Two protons are released from rest when they are 0.750 {/rm nm} apart.

a) What is the maximum speed they will reach?

b) What is the max acceleration they will reach?

Given:

m = 1.673 * 10^-27 kg

Q = q = 1.602 * 10^-19 C

r = 0.75 nm

= 0.75 * 10^-9 m

A.

Energy, U = (kQq) / r

Ut = 1/2 mv^2 + 1/2 mv^2

1.673 * 10^-27 * v^2 = (8.99 * 10^9 * (1.602 * 10^-19) ^2) / 0.75 * 10^-9

v = 1.356 * 10^4 m/s

B.

F = (kQq) / r^2

F = m * a

1.673 * 10^-27 * a = ((8.99 * 10^9 * (1.602 * 10-19) ^2) / (0.075 * 10^-9) ^2

a = 2.45 * 10^17 m/s^2.