Question

If m_1,m₂ and m₃ be the masses of three particles v_12, v₂₃ and v₁₃ be their relative velocities, prove that the total kinetic energy (E) of the system about the centre of mass is given by

m₁m₂v²₁₂₊ m₂m₃v²₂₃₊ m₁m₃v²₁₃ / m₁+ m₂ + m₃

Answer 1

$\vec v_{12}=\vec{v_1}-\vec{v_2}$

$\vec v_{23}=\vec{v_2}-\vec{v_3}$

$\vec v_{13}=\vec{v_1}-\vec{v_3}$

$m_1m_2\vec v_{12}^2+m_2m_3\vec v_{23}^2+m_1m_3\vec v_{13}^2=$

$m_1m_2(\vec{v_1}-\vec{v_2})^2+m_2m_3(\vec{v_2}-\vec{v_3})^2+m_1m_3(\vec{v_1}-\vec{v_3})^2=$

$\text{(After expanding the brackets and grouping the terms )}$

$(m_1\vec v_1^2+m_2\vec v_2^2+m_3\vec v_3^2)(m_1+m_2+m_3)-$

$(m_1\vec v_1+m_2\vec v_2+m_3\vec v_3)^2\ (1)$

$\text{Divide the resulting expression (1) by }(m_1+m_2+m_3)$

$\text{We also take into account the fact that:}$

$T= \frac{m_1\vec v_1^2}{2}+ \frac{m_2\vec v_2^2}{2}+ \frac{m_3\vec v_3^2}{2}$

$(m_1\vec v_1^2+m_2\vec v_2^2+m_3\vec v_3^2)= 2T$

$\text{where }T - \text{total kinetic energy of the system}$

$(m_1\vec v_1+m_2\vec v_2+m_3\vec v_3)^2$

$P = m_1\vec v_1+m_2\vec v_2+m_3\vec v_3$

$\text{where } P\text{ center of gravity impulse}$

$T_c =\frac{1}{2}(m_1+m_2+m_3)v_c^2$

$T_c -\text{kinetic energy of the center of gravity}$

$P =(m_1+m_2+m_3)v_c$

$(m_1\vec v_1+m_2\vec v_2+m_3\vec v_3)^2= 2(m_1+m_2+m_3)T_c$

$\frac{m_1m_2\vec v_{12}^2+m_2m_3\vec v_{23}^2+m_1m_3\vec v_{13}^2}{m_1+m_2+m3}=2T-2T_c$

$\text {by Koenig's theorem}$

$T = T_c+T_r$

$T_r -\text{ relative kinetic energy of the system}$

$\frac{m_1m_2\vec v_{12}^2+m_2m_3\vec v_{23}^2+m_1m_3\vec v_{13}^2}{m_1+m_2+m3}=2T_r$

$\text{hence the hypothesis is false}$

$\text{Answer: the proposed proof formula is false}$