Question

A bullet moving with velocity $‘v^{\prime }$ penetrates a sandbag up to distance ${}^{\prime }{x}^{\prime }$. The same bullet with velocity ${}^{\prime }2v^{\prime }$ can penetrate bag up to a distance of:

• A. $x$
• B. $2x$
• C. $3x$
• D. $4x$

Answer 1

The correct option is D $4x$

The bullet moves a distance $x$ in the sandbag, let the retardation due to the sand be $a$, then from the third equation of motion.
$v^2=u^2–2ax$
The final velocity is zero, and the initial velocity is $u$, then
$v^2=2ax$
$a=\frac{v^2}{2x}$
Then for the velocity $2v$, from the same equation
$(2v{)}^2=2aS$
$4v^2=2aS$
Now, substituting the value of $a$ we get,
$S=4x$
Therefore, the bullet will move a distance of $4x$ in the sandbag with a velocity of $‘2v^{\prime }$
Therefore, option d is correct.