Question

[Key Concept: With a constant resisting force, penetration depth is proportional to the kinetic energy lost in the bullet–block relative motion.]

Case-1 : When block is fixed

$$\begin{matrix} & & \omega_{Net} = \Delta KE \\ & & \omega_{Res} = K_{F} - K_{I} \\ & & \ - F_{R} \cdot \text{ }S = O - \frac{1}{2}{mv}_{0}^{2} \\ & & \end{matrix}$$

${\text{ }F}_{R} = \frac{1}{2}\frac{{mv}_{0}^{2}}{\text{ }S}$ $\text{(}\text{i}\text{)}$

F_R=Constant resistance foffered by block.

$V_{0} =$ Initial speed of bullet

Case-2 : When block is free to move

Stepe-1 : Linear momentum conservation of bullet + block system, as no consntant force is acting

$$\begin{array}{r} {mv}_{0} = (m + M)v \Rightarrow v = \frac{{mv}_{0}}{\text{ }m + M}\#(ii) \end{array}$$

Step-2 : Work energy theorem

$$W_{\text{net~}} = k_{F} - K_{I}$$

$$\begin{matrix} & & \\ & \ - F_{R} \cdot {\text{ }S}^{'} = \frac{1}{2}(\text{ }m + M)v^{2} - \frac{1}{2}{mv}_{0}^{2} & \text{(iii)} \end{matrix}$$

On solving equation (i), (ii) and (iii)

$$\begin{matrix} & \ - F_{R} \cdot S^{'} = \frac{1}{2}(m + M)\left\lbrack \frac{mv_{0}}{m + M} \right\rbrack^{2} - \frac{1}{2}mv_{0}^{2} \\ & \ - \frac{1}{2}\frac{mv_{0}^{2}}{S} \cdot S^{'} = \frac{1}{2}\frac{m^{2}v_{0}^{2}}{m + M} - \frac{1}{2}mv_{0}^{2} \\ & \ - \frac{S^{'}}{S} = \frac{m}{m + M} - 1 \\ & \ \Rightarrow \frac{S^{'}}{S} = 1 - \frac{m}{m + M} = \frac{M}{m + M} \\ & \ \Rightarrow S^{'} = \frac{Ms}{m + M} \end{matrix}$$