Chapter 2, Problem 1P

Calculate the momentum of a proton moving with a speed of (a) 0.010c, (b) 0.50c, (c) 0.90c. (d) Convert the answers of (a)–(c) to MeV/c.

To determine

The momentum of a proton moving with a speed $0.010c$

Answer to Problem 1P

The momentum of a proton moving with a speed $0.010c$ is $5.01\times {10}^{-21}\text{kg}\cdot \text{m}/\text{s}$

Explanation of Solution

Write the equation for the momentum.

    $p=\frac{mv}{{\left[1-\left(v^2/c^2\right)\right]}^{1/2}}$        (I)

Here $m$ is the mass of the proton, $v$ is the velocity of the proton, $p$ is the momentum of the proton and $c$ is the velocity.

Conclusion:

Substitute $1.67\times {10}^{-27}\text{ kg}$ for $m$, and $0.01c$ for $v$

    $\begin{array}{c}p=\frac{\left(1.67\times {10}^{-27}\text{ kg}\right)\left(0.01c\right)}{{\left[1-{\left(0.01c/c\right)}^2\right]}^{1/2}}\\ =5.01\times {10}^{-21}\text{kg}\cdot \text{m}/\text{s}\end{array}$

The momentum of a proton moving with a speed $0.010c$ is $5.01\times {10}^{-21}\text{kg}\cdot \text{m}/\text{s}$

To determine

The momentum of a proton moving with a speed $0.5c$

Answer to Problem 1P

The momentum of a proton moving with a speed $0.5c$ is $2.89\times {10}^{-19}\text{kg}\cdot \text{m}/\text{s}$

Explanation of Solution

Write the equation for the momentum.

    $p=\frac{mv}{{\left[1-\left(v^2/c^2\right)\right]}^{1/2}}$        (I)

Here $m$ is the mass of the proton, $v$ is the velocity of the proton, $p$ is the momentum of the proton and $c$ is the velocity.

Conclusion:

Substitute $1.67\times {10}^{-27}\text{ kg}$ for $m$, and $0.5c$ for $v$

    $\begin{array}{c}p=\frac{\left(1.67\times {10}^{-27}\text{ kg}\right)\left(0.5c\right)}{{\left[1-{\left(0.5c/c\right)}^2\right]}^{1/2}}\\ =2.89\times {10}^{-19}\text{kg}\cdot \text{m}/\text{s}\end{array}$

The momentum of a proton moving with a speed $0.5c$ is $2.89\times {10}^{-19}\text{kg}\cdot \text{m}/\text{s}$

To determine

The momentum of a proton moving with a speed $0.9c$

Answer to Problem 1P

The momentum of a proton moving with a speed $0.9c$ is $1.03\times {10}^{-18}\text{kg}\cdot \text{m}/\text{s}$

Explanation of Solution

Write the equation for the momentum.

    $p=\frac{mv}{{\left[1-\left(v^2/c^2\right)\right]}^{1/2}}$        (I)

Here $m$ is the mass of the proton, $v$ is the velocity of the proton, $p$ is the momentum of the proton and $c$ is the velocity.

Conclusion:

Substitute $1.67\times {10}^{-27}\text{ kg}$ for $m$, and $0.5c$ for $v$

    $\begin{array}{c}p=\frac{\left(1.67\times {10}^{-27}\text{ kg}\right)\left(0.9c\right)}{{\left[1-{\left(0.9c/c\right)}^2\right]}^{1/2}}\\ =1.03\times {10}^{-18}\text{kg}\cdot \text{m}/\text{s}\end{array}$

The momentum of a proton moving with a speed $0.9c$ is $1.03\times {10}^{-18}\text{kg}\cdot \text{m}/\text{s}$

To determine

Convert the answer from subpart (a) to (c) into $\text{MeV}/\text{c}$

Answer to Problem 1P

$5.01\times {10}^{-21}\text{kg}\cdot \text{m}/\text{s}$ into $\text{MeV}/\text{c}$ is $9.38\text{MeV}/c$, $2.89\times {10}^{-19}\text{kg}\cdot \text{m}/\text{s}$ into $\text{MeV}/\text{c}$ is $540\text{MeV}/c$, $1.03\times {10}^{-18}\text{kg}\cdot \text{m}/\text{s}$ into  $\text{MeV}/\text{c}$ is $1930\text{MeV}/c$

Explanation of Solution

Write the equation for the momentum.

    $p=\frac{mv}{{\left[1-\left(v^2/c^2\right)\right]}^{1/2}}$        (I)

Here $m$ is the mass of the proton, $v$ is the velocity of the proton, $p$ is the momentum of the proton and $c$ is the velocity.

To convert $\text{kg}\cdot \text{m/s}$ to $\text{MeV}/\text{c}$ multiple the term with $\frac{3\times {10}^8}{1.6\times {10}^{-19}}$

Conclusion:

Multiple  $\frac{3\times {10}^8}{1.6\times {10}^{-19}}$ with expression $p=\frac{\left(1.67\times {10}^{-27}\text{ kg}\right)\left(0.01c\right)}{{\left[1-{\left(0.01c/c\right)}^2\right]}^{1/2}}$ from subpart (a)

    $\begin{array}{c}p=\frac{\left(1.67\times {10}^{-27}\text{ kg}\right)\left(0.01c\right)}{{\left[1-{\left(0.01c/c\right)}^2\right]}^{1/2}}\times \frac{3\times {10}^8}{1.6\times {10}^{-19}}\\ =9.38\text{MeV/c}\end{array}$

Multiple  $\frac{3\times {10}^8}{1.6\times {10}^{-19}}$ with expression $p=\frac{\left(1.67\times {10}^{-27}\text{ kg}\right)\left(0.5c\right)}{{\left[1-{\left(0.5c/c\right)}^2\right]}^{1/2}}$ from subpart (a)

    $\begin{array}{c}p=\frac{\left(1.67\times {10}^{-27}\text{ kg}\right)\left(0.5c\right)}{{\left[1-{\left(0.5c/c\right)}^2\right]}^{1/2}}\times \frac{3\times {10}^8}{1.6\times {10}^{-19}}\\ =540\text{MeV/c}\end{array}$

Multiple  $\frac{3\times {10}^8}{1.6\times {10}^{-19}}$ with expression $p=\frac{\left(1.67\times {10}^{-27}\text{ kg}\right)\left(0.9c\right)}{{\left[1-{\left(0.9c/c\right)}^2\right]}^{1/2}}$ from subpart (a)

    $\begin{array}{c}p=\frac{\left(1.67\times {10}^{-27}\text{ kg}\right)\left(0.9c\right)}{{\left[1-{\left(0.9c/c\right)}^2\right]}^{1/2}}\times \frac{3\times {10}^8}{1.6\times {10}^{-19}}\\ =1930\text{MeV/c}\end{array}$

Therefore $5.01\times {10}^{-21}\text{kg}\cdot \text{m}/\text{s}$ into $\text{MeV}/\text{c}$ is $9.38\text{MeV}/c$, $2.89\times {10}^{-19}\text{kg}\cdot \text{m}/\text{s}$ into $\text{MeV}/\text{c}$ is $540\text{MeV}/c$, $1.03\times {10}^{-18}\text{kg}\cdot \text{m}/\text{s}$ into  $\text{MeV}/\text{c}$ is $1930\text{MeV}/c$