As the outlaws escape in their getaway car, which goes,$\frac{\mathbf{3}}{\mathbf{4}}\mathit{c}$the police officer fires a bullet from the pursuit car, which only goes$\frac{\mathbf{1}}{\mathbf{2}}\mathit{c}$(Fig. 12.3). The muzzle velocity of the bullet (relative to the gun)$\frac{\mathbf{1}}{\mathbf{3}}\mathit{c}$is. Does the bullet reach its target (a) according to Galileo, (b) according to Einstein?

Given data:

The velocity of the bullet relative to the gun is$v_{bg}=\frac{1}{3}c$.

The velocity of the gun fitted to the car is$v_{gE}=\frac{1}{2}c$.

(a)

Write the equation for the velocity of the bullet relative to the ground.

${\mathit{v}}_{\mathbf{b}\mathbf{E}}\mathbf{=}{\mathit{v}}_{\mathbf{b}\mathbf{g}}\mathbf{+}{\mathit{v}}_{\mathbf{g}\mathbf{E}}$

Substitute the value of$v_{bg}$and$v_{gE}$in the above expression.

$$\begin{gathered} v_{bE}=\left(\frac{1}{3}c\right)+\left(\frac{1}{2}c\right) \\ {v}_{bE}=\frac{2c+3c}{6} \\ {v}_{bE}=\frac{5}{6}c \end{gathered}$$

The velocity of the gateway is given as$v_G=\frac{3}{4}c$.

Now, to make the same denominator values of$v_{bE}$and $v_G$, multiply and divide by 2 in$v_{bE}$value and multiply and divide by 3 in $v_G$value. Hence,

$$\begin{gathered} v_{bE}=\frac{5}{6}c\times \frac{2}{2}=\frac{10}{12}c \\ {v}_G=\frac{3}{4}c\times \frac{3}{3}=\frac{9}{12}c \\ {v}_G<v_{bE} \end{gathered}$$

As the velocity of the gateway car is less than the velocity of the bullet relative to the ground, the bullet reaches its target, according to Galileo.

Therefore, according to Galileo, the bullet reaches its target.

(b)

According to Einstein, the velocity of a bullet with respect to ground will be,

$v_{bE}=\frac{v_{bg}+v_{gE}}{1+{\displaystyle \frac{v_{bg}{v}_{gE}}{c^2}}}$

Substitute the value of$v_{bg}$and $v_{gE}$in the above expression.

$$\begin{gathered} v_{bE}=\frac{\left({\displaystyle \frac{1}{3}}c\right)+\left({\displaystyle \frac{1}{2}}c\right)}{1+{\displaystyle \frac{\left(\frac{1}{3}c\right)+\left(\frac{1}{2}c\right)}{c^2}}} \\ {v}_{bE}=\frac{\left({\displaystyle \frac{5}{6}}c\right)}{\left({\displaystyle \frac{7}{6}}\right)} \\ {v}_{bE}=\frac{5}{7}c \end{gathered}$$

Now, to make the same denominator values of$v_{bE}$and$v_G$, multiply and divide by 4 in$v_{bE}$value and multiply and divide by 7 in $v_G$value. Hence,

$$\begin{gathered} v_{bE}=\frac{5}{7}c\times \frac{4}{4}=\frac{20}{28}c \\ {v}_G=\frac{3}{4}c\times \frac{7}{7}=\frac{21}{28}c \\ {v}_G>v_{bE} \end{gathered}$$

As the velocity of the gateway car is greater than the velocity of the bullet relative to the ground, the bullet does not reach its target, according to Einstein.

Therefore, according to Einstein, the bullet does not reach its target.