Question

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

• A. $11$
• B. $20$
• C. $21$
• D. Infinite

Answer 1

The correct option is A $11$

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

$\frac{1}{2}m[v^2-{\left(\frac{19}{20}v\right)}^2]n=\frac{1}{2}m[0-v^2]$
Solving for $n$, $n=10.25$
So, $11$ planks.