# 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).

• Additional constrain: the problem must be one dimensional. As to why, it is the only way both momentum and kinetic energy can be conserved. (And conserving kinetic energy is what "elastic" means.) – dmckee Oct 23 '13 at 19:58

In any collision, momentum is conserved. This means

$$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$$

For a perfectly elastic collision, kinetic energy is also conserved

$$m_1u_1^2 + m_2u_2^2 = m_1v_1^2 + m_2v_2^2$$

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

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

when $m_1=m_2$, these reduce to $$v_1 = u_2;\\ v_2 = u_1;$$

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

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.

This is a very interesting question. The problem is simpler in kinematics. However, if we view it as a problem in dynamics, invoking forces and Newton's laws, then the question becomes a natural consequence and the answers become rather complicated.

The question is one of reference frames, in kinematics. The collision is viewed from the laboratory frame of reference where ball B is at rest (this is not spelt out, but implicit, in the question) before collision. Since the two balls are identical, Ball A transfers all of its momentum to B and comes to rest after collision.Ball B starts moving with the momentum gained from A with the same velocity with which ball A was moving before collision.

If the collision is viewed from the center of mass (CM) reference frame, then and only then, the use of concept of force and Newton's third law applicability, become useful and meaningful. In CM frame, B will not be at rest before collision and A will not stop after collision. In CM frame both A and B move with equal speeds in opposite directions along the same line, before collision. After collision each reverses its direction of motion keeping the speeds the same and move away from each other along the same straight line.

Here the concept of force, and Newton's laws become helpful and the solution becomes one of dynamics, which is not as simple as the kinematics solution.

In the case of unequal masses with A having higher mass, your logic that A coninues to move after collision in the same direction (with reduced speed) is right (lab frame).

## protected by AccidentalFourierTransformAug 11 '18 at 13:55

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