Chapter 26, Problem 19P

(a)

To determine

The percentage change in momentum for the given values.

Answer to Problem 19P

  $\text{310\%}$ .

Explanation of Solution

Given:

Speeds: $0.45c,0.90c,0.98c$

Formula used:

It is known that the momentum of proton or particle:

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

Where,

  $m_0$ is rest mass

  $v$ is velocity

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

Calculation:

The proton’s momenta are found using $p=\frac{m_{0}v}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}$

Calculation for speeds: $0.45c$

  $p_1=\frac{m_0\left(0.45c\right)}{\sqrt{1-{\left(0.45\right)}^2}}=0.5039c{m}_0$

Calculation for speeds: $0.90c$

  $p_2=\frac{m_0\left(0.90c\right)}{\sqrt{1-{\left(0.90\right)}^2}}=2.065c{m}_0$

Calculation for speeds: $0.98c$

  $p_3=\frac{m_0\left(0.98c\right)}{\sqrt{1-{\left(0.98\right)}^2}}=4.925c{m}_0$

Calculation for the percentage change in momentum of a proton that accelerates from $$ to $\text{0}\text{.90}\text{c}$ :

  $\frac{{\text{p}}_{\text{2}}{\text{-p}}_{\text{1}}}{{\text{p}}_{\text{1}}}\text{=}\frac{\text{2}\text{.065}{\text{cm}}_{\text{0}}\text{- 0}\text{.5039}{\text{cm}}_{\text{0}}}{\text{0}\text{.5039}{\text{cm}}_{\text{0}}}\text{=3}\text{.10}\text{=}\text{310\%}$

(b)

To determine

The percentage change in momentum for given values.

Answer to Problem 19P

  $140\text{\%}$ .

Explanation of Solution

Given:

Speeds: $0.45c,0.90c,0.98c$

Formula used:

It is known that the momentum of proton or particle:

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

Where,

  $m_0$ is rest mass

  $v$ is velocity

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

Calculation:

The proton’s momenta are found using $p=\frac{m_{0}v}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}$

Calculation for speeds: $0.45c$

  $p_1=\frac{m_0\left(0.45c\right)}{\sqrt{1-{\left(0.45\right)}^2}}=0.5039c{m}_0$

Calculation for speeds: $0.90c$

  $p_2=\frac{m_0\left(0.90c\right)}{\sqrt{1-{\left(0.90\right)}^2}}=2.065c{m}_0$

Calculation for speeds: $0.98c$

  $p_3=\frac{m_0\left(0.98c\right)}{\sqrt{1-{\left(0.98\right)}^2}}=4.925c{m}_0$

Calculation for the percentage change in momentum of a proton that accelerates from $\text{0}\text{.90}\text{c}$ to $\text{0}\text{.98}\text{c}$ :

  $\frac{{\text{p}}_{\text{3}}{\text{-p}}_{\text{2}}}{{\text{p}}_{\text{2}}}\text{=}\frac{\text{4}\text{.925}{\text{cm}}_{\text{0}}\text{- 2}\text{.065}{\text{cm}}_{\text{0}}}{\text{2}\text{.065}{\text{cm}}_{\text{0}}}\text{=1}\text{.40}\text{=}140\text{\%}$