Chapter 26, Problem 38P

To determine

The kinetic energy and the momentum of a proton.

Answer to Problem 38P

The kinetic energy is $29.5MeV\left(4.7\times {10}^{-12}J\right)$ .

The momentum is $1.3\times {10}^{-19}kg\cdot m/s$ .

Explanation of Solution

Given info:

Rest mass, $m_0=1.67\times {10}^{-27}kg$

Speed, $v=7.35\times {10}^{7}m/s$

Formula used:

Kinetic energy,

  $KE=\left(m-m_0\right)c^2=m_0{c}^2\left[\frac{1}{{\left[1-{\left(\frac{v}{c}\right)}^2\right]}^{\frac{1}{2}}}-1\right]$

Where,

  $m_0$ is rest mass

  $m$ is relativistic mass

  $v$ is velocity

  $c$ is speed of light, $3.00\times {10}^{8}m/s$

The momentum of particle:

  $p=\frac{m_{0}v}{{\left[1-{\left(\frac{v}{c}\right)}^2\right]}^{\frac{1}{2}}}$

Where,

  $m_0$ is rest mass

  $v$ is velocity

  $c$ is speed of light, $3.00\times {10}^{8}m/s$

Calculation:

The speed of the proton in terms of speed of light is:

  $v=\frac{\left(7.35\times {10}^{7}m/s\right)}{\left(3.00\times {10}^{8}m/s\right)}c=0.245c$

The kinetic energy is

  $KE=\left(m-m_0\right)c^2=m_0{c}^2\left[\frac{1}{{\left[1-{\left(\frac{v}{c}\right)}^2\right]}^{\frac{1}{2}}}-1\right]$

  $KE=\left(938MeV\right)\left(\left\{\frac{1}{{\left[1-{\left(0.245\right)}^2\right]}^{\frac{1}{2}}}\right\}-1\right)$

  $\begin{array}{l}KE=29.5MeV\\ KE=4.7\times {10}^{-12}J\end{array}$

The proton’s momentum is

  $p=\frac{m_{0}v}{{\left[1-\frac{v^2}{c^2}\right]}^{\frac{1}{2}}}$

  $p=\left(1.67\times {10}^{-27}kg\right)\left(7.35\times {10}^{7}m/s\right)\left\{\frac{1}{{\left[1-{\left(0.245\right)}^2\right]}^{\frac{1}{2}}}\right\}$

  $=1.3\times {10}^{-19}kg\cdot m/s$

Conclusion: The kinetic energy is $4.7\times {10}^{-12}J$ and momentum is $1.3\times {10}^{-19}kg\cdot m/s$ .