Chapter 39, Problem 83PQ

(a)

To determine

The speed of the first piece inspected by an observer in the lab.

Answer to Problem 83PQ

The speed of the first piece inspected by an observer in the lab is $0.990c$.

Explanation of Solution

Write the expression for the Lorentz constant.

    ${\gamma }_1=\frac{1}{\sqrt{1-\frac{v_1^2}{c^2}}}$

Here, Lorentz constant is ${\gamma }_1$, speed of the first piece inspected by an observer in the lab is $v_1$ and speed of light is $c$.

Write the expression for the momentum in fixed frame.

    ${\overrightarrow{p}}_1=m_1{v}_1$

Here, momentum in fixed frame is ${\overrightarrow{p}}_1$, rest mass is $m_1$ and velocity is $v_1$.

Write the expression for the momentum with respect to laboratory frame.

    ${\overrightarrow{p}}_1^{\text{'}}={\gamma }_1{\overrightarrow{p}}_1$

Here, momentum with respect to laboratory frame is $\overrightarrow{p}{\text{'}}_1$.

Substitute $m_1{v}_1$ for ${\overrightarrow{p}}_1$ and $\frac{1}{\sqrt{1-\frac{v_1^2}{c^2}}}$ for ${\gamma }_1$ in the above equation to find ${\overrightarrow{p}}_1^{\text{'}}$.

    $\begin{array}{c}{\overrightarrow{p}}_1^{\text{'}}=\left(\frac{1}{\sqrt{1-\frac{v_1^2}{c^2}}}\right)m_1{v}_1\\ =\frac{m_1{v}_1}{\sqrt{1-\frac{v_1^2}{c^2}}}\end{array}$

Conclusion:

Substitute $1.45\text{MeV}/{\text{c}}^2$ for $m_1$ and $-10.0\text{MeV}/\text{c}$ for $\overrightarrow{p}{\text{'}}_1$ in the above equation to find $v_1$.

    $\begin{array}{c}-10.0\text{MeV}/c=\frac{\left(1.45\text{MeV}/c^2\right)v_1}{\sqrt{1-\frac{v^2{}_1}{c^2}}}\\ \frac{v_1}{\sqrt{1-\frac{v_1^2}{c^2}}}=-\frac{-10.0\text{MeV}/c}{\left(1.45\text{MeV}/c^2\right)}\\ =6.896c\end{array}$

Take square both sides.

    $\begin{array}{c}{\left(\frac{v_1}{\sqrt{1-\frac{v^2{}_1}{c^2}}}\right)}^2={\left(-6.896c\right)}^2\\ \frac{v_1^2}{1-\frac{v_1^2}{c^2}}=47.56c^2\\ \frac{v_1^2}{\frac{c^2-v_1^2}{c^2}}=47.56c^2\\ \frac{c^2{v}_1^2}{c^2-v_1^2}=47.56c^2\end{array}$

Further solve the above equation.

    $\begin{array}{c}{c}^2{v}_1^2=47.56c^2\left(c^2-v_1^2\right)\\ c^2{v}_1^2=47.56c^4-47.56c^2{v}_1^2\\ c^2{v}_1^2+47.56c^2{v}_1^2=47.56c^4\\ v_1^2+47.56v_1^2=47.56c^2\end{array}$

Further solve the above equation.

    $\begin{array}{c}{v}_1^2\left(1+47.56\right)=47.56c^2\\ v_1^2\left(48.56\right)=47.56c^2\\ v_1^2=\frac{47.56c^2}{48.56}\\ v_1^2=0.9794c^2\end{array}$

Further solve the above equation.

    $\begin{array}{c}{v}_1=\sqrt{0.9794c}\\ =0.990c\end{array}$

Therefore, the speed of the first piece inspected by an observer in the lab is $0.990c$.

(b)

To determine

The speed of the second piece inspected by an observer in the lab.

Answer to Problem 83PQ

The speed of the second piece inspected by an observer in the lab is $0.989c$.

Explanation of Solution

Write the expression for the Lorentz constant.

    ${\gamma }_2=\frac{1}{\sqrt{1-\frac{v_2^2}{c^2}}}$

Here, Lorentz constant is ${\gamma }_2$, speed of the first piece inspected by an observer in the lab is $v_2$ and speed of light is $c$.

Write the expression for the momentum in fixed frame.

    ${\overrightarrow{p}}_2=m_2{v}_2$

Here, momentum in fixed frame is ${\overrightarrow{p}}_2$, rest mass is $m_2$ and velocity is $v_2$.

Write the expression for the momentum with respect to laboratory frame.

    ${\overrightarrow{p}}_2^{\text{'}}={\gamma }_2{\overrightarrow{p}}_2$

Here, momentum with respect to laboratory frame is ${\overrightarrow{p}}_2^{\text{'}}$.

Substitute $m_2{v}_2$ for ${\overrightarrow{p}}_2$ and $\frac{1}{\sqrt{1-\frac{v_2^2}{c^2}}}$ for ${\gamma }_1$ in the above equation to find ${\overrightarrow{p}}_2^{\text{'}}$.

    $\begin{array}{c}{\overrightarrow{p}}_2^{\text{'}}=\left(\frac{1}{\sqrt{1-\frac{v_2^2}{c^2}}}\right)m_2{v}_2\\ =\frac{m_2{v}_2}{\sqrt{1-\frac{v_2^2}{c^2}}}\end{array}$

Conclusion:

Substitute $0.850\text{MeV}/{\text{c}}^2$ for $m_2$ and $5.75\text{MeV}/\text{c}$ for $\overrightarrow{p}{\text{'}}_2$ in the above equation to find $v_2$.

    $\begin{array}{c}5.75\text{MeV}/c=\frac{\left(0.850\text{MeV}/c^2\right)v_2}{\sqrt{1-\frac{v_2^2}{c^2}}}\\ \frac{v_2}{\sqrt{1-\frac{v_2^2}{c^2}}}=-\frac{5.75\text{MeV}/c}{\left(0.850\text{MeV}/c^2\right)}\\ =6.764c\end{array}$

Take square both sides.

    $\begin{array}{c}{\left(\frac{v_2}{\sqrt{1-\frac{v_2^2}{c^2}}}\right)}^2={\left(6.764c\right)}^2\\ \frac{v_2^2}{1-\frac{v^2{}_2}{c^2}}=45.76c^2\\ \frac{v_2^2}{\frac{c^2-v_2^2}{c^2}}=45.76c^2\\ \frac{c^2{v}_2^2}{c^2-v_2^2}=45.76c^2\end{array}$

Further solve the above equation

    $\begin{array}{c}{c}^2{v}_2^2=45.76c^2\left(c^2-v_2^2\right)\\ c^2{v}_2^2=45.76c^4-45.76c^2{v}_2^2\\ c^2{v}_2^2+45.76c^2{v}_2^2=45.76c^4\\ v_2^2+45.76v_2^2={45.76}_{}{c}^2\end{array}$

Further solve the above equation

    $\begin{array}{c}{v}_2^2\left(1+45.76\right)=45.76c^2\\ v_2^2\left(46.76\right)=45.76c^2\\ v_2^2=\frac{45.76c^2}{46.76}\\ v_2^2=0.9786c^2\end{array}$

Further solve the above equation

    $\begin{array}{c}{v}_2=\sqrt{0.9786c}\\ =0.989c\end{array}$

Therefore, the speed of the second piece inspected by an observer in the lab is $0.989c$.