Question

Using second law of motion, derive the relation between force and acceleration. A bullet of $10\mathrm{g}$ strikes a sand-bag at a speed of ${10}^3\mathrm{m}/\mathrm{s}$ and gets embedded after travelling $5\mathrm{cm}$. Calculate

1. The resistive force exerted by the sand on the bullet.
2. The time taken by the bullet to come to rest.

Answer 1

Step 1: Given parameters

Mass, $\mathrm{m}=10\mathrm{g}=\frac{10}{1000}\mathrm{kg}$

Initial velocity, $\mathrm{u}={10}^3\mathrm{m}/\mathrm{s}$

Final velocity, $\mathrm{v}=0$

Distance, $\mathrm{s}=5\mathrm{cm}=\frac{5}{100}\mathrm{m}$

Step 2: To find

We have to determine the resistive force exerted by the sand on the bullet and the time taken by the bullet to come to rest.

Step 3: Finding the resistive force exerted by the sand on the bullet

By using the formula,

${\mathrm{v}}^2-{\mathrm{u}}^2=2\mathrm{as}$

By substituting the given values, we get

$$\begin{gathered} 0-{\left({10}^3\right)}^2=2\times \mathrm{a}\times \frac{5}{100} \\ -{\left({10}^3\right)}^2=\frac{10}{100}\mathrm{a} \\ {-\left({10}^3\right)}^2=0.1\mathrm{a} \\ \mathrm{a}=-\frac{{10}^6}{0.1} \\ =-{10}^7{\mathrm{ms}}^{-2} \end{gathered}$$

The resistive force exerted by:

$$\begin{gathered} \mathrm{F}=\mathrm{m}\times \mathrm{a} \\ =\frac{10}{1000}\times -{10}^7 \\ =-{10}^5\mathrm{N} \end{gathered}$$

Step 4: Finding the time taken by the bullet to come to rest.

We know the formula,

$$\begin{gathered} \mathrm{v}=\mathrm{u}+\mathrm{at} \\ 0={10}^3+(-{10}^7)\times \mathrm{t} \\ {10}^7\mathrm{t}={10}^3 \\ \mathrm{t}=\frac{{10}^3}{{10}^7} \\ ={10}^{-4}\mathrm{s} \end{gathered}$$

Therefore, the force exerted is $-{10}^5\mathrm{N}$ and the time is taken be ${10}^{-4}\mathrm{s}$.