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
$\left(u_1\right)=100m/s$. Initial number of planks $\left(n_1\right)=2$
Initial stopping distance $\left(s_1\right)=n_{1}x=2x$ (where $x$ is the thickness of one plank).
Initial velocity of bullet in second case $\left(u_2\right)=200m/s$. Relation for the stopping distance $(s)$ is $v^2=u^2+2as$
Since the bullet is just able to penetrate the planks, therefore its final velocity $v=0$. Thus, $2as=-u^2$ or $s\propto u^2$
$\therefore \frac{s_1}{s_2}={\left(\frac{u_1}{u_2}\right)}^2=\frac{1}{4}$
or $s_2=4s_1=4\times 2x=8x$
Thus final number of planks $\left(n_2\right)=\frac{s_2}{x}=\frac{8x}{x}=8$