Question

A bullet moving with a speed of $100m/s$ 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

• A. $6$
• B. $10$
• C. $4$
• D. $8$

Answer 1

The correct option is D

$8$

Step 1: Given data

Initial velocity in first case,$u_1$$=100m/s$

Let the thickness of each plank be $x$

Distance traveled by the bullet in the first case, $S_1$$=noofplanks\times x=2xm$

Initial velocity in the second case,$u_2$$=200m/s$

Let the Distance traveled by the bullet in the second case, $S_2$$=nxm$

Where n is a number of planks penetrated by a bullet in the second case.

Note that the penetration of the bullet signifies that the final speed is zero in both conditions

Step 2: Formula used

$v^2=u^2+2as$ where $v$is the final velocity, $u$is the initial velocity, $a$is the acceleration.

Step 3: Find the relation between distance traveled and initial velocity

Now Putting v=0, we get

$s=\frac{-u^2}{2a}$

If the acceleration is constant, the distance traveled,$s$ is proportional to the square of initial velocity $u$

$s$∝$u^2$

Step 4: Find the distance traveled when the speed is doubled

$$\begin{gathered} S_1/S_2={u_1}^2/{u_2}^2 \\ \Rightarrow S_2=({u_2}^2/{u_1}^2)*S_1 \\ \Rightarrow S_2=({200}^2/{100}^2)*2x \\ \Rightarrow S_2=8x \end{gathered}$$

Comparing it with $S_2=nx$

We get $n=8$

Hence the number of such planks penetrated is $8planks$. Option D is correct.