Q 11-02-100JEE MainJEE Main 2022 (27 Jul, Shift 2)Medium
The velocity of the bullet becomes one third after it penetrates $4$ cm in a wooden block. Assuming that bullet is facing a constant resistance during its motion in the block. The bullet stops completely after travelling at $(4 + x)$ cm inside the block. The value of $x$ is
Answer: (C) $0.5$
With constant retardation $a$:
$$u^2 - \left(\frac u3\right)^2 = 2a(4) \Rightarrow \frac89u^2 = 8a \Rightarrow a = \frac{u^2}{9}$$
For the rest of the motion, from $\dfrac u3$ to $0$:
$$\frac{u^2}{9} = 2a\,x = \frac{2u^2}{9}x \Rightarrow x = 0.5\ \text{cm}$$
Solution by Sreeraj P, M.Sc Physics