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Force and Laws of Motion — Science STD 9 — Question

Gujarat BoardEnglish MediumSTD 9ScienceForce and Laws of Motion5 Marks
Question
Using second law of motion, derive the relation between force and acceleration. A bullet of 10g strikes a sand-bag at a speed of $103m s^{-1}$ and gets embedded after travelling $5\ cm.$ Calculate
  1. The resistive force exerted by the sand on the bullet.
  2. The time taken by the bullet to come to rest.
✓

Answer

If a body of mass $( m ),$ moving at velocity $( u )$ accelerates uniformly at $(a)$ for time $T$,
so that its velocity changes to $v$, then initial momentum $p _1= mu$ and final momentum $p _2= mv$
$ \therefore$ Change in momentum $= p _2- p _1= mv - mu = m (v -u)$ According to the second law of mation,force
$\text{F}\propto\frac{\text{Change in momentum}}{\text{Time}}$
$\Rightarrow\text{F}\propto\frac{\text{p}_2-\text{p}_1}{\text{t}}$
$\Rightarrow\text{F}\propto\frac{\text{m(v}-\text{u})}{\text{t}}$
$\Rightarrow\text{F}\propto\text{ma}$
$\Big[\text{where},\frac{\text{v}-\text{u}}{\text{t}}=\text{a}\Big]$
$\Rightarrow\text{F}=\text{kma}$ Here, $\text{k}=1$
$\therefore \text{F}=\text{ma}$ Given, $\text{m}=10\text{g}=\frac{10}{1000}\text{kg}=0.01\text{kg},\text{u}=10^3\text{m/s},\text{v}=0 [\because 1kg = 1000g] $ and $\text{s}=5\text{cm}=\frac{5}{100}\text{m}=0.05\text{m}[\because 1m = 100cm]$
  1. From thrid equation of motion, $\text{v}^2=\text{u}^2+2 \text {as}$
$\Rightarrow\text{v}^2-\text{u}^2+2\text{as}$
$\Rightarrow\text{a}=\frac{\text{v}^2-\text{u}^2}{2\text{s}}=\frac{(0)^2-(10^3)^2}{2\times0.05}$
$\text{a}=\frac{-10^6}{0.1}=-10^7\text{m/s}^2$
Now, force applied by the bullet,
F = ma
$= 0.01 \times 10^7 = -10^5 N$ [negative sing show against the direction of motion]
The resistive force exerted by the sand on the bullet $= 10^5 N$
  1. From first equation of motion, $\text{v = u + at}$
$\Rightarrow\text{at = v}-\text{u}$
Now, time taken by bullet to come to rest, $\text{t}=\frac{\text{v}-\text{u}}{\text{a}}$
$\Rightarrow\text{t}=\frac{0-10^3}{-10^7}$
$\Rightarrow\text{t}=\frac{-10^3}{-10^7}=10^3\times10^{-7}$
$\Rightarrow\text{t}=10^{-4}\text{s}$
Hence, force $(F) = 10^5 N$ and time $(t) = 10^{-4} s.$

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