Question

A rifle bullets loses ${\left(\frac{1}{20}\right)}^{th}$ 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 as the bullet moves through the block is $F$ and displacement inside one block is $x$.
Using work energy theorem for one block,
$-F.x=\frac{1}{2}m[{\left(\frac{19}{20}v\right)}^2-v^2]...\left(1\right)$
Applying work energy theorem for $n$ blocks,
$-F.nx=\frac{1}{2}m\left[v^2\right]$

Using value of $Fx$ from ...(1)
$\frac{1}{2}m[v^2-{\left(\frac{19}{20}v\right)}^2]n=\frac{1}{2}m[0-v^2]$
$n=\frac{400}{39}$
After solving, we get $n=10.25$
So, $11$ planks

Why this question? "Work done by sum of all forces acting on a body is change in its kinetic energy." "Applications of work energy theorem are prominent in NEET examination"