Mass of metal ball = 2 kg;
Speed of metal ball (v$_1$) = 36 km/h = 10 m/s and
mass of stationary ball = 3 kg.
Applying law of conservation of momentum,
$m_1v_1+m_2v_2=(m_1+m_2)v$
or, $v=\frac{m_1v_1+m_2v_2}{m_1+m_2}=\frac{(2 \times 10)+(3 \times 0)}{2+3}=\frac{20}{5}$
= 4 m/s
Therefore loss of energy
$ \, \, \, \, \, \, =\bigg[\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2\bigg]-\frac{1}{2}+(m_1+m_2)v^2$
=$\bigg[\frac{1}{2} \times 2 \times (10)^2 +\frac{1}{2} \times 3 (0)^2\bigg] -\frac{1}{2} \times (2+3)\times(4)^2$
= 100 - 40 = 60 J.