Backspin!
Those shots in which the cue ball "draws" backwards after hitting the target ball involve backspin. Without backspin, the cue ball cannot reverse direction.
Consider what happens when the cue ball is not spinning at all when it hits the target ball. The cue ball will come to a dead stop if it hits the target ball straight on. Think of Newton's cradle. The cue ball will continue moving forward (but at an angle) if a non-spinning cue ball hits the target ball obliquely.
The cue ball always moves forward after striking the target ball if the cue ball is rolling without slipping whilst hitting the target ball. A rolling cue ball will initially stop if it hits the target ball straight on. The cue ball will still be spinning, however, and this spin will soon make the cue ball start moving forward again. When a rolling cue ball hits the target ball obliquely, the collision will change the cue ball's direction and the spin will accentuate the forward motion.
The only way to combat these effects is to have the cue ball spinning backwards when it strikes the target ball. A backspinning cue ball that hits a target ball straight on will initially stop, but now the backspin will make the cue ball reverse direction.
So how can one make the cue ball have backspin? The answer is simple: Strike the ball below center. How much below depends on the distance to the target ball. This is easy if the target ball is close to the cue ball: Strike the cue ball a bit below center. You'll need to strike the cue ball a bit further below center if the target ball is further away. When the target ball is very far away (across the length of the table), it's very hard to have the cue ball spinning backwards at the point of collision.
You need to take care in your shot and how far from off-center you hit the cue ball. Hit the cue ball too far off-center and you'll hear a nasty "clink" sound. You've just miscued; the cue ball won't move anything like you planned. And maybe you've even ripped the table, bad move!
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.
Best Answer
The answer is that there is no simple answer. The way that energy and momentum get split up in the aftermath of a collision depends on the details of the collision itself, and there is nothing in the conservation laws themselves that influences this.
The simplest case is in one-dimensional collisions, where both objects are constrained to move along the same line. Even in this simple system the ensuing dynamics are not completely determined, and they depend on how 'elastic' or 'inelastic' the collision is; that is, on how much kinetic energy is lost to other energetic channels. You can and should experiment with this: take carts on an air rail and make them collide with each other, both elastically (metal-on-metal should be fine, or add springs if not) and inelastically (use blue-tack to make them stick, or add e.g. a ball of paper that can be crumpled). You will find that the details of the collision affect the outcome, even for the same initial velocities.
When more dimensions are involved, the details of the collision become much more important, even for fully elastic collisions. This is again something you should experiment with, using a pool table or an air table or something similar. In general, the precise point of contact has a large influence on the collision outcome, and it is in such details that the balls 'decide' which way they're going, and how fast.