Question

A rifle bullet loses $\dfrac{1}{{20}}{\text{th}}$ of its velocity in passing through a plank. Assuming constant resistive force, the least number of such planks required just to stop the bullet is:
A. ${\text{15}}$
B. ${\text{10}}$
C. $11$
D. $20$

Answer 1

We are asked to find the minimum number of planks that would be required to just stop the bullet. The information of velocity of bullet after passing one plank is given, you will need to use here the third equation of motion. Apply the third equation of motion to form equations with which you can find the least number of planks required.

Complete step by step answer:
Given, a rifle bullet loses $\dfrac{1}{{20}}{\text{th}}$ of its velocity in passing through a plank.We will use here the third equation of motion. From third equation of motion we have,
${v^2} - {u^2} = 2as$ ……………………(i)
where $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration and $s$ is the distance travelled.
Here, let the initial velocity of the bullet be $u$.Let the thickness of each plank be $d$. Let the acceleration provided by each plank be $a$.It is given in the question that the bullet stops, so final velocity of bullet $v = 0$. So, applying the third equation of motion ( using equation (i)) we get,
${0^2} - {u^2} = 2as$ ……………(ii)
Let $n$ be the number of planks required to stop the bullet, so total thickness of all planks will be, $s = nd$
Putting this value in equation (ii), we get
${0^2} - {u^2} = 2and$
$\Rightarrow - {u^2} = 2and$
$\Rightarrow n = - \dfrac{{{u^2}}}{{2ad}}$ ……………(iii)
It is said after passing each plank the velocity of bullet reduces by $\dfrac{1}{{20}}{\text{th}}$ of its initial velocity. Therefore, velocity after passing first plank will be,
$v' = u - \dfrac{1}{{20}}u$
$\Rightarrow v' = \dfrac{{19}}{{20}}u$
Now, applying third equation of motion after the bullet passes first plank we get, (using equation (i))
${v'^2} - {u^2} = 2ad$
Putting the value of $v'$ we get,
${\left( {\dfrac{{19}}{{20}}u} \right)^2} - {u^2} = 2ad$
$\Rightarrow \dfrac{{361}}{{400}}{u^2} - {u^2} = 2ad$
$\Rightarrow - \dfrac{{39}}{{400}}{u^2} = 2ad$
$\Rightarrow 2ad = - \dfrac{{39}}{{400}}{u^2}$
Substituting the value of $2ad$ in equation (iii), we get
$n = - \dfrac{{{u^2}}}{{\left( { - \dfrac{{39}}{{400}}{u^2}} \right)}}$
$\Rightarrow n = \dfrac{1}{{\left( {\dfrac{{39}}{{400}}} \right)}}$
$\Rightarrow n = \dfrac{{400}}{{39}}$
$\therefore n = 10.26 \approx 11$
Therefore, the least number of planks required to stop the bullet is $11$.

Hence, the correct answer is option C.

Note: There are three equations of motion that one should always remember. These are,
-First equation of motion:- $v = u + at$
-Second equation of motion:- $s = ut + \dfrac{1}{2}a{t^2}$
-Third equation of motion :- ${v^2} - {u^2} = 2as$
Also remember, while using these equations check what are the information given in the question and which equation would help us to get the required answer.