Chapter 39, Problem 39.49P

A proton in a high-energy accelerator moves with a speed of c/2. Use the work-kinetic energy theorem to find the work required to increase its speed to (a) 0.750c and (b) 0.995c

To determine

The work required to increase the speed of the particle to $0.750c$.

Answer to Problem 39.49P

The work required to increase the sped of the particle is $\text{335}\text{MeV}$.

Explanation of Solution

Given Info: The initial speed of the particle is $\frac{c}{2}$.

The rest mass of the proton is $1.67\times {10}^{-27}\text{kg}$.

The speed of the light is $3\times {10}^8\text{m}/\text{s}$.

The final kinetic energy of the proton is,

$K_{\text{f}}=\left(\frac{1}{\sqrt{1-{\left(\frac{u_{\text{f}}}{c}\right)}^2}}-1\right)m{\left(c\right)}^2$

Here,

$u_{\text{f}}$ is the final velocity

$m$ is the mass of the proton.

$c$ is the speed of the light.

The initial kinetic energy of the proton is,

$K_{\text{i}}=\left(\frac{1}{\sqrt{1-{\left(\frac{u_{\text{i}}}{c}\right)}^2}}-1\right)m{\left(c\right)}^2$

Here,

$u_{\text{i}}$ is the initial velocity.

Form the work energy theorem.

$W=K_{\text{f}}-K_{\text{i}}$

Substitute $\left(\frac{1}{\sqrt{1-{\left(\frac{u_{\text{f}}}{c}\right)}^2}}-1\right)m{\left(c\right)}^2$ for $K_{\text{f}}$ and $\left(\frac{1}{\sqrt{1-{\left(\frac{u_{\text{i}}}{c}\right)}^2}}-1\right)m{\left(c\right)}^2$ for $K_{\text{i}}$ in the above equation.

$\begin{array}{c}W=\left(\frac{1}{\sqrt{1-{\left(\frac{u_{\text{f}}}{c}\right)}^2}}-1\right)m{\left(c\right)}^2-\left(\frac{1}{\sqrt{1-{\left(\frac{u_{\text{i}}}{c}\right)}^2}}-1\right)m{\left(c\right)}^2\\ =\left(\frac{1}{\sqrt{1-{\left(\frac{u_{\text{f}}}{c}\right)}^2}}\right)m{\left(c\right)}^2-m{c}^2-\left(\frac{1}{\sqrt{1-{\left(\frac{u_{\text{i}}}{c}\right)}^2}}\right)m{\left(c\right)}^2+m{c}^2\\ =\left(\frac{1}{\sqrt{1-{\left(\frac{u_{\text{f}}}{c}\right)}^2}}\right)m{\left(c\right)}^2-\left(\frac{1}{\sqrt{1-{\left(\frac{u_{\text{i}}}{c}\right)}^2}}\right)m{\left(c\right)}^2\end{array}$

Substitute $\frac{c}{2}$ for $u_{\text{i}}$ and $0.750c$ for $u_{\text{f}}$ in the above equation.

$\begin{array}{c}W=\left(\frac{1}{\sqrt{1-{\left(\frac{0.75c}{c}\right)}^2}}\right)m{\left(c\right)}^2-\left(\frac{1}{\sqrt{1-{\left(\frac{c}{2c}\right)}^2}}\right)m{\left(c\right)}^2\\ =\left(\frac{1}{0.661}\right)m{c}^2-\left(\frac{1}{0.75}\right)m{c}^2\\ =0.357\left(m{c}^2\right)\end{array}$

Substitute $1.67\times {10}^{-27}\text{kg}$ for $m$ and $3\times {10}^8\text{m}/\text{s}$ for $c$ in the above equation.

$\begin{array}{c}W=0.357\left(1.67\times {10}^{-27}\text{kg}{\left(3\times {10}^8\text{m}/\text{s}\right)}^2\right)\\ =5.37\times {10}^{-10}\text{J}\times \left(\frac{\text{1}\text{eV}}{1.602\times {10}^{-19}\text{J}}\right)\\ =335.20\times {10}^6\text{}\text{eV}\\ \cong \text{335}\text{MeV}\end{array}$

Conclusion:

Therefore, the work required to increase the sped of the particle is $\text{335}\text{MeV}$.

To determine

The work required to increase the speed of the particle to $0.995c$.

Answer to Problem 39.49P

The work required to increase the sped of the particle is $8.31\text{GeV}$.

Explanation of Solution

Given Info: The initial speed of the particle is $\frac{c}{2}$.

The rest mass of the proton is $1.67\times {10}^{-27}\text{kg}$.

The speed of the light is $3\times {10}^8\text{m}/\text{s}$.

The work done is,

$W=\left(\frac{1}{\sqrt{1-{\left(\frac{u_{\text{f}}}{c}\right)}^2}}\right)m{\left(c\right)}^2-\left(\frac{1}{\sqrt{1-{\left(\frac{u_{\text{i}}}{c}\right)}^2}}\right)m{\left(c\right)}^2$

Substitute $\frac{c}{2}$ for $u_{\text{i}}$ and $0.995c$ for $u_{\text{f}}$ in the above equation.

$\begin{array}{c}W=\left(\frac{1}{\sqrt{1-{\left(\frac{0.995c}{c}\right)}^2}}\right)m{\left(c\right)}^2-\left(\frac{1}{\sqrt{1-{\left(\frac{c}{2c}\right)}^2}}\right)m{\left(c\right)}^2\\ =\left(\frac{1}{0.099}\right)m{c}^2-\left(\frac{1}{0.75}\right)m{c}^2\\ =8.85\left(m{c}^2\right)\end{array}$

Substitute $1.67\times {10}^{-27}\text{kg}$ for $m$ and $3\times {10}^8\text{m}/\text{s}$ for $c$ in the above equation.

$\begin{array}{c}W=8.85\left(1.67\times {10}^{-27}\text{kg}{\left(3\times {10}^8\text{m}/\text{s}\right)}^2\right)\\ =1.33\times {10}^{-9}\text{J}\times \left(\frac{\text{1}\text{eV}}{1.602\times {10}^{-19}\text{J}}\right)\\ =8.303\times {10}^9\text{}\text{eV}\\ \approx \text{8}\text{.31}\text{GeV}\end{array}$

Conclusion:

Therefore, the work required to increase the sped of the particle is $8.31\text{GeV}$.