Question

Two 2.5-mm-diameter beads, C and D, are 11 mm apart, measured between their centers. Bead C has mass 1.0 g and charge 1.8 nC .Bead D has mass 2.5 g and charge -1.0 nC.

If the beads are released from rest, what is the speed vC of C at the instant the beads collide?

Express your answer to two significant figures and include the appropriate units.

What is the speed vD of D at the instant the beads collide?

Express your answer to two significant figures and include the appropriate units.

Answer 1

Initial potential energy(U_i) = kq_Cq_D/r
where K is constant , q_C is charge on bead C = 1.8 nC
q_D is charge on bead D = -1.0nC
U_i = -1.473*10^-6 J
Final potential energy when they collide
U_f = kq_Cq_D/r
now the distance between them (r) = d_C + d_D = 2.5 + 2.5 = 5 mm
where d is diameter of bead C and D
U_f = 9*10⁹*(1.8*10^-9)*(-1*10^-9) /(5*10^-3) = -3.24*10^-6 J
Change in energy = U_i - U_f = [-1.473 - (3.24)]*10^-6 = 1.767*10^-6 J
From the conservation of energy
Change in potential energy = increase in kinetic energy
1.767*10^-6 = (1/2)m_CV_C² + (1/2)m_DV_D²   
where m is masss of beads
2*1.767*10^-6 = (1*10^-3)V_C² + (2.5*10^-3)V_D²
3.534*10^-3 = V_C² + 2.5V_D² ---------------(1)
Now the initial momentum of the system = 0
Final momentum of the system = m_CV_C - m_DV_D
Applying conservation of momentum
m_CV_C - m_DV_D = 0
m_CV_C = m_DV_D
1*V_C = 2.5V_D
V_C = 2.5V_D
using this in equation 1
3.534*10^-3 = (2.5V_D)² + 2.5V_D²
V_D = 0.022 m/s
V_C = 2.5V_D = 0.0552 m/s