Electric Potential Energy Video Lessons

Concept

# Problem: 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 vC at the instant the beads collide? What is the speed vD at the instant the beads collide?

###### FREE Expert Solution

Electrostatic potential energy:

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

Momentum:

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

Kinetic energy:

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

Law of conservation of energy:

$\overline{){{\mathbf{K}}}_{{\mathbf{i}}}{\mathbf{+}}{{\mathbf{U}}}_{{\mathbf{i}}}{\mathbf{+}}{{\mathbf{W}}}_{{\mathbf{nc}}}{\mathbf{=}}{{\mathbf{K}}}_{{\mathbf{f}}}{\mathbf{+}}{{\mathbf{U}}}_{{\mathbf{f}}}}$ Wnc 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.

pi = pf

0 = mcvc - mdvd

mcvc = mdvd

(1.0g)vc = (1.7g)vd

vc = 1.7vd

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###### Problem Details

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 vC at the instant the beads collide? What is the speed vD at the instant the beads collide?