Question

Figure 37-16 shows a ship (attached to reference frame ${\mathbf{S}}^{\mathbf{\text{'}}}$) passing us (standing in reference frame ${\mathbf{S}}^{}$) with velocity $\overrightarrow{\mathbf{v}}\mathbf{=}\mathbf{0.950}\mathbf{c}\widehat{\mathbf{i}}$. A proton is fired at speed $\mathbf{0.980}\mathbf{c}$ relative to the ship from the front of the ship to the rear. The proper length of the ship is $\mathbf{760}\mathbf{\text{\hspace{0.17em}m}}$ . What is the temporal separation between the time the proton is fired and the time it hits the rear wall of the ship according to (a) a passenger in the ship and (b) us? Suppose that, instead, the proton is fired from the rear to the front. What then is the temporal separation between the time it is fired and the time it hits the front wall according to (c) the passenger and (d) us?

Short answer

(a) The temporal separation of a passenger in the ship is $2.6\times {10}^{-6}\text{\hspace{0.17em}s}$.

(b) Our temporal separation is $1.05\times {10}^{-6}\text{\hspace{0.17em}s}$.

(c) The temporal separation of a passenger in the ship is $2.6\times {10}^{-6}\text{\hspace{0.17em}s}$.

(d) Our temporal separation is $2.63\times {10}^{-5}\text{\hspace{0.17em}s}$.

Step-by-step solution

Identification of given data

The given data can be listed below as:

The velocity of the ship is $\overrightarrow{\mathrm{v}}=0.950\mathrm{c}\widehat{\mathrm{i}}$.

The velocity of the proton is ${\mathrm{v}}_1=0.980\mathrm{c}$.

The ship’s length is $\mathrm{l}=760\text{\hspace{0.17em}m}$.

Significance of the relativity

The relativity mainly refers to the behaviour of a particular object is time and space. This theory is also beneficial for predicting the existence of a particle and light’s bending.

(a) Determination of the temporal separation of a passenger in the first case

The equation of the temporal separation of a passenger is expressed as:

${\mathrm{t}}_1=\frac{\mathrm{l}}{{\mathrm{v}}_1}$

Here, ${\mathrm{t}}_1$ is the temporal separation of a passenger, $l$ is the ship’s length and ${\mathrm{v}}_1$ is the velocity of the proton.

Substitute the values in the above equation.

$\begin{array}{c}{\mathrm{t}}_1=\frac{760\text{\hspace{0.17em}m}}{0.980(3\times {10}^8\text{\hspace{0.17em}m/s})}\\ =\frac{760\text{\hspace{0.17em}m}}{(2.9\times {10}^8\text{\hspace{0.17em}m/s})}\\ =2.6\times {10}^{-6}\text{\hspace{0.17em}s}\end{array}$

Thus, the temporal separation of a passenger in the ship is $2.6\times {10}^{-6}\text{\hspace{0.17em}s}$.

(b) Determination of the temporal separation of us in the first case

As in our case, the length of the ship is taken negative. The equation of our temporal separation is expressed as:

${\mathrm{t}}_2=\frac{1}{\sqrt{1-{\left(\frac{{\mathrm{v}}_1}{\mathrm{c}}\right)}^2}}\left({\mathrm{t}}_1+\frac{{\mathrm{v}}_1(-\mathrm{l})}{{\mathrm{c}}^2}\right)$

Here, $t_2$is the temporal separation of us and ${\mathrm{v}}_1$is the velocity of the proton.

Substitute the values in the above equation.

$\begin{array}{c}{\mathrm{t}}_2=\frac{1}{\sqrt{1-{\left(\frac{0.980\mathrm{c}}{\mathrm{c}}\right)}^2}}\left(2.6\times {10}^{-6}\text{\hspace{0.17em}s}+\frac{\left(0.980\mathrm{c}\right)(-760\text{\hspace{0.17em}m})}{{\mathrm{c}}^2}\right)\\ =\frac{1}{\sqrt{1-{\left(0.980\right)}^2}}\left(2.6\times {10}^{-6}\text{\hspace{0.17em}s-}\frac{\left(744.8\text{\hspace{0.17em}m}\right)}{3\times {10}^8\text{\hspace{0.17em}m/s}}\right)\\ =\frac{1}{\sqrt{1-0.9604}}(2.6\times {10}^{-6}\text{\hspace{0.17em}s-2.4}\times {\text{10}}^{-6}\text{\hspace{0.17em}s})\\ =\frac{1}{0.19}(0.2\times {10}^{-6}\text{\hspace{0.17em}s})\\ =1.05\times {10}^{-6}\text{\hspace{0.17em}s}\end{array}$

Thus, our temporal separation is $1.05\times {10}^{-6}\text{\hspace{0.17em}s}$.

(c) Determination of the temporal separation of a passenger in the second case

According to the observers of ship, there is no difference of firing proton from front to back that significantly makes no difference. Hence, the temporal separation of the passenger is the same as previous.

Thus, the temporal separation of a passenger in the ship is $2.6\times {10}^{-6}\text{\hspace{0.17em}s}$.

(d) Determination of the temporal separation of us in the second case 

The equation of our temporal separation is expressed as:

${\mathrm{t}}_3=\frac{1}{\sqrt{1-{\left(\frac{{\mathrm{v}}_1}{\mathrm{c}}\right)}^2}}\left({\mathrm{t}}_1+\frac{{\mathrm{v}}_1\left(\mathrm{l}\right)}{{\mathrm{c}}^2}\right)$

Here, ${\mathrm{t}}_3$ is the temporal separation of us.

Substitute the values in the above equation.

$\begin{array}{c}{\mathrm{t}}_3=\frac{1}{\sqrt{1-{\left(\frac{0.980\mathrm{c}}{\mathrm{c}}\right)}^2}}\left(2.6\times {10}^{-6}\text{\hspace{0.17em}s}+\frac{\left(0.980\mathrm{c}\right)\left(760\text{\hspace{0.17em}m}\right)}{{\mathrm{c}}^2}\right)\\ =\frac{1}{\sqrt{1-{\left(0.980\right)}^2}}\left(2.6\times {10}^{-6}\text{\hspace{0.17em}s+}\frac{\left(744.8\text{\hspace{0.17em}m}\right)}{3\times {10}^8\text{\hspace{0.17em}m/s}}\right)\\ =\frac{1}{\sqrt{1-0.9604}}(2.6\times {10}^{-6}\text{\hspace{0.17em}s+2.4}\times {\text{10}}^{-6}\text{\hspace{0.17em}s})\\ =\frac{1}{0.19}(5\times {10}^{-6}\text{\hspace{0.17em}s})\\ =2.63\times {10}^{-5}\text{\hspace{0.17em}s}\end{array}$

Thus, our temporal separation is $2.63\times {10}^{-5}\text{\hspace{0.17em}s}$.