Question

A bullet moving with a speed of $100\,ms^{-1}$ can just penetrate two planks of equal thickness. Then the number of such planks penetrated by the same bullet when the speed is doubled will be _________.

Answer 1

The amount of planks penetrated is to be considered as distance travelled by the given bullet. There is a relation established between the distance travelled and speed of the bullet, which can be formulated using equations of motion. There are three equations of motions but here we use the equation that can relate the quantities of speed and distance.

Complete step by step answer:
We start by noting down the given quantities; a bullet is given and it travels at a speed of $100\;m{s^{ - 1}}$, this can be considered as the initial velocity (${u_1}$). This bullet penetrates only two planks and considers it as two units, so the distance covered in the initial speed can be: ${s_1} = 2$. Now we are given that the speed in the second condition becomes twice the initial speed of the first condition so the initial speed in the second condition: ${u_2} = 2 \times {u_1}$. Note that the penetration of the bullet signifies that the final speed is zero in both conditions: ${v_1},{v_2} = 0$. We are to find the distance covered in the second condition so let it be represented as: ${s_2} = ?$

To properly relate all the conditions given we make use of equations of motion. The equation of motion we use here needs to have distance as well as initial and final velocities.
So we use $\Rightarrow {v^2} = {u^2} + (2 \times a \times s)$
Here the acceleration becomes negative since penetration into planks take place so equation is;
$\Rightarrow {v^2} = {u^2} - (2 \times a \times s)$
But again we know that velocity is zero during penetration of bullets so;
$\Rightarrow 0 = {u^2} - (2 \times a \times s)$
$\Rightarrow {u^2} = (2 \times a \times s)$
From the above equation we can draw a relation between initial velocity and distance, it is as follows:
$\Rightarrow {u^2}\;\alpha \;s$

It can also be written as:
$\Rightarrow {u_1}^2\;\alpha \;{s_1} \to (i)$ and $\Rightarrow {u_2}^2\;\alpha \;{s_2} \to (ii)$
The proportionalities from $(i)$ and $(ii)$ can be related as follows:
$\Rightarrow \dfrac{{{u_1}^2}}{{{u_2}^2}}\; = \;\dfrac{{{s_1}}}{{{s_2}}}$
The final step is to substitute the values that we received from the question:
$\Rightarrow \dfrac{{{u_1}^2}}{{2 \times {u_1}^2}}\; = \;\dfrac{{{s_1}}}{{{s_2}}}$
$\Rightarrow {\left( {\dfrac{{100}}{{200}}} \right)^2}\; = \;\dfrac{2}{{{s_2}}}$
Rearranging the equation we get;
$\Rightarrow \;{s_2} = \;2 \times {\left( {\dfrac{{200}}{{100}}} \right)^2}$
Simplifying;
$\Rightarrow \;{s_2} = \;2 \times {\left( 2 \right)^2}$
$\therefore \;{s_2} = \;8\;units$

Therefore the number of planks that were penetrated by the same bullet when the initial speed was doubled resulted to be $8$ planks.

Note: We dealt with speed and velocity here in this question, interchangeably. But there is a specific difference we must keep in mind while dealing with these two terms. The velocity is the ‘displacement’ that occurs on a body at a particular unit of time, while, in unit time the ‘distance’ covered is speed. Speed is scalar due to its lack of direction but velocity is vector since it possesses magnitude and direction.