Answer: CALC A proton with mass m moves in one dimension.

QUESTION:

A proton with mass m moves in one dimension. The potential-energy function is \(U(x)=\alpha / x^{2}-\beta / x\) where \(\alpha \text { and } \beta\) are positive constants. The proton is released from rest at \(x_{0}=\alpha / \beta\)(a) Show that \(U(x)\)can be written as

\(U(x)=\frac{a}{x}\)

                U(x)=\frac{\alpha}{x_{0}^{2}}\left[\left(\frac{x_{0}}{x}\right)^{2}-\frac{x_{0}}{x}\right]

Graph . Calculate \(U\left(x_{0}\right)\) and thereby locate the point \(x_{0}\)on the graph.

(b) Calculate  the speed of the proton as a function of position. Graph and give a qualitative description of the motion.

(c) For what value of x is the speed of the proton a maximum? What is the value of that maximum speed?

(d) What is the force on the proton at the point in part (c)?

(e) Let the proton be released instead at \(x_{1}=3 \alpha / \beta\) Locate the point  \(x_{1}\) on the graph of . Calculate  and give a qualitative description of the motion.

(f) For each release point \(\left(x=x_{0} \text { and } x=x_{1}\right)\) what are the maximum and minimum values of x reached during the motion?

Equation Transcription:

Text Transcription:

U(x)=\alpha / x^{2}-\beta / x

\alpha \text { and } \beta

x_{0}=\alpha / \beta

U(x) = frac{a}{x}

U(x)=\frac{\alpha}{x_{0}^{2}}\left[\left(\frac{x_{0}}{x}\right)^{2}-\frac{x_{0}}{x}\right]

U(x)

U\left(x_{0}\right)

X_{0}

v(x)

v(x)

\(x_{1}=3 \alpha / \beta\)

x_{1}

\left(x=x_{0} \text { and } x=x_{1}\right)