# Problem: A proton follows the path shown in (Figure 1). Its initial speed is v0 = 2.4x106 m/s .What is the proton's speed as it passes through point P?

###### FREE Expert Solution

Law of conservation of energy:

$\overline{){{\mathbf{K}}}_{{\mathbf{i}}}{\mathbf{+}}{{\mathbf{U}}}_{{\mathbf{i}}}{\mathbf{+}}{{\mathbf{W}}}_{{\mathbf{nc}}}{\mathbf{=}}{{\mathbf{K}}}_{{\mathbf{f}}}{\mathbf{+}}{{\mathbf{U}}}_{{\mathbf{f}}}}$

Kinetic energy, K:

$\overline{){\mathbf{K}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}{{\mathbf{mv}}}^{{\mathbf{2}}}}$

Potential energy, U:

$\overline{){\mathbf{U}}{\mathbf{=}}{\mathbf{-}}\frac{\mathbf{k}{\mathbf{q}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{\mathbf{r}}}$

Wnc = 0

The law of conservation of energy is now written as:

$\begin{array}{lcl}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{m}{{\mathbf{v}}_{\mathbf{0}}}^{\mathbf{2}}{\mathbf{-}}\frac{{\mathbf{kq}}_{\mathbf{1}}\mathbf{e}}{{\mathbf{r}}_{\mathbf{0}}}& \mathbf{=}& \frac{\mathbf{1}}{\mathbf{2}}{\mathbf{m}}{{\mathbf{v}}_{\mathbf{f}}}^{{\mathbf{2}}}{\mathbf{-}}\frac{{\mathbf{kq}}_{\mathbf{1}}\mathbf{e}}{{\mathbf{r}}_{\mathbf{f}}}\\ \frac{\mathbf{1}}{\mathbf{2}}{\mathbf{m}}{{\mathbf{v}}_{\mathbf{f}}}^{{\mathbf{2}}}& \mathbf{=}& {{\mathbf{kq}}}_{{\mathbf{1}}}{\mathbf{e}}{\mathbf{\left(}}\frac{\mathbf{1}}{{\mathbf{r}}_{\mathbf{f}}}{\mathbf{-}}\frac{\mathbf{1}}{{\mathbf{r}}_{\mathbf{0}}}{\mathbf{\right)}}{\mathbf{+}}\frac{\mathbf{1}}{\mathbf{2}}{{{\mathbf{mv}}}_{{\mathbf{0}}}}^{{\mathbf{2}}}\end{array}$

96% (119 ratings) ###### Problem Details

A proton follows the path shown in (Figure 1). Its initial speed is v0 = 2.4x106 m/s . What is the proton's speed as it passes through point P?