Question

A proton’s speed as it passes point A is 50,000 m/s. It follows the trajectory shown in FIGURE P25.43. What is the proton’s speed at point B? A proton’s speed as it passes point A is 50,000 m/s. It follows the trajectory shown in FIGURE P25.43. What is the proton’s speed at point B?

Short answer

The value of proton’s speed at the point B is$1.01\times {10}^{-5}m/s$

Step-by-step solution

Speed of photon

The potential energy and kinetic energy at the A point

$$\begin{gathered} Kinecticenergyis{K}_A=\frac{1}{2}m{v^2}_A \\ potentialenergyis{U}_A=e{V}_A \end{gathered}$$

The potential energy and kinetic energy at the B point

$$\begin{gathered} Kinecticenergyis{K}_B=\frac{1}{2}m{v^2}_B \\ potentialenergyis{U}_B=e{V}_B \end{gathered}$$

$$\begin{gathered} e{V}_A+\frac{1}{2}m{v^2}_A=e{V}_B+\frac{1}{2}m{v^2}_B \\ {v}_B=\sqrt{{v^2}_A+\frac{2\times e\left(V_A-V_B\right)}{m}} \\ {v}_A=5000m/s \\ e=1.6\times {10}^{-19}C \\ {V}_A=30V \\ {V}_B=-10V \\ {v}_B=1.01\times {10}^{5}m/s \end{gathered}$$

Calculation 

On the basis of conservation theory

$K_A+U_A=K_B+U_B$

Applying all data

role="math" localid="1648296227023" $\begin{array}{rcl}{V}_B& =& \sqrt{{V_A}^2+\frac{2e(V_A-V_B)}{m}}\\ & =& \sqrt{{(50000m/s)}^2+\frac{2(1.6\times {10}^{-19}C)(30V-(-10V\left)\right)}{(1.67\times {10}^{-27}kg)}}\\ & =& 1.008\times {10}^{5}m/s\end{array}$

Therefore, the speed of the proton is $1.01\times {10}^{5}m/s$