JEE MAIN - Physics (2024 - 27th January Evening Shift - No. 11)
A bullet is fired into a fixed target looses one third of its velocity after travelling $$4 \mathrm{~cm}$$. It penetrates further $$\mathrm{D} \times 10^{-3} \mathrm{~m}$$ before coming to rest. The value of $$\mathrm{D}$$ is :
23
32
42
52
Explanation
$$\begin{aligned} & v^2-u^2=2 a S \\ & \left(\frac{2 u}{3}\right)^2=u^2+2(-a)\left(4 \times 10^{-2}\right) \\ & \frac{4 u^2}{9}=u^2-2 a\left(4 \times 10^{-2}\right) \\ & -\frac{5 u^2}{9}=-2 a\left(4 \times 10^{-2}\right) \ldots(1) \\ & 0=\left(\frac{2 u}{3}\right)^2+2(-a)(x) \\ & -\frac{4 u^2}{9}=-2 a x \ldots(2) \end{aligned}$$
$$(1)/(2)$$
$$\begin{aligned} & \frac{5}{4}=\frac{4 \times 10^{-2}}{\mathrm{x}} \\ & \mathrm{x}=\frac{16}{5} \times 10^{-2} \\ & \mathrm{x}=3 \cdot 2 \times 10^{-2} \mathrm{~m} \\ & \mathrm{x}=32 \times 10^{-3} \mathrm{~m} \end{aligned}$$
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