Chapter 26, Problem 61GP

(a)

To determine

To find: Momentum of proton

Answer to Problem 61GP

  $1.32\times {10}^{-18}\text{kg}\cdot \text{m}/\text{s}$

Explanation of Solution

Given data:

Speed of proton, $v=935c$

Formula used:

Relativistic momentum formula

  $p=\gamma mv=\frac{mv}{\sqrt{1-v^2/c^2}}=\frac{1}{c}\frac{m{c}^2(v/c)}{\sqrt{1-v^2/c^2}}$

Calculation:

Relativistic momentum of proton$\begin{array}{c}p=\frac{1}{c}\frac{m{c}^2(v/c)}{\sqrt{1-v^2/c^2}}\\ =\frac{1}{c}\frac{(938.3\text{MeV})(0.935)}{\sqrt{1-{0.935}^2}}\\ =2473.7\text{MeV}/\text{c}\end{array}$

  $\begin{array}{l}\approx 2.47\text{GeV}/\text{c}\\ =(2.47\text{GeV}/\text{c})\left(\frac{1c}{3.00\times {10}^8\text{m}/\text{s}}\right)\left(\frac{1.602\times {10}^{-10}\text{J}/\text{GeV}}{1\text{GeV}}\right)\end{array}$

  $=1.32\times {10}^{-18}\text{kg}\cdot \text{m}/\text{s}$

Conclusion:

Momentum of proton is $1.32\times {10}^{-18}\text{kg}\cdot \text{m}/\text{s}$

(b)

To determine

To find: Total momentum of two proton

Explanation of Solution

Both proton are moving with same speed but in opposite direction. So the momentum of both proton equal in magnitude but opposite in sign.

Therefore, the sum of both momentum is zero.

(c)

To determine

To find: Momentum of one proton seen by other proton

Answer to Problem 61GP

  $7.37\times {10}^{-18}\text{kg}\cdot \text{m}/\text{s}$

Explanation of Solution

Given data:

Speed of proton, $v=935c$

Formula used:

Relativistic momentum formula

  $p=\gamma mv=\frac{mv}{\sqrt{1-v^2/c^2}}=\frac{1}{c}\frac{m{c}^2(v/c)}{\sqrt{1-v^2/c^2}}$

Calculation:

From relativistic addition of velocities

  $u=\frac{v+u^{\prime }}{1+v{u}^{\prime }/c^2}$

Since the both protons are moving opposite directions $u=0.935c$ and $v=-0.935$

Then, the speed of one proton relative to the other

  $u^{\prime }=\frac{u-v}{\left(1-\frac{vu}{c^2}\right)}$

  $u^{\prime }=\frac{0.935c-(-0.935c)}{\left(1-\frac{(-0.935c)(0.935c)}{c^2}\right)}$

  $u^{\prime }=0.9977c$

Relativistic momentum of proton seen by other proton

  $\begin{array}{c}p=\frac{1}{c}\frac{m{c}^2(u\text{'}/c)}{\sqrt{1-u{\text{'}}^2/c^2}}\\ =\frac{1}{c}\frac{(938.3\text{MeV})\left(\frac{0.9977c}{c}\right)}{\sqrt{1-\frac{{\left(0.9977c\right)}^2}{c^2}}}\\ =13810.6\text{MeV}/c\\ \approx 13.8\text{GeV}/c\\ =(13.8\text{GeV}/c)\left(\frac{1c}{3.00\times {10}^8\text{m}/\text{s}}\right)\left(\frac{1.602\times {10}^{-10}\text{J}/\text{GeV}}{1\text{GeV}}\right)\\ =7.37\times {10}^{-18}\text{kg}\cdot \text{m}/\text{s}\end{array}$

Conclusion:

Momentum of one proton seen by other proton is $7.37\times {10}^{-18}\text{kg}\cdot \text{m}/\text{s}$ .