Centre of Mass and Collision Part 1

Given Mass of large block of wood, $M=5.99\,\text{kg}$ Mass of bullet, $m=10\,\text{g}$ Height at which their centre of mass rise, $h=9.8\,\text{cm}$ From the law of conservation of energy, Energy of the system when bullet gets embedded = Energy of the system till it momentarily comes to rest. $\Rightarrow \;\; \;\frac{1}{2}(M+m)v_{1}^{2}=(M+m)gh$ where, $v_{1}=$ velocity of bullet $+$ block system $\Rightarrow \;\; v_{1}=\sqrt{2gh}$.....(i) According to law of conservation of momentum, Momentum before collision $=$ Momentum after collision. $\Rightarrow \;\; mv=(M+m)v_{1}$ [where, $v=$ velocity of bullet before collision] $\Rightarrow \;\; mv=(M+m)\sqrt{2gh}$ [using Eq. (i)] $\Rightarrow \;\; v=\left(\;\frac{M+m}{m}\right)\sqrt{2gh}$ $\Rightarrow \;\; v=\;\frac{5.99+0.01}{10\times 10^{-3}}\times \sqrt{2\times 9.8\times 9.8\times 10^{-2}}$ $\Rightarrow \;\; v=831.55\,\text{ms}^{-1}$