# Question #22770

Nov 18, 2016

Let $m$ be the mass of bullet and $v$ be its initial velocity before it strikes 1st plank of thickness $d$. It loses $\frac{1}{40} t h$ of its velocity in piercing the 1st plank and the remaining velocity is $v - \frac{v}{40} = \frac{39 v}{40}$

If the average resisting force acting within plank be F then by law of conservation of energy we can write

$\frac{1}{2} m \left({v}^{2} - {\left(39 v\right)}^{2} / {40}^{2}\right) = F \times d$

$\implies \frac{1}{2} \times m \times \frac{79 {v}^{2}}{40} ^ 2 = F \times d \ldots . . \left(1\right)$

If n such planks are required to stop the bullet (that means to gain zero kinetic energy) then

$\frac{1}{2} m {v}^{2} = F \times n \times d$

Dividing (2) by (1) we get

$n = {40}^{2} / 79 = 20.25$

As n represents number of planks required to stop the bullet , it should be a whole number just greater than 20.25.

So n should be $= 21$