A body of mass $5\times {10}^3$kg moving with speed $2 {\mathrm{ms}}^{-1}$ collides with a stationary  body of mass $15\times {10}^3$ kg  inelastically and sticks to it. Then, loss in kinetic energy of the system will be [AIIMS 2019]

Given, mass of body,  ${\mathrm{m}}_1= 5\times {10}^3 \mathrm{kg}$

Velocity, ${\mathrm{v}}_1=2 {\mathrm{ms}}^{-1}$

and mass of another body, ${\mathrm{m}}_2=15\times {10}^3 \mathrm{kg}$

For perfectly inelastic collision, e = 0.
$\therefore$ Loss in kinetic energy of system,

                     $\Delta E_K=\frac{1}{2}\frac{m_1{m}_2}{m_1+m_2}\times v_1^2$

                              $=\frac{1}{2}\times \frac{5\times {10}^3\times 15\times {10}^3}{5\times {10}^3+15\times {10}^3}\times (2{)}^2$

                              $= 7.5\times {10}^3\mathrm{J}$

                               = 7.5 kJ