Question

A bullet travelling horizontally loses $\dfrac{1}{20}^{th}$ 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

Hint: Begin by determining the final velocity after the bullet pierces the wooden plank. Plug this into an appropriate equation of motion relating the initial, final velocities and the distance travelled and acceleration. Remember that the bullet loses velocity, so it is essentially retarding, so use the proper sign for acceleration. Then, use the same equation of motion to describe the motion of the bullet where its final velocity becomes zero after piercing through a number of planks. Use the two equations at hand to arithmetically arrive at the number of planks to an approximation and choose the nearest option.

Complete step by step solution:
We are required to determine the number of planks that are required to stop a bullet whose velocity is reduced by $\dfrac{1}{20}$ while piercing through a single plank.
Let the thickness of each wooden plank be $t$. When the bullet travels through this distance t, it loses $\dfrac{1}{20}^{th}$ of its initial velocity, i.e.,
If u is the initial velocity of the bullet, then $v = u - \dfrac{1}{20}u = \dfrac{20u-1u}{20} = \dfrac{19u}{20}$.
Since it is essentially slowing down, it will experience a retardation of $-a$
Plugging this into the following kinematic equation of motion:
$v^2 =u^2-2as \Rightarrow \dfrac{19^2u^2}{20^2}=u^2 -2at$
$\Rightarrow 2at = u^2 - \dfrac{361u^2}{400} = \dfrac{400u^2-361u^2}{400} $
$\Rightarrow 2at = \dfrac{39u^2}{400}$
Suppose the bullet has to pass through, say n blocks in order to come to a complete stop, i.e., $v=0$, it has to travel through a thickness of $s=nt$. Plugging this into the same kinematic equation of motion we get:
$v^2=u^2-2as \Rightarrow 0 = u^2-2ant \Rightarrow 2ant = u^2 \Rightarrow n = \dfrac{u^2}{2at}$
Substituting for 2at from the first equation of motion we get:
$n = \dfrac{u^2}{\left(\dfrac{39u^2}{400}\right)} = \dfrac{u^2 \times 400}{39u^2} = \dfrac{400}{39} = 10.25 \approx 11$
Therefore, the correct choice would be C. 11

Note: In addition to the above discussed equation of motion, it is advisable to keep in mind a couple more equations that describe a variety of quantities associated with the motion of objects:
1. $v = u +at$, where t is the time taken by the object.
2. $s=ut +\dfrac{1}{2}at^2$
The above equations can be used to describe the motion of an object under various circumstances, correlating the quantities in the equations, and are most often used to determine unknown quantities from a limited description of any object’s trajectory/motion.