A block m₁ strikes a stationary block m₃ inelastically. Another block m₂ is kept on m₃. Neglecting the friction between all contacting surfaces, the fractional decrease of K. E. of the system in collision is:

Since the impact between m₁ and m₃ is inelastic, m₁ and m₃ will move together towards right and m₂ does not move due to the absence of friction. The velocity of the combined mass
$={\mathrm{v}}^{\text{'}}=\frac{{\mathrm{m}}_1\mathrm{v}}{{\mathrm{m}}_1+{\mathrm{m}}_3}$

$\Rightarrow \frac{\left|∆\mathrm{KE}\right|}{\mathrm{kE}}=\frac{{\displaystyle \frac{1}{2}{\mathrm{m}}_1{\mathrm{v}}^2-\frac{1}{2}\left({\mathrm{m}}_1+{\mathrm{m}}_3\right){\left({\mathrm{v}}^{\text{'}}\right)}^2}}{\frac{1}{2}{\mathrm{m}}_1{\mathrm{v}}^2}=1-\left(\frac{{\mathrm{m}}_1+{\mathrm{m}}_3}{{\mathrm{m}}_1}\right){\left(\frac{{\mathrm{v}}^{\text{'}}}{\mathrm{v}}\right)}^2$

$=1-\left(\frac{{\mathrm{m}}_1+{\mathrm{m}}_3}{{\mathrm{m}}_1}\right){\left(\frac{{\mathrm{m}}_1}{{\mathrm{m}}_1+{\mathrm{m}}_3}\right)}^2=1-\frac{{\mathrm{m}}_1}{{\mathrm{m}}_1+{\mathrm{m}}_3}$

$=\frac{{\mathrm{m}}_3}{{\mathrm{m}}_1+{\mathrm{m}}_3}$