Question

[Key Concept: Energy conservation with the release point fixing the total energy; maximum speed occurs at the potential minimum.]

a) $U(x) = \frac{\alpha}{x^{2}} - \frac{\beta}{x} = \frac{\alpha}{x_{0}^{2}}\left\lbrack \frac{x_{0}^{2}}{x^{2}} - \frac{\beta x_{0}^{2}}{x\alpha} \right\rbrack$

$$U(x) = \frac{\alpha}{x_{0}^{2}}\left\lbrack \left( \frac{x_{0}}{x} \right)^{2} - \frac{x_{0}}{x} \right\rbrack\ \because x_{0} = \frac{\alpha}{\beta}$$

b)$\ K_{i} + U_{i} = K_{f} + U_{f}$

$$O + \left( \frac{\alpha}{x_{0}^{2}} - \frac{\beta}{x_{0}} \right) = \frac{1}{2}{mv}^{2} + \frac{\alpha}{x_{0}^{2}}\left( \left( \frac{x_{0}}{x} \right)^{2} - \frac{x_{0}}{x} \right)$$

$$ \begin{matrix} & \frac{1}{2}mv^{2} = \frac{\alpha}{x_{0}^{2}}\left\lbrack \frac{x_{0}}{x} - \left( \frac{x_{0}}{x} \right)^{2} \right\rbrack + \left( \frac{\alpha}{x_{0}^{2}} - \frac{\beta}{x_{0}} \right) \\ & \ = \frac{\alpha}{x_{0}^{2}}\left\lbrack \frac{x_{0}}{x} - \left( \frac{x_{0}}{x} \right)^{2} \right\rbrack + \left( \frac{\alpha}{x_{0}^{2}} - \frac{\alpha}{x_{0}^{2}} \right) \\ & \end{matrix}$$

$$\because\beta = \frac{\alpha}{x_{0}}$$

$$v = \left\lbrack \frac{2\alpha}{mx_{0}^{2}}\left\{ \frac{x_{0}}{x} - \left( \frac{x_{0}}{x} \right)^{2} \right\} \right\rbrack^{1/2}$$

c) for $v_{\text{max~}} \rightarrow U_{\text{min~}}$

$$\begin{matrix} & \frac{dU}{dx} = 0 = \alpha\frac{( - 2)}{x^{3}} + \frac{\beta}{x^{2}} = 0 \\ & \ \Rightarrow - \frac{2\alpha}{x} + \beta = 0 \\ & x = \frac{2\alpha}{\beta} = 2x_{0} \end{matrix}$$

$$ {so\ v_{\text{max~}} = \left\{ \frac{2\alpha}{{mx}_{0}^{2}}\left\lbrack \frac{x_{0}}{2x_{0}} - \left( \frac{x_{0}}{2x_{0}} \right)^{2} \right\rbrack \right\}^{1/2}}$$

$$ {v_{\max} = \left\{ \frac{2\alpha}{{mx}_{0}^{2}}\left\lbrack \frac{1}{2} - \frac{1}{4} \right\rbrack \right\}^{1/2} = \sqrt{\frac{\alpha}{2{mx}_{0}^{2}}}}$$

d) $T.E. = K_{i} + U_{i} = 0 + \left( \frac{\alpha}{x_{0}^{2}} - \frac{\beta}{x_{0}} \right) = 0 + \frac{\alpha}{x_{0}^{2}} - \frac{\alpha}{x_{0}^{2}} = 0$