Particle 1 has a mass of m 1 = 3.10 × 10 -6 kg, while particle 2 has a mass of m

Question

Particle 1 has a mass of m1 = 3.10 × 10-6 kg, while particle 2 has a mass of m2 = 6.20 × 10-6 kg. Each has the same electric charge. These particles are initially held at rest, and the two-particle system has an initial electric potential energy of 0.160 J. Suddenly, the particles are released and fly apart because of the repulsive electric force that acts on each one (see the figure). The effects of the gravitational force are negligible, and no other forces act on the particles. At one instant following the release, the speed of particle 1 is measured to be v1 = 164 m/s. What is the electric potential energy at this instant?

Explanation / Answer

Here,

m1 = 3.1 *10^-6 Kg

m2 = 6.20 *10^-6 Kg

as there is no external force acting on the particles

Using conservation of momentum

m1 * v1 - m2 * v2 = 0

3.1 *10^-6 * 164 - 6.2 *10^-6 * v2 = 0

v2 = 82 m/s

let the final electric potential energy is Uf

Using conservation of energy

Uf + 0.5 * 3.1 *10^-6 * 164^2 + 0.5 * 6.2 *10^-6 * 82^2 = 0.160

solving for Uf

Uf = 0.0974 J

the final electrical potential energy of the system is 0.0974 J

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects