy is conserved. The correct interpretation is that if the particle passes through the origin, then KE + PE must exceed A. Understanding these energy relationships is crucial for analyzing particle motion effectively.
I_Try_Math
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Homework Statement
(a) Sketch a graph of the potential energy function ## U(x) = \frac {kx^2} {2} + Ae^{-\alpha x^2} ## where k, A, and, ##\alpha## are constants. (b) What is the force corresponding to this potential energy? (c) Suppose a particle of mass m moving with this potential energy has a velocity ##v_a## when its position is x=a. Show that the particle does not pass through the origin unless ## A \leq \frac {mv^2_a + ka^2} {2(1 - e^{-\alpha a^2})} ##
Relevant Equations
## U(x) = \frac {kx^2} {2} + Ae^{-\alpha x^2} ##
For part (c) my understanding is that in this case for the particle to cross the origin it must have at least as much kinetic energy as there is potential energy at ##U(0) = A##. Given the potential energy and kinetic energy at any position x=a is equal to ## U(a) = \frac {ka^2} {2} + Ae^{-\alpha a^2} ## and ## \frac 1 2mv^2_a ## respectively, this implies that:



##A \leq \frac 1 2mv^2_a - \frac {ka^2} {2} - Ae^{-\alpha a^2} ##

##A + Ae^{-\alpha a^2} \leq \frac 1 2mv^2_a - \frac {ka^2} {2} ##

##A(1 + e^{-\alpha a^2}) \leq \frac 1 2(mv^2_a - ka^2) ##

##2A(1 + e^{-\alpha a^2}) \leq mv^2_a - ka^2)##

##A \leq \frac {mv^2_a - ka^2} {2(1 + e^{-\alpha a^2})}##

Can't tell where I'm making a mistake.
 
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Why do you start with KE-PE rather than KE+PE?
 
Hill said:
Why do you start with KE-PE rather than KE+PE?
I guess I'm just getting confused. So supposing the particle goes through the origin, ##KE + PE \gt A##, correct?
 
I_Try_Math said:
I guess I'm just getting confused. So supposing the particle goes through the origin, ##KE + PE \gt A##, correct?
The total energy is conserved: ##KE(a)+PE(a)=KE(0)+PE(0)=KE(0)+A##
 
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