Question

Chapter 7, Problem 1. A proton (mass m = 1.67 × 10-27 kg) is being accelerated along a straight line at 5.40 × 1013 m/s2 in a machine. If the proton has an initial speed of 6.20 × 105 m/s and travels 2.40 cm, what then is (a) its speed and (b) the increase in its kinetic energy?

Answer 1

given

m = 1.67*10^-27 kg
a = 5.40*10^13 m/s^2
vi = 6.2*10^5 m/s
d = 2.40 cm = 0.024 m

a) vf = ?

use, vf^2 - vi^2 = 2*a*d

vf^2 = vi^2 + 2*a*d

vf = sqrt(vi^2 + 2*a*d)

= sqrt((6.20*10^5)^2 + 2*5.4*10^13*0.024)

= 1.73*10^6 m/s <<<<<<<<<=============Answer

b) increas in kinetic energy = (1/2)*m*vf^2 - (1/2)*m*vi^2

= (1/2)*m*(vf^2 - vi^2)

= (1/2)*1.67*10^-27*( (1.73*10^6)^2 - (6.2*10^5)^2 )

= 2.18*10^-15 J <<<<<<<<<=============Answer