Unless I made some mistakes, impulse is equal to momentum and not to
change of momentum. Where did I go wrong, or, what is the final word?
I'm not sure I completely understand your notation in your reasoning preceding the statements quoted above but 3rd line certainly doesn't look correct:
Acceleration = a=[1]Δv change of velocity (m) v/s
The correct equation is
$$\bar a = \frac{\Delta v}{\Delta t} $$
where $\bar a$ is the average acceleration. With this correction, the final equation becomes
$$J = (m)\bar a \Delta t = (m) \Delta v = \Delta p$$
This seems so straightforward that I suspect I don't understand what you're actually trying to show.
what is change of velocity then? if a football is at rest and I kick
it and it aquires v=10m/s, haven't I accelerated it over a period of
time? isn't that difference of Δv=+10m/s acceleration?
Yes, you accelerated it over a period of time and no, the difference in velocity is not acceleration.
(Average) acceleration is, as I wrote above, the ratio of the change in velocity to the elapsed time over which the change occurred.
So, if the change in velocity is
$$\Delta v = 10 \frac{m}{s}$$
one does not know the average acceleration unless one also knows the elapsed time $\Delta t$ since the average acceleration is given by
$$\bar a = \frac{\Delta v}{\Delta t}$$
Clearly, the average acceleration is inversely proportional to the elapsed time so, the smaller the elapsed time, the larger the average acceleration.
To be concrete, let us say that the foot was in contact with the football for $100 ms$. Then, the average acceleration of the football is
$$\bar a = \frac{10 \frac{m}{s}}{0.1s} = 100 \frac{m}{s^2} $$
The reason your expectations of kinetic energy loss are violated is because you picked a non-inertial reference frame.
In fact, you didn't notice, but based on your assumptions, momentum isn't even conserved.
Let's look more closely at your first equation:
$$m_e * 0 + m_i v_i = m_e * 0 + m_f v_f$$
Let's say the ball's mass is 5 kg. If the ball is falling down towards the ground, the initial velocity before the collision should be negative (assuming we have adopted a coordinate system where "up" is positive). Let's say the ball was moving at 10 m/s. Here is our equation so far:
$$m_e * 0 + (5 kg) (-10 m/s) = m_e * 0 + (5 kg) v_f$$
Let's simplify, given your assumption that the Earth doesn't move:
$$(5 kg) (-10 m/s) = (5 kg) v_f$$
Which gives us
$$ v_f = -10 m/s $$
Whaaa? The final velocity is negative? That means . . . after this "collision", the ball is still falling downward at the same speed! If we assume, as you did, that the velocity was positive, then the momentum of the system can't be conserved! Clearly, something is wrong here.
Here's the trouble: You began in the reference frame of the Earth. Once the ball hit the Earth, the Earth's reference frame is no longer inertial! You either have to introduce a fictitious force on the ball to account for the fact that the Earth accelerated (a tiny bit, yes, but an important tiny bit), or you need to find an inertial reference frame.
So, Let's pick the frame of reference in which the Earth starts out at rest. When the Earth moves, we'll let our frame of reference stay where it is, so as to allow it to remain inertial:
$$m_e * 0 + (5 kg) (-10 m/s) = m_e * v_{e, f} + (5 kg) v_f$$
$$-50 kg m/s = m_e * v_{e, f} + (5 kg) v_f$$
NOW we can introduce a coefficient of elasticity and demand that energy be either conserved or not, and find the final velocity of the Earth and the ball as they rebound from each other.
Note that although $v_{e, f}$ will be incredibly tiny, due to the enormous mass of the Earth $m_e$, the final momentum of the Earth will be non-negligible - in fact, the final momentum of the Earth must be comparable to the final momentum of the ball in order for momentum to be conserved!
So, to sum up: You picked a non-inertial reference frame, and didn't account for it, so your reliance on the laws of physics was betrayed.
Best Answer
You're right about the stopping time, if you continuously apply a constant force, this will indeed be true.
The stopping distance will probably not be the same, as the pingpong-ball is moving much faster initially (why?). Can you determine the velocity of the ball as a function of time? How will you use this velocity for determining the stopping distance?