Question

A bullet when fired at a target has its velocity decreased to $50\%$ after penetrating $30cm$ into it. Then the additional thickness it will penetrate before resting is

• A. $10cm$
• B. $30cm$
• C. $40cm$
• D. $60cm$

Answer 1

The correct option is A

$10cm$

Step 1: Given data and assumptions

Let the initial velocity of the bullet be $vm{s}^{-1}$.

The velocity decreased by $50\%$ after penetrating the target by, $S=30cm$

The new velocity after penetrating $30cm$will be $\frac{v}{2}m{s}^{-1}$

Let the mass of the bullet be taken as $\mathrm{m}kg$.

Step 2: Calculating additional thickness of the target

The velocity is retarding, so the resisting force will be taken as negative.

The mass of the bullet is taken as $\mathrm{m}kg$.

The decrease in work done can be written as,

$$\begin{gathered} \frac{1}{2}\mathrm{m}{\left(\frac{v}{2}\right)}^2-\frac{1}{2}\mathrm{m}{v}^2=-FS\left[K.E=\frac{1}{2}m{v}^2\right] \\ \Rightarrow \frac{\mathrm{m}{v}^2}{8}-\frac{\mathrm{m}{v}^2}{2}=-FS \\ \Rightarrow \frac{\mathrm{m}{v}^2-4\mathrm{m}{v}^2}{8}=-FS \\ \Rightarrow -\frac{3}{8}\mathrm{m}{v}^2=-FS \\ \Rightarrow \frac{1}{2}\left(-\frac{3}{4}\right)\mathrm{m}{v}^2=-FS \\ \Rightarrow \frac{1}{2}\mathrm{m}{v}^2=\frac{4}{3}FS\left[equation1\right] \end{gathered}$$

Let the total distance travelled by the bullet be $S\text{'}$.

The bullet comes at rest after covering the total distance, so we will take the velocity as zero.

$$\begin{gathered} \frac{1}{2}\mathrm{m}{\left(0\right)}^2-\frac{1}{2}\mathrm{m}{v}^2=-FS\text{'} \\ \Rightarrow -\frac{1}{2}\mathrm{m}{v}^2=-FS\text{'} \\ \Rightarrow \frac{1}{2}\mathrm{m}{v}^2=FS\text{'}\left[equation2\right] \end{gathered}$$

Now from equations 1 and 2,

$$\begin{gathered} \frac{4}{3}FS=FS\text{'} \\ \Rightarrow \frac{4}{3}S=S\text{'} \\ \Rightarrow \frac{4}{3}\times 30cm=S\text{'}\left[S=30cm\right] \\ \Rightarrow S\text{'}=40cm \end{gathered}$$

The additional thickness penetrated by the bullet is$S\text{'}-S=40cm-30cm=10cm$.

Hence, option (a) is correct.