Question

A rifle bullet loses 1/20 th of its velocity in passing
through a plank. Find the least number of such
planks required to just stop the bullet​.​.

Answer 1

$\huge\red{\underline{{\boxed{\textbf{QUESTION}}}}}$

A rifle bullet loses 1/20 th of its velocity in passing through a plank. Find the least number of such planks required to just stop the bullet.

$\huge\red{\underline{{\boxed{\textbf{ANSWER}}}}}$

Let us assume that the thickness of the plank be s and acceleration provided by each plank is a. Using equation of motion we have,

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

As the bullet stops finally, therefore above equation can be rewritten as,

$0 - u {}^{2} = 2as$

Now, let the number of planks required be n, therefore the equation can be written as,

$- u {}^{2} = 2asn$

or,

$n = \frac{ - u {}^{2} }{2as} ........\: (1)$

Now, question says that the bullet loses its speed by 1/20 on passing through plank, so the final speed of the bullet when it leaves one plank is,

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

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

Again using equation of motion we have,

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

Substituting for v in above equation we get,

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

$2as = ( - \frac{39}{400} {u}^{2} )......(2)$

substituting equation (2) in equation (1) we get,

$n = \frac{ - {u}^{2} }{ \frac{39}{400 } {u}^{2} }$

$n = 10.26$

So, the number of planks required is 11 because 0.26 can be considered as one plank.

@HarshPratapSingh