Question

A rifle with a barrel length of $60$cm fires a $10$g bullet with a horizontal speed of $400$m/s. The bullet strikes a block of wood and penetrates to a depth of $12$cm.

a. What resistive force (assumed to be constant) does the wood exert on the bullet?

b. How long does it take the bullet to come to rest?

Short answer

a). $F=6700N=6.7kN$pointing out of the wood

b).$t=0.0006s=600\mu s$take the bullet to come to rest

Step-by-step solution

Step 1. Given Information

We have, rifle with barrel length $l=60$cm

Weight $m=10$g

The horizontal speed of $v=400$m/s

Depth $d=12$cm

Step 1. Part a).  Find the resistive force

Rearranging the following well-known equation of kinematics we get

$$\begin{gathered} {v_f}^2={v_0}^2+2ad \\ \Rightarrow {v_0}^2=-2ad \end{gathered}$$since$v_f=0$

From Newton's second law we have $a=\frac{F}{m}$, which we can substitute in the previous equation find

$$\begin{gathered} {v_0}^2=-\frac{2Fd}{m} \\ \Rightarrow F=\frac{m{v_0}^2}{2d} \end{gathered}$$

Substitute our values, we get

localid="1649521147287" $F=-\left(\frac{0.01\times {400}^2}{2\times 0.12}\right)\approx -6.7kN$

Step 1. Part b).  Find the time

Knowing the definition of acceleration, we can find time taken for the bullet to come to rest.

$$\begin{gathered} a=\frac{v_f-v_0}{t} \\ \Rightarrow t=-\frac{v_0}{a} \\ \Rightarrow t=-\frac{m{v}_0}{F} \\ \Rightarrow t=\left(-\left(\frac{0.01\times 400}{-6700}\right)\right)=0.0006s \\ \Rightarrow t=\left(600\right)\mu s \end{gathered}$$ Since$v_f=0$