Question

A lead bullet of mass $2kg$, traveling with a velocity of $20m·s^{-1}$, comes to rest after penetrating $20m$ in a still target. Find the resistive force applied by the target and acceleration of the bullet.

Answer 1

Step 1: Given data

1. The initial velocity of the bullet, $u=20m·s^{-1}$.
2. As the bullet comes to rest thus the final velocity of the bullet, $v=0m·s^{-1}$.
3. Displacement before the bullet stops, $s=20m$.

Step 2: Determine the acceleration of the bullet

1. Let the acceleration of the bullet be $a$.
2. It is known that, $v^2=u^2+2as$.

$$\begin{gathered} \Rightarrow a=\frac{v^2-u^2}{2s} \\ \Rightarrow a=\frac{{\left(0\right)}^2-{\left(20\right)}^2}{2·20} \\ \Rightarrow a=-\frac{400}{40} \\ \Rightarrow a=-10 \end{gathered}$$

Thus, the acceleration is $-10m·s^{-2}$, negative sign implies that the velocity is decreasing.

Step 3: Determine the resistive force applied by the target

The acceleration of the bullet is $a=-10m·s^{-2}$.

The mass of the bullet is $m=2kg$.

Thus, the force applied by the target can be given by, $F=ma$

$$\begin{gathered} \Rightarrow F=2\times \left(-10\right)N \\ \Rightarrow F=-20N \end{gathered}$$

A negative sign indicates that the force is resistive.

Therefore, the resistive force is $20N$.

Hence, the resistive force applied by the target is $20N$ and the acceleration of the bullet is $-10m·s^{-2}$.