Chapter 2, Problem 29P

(a)

To determine

The mass of the original object.

Answer to Problem 29P

The mass of the original object is $\frac{3.65\text{MeV}}{c^2}$

Explanation of Solution

The formula to calculate energy of the first fragment is,

    $E_1^2=p_1^2{c}^2+{\left(m_1{c}^2\right)}^2$        (I)

Here, $m_1$ is the mass of the fragment, $p_1$ is the linear momentum of the first fragment, and $c$ is the speed of light.

The formula to calculate energy of the second fragment is,

  $E_2^2=p_2^2{c}^2+{\left(m_2{c}^2\right)}^2$

Here, $m_2$ is the mass of the fragment, $p_2$ is the linear momentum of the second fragment, and $c$ is the speed of light.

Conclusion:

Substitute $\frac{\text{1}\text{.75MeV}}{c}$ for $p_2$, and $\frac{\text{1}\text{.00MeV}}{c^2}$ for $m_2$ in expression (I) to calculate the value for $E_2$,

    $\begin{array}{c}{E}_2^2={\left(\frac{\text{1}\text{.75MeV}}{c}\right)}^2{c}^2+{\left(\left(\frac{\text{1}\text{.00MeV}}{c^2}\right)c^2\right)}^2\\ E_2=\sqrt{{\left(\text{1}\text{.75MeV}\right)}^2+{\left(\text{1}\text{.00MeV}\right)}^2}\\ =2.02\text{MeV}\end{array}$

Thus the total energy of the first fragment is $2.02\text{MeV}$.

Substitute $\frac{\text{2}\text{.00MeV}}{c}$ for $p_2$, and $\frac{\text{1}\text{.50MeV}}{c^2}$ for $m_2$ in expression (I) to calculate the value for $E_2$,

    $\begin{array}{c}{E}_2^2={\left(\frac{\text{2}\text{.00MeV}}{c}\right)}^2{c}^2+{\left(\left(\frac{\text{1}\text{.50MeV}}{c^2}\right)c^2\right)}^2\\ E_2=\sqrt{{\left(\text{2}\text{.00MeV}\right)}^2+{\left(\text{1}\text{.50MeV}\right)}^2}\\ =2.50\text{MeV}\end{array}$

Thus the total energy of the first fragment is $2.50\text{MeV}$.

The total energy of the original object is, by substituting $2.50\text{MeV}$ and $2.02\text{MeV}$,

  $\begin{array}{c}E=E_1+E_2\\ =2.02\text{MeV}+2.50\text{MeV}\\ =4.52\text{MeV}\end{array}$ 

The total linear momentum of the original object is,

  $\begin{array}{c}{p}^2=p_1^2+p_2^2\\ ={\left(\frac{\text{2}\text{.00MeV}}{c}\right)}^2+{\left(\frac{\text{1}\text{.75MeV}}{c}\right)}^2\\ =\frac{1}{c^2}\left({\left(\text{2}\text{.00MeV}\right)}^2+{\left(\text{1}\text{.75MeV}\right)}^2\right)\end{array}$

The equation for energy of the object is,

  $E^2=p^2{c}^2+{\left(m{c}^2\right)}^2$     

Here, $m$ is the mass of the object.

Substitute $\text{4}\text{.52MeV}$ for $E$ and $\frac{1}{c^2}\left({\left(\text{2}\text{.00MeV}\right)}^2+{\left(\text{1}\text{.75MeV}\right)}^2\right)$ for $p^2$ in above equation $E^2=p^2{c}^2+{\left(m{c}^2\right)}^2$

  $\begin{array}{c}{\left(\text{4}\text{.52MeV}\right)}^2=\frac{1}{c^2}\left({\left(\text{2}\text{.00MeV}\right)}^2+{\left(\text{1}\text{.75MeV}\right)}^2\right)c^2+{\left(m{c}^2\right)}^2\\ ={\left(\text{1}\text{.75MeV}\right)}^2+{\left(\text{2}\text{.00MeV}\right)}^2+{\left(m{c}^2\right)}^2\\ =\frac{3.65\text{MeV}}{c^2}\end{array}$

Therefore the mass of the original object is $\frac{3.65\text{MeV}}{c^2}$

(b)

To determine

The speed of the original object.

Answer to Problem 29P

The speed of the original object is $0.589c$

Explanation of Solution

Write the expression for the total energy of the object is,

  $E=\frac{m{c}^2}{\sqrt{1-\frac{u^2}{c^2}}}$        (I)

Here, $u$ is the speed of the original object, $m$ is the mass of the object, and $c$ is the speed of light.

Conclusion:

Substitute $\text{4}\text{.52MeV}$ for $\text{E}$, and $\frac{3.65\text{MeV}}{c^2}$ for $\text{m}$ expression (I) and solve for $u$

  $\begin{array}{c}4.52\text{MeV}=\frac{\left(\frac{3.65\text{MeV}}{c^2}\right)c^2}{\sqrt{1-\frac{u^2}{c^2}}}\\ 1-\frac{u^2}{c^2}=0.652\\ \frac{u^2}{c^2}=\sqrt{0.348c^2}\\ =0.589c\end{array}$

Therefore the speed of the original object is $0.589c$