Two point charges are moving to the right along the

Chapter 23, Problem 92CP

(choose chapter or problem)

СP Two point charges are moving to the right along the x-axis. Point charge 1 has charge \(q_1=2.00 \mu \mathrm{C}\), mass \(m_1=\) \(6.00 \times 10^{-5} \mathrm{~kg}\), and speed \(v_1\). Point charge 2 is to the right of \(q_1\) and has charge \(q_2=-5.00 \mu \mathrm{C}\), mass \(m_2=3.00 \times 10^{-5} \mathrm{~kg}\), and speed \(v_2\). At a particular instant, the charges are separated by a distance of \(9.00 \mathrm{~mm}\) and have speeds \(v_1=400 \mathrm{~m} / \mathrm{s}\) and \(v_2=1300 \mathrm{~m} / \mathrm{s}\). The only forces on the particles are the forces they exert on each other.

(a) Determine the speed \(v_{\mathrm{cm}}\) of the center of mass of the system.

(b) The relative energy \(E_{\mathrm{rel}}\) of the system is defined as the total energy minus the kinetic energy contributed by the motion of the center of mass:

\(E_{\text {rel }}=E-\frac{1}{2}\left(m_1+m_2\right) v_{\mathrm{cm}}^2\)

where \(E=\frac{1}{2} m_1 v_1^2+\frac{1}{2} m_2 v_2^2+q_1 q_2 / 4 \pi \epsilon_0 r\) is the total energy of the system and r is the distance between the charges. Show that \(E_{\text {rel }}=\frac{1}{2} \mu v^2+q_1 q_2 / 4 \pi \epsilon_0 r\), where \(\mu=m_1 m_2 /\left(m_1+m_2\right)\) is called the reduced mass of the system and \(v=v_2-v_1\) is the relative speed of the moving particles.

(c) For the numerical values given above, calculate the numerical value of \(E_{\text {rel }}\).

(d) Based on the result of part (c), for the conditions given above, will the particles escape from one another? Explain.

(e) If the particles do escape, what will be their final relative speed when \(r \rightarrow \infty\)? If the particles do not escape, what will be their distance of maximum separation? That is, what will be the value of r when v = 0?

(f) Repeat parts (c)-(e) for \(v_1=400 \mathrm{~m} / \mathrm{s}\) and \(v_2=1800 \mathrm{~m} / \mathrm{s}\) when the separation is 9.00 mm.