Motion In One Dimension Question 5
Question 5 - 2024 (27 Jan Shift 2)
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 :
[We changed options. In official NTA paper no option was correct.]
(1) 2
(2) 5
(3) 3
(4) 32
Show Answer
Answer: (4)
Solution:
$v^{2}-u^{2}=2 a S$
$\left(\frac{2 \mathrm{u}}{3}\right)^{2}=\mathrm{u}^{2}+2(-\mathrm{a})\left(4 \times 10^{-2}\right)$
$\frac{4 u^{2}}{9}=u^{2}-2 \mathrm{a}\left(4 \times 10^{-2}\right)$
$-\frac{5 \mathrm{u}^{2}}{9}=-2 \mathrm{a}\left(4 \times 10^{-2}\right) \ldots$
$0=\left(\frac{2 \mathrm{u}}{3}\right)^{2}+2(-\mathrm{a})(\mathrm{x})$
$-\frac{4 \mathrm{u}^{2}}{9}=-2 \mathrm{ax}$
(1) $/(2)$
$\frac{5}{4}=\frac{4 \times 10^{-2}}{\mathrm{x}}$
$x=\frac{16}{5} \times 10^{-2}$
$\mathrm{x}=3 \cdot 2 \times 10^{-2} \mathrm{~m}$
$\mathrm{x}=32 \times 10^{-3} \mathrm{~m}$
