(a)
Determine the system's initial configuration at ri = infinite particle separation and the system's final configuration at the point of closest approach.
Since the two-particle system is not being affected by any outside forces, we may treat it as an isolated system for momentum and use the momentum conservation law.
m1v1 + m1v2 = (m1+m2)v
The second particle's starting velocity is zero, so:
m1v1 = (m1+m2)v
After substituting the values we get,
v = 6i m/s
(b)
Since the two particle system is also energy-isolated, we may use the energy-conservation principle.
dK + dU = 0
Ki +Ui = Kf + Uf
Substituting the values,
1/2m1v1^2i + 1/2 m2v2^2i + 0 = 1/2m1v1^2f + 1/2m2v2^2f +ke q1q2/rf
The second particle's initial speed is 0 (v2 = 0). Additionally, both the first and second particle's final velocity have the same value, v. Put these values in place of the preceding expression:
1/2m1v1^2i = 1/2m1v1^2 + 1/2m2v2^2 +ke q1q2/rf
After solving we get,
rf = 2ke q1q2 / m1v1^2 - (m1+m2)v^2
Substituting the values we get,
rf = 3.64m
(c)
v1f = (m1-m2 / m1 + m2) v1i
v1f = -9i m/s
(d)
v2f = (2m1/ m1 +m2) v1i
After substituting the values,
v2f = 12i m/ s
Question :
Review. From a large distance away, a particle of mass 2.00 g and charge 15.0 \muμC is fired at 21.0 m/s straight toward a second particle, originally stationary but free to move, with mass 5.00 g and charge 8.50 \muμC. Both particles are constrained to move only along the x axis. (a) At the instant of closest approach, both particles will be moving at the same velocity. Find this velocity. (b) Find the distance of closest approach. After the interaction, the particles will move far apart again. At this time, find the velocity of (c) the 2.00-g particle and (d) the 5.00-g particle. \hat{i}
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