# Velocities in an elastic collision

My teacher keeps saying that, in a head-on elastic collision of two bodies of equal masses, the velocities get exchanged. But consider two balls of equal mass moving with the same velocity in the opposite direction. In this case, after the collision, both the bodies come to rest, i.e. no velocity exchange occurs. I consulted a couple of books which all repeat what my teacher said, but I found neither an explanation for this assertion nor the clarification of my consfusion. It would be great if someone helped.

Your assumption that both bodies come to rest is false.

Recall that in an elastic collision with two balls $$A$$ and $$B$$ with initial velocities $$v_{A,\;i}$$ and $$v_{B,\;i}$$, the final velocities are given by: $$v_{A,\;f} = \dfrac{m_A - m_B}{m_A+m_B} v_{A,\;i} + \dfrac{2 m_B}{m_A+m_B}v_{B,\;i}$$ $$v_{B,\;f} = \dfrac{2m_A}{m_A+m_B} v_{A,\;i} - \dfrac{m_A-m_B}{m_A+m_B}v_{B,\;i}$$

For the particular case where $$v_{B,\;i}=-v_{A,\;i}$$ and $$m_A=m_B$$, you get,

$$v_{A,\;f} = v_{B,\;i}$$ $$v_{B,\;f} = v_{A,\;i}$$

As you can see, the velocities are indeed exchanged. You must have made a mistake in your calculations.

Do note that in an inelastic collision, it would be possible for the two masses to both end up at rest.

In the context of problems with colliding balls, an elastic collision is the collision in which the kinetic energy of the balls is conserved, while an inelastic collision is a collision where the kinetic energy is not conserved (e.g., it could be transferred to heat, if the two balls stick together). In both cases the total momentum is conserved.

We thus have the following equations for an elastic collision: $$m_Av_A + m_B v_B = m_A v_A' + m_B v_B'\text{ (momentum conservation)},\\ \frac{m_Av_A^2}{2} + \frac{m_Bv_B^2}{2} = \frac{m_Av_A'^2}{2} + \frac{m_Bv_B'^2}{2}\text{ (energy conservation)},$$ where $$v_A, v_B$$ are the velocities of the balls before the collision, while $$v_A', v_B'$$ are those after the collision. Assuming the velocities before the collision know, we can solve for the velocities after the collision (see the answer by @user256872). For the balls of equal mass this gives: $$v_A'=v_B,\\ v_B'=v_A$$ (There exists also the trivial solution $$v_A'=v_A, v_B'=v_B$$, which corresponds to no collision.)

To summarize: the claim about the balls exchanging velocities is based on a) proper understanding of what elastic collision is, and b) doing the math.

They could only come to rest after collision if they were, for instance, wet balls of clay with similar kinetic energy, which is not an elastic collision. If you roll two billiard balls of similar mass straight towards each other they do not stop at collision, they will exchange momentum, minus small losses due to material deformation, sound, heat, friction, or others as it is not a perfectly elastic collision.