Electrostatic potential energy:

$\overline{){\mathbf{U}}{\mathbf{=}}\frac{\mathbf{k}{\mathbf{q}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{\mathbf{r}}}$

Momentum:

$\overline{){\mathit{p}}{\mathbf{=}}{\mathit{m}}{\mathit{v}}}$

Kinetic energy:

$\overline{){\mathbf{K}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}{{\mathbf{mv}}}^{{\mathbf{2}}}}$

Law of conservation of energy:

$\overline{){{\mathbf{K}}}_{{\mathbf{i}}}{\mathbf{+}}{{\mathbf{U}}}_{{\mathbf{i}}}{\mathbf{+}}{{\mathbf{W}}}_{{\mathbf{nc}}}{\mathbf{=}}{{\mathbf{K}}}_{{\mathbf{f}}}{\mathbf{+}}{{\mathbf{U}}}_{{\mathbf{f}}}}$ W_{nc} is the work done by non-conservative forces like friction.

We'll use conservation of momentum to determine the relationship in the velocities of the balls. Take bead C is moving in the positive direction and bead D is moving in the negative direction.

p_{i} = p_{f}

0 = m_{c}v_{c} - m_{d}v_{d}

m_{c}v_{c} = m_{d}v_{d}

(1.0g)v_{c} = (1.7g)v_{d}

v_{c} = 1.7v_{d}

Two 2.0-mm diameter beads, C and D, are 12 mm apart, measured between their centers. Bead C has mass 1.0 g and charge 2.5 nC. Bead D has mass 1.7 g and charge -1.0 nC. If the beads are released from rest, what is the speed v_{C} at the instant the beads collide? What is the speed v_{D} at the instant the beads collide?

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