Physics - Kinematics Question with Solution | TestHub

When the bullet just enters the target, its equation of motion is given by

$v^2-u^2=2aS ...\left(1\right)$

Equation (1) implies that

$$\begin{gathered} {\left(\frac{2u}{3}\right)}^2=u^2+2\left(-a\right)\left(4\times {10}^{-2}\right) \\ \Rightarrow \frac{4u^2}{9}=u^2-2a\left(4\times {10}^{-2}\right) \\ \Rightarrow \frac{5u^2}{9}=2a\left(4\times {10}^{-2}\right) \dots \left(2\right) \end{gathered}$$

When the bullet stops inside the target, its equation of motion can be written as

$0={\left(\frac{2u}{3}\right)}^2+2\left(-a\right)\left(x\right) ...\left(3\right)$

Equation (3) implies that

$\frac{4u^2}{9}=2ax ...\left(4\right)$

Dividing equation (2) by equation (4), we have

$$\begin{gathered} \frac{5}{4}=\frac{4\times {10}^{-2}}{x} \\ \Rightarrow x=\frac{16}{5}\times {10}^{-2} \\ \Rightarrow x=3.2\times {10}^{-2} \mathrm{m} \\ =32\times {10}^{-3} \mathrm{m} \end{gathered}$$

Hence, $D=32$.