Two particles each have a mass of 5.5 10-3 kg. One has a charge of +4.8 10-6 C, and the other has a charge of -4.8 10-6 C. They are initially held at rest at a distance of 0.88 m apart. Both are then released and accelerate toward each other. How fast is each particle moving when the separation between them is one-third its initial value?

To solve this problem, we can use the principles of conservation of energy and the electric potential energy.

First, we can find the initial electric potential energy between the two charges using the formula:

\(U_{initial} = \frac{k \cdot q_1 \cdot q_2}{r_{initial}}\)

where \(k\) is the Coulomb's constant, \(q_1\) and \(q_2\) are the charges, and \(r_{initial}\) is the initial distance between the charges.

Plugging in the given values:
\(U_{initial} = \frac{9 \times 10^9 \cdot (4.8 \times 10^{-6})^2}{0.88}\)

We can calculate the value of \(U_{initial}\) to be approximately \(2.204 \times 10^{-4}\) Joules.

Now, as the particles move closer together, their potential energy decreases and converts into kinetic energy.

When the separation between the particles becomes one-third of its initial value, the distance between them is \(r_{final} = \frac{1}{3} \cdot r_{initial}\).

The final electric potential energy \(U_{final}\) is given by:

\(U_{final} = \frac{k \cdot q_1 \cdot q_2}{r_{final}}\)

Plugging in the values:

\(U_{final} = \frac{9 \times 10^9 \cdot (4.8 \times 10^{-6})^2}{(1/3) \cdot 0.88}\)

Simplifying the expression, we find that \(U_{final} = 3 \times 10^{-3}\) Joules.

Since the total energy is conserved, the initial potential energy \(U_{initial}\) is equal to the final kinetic energy \(K_{final}\) of both particles. Therefore:

\(K_{final} = U_{initial} = 2.204 \times 10^{-4}\) Joules

To find the final velocity of each particle, we use the equation for kinetic energy:

\(K_{final} = \frac{1}{2} m v^2\)

Solving for \(v\), we get:

\(v = \sqrt{\frac{2K_{final}}{m}}\)

where \(m\) is the mass of each particle.

Plugging in the values, we have:

\(v = \sqrt{\frac{2 \cdot 2.204 \times 10^{-4}}{5.5 \times 10^{-3}}}\)

Calculating this expression, we find that the final velocity of each particle when the separation between them is one-third of its initial value is approximately 0.119 m/s.