# In an elastic collision, why is the velocity of separation equal to the velocity of approach?

For two bodies in an elastic collision, the magnitude of the difference in their velocities along the line of collision remains the same before and after the collision. Or,

$$v_1-v_2=v_2^{'}-v_1^{'}$$

where primes ($$'$$) indicate velocities after the collision. (The LHS is called the velocity of approach, the RHS is the velocity of separation.)

This result can be derived using the conservation of linear momentum and energy. Is there a more intuitive explanation?

An easy way to understand this is that the total kinetic energy could be divided into two part:$$E_k=\frac{1}{2}(m_1+m_2)v_c^2+\frac{1}{2}\frac{m_1m_2}{m_1+m_2}v_r^2$$ the first term is the kinetic energy of the mass center and the second term is the relative kinetic energy, where $$v_r=v_1-v_2$$ is the relative velocity and the relative kinetic energy is defined as the kinetic energy you see when you are in the center of mass frame.

The conservation of momentum require $$v_c$$ to be invariant so $$v_r^2$$ doesn't change as well.

• How did you get the second term? Nov 7, 2020 at 16:21
• @ijm $$v_c=\frac{m_1v_1+m_2v_2}{m_1+m_2}$$ Nov 7, 2020 at 16:56