Two protons are released from rest when they are 0.750 nm apart. (a) What is the maximum speed they will reach? When does this speed occur? (b) What is the maximum acceleration they will achieve? When does this acceleration occur?

Solution 9E Step 1: Introduction: In this question, we need to find the maximum speed reached by the protons and when protons to achieve maximum speed In the second part, we need to find the maximum acceleration reached by the protons and when the protons to achieve maximum acceleration Data given 9 Distance between the protons d = 0.750 nm = 0.75 × 10 m Charge of proton e = p.6 × 10 19C Mass of proton m = p.67 × 10 27 kg Step 2 : We shall find the potential energy of the protons when they are released It is given by U = K× (e /d)p2 f Here U fis the final potential energy K coulomb s constant = 8.99 × 10 Nm /C 2 2 Substituting values we get 9 2 19 2 9 U =f8.99 × 10 Nm × ((1.6 × 10 C) /0.75 × 10 m) U = 8.99 × 10 Nm × ((2.56 × 10 38C/0.75 × 10 m)9 f 9 2 29 U f 8.99 × 10 Nm × 3.4134 × 10 C/m 19 U f 3.06 × 10 J Step 3 : We need to find the velocity of the protons We have potential given 2 U f mv f Rearranging to find velocity, we have vf= U fm p Substituting values we get vf= 3.06 × 10 19J/1.67 × 10 27kg v = 13558 m/s f v = 1.36 km /s f Hence we have maximum velocity as 1.36 km /s