1

Suppose two particles of masses $m_1$ and $m_2$ with initial velocity $v_1$ and $v_2$ have an elastic collision. Let $v_1'$ and $v_2'$ be their new velocities. We know that $$v_1'= \frac{(m_1 - m_2)v_1 + 2m_2 v_2}{m_1+ m_2}, \quad v_2'= \frac{(m_2 - m_1)v_2 + 2m_1 v_1}{m_1+ m_2}.$$ Now we compute the average of the velocities of particle 1 before and after collision: $$\bar v_1= \frac{v_1' + v_1}{2} = \frac{(m_1 - m_2)v_1 + 2m_2 v_2 + m_1v_1+ m_2v_1}{2(m_1+ m_2)} =\frac{m_1v_1+ m_2v_2}{m_1+ m_2}.$$ In some sense it means the average velocity of particle 1 with respect to time (before and after) equals to the average velocity with respect to space (of particle 1 and particle 2).

Is this a coincidence? Or does it has some physical meanings?

Thanks!

Pengfei
  • 113
  • 1
    The first average is not an "average with respect to time" and the final expression is not an "average with respect to space." – Triatticus Nov 04 '19 at 18:24
  • 1
    Note that If you put a coordinate system co-moving with the center of mass, then $m_1 r_1 + m_2 r_2 = 0$. Take the derivative of the above and see that $m_1 v_1 + m_2 v_2 = 0$. So without loss of generality, the average is zero. – John Alexiou Nov 04 '19 at 19:38

1 Answers1

2

The averages you have calculated give the velocity of the centre of mass.

Consider the collision in the centre of mass frame, i.e. in the frame where the total momentum is zero. In that frame the collision looks like this:

COM frame

In this frame the average velocities of the two masses are both zero.

Now suppose we switch to a frame in which the COM is moving with some velocity $u$. In this frame the initial and final velocities of mass one become $v_1 + u$ and $-v_1 + u$ respectively, so the average is just the COM velocity $u$. The same argument tells us that the average velocity of mass two is also the COM velocity $u$. So what we find is that the average velocities of the two particles are aways equal and always equal to the COM velocity.

John Rennie
  • 355,118