Question

A bullet travelling horizontally loses ${\frac{1}{20}}^{th}$ of its velocity while piercing a wooden plank. Number of such planks required to stop the bullet is:

• A. $6$
• B. $9$
• C. $11$
• D. $13$

Answer 1

The correct option is C $11$

Let $s$ be the length of each block.

After piercing it is decelerating so acceleration is negative.

So, the equation of motion:

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

$⟹\frac{{19}^2{u}^2}{{20}^2}=u^2-2as$

or, $2as=\frac{39}{400}{u}^2$

Now let's assume the bullet will pass through $n$ number of blocks. So the net distance traveled by the bullet will be $ns$ and finally, it will come to rest.

So, using the equation of motion:

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

$\Rightarrow$ $0=u^2-2ans$

$\Rightarrow$ $n=\frac{u^2}{2as}$

Substituting value of $2as$, we get:

$n=\frac{400}{39}=10.3$

Hence, a number of planks required is $11$.