Question

A bullet of mass 5 g leaves a rifle with a velocity of 2 m/s. If it strikes a wooden target and penetrates 20 cm into it, what is the average resistance offered by the target?

Answer 1

Step 1: Given data

Mass of bullet is, $m=5g=0.005kg$

The velocity of the rifle (initial velocity) is, $u=2m/s$

The distance through which the bullet penetrates the target is $S=20cm=0.2m$

Step 2: Calculating acceleration

The third equation of motion is given as,

$v^2-u^2=2aS$

where, $v$ is final velocity, $u$ is initial velocity, $a$ is acceleration and $S$ is distance.

Since the bullet finally comes to rest, the final velocity will be zero, that is, $v=0m/s$

Substituting the known values, we get,

$$\begin{gathered} {(0m/s)}^2-{(2m/s)}^2=2\times a\times 0.2m \\ a=\frac{-4}{0.4} \\ a=-10m/s^2 \end{gathered}$$

Step 3: Calculating resistance offered by the target

The force (resistance) offered by the target is given as,

$F=ma$

where $m$ is mass and $a$ is acceleration.

Substituting the known values, we get,

$$\begin{gathered} F=0.005kg\times 10m/s^2 \\ F=0.05N \\ F=5000dynes[1N={10}^{5}dynes] \end{gathered}$$

Therefore, the resistance offered by the target is $5000dynes$.