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April 16, 2014

Homework Help: physics

Posted by Ann on Saturday, February 3, 2007 at 3:48pm.

One particle has a mass of 3.00*10-3 kg and a charge of +7.60 µC. A second particle has a mass of 6.00*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 0.100 m, the speed of the 3.00*10-3 kg particle is 140 m/s. Find the initial separation between the particles.

You can so,ve this problem by using conservation of energy.

First you find the speed of the aother particle. You use the fact that momentum is conserved. It was zero when the particles were released so it's still zero.

The fact that the total mometum is zero leads to the conclusion that second particle is moving at half the speed of the first particle (and in the opposite direction), because it's mass is twice that of the first particle.

You now know the speeds of both particles and from that you can calculate the kinetic energy. From the distance of the two particles you calculate the potential energy:

q1*q2/(4 pi epsilon_0 r)

The total energy is the sum of the potential and kinetic energy. When the particles were just released they both had a velocity of zero. So at that time the total energy was the potential energy they had. But since energy is conserved that potential energy was equal to the total energy you just computed. This means that the distance r_int is such that:

q1*q2/(4 pi epsilon_0 r_int) = total energy

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