A bullet fired into a wooden block loses half of its velocity after penetrating $40 \mathrm{~cm} .$ It comes…
- $\frac{22}{3} \mathrm{~cm}$
- $\frac{40}{3} \mathrm{~cm}$
- $\frac{20}{3} \mathrm{~cm}$
- $\frac{22}{5} \mathrm{~cm}$
Solution
For latter part of penetration $(0)^{2}-\left(\frac{\mathrm{u}}{2}\right)^{2}=2 \mathrm{aS}^{\prime}, \mathrm{S}^{\prime}=-\frac{\mathrm{u}^{2}}{8 \mathrm{a}}$
$\mathrm{S}^{\prime}=-\frac{\mathrm{u}^{2}}{8}\left(\frac{8 \mathrm{~S}}{-3 \mathrm{u}^{2}}\right) \quad$ (Using (i))
$\mathrm{S}^{\prime}=\frac{\mathrm{S}}{3}$ or $\mathrm{S}^{\prime}=\frac{40}{3} \mathrm{~cm}$
Asked in: JEE Mains - Motion In One Dimension - Test 2