Question: Two point charges are moving to the right along the x-axis. Point charge 1 has charge ql = 2.00 µC mass ml = 6.00 X
Two point charges are moving to the right along the x-axis. Point charge 1 has charge ql = 2.00 µC mass ml = 6.00 X 10-5 kg, and speed u1 Point charge 2 is to the right of ql and has charge q2 = -5.00 µC, mass m2 = 3.00 X 10-5 kg, and speed u2. At a particular instant, the charges are separated by a distance of 9.00 mm and have speeds u1 = 400 m/s and u2 = 1300 m/s. The only forces on the particles are the forces they exert on each other.
(a) Determine the speed Ucm of the center of mass of the system.
(b) The relative energy Erel of the system is defined as the total energy minus the kinetic energy contributed by the motion of the center of mass:
Where E = 1/2m1U2/1 + 1/2m2U2/2 + q1q2/4πє0r is the total energy of the system and r is the distance between the charges. Show that Erel = 1/2µu2 + q1q2/4πє0r where µ = m1m2/ (ml + m2) is called the reduced mass of the system and V = V2 - VI is the relative speed of the moving particles.
(c) For the numerical values given above, calculate the numerical value of Erel.
(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 → ∞? If the particles do not escape, what will be their distance of maximum separation? That is, what will be the value of r when U = 0?
(f) Repeat parts (c)-(e) for U1 = 400 m/s and U2 = 1800 m/s when the separation is 9.00 mm.
E = E -(m, + m2)um Ee
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