A proton's speed as it passes point A is 50000 m/s . It follows the trajectory shown in the figure.(Figure 1)What is the proton's speed at point B?
Work done by a proton is, W = qV Here, W is the work done, q is the charge on proton and V is the potential. W = 9V = (1.602 x 10-19 C)(40) = 6.408x10-18 J The above work done is equal to increased kinetic energy because proton is moving from higher potential to lower potential its energy will increase. (K.E), = 64.08x10-19 J Let (K.E), is the final kinetic energy and (K.E), is the initial energy. (K.E), = (KE), +(K.E), 1 mv = 1 mv +6.408x10-16 VA=v24 (2)(6.408x10-18) m Here m is the mass of proton v} = (50000 m/s) + 4 12 (2)(64.08x10-19) 1.673x10-2 = (50000 m/s) +7.6605x10° Vj = V1.01605x1010 =100799.306