Question

A proton initially travels at a speed of 3000 m/s. After it passes through a region in which there is an electric field, the proton’s speed is 5000 m/s. (a) What is the initial kinetic energy of the proton? (b) What is the final kinetic energy of the proton? (c) What is the change in kinetic energy of the proton?

Short answer

(a) Initial kinetic energy of the proton is $7.51\mathrm{x}{10}^{-21}\mathrm{J}$.

(b) Final kinetic energy of the proton is$20.87\times {10}^{-21}\mathrm{J}$

(c) Change in the kinetic energy of the proton is $13.36\times {10}^{-21}\mathrm{J}$.

Step-by-step solution

The kinetic energy of an electric particle

The kinetic energy of any particle relies upon the ‘mass’ and ‘velocity’ of the particle.

The formula for the kinetic energy (K.E.) of the particle having mass ‘m’ and velocity ‘v’ is given by,

$K.E.=\frac{1}{2}m{v}^2$

Identification of given data

• The initial speed of the proton is $u=3000m/s$.
• The final speed of the proton is, $\mathrm{v}=5000\mathrm{m}/\mathrm{s}$.

(a) The initial kinetic energy of the proton

The mass of the proton is given by,

$m=1.67\times {10}^{-27}kg$

The formula for the initial kinetic energy of the proton is given by,

role="math" localid="1657080291356" $$\begin{gathered} {\mathrm{E}}_1=\frac{1}{2}{\mathrm{mu}}^2 \\ {\mathrm{E}}_1=\frac{1}{2}\times 1.67\times {10}^{-27}\mathrm{kg}{\left(3000\mathrm{m}/\mathrm{s}\right)}^2\times \frac{1\mathrm{J}}{1\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2} \\ {\mathrm{E}}_1=7.51\times {10}^{-21}\mathrm{J} \end{gathered}$$

Hence, the initial kinetic energy of the proton is $7.51\times {10}^{-21}\mathrm{J}$.

(b) The final kinetic energy of the proton

The formula for the final kinetic energy of the proton is given by,

$$\begin{gathered} {\mathrm{E}}_2=\frac{1}{2}m{v}^2 \\ {\mathrm{E}}_2=\frac{1}{2}\times 1.67\times {10}^{-27}\mathrm{kg}{\left(5000\mathrm{m}/\mathrm{s}\right)}^2\times \frac{1\mathrm{J}}{1\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2} \\ {\mathrm{E}}_2=20.87\times {10}^{-21}\mathrm{J} \end{gathered}$$

Hence, the final kinetic energy of the proton is $20.87\times {10}^{-21}\mathrm{J}$.

(c) The change in the kinetic energy of the proton

The formula for the change in the kinetic energy of the proton is given by,

$$\begin{gathered} ∆\mathrm{E}={\mathrm{E}}_2-{\mathrm{E}}_1 \\ ∆\mathrm{E}=20.87\times {10}^{-21}\mathrm{J}-7.51\times {10}^{-21}\mathrm{J} \\ ∆\mathrm{E}=13.36\times {10}^{-21}\mathrm{J} \end{gathered}$$

Hence, the change in the kinetic energy of the proton is $13.36\times {10}^{-21}\mathrm{J}$.