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 $4 cm$. It penetrates further $D \times 10^{-3} m$ before coming to rest. The value of $D$ 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:
$v^{2}-u^{2}=2 a S$
$\left(\frac{2 u}{3}\right)^{2}=u^{2}+2(-a)\left(4 \times 10^{-2}\right)$
$\frac{4 u^{2}}{9}=u^{2}-2 a\left(4 \times 10^{-2}\right)$
$-\frac{5 u^{2}}{9}=-2 a\left(4 \times 10^{-2}\right) \ldots$
$0=\left(\frac{2 u}{3}\right)^{2}+2(-a)(x)$
$-\frac{4 u^{2}}{9}=-2 ax$
(1) $/(2)$
$\frac{5}{4}=\frac{4 \times 10^{-2}}{x}$
$x=\frac{16}{5} \times 10^{-2}$
$x=3 \cdot 2 \times 10^{-2} m$
$x=32 \times 10^{-3} m$
