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A bullet fired into a wooden block loses half of its velocity after penetrating 40 cm. It comes to rest after penetrating a further distance of
Explanation
For first part of penetration, by equation of motion $(u/2)^{2} - (u)^{2} = 2aS$ or $a = -3u^{2}/8S$ For latter part of penetration $(0)^{2} - (u/2)^{2} = 2aS'$, $S' = u^{2}/8a$ $S' = -u^{2}/8 (8S/-3u^{2})$ (Using (i)) $S' = S/3$ or $S' = 40/3 cm$