A proton $\left(\mathrm{mass}m=1.67\times {10}^{-27}\mathrm{kg}\right)$is being accelerated along a straight line at $\mathbf{3}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{15}}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$in a machine. If the proton has an initial speed of $\mathbf{2}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$and travels 3.5 cm, what then is (a) its speed and (b) the increase in its kinetic energy?

1. Mass of the proton is, $\mathrm{m}=1.67\times {10}^{-27}\mathrm{kg}$.
2. Acceleration is, $\mathrm{a}=3.6\times {10}^{15}\mathrm{m}/{\mathrm{s}}^2$.
3. The initial speed is, ${\mathrm{v}}_{\mathrm{i}}=2.4\times {10}^7\mathrm{m}/\mathrm{s}$.
4. Distance traveled is, $\mathrm{d}=3.5\mathrm{cm}=0.035\mathrm{m}$.

Using the third kinematic equation, we can find the final speed of the proton from a given mass, acceleration, and initial speed of the proton. Also, we can use this final velocity and given initial velocity to calculate the increase in its kinetic energy.

From newtons law of motion equations, the final velocity can be calculated as,

${\mathrm{v}}_{\mathrm{f}}^2={\mathrm{v}}_{\mathrm{i}}^2+2\mathrm{ad}$

Substitute the given values in the above equation, and we get,

$$\begin{gathered} {\mathrm{v}}_{\mathrm{f}}^2={\left(2.4\times {10}^7\frac{\mathrm{m}}{\mathrm{s}}\right)}^2+2\left(3.6\times {10}^{15}\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)\left(0.035\mathrm{m}\right) \\ {\mathrm{v}}_{\mathrm{f}}^2=5.76\times {10}^{14}{\mathrm{m}}^2/{\mathrm{s}}^2+2.52\times {10}^{14}{\mathrm{m}}^2/{\mathrm{s}}^2 \\ {\mathrm{v}}_{\mathrm{f}}^2=8.41\times {10}^{14}{\mathrm{m}}^2/{\mathrm{s}}^2 \\ {\mathrm{v}}_{\mathrm{f}}=2.9\times {10}^7\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the final speed of the proton is $2.9\times {10}^7\mathrm{m}/\mathrm{s}$.

The expression for kinetic energy is,

$\mathrm{K}=\frac{1}{2}{\mathrm{mv}}^2$

Initial kinetic energy can be calculated as,

${\mathrm{K}}_{\mathrm{i}}=\frac{1}{2}{\mathrm{mv}}_{\mathrm{i}}^2$

Substitute the given values in the above equation, and we get,

$\begin{array}{rcl}{\mathrm{K}}_{\mathrm{i}}& =& \frac{1}{2}\left(1.67\times {10}^{-27}\mathrm{kg}\right){\left(2.4\times {10}^7\frac{\mathrm{m}}{\mathrm{s}}\right)}^2\\ & =& 4.8\times {10}^{-13}.\left(1\mathrm{kg}\times 1\frac{{\mathrm{m}}^2}{{\mathrm{s}}^2}\times \frac{1\mathrm{J}}{1\mathrm{N}.\mathrm{m}}\times \frac{1\mathrm{N}}{1\mathrm{kg}.{\displaystyle \frac{\mathrm{m}}{{\mathrm{s}}^2}}}\right)\\ & =& 4.8\times {10}^{-13}\mathrm{J}\end{array}$

Final kinetic energy can be calculated as,

${\mathrm{K}}_{\mathrm{f}}=\frac{1}{2}{\mathrm{mv}}_{\mathrm{f}}^2$

Substitute the given values in the above equation, and we get,

$\begin{array}{rcl}{\mathrm{K}}_{\mathrm{f}}& =& \frac{1}{2}\left(1.6\times {10}^{-27}\mathrm{kg}\right){\left(2.9\times {10}^7\frac{\mathrm{m}}{\mathrm{s}}\right)}^2\\ & =& 6.9\times {10}^{-13}.\left(1\mathrm{kg}\times 1\frac{{\mathrm{m}}^2}{{\mathrm{s}}^2}\times \frac{1\mathrm{J}}{1\mathrm{N}.\mathrm{m}}\times \frac{1\mathrm{N}}{1\mathrm{kg}.{\displaystyle \frac{\mathrm{m}}{{\mathrm{s}}^2}}}\right)\\ & =& 6.9\times {10}^{-13}\mathrm{J}\end{array}$

Therefore, an increase in the kinetic energy of the proton can be calculated as,

$∆\mathrm{K}={\mathrm{K}}_{\mathrm{f}}-{\mathrm{K}}_{\mathrm{i}}$

Substitute the given values in the above equation, and we get,

$$\begin{gathered} ∆\mathrm{K}=6.9\times {10}^{-13}\mathrm{J}-4.8\times {10}^{-13}\mathrm{J} \\ ∆\mathrm{K}=2.1\times {10}^{-13}\mathrm{J} \end{gathered}$$

Therefore, an increase in the kinetic energy of the proton is $2.1\times {10}^{-13}\mathrm{J}$.