I_Try_Math
- 119
- 25
- 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.
##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.