# A bullet of 10 g strikes a sand-bag at a speed of 10^3 m/s and gets embedded after travelling 5 cm. Calculate: (i) the acceleration of the bullet (ii) the resistive force exerted by the sand on the bullet. (iii)the time taken by the bullet to come to rest..

Given,

Mass of the bullet, m = 10 g = 10

^{-2}kg

Speed of the bullet, u = 10

^{3}m/s

Distance travelled by the bullet before stopping (v=0), d = 5 cm = 0.05 m

(i)Let the acceleration of the bullet be a.

Using the kinematic equation for motion under uniform acceleration:

${v}^{2}={u}^{2}+2ad\phantom{\rule{0ex}{0ex}}0={\left({10}^{3}\right)}^{2}+2\times a\times 0.05\phantom{\rule{0ex}{0ex}}a=-\frac{{10}^{6}}{2\times 0.05}=-{10}^{7}m/{s}^{2}$

(ii) Let the force exerted by the sand on the bullet be F

Using Newton's 2nd law of motion

$F=ma\phantom{\rule{0ex}{0ex}}F={10}^{-2}\times \left(-{10}^{7}\right)=-{10}^{5}N\phantom{\rule{0ex}{0ex}}Forceisnegativewhichmeansthatitisresistiveorretardinginnature.$

(iii) Let t be the time taken by the bullet to come to rest.

Using the equation

$v=u+at\phantom{\rule{0ex}{0ex}}0=u+at\phantom{\rule{0ex}{0ex}}-{10}^{3}=-{10}^{7}\times t\phantom{\rule{0ex}{0ex}}t=\frac{{10}^{3}}{{10}^{7}}={10}^{-4}s$

Regards

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