A block of mass 0.5 kg  is moving with speed of 2 m/s on a smooth surface.  It strikes another mass of 1 kg and then they move together as a single body.   The Energy loss during the Collision is,

 $$\begin{gathered} apply law of conservation of linear momentum \\ {p}_{before}=p_{after} \\ mu = (m+M)V \\ m=0.5kg;M=1kg;u=2m/s \\ 0.5\left(2\right)=(0.5+1)V \\ V=\frac{1}{1.5}m/s \\ K{E}_{initial}=\frac{1}{2}m{v}^2=\frac{1}{2}x0.5x{2}^2=1 joule \\ K{E}_{final}=\frac{1}{2}m{v}^2=\frac{1}{2}(1.5){\left(\frac{1}{1.5}\right)}^2=0.33 joule \\ loss of KE=K{E}_{initial}-K{E}_{final}=1-0.33=0.67 joule \end{gathered}$$