Two protons at rest and separated by \(1.00 \mathrm{~mm}\) are released simultaneously. What is the speed of either at the instant when the two are \(10.0 \mathrm{~mm}\) apart?

We have two protons initially separated by a distance of \(1.00 \mathrm{~mm}\), and we want to find their speed when the separation distance increases to \(10.0 \mathrm{~mm}\). We know the charges of both protons are equal, \(q= 1.6 \times 10^{-19} \mathrm{C}\), and their masses are also the same, \(m= 1.67 \times 10^{-27} \mathrm{kg}\). The initial distance between the protons is \(r_{initial} = 1.00 \times 10^{-3} \mathrm{m}\) and the final distance is \(r_{final} = 10.0 \times 10^{-3} \mathrm{m}\). The protons are under the influence of the Coulomb force, which can be calculated using the equation \(F_c = \dfrac{kq^2}{r^2}\), where \(k\) is the Coulomb's constant (\(k=8.99 \times 10^9 \mathrm{N m^2/C^2}\)) and \(r\) is the distance between the protons.

The mechanical energy will be conserved in this problem since there are no non-conservative forces acting on the protons. We can write the conservation of mechanical energy equation as follows: \(U_{initial} + K_{initial} = U_{final} + K_{final}\) Since both protons initially start at rest, the initial kinetic energy (\(K_{initial}\)) will be zero. The final kinetic energy is what we're looking for, \(K_{final} = \dfrac{mV^2}{2}\), where \(V\) is the final speed of either proton.

The initial and final electric potential energy can be calculated using the formula \(U = \dfrac{kq^2}{r}\) We need to find \(U_{initial}\) and \(U_{final}\) using \(r_{initial}\) and \(r_{final}\), respectively. \(U_{initial} = \dfrac{(8.99 \times 10^9 \mathrm{N m^2/C^2})(1.6 \times 10^{-19} \mathrm{C})^2}{1.00 \times 10^{-3} \mathrm{m}} = 2.30 \times 10^{-17} \mathrm{J}\) \(U_{final} = \dfrac{(8.99 \times 10^9 \mathrm{N m^2/C^2})(1.6 \times 10^{-19} \mathrm{C})^2}{10.0 \times 10^{-3} \mathrm{m}} = 2.30 \times 10^{-18} \mathrm{J}\)

Next, we plug in \(U_{initial}\), \(U_{final}\), and \(K_{initial}\) into the conservation of mechanical energy equation from Step 2: \(2.30 \times 10^{-17} \mathrm{J} + 0 = 2.30 \times 10^{-18} \mathrm{J} + \dfrac{mV^2}{2}\) Subtract \(2.30 \times 10^{-18} \mathrm{J}\) from both sides: \(2.07 \times 10^{-17} \mathrm{J} = \dfrac{mV^2}{2}\) Now we need to solve for \(V\): \(V^2 = \dfrac{2(2.07 \times 10^{-17} \mathrm{J})}{m} \Rightarrow V = \sqrt{\dfrac{4.14 \times 10^{-17} \mathrm{J}}{1.67 \times 10^{-27} \mathrm{kg}}}\)

Finally, plug in the expression from Step 4 to find the final speed of either proton: \(V = \sqrt{\dfrac{4.14 \times 10^{-17} \mathrm{J}}{1.67 \times 10^{-27} \mathrm{kg}}} = 1.57 \times 10^5 \mathrm{m/s}\) The speed of either proton when they are \(10.0 \mathrm{~mm}\) apart is approximately \(1.57 \times 10^5 \mathrm{m/s}\).