Vi = 2.3 x 10" mis q = - 8nc Vf = ? speed x = 2." initial separation y = 4. 14 final separation Speed Stay the Same ? KEi + PEI = PE + KEf Imv, + kae = 1 mvp + Kave . 10 - 9 8.35 x 10- 28 VF + 1. 73 * 10 12 (9. 109 N. m2/62) (- 8 n6 ) (1.6 - 10-19 0) - (1.67 *10 -27 kg ) ( 2.3 x 10" mis) 2 + 7 (1.67 x 10 -27 kg ) VE2 + 10.2 2.4cm (0.01cm) ( m ( 9 . 109 N . m=1(2 ) ( - 8 nc ) (1 .6 x 10-19 c) - 4.09 x 10-7 = 6 8. 35 x 10- 28 VF + 1.75 x 10 12 Im (4.1 cm ) ( 0 .01 cm ) (1.6 * 10 -19 0 ) - 4.69 x 10 - 7 = 8. 25 . 10- 28 VF + 1.75 x 1012 - 1. 75 X 10 12 - 1. 75 x 1012 1. 75 x 10 12 = 8. 35 . 10-28 VF 8. 35 x 10-28 8. 85- 10-28 2.09 x 10 39 = VF 2 4.48. 1019 = Vf 2.4 cm (0.01Cm - (1. 67 x 10 - 27 Rg ) ( 2 . 3 x 10 # m 15 ) 2 + ( 9 x 10 - ( N. my/ 62 ) ( - 8 x10 - 9) (1. 6x 10-19 6) = 1 (1.67 x 10- 27 Rg ) VE + ( 9* 109 N. m=/63)(-8x 109)(1-4 x 10-#90) Im ( 4 .1 cm ) ( 0.01 cm ) (16 * 10 -PC ) 3.93 x 10-15 3.93 x 10-1 =1 (1-67 * 10-27 kg) (VF) + (9 * 109 N. m2/c = (-8x 10-9) (1.6 * 10-tag) VF =1. 4 x 10-15 4. 1 0 m ( D.01 CM) ( 1.6 * 10- 19 6)A proton is flying through space with an initial velocity v. = 2.3 x 10 m/s, as shown below. It interacts with a fixed negative charge q = - 8 nC along the path shown. What is its speed in m/s when it reaches the point P, given that x = 2.4 cm and y = 4.1 cm? Give your answer to 3 significant digits. (note the drawing may not be to scale) P y Vo X