Q 11-02-063JEE MainJEE Main 2024 (27 Jan, Shift 2)Medium
A bullet is fired into a fixed target and loses one third of its velocity after travelling $4\ \text{cm}$. It penetrates further $D\times10^{-3}\ \text{m}$ before coming to rest. The value of $D$ is:
Answer: (A) 32
With uniform retardation $a$, the speed drops from $v$ to $\frac{2v}{3}$ in $4\ \text{cm}$:
$$v^2 - \frac{4v^2}{9} = 2a(0.04) \;\Rightarrow\; \frac{5v^2}{9} = 0.08a$$
It then stops in a further distance $x$:
$$\frac{4v^2}{9} = 2ax$$
Dividing, $\dfrac{x}{0.04} = \dfrac45 \Rightarrow x = 0.032\ \text{m} = 32\times10^{-3}\ \text{m}$, so $D = 32$.
Solution by Sreeraj P, M.Sc Physics