Chapter 26, Problem 29P

To determine

The kinetic energy and momentum of a proton.

Answer to Problem 29P

The kinetic energy: $\text{1}\text{.51}{\text{×10}}^{\text{-10}}\text{J}$

The momentum of a proton: $8.70\times {10}^{-19}kg\cdot m/s$

Explanation of Solution

Given:

Speed of proton is $\left(2.60\times {10}^{8}m/s\right)$

Formula used:

It is known that:

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

Where,

  $m_0$ is rest mass

m is relativistic mass

v is velocity

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

It is known that the momentum of proton:

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

Calculation:

It is known that the mass of a proton is $=1.67\times {10}^{-27}kg$

Also, the speed of the proton in terms of speed of light is

  $v=\frac{\left(\text{2}{\text{.60×10}}^{\text{8}}\text{m/s}\right)}{\left(\text{3}{\text{.00×10}}^{\text{8}}\text{m/s}\right)}=0.867c$

The kinetic energy is

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

  $\text{=1}\text{.67}\times {\text{10}}^{-27}\left[\left\{\frac{\text{1}}{{\left[\text{1}-{\left(\text{0}\text{.867}\right)}^{\text{2}}\right]}^{\frac{\text{1}}{\text{2}}}}\right\}\text{-1}\right]$

  $\text{=}\text{1}\text{.51}{\text{×10}}^{\text{-10}}\text{J}$

It is known that the momentum of proton:

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

  $=\left(1.67\times {10}^{-27}kg\right)\left(2.60\times {10}^{8}m/s\right)\left\{\frac{1}{{\left[1-{\left(0.867\right)}^2\right]}^{\frac{1}{2}}}\right\}$

  $=8.70\times {10}^{-19}kg\cdot m/s$

Conclusion:

The kinetic energy is, $\left(\text{1}\text{.51}{\text{×10}}^{\text{-10}}\text{J}\right)$

The momentum of a proton $8.70\times {10}^{-19}kg\cdot m/s$ .