Question

A proton (mass m 1.67 x 1027 kg) is being accelerated along a straight line at 1.4 x 1015 m/s2 in a machine. The proton has an initial speed of 2.4 x 107 m/s and travels 3.0 cm
(a) What is its speed? 25195237.6452 m/s
(b) What is the increase in its kinetic energy? 163660000000 ×

Answer 1

Answer:

Part a)

$v_f = 2.569 \times 10^7 m/s$

Part b)

$\Delta K = 7.014 \times 10^{-14} J$

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

$v_i = 2.4 \times 10^7 m/s$

$d = 3 cm$

$a = 1.4 \times 10^{15} m/s^2$

so we will have

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - (2.4 \times 10^7)^2 = 2(1.4 \times 10^{15})(0.03)$

$v_f^2 = 6.6 \times 10^{14}$

$v_f = 2.569 \times 10^7 m/s$

Part b)

Now increase in kinetic energy is given as

$\Delta K = \frac{1}{2}m(v_f^2 - v_i^2)$

$\Delta K = \frac{1}{2}(1.67 \times 10^{-27})[(2.569 \times 10^7)^2 - (2.4 \times 10^7)^2]$

$\Delta K = 7.014 \times 10^{-14} J$