Question

A bullet moving with a velocity of 100 m / s can just penetrate two planks of equal thickness. The number of such planks penetrated by the same bullet, when the velocity is doubled, will be

• A. 4
• B. 6
• C. 8
• D. 10

Answer 1

Answer: (C)

Given initial velocity of bullet in first case
$\left(u_{1}\right)=100 m / s$. Initial number of planks $\left(n_{1}\right)=2$
Initial stopping distance $\left(s_{1}\right)=n_{1} x=2 x$ (where $x$ is the thickness of one plank).
Initial velocity of bullet in second case $\left(u_{2}\right)=200 m / s$. Relation for the stopping distance $(s)$ is $v^{2}=u^{2}+2 a s$
Since the bullet is just able to penetrate the planks, therefore its final velocity $v=0$. Thus, $2 a s=-u^{2}$ or $s \propto u^{2} $
$\therefore \frac{s_{1}}{s_{2}}=\left(\frac{u_{1}}{u_{2}}\right)^{2}=\frac{1}{4}$
or $s_{2}=4 s_{1}=4 \times 2 x=8 x$
Thus final number of planks $\left(n_{2}\right)=\frac{s_{2}}{x}=\frac{8 x}{x}=8$