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Answer, showing all work

Bhtps/www.webassign.net/web/student/Assignment-Responees/s.bmitdep- 13246876 10 points SerPSED 25 P 055 Notes O Ask Your From a large distance away, a particle of mass 1.80 g and charge 15.0 C is fired at 18.01 m/s straight toward a second particle, originally stationary but free to move, with mass 4.50 g and charge 8.50 uC. Both particles are constrained to move only along the x axis. (o) At the instant of closest approach, both particles will be moving at the same velocity. Find this velocty i m/s (b) Find the distance of closest approach. After the interaction, the particles will move far apart again. At this time, find the velocity of the following. (c) the 1.80-g particle i m/s (d) the 4.50-g particle m/'s Read It

Explanation / Answer

a) from the momentum conservation, we have
m1vo = (m1 + m2)V
  V = m1vo/(m1 +m2)

= 5.14 m/s i^

b) from the law of conservation of energy, we have
m1vo2/2 = kq1q2/d + (m1 +m2)V2/2

0.0018*18*18/2  = [9*10^9*15*10^-6*8.5*10^-6] / d +{ [0.0018+0.0045][5.14*5.14]/2}

thus, the diatnce is, d = rc= 5.5 m

d)
velocity of 1.8 g particle is,

v1' = (m1 -m2)vo/(m1 + m2) = -7.71 m/s i^

velocity of 4.5 g particle is,
v2' = 2m1vo/(m1 +m2) = 10.285 m/s i^

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