Question

An armored car $2\text{m}$ long and $3\text{m}$ wide is moving at $10\text{m/s}$ when a bullet hits it in a direction making an angle ${\tan }^{-1}\left(\frac{3}{4}\right)$ with the car as seen from the street. The bullet enters one corner of the car and passes out of the diagonally opposite corner. Neglecting any interaction between the bullet and the car, find the time for bullet to cross through the car.

• A. $t=0.35\text{s}$
• B. $t=0.2\text{s}$
• C. $t=0.6\text{s}$
• D. $t=\frac{2}{13}\text{s}$

Answer 1

The correct option is B $t=0.2\text{s}$


In this problem both objects i.e car and bullet are moving. So we will solve it from frame of reference of car by applying concept of relative velocity.

CONCEPT:-When bullet travels though car, then X-displaceement and Y-displacement of bullet should be equal to length & width of car respectively.

Let $\text{v}$ be the speed of the bullet w.r.t ground.
Now taking the relative velocity of bullet in X & Y direction w.r.t car
${\text{v}}_{bc}={\text{v}}_b-{\text{v}}_c$, where ${\text{v}}_c$= velocity of car w.r.t ground

$\therefore$$({\text{v}}_{bc}{)}_x=(\text{v}\cos \theta -10)$
Similarly relative velocity of bullet in Y-direction w.r.t car:
$({\text{v}}_{bc}{)}_y=\text{v}\sin \theta$
For crossing through the car, the relative displacement is $X=2\text{m},Y=3\text{m}$
$(\text{v}\cos \theta -10)\times t$= $2$..........(1)
$\left(\text{v}\sin \theta \right)\times t$= $3$.................(2)
Putting $\sin \theta$=$\frac{3}{5}$ & $\cos \theta$=$\frac{4}{5}$ in equations (1) and (2) ;

We get $t$=$0.2\text{s}$