Question

A wooden boxcar is moving along a straight railroad track at speed ${\mathit{v}}_{\mathbf{1}}$. A sniper fires a bullet (initial speed ${\mathit{v}}_{\mathbf{2}}$) at it from a high-powered rifle. The bullet passes through both lengthwise walls of the car, its entrance and exit holes being exactly opposite each other as viewed from within the car. From what direction, relative to the track, is the bullet fired? Assume that the bullet is not deflected upon entering the car, but that its speed decreases by$\mathbf{20}\mathbf{\%}\mathbf{.}\mathbf{Take}{\mathbf{v}}_{\mathbf{1}}\mathbf{=}\mathbf{85}\mathbf{}\mathbf{km}\mathbf{/}\mathbf{h}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{v}}_{\mathbf{2}}\mathbf{=}\mathbf{650}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$. (Why don’t you need to know the width of the box car?)

Short answer

Bullet was fired at $93°$measured from the direction of motion.

Step-by-step solution

Given information

It is given that,

$$\begin{gathered} v_1=85\mathrm{km}/\mathrm{h} \\ {\mathrm{v}}_2=650\mathrm{km}/\mathrm{h} \end{gathered}$$

Speed reduction of 20%

Determining the concept

This problem deals with the relative velocity. Relative velocity is the velocity of an object in the rest frame of another object. To find the direction of bullet fired use the relative velocity of the bullet.

Determining the velocity

Velocity of box,

${\stackrel{\rightharpoonup }{v}}_{cg}=v_1\hat{i}$

Bullet has velocity relative to ground is given as below,

${\stackrel{\rightharpoonup }{V}}_{0bg}=V_{2}cs\theta \hat{i}+v_2\sin \theta \hat{j}$

Velocity of bullet in car,

${\stackrel{\rightharpoonup }{v}}_{bc}=v_3\hat{\mathrm{j}}$

As speed reduced to 20%, so new velocity is,

$$\begin{gathered} {\stackrel{\rightharpoonup }{v}}_{bg}=0.8v_2\cos \theta \hat{\mathrm{i}}+0.8{\mathrm{v}}_2\sin \theta \hat{\mathrm{j}} \\ 0.8{\mathrm{v}}_2\cos \theta \hat{i}+0.8v_2\sin \theta \hat{\mathrm{j}}=v_3\hat{\mathrm{j}}+v_1\hat{\mathrm{i}} \end{gathered}$$

Determining the direction of bullet

To find the direction, equate x components,

$$\begin{gathered} 0.8{\mathrm{v}}_2\cos \theta =v_1 \\ \cos \theta =\frac{v_1}{0.8{\mathrm{v}}_2} \\ \mathrm{So}, \\ \cos \theta =\frac{85\mathrm{km}/\mathrm{h}}{0.8\times 650\mathrm{m}/\mathrm{s}}\times \frac{1000\mathrm{m}/\mathrm{km}}{3600\mathrm{s}/\mathrm{h}} \\ \mathrm{So}\theta =87° \\ \mathrm{So},87°\mathrm{is}\mathrm{the}\mathrm{direction}\mathrm{of}{\stackrel{\rightharpoonup }{\mathrm{v}}}_{\mathrm{bg}} \end{gathered}$$

Since, the direction in which bullet was fired had been found, it would give us the answer as,$78=93°$.