Q.A bullet of mass $m_1$ travelling with a velocity u strikes a stationary wooden block of mass $m_2$ and gets embedded into it. Determine the expression for loss in the kinetic energy of the system. Is this violating the principle of conservation of energy? If not, how can you account for this loss?

[!TLDR] $\Delta(KE)=\frac{1}{2}\frac{m_1m_2}{m_1+m_2}u^2$; this does NOT violate energy conservation -- the 'lost' kinetic energy converts into heat, sound and deformation energy inside the block and bullet, so TOTAL energy is still conserved. [!ANSWER] $\Delta(KE)= …

This is a perfectly inelastic head-on collision (bullet embeds in the block, so they move off together with a common final velocity v). By conservation of momentum, $m_1u+m_2(0)=(m_1+m_2)v\Rightarrow v=\frac{m_1u}{m_1+m_2}$. Using the general perfectly-inelastic kinetic-energy-loss formula from 4.8.5 (with $u_2=0$ here), $\Delta(KE)=(KE)_{\text{initial}}-(KE)_{\text{final}}=\frac{1}{2}m_1u^2-\frac{1}{2}(m_1+m_2)v^2=\frac{1}{2}\frac{m_1m_2}{m_1+m_2}u^2$ (setting $u_2=0$ in the general reduced-mass formula $\frac{1}{2}\frac{m_1m_2}{m_1+m_2}(u_1-u_2)^2$). This loss in KINETIC energy does NOT violate the principle of conservation of TOTAL energy: the 'lost' kinetic energy is not destroyed, but CONVERTED into other forms -- heat generated by friction as the bullet forces its way into the wood, sound of the impact, and permanent deformation/damage of the wood fibres and the bullet itself (plastic deformation energy). Total energy (kinetic + heat + sound + deformation/strain energy) of the bullet-block …

Apply momentum conservation to find the common final velocity, then use the perfectly-inelastic-collision kinetic-energy-loss formula (with the second body's initial velocity set to …

Concluding that kinetic energy loss means energy is not conserved -- conflating conservation of KINETIC energy (which is not generally conserved in inelastic …