newtonian mechanics - Perfect elastic collision and velocity transfer

So my teacher told me that when you have two identical balls in a perfectly elastic collision, the first ball A will collide with B and afterwards A will stop and B continue. Why is this? Doesn't Newton's 3rd law imply both balls would get an equal force into opposite direction during the collision? And if A was heavier than B, does A continue in the same direction after elastically colliding with B (that's the only logical result I can think of if this is true).

Answer

In any collision, momentum is conserved. This means

\begin{equation} m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2 \end{equation}

For a perfectly elastic collision, kinetic energy is also conserved

\begin{equation} m_1u_1^2 + m_2u_2^2 = m_1v_1^2 + m_2v_2^2 \end{equation}

Solving these equations simultaneously ($v_1$ and $v_2$ are the variables)

\begin{equation} v_1 = \frac{u_1(m_1-m_2) + 2m_2u_2}{m_1+m_2};\\ v_2 = \frac{u_2(m_2-m_1) + 2m_1u_1}{m_1+m_2}; \end{equation}

when $m_1=m_2$, these reduce to \begin{equation} v_1 = u_2;\\ v_2 = u_1; \end{equation}

You can check out what happens for other cases as well ($m_1 >> m_2$ or $u_2 = 0$, etc.)

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EDIT: If you look at it from the point of view of forces, you will see the same force act on both objects, in opposite directions. This will cause an acceleration depending on the mass of the object ($F=ma$), but only for the tiny instant that the two are in contact. Now, for example, considering equal masses, the force would decelerate the first object to some velocity, and accelerate the second object to the same velocity (because both have equal masses, and the force acts for an equal amount of time). From the momentum equations, we find that the velocities are swapped.

Important point to remember: Force is not velocity. The same force can produce different accelerations and hence different velocities for different masses.