Chapter 26, Problem 87P

(a)

To determine

The momentum magnitude of the original particle.

Answer to Problem 87P

The momentum magnitude of the original particle is $409\text{MeV}/c$.

Explanation of Solution

Write the expression for the magnitude of momentum for each particle.

  $p=\gamma mv$                                                                                                         (I)

Here, $p$ is the magnitude of momentum of each pion, $\gamma$ is the Lorentz factor, $m$ is the rest mass of the pion and $v$ is the speed of the pion.

Write the expression for Lorentz factor.

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

Here, $c$ is the speed of light in vacuum.

Multiply denominator and numerator of equation (I) by $c^2$ in to obtain new equation for $p$.

  $p=\frac{\gamma m{c}^{2}v}{c^2}$

In problem it is given that the particles are moving at right angle to each other.

Calculate the vector sum of their momenta.

  $\begin{array}{c}p=\sqrt{{\left(\frac{\gamma m{c}^{2}v}{c^2}\right)}^2+{\left(\frac{\gamma m{c}^{2}v}{c^2}\right)}^2}\\ =\sqrt{2}\frac{\gamma m{c}^{2}v}{c^2}\end{array}$                                                                               (II)

Write the expression for the rest mass energy of the particles.

  $E_0=m{c}^2$

Here, $E_0$ is the rest energy.

Substitute $E_0$ for $m{c}^2$ in equation (II) to get $p$.

  $p=\sqrt{2}\frac{\gamma E_{0}v}{c^2}$

Substitute ${\left(1-\frac{v^2}{c^2}\right)}^{-1}$ for $\gamma$ in above equation to get final expression for $p$.

  $p=\sqrt{2}\frac{\left({\left(1-\frac{v^2}{c^2}\right)}^{-1}\right)E_{0}v}{c^2}$                                                                                      (III)

Conclusion:

Substitute $0.900c$ for $v$, $140.0\text{MeV}$ for $E_0$ in above equation to get $p$.

  $\begin{array}{c}p=\sqrt{2}\frac{\left({\left(1-\frac{{\left(0.900c\right)}^2}{c^2}\right)}^{-1}\right)\left(140.0\text{MeV}\right)\left(0.900c\right)}{c^2}\\ =409\text{MeV}/c\end{array}$

According to momentum conservation principle, momentum of original particle is equal to vector sum of the momenta of the pions.

Therefore, the momentum magnitude of the original particle is $409\text{MeV}/c$.

(b)

To determine

The kinetic energy of the original particle.

Answer to Problem 87P

The kinetic energy of the original particle is $147\text{MeV}$.

Explanation of Solution

According to conservation of energy, total energy of the original particle must equal to the sum of the total energy of two pions.

Write the expression for the total energy pions.

  $E=E_1+E_2$                                                                                                             (IV)

Here, $E$ is the sum of energy of the two pions $E_1$ is the total energy first pion and $E_2$ is the total energy of second pion.

Write the expression for the total energy of first pion.

  $E_1=\left({\left(1-\frac{v^2}{c^2}\right)}^{-1}\right)E_0$

Write the expression for the total energy of second pion.

  $E_2=\left({\left(1-\frac{v^2}{c^2}\right)}^{-1}\right)E_0$

Substitute $\left({\left(1-\frac{v^2}{c^2}\right)}^{-1}\right)E_0$ for $E_1$ and $\left({\left(1-\frac{v^2}{c^2}\right)}^{-1}\right)E_0$ for $E_2$ in equation (IV) to get $E$.

  $\begin{array}{c}E=\left({\left(1-\frac{v^2}{c^2}\right)}^{-1}\right)E_0+\left({\left(1-\frac{v^2}{c^2}\right)}^{-1}\right)E_0\\ =2\left({\left(1-\frac{v^2}{c^2}\right)}^{-1}\right)E_0\end{array}$                                                                    (V)

Write the expression for the total energy of original particle.

  $E^2={E^{\prime }}_0^2+{\left(pc\right)}^2$

Here, $E$ is the total energy of original particle,${E^{\prime }}_0$ is the rest mass energy of the original particle and $p$ is the momentum of the original particle.

Rearrange above equation to get $E_0$.

  ${E^{\prime }}_0=\sqrt{E^2-{\left(pc\right)}^2}$                                                                                                 (VI)

Write the expression for the kinetic energy of original particle.

  $K=E-{E^{\prime }}_0$                                                                                                            (VII)

Here, $K$ is the kinetic energy of original particle.

Conclusion:

Substitute $0.900c$ for $v$ and $140.0\text{MeV}$ for $E_0$ in equation (V) to get $E$.

  $\begin{array}{c}E=2\left({\left(1-\frac{{\left(0.900c\right)}^2}{c^2}\right)}^{-1}\right)\times 140.0\text{MeV}\\ =\text{642}\text{MeV}\end{array}$

According to conservation of energy, $642\text{MeV}$ is the total energy of original particle.

Substitute $642\text{MeV}$ for $E$ and $\frac{409}{c}\text{MeV}$ for $p$ in equation (VI) to get ${E^{\prime }}_0$.

  $\begin{array}{c}{E^{\prime }}_0=\sqrt{{\left(642\text{MeV}\right)}^2-{\left(\frac{409}{c}\text{MeV}\times c\right)}^2}\\ =\sqrt{{\left(642\text{MeV}\right)}^2-{\left(409\text{MeV}\right)}^2}\\ =495\text{MeV}\end{array}$

Substitute $642\text{MeV}$ for $E$ and $495\text{MeV}$ for ${E^{\prime }}_0$ in equation (VII) to get $K$.

  $\begin{array}{c}K=642\text{MeV}-495\text{MeV}\\ \text{=147}\text{MeV}\end{array}$

Therefore, the kinetic energy of the original particle is $147\text{MeV}$.

(c)

To determine

The mass of original particle in units of $\text{MeV}/c^2$.

Answer to Problem 87P

The mass of original particle in units of $\text{MeV}/c^2$ is $495\text{MeV}/c^2$.

Explanation of Solution

Write the expression for the rest mass of the particle in terms of rest energy.

  $m=\frac{E_0}{c^2}$

Conclusion:

Substitute $495\text{MeV}$ for $E_0$ in above equation to get $m$.

  $\begin{array}{c}m=\frac{495\text{MeV}}{c^2}\\ =495\text{MeV}/{\text{c}}^2\end{array}$

Therefore, the mass of original particle in units of $\text{MeV}/c^2$ is $495\text{MeV}/c^2$.