In Problems 74 through 76 you are given the equation(s) used to solve a problem. For each of these,

a. Write a realistic problem for which this is the correct equation(s).

b. Finish the solution of the problem.

75.$\frac{1}{2}(1.67\times {10}^{-27}kg){(2.5\times {10}^{6}m/s)}^2+0=\frac{1}{2}(1.67\times {10}^{-27}kg){v_i}^2+\frac{(9.0\times {10}^{9}N{m}^2/C^2)(2.0\times {10}^{-9}C)(1.60\times {10}^{-19}C)}{0.0010m}$

Given equation :

$\frac{1}{2}(1.67\times {10}^{-27}kg){(2.5\times {10}^{6}m/s)}^2+0=\frac{1}{2}(1.67\times {10}^{-27}kg){v_i}^2+\frac{(9.0\times {10}^{9}N{m}^2/C^2)(2.0\times {10}^{-9}C)(1.60\times {10}^{-19}C)}{0.0010m}$

Theory used :

Coulomb's law states that the electrical force between two charged objects is directly proportional to the product of the quantity of charge on the objects and inversely proportional to the square of the separation distance between the two objects.

a. Real life problem :

"Find the speed of a proton launched from a distance of $1mm$to a charge of $2nC$with a speed of $2.5\times {10}^{6}m/s$while at a distance of $1mm$from the charge."

b. We can derive the following numerical equality from the law of conservation of energy, which is presented in the exercise:

$\frac{1.67\times {10}^{-27}(2.5\times {10}^6)}{2}=\frac{1.67\times {10}^{-27}{v_i}^2}{2}+\frac{9\times {10}^9\times 2\times {10}^{-9}\times 1.67\times {10}^{-19}}{0.001}$

We can simplify this by conducting calculations, as shown below, to make it easier:

$5.219\times {10}^{-15}=8.35\times {10}^{-28}{v_i}^2+2.88\times {10}^{-15}$

Now we can calculate the speed :

$v_i=\sqrt{\frac{(5.219-2.88)\times {10}^{-15}}{8.35\times {10}^{-28}}}=\boxed{\mathbf{1}\mathbf{.}\mathbf{67}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}}$