What is the impact speed of a proton colliding with a charged plastic bead?

[Sarah88]

Homework Statement

A 2.0-mm-diameter plastic bead is charged to –1.0 nC. A proton is fired at the bead from far away with a speed of 1.0 x 10^6 m/s, and it collides head-on. What is the impact speed?

Homework Equations

Conservation of energy: Kf + qVf= Ki + qVi

Possibly U elect/ Electric potential energy= K* (q*q'/r)

K= 1/2*m*v^2

V= K* (q/r)

The Attempt at a Solution

Using the equation Ki+ qVi = Kf + qVf (instead of qVf, used K*(q*q'/r))

1/2*1.67*10^-27 kg (mass of proton)*(1*10^6 m/s)^2 + 0 (due to far away distance, being infinitity)= 1/2 * (1.67*10^-27 kg + mass of plastic bead) * (Vfinal)^2 + 8.99*10^9 Nm^2/C^2 (1.60*10^-19 C + -1*10^-9 C/ r)

In terms of the mass of the plastic bead, is there an equation which allows one to find the mass of the plastic bead if the volume is known? Also, in terms of r (in the second part of the equation/final distance) at first 0 seems logical but this would give an undefined answer, therefore would r be the diameter or radius of the plastic bead? Thank you!

 

[Berko]

Yes, the impact is when the proton is at the surface of the bead. Also, the bead is so much more massive than the proton, you can consider it to be stationary.

 

[Sarah88]

Perfect, thank you for your help!