Question

One particle has a mass of 3.46 x 10-3 kg and a charge of +6.83 μC. A second particle has a mass of 7.68 x 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 0.112 m, the speed of the 3.46 x 10-3 kg-particle is 191 m/s. Find the initial separation between the particles. PLEASE get it right I've done this so many times :(

Answer 1

Given,

The mass of first particle, m ₁ = 3.46 * 10^-3 kg

The mass of second particle, m ₂ = 7.68 * 10^-3 kg

The speed, = 191 m/s

Charge q = 6.83 $\mu$ C = 6.83 * 10^-6C

Let the initial separation be 'd' meters.

Since the initial momentum of the system is 0, the final momentum will also be 0.

Therefore, the particles moves in opposite direction so that the magnitude of each of  the particle's momentum be equal.

Let the speed of 3.46 * 10^-3 kg particle at the given final moment be v m/s.

By equating the magnitudes of momentum:

7.68 *  0.001 * v = 3.46 * 0.001 * 191

v = 86.049 m/s

Since, there is no energy loss.

The initial electrical potential energy = final electrical P.E + final K.E

9 * 10⁹ *  (6.83 * 10^-6)² /d = (9 * 10⁹ *  (6.83 * 10^-6)² / 0.112 )+ (0.5 * 3.46 * 10^-3 * 191² ) + 0.5 * 7.68 * 10^-3 * 86.049²

419.84 * 10^-3 / d = 3748.57 * 10^-3 + 63112.13 * 10^-3 + 27772.16 * 10^-3

419.84 * 10^-3 / d = 94632.86 * 10^-3

d = 0.0044365 m

d = 0.44365 cm