Question

Two small discs of masses $m_1$ and $m_2$ interconnected by a weightless spring rest on a smooth horizontal plane. The discs are set in motion with initial velocities $v_1$ and $v_2$ whose directions are mutually perpendicular and lie in a horizontal plane. Find the total energy $\tilde{E}$ of this system in the frame of the centre of inertia

Answer 1

As initially $U=\tilde{U}=0$, so, $\tilde{E}=\tilde{T}$
Total kinetic energy in center of mass frame is
$\tilde{T}=\frac{1}{2}\mu |\overrightarrow{v_1}-\overrightarrow{v_2}|$,
As $\overrightarrow{v_1}\perp \overrightarrow{v_2}$
Thus $\tilde{T}=\frac{1}{2}\frac{m_1{m}_2}{m_1+m_2}(v_1^2+v_2^2)$