A bullet fired into a fixed target loses half | vedclass

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(B) Let the initial velocity be $v$ and the constant retardation be $a$.After penetrating $s_1 = 3\,cm$,the velocity becomes $v/2$.Using the equation

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$1$

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(B) Let the initial velocity be $v$ and the constant retardation be $a$.After penetrating $s_1 = 3\,cm$,the velocity becomes $v/2$.Using the equation of motion $v_f^2 = v_i^2 + 2as$:$(v/2)^2 = v^2 - 2a(3) \implies v^2/4 = v^2 - 6a \implies 6a = 3v^2/4 \implies a = v^2/8$.Now,let the total distance penetrated be $s_2$ until the final velocity becomes $0$.$0^2 = v^2 - 2a(s_2) \implies v^2 = 2(v^2/8)s_2 \implies v^2 = (v^2/4)s_2 \implies s_2 = 4\,cm$.The additional distance covered is $s_2 - s_1 = 4\,cm - 3\,cm = 1\,cm$.

$A$ 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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