(B) Let the initial velocity of the bullet be $u$.After penetrating $3\,cm$,its velocity becomes $u/2$.Using the equation of motion $v^2 = u^2 - 2as$:

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(B) Let the initial velocity of the bullet be $u$.After penetrating $3\,cm$,its velocity becomes $u/2$.Using the equation of motion $v^2 = u^2 - 2as$:$(u/2)^2 = u^2 - 2a(3)$$u^2/4 = u^2 - 6a$$6a = 3u^2/4$$a = u^2/8$Let the bullet penetrate an additional distance $x$ before coming to rest.For this motion,initial velocity is $u/2$,final velocity is $0$,and acceleration is $-a = -u^2/8$.Using $v^2 = u^2 - 2as$:$0^2 = (u/2)^2 - 2(u^2/8)x$$0 = u^2/4 - (u^2/4)x$$u^2/4 = (u^2/4)x$$x = 1\,cm$.

Class 11 Physics Motion in a Straight Line Uniformly Accelerated Motion

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

AIEEE 2005 All AIEEE

A
$1.5$
B
$1$
C
$3$
D
$2$

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Motion in a Straight Line

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Uniformly Accelerated Motion

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

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