Question

A rifle bullets losses ${\left(\frac{1}{20}\right)}^{\text{th}}$ of its velocity in passing through a plank. Assuming that the plank exerts a constant retarding force, the least number of such planks required to just stop the bullet is

Answer 1

Let the retarding force exerted by one plank be $F$ and displacement inside the plank be $x$.
So, using work energy theorem for one plank,
Work done = $\Delta KE$ of the bullet
$-F.x=\frac{1}{2}m[{\left(\frac{19}{20}v\right)}^2-v^2]\dots \dots \left(1\right)$
Applying work energy theorem for $n$ planks, $-F.nx=\frac{1}{2}m[0-v^2]\dots \left(2\right)$

Using value of $Fx$ from (1) in (2)
$\frac{1}{2}m[{\left(\frac{19}{20}v\right)}^2-v^2]n=\frac{1}{2}m[0-v^2]$
Solving, we get $n=10.25$
So, $11$ planks are needed to just stop the bullet.