Question

If a bullet looses 1/nth of it's velocity while passing through a plank then minimum number of such planks required to just stop the bullet​

Answer 1

here is yor answer...

source : quora

Answer 2

The number of planks required to just stop the bullet is n²/(2n-1)

Explanation:

Bullet looses 1/nth of its velocity while passing through plank

Let the resistance offered by the plank be a

if the initial velocity of bullet be u and final velocity be v and the depth of plank d

Then

By the second equation of motion

$v^2=u^2-2as$

$v^2=u^2-2ad$

But

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

$\implies v=u(1-\frac{1}{n})$

Therefore,

$u^2(1-\frac{1}{n})^2=u^2-2ad$

$\implies 2ad=u^2[1-(1-\frac{1}{n})^2]$

$\implies 2ad=u^2[1-(1-\frac{2}{n}+\frac{1}{n^2}]$

$\implies 2ad=u^2[\frac{2}{n}-\frac{1}{n^2}]$  .............. (1)

Let there are N planks

Then, after passing through these N planks, the velocity of bullet will be zero

Thus,

$0^2=u^2-2aNd$

$\implies 2aNd=u^2$  ................... (2)

Dividing eq (1) by eq (2)

$\frac{1}{N}=\frac{2}{n}-\frac{1}{n^2}$

$\implies \frac{1}{N}=\frac{2n-1}{n^2}$

[tex]\implies N=\frac{n^2}{2n-1}

Hope this answer is helpful.

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