A proton (mass m = 1.67 multiplied by 10-27 kg) is being accelerated along a straight line at 2.2 multiplied by 1015 m/s2 in a machine. The proton has an initial speed of 2.4 multiplied by 107 m/s and travels 3.8 cm.

Question

(a) What is its speed?
m/s
(b) What is the increase in its kinetic energy? j

Henry · Answer

a. V^2 = Vo^2 + 2ad,
V^2 = (2.4*10^7)^2 + 4.4*10^15*0.038,
V^2=5.76*10^14+1.67*10^14=7.43*10^14,
V = 2.73*10^7.

b. KE = 0.5m*V^2,
KE1=(1.67/2)(2.4*10^7)^2=4.81*10^14J.
KE2=(1.67/2)*(2.73*10^7)^2=6.22*10^14J.

Increase = KE2 - KE! =
6.22*10^14 - 4.81*10^14 = 1.41*10^14J.