Two particles each have a mass of 6.7 x 10-2 kg. One has a charge of +7.3 x 10-6 C, and the other has a charge of -7.3 x 10-6 C. They are initially held at rest at a distance of 1.1 m apart. Both are then released and accelerate toward each other. How fast is each particle moving when the separation between them is one-half its initial value?

To solve this problem, we can use the principles of electrostatics and conservation of mechanical energy.

Step 1: Calculate the initial potential energy of the system.
The potential energy between two charged particles can be calculated using the formula:
U(initial) = (k * |q1 * q2|) / r
where k is the Coulomb's constant (8.99 x 10^9 N m^2/C^2), q1 and q2 are the charges of the two particles, and r is the initial separation between them.

U(initial) = (8.99 x 10^9 N m^2/C^2) * |(7.3 x 10^-6 C)^2| / (1.1 m)
U(initial) = 5.63 x 10^-3 J

Step 2: Calculate the final potential energy of the system.
When the separation between the particles is halved, the potential energy can be calculated using the same formula but with the new separation distance (r/2):
U(final) = (8.99 x 10^9 N m^2/C^2) * |(7.3 x 10^-6 C)^2| / (0.5 x 1.1 m)
U(final) = 2.54 x 10^-2 J

Step 3: Apply the conservation of mechanical energy.
According to the conservation of mechanical energy, the initial total mechanical energy of the system (potential energy + kinetic energy) is equal to the final total mechanical energy.

Initial mechanical energy = U(initial) + kinetic energy of both particles
Final mechanical energy = U(final) + kinetic energy of both particles

Since the particles are initially at rest, their kinetic energy is zero.

U(initial) + 0 = U(final) + 0
5.63 x 10^-3 J = 2.54 x 10^-2 J + 0

Step 4: Solve for the velocity of each particle.
The kinetic energy of a particle can be calculated using the formula:
Kinetic energy = (1/2) * m * v^2
where m is the mass of the particle and v is its velocity.

Since the two particles have the same mass, we can assume that the final velocities of both particles are equal.

(1/2) * (6.7 x 10^-2 kg) * v^2 = 5.63 x 10^-3 J - 2.54 x 10^-2 J

Simplifying and solving for v:
(1/2) * (6.7 x 10^-2 kg) * v^2 = -1.98 x 10^-2 J

v^2 = (-1.98 x 10^-2 J) / (1/2) * (6.7 x 10^-2 kg)
v^2 = -5.92 m^2/s^2

Taking the square root of both sides:

v = √(-5.92 m^2/s^2)
v = ±2.43 m/s

Therefore, when the separation between the particles is halved, each particle is moving at a speed of approximately ±2.43 m/s.

To find the speed of each particle when the separation between them is one-half its initial value, we can use the law of conservation of energy.

The initial potential energy of the system is given by the equation:

U_initial = k * (q1 * q2) / r_initial,

where U_initial is the initial potential energy, k is the electrostatic constant (9 x 10^9 Nm^2/C^2), q1 and q2 are the charges of the particles, and r_initial is the initial separation between the particles.

The final potential energy of the system, when the separation between the particles is one-half its initial value, is given by:

U_final = k * (q1 * q2) / r_final,

where U_final is the final potential energy, and r_final is the final separation between the particles.

Since energy is conserved, the initial potential energy is equal to the final kinetic energy:

U_initial = U_final = (1/2) * m1 * v1^2 + (1/2) * m2 * v2^2,

where m1 and m2 are the masses of the particles, and v1 and v2 are their respective velocities.

To solve for the velocities, we rearrange the equation:

(1/2) * m1 * v1^2 + (1/2) * m2 * v2^2 = k * (q1 * q2) / r_initial.

Substituting known values:

(1/2) * (6.7 x 10^-2 kg) * v1^2 + (1/2) * (6.7 x 10^-2 kg) * v2^2 = (9 x 10^9 Nm^2/C^2) * (7.3 x 10^-6 C)^2 / (1.1 m).

Simplifying the equation:

(1/2) * (6.7 x 10^-2 kg) * v1^2 + (1/2) * (6.7 x 10^-2 kg) * v2^2 = 4.95 x 10^-4 J.

Now, plug in the value for the final separation, which is one-half of the initial distance (1.1 m / 2 = 0.55 m):

(1/2) * (6.7 x 10^-2 kg) * v1^2 + (1/2) * (6.7 x 10^-2 kg) * v2^2 = 4.95 x 10^-4 J,

(1/2) * (6.7 x 10^-2 kg) * v1^2 + (1/2) * (6.7 x 10^-2 kg) * v2^2 = (9 x 10^9 Nm^2/C^2) * (7.3 x 10^-6 C)^2 / (0.55 m).

Next, we need to solve this system of equations to find the values of v1 and v2.

Solving these equations will give you the velocities of each particle when the separation between them is one-half its initial value.