JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion
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 \(D\) is:
- A \(20\)
- B \(50\)
- C \(32\)
- D \(42\)
Answer & Solution
Correct Answer
(C) \(32\)
Step-by-step Solution
Detailed explanation
\( v^2-u^2=2 a S \) \( \left(\frac{2 u}{3}\right)^2=u^2+2(-a)\left(4 \times 10^{-2}\right) \) \( \frac{4 u^2}{9}=u^2-2 a\left(4 \times 10^{-2}\right) \) \( -\frac{5 u^2}{9}=-2 a\left(4 \times 10^{-2}\right) \ldots(1) \) \( 0=\left(\frac{2 u}{3}\right)^2+2(-a)(x) \)…
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