Question

A bullet loses $\left(\frac{1}{n}\right)$ of the velocity while passing through a plank, then minimum number of plates required to stop bullet is:

Answer 1

Let the thickness of one plank $=d$

and the acceleration provided by the plank $=a$

$v^2=v{o}^2+2ad$

If n planks are required to stop the bullet, then

$0^2=v{o}^2+2a\times nd$

$2and=-v{o}^2$

$n=v{o}^2/(-2ad)$

$v=vo-vo/20=19vo/20$ in passing through one plank

$(19vo/20{)}^2=v{o}^2+2ad$

$361/400\times v{o}^2=v{o}^2+2ad$

$-2ad=v{o}^2(1-361/400)$

$-2ad=v{o}^2\times 39/400$

Substituting this value of $-2ad$

$n=v{o}^2/(v{o}^2\times 39/400)=400/39$

The minimum number of planks needed = smallest integer greater than $400/39=11.$