A bullet fired into a wooden block loses half of its velocity after penetrating $40 \mathrm{~cm} .$ It comes…

A bullet fired into a wooden block loses half of its velocity after penetrating $40 \mathrm{~cm} .$ It comes to rest after penetrating a further distance of
  1. $\frac{22}{3} \mathrm{~cm}$
  2. $\frac{40}{3} \mathrm{~cm}$
  3. $\frac{20}{3} \mathrm{~cm}$
  4. $\frac{22}{5} \mathrm{~cm}$

Solution

For first part of penetration, by equation of motion $\left(\frac{\mathrm{u}}{2}\right)^{2}-(\mathrm{u})^{2}=2 \mathrm{aS}$ or $\mathrm{a}=-\frac{3 \mathrm{u}^{2}}{8 \mathrm{~S}} \quad \ldots(\mathrm{i})$
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

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