Question

A proton with a speed of \(1.23 \cdot 10^{4} \mathrm{~m} / \mathrm{s}\) is moving from infinity directly toward a second proton. Assuming that the second proton is fixed in place, find the position where the moving proton stops momentarily before turning around.

Short answer

Question: Determine the position where a moving proton stops momentarily before turning around due to the electrostatic repulsion from a second fixed proton. Given the initial speed of the moving proton as \(1.23 \times 10^{4}\) m/s. Answer: The distance between the protons when the moving proton stops momentarily is \(r_2 = \frac{k_{e}e^{2}}{K_{1}} = \frac{(8.99 \times 10^9 \mathrm{Nm^{2}/C^{2}})(1.6 \times 10^{-19}\mathrm{C})^2}{\frac{1}{2}(1.67 \times 10^{-27}\mathrm{kg})(1.23 \times 10^{4}\mathrm{m/s})^2}\). Calculate the value of \(r_2\) to find the position where the moving proton stops.

Step-by-step solution

Calculate the initial kinetic energy of the moving proton

First, let's find the initial kinetic energy (\(K_{1}\)) of the moving proton using the equation: \(K_{1} = \frac{1}{2}m_{p}v^{2}\), where \(m_{p}\) is the mass of a proton and \(v\) is the initial speed. The mass of a proton is \(m_{p} = 1.67 \times 10^{-27} \mathrm{kg}\), and we are given its initial speed as \(v = 1.23 \times 10^{4} \mathrm{m/s}\). Plugging these values into the equation, we get: \(K_{1} = \frac{1}{2}(1.67 \times 10^{-27}\mathrm{kg})(1.23 \times 10^{4}\mathrm{m/s})^2\).

Applying the conservation of mechanical energy principle

The total mechanical energy before and after the moving proton stops momentarily will be conserved: \(E_{initial} = E_{final}\). Initially, the moving proton has kinetic energy (\(K_{1}\)), and its potential energy (\(U_{1}\)) is zero since it is assumed to be at an infinite distance from the fixed proton. Finally, the moving proton has zero kinetic energy (\(K_{2}\)) when it stops, and its potential energy (\(U_{2}\)) is given by the electrostatic potential energy equation: \(U_{2} = \frac{k_{e}e^{2}}{r_{2}}\), where \(k_{e}\) is the electrostatic constant, \(e\) is the charge of a proton, and \(r_{2}\) is the distance between the protons when the moving proton stops. Using the conservation of mechanical energy principle, we get: \(K_{1} + U_{1} = K_{2} + U_{2}\). Since \(U_{1} = 0\) and \(K_{2} = 0\), this simplifies to: \(K_{1} = U_{2}\).

Solve for the distance between the protons when the moving proton stops

Now we can set \(K_{1} = \frac{k_{e}e^{2}}{r_{2}}\). Plugging the values for \(K_{1}\), \(k_{e} = 8.99 \times 10^9 \mathrm{Nm^{2}/C^{2}}\), and \(e = 1.6 \times 10^{-19} \mathrm{C}\) into the equation, we can solve for \(r_{2}\): \(r_{2} = \frac{k_{e}e^{2}}{K_{1}} = \frac{(8.99 \times 10^9 \mathrm{Nm^{2}/C^{2}})(1.6 \times 10^{-19}\mathrm{C})^2}{\frac{1}{2}(1.67 \times 10^{-27}\mathrm{kg})(1.23 \times 10^{4}\mathrm{m/s})^2}\). Compute the value of \(r_{2}\) to find the position where the moving proton stops momentarily before turning around.