Chapter 9, Problem 46P

(a)

To determine

The mass of the original object.

Answer to Problem 46P

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

Explanation of Solution

Given info: The first mass of first fragment is $\frac{1.00\text{MeV}}{c^2}$, the momentum of the first fragment is $\frac{1.75\text{MeV}}{c}$ in positive $x$ direction, the first mass of second fragment is $\frac{1.50\text{MeV}}{c^2}$ and the momentum of the second fragment is $\frac{2.00\text{MeV}}{c}$ in positive $y$ direction.

The formula to calculate energy of the first fragment is,

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

Here,

$m_1$ is the mass of the first fragment.

$p_1$ is the linear momentum of the first fragment.

$c$ is the speed of light.

Substitute $\frac{1.75\text{MeV}}{c}$ for $p_1$ and $\frac{1.00\text{MeV}}{c^2}$ for $m_1$ in above equation to find $E_1$.

$\begin{array}{c}{E}_1^2={\left(\frac{1.75\text{MeV}}{c}\right)}^2{c}^2+{\left(\left(\frac{1.00\text{MeV}}{c^2}\right)c^2\right)}^2\\ ={\left(1.75\text{MeV}\right)}^2+{\left(1.00\text{MeV}\right)}^2\\ E_1=\sqrt{{\left(1.75\text{MeV}\right)}^2+{\left(1.00\text{MeV}\right)}^2}\\ \cong 2.02\text{MeV}\end{array}$

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

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 second fragment.

$p_2$ is the linear momentum of the second fragment.

Substitute $\frac{2.00\text{MeV}}{c}$ for $p_2$ and $\frac{1.50\text{MeV}}{c^2}$ for $m_2$ in above equation to find $E_2$.

$\begin{array}{c}{E}_1^2={\left(\frac{2.00\text{MeV}}{c}\right)}^2{c}^2+{\left(\left(\frac{1.50\text{MeV}}{c^2}\right)c^2\right)}^2\\ ={\left(2.00\text{MeV}\right)}^2+{\left(1.50\text{MeV}\right)}^2\\ E_2=\sqrt{{\left(2.00\text{MeV}\right)}^2+{\left(1.50\text{MeV}\right)}^2}\\ =2.50\text{MeV}\end{array}$

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

The total energy of the original object is,

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

Thus, the total energy of the original object is $4.52\text{MeV}$.

The total linear momentum of the original object is,

$\begin{array}{c}{p}^2=p_1^2+p_2^2\\ ={\left(\frac{1.75\text{MeV}}{c}\right)}^2+{\left(\frac{2.00\text{MeV}}{c}\right)}^2\\ =\frac{1}{c^2}\left({\left(1.75\text{MeV}\right)}^2+{\left(2.00\text{MeV}\right)}^2\right)\end{array}$

The equation for the energy of the object is,

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

Here,

$m$ is the mass of the original object.

Substitute $4.52\text{MeV}$ for $E$ and $\frac{1}{c^2}\left({\left(1.75\text{MeV}\right)}^2+{\left(2.00\text{MeV}\right)}^2\right)$ for $p^2$ in above equation to find $m$.

$\begin{array}{c}{\left(4.52\text{MeV}\right)}^2=\frac{1}{c^2}\left({\left(1.75\text{MeV}\right)}^2+{\left(2.00\text{MeV}\right)}^2\right)c^2+{\left(m{c}^2\right)}^2\\ ={\left(1.75\text{MeV}\right)}^2+{\left(2.00\text{MeV}\right)}^2+{\left(m{c}^2\right)}^2\\ m=\sqrt{\frac{13.36\text{MeV}}{c^4}}\\ =\frac{3.65\text{MeV}}{c^2}\end{array}$

Conclusion:

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 46P

The speed of the original object is $0.589c$.

Explanation of Solution

Given info: The first mass of first fragment is $\frac{1.00\text{MeV}}{c^2}$, the momentum of the first fragment is $\frac{1.75\text{MeV}}{c}$ in positive $x$ direction, the first mass of second fragment is $\frac{1.50\text{MeV}}{c^2}$ and the momentum of the second fragment is $\frac{2.00\text{MeV}}{c}$ in positive $y$ direction.

The equation for the total energy of the object is,

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

Here,

$u$ is the speed of the original object.

Substitute $4.52\text{MeV}$ for $E$ and $\frac{3.65\text{MeV}}{c^2}$ for $m$ in above equation to find $u$.

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

Further, solve for $u$.

$\begin{array}{c}\frac{u^2}{c^2}=1-0.652\\ u=\sqrt{0.348c^2}\\ =0.589c\end{array}$

Conclusion:

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