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

Part a)

`v_f = 2.569 * 10^7 m/s`

Part b)

`\Delta K = 7.014 * 10^(-14) J`

Step-by-step 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 * 10^7 m/s`

`d = 3 cm`

`a = 1.4 * 10^(15) m/s^2`

so we will have

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

`v_f^2 - (2.4 * 10^7)^2 = 2(1.4 * 10^(15))(0.03)`

`v_f^2 = 6.6 * 10^(14)`

`v_f = 2.569 * 10^7 m/s`

Part b)

Now increase in kinetic energy is given as

`\Delta K = (1)/(2)m(v_f^2 - v_i^2)`

`\Delta K = (1)/(2)(1.67 * 10^(-27))[(2.569 * 10^7)^2 - (2.4 * 10^7)^2]`

`\Delta K = 7.014 * 10^(-14) J`