A proton of mass m, is accelerated along a straight line at a rate of a = 3.6 x 10^15 m/s^2 in a machine. The proton had an initial speed of v; = 2.4 x 10^7m/s and traveled for d= 0.035 m with the given acceleration. Givens: m2 = 1.67 x 10^-27 kg, a = 3.6 x 10^15 m/s^2, v; = 2.4 x 10^7m/s, d=0.035 m
(a) Use the 1D kinematic equations for constant acceleration motion to derive a symbolic expression for the speed of the proton, vi(mp, a, Vi, d), after it has traveled a distance d. (Chapter 2 physics!)
(b) Calculate a numerical value for us if d = 0.035 m.
(c) With the initial speed, Vi, taken as given, and your expression for Uf from part (a), derive a symbolic expression for the increase in the proton's kinetic energy, AK(mp, a, d). Simplify your expression as much as possible. (Hint: Your final answer should simplify to the product of three parameters.)
(d) By Newton's 2nd law the net force, F, acting on the proton must be F = mya. Rewrite your symbolic expression for AK in terms of F and d. In this form, your equation should look like a theorem from the physics of motion. What is the name of the theorem?
(e) Using your result from part (c) or from part (d), calculate a numerical value for AK
Check: If my = 1.67 x 10^-27 kg, a = 1.8 x 10^15 m/s^2, v; = 1.2 x 10^7m/s, and d = 0.050 m, then (b) 1.8 x 10^7m/s, (e) 1.5 x 10^-13 J.