Question

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?

Short answer

Based on the step-by-step solution provided, create a short answer question: Question: Two protons are initially at rest and separated by \(1.00 \mathrm{~mm}\). Determine their speed when they are \(10.0 \mathrm{~mm}\) apart. Answer: The speed of either proton when they are \(10.0 \mathrm{~mm}\) apart is approximately \(1.57 \times 10^5 \mathrm{m/s}\).

Step-by-step solution

Define the given variables and identify the forces involved.

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.

Apply conservation of energy.

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.

Calculate the initial and final electric potential energy.

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. \(U_{initial} = \dfrac{(8.99 \times 10^9 \mathrm{N m^2/C^2})(1.6 \times 10^{-19} \mathrm{C})^2}{1.00 \times 10^{-3} \mathrm{m}} = 2.30 \times 10^{-17} \mathrm{J}\) \(U_{final} = \dfrac{(8.99 \times 10^9 \mathrm{N m^2/C^2})(1.6 \times 10^{-19} \mathrm{C})^2}{10.0 \times 10^{-3} \mathrm{m}} = 2.30 \times 10^{-18} \mathrm{J}\)

Solve for the final kinetic energy.

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 \(V\): \(V^2 = \dfrac{2(2.07 \times 10^{-17} \mathrm{J})}{m} \Rightarrow V = \sqrt{\dfrac{4.14 \times 10^{-17} \mathrm{J}}{1.67 \times 10^{-27} \mathrm{kg}}}\)

Calculate the final speed.

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 \(1.57 \times 10^5 \mathrm{m/s}\).