A bullet fired into a fixed target loses half | vedclass

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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 $v^2 = u^2 - 2as$

Vedclass · Accepted Answer

$1$

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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 $v^2 = u^2 - 2as$,where $a$ is the retardation due to constant resistance:$(u/2)^2 = u^2 - 2a(3)$$u^2/4 = u^2 - 6a$$6a = 3u^2/4$$a = u^2/8$Now,let the bullet penetrate an additional distance $x$ before coming to rest $(v=0)$.Using the equation of motion $v^2 = u^2 - 2as$ for this segment,with initial velocity $u/2$:$0^2 = (u/2)^2 - 2ax$$0 = u^2/4 - 2(u^2/8)x$$u^2/4 = (u^2/4)x$$x = 1\, cm$.Therefore,the bullet will penetrate $1\, cm$ further.

$A$ 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?.......$cm$

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