There's nothing wrong with any of these other answers, but for another perspective, if you have a constant force acting on an object starting with zero velocity, then it will accelerate with constant acceleration $\frac{F}{m}$, and thus, after $t$ time, will have velocity $v=\frac{F}{m}t$. This means that the kinetic energy that it has acquired will be given by $\frac{1}{2}mv^{2} = \frac{F^{2}t^{2}}{2m}$.
Since the power is the rate of energy consumption, we have:
$$P = {\dot E} = \frac{F^{2}t}{m}$$
so, it should be obvious that the power increases with time. It should also be clear that our expression for $P$ is equal to $Fv$.
First of all, I have to correct your terminology: acceleration doesn't cause force or vice versa. Acceleration is proportional to force, meaning, if acceleration exists, a force exists too.
Force, in its most basic form, is defined as a change in momentum, and that's how Newton stated it
$F=\frac{dP}{dt}=\frac{d(mv)}{dt}=m\frac{dv}{dt}=ma$
The wall does not "stop" the ball, but a collision happens, and the collision between the wall and the ball just reflects the ball doing nothing to the wall because the mass of the wall is very high compared to the ball.
Theoretically, when the the ball hits the wall, it gets repelled back because the wall is much stronger than the ball. Macroscopically, the wall absorbs the hit of the ball, and vibrates, and the vibrations damp so fast because the wall is very rigid, and realistically you don't even notice the vibrations.
How do you calculate that? You have to solve the classical collision problem. The mass of the ball is M and the mass of the wall tends to infinity.
Best Answer
You might want to use the idea of impulse, $J$, defined as $$J=\int_{t_a}^{t_b} F\,\mathrm{d}t=\Delta p=mv_f-mv_0$$ In your first case, $F$ is not time dependent, and so you have $$J=Ft_b-Ft_a=mv_f-mv_0$$ You should be able to solve this.
In the second case, $F$ may or may not be time dependent. The equation for impulse can be changed to $$v_f=\frac{\int_{t_a}^{t_b} F\,\mathrm{d}t+mv_0}{m}$$ If $t_a=t_b$ - your second case - then $v_f=v_0$. Therefore, this proves that a force applied for an instantaneous amount of time produces an acceleration lasting for an instantaneous amount of time - which causes no change in velocity whatsoever.