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 $11$

If the bullet looses 1/20th of its velocity then,

$v=u-\frac{1}{20}u$

$v=\frac{19u}{20}$

Since, $v^2=u^2+2as$

${\left(\frac{19u}{20}\right)}^2=u^2-2as$

$2as=u^2-\frac{19u}{20}...........\left(1\right)$

Now initial velocity =0

$\therefore 0^2=u^2-2as\left(n\right)$

$u^2=2as\left(n\right)$

Now putting value of 2as from (1)

$u^2=n[u^2-{\left(\frac{19u}{20}\right)}^2]$

$u^2=n{u}^2\left[\frac{{20}^2-{19}^2}{{20}^2}\right]$

$n=\frac{{20}^2}{{20}^2-{19}^2}$

$n=\frac{400}{400-361}$

$n=\frac{400}{39}$

$n=10.26\approx 11planks$