Question

Two small discs of masses $m_1$ and $m_2$ are connected by a weightless spring resting on a smooth horizontal plane. The discs are set in motion with initial velocities ${\nu }_1$ and ${\nu }_2$, whose directions are mutually perpendicular and in the same horizontal plane. Find the total energy $E$ of the system with reference to the frame fixed to the centre of mass.

Answer 1

Let ${\overline{\nu }}_{cm}$ be the centre of mass. Velocity of $m_1$ relative to the centre of mass is $({\overrightarrow{\nu }}_1-{\overrightarrow{\nu }}_{cm})$ and that of $m_2$ relative to the centre of mass is $({\overrightarrow{\nu }}_2-{\overrightarrow{\nu }}_1)$
velocity of centre of mass ${\overrightarrow{\nu }}_{cm}=\frac{m_1{\overrightarrow{\nu }}_1+m_2{\overrightarrow{\nu }}_2}{m_1+m_2}$
$({\overrightarrow{\nu }}_1-{\overrightarrow{\nu }}_{cm})=\frac{m_2}{m_1+m_2}({\overrightarrow{\nu }}_1-{\overrightarrow{\nu }}_{cm})$
Hence, $|{\overrightarrow{\nu }}_1-{\overrightarrow{\nu }}_{cm}|=\frac{m_2}{m_1+m_2}({\overrightarrow{\nu }}_1^2-{\overrightarrow{\nu }}_2^2{)}^{\frac{1}{2}}$
and $({\overrightarrow{\nu }}_1-{\overrightarrow{\nu }}_{cm})=\frac{m_2}{m_1+m_2}({\overrightarrow{\nu }}_1-{\overrightarrow{\nu }}_2)$
Hence, $|{\overrightarrow{\nu }}_2-{\overrightarrow{\nu }}_{cm}|=\frac{m_1}{m_1+m_2}({\overrightarrow{\nu }}_1^2+{\overrightarrow{\nu }}_2^2{)}^{\frac{1}{2}}$
total energy with respect to the centre of mass,
$E=\frac{1}{2}{m}_1|{\overrightarrow{\nu }}_1-{\overrightarrow{\nu }}_{cm}{|}^2+\frac{1}{2}{m}_1|{\overrightarrow{\nu }}_2-{\overrightarrow{\nu }}_{cm}{|}^2$
$=\frac{1}{2}\left(\frac{m_1{m}_2}{m_1+m_2}\right)({\nu }_1^2+{\nu }_2^2)$
Where $\mu =\frac{m_1{m}_2}{m_1+m_2}$ is called reduced mass.