Question

A metal ball of mass 2 kg moving with speed of 36 km/h has a head on collision with a stationary ball of mass 3 kg. If after collision both the balls move as a single mass, what will be the loss in kinetic energy due to collision?

Answer 1

Dear Student,

$$\begin{gathered} {\mathrm{u}}_1=36\mathrm{km}/\mathrm{h} \\ {\mathrm{u}}_1=36\times \frac{5}{18}\mathrm{m}/\mathrm{s}=10\mathrm{m}/\mathrm{s} \\ {\mathrm{u}}_2=0 \\ {\mathrm{m}}_1=2\mathrm{kg} \\ {\mathrm{m}}_2=3\mathrm{kg} \\ {\mathrm{m}}_1{\mathrm{u}}_1+{\mathrm{m}}_2{\mathrm{u}}_2=\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)\mathrm{v} \\ 2\times 10+3\times 0=\left(2+3\right)\mathrm{v} \\ \mathrm{v}=\frac{20}{5}=4\mathrm{m}/\mathrm{s} \\ \mathrm{Loss}\mathrm{in}\mathrm{kinetic}\mathrm{energy}=\frac{1}{2}{\mathrm{m}}_1{{\mathrm{u}}_1}^2-\frac{1}{2}\left({\mathrm{m}}_1+{\mathrm{m}}_2\right){\mathrm{v}}^2 \\ \mathrm{Loss}\mathrm{in}\mathrm{kinetic}\mathrm{energy}=\frac{1}{2}\times 2\times {10}^2-\frac{1}{2}\times 5\times 4^2 \\ \mathrm{Loss}\mathrm{in}\mathrm{kinetic}\mathrm{energy}=100-40=60\mathrm{J} \end{gathered}$$