Question

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

Answer 1

11 planks 100% sure

Answer 2

$\textsf{Given:}$

• rifles bullet loses 1/20th of its velocity in passing through a plank

$\textsf{To find:}$

• the least number of such planks required to just stop the bullet

$\textsf{Solution:}$

For one plank, we know by work energy theorem that:

     $\boxed{\textsf{W = Fsd = \frac{1}{2}m[Vf^{2} - Vi^{2} ] }}$

So, according to the question, putting the given values in the formula, we get:

     $\textsf{Fsd = }$ $\frac{1}{2} m [(\frac{19u}{20})^{2} - u^{2} ]$  _____(1)

Similarly, for 'n' planks we have:

    $\textsf{nFsd = }$ $0 - \frac{1}{2} mu^{2}$ _____(2)

[final velocity is zero since the bullet finally comes to rest]

Putting (1) in (2), we get:

    $\textsf{n}$ $(\frac{1}{2} m [(\frac{19u}{20})^{2} - u^{2}])$  =  $- \frac{1}{2} mu^{2}$

    $\textsf{n}$ $(\frac{m}{2} (\frac{361u^{2} - 400u^{2} }{400} )= - \frac{m}{2} u^{2}$

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

    $\textsf{n =}$ $\frac{400}{39} = 11$

$\boxed{\textsf{Hence, the least number of planks required is 11.}}$