Question

In: Physics

One particle has a mass of 3.84 x 10-3 kg and a charge of +8.90 μC....

One particle has a mass of 3.84 x 10-3 kg and a charge of +8.90 μC. A second particle has a mass of 8.33 x 10-3 kg and the same charge. The two particles are initially held in place and then released. The particles fly apart, and when the separation between them is 0.158 m, the speed of the 3.84 x 10-3 kg-particle is 190 m/s. Find the initial separation between the particles.

Solutions

Expert Solution

Given Data

m1 = 3.84 x 10-3 kg,

q = + 8.90 uC = +8.90 x 10-6 C,

m2 = 8.33 x 10-3 kg,

d = 0.158 m,

v1 = 190 m/s.

To Find the initial separation d0

Solution:-

initial energy = kq2/d0

final energy = m1v12/2 + m2v22/2 + kq2/d

momentum conservation: 0 = m1v1 + m2v2, so v2 = -m1v1/m2 = -(3.84x 10-3 kg) * 190 / (8.33 x 10-3 kg)

                                                                        = - 87.58 m/s

kq2/d0 = m1v12/2 + m2v22/2 + kq2/d

d0 = [1/d + (m1v12 + m2v22)/(2kq2)]-1

    = [1/0.158 + { (3.84 x 10-3 kg)*1902 + (8.33x 10-3 kg)*87.582) / { (2*(9*109)*(8.90 x 10-6)2 }]-1

    = 0.00673 m


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