Question

Two small discs of masses m1 and m2 interconnected by a weightless spring rest on a smooth horizontal plane. The discs are set in motion with initial velocities v1 and v2 whose directions are mutually perpendicular and lie in a horizontal plane. Find the total energy E of this system in the frame of the centre of inertia

Answer 1

Initially $U=\tilde{U}=0$

so $\tilde{E}=\tilde{T}$

$\tilde{T}=\frac{1}{2} \mu v_{rel}^2=\frac{1}{2}\mu|\vec{v_1}-\vec{v_2}|^2$

Therefore from the equation;

$\tilde{T}=\frac{1}{2} \mu |\vec{v_1}-\vec{v_2}|$

$\vec{v_1}-\vec{v_2}$

Hence;

$\tilde{T}=\frac{1}{2}\frac{m_1m_2}{m_1+m_2}(v^2_1+v^2_2)$

where $\mu=\frac{m_1m_2}{m_1+m_2}$ refers to reduced mass