If the kinetic energy of both the ball decreases, how can their
velocities be equal?? The front one ( B ),from the beginning, had low KE; if
it decreases during the deformation, how can its velocity be equal to
the velocity of the rear ball ( A )?
In order to get a clear picture, let's consider the extreme case when the velocity of B = 0
Let's make a concrete example with numbers $m_A = 1, m_B = 2, M = 3$:
Suppose that:
$v_a = 6m/s$ and $v_b, p, E_k = 0 \rightarrow E_k = 0.5 * 6^2 = 18, p = 1 * 6 = 6, v_{cm} = p/M = 2$
Kinetic energy and momentum are conserved only in elastic collisions, but if the bodies stick together the collision is inelastic an only momentum is conserved:
After the collision velocity of A would be anyway lower as KE should be distributed among more mass, but some KE is lost in the crash. How much?
Momentum is conserved: $ p_{ab} = 6$ , from this datum you can calculate its velocity which now coincides with the velocity of center of mass:
$$v_{ab} = v_{cm}= \frac{6}{3} = 2$$ and $E_{AB} = 0.5 * 2^2 *3 = 6 \rightarrow E_A = 2 + E_B = 4$.
Some energy has been transferred to B (4 J), but two thirds of the kinetic energy(12 J) have been changed to other forms of energy. The general law of 'conservation of energy' has not been, anyway, violated
Velocity of center of mass is the same, although KE has changed. Note that momentum is conserved because we are assuming that on the surface of contact there is no friction.
I hope your main question has got an answer by now, velocities can and must be equal because AB is now one single body: the rear ball has decreased and the front ball has increased its v and the two values level out.
(This is not due to the loss of KE, even if it had been conserved the two bodies would have levelled their v to 3.464, but this would violate the principle of conservation of momentum that would have increased to 10.4)
As to the queries in your comments: when the bodies have reached the maximum deformation they will move at final and same v. It is impossible to determine how much of the amount of KE lost and transformed will be absorbed by each body, as this depends on the material they are made of: the more a body is deformable the more energy it will absorb
.. But what about the case when the front ball is moving?
It makes no difference! Just think of communicating vessels, once two bodies are joined and become a single body... energy, velocity and momentum level out and are unified.
There is no direct relationship between the forces involved and degree of elasticity.
As an example, let's roll two clay balls at each other. The collision is completely inelastic as they stick to each other. If we surround the clay balls with a hard plastic shell, the collision becomes almost completely elastic. The time of interaction is tiny, so the forces felt are very high.
A different possibility is to place a light spring between the two balls. Again the interaction is very elastic, but this time the forces involved are much lower than the clay ball collision.
What you can say is that the impulse (change in momentum) is greater in the elastic collision, and (assuming mass and speed are held constant) determines the speed of the object ball after the impact. High force or low force, the object ball will be moving more quickly after the elastic collision.
Best Answer
I think I understand your question. How is it possible, that even though the first ball deforms and stores elastic energy during the collision, it suddenly ends up with no motion after. The short answer is that some elastic energy is temporarily stored in the left-most ball during the collision, but somehow during the collision, the combination of forces, compression and relative motion makes it so all this stored energy is finally transmitted to kinetic energy in the right-most ball. The quickest way to argue why it must be so is by assuming conservation of energy and momentum, as is typically done in any introductory mechanics course. However, it seems that you would like to "see" the process unfold during the process during the collision.
Let us model the balls as elastic springs obeying Hooke's law. Let $x_1,x_2$ be the positions, $v_1,v_2$ the velocities, $\Delta x_1, \Delta x_2$ the widths, and $m_1,m_2$ the masses of the left-most and right-most ball, respectively.
We assume the balls have equal size and spring constant so that the equilibrium length of both is $\Delta x_0$ and the spring constant is $k$. At the start of the collision, we set $x_1=0$ so that $x_2=\frac{\Delta x_0}{2} + \frac{\Delta x_0}{2}$ and the initial velocities are $v_1 = v_0$, $v_2=0$. During the collision the balls deform, store elastic energy, and exert a force on each other.
Due to N3, the force $F_{21}$ exerted by ball 2 on ball 1 is equal to $-F_{12}$, where $F_{12}$ is the force exerted by ball $1$ on ball $2$. Since the forces are related to the deformation of the balls, we have
$$ F_{12} = -k_1 (\Delta x_1 - \Delta x_0) $$
$$ F_{21} = k_2 (\Delta x_2 -\Delta x_0) $$
Equating $F_{12} = -F_{21}$, we find
$$ k (\Delta x_1 - \Delta x_0) = k (\Delta x_2 -\Delta x_0) $$ $$ \Delta x_1 = \Delta x_2 \equiv \Delta x $$
Additionally, we have
$$ x_2-x_1 = \frac{\Delta x_1}{2} + \frac{\Delta x_2}{2} = \Delta x. $$
From N2, the equations of motion are
$$ a_1 = \frac{k}{m_1} (\Delta x_2 - \Delta x_0) = \frac{k}{m_1} (x_2-x_1 - \Delta x_0) $$ $$ a_2 = -\frac{k}{m_2} (\Delta x_2 - \Delta x_0) = -\frac{k}{m_2} (x_2-x_1 - \Delta x_0). $$
where $a_1, a_2$ are the accelerations of balls 1 and 2, respectively. I don't know if these equations can be solved analytically, but for the purposes of this post, I wrote a midpoint Euler script to solve them numerically. Below is the result from setting $k = 5 N/m$, $\Delta x_0=1 m$, $v_0=3 m/s$, $m_1 = m_2 = 1 kg$:
As we see from the third column (since $\Delta x_1 = \Delta x_2 = \Delta x = x_2-x_1$), both balls are compressed, and thus store elastic energy during the collision. The first ball does move during the collision and ends up at approximately $x_1\sim 0.6m$ after the collision. All the kinetic energy from the first ball is given onto the second, as seen in the middle figure. I made a simple animation of how it looks when two balls of equal mass collide.
The fact that the first ball comes to a complete stop is due to the masses being equal. If we increase the mass of the first ball to $m_1 = 2kg$ (keeping $m_2=1kg$), we get the following result
where, as you can see, the both balls continue along after the collision. For completeness, if we double the mass of the second ball and keeping the first at its original mass, i.e., $m_1=1 kg$, $m_2=2kg$, we get the following:
where the first ball hits the second and turns around.