2.Motion in Straight Line
hard

A bullet fired into a fixed target loses half of its velocity after penetrating $3\,cm$. How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion?........$cm$

A

$1.5$

B

$1$

C

$3$

D

$2$

(AIEEE-2005)

Solution

(b) Let initial velocity of the bullet $= u$

After penetrating $3\, cm$ its velocity becomes $\frac{u}{2}$

From ${v^2} = {u^2} – 2as$

${\left( {\frac{u}{2}} \right)^2} = {u^2} – 2a\,(3)$

$⇒$ $6a = \frac{{3{u^2}}}{4}$ $⇒$  $a = \frac{{{u^2}}}{8}$

Let further it will penetrate through distance $x$ and stops at point $C$.

For distance $BC$, $v = 0,\,u = u/2,\,s = x,\,a = {u^2}/8$

From ${v^2} = {u^2} – 2as$ $⇒$ $0 = {\left( {\frac{u}{2}} \right)^2} – 2\left( {\frac{{{u^2}}}{8}} \right)\,.\,x$ $⇒$ $x = 1\,cm$.

Standard 11
Physics

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