The linear momentum remains conserved before and after the collision, therefore, 

$$\begin{gathered} \Rightarrow 2 \mathrm{kg}\times \left(36\times \frac{1000}{3600} \mathrm{m} {\mathrm{s}}^{-1}\right)+3 \mathrm{kg}\times 0=\left(3 \mathrm{kg}+2 \mathrm{kg}\right)v \\ v=\frac{20}{5}=4 \mathrm{m} {\mathrm{s}}^{-1} \end{gathered}$$

The loss in kinetic energy is, $$\begin{gathered} ∆KE=K{E}_i-K{E}_f \\ =\left[\frac{1}{2}\left(2 \mathrm{kg}\right){\left(10 \mathrm{m} {\mathrm{s}}^{-1}\right)}^2+0\right]-\left[\frac{1}{2}\left(2 \mathrm{kg}+3 \mathrm{kg}\right){\left(4 \mathrm{m} {\mathrm{s}}^{-1}\right)}^2\right] \\ =100-40=60 \mathrm{kg} {\mathrm{m}}^2 {\mathrm{s}}^{-2} \\ =60 \mathrm{J} \end{gathered}$$