Question

One particle has a mass of 3.00 × 10⁻³ kg and a charge of 8.00 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 125 m/s. Find the distance of the initial separation between the particles. (Hint: The momentum is conserved.)

Answer 1

The initial separation between the particles is 0.1 m.

Step-by-step explanation:

To solve this problem, we can use the principle of conservation of momentum. The initial momentum of the system is zero since both particles are initially at rest. When the particles fly apart, their momenta will be equal and opposite. We can use the formula for momentum, p = mv, to calculate the initial momentum of the 3.00 × 10⁻³ kg particle:

p = (3.00 × 10⁻³ kg)(125 m/s) = 0.375 kg*m/s

Since the momenta of the two particles are equal and opposite, the final momentum of the 6.00 × 10⁻³ kg particle will be:

p = -0.375 kg*m/s

Using the formula for momentum again, we can find the initial separation between the particles:

p = mv

-0.375 kg*m/s = (6.00 × 10⁻³ kg)(v)

v = -0.375 kg*m/s / 6.00 × 10⁻³ kg

v = -62.5 m/s

The negative sign indicates that the particles are moving in opposite directions. Finally, we can use the formula for distance traveled, d = vt, to calculate the distance of the initial separation:

d = (-62.5 m/s)(t)

Since the particles start from rest and move apart, the final velocity of the 3.00 × 10⁻³ kg particle is 0 m/s. Using this information, we can find the time it takes for the particles to reach their final separation:

d = (-62.5 m/s)(t)

0.1 m = (-62.5 m/s)(t)

t = 0.1 m / (-62.5 m/s)

t = -0.0016 s

The negative sign indicates that the particles are moving in opposite directions. Therefore, the initial separation between the particles is

d = (-62.5 m/s)(-0.0016 s)

d = 0.1 m