Question

The average force necessary to stop a bullet of mass $20g$ moving with a speed of $250m/s$, as it penetrates into the wood for a distance of $12cm$ is

Answer 1

Step1: Given data

Mass of bullet$\left(m\right)$=$20g=20\times {10}^{-3}kg$

Speed of bullet$\left(u\right)$=$250m/s$

The distance at which the bullet penetrates $\left(s\right)$=$12cm=0.12mas1cm=\frac{1}{100}m$

Final speed $\left(v\right)=0$

Step2: Formula used

To compute the acceleration, use the third equation of motion $v^2=u^2+2as$

Where $v=finalspeed,u=initialspeed,a=acclerationands=dis\tan ce$.

Step3: Determine the acceleration

Substitute the known values in the formula, $v^2=u^2+2as$

$$\begin{gathered} {\left(0\right)}^2={\left(250\right)}^2+2a\times \left(0.12\right) \\ -62500=0.24\times a \\ a=-\frac{62500}{0.24}m/s^2 \end{gathered}$$

Acceleration is negative, it means it is acting in opposite direction.

Step4: Compute the force needed to stop the bullet

We know that $F=ma$. So,

$$\begin{gathered} F=\left(20\times {10}^{-3}\right)\times \frac{62500}{0.24} \\ F=5208.3N \end{gathered}$$

Thus, the required force to stop the bullet is $5208.3N$.