Answer:
Part a)
[tex]v_f = 2.569 \times 10^7 m/s[/tex]
Part b)
[tex]\Delta K = 7.014 \times 10^{-14} J[/tex]
Explanation:
Part a)
As we know that proton is accelerated uniformly so we can use kinematics here to find the final speed
so we know that
[tex]v_i = 2.4 \times 10^7 m/s[/tex]
[tex]d = 3 cm[/tex]
[tex]a = 1.4 \times 10^{15} m/s^2[/tex]
so we will have
[tex]v_f^2 - v_i^2 = 2 a d[/tex]
[tex]v_f^2 - (2.4 \times 10^7)^2 = 2(1.4 \times 10^{15})(0.03)[/tex]
[tex]v_f^2 = 6.6 \times 10^{14}[/tex]
[tex]v_f = 2.569 \times 10^7 m/s[/tex]
Part b)
Now increase in kinetic energy is given as
[tex]\Delta K = \frac{1}{2}m(v_f^2 - v_i^2)[/tex]
[tex]\Delta K = \frac{1}{2}(1.67 \times 10^{-27})[(2.569 \times 10^7)^2 - (2.4 \times 10^7)^2][/tex]
[tex]\Delta K = 7.014 \times 10^{-14} J[/tex]
