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 law of conservation of linear momentum. Here are the steps to calculate the final velocity of each particle:

Step 1: Calculate the initial potential energy
The initial potential energy (U_initial) between the two particles is given by the equation:

U_initial = k * (q1 * q2) / r_initial

Where,
k = Coulomb's constant = 8.99 × 10^9 N m²/C²
q1 = charge of the first particle = +4.8 × 10^-6 C
q2 = charge of the second particle = -4.8 × 10^-6 C
r_initial = initial separation distance = 0.88 m

Substituting the values:

U_initial = (8.99 × 10^9 N m²/C²) * ((4.8 × 10^-6 C) * (-4.8 × 10^-6 C)) / (0.88 m)
U_initial = -191.92 J (negative sign indicates that the potential energy is attractive)

Step 2: Calculate the final potential energy
The final potential energy (U_final) between the two particles, when the separation distance is one-third its initial value, can be calculated using the formula:

U_final = k * (q1 * q2) / r_final

Where,
r_final = one-third of the initial separation distance = 0.88 m / 3 = 0.293 m

Substituting the values:

U_final = (8.99 × 10^9 N m²/C²) * ((4.8 × 10^-6 C) * (-4.8 × 10^-6 C)) / (0.293 m)
U_final = -1,204.51 J

Step 3: Apply conservation of energy
According to the law of conservation of energy, the initial potential energy will be converted into the final kinetic energy of the particles:

U_initial = KE_particle1 + KE_particle2

Where,
KE_particle1 = kinetic energy of particle 1
KE_particle2 = kinetic energy of particle 2

Substituting the values:

-191.92 J = (1/2) * m1 * v1² + (1/2) * m2 * v2²

Since both particles have the same mass (m1 = m2 = 5.5 × 10^-3 kg), we can write:

-191.92 J = (1/2) * (5.5 × 10^-3 kg) * v1² + (1/2) * (5.5 × 10^-3 kg) * v2²

Step 4: Apply conservation of linear momentum
According to the law of conservation of linear momentum, the initial momentum of the system will be conserved during the interaction:

p_initial = p_final

Where,
p_initial = initial momentum of the system = m1 * v1_initial + m2 * v2_initial
p_final = final momentum of the system = m1 * v1_final + m2 * v2_final

Since both particles start from rest, the initial momentum of the system is zero. Therefore, the final momentum of the system is also zero. This can be written as:

m1 * v1_final + m2 * v2_final = 0

Substituting the values:

(5.5 × 10^-3 kg) * v1_final + (5.5 × 10^-3 kg) * v2_final = 0

Step 5: Solve the equations
We now have two equations:

Equation 1: -191.92 J = (1/2) * (5.5 × 10^-3 kg) * v1² + (1/2) * (5.5 × 10^-3 kg) * v2²
Equation 2: (5.5 × 10^-3 kg) * v1_final + (5.5 × 10^-3 kg) * v2_final = 0

We can solve these two equations simultaneously to find the values of v1 and v2.