A body of mass 4 kg moving with velocity $12 {\mathrm{ms}}^{-1}$ collides with another body of mass 6 kg at rest. If two bodies stick together after collision, then the loss of kinetic energy of system is

Loss of kinetic energy $=\frac{1}{2}{m}_{{\mu }_1}^2-\frac{m_1^2{u}_1^2}{2\left(m_1+m_2\right)}=\frac{m_1{m}_2{u}_1^2}{2\left(m_1+m_2\right)}$

Given, ${\mathrm{m}}_1=4 \mathrm{kg}, {\mathrm{u}}_1=12 {\mathrm{ms}}^{-1},$

            ${\mathrm{m}}_2=6 \mathrm{kg} \mathrm{and} {\mathrm{u}}_2=\mathrm{O}$

$\therefore$       $∆\mathrm{KE}=\frac{1}{2}\frac{4\times 6}{(4+6)}(12{)}^2=\frac{1}{2}\times \frac{24}{10}\times (12{)}^2$

                     $=\frac{12}{10}\times 144=172.8 \mathrm{J}$