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A rifle bullet loses $1 / 20^{\text {th }}$ of its velocity in passing through a plank. The least number of such planks required just to stop the bullet is
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1222 Upvotes
Verified Answer
The correct answer is:
11
$\mathrm{v}=\left(u-\frac{u}{20}\right)=\frac{19}{20} u$
from $v^2=u^2-2 a s$
$\Rightarrow\left(\frac{19}{20} u\right)^2=u^2-2 a s \Rightarrow \frac{400}{39}=\frac{u^2}{2 a s}$
Now if the $n$ planks are arranged just to stop the bullet then again from
$\begin{aligned}
& 0=u^2-2 a n s \\
& \Rightarrow n=\frac{u^2}{2 a s}=\frac{400}{39} \\
& \Rightarrow n=10.25
\end{aligned}$
As the planks are more than 10 so we can consider $n=11$
from $v^2=u^2-2 a s$
$\Rightarrow\left(\frac{19}{20} u\right)^2=u^2-2 a s \Rightarrow \frac{400}{39}=\frac{u^2}{2 a s}$
Now if the $n$ planks are arranged just to stop the bullet then again from
$\begin{aligned}
& 0=u^2-2 a n s \\
& \Rightarrow n=\frac{u^2}{2 a s}=\frac{400}{39} \\
& \Rightarrow n=10.25
\end{aligned}$
As the planks are more than 10 so we can consider $n=11$
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