Chapter 2, Problem 30P

To determine

Coordinates (x, t) in System K and Kʹ at time of bullet firing and time when bullet reached at target.

Answer to Problem 30P

The coordinates in system K at time of bullet firing is $(1\text{ m, }3\text{ ns)}$ and at time of target attack is $\text{(}121\text{ m, }411\text{ ns)}$ and in system Kʹ are $(0.47\text{ m, }0.55\text{ ns)}$ and $\text{(}37.3\text{ m, }147.2\text{ ns)}$ respectively.

Explanation of Solution

Let’s coordinates at the timing of bullet firing are as follows:

In system K $\left(x_1,t_1\right)$ and $\left(x{\prime }_1,t{\prime }_1\right)$ in system Kʹ.

Let’s coordinates at the timing of target attack are as follows:

In system K $\left(x_2,t_2\right)$ and $\left(x{\prime }_2,t{\prime }_2\right)$ in system Kʹ.

System Kʹ is moving along the x-axis of the system K, therefore coordinates on x-axis in both system will change. Thus $\left(y=y\prime \right)$ and $\left(z=z\prime \right)$

In system K coordinates at time of bullet firing are as follows:

$x_1=1\text{ m, }{t}_1=3\text{ ns }$        (I)

In system K coordinates at time of target attacking are as follows:

$\begin{array}{c}{x}_2=1\text{ m}+120\text{ m }\\ =121\text{ m}\end{array}$        (II)

And

$t_2=t_1+\frac{\text{distnace tarvel by bullet }}{\text{speed of the bullet}}$

Substitute the distance travelled equal to 120 m and speed of the bullet 0.98c in above equation to get the time coordinate at time of target attack in system K. Thus,

$\begin{array}{c}{t}_2=3\times {10}^{-9}\text{ s}+\frac{\text{120 m}}{\text{0}\text{.98}\times \text{3}\times {10}^8\text{ m / s}}\\ =3\times {10}^{-9}\text{ s}+408\times {10}^{-9}\text{ s}\\ =411\text{ ns}\end{array}$        (III)

Use Lorentz transformations to get the coordinates in the system Kʹ. Thus,

$\begin{array}{l}t{\prime }_1=\gamma \left(t_1-\frac{v{x}_1}{c^2}\right)\\ t{\prime }_2=\gamma \left(t_2-\frac{v{x}_2}{c^2}\right)\\ x{\prime }_1=\gamma \left(x_1-v{t}_1\right)\\ x{\prime }_2=\gamma \left(x_2-v{t}_2\right)\end{array}$        (IV)

Where, $\gamma$ is the relativistic factor,

$y=\frac{1}{\sqrt{1-v^2/c^2}}$

Where, v is the velocity of the system Kʹ and c is the velocity of the light.

Substitute $v=0.8c$ in $y=\frac{1}{\sqrt{1-v^2/c^2}}$ to get relativistic factor. Thus

$\begin{array}{c}y=\frac{1}{\sqrt{1-{\left(0.8c\right)}^2/c^2}}\\ =\frac{1}{\sqrt{1-0.64}}\\ =\frac{10}{6}\\ =\frac{5}{3}\end{array}$        (V)

Substitute $t_1=3\text{ ns, }{x}_1=1\text{ m, }\gamma =\frac{5}{3},c=3\times {10}^8\text{ m / s,}$ and $v=0.8c$ in equation $t{\prime }_1=\gamma \left(t_1-\frac{v{x}_1}{c^2}\right)$ to get $t{\prime }_1$. Thus,

$\begin{array}{l}t{\prime }_1=\frac{5}{3}\left(3\text{ ns }-\frac{0.8c\times 1\text{ m}}{c^2}\right)\\ \text{=}\frac{5}{3}\left(3\text{ ns }-\frac{0.8\times 1\text{ m}}{3\times {10}^8\text{ m / s}}\right)\\ =\frac{5}{3}\left(3\text{ ns }-2.67\times {10}^{-9}\text{ s}\right)\\ =\frac{5}{3}\left(3\text{ ns }-2.67\text{ ns}\right)\end{array}$

Simplify the above equation

$\begin{array}{c}t{\prime }_1=\frac{5}{3}\left(3\text{ ns }-2.67\text{ ns}\right)\\ =\frac{5}{3}\times 0.33\text{ ns}\\ =0.55\text{ ns}\end{array}$

Substitute $t_2=411\text{ ns, }{x}_2=121\text{ m, }\gamma =\frac{5}{3},c=3\times {10}^8\text{ m / s,}$ and $v=0.8c$ in equation $t{\prime }_2=\gamma \left(t_2-\frac{v{x}_2}{c^2}\right)$ to get $t{\prime }_2$. Thus,

$\begin{array}{l}t{\prime }_2=\frac{5}{3}\left(411\text{ ns }-\frac{0.8c\times 121\text{ m}}{c^2}\right)\\ \text{=}\frac{5}{3}\left(\text{411 ns }-\frac{0.8\times 121\text{ m}}{3\times {10}^8\text{ m / s}}\right)\\ =\frac{5}{3}\left(411\text{ ns }-322.67\times {10}^{-9}\text{ s}\right)\\ =\frac{5}{3}\left(411\text{ ns }-322.67\text{ ns}\right)\end{array}$

Simplify the above equation

$\begin{array}{c}t{\prime }_1=\frac{5}{3}\left(411\text{ ns }-322.67\text{ ns}\right)\\ =\frac{5}{3}\times 88.33\text{ ns}\\ =147.2\text{ ns}\end{array}$

Substitute $t_1=3\text{ ns}=3\times {10}^{-9}\text{ s, }{x}_1=1\text{ m, }\gamma =\frac{5}{3},c=3\times {10}^8\text{ m / s,}$ and $v=0.8c$ in equation $x{\prime }_1=\gamma \left(x_1-v{t}_1\right)$ to get $x{\prime }_1$. Thus,

$\begin{array}{c}x{\prime }_1=\frac{5}{3}\left(1\text{ m}-0.8\times 3\times {10}^8\text{ m / s}\times 3\times {10}^{-9}\text{ s}\right)\\ =\frac{5}{3}\left(1\text{ m}-0.72\text{ m}\right)\\ =0.47\text{ m}\end{array}$

Substitute $t_2=411\text{ ns}=411\times {10}^{-9}\text{ s, }{x}_2=121\text{ m, }\gamma =\frac{5}{3},c=3\times {10}^8\text{ m / s,}$ and $v=0.8c$ in equation $x{\prime }_2=\gamma \left(x_2-v{t}_2\right)$ to get $x{\prime }_2$. Thus,

$\begin{array}{c}x{\prime }_2=\frac{5}{3}\left(121\text{ m}-0.8\times 3\times {10}^8\text{ m / s}\times 411\times {10}^{-9}\text{ s}\right)\\ =\frac{5}{3}\left(121\text{ m}-98.64\text{ m}\right)\\ =37.3\text{ m}\end{array}$

Conclusion:

Therefore, the coordinates in system K at time of bullet firing is $(1\text{ m, }3\text{ ns)}$ and at time of target attack is $\text{(}121\text{ m, }411\text{ ns)}$ and in system Kʹ are $(0.47\text{ m, }0.55\text{ ns)}$ and $\text{(}37.3\text{ m, }147.2\text{ ns)}$ respectively.