A metal ball of mass \(2 \mathrm{~kg}\) moving with a speed of \(10 \mathrm{~ms}^{-1}\) had a head-on collision with a stationary ball of mass \(3 \mathrm{~kg}\). If after collision, both the balls move together, then the loss in kinetic energy due to collision is

Apply conservation of momentum,
\( \begin{array}{l} \mathrm{m}_{1} \mathrm{v}_{1}=\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \mathrm{v} \\ \mathrm{v}=\frac{\mathrm{m}_{1} \mathrm{v}_{1}}{\left(\mathrm{~m}_{1}+\mathrm{m}_{2}\right)} \end{array} \)
Here \(\mathrm{v}_{1}=36 \mathrm{~km} / \mathrm{hr}=10 \mathrm{~m} / \mathrm{s}\),
\( \begin{array}{l} \mathrm{m}_{1}=2 \mathrm{~kg}, \mathrm{~m}_{2}=3 \mathrm{~kg} \\ \mathrm{v}=\frac{10 \times 2}{5}=4 \mathrm{~m} / \mathrm{s} \end{array} \)
K.E. (initial) \(=\frac{1}{2} \times 2 \times(10)^{2}=100 \mathrm{~J}\)
K.E. (Final) \(=\frac{1}{2} \times(3+2) \times(4)^{2}=40 \mathrm{~J}\)
Loss in K.E. \(=100-40=60 \mathrm{~J}\)
Alternatively use the formula
\(-\Delta \mathrm{E}_{\mathrm{k}}=\frac{1}{2} \frac{\mathrm{m}_{1} \mathrm{~m}_{2}}{\left(\mathrm{~m}_{1}+\mathrm{m}_{2}\right)}\left(\mathrm{u}_{1}-\mathrm{u}_{2}\right)^{2}\)